Master of Science in

Computacional Mechanics
Erasmus Mundos
Master Course

INTRODUCTION TO
THE FINITE ELEMENT METHOD

Lectures

EUGENIO OÑATE
PEDRO DÍEZ
FRANCISCO ZÁRATE
ANTONIA LARESE

October 2008
Contents
CHAPTER 1
1 INTRODUCTION TO THE FINITE ELEMENT METHOD 1
1.1 WHAT IS THE FINITE ELEMENT METHOD? . . . . . . . . 1
1.2 ANALYTICAL AND NUMERICAL METHODS . . . . . . . . 1
1.3 WHAT IS A FINITE ELEMENT? . . . . . . . . . . . . . . . . 2
1.4 STRUCTURAL MODELLING AND FEM ANALYSIS . . . . . 3
1.4.1 Classification of the problem . . . . . . . . . . . . . . . . 3
1.4.2 Structural model . . . . . . . . . . . . . . . . . . . . . . 3
1.4.3 Structural analysis by the FEM . . . . . . . . . . . . . . 5
1.4.4 Verification and validation of FEM results . . . . . . . . 6
1.5 DISCRETE SYSTEMS. BAR STRUCTURES . . . . . . . . . . 9
1.5.1 Basic concepts of matrix analysis of bar structures . . . . 10
1.5.2 Analogy with the matrix analysis of other discrete systems 13
1.5.3 Basic steps for matrix analysis of discrete systems . . . . 15
1.6 DIRECT ASSEMBLY OF THE GLOBAL STIFFNESS MATRIX 17
1.7 DERIVATION OF THE MATRIX EQUILIBRIUM EQUATI-
ONS FOR THE BAR ELEMENT USING THE PRINCIPLE
OF VIRTUAL WORK . . . . . . . . . . . . . . . . . . . . . . . 19
1.8 DERIVATION OF THE BAR EQUILIBRIUM EQUATIONS
VIA THE MINIMUM TOTAL POTENTIAL ENERGY PRIN-
CIPLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
1.9 PLANE PIN-JOINTED FRAMEWORKS . . . . . . . . . . . . 22
1.10 TREATMENT OF PRESCRIBED DISPLACEMENTS AND
COMPUTATION OF REACTIONS . . . . . . . . . . . . . . . . 24
1.11 INTRODUCTION TO THE FINITE ELEMENT METHOD
FOR ANALYSIS OF CONTINUUM SYSTEMS . . . . . . . . . 27

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Contents
CHAPTER 2

2 FEM ANALYSIS OF 1D PROBLEMS. APPLICATION TO

THE POISSON EQUATION 33
2.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . 33
2.2 THE POISSON EQUATION . . . . . . . . . . . . . . . . . . . 34
2.3 WEIGHTED RESIDUAL METHOD . . . . . . . . . . . . . . . 40
2.3.1 Approximation of the unknown. Weighted Residuals . . 42
2.3.2 Application of the WRM for the solution of the 1D heat
conduction equation . . . . . . . . . . . . . . . . . . . . 43
2.3.3 Global definition of the shape functions . . . . . . . . . . 45
2.3.4 Point Collocation Method . . . . . . . . . . . . . . . . . 48
2.3.5 Subdomain collocation method . . . . . . . . . . . . . . 51
2.3.6 Galerkin method . . . . . . . . . . . . . . . . . . . . . . 53
2.3.7 Least square method . . . . . . . . . . . . . . . . . . . . 55
2.4 GENERAL SOLUTION PROCEDURE . . . . . . . . . . . . . 59
2.4.1 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . 61
2.5 INTEGRABILITY CONDITION . . . . . . . . . . . . . . . . . 62
2.6 WEAK FORM OF THE WEIGHTED RESIDUAL METHOD . 62
2.6.1 Natural boundary condition . . . . . . . . . . . . . . . . 64
2.6.2 Discretization of the weak form . . . . . . . . . . . . . . 64
2.7 DERIVATION OF THE PRINCIPLE OF VIRTUAL WORK
FROM THE WEIGHTED RESIDUAL METHOD . . . . . . . 71
2.8 THE PVW IN POISSON PROBLEMS . . . . . . . . . . . . . . 73
2.9 THE FEM IN 1D POISSON PROBLEMS . . . . . . 73
2.9.1 Discretization of the problem. Local definition of the
shape functions . . . . . . . . . . . . . . . . . . . . . . . 75
2.9.2 Derivation of the algebraic equation systems. Solution
of the 1D Poisson problem using one 2-noded 1D element 77
2.9.3 Solution of the 1D Poisson problem using two 2-noded
elements . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
2.10 GENERALIZATION OF THE SOLUTION FOR A MESH OF
TWO-NODED ELEMENTS . . . . . . . . . . . . . . . . . . . . 85

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CHAPTER 3

3 1D FINITE ELEMENTS FOR AXIALLY LOADED RODS 87

3.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . 87
3.2 AXIALLY LOADED ROD . . . . . . . . . . . . . . . . . . . . . . 87
3.3 AXIALLY LOADED ROD OF CONSTANT CROSS SECTIONAL
AREA. DISCRETIZATION IN ONE LINEAR ROD ELEMENT . 89
3.4 DERIVATION OF THE DISCRETIZED EQUATIONS FROM
THE GLOBAL DISPLACEMENT INTERPOLATION FIELD . . 94
3.5 AXIALLY LOADED ROD OF CONSTANT CROSS
SECTIONAL AREA. DISCRETIZATION IN TWO LINEAR ROD
ELEMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
3.6 GENERALIZATION OF THE SOLUTION WITH N LINEAR
ROD ELEMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . 102
3.7 EXTRAPOLATION OF THE SOLUTION FROM TWO DIF-
FERENT MESHES . . . . . . . . . . . . . . . . . . . . . . . . . . 106
3.8 MATRIX FORMULATION OF THE ELEMENT EQUATIONS . 108
3.8.1 Shape function matrix . . . . . . . . . . . . . . . . . . . . 109
3.8.2 Strain matrix . . . . . . . . . . . . . . . . . . . . . . . . . 109
3.8.3 Constitutive matrix . . . . . . . . . . . . . . . . . . . . . . 109
3.8.4 Principle of Virtual Work . . . . . . . . . . . . . . . . . . 110
3.8.5 Stiffness matrix and equivalent nodal force vector . . . . . 110
3.9 SUMMARY OF THE STEPS FOR THE ANALYSIS OF A STRUC-
TURE USING THE FEM . . . . . . . . . . . . . . . . . . . . . . 112
3.10 ADVANCED ROD ELEMENTS AND REQUIREMENTS FOR
THE NUMERICAL SOLUTION . . . . . . . . . . . . . . . . . . 113
3.10.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . 113
3.11 ONE DIMENSIONAL Co ELEMENTS. LAGRANGE
ELEMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
3.12 ISOPARAMETRIC FORMULATION AND NUMERICAL INTE-
GRATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
3.12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . 117
3.12.2 The concept of parametric interpolation . . . . . . . . . . 118
3.12.3 Isoparametric formulation of the two-noded rod element . 120
3.12.4 Isoparametric formulation of the 3-noded quadratic rod el-
ement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

3
3.13 NUMERICAL INTEGRATION . . . . . . . . . . . . . . . . . . . 123
3.14 STEPS FOR THE COMPUTATION OF MATRICES AND VEC-
TORS FOR AN ISOPARAMETRIC ROD ELEMENT . . . . . . 125
3.14.1 Interpolation of the axial displacement . . . . . . . . . . . 126
3.14.2 Geometry interpolation . . . . . . . . . . . . . . . . . . . . 126
3.14.3 Interpolation of the axial strain . . . . . . . . . . . . . . . 126
3.14.4 Computation of the axial force . . . . . . . . . . . . . . . . 127
3.14.5 Element stiffness matrix . . . . . . . . . . . . . . . . . . . 127
3.14.6 Equivalent nodal force vector . . . . . . . . . . . . . . . . 128
3.15 BASIC ORGANIZATION OF A FINITE ELEMENT PROGRAM 129
3.16 SELECTION OF ELEMENT TYPE . . . . . . . . . . . . . . . . 129
3.17 REQUIREMENTS FOR CONVERGENCE OF THE SOLUTION 133
3.17.1 Continuity condition . . . . . . . . . . . . . . . . . . . . . 133
3.17.2 Derivativity condition . . . . . . . . . . . . . . . . . . . . 133
3.17.3 Integrability condition . . . . . . . . . . . . . . . . . . . . 133
3.17.4 Rigid body condition . . . . . . . . . . . . . . . . . . . . . 134
3.17.5 Constant strain condition . . . . . . . . . . . . . . . . . . 134
3.18 ASSESSMENT OF CONVERGENCE REQUIREMENTS.
THE PATCH TEST . . . . . . . . . . . . . . . . . . . . . . . . . 135
3.19 OTHER REQUIREMENTS FOR THE FINITE ELEMENT AP-
PROXIMATION . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
3.19.1 Compatibility condition . . . . . . . . . . . . . . . . . . . 137
3.19.2 Condition of complete polynomial . . . . . . . . . . . . . . 137
3.19.3 Stability condition . . . . . . . . . . . . . . . . . . . . . . 138
3.19.4 Geometric invariance condition . . . . . . . . . . . . . . . 139
3.20 SOME REMARKS ON THE COMPATIBILITY AND EQUILIB-
RIUM OF THE SOLUTION . . . . . . . . . . . . . . . . . . . . . 139
3.21 CONVERGENCE REQUIREMENTS IN ISOPARAMETRIC EL-
EMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
3.22 ERROR TYPES IN THE FINITE ELEMENT SOLUTION . . . 142
3.22.1 Discretization error . . . . . . . . . . . . . . . . . . . . . . 142
3.22.2 Error in the geometry approximation . . . . . . . . . . . . 143
3.22.3 Error in the computation of the element integrals . . . . . 144
3.22.4 Errors in the solution of the global equation system . . . . 144
3.22.5 Errors associated with the constitutive equation . . . . . . 146

ii

4
 ii CONTENTS

Contents
CHAPTER 4

Contents
4 FEM ANALYSIS OF THE 2D POISSON EQUATION 147
Contents
4.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . 147
4.2 STEADY-STATE POISSON EQUATION IN 2D . . . . . . . . 147
4 4.3
FEMPROBLEM
ANALYSIS OF THEUSING
SOLUTION 2D POISSONTHE FINITE EQUATION ELEMENT 147
4.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . 148
METHOD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
4.3.1 Integral
4.2 STEADY-STATE form of the GWR method . . . . . . . . . . . . 149
4 FEM ANALYSIS OFPOISSON THE 2DEQUATION POISSONIN 2D . . . . . . . . 147
EQUATION 147
4.3 4.3.2
PROBLEMFEM Discretization
SOLUTION .
USING . . .THE. . . . .
FINITE . . .ELEMENT
. . . . . . . . . 150
4.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . 147
4.4 THREE-NODED
METHOD . . . . TRIANGULAR . . . ELEMENT
. . . . . . EQUATION . . . . . IN . .2D .. .. .. .. .. .. .. .. .. .. 153
4.2 STEADY-STATE POISSON . . . . . . . .148 147
4.4.1
4.3.1 Elemental
Integral stiffness
form of the matrix
GWR . .
method . . . .. .. .. .. .. .. .. .. .. .. .. .. 154 149
4.3 4.4.2
PROBLEMThe SOLUTION
equivalent nodalUSING flow THE FINITE
vector .. .. .. ..ELEMENT
. . . . . . . . . 156
METHOD . . . . . . . . . . . . . . . . . . . . . .. .. .. .. .. .. .. .. ..150
4.3.2 FEM Discretization . . . . . . . . 148
4.4 THREE-NODED TRIANGULAR ELEMENT . . . . . . . . . . 153
5 FEM4.3.1
ANALYSIS
4.4.1
Integral OF formTHE
Elemental stiffness
of the3D GWR
matrix POISSON
method EQUATION
. . . . . . . . . . . . . . . . .163
. . . . . . . . . . . . 149
154
CHAPTER 5
4.3.2 FEM Discretization
5.1 INTRODUCTION . . . . .. .. .. .. .. .. . . . . . . . . . . . . . . . . . . . . . . . . . .. .. .163
150
4.4.2 The equivalent nodal flow vector . . . . . . . . . . . . . 156
4.4 THREE-NODED POISSONS
5.2 STEADY-STATE TRIANGULAR EQUATIONELEMENT IN 3D . . . . . . . . . . . . . .. .. .163153
5 5.3 4.4.1
FEMSOLUTION Elemental
ANALYSIS USINGOFstiffness
THE FINITE
THE matrix
3D POISSON . . . . .EQUATION
ELEMENT . . METHOD
. . . . . . . . .. .. .163 154
164
4.4.2
5.1 5.3.1 The equivalent
Weak
INTRODUCTION integral. form .nodal
. . .. flow
.. .. .. vector
.. .. .. .. ... ... ... ... ... ... ... ... ... ... ........ .164
156
163
5.2 5.3.2 FEM Discretization
STEADY-STATE POISSONS . . EQUATION
. . . . . . . .IN. .3D. .. .. .. .. .. .. .. .. 165 163
5 FEM
5.4 ANALYSIS
FOUR-NODED OF THE
TETRAHEDRAL 3D POISSON ELEMENT EQUATION. . . . . . . . . . 163
167
5.3 SOLUTION USING THE FINITE ELEMENT METHOD . . . 164
5.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . 163
5.3.1 Weak integral form . . . . . . . . . . . . . . . . . . . . . 164
6 5.2
MATRIX FORMULATION
STEADY-STATE POISSONS FOR THE FEM
EQUATION INSOLUTION
3D . . . . . .OF . . 163
5.3.2 FEM Discretization . . . . . . . . . . . . . . . . . . . . . 165
THESOLUTION
5.3 POISSON USING EQUATION THE FINITE ELEMENT METHOD . . 173 . 164
5.4 FOUR-NODED TETRAHEDRAL ELEMENT . . . . . . . . . . 167
5.3.1 Weak integral. form
6.1 INTRODUCTION . . . .. .. .. .. .. .. . . . . . . . . . . . . . . . . . . . . . . . . . .. .. .173 164
6 6.2 MATRIX
MATRIX
CHAPTER 6
FORMULATION
5.3.2 FORMULATION
FEM Discretization FOR .OF. .THE.THE . .POISSON
. FEM
. . . .SOLUTION.EQUATION
. . . . . . .OF .. .173
165
6.3
THEWEAK
5.4 INTEGRAL
FOUR-NODED
POISSON FORM . . . . ELEMENT
TETRAHEDRAL
EQUATION . . . . . . . . . . . . . . . . . . . . . .. .. .173 175
167
6.4
6.1 MATRIX FORM OF. THE
INTRODUCTION . . . FEM. . . EQUATIONS
. . . . . . . . . .. .. .. .. .. .. .. .. .. 175 173
6 MATRIX
6.5 THE FORMULATION
MATRIX EQUILIBRIUM FOR
6.2 MATRIX FORMULATION OF THE POISSON EQUATION . 173 THE
EQUATIONS FEM SOLUTION
OBTAINED OF
THE POISSON
6.3 FROM
WEAK THE EQUATION
NODAL
INTEGRAL FLUXES
FORM . . .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. 177 173
175
6.1 INTRODUCTION
6.4 6.5.1
MATRIX OneFORM OF. THE
dimensional . 2-noded
. . FEM
. . . element
.EQUATIONS
. . . . . . . . . . . . ... ... ... ... ... ... .........177 173
175
6.2 MATRIX
6.5 6.5.2 2D FORMULATION
THE MATRIX three-noded
EQUILIBRIUM triangularOF EQUATIONS
THE
element POISSON . .EQUATION
. . OBTAINED . . . . . . . . .180 173
6.3 WEAK
FROM THE INTEGRAL
NODALFORM FLUXES . . .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..177 175
6.4 MATRIX
6.5.1 One FORM OF THE
dimensional FEM element
2-noded EQUATIONS . . . . .. .. .. .. .. .. .. .. ..177 175
6.5 THE
6.5.2 MATRIX EQUILIBRIUM
2D three-noded triangularEQUATIONS
element . . OBTAINED . . . . . . . . . . 180
FROM THE NODAL FLUXES . . . . . . . . . . . . . . . . . . 177
6.5.1 One dimensional 2-noded element . . . . . . . . . . . . . 177
6.5.2 2D three-noded triangular element . . . . . . . . . . . . 180

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CHAPTER 7

7 SHAPE FUNCTIONS FOR 2D AND 3D ELEMENTS WITH

C ◦ CONTINUITY 183
7.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . 183
7.2 DERIVATION OF THE SHAPE FUNCTIONS FOR Co TWO
DIMENSIONAL ELEMENTS . . . . . . . . . . . . . . . . . . . 183
7.2.1 Complete polynomials in two dimensions. Pascal’s triangle183
7.2.2 Shape functions of C o rectangular elements. Natural
coordinates in two dimensions . . . . . . . . . . . . . . . 184
7.3 LAGRANGE RECTANGULAR ELEMENTS . . . . . . . . . . 186
7.3.1 Four-noded Lagrange rectangle . . . . . . . . . . . . . . 186
7.3.2 Nine-noded quadratic Lagrange rectangle . . . . . . . . . 186
7.3.3 Sixteen-noded cubic Lagrange rectangle . . . . . . . . . . 190
7.3.4 Other Lagrange rectangular elements . . . . . . . . . . . 190
7.4 SERENDIPITY RECTANGULAR ELEMENTS . . . . . . . . . 191
7.4.1 Eigth-noded quadratic Serendipity rectangle . . . . . . . 192
7.4.2 Twelve-noded cubic Serendipity rectangle . . . . . . . . . 194
7.4.3 Seventeen-noded quartic Serendipity rectangle . . . . . . 195
7.5 SHAPE FUNCTIONS FOR C o CONTINUOUS TRIANGU-
LAR ELEMENTS . . . . . . . . . . . . . . . . . . . . . . . . . 196
7.5.1 Area coordinates . . . . . . . . . . . . . . . . . . . . . . 196
7.5.2 Derivation of the shape functions for C o continuous tri-
angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198
7.5.3 Shape functions for the 3-noded linear triangle . . . . . . 198
7.5.4 Shape functions for the six-noded quadratic triangle . . . 199
7.5.5 Shape functions for the ten-noded cubic triangle . . . . . 200
7.5.6 Natural coordinates for triangles . . . . . . . . . . . . . . 201
7.6 ANALYTIC COMPUTATION OF INTEGRALS OVER
RECTANGLES AND STRAIGHT SIDE TRIANGLES . . . . . 201
7.7 SHAPE FUNCTIONS FOR 3D ELEMENTS . . . . . . . . . . . 205
7.8 RIGHT PRISMS . . . . . . . . . . . . . . . . . . . . . . . . . . 205
7.8.1 Right prisms of the Lagrange family . . . . . . . . . . . 206
7.8.2 Serendipity prisms . . . . . . . . . . . . . . . . . . . . . 209
7.9 STRAIGHT EDGED TETRAHEDRA . . . . . . . . . . . . . . 215
7.9.1 Shape functions for the 10-noded quadratic tetrahedron . 219

6
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7.9.2 Shape functions for the 20-noded quadratic tetrahedron . 220

7.10 COMPUTATION OF THE ELEMENT INTEGRALS . . . . . . 221
7.10.1 Analytical computation of element integrals . . . . . . . 221

Contents
CHAPTER 8

8 2D AND 3D ISOPARAMETRIC ELEMENTS 225

8.1 ISOPARAMETRIC QUADRILATERAL ELEMENTS . . . . . 225
8.1.1 Stiffness matrix and load vector of isoparametric quadri-
lateral . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227
8.1.2 A comparison between the 8- and 9-noded isoparametric
quadrilaterals . . . . . . . . . . . . . . . . . . . . . . . . 230
8.2 ISOPARAMETRIC TRIANGULAR ELEMENTS . . . . . . . . 231
8.3 NUMERICAL INTEGRATION IN TWO DIMENSIONS . . . . 234
8.3.1 Numerical integration in quadrilateral domains . . . . . . 234
8.3.2 Numerical integration over triangles . . . . . . . . . . . . 235
8.4 NUMERICAL INTEGRATION OF THE ELEMENT MATRI-
CES AND VECTORS . . . . . . . . . . . . . . . . . . . . . . . 236
8.5 COMPUTER PROGRAMMING OF K(e) AND f (e) . . . . . . . 238
8.6 OPTIMAL POINTS FOR COMPUTING GRADIENT FLUXES239
8.7 SELECTION OF THE QUADRATURE ORDER . . . . . . . . 243
8.8 THE PATCH TEST FOR 2D ELEMENTS . . . . . . . . . . . . 247
8.9 3D ISOPARAMETRIC ELEMENTS . . . . . . . . . . . . . . . 248
8.10 NUMERICAL INTEGRATION IN THREE DIMENSIONS . . . 252
8.10.1 Hexahedral elements . . . . . . . . . . . . . . . . . . . . 252
8.10.2 Tetrahedral elements . . . . . . . . . . . . . . . . . . . . 253
8.11 NUMERICAL INTEGRATION OF THE ELEMENT
MATRICES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253
8.11.1 Isoparametric hexahedral elements . . . . . . . . . . . . 253
8.11.2 Isoparametric tetrahedral elements . . . . . . . . . . . . 255
8.11.3 Selection of the quadrature order . . . . . . . . . . . . . 256
8.12 PERFORMANCE OF 3D SOLID ELEMENTS . . . . . . . . . 256

7
Contents
CHAPTER 9

9 2D SOLIDS. LINEAR TRIANGULAR AND RECTANGU-

LAR ELEMENTS 259
9.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . 259
9.2 TWO DIMENSIONAL ELASTICITY THEORY . . . . . . . . . 260
9.2.1 Displacement field . . . . . . . . . . . . . . . . . . . . . 260
9.2.2 Strain field . . . . . . . . . . . . . . . . . . . . . . . . . 261
9.2.3 Stress field . . . . . . . . . . . . . . . . . . . . . . . . . . 262
9.2.4 Stress-strain relationship . . . . . . . . . . . . . . . . . . 262
9.2.5 Virtual work expression . . . . . . . . . . . . . . . . . . 270
9.3 FINITE ELEMENT FORMULATION. THREE-NODED
SOLID TRIANGULAR ELEMENT . . . . . . . . . . . . . . . . 270
9.3.1 Discretization of the displacement field . . . . . . . . . . 271
9.3.2 Discretization of the strain field . . . . . . . . . . . . . . 273
9.3.3 Discretization of the stress field . . . . . . . . . . . . . . 275
9.3.4 Discretized equilibrium equations . . . . . . . . . . . . . 275
9.3.5 Stiffness matrix and equivalent nodal force vectors for
the 3-noded solid triangular element . . . . . . . . . . . 278
9.4 THE FOUR NODED SOLID RECTANGULAR ELEMENT . . 282
9.4.1 Basic formulation . . . . . . . . . . . . . . . . . . . . . . 282
9.4.2 Some remarks on the behaviour of the 4-noded solid rect-
angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286
9.5 PERFORMANCE OF THE 3-NODED SOLID TRIANGLE AND
THE 4-NODED SOLID RECTANGLE . . . . . . . . . . . . . . 289
9.6 GENERAL PERFORMANCE OF TRIANGULAR AND
RECTANGULAR ELEMENTS . . . . . . . . . . . . . . . . . . 291
9.7 CONCLUDING REMARKS . . . . . . . . . . . . . . . . . . . . 293

10 REFERENCES 295

8
Chapter 1

INTRODUCTION TO THE FINITE

ELEMENT METHOD

1.1 WHAT IS THE FINITE ELEMENT METHOD?

The finite element method (FEM) is a procedure for the numerical solution of
the equations which govern problems found in nature. Usually the behaviour of
nature can be described by equations expressed in differential or integral form.
For this reason the FEM is understood in mathematical circles as a numerical
technique for solving partial differential or integral equations. Generally, the
FEM allows to obtain the evolution in space and/or time of one or more
variables representing the behaviour of a physical system.
When referred to the analysis of structures the FEM is a powerful method
for computing the displacements, stresses and strains in a structure under a
set of loads. This is precisely what we aim to study in this book.

1.2 ANALYTICAL AND NUMERICAL METHODS

The conceptual difference between analytical and numerical methods is that
the former search for the universal mathematical expressions representing the
general and “exact” solution of a problem governed typically by mathematical
equations. Unfortunately exact solutions are only possible for a few particular
cases which frequently represent coarse simplifications of reality.
On the other hand, numerical methods, such as the FEM aim to providing
a solution, in the form of a set of numbers, to the mathematical equations
governing a problem. The strategy followed by most numerical methods is to
transform the mathematical equations into a set of algebraic equations which
depend on a finite set of parameters. These equations in the practical case
involve many thousands (or even millions) of unknowns and therefore the final
system of algebraic equations can only be solved with the help of computers.
This explains why even though many numerical methods have been known

1
since the XVIII century, their development and popularity has occurred in
tandem to the progress of modern computers in the XX century. The term
numerical method is synonym of computational method in this text.
Numerical methods represent, in fact, the return of numbers as the true
protagonists in the solution of a problem. The loop initiated by Pythagoras
some 25 centuries ago has been closed in the last few decades with the evidence
that, with the help of numerical methods, we can find precise answers to any
problem in science and engineering.
We should keep in mind that numerical methods for structural engineering
are inseparable from mathematics, material modelling and computer science.
Nowadays it is unthinkable to attempt the development of a new numerical
method for the solution of a structural problem without referring to those disci-
plines. As an example, any numerical method for solving large scale structural
problems has to take into account the hardware environment where it will be
implemented (most probably using parallel computing facilities). Also a mod-
ern computer program for structural analysis should be able to incorporate the
continuous advances in the modelling of new materials.
The word which perhaps best synthesizes the immediate future of numer-
ical methods is “computational multiphysics”. The solution of problems will
not be attempted from the perspective of a single physical field and it will
involve all the couplings which characterize the complexity of reality. For in-
stance, the design of a structural component for a vehicle (an automobile,
an aeroplane, etc.) will take into account the manufacturing process and the
function which the component will play throughout its life time. Structures in
civil engineering will be studied considering the surrounding environment (soil,
water, air). Similar examples are found in mechanical, naval and aeronauti-
cal engineering and indeed in practically all branches of engineering science.
Accounting for the non-deterministic character of data will be essential for esti-
mating the probability that the new products and processes conceived by men
behave as planned. The huge computational needs resulting from a “stochastic
multiphysics” viewpoint will demand better numerical methods, new material
models and, indeed, faster computers.
It is only through the integration of a deep knowledge of the physical and
mathematical basis of a problem and of numerical methods and informatics,
that effective solutions will be found for the mega-structural problems of the
twenty-first century.

1.3 WHAT IS A FINITE ELEMENT?

A finite element can be visualized as a small portion of a continuum (i.e. a
structure). The word “finite” distinguishes such a portion from the “infinites-
imal” elements of a differential calculus. The geometry of the continuum is
considered to be formed by the assembly of a collection of non-overlapping do-
mains with simple geometry termed finite elements (i.e. triangles and quadri-

2
laterals in 2D, or tetrahedra and hexahedra in 3D). It is usually said that a
“mesh” of finite elements “discretizes” the continuum (Figure 1.1). The space
variation of the problem parameters (i.e. the displacements in a structure) is
expressed within each element by means of a polynomial expansion. Given
that the “exact” analytical variation of such parameters is more complex (and
generally unknown), the FEM only provides an approximation to the exact
solution.

1.4 STRUCTURAL MODELLING AND FEM ANALYSIS

1.4.1 Classification of the problem
The first step in the solution of a problem is the identification of the problem
itself. Hence, before we can analyze a structure we must ask ourselfs the fol-
lowing questions: Which are the more relevant physical phenomena influencing
the structure? Is the problem of static or dynamic type? Are the kinematics
or the material properties linear or non-linear? Which are the key results re-
quested? What is the level of accuracy sought? The answers to these questions
are essential for selecting a structural model and the adequate computational
method.

1.4.2 Structural model

Computational methods, such as the FEM, are applied to “models” of a real
problem, and not to the actual problem itself. Even experimental methods
in structural laboratories make use of scale physical models, unless the actual
structure is tested in real size, which rarely occurs. Models can be developed
once the physical nature of a problem is clearly understood. In the derivation
of a model we should aim to exclude superfluous details and include all the
relevant features of the problem under consideration so that the model can
describe reality with enough accuracy.
A structural model must include three fundamental aspects. The geome-
tric description of the actual structure by means of its specific geometrical
components (points, lines, surfaces, volumes), the mathematical expression of
the basic physical laws governing the behaviour of the structure (i.e. force-
equilibrium equations) usually written in terms of differential and/or integral
equations, and the specification of the properties of the materials in the struc-
ture. Note that the same structure can be analyzed using different structural
models depending on the accuracy and/or simplicity sought in the analysis.
As an example, a beam can be modelled using the more general 3D elasticity
theory, the 2D plane stress theory or the simpler beam theory. Each model
provides a different set out for the analysis of the actual structure. We should
bear in mind that the solution taking as its starting point an incorrect struc-
tural model, even if obtained with the most accurate numerical method, will

3
 Fig. 1.1: Discretization of different solids and structures into finite elements

be a wrong solution, far from correct physical values.

In this book we will study the analysis by the FEM of a number of struc-
tural models, each one applicable to one or more practical structural types.
The material properties in all cases will be considered to be linear elastic.

4
Furthermore the analysis will be restricted to linear kinematics and to static
loading. The structures are therefore analyzed under linear static conditions
conditions.
Despite their simplicity, these assumptions are applicable to most of the situ-
ations found in the everyday practice of structural analysis and design.
The structural models considered in this book are classified as solid models
(2D/3D solids and axisymmetric solids), beam and plate models and shell mod-
els (faceted shells, axisymmetric shells and curved shells). Figure 1.2 shows
the general features of a typical member of each structural model family. The
structures that can be analyzed with these models cover the main problems
in structural engineering and include frames, dams, retaining walls, tunnels,
bridges, cylindrical tanks, shell roofs, ship hulls, mechanical parts, airplane
fuselages, vehicle components, etc.

1.4.3 Structural analysis by the FEM

The geometry of a structure is discretized when it is split into a mesh of finite
elements of a certain accuracy. Clearly the discretization introduces another
approximation. With respect to reality we have therefore to account for two
error sources from the outset: the modelling error and the discretization error.
The former can be reduced by improving the structural model which describes
the actual behaviour of the structure, as previously explained. The discretiza-
tion error, on the other hand, can be reduced by using a finer mesh (i.e. more
elements), or else by increasing the accuracy of the finite elements chosen us-
ing higher order polynomial expansions for approximating the displacement
field within each element. We have to recall that even if we could reduce the
discretization error to zero, we would not be able to reproduce accurately the
actual behaviour of the structure, unless the structural model was perfect.
Additionally, the use of computers introduces numerical errors associated
with the ability of the computer to represent data accurately with numbers of
finite precision. The numerical error is usually small, although it can be large
in some problems, such as when some parts of the structure have very different
physical properties. The sum of discretization and numerical errors contribute
to the error of the computational method.
Figures 1.3(a) and (b) show schematically the discretization of some geo-
metrical models of structures using specific finite elements. Figure 1.4 shows
the actual image of a car panel, the geometrical definition of the panel surface
by means of NURBS (non-uniform rational b-splines) patches using computer-
aided design (CAD) tools, the discretization of the surface by a mesh of 3-noded
shell triangles and some numerical results of the FEM analysis. The differences
between the real structure of the panel, the geometrical description and the
analysis mesh can be seen clearly. A similar example of the FEM analysis of
an office building is shown in Figure 1.5.

5
 Fig. 1.2: Structural models for some structures

1.4.4 Verification and validation of FEM results

Developers of structural finite element computer codes, analysts who use the
codes and decision makers who rely on the results of the analysis face a critical
question: How should confidence in modelling and computation be critically
assessed? Validation and verification of FEM computations are the primary
methods for building and quantifying this confidence. In essence, validation
is the assessment of the accuracy of both the structural and computational
models by comparison of the numerical results with experimental data. Ex-
periments can be performed in laboratory using scale models of a structure

6
 Fig. 1.3: (a) Discretization of structural models into finite elements

or on actual structures. The correct definition of the experimental tests and

the reliability of the experimental results are crucial issues in the validation
process.
In verification, on the other hand, the relationship between the numerical
results to the real world is not an issue. The verification of FEM computations
is usually made by comparing the numerical results for simple benchmark
problems with “exact” solutions obtained analytically, or using more accurate
numerical methods.
Figure 1.6 shows a scheme of the validation and verification procedures.

7
Fig. 1.3: (b) Discretization of structural models in one-, two- and three-dimensional
finite elements

The verification process is typically performed first in order to evaluate and re-
duce the possible sources of numerical error (i.e. discretization error, numerical
errors, etc.). These errors can be appraised using error estimation techniques.
A more accurate numerical solution can be found with a finer discretization
or by using higher order elements (Chapter 0). The subsequent experimental
validation provides insight on the capacity of the overall structural model to

8
Fig. 1.4: (a) Actual geometry of an automotive panel. (b) CAD geometrical de-
scription by NURBS patches. (c) Finite element mesh of 3-noded shell triangles
discretizing the panel geometry. (d) FEM numerical results of the structural analy-
sis showing the equivalent strain distribution (Images by courtesy of Quantech ATZ
SA, www.quantech.es)

reproduce the behaviour of a real structure with enough precision. Although

both the accuracy of the structural model and the computational method are
assessed in a validation exercise, a large validation error for an already verified
code typically means that the structural model chosen is not adequate and
that a more refined structural model should be used. In summary, verification
serves to check that we are solving structural problems accurately, while vali-
dation tell us that we are solving the right problem. The issue of verification
and validation of FEM codes is treated extensively in [].
In the following sections we will revisit the basic concepts of the matrix
analysis of bar structures, considered here as a particular class of the so-called
“discrete systems”. Then we will summarize the general steps in the analysis of
“continuous” structures by the FEM. The interest of classical matrix structural
analysis is that it provides a general solution framework which reassembles very
closely that followed in the FEM.

1.5 DISCRETE SYSTEMS. BAR STRUCTURES

The solution of a many technical problems requires the analysis of a network
system formed by different “elements”connected by their extremities or joints,
and subjected to a set of “loads” which are usually external to the system.
Examples of such systems, which we will call discrete systems, are common in
structural engineering (pin-jointed bar structures, frames, grillages, etc.) and
in many other different engineering problems, e.g.: hydraulic piping networks,

9
Fig. 1.5: FEM analysis of the Agbar tower (Barcelona). Actual structure and dis-
cretization into shell and 3D beam elements. Deformed mesh (amplified) under wind
loads (Images courtesy of Compass Ingenierı́a y Sistemas SA, www.compassis.com)

electric networks, transport planning networks, production organization sys-

tems (PERT, etc) amongst others. Figure 1.7 shows some of these discrete
systems.
Most discrete systems can be studied using matrix analysis procedures
which have a very close resemblance to the Finite Element Method (FEM). In
Appendix I the basic concepts of matrix algebra are summariced. An outline
of the matrix analysis techniques for bar structures and other discrete systems
such as electric and hydraulic networks is presented in the next section.

1.5.1 Basic concepts of matrix analysis of bar structures

Matrix analysis is the most popular technique for the solution of bar structures
[L3], [P14]. Matrix analysis also provides a general methodology for the FEM
analysis of other structural problems. A good knowledge of this technique is

10


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Fig. 1.6: Schematic view of the verification and validation processes in the FEM

essential for the study of this book.

Fig. 1.7: Some discrete systems

The matrix equations for a bar structure are obtained from the study of
the “equilibrium” of the different individual bars. We will consider first the
(e)
case of an isolated bar, e, of length l(e) subjected to axial end forces R1 and
(e)
R2 only (Figure 1.8). Strength of Materials defines the strain at any point

11
 Fig. 1.8: Deformation of a bar subjected to axial forces

along the bar by the relative elongation [T7,10], i.e.

(e) (e)
∆l(e) u −u
ε = (e)
= 2 (e) 1 (1.1)
l l
(e) (e)
where u1 and u2 are the displacements of the end points 1 and 2, respectively.
In Eq.(1.1) and the following the superindex denotes values associated to an
individual bar.
The axial stress σ is related to the strain ε by Hooke’s law [T6,7] as
(e) (e)
u2 − u1
σ = E (e) ε = E (e) (1.2)
l(e)
where E (e) is the Young modulus of the material. The axial force N at each
section is obtained by integrating the stress over the cross sectional area. The
axial force is transmitted to the adjacent bars through the joints. In the case
of a homogeneous material we have
(e) (e)
(e) (e) u2 − u1
N = A σ = (EA) (1.3)
l(e)
The equilibrium equation for the bar of Figure 1.8 is simply
(e) (e)
R1 + R2 = 0 (1.4a)
(e) (e)
with (e) (e) u2 − u1
R2 = N = (EA) = k (e) (u2 − u1 ) (1.4b)
l(e)
³ ´(e)
EA
where k (e) = l
. Eqs.(1.4) can be written in matrix form as
( (e)
) · ¸ ( (e) )
R1 1 −1 u1
q(e) = (e) = k (e) (e) = K(e) a(e) (1.5a)
R2 −1 1 u2

where · ¸
1 −1
K(e) = k (e) (1.5b)
−1 1

12
K(e) is the stiffness matrix of the bar, which depends on the geometry of the
(e) (e)
bar (l(e) , A(e) ) and its mechanical properties (E (e) ) only; a(e) = [u1 , u2 ]T and
(e) (e)
q(e) = [R1 , R2 ]T are the joint displacements and the joint equilibrium force
vectors for the bar, respectively.
The effect of a uniformly distributed external axial load of intensity b(e)
can easily be taken into account by adding one half of the total external load
to each axial force at the bar ends. The equilibrium equation now reads
( (e)
) · ¸ ( (e) ) ½ ¾
(e) R1 (e) 1 −1 u1 (bl)(e) 1
q = (e) = k (e) − = K(e) a(e) − f (e)
R2 −1 1 u2 2 1
½ ¾
(1.6)
(bl)(e) 1
where f (e) = is the vector of joint forces due to the distributed
2 1
loading.
The equilibrium equations for the whole structure are obtained by imposing
the equilibrium of forces at each of the n joints. This condition can be written
as
ne
X (e)
Ri = Rj , j = 1, n (1.7)
e=1

The sum on the left hand side (l.h.s.) of Eq.(1.7) extends over all bars ne
sharing the joint point with global number j and Rjext represents the external
(e)
load acting on that joint. The values of the bar end forces Ri of Eq.(1.7)
are expressed in terms of the joint displacements using Eq.(1.6). This process
leads to the following global equilibrium equation
    
K11 K12 ······ K1n   u1   
 f1 
K    
 21 K22 ······ K2n 




u 

2





f2 


 .  . .
 ..  .. = .
.  (1.8a)
 
 ..  .. 
 
 .. 
 .  
 
 . 

 . 

 







Kn1 Kn2 · · · · · · Knn un fn

Ka = f (1.8b)
where K is the global stiffness matrix of the structure and a and f are the
global joint displacement vector and the global joint force vector respectively.
The derivation of Eq.(1.8a) is termed the assembly process. The solution of
Eq.(1.8a) yields the displacements at all joint points from which the values of
the internal axial forces in the bars can be computed.

1.5.2 Analogy with the matrix analysis of other discrete systems

The steps between Eqs.(1.1) and (1.8) are very similar for many discrete sys-
tems. For instance, the study of a single resistance element 1-2 in an electric
network (Figure 1.9a) yields the following relationship between the currents

13
entering the resistance element and the voltages at the end points (Ohm’s law)

(e) (e) 1 (e) (e) (e) (e)

I1 = − I2 = (e)
(V1 − V2 ) = k (e) (V1 − V2 ) (1.9)
R

Fig. 1.9: a) Electrical resistance, b) Fluid carrying pipe. Equations of equilibrium

We note that this equation is identical to Eq.(1.4) for the bar element if
the current intensities and the voltages are replaced by the joint forces and the
³ ´(e)
joint displacements, respectively, and 1/R(e) by EA l
. Indeed, if uniformly
(e)
distributed external currents b are supplied along the length of the element,
the force term f (e) of Eq.(1.6) is found. The “assembly rule” is the well known
Kirchhoff’s law stating that the sum of all the current intensities arriving at a
joint must be equal to zero, i.e.
ne
X (e)
Ii = Ij , j = 1, n (1.10)
e=1

where Ij is the external current intensity entering joint j. Note the analogy
between Eqs.(1.10) and (1.7).
The same analogy can be found in the study of fluid carrying pipe networks.
The equilibrium equation relating fluid flow q and hydraulic head h at the ends
of a single pipe element can be written as (Figure 1.5b)

(e) (e) (e) (e)

q1 = − q2 = k (e) (h1 − h2 ) (1.11)

where k (e) is a parameter which is a function of the pipe roughness and the
hydraulic head. This implies that the terms of the stiffness matrix K(e) for
(e)
a pipe element are known functions of the joint heads hi . The equilibrium
equation for each pipe element is written in an identical manner to Eq.(1.6)
(e) (e) (e) (e)
where ui and Ri are replaced by hi and qi , respectively and b(e) represents
the input of a uniformly distributed flow source along the pipe length.

14
 The assembly rule simply states that at each of the n pipe joint the sum of
the flow contributed by the adjacent pipe elements should equal the external
flow source, i.e.
ne
X (e)
qi = qj , j = 1, n (1.12)
e=1

The global equilibrium equations are assembled similarly as for the bar
element yielding the system of Eqs.(1.8). In the general problem matrix K
will be a function of the nodal hydraulic head via the k (e) parameter. Iterative
techniques for the solution of the resulting non-linear system of equations are
needed in this case.

1.5.3 Basic steps for matrix analysis of discrete systems

What we have seen thus far leads us to conclude that the analysis of a discrete
system (i.e. a bar structure) involves the following steps:

a) Definition of a network of discrete elements (bars) connected among

themselves by joints adequately numbered. Each element e has known
geometrical and mechanical properties. All these characteristics consti-
tute the problem data and should be defined in the simplest possible way
(preprocessing step).

b) Computation of the stiffness matrix K(e) and the joint force vector f (e)
for each element of the system.

c) Assembly and solution of the resulting global matrix equilibrium equa-

tion (Ka = f ) to compute the unknown parameters at each joint (i.e.
the displacements for the bar system).

d) Computation of other relevant parameters for each element (i.e. the

axial strains and forces) in terms of the joint parameters.

The results of the analysis should be presented in a clear graphical form to

facilitate the assessment of the system’s performance (postprocessing step).

Fig. 1.10: Analysis of a simple three-bar structure under an axial load

15
Example: 1.1 Compute the displacements and axial forces in the three bar structure of
Figure 1.10 subjected to an horizontal force P acting at its right hand end.

- Solution

The equilibrium equations for each joint are (see Eq.(1.5))

( ) · ¸( )
(1) (1)
R1 (1) 1 −1 u1
Bar 1 = k
(1)
R2 −1 1 (1)
u2

( ) · ¸( )
(2) (2)
R1 1 −1 u1
Bar 2 = k (2)
(2)
R2 −1 1 (2)
u2
( ) · ¸( )
(3) (3)
R1 (3) 1 −1 u1
Bar 3 = k
(3)
R2 −1 1 (3)
u2

EA 2EA
with k (1) = k (2) = l and k (3) = l .
The compatibility equations between local and global displacements are

(1) (1) (2)

u1 = u1 ; u2 = u3 ; u1 = u2

(2) (3) (3)

u2 = u3 ; u1 = u3 ; u2 = u4

Applying the assembly equation (1.7) to each of the four joints we have

3
X (e)
joint 1: Ri = −R1
e=1

3
X (e)
joint 2: Ri = −R2
e=1

3
X (e)
joint 3: Ri = 0
e=1

3
X (e)
joint 4: Ri = P
e=1

(e)
Substituting the values of Ri from the equilibrium equation of each bar, we obtain

(1) (1)
joint 1 : k (1) (u1 − u2 ) = −R1
(2) (2)
joint 2 : k (2) (u1 − u2 ) = −R2
(1) (1) (2) (2) (3) (3)
joint 3 : k (1) (−u1 + u2 ) + k (2) (−u1 + u2 ) + k (3) (u1 + u2 ) = 0
(3) (1)
joint 4 : k (3) (−u1 + u2 ) = P

16
 Above equations can be written in matrix form using the displacement compatibility
conditions as

1 2 3 4
     
1 k (1) 0 −k (1) 0  u1 
   −R1 
 
2  0 k (2) −k (2) 0  u2 −R2
 −k (1) =
3 −k (2) (k (1) + k (2) + k (3) ) −k (3)   u3 
   0 
 
4 0 0 −k (3) k (3) u4 P

Substituting the values of k (e) for each bar and imposing the boundary conditions
u1 = u2 = 0, the previous system can be solved to give

Pl Pl P
u3 = ; u4 = ; R1 = R2 =
2EA EA 2

The axial forces in each bar are finally obtained as

EA P
Bar 1 : N (1) = (u3 − u1 ) =
l 2

EA P
Bar 2 : N (2) = (u3 − u2 ) =
l 2
2EA
Bar 3 : N (3) = (u4 − u3 ) = P
l

1.6 DIRECT ASSEMBLY OF THE GLOBAL STIFFNESS MA-

TRIX
The stiffness contribution of each individual bar can be directly assembled
in the global stiffness matrix by the following procedure. Consider a bar e
connecting two joints with global numbers i and m. Each term (i, m) of the bar
stiffness matrix contributes to the same position (i, m) of the global stiffness
matrix (Figure 1.11). Thus, the global stiffness matrix terms can be directly
computed by sistematically adding the contributions from the different bars
using information from the nodal numbers. This assembly process can be
programmed in a simple and general form [H5].

17
 Fig. 1.11: Contributions to the global stiffness matrix from an individual bar

Example: 1.2 Obtain the bandwidth of the stiffness matrix of the structure of Figure 1.4
with the node numbering indicated below.

- Solution

Numbering a)

The local numbering of each bar element is always taken from left to right
 (1) (1)

k11 k12 0 0
 (1) (1) (2) (3) (2) (3) 
k (k22 + k22 + k11 ) k21 k12 
K(e) =  21 (2) (2) 
 0 k12 k11 0 
(3) (3)
0 k21 0 k22

Numbering b)

18
  (1) (1)

k11 0 0 k12
 (3) (3) 
 0 k22 0 k21 
K(e) =  (2) (2) 
 0 0 k11 k12 
(1) (2) (1) (2) (3)
k21 0 k21 (k11 + k22 + k11 )

It is observed that in numbering (a) the bandwidth is 4, whereas in numbering (b)

the banded structure is lost and the bandwidth coincides with the number of terms of
the main diagonal (=6). These differences, although of little relevance in this simple
example, can be very significant in practical problems.

1.7 DERIVATION OF THE MATRIX EQUILIBRIUM EQUATI-

ONS FOR THE BAR ELEMENT USING THE PRINCIPLE
OF VIRTUAL WORK
One of the main steps in the matrix analysis of bar structures is the derivation
of the matrix equations for the single bar element. These equations express
the equilibrium between the loads acting at the bar joints and the displace-
ments of the joint points (Eq.(1.5)). For the simple axially loaded bar these
equations can be directly obtained using concepts from Strength of Materials.
For complex structures more general procedures are needed. Among these,
the Principle of Virtual Work (PVW) is the more powerful and widespread
technique. This well known principle states that: “A structure is in equilib-
rium under a set of external loads if after imposing to the structure arbitrary
(virtual) displacements compatible with the boundary conditions, the work
performed by the external loads on the virtual displacements equals the work
performed by the actual stresses on the strains induced by the virtual displace-
ments”.
The PVW is a necessary and sufficient condition for the equilibrium of the
whole structure or any of its parts [T7], [Z6]. Next, we will apply this technique
to the axially loaded bar of Figure 1.2. The PVW in this case is written as
Z Z Z
(e) (e) (e) (e)
δεσdV = δu1 R1 + δu2 R2 (1.13)
V (e)

(e) (e)
where δu1 and δu2 are, respectively, the virtual displacements of ends 1 and
2 of a bar with volume V (e) , and δε is the corresponding virtual strain which
(e) (e)
can be obtained in terms of δu1 and δu2 as
(e) (e)
δu2 − δu1
δε = (1.14)
l(e)

19
 Substituting the values of σ and δε of Eqs.(1.2) and (1.14) into (1.13) and
integrating the stresses over the cross sectional area of the bar gives
Z
1 h (e) (e)
i 1 h (e) (e)
i
(e) (e) (e) (e)
δu2 − δu1 (EA)(e) u2 − u 1 dx = δu1 R1 + δu2 R2
l(e) l(e) l(e)
(1.15)
(e)
Integrating over the bar length, assuming the Young modulus E and the
area A(e) to be constant, yields
µ ¶(e) h i µ ¶(e) h i
EA (e) (e) (e) EA (e) (e) (e)
u1 − u2 δu1 + u2 − u1 δu2 =
l l
(e) (e) (e) (e)
= δu1 R1 + δu2 R2 (1.16)
Since the virtual displacements are arbitrary, the satisfaction of Eq.(1.16)
(e) (e)
for any value of δu1 and δu2 requires that the terms multiplying each virtual
displacement at each side of the equation should be identical. This leads to
the following system of two equations
µ ¶(e) h i
(e) EA (e) (e) (e)
For δu1 : u1 − u2 = R1 (1.17a)
l
µ ¶
EA (e) h (e)
(e) (e)
i
(e)
For : δu2 u2 − u1 = R2 (1.17b)
l
which are the equilibrium equations we are looking for.
We can check that these equations, written in matrix form, coincide with
Eqs.(1.5) directly obtained using more physical arguments. The effect of a
distributed load can easily be taken
R
into account by adding to the right hand
side (r.h.s.) of Eq.(1.13) the term l(e) δub dx. Assuming a linear distribution of
the virtual displacements in terms of the end values, the expression of Eq.(1.6)
is recovered.
The PVW will be used throughout this book to derive the matrix equilib-
rium equations for the different structures studied.

1.8 DERIVATION OF THE BAR EQUILIBRIUM EQUATIONS

VIA THE MINIMUM TOTAL POTENTIAL ENERGY PRIN-
CIPLE
The equilibrium equations for a structure can also be derived via the prin-
ciple of Minimum Total Potential Energy (MTPE). The resulting equations
are identical to those obtained via the PVW. The applications of the MTPE
principle are generally limited to elastic materials for which simple forms of
the total potential energy can be derived. The PVW is more general as it
is applicable to non linear problems (including both material and geometrical

20
non linearities) and it is usually chosen as the starting variational form for
deriving finite element equations.
(e)
The total potential energy for a single bar e under point forces Ri at the
two ends is
(e) 1Z X2
(e) (e)
Π = εN dx − ui R i (1.18)
2 l (e)
i=1

Substituting into Eq.(1.18) the expression for the elongation ε and the axial
forces in terms of the end displacements, i.e.
(e) (e) (e) (e)
u −u (e) u2 − u1
ε = 2 (e) 1 , N = (EA) (1.19)
l l(e)
gives
   
1 Z  u2 − u1 
(e) (e) (e) (e) ³ ´
(e) (e)  u2 − u1  (e) (e) (e) (e)
Π = (EA) dx − u 1 R1 + u 2 R 2
2 l(e) l(e) l(e)
(1.20)
The MTPE principle states that a structure is in equilibrium for values of
the displacement making Π stationary. The MTPE also holds for the equilib-
rium of any part of the structure. The equilibrium condition is written for the
single bar as
∂Π(e)
(e)
=0 i = 1, 2 (1.21)
∂ui
i.e.  
1 Z
(e) (e)
∂Π(e) u − u (e)
(e)
= − (e) (EA)(e)  2 (e) 1  dx − R1 = 0
∂u1 l l (e) l
  (1.22)
1 Z
(e) (e)
∂Π(e) (e)  2 u − u 1  (e)
(e)
= (e) (EA) dx − R2 = 0
∂u2 l l(e) l(e)
For a linear material, the above equations simplify to
µ ¶
EA (e) h (e) (e)
i
(e)
u1 − u2 = R1
l (1.23)
µ ¶
EA (e) h (e) (e)
i
(e)
u2 − u1 = R2
l
Note the coincidence between the above end force-displacement equilibrium
equations and those obtained via the PVW (Eqs.(1.17)).
Eq.(1.20) can be rewritten as
l
Π(e) = [a(e) ]T K(e) a(e) − [a(e) ]T q(e) (1.24)
2
where K(e) , a(e) and q(e) are respectively the stiffness matrix, the joint dis-
placement vector and the joint equilibrium force vector for the single bar (see
Eqs.(1.5)).

21
 The stationarity of Π(e) with respect to the joint displacements gives
∂Π(e)
=0 → K(e) a(e) = q(e) (1.25)
∂a(e)
Eq.(1.25) is the same matrix equilibrium equation between the forces and
the displacements at the bar joints obtained in the the previous section (see
Eq.(1.5a)).
The total potential energy for a bar structure can be written in a form
analogous to Eq.(1.24) as
1
Π = aT Ka − aT f (1.26)
2
where K, a and f are respectively the stiffness matrix, the joint displacement
vector and the external joint force vector for the whole structure. The station-
arity of Π with respect to a gives
∂Π
=0 → Ka = f (1.27)
∂a
Eq.(1.27) is the global matrix equilibrium equation relating the displace-
ments and the external forces at all the joints of the structure. The global
matrix equations can be obtained by assembly of the contributions from the
individual bars, as previously explained.

Fig. 1.12: Forces and displacements at the end points of a plane pin-jointed bar

1.9 PLANE PIN-JOINTED FRAMEWORKS

We will briefly treat the case of plane pin-jointed frameworks as an extension
of the concepts previously studied. Each joint has now two degrees of free-
dom (d.o.f.) corresponding to the displacements along the two cartesian axes.

22
Eqs.(1.4) relating the joint displacements and the axial forces in local axes still
holds. However, the sum of the joint forces for the different bars sharing a
joint requires the force-displacement relationships to be expressed in a global
cartesian system x, y.
Let us consider a bar 1-2 inclined an angle α with respect to the global axis
x, as shown in Figure 1.12. For joint 1 we have
0(e)
R1 = Rx(e)1 cos α + Ry(e)
1
sen α
(1.28)
0(e) (e) (e)
u1 = u1 cos α + v1 sen α
where the primes denote the components in the direction of the local axis x0 .
In matrix form
½ ¾(e)
0(e) Rx1 (e)
R1 = [cos α, sen α] = L(e) q1
Ry1
½ ¾(e)
(1.29)
0(e) u1 (e)
u1 = [cos α, sen α] = L(e) u1
v2
(e) (e)
where u1 and q1 contain the two displacements and the two forces of joint 1
expressed in the global cartesian system x, y, respectively and L(e) = [cos α, sen α].
Analogous expressions can be found for node 2 as
0(e) (e) 0(e) (e)
R2 = L(e) q2 and u2 = L(e) u2 (1.30)
with h iT h iT
(e) (e) (e) (e)
q2 = Rx(e)2 , Ry(e)
2
and u2 = u2 , v2
From Figure 1.6, we deduce
h i µ ¶(e)
0(e) 0(e) (e) 0(e) 0(e) (e) EA
R1 = − R2 = k u1 − u2 with k = (1.31)
l
Multiplying Eq.(1.31) by [L(e) ]T and using Eqs.(1.29) and (1.30) the fol-
lowing two equations are obtained
h iT h iT
(e) (e) (e)
q1 = L(e) k (e) L(e) u1 − L(e) k (e) L(e) u2
h iT h iT (1.32)
(e) (e) (e) (e) (e)
q2 = − L k L(e) u1 + L (e) (e)
k L (e)
u2

In matrix form  (e)   (e) (e)  (e) 

 q1 
  K11 K12  u1 
 
= 


 (1.33a)

 (e) 
 (e) (e)  (e) 

q2 K21 K22 u2
where
h iT
(e) (e) (e) (e)
K11 = K22 = −K12 = −K21 = L(e) k (e) L(e) =
· ¸
cos2 α sin α cos α (1.33b)
= k (e)
sin α cos α sin2 α

23
 The assembly of the contributions of the individual bar members into the
global stiffness matrix follows precisely the steps explained in Section 1.2. Note
that each joint contributes a 2 ×2 matrix as shown in Figure 1.13. An example
of the assembly process is presented in Figure 1.14.

 (e) (e) 
K11 K12
K(e) = 



(e) (e)
K21 K22

µ ¶(e) · ¸
(e) (e) (e) (e) EA cos2 α sinα cos α
K11 = K22 = −K12 = −K21 =
l sinα cos α sin2 α

i m
 (e) (e)
  u 
Rxi
 
K11 K12 i
i
   vi   R 
     yi 
 = 
  
u 

 Rxm 
m
m K21
(e) (e)
K22 vm Rym

Fig. 1.13: Contributions to the global stiffness matrix from a general member of a
pin-jointed framework

1.10 TREATMENT OF PRESCRIBED DISPLACEMENTS AND

COMPUTATION OF REACTIONS
In this book we will not enter into the details of techniques for solving the
system of algebraic equations Ka = f . This is a problem typical of matrix
algebra and many well known solution procedures are available (i.e.: Gauss
reduction, Choleski, modified Choleski, Frontal; Profile, etc.) [H4], [P13],
[R2]. We will just treat here briefly the problem of prescribed displacements
and the computation of the corresponding reactions, as these are issues of
general interest for the study of this book.

24
 1 2
  2 3
K(1) K(1) 1  
K (1)
= 
11 12
 ; K(2)
11 K(2)
12
2
K (1)
K (1)
2 K (2)
=  
21 22
K(2)
21 K(2)
22 3

1 2 3
   (e)

   a1  
 f 

1 K (1)
11 K (1)
12 0 








1 


 
  − − − 
 
 − − − 

 (1)  (e)
Ka = 2  K21 K(1)
22 + K11
(2)
K12   a2  =  f2
(2)

=f
  
− − −
 



 − − −



3 
 
 

0 K(2) K(2)  (e) 
21 22
a 3 f3
(e) (e)
ai = [ui , vi ]T , fi = [Rxi , Ryi ]T , Kij as in ec.(1.33b)

Fig. 1.14: Plane pin-jointed framework. Equation of global equilibrium

Let us consider the following system of equations

k11 u1 + k12 u2 + k13 u3 + ... + k1n un = f1

k21 u1 + k22 u2 + k23 u3 + ... + k2n un = f2
k31 u1 + k32 u2 + k33 u3 + ... + k3n un = f3 (1.34)
.. .. .. .. ..
. . . . .
kn1 u1 + kn2 u2 + kn3 u3 + . . . + knn un = fn

where fi are external forces (which can be equal to zero) or reactions in points
where the displacement is prescribed.
Let us assume that a displacement, for example u2 , is prescribed to the

25
value u2 , i.e.
u2 = u2 (1.35)
There are two basic procedures to introduce this condition in the above
system of equations:

a) The second row and column of Eq.(1.34) are eliminated and the values
of fi in the right hand side are substituted by fi − ki2 u2 . That is, the
system of n equations with n unknowns is reduced in one equation and
one unknown as follows
k11 u1 + k13 u3 + ... + k1n un = f1 − k12 u2
k31 u1 + k33 u3 + ... + k3n un = f3 − k32 u2
.. .. .. .. .. (1.36)
. . . . .
kn1 u1 + kn3 u3 + . . . + knn un = fn − kn2 u2

Once the values of u1 , u3 , . . . , un are obtained, the reaction f2 is com-

puted by the following equation (in the case that the external force acting
at node 2 is equal to zero)

f2 = k21 u1 + k22 u2 + k23 u3 + . . . + k2n un (1.37)

If u2 is zero, the procedure remains the same, although the values of fi

are not modified and f2 is obtained by Eq.(1.37) with u2 = 0.

b) An alternative procedure which does not require the original system of

equations to be modified substantially, is to add a very large number to
the term of the main diagonal corresponding to the prescribed displace-
ments. The force term in the modified row is substituted by the value
of the prescribed displacement multiplied by the large number chosen.
Thus, if we have u2 = u2 we substitute k22 by k22 +1015 k22 (for instance),
and f2 by 1015 k22 × u2 . The final system of equations is
k11 u1 + k12 u2 + k13 u3 + ... + k1n un = f1
k21 u2 + (1 + 1015 )k22 u2 + k23 u3 + ... + 15
k2n un = 10 k22 u2
k31 u1 + k32 u2 + k33 u3 + ... + k3n un = f3
.. .. .. .. ..
. . . . .
kn1 u1 + kn2 u2 + kn3 u3 + . . . + knn un = fn

(1.38)
In this way, the second equation is equivalent to

1015 k22 u2 = 1015 k22 u2 or u2 = u2 (1.39)

which is the prescribed condition. The value of the reaction f2 is com-

puted “a posteriori” by Eq.(1.37).

26
1.11 INTRODUCTION TO THE FINITE ELEMENT METHOD
FOR ANALYSIS OF CONTINUUM SYSTEMS
Most problems in science and engineering are of continuous nature and can
not be naturally modelled by a collection of discrete element. Examples of
“continuous” problems in structural analysis are standard in civil, mechanical,
aeronautical and naval engineering. Amongst the more common we can list:
plates, foundations, roofs, containers, bridges, dams, airplane fuselages, car
bodies, ship hulls, mechanical components, etc. (Figure 1.15).

Fig. 1.15: Continuous structures: a) Dam, b) Shell, c) Bridge, d) Plate

Although a continuous system is inherently three-dimensional (3D), its

behaviour can be accurately described in some cases by one- (1D) or two-
dimensional (2D) structural models. This occurs, for instance, in the analysis
of plates in bending, where only the deformation of the plate mid-plane is
considered. Other examples are the case of prismatic bodies analyzed with 2D
or axisymmetric models.
The analytical solution of a “continuous” system is very difficult (generally
impossible), due to the complexities of the geometry, the boundary conditions,
the material properties, the type of loading, etc. This explains the need for
computational methods to analyse continuous systems.
The FEM is the simpler and more powerful computational procedure for the
analysis of continuous systems with arbitrary geometry and general material

27
properties subjected to any type of loading.
The FEM allows one the behaviour of a continuous system with an infinite
number of d.o.f. to be modelled by that of another one with approximately
the same geometrical and mechanical properties, but with a finite number of
d.o.f. The latter are related to the external forces by a system of algebraic
equations expressing the equilibrium of the system. We will find that the
basic finite element methodology is analogous to the matrix analysis techniques
studied for bar structures. These analogies between matrix analysis can be
clearly visualized in the analysis of the bridge shown in Figure 1.16. Without
entering into too the details, the basic steps in the finite element analysis are
the following:

Fig. 1.16: Analysis of a bridge by the finite element method

Step 1 : Starting with the geometrical description of the bridge, its supports
and the loading, the first step is to select a structural model. For example, we

28
could use a stiffened plate model (Chapter 13), a facet shell model (Chapter
10), or a 3D solid model (Chapter 7). The material properties must also be
defined, as well as the scope of the analysis (small or large displacements, static
or dynamic analysis, etc.). As mentioned earlier in this book we will focus on
linear static analysis only.

Step 2 : The structure is subdivided into a mesh of non-intersecting do-

mains termed finite elements (discretization process). The problem variables
(displacements) are interpolated within each element in terms of their values
at a known set of points of the element called nodes. The number of nodes
defines the approximation of the solution within each element. Some nodes are
placed at the element boundaries and they can be interpreted as linking points
between adjacent elements. However, nodes in the interior of the elements are
needed for higher-level approximations and, hence, the nodes do not have a
physical meaning as the connecting joints in bar structures. The mesh can
include elements with different geometries, such as 2D plate elements coupled
with 1D beam elements. The discretization process is an essential part of the
preprocessing step which includes the definition of all the analysis data. The
preprocessing step typically consumes a considerable amount of human effort.
The use of efficient preprocessing tools is therefore essential for the analysis of
practical structures in competitive times [F2,F,H,G].

Step 3 : The stiffness matrices K(e) and the load vectors f (e) are obtained
for each element. The computation of K(e) and f (e) is more complex than
for bar structures and it usually requires the evaluation of integrals over the
element domain.

Step 4 : The element stiffness and the load terms are assembled into the
overall stiffness matrix K and the load vector f for the structure.

Step 5 : The global system of linear simultaneous equations Ka = f is solved

for the unknown displacement variables a.

Step 6 : Once the displacements a are computed, the strains and the stresses
are evaluated within each element. Reactions at the nodes restrained against
movement are also computed.

Step 7 : Solving steps 3-6 requires a computer implementation of the FEM

by means of a standard or specially developed program.

Step 8 : After a successful computer run, the next step is the interpretation
and presentation of results. Results are presented graphically to aid their in-
terpretation and checking (postprocessing step). The use of specialized graphic
software is essential in practice [F2,F,G,H].

29
 Step 9 : Having assessed the finite element results, the analyst may con-
sider several modifications which may be introduced at various stages of the
analysis. For example, it may be found that the structural model selected is
inappropriate and hence it should be adequately modified. Alternatively, the
finite element mesh chosen may turn out to be too coarse to capture the ex-
pected stress distributions and must therefore be refined or a different, more
accurate element used. Round-off problems arising from ill-conditioned equa-
tions, the equation solving algorithm and the computer word length employed
in the analysis may cause difficulties and can require the use of double-precision
arithmetic or some other techniques. Input data errors which occur quite fre-
quently must be also corrected.

All these possible modifications are indicated by the feedback loop shown
in Figure 1.17 taken from [H5].
From the structural engineer’s point of view, the FEM can be considered as
an extension to continuous systems of the matrix analysis procedures for bar
structures. The origins of the FEM go back to the early 1940’s with the first
attempts to solve problems of 2D elasticity using matrix analysis techniques
by subdividing the continuum into bar elements [H11,M6]. In 1946 Courant
[C16] introduced for the first time the concept of “continuum element” to
solve 2D elasticity problems using a subdivision into triangular elements with
an assumed displacement field. The arrival of digital computers in the 1960’s
contributed to the fast development of matrix analysis based techniques, free
from the limitations imposed by the need to solve large systems of equations.
It was during this period what the FEM rapidly establish itself as a powerful
approach to solve many problems in mathematics and physics. It is interest-
ing that the first applications of the FEM were related to structural analysis
and, in particular, to aeronautical engineering [A10], [T12]. It is acknowledged
that Clough first used the name “finite elements” in relation to the solution
of 2D elasticity problems in 1960 [C8]. Since then the FEM had a tremen-
dous expansion in its application to many different fields. Supported by the
continuous upgrading of computers and by the increasing complexity of many
areas in science and technology, today the FEM enjoys a unique position as a
powerful technique for solving the most difficult problems in engineering and
applied sciences.
It would be an impossible task to list here all the significant published work
since the origins of the FEM. Only in 2006, scientific publications in this field
were estimated to number in excess of 25,000. The reader interested in the
historical aspects of the FEM should consult the reference list in Zienkiewicz
and Taylor [Z15] and the Encyclopedia of Computational Mechanics [].
Within the fields of engineering, applied mathematics and physics the prob-
lems to which the FEM is applied are basically the following:

1. Stationary equilibrium problems: those problems where the properties of

30
 Fig. 1.17: Flow chart of the analysis of a structure by the FEM

the system do not vary with time.

2. Eigenvalue problems. They are an extension of the stationary equilib-

rium problems for which some critical values of certain parameters are
obtained.

31
Chapter 2

FEM ANALYSIS OF 1D PROBLEMS.

APPLICATION TO THE POISSON
EQUATION

2.1 INTRODUCTION
The behaviour of continuum systems can in general be expressed in terms of
differential equations with their adequate boundary conditions. The objective
of this chapter is to present an overview of the solution of one-dimensional
(1D) partial differential equations with the FEM.
There are two general procedures for solving a differential equation: a) the
direct integration of the equation, which yields the so called analytical solution
(exact method ), and the approximate solution using numerical methods.
Numerical solution procedures for partial differential equations (PDE) be
classified as: a) those which are applied on the original PDE (for example,
the finite difference method), and b) those which work with an equivalent
integral expression. The FEM belongs to this second class of methods. The
two standard approaches of this kind are: a) variational methods, and b)
residual formulations.
The variational method is based on the search for the solution to the prob-
lem by solving an integral equation which represents a general property of
the system. A typical example is the integral expression obtained from the
minimum energy principle.
The residual formulations are based on Weighted Residual (WR) methods
such as the Point Collocation method, the Subdomain Collocation method,
the Galerkin method, the Minimum Least Squares method, etc.
The FEM can therefore be viewed as a procedure for solving the PDEs
governing a physical problem via a residual or variational formulation. Typ-
ically, residual methods are more general and advantageous than variational
methods and they will be the focus of this chapter.
This chapter presents different weighted residual techniques for solving a

33
PDE with the FEM. The model problem considered is the Poisson equation in
1D.

2.2 THE POISSON EQUATION

The Poisson equation expresses in mathematical form the behaviour of many
physical problems. For this reason it has been chosen here as the model prob-
lem for introducing the basis of the FEM.
The simplest form of the Poisson equation is as follows:

d2 φ 
In 1D: k 2 + Q(x) = 0 

dx 





d2 φ d2 φ 
In 2D: k + k + Q(x, y) = 0 k∇2 φ + Q = 0 (2.1)
dx2 dy 2 





2
dφ 2
dφ 2
dφ 

In 3D: k 2 + k 2 + k 2 + Q(x, y, z) = 0 

dx dy dz

where φ is the problem unknown, k is a parameter expressing a physical prop-

erty of the system, Q is a source term and ∇2 is the Laplace operator, i.e.
∂2 ∂2
∇2 = ∂x 2 + ∂y 2 in 2D. In general the parameter k can be different for each of

the space directions. The simplest isotropic form is chosen here.

Table 2.1 presents the meaning of the terms in Eq.(2.2) for some physical
problems.

Unknown, φ k Q
Heat conditions temperature heat internal heat
conductivity source
Flow through pressure permeability water
porous media head source
1D elasticity displacement Young modulus body force
x Area
Potential flow velocity density –
potential
Magnetostatics magnetic potential reductivity –
Torsion warping function shear modulus –
Torsion stress function (shear modulus) -1 twist
Gass diffusion concentration diffusivity –
Reynolds film pressure (film thickness)3/ lubricant
lubrication viscosity supply

Table 2.1: Meaning of the terms in the Poisson equation for some problems

34
 The form of the Poisson equation given in Eq.(2.2) assumes that k is con-
stant. In fact k can be a function of the position and even of the problem
unknown φ and its derivatives, as it happens in non-linear problems.
We present next examples of the derivation of the Poisson equation for
three specific problems: 1) heat transfer in a bar, 2) bar under axial forces and
3) seepage in a porous media.

Figure 2.1: Heat conduction in a bar. Heat flow balance in an infinitesimal

domain.

Example 2.1 Steady state heat conduction in a bar

Solution

Let us consider the bar of Figure 2.1 representing a 1D domain through which heat
is transfered via conduction effects in a steady manner. The temperature φ̄ at x = 0
is known, i.e.
φ = φ̄|x=0 (a)
and also the heat flux q̄ at x = l
q = q̄|x=l (b)

The balance of heat flux in a differential domain of the bar in expressed as (Figure
2.2b)
(q + dq) − q =0
| {z } |{z}
out going flux in-coming flux

dq = 0 (c)

The relationship between the heat flux q and the temperature is expressed by Fourier’s
law
dφ
q = −k (d)
dx
where k is the thermal conductivity parameter.

35
Figure 2.2: a) Bar under a heat source Q(x); b) Heat balance in an infinitesimal
domain.

Derivation of (d) gives µ ¶

dq d dφ
dq = dx = − k dx
dx dx dx
Substituting this equation into (c) gives
µ ¶
d dφ
k =0 heat balance equation (e)
dx dx

Note that the mathematical problem is well defined as we know

1. The governing differential equation (e)

2. The boundary conditions (a) and (b)

Let us know assume that the bar is subjected to an external heat source per unit
length Q (Figure 2.2a). A similar heat balance procedure on an infinitesimal domain
(Figure 2.2b) leads to the following equations

(q + dq) − q − Qdx = 0 (f )

dq
−Q=0 (g)
dx
Substituting (d) into (g) gives
µ ¶
d dφ
k +Q=0 (h)
dx dx

Expressions (e) and (h) are two forms of the Poisson equation. The form (e) (with
Q = 0) is usually known as the Laplace equation.
Note that for k being constant, the simplified form of the Poisson equation of Eq.(2.1)
is obtained.

36
Figure 2.3: Bar under axial forces. Equilibrium of force in an infinitesimal
domain.

The boundary conditions (a) and (b) completing the definition of the problem can be
summarized as


 Prescribed temperature: φ − φ̄ = 0 at x = 0 (i)


 
Boundary conditions  q − q̄ = 0 at x = l

 or (j)

 Prescribed heat flux
  dφ
k dx + q̄ = 0 at x = l

Condition (i) are known in mathematics a Dirichlet boundary condition, while condi-
tion (j) is known as Neumann boundary condition.

Example 2.2 Bar under axial forces

Solution

Let us consider the clamped bar of Figure 2.3, under an horizontal force x̄2 acting in
the free end and distributed axial forces b(x).
The governing equations of the problem are obtained following a similar process as in
Example 2.1. The equilibrium of axial forces is established in an infinitesimal domain
(Figure 2.3b), i.e.
dN
+ b(x) = 0 (a)
dx

The stress-strain relationship is expressed by Hooke’s law

du
σ = Eε = E (b)
dx
|{z}
ε

The axial force N is obtained by integrating the stress over the transverse cross-section
as
N = σA (c)

37
In (a) and (b) σ is the normal stress, E is the Young’s modulus, ε is the axial strain,
u the horizontal displacement and A the area of the transverse cross section.
From (b) and (c)

du
N = σA = EA (d)
dx
Substituting (d) into (a) gives
µ ¶
d du
EA +b=0 (e)
dx dx

The above equation expresses the equilibrium (or balance) of forces at each point of
the bar. It is therefore named the equilibrium equation for the bar.
The boundary conditions are

Prescribed displacement: u = 0 at x = 0

 N = x2 in x = l (f )
Prescribed force ó

EA du
dx − X2 = 0 in x = l

Eqs.(e) and (f) define the governing equations of the problem.

Note the analogy of Eq.(e) with Eq.(h) of the previous example. Both are Poisson
equations in 1D. The following lines show the analogies between the structural and
thermal problems.

Thermal problem – Axially loaded bar

temperature φ ←→
u displacement
conductivity k ←→
EA axial stiffness
heat source Q ←→
b distributed force
prescribed heat flow q̄ ←→
−X2 point force

The only difference in the analysis between the equation parameters in the thermal
problem and the axially loaded bar is the negative sign in the point force X2 . This
is a consequence of the proportionality between the displacement gradient and the
stress in structures, whereas in thermal problems the heat flux goes in the opposite
direction of the gradient, i.e.

du du
N = EA
|{z} =k
dx dx
k
dφ
q = −k
dx

38
Figure 2.4: a) Water flow through a porous bar.; b) Balance of water fluxes in
an infinitesimal domain

Example 2.3 Water flow in a porous medium

Solution

The balance of water flow in the infinitesimal doman of Figure 2.4 gives

(q + dq) − Qdx − q = 0

dq
−Q=0 (a)
dx
In (a) q and Q represent the internal water flux and the external source of water per
unit length, respectively.
The relationship between the water flow q and the pressure p is expressed by means
of Darcy law
dp
q = −k (b)
dx
where k is the permeability of the porous medium.
Note the analogy of above equations with the equivalent ones in the thermal and
structural problems. Substituting (b) into (a) gives
µ ¶
d dp
k +Q=0 (c)
dx dx

Boundary conditions

Prescribed pressure p − p̄ = 0 at x = 0

Prescribed flux q − q̄ = 0 at x = l (d)

or
dp
k dx + q̄ = 0 at x = l

39
 There is a perfect analogy between the governing equations in the thermal and porous
media flow problems, as expressed in the following lines:

Thermal problem – Porous media flow

temperature φ ←→
p water pressure
conductivity k ←→
k permeability
heat source Q ←→
Q water source
heat flux q ←→
q water flux

2.3 WEIGHTED RESIDUAL METHOD

The weighted residual method (WRM) is based on transforming the differential
equation which governs the problem in an equivalent integral expressions.
Let A(φ) be the governing differential equation (i.e. the Poisson equation)
µ ¶
d dφ
A(φ) = k + Q = 0 in Ω (2.2)
dx dx

where Ω is the analysis domain.

Also, let B(φ) be the equation expressing the boundary conditions

 φ − φ̄ = 0 in Γφ
B(φ) : (2.3)

k dφ
dx
+ q̄ = 0 in Γq

where Γφ is the Dirichlet boundary where the unknown function is prescribed

and Γq is the Neumann boundary where the flux incoming or outgoing the
domain is prescribed. The unknown φ and the parameters k, Q and q have
adequate meanings for each physical problem, as explained in the previous
section.
For instance, for the problem of Figure 2.1, we have
Ω: 0 ≤ x ≤ l (the bar length)
Γφ : x = 0 (left hand end point)
Γq : x = l (right hand end point)
The integral form equivalent to the governing equations given above is
obtained by multiplying the differential expressions A and B by arbitrary
weighting functions and integrating over the domain of each equation. Thus
Z I
W (x)A(φ)dΩ + W̄ (x)B(φ)dΓ = 0 (2.4)
Ω Γ

where W (x) and W̄ (x) are the weighting functions.

40
 It is clear that if above integral equation is satisfied for each pair of weight-
ing functions W and W̄ , the following conditions must be satisfied
A(φ) = 0 in Ω

B(φ) = 0 in Γ
Conversely, satisfaction of Eq.(2.2) and (2.3) indicates necessarily that the
integral equation (2.4) is satisfied for any weighting function.
This heuristic proof indicates that Eq.(2.4) is a necessary and sufficient
condition for the satisfaction of the governing equations. In other words, func-
tion φ satisfying the differential equations (2.2) and (2.3) also satisfies the
equivalent integral form (2.4). The solution of the problem via Eq.(2.4) or via
(2.2) and (2.3) are the starting point for method such as finite difference or
finite point procedures based on the satisfaction of the differential equations
in a finite set governing of points in the analysis domain. On the other hand,
Eq.(2.4) is the basis of the so-called integral methods, such as the finite element
method (FEM).
An interesting feature of Eq.(2.4) is the additive property of the integral.
This, if the integrals of Eq.(2.4) are computable, the integral from of Eq.(2.4)
can be written as
XZ XZ
W A(φ)dx + W̄ B(φ)dΓ = 0 (2.5)
e Ωe e Γe

where the sum extends over the collection of non-interesting domains (ele-
ments) covering the domain Ω and its boundary Γ. Eq.(2.5) is the basis of the
assembly process in the FEM.

2.3.1 Approximation of the unknown. Weighted Residuals

The numerical solution of the problem sought for an approximate value of the
unknown φ̂ such that
φ(x) ∼
= φ̂(x) (2.6)
The usual way for expressing the approximate solution is via a linear com-
bination of functions as
Xn
φ̂(x) = Ni (x)ai (2.7)
i=0

where ai are the unknown parameters and Ni (x) are functions of the indepen-
dent variable x. Typical choices for Ni , are
1. Monomials (Ni (x) = xi ):
n
X n
X
i
φ̂(x) = ai x = a0 + ai xi (2.8a)
i=0 i=1

41
 2. Fourier series functions (Ni (x) = sin ix, cos ix):
n n p p
X X X X
φ̂(x) = ai cos ix + βi sin ix = a0 + ai cos ix + βi sin ix
i=0 i=0 i=1 i=1
(2.8b)

3. Exponential functions (Ni (x) = ebix ):

n
X
φ̂(x) = ai ebix (2.8c)
0

Substituting the approximate function φ̂ into the integral expression (2.4) gives
Z I
W A(φ̂)dΩ + W̄ B(φ̂)dΓ = 0 (2.9)
Ω Γ

The above expression is an approximation of the integral form (2.4) and it

is usually called weighed residual expression. This name comes after observing
that A(φ̂) and B(φ̂) are the “residuals” of the approximate solution in the do-
main and the boundary, respectively. Thus, substituting φ̂ into the governing
equations we have
A(φ̂) = rΩ in Ω

B(φ̂) = rΓ in Γ
Hence, Eq.(2.9) can be written as
Z I
W rΩ dΩ + W̄ rΓ dΓ = 0 (2.10a)
Ω Γ

Obviously, if φ̂ is the exact solution, i.e. of φ = φ̂, then rΩ = 0 in Ω and

rΓ = 0 in Γ.
The value of the residuals rΩ and rΓ indicate the error in the satisfaction
of the governing differential equations due to the choice of the approximate
function φ̂.
Expressions (2.9) or (2.10a) can be therefore be interpreted as the integral
of the residuals of the differential equations weighted by the functions w(x)
and w̄(x). This explains the name of the weighted residual method.
The approximate solution is found by choosing a finite set of weighted
functions. Thus, by choosing a number of weighting functions equal to the
terms of the expansion (3), the following system of equations is obtained
Z X I X
Wi A( Nj aj )dΩ + W̄i B( Nj aj )dΓ = 0; i = 1, n (2.11)
Ω j Γ j

42
 The above expression can be written in matrix form, after an adequate
ordering, as
Ka = f (2.12)
where K is a square matrix which elements depend on the geometrical and
physical properties of the problem, a is the vector containing the n unknown
parameters and f is a vector depending on the prescribed value of the fluxes
and the unknown function in the boundary.
Solution of the algebraic equation system (2.12) yields the value of the
unknowns ai . The derivation of Eq.(2.12) for some specific problems will be
presented in the following sections.

2.3.2 Application of the WRM for the solution of the 1D heat conduc-
tion equation
The more typical particular cases of the WRM are

1. Point collocation method

2. Subdomain collocation method

3. Galerkin methoed

4. Least square method

Let us consider the solution of the 1D Poisson equation for k = constant,

i.e.
d2 φ
A(φ) = k +Q=0 in Ω (2.13)
dx2
More specifically, let us consider the solution of a heat conduction problem
in 1D, where k is the heat conductivity, Q is the external heat source per unit
length and φ is the temperature.
In the example to be solved (Figure 2.5) we will take

k=1

 1 for 0 ≤ x < l/2
Q: (2.14)

0 for l/2 < x ≤ l
Boundary conditions
½
φ = 0 in x = 0
B(φ) : (2.15)
φ = 0 in x = l
The solution process via the WRM follows the steps previously explained,
i.e.

43
Figure 2.5: Heat conduction in a bar of length l. Piece-wise distribution of the
heat source Q and prescribed temperature at the ends

• We define the governing differential equations of the problem

A(φ) = 0 in the domain Ω

• and the boundary conditions

B(φ) = 0 in the boundary Γ of Ω

The solution process is as follows

1. The unknown φ(x) is approximated by a function φ̂(x) as

n
X
φ∼
= φ̂ = Ni (x)ai (2.16)
i=1

where Ni (x) is a known “shape functions” and ai i = 1, n are n the

unknown parameters. Substituting Eq.(2.16) into Eqs.(2.15) yields the
residuals (errors) in the domain and the boundary, i.e.
A(φ̂) = rΩ 6= 0 in Ω
B(φ̂) = rΓ 6= 0 in Γ

2. We impose that the integral of the residuals, weighted with n test func-
tions Wi and W̄i is zero, i.e.
Z Z
Wi (x)A(φ̂)dΩ + W̄i (x)B(φ̂)dΓ = 0 i = 1, 2, . . . , n (2.17)
Ω Γ

3. the parameters ai are found by solving the algebraic system of n equa-

tions with n unknowns resulting from Eq.(2.17) and written as
Ka = f

44
2.3.3 Global definition of the shape functions
The shape functions Ni (x) can be defined “globally” over the whole domain Ω,
or else “locally” in subdomains Ωi . Next we will consider the global definition
of Ni (x).
For the solution of Eqs.(2.13)–(2.15) we will assume first an approximation
of φ using Fourier series
n
X πxi
φ(x) ' φ̂(x) = ai sin ; i = 1, 2, . . . , n (2.18)
i=1
l

πxi
Ni (x) = sin (2.19)
l
From the expression of Ni (x) we observe that

• they are continuous over the domain Ω,

• they satisfy the boundary conditions (2.15), i.e.

¾
Ni (0) = sin(0) = 0
∀i
Ni (l) = sin(iπ) = 0
Hence
P
n
φ̂(0) = ai Ni (0) = 0
i=1
Pn
φ̂(l) = ai Ni (l) = 0
i=1

This gives

B(φ̂) = rΓ (x) = 0 in the boundary Γ (x = 0, x = l) (2.20)

Also, from Eqs.(2.13) and (2.18)

µn ¶ µn ¶
P πxi d2
P πxi
A(φ̂) = A ai sin l = k dx2 ai sin l + Q =
i=1 i=1
(2.21)
n h
P ¡ iπ ¢2 i
= −ai l
sin πxi
l
+Q
i=1

Eq.(2.18) are re-written (for rΓ = 0) as

Z l à !
d2 φ̂
Wi (x) + Q dx = 0; i = 1, 2, . . . , n
0 dx2
| {z }
rΩ

45
 Substituting Eq.(2.18) into above integral gives

Z l "X
n µ ¶2 #
πj πx
− aj sin j + Q Wi (x)dx = 0; i = 1, 2, . . . , n
0 1
l l
 
n 
Z Z l
P 
³ π ´2 l
πx  
 j j
a Wi (x) sin j dx = Wi (x)Qdx, i = 1, 2, . . . , n
j=1  l 0 l  0
| {z } | {z }
Kij fi
(2.22)
In compact form

n
X
Kij aj = fi : i = 1, 2, . . . , n (2.23)
j=1

which leads to a system of n linear equation with n unknowns.

For i = 1

Z l µ ¶2 Z l
¡ π ¢2 πx 2π 2πx
a1 l
W1 sin dx + a2 W1 sin dx + . . .
0 l l 0 l
Z Z l
¡ nπ ¢2 l nπx
· · · + an l W1 sin dx = W1 Qdx
0 l 0

For i = 2

Z l µ ¶2 Z l
¡ π ¢2 πx 2π 2πx
a1 l
W2 sin dx + a2 W2 sin dx + . . .
0 l l 0 l
Z Z l
¡ nπ ¢2 l nπx
· · · + an l W2 sin dx = W2 Qdx
0 l 0

For i = n

³ π ´2 Z l µ ¶2 Z l
πx 2π 2πx
a1 Wn sin dx + a2 Wn sin dx + . . .
l 0 l l 0 l
³ nπ ´2 Z l
nπx
Z l
· · · + an Wn sin dx = Wn Qdx
l 0 l 0

46
 The above equations can be written in matrix form as
Z l Z l Z l 
πx 2πx 2 nπx
 W1 sin l dx 4 W1 sin
l
dx . . . n W1 sin
l
dx   
 Z0 Z 0 Z 0   a1 
 l l l  
π2 
πx 2πx
dx . . . n2
nπx
dx    
 W2 sin dx 4 W2 sin W2 sin  a2
 0 l 0 l 0 l  .. =
l2  .
. .
. .
.  . 

 . . .  
Z l Z Z  an 
 πx l
2πx l
nπx 
Wn sin dx 4 Wn sin dx . . . n2 Wn sin dx
0 l 0 l 0 l
Z l 

 

 QW1 dx  


 


 Z 0
l 


 QW dx  
2
= 0

 .. 


 . 

Z l
 


 


 QWn dx 
0

or
j
 ↓    
K11 K12 ... K1n  a1   f1 
K  
   
 21 K22 ... K2n 
 a2 f2

 . .. ..  = (2.24)
i → .. . Kij .  ai   fi 
     
Kn1 Kn2 ... Knn | a{zn } | f{zn }
| {z }
K a f

Ka = f (2.25)

where a term Kij of K can be expressed as

µ ¶2 Z l
jπ πx
Kij = Wi (x) sin j dx (2.26)
l 0 l

Similarly a term fi of f is
Z l
fi = Wi (x)Q(x)dx (2.27)
0

Expressions (2.26) and (2.27) are a generic form of the terms of K and f ,
respectively, which favours the programming of all the expressions.
The remaining step is the evaluation of all matrices and vectors after defin-
ing the weighted functions Wi . The selection of Wi . yields different forms of
the Weighted Residual Method.

47
2.3.4 Point Collocation Method
This method consists in defining n points in the analysis domain and choosing
the Dirac Delta functions as the weighting function. Thus,

Wi (x) = δ(x − xi ), i = 1, 2, . . . , n (2.28)

Function δ has the property that for x 6= xi , δi (x) = 0 and satisfies the
conditions Z ∞
δ(x − xi )dx = 1 (2.29a)
−∞
Z l
f (x)δ(x − xi )dx = f (xi ) (2.29b)
0

Thus
Z l Z l ¯
¯
Wi (x)A(φ̂)dx = δ(x − xi )A(φ̂)dx = A(φ̂)¯¯ = 0, i = 1, 2, . . . , n
0 0 x=xi
(2.30)
Choosing the Dirac Delta functions as the weighting function, it is equiva-
lent to making A(φ̂) equal to zero at each point x = xi (i = 1, . . . , n). Hence,
the procedure is equivalent to imposing the vanishing of the residual at each
one of the sampling points chosen. From Eq.(2.30) we deduce that there is no
need to perform an integration for obtaining the system of algebraic equations.
We will solve the previous problem for one and two terms of the Fourier
series expansions (2.18).
Solution with Wi = δi for n = 1

φ̂ = N1 (x)a1
We take
xi = x1 = l/2
From Eq.(2.28)
W1 (x) = δ(x − l/2)
From Eq.(2.27)
Z l µ ¶
l 1
fi = f1 = δ(x − l/2)Q(x)dx = Q =
0 2 2
This value is deduced from Figure 2.5. At a point infinitely close to x = l/2
from the left Q takes a unit value, where Q = 0 for a point to the right of
x = l/2. This difficulty is overcome by taking Q = l/2 for x = l/2.
On the other hand
³ π ´2 Z l µ x − l ¶ πx ³ π ´2 πl ³ π ´2
K11 = δ sin dx = sin =
l 0 2 l l l2 l

48
 From (2.24)
K11 a1 = f1

π2 1
2
a1 = =⇒ a1 = l2 /2π 2
l 2
The approximate function sought is therefore

l2 π πx
φ̂(x) = 2
sin x −→ φ̂ = 0, 0507 l2 sin
2π | {zl }
|{z} l
a1 N1 (x)

We can verify that the residual rΩ vanishes at x1 = l/2

¯ µ ¶¯ ¯
¯ d2 φ̂ d2 l2 π ¯¯ ¯
A(φ̂)¯¯ = 2 +Q= 2 sin x ¯ + Q(x)¯
¯ =
x=x1 = 2l dx dx 2π 2 l x=l/2 x=l/2

µ ¶
l2 π 2 π l 1 1
= − 2 2 sin x|x=l/2 + Q =− + =0
2π l l 2 2 2

Solution with Wi = δi for n = 2

φ̂(x) = N1 (x)a1 + N2 (x)a2

Taking

x1 = l/4
x2 = 3/4l

W1 (x) = δ(x − l/4)

W2 (x) = δ(x − 3/4l)

From Eq.(2.23) we obtain

For i = 1

K11 a1 + K12 a2 = f1
For i = 2
K21 a1 + K22 a2 = f2
The algebraic equation system is
· ¸½ ¾ ½ ¾
K11 K12 a1 f1
=
K21 K22 a2 f2

49
For i = 1
Z l Z l µ ¶ √ 2
π2 πx π2 l πx π2 πl 2π
K11 = l2
W1 sin dx = 2 δ x− sin dx = 2 sin =
0 l l 0 4 l l l4 2 l2
Z l Z l µ ¶
4π 2 2πx 4π 2 l 2πx 4π 2 2π 4π 2
K12 = l2
W1 sin dx = 2 δ x− sin dx = 2 sin l = 2
0 l l 0 4 l l 4l l
Z l Z l µ ¶ µ ¶
l l
f1 = Q(x)W1 (x)dx = Qδ x − dx = Q =1
0 0 4 4

For i = 2
Z l Z l µ ¶ √ 2
π2 πx π2 3l πx π2 3 πl 2π
K21 = l2
W2 sin dx = 2 δ x− sin dx = 2 sin =
0 l l 0 4 l l 4 l 2 l2
Z l Z l µ ¶
4π 2 2πx 4π 2 3l 2πx 4π 2 2π 3l 4π 2
K22 = l2
W2 sin dx = 2 δ x− sin dx = 2 sin =− 2
0 l l 0 4 l l 4 l l
Z l Z l µ ¶ µ ¶
3l 3l
f2 = Q(x)W2 (x)dx = Qδ x − dx = Q =0
0 0 4 4

Therefore "√ #½ ¾ ½ ¾ ½ √
π2 2
4 a1 1 a1 = l2 / 2π 2
√2 = ⇒
l2 2
2
−4 a2 0 a2 = l2 /8π 2

The approximate solution is

1 l2 πx l2 2πx
φ̂(x) = √ 2 sin + 2 sin
2π l 8π l
| {z } | {z } |{z} | {z }
a1 N1 (x) a2 N2 (x)

We can verify that

A(φ̂) = 0| x1 =l/4
x2 = 3
4l

½ ¾
d2 φ̂ 1 l2 π 2 π 1 l2 4π 2 2π
A(φ̂) = + Q(x) = − √ 2 2 sin x + sin x + Q(x) |x1 ,x2
dx2 2π l l 8 π 2 l2 l
à √ !
1 2 1
A(l/4) = − √ + 1 + Q(l/4) = −1 + 1 = 0
2 2 2
à √ ! µ ¶
1 2 1 3
A(3/4l) = − √ + (−1) + Q l = −0 + 0 = 0
2 2 2 4

50
2.3.5 Subdomain collocation method
The method consists in subdividing the domain Ω into as many subdomains
as unknown parameters ai . The test functions Wi are then chosen so that they
take a unit value in each subdomain Ωi and zero in the rest. Thus,

Wi (x) = 1 ∀x ∈ Ωi
(2.31)
Wi (x) = 0 ∀x 6∈ Ωi

Applying the WRM gives

Z
A(φ̂)dx = 0, i = 1, 2, . . . , n (2.32)
Ωi

The previous equation is equivalent to making zero the integral of the

residual in each subdomain Ωi .
As in the previous case we will found φ̂(x) for one and two terms of the
Fourier series expansion (2.18)
Solution for n = 1

φ̂(x) = N1 (x)a1
Ωi ≡ Ω
i.e., the subdomain Ωi is taken to coincide with the whole domain Ω : [0, l]

W1 = 1 0≤x≤l

From Eq.(2.32) we deduce

K11 a1 = f1
where Z Z
l l
π2 πx π2 πx 2π
K11 = 2 W1 sin dx = 2 sin dx =
l 0 l l 0 l l

Z l Z l Z l/2 Z l
f1 = W1 Q(x)dx = Q(x)dx = Q(x) dx + Q(x)dx = l/2
0 0 0 | {z } l/2
1

Therefore

K 11 f1
z}|{ z}|{
2π l l2
a1 = =⇒ a1 =
l 2 4π

51
and
l2 πx
φ̂(x) = sin
4π | {z l }
|{z}
a1 N1 (x)

Solution for n = 2

φ̂(x) = N1 (x)a1 + N2 (x)a2

We take as subdomains the two intervals

[0, l/2] and [l/2, l]

hence


1 0 ≤ x ≤ l/2
W1 (x) :


0 l/2 < x ≤ l



0 0 ≤ x ≤ l/2
W2 (x) :


1 l/2 < x ≤ l

The above is equivalent to imposing that the integral of the residual is zero
in each of the two subdomains chosen.
In Eq.(2.23) we have for n = 2, i = 1, 2
¯ · ¸½ ¾ ½ ¾
i = 1; K11 a1 + K12 a2 = f1 ¯¯ K11 K12 a1 f1
=⇒ =
i = 2; K21 a1 + K22 a2 = f2 ¯ K21 K22 a2 f2

For i = 1
Z l Z l/2
π2 πx π2 πx π
K11 = l2
W1 sin dx = 2 sin dx =
0 l l 0 l l
Z l Z l/2
4π 2 2πx 4π 2 2πx 4π
K12 = l2
W1 sin dx = 2 sin dx =
0 l l 0 l l
Z l Z l/2
l
f1 = W1 Qdx = Qdx =
0 0 2

52
For i = 2
Z l Z l
π2 πx π2 πx π
K21 = l2
W2 sin dx = 2 sin dx =
0 l l l/2 l l
Z l Z l
4π 2 2πx 4π 2 2πx 4π
K22 = l2
W2 sin dx = 2 sin dx = −
0 l l l/2 l l
Z l Z l
f2 = W2 Qdx = Qdx = 0
0 l/2

Hence · ¸½ ¾ ½ ¾ ½
π 1 4 a1 l/2 a1 = l2 /4π
= ⇒
l 1 −4 a2 0 a2 = l2 /16π
The approximate solution is now

l2 πx l2 2πx
φ̂(x) = sin + sin
4π | {z l } |{z}
|{z} 16π | {z l }
a1 N1 (x) a2 N2 (x)

2.3.6 Galerkin method

This popular method consists in taking as weighting functions the approxima-
tion functions Ni , i.e.
Wi (x) ≡ Ni (x) (2.33)
The WRM is written as
Z Z
Ni A(φ̂)dΩ + Ni B(φ̂)dΓ = 0 (2.34)
Ω Γ

In the problem we are solving the integral over the boundary Γ is zero. Next
we will solve the problem by taking one and two terms of the approximation
(2.18) as in the previous cases.
Solution for n = 1
φ̂(x) = N1 (x)a1
πx
W1 (x) = N1 (x) = sin
l
Hence
Z l Z l
π2 πx π2 πx π2
K11 = 2 W1 sin dx = sin2 dx =
l 0 l l 0 l 2l
Z l Z l/2
πx
f1 = W1 Qdx = sin 1dx = l/π
0 0 l

53
 Therefore
π2 l 2l2
a1 = → a1 = 3
2l
|{z} π
|{z} π
K11 f1
The final result is
2l2 πx
φ̂(x) = 3
sin
π | {z l }
|{z}
a1 N1 (x)

Solution for n = 2

φ̂(x) = N1 (x)a1 + N2 (x)a2

πx
W1 (x) = N1 (x) = sin
l
2πx
W2 (x) = N2 (x) = sin
l
The terms of the system (2.23) are
For i = 1
Z l
R
π2 l πx π2 2 πx π2
K11 = l2 0 W1 dx sin l = l2 sin dx =
0 l 2l
Z l
4π 2
Rl 2πx 4π 2 πx 2πx
K12 = l2 0
W1 sin l
dx = l2
sin sin dx = 0
0 l l
Z l/2
Rl πx l
f1 = 0
W1 Qdx = sin 1dx =
0 l π
For i = 2
Z l Z l
π2 πx π2 2πx πx
K21 = l2
W2 sin dx = 2 sin sin dx = 0
0 l l 0 l l
Z l Z l
4π 2 2πx 4π 2 2πx π2
K22 = l2
W2 sin dx = 2 sin2 dx = 2
0 l l 0 l l
Z l/2
Rl 2πx l
f2 = 0
W2 Qdx = sin 1dx =
0 l π
The resulting matrix expression is
· ¸½ ¾ ½ ¾ ½
π 2 1/2 0 a1 l/π a1 = 2l2 /π 3
= ⇒
l 0 2 a2 l/π a2 = l2 /2π 3
The approximate solution is
2l2 πx l2 2πx
φ̂(x) = 3
sin + 3
sin
π | {z l } |{z}
|{z} 2π | {z l }
a1 N1 (x) a2 N2 (x)

54
2.3.7 Least square method
The least square (LSQ) method consists in finding the unknown parameters
by minimizing the integral of the residual squared in the analysis domain. The
minimization process implies the derivation of the integral with respect to each
unknown. This leads to an algebraic system of n equations with n unknowns,
as in the previous methods.
The LSQ method can be interpreted as a particular case of the WRM with
W = A(φ) and Ŵ = B(φ).
The expression to be minimized is
Z Z
2
I= [A(φ̂)] dΩ + [B(φ̂)]2 dΓ (2.35)
Ω Γ

Making
∂I ∂I ∂I
δI = δa1 + δa2 + · · · + δan = 0 (2.36)
∂a1 ∂a2 ∂an
Satisfaction of this equation implies

 ∂I

 =0

 ∂a1



 ∂I
=0
∂a2 (2.37)

 ..

 .



 ∂I = 0

∂an

For the 1D heat conduction problem in hand the minimization process can
be written as
Z l
∂
[A(φ̂)]2 dx = 0; i = 1, 2, . . . , n
∂ai 0
or

Z l" #2 Z l
"
# " #
∂ d2 φ̂ d2 φ̂ ∂ d2 φ̂
+Q dx = 2 +Q dx
∂ai 0 dx2 0 dx2 ∂ai dx2
Z l " #" #
∂ d2 φ̂ d2 φ̂
= 2 + Q dx
0 ∂ai dx2 dx2
| {z }| {z }
Wi (x) A(φ̂)
Z l
= Wi (x)A(φ̂)dx = 0 i = 1, 2, . . . , n
0

55
 The outcome of the minimization process can be interpreted as a form of
the WRM with
h 2 i · µn ¶¸
∂ d φ̂ ∂ d2
P jπx
Wi = 2 ∂ai dx2 = 2 ∂ai dx2 aj sin l =
1
h ¡ ¢2 ¡ 2π ¢2
= 2 ∂ai −a1 πl sin 1πx
∂
l
− a 2 l
sin 2πx
l
− ...
¡ iπ ¢2 ¡ ¢ i
π 2
−ai l sin iπx l
− · · · − a n n l
sin nπx
l
=
¡ iπ ¢2
= −2 l sin iπx l

Solution for n = 1

φ̂(x) = N1 (x)a1
2π 2 πx
W1 = − 2
sin ; i=1
l l
After a little algebra we find
Z Z
π2 l πx π 2 l 2π 2 πx π4
K11 = 2 W1 sin dx = 2 − 2 sin2 dx = − 3
l 0 l l 0 l l l
Z l Z l/2
2π 2 πx 2π
f1 = W1 Qdx = − 2
sin dx = −
0 0 l l l
Finally we obtain

π4 π 2l2
− a1 = −2 ⇒ a1 =
l3 l π3
Hence
2l2 πx
φ̂(x) = sin
π 3 | {z l }
|{z}
a1 N1 (x)

Solution for n = 2

φ̂(x) = N1 (x)a1 + N2 (x)a2

π2 πx
W1 = −2 2
sin ; i=1
l l
8π 2 2πx
W2 = − 2 sin ; i=2
l l

56
 After some algebra we find
For i = 1
Z l Z l
π2 πx π2 2π 2 2 πx π4
K11 = l2
W1 sin dx = 2 − sin dx = −
0 l l 0 l2 l l3
Z l Z l
4π 2 2πx 4π 2 2π 2 πx 2πx
K12 = l2
W1 sin dx = 2 − 2
sin sin dx = 0
0 l l 0 l l l
Z l Z l/2
2π 2 πx 2π
f1 = W1 Qdx = − 2
sin 1dx = −
0 0 l l l
For i = 2
Z l Z l
π2 πx π2 8π 2 2πx πx
K21 = l2
W2 sin dx = 2 − 2
sin sin dx = 0
0 l l 0 l l l
Z l Z l
4π 2 2πx 4π 2 8π 2 2 2πx π4
K22 = l2
W2 sin dx = 2 − sin dx = −16
0 l l 0 l2 l l3
Z l Z l/2
8π 2 2πx 8π
f2 = W2 Qdx = − 2
sin 1dx = −
0 0 l l l
The final system of equations is
· ¸½ ¾ ½ ¾ ½
π 4 −1 0 a1 −2π/l a1 = 2l2 /π 3
= ⇒
l3 0 −16 a2 −8π/l a2 = l2 /2π 3
The approximate solution is
2l2 πx l2 2πx
φ̂(x) = 3
sin + 3
sin
π l 2π l
Note that the solution coincides with that obtained with the Galerkin
method. This coincidence is fortuitous and it can not be generalized to other
problems.
Figures 2.6 and 2.7 show the results obtained with the different WRM pre-
sented using one and two terms of the Fourier series used for the approximation
(Eq.2.18). Note the best accuracy of the Galerkin method, which explains its
popularity of this method for application with the FEM.

2.4 GENERAL SOLUTION PROCEDURE

Let is consider the equation
µ ¶ Ω
z }| {
d dφ
A(φ) = k +Q=0 en 0 ≤ x ≤ l (2.38)
dx dx

57
Figure 2.6: Exact and approximate solution using Fourier series (one term) for
the 1D heat conduction problem of Figure 2.4

which governs the heat conduction problem of Figure 2.8 with the boundary
conditions

 φ − φ̄ = 0; en x = 0 (Γφ )
B(φ) (2.39)

k dφ
dx
+ q̄ = 0; en x = l (Γq )

The unknown φ is approximated by

n
X
φ ' φ̂ = Ni (x)ai (2.40)
i=1

58
Figure 2.7: Exact and approximate solution using Fourier series (two terms)
for the 1D heat conduction problem of Figure 2.4

Application of the WRM gives

 
Z l " Ã ! #  Ã !
d dφ̂  dφ̂ 
Wi k + Q dx + 
 W̄ i k + q̄ 
 =0 (i = 1, 2, . . . , n)
0 dx dx  dx 
| {z } | {z }
A(φ̂) B(φ̂) x=l
(2.41)
Note that we have not included in above expression the term in the Dirichlet
boundary Γ(x=0) . The reason is that in the following we will choose functions
φ̂ that satisfy the Dirichlet boundary condition.

59
 Figure 2.8: The 1D heat conduction problem in a bar

Introducing Eq.(2.40) into (2.41) gives

Z l · µ ¶¸
d dN1 dN2 dNn
Wi k a1 + a2 + · · · + an dx+
0 dx dx dx dx
· µ ¶ ¸
dN1 dN2 dNn
+ W̄i k a1 + a2 + · · · + an + q̄ +
dx dx dx x=l

Z l
+ Wi Qdx = 0 (i = 1, 2, . . . , n)
0

The above equations is a system of n equations with n unknowns that can

be written in matrix form as
    
K11 K12 . . . K1n   a1 
   f1 

 K21 K22 . . .  a2 
    f2 
 K2n 


 ..  .. = . (2.42)
 . .
  
  .. 
 

Kn1 Kn2 . . . Knn an  fn 
| {z } | {z } | {z }
K a f

where Zµ ¶ · ¸
l
d dNj dNj
Kij = Wi k dx + W̄i k (2.43)
0 dx dx dx x=l
Z l
fi = − Wi Qdx − [W̄i q̄]x=l
0
The selection of the test functions Wi leads to the different modalities of
the WRM. For Wi = Ni we obtain the Galerkin method. Note that in general

Kij 6= Kji
and, therefore, matrix K is non symmetrical.
Application of the Galerkin method in the example of previous section lead
to a symmetric expression of K. This a consequence of the particular form of
the Fourier expansion chosen for the approximation.

60
2.4.1 Conclusions
We point out some key aspects of the solution process explained in the previous
section.

1. The unknown function is approximated by the sum of product of known

functions Ni (x) and unknown parameters ai
n
X
φ(x) ∼
= φ̂(x) = Ni (x)ai (2.44)
i=1

2. The governing equations of the problem are written in an integral form

via the WRM as
Z Z
Wi A(φ̂)dx + W̄i B(φ̂)dx = 0 (2.45)
Ω Γ

3. Substituting the approximation of the unknown into the previous integral

form leads to a system of algebraic equations

Ka = f

Solution of above system gives the value of the parameters ai , which define
the approximate expression of the unknown function φ.

2.5 INTEGRABILITY CONDITION

Integration of the mth-derivative of a function f (x) requires the continuity
R m
of its first m − 1 derivatives. In other words, the integral Ω d dxfm(x) dx will be
m−1 f (x)
compatible if the derivative d dxm−1 is continuous.
This property is clarified in the example of Figure 2.9. The figure shows a
linear function f (x). The integral exists in the interval [0, xi + 1] and coincides
with the area of the triangular region shown. Figure 2.9b shows the first
0
derivative f (x). Note that this function is discontinuous and its integral is
equal to the sum of the areas of the two rectangular regions shown. Figure
00
2.9c shows f (x). The function is not integrable and this is a consequence
0
of f (x) not being continuous, in agreement with the integrability criterium
stated above.
A function is C m continuous if the function and its m first derivatives are
continuous. This C 1 continuous function implies that both the function and its
first derivatives are continuous, C 0 continuity indicates that just the function
is continuous. C −1 continuity implies that the function is discontinuous.

61
 Figure 2.9: Integral of a linear function and its two first derivative

2.6 WEAK FORM OF THE WEIGHTED RESIDUAL METHOD

Let us consider once more the 1D Poisson equation
µ ¶
d dφ
k +Q=0
dx dx
with boundary conditions
φ − φ̄ = 0 en x = 0
dφ
k + q̄ = 0 en x = l
dx
The integral form is obtained by multiplying the above equations by test
functions and integrating over the domain. Thus
Z l · µ ¶ ¸ · µ ¶¸
d dφ dφ
W k + Q dx + W̄ k + q̄ =0 (2.46)
0 dx dx dx l

From this expression we note

• The derivative of φ is of order 2. This requires C 1 continuity for φ and,

therefore, the shape functions must also be C 1 continuous.
• The derivative of k appears in the integral. This requires that is con-
tinuous (C 0 continuity). This is an obvious inconvenience for problems
where the analysis domain contains material with different properties.
• No derivative is applied on W . This implies that W can be discontinuous
(C −1 continuity). Above requirements are clearly non-symmetrical with
request to φ and W . Thus, while continuity of the first derivative is
required for φ, no continuity requirement is demanded for φ. This lack of
symmetry leads generally to a non symmetric expression of the algebraic
equation system resulting from the discretization process.

62
 This “problem” can be overcome by an integration by parts of the second
derivative term in the integral expression of the WRM. Recalling the integra-
tion by parts rule Z Z
l l
udv = [uv]l0 − vdu
0 0
and applying this rule to the term involving the second derivative of φ in the
expression of the WRM (Eq.(2.46)) gives (for u = W and v = k dφ dx
)
Z l µ ¶ · ¸l Z l
d dφ dφ dW dφ
W k dx = W k − k dx (2.47a)
0 dx dx dx 0 0 dx dx
Substituting this expression into the WRM gives
Z l Z l · ¸l · µ ¶¸
dW dφ dφ dφ
− k dx + W Qdx + W k + W̄ k + q̄ = 0 (2.47b)
0
dx dx 0
dx 0 dx l

Note that the continuity requirements for φ, W and k have changed. Thus,
the first derivatives of φ and W are just now involved, which simply requires
C 0 continuity for these two functions.
Also, the derivative of the k has disappeared and k can now be a discon-
tinuous function. The changes in the continuity requirement are summarized
in Table 2.2.

Continuity requirements
Original integral form Weak integral form
φ C C0
−1
W C C0
k C0 C −1

Table 2.2: Continuity requirements for φ, W and k in the original integral

form and the weak form

We recall that C −1 continuity means that the function can be discontinuous.

Eq.(2.47b) is called the weak integral form (or simple the weak form). This
name comes from the fact that we have restricted the field for selecting the
test functions W . Note that the original integral form we allowed any choice
for W , while the weak form is restricted to functions W which are continuous.

2.6.1 Natural boundary condition

Let us choose for convenience W̄ = −W in the weak form (2.47b). Simplifica-
tion and rearranging of terms leads to
Z l Z l · ¸
dW dφ dφ
k dx = W Qdx − W k − [W q̄]l (2.48)
0 dx dx 0 dx 0

63
 Recalling that q = −k dφ
dx
gives
Z l Z l
dW dφ
k dx = W Qdx + [W q]0 − [W q̄]l (2.49)
0
dx dx 0

We note the following points:

1. The variable φ has disappeared from the Neumann boundary where the
heat flux is prescribed ((x = l).

2. If the heat flux at x = l is prescribed to a zero value (q̄ = 0), then all
the Neumann boundary terms disappear from the weak form (2.49). The
conditions of zero prescribed flux at the boundary are typically termed
natural boundary condition.

3. q0 is the incoming flux into the domain by the left end (x = 0) where the
value of φ is prescribed. Thus q0 can be interpreted as an end “reaction”
(using a terminology from structural mechanics) corresponding to the
prescribed value of the unknown. The value of q0 can be computed “a
posteriori”, once the approximate solution for φ has been found.

2.6.2 Discretization of the weak form

The weak form will be in the following the starting expression for deriving the
discretized equation in the FEM. Thus, substituting the approximation for φ
defined as n
X
φ∼= φ̂ = Nj aj
j=1

into the weak form (2.49) and choosing n test functions Wi , i = 1, 2, . . . , n

gives
Z n Z
dWi X dNj
l l
k aj dx = Wi Qdx+ [Wi q]x=0 − [Wi q̄]x=l
0 dx dx 0
j=1 (2.50)

i = 1, 2, . . . n

The algebraic system of n equations and n unknowns is obtained by giving

values to Wi in above expression from i = 1 to n.
For i = 1

Z l µ ¶ Z l
dW1 dN1 dN2 dNn
k a1 + a1 + · · · + an dx = W1 Qdx+(W1 q)0 −(W1 q̄)l
0 dx dx dx dx 0
(2.51)

64
For i = 2

Z l µ ¶ Z l
dW2 dN1 dN2 dNn
k a1 + a2 + · · · + an dx = W2 Qdx+(W2 q)0 −(W2 q̄)l
0 dx dx dx dx 0

For i = n
Z l µ ¶ Z l
dWn dN1 dN2 dNn
k a1 + a2 + · · · + an dx = Wn Qdx+(Wn q)0 −(Wn q̄)l
0 dx dx dx dx 0

Above equations can be written in matrix form as

    
K11 K12 . . . K1n   a1 
   f1 
 K21 K22 . . . K2n    a2 
   f2 
 
 .. ..  .. = ..
 . . 
 .



 .

Kn1 Kn2 . . . Knn  an   fn 

Ka = f
where
Z l
dWi dNj
Kij = k dx
Z l 0 dx dx (2.52)
fi = Wi Qdx + (Wi q)0 − (Wi q̄)l
0
We are still free to choose an appropriate expression for Wi . Choosing the
Galerkin form (Wi = Ni ) gives
Z l
dNi dNj
Kij = k dx
Z0 l dx dx (2.53)
fi = Ni Qdx + (Ni q)0 − (Ni q̄)l
0

Note that matrix K is symmetrical, which is advantageous for the compu-

tations. This symmetry is one of the key properties of the so-called Galerkin
weak form.
We point out again that fi contains the incoming flux q0 at the boundary
where φ is prescribed. The flux is an unknown that can be computed “a
posteriori” once the value of the unknown parameters have been found. This
process is explained in the example given next.

65
Example 2.4 Solution of 1D Poisson equation via the Galerkin weak form and a polyno-
mial approximation

Solution

Let us consider the Poisson equation

d2 φ
A(φ) : +3=0 en Ω (a)
dx2

The boundary conditions are

½
φ−1 = 0, x = 0 (Γ)φ
B(φ) : dφ (b)
dx + 2 = 0, x = l (Γq )

From (a) and (b) we deduce k = 1, Q = 3, φ̄ = 1 and q̄ = 2.

We define a polynomic approximation of the unknown such that the boundary con-
ditions for the temperature prescribed at x = 0 is satisfied, i.e.
n
X
φ(x) ∼
= φ̂(x) = 1 + aj xj (c)
j=1

Note that x = 0, φ = 1.
From (c) we deduce Nj = xj (i = 1, 2, . . . , n).
Let us find now the solution for n = 2

φ̂(x) = 1 + a1 x + a2 x2
(d)
N1 (x) = x, N2 (x) = x2

where a1 and a2 are the unknowns.

The Galerkin weak form leads to an algebraic system with two equations that can be
expressed as · ¸½ ¾ ½ ¾
K11 K12 a1 f1
= (e)
K21 K22 a2 f2

where
Z l Z l
dN1 dN1
K11 = dx = dx = l
0 dx dx 0

Z l Z l
dN1 dN2
K12 = = 2xdx = l2 = K21
0 dx dx 0

Z l Z l
dN2 dN2 4 3
K22 = dx = 4x2 = l (f )
0 dx dx 0 3
Z l Z l
3 2
f1 = N1 Qdx + (N1 q)0 − (N1 q̄)l = 3xdx − 2l = l − 2l
0 0 2
Z l Z l
f2 = N2 Qdx + (N2 q)0 − (N2 q̄)l = 3x2 dx − 2l2 = l3 − 2l2
0 0

66
Substituting into (e) gives
· ¸ ½ ¾ ½ 3l2 ¾
l l2 a1 2 − 2l
4 3 = (g)
l2 l a2 l3 − 2l2
| {z3 } | {z } | {z }
K a f

Solution of the system gives a1 = 3l − 2, a2 = −3/2.

The approximate solution is

3
φ̂(x) ≡ φ(x) = 1 + (3l − 2)x − x2 (h)
2

Let us find out as an exercise the analytical solution. Integration of the original
differential equation gives

dφ
= −3x + A
dx
3
φ = − x2 + Ax + B
2

Constants A and B are found by using the boundary conditions

(φ)0 = 1 =⇒ B = 1
µ ¶
dφ
|l = − 2 =⇒ −3l + A = −2; A = 3l − 2
dx

The exact solution is therefore

3
φ = − x2 + (3l − 2)x + 1 (i)
2

Note that the approximate solution coincides with the exact one. This is a conse-
quence of the quadratic approximation chosen that coincides with the exact one.
The incoming flux q0 is found as
¯ ¯
dφ ¯¯ ¯
¯
q0 = −k ¯ = −(a 1 + 2a x)
2 ¯ = −a1 = 2 − 3l
dx x=0 x=0

This shows that the flux at the Dirichlet boundary where the unknowns is prescribed
can be computed “a posteriori” in terms of the parameters of the approximation.
Note the analogy with the computation of the reaction at the support of a structure
from the displacements at the joints of the structure.

67
Example 2.5 Solve the 1D heat conduction equation accounting for heat conduction loses
via the Galerkin weak form and a polynomial approximation.

Solution

Let us consider the heat balance equation in a bar where heat is generated due to a
distributed source Q and heat is lost along the bar length due to the difference between
the temperature of the bar and that of the external medium (heat conduction loss).
The heat balance equation is an infinitesimal element of the bar is written as

q + Qdx − (q + dq) − h(φ − φe )dx = 0

³ ´
Simplifying and using Fourier’s law q = −k dφ
dx , we obtain

µ ¶
d dφ
k + Q − h(φ − φe ) = 0 (a)
dx dx

where h is the convection parameter and φe the external temperature.

The third term expresses the heat loss due to convection along the bar.
Let us assume a problem with k = 1, Q = 0, h = 1 and φe = 0, i.e.

d2 φ
A(φ) : −φ=0 in Ω (b)
dx2
with the following boundary conditions

φ = 0; x=0 (Γφ
B(φ) : dφ (c)
 − 20 = 0 ; x=l (Γq
dx
The integral expression of the previous equation is obtained in the usual way as
Z l µ ¶ · µ ¶¸
d2 φ dφ
Wi − φ dx + W̄ i − 20 =0 (d)
0 dx2 dx x=l

The weak form is obtained integrating by parts the second derivative term. This
Z l Z l · ¸l
d2 φ dWi dφ dφ
Wi dx = − dx + Wi (e)
0 dx 0 dx dx dx 0

Substituting (e) into (d) and making W̄i = −Wi , gives

Z l · ¸l Z l · µ ¶¸
dWi dφ dφ dφ
− dx + Wi − Wi φdx − Wi − 20 =0 (f )
0 dx dx dx 0 0 dx l

After little algebra we obtain

Z lµ ¶
dWi dφ
+ Wi φ dx = (Wi q)0 + 20(Wi )l (i = 1, 2, . . . , n) (g)
0 dx dx

68
where ¯
dφ ¯¯
q0 = −k ¯
dx 0

Note that the effect of convection leads to a new term is the l.h.s. of (g) which
obviously affects the stiffness matrix. Also note that φe 6= 0 introduces and additional
term in the flux vector.
Let us assume the following expansion

P
2
φ∼
= φ̂ = aj xj = a1 x + a2 x2
j=1 (h)
N1 = x; N2 = x2

Substituting Eq.(h) into (g) and applying the Galerkin method (Wi = Ni ) gives

Z là !
dNi dφ̂
+ Ni φ̂ dx = (Ni q)0 + 20(Ni )l ; (i = 1, 2) (i)
0 dx dx

Substituting the approximation for φ gives

   
Z l X2 X 2
 dN i  dN j
aj  + Ni ( Nj aj ) dx = (Ni q)0 + 20(Ni )l ; (i = 1, 2) (j)
0 dx j=1 dx j=1

The expressions for the stiffness matrix and the flux vector are

Z lµ ¶
dNi dNj
Kij = + Ni Nj dx (k)
0 dx dx

dNi dNj
where dx dx is the conduction term and Ni Nj is the term due to convection.

fi = (Ni q)0 + 20(Ni )l (l)

The numerical expressions for Kij and fi are

Z 1
K11 = (1 + x2 )dx ⇒ K11 = 4/3
0
Z 1
K12 = (2x + x3 )dx ⇒ K12 = 5/4 = K21
0
Z 1
K22 = (4x2 + x4 )dx ⇒ K22 = 23/15
0
f1 = (xq)0 + 20x|l = 20
f2 = (x2 q)0 + 20x2 |l = 20

The system of equations is written in matrix form as

· ¸½ ¾ ½ ¾
4/3 5/4 a1 20
=
5/4 23/15 a2 20

69
 From above we obtain
½
4/3a1 + 5/4a2 = 20 a1 = 11, 7579
⇒
5/4a1 + 23/15a2 = 20 a2 = 3, 4582

The solution is
a1 a2
z }| { z }| {
φ̂ = 11, 7579 x + 3, 4582 x2

The heat flux q0 at x = 0 is

¯
dφ̂ ¯¯
q0 = − = −11, 7579
dx ¯x=0

Let us verify the approximation of the flux computed at x = 1 with the exact value
¯
dφ̂ ¯
ql = − ¯¯ = −11, 7579 − 2 × 3, 4582 = −18, 67
dx x=1

The minus sign denotes that the flux goes out the domain, as expected.
Using a single term in the approximation (h), gives φ̂ = 15 and ql = −15.
We can plot the approximate solution for comparison with the exact solution. The
following tables³show´ the approximate and exact values of φ at x = 0, 5 and x = 1,
and of the flux dφdx at x = 1.

φ̂ φ (exact)
n=1 n=2
x = 0, 5 7,5 6,7435 6,7540
x=1 15 15,2161 15,2319

¯ ¯
¯
dφ̂ ¯
¯
dφ ¯
dx ¯ dx ¯
(exact)
1 1
n=1 n=2
-15 -18,67 -20
¯
¯
dφ̂ ¯
Table 2.3: Comparison of the approximate and exact values for φ y dx ¯
1
(Example 2.5)
Note that the accuracy for φ is greater than that for the flux at the end point. This is
due to the computation of the flux as the first derivative of the approximate function
φ̂. Recall that the derivatives of an approximate function introduces an additional
error. Consequently, the fluxes computed from the numerical solution are an order
less accurate than the values of φ̂. This conclusion also extends to the FEM.

70
2.7 DERIVATION OF THE PRINCIPLE OF VIRTUAL WORK
FROM THE WEIGHTED RESIDUAL METHOD
Let us consider the bar under traction forces of Figure 2.3.
The governing equation is
µ ¶
d du
A(u) : EA +b=0 (2.54)
dx dx

and the boundary conditions


 Prescribed displacement u = 0 x = 0
B(u) :

Prescribed forces EA du
dx
− H̄ = 0 x = l

The equivalent integral form is derived via the WRM

Z l · µ ¶ ¸ · µ ¶¸
d du du
W EA + b dx + W̄ EA − H̄ =0 (2.55)
0 dx dx dx x=l

The weak form is obtained integrating by parts the term

Z l · µ ¶¸ Z l · ¸l
d du dW du du
W EA dx = − EA dx + W̄ EA (2.56)
0 dx dx 0 dx dx dx 0

Substituting above expressions into the WRM and making W̄ = −W , gives

Z l · ¸l · µ ¶¸ Z l
dW du du du
− EA dx + W EA − W EA − H̄ + W bdx = 0
0 dx dx dx 0 dx l 0
(2.57)
Let us recall that
du
N = σA = E²A = EA (2.58)
dx
Hence, the weak form can be written as
Z l Z l
dW du
(EA) dx = W bdx + (W H̄)l − (W N )0 (2.59)
0 dx | {z dx } 0
N

Let us denote, for convenience, the arbitrary test functions as virtual dis-
placement δu. Hence,
W (x) ≡ δu(x) (2.60)
The term (W N )0 is eliminated by imposing that the virtual displacement
δu satisfies the kinematic boundary conditions, i.e.

δu(x)|x=0 = 0 (2.61)

71
 Substituting W by δu into the weak form and noting (2.60) leads to
Z l Z l
d(δu)
N dx = δubdx + (δuH̄)l (2.62)
0 dx 0

The virtual strain is defined as

d(δu)
δε = (2.63)
dx
Substituting this expression into the weak form leads finally to
Z l Z l
δ²N (x)dx = δubdx + δul H̄ (2.64)
0 0

The terms δub can be interpreted as the work (per unit length) performed
by the distributed force b over the virtual displacement δu. Also, the term
δul H̄ is the work of the force H̄ over the virtual displacement of the bar end.
Finally the integral of the l.h.s. of Eq.(2.63) represents the work of the axial
forces over the virtual strains (elongations) along the bar.
The previous integral expression is known in practice as the Principle of
Virtual Work (PVW). The definition of PVW is a follows:
“A body is in equilibrium if, for any virtual displacement δu that satisfies
the kinematic boundary conditions, the work performed by the stresses over
the virtual strains equals the work performed by the external forces over the
virtual displacement”.
As the starting point for the derivation of the PVW is the equilibrium
equation, the PVW is an equilibrium expression. Also, starting from the PVW
and integrating by parts the l.h.s. and taking into account that the virtual
displacement are arbitrary, we recover the original governing equations. The
PVW is therefore a necessary and sufficient condition for the equilibrium of a
structure
The PVW has advantages versus the governing differential equations ex-
pressing the equilibrium of a body, as it has a simple physical meaning. The
PVW is the starting point for the solution via FEM of most problems in struc-
tural mechanics.

2.8 THE PVW IN POISSON PROBLEMS

The PVW has an analogous expression in Poisson problems. Thus making
W (x) = δφ(x)
N = −q
d(δφ) ¡ dφ ¢
dx
= δ dx
= δg
b(x) = Q(x)
H̄ = − q̄

72
 In above δφ or δg are the virtual variables (i.e. temperature, pressure, etc.)
and the virtual gradient, respectively. The PVW is written in this case as,
Z l Z l
δg(−q)dx = δφQdx − (δφ)l q̄
0 0

The minus sign in the above expression comes from the fact that the flux
direction is the opposite to that of the gradient of φ.

(e) (e)
x1 , x2 : abscissae of the coordinates of nodes 1 and 2 of element (e)

Figure 2.10: Discretization of a bar into 1D two-noded finite element

2.9 THE FEM IN 1D POISSON PROBLEMS

We will study next the general solution of 1D Poisson problems with the FEM.
Let us consider a discretization of the analysis domain (the 1D domain)
in non-overlapping sub-domains, termed hereafter “finite elements” (Figure
2.10).
Among the many choice for approximately the unknown function φ we will
choose the simplest one, using polynomial functions locally defined within each
finite element as
n
X
φ(x) ' φ̂(x) = ao + a1 x + a2 x2 + · · · + an xn = a i xi (2.64)
i=0

73
where n is the number of points within the element where the value of φ is
assumed to be known. Such points are called nodes. Also, ao , a1 , . . ., an are
constant parameters depending on the value of φ at the nodes only. The above
expression is typically written as

n
X
(e) (e) (e) (e) (e) (e)
φ̂(x) = N1 (x)φ1 + N2 (x)φ2 + · · · + Nn(e) (x)φ(e)
n = Ni (x)φi
i=1
(2.65)
(e) (e)
where N1 (x), . . ., Nn (x) are the polynomial functions that interpolate φ
(e)
within the element domain (they are called shape functions) and φi is the
value of the approximate function φ at the node i. In the follows for simplicity
we will skip the hat over φ̂. This means that we will make no distinction
between the exact and approximate expressions of the unknown φ.
(e)
The shape function Ni (x) interpolates within the element the unknown
corresponding to node i only and, therefore, it is called the shape function of
(e)
node i. It is easy to note that Ni (x) takes the unit value at the node and
zero at the other nodes.
Substituting the approximate expression from φ(x) for each element into
the integral form of the WRM leads, after simple algebra, to a system of
algebraic equations expressing the equilibrium of the problem in terms of the
nodal values of unknown φ at the nodes of the finite element mesh. Such
equations can be written in matrix form as

Ka = f (2.66)

with
     
K11 K12 . . . K1n 
 φ 1 
 
 f1 
 .. ..  
 φ2    f2 
 
 . . 
K= .. 
..  ; a = .
..  ; f = .
..  (2.67)
 . . 
  
 

φ   f 
 
KN 1 KN 2 . . . K N N N N

where N is the total number of nodes in the finite element mesh.

By analogy with matrix analysis of structures, K is called the stiffness
matrix of the finite element mesh and a and f are called the nodal unknown
vector and the vector of equivalent nodal fluxes, respectively. K and f can be
obtained by assembling the individual contributions from each element as in
matrix analysis of bar structures. The solution of the global system of algebraic
equations gives the value of φ at the nodes from which the gradient and the
flux within each element can be found.
For the sake of clarity we will study in a subsequent section the solution
of the 1D Poisson problem using two meshes of one and two two-noded linear
finite elements, respectively.

74
2.9.1 Discretization of the problem. Local definition of the shape func-
tions
Let us define the distribution of the unknown function φ within each element.
The bar is discretized into two-noded 1D finite element. Within each ele-
ment the unknown function φ(x) is approximated using a linear polynomial as
(Figure 2.11)
1
X
φ(e) (x) ∼
= φ̂(e) (x) = αi xi = α0 + α1 x (2.68)
i=0

Taking into account that at the end node of each element the unknown
function values coincide with those of the nodal unknowns we can write
(e) (e)
φ̂(e) (x1 ) = φ1 = α0 + α1 x1
(2.69)
(e) (e) (e)
φ̂ (x2 ) = φ2 = α0 + α1 x2

Solving the above equation system with two unknowns gives

(e) (e)
(e) φ2 −φ1 (e)
α0 = φ1 − (e) (e) x1
x2 −x1

(e) (e)
φ2 −φ1
α1 = (e) (e)
x2 −x1

Substituting α0 and α1 into (2.68) gives

à ! à !
(e) (e) (e) (e)
(e) φ − φ (e) φ − φ
φ̂(e) (x) = φ1 − 2(e) 1
x
(e) 1
+ 2
(e)
1
(e)
x
x2 − x1 x2 − x1
After some rearranging we obtain
(e) (e)
x2 − x (e) x − x1 (e)
φ̂(e) (x) = (e) (e)
φ1 + (e) (e)
φ2
x2 − x1 x2 − x1
| {z } | {z }
N1 (x)
(e) (e)
N2 (x) (2.70)

(e) (e) (e) (e) P

2
(e) (e)
= N1 (x)φ1 + N2 (x)φ2 = Ni (x)φi
i=1

(e)
from which the expression of the shape functions Ni is deduced as
(e) (e)
(e) x2 −x x2 −x
N1 (x) = (e) (e) = l(e)
x2 −x1
(2.71)
(e) (e)
(e) x−x1 x−x1
N2 (x) = (e) (e) = l(e)
x2 −x1

75
 Note that
( (
(e) (e)
(e) N1 (x1 ) = 1 (e) N1 (x2 ) = 0
x= x1 → (e) x= x2 → (e)
N2 (x1 ) = 0 N2 (x2 ) = 1

The above expressions show that the shape functions of a node take a unit
value at that node and zero at the other node.
Figure 2.11 shows the graphic representation of the shape functions of the
two-noded bar element.

Figure 2.11: Linear shape functions for the 1D two-noded bar element

The interpolation (2.70) allows us to obtain the value of the unknown φ

(e) (e)
at any point of the element in terms of the nodal values φ1 and φ2 , which
become now the problem unknowns.
The linear interpolation chosen within each element provides a linear-wise
approximation of φ over the analysis domain. If the nodal values coincide
with the exact ones (which only occurs in exceptional cases), we will have the
situation of Figure 2.12a. Clearly the coincidence of the approximate and the
exact solution will only occur when the exact solution varies also linearly. In
the most general case the nodal unknown values will not coincide with the
exact ones (Figure 2.12b), neither the exact solution will vary linearly along
the analysis domain. The error can be corrected with a finer discretization or
using higher order finite elements.

2.9.2 Derivation of the algebraic equation systems. Solution of the 1D

Poisson problem using one 2-noded 1D element
Let us retake the problem of the heated bar of Figure 2.8. The governing
equations for the heat conduction problem are
µ ¶
d dφ
A(φ) = k +Q=0 0≤x≤l
dx dx

76
Figure 2.12: Approximation of the solution via linear finite elements. a)
Nodaly exact solution; b) Nodally approximate solution

½
φ − φ̄ = 0 x=0
B(φ) :
k dφ
dx
+ q̄ = 0 x=l
The analytical solution of this problem (for k constant) is obtained as
follows:
k dφ
dx
= − Qx + A

Q 2 Ax
φ = − 2k
x + k
+B
From the boundary conditions we deduce

B =¯ φ̄
¯
k dφ ¯ = − Ql + A
dx ¯
l
A = − Ql − q̄

Substituting into the expression for φ gives

Q 2 x
φ=− x + (Ql − q̄) + φ̄ (2.72)
2k k

77
Figure 2.13: Solution of the 1D problem of Figure 2.8 using one 2-noded linear
element

The weak form is written as

Z l Z l ¯ ¯
dWi dφ̂ ¯ ¯
k dx = Wi Qdx + Wi q ¯¯ −Wi q̄ ¯¯ (2.73)
0 dx dx 0 x=0 x=l

We will now solve the FEM problem with a single 2-noded element (Figure
2.13). The temperature φ is approximated within the element as
φ̂(x) = N1 (x)φ1 + N2 (x)φ2 (2.74)
where the upper index e has been omitted as we only have one element.
Let us recall that φ1 and φ2 are temperatures at nodes 1 and 2, respectively.
Derivating the above expression gives

dφ̂(x) dN1 (x) dN2 (x)

= φ1 + φ2 (2.75)
dx dx dx
Substituting (2.74) into the weak form (2.73) and applying the Galerkin WR
method (Wi = Ni ), gives
Z l · ¸ Z l
dNi dN1 dN2
k φ1 + φ2 dx = Ni (x)Qdx+[Ni (x)q]0 −[Ni (x)q̄]l , i = 1, 2
0 dx dx dx 0

For i = 1 Z · ¸ Z l
l
dN1 dN1 dN2
k φ1 + φ2 dx = N1 Qdx + q0
0 dx dx dx 0
as
N1 (0) = 1
N1 (l) = 0

78
For i = 2
Z l · ¸ Z l
dN2 dN1 dN2
k φ1 + φ2 dx = N2 Qdx − q̄
0 dx dx dx 0
as

N2 (0) = 0
N2 (l) = 1

The above equations define a system of two equations with two unknowns
that can be written as
· ¸½ ¾ ½ ¾
K11 K12 φ1 f1
=
K21 K22 φ2 f2
where the elements of the stiffness matrix are
Z l
dNi dNj
Kij = k dx
0 dx dx
Similarly we obtain for the components of the equivalent flux vector
Z l
f1 = N1 Qdx + q0
0

Z l
f2 = N2 Qdx − q̄
0
The above expressions are completely general and applicable to any 1D
heat conduction element. It is easy to particularize the expressions for the
2-noded element.
Note that
dN1 −1 dN2 1
= (e) , = (e)
dx l dx l
Thus gives
(e) (e) k
K11 = K22 =
l
(e) (e) −k
K21 = K12 =
l
Ql
f1 = + q0
2
Ql
f2 = − q̄
2
The system of algebraic equations to be solved in this simple case is
· ¸ ½ ¾ ½ Ql ¾
k 1 −1 φ1 + q0
= Ql 2 , φ1 = φ̄ (2.76)
l −1 1 φ2 2
− q̄

79
 The approximate solution for the end value φ2 is
Ql2 q̄l
φ2 =− + φ̄ (2.77)
2k k
Note that the value φ2 coincides with that given by the exact analytical
solution. This is remarkable as the exact parabolic solution differs from the
linear approximation chosen for the element. This coincidence only occurs for
some special 1D problems.
The end flux “q0 ” (considered here as a nodal “reaction”) can be obtained
by substituting φ2 into the first equation giving
q0 = q̄ − Ql (2.78)
Note that the end flux satisfies the flux balance equation, i.e.
(q + Ql) − q̄ =0
| 0 {z } |{z}
incoming flux outgoing flux

2.9.3 Solution of the 1D Poisson problem using two 2-noded elements

We will now solve the problem using a discretization of the bar into two linear
elements as shown in Figure 2.14. The approximation of φ over the whole bar
can be written as
φ ' φ̂ = N1 φ1 + N2 φ2 + N3 φ3 (2.79)
where Ni are called global shape function. From Figure 2.14 we deduce
(1)
¾
N1 = N1 0 ≤ x ≤ l/2
N1 = 0 l/2 < x ≤ l
)
(1)
N 2 = N2 0 ≤ x ≤ l/2
(2) (2.80)
N 2 = N1 l/2 < x ≤ l
¾
N3 = 0 0 ≤ x < l/2
(2)
N3 = N2 l/2 ≤ x ≤ l
We can see that each global shape function Ni coincides with a local shape
(e)
function Nj within each element. Note also that the global shape function of
a node taking a unit value at the node and zero at the other node in the mesh.
This means that the “influence domain” of each node in the mesh extends only
over the element that share the node.
The discretized weak form is written now as

Z l · ¸ Z l
dNi dN1 dN2 dN3
k φ1 + φ2 + φ3 dx = Ni Qdx + [Ni q]0 − [Ni q̄]l
0 dx dx dx dx 0

i = 1, 2, 3 “global” numbering
(2.81)

80
Figure 2.14: Discretization of a bar with two 1D linear elements. Global shape
functions

We particularize above expression for different values of i.

For i = 1

Z " # Z l/2
l/2 (1) (1) (1)
dN1 dN1 dN2 (1)
k φ1 + φ2 dx = N1 Qdx + [N1 q]0 − [N1 q̄]l =
0 dx dx dx 0

Z l/2
(1)
= N1 Qdx +q0
|0 {z }
(1)
f1
(2.82a)
dN3
Note that it makes no sense to include φ
within the bracket term in
dx 3
the l.h.s., as N3 takes a zero value within the first element (0 ≤ x < l/2).
For the derivation of the r.h.s. of (2.82a) we have used that N1 = 1 at
x = 0 and N1 = 0 at x = l.

81
For i = 2

Z " # Z l " #
l/2 (1) (1) (1) (1) (2) (2)
dN2 dN1 dN2 dN1 dN1 dN2
k φ1 + φ2 dx + k φ2 + φ3 dx=
0 dx dx dx l/2 dx dx dx

Z l/2 Z l
(1) (2)
= N2 Qdx + N1 Qdx
|0 {z } |
l/2
{z }
(1) (2)
f2 f1
(2.82b)
Note that the boundary terms are zero as N2 = 0 at x = 0 and x = l.
For i = 3
Z " # Z l
l (2) (2) (2)
dN2 dN1 dN2 (2)
k φ2 + φ3 dx = N2 Qdx+
l/2 dx dx dx l/2

Z l (2.82c)
(2)
+[N3 q]0 − [N3 q̄]l = N2 Qdx −q̄
l/2
| {z }
(2)
f2

The above expressions can be written in matrix form as

 (1) (1)
   (1) 
K11 K12 0 φ
 1 = φ̄  
 f + q 0 
   
1
 
 
 
 

    
 (1) (1) (2) (2)  (1) (2)
K21 K22 + K11 K12  φ2 = f2 + f1
    
 










  
 

0
(2) (2)
K21 K22 | {z } φ3
(2)
f2 − q̄
| {z } | {z }
K a f

with
Z (e) (e)
(e) dNi dNj
Kij = k dx
l(e) dx dx
Z
(e) (e)
fi = Ni Qdx
l(e)

We see that the global stiffness matrix can be obtained by assembling the
individual element matrices K(e) , as we did for the assembly of the stiffness ma-
trices in bar structures. Also note the banded structure of K. The bandwidth
size depends on the numbering of the nodes in the finite element mesh
The components of the global equivalent nodal flux vector can be also
obtained by assembling the contributions from each element. Note that at the
end nodes 1 and 3 we should add the global flux components q0 and q̄. As

82
usual the sign of the end flux is positive is heat comes into the body, whereas
it is negative for heat fluxes that originate heat loses in the body.
Note again the coincidence of the assembly process for K and f with that
followed for the matrix analysis of bar structures.
The expressions of K(e) and f (e) given above are general and applicable to
any 1D heat conduction element. We particularize next these expressions for
the 2-noded element. From the expression of the shape function we obtain

(e)
dN1 (x) 1
= − l(e)
dx
(e)
dN2 (x) 1
= l(e)
dx
(e)
Substituting the above expressions into Kij , we obtain
Z µ ¶µ ¶ µ ¶(e)
(e) 1 1 k (e)
K11 = − (e) − (e) kdx = = K22
l(e) l l l
where k has been assumed to be constant within each element. Also
Z µ ¶ µ ¶ µ ¶(e)
(e) (e) 1 1 k
K12 = K21 = − (e) k (e) dx = −
l(e) l l l

Assuming a uniform distribution of Q over the bar gives

Z (e) Z (e) Z
(e) x2 − x x2 x
f1 = Qdx = Qdx − Qdx =
l(e) l(e) l(e) l(e) l(e) l(e)
· ¸l(e)
(e) Q x2 Ql(e)
= Q[x]l0 − (e) =
l 2 0 2
Z (e) · ¸l(e)
(e) x − x1 Q x2 Ql(e)
f2 = dx = =
l(e) l(e) l(e) 2 0 2

The global matrix equation is therefore written as

 
 ¡ ¢(1) ¡ ¢  φ1 
   (1) 
k
− k (1)
0 
 
  Ql
+ q 
   0 
¡ l k ¢(1) ¡ k ¢(1) l ¡ k ¢(2) ¡ k ¢(2)  Ql
2
(1) Ql (2)
 −l + −  φ2 = + 2
l ¡ k ¢(2)l ¡ k l¢(2) 




 2
 Ql(2) 

0 − l l

   2
− q̄
φ3

83
 For k constant over the bar and an uniform mesh (l(e) = l/2) we obtain
 
  φ1 
 
  Ql 
1 −1 0   
  + q0 
2k  4
−1 2 −1 φ2 = Ql
2
, φ1 = φ̄
l 
    Ql − q̄ 
0 −1 1   
   4
φ3
The solution of the above system gives
Ql2 q̄l
φ3 = 2k
− k
+ φ̄

3Ql2 q̄l (2.83)

φ2 = 8k
− 2k
+ φ̄

φ0 = q̄ − Ql
Note that the solution for φ2 and φ3 coincides again with the exact values
(eq. (2.72)). Also, we obtain again q0 = q̄ − Ql which balances the sum of heat
fluxes over the bar.

2.10 GENERALIZATION OF THE SOLUTION FOR A MESH

OF TWO-NODED ELEMENTS
We summarize next the steps to be followed for solving the 1D Poisson problem
with 2-noded linear elements.
1. Obtain the stiffness matrix K(e) and the equivalent nodal flux vector f (e)
for each element.
For the 2-noded element the expressions are
 (e) (e)  ( )
K11 K12 (e)
f1
K(e) =   f (e) = (e)
(e) (e) f2
K21 K22
with

Z (e) (e) µ ¶(e)

(e) dNi (x) dNj (x) k
Kij = k dA = (−1)i+j
l(e) dx dx l
(2.84)
Z (e) (e)
(e) (e) (Ql)
fi = Ni Qdx =
l(e) 2

Note that K(e) has a many rows and columns as degrees of freedom
(d.o.f.) has the element. Recall that for the 1D Poisson problem we just
have a single d.o.f. per node.

84
2. Obtain the global stiffness matrix K and the global equivalent nodal flux
vector f by assembling the contributions for each element. For a mesh
of n 2-noded elements, the resulting matrix expression can be written as

 
 (1)     f (1) + q 
(1)  φ1  
 1 0 

K11 K12 
 
 
 (1) (2)  
 (1)  
 φ 
 
 f + f 

(1)
K21 K22 + K
(2)
K
(2)
O  

2 
 
 2 1 

 11 12   φ 3
  f
(2)
+ f
(3) 
 (2) (2) (3)  2 1
 K 21 K22 + K 11 .
 .   = ..
 (n−1) (n)   .   . 

+ K11 K12  
(n)   
 O K12 
 
 
 

(n) (n)   φ n 
 
 f
(n−1)
+ f
(n) 

K21 K22   
 2 1 

| {z φn+1
} | {z }  (n) 
f2 − q̄
K a | {z }
f

The assembling process is identical to that followed for assembly the

global equilibrium equations in the matrix analysis of discrete systems
(Chapter 1).

3. Solve the global equations system Ka = f for the values of the nodal
unknowns φi . Obtain the gradient of φ and the heat fluxes within each
element in terms of the nodal unknowns φi .

85
 3. Propagation problems. They include all non-stationary problems where
the evolution of some variables depends on time.

Table I lists different applications of the FEM in different engineering dis-

ciplines related to the three types of problems mentioned above.

STATIONARY EIGENVALUE PROPAGATION

Discipline PROBLEMS PROBLEMS PROBLEMS
Civil, Aerospace Static analysis Structural Seismic and
and Naval of structures. stability dynamic analysis
Engineering Stationary porous problems of structures.
media flow Fluid dynamics
analysis
Mechanical Stress analysis Modal and Dynamic analysis
Engineering of mechanical vibration of mechanical
systems. analysis systems.
Stationary heat of mechanical Transient heat
transfer components transfer
Geothecnical Stress analysis in Modal and Wave propagation
Engineering soils and rocks. vibration in soils
Flow through analysis of and rocks
porous media soils
Nuclear Stress and heat Modal and Seismic and
Engineering transfer analysis vibration dynamic analysis
in nuclear analysis of nuclear of nuclear structures.
structures. structures Transient heat
Fracture mechanics transfer
Electrical and Stationary analysis Frequency Transient
Telecommunication of Maxwell analysis solution
Engineering equations of Maxwell of Maxwell
equations equations

From the practical point of view of the analyst it should never be forgotten
that the FEM is a very powerful technique to obtain approximate solutions for
continuous problems. In the hands of a careful and expert user the FEM is
an indispensable tool for the analysis of complex problems which cannot be
studied otherwise. However, being an approximate method it involves a certain
error in the numerical values and users should always consider FEM results
with a critical eye. In this course we will try to facilitate the understanding
of the theoretical and applied aspects of the FEM for analysis of continuum
systems.

32
Chapter 3

1D FINITE ELEMENTS FOR AXIALLY

LOADED RODS

3.1 INTRODUCTION
The objective of this chapter is to introduce the basic concepts of the FEM in
its application to the analysis of simple one-dimensional (1D) axially loaded rods
via the principle of virtual work (PVW).
The organization of the chapter is as follows. In the first section the analysis
of axially loaded rods using 2-noded rod elements is presented. Particular em-
phasis is put in the analogies with the solution of the same problem using the
standard matrix analysis techniques studied in the Chapter 1 for bar structures.
Here some examples of application are given. Then the chapter the matrix finite
element formulation which will be adopted throughout this course is presented.
In the final part of the chapter the general derivation of the one-dimensional
(1D) shape functions is presented. Such functions are very useful for obtaining
the shape functions of two- (2D) and three- (3D) dimensional elements in the
next chapters. The concepts of isoparametric element and numerical integration
are explained. These concepts are essential for the study of high-order 2D and 3D
elements. Finally, the requirements for the convergence of the numerical solution
are discussed, together with a description of the more usual solution errors.

3.2 AXIALLY LOADED ROD

Let us consider a rod of length l subjected to a uniformly distributed axial load
per unit lentgth b(x) and a set of axial point loads Xi acting at p different points
xi (Figure 3.1). The rod can also have prescribed displacements uj at m points xj .
The displacement of the rod points produces the corresponding strains ε(x) = du dx
and stresses σ in the rod which are related by Hooke’s law, i.e.

du
σ = Eε = E (3.1)
dx

87
 Fig. 3.1: Axially loaded rod

where E is the Young modulus of the material.

In the equilibrium configuration the stresses and the external forces satisfy
the Principle of Virtual Work (PVW) defined in Section 1.7. The PVW for the
rod is written as
ZZZ Z l Xp
δε σ dV = δu b dx + δui Xi (3.2)
V 0 i=1

where δu and δε are the virtual displacement and the virtual strain of an arbitrary
point of the rod center line δui is the virtual displacement of the point where the
point load Xi acts, and V is the rod volume. The left- and right-hand sides of
Eq.(3.2) represent the internal and external virtual work carried out by the actual
stresses and the external loads, respectively.
Eq.(3.2) can be rewritten after integration over the cross section of area A
(note that dV = dA · dx) and using Eq.(3.1) as
Z l Z l p
X
du
δε EA = δu b dx + δui Xi (3.3)
0 dx 0 i=1

It can be proved [T7] that the equilibrium solution of the rod problem is
reduced to finding a displacement field u(x) satisfying Eq.(3.3) and the displace-
ment boundary conditions (kinematic conditions). The approximate solution
using the FEM is set as follows: find an alternative displacement field û(x) which
approximates u(x) and which also satisfies Eq.(3.3) and the kinematic conditions.
Among the different options available to express the approximate displacement
field û(x) we will choose the simplest one using polinomials locally defined for each
element. Thus, after discretization we can write for each element
n
X
2 n
u(x) ' û(x) = ao + a1 x + a2 x + · · · + an x = ai xi (3.4)
i=1

In Eq.(3.4) n is the number of points of the element where the displacement

is assumed to be known. These points are called nodes. The parameters ao , a1 ,
. . ., an depend on the nodal displacements only. In the following we will skip the
hat over the approximate solution and write Eq.(3.4) in the form
n
X
(e) (e) (e) (e) (e) (e)
u(x) = N1 (x)u1 + N2 (x)u2 + · · · + Nn(e) (x)u(e)
n = Ni (x)ui (3.5)
i=1

88
 (e) (e)
where N1 (x), . . ., Nn (x) are the polinomial interpolating functions defined over
(e)
the domain of each element e (termed hereafter shape functions) and ui is the
(e)
value of the (approximate) displacement of node i. The shape function Ni (x)
interpolates within each element the displacements of node i only and for this
(e)
reason it is called the shape function of node i. Note from Eq.(3.5) that Ni (x)
is equal to one at node i and equal to zero at all other nodes. These concepts
will be extended in the next section.
Substituting the displacement approximation for each element in the PVW
allows us to express the equilibrium equations in terms of the nodal displacements
of the finite element mesh. These algebraic equations can be written in the
standard matrix form
Ka = f (3.6)
where, by analogy with bar systems, K is termed the stiffness matrix of the
finite element mesh, and a and f are the vectors of nodal displacements and of
equivalent nodal forces, respectively. Both K and f are obtained by assembling the
contributions from the individual elements as in matrix analysis of bar structures.
Solving Eq.(3.6) yields the values of the displacements at all the nodes in the mesh
from which the strains and stresses within each element can be found.
These concepts will be illustrated in the next section for the analysis of an
axially loaded rod with constant cross sectional area using two meshes of one and
two linear rod elements, respectively.

3.3 AXIALLY LOADED ROD OF CONSTANT CROSS SECTIONAL

AREA. DISCRETIZATION IN ONE LINEAR ROD ELEMENT
Let us consider the bar in Figure 3.2. The bar is first discretized in a single
element with two nodes which define a linear interpolation of the displacement
field as
u(x) = ao + a1 x (3.7)
(1) (1)
Naturally, u(x) must take the values u1 and u2 at nodes 1 and 2, i.e.

(1) (1) (1) (1)

u(x1 ) = u1 and u(x2 ) = u2 (3.8)

(1) (1)
where x1 and x2 are the coordinates of nodes 1 and 2, respectively. The
superindex 1 in Eq.(3.8) denotes that all the parameters refer to the element
number one.
From Eqs.(3.8) and (3.7) the following system of equations is obtained

(1) (1)
u1 = ao + a1 x1
(3.9)
(1) (1)
u2 = ao + a1 x2

89
Fig. 3.2: Axially loaded rod of constant cross section. Discretization in a single 2-
noded rod element

from which the parameters ao and a1 are found as

(1) (1) (1) (1) (1) (1)
u1 − u2 x 2 u1 − x 1 u2
ao = (1) (1)
and a1 = (1) (1)
(3.10)
x1 − x2 x2 x1
Substituting Eq.(3.10) into (3.7) allows us to rewrite the latter as
(1) (1) (1) (1)
u = N1 (x)u1 + N2 (x)u2 (3.11)
(1) (1)
where N1 and N2 are the shape functions of nodes 1 and 2, respectively given
by
(1) (1)
(1) x2 − x (1) x − x1
N1 (x) = ; N2 (x) = (3.12)
l(1) l(1)
(1) (1)
where l(1) = x2 −x1 is the element length. It is deduced from Eq.(3.12) that the
(1)
shape functions Ni (i = 1, 2) vary linearly within the element and take the value
one at node i and the value zero at the other node. This is a general property
which is a consequence of the local definition of the finite element interpolation
chosen. This lets us anticipate in most cases the geometry of the shape functions,
as will be frequently seen throughout this book.
Before we proceed any further it is important to clarify the differences between
local and global numbering. Table 3.1 shows an example of both numberings for

90
the nodes, the coordinates and the nodal displacements for the example in Figure
3.2.

nodes coordinates displacement

Element local global local global local global
1 1 x(1)
1 x1 u(1)
1 u1
1
2 2 x(1)
2 x2 u(1)
2 u2

Table 3.1: Local and global parameters for the example in Figure 3.2

Note that since we have taken in this case a single element, the local and
global numbers coincide.
The derivatives of the shape functions are computed as
(1) (1)
dN1 1 dN2 1
= − and = (3.13)
dx l(1) dx l(1)
In this way the axial strain can be obtained at each point within the element
as
(1) (1)
du dN1 (1) dN2 (1) 1 (1) 1 (1)
ε = = u1 + u2 = − u1 + u2 (3.14)
dx dx dx l(1) l(1)

Obviously, the linear shape functions yield a constant strain (and stress) field
over the element.
The forces between elements are transmited across the nodes only. These
forces are termed equilibrating nodal forces and can be obtained for each element
using the PVW. For the single element of Figure 3.2 we can write
Z x2
(1) Z x2
(1)
(1) (1) (1) (1)
δε EA · ε dx = δu · b dx + δu1 X1 + δu2 X2 (3.15)
(1) (1)
x1 x1

(1) (1) (1) (1)

where δu1 , δu2 , X1 and X2 are the virtual displacements and the equili-
brating nodal forces of nodes 1 and 2 of the element, respectively. The virtual
displacement can also be linearly interpolated in terms of the nodal values as
(1) (1) (1) (1)
δu = N1 δu1 + N2 δu2 (3.16)

The virtual strain is now expressed in terms of the virtual nodal displacements
as
(1) (1)
d dN1 (1) dN2 (1)
δε = (δu) = δu1 + δu2 (3.17)
dx dx dx

91
 Eq.(3.15) is rewritten, after substitution of (3.11), (3.16) and (3.17), as
Z x(1) " # " #
(1) (1) (1) (1)
2 dN1 (1) dN2 (1) dN1 (1) dN2 (1)
δu1 + δu2 (EA) δu1 + δu2 dx −
(1)
x1 dx dx dx dx
Z x(1) " #
2
(1) (1) (1) (1) (1) (1) (1) (1)
− N1 δu1 + N2 δu2 b dx = δu1 X1 + δu2 X2 (3.18)
(1)
x1

Grouping terms gives

"Z (1) Ã !
x2 (1) (1) (1) (1)
(1) dN1 dN (1) dN1 dN (1)
δu1 (EA) 1 u1 + (EA) 2 u2 dx −
(1)
x1 dx dx dx dx
Z x2 (1)
# " Z x2(1)
Ã
(1) (1)
(1) (1) (1) dN2 dN1 (1)
− N1 bdx − X1 + δu2 (EA) u1 +
(1)
x1
(1)
x1 dx dx
! Z x(1) #
(1) (1)
dN2 dN2 (1) 2
(1) (1)
+ (EA) u2 dx − N2 bdx − X2 =0 (3.19)
dx dx x1
(1)

Since the virtual displacements are arbitrary, satisfying of Eq.(3.19) leads to

the following system of two equations
Z x(1) Ã !
(1) (1) (1) (1)
2 dN1 dN1 (1) dN1 dN2 (1)
(EA) u + (EA) u dx−
(1)
x1 dx dx 1 dx dx 2
Z x(1)
2
(1) (1)
− N1 bdx − X1 = 0
(1)
x1
(3.20)
Z (1)
à !
x2 (1) (1) (1) (1)
dN2 dN (1) dN2 dN (1)
(EA) 1 u1 + (EA) 2 u2 dx
x1
(1) dx dx dx dx
Z x(1)
2
(1) (1)
− N2 bdx − X2 = 0
(1)
x1

From Eq.(3.20) the values of the equilibrating nodal forces are obtained. In
matrix form
   
³ dN (1) dN
(1) ´ ³
dN
(1)
dN
(1) ´
( )
Z (1) 1
(EA) 1 1
(EA) 2
 x2    u(1)
  dx dx dx dx  dx 1
−
 (1) ³ dN (1) dN1
(1) ´ ³
dN2
(1)
dN2
(1) ´  (1)
u2
x1 2
(EA) (EA)
( dx) dx ) dx
( dx
Z x(1) (1) (1)
2
N1 X1
− (1) bdx = (1) (3.21)
(1)
x1 N2 X2
or
K(1) a(1) − f (1) = q(1) (3.22a)

92
with
Z (1)
(1)
x2
dNi (1) dNj (1)
Kij = (EA) dx (3.22b)
(1)
x1 dx dx
Z (1)
x2
(1) (1)
fi = Ni b dx i, j = 1, 2
(1)
x1

(1) (1) T (1) (2) T

a(1) = [u1 , u2 ] ; q(1) = [X1 , X2 ]
Eq.(3.22a) expresses the balance between the equilibrating nodal forces q(1) ,
the distributed forces f (1) , and the nodal displacements a(1) . In Eq.(3.22a) K(1)
is the element stiffness matrix and f (1) is the equivalent nodal force vector for the
element.
If the Young modulus, the cross sectional area and the distributed loading are
constant over the element, the following is obtained
· ¸ ½ ¾
(1) EA (1) 1 −1 (1) (bl)(1) 1
K = ( ) ; f = (3.23)
l −1 1 2 1

Note the coincidence between these expressions and those obtained for the
axially loaded bar in Chapter 1. This coincidence could have been anticipated if
we had observed that in both cases the same linear displacement field is assumed.
This obviously leads, via the PVW, to the same expressions for the element
stiffness matrix and the nodal load vector.
The global equilibrium equations are obtained by the same nodal load bal-
ancing procedure explained for bar structures in the previous chapter. Thus, for
each of the N nodes in the mesh we have
X (e)
Xi = Xjext , j = 1, N (3.24)
e

where the sum is extended over all the elements sharing the node with global
(e)
number j, Xi is the equilibrating nodal force contributed by each element, Xjext
is the external point load acting at the node.
For the single element mesh considered, Eq.(3.24) is written as (see Figure
3.2)
(1)
node 1 : X1 = R1
(1)
node 2 : X2 = P
Substituting the values of the equilibrating nodal forces from Eq.(3.21) and
making use of Table 3.1 the global equilibrium equations are obtained as
 

 bl 
EA
· ¸ ½ ¾ R1 +  
1 −1 u1 2
( ) =
l −1 1 u2  
 P + bl 
 
2

93
or
Ka = f (3.25)
where, as usual, K, a and f are, respectively, the global stiffness matrix, the vector
containing the displacements of all nodes in the mesh and the global equivalent
nodal force vector. Eq.(3.25) is solved after imposing the condition u1 = 0, to
give
l bl
u2 = (P + ) ; R1 = − (P + bl) (3.26)
EA 2
The axial strain ε and the axial stress N in the element are given by
(1) (1) (1)
(1) dN1 (1) dN2 (1) u1 P + bl2
ε = u1 + u2 = − (1)
=
dx dx l1 EA
bl
N (1) = (EA)(1) ε(1) = P + (3.27)
2
The exact solution for this simple problem is [T7]
· ¸
1 bx2
u = − + (P + bl) x
EA 2 (3.28)
1
ε = [P + b(l − x)]
EA
The finite element and the exact solutions are compared in Figure 3.3 for
P = 0 and b = 1T /m. Note that the value of the end displacement is the
exact solution. This is a coincidence and it should be considered as an exception
which only occurs for a few occasions only.1 Within the rod the single element
approximation yields a linear displacement field very different from the exact
quadratic solution. Also note that the constant axial stress value obtained differs
substantially from the linear exact solution. As expected, the numerical solution
improves as the mesh is refined and this is shown in Section 3.5 for a mesh of two
elements.

3.4 DERIVATION OF THE DISCRETIZED EQUATIONS FROM

THE GLOBAL DISPLACEMENT INTERPOLATION FIELD
It is interesting that a general expression for the displacement interpolation field
for the whole mesh can be obtained by simple superposition of the local ap-
proximations for each element. This lets us define global shape functions which
naturally should coincide with the original local expressions within each element.
The use of global shape functions leads to identical results as with the simpler
local functions. However, its study has an academic interest.
1
It has been proved [Z7] that the finite element solution coincides with the exact one for 1D
problems if the interpolation chosen satisfies exactly the homogeneous form of the differential
2
equation of equilibrium. This is written for the rod problem as ddxu2 = 0, which is obviously
satisfied by the linear approximation chosen.

94
Fig. 3.3: Axially loaded rod under uniformly distributed loading. Exact and approxi-
mate solutions using one and two linear rod elements

The conceptual differences between global and local shape functions will be
clarified by repeating the single rod element problem using a global interpolation
for the displacement field.
The axial displacement can be written in the single element mesh as

u(x) = N1g (x) u1 + N2g (x) u2 (3.29)

where N1g (x) and N2g (x) are the global shape functions of nodes 1 and 2, respec-
tively, and u1 and u2 are the displacements of these nodes. Note that for the
global displacements we skip the superindex e. We deduce from Eq.(3.29) that
the global function of a node takes the value one at that node and zero at all other
nodes. This provides the relationship between global and local shape functions
as
(e)
Nig (x) = Ni (x) if x belongs to element e
(3.30)
= 0 if x does not belong to element e

For the single element case considered, the global and local shape functions
coincide (Figure 3.2). Thus,
(1) (1)
N1g (x) = N1 (x) and N2g (x) = N2 (x) (3.31)

The axial strain in the rod of Figure 3.2 can be obtained as

du dN1g dN2g
ε = = u1 + u2 (3.32)
dx dx dx

95
 The virtual displacement and the virtual axial strain are expressed as

dN1g dN2g
δu = N1g δu1 + N2g δu2 , δε = δu1 + δu2 (3.33)
dx dx
The PVW is written for the rod as
Z l· g ¸ · g ¸
dN1 dN2g dN1 dN2g
δu1 + δu2 (EA) u1 + u2 dx −
0 dx dx dx dx
Z l
− [N1g δu1 + N2g δu2 ] b dx = δu1 R1 + δu2 P (3.34)
0

After eliminating the virtual displacements, Eq.(3.34) leads to

  g g g g 
Z l dN1 (EA) dN1 dN1 (EA) dN2 ½ ¾
  dx dx dx dx   u1
  dN g dx −
0 2 dN1g dN2g dN2g  u2
(EA) (EA)
dx dx dx dx
Z l ½ g¾ ½ ¾
N1 R1
− g b dx = (3.35)
0 N 2 P

The following relationships are important for the computation of the integrals
in Eq.(3.35) 
(1)
N1g = N1 





N2g
= N2
(1) 



dN1  0 ≤ x ≤ l (3.36)
g (1)
dN1
= 
dx dx  


(1) 
dN2 g
dN2  

=
dx dx
Using Eq.(3.36) we obtain
Z Z Ã !2
l(1) (1)
l
dN1g dN1g dN1 1
dx = dx =
0 dx dx 0 dx l
Z Z l(1) (1) (1)
l
dN1g dN2g dN1 dN2 1
dx = dx = −
0 dx dx 0 dx dx l
Z Z l(1) Ã (1) !2
l
dN2g dN2g dN2 1
dx = dx =
0 dx dx 0 dx l
Z Z l(1) Z l(1) (3.37)
l
(1) (1) l
Nig dx = N1 dx = N2 dx =
0 0 0 2
Substituting Eq.(3.37) into the PVW expression (3.34) yields the global equi-
librium equation (3.25) directly. Recall that in the previous section this equation

96
 Fig. 3.4: Axially loaded rod. Discretization in two linear rod elements

was obtained from the assembly of the element contributions. From this point
onwards the solution process is identical to that explained in Eqs.(3.25)-(3.27)
and it will not be repeated here.
In the next section the same problem is solved using a mesh of two linear
elements.

3.5 AXIALLY LOADED ROD OF CONSTANT CROSS

SECTIONAL AREA. DISCRETIZATION IN TWO LINEAR
ROD ELEMENTS
The same rod as for the previous example is discretized now in two linear rod
elements as shown in Figure 3.4 where the differences between local and global
shape functions are also shown.

97
 The discretized equilibrium equations will be obtained first using the local
description of the shape functions.
The displacements within each element are interpolated as

Element 1 Element 2
¯
¯ (3.38)
(1) (1)
u(x) = N1 (x)u1 + N2 (x)u2
(1) (1) ¯ u(x) = N1(2) (x)u(2) (2) (2)
1 + N2 (x)u2
¯

The shape functions and their derivatives are

¯
1 ¯¯
(1) (1) (2) (2)
(1) x2 − x dN1 (2) x2 − x dN1 1
N1 = ; = − (1) ¯ N1 = ; = − (2)
l (1) dx l ¯ l (2) dx l
¯ (3.39)
x − x2
(1)
dN2
(1)
1 ¯ x − x1
(2)
dN2
(2)
1
(1) ¯ (2)
N2 = ; = (1) ¯ N2 = ; = (2)
l (1) dx l ¯ l (2) dx l

The axial strain in each element is

(1) (1) ¯ (2) (2)
du dN1 (1) dN2 (1) ¯¯ du dN1 (2) dN2 (2)
ε= = u + u ε = = u + u
dx dx 1 dx 2 ¯ dx dx 1 dx 2
(3.40)
The matrix form of the discretized equilibrium equations is obtained using
the PVW as explained in the previous section for the single element case (see
Eqs.(3.12)-(3.20)). We find that

q(1) = K(1) a(1) − f (1) ; q(2) = K(2) a(2) − f (2) (3.41a)

where
 (1) (1) (1) (1) 
Z x(1) dN 1 dN 1 dN 1 dN 2
2  dx dx dx dx 
K =(1)
(EA)  (1) (1) (1)
 dx
(1) 
x1
(1)
dN2 dN1 dN2 dN2
Z x(1) dx dx dx dx
2 h iT
(1) (1)
f (1) = N1 , N 2 b(1) dx
(1)
x
h1 iT h iT
q(1) = X1(1) , X2(1) , a(1) = u(1) 1 , u2
(1)

 (2) (3.41b)
(2) (2) (2) 
Z x2 (1)
dN 1 dN 1 dN 1 dN 2
 dx dx dx dx 
K =(2)
EA  (2) (2) (2) (2) 
 dx
x1
(1)
dN2 dN1 dN2 dN2
Z x(2) dx dx dx dx
2 h iT
(2) (2)
f (2) = N1 , N 2 b(2) dx
(2)
x1
h iT h iT
q(2) = X1(2) , X2(2) , a(2) = u(2)
1 , u
(2)
2

98
are respectively the stiffness matrices, the equivalent nodal force vectors, the
equilibrating nodal force vectors and the nodal displacement vectors of elements
1 and 2.
The integrals in Eqs.(3.41b) are computed keeping in mind the relationships
between the local and global numbering of the element parameters summarized
in Table 3.2.
nodes coordinates displacement
Element
local global local global local global

1 1 x(1)
1 x1 u(1)
1 u1
1
2 2 x(1)
2 x2 u(1)
2 u2

1 2 x(2)
1 x2 u(2)
1 u2
2
2 3 x(2)
2 x3 u(2)
2 u32

Table 3.2: Local and global parameters for the example of Figure 3.4

Substituting Eq.(3.39) into (3.41) and using Table 3.2 the following expres-
sions are obtained for the case of homogeneous material and uniformly distributed
loading:
µ ¶(1) · ¸ µ ¶(2) · ¸
(1) EA 1 −1 (2) EA 1 −1
K = ; K =
l −1 1 l −1 1 (3.42)
(1) £ ¤T
(1) (bl) (2) (bl) £
(2) ¤T
f = 1, 1 ; f = 1, 1
2 2
The equilibrium of nodal forces is written as (see Eq.(3.24) and Figure 3.4)
(1)
Node 1 : X1 = R1
(1) (2)
Node 2 : X2 + X1 = 0 (3.43)
(2)
Node 3 : X2 = P
(e)
Substituting Xi from Eq.(3.42) into (3.43) the following matrix equilibrium
equation is obtained
 
 ¡ ¢   bl
EA (1)
−
¡ ¢
EA (1)
0
    + R1 


 u  
 4 

 ¡ EA ¢(1) h¡ EA ¢(1) ¡ EA ¢(2) i ¡ EA ¢(2) 
l l 1
bl
− 
 l l
+ l l  u2 = (3.44a)
¡ EA ¢(2) ¡ EA ¢(2) u3   
 2 


0 − l l  bl + P 
 
4
Ka = f (3.44b)

99
 The element stiffness matrix is deduced from Eqs.(3.42) and (3.44) as
 (e) (e) (e) (e)

" (e) (e)
# Z x2(e) dN1 dN1 dN1 dN 2
K11 K12 
(e)  dx

K (e)
= = (EA) dx dx dx  dx =
(e) (e) (e)  dN (e) dN (e) dN (e) dN (e) 
K21 K22 x1 2 1 2 2
dx dx dx dx
µ ¶(e) · ¸
EA 1 −1
= (3.45)
l −1 1

Note that the assembly process is identical to that explained in the previous
chapter for bar structures.
The same assembly process applies to the global vector of equivalent nodal
forces. The elemental expression of this vector is
( (e) ) Z x(e) ( (e) ) µ ¶(e) ( )
f1 2 N 1 bl 1
f (e) = = b (e)
dx = (3.46)
f2
(e) (e)
x1 N2
(e) 2 1
¡ ¢(1) ¡ EA ¢(2)
Substituting EAl
= l = 2EA
l
into Eq.(3.44a) and solving the equa-
tion system we find that
µ ¶
l 3bl
u1 = 0 ; u2 = P+
2EA 4 (3.47)
l
u3 = (2P + bl) ; R1 = −(P + bl)
2EA
The axial strain and the axial force are constant within each element and are
obtained as
Element 1 Element 2
µ ¶(1) ¯ µ ¶(2) µ ¶
(1) du u2 P + 3bl
4
¯
¯ (2) du u3 − u2 1 bl
ε = = (1) = ¯ ε = = = +P
dx l EA ¯ dx l(2) EA 4
¯
3bl ¯ bl
¯
N (1) = (EA)(1) ε(1) = P + ¯ N (2) = (EA)(2) ε(2) = +P
4 ¯ 4
(3.48)
The distribution within each element of the displacement u and the constant
axial force N is shown in Figure 3.3. Note that for the same reasons explained
in Section 3.3 the nodal displacements coincide with the exact values. Some im-
provement in the approximation of the global displacement field is also observed.
However, the error in the axial force is still considerable and its reduction requires
a finer discretization and the nodal smoothing of the constant axial forces over
each element. This can be done by simply averaging the nodal forces.
The results obtained for the axial forces (and strains) are more innacurate
than those for the displacement field. This is a general rule which is explained
by the fact that the strains and the stresses are computed from the derivatives of

100
the approximate displacement field. This, naturally, increases the solution error
for those values.
The same problem is now solved using the global description of the shape
functions.
The axial displacement can be expressed globally over the two elements mesh
as (see Figure 3.4)

u(x) = N1g (x)u1 + N2g (x)u2 + N3g (x)u3 (3.49)

and the axial strain is given by

du dN1g dN2g dN3g

ε = = u1 + u2 + u3 (3.50)
dx dx dx dx

The discretized form of the PVW is written using above equations as

Z lµ ¶
dN1g dN2g dN3g
δu1 + δu2 + δu3 (EA)
0 dx dx dx
µ g ¶
dN1 dN2g dN3g
u1 + u2 + u3 dx− (3.51)
dx dx dx
Z l³ ´
− N1g δu1 + N2g δu2 + N3g δu3 b dx = δu1 R1 + δu3 P
0

This leads, after eliminating the virtual displacements, to the following matrix
system of equations

   
dN1g dN1g dN1g dN2g dN1g dN3g
 
Z l  dxg dxg dxg dxg dxg dxg   u1 
  dN dN dN2 dN2 dN2 dN3  
  2 1  (EA) dx
  dx dx dxg dxg dxg dxg   u2  −
  g
0   u3
dN3 dN1g dN3 dN2 dN3 dN3
(3.52)
dx dx dx dx dx dx
   
Z l N1g  R1 
g
− N2 b dx = 0
0 N g   
P
3

The computation of the integrals in Eq.(3.52) requires a correspondence be-

tween the global and local shape functions. The following relationships are de-

101
duced from Table 3.2 and Figure 3.4
 
N1g = N1
(1)

 N1g =0 


 

(1)  dN1g 
dN1 g
dN1  

 =0




=  dx 
dx dx  
 (2)



g (1) 
 N2g = N1 

N2 = N2  l  l
g (1) 0 ≤ x ≤ ; dN2g dN1
(2) ≤ x ≤ l (3.53)
dN2 dN2   2 = 
 2
=  
dx dx  


dx
g
dx
(2)




N3g
=0 
 N3 = N2 


 


 (2) 

dN̄3 

 dN3g dN2 


= 0 =
dx dx dx
Making use of the expressions (3.53) in (3.52) the global equilibrium equation
is directly obtained. The reader can easily verify the coincidence of this equation
with Eq.(3.44a) obtained by assembly of the element contributions.
This example clearly shows that the use of the global shape functions is less
systematic and requires more detailed computations than the element by element
approach. These differences are even more apparent for finer meshes. As a
consequence, the assembly of the global equations from the elemental expressions
derived via the local shape functions is recommended in practice.

3.6 GENERALIZATION OF THE SOLUTION WITH N LINEAR

ROD ELEMENTS
The solution process explained in the previous sections can be easily generalized
for a discretization using a mesh of n 2-noded (linear) rod elements. The stiffness
matrix and the equivalent nodal force vector for each element are given by
 (e) (e) (e) (e) 
dN1 dN1 dN1 dN2
Z x(e)
2
 dx dx dx dx 
K(e) = (EA)(e)   dx
(e) (e) (e) (e) (e)
x1 dN2 dN1 dN2 dN2
dx dx dx dx (3.54)
( )
R x(e) (e)
N1
f (e) = 2
(e) (e) b(e) dx
x1 N2

which after substitution of the shape functions and their derivatives

(e) (e)
(e) x2 − x dN1 1
N1 = ; = −
l(e) dx l(e)
(3.55)
(e) (e)
(e) x − x1 dN2 1
N2 = ; =
l(e) dx l(e)

102
gives (for homogeneous material and uniformly distributed loading)
µ ¶(e) · ¸ ½ ¾
(e) EA 1 −1 (e) (bl)(e) 1
K = ; f = (3.56)
l −1 1 2 1
The assembly process leads, after some algebra, to the following global matrix
equation
 (1)  
k £ −k (1) ¤ 0 ... 0  u1 
 
−k (1) k (1) + k (2)
£ −k (2) ¤ ... 0  
 u2  

  
 

 0 −k (2)
k (2)
+ k (3)
. . . 0  u 3 

 ..    .. 
 (3)
 0 0 −k ... .  . =
 . . . .  . 

 .. .. .. ..   .. 
 £ ... ¤ 




 0 0 ... k (n−1)
+k (n)
−k 
(n)  
un−1  

 

(n) (n) un
0 0 ... −k k
| {z } | {z }
K a
 (bl)(1)


 + P1 


 (1)
2
(2) 


 (bl)
+ (bl)
+ P 


 2 2 2 


 (bl)(2) (bl) (3) 


 2
+ 2
+ P 3 
 µ ¶(e)
.. (e) EA
= . with k = (3.57)

 . 
 l

 .. 


 


 (bl)(n−1) (bl) (n) 


 + + P n−1 


 2
(n)
2 

(bl)
2
+ P n
| {z }
f
Note that matrix K depends on the geometrical (l and A) and material (E)
parameters of each element, while vector f depends on the intensity of the dis-
tributed load b(e) , the element length and the point loads Pi acting at the nodes.
The unknown reactions at the prescribed nodes are treated as point loads and
they are computed “a posteriori” following standard procedures of structural ma-
trix analysis (see Section 1.4) [L3], [T9].

Example: 3.1 Analyse the axially loaded rod with exponentially varying cross sectional area
of Figure 3.5 using three meshes of one, two and three linear rod elements.

-- Solution
x
The change in cross sectional area is defined by A = Ao e− l where Ao is the cross
sectional area at the clamped end and l is the rod length. The rod is subjected to an
axial force acting at the free end. The exact solution for this simple problem is

F F x σ F x
σ = = el , ε = = el
A Ao E EAo

103
Fig. 3.5: Axially loaded road with exponentialy varying cross section. Discretization
in 3 meshes of two-noded rod elements

Z x Z x
F F l x/l
u(x) = εdx = ex/l dx = (e − 1)
0 0 EAo EAo

Fl Fl
u(l) = (e − 1) = 1.71828 ; R1 = − F
EAo EAo

Two options are possible for the finite element solution: a) to use the exact expression
of the cross sectional area, and b) to assume a constant cross sectional area within each
element. The second option has been chosen here for simplicity. The reader is encouraged
to repeat this problem as an exercise using the first alternative.

One elements solution

The cross sectional area is assumed to be constant and equal to A = Ao e−1/2 . The
element stiffness matrix is directly given by Eq.(3.56), i.e.
· ¸ · ¸
EAo −1/2 1 −1 EAo 1 −1
K(1) = e = 0.60653
l −1 1 l −1 1

The equilibrium equation is deduced from Eq.(3.57) (noting that the distributed loads
b(e) are zero) as
· ¸ ½ ¾ ½ ¾
EAo 1 −1 u1 R1
0.60653 = ; u1 = 0
l −1 1 u2 F

104
which when solved gives

1 Fl Fl
u2 = = 1.6487 ; R1 = − F
0.60653 EAo EAo

The percentage of error with respect to the exact solution is 4.21%. This can be consid-
ered acceptable given the simplicity of the mesh.

Two elements solution

Now A(1) = Ao e−1/4 and A(2) = Ao e−3/4 . The equilibrium equations for each element
(Figure 3.5) are obtained as explained in Section 3.5.

Element 1 ( )
³ EA ´ · ¸ ½ ¾ (1)
o 1 −1 u1 X1
1.5576 = (1)
l −1 1 u2 X2

Element 2 ( )
³ EA ´ · 1 −1
¸ ½ ¾
u2
(2)
X1
o
0.9447 = (2)
l −1 1 u3 X2

After global assembly we have

     
1.5576 −1.5576 0 u1  R1 
EAo  u1 = 0
−1.5576 2.5023 −0.9447 u2 = 0
l    
0 −0.9447 0.9447 u3 F

which gives
Fl
u3 = 1.7005 (Error = 1.04%)
EAo
Fl
u2 = 0.377541 u3 = 0.6419
EAo
R1 = − F

Three elements mesh

For the three elements mesh (Figure 3.5) A(1) = Ao e−1/6 , A(2) = Ao e−1/2 and A(3) =
Ao e−5/6 . The equilibrium equations for each element are

Element 1 · ¸ ½ ¾ ( )
(1)
EAo 1 −1 u1 X1
2.5394 = (1)
l −1 1 u2 X2

Element 2 · ¸ ½ ¾ ( )
(2)
EAo 1 −1 u2 X1
1.8196 = (2)
l −1 1 u3 X2

Element 3 · ¸ ½ ¾ ( )
(3)
EAo 1 −1 u3 X1
1.3028 = (3)
l −1 1 u4 X2

105
 The global equilibrium equation after assembly is
     
2.5394 −2.5394 0 0 
 u1   R1 
EAo −2.5394        u1 = 0
 4.3590 −1.8196 0  u2 0
 =
l 0 −1.8196 3.1234 −1.3038 u3 
  0
     
0 0 −1.3038 1.3038 u4 F

and the solution is

Fl
u4 = 1.71036 (Error = 0.46%)
EAo
Fl
u3 = 0.55156 u4 = 0.9432
EAo
Fl
u2 = 0.230241 u4 = 0.3938
EAo
R1 = −F

Once the nodal displacements have been obtained, the axial strain and the axial stress
can be computed for each element. For example, at the central point of element number
2 in the three elements mesh we have
³ l(2) ´ (2) ³ l(2) ´ (2)
(2) (2)
u(l/2) = N1 x(2) = u1 + N2 x(2) = u2 =
2 2
1 1 Fl
= 0.23024u4 + 0.55156u4 = 0.3909u4 = 0.6686
2 2 EAo

Fl
(Exact value = 0.6487 . Error : 3.07%).
EAo

³ dN (2) ´ ³ dN (2) ´ F
1 (2) 2 (2)
ε(l/2) = u1 + u2 = 1.6491
dx x(2) = l(2) dx x(2) = l(2) EA o
³ 3 2
3 ´ 2
u4
= − 0.2302 + 0.5516 u4 = 0.9642
l l l
F F
σ(l/2) = EεA = 1.649 (Exact value: 1.6487 . Error: 0.02% )
Ao Ao
The convergence of the end displacement value with the number of elements is shown in
Figure 3.6. We see that the simple assumption of constant cross sectional area leads to
percentage errors of less than 1% for meshes finer than two elements.
The displacement and stress distribution along the rod for the three meshes are plotted
in Figure 3.7 together with the exact solution. Note that the nodal displacements and
even the linear displacement field within each element are very accurate for the three
meshes. However, the convergence of the constant stress field for each element to the
exact exponential solution is quite slow.

3.7 EXTRAPOLATION OF THE SOLUTION FROM TWO DIF-

FERENT MESHES
Expanding in Taylor series the displacement in the vecinity of a node i gives
³ ∂u ´ ³ ∂ 2u ´
u = ui + (x − xi ) + (x − xi )2 + · · · (3.58)
∂x i ∂x2 i

106
Fig. 3.6: Axially loaded rod with varying cross sectional area. Convergence of the end
displacement value and the stress distribution with the number of elements

Fig. 3.7: Nodal displacements and stress distribution along the rod for the three
meshes

If the shape functions Ni (x) are polynomials of pth degree it is obvious that
only the first p terms of the Taylor expansion can be approximated exactly as the
derivatives of order p + 1, p + 2, etc. are zero. The error of this approximation is
then of the order of the first term disregarded in the above expansion, i.e.
error = uexact − uapprox = O(x − xi )p+1 ' O(lp+1 ) (3.59)
where O(lp+1 ) is read as “of the order of lp+1 ” and l is the element length.

107
 Let us now consider two solutions u1 and u2 obtained with two meshes of
uniform element sizes l and l/d, respectively. We can write

uexact − u1 = O(lp+1 )
h l i (3.60)
uexact − u2 = O ( )p+1
d
The approximate value of uexact is obtained in terms of u1 and u2 from
Eqs.(3.60) as
(dp+1 )u2 − u1
uexact ∼
= (3.61)
(dp+1 ) − 1
This technique is known as Richardson extrapolation [R2]. For the rod of
Figure 3.5 we obtain for the end displacement value:

1. Extrapolated solution from meshes 1 and 2 (d = 2)

4u2 − u1 Fl
u(l) = = 1.7178 (Error: 0.03%)
3 EAo
2. Extrapolated solution using meshes 1 and 3 (d = 3)

9u2 − u1 Fl
u(l) = = 1.71807 (Error: 0.012%)
8 EAo
3. Extrapolated solution using meshes 2 and 3 (d = 1.5)
(9.5)2 u2 − u1 Fl
u(l) = = 1.71825 (Error: 0.002% )
(1.5)2 − 1 EAo

We see that Richardson’s extrapolation is an effective technique to improve the

displacement solution obtained from two meshes using elements of the same type.
This simple procedure is also applicable for 2D and 3D problems. Obviously, the
enhanced nodal displacement values can be used to obtain an improved solution
for the stress field. Unfortunately the improvement is not so relevant as for the
nodal displacements.

3.8 MATRIX FORMULATION OF THE ELEMENT EQUATIONS

The methodology explained in the previous sections is very useful for introducing
the basic steps of the FEM. However, for problems with more than one displace-
ment variable per node a matrix formulation is much more convenient as it allows
all variables and algebraic operations to be grouped together in a compact form.
The matrix formulation also provides a systematic finite element methodology
for all the structural problems treated in this book. The basic concepts of the
matrix formulation will be presented next.

108
 Most expressions used henceforth will be referred to an individual
element only. Therefore, superindex “e” denoting element values
will be omitted hereafter for simplicity, with the exception of a few
significative element parameters such as the main geometrical di-
mensions (l(e) , A(e) and V (e) ), the nodal displacement vector a(e) ,
the nodal coordinates vector x(e) , the nodal force vectors (f (e) , q(e) ),
the stiffness matrix K(e) , and other relevant elemental parameters.
All other parameters, vectors and matrices appearing in the text
should be interpreted, unless otherwise mentioned, as belonging to
an individual element.

3.8.1 Shape function matrix

Let us consider a general 2-noded rod element. The displacement field is expressed
within the element as
u = N1 u 1 + N2 u 2 (3.62)

Eq.(3.62) is written in matrix form as

½ ¾
u1
u = {u} = [N1 , N2 ] = N a(e) (3.63)
u2

where ½ ¾
(e) u1
N = [N1 , N2 ] ; a = (3.64)
u2

are the shape function matrix and the nodal displacement vector of the element.
As mentioned above, the superindex “e” denoting element values has been omit-
ted from most terms in Eqs.(3.62)-(3.64).

3.8.2 Strain matrix

The strain vector is written as
½ ¾
dN1 dN2 dN1 dN2 u1
ε = {ε} = { u1 + u2 } = [ , ] = B a(e) (3.65)
dx dx dx dx u2

where
dN1 dN2
B = [ , ] (3.66)
dx dx
is the strain matrix of the element.

109
3.8.3 Constitutive matrix
The stress vector is written as
σ = [N ] = (EA) ε = [EA] ε = DBa(e) (3.67)
where
D = [EA] (3.68)
is the matrix of mechanical properties of the material, also called hereafter con-
stitutive matrix.
For the axial rod problem vectors ε and σ and matrix D have a single compo-
nent only. In general σ and ε will have t components. Thus, if n is the number
of nodes of an individual finite element and d the number of d.o.f. of each node,
the dimensions of the vectors and matrices in the constitutive equation are
σ = D· B · a(e) (3.69)
t×1 t×t [t×(n×d)] [(n×d)×1]

3.8.4 Principle of Virtual Work

The PVW for an individual element is written in matrix form as
Z Z
T
δεε σ dx =
T
δuT b dx + [δa(e) ] q(e) (3.70)
l(e) l(e)

Note that the PVW is a scalar equation, i.e. both sides of Eq.(3.70) are
numbers representing the internal and external virtual work, respectively. This
explains the organization of the terms in Eq.(3.70), as a scalar number is obtained
as product of a row vector times a column vector, i.e. if s is a scalar number we
can write
 
 b1 
 

 b2 
s = a1 b1 + · · · + an bn = [a1 , a2 , . . . , an ] .. = aT b (3.71)

  . 


b 
n

Naturally, if vectors ε and σ have a single term, as in the axially loaded rod
problem, the vector product (3.71) reduces to multiplying two numbers. However,
for most of the problems studied in this book, vectors ε and σ have several
components and the matrix form of the PVW will be used.

3.8.5 Stiffness matrix and equivalent nodal force vector

From Eqs.(3.63) and (3.65) we have
T
[δu]T = [δa(e) ] NT
T
(3.72)
[δεε]T = [δa(e) ] BT

110
 Substituting now Eqs.(3.65), (3.67) and (3.72) into the PVW written for a
single element gives
Z Z
(e) T (e) T
[δa ] B D B a dx −
T
[δa(e) ] NT b dx = δa(e) q(e) (3.73)
l(e) l(e)

where b = {b} is the vector of distributed loads acting on the element. Elimina-
ting the virtual displacements from Eq.(3.73) yields
·³Z ´ Z ¸
(e) T (e) (e)
[δa ] B D B dx a
T
− N b dx − q
T
= 0 (3.74)
l(e) l(e)

As the virtual displacements are arbitrary, satisfaction of Eq.(3.74) implies

³Z ´ Z
(e)
B D B dx a
T
− NT b dx = q(e) (3.75)
l(e) l(e)

Eq.(3.75) is a system of algebraic equations which can be written in the stan-

dard form

K(e) a(e) − f (e) = q(e) (3.76a)

where
Z
(e)
K = BT D B dx
l(e)
Z (3.76b)
(e)
f = N b dx
T

l(e)

are respectively the stiffness matrix and the equivalent nodal force due to dis-
tributed loading for the element. Vector q(e) in Eq.(3.76a) is the equilibrating
nodal force vector for the element which is essential for the global assembly pro-
cess.
Note that the derivation of K(e) and f (e) is completely general . Expressions
(3.76) will frequently appear throughout the text and will be particularized for
each element type studied.
The explicit form of K(e) and f (e) for the 2-noded axially loaded rod element
is found by substituting into Eqs.(3.76b) the adequate expressions for B, D, N
and b. In this case we have
hx − x x − x i
2 1
N = [N1 , N2 ] = [N1 , N2 ] = (e)
, (e)
l l
h dN dN i h 1 1 i (3.77)
1 2
B = [B1 , B2 ] = , = − (e) , (e)
dx dx l l

D = [EA] and b = {b}

111
Substituting Eqs.(3.77) into (3.76b) gives
Z ½ 1
¾ h 1 1 i ³ EA ´(e) · 1 −1¸
(e) − l(e)
K = 1 (EA) − (e) , (e) dx =
l(e) l(e) l l l −1 1
½ ¾ ½ ¾ (3.78)
Z
x2 − x ³ b ´(e) (bl)(e) 1
f (e) = dx =
l(e) x − x1 l 2 1

Note the coincidence of these expressions with those obtained in Eq.(3.56) by

a more elaborate procedure.
We remark finally that the element stiffness matrix and the equivalent nodal
force vector can be obtained from the corresponding sub-matrices and subvectors.
Thus, from Eqs.(3.77) and (3.76b) we have
 .. 
½ T¾
Z Z BT1 DB1 . BT1 DB2
B1  
K(e) = D [B1 , B2 ] dx = . . . . . . . . . . . . . . . . . . . . . dx
l(e) B T
2 l(e) ..
BT2 DB1 . BT2 DB2
Z ½ T¾ Z ½ T ¾
N1 N1 b
f (e) = b dx = dx (3.79)
l(e) N2 l(e) N2 b
T T

(e)
Matrix Kij relating nodes i and j of element e is
Z
(e)
Kij = BTi D Bj dx ; i, j = 1, 2 (3.80)
d×d l(e) (d×t)(t×t)(t×d)

and the equivalent nodal force vector of node i of element e is

Z
(e)
fi = b dx i = 1, 2 (3.81)
(d×1) l(e) (d×d)

Recall that d is the number of d.o.f. of each node (i.e. d = 1 for the axially
loaded rod). For the particular case of the 2-noded rod element we have
Z ³ EA ´(e)
(e) dNi dNj
Kij = EA dx = (−1)i+j
l(e) dx dx l
(3.82)
Z (e)
(e) (bl)
fi = Ni b dx =
l(e) 2

from which the expressions of K(e) and f (e) of Eq.(3.78) can be obtained.
The computation of the element stiffness matrix K(e) and the equivalent nodal
(e) (e)
force vector f (e) from the nodal contributions Kij and fi is a simple and eco-
nomical procedure and it also facilitates the organization of a computer program.
We will be able to verify this on many occasions throughout the text.

112
3.9 SUMMARY OF THE STEPS FOR THE ANALYSIS OF A
STRUCTURE USING THE FEM
Let us summarize the main steps to be followed in the finite element analysis of
a structure.

Step 1. Discretize the structure into a mesh of finite elements.

Step 2. Compute for each element the stiffness matrix and the equivalent nodal
force vector using expressions of the type
Z Z
(e) (e)
K = B DB dx ; Kij =
T
BTi DBj dx
l(e) l(e)
Z Z
(e) T (e)
f = N b dx ; fi = NTi b dx (3.83)
l(e) l(e)

For two and three dimensional structures the element integrals are com-
puted over the element area or volume.

Step 3. Assemble the stiffness matrix and the equivalent nodal force vector for
each element into the global system

Ka = f (3.84a)

K = A K(e) ; f = A f (e) (3.84b)

e e

where A denotes the global assembly process of all the individual matrices
e
and vectors for each element in the mesh.

Step 4. The nodal displacement values are computed by solving the equation sys-
tem (3.84a) where the prescribed displacement must be adequately imposed,
i.e.
a = K−1 f (3.85)

Also, the nodal reactions are obtained at the prescribed nodes.

Step 5. The strains and stresses are computed within each element from the nodal
displacement values as

ε=Ba ; σ = DBa (3.86)

Details of each of above steps and of the precise form of the element vectors
and matrices will be given in the next chapters for each of the problems
studied.

113
3.10 ADVANCED ROD ELEMENTS AND REQUIREMENTS FOR
THE NUMERICAL SOLUTION
3.10.1 INTRODUCTION
The analysis of the simple axially loaded rod problem using 2-noded rod elements
studied in the previous chapter is of big interest as it summarises the basic steps
for the analysis of a structure by the FEM. However, a number of important
questions still remain unanswered, such as: Can higher order rod elements be
effectively used? What are their advantages versus the simpler 2-noded rod ele-
ment? Can it be guaranteed that the numerical solution converges to the exact
one as the mesh is refined? What are the conditions influencing the error in the
numerical solution? The reader who faces the application of the FEM for the first
time will certainly come across these and similar questions. In the following sec-
tions we will see that there are not definitive answers for many of the questions,
and in some cases only some practical hints are possible. For simplicity we will
mostly refer to the axially loaded rod problem as it allows a simple explanation
of topics which are of general applicability to more complex problems.

3.11 ONE DIMENSIONAL Co ELEMENTS. LAGRANGE

ELEMENTS
In the previous sections we introduced the basic concepts of the FEM using sim-
ple 2-noded one-dimensional elements with linear shape functions. Obviously,
the polynomial interpolation guarantees that the axial displacement is continu-
ous within the element and between adjacent elements. Elements satisfying this
condition are termed “C o continuous”. Additionally, we could require continuity
of the first derivative of the displacement and the approximation is then called
“C 1 continuous”. In general, an element is“C m continuous” if the displacement
field has continuous the n − 1 first derivatives. In Section 3.17.3 we will come
back to this subject. In this section we will derive the shape functions for C o
continuous 1D elements. These ideas will be very useful for deriving the shape
functions of 2D elements.
The approximation of a displacement unknown in 1D elements can be written
as
u(x) = αo + α1 x + α2 x2 + · · · (3.87)
where αo , α1 , etc., are constant parameters.
Let us choose a first degree polynomial

u(x) = αo + α1 x (3.88)

The parameters αo and α1 can be obtained from the value of u(x) at two element
points. This requires the element associated with the interpolation (3.2) to have

114
Fig. 3.8: Definition of the natural coordinate system ξ. Actual and normalized ge-
ometries for a 2-noded element

two nodes. For a 2-noded element of length `(e) with node 1 at x = x1 and node
2 at x = x2 (Figure 3.8), we have
u(x1 ) = u1 = αo + α1 x1
(3.89)
u(x2 ) = u2 = αo + α1 x2
where u1 and u2 are the values of the axial displacement at the two nodes. Sub-
stituting the values of αo and α1 obtained from Eq.(3.89) into Eq.(3.88) gives
u(x) = N1 (x)u1 + N2 (x)u2 (3.90a)
where
(x2 − x) (x − x1 )
N1 (x) = (e)
; N2 (x) = (3.90b)
l l(e)
are the element shape functions. Note the coincidence with the expressions ob-
tained in Section 3.3 (see Eq.(3.12)).
The shape functions for C o continuous 1D elements can be simply derived from
the expressions of Lagrange polynomials. A n − 1th degree Lagrange polynomial
`ni (x) is defined in terms of n points with coordinates x1 , x2 , · · · xn as follows
`ni (x) = (x − x1 )(x − x2 ) · · · (x − xi−1 )(x − xih ) · · · (x − xn ) (3.91)
Note that `ni (xi ) = yi (6= 0) and `ni (xj ) = 0 for j = 1, 2, · · · n(j 6= i). There-
fore if the points coincide with the element nodes and the non-zero value yi is
normalized to the unity, the resulting Lagrange polynomial coincides with the
shape function Ni (x) of the corresponding 1D element. This explains why C o
continuous 1D elements are also called “Lagrange” elements.
The shape function Ni or a Lagrange element with n nodes can be obtained
directly by the following expression
`ni (x)
Ni (x) = (3.92a)
`ni (xi )

115
which can also be written as
n
Y ³x−x ´
j
Ni (x) = (3.92b)
xi − xj
j=1(j6=i)

For a two-noded element we find again that

x − x2 x2 − x
N1 = =
x1 − x2 l(e)
(3.93)
x − x1 x − x1
N2 = =
x2 − x1 l(e)
A natural coordinate ξ is introduced for convenience as (Figure 3.8)
x − xc
ξ = 2 (3.94)
l(e)
where xc is the cartesian coordinate of the element midpoint. Eq. (3.8) gives

ξ = −1 at the left-hand end of the element

ξ = 0 at the element mid point
ξ = 1 at the right-hand end of the element

Eq. (3.94) transforms the actual element geometry into a normalized geometry
of length equal to 2. The shape functions can now be written in terms of the
natural coordinate ξ. By analogy with Eq.(3.92b) we write
n
Y ¡ ξ − ξj ¢
Ni (ξ) = (3.95)
ξi − ξj
j=1(j6=i)

For a linear Lagrange element with two nodes at ξ = −1 and ξ = +1 we

obtain
ξ − ξ2 1
N1 = = (1 − ξ)
ξ1 − ξ2 2
(3.96)
ξ − ξ1 1
N2 = = (1 + ξ)
ξ2 − ξ1 2
For a quadratic Lagrange element with three nodes at ξ1 = −1, ξ = 0 and
ξ = +1 (Figure 3.9) the shape functions are
(ξ − ξ2 )(ξ − ξ3 ) 1
N1 = = ξ(ξ − 1)
(ξ1 − ξ2 ) 2
(ξ − ξ1 )(ξ − ξ3 )
N2 = = (1 − ξ 2 ) (3.97)
(ξ2 − ξ1 )(ξ2 − ξ3 )
(ξ − ξ1 )(ξ − ξ2 ) 1
N3 = = ξ(1 + ξ)
(ξ3 − ξ1 )(ξ3 − ξ2 ) 2

116
 Fig. 3.9: Quadratic and cubic 1D elements with C o continuity

Finally, for a cubic element with four nodes at ξ1 = −1, ξ2 = −1/3, ξ3 = 1/3
and ξ4 = +1 (Figure 3.9) the shape functions are
µ ¶
(ξ − ξ2 )(ξ − ξ3 )(ξ − ξ4 ) 9 2 1
N1 = = − ξ − (ξ − 1)
(ξ1 − ξ2 )(ξ1 − ξ3 )(ξ1 − ξ4 ) 16 9
µ ¶
(ξ − ξ1 )(ξ − ξ3 )(ξ − ξ4 ) 27 1
N2 = = ξ− (ξ 2 − 1)
(ξ2 − ξ1 )(ξ2 − ξ3 )(ξ2 − ξ4 ) 16 3
µ ¶ (3.98)
(ξ − ξ1 )(ξ − ξ2 )(ξ − ξ4 ) 27 1
N3 = = − ξ+ (ξ 2 − 1)
(ξ3 − ξ1 )(ξ3 − ξ2 )(ξ3 − ξ4 ) 16 3
µ ¶
(ξ − ξ1 )(ξ − ξ2 )(ξ − ξ3 ) 9 2 1
N4 = = (ξ + 1) ξ −
(ξ4 − ξ1 )(ξ4 − ξ2 )(ξ4 − ξ3 ) 16 9

The cartesian expressions of the above shape functions can be obtained from
the transformation (3.94). However, only the normalized forms are necessary in
practice.
The reader is encouraged to derive by himself the expressions of the shape
functions of higher order 1D Lagrange elements.
The shape functions of C 1 continuous 1D elements will be derived in Chapter
7.

117
3.12 ISOPARAMETRIC FORMULATION AND NUMERICAL IN-
TEGRATION
3.12.1 Introduction
We will now introduce two key concepts which have been essential for the de-
velopment of the FEM. The first one is that of isoparametric interpolation. The
basic idea is to interpolate the element geometry from the coordinates of the
nodes. Such an interpolation is essential to obtain a general relationship between
the natural and cartesian coordinates.
The second concept is that of numerical integration. In most cases the exact
analytical computation of the element integrals is not possible and numerical
integration is the only option to evaluate them in a simple and precise way.
The application of these two techniques to C o continuons 1D elements is
presented in the next sections. However, the advantages of these procedures will
become clearer when dealing with 2D and 3D elements.

3.12.2 The concept of parametric interpolation

Let us recall the displacement interpolation for a 2-noded axial rod element:

u(ξ) = N1 (ξ)u1 + N2 (ξ)u2 (3.99)

Note that in Eq.(3.99) we have used the expression of the shape funtions in
terms of the natural coordinate ξ. With a few exceptions this will be the usual
procedure throughout the text.
The axial strain is obtained from Eq.(3.99) as

du dN1 (ξ) dN2 (ξ)

ε = = u1 + u2 (3.100)
dx dx dx
The cartesian derivatives of the shape functions are therefore needed to com-
pute the strain. This would be an easy task if the shape functions were expressed
in terms of the cartesian coordinate x. However, as this will not generally be the
case, some transformations are necessary. For the 1D problem we have

dN1 (ξ) dN1 (ξ) dξ d ³1 − ξ ´ dξ 1 dξ

= = = −
dx dξ dx dξ 2 dx 2 dx
(3.101)
dN2 (ξ) dN2 (ξ) dξ d ³1 + ξ ´ dξ 1 dξ
= = =
dx dξ dx dξ 2 dx 2 dx

and the strain is obtained by

1 ³ dξ ´ 1 ³ dξ ´
ε = − u1 + u2 (3.102)
2 dx 2 dx

118
 dξ
Eq.(3.102) involves the evaluation of dx . This obviously requires an explicit
relationship between ξ and x which can be obtained using a parametric interpo-
lation of the element geometry. This expresses the coordinate of any point within
the element in terms of the coordinates of m element points x1 , x2 , · · · , xm by the
following interpolation
x = N̂1 (ξ)x1 + N̂2 (ξ)x1 + · · · + N̂m (ξ)xm (3.103)
In Eq.(3.103) N̂i (ξ) are geometry interpolation functions which satisfy the
same requirements as the displacement shape functions; i.e. N̂i (ξ) takes the
value one at point i and zero at the other m − 1 points for which the coordinates
are known. Hence, the expression of N̂i (ξ) can be obtained simply by changing
n for m in Eq.(3.95), where ξi are now the natural coordinates of the geometry
interpolating points.
It is important to note that Eq. (3.103) yields precisely the relationship we
are looking for between the coordinates ξ and x. This expression can also be
interpreted as transformation between coordinates ξ and x, such that every point
in the normalized space [-1,1] is mapped onto another point in the cartesian space
[x1 , x2 ]. It is essential that this mapping be unique and this generally depends
on the element geometry.

Example: 3.2 Parametric interpolation of a cubic polinominal.

- Solution

Let us consider the polynomial y = x3 − 2x2 − x + 4 plotted in Figure 3.10. Such a

function can represent, for instance, the geometry of a curved beam or the boundary of
a curved 2D element. We will assume that the coordinates of the three points at x1 = 0,
x2 = 1, and x3 = 2 are known.
The coordinates of the three points will be used to build up a quadratic approximation
using a 3-noded 1D element. The relationship between the cartesian (x, y) coordinates
and the natural coordinate ξ is obtained as a particular case of Eq.(3.103), i.e.
3
X 1 1
x= Ni (ξ)xi = ξ(ξ − 1)x1 + (1 − ξ 2 )x2 + ξ(ξ + 1)x3 = 1 − ξ
i=1
2 2
3
X 1 1
y= Ni (ξ)yi = ξ(ξ − 1)y1 + (1 − ξ 2 )y2 + ξ(1 + ξ)y3 = ξ 2 − ξ + 2
i=1
2 2

Figure 3.10 shows the approximating quadratic function. Note the error with respect to
the “exact” cubic function. Also note that this error is much larger outside the interval
[0,2] which includes the three points selected.
The accuracy can be dramatically improved by using a cubic approximation in terms of
the coordinates of four known points at x1 = 0, x2 = 2/3, x3 = 4/3, and x4 = 2.0, with
y(x1 ) = 4.0, y(x2 ) = 74/27, y(x3 ) = 40/27, and y(x4 ) = 2.0, respectively. A cubic 1D
element is now used giving
4
X 4
X
x= Ni (ξ)xi ; y= Ni (ξ)yi
i=1 i=1

119
 Fig. 3.10: Parametric interpolation of a cubic polynomial

where Ni are the cubic shape function of Eq.(3.97). After some algegra, the following is
obtained (easily)

x=1+ξ ; y = (1 + ξ)3 − 2(1 + ξ)2 − (1 + ξ) + 4

As expected, the cubic field chosen exactly approximates the original cubic function.

Example 3.2 shows that important errors in the geometry approximation can
occur unless a correct interpolation of the geometry is chosen. These errors are
undesirable and should be avoided or, at least, minimized.
Two types of points must therefore be considered in an element: a) the points
used for interpolating the displacement field (nodes) by the shape functions Ni (ξ);
and b) the points chosen to approximate the element geometry via the geometry
interpolation function N̂i (ξ). These two sets of points can coincide depending on
the problem. Complex structures might require a higher order interpolation of
the geometry, whereas a simple geometry can be exactly approximated using a
linear field for N̂i , independently of the interpolation used for the displacement
field.
If the number of geometry points m is greater than that of element nodes, the
geometry interpolation functions N̂i will be polynomials of a higher order than
the displacement shape functions, and the element is termed superparametric. If
n coincides with the number of nodes, then Ni ≡ N̂i and the element is isopara-
metric. Finally, if the number of geometry points is less than that of nodes, the
element is called subparametric.

120
 In practice it is usual to choose an isoparametric formulation. However, it is
important to have a clear picture of the two other options which are useful in
some cases.
The isoparametric concept originates from the work of Taig [T1,2] who derived
the first 4-noded isoparametric quadrilateral. Irons [I3,4] extended these ideas to
formulate high order isoparametric elements. In the following chapters we will
extend this concept for 2D and 3D elements.

3.12.3 Isoparametric formulation of the two-noded rod element

The geometry of the linear rod element is expressed in terms of the coordinates
of the two nodes as
x(ξ) = N1 (ξ)x1 + N2 (ξ)x2 (3.104)
where N1 and N2 are the same linear shape functions used for interpolating the
displacement field (see Eq.(3.96)).
From Eq.(3.104) we obtain
dx dN1 dN2 1 1 l(e)
= x1 + x2 = − x1 + x2 = (3.105)
dξ dξ dξ 2 2 2
and
l(e) dξ 2
dx = dξ y = (e) (3.106)
2 dx l
Substituting Eq.(3.106) into (3.101) gives
dN1 2 dN1 1
= (e) = − (e)
dx l dξ l
(3.107)
dN2 2 dN2 1
= (e) = (e)
dx l dξ l
and from Eqs.(3.107) and (3.100)
h i
ε = −1/l(e) , 1/l(e) a(e) = B a(e) (3.108a)

with h i
(e) (e)
B = −1/l , 1/l (3.108b)
which naturally coincides with the expressions previously derived by a more direct
procedure. The systematic approach chosen here is useful in order to understand
the application of the isoparametric concept.
The stiffness matrix and the equivalent nodal force vector are expressed in
the natural coordinate system combining Eqs.(3.106) and (3.83b) as
Z +1 Z +1
(e) T l(e) (e) T l(e)
K = B (EA)B dξ , f = N b dξ (3.109)
−1 2 −1 2

For homogeneous material and uniformly distributed loading the computation

of the above integrals is simple, leading to the expressions (3.85).

121
3.12.4 Isoparametric formulation of the 3-noded quadratic rod element
We will now study the 3-noded rod element of Figure 3.9 with quadratic shape
functions. The axial displacement is expressed by

u = N1 (ξ)u1 + N2 (ξ)u2 + N3 (ξ)u3 (3.110)

where the shape functions N1 (ξ), N2 (ξ) and N3 (ξ)are given by Eq.(3.97).
The x coordinate of a point within the element is written in the isoparametric
formulation as
x = N1 (ξ)x1 + N2 (ξ)x2 + N3 (ξ)x3 (3.111)
The axial strain is obtained by
 
3
X · ¸ u1 
du dNi dN1 dξ dN2 dξ dN3 dξ
ε= = ui = , , u2 = B a(e)
dx dξ dξ dx dξ dx dξ dx  
i=1 u3
(3.112)
From Eq.(3.97)
dN1 1 dN2 dN3 1
= ξ− ; = − 2ξ ; =ξ+ (3.113)
dξ 2 dξ dξ 2
and the strain matrix B is
³ dξ ´ h 1 1 i
B = (ξ − ), −2ξ, (ξ + ) (3.114)
dx 2 2
dx
The derivative dξ
is computed from Eq.(3.111) as

dx dN1 dN2 dN3 1

= x1 + x2 + x3 = (ξ − ) x1 −
dξ dξ dξ dξ 2
(e)
(3.115)
1 l
− 2ξx2 + (ξ + ) x3 = + ξ (x1 + x3 − 2x2 )
2 2
and dξ 2
= (e) (3.116)
dx l + 2ξ(x1 + x3 − 2x2 )
Eq.(3.116) provides a relationship between dx and dξ in terms of the three
nodal coordinates. In the (usual) case that the central node is located at the
element midpoint, we have
dξ 2
= (e) (3.117)
dx l
and dx 2 l(e)
= (e) y dx = dξ (3.118)
dξ l 2
In this latter case the strain matrix of Eq.(3.114) is simply
2 h 1 1 i
B = (e) (ξ − ), −2ξ, (ξ + ) (3.119)
l 2 2

122
 The expression of B for an arbitrary position of the central node is obtained
by substituting Eq.(3.116) into (3.114).
The element stiffness matrix and the equivalent nodal force vector are obtained
from the PVW as explained the 2-noded element. It can easily be found that the
element stiffness matrix has once again the general form
Z
(e)
K = BT (EA) B dx (3.120)
l(e)

Substituting the above expressions of dx and B in terms of ξ into Eq.(3.120)

leads to (in the case of the mid-node being central in the element)
 
Z +1 (ξ − 12 ) · ¸ (e)
(e) 2 2 1 1 l
K = −2ξ (EA) (e) (ξ − ), −2ξ, (ξ + ) dξ (3.121)
−1 l
(e)   l 2 2 2
(ξ + 21 )

The computation of the above integral is straightforward if both E and A are

constant over the element, giving
 
³ EA ´(e) 14 −16 2
K(e) = −16 32 −16 (3.122)
6l
2 −16 14

The equivalent nodal force vector for a distributed loading of intensity b is

 
Z Z +1  21 ξ(ξ − 1)
l(e)
f (e) = NT b dx = 1 − ξ2 b dξ (3.123)
l(e) −1  1 ξ(1 + ξ) 2
2

For a uniformly distributed loading

 
(bl) (e) 1
f (e) = 4 (3.124)
6  
1

We note that in this case the central node absorbes four times more load-
ing than the end nodes. This result which is not the obvious one, is a natural
consequence of the PVW and the quadratic approximation chosen.
The expressions of K(e) and f(e) for an arbitrary position of the central node are
obtained by making use of Eq.(3.116). Note that in this case rational polynominal
terms in ξ are involved and, therefore, the analytical computation of the element
integrals is not so simple.
The isoparametric formulation of higher order rod elements follows the rules
explained for the quadratic element. The increasing complexity of the element
integrals can be overcome by using numerical integration as explained in the next
section.

123
3.13 NUMERICAL INTEGRATION
We have seen that in some cases the exact analytical computation of the integrals
appearing in K(e) and f (e) can be difficult and sometimes impossible. This is
typically the case for 2D and 3D isoparametric elements, due to the complexity
of the rational terms involved in the integral expression. Numerical integration
appears here as the only option to compute the element integrals in a simple and
accurate way.
To enter into the mathematics of numerical integration falls outside the scope
of this text. For simplicity we will only consider here the Gauss quadrature [R2]
as this is the more popular numerical integration procedure used in the FEM.
We will introduce here the basic ideas for 1D problems which will be extended
for 2D and 3D problems in subsequent chapters.
Let us assume that the integral of a function f (x) in the interval [-1,1] is
required Z +1
I = f (ξ) dξ (3.125)
−1
The Gauss integration rule, or Gauss quadrature, expreses the value of the
above integral as a sum of the products of the function values at a number of
known points by prescribed weights. For a quadrature of order qth
q
X
I ' Iq = f (ξi )Wi (3.126)
i=1

where Wi is the weight corresponding to the ith sampling point located at ξ = ξi

and q the number of sampling points. A Gauss quadrature of qth order integrates
exactly a polynomial function of degree 2q − 1 [R2]. The error in the computation
of the integral is of the order 0(42q ), where 4 is the spacing between the sampling
points. The coordinates of the sampling points and the corresponding weights
for the first six Gauss quadratures are shown in Table 3.3.
Note that the sampling points are all located within the normalized domain
[-1,1]. This is useful for the computation of the element integrals expressed in
terms of the natural coordinate ξ. The popularity of the Gauss quadrature derives
from the fact that it requires the minimum number of sampling points to achieve
a prescribed error in the computation of the integral. Thus, it minimizes the
number of times the integrand function is computed. The reader can find further
details in [R1,2] and [P13].

Example: 3.3 Applications of the Gauss quadrature.

-- Solution

Let us consider the polynomial

f (x) = 1 + x + x2 + x3 + x4

124
 q ξq Wq
1 0.0 2.0
2 ±0.5773502692 1.0
±0.774596697 0.5555555556
3
0.0 0.8888888889
±0.8611363116 0.3478548451
4
±0.3399810436 0.6521451549
±0.9061798459 0.2369268851
5 ±0.5384693101 0.4786286705
0.0 0.5688888889
±0.9324695142 0.1713244924
6 ±0.6612093865 0.3607615730
±0.2386191861 0.4679139346
±0.9491079123 0.1294849662
±0.7415311856 0.2797053915
7
±0.4058451514 0.3818300505
0.0 0.4179591837
±0.9602898565 0.1012285363
±0.7966664774 0.2223810345
8
±0.5255324099 0.3137066459
±0.1834346425 0.3626837834
Table 3.3: Coordinates and weights of Gauss quadratures

The exact integral of f (x) over the interval −1 ≤ x ≤ 1 is

Z +1
2 2
I = f (x)dx = 2 + + = 3.0666
−1 3 5

- First order Gauss quadrature:

q = 1 , x1 = 0 , W 1 = 2 ; I = W1 f (x1 ) = 2

- Second order quadrature:

(
x1 = −0.57735 , W1 = 1
q=2
x2 = +0.57735 , W2 = 2

I = W1 f (x1 ) + W2 f (x2 ) = 0.67464 + 2.21424 = 2.8888

- Third order quadrature:

 
 x1 = −0.77459
 , W1 = 0.5555

q=3 x2 =0.57735 , W2 = 0.8888

 

x3 = +0.77459 , W3 = 0.5555

125
 I = W1 f (x1 ) + W2 f (x2 ) + W3 f (x3 ) = 0.7204 × 0.5555 +

+ 1.0 × 0.8888 + 3.19931 × 0.5555 = 3.0666 Exact value

We see that the exact integration of a fourth order polynominal requires a third order
Gauss quadrature as expected.

3.14 STEPS FOR THE COMPUTATION OF MATRICES AND

VECTORS FOR AN ISOPARAMETRIC ROD ELEMENT
We will now present the basic steps for computing the stiffness matrix and the
equivalent nodal vector for an isoparametric rod element with n nodes. The steps
have been arranged so as to facilitate their implementation within a computer
program.

3.14.1 Interpolation of the axial displacement

The axial displacement within the element is expressed as

u = N1 u1 + N2 u2 + . . . + Nn u n =


 u1  
Xn  u2 
 
= Ni ui = [N1 , N2 , . . . , Nn ] .. = N a(e) (3.127)
.
 
i=1
u 
 
n

3.14.2 Geometry interpolation

The coordinate x is interpolated using the isoparametric form as
n
X
x = N1 x 1 + N2 x 2 + . . . + Nn x n = Ni x i =
i=1
 

 x1 
  
x2
= [N1 , N2 , . . . , Nn ] = N x(e) (3.128)

. . . 
  
xn

126
3.14.3 Interpolation of the axial strain
The axial strain is expressed in terms of the nodal displacements as
Xn
du dN1 dN2 dNn dNi
ε = = u1 + u2 + . . . + un = ui =
dx dx dx dx i=1
dx
 

 u1  
h dN dN i 
 

1 2 dN n u 2
= , ,..., .. = B a(e) (3.129)
dx dx dx   . 


u  
n

The cartesian derivative of the shape functions is obtained by

dNi dNi dξ
= (3.130)
dx dξ dx
From Eq.(3.128) we deduce
Xn
dx dNi
= xi = J (e) (3.131)
dξ i=1
dξ
and, therefore
dξ 1
dx = J (e) dξ ; = (e) (3.132)
dx J
and dNi 1 dNi
= (e) (3.133)
dx J dξ
Substituting Eq.(3.133) into the expression of B gives
· ¸ · ¸
dNi dN2 dNn 1 dNi dN2 dNn
B = , ,..., = (e) , ,..., (3.134)
dx dx dx J dξ dξ dξ
We point out that J (e) = dx
dξ
is the Jacobian of the 1D coordinate transforma-
tion x → ξ. In the following chapters we will find that for 2D and 3D problems
J (e) is a matrix whose determinant relates the infinitessimal areas (for 2D) and
volumes (for 3D) in the cartesian and natural coordinate systems.

3.14.4 Computation of the axial force

The axial force N is obtained in terms of the nodal displacements by
N = (EA) ε = D B a(e) (3.135a)
with
D = [EA] (3.135b)

3.14.5 Element stiffness matrix

The PVW leads to the following general expression for the element stiffness matrix
Z Z +1
(e) T
K = B D B dx = BT D B J (e) dξ (3.136)
l(e) −1

127
From Eqs.(3.134) and (3.135b) we deduce
Z +1
(e) 1 dNi dNj
Kij = (e)
(EA) dξ (3.137)
−1 J dξ dξ
The simplicity of the above integral depends on the expression of the shape
functions and of J (e) . In general, the computation of K(e) is perfomed using the
Gauss quadrature which evaluates (3.137) as exactly as possible. For a qth order
quadrature we can write.
q · ¸
(e)
X 1 dNi dNj
Kij = (e) dξ
(EA) Wr (3.138)
r=1
J dξ r

where [·]r denotes values computed at the sampling point ξ = ξr .

3.14.6 Equivalent nodal force vector

For a distributed loading of intensity b(x) we have
Z Z +1
(e) T
f = N b dx = NT · b J (e) dξ (3.139)
l(e) −1

The computation of Eq.(3.139) can be standarized by using numerical inte-

gration as
q
(e)
X
fi = [Ni b J (e) ]r Wr (3.140)
r=1

(e)
Example: 3.4 Compute the term K11 of the stiffness matrix for the 3-noded rod element of
Figure 3.9 using an isoparametric formulation and numerical integration.

- Solution
(e)
The term K11 is obtained from Eq.(3.137) as

Z +1
(e) 1 dN1 dN1
K11 = (EA) dξ
−1 J (e) dξ dξ

The expression of N1 for the 3-noded rod element is (see Eq. (3.97))

1 dN1 1
N1 = (ξ − 1)ξ and =ξ−
2 dξ 2

l(e)
Assuming that node 2 is centered in the element J (e) = .
2
dN1 (e)
Substituting and J (e) into K11 we have
dξ
Z +1
(e) 2EA 1
K11 = (e)
(ξ − )2 dξ
−1 l 2

128
 The integrand is a quadratic function and hence, the exact integral requires a Gauss
quadrature of 2nd order (q = 2). From Eq.(3.138) and Table 3.1 we obtain

2 h
(e)
X 2EA 1 2i h 2EA 1 2i
K11 = (ξ − ) W r = (ξ − ) √ +
r=1
l(e) 2 r l(e) 2 ξ=− 33
h 2EA 1 i 7 EA (e)
+ (e) (ξ − )2 √ = ( )
l 2 ξ= 33 3 l

The same procedure can be followed for computing the rest of terms of K(e) .

3.15 BASIC ORGANIZATION OF A FINITE ELEMENT PRO-

GRAM
The steps presented in the previous section for computing the stiffness matrix and
the equivalent nodal force vector of the simple rod element are general and almost
identical to those required for more complex 2D and 3D elements. Also, these
steps define naturally the basic subroutines of a computer program for structural
analysis using the FEM. Here we will just introduce the basic format of a finite
element program for structural analysis.
Figure 3.11 shows the flow chart of a finite element program for the analysis of
axially loaded rods. The first subroutine deals with the reading of the geometrical
and mechanical data required for the analysis (subroutine INPUT). Then, the
stiffness matrix and the equivalent nodal force vector are computed for each
element in subroutines STIFFNESS and LOAD, respectively. The next step is
the assembly and solution of the global system of algebraic equilibrium equations
to obtain the nodal displacement values in subroutine SOLVE. Finally, the strains
and stresses are computed at selected points within each element in subroutine
STRESS. Note the analogy of the program skeleton with that of a program for
matrix analysis of bar structures [L3], [L5].

3.16 SELECTION OF ELEMENT TYPE

The first task in the analysis of a structure by the FEM is to select the element
to be used. This is an important decision and, moreover, not a simple one, as
there are many elements available for solving the same structural problem, and
each one has different advantages and disadvantages with regard to simplicity,
accuracy, cost, etc.
In most cases, the selection of an element for a particular problem is made by
the analyst responsible for the computations. This decision should be based on:
1) the characteristics of the structure to be analysed; 2) the elements available in
comercial on in-house computer programs and the type of computer to be used;

129
 

Input:



• Element type


SUBROUTINE INPUT 
• Mesh topology and nodal

Input data defining coordinates
geometry and mechanical 
• Material properties


properties. 
• Boundary conditions
¯ 

¯ 
• Coordinates and weights of
¯ 
¯ Gauss quadrature
¯
¯ 
¯
¯ 
 Compute at each Gauss point r :
¯ 

¯ 
 • Material properties (EA)



 • Derivatives ∂N i
SUBROUTINE STIFFNESS  (e) P ∂N∂ξ

Evaluate the stiffness matrix •J = i

∂ξ i
x

 i
for each element 
 •B
¯ 
Compute:
¯ 
 P (e) T
¯ 

¯ • K(e) = [J B (EA)B]r Wr
¯ r
¯
¯
¯ 
¯
¯ 
 Compute at each Gauss point r :



 • Shape functions
P ∂Ni Ni
SUBROUTINE LOAD  (e)
•J = ∂ξ i
x
Evaluate the equivalent nodal i


force vector for each element 
 Compute: P
¯ 

¯ • f (e) = [J (e) NT b]r Wr
¯ r
¯
¯

SUBROUTINE SOLVE  Gaussian elimination [R2]
Assembly and solution of Frontal method [H5]

Ka = f Profile solver [Z6], etc.
¯
¯
¯
¯
¯
SUBROUTINE STRESS ½
ε = Ba
Evaluate strains and stresses
σ = DBa
for each element.
¯
¯
¯
STOP

Fig. 3.11: Flow chart of a finite element program for analysis of axially loaded rods

and 3) the experience of the analyst in the solution of similar structures by the
FEM.
Several rules for the selection of the best element for each particular structural

130
problem will be given throughout this text. Nevertheless a few rules of “thumb”
can be summarized at this stage. These are:

1. The element chosen must be robust. This simply means that there should
be no danger of obtaining a spurious solution due to intrinsic bad behaviour
of the element under general geometrical or mechanical conditions. A test
for robustness of the element is provided by the patch test studied in a next
section.
2. The mesh should account for the probable stress gradients in the solution,
i.e. the mesh should be finer in zones where stress gradients are expected to
be higher. Here the use of error estimators and adaptive mesh refinement
procedures is recommended.
3. The element should be as accurate as possible. The debate between using
few elements of high order, or a finer mesh of simpler low order elements is
still open in FEM practice. We note that the growing popularity of adaptive
mesh refinement strategies, and the continuing increase in computer power
is favouring the use of low order elements.

The choice of low or high order order elements is schematically represented

in Figure 3.12 showing the approximation of a third degree polynomial function
representing the solution of an axial rod problem using different elements. Note
that a large number of simple 2-noded elements is required, whereas a single
4-noded cubic element provides the exact solution.
An indicator to decide between two elements is the ratio between the accuracy
and the number of nodal variables. This requires a definition of “accuracy”, which
is not obvious if the exact solution is not known “a priori”. Another guideline is
that in case of doubt between two elements of different order, the analyst should
always choose the simplest one (which is generally the low order one).
A comparison between the quadratic and linear rod elements is presented
next.

Example: 3.5 Solve the problem of Figure 3.2 using a single 3-noded quadratic rod element.

- Solution

Since we have only one element the global equilibrium equation is written from Eqs.(3.122)
and (3.124) as
     bl 
EA 
14 −16 2 u1   6 + R
−16 32 −16 u2 = 2bl
, u1 = 0
6l    bl 3 
2 −16 14 u3 6 + P

Solving the above system with the condition u1 = 0, gives

3bl2 Pl bl2 Pl
u2 = + , u3 = +
8EA 2A 2EA EA

131
Fig. 3.12: Approximation of a cubic solution with different rod elements. For simplic-
ity the finite element solution has been assumed to be exact at the nodes

It is easy to check that these values coincide with the exact solution (3.114) at the nodes.
The displacement field within the element is

1
u = (1 − ξ 2 )u2 + (1 + ξ)ξu3
2

132
 2x−l
Substituting the values for u2 and u3 and making the change of variable (ξ = l )
gives
1 bx2
u = (− + (P + bl)x)
EA 2

which coincides with the exact solution (3.28). This could have been anticipated as the
assumed displacement field contains the quadratic solution. The axial strain and stress
fields within the element are

1 ³ ´
ε = P + b(l − x)
EA

σ = P + b(l − x)

which again coincide with the exact solution.

This example shows that the quadratic rod element has a better perfomance
than the linear one (Figure 3.3). However, this is not a general rule in favour of
quadratic elements, as in many cases (particularly for 3D problems) their greater
accuracy is counterbalanced by an excesive computing cost.

3.17 REQUIREMENTS FOR CONVERGENCE OF THE SOLU-

TION
The finite element approximation must satisfy certain conditions which guarantee
that as the mesh is refined the numerical solution converges to the exact values.
It is very important that the elements satisfy these conditions which are the basis
for the success of mesh refinement strategies.

3.17.1 Continuity condition

The displacement must be continuous within each element. This condition is
automatically satisfed by using polynomial approximations for the displacement
field. The issue of continuity of displacements along the element interfaces is
treated in Section 3.19.1.

3.17.2 Derivativity condition

Obviously, the derivatives of the polynomial approximation should exist up to
the order of the derivatives appearing in the element integrals.
For instance, for the axially loaded rod problem the element expressions de-
rived from the PVW contain first order derivatives only. This requires that the
shape functions should be at least first order polynomials.

133
 Fig. 3.13: Integral of a bilinear function and its two first derivatives

3.17.3 Integrability condition

Logically, the integrals appearing in the element expressions must have a primitive
function. This condition can be explained by considering the simple example of
Figure 3.13 where a continuous function f (x) and its two first derivatives are
represented. We see that the integral of f (x) in the interval considered exists
and it is equal to the area shown in the figure. Also, the integral of f 0 (x) exists,
although it is not a continuous function. Finally, we observe that the second
derivative f 00 (x) has two singular points, due to the discontinuity of f 0 (x), and it
is not integrable. Therefore, the general rule deduced from this simple example
is the following. The derivative of a function is integrable if its m − 1 first
derivatives are continuous (C m−1 continuity). It can be thus stated that if mth
order derivatives of the displacement field appear in the PVW, the displacement
field (and therefore the shape functions) must be C m−1 continuous. This condition
ensures that the strains at the interface between elements are finite (even though
they are indeterminate).
As an example let us consider the axial rod problem. The PVW (Eq.(3.3))
contains only first derivatives of u and hence just displacement continuity (C o
continuity) is required. Such a continuity is guaranteed within each element by
the polynomial approximation chosen, and between elements by the coincidence
of the nodal interelemental values.

3.17.4 Rigid body condition

The displacement field closen should not allow straining of an element to occur
when the nodal displacements are caused by a rigid body displacement.
This physical condition is satisfied for a single element if the sum of the shape
functions at any point is equal to one. To prove this, let us consider the simple
2-noded axially loaded rod element with equal prescribed nodal displacements ū.

134
Within the element we have

u = N1 ū + N2 ū = (N1 + N2 ) ū (3.141)

and for u = ū then N1 + N2 = 1 must be satisfed.

3.17.5 Constant strain condition

The displacement function has to be such that if nodal displacements are com-
patible with a constant strain field, such constant strain will in fact be obtained.
Clearly, as elements get smaller, nearly constant strain conditions will prevail
in them. It is therefore desirable that a finite size element should be able to
reproduce a constant strain condition.
The constant strain criterion incorporates the rigid body requeriment, as a
rigid body displacement is a particular case of a constant (zero) strain field.
Strictly both criteria need only be satisfied in the limit as the size of the elements
tends to zero. However, satisfaction of these criteria on elements of finite size
leads to a convergent and more accurate solution.

3.18 ASSESSMENT OF CONVERGENCE REQUIREMENTS.

THE PATCH TEST
The patch test was first introduced by Irons [I5,I7] and has since then provided a
necessary and sufficient condition for convergence. The test is based on selecting
an arbitrary patch of elements and imposing upon it nodal displacements corre-
sponding to any state of constant strain. If nodal equilibrium is achieved without
imposing external nodal forces, and if a state of constant stress is obtained, then
clearly the constant strain criterion of the previous section is satisfied and fur-
thermore, displacement continuity is guaranteed, since no external work is lost
through the interelement interfaces.
The patch test also includes the satisfaction of the rigid body condition by
simply imposing a nodal displacement field corresponding to a zero strain value.
An alternative patch test is to prescribe a known linear displacement field at
the boundary of the patch nodes only. It is then verified that the displacement
solution at the interior nodes coincides with the exact values and that a constant
strain field is obtained throughout the patch.
An interesting feature of the patch test is its ability to asess the convergence
of elements with shape functions which are discontinuous along the element inter-
faces between adjacent elements. This issue will be discussed further in Section
3.19.1.
The application of the patch test to the simple 2-noded rod element is shown
in the next example.

135
 Fig. 3.14: Example of patch test in 2-noded rod elements

Example: 3.6 Apply the patch test to the three element patch of 2-noded rod elements shown
in Figure 3.14. All elements have equal length and the same material properties.

-- Solution

(a) Constant strain condition

We will assume a displacement field u = l + x giving a constant strain field in the whole
mesh, i.e. ε = du
dx = 1. The following displacements are prescribed at the end nodes of
the patch:
u3 = l + 3l = 4l
u6 = l + 6l = 7l

We now look for the solutioon for the nodal displacements u4 and u5 . The equation
system to be solved is
     
1 −1 0 0 u3 
  
R3  u3 = 4l
EA      
−1 2 −1 0 u4 0
=
l  0 −1 2 −1 u5 
  
 0
    
0 0 −1 1 u6 R6 u6 = 7l

which gives u4 = 5l and u5 = 6l. These values coincide with the exact ones given by the

136
 prescribed field. It can also be checked that

1 1
ε(3) = − u3 + u4 = 1
l l
(4) 1 1
ε = − u4 + u5 = 1
l l
(5) 1 1
ε = − u5 + u6 = 1
l l

which corresponds to the exact constant field imposed. Therefore, the element satisfies
the patch test.

(b) Rigid body condition

A particular case of the previous example is to study the perfomance of the patch subjeted
to a the constant displacement field u = 3l, corresponding to a rigid body movement of
the patch. The FEM solution yields u4 = u5 = 3l, which is the correct answer giving a
zero strain field over the patch.

3.19 OTHER REQUIREMENTS FOR THE FINITE ELEMENT

APPROXIMATION
Next, we will consider some requirements which, in fact, are not strictly necessary
for the convergence of the finite element solution. However, their fulfilment is
always desirable since, otherwise, the convergence and accuracy of the solution
can deteriorate in some cases.

3.19.1 Compatibility condition

The elements must be compatible. This implies that the displacement field for
C 0 elements, or its first derivative field for C 1 elements, must be continuous along
interelemental boundaries. This is a consequence of the more general integrabil-
ity condition of Section 3.18.3. Elements satisfying the compatibility condition
are termed compatible or conforming. These elements, when integrated exactly,
always converge to the exact solution from the stiffer side.
The compatibility condition is usually satisfied when the displacement field
is defined by a polynomial taking a unique value at the nodes. This, however,
is not sufficient in some particular cases, such as in some C 1 thin plate bending
elements based in Kirchhoff’s theory where a discontinuity of the gradient of the
deflection occurs at the element sides. These elements are termed incompatible or
non-conforming. Incompatible elements can still converge to the exact solution
if the patch test is satisfied. This guarantees that the compatibility condition is
fulfilled in the limit as the mesh is refined.
It is interesting that non-conforming elements can be competitive in practice.
The reason is that the interelemental discontinuities introduce a greater flexibility

137
in the element which tends to counterbalance the intrinsic rigidity of the finite
element approximation. This leads in some occassions to very good solutions
with relatively coarse meshes.
In summary, the non-conformity is an undesirable deficiency which, however,
does not automatically invalidate an element. The patch test is the critical proof
for acceptance of the element for practical purposes. Although incompatible
elements can sometimes be very attractive, they should be looked upon with
caution since they can have unexpected features. For instance, some incompatible
solid elements show a spurious dependency with the Poisson’s ratio which varies
with the mesh size [N1].

3.19.2 Condition of complete polynomial

This condition can be explained by recalling that the finite element approximation
can reproduce only a finite number of the Taylor expansion terms of the exact
solution, which can written in the vicinity of a point xi as
³ du ´ ³ d2 u ´ ³ dn u ´
2
u(x) = u(xi ) + (x − xi ) + 2
(x − xi ) + · · · + n
(x − xi )n (3.142)
dx i dx i dx i
It is obvious that the finite element approximation written as
ū(x) = ao + a1 x + a2 x2 + · · · + am xm (3.143)
can only reproduce exact results up to the nth term of the Taylor expansion
(3.142) when ū(x) contains all the terms of the polynomial of mth degree. In
such a case the approximation error is of the order Ohm+1 and this can be used
to derive appropiate extrapolation rules (see Section 3.7).
Therefore, the finite element approximation depends on the higher complete
polynomial included in the shape functions. The approximation will be “optimal”
if the shape functions are complete polynomials. Unfortunately this is not always
possible, and in many cases the shape functions contain incomplete polynomial
terms which do not contribute to a higher approximation of the element.

Example: 3.7 Complete and incomplete polynomials and approximations.

-- Solution

a) Complete approximations of 2nd degree.

1D : ū(x) = ao + a1 x + a2 x2
2D : ū(x, y) = ao + a1 x + a2 y + a3 xy + a4 x2 + a5 y 2

b) Incomplete approximations of 3rd degree.

1D : ū(x) = ao + a1 x + a2 x3
2D : ū(x, y) = ao + a1 x + a2 y + a3 x2 + a4 y 2 + a5 x3

138
 The terms of a complete polynomial of several variables can be deduced from
the Pascal triangle [R2]. This subject will be treated when studying the shape
functions of 2D and 3D elements.
In conclusion, it is desirable for the shape functions be complete polynomi-
als, or, if this is not possible, that they contain to a small number of incomplete
polynomials. Note that an incomplete approximation does not preclude the con-
vergence of the element.

3.19.3 Stability condition

The analysis of a bar structure requires prescribing enough displacements to pre-
vent the appearance of unstable mechanisms. Lack of stability is usually detected
by the existance of one or more mechanisms which correspond to the same num-
ber of zero eigenvalues in the stiffness matrix and the associated so-called rigid
body modes. The same concept applies to the stability of an element. In conse-
quence, the stiffness matrix of an individual element (and also that of a patch
of elements) must have the correct rank [R2]. This means that the number of
zero eigenvalues of a single isolated element free of external constrains must be
equal to the number of rigid body displacements of the element. The element
is considered as stable if these zero eigenvalues disappear after prescribing the
appropriate d.o.f. Element stability is generally guaranteed if the stiffness matrix
is integrated exactly. However, the inexact computation of some terms of the
stiffness matrix (by using reduced integration, for instance) can introduce un-
desirable internal mechanisms in addition to those of rigid body motion. These
mechanisms should be avoided since they can lead to instability of the solution.
The existence of internal mechanisms is not always a reason to exclude an ele-
ment as, in some cases, these mechanisms can not propagate themselves through-
out the mesh. Therefore, eigenvalue tests must be performed to assess the exis-
tence of spurious mechanisms in an individual element, and also in patches of two
or more element assemblies, in order to detect the capability of these mechanisms
to propagate in a mesh.

3.19.4 Geometric invariance condition

An element should not have preferred directions. This means that the elements
must have what is usually called “geometric-invariance”, also known as frame-
invariance and geometric or spatial isotropy.
The lack of geometric-invariance is detected if different displacements or stresses
are obtained when the element position is changed in space without changing the
relative direction of the loading. In general, an element is geometric-invariant if
all the displacement d.o.f. are interpolated with the same polynomial and this is
not sensitive to the interchange of the coordinates. This can be achieved by using
complete polynomial interpolations expressed in the natural coordinate system
and an isoparametric formulation [C15].

139
 Geometric-invariance can be lost in an element by under integration of some
of the terms in the stiffness matrix, such as in the case of selective integration
procedures. The lack if geometric-invariance is a defect to be avoided. However,
this does not necessarily destroies the convergence of the element.

3.20 SOME REMARKS ON THE COMPATIBILITY AND

EQUILIBRIUM OF THE SOLUTION
We should keep in mind that the finite element solution is approximate and in
general does not satisfy the equilibrium and compatibility requirements of the
exact solution. In most cases we will find that:

1. The solution is compatible within the elements. This is always guaranteed

by using continuous polynomial approximations.

2. The solution can be incompatible along the interelemental boundaries. As

previously explained interelemental continuity can be violated if the patch
test is satisfied and this guarantees compatibility in the limit case of infinite
refinement. Also, incompatible elements can sometimes produce excellent
answers.

3. Usually there is not equilibrium of stresses along interelemental boundaries.

Compatibility is always satisfied at the nodes, since they are the points
where displacement continuity is enforced due to the polynomial form of
the shape functions.

4. Equilibrium of forces is satisfed at the nodes, since these are the points
where equilibrium is enforced during the assembly process and, therefore,
at each node Ka − f = 0 it is satisfied.

5. Usually there is not equilibrium of stresses along interelemental boundaries.

Nodal stresses σ can be directly obtained for each element in terms of the
nodal displacements, or by extrapolating the values computed at the Gauss
points within the element. Stresses at interface nodes are different for each
element and the global stress field is discontinuous. A continuous stress field
can be obtained by smoothing the discontinuous nodal values (for instance
by simple nodal averaging) as shown in Figure 3.15. Also, the stresses
computed at the free boundaries are usually not zero, although they are
much smaller than the values inside the mesh. This incompability of the
stress field is a consequence of the displacement formulation, where only
displacement continuity is required and the stresses can be discontinuous.
The stress discontinuity does not violate the convergence requirements and
it is usually corrected as the mesh is refined.

140
 Fig. 3.15: Example of smothing of nodal stresses in linear rod elements

6. The stresses are not in equilibrium within elements. The finite element val-
ues approximates the exact solution in an average integral form (by means
of the PVW) [Z15]. Therefore, the differential equations of equilibrium in
stresses are only approximately satisfied pointwise. An exception to this
rule is for elements with linear shape functions, where the strain and stress
fields are constant and the equilibrium differential equations (involving first
derivatives of stresses only) are automatically satisfied. This lack of equi-
librium is also corrected as the mesh is refined and it does not preclude the
convergence of the numerical solution.

3.21 CONVERGENCE REQUIREMENTS IN ISOPARAMETRIC

ELEMENTS
Isoparametric elements are based on the interpolation of the geometry field in
terms of the nodal coordinate values using the same shape functions as for the
displacement field. The coordinate transformation changes the derivatives of any
function by a jacobian relation. For 1D problems we have
µ ¶−1
du du dx
= (3.144)
dx dξ dξ

Therefore the PVW can be expressed in terms of the natural coordinate ξ

with the maximum order of differentiation unchanged.
It follows immediately that if the displacement shape functions are so chosen
in the natural coordinate system as to observe the usual rules of convergence then
convergence of isoparametric elements will occur.
Furthermore, C ◦ isoparametric elements always satisfy the patch test. The
proof of this is simple; let us prescribe the following linear displacement field

u = a1 + a2 x (3.145)

over a mesh of linear rod elements. The nodal displacements will take the values

ui = a1 + a2 xi ; i = 1, 2 (3.146)

141
 P
2
Inside the element u = Ni ui . Hence, making use of the prescribed field
i=1

2
X 2
X 2
X
u = N1 (a1 + a2 xi ) = a1 Ni + a 2 N i xi (3.147)
i=1 i=1 i=1

Since the element is isoparametric we have

X2
x = N i xi (3.148)
i=1

From Eqs.(3.147) and (3.148) we deduce that the displacement field will co-
incide with the prescribed one (3.145) if
2
X
Ni = 1 (3.149)
i=1

is satisfied for any value of the natural coordinate ξ between −1 and +1. Note
that Eq.(3.149) is the usual rigid body requirement for the shape functions (see
Section 3.17.4). We conclude that the constant derivative condition is satisfied
for C ◦ isoparametric elements.

3.22 ERROR TYPES IN THE FINITE ELEMENT SOLUTION

We recall that the finite element solution is approximate. This automatically
implies that some kind of error in the numerical solution is unavoidable. Next,
we will study the more usual sources of error.

3.22.1 Discretization error

This error is intrinsic to the polynomial form of the finite element approximation.
We showed in Section 3.19.2 that the error involved in the approximation is of
the order of the first term in the Taylor expansion of the solution not included
in the complete shape function polynomial. Strang and Fix [S14] proposed the
following general expression to estimate this error for 1D problems
¯ p+1 ¯
¯ ∂ uexact ¯
e(error) = uaprox − uexact ≤ Ch p+1
Max ¯¯ ¯ (3.150)
∂xp+1 ¯
where Max denotes the maximum value of the derivative over the element, C is
a constant parameter depending on the element type, n is the maximum charac-
teristic element dimension (i.e. the length in rod elements) and p the degree of
the highest complete polynomial contained in the shape functions.
Eq.(3.150) shows that convergence is guaranteed if C and the n+1th derivative
of the solution are bounded. Note that in this case the error will tend to zero as
the element size diminishes.

142
 Fig. 3.16: Continuous rod elements of different sizes

The application of this concept to the 1D linear rod element gives for uniformly
spaced meshs
∂2u
e ∝ h2 (3.151)
∂x2
which implies that the error is proportional to the strain (or stress) gradient.
Therefore, smaller elements should be used in zones whese this gradient is ex-
pected to be higher. The reduction of the error by diminishing the element size
is known in the mesh refinement literature as the h method.
The error can also be reduced by increasing the approximation order of the
elements, while keeping their sizes constant. This results in a larger value of the
exponent p in Eq.(3.150). This approach is known as the p method [Z15].
Eq.(3.150) assumed a mesh of equal element sizes. The effect of using ele-
ments of different sizes has been studied for the analysis of an axially loaded rod
using linear elements of two different sizes. The error in the satisfaction of the
differential equilibrium equation at the jth node is
µ 3 ¶
h ∂ u h2 ³ 1 + a3 ´ ∂ 4 u
e = − (1 − a) + (xj ) + · · · (3.152)
3 ∂x3 j 12 1 + a ∂x4

where h and a are the lengths of two adjacent elements (see Figure 3.16). Eq.(3.155)
shows that the error is of the order h2 for a uniform mesh (a ' 1), whereas a
higher error of order h is obtained when the element sizes are very different
(a 6= 1). This indicates that drastic changes in the sizes of contiguous elements
in a mesh should be avoided.
The same concepts apply for 2D and 3D problems. The estimation of the
discretization error in two dimensions involves the Taylor expansion
h ∂u ∂u i
u(x + h, y + k) = u(x, y) + h +k +
∂x ∂y
1 h 2 ∂ 2u ∂2u 2 i
2∂ u
+ h + 2hk +k + · · · (3.153)
2! ∂x2 ∂x∂y ∂y 2

where u is the exact solution and h and k are a measure of the element sizes in
the x and y directions respectively.
It can be shown that the discretization error for 2D linear elements, like the
3-noded triangle (Chapter 4), is proportional to the underlined term on the right-
hand side of Eq.(3.67) [S14]. The second derivatives in Eqs.(3.153) can be related
to strain (or stress) gradients. Thus, for a constant strain field the error is very
small.

143
 It is interesting that the discretization error can also be expressed in terms
of the ratio hk .This is a measure of the relative dimensions of the element and it
is known as the element aspect ratio. For an equilateral element its aspect ratio
should be equal to one. However, it will take a large value for a long triangular
element. It is recommended to keep the element aspect ratio as close to unity as
possible through the mesh.

3.22.2 Error in the geometry approximation

In many cases the interpolation of the geometry is unable to reproduce exactly
the real shape. This can be due to a geometry approximation of a lower order
than the exact one, or, what is more usual, to the ignorance of the exact analytical
form of the geometry defined by the coordinates of a number of points. In both
cases, there will be an error in the approximation of the geometry. This error
can be reduced by refining the mesh, or by using higher order superparametric
approximations. A compromise between these two options is sought in practice
and usually isoparametric elements are used. This unavoidably introduces an
error in the geometry approximation in some cases. An exception is the case
of structures with linear boundaries where the geometry can always be exactly
approximated.

3.22.3 Error in the computation of the element integrals

The exact numerical computation of the element integrals implies the use of an
appropriate quadrature. Otherwise, an error occurs due to the underestimation of
the integral value. In many cases, the exact numerical integration is not possible
due to the rational terms appearing in the element integrals. Also, the approxi-
mation of the exact value may require a large number of integration points, which
may be very expensive. In such cases, it is usual to accept a certain error in the
computation of the element integrals.
Paradoxically enough, this error can, on occasions, be beneficial. Usually by
subintegrating the rigidity terms the element becomes more flexible, and this
balances the stiffening introduced by the approximation of the displacement and
geometry fields. This explains why sometimes good results can be obtained with
coarse meshes. In the following chapters we will see that the “reduced integration”
quadrature is sometimes used to guarantee the correct solution. Note, however,
that the inexact computation of the stiffness matrix can modify its correct rank
and introduce spurious mechanisms. Reduced integration is therefore a technique
which should be used with extreme care.

3.22.4 Errors in the solution of the global equation system

Three type of errors are typical in the solution of the global system of FEM equa-
tions using a direct solution method (i.e. Gaussian elimination, Choleski, Frontal

144
method, etc.): errors due to the ill-conditioning of the equations; truncation
errors and round-off errors [R5].
The equation system Ka = f is ill-conditioned if small changes in the terms of
K or f induce large changes in the solution a. The main reason for ill conditioning
is the existance of an element, or a group of elements, of large stiffness connected
to elements of much smaller stiffness. The behaviour of such a structure can be
artificially altered and, unless the computer can store a sufficiently large number
of digits, the stiffness matrix behaves as singular or quasi-singular. The error
associated with ill-conditioning of the equation system therefore depends on the
digit storage capacity of the computer, i.e. in the truncation and round-off errors
which are the main contributors to the total error in the solution.
The truncation error is the most important. A computer using d digits to
represent a number in simple precission can only store the first d digits of each
term of K and f. It is then possible that essential information for the correct
solution is lost by truncating a number.
The round-off error is due to the adjustment automatically performed by the
computer on the last digit of each number during computations. Experience
shows that this error is less important than the truncation error. Nevertheless,
unnecessary round-off errors, such as those in some parameters like the coordi-
nates and weights of the numerical quadrature, should be avoided by defining
these parameters with the maximum number of digits allowed by the computer.

Example: 3.8 Study the influence of truncation error in the solution of the spring system
shown in Figure 3.17.

-- Solution

The stiffness matrix of the structure and its inverse are

 
  1 1 1
K1 −K1 +
 K1 K 2 K2 
 
K=  ; K−1 = 
−K1 K1 + K2  1 1 
K2 K2

If K1 >> K2 , K1 dominates in K. However, K2 dominates K−1 and therefore the value

of the solution. The computation of K−1 is only correct if the terms in K are evaluated
in a way such that K2 is not lost during the solution. Thus, if K1 = 50 and K2 = 0.0023
the computer must retain at least six digits and K1 must be represented as 50.0000 so
that the last digit of K2 is retained in the term K1 + K2 . If only four digits are retained
the sum K1 + K2 will give 50.00 and K will be singular. This problem is ill conditioned
since the solution is sensitive to the changes (truncation) in the sixth digit of the term
K1 + K2 .
Also, if the system Ka=f is solved using Gauss ellimination, the ellimination of the dis-
placement u1 changes the last diagonal term to (K1 + K2 ) − K1 . We see that information
for a correct solution can again be lost if K1 >> K2 .

145
 Fig. 3.17: Spring system with two degrees of freedom

A way to avoid truncation errors and to improve the solution is to use double
precision throughout the solution process (i.e. for computing the terms of K and
f and during the solution of the equation system).
An indicator of how sensitive the system Ka = f is to truncation and round-off
errors is the condition number of K. An estimation of the number of significative
figures exactly computed in the solution process is [B4]

s ≈ t − log10 [cond(K)] (3.154)

where t is the maximum number of digits which can be stored by the computer
and cond (K) is the condition number of K defined as

λmax
cond(K) ' (3.155)
λmin
where λmax and λmin are respectively the larger and smaller eigenvalues of K.
Although Eq.(3.154) is only approximate, it indicates that the accuracy of the
solution decreases as the condition number increases.
It is therefore desirable that the condition number of K should be as low as
possible. This can be achieved by an adequate scaling of the terms of K [B4],
[R2], [R6].

3.22.5 Errors associated with the constitutive equation

The survey of the error sources in the finite element solution of a structure would
be incomplete without referring to the errors arising from a wrong definition
of the material properties. The importance of the evaluation of the relevant
parameters in the constitutive equation is obvious. For example, in a structure
with homogeneous and isotropic material the displacements are proportional to
the Young modulus, although the stresses are not affected by this value and they
depend only on the Poisson’s ratio. For a structure with orthotropic or anisotropic
materials both the displacements and the stresses depend on the Young modulus
and the Poisson’s ratio. An incorrect definition of the material parameters can
lead to larger errors than those induced by all the error sources mentioned in the
previous sections.

146
Chapter 4

FEM ANALYSIS OF THE 2D POISSON

EQUATION

4.1 INTRODUCTION
In this chapter the formulation of the FEM applied to the two-dimensional (2D)
form of the Poisson equation is described.
Initially the main assumptions are presented; then, the FEM solution with
the simple three-noded triangular element is presented.

4.2 STEADY-STATE POISSON EQUATION IN 2D

The steady-state Poisson equation for a 2D problem is
µ ¶ µ ¶
∂ ∂φ ∂ ∂φ
A(φ) : = kx + ky +Q=0 in Ω (4.1)
∂x ∂x ∂y ∂x

where φ(x, y) is the unknown function, kx and ky are the conductivities in the
x and y directions and Q(x, y) is a heat source over the domain Ω.
This equation can be obtained by improving flux balance in a differential
domain (Figure 4.1). There exist two kinds of boundary conditions


 Prescribed variable (Dirichlet condition)


 φ − φ̄ = 0 on Γφ
B(φ) : (4.2a)



 Prescribed normal heat flow (Neumann condition)

qn = q̄n + α(φ − φa ) on Γq

The normal flux qn is obtained projecting the boundary flow over the bound-
ary normal
qn = nT q = nx qx + ny qy

147
 Figure 4.1: Problem definition for the Poisson equation in a 2D domain

Substituting qx = −kx ∂φ
∂x
and qy = −ky ∂φ
∂y
in the previous equations gives
the prescribed heat flux condition along the boundary (Neumann condition) as
follows:
∂φ ∂φ
kx nx + ky ny + q̄n + α(φ − φ̄a ) = 0 en Γq (4.2b)
∂x ∂y

where q̄n is the normal heat flow prescribed flow at the domain’s boundary Γq
(q̄n is positive if the flow is in the outgoing normal direction n). The last term
in (4.2b) express the heat flow at the boundary due the temperature difference
in the boundary (φ) and the temperature of the external media (φa ); α is the
convection coefficient.
Note that
Γ = Γφ ∪ Γ q (4.3)

The boundary split in two zones Γφ and Γq is conceptual, because these

conditions can appear in different zones along the boundary.

4.3 PROBLEM SOLUTION USING THE FINITE ELEMENT

METHOD
We solve the problem defined by Eqs.(4.1) and (4.2) in a similar way as we
did for the 1D case. First we apply the Galerkin’s weighted residual (GWR)
method on the Poisson equation that defines the balance of heat flow over the
domain and the equation that defines the Neumann the boundary conditions.

148
 Figure 4.2: Normal vector to the boundary

4.3.1 Integral form of the GWR method

The equivalent integral form for the differential equations are obtained applying
the GWR method. In this manner, multiplying the equations (4.1) and (4.2)
by the weighted functions W and W̄ and integrating over the domain and the
boundary Γq respectively gives:
ZZ · µ ¶ µ ¶ ¸
∂ ∂φ ∂ ∂φ
W kx + ky + Q dxdy+
Ω ∂x ∂x ∂y ∂y
Z · ¸ (4.4)
∂φ ∂φ
+ W̄ kx nx + ky ny + q̄n + α(φ − φa ) dΓ = 0
Γq ∂x ∂y
The weak form is obtained integrating by parts the terms with second deriva-
tives in (4.4)
ZZ µ ¶ Z Z µ ¶
∂ ∂φ ∂ ∂φ
W kx dxdy = dy W kx dx =
Ω ∂x ∂x ∂x ∂x
ZZ Z (4.5)
∂W ∂φ ∂φ
=− kx dxdy + W kx dy
Ω ∂x ∂x Γ=Γq +Γφ ∂x
³ ´
A similar operation is applied for the integral of ∂y∂
kx ∂φ
∂y
to obtain:
ZZ · µ ¶ µ ¶ ¸
∂ ∂φ ∂ ∂φ
W kx + ky + Q dxdy =
Ω ∂x ∂x ∂y ∂y
ZZ · ¸ ZZ
∂W ∂φ ∂W ∂φ
=− kx + ky dxdy + W Qdxdy+ (4.6)
Ω ∂x ∂x ∂y ∂y Ω

Z · ¸
∂φ ∂φ
+ W kx dy + ky dx
Γq +Γφ ∂x ∂y
Considering that (Figure 4.2)
dy = dΓ cos α = dΓnx
(4.7)
dx = dΓ cos β = dΓny

149
and adopting W̄ = −W which eliminates the derivatives of in the Neumann
boundary when substituting in (4.6) gives:
ZZ · ¸ ZZ
∂W ∂φ ∂W ∂φ
− kx + ky dxdy + W Qdxdy+
Ω ∂x ∂x ∂y ∂y Ω

Z · ¸
∂φ ∂φ
+ W kx nx + ky ny dΓ− (4.8)
Γq +Γφ ∂x ∂y
Z · ¸
∂φ ∂φ
− W kx nx + ky ny + q̄n + α(φ − φa ) dΓ = 0
Γq ∂x ∂y
where the underlined terms are canceled together.
The resulting integral form is:
ZZ · ¸ ZZ
∂W ∂φ ∂W ∂φ
− kx + ky dΩ + W QdΩ+
Ω ∂x ∂x ∂y ∂y Ω

Z · ¸ Z
∂φ ∂φ
+ W kx nx + ky ny ) dΓ − W [q̄n + α(φ − φa )] dΓ = 0
Γφ ∂x ∂y Γq
(4.9)
Recalling that
∂φ ∂φ
qn = qx nx + qy ny = −kx nx − ky ny (4.10)
∂x ∂y
and ordering the terms for φ the weak form is finally obtained. This equation
allows to recover the C ◦ continuity condition for W and φ:
ZZ h i R
∂W
kx ∂φ + ∂W
k ∂φ
y ∂y dΩ + Γq αW φdΓ =
Ω
∂x ∂x ∂y

ZZ Z Z (4.11)
= W QdΩ − W [q̄n − αφa ]dΓ − W qn dΓ
Ω Γq Γφ

The weak form (4.11) is the starting expression for deriving the FEM equations.

4.3.2 FEM Discretization

The last integral in Eq.(4.11) is due the normal heat flow qn crossing the bound-
ary Γφ where φ is a known value. Note that the sign of the integral is negative,
which means that the normal flux qn takes heat from the domain toward outside
the boundary. As in the 1D problem, this flux can be considered as a “reaction”
that can be computed once the value of φ is solved over Ω.
Let us assume a discretization of the domain Ω using 2D finite elements with
n nodes (Figure 4.3). The approximation of φ inside each element is defined as:
n
X
φ∼
(e) (e)
= φ̂ = Ni (x, y)φi (4.12)
i=1

150
Figure 4.3: Discretization of a two-dimensional domain by tree-noded triangular
elements.
(e)
where Ni (x, y) are the 2D shape functions function and φ(e) are the nodal
values of the approximated solution.
The system of equations for the discretized domain is obtained substituting
the approximation (4.12) into the weak form (4.11) and choosing N weighting
functions Wi , where N is the total number of nodes in the mesh
ZZ " # Z
∂Wi ∂ φ̂ ∂Wi ∂ φ̂
kx + ky dΩ + αWi φ̂dΩ =
Ω ∂x ∂x ∂y ∂y Γq

ZZ Z Z
= Wi QdΩ − Wi [q̄n − αφa ]dΓ − Wi qn dΓ i = 1, . . . , N
Ω Γq Γφ
(4.13)
The Galerkin form is obtained by making Wi = Ni . Assuming that all the
functions that appear in (4.13) are integrable and having in mind the additive
property of the integrals, the Galerkin form can be expressed as a function of
the element contributions:
ZZ " Ã n ! Ã n !#
P (e)
∂Ni X ∂Nj(e) (e) ∂Ni
(e) X ∂Nj(e) (e)
kx φ + ky φ dΩ+
e Ω(e) ∂x ∂x j ∂y ∂y j
à n
j=1 ! j=1
XZ (e)
X (e) (e)
+ αNi Nj φ j dΓ =
(e)
e Γq j=1
X ZZ (e)
XZ (e)
XZ (e)
= Ni QdΩ − Ni [q̄n − αφa ]dΓ − Ni qn dΓ
(e) (e)
e Ω(e) e Γq e Γφ
(4.14)
where the sum applies are over all the elements discretizing the domain Ω.
Expression (4.14) can be written in matrix form as
Ka = f (4.15)

151
 The terms of K and f are obtained by assembling the elemental contributions
(e)
as usual. Because the shape function forms Ni have a zero value outside the
element (e), we deduce that:
ZZ Ã (e) (e)
!
(e) (e)
(e) ∂N i
∂N j ∂N i
∂Nj
Kij = kx + ky dΩ+
Ω(e) ∂x ∂x ∂y ∂y
(4.16)
Z
(e) (e) (e)
+ αNi Nj dΓ = Kdij + Kc(e) ij
(e)
Γq

ZZ Z Z
(e) (e) (e) (e)
fi = Ni QdΩ − Ni [q̄n − αφa ]dΓ − Ni qn dΓ (4.17)
(e) (e)
Ω(e) Γq Γφ

(e) (e)
The terms Kdij and Kcij in (4.16) represent the elemental contributions due
to the conductivity and convection in the element stiffness matrix, respectively.
These terms are obtained by:
ZZ " (e) (e)
#
(e) (e)
(e) ∂N i
∂N j ∂N i
∂N j
Kdij = kx + ky dΩ (4.18)
Ω (e) ∂x ∂x ∂y ∂y
Z
(e) (e) (e)
Kcij = αNi Nj dΓ (4.19)
(e)
Γq
(e) (e)
The independent computation of Kdij and Kcij simplifies the evaluation of
the element stiffness matrix coefficients.
(e)
Similarly, the components of vector fi are:
(e) (e)
fi = fQi + fc(e)
i
+ qni (4.20)
where ZZ
(e) (e)
fQi = Ni QdΩ (4.21)
Ω(e)
is the contribution of the heat source over the domain
Z
(e) (e)
fi = − Ni [q̄n − αφa ]dΓ (4.22)
(e)
Γq

is the contribution of the heat flow trough the Neumann boundary where the
flow is prescribed and Z
(e) (e)
qni = − Ni qn dΓ (4.23)
(e)
Γφ

is the “reaction” flow term over the boundary Γφ where the variable φ is pre-
scribed. This flow is evaluated “a posteriori ” once the nodal values of φ have
been calculated.
Recall that the element boundary integral of (4.17) only contributes to the
nodal force vector when the element has one or more sides over Γq or Γφ .

152
4.4 THREE-NODED TRIANGULAR ELEMENT
Let us assume a three-noded triangular element discretization over the domain,
as shown in Figure 4.3.
Each element node has one degree of freedom (d.o.f.) which corresponds
to the nodal unknown value of φ. The unknown variation is expressed by the
following polynomial:
φ(x, y) = α0 + α1 x + α2 y (4.24)
To find the coefficients α0 , α1 and α2 it is necessary to substitute the nodal
coordinates into Eq.(4.24) as follows
(e) (e) (e)
φ1 = α0 + α1 x1 + α2 y1

(e) (e) (e)

φ2 = α0 + α1 x2 + α2 y2 (4.25)

(e) (e) (e)

φ3 = α0 + α1 x3 + α3 y3
Solving these equations and substituting the values of αi we obtain
1 (e) (e) (e) (e) (e) (e) (e) (e)
φ(x, y) = 2A(e)
[(a1 + b1 x + c1 y)φ1 + (a2 + b2 x + c2 y)φ2 +
(4.26)
(e) (e) (e) (e)
+(a3 + b3 x + c3 y)φ3 ]
where
(e) (e) (e) (e) (e)
ai = xj yk − xk yj

(e) (e) (e)

bi = yj − yk (4.27)

(e) (e) (e)

ci = xk − xj
Indexes i, j, k go from 1 to 3 (Figure 4.3). Permuting the indexes allows to
find the expression for each parameter.
The element area A(e) is obtained by the determinant of the coefficients in
Eq.(4.25) as follows ¯ ¯
¯1 x(e) y (e) ¯
¯ i i ¯
¯ (e) ¯
2A(e) = ¯1 x(e)
j yj ¯ (4.28)
¯ (e) ¯¯
¯1 x(e)
k yk
Eq.(4.26) can be compacted as
3
X (e) (e) (e) (e) (e) (e) (e) (e)
φ= Ni φi = N1 φ1 + N2 φ2 + N3 φ3 (4.29)
i=1

which allows identifying the element shape functions by:

(e) 1 (e) (e) (e)
Ni (x, y) = (a + bi x + ci y); i = 1, 2, 3 (4.30)
2A(e) i

153
 Figure 4.4: Shape functions for the three-noded triangular element.

In the 3-noded triangular element these shape functions are expressed by

linear polynomials with coefficients depending on the nodal coordinates. The
nodal shape functions take a unit value at the node and zero at the other nodes,
as usual (Figure 4.4).
As for the 1D case, the global shape functions of a node has a unit value
at the node and zero at any other node. Also, the influence domain for this
function corresponds to all the elements surrounding the node.

4.4.1 Elemental stiffness matrix

From Eq.(4.30) we deduce
(e) (e)
∂Ni bi ∂Ni ci
= ; = (4.31)
∂x 2A(e) ∂y 2A(e)
Substituting Eqs.(4.30) and (4.31) into (4.18) and (4.19) gives
ZZ
(e) 1 1
Kdij = (e)2
[kx bi bj + ky ci cj ]dΩ = [kx bi bj + ky ci cj ] (4.32)
Ω(e) 4A 4A(e)
(e)
and the Kd matrix is obtained by
 (e)
kx b1 b1 + ky c1 c1 kx b1 b2 + ky c1 c2 kx b1 b3 + ky c1 c3
(e) 1 
Kd = kx b2 b1 + ky c2 c1 kx b2 b2 + ky c2 c2 kx b2 b3 + ky c2 c3  (4.33)
4A(e)
kx b3 b1 + ky c3 c1 kx b3 b2 + ky c3 c2 kx b3 b3 + ky c3 c3

The evaluation of the integral defined in (4.19) yields the convection stiffness
matrix as
 (e) (e) (e) (e)

(e) (e) (e)
Z N1 N1 N1 N2 N1 N3
 
K(e)
c = α N2(e) N1(e) N2(e) N2(e) N2(e) N3(e)  dΓ (4.34)
(e)
Γq (e) (e) (e) (e) (e) (e)
N3 N1 N3 N2 N3 N3

It is important to note that the above integral is computed over the length of
the sides of the triangular element laying on the Neumann boundary. Therefore
the shape function for the third node which does not belong to a side takes a
zero value over that side. This allows us to simplify expression (4.34) depending
on the element’s side where the boundary integral is computed.

154
 (e)
Hence, for side 1-2, N3 = 0 and
 (e) (e) (e) (e) 
Z N1 N1 N1 N2 0
Kc(e)
12
= α N2(e) N1(e) N2(e) N2(e) 0 dΓ (4.35)
(e)
l12
0 0 0
(e)
For side 2-3, N1 = 0 and
 
Z 0 0 0
(e) (e) (e) (e)
Kc(e)
23
= α  0 N2 N2 N2 N3  dΓ (4.36)
(e)
l23 (e) (e) (e) (e)
0 N3 N2 N3 N3
(e)
For side 1-3, N2 = 0 and
 (e) (e) (e) (e) 
Z N1 N1 0 N1 N3
Kc(e)
13
= α 0 0 0  dΓ (4.37)
(e)
l13 (e) (e) (e) (e)
N1 N3 0 N3 N3
Considering that α is constant over the element sides, the above expressions
can be integrated to obtain:
Z (
(e)
(e) (e) lij /3; i = j
Ni Nj dΓ = (e) (4.38)
(e)
lij lij /6; i 6= j
(e)
where lij is the length of the (e) element’s side i − j; this gives:
     
(e) 2 1 0 (e) 0 0 0 (e) 2 0 1
(e) αl12   (e) αl23   (e) αl13 0 0 0 
Kc12 = 1 2 0 ; Kc23 = 0 2 1 ; Kc13 =
6 6 6
0 0 0 0 1 2 1 0 2
(4.39)
The total element stiffness matrix is obtained as:
(e)
K(e) = Kd + K(e) (e) (e)
c12 + Kc23 + Kc13 (4.40)
(e)
We recall again that the contribution of matrices Kcij is only relevant if the
element side i − j belongs to the external Neumann boundary Γq . Also note
that we have obtained explicit expressions for the element stiffness matrix due
the linear form of the shape functions.

4.4.2 The equivalent nodal flux vector

The generic expression of f (e) for the 3-noded triangular element is:
 
(e)
f1 
 
(e) (e)
f = f2 (4.41)
f (e) 
 
3

155
 The heat source terms are deduced from Eq.(4.21)
 (e)   
(e)
 f
 Q1   Z Z 
 1 
N 
(e) (e) (e)
fQ = fQ2 = N2 QdΩ (4.42)

 (e)   N (e) 
Ω(e)  
fQ3 3

For a constant value of Q over the element, it is easy to obtain:

 
(e) 1
(e) QA
fQ = 1
3  
1
Hence, the total element flux is split in equal parts over the three nodes, as
expected.
From equation (4.22) the nodal flux contribution due convection is evaluated
as follows  
(e)
Z   1 
N 
(e) (e)
fc = − N 2 [q̄n − αφa ]dΓ (4.43)
Γq  N (e) 
(e)

As for the convection stiffness matrix, the evaluation of the above integral
requires to consider the element’s side where the integral is computed. Therefore
for each side we have:

(e)
Side 1-2 (N3 = 0)
 (e) 
Z  N1 
fc(e) =− (e)
N2  [q̄n − αφa ]dΓ (4.44)
12 (e) 
l12
0

(e)
Side 1-3 (N2 = 0)
 (e) 
Z  N1 
fc(e) =− 0 [q̄n − αφa ]dΓ (4.45)
l13  (e) 
13 (e)
N3

(e)
Side 2-3 (N1 = 0)
 
Z  0 
(e)
fc(e) =− N2 [q̄n − αφa ]dΓ (4.46)
(e) 
(e) 
23
l23
N3
Assuming that q̄n , α and φa are constant over the side gives
 
(e)
l12 1
(e)
fc12 = − [q̄n − αφa ] 1 (4.47)
2  
0

156
 (e) (e)
with similar expressions for fc13 and fc23 .
The last contributions to the nodal flux vector are the reactions at the
boundary where φ is prescribed (Eq.4.23)). These fluxes are added directly to
the boundary nodes and are evaluated “a posteriori” once the nodal values of
φ have been computed.

Example 4.1 Heat conduction into a square domain

Solution

Consider the square domain shown in Figure 4.5. A known amount of

heat is generated internally due to a heat source and conduction effects.
The goal is to compute the temperature distribution over the domain.
We assume that kx = ky = k = 2, 0 [cal/cm ◦ C] and Q =2,4 [cal/cm2 s].
Due to symmetry conditions only an eighth section of the domain is ana-
lyzed. The boundary conditions are:

• Along the external boundary 4-6 α = 0, φ̄ = 0

• Along the boundaries 1-6 and 1-4, the symmetry condition implies
that

∂φ
=0
∂n
This means that the temperature gradient in the boundary’s normal di-
rection is zero. The physical meaning of this condition corresponds to an
isolated boundary. Because of the conditions applied, the heat flow only
goes out the domain through the external side 1-4.
The analysis domain is discretized into a mesh of four 3-noded triangular
elements as shown in Figure 4.5.
Because α = 0
 
b1 b1 + c1 c1 , b1 b2 + c1 c2 , b1 b3 + c1 c3
(e) k b2 b1 + c2 c1 , b2 b2 + c2 c2 , b2 b3 + c2 c3 
K(e) = Kk =
4A(e)
b3 b1 + c3 c1 , b3 b2 + c3 c2 , b3 b3 + c3 c3

with
(e) (e)
bi = yj − yk
(e) (e)
c i = xk − xj
Following the element numbering shown in Figure 4.5b, for elements 1, 2
and 4 we have:
   
6, 25 −6, 25 0 1 −1 0
2, 0
K(1) = K(2) = K(4) = 2,52 −6, 25 6, 25 + 6, 25 −6, 25 = −1 2 −1
4 2 0 −6, 25 6, 25 0 −1 1

157
Figure 4.5: a) Heat conduction in a square domain. b) Mesh of 3-noded trian-
gular elements using the symmetry simplification.

For the element number 3:

   
6, 25 0 −6, 25 1 0 −1
2, 0
K(3) = 6,25  0 6, 25 −6, 25 =  0 1 −1
4 2 −6, 25 −6, 25 12, 50 −1 −1 2

158
 bi ci
(e) (e) (e) (e)
b1 = y2 − y3 = −2, 5 c 1 = x3 − x2 = 0
(e) (e) (e) (e)
b2 = y3 − y1 = 2, 5 c2 = x1 − x3 = −2, 5
(e) (e) (e) (e)
b3 = y1 − y2 = 0 c3 = x2 − x1 = 2, 5

Table 4.1: Coefficients bi and ci for elements 1, 2 and 4

bi ci
(3) (3) (3) (3)
b1 = y2 − y3 = 0 c1 = x3 − x2 = −2, 5
(3) (3) (3) (3)
b2 = y3 − y1 = 2, 5 c 2 = x1 − x3 = 0
(3) (3) (3) (3)
b3 = y1 − y2 = −2, 5 c3 = x2 − x1 = 2, 5

Table 4.2: Coefficients bi and ci for element 3

The elemental nodal flux vectors are

 
1
(e) QA(e)
f (e) = fQ = 3
1
 
1
   
1 1
2,4×2,52
f (e) = 3×2
1 = 2, 5 1
   
1 1

Assembly of the global stiffness matrix

The first steep is to identify the position of each coefficient of the elemen-
tal stiffness matrix K through the correspondence between the local and
global nodal numbers. From Figure 4.5 we deduce

1 2 3
(1) (1) (1) 
1 K11 K12 K13
(1) (1) (1)
K(1) = 2 K21 K22 K23 
(1) (1) (1)
3 K31 K32 K33

2 4 5
(2) (2) (2) 
2 K11 K12 K13
(2) (2) (2)
K(2) = 4 K21 K22 K23 
(2) (2) (2)
5 K31 K32 K33

159
 2 5 3

(3) (3) (3) 
2 K11 K12 K13
(3) (3) (3)
K(3) = 5 K21 K22 K23 
(3) (3) (3)
3 K31 K32 K33

3 5 6

(4) (4) (4) 
3 K11 K12 K13
(4) (4) (4)
K(4) = 5 K21 K22 K23 
(4) (4) (4)
6 K31 K32 K33

The global stiffness matrix is:

1 2 3 4 5 6
 
(1) (1) (1)
1 K11 K12 K13 0 0 0 
 
 (1) (2) (3) (1) (3) (2) (2) (3) 
2 K22 + K11 + K11 K23 + K13 K12 K13 + K12 0  
 
3  (1) (3)
K33 + K33 + K11
(4)
0
(3)
K32 + K12
(4) (4) 
K13 
K=  
 (2) (2) 
4 K22 K23 0 
 
 
5
(2) (3) (4) (4) 
 Symm. K33 + K22 + K22 K23 
(4)
6 K33

Finally the elemental nodal flow are assembled into the vector f as follows
 
    f (1) = 2, 5 
1 −1 0 0 0 0  φ1  
 1 


 
 
 

    
 
 (1) 

     (2) (3) 

−1 4 −2 −1 0 0  
 φ 
 
 f 2 + f1 + f 1 = 7, 5 

  
2 
 
 

   
 
 

      (1) 

 0 −2 4 0 −2 0  
 φ 
 
 f + f
(3)
+ f
(4)
= 7, 5 

  3 3 3 1
 
   =  
 0 −1 0 2 −1 0  φ4   (2)  φ4 = 0
  






 f2 = 2, 5 + R4 


    
  










0 0 −2 −1 4 −1 
 φ 
 
 (2) (3) (4) 
 φ5 = 0
   5 
 
 f 3 + f2 + f 2 = 7, 5 + R 5

   
 
 


     

0 0 0 0 −1 1 φ6 
f (4) = 2, 5 + R 
 φ6 = 0
| {z } | {z } 3 6
K a | {z }
f

where R4 , R5 y R6 are the unknown fluxes at the nodes 4, 5 and 6 respectively.

Note that the value of φ is known at those nodes (φ4 = φ5 = φ6 = 0).

160
 Solution of the system of equations gives
◦
φ1 = 9, 3748 C
◦
φ2 = 6, 8748 C
◦
φ3 = 5, 3123 C
φ4 = 0 ◦ C; R4 = −9, 3748 cal/s/cm2
φ5 = 0 ◦ C; R5 = −18, 124 cal/s/cm2
φ6 = 0 ◦ C; R6 = −2, 5 cal/s/cm2

Let us compare the total flux generated in the domain (QA) with the flow
that goes out through the domain boundary

R4 + R5 + R6 = 29, 9988
QA = 30

Hence, the total flux generated in the domain coincides with the total flux
that goes out through the boundary. This proves the global equilibrium of
nodal fluxes which is a consequence of the flux balance (equilibrium) approach
followed in the FEM.

161
Chapter 5

FEM ANALYSIS OF THE 3D POISSON

EQUATION

5.1 INTRODUCTION
In this chapter the formulation of the FEM applied to the 3D form of the Poisson
equation is described. This chapter completes the study of problems with one
variable per node.

5.2 STEADY-STATE POISSON EQUATION IN 3D

The steady-state Poisson equation for a three-dimensional (3D) problem is
µ ¶ µ ¶ µ ¶
∂ ∂φ ∂ ∂φ ∂ ∂φ
A(φ) : = kx + ky + kz + Q = 0 in Ω (5.1)
∂x ∂x ∂y ∂y ∂z ∂z

where φ(x, y, z) is the unknown function, kx , ky and kz are the conductivities

in the x, y, z directions and Q is a heat source over the domain Ω.
As for the 1D and 2D cases this equation is obtained by considering the
balance of heat fluxes in a differential domain (Figure 5.1).
The boundary conditions are:


 Prescribed variable condition (Dirichlet condition):



 φ − φ̄ = 0 on Γφ

B(φ) : (5.2)

 Prescribed normal flux condition (Neumann condition):

 ∂φ ∂φ ∂φ


 kx nx + ky ny + kz nz + q̄n + α(φ − φ̄a ) = 0 on Γq
∂x ∂y ∂z

As for the 2D case, q̄n is the prescribed flux in the normal direction to the
boundary Γq , φa is the external temperature, α is the convection coefficient and
n = [nx , ny , nz ]T is the unit normal to Γq and Γ = Γφ ∪ Γq .

163
 Figure 5.1: Three dimensional definition of the Poisson equation.

5.3 SOLUTION USING THE FINITE ELEMENT METHOD

5.3.1 Weak integral form
The equivalent integral form for the differential equations (5.1) and (5.2) are
obtained applying the GWR method as
ZZZ · µ ¶ µ ¶ µ ¶ ¸
∂ ∂φ ∂ ∂φ ∂ ∂φ
W kx + ky + kz + Q dΩ+
Ω ∂x ∂x ∂y ∂y ∂z ∂z
ZZ · ¸
∂φ ∂φ ∂φ
+ W̄ kx nx + ky ny + kz nz + q̄n + α(φ − φa ) dΓ = 0
Γq ∂x ∂y ∂z
(5.3)
where W and W̄ are arbitrary weighting functions.
Integrating by parts the terms with second derivatives of φ in the first inte-
gral and choosing W̄ = −W gives:
ZZZ µ ¶
∂W ∂φ ∂W ∂φ ∂W ∂φ
− kx + ky + kz dΩ+
Z Z ZΩ ∂x ∂x Z Z ∂y ∂y µ ∂z ∂z ¶
∂φ ∂φ ∂φ
+ W QdΩ + W kx nx + ky ny + kz nz dΓ (5.4)
Ω Γq +Γφ ∂x ∂y ∂z
ZZ · ¸
∂φ ∂φ ∂φ
− W kx nx + ky ny + kz nz + q̄n + α(φ − φa ) dΓ = 0
Γq ∂x ∂y ∂z

where the underlined terms cancel together.

164
 Figure 5.2: Solid domains discretized with finite elements.

Reordering the terms the weak form is finally expressed as

ZZZ h i ZZ
∂W ∂φ ∂W ∂φ ∂W ∂φ
k
∂x x ∂x
+ ∂y ky ∂y + ∂z kz ∂z dΩ + W αφdΓ =
Ω Γq

ZZZ ZZ ZZ (5.5)
= W QdΩ − W [q̄n − αφa ]dΓ − W qn dΓ
Ω Γq Γφ

In the above equation the use of the normal flux definition has been consid-
ered, i.e.
∂φ ∂φ ∂φ
qn = qT n = qx nx + qy ny + qz nz = −kx nx − ky ny − kz nz (5.6)
∂x ∂y ∂z
The weak form (5.5) is the starting expression for deriving the FEM equa-
tions.

5.3.2 FEM Discretization

Consider the 3D domain discretizations shown in Figure 5.2 using finite elements
with n nodes. The approximation of the unknown φ within each element is
expressed as
Xn
∼
φ = φ̂ =
(e)
Nj (x, y, z)φj
(e)
(5.7)
j=1
(e) (e)
where Nj are the shape functions of the 3D element and φj are the nodal
values of the approximation function.

165
 Substituting the above approximation into the weak form (5.5) and selecting
the same number of weighting functions as nodal unknowns, the approximate
expression of the weak form is obtained
ZZZ " # ZZ
∂Wi ∂ φ̂ ∂Wi ∂ φ̂ ∂Wi ∂ φ̂
kx + ky + kz dΩ + Wi αφ̂dΓ =
Ω ∂x ∂x ∂y ∂y ∂z ∂z Γq

ZZZ ZZ ZZ (5.8)
= Wi αφ̂dΓ − Wi [q̄n − αφa ]dΓ − Wi qn dΓ
Ω Γq Γφ
i = 1, 2, . . . , N
where N is total number of nodes in the mesh.
The Galerkin symmetric form is obtained by choosing Wi = Ni . Assuming
that all the functions in the last expression can be integrated and using the
additive property of the integral, the weak form can be expressed as the sum of
the element contributions as
(Z Z Z " Ã n ! Ã n !
P (e)
∂Ni X ∂Nj(e) (e) ∂Ni
(e) X ∂Nj(e) (e)
kx φj + ky φj +
e Ω(e) ∂x j=1
∂x ∂y j=1
∂y
à !# ZZ à n ! )
(e)
∂Ni P
n (e)
∂Nj (e) (e)
X (e) (e)
+ ∂z
kz ∂z
φj dΩ + Ni α Nj φj dΓ =
(e)
j=1 Γq j=1

(Z Z Z ZZ ZZ )
P (e) (e) (e)
= Ni QdΩ − Ni (q̄ − αφa )dΓ − Ni qdΓ
(e) (e)
e Ω(e) Γq Γφ
i = 1, 2, . . . , N
(5.9)
where the sum is performed over all the elements in the mesh.
The above integral expression can be interpreted as the i-th row of the
algebraic equation system of the discretization, i.e.
Ki1 φ1 + Ki2 φ2 + · · · + KiN φN = fi , i = 1, 2, . . . , N (5.10)
The global equation system is therefore written as
Ka = f
or
1 2 N
    
i=1 K11 K12 ... K1N   φ1   
 f1  
 










  
 
 

i=2   K21 K22 ... K2N   φ 2   f 2 
..  .. .. ..  .. = .. (5.11)
.  . . .  .   . 
..  .. ... ..  

 .





 .. 


.  . . 

.. 



 . 

i = N KN 1 KN 2 ... KN N    
φN fN
| {z } | {z } | {z }
K a f

166
 As for the 1D and 2D cases, the K matrix and the f vector can be obtained
by assembling the stiffness matrices K(e) and the elemental nodal flux vectors
f (e) for each element given by:
 (e) (e)

K11 . . . K 1n
K(e)  .. 
=  ... ..
. .  (5.12)
n×n (e) (e)
Kn1 . . . Knn
 (e) 

f1  
f1 (e) .
..
= (5.13)
n×1   (e)  
fn
where n is the number of nodes in the element and
(e)
Kd
ij
z à }| ! {
ZZZ (e) (e) (e) (e) (e) (e)
(e) ∂Ni ∂Nj ∂Ni ∂Nj ∂Ni ∂Nj
Kij = kx + ky + kz dΩ +
Ω(e) ∂x ∂x ∂y ∂y ∂z ∂z
ZZ
(e) (e) (e) (e)
+ αNi Nj dΓ = Kkij + Kcij
(e)
Γq
| {z }
(e)
Kcij
(5.14)
(e) (e)
Kdij is the conduction stiffness term and Kcij
is the convection stiffness term.
The second integral only appears in the Neumann boundary part where the
convection condition is prescribed.
For the flux vector
ZZZ ZZ
(e) (e) (e)
fi = Ni QdΩ − Ni (q̄ − αφa )dΓ −qn(e)i (5.15)
(e) (e)
Ω Γq
| {z }| {z }
(e) (e)
fQ fci
i

(e) (e)
where fQi is the nodal flux term related to the heat source Q, fci is the nodal
flux term
Z Z related to the prescribed heat flux at the Neumann boundary and
(e) (e)
q ni = Ni qn dΓ is the reaction flux in the node where the value of φ is
(e)
Γq
(e)
prescribed. Again, the term fci will be considered only in nodes that belong
to the Neumann boundary.
In the next section the above expressions will be particularized the four-
noded tetrahedral element.

5.4 FOUR-NODED TETRAHEDRAL ELEMENT

In this section the four-noded tetrahedral element will be considered. This
element is also know as the linear tetrahedron (Figure 5.3).

167
 Figure 5.3: Four-noded tetrahedral element.

The linear tetrahedron is the simplest 3D element, similarly as the three-

noded triangular element is the simplest element for 2D problems. Both element
families have similar properties.
The approximation of the unknown field inside each tetrahedron is expressed
by the following linear polynomial interpolation
φ(e) = α0 + α1 x + α2 y + α3 z (5.16)
From the above expression we can deduce
(e) (e) (e) (e)
φ1 = α0 + α1 x2 + α2 y1 + α3 z1
(e) (e) (e) (e)
φ2 = α0 + α1 x2 + α2 y2 + α3 z2
(e) (e) (e) (e) (5.17)
φ3 = α0 + α1 x3 + α2 y3 + α3 z3
(e) (e) (e) (e)
φ4 = α0 + α1 x4 + α2 y4 + α3 z4
Solving this equation system and substituting the expression for αi into
Eq.(5.16) and reordering terms yields:
4
X (e) (e) (e) (e) (e) (e)
φ= Ni (x, y, z)φi = N1 (x, y, z)φ1 + · · · + N4 (x, y, z)φ4 (5.18)
i=1

where
(e) 1 (e) (e) (e) (e)
Ni (x, y, z) = (a + bi (x) + ci y + di z) (5.19)
6Ω(e) i
with ¯ ¯ ¯ ¯
¯x(e) (e)
yj zj ¯¯
(e) ¯1 y (e) z (e) ¯
¯ j ¯ j j ¯
(e) ¯ (e) ¯ (e) ¯ ¯
ai = ¯x(e) (e)
yk zk ¯ ; bi = − ¯1 yk zk(e) ¯
(e)
¯ k(e) (e) (e) ¯ ¯ ¯
¯ xl yl zl ¯ ¯1 yl(e) zl(e) ¯
¯ ¯ ¯ ¯ (5.20)
¯x(e) 1 z (e) ¯ ¯x(e) y (e) 1¯
¯ j j ¯ ¯ j j ¯
(e) ¯ (e) ¯ (e) ¯ (e) (e) ¯
ci = ¯xk 1 zk(e) ¯ ; di = − ¯xk yk 1¯
¯ (e) (e) ¯ ¯ (e) (e) ¯
¯xl 1 zl ¯ ¯xl yl 1¯

168
 The rest of the constants can be obtained by rotating the indexes (i, j, k, l)
using the right-hand rule
The element volume Ω(e) can be found from the following determinant
¯ ¯
¯1 x(e) y (e) z (e) ¯
¯ i i i ¯
¯ (e) (e) (e) ¯
(e) ¯1 x j yj z j ¯
6Ω = ¯ (e) ¯ (5.21)
¯1 x(e) y
(e)
z ¯
¯ k k k ¯
¯1 x(e) y (e) z (e) ¯
l l l
(e)
The three dimensional shape functions forms are hyper-planes and Ni can
not be visualized as for the 2D problem. Note however that, as usual, the shape
(e)
function Ni take a unit value at the node and zero at the rest of the nodes.
Another difficulty related to 3D elements is the visualization of the results.
There are some methodologies to help with this task, like the previous subdivi-
sion in hexahedra followed by an automatic division in tetrahedra.
From the shape functions expression we deduce:
(e)
∂Ni bi ∂Ni ci ∂Ni di
= ; = ; = (5.22)
∂x 6Ω(e) ∂y 6Ω(e) ∂z 6Ω(e)
(e)
Substituting into Kij gives
(e) 1
Kdij = (kx bi bj + ky ci cj + kz di dj ) (5.23)
36Ω(e)
and
 
kx b1 b1 + ky c1 c1 + kz d1 d1
. . . kx b1 b4 + ky c1 c4 + · · · + kz d1 d4
1 kx b2 b1 + ky c2 c1 + kz d2 d1
. . . kx b2 b4 + ky c2 c4 + · · · + kz d2 d4 
(e)
Kd =  
36Ω(e) kx b3 b1 + ky c3 c1 + kz d3 d1
. . . kx b3 b4 + ky c3 c4 + · · · + kz d3 d4 
kx b4 b1 + ky c4 c1 + kz d4 d1
. . . kx b4 b4 + ky c4 c4 + · · · + kz d4 d4
(5.24)
is the conductivity stiffness matrix for the element.
The general expression for the element convection stiffness matrix is
 (e) (e) (e) (e)

ZZ N1 N1 . . . N 1 N4
 .. ... .. 
K(e)
c =  . .  αdΓ (5.25)
(e)
Γq (e) (e) (e) (e)
N4 N1 ... N 4 N4
The tetrahedron boundaries correspond to the element faces (Figure 5.3).
(e)
In that way the shape function Ni will have a null value if the i-th node does
not belong to the face where the integral is evaluated. From the shape function
expression it is easy to obtain:


 αA(e)
ZZ 
 ; i=j
 6
(e) (e)
Kc(e) = αNi Nj dΓ = (5.26)
ij (e)
Γq 


 αA (e)
 ; i 6= j
12

169
where A(e) is the tetrahedron’s face area where the convection boundary is
prescribed.
As an example, for the face 1-2-4 of the tetrahedron shown in Figure 5.3,
(e)
N3 = 0 and
 (e) (e) (e) (e) (e) (e)

N1 N1 N1 N2 0 N1 N4
 (e) (e) (e) (e) (e) (e) 
ZZ  N2 N1 N2 N2 0 N2 N4 
 
 
K(e)
c124 = α   dΓ (5.27)
(e)
Γq124  0 0 0 0 
 
 
(e) (e) (e) (e) (e) (e)
N4 N1 N4 N2 0 N4 N4
If α is constant over the face 1-2-4
 
1 1/2 0 1/2
(e)
αA124 1/2 1 0 1/2
K(e)
c124 =
  (5.28)
6  0 0 0 0 
1/2 1/2 0 1
(e)
The same procedure is followed to evaluate Kc over the other faces of the
tetrahedron.
As a conclusion, the element stiffness matrix can be written as:
(e)
K(e) = Kd + K(e) (e) (e) (e)
c123 + Kc124 + Kc134 + Kc234 (5.29)
The equivalent nodal flux vector for a uniformly distributed heat source is
 (e)   (e) 

f   N1 
  ZZZ 
 Q(e)1 
  (e) 
 

(e) fQ2 N2
fQ = (e) = (e) QdΩ (5.30)

f 
 Ω N 

 (e) 
 Q 3
  (e) 
 3

fQ4 N4
If Q is constant over the element
 

 1
QΩ(e)  
(e) 1
fQ = (5.31)
4 
 1
 
1
Note that in this case the total flux over the element is uniformly distributed
in all the element’s nodes, similarly as for the 3-noded triangle.
The vector associated to the ijl face where the normal flux is prescribed
(Figure 5.3) is
 (e)   
(e)

N i   
 N i 

ZZ 
 (e)  ZZ 
 (e)  
Nj Nj
fc(e) = − (q̄ n − αφ a ) (e) dΓ = − (q̄ n − αφ a ) dΓ (5.32)
ijl (e)
Γqijl 
N 
 (e)
Aijl 
 0  

 (e) 
k
 
 (e)  
Nl Nl

170
 (e) (e)
where Aijl is the ijl side area where Nk = 0. If q̄n , α and φa are constant over
the face, then  

 1
A
(e)  
(e) ijl 1
fcijl = − (q̄n − αφa ) (5.33)
3 
 0
 
1
(e)
In a similar form, the expression for fc can be obtained for the other faces
where the normal flux is prescribed.
As a conclusion, the element equivalent flux vector can be written as:
(e)
f (e) = fQ + fc(e)
123
+ fc(e)
124
+ fc(e)
134
+ fc(e)
234
(5.34)

where  

0
(e)
A234  

1
f234 =− (q̄n − αφa ) , etc. (5.35)
3 
1
 

1

171
Chapter 6

MATRIX FORMULATION FOR THE FEM

SOLUTION OF THE POISSON EQUATION

6.1 INTRODUCTION
In previous sections the Poisson equation has been analysed for 1D, 2D and 3D
problems with the FEM. In this section, the FEM solution will be presented in
a general form using a matrix formulation. This approach is useful to treat in
a similar manner many other problems solved with the FEM.

6.2 MATRIX FORMULATION OF THE POISSON EQUATION

The heat flow balance equation over an infinitesimal domain for 1D, 2D or 3D
problems can be expressed in a matrix form as

∇T q − Q = 0 (6.1)

where ∇ is the gradient vector operator and q is flux vector given by

£∂¤
1D: ∇ = ∂x , q = [q]
h iT
∂ ∂
2D: ∇ = ∂x , ∂y , q = [qx , qy ]T (6.2)
h iT
∂ ∂ ∂
3D: ∇ = ∂x , ∂y , ∂z , q = [qx , qy , qz ]T

The extended form of Eq.(6.1) corresponds to the heat flow balance equation
obtained in previous chapters. For example, Eq.(6.1) has the following form for
2D problems: · ¸½ ¾
∂ ∂ qx ∂qx ∂qy
, −Q= + −Q=0 (6.3)
∂x ∂y qy ∂x ∂y
The gradient vector is defined as

g = ∇φ (6.4)

173
where £ ∂φ ¤
1D: g = [g] = ∂x h iT
T ∂φ ∂φ
2D: g = [gx , gy ] = ,
∂x ∂y (6.5)
h iT
∂φ ∂φ ∂φ
3D: g = [gx , gy , gz ]T = , ,
∂x ∂y ∂z

The relationship between the heat flow vector and the gradient vector is
defined by Fourier’s law
q = −Dg = −D∇φ (6.6)
where D is the conductivity matrix given by

1D: D = [k]
· ¸
k 0
2D: D = x
0 ky
(6.7)
 
kx 0 0
3D: D = 0 ky 0 

0 0 kz

Substituting Eq.(6.6) into (6.1) the matrix form of the flux balance equation
is obtained as
∇T (D∇φ) + Q = 0 en Ω (6.8)

Expanding this equation, the usual form of the Poisson equation is obtained.
For 2D problems
 
· ¸· ¸  ∂φ  µ ¶ µ ¶
∂ ∂ kx 0 ∂x ∂ ∂φ ∂ ∂φ
, ∂φ +Q= kx + ky + Q = 0 (6.9)
∂x ∂y 0 ky   ∂x ∂x ∂y ∂y
∂y

The boundary conditions can be described in a general way as follows.

Prescribed variable condition

φ − φ̄ = 0 on Γφ (6.10)

Prescribed flux condition

nT D∇φ + q̄n + α(φ − φq ) = 0 on Γq (6.11)

where n is the external normal to the boundary Γq and the rest of the terms
have the usual meaning.

174
6.3 WEAK INTEGRAL FORM
The equivalent weak integral form of the balance equation and the boundary
conditions can be obtained from the GWR method as
Z I
T
W [∇ D∇φ + Q] + W̄ [nT D∇φ + q̄n + α(φ − φa )]dΓ = 0 (6.12)
Ω Γq

Integrating by parts the term with second derivates in the first integral and
selecting W̄ = −W , the weak form is obtained as
Z Z Z
T
− (∇ W )D∇φdΩ + W QdΩ − W [q̄n − α(φ − φa )]dΩ +
Ω Ω Γq
Z
W nT D∇φdΩ = 0 (6.13)
Γφ

Having in mind the definition of the heat flow and selecting W = δφ as the
virtual variation of the unknown and ∇ = δg as the virtual gradient vector, the
weak form can be rewritten as
Z Z Z Z
T
δg qdΩ + δφQdΩ − δφ[q̄n − α(φ − φa )]dΓ − δφqn dΓ = 0 (6.14)
Ω Ω Γq Γφ

Eq.(6.14) can be named as the virtual variation principle and it is equivalent

to the virtual work principle for solid mechanics. The virtual field δφ satisfies
the boundary conditions at the Dirichlet boundary. Therefore if φ = 0 on Γφ
then also δφ = 0 on Γφ and the last integral in (6.14) can be omitted.
The integral expression in (6.13) (or the equivalent one in (6.14)) is the
starting point for deriving the FEM discretized equations.

6.4 MATRIX FORM OF THE FEM EQUATIONS

The finite element approximation is expressed for a n-noded element in matrix
form as
 (e) 

φ1  

 


 


 (e) 

Xn 
 2 
φ 
(e) (e) (e) (e)
φ= Ni φi = [N1 , N2 , . . . , Nn(e) ] = N(e) a(e) (6.15)

 . 

i=1 
 .. 

 


 


 
(e) 
φn
with
(e) (e) (e)
N(e) = [N1 , N2 , . . . , Nn ]
(6.16)
(e) (e) (e) (e)
a = [φ1 , φ2 , . . . , φn ]T

175
 The N(e) matrix is the shape function matrix and a(e) is the nodal variable
vector for the element.
The gradient vector is expressed as a function of the elemental nodal vari-
ables by
(e)
g = ∇φ = [∇N(e) a(e) ] = [∇N(e) ]a(e) = Bi a(e) (6.17)
where B(e) is the gradient matrix for the element defined by
(e) (e) (e)
B(e) = ∇N(e) = ∇[N1 , N2 , . . . , Nn ] =
(6.18)
(e) (e) (e) (e) (e)
= [∇N1 , ∇N2 , . . . , ∇Nn ] = [B1 , B2 , . . . , B(e)
n ]

(e) (e)
where Bi = ∇Ni is the gradient matrix for the i-th node of element e. Note
(e)
that B(e) is formed by as many Bi sub-matrices as nodes has the element. For
example, for a 3-noded triangular element
 ¯ ¯ 
∂N1 ¯¯ ∂N2 ¯¯ ∂N3  (e)  ( )
(e) (e) (e)
∂Ni
 ∂x ¯ ∂x ¯ ∂x  1 (e)
bi
(e) 
B =  (e) ¯ ¯ ¯  ;
(e)  ∂x 
Bi = ∂N (e) = (6.19)
∂N1 ¯ ∂N2 ¯¯ ∂N3  2A(e) c(e)
(e) (e)
i
i
∂y ¯ ∂y ¯ ∂y ∂y

The heat flux vector is expressed in terms of the nodal variables as:

q = −Dg = −DB(e) a(e) (6.20)

Substituting Eqs.(6.17) and (6.18) into the integral expression ((6.13)) and
selecting the Galerkin form with Wi = δφ = Ni the discretized equations system
is obtained in the standard form Ka = f. The K matrix and the f vector are
obtained by assembling the element contributions as usual.
The element stiffness matrix expression is
Z Z
(e) (e)T (e)
K = B DB dΩ + α[N(e) ]T N(e) dΓ (6.21)
(e)
Ω(e) Γq

with Z Z
(e) (e) (e) (e) (e)
Kij = [Bi ]T DBj dΩ + αNi Nj dΓ (6.22)
(e)
Ω(e) Γq

The expression for K(e) holds for 1, 2 and 3 dimensional problems. For each
case the integrals are computed over the corresponding elemental domain and
the element boundaries belonging to the external Neumann boundary were the
normal flux is prescribed.
The equivalent nodal flux vector can be written in matrix form as:
Z I
(e) (e) T
f = [N ] QdΩ − [N(e) ]T [q̄n − αφa ] − r(e) (6.23)
(e)
Ω(e) Γq

176
with Z I
(e) (e) (e)
fi = Ni QdΩ − Ni [q̄n − αφa ]dΓ − qn(e)i (6.24)
(e)
Ω(e) Γq

As usual, the integrals are evaluated over the corresponding elemental do-
main.
The r(e) vector corresponds to the unknown “reaction” flux in the nodes
were the φ variable is prescribed. These fluxes are computed “a posteriori”
once the solution for the nodal value of φ is found.
It is important to note that the matrix formulation provides a general frame-
work for solving any Poisson problem in 1D, 2D and 3D. The matrix formulation
presented in this section is applied in similar way to solid mechanics and fluid
mechanics problems solved using the FEM.

6.5 THE MATRIX EQUILIBRIUM EQUATIONS OBTAINED

FROM THE NODAL FLUXES
The usual procedure to obtain the global equations system is applying the
concept of nodal flux balance (equilibrium). The process is similar to using
Kirchhoff’s law in electrical networks or in hydraulic networks. Expressing
the fluxes in the element “sides” as a function of the nodal variables and the
external actions via the balance equilibrium equation for each element, and
imposing flux balance at all nodes yields the global equilibrium equations as
Ka = f. The “flux addition” procedure allows to obtain K and f by assembling
the expressions for K(e) and f(e) for each element.
The flux balance expression will be described for the 1D and 2D elements
shown in Figure 6.1.

6.5.1 One dimensional 2-noded element

We will start with the simplest 1D element under a uniform flux Q and two
(e) (e)
nodal fluxes q1 y q2 applied at nodes 1 and 2 respectively.
Let us consider the global flux balance expression over the element written
in the weak integral form, or the equivalent expression derived from the virtual
variation principle
Z Z
T (e) (e) (e) (e)
δg q dx + δφQ dx − δφ1 q1 − δφ2 q2 = 0 (6.25)
l(e) l(e)

(e)
The negative signs in the above equation are due the flux definition for q1
(e)
and q2 which go out the element (the heat goes out the element).
Note that δφ are virtual variations of φ and δg is the virtual gradient field
(δg = ∇δφ).

177
Figure 6.1: Flux balance in a 1D 2-noded element (a) and in a 2D 3-noded
triangle (b)

Eq.(6.25) allows expressing the “virtual work” produced by the nodal fluxes
in terms of the elemental variables as:
Z Z
(e) T (e)
[δa ] q = δgqdΩ + δφQdΩ (6.26)
l(e) l(e)

with
(e) (e) (e) (e)
δa(e) = [δφ1 , δφ2 ]T , q(e) = [q1 , q2 ]T (6.27)
where q(e) is the so called vector of the interelemental balancing nodal fluxes.
Let us substitute the FEM approximation into Eq.(6.26). The φ and δφ
fields can be expressed inside each element in a matrix form by
2
X (e) (e)
φ= Ni φi = N(e) a(e) (6.28)
i=1

2
X (e) (e)
δφ = Ni δφi = N(e) δa(e) (6.29)
i=1

with
(e) (e) (e) (e)
N(e) = [N1 , N2 ], a(e) = [φ1 , φ2 ]T (6.30)

178
and
(e) (e)
δa(e) = [δφ1 , δφ2 ]T (6.31)
The virtual gradient is expressed in terms of the nodal virtual variables by

δg = ∇δφ = [∇N(e) ]δa(e) = B(e) δa(e) (6.32)

with ¸ · h i
∂ (e) (e) (e) (e) (e)
∂Ni
∇= , B = [B1 , B2 ], Bi = (6.33)
∂x ∂x

Substituting the above expression into (6.26) and extracting δa(e) in the
second term gives
½ Z Z ¾
(e) T (e) (e) T (e) T (e) (e) (e) T
[δa ] q = [δa ] −[ [B ] DB dΩ]a + [N ] QdΩ] (6.34)
l(e) l(e)

(e)
Having in mind that the virtual variables δφi are arbitrary, the above
equation must be true for any value of δa(e) ; this gives
·Z ¸ Z
(e) T
q (e)
=− [B ] DB dx a (e) (e)
+ [N(e) ]T Qdx (6.35)
l(e) l(e)

or
q(e) = −K(e) a(e) + f(e) (6.36)
where Z
(e)
K = [B(e) ]T DB(e) dx (6.37)
l(e)
Z
(e)
f = [N(e) ]T Qdx (6.38)
l(e)
are the elemental stiffness matrix and the nodal flux vector due the uniform
distributed Q source.
Eq.(6.36) is the flux balance equation for the element and express the equi-
librium (balance) between the interelemental balancing nodal fluxes, the nodal
flux dues the Q source and the fluxes due the gradient of φ inside the element.
Note the analogy between Eq.(6.36) and the equation expression the equi-
librium of nodal forces for a discrete system (Eq.(1.6)). This analogy also holds
for the assembling process.
Eq.(6.36) is the key to obtain the global equilibrium equation in terms of the
nodal variables. The process is similar to that explained for discrete systems.
For each node the equilibrium of fluxes is considered taking into account the
contribution from all the elements sharing the node.
Therefore the expression for balance of nodal fluxes is written as
X (e)
qi = qjext j = 1, 2, . . . N (6.39)
e

179
where the sum is over all the elements that share the node with global number
j, qjext is the external flux that goes into the j node (in general qjext = 0) and N
is the total number of nodes in the mesh.
(e)
Expressing the values of qi in terms of the nodal variables a(e) and the
external fluxes f(e) via Eq.(6.36), the global system of equations is obtained in
the usual form Ka = f. The stiffness matrix K and the nodal flux vector f are
assembled from the element contributions following the same rules as for the
discrete systems.
It is important to understand that this assembling process coincides with
that explained in the previous chapter using the weak integral form and the
additive properties of the integrals defined over the analysis domain.
To clarify this concept the assembling process will be repeated for the three-
noded triangular element.

6.5.2 2D three-noded triangular element

Consider the three-noded triangular element shown in Figure 6.1b. The element
is subjected to a uniform distributed source Q, to the normal fluxes through
the boundaries sides Γq and to the interelemental fluxes. The flux balance
expression for all the fluxes and the inner gradients can be expressed by the
virtual variation principle as:
ZZ ZZ Z
(e) T
[δg ] qdΩ + δφQdΩ − δφq̄n dΓ − [δa(e) ]T q(e) = 0 (6.40)
(e)
Ω(e) Ω(e) Γq

where
δg(e) = [δgx , δgy ]T = ∇(δφ) (6.41)
½∂¾
T
q = [qx , qy ] = −Dg = −D∇φ, ∇= ∂x
∂ (6.42)
∂x

(e) (e) (e)

δa(e) = [δφ1 , δφ2 , δφ3 ]T (6.43)

(e) (e) (e)

q(e) = [q1 , q2 , q3 ]T (6.44)
(e)
In the above expressions qi is the interelemental nodal flux that goes out
through the i-th node. This flux can be expressed as the sum of the fluxes in
the x and y directions.
It is important to note that the third integral in the expression (6.40) only
should be considered if the element has sides on the Neumann boundary were
the flux is prescribed to q̄n . For elements laying in the interior domain, the
fluxes along the sides cancel out in the assembly process, as these fluxes are
identical but with opposite sign for the two elements sharing a side. Also, if
any convection with the external environment exist it is necessary to add the
term α(φ − φa ) to q̄n . For simplicity reasons we do not consider this term here.

180
 The FEM discretization is written in matrix form as
3
X (e) (e)
φ= Ni φi = N(e) a(e) (6.45)
i=1

g = B(e) a(e) , δg = B(e) δa(e) (6.46)

q = −Dg = −DB(e) a(e) (6.47)
with
(e) (e) (e)
N(e) = [N1 , N2 , N3 ] (6.48)
 (e)

∂Ni
(e) (e) (e) (e) (e)  ∂x(e) 
B(e) = [B1 , B2 , B3 ], Bi = ∇Ni = ∂Ni
(6.49)
∂y
· ¸
k 0
D= x (6.50)
0 ky
Substituting these expressions into the integral form and extracting δa(e) we
obtain
½· Z Z
(e) T (e)
[δa ] q = [δa ](e) T
− [B(e) ]T DB(e) dΩ]a(e) +
Ω(e)

ZZ I #) (6.51)
+ [N(e) ]T QdΩ − [N(e) ]T q̄n dΓ
(e)
Ω(e) Γq

This gives
q(e) = −K(e) a(e) + f (e) (6.52)
with ZZ
(e) (e) (e)
Kij = [Bi ]T DBj dΩ
Ω(e)
ZZ I (6.53)
(e) (e) (e)
fi = Ni QdΩ − Ni q̄n dΓ
(e)
Ω(e) Γq

The assembling process follows the usual steps of imposing the equilibrium
of nodal fluxes as X (e)
qi = qjext j = 1, 2, . . . N (6.54)
e
(e)
Substituting in (6.54) the expressions for qi from Eq.(6.53) we obtain the
global equation systems of equations in matrix form Ka = f were matrix K
and vector f are formed by assembling the contributions if K(e) and f (e) for
each element in the standard way.

181
Chapter 7

SHAPE FUNCTIONS FOR 2D AND 3D

ELEMENTS WITH C ◦ CONTINUITY

7.1 INTRODUCTION
This chapter extends the concepts previously studied for the analysis of 2D
problems with the FEM using triangular and quadrilateral elements. The use
of higher order elements with arbitrary geometry including curved sides is
now considered. The isoparametric formulation and numerical integration is
mandatory in these cases. These topics are explained in Chapter 8.
The chapter is organized as follows. In the first sections we detail the
systematic derivation of the shape functions for rectangular and triangular el-
ements of different order of approximation. Next, some rules for the analytical
computation of the element integrals over rectangles and straight side triangles
are given. The same topics are then explained for 3D elements of tetrahedral
and prismatic shape.

7.2 DERIVATION OF THE SHAPE FUNCTIONS FOR Co TWO

DIMENSIONAL ELEMENTS
Next, we will derive nex the shape functions for different triangular and rect-
angular elements with C o continuity. The possibilities of distorting these ele-
ments into arbitrary shapes including curved sides will be treated in Chapter
8 using the concept of isoparametric interpolation.

7.2.1 Complete polynomials in two dimensions. Pascal’s triangle

We recall that the chosen displacement field can only reproduce exactly a
polynomial solution of an order equal to or less than that of the complete
polynomial contained in the shape functions. Consequently, the finite element
solution will improve as the degree of such a complete polynomial increases.

183
 Fig. 7.1: Pascal’s triangle in two dimensions

A 2D complete polynomial of nth degree can be written as

p
X
f (x, y) = αi xj y k ; j+k ≤n (7.1)
i=1

where the number of terms is

p = (n + 1)(n + 2)/2 (7.2)

For a linear polynomial (p = 3)

f (x, y) = α1 + α2 x + α3 y (7.3a)

and for a quadratic polynomial (p = 6)

f (x, y) = α1 + α2 x + α3 y + α4 xy + α5 x2 + α6 y 2 (7.3b)

The terms of a 2D complete polynomial can be readily identified by means

of the Pascal’s triangle (Figure 7.1).
The shape functions of many elements contain incomplete polynomial terms.
For instance, the 4-noded rectangle includes the term α4 xy from the quadratic
polynomial (Section 7.3.1). These terms involve nodal variables which in gen-
eral do not contribute to increasing the order of the approximation. This
should be taken into account when designing a new element.

7.2.2 Shape functions of C o rectangular elements. Natural coordi-

nates in two dimensions
A local coordinate system ξ, η is defined for each element in order to facilitate
the derivation of the shape functions. Such a natural or intrinsic coordinate
system is normalized so that the element sides are located at ξ = ±1 and
η = ±1 as shown in Figure 7.2. The natural coordinate ξ was introduced for
1D rod elements in Chapter 2. It is easy to deduce from Figure 7.2 that
x − xc y − yc
ξ= ; η= (7.4)
a b

184
Fig. 7.2: Geometry of a generic rectangular element. Cartesian and natural coor-
dinates

where xc and yc are the coordinates of the element centroid. Thus

dξ 1 dη 1
= ; = (7.5)
dx a dy b
The differential of area in both systems is related by
dx dy = ab dξ dη (7.6)
The integration of a function f (x, y) over a rectangular element can be
expressed in the natural coordinate system by
ZZ Z +1 Z +1
f (x, y) dx dy = g(ξ, η)ab dξ dη (7.7)
A(e) −1 −1
The shape functions when expressed in the natural coordinates must sat-
isfy the same requirements as in cartesian coordinates. Therefore, the shape
functions of C o continuous elements must satisfy:

a) Condition of nodal compatibility

1 i=j
Ni (ξj , ηj ) = (7.8)
0 i 6= j

b) Rigid body condition

n
X
Ni (ξ, η) = 1 (7.9)
i=1
Two element families can be clearly identified within the C o continuous
rectangular elements, i.e. the Lagrange family and the Serendipity family.
The derivation of the shape functions for each of these two element families is
presented next.

185
7.3 LAGRANGE RECTANGULAR ELEMENTS
The shape functions of Lagrange elements can be obtained by the simple prod-
uct of the two one dimensional (1D) Lagrange polynomials corresponding to
the natural coordinates ξ and η of the node. Thus, if lIi (ξ) is the Lagrange
polynomial of order I in the ξ direction for node i and lJi (η) is the Lagrange
polynomial of order J in the η direction for node i, the shape function of node
i is
Ni (ξ, η) = lIi (ξ) lJi (η) (7.10)
The 1D Lagrange polynomials for each node can be obtained by Eq.(3.9)
which can be indistinctly used for the coordinates ξ or η. Figure 7.3 shows some
of the more popular Lagrange rectangular elements. Note that the number of
nodes in each of the two directions ξ and η are the same along any nodal line.
This is a particular feature of Lagrange elements.
The polynomial terms contained in the shape functions can be directly
obtained from the Pascal’s triangle as shown in Figure 7.3. Note that the
shape functions are not complete polynomials and all contain some incomplete
polynomial terms.
The derivation of the shape functions for the more popular Lagrange rect-
angular elements is presented next.

7.3.1 Four-noded Lagrange rectangle

This is the simplest element of the Lagrange family. We will derive its shape
functions again using natural coordinates (Figure 7.4).
Let us consider a generic node i. The 1D shape functions corresponding to
the local directions ξ and η coincide with the shape functions of the 2-noded
bar element. Thus, it is easy to find
1 1
l1i (ξ) = (1 + ξξi ) ; l1i (η) = (1 + ηηi ) (7.11)
2 2
where ξi and ηi take the values given in the table of Figure 7.4. The shape
function of any node can be written in compact form as
1
Ni (ξ, η) = l1i (ξ)l1i (η) = (1 + ξξi )(1 + ηηi ) (7.12)
4
Figure 4.16 shows in graphic form the shape function of node 1. It is easy
to verify that the shape functions (7.12) satisfy the conditions (7.8) and (7.9).

7.3.2 Nine-noded quadratic Lagrange rectangle

The shape functions of the 9-noded Lagrange rectangle (Figure 7.5) are ob-
tained by the product of two 1D quadratic polynomials in ξ and η. These

186
Fig. 7.3: Some Lagrange rectangular elements. Polynomial terms contained in the
shape functions

polynomials are obtained from the shape functions of the quadratic rod ele-
ment (Eq.(3.11)). Thus, for node 1

1 1
l21 (ξ) = (ξ − 1)ξ ; l21 (η) = (η − 1)η (7.13)
2 2

187
 Fig. 7.4: Four-noded Lagrange rectangular element

and the shape function is

1
N2 (ξ, η) = l21 (ξ)l21 (η) = (ξ − 1)(η − 1)ξ η (7.14)
4
Following a similar procedure for the rest of the nodes, the shape functions
can be written in compact form as

a) Corner nodes

1
Ni = (ξ 2 + ξξi )(η 2 + ηηi ) ; i = 1, 3, 5, 7 (7.15)
4

b) Mid-side nodes

1 1
Ni = ηi2 (η 2 − ηηi )(1 − ξ 2 ) + ξi2 (ξ 2 − ξξi )(1 − η 2 ) ; i = 2, 4, 6, 8 (7.16)
2 2

c) Central node
N9 (ξ, η) = (1 − ξ 2 )(1 − η 2 ) (7.17)

188
 Fig. 7.5: Nine-noded quadratic Lagrange rectangle

We can verify that these shape functions satisfy Eqs.(7.8) and (7.9).
Figure 7.5 shows the shape function of three characteristic nodes. These
functions contain the polynomial terms shown in Figure 7.3. Note that the
9-noded Lagrange rectangle contains all the terms of a complete quadratic
polynomial plus three additional terms of the cubic and quartic polynomials
(x2 y, xy 2 , x2 y 2 ). Therefore, the approximation is simply of quadratic order.

189
Fig. 7.6: Sixteen-noded cubic Lagrange rectangular element. Shape functions in
natural coordinates

7.3.3 Sixteen-noded cubic Lagrange rectangle

This element has four nodes in each of the two directions ξ and η. The shape
functions are obtained by the product of two cubic 1D Lagrange polynomials in
the ξ and η directions deduced from Eq.(3.12). Figure 7.6 shows the resulting
expressions of the shape functions. Figure 7.3 shows that the shape function
are complete cubic polynomials and contain the following additional terms:
x3 y, x2 y 2 , xy 3 , x3 y 2 , x2 y 3 and x3 y 3 from the quartic, quintic and sextic
polynomials. The shape functions satisfy Eqs.(7.8) and (7.9).

7.3.4 Other Lagrange rectangular elements

The shape functions of higher order Lagrange rectangular elements with 5, 6,
7, etc. nodes in each of the ξ and η directions are obtained by the product of

190
Fig. 7.7: Lagrange rectangular elements with different number of nodes along each
local direction

fourth, fifth, sixth, etc. degree 1D Lagrange polynomials in ξ and η, similar

to the linear, quadratic and cubic elements previously studied. It is easy to
verify that the shape functions of a Lagrange element with n nodes in each of
the two local directions ξ and η contain a complete nth degree polynomial and
n(n + 1)/2 terms of incomplete polynomials of a higher degree which can be
deduced from the Pascal’s triangle.
Lagrange elements can have different number of nodes in each local direc-
tion ξ or η (Figure 7.7). The shape functions in this case are obtained by
the product of the adequate 1D polynomials in ξ and η corresponding to the
number of nodes in each direction. The shape functions now contain a com-
plete 2D polynomial of a degree equal to the smallest degree of the two 1D
polynomials in each local direction. Therefore, the degree of approximation of
the element does not change by simply increasing the number of nodes in one
of the two local directions only. This explains why these elements are not very
popular and they are only occasionaly used as a transition between elements
of two different orders.

7.4 SERENDIPITY RECTANGULAR ELEMENTS

Serendipity elements are obtained as follows. First the number of nodes defin-
ing a 1D polynomial of a given degree along each side is chosen. Then, the
minimum number of nodes within the element is added so that a complete
and symmetrical 2D polynomial of the same degree as the 1D polynomial cho-
sen along the sides is obtained. Figure 7.8 shows some of the more popular
Serendipity elements and the polynomial terms contained in the shape func-
tions. Note that the simplest element of the Serendipity family, i.e. the 4-noded
rectangle, coincides with the same element of the Lagrange family. Also note
that the quadratic and cubic elements of 8 and 12 nodes, respectively, do not
have interior nodes, whereas the 17 node element requires a central node to
guarantee the complete quartic approximation, as is explained next.
The derivation of the shape functions of Serendipity elements are not as
straightforwad as those of Lagrange elements. In fact, some ingenuity is needed
and this explains the name Serendipity, after the ingenuous discoveries of the
Prince of Serendip quoted in the romances of Horacio Walpole in the eighteenth
century.

191
Fig. 7.8: Some Serendipity elements and terms contained in their shape functions

7.4.1 Eigth-noded quadratic Serendipity rectangle

The shape functions of the side nodes are readily obtained as the product of
a second degree polynomial in ξ (or η) and another one in η (or ξ). It can
be checked that this product contains the required complete quadratic terms

192
(Figure 7.9). For these nodes we obtain
1
Ni (ξ, η) = (1 + ξξi )(1 − η 2 ) ; i = 4, 8
2 (7.18)
1
Ni (ξ, η) = (1 + ηηi )(1 − ξ 2 ) ; i = 2, 6
2
Unfortunately this strategy can not be applied for the corner nodes, since
in this case the product of two quadratic polynomials will yield a zero value at
the center and thus the criterion of Eq.(7.9) would be violated. Consequently,
a different procedure is followed as detailed below.

Step 1 . The shape function of the corner node is initially assumed to be

bi-linear, i.e. for node 1 (Figure 7.9) we have
1
N1L = (1 − ξ)(1 − η) (7.19)
4
This shape function is equal to one at the corner node and takes the value
zero at all the other nodes, except for the two nodes 2 and 8 adjacent to node
1 where it takes the value 1/2.

Step 2 . The shape function is made zero at node 2 by subtracting from the
bi-linear function one half of the quadratic shape function (7.19) of node 2:
1
N 1 (ξ, η) = N1L − N2 (7.20)
2

Step 3 . Function N 1 still takes the value 1/2 at node 8. Hence, the final step
is to substract from N 1 one half of the quadratic shape function of node 8
(Eq.(7.18)), i.e.
1 1
N1 (ξ, η) = N1L − N2 − N8 (7.21)
2 2
It can be verified that the resulting function N1 satisfies the conditions (7.8)
and (7.9) and contains the desired (quadratic) polynomial terms. Therefore,
it is the shape function of node 1 we are looking for.
Following the same procedure for the rest of the corner nodes yields the
following general expression
1
Ni (ξ, η) = (1 + ξξi )(1 + ηηi )(ξξi + ηηi − 1) ; i = 1, 3, 5, 7 (7.22)
4
Figure 7.9 shows that the shape functions of the 8-noded Serendipity ele-
ment contain a complete quadratic polynomial and two terms x2 y and xy 2 of
the cubic polynomial. Therefore, this element has the same approximation as
the 9-noded Lagrange element and it has a node less. This makes the 8-noded
quadrilateral in principle more competitive for practical purposes (see Section
8.1.2 for further details).

193
Fig. 7.9: 8-noded quadratic Serendipity rectangle. Derivation of the shape functions
for a mid-side node and a corner node

7.4.2 Twelve-noded cubic Serendipity rectangle

This element has four nodes along each side and a total of twelve side nodes
which define the twelve term polynomial approximation shown in Figure 7.8.
The shape functions are derived following the same procedure explained for
the 8-noded element. Thus, the shape functions for the side nodes are obtained
by the simple product of two Lagrange cubic and linear polynomials. For the
corner nodes the starting point is again the bilinear approximation. This initial
shape function is forced to take a zero value at the two side nodes adjacent to
the corner node by subtracting the shape functions of those nodes weighted
by the factors 2/3 and 1/3. Figure 7.10 shows the expression of the shape
functions which can be derived by the reader as an exercise.
It is simple to check that the element satisfies conditions (7.8) and (7.9).

194
Fig. 7.10: Shape functions for the cubic (12 nodes) and quartic (17 nodes) Serendip-
ity rectangles

Figure 7.8 shows that the shape functions contain a complete cubic approx-
imation plus two terms (x3 y, xy 3 ) of the quartic polynomial. This element
compares very favourably with the 16-noded Lagrange element, since both
have a cubic approximation but the Serendipity element has fewer nodes (12
nodes versus 16 nodes for the cubic Lagrange rectangle).

7.4.3 Seventeen-noded quartic Serendipity rectangle

The quartic Serendipity rectangle has five nodes along each side and a total
of seventeen nodes (sixteen side nodes plus a central node, Figure 7.10). The
central node is necessary to introduce the “bubble” function (1 − ξ 2 )(1 − η 2 ),
contributing the term ξ 2 η 2 to complete a quartic approximation as shown in
Figure 7.8.
The derivation of the shape functions follows a procedure similar to that
for the 8 and 12 node Serendipity elements. The shape functions for the side
nodes are obtained by the product of a quartic and a linear polynomial. An
exception are nodes 3, 7, 11 and 15 for which the function 1/2 (1 − ξ 2 )(1 − η 2 )
is subtracted from that product so that the resulting shape function takes a
zero value at the central node. The starting point for the corner nodes is
the bilinear function to which a proportion of the shape functions of the side
nodes is subtracted so that the resulting shape function takes a zero value at

195
these nodes. Finally, the shape function for the central node coincides with
the bubble function. Figure 7.10 shows the shape functions for this element.
Figure 7.8 shows that the shape functions contain a complete quartic ap-
proximation plus two additional terms (x4 y and yx4 ) from the quintic polyno-
mial. The corresponding quartic Lagrange element has 25 nodes (Figure 7.2)
and hence the 17-noded Serendipity rectangle is in principle more competitive
for practical purposes.

7.5 SHAPE FUNCTIONS FOR C o CONTINUOUS TRIANGU-

LAR ELEMENTS
The shape functions for the more popular C o continuous triangular elements
are complete polynomials whose terms can be readily identified by the Pas-
cal’s triangle. This also defines the position of the nodes within the element.
We recall, for instance, that the shape function of the 3-noded triangle is lin-
ear. Similarly, the six and ten-noded triangles define the following complete
quadratic and cubic approximations

6-noded triangle

φ = αo + α1 x + α2 y + α3 xy + α4 x2 + α5 y 2 (7.23)

10-noded triangle

φ = αo + α1 x + α2 y + α3 xy + α4 x2 + α5 y 2 + α6 x3 + α7 x2 y + α8 xy 2 + α9 y 3
(7.24)
The αi parameters are obtained by the same procedure as described for
the 3-noded triangle in Chapter 4. This method has obvious difficulties for
high order elements and it is simpler to apply the technique based on area
coordinates that is explained below.

7.5.1 Area coordinates

Let us join a point P within a triangle of area A with the three vertices 1, 2,
3 (Figure 7.11). This defines three sub-areas A1 , A2 and A3 corresponding to
the triangles P 13, P 12 and P 23, respectively (note that A1 + A2 + A3 = A).
The area coordinates are defined as
A1 A2 A3
L1 = ; L2 = ; L3 = (7.25)
A A A
Obviously
A1 + A2 + A3 A
L1 + L2 + L3 = = =1 (7.26)
A A

196
 Fig. 7.11: Area coordinates for a triangle

The position of point P can be defined by any two of these three coordi-
nates. The area coordinates of a node can be interpreted as the ratio between
the distance from point P to the opposite side divided by the distance from
the node to that side (Figure 7.11). Thus, area coordinates of the centroid
are L1 = L2 = L3 = 1/3. Area coordinates are also known as barycentric,
triangular, and trilinear coordinates and they are typical of geometry treatises
[F3]. In the FEM context area coordinates have proved to be very useful for
deriving the shape functions of triangular finite elements.
Area coordinates are also of interest to define a parametric interpolation
of the element geometry. For a straight side triangle the following relationship
between the area and cartesian coordinates can be written

x = L1 x1 + L2 x2 + L3 x3
(7.27)
y = L1 y1 + L2 y2 + L3 y3

This equation system is completed with Eq.(7.26) so that L1 , L2 and L3

can be obtained in terms of the cartesian coordinates by

1
Li = (ai + bi x + ci y) (7.28)
2A(e)

where A(e) is the area of the triangle and ai , bi , ci coincide with the values given
in Eq.(4.27) for the 3-noded triangle. It is concluded that the area coordinates
coincide precisely with the shape functions of the 3-noded triangular element.

197
7.5.2 Derivation of the shape functions for C o continuous triangles
The shape functions for triangles containing complete M th degree polynomials
can be obtained in terms of the area coordinates as follows. Let us consider a
node i characterized by the position (I, J, K) where I, J and K are the powers
of the area coordinates L1 , L2 and L3 , respectively in the expression of the
shape function of the node. Thus, I + J + K = M and the shape function of
node i is given by
Ni = lIi (L1 ) lJi (L2 ) lK
i
(L3 ) (7.29)
where lIi (L1 ) is the 1D Lagrange Ith degree polynomial in L1 which takes the
value one at node i (Eq.(3.6a)), i.e.
³ ´
Y L1 − Lj1
lIi (L1 ) = ³ ´ (7.30)
j=1,I Li1 − Lj1
j6=i

with identical expressions for lJi (L2 ) and lK

i
(L3 ). In Eq.(7.30) Li1 is the value
of L1 at node i.
The main difficulty in applying Eq.(7.29) is to deduce the values of I, J, K
for each node. This can be done simply by noting that: a) the shape function
of a corner node depends on a single area coordinate only and thus the corre-
sponding I, J or K power for that node is equal to M ; b) all nodes located on
the lines L1 = constant have the same value for I and the same occurs with L2
and J and L3 and K; and c) the values of I, J and K associated with L1 , L2
and L3 , respectively, decrease progressively from the maximum value for the
lines Li = 1 at the corner nodes, to a value equal to zero at the lines Li = 0
which coincide with the opposite side to each corner node i (Figure 7.12).
This procedure will be clarified next with some examples of its application.

7.5.3 Shape functions for the 3-noded linear triangle

The shape functions of the 3-noded triangle are linear polynomials (M = 1).
The area coordinates and the values of I, J, K for each node can be seen in
Figure 7.12.

Node 1
Position (I, J, K) : (1, 0, 0)
Area coordinates: (1, 0, 0)

N1 = l11 (L1 ) = L1 (7.31)

It is straight-forward find N2 = L2 and N1 = L3 as expected.

198
Fig. 7.12: Linear, quadratic and cubic triangular elements. Area coordinates and
values of (I, J, K) for each node

7.5.4 Shape functions for the six-noded quadratic triangle

The shape functions for this element are complete quadratic polynomials (M =
2). The area coordinates and the values of I, J and K at each node are shown
in Figure 7.12.

Node 1
Position (I, J, K) : (2, 0, 0)
Area coordinates: (1, 0, 0)
(L1 − 1/2)L1
N1 = l21 (L1 ) = = (2L1 − 1)L1 (7.32)
(1 − 1/2) 1
Node 4
Position (I, J, K) : (1, 1, 0)
Area coordinates: (1/2, 1/2, 0)
L1 L2
N4 = l12 (L1 ) l12 (L2 ) = = 4L1 L2 (7.33)
1/2 1/2

199
Fig. 7.13: Shape functions of a corner node and a side node for a quadratic triangle

Following the same procedure for the rest of nodes we find (that)
N1 = (2L1 − 1)L1 ; N2 = (2 L2 − 1)L2 ; N3 = (2L3 − 1)L3
(7.34)
N4 = 4L1 L2 ; N5 = 4 L2 L3 ; N6 = 4L1 L3
Figure 7.13 shows two characteristic shape functions for this element.

7.5.5 Shape functions for the ten-noded cubic triangle

The shape functions for this element are complete cubic polynomials (M = 3).
Figure 7.12 shows the area coordinates and the values of I, J and K at each
node:

Node 1
Position (I, J, K) : (3, 0, 0)
Area coordinates: (1, 0, 0)
(L1 − 2/3) (L1 − 1/3) L1 1
N1 = l31 (L1 ) = = L1 (3L1 − 1) (3L1 − 2) (7.35)
(1 − 2/3) (1 − 1/3) 1 2
Node 4
Position (I, J, K): (2, 1, 0)
Area coordinates: (2/3, 1/3, 0)
(L1 − 1/3) L1 L2 9
N4 = l22 (L1 ) l12 (L2 ) = · = (3L1 − 1) L1 L2 (7.36)
(2/3 − 1/3) 2/3 1/3 2
The same procedure applied to the rest of the nodes gives
1 1
N1 = L1 (3L1 − 1)(3L1 − 2) ; N2 = L2 (3L2 − 1)(3L2 − 2)
2 2
1 9
N3 = L3 (3L3 − 1)(3L3 − 2) ; N4 = (3L1 − 1)L1 L2 )
2 2
9 9
N5 = (3L2 − 1)L1 L2 ; N6 = (3L2 − 1)L2 L3 (7.37)
2 2
9
N7 = (3L3 − 1) L2 L3 ; N8 = 92 (3L3 − 1)L3 L1
2
9
N9 = (3L2 − 1)L3 L1 ; N10 = 27L1 L2 L3
2

200
 Fig. 7.14: Natural coordinates for a triangular element

A similar technique can be employed to derive the shape functions of higher

order triangular elements.

7.5.6 Natural coordinates for triangles

It is very convenient to define a normalized coordinate system α, β (also called
natural coordinate system), such that the element has the sides over the lines
α = 0, β = 0 and 1 − α − β = 0 as shown in Figure 7.14. The shape functions
for the 3-noded triangle can then be written as

N1 = 1 − α − β ; N2 = α , N3 = β (7.38)

It is deduced that the area coordinates L2 and L3 coincide with the natural
coordinates α and β, respectively and L1 = 1 − α − β.
These coincidences allow us to express the shape functions of triangular
elements in terms of the natural coordinates. This is particularly useful for
deriving isoparametric triangular elements.

7.6 ANALYTIC COMPUTATION OF INTEGRALS OVER

RECTANGLES AND STRAIGHT SIDE TRIANGLES
The integration of polynomials over 2D elements can be difficult and, in gen-
eral, numerical integration is used. However, some interesting analytical ex-
pressions in terms of the local cartesian coordinates x, y shown in Figure 7.15
exist for rectangles and straight side triangles. A typical integral term such as
ZZ
C ij = D xm y n dA (7.39)
A(e)

can be directly integrated by the following expressions:

201
Fig. 7.15: Local coordinate system x̄, ȳ for the analytical computation of integrals
over triangular and rectangular elements

Straight side triangular element

h i m!n!
C ij = D cn+1 am+1 − (−b)m+1 (7.40)
(m + n + 2)!

Rectangular element
(2a)m+1 (2b)n+1
C ij = D (7.41)
(m + 1) (n + 1)
In the above m and n are integers and a, b and c are typical element di-
mensions (Figure 7.15). Once K and f have been found in the local coordinate
system x, y using the above expressions, they can be transformed into the
global axes using the standard relationship (see Chapter 1)

Kij = TT Kij T , fi = TT f i (7.42)

where T is the 2 × 2 coordinate transformation matrix

· ¸
cos (xx) cos (xy)
T= (7.43)
cos (yx) cos (yy)

with (xx) being the angle between the local x axis and the global x axis, etc.
Other simple analytical forms in terms of area coordinates can be found
for area integrals over straight side triangles. A typical stiffness matrix term
involves the cartesian derivatives of the shape functions. This is expressed in
terms of area coordinates by the standard chain rule

∂N1 (L1 , L2 , L3 ) ∂N1 ∂L1 ∂N1 ∂L2 ∂N1 ∂L3

= + + (7.44)
∂x ∂L1 ∂x ∂L2 ∂x ∂L3 ∂x
As the element sides are assumed to be straight, Eq.(7.28) leads to

∂Li bi ∂Li ci
= and = (7.45)
∂x 2A(e) ∂y 2A(e)

202
where bi and ci are given by Eq.(4.27). Combining Eqs.(7.44) and (7.45) gives
3 3
∂Ni 1 X ∂Ni ∂Ni 1 X ∂Ni
= bi ; = ci (7.46)
∂x 2A(e) i=1 ∂Li ∂y 2A(e) i=1 ∂Li

Thus, the element integrals can be easily expressed in terms of area coor-
dinates and they can be directly computed by the following expressions
ZZ
k! l! m!
Lk1 Ll2 Lm 3 dA = 2A
(e)
(7.47a)
A(e) (2 + k + l + m)!
I
k! l!
Lki Llj ds = l(e) (7.47b)
l(e) (1 + k + l)!
If one of the area coordinates is missing in the integrand, the corresponding
power is omitted in the denominator of Eqs.(7.47) and it is made equal to a
unit value in the numerator.
The use of the natural coordinates α and β does not introduce any addi-
tional difficulty. Exact expressions for the integrals over straight side triangles
can be found as
ZZ
2A(e) Γ(m + 1) Γ(n + 1)
I= αm β n dA = (7.48)
A(e) Γ (3 + m + n)

where Γ is the Gamma function [R2]. When m and n are possitive integers

m! n!
I = 2A(e) (7.49)
(2 + m + n)!

Note that this is just a particular case of Eq.(7.47a) when one of the area
coordinates is missing. Similarly, it is deduced from (7.47b) that
I
m!
αm ds = l(e) (7.50)
l(e) (2 + m)!

For curved side elements the computation of the cartesian derivatives of the
shape functions requires a parametric formulation. This generally introduces
rational terms in the integrals and makes the use of numerical integration
unavoidable.

(e)
Example: 7.1 Compute the stiffness matrix K11 for an ortrothopic quadratic triangle with
straight sides and unit thickness.

-- Solution

The first step is to obtain the cartesian derivatives of the shape function N1 expressed
in terms of the area coordinates as

203
 ∂N1 ∂N1 ∂L1 ∂N1 ∂L2 ∂N1 ∂L3
= + +
∂x ∂L1 ∂x ∂L2 ∂x ∂L3 ∂x

∂N1 ∂N1 ∂L1 ∂N1 ∂L2 ∂N1 ∂L3

= + +
∂y ∂L1 ∂y ∂L2 ∂y ∂L3 ∂y

From Eq.(7.32) we deduce

∂N1 ∂N1 ∂N1

= 4L1 − 1 ; = =0
∂L1 ∂L2 ∂L3

and from Eq.(7.45)

∂Li bi ∂Li ci
= ; =
∂x 2A(e) ∂y 2A(e)

Therefore
∂N1 b1 ∂N1 c1
= (4L1 − 1) ; = (4L1 − 1)
∂x 2A(e) ∂y 2A(e)
and  
b 0
(4L1 − 1)  1
B1 = 0 c1 
2A(e) c1 b1

(e)
Matrix K11 is thus obtained by
 
ZZ · ¸ d d12 0
(e) t b1 0 b1  11
K11 = T
B1 D B tdA = d12 d22 0 ×
A(e) (2A (e) ) 2 0 c1 c1
0 0 d33

 
b1 0 ZZ
2
×0 c1  (4L1 − 1) dA
A (e)
c1 b1

We deduce from Eq.(7.47a)

ZZ " #
2 (e) 16 · 2! 8 · 1! 1
(4L1 − 1) dA = 2A − + = A(e)
A(e) 4! 3! 2

which leads to
· 2 ¸
(e) 1 b1 d11 + c21 d33 b1 c1 (d33 + d12 )
K11 =
4A (e) b1 c1 (d33 + d12 ) b21 d33 + c21 d22

(e)
The rest of the Kij matrices are obtained following an identical procedure. The
complete expression of the stiffness matrix for the quadratic triangle can be found in
[C15] and [W2].

204
 Fig. 7.16: Natural coordinate system for a right prism

7.7 SHAPE FUNCTIONS FOR 3D ELEMENTS

We will study next the derivation of two families of 3D solid elements: hexa-
hedra and tetrahedra of different orders. In common with 2D solid elements,
3D hexahedra can be of Lagrange or Serendipity types. The shape functions
of hexahedra elements are obtained for a right prismatic geometry in terms of
the three natural coordinates following similar rules as for 2D rectangles. The
extension of hexahedra elements to model arbitrary 3D geometries of irregular
shapes, including curved surfaces, is straightforward using an isoparametric
formulation (Chapter 8).
The shape functions for Lagrange right prisms are simply obtained by the
product of three Lagrange polynomials in 1D. The derivation of 3D Serendipity
prismatic elements is more cumbersome, although again the same rules given
for the 2D case apply (Chapter 8).
Like triangles, the shape functions for tetrahedra are complete polynomials
and they are more easily expressed in terms of volume and natural coordinates.

7.8 RIGHT PRISMS

Let us consider a right prism with the natural coordinates ξ, η, ζ defined as
shown in Figure 7.16. For an element with edges 2a × 2b × 2c we have
(x − xc ) (y − yc ) (z − zc )
ξ= ; η= ; ζ= (7.51)
a b c
where (xc , yc , zc ) are the coordinates of the element centroid. From Eq.(7.51)
it is deduced that
dξ 1 dη 1 dζ 1
= ; = ; = (7.52)
dx a dy b dz c

205
and a differential of volume is expressed by
dx dy dz = abc dξ dη dζ (7.53)
The integration of a function f (x, y, z) over the element is expressed in the
natural coordinate system as

ZZZ Z +1 Z +1 Z +1
f (x, y, z) dV = f (ξ, η, ζ)abc dξ dη dζ (7.54)
V (e) −1 −1 −1

Since the element is a right prism, the cartesian derivatives of the shape
functions are directly obtained by
∂Ni 1 ∂Ni ∂Ni 1 ∂Ni ∂Ni 1 ∂Ni
= ; = ; = (7.55)
∂x a ∂ξ ∂y b ∂η ∂z c ∂ζ

The shape functions must satisfy the standard conditions

(
1 if i = j
Ni (ξj , ηj , ζj ) = (7.56a)
0 if i 6= j
and n
X
Ni (ξ, η, ζ) = 1 (7.56b)
i=1

7.8.1 Right prisms of the Lagrange family

The shape functions of Lagrange right prisms are obtained by multilying three
1D polynomial Lagrange polynomials as

Ni (ξ, η, ζ) = lIi (ξ) lIi (η) lIi (ζ) (7.57)

where lIi (ξ) is the Lagrange polynomial of Ith degree passing by node i, etc.
For the same reasons mentioned in Section 7.3.4 it is usual, to choose the same
polynomial approximation in each of the three directions ξ, η and ζ.
The terms contained in the shape functions of prismatic elements are
deduced from the Pascal’s tetrahedron. Figure 7.17 shows the linear and
quadratic elements of this family whose shape functions are derived next.

Linear right prism of the Lagrange family

The simplest Lagrange prismatic element is the 8-noded linear prism shown in
Figure 7.17.
The shape function of a node is obtained by multilying the three linear
polynomials in ξ, η and ζ, corresponding to the node. In general form
1
Ni (ξ, η, ζ) = (1 + ξi ξ) (1 + ηi η) (1 + ζi ζ) (7.58)
8

206
Fig. 7.17: Linear and quadratic right prisms of the Lagrange family. Polynomial
terms contained in the shape functions deduced from the Pascal’s tetrahedron

Note that:
1. The shape functions contain a complete linear polynomial in ξ, η, ζ and
the incomplete quadratic and cubic terms ξη, ξζ, ηξ and ξηζ, ξηζ (see
Figure 7.18).

2. The shape functions satisfy the conditions (7.56).

The linear hexahedron is a popular element due to the small number of
nodal variables, which is attractive for practical 3D analysis.
The linear hexahedron performs very well for problems with uniform gradi-
ents, whereas its accuracy is poor for problems where strains gradient changes
occur.

207
 Fig. 7.18: Shape functions for the linear right prism

Fig. 7.19: Quadratic right prism of the Lagrange family. Shape functions for a
corner node and a mid-side node

208
Quadratic right prism of the Lagrange family

The quadratic Lagrange prism has 27 nodes (see Figure 7.19). The shape
functions are obtained by the product of three 1D quadratic Lagrange poly-
nomials. Figure 7.19 shows the derivation of the shape functions for a corner
node and a mid-side node. It is simple to extrapolate this procedure to obtain
the following general expressions

Corner nodes
1 2 1, 3, 5, 7
Ni = (ξ + ξξi )(η 2 + ηηi )(ζ 2 + ζζi ) ; i = 19, 21, 23, 25 (7.59)
8
Mid-side nodes

1 1
Ni = ηi2 (η 2 − ηηi )ζ 2 (ζ 2 − ζζi )(1 − ξ 2 ) + ζi2 (ζ 2 − ζζi )+ 2, 4, 6, 8
4 4 ; i= 10, 12, 14, 16
1 2 2 20, 22, 24, 26
+ξi (ξ − ξξi )(1 − η ) + ξi (ξ − ξξi )ηi2 (η 2 − ηηi )(1 − ζ 2 )
2 2 2
4
(7.60)
Face nodes

1 1
Ni = (1 − ξ 2 )(1 − η 2 )(ζ + ζi ζ 2 ) + (1 − η 2 )(1 − ζ 2 )(ξ + ξi ξ 2 )+
2 2 9, 11, 13
; i = 15,
1 2 2 2
17, 27
+ (1 − ξ )(1 − ζ )(η + ηi η )
2
(7.61)
Central internal node
N18 = (1 − ξ 2 )(1 − η 2 )(1 − ζ 2 ) (7.62)

Other hexahedral elements of the Lagrange family

The next members of the Lagrange family are the 64-noded cubic prism (4
nodes along each edge) and the 125-noded quartic prism. Their shape func-
tions are obtained by the product of three 1D cubic and quartic polynomials
respectively. These elements in their right form are in general not competitive
versus the analogous Serendipity element which involve lewer nodal variables.

7.8.2 Serendipity prisms

Serendipity prisms are obtained by extension from the corresponding 2D rect-
angular Serendipity elements. Figure 7.20 shows the first two elements in the
family, i.e. the 8- and 20-noded right prisms. Note that the 8-noded prism is
common to the Lagrange and Serendipity families and hence its shape func-
tions coincide with those given in (7.58).

209
Fig. 7.20: 8-noded and 20-noded Serendipity prisms. Polynomial terms contained
in the shape functions deduced from the Pascal’s tetrahedron

20-noded quadratic Serendipity prism

The shape functions are obtained using similar criteria as for the 8-noded rect-
angle (Section 7.4.1). The shape functions for the side nodes are obtained by
multilying a 1D quadratic Lagrange polynomial and two 1D linear polynomials
expressed in the natural coordinates. For the corner nodes, a two step proce-
dure is followed. The first step involves the derivation of the trilinear function
corresponding to the node. This function is subsequently modified so that it
takes a zero value at the side nodes, by subtracting one half of the values of the
shape function of the side nodes adjacent to the relevant corner node under
consideration. See Section 7.4.1 for details.
Figure 7.21 shows the derivation of the shape function for a side node (20)
and a corner node (13). The element shape functions are written in compact
form as

Corner nodes
1 1, 3, 5, 7
Ni = (1 + ξi ξ)(1 + ηi η)(1 + ζi ζ)(ξi ξ + ηi η + ζi ζ − 2) ; i = 13, 15, 17, 19 (7.63)
8

210
Fig. 7.21: 20-noded quadratic Serendipity prism. Shape functions for a side node
and a corner node

Side nodes
1
Ni = (1 − ξ 2 )(1 + ηi η)(1 + ζi ζ) ; i = 2, 6, 14, 18
4
1
= (1 − η 2 )(1 + ζi ζ)(1 + ξi ξ) ; i = 4, 8, 16, 20 (7.64)
4
1
= (1 − ζ 2 )(1 + ηi η)(1 + ξi ξ) ; i = 9, 10, 11, 12
4
Note that:
1. The shape functions contain a complete quadratic polynomial plus the
terms ξη 2 , ξ 2 η, ξ 2 ζ, ξ.ζ 2 , ζ 2 η, η 2 ζ, ξηζ, ξ 2 ηζ, ξη 2 ζ and ξηζ 2 . The shape
functions satisfy the conditions in (7.56).

2. The 20-noded Serendipity prism has the same quadratic approximation

as the 27-noded Lagrange prism. This means savings of 21 nodal vari-

211
 Fig. 7.22: Shape functions for the 32-noded cubic Serendipity prism

ables per element, which explains the popularity of the 20-noded right
prism for practical applications.

32-noded cubic Serendipity prism

This element has 8 corner nodes and 24 nodes along the edges as shown in Fig-
ure 7.22. The 12 face nodes define a cubic polynomial over each face as for the
corresponding quadrilateral element (see Section 7.4.2). The shape functions
for the side nodes are obtained by multilying a 1D cubic Lagrange polynomial
and two 1D linear polynomials expressed in the natural coordinates. For the
corner nodes the starting point is the trilinear shape function, from which a
proportion of the shape functions of the side nodes are subtracted so that the
final shape function takes a zero value at these nodes. The reader is invited to
derive the expressions of the shape functions for this element shown in Figure
7.22.
The shape functions contain a complete cubic polynomial (20 terms) plus
the following twelve terms: ξ 3 η, ξη 3 , η 3 ζ, ηζ 3 , ξζ 3 , ζ 3 ξ, ξ 2 ηζ, ξη 2 ζ, ξηζ 2 , ξ 3 ηζ,
ξη 3 ζ, ξηζ 3 . Therefore, the cubic Serendipity prism has the same approximation
as the analogous 64-noded Lagrangian prism, but with a substantial reduction

212
in the number of nodal variables. This element is less popular however than
the 20-noded prism.

(e)
Example: 7.2 Compute the matrix K11 for the 8-noded right prism of Figure 7.8 for a
homogeneous isotropic material.

-- Solution

The cartesian derivatives of the shape function N1 are obtained using Eqs.(7.58) as
(note that the element sides are straight)

· ¸
∂N1 ∂N1 ∂ξ 1 ∂ 1 1
= = (1 − ξ)(1 − η)(1 − ξ) = − (1 − η)(1 − ζ)
∂x ∂ξ ∂x a ∂ξ 8 8a

and

∂N1 ∂Ni ∂y 1
= = − (1 − ξ)(1 − ζ)
∂y ∂y ∂η 8b

∂N1 ∂Ni ∂z 1
= = − (1 − ξ)(1 − η)
∂z ∂z ∂ζ 8c

The strain matrix B1 is

 
1
− (1 − η)(1 − ζ) 0 0
 a 
 
 1 
 0 − (1 − ξ)(1 − ζ) 0 
 b 
 
 1 

1 0 0 − (1 − ξ)(1 − η) 
c 
B1 =  
8 1 1 
 − (1 − ξ)(1 − ζ) − (1 − η)(1 − ζ) 0 
 b a 
 
 1 1 
− (1 − ξ)(1 − η) 0 − (1 − η)(1 − ζ)
 c a 
 1 1 
0 − (1 − ζ)(1 − η) − (1 − ξ)(1 − ζ)
c b

(e)
Matrix K11 is obtained by

ZZZ Z +1 Z +1 Z +1
(e)
K11 = BT1 DB1 dV = BT1 DB1 abc dξ dη dζ
V (e) −1 −1 −1

213
Denoting ξ = 1 − ξ, η = 1 − η and ζ̄ = 1 − ζ we can write
 
1
− ηξ 0 0
   
d11 d12 d13  a 1 
 
  0 − ξζ 0 
d12 d22 d23 0  b 
  1 
  0 0 − ξη 
1d13 d23 d33 
D B1 =   c   =

8  1 1 
d44  − ξζ − ηζ 0 
  b a 
0 d55  
  1 1 
− ξ η̄ 0 − ηζ 
d66  c a 
1 1 
0 − ξη − ξζ
c b

 
d11 d13
− a ηζ − db12 ξζ − ξη 
 c 
 d12 d22 d23 
− ξη 
 a ηζ −
b
ξζ −
c 
 
 d d23 d33 
 13 
− ηζ − ξζ − ξη 
1 a b c 
=  
8  d44 d44 
− 0 
 b ξζ −
a
ηζ 
 
 d55 d55 
− ηζ 
 c ξη 0 −
a 
 
 d66 d66 
0 − ζη − ξζ
c b

Multiplying the previous equation by BT1 gives

BT1 DB1 =
µ µ ¶ µ ¶ 
d11 d44 d55 ξ3 ξ4
 a2 η 1 + b2 ξ 2 + c2 ξ 1 ) d12 + d44 ab d13 + d55 ac 
 µ ¶ µ ¶ 
1 



= d22
b2 ξ 2 + d44
a2 η 1 + d66
c2 ξ 1 d23 + d66 ξbc5
64 
 µ

¶
 
d33 d55 d66
Symmetrical c2 ξ 1 + a2 η 1 + b2 ξ 2

with ξ¯1 = ξ¯η̄, ξ¯2 = ξ¯2 ζ̄ 2 , ξ¯3 = ξ¯η̄ ζ̄ 2 , ξ¯4 = ξ¯η̄ 2 ζ̄, ξ¯5 = ξ¯2 η̄ ζ̄, η̄1 = η̄ 2 ζ̄ 2 .

Taking into account that

Z +1 Z +1 Z +1
2 2 8
ξ dξ = η 2 dη = ζ =
−1 −1 −1 3
Z +1 Z +1 Z +1
ξ¯ dξ = η̄ dη = ζ̄ dζ = 2
−1 −1 −1

214
 we finally obtain
 µ ¶ 
1 d11 d44 d55 1 1
9 a2 + b2 + c2 6ab (d12 + d44 ) 6ac (d13 + d55 ) 
 µ ¶ 
(e) V(e) 
K11 =  1 d22
+ d44
+ d66 1
(d + d ) 
8  
9 b2 a2 c2 6bc 23 66
 µ ¶
 
1 d33 d55 d66
Symmetrical 9 c 2 + a 2 + b2

(e)
The rest of the Kij matrices are obtained following a similar procedure.

7.9 STRAIGHT EDGED TETRAHEDRA

Straight edged tetrahedral elements are a direct 3D extension of straight sided
triangles. Their shape functions are also complete polynomials whose terms
can easily be deduced from the Pascal’s tetrahedron as shown in Figure 7.23.
The shape functions of tetrahedral elements can be written in terms of
volume coordinates and/or natural coordinates. The volume coordinates are
identified by L1 , L2 , L3 and L4 and have a similar meaning to the area coor-
dinates in triangles. Each coordinate Li is now defined as the ratio between
the volume of the tetrahedron formed by a point inside the element P and the
face opposite node i, and the total volume (see Figure 7.24). Thus
Volume P jkl
Li = ; i = 1, 2, 3, 4 (7.65)
V (e)
Obviously, the following expression holds

L1 + L2 + L3 + L4 = 1 (7.66)

Volume coordinates can be used to define a linear interpolation of the

element geometry as
4
X 4
X X4
x= Li xi , y = Li yi , z = Li zi (7.67)
i=1 i=1 i=1

Eqs. (7.65) and (7.66) allow us to eliminate Li in terms of the cartesian

coordinates as l
Li = (ai + bi x + ci y + di z) = Ni (7.68)
6V (e)
where ai , bi , ci , di coincide with the values given in Eq.(5.20) for the shape
functions of the linear tetrahedron.
Eq.(7.68) allows us to obtain the cartesian derivatives of the volume coor-
dinates as
∂Li l ∂Li l ∂Li l
= bi ; = c i ; = di (7.69)
∂x 6V (e) ∂y 6V (e) ∂z 6V (e)

215
Fig. 7.23: Straight edged tetrahedral elements: linear (4 nodes), quadratic (10
nodes), cubic (20 nodes). Polynomial terms contained in the shape functions

The natural coordinates α, β, γ define a normalized straight tetrahedron

with faces for α = 0, β = 0, γ = 0 and 1 − α − β − γ = 0 (Figure 7.25). For a
tetrahedron with edges a, b, c we have

x − xi y − yi z − zi
α= ; β= ; γ= (7.70)
a b c

where i is the node taken as origin of the natural coordinate system. It is

deduced from Eq.(7.70)

dα 1 dβ 1 dγ 1
= ; = ; = (7.71)
dx a dy b dz c

216
 Fig. 7.24: Volume coordinates in a tetrahedron

Fig. 7.25: Natural coordinate system α, β, γ in a tetrahedron

A differential of volume can be expressed as

dx dy dz = abc dα dβ dγ (7.72)

The integral of a function f (x, y, z) over the element can be written in the
natural coordinate system as
ZZZ Z 1 Z 1−α Z 1−β−γ
f (x, y, z) dx dy dz = f (α, β, γ)abc dα dβ dγ (7.73)
V (e) 0 0 0

The shape functions of the linear tetrahedron can be expressed simply in

terms of the natural coordinates α, β, γ as

N1 = 1 − α − β − γ ; N2 = α; N3 = β; N4 = γ (7.74)

Note that the shape functions in the natural coordinate system satisfy
Eqs.(7.56). Also from Eqs.(7.74) and (7.71) we obtain
∂Ni 1 ∂Ni ∂Ni 1 ∂Ni ∂Ni 1 ∂Ni
= ; = ; = (7.75)
∂x a ∂α ∂y b ∂β ∂z c ∂γ

217
Fig. 7.26: 10-noded quadratic tetrahedron. Generalized coordinates I, J, K, L and
natural coordinates α, β, γ for each node

The relationship between the volume and natural coordinates is readily

deduced from Eqs.(7.68) and (7.74) as
L1 = 1 − α − β − γ; L2 = α; L3 = β L4 = γ (7.76)
Volume coordinates allow us to express the element shape functions of
tetrahedral elements by the product of four 1D Lagrange polynomials, in a
similar way as explained for triangular elements. Thus the shape function of
a node i with generalized coordinates (I, J, K, L) is given by
i
Ni = lIi (L1 ) lIi (L2 ) lK (L3 ) lLi (L4 ) (7.77)
where the value of the generalized coordinates I, J, K, and L coincides with the
power of each volume coordinate in the expression of Ni . There I +J +K +L =
M where M is the degree of the complete polynomial contained in Ni . Also,
lIi (Lj ) is the Lagrange polynomial of Ith degree in Lj passing by node i (see
Eq.(3.5)). Figures 7.26 and 7.27 show the values of the generalized coordinates
I, J, K, L for two typical tetrahedral elements.
The expression of the shape functions in terms of natural coordinates α, β, γ
is straightforward by making use of the transformations (7.76).

218
Fig. 7.27: 20-noded cubic tetrahedron. Generalized coordinates I, J, K, L and na-
tural coordinates for each node

7.9.1 Shape functions for the 10-noded quadratic tetrahedron

The nodal values of the generalized coordinates I, J, K, L and of the natural
coordinates are shown in Figure 7.26. The values of the volume coordinates
Li are deduced from Eq.(7.76). The shape functions are derived next using
Eq.(7.77).

219
Node 1
Position (I, J, K, L) : (2, 0, 0, 0). Volume coordinates : (1, 0, 0, 0)
µ ¶
1
L1 − 2 L1
1
N1 = l2 (L1 ) = = (2L1 − 1)L1 (7.78)
1 − 12

Node 2
Position (I, J, K, L) : (1, 1, 0, 0). Volume coordinates : ( 12 , 12 , 0, 0)
L1 L2
N2 = l12 (L1 ) l12 (L2 ) = 1 1 = 4L1 L2 (7.79)
2 2
The same procedure gives
N3 = (2L2 − 1) ; N7 = 4 L2 L4
N4 = 4L2 L3 ; N8 = 4 L3 L4
(7.80)
N5 = (2 L3 − 1)L − 3 ; N9 = 4 L1 L4
N6 = 4 L1 L3 ; N10 = (2 L4 − 1) L4
The expression of Ni in terms of the natural coordinates is obtained using
Eq.(7.76). The cartesian form of Ni for a straight edged tetrahedron is directly
obtained from Eq.(7.68).
The shape functions of this element contain all the terms of a quadratic
polynomial (Figure (7.23)) and they satisfy Eqs.(7.56).

7.9.2 Shape functions for the 20-noded quadratic tetrahedron

The nodal values of the generalized coordinates I, J, K, L and α, β, γ are shown
in Figure 7.27. From Eqs.(7.77) we obtain:

Node 1
Position (I, J, K, L) : (3, 0, 0, 0). Volume coordinates : (3, 0, 0, 0)
µ ¶µ ¶
2 1
L1 − 3 L1 − 3 L1
1 1
N1 = l3 (L1 ) = µ ¶µ ¶ = L1 (3L1 − 1) (3L1 − 2) (7.81)
2
1 − 23 1 − 31 1

Node 2
Position (I, J, K, L) : (2, 1, 0, 0). Volume coordinates : ( 32 , 13 , 0, 0)
µ ¶
1
L1 − 3 L − 1
2 2 L2 9
N2 = l2 (L1 ) L1 (L2 ) = µ ¶ 1 = (3L1 − 1) L1 L2 (7.82)
2 1 2 2
3
−3 3 3

220
Fig. 7.28: Local coordinate system x̄ȳz̄ for the analytical computation of volume
integrals over right prisms

Following a similar procedure gives

9 9 9
N3 = (3L2 − 1)L1 L2 ; N9 = (3L1 − 1)L1 L3 ; N15 = (3L1 − 1)L1 L4
2 2 2
1
N4 = L2 (3L2 − 1)(3L2 − 2) ; N10 = 27 L1 L2 L3 ; N16 = 27L1 L2 L4
2
9 9 9
N5 = (3L2 − 1)L2 3 ; N11 = (3L2 − 1)L2 L4 ; N17 = (3L4 − 1)L4 L2
2 2 2
9 9
N6 = (3 L3 − 1)L3 L2 ; N12 = 27 L2 L3 L4 ; N18 = (3L4 − 1)L4 L3
2 2
1 9 9
N7 = L3 (3L3 − 1)(3L3 − 2) ; N13 = (3L3 − 1)L3 L4 ; N19 = (3L4 − 1)L4 L1
2 2 2
9 1
N8 = (3L3 − 1)L3 L1 ; N14 = 27L1 L3 L4 ; N20 = L4 (3L4 − 1)(3L4 − 2)
2 2
(7.83)
Eqs.(7.76) allows us to express Ni in terms of the natural coordinates. The
cartesian form of Ni for a straight edged tetrahedron is obtained substituting
Eq.(7.68) into above expressions. Note that the shape functions are complete
cubic polynomials and they satisfy Eq.(7.56).

7.10 COMPUTATION OF THE ELEMENT INTEGRALS

7.10.1 Analytical computation of element integrals
In general, the computation of the element integrals is carried out via numerical
integration. Useful analytical rules can, however, be derived for straight edged
tetrahedra or right prisms as shown below.
Volume integrals over right prisms can be computed exactly using the local
coordinate system xȳz shown in Figure 7.28. For instance, the integral over a
face with x = constant, is first obtained by Eq.(7.41) and then a simple line

221
integral in the z direction is performed. The resulting local stiffness matrix
K is next transformed to the global axes xyz by a transformation similar to
Eq.(7.42) taking into account the third cartesian axis. This procedure is not
applicable to non-regular or hexahedra curved sides for which the use of an
isoparametric formulation and numericaly integration is essential.
The integrals over straight-sided tetrahedra have simpler expressions. Thus,
the volume integral of a polynomial term expressed in volume coordinates is
given by
ZZZ
k! l! m! n!
Lk1 Ll2 Lm n
3 L4 dV = 6V
(e)
(7.84)
V (e) (k + l + m + n + 3)!

Similarly, the surface integrals over the element faces can be obtained by
Eqs.(7.40) or (7.38).
The use of natural coordinates does not introduce any additional difficulty.
The volume integrals are computed by
ZZZ
k! l! m!
αk β l γ m dV = 6V (e) (7.85)
V (e) (k + l + m + 3)!
and the surface integrals can be obtained by Eq.(7.48). Recall that if any of
the coordinates is missing in the integrals of Eqs.(7.84) and (7.85) then the
corresponding power is made equal to one in the numerator and to zero in the
denominator of the corresponding right-hand side.
Curved-side tetrahedra require an isoparametric formulation and numerical
integration (Chapter 8).

(e)
Example: 7.3 Compute the submatrix K11 of a 20-noded quadratic tetrahedron with
straight sides.

-- Solution

We compute first the cartesian derivatives of the shape function N1 expressed in terms
of volume coordinates. For instance,

∂N1 ∂N1 ∂L1 ∂N1 ∂L2 ∂N1 ∂L3 ∂N1 ∂L4

= + + +
∂x ∂L1 ∂x ∂L2 ∂x ∂L3 ∂x ∂L4 ∂x

Since N1 = L1 (2L1 − 1) we have from Eq.(7.68)

∂N1 bi
= (4L1 − 1)
∂x 6V (e)

Following a similar procedure gives

∂N1 ci ∂N1 di
= (4L1 − 1) and = (4L1 − 1)
∂y 6V (e) ∂z 6V (e)

222
The strain matrix B1 is written as
 
bi 0 0
0 ci 0
 
 
(4L1 − 1) 

0 0 di  (4L1 − 1)
=
B1 =  B1
6V (e)
 ci bi 0  6V (e)
 
di 0 bi 
0 di ci

(e)
and matrix K11 is given by
ZZZ ZZZ
(e) 1 T T 2
K11 = BT1 DB1 dV = 2 B1 DB1 (4L1 − 1) dV
V (e) 36(V (e) ) V (e)

Making use of Eq.(7.84) gives finally

(d b2 + d c2 + d d2 ) b1 c1 (d12 + d44 ) (d13 b1 d1 + d55 d1 c1 )


11 1 44 1 55 1

(e) 1  
K11 =  (d22 c21 + d44 b21 + d66 d21 ) c1 d1 (d23 + d66 ) 
60V (e)
Symmetrical (d33 d21 + d55 b21 + d66 c21 )

(e)
The rest of the Kij matrices are obtained in a similar manner.

223
Chapter 8

2D AND 3D ISOPARAMETRIC ELEMENTS

8.1 ISOPARAMETRIC QUADRILATERAL ELEMENTS

We recall that the term “isoparametric” means that the standard shape func-
tions are used to interpolate the element geometry in terms of the nodal coor-
dinates. Thus, the geometry of a 2D isoparametric quadrilateral with n nodes
is expressed as
Xn Xn
x= Ni (ξ, η) xi ; y = Ni (ξ, η) yi (8.1)
i=1 i=1

where Ni (ξ, η) are the standard displacement shape functions. Eqs.(8.1) relate
the cartesian and the natural coordinates at each point. Such a relationship
must be unique and this is satisfied if the Jacobian matrix of the coordinate
transformation x, y → ξ, η has a constant sign over the element [S14].
It can be shown that this condition is satisfied for linear quadrilateral el-
ements if all internal angles between two element sides are greater than 180◦
[S14]. For quadratic elements it is additionally required that the side nodes
are located within the “middle third” of the distance between adjacent corners
[J5]. There are no practical rules for higher order quadrilateral elements and
the constant sign of the Jacobian is the only possible verification in this case.
Figure 8.1 shows some examples of 2D isoparametric elements.
Most of the ideas of the isoparametric formulation originated from the work
of Taig [T1, T2], who derived 4-noded isoparametric quadrilaterals. These
concepts were generalized to more complex 2D and 3D elements by Irons [I2,7].
Eq.(8.1) allows us to obtain a relationship between the derivatives of the
shape functions with respect to the cartesian and the natural coordinates. In
general, Ni is expressed in terms of the natural coordinates ξ and η and the
chain rule of derivation yields
∂Ni ∂Ni ∂x ∂Ni ∂y
= +
∂ξ ∂x ∂ξ ∂y ∂ξ
(8.2)
∂Ni ∂Ni ∂x ∂Ni ∂y
= +
∂η ∂x ∂η ∂y ∂η

225
 Fig. 8.1: Some two-dimensional isoparametric elements

In matrix form
   
∂y    
 ∂N i  ∂x  ∂Ni   ∂Ni 
∂ξ  ∂ξ ∂ξ  ∂x ∂x
= = J(e) (8.3)
 ∂Ni  ∂x ∂y   ∂Ni   ∂Ni 
∂η ∂η ∂η ∂y ∂y
| {z }
J(e)

where J(e) is the Jacobian matrix of the transformation between the cartesian
and natural coordinates. The superindex e in J denotes that this matrix is
always computed at element level. We deduce from Eq.(8.3)
  " #−1  ∂Ni   ∂y  
 ∂Ni    ∂y   ∂Ni 
∂x ∂ξ 1  ∂η −
∂ξ  ∂ξ
= J(e) = ¯¯ ¯ (8.4)
 ∂Ni   ∂Ni  ¯J(e) ¯¯ − ∂x ∂x  ∂Ni 
∂y ∂η ∂η ∂ξ ∂η
¯ ¯
¯ ¯
where ¯J(e) ¯ is the determinant of the Jacobian matrix. This determinant also
relates the differential of area in the two coordinate systems, i.e.
¯ ¯
¯ ¯
dx dy = ¯J(e) ¯ dξ dη (8.5)

226
 The terms of the Jacobian matrix are computed using the isoparametric
approximation (8.1), i.e.
n n
∂x X ∂Ni ∂x X ∂Ni
= xi ; = xi ; etc. (8.6)
∂ξ i=1
∂ξ ∂η 1=1
∂η

and  ∂N
n ix ∂Ni y 
X i
∂ξ i 
J(e) =  ∂ξ (8.7)
∂Ni x ∂Ni y
i=1
∂η i ∂η i
For a rectangular element
· ¸
(e) a 0
J = and |J(e) | = ab (8.8)
0 b

8.1.1 Stiffness matrix and load vector of isoparametric quadrilateral

Substituting the cartesian derivatives of the shape functions from Eq.(8.4)
into (6.19) yields the general form of the gradient matrix for an isoparametric
quadrilateral element in terms of the natural coordinates by
 
 ∂N
 ∂x 
i 
  ½ ¾
1
Bi (ξ, η) = = ¯¯ ¯ bi (8.9)

 ∂N 
 J (e) ¯ ci
 i  ¯ ¯
∂y
where
∂y ∂Ni ∂y ∂Ni ∂x ∂Ni ∂x ∂Ni
bi = − ; ci = − (8.10)
∂η ∂ξ ∂ξ ∂η ∂ξ ∂η ∂η ∂ξ
The element stiffness matrix for a 2D Poisson element can be written in
the normalized natural coordinate space as
Z Z Z+1Z+1 ¯ ¯
(e) ¯ (e) ¯
Kij = BTi D Bj dxdy = BTi (ξ, η) D Bj (ξ, η) ¯ J ¯ dξ dη =
A(e)
−1 −1
Z+1Z+1
1
= (kx bi bj + ky ci cj ) ¯¯ ¯ dξdη =
¯
(e)
−1 −1
¯J ¯
Z+1Z+1
= Gij dξdη (8.11a)
−1 −1

with
1
Gij = (kx bi bj + ky ci cj ) ¯¯ ¯
¯ (8.11b)
¯J(e) ¯

227
 (e)
Eq.(8.11a) shows that the integrand of Kij contains rational functions in
ξ and η. An exception to this rule is when the determinant of the Jacobian
matrix is constant. This only occurs for rectangular elements (and for straight
side triangles). In these cases the element integrals contain simple polynomials
and the analytical expressions of Section 7.6 can be applied. However, for
(e)
general quadrilateral shapes the analytical integration of Kij is difficult (and
in most cases impossible!) and the only option is to use numerical integration.
A similar procedure will be followed to compute the equivalent nodal force
vectors for isoparametric elements. For example, for the case of a heat source
ZZ Z +1 Z +1 ¯ ¯
(e) ¯ (e) ¯
fi = Ni Q dx dy = Ni Q¯J ¯t dξ dη (8.12)
A(e) −1 −1

Numerical integration is also typically used to compute integrals such as

that of Eq.(8.12).

Example: 8.1 Formulate an isoparametric quadrilateral of 4 nodes.

Solution

The actual and normalized geometries of the element are shown in the Figure 8.2
below.
The isoparametric description of the geometry is written as

½ ¾ X4 ½ ¾
x x
x= y = Ni (ξ, η) y i (8.13)
i
i=1

where Ni (ξ, η) = 14 (1 + ξξi )(1 + ηi η).

Fig. 8.2: Four-noded isoparametric quadrilateral. Actual and normalized geometry

228
The above expression maps the natural coordinates of each element point on to the
cartesian space. For instance, the cartesian position of the central point O with
natural coordinates (0,0) is
4
X ½ ¾ ½ ¾
xi 1 x1 + x2 + x3 + x4
xO = Ni (0, 0) y = (8.14a)
i=1
i 4 y1 + y2 + y3 + y4

Point A at the center of side 1–2 with natural coordinates (0,-1) is located
in the cartesian space at
X 4 ½ ¾ ½ ¾
xi 1 x1 + x2
xA = Ni (0, −1) y = (8.14b)
i=1
i 2 y1 + y2

We can observe that the isoparametric description is useful to express all the element
expressions in the natural coordinate system.
The Jacobian matrix of Eq.(8.7) is given by
 
4
ξi ξi
X  4 (1 + ηηi )xi 4 (1 + ηηi )yi 
J(e) =  ηi ηi  (8.15)
i=1 (1 + ξξi )xi (1 + ξξi )yi
4 4
The cartesian derivatives of the shape functions are obtained from Eq.(8.4) as
    
 ∂Ni  ηj ξj  ξ 
  1 X  4 (1 + ξξj )yj − 4 (1 + ηηj )yj   4 (1 + ηηi )
4 i
∂x = (e)   ηi
 ∂Ni 
  |J | j=1 − ηj (1 + ξξj )xj ξj (1 + ηηj )xj 
 (1 + ξξi )
∂y 4 4 4
(8.16)
From above equations we deduce
 
 ∂Ni  ½ ¾
 
∂x = b̂i (8.17a)

 ∂N i
 ĉi
∂y
with
α1i + α2i ξ + α3i η + α4i ξη α5i + α6i ξ + α7i η + α8i ξη
b̂i = , ĉi = (8.17b)
β1i + β2i ξ + β3i η + β4i ξη β5i + β6i ξ + β7i η + β8i ξη
where α1i · · · α8i , β1i · · · β8i are nodal parameters depending on the nodal coordinates.
Clearly, the element strain matrix and the stiffness matrix now contains rational poly-
nomials making exact integration over an arbitrary quadrilateral domain is extremely
difficult. This problem can be overcome by using numerical integration (Section 8.3).
The above equations simplify considerably for rectangular shapes. For a rectangle of
size 2a × 2b we have
· ¸
(e) a 0
J = 0 b , |J(e) | = ab and dx dy = ab dξ dη (8.18)

229
 The cartesian derivatives of the shape functions are now simply given by
   
 ∂Ni   ξi
   (1 + ηηi ) 
∂x 4b
= η (8.19)
 ∂Ni 
   i (1 + ξξi )

∂y 4a

The element stiffness matrix is expressed in the natural system by

Z +1 Z +1
(e)
Kij = BTi DBi ab dξ dη (8.20)
−1 −1

(e)
The integral expression of Kij now contains polynomial expressions in ξ, η, ξ 2 , η 2 and
ξη which can be directly computed noting that
Z +1 Z +1 ·¸
2 2 4 4
C[1, ξ, η, ξ , η , ξη] dξ dη = C 4, 0, 0, , , 0 (8.21)
−1 −1 3 3
where C is a constant parameter.

8.1.2 A comparison between the 8- and 9-noded isoparametric quadri-

laterals
It is interesting to assess the circumstances under which the linearly distorted
8 and 9-noded quadrilaterals can fully represent any quadratic cartesian ex-
pansion. The straight-sided element geometry is exactly approximated by the
bilinear (subparametric) interpolation
4
X
x= Ni x i (8.22)
i=1

where Ni = 14 (1+ξξi )(1+ηηi ) are the shape functions of the 4-noded rectangle.
We wish to be able to reproduce

u = α1 + α2 x + α3 y + α4 x2 + α5 xy + α6 y 2 (8.23)

Noting that the bilinear form contains terms such as 1, ξ, η and ξη and by
substituting Eq.(8.22) into (8.23) the above can be written as

u = β1 + β2 ξ + β3 η + β4 ξ 2 + β5 ξη + β6 η 2 + β7 η 2 + β8 ξ 2 + β9 ξ 2 η 2 (8.24)

where β1 to β9 depend on the values of α1 to α6 .

We shall now try to match the terms arising from the quadratic expansion
of the 8-noded Serendipity shown in Figure 7.8. Noting the terms occuring in
the Pascal’s triangle of Figure 7.8, the interpolation can be written as

u = b1 + b2 ξ + b3 η + b4 ξ 2 + b5 ξη + b6 η 2 + b7 ξη 2 + b8 ξ 2 η (8.25)

230
 It is evident that for arbitrary values of η1 to η9 it is impossible to match
the coefficients b1 to b8 due to the absence of the term ξ 2 η 2 in Eq.(8.25).
For the 9-noded Lagrange element (Figure 7.5) the expansion similar to
Eq.(8.24) gives
u = b1 + b2 ξ + b3 η + b4 ξ 2 + b5 ξη + b6 η 2 + b7 ξη 2 + b8 ξ 2 η + b9 ξ 2 η 2 (8.26)
and the matching of the coefficients in Eqs.(8.24) and (8.26) can be made
directly.
We conclude that the 9-noded elements can better represent quadratic
cartesian polynomials on linearly distorted shapes and therefore are gener-
ally preferable in modelling smooth solutions. Figure 8.3 shows an example
of this for the analysis, with 8- and 9-node elements, respectively of a simple
beam solution where exact answers are quadratic. With no distorsion both el-
ements with a full (3 × 3) integration rule give exact results but when distorted
only the 9-node element does so, the 8-noded element giving a significant stress
fluctuation.
A similar argument leads to the conclusion that in 3D, again only 27-noded
Lagrangian elements are capable of reproducing fully a quadratic function in
cartesian coordinates when trilinearity distorted.

8.2 ISOPARAMETRIC TRIANGULAR ELEMENTS

The isoparametric interpolation for triangular elements is written as
n
X n
X
x= Ni (L1 , L2 , L3 ) xi ; y= Ni (L1 , L2 , L3 ) yi (8.27)
i=1 i=1

The computation of the cartesian derivatives of Ni and the Jacobian matrix

is inmediate for straight side triangles (see Example 8.2) giving
· ¸(e)
(e) x2 − x1 y2 − y1
J = and |J(e) | = 2A(e) (8.28)
x3 − x1 y3 − y1
The computation of the element integrals in this case is simple via the
analytic expressions of Section 7.6.
For curved side triangles it is convenient to use the natural coordinates α
and β defined in Figure 7.14. This implies that L2 and L3 will be replaced by
α and β, respectively, and L1 by 1 − α − β in all expressions. The compu-
tation of the cartesian derivatives of Ni follows precisely the steps described
in the previous section, simply changing the coordinates ξ and η for α and β,
respectively. For instance
   ∂y  
 ∂Ni  ∂y   ∂Ni 
  1  ∂β −  
∂x ∂α  ∂α
= (e)   ∂N (8.29)
 ∂Ni 
  |J | − ∂x − ∂x   i 
∂y ∂β ∂α ∂β

231
Fig. 8.3: Quadratic Serendipity and Lagrangian 8- and 9-noded elements in regular
and distorted form. Elastic deflection of a beam under constant moment. Note poor
results of 8-noded element

 ∂N ∂Ni 
i
n
X xi yi
 ∂α ∂α 
J(e) =  ∂N ∂Ni  (8.30)
i
i=1 xi yi
∂β ∂β
n n
∂x X ∂Ni (α, β) ∂x X ∂Ni (α, β)
= xi ; = xi ; etc. (8.31)
∂α i=1
∂α ∂β i=1
∂β
The element stiffness matrix is obtained by an expression analogous to
Eq.(8.11a) by
Z 1 Z 1−β ¯ ¯ Z 1 Z 1−β
(e) ¯ (e) ¯
Kij = BTi D Bj ¯J ¯dαdβ = Gij (α, β) dαdβ (8.32)
0 0 0 0

232
where all the terms in Bi and Gij are deduced from Eqs. (8.11) simply sub-
stituting ξ and η for α and β, respectively.
For curved side triangles the integrand of Eq.(8.32) is a rational polynomial
and numerical integration is needed.

Example: 8.2 Derive the expression of the Jacobian matrix for an isoparametric triangle
with straight sides.

Solution

As the element sides are straight, a linear interpolation for the geometry will suffice,
i.e.
x = L1 x1 + L2 x2 + L3 x3
(8.33)
y = L1 y1 + L2 y2 + L3 y3
where xi , yi i = 1, 2, 3 are the coordinates of the three vertex nodes.
Recall that L1 = 1 − L2 − L3 . Substituting this into the above gives
x = x1 + (x2 − x1 )L2 + (x3 − x1 )L3
(8.34)
y = y1 + (y2 − y1 )L2 + (y3 − y1 )L3
The Jacobian matrix is obtained from
      
 ∂Ni  ∂x ∂y   ∂Ni   ∂Ni 
∂L2  ∂L2 ∂L2  ∂x ∂x
= = J(e) (8.35)
 ∂Ni  ∂x ∂y  ∂Ni   ∂Ni 
∂L3 ∂L3 ∂L3 ∂y ∂y
The Jacobian matrix is deduced from the above two equations as
 ∂y  ·
∂x ¸
∂L ∂L2  x2 − x1 y2 − y1
J(e) = 2 = (8.36)
∂x ∂y x3 − x1 y3 − y1
∂L3 ∂L3
¯ ¯
¯ (e) ¯
The cartesian derivatives are obtained by (noting that ¯J ¯¯ = 2A(e) )
¯

   
 ∂Ni  1
·
y3 − y1 y1 − y2
¸  ∂Ni 
∂x ∂L2
= (8.37)
∂N
 i  2A (e) ∂N
x1 − x3 x2 − x1  i 
∂y ∂L3
Let us verify the above expression for the simple 3-noded triangle. For instance, for
∂N1 ∂N1
i = 1, = = −1 (as N1 = L1 = 1 − L2 − L3 ) and
∂L2 ∂L3
 
 ∂N1  1
½
y2 − y3
¾
∂x
= (8.38)
 ∂N1  2A(e) x3 − x2
∂y
Note the coincidence with the expression obtained using Eqs.(4.30) directly.
The expression for the cartesian derivatives obtained is completely general for straight
side triangles of any approximation order (i.e. quadratic, cubic, etc.).

233
Fig. 8.4: Gauss quadratures over quadrilateral elements, a) 1 × 1, b) 2 × 2, c) 3 × 3,
d) 4 × 4 integration points

8.3 NUMERICAL INTEGRATION IN TWO DIMENSIONS

It has been shown in the previous section that all element integrals can be
written in the natural coordinate space making use of the isoparametric for-
mulation. The numerical integration by a Gauss quadrature will be considered
next.

8.3.1 Numerical integration in quadrilateral domains

The integral of a term g(ξ, η) over the normalized isoparametric quadrilateral
domain can be evaluated using a 2D Gauss quadrature by

Z " nq #
+1Z +1 Z +1 X np nq
X X
g(ξ, η) dξ dη = dξ g(ξ, ηq )Wq = g(ξp , ηq )Wp Wq
−1 −1 −1 q=1 p=1 q=1
(8.39)
where np and nq are the number of integration points along each natural co-
ordinate ξ and η respectively; ξp and ηq are the natural coordinates of the pth
integration point and Wp , Wq are the corresponding weights.
The coordinates and weights for each natural direction are directly deduced
from those given in Table 3.3 for the 1D case. Recall that a 1D quadrature
of qth order integrates exactly a polynomial of degree q ≤ 2n − 1. Figure 8.4
shows the more usual quadratures for quadrilateral elements.

Example: 8.3 Integrate numerically the function f (ξ, η) = ξ 2 η 2 over a rectangular element
with dimensions 2a × 2b.

Solution

Since the element is rectangular |J(e) | = ab (see Eq.(8.18)).

The integrand is a quadratic function in ξ and η and hence a 2 × 2 quadrature is

234
 needed (Figure 8.4b). Thus

ZZ Z +1 Z +1 2 X
X 2
2 2
I = ξ η dA = ab ξ 2 η 2 = ab (ξ 2 η 2 )p,q Wp Wq =
A −1 −1 p=1 q=1
" √ √ √ √ √ √ √ # √
3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 4
= ab (− ) (− ) + (− ) ( ) +( ) (− ) +( ) ( ) = ab
3 3 3 3 3 3 3 3 9

which is the exact solution.

8.3.2 Numerical integration over triangles

The Gauss quadrature for triangles is written as
Z 1 Z 1−L3 np
X
f (L1 , L2 , L3 ) dL2 dL3 = f (L1p , L2p , L3p ) Wp (8.40)
0 0 p=1

where np is the number of integration points: L1p , L2p , L3p and Wp are the area
coordinates and the corresponding weight for the pth integration point.
Figure 8.5 shows the more usual coordinates and weights; the term “ac-
curacy” in the figure refers to the highest degree polynomial which is exactly
integrated by each quadrature. Figure 8.5 is also of direct application for
computing the integrals defined in terms of the natural coordinates α and β,
simply recalling that L2 = α, L3 = β and L1 = 1 − α − β.
It is important to note that the weights in Figure 8.5 are normalized so that
their sum is 1/2. In many references this value is changed to the unity and
this requires the sum of Eq.(5.91) to be multiplied by 1/2 so that the element
area is correctly computed in those cases.

Example: 8.4 Compute the area of a triangular element with straight sides by numerical
integration.

Solution
ZZ Z 1 Z 1−β X
(e) |J (e) |
A = dx dy = |J (e) | dα dβ = |J (e) | Wp = (8.41)
A(e) 0 0 p
2

which corresponds with the value obtained in Eq.(8.28).

For further information on numerical integration over triangular domains

see refs. [C15], [C19], [D10], [H20] y [Z6].

235
 Figure n Accuracy Points L1 L2 L3 Wi
(a) 1 Lineal a 1/3 1/3 1/3 1/2
a 1/2 1/2 0 1/6
(b) 3 Quadratic b 0 1/2 1/2 1/6
c 1/2 0 1/2 1/6
a 1/3 1/3 1/3 γ1
(c) 4 Cubic b 0.6 0.2 0.2 γ2
c 0.2 0.6 0.2 γ2
d 0.2 0.2 0.6 γ2
a α1 β1 β1 γ3
b β1 α1 β1 γ3
(d) 6 Quartic c β1 β1 α1 γ3
d α2 β2 β2 γ4
e β2 α2 β2 γ4
f β2 β2 α2 γ4
27
α1 = 0.8168475730 ; β1 = 0.0915762135 ; γ1 = − ; 2γ3 = 0.1099517437
96
25
α2 = 0.1081030182 ; β2 = 0.4459484909 ; γ2 = ; 2γ4 = 0, 2233815897
96

Fig. 8.5: Coordinates and weights for the Gauss quadrature in triangular elements

8.4 NUMERICAL INTEGRATION OF THE ELEMENT MATRI-

CES AND VECTORS
The stiffness matrix of an isoparametric quadrilateral element is computed
using numerical integration as
ZZ Z +1 Z +1 ¯ ¯
(e) T ¯ ¯
Kij = Bi DBj dx dy = BTi D Bj ¯J(e) ¯ dξ dη =
A(e) −1 −1
XXh ¯
np ¯i
nq np nq
X X (8.42)
T ¯ (e) ¯
= Bi DBj ¯J ¯t Wṗ Wq̇ = [Gij ]p,q Wp Wq
p,q
p=1 q=1 p=1 q=1

where Gij is given in Eq.(8.11b). For a triangular element we deduce from

Eq.(8.32)
Z 1 Z 1−β np
X np
X
(e)
Kij = BTi DBj |J(e) | dα dβ = T (e)
[Bi DBj |J |]p Wp = [Gij ]p
0 0 p=1 p=1
(8.43)

236
where the position of the integration points and the corresponding weights is
obtained from Figure 8.5.
The numerical integration of the stiffness matrix requires the evaluation of
the Jacobian J(e) and its determinant, the strain matrix Bi and the constitutive
matrix D at each integration point. Similarly, the numerical integration of the
equivalent nodal flux vector due to a heat source for isoparametric quadrilateral
elements gives
Z +1 Z +1 ¯ ¯ np nq ³
X X ¯ ¯´
(e) ¯ (e) ¯ ¯ (e) ¯
fbi = Ni Q ¯J ¯dξdη = Ni Q¯J ¯ Wṗ Wq̇ (8.44)
−1 −1 p,q
p=1 q=1

For triangular elements the double sum is replaced by the single sum of
Eq.(8.43).
The computation of the equivalent nodal flux vector due to surface fluxes
deserves a special comment. Let us recall that this vector has the following
expression (see Eq.(4.57))
I
(e)
fi = Ni (αφa − q̄n )dS (8.45)
l(e)

where l(e) is the loaded element boundary. In general this boundary represents
a line ξ = constant or η = constant in the natural coordinate space (see
Figure 8.6). Therefore, the differential of length over the side η = 1 for an
isoparametric quadrilateral element is computed by
"r #
³ ´2 ³ ´2
dx dy
(ds)η=1 = (dx2 + dy 2 )η=1 = dξ
+ dξ
dξ =
η=1 η=1
"v
uà !2 à ! 2 #
u Pn dN Pn
= t i=1 dξ xi
i
+ dNi
i=1 dξ yi dξ = c(ξ) dξ
η=1 η=1
(8.46)
Substituting Eq.(8.46) into (8.45) yields a line integral which is a function
of the natural coordinate ξ only. This is computed using a 1D quadrature by
I Z +1 np
X
(e)
fi = Ni (αφa − q̄n )c(ξ)dξ = g(ξ)dξ = g(ξp )Wp (8.47)
l(e) −1 p=1

(e)
For three-noded triangles and four-noded quadrilaterals, fi has the fol-
lowing expression for a uniform boundary flux

(e) l(e)
fi = (αφa − q̄n ) (8.48)
2

237
 Fig. 8.6: Normal flux across a boundary of a 2D element

8.5 COMPUTER PROGRAMMING OF K(e) AND f (e)

Previous sections provided all the necessary expressions for programming the
computation of the stiffness matrix and the equivalent nodal force vector for
each element. The basic steps involved in programming the computation of
K(e) and f (e) for quadrilateral elements are given next.
Figure 8.7 shows the flow diagram for computing of K(e) as deduced from
Eq. (8.42).
The evaluation of the constitutive matrix D can be taken out of the nu-
merical integration loop if the material properties are homogeneous over the
element. The case of D variable is treated simply by using a standard inter-
polation within the element in terms of the nodal values as

n
X
D= Ni D i (8.49)
i=1

Eq.(8.49) is used to obtain the values of D at the Gauss points within the
numerical integration loop.

Figure 8.8 shows the flow chart for computing the flux vector f (e) for the
case of a heat source Q as deduced from Eq.(8.44).
The computation of the heat source Q can be taken out from the numerical
integration loop if it is constant over the element. A variable heat source can
be accounted for by interpolating the known nodal values, in the same way as
was done for matrix D in Eq. (8.49).
The flow charts in Figures 8.7 and 8.8 are completely general and applicable
to most problems solved with the FEM.

238
 Fig. 8.7: Flow chart for the computation of K(e)

8.6 OPTIMAL POINTS FOR COMPUTING GRADIENT FLUXES

The gradient (and the fluxes) are obtained from the derivatives of the unknown
function φ. Therefore, their approximation is always of a lower order than that
for φ. In general, if the shape functions are complete polynomials of pth degree
the approximation for the fluxes will be a polynomial of degree p − 1 or p − 2,
depending if they are obtained as first or second derivatives of the φ field,
respectively.
It can be proved that the fluxes from the finite element solution can be
considered a least square interpolation of the exact flux field [Z6]. Naturally,
the exact flux field is unknown. However, an enhanced stress distribution can
be found by the following property of the Gauss quadrature: A nth degree

239
 Fig. 8.8: Flow diagram for computing f (e) for the case of body forces

polynomial and a n − 1th degree polynomial, obtained by least square fitting

of the first one, take the same values at the points of the Gauss quadrature of
order nth. Hence, we can obtain an approximation of the fluxes and the gra-
dients that is one order higher higher by computing these at the Gauss points.
This important property will be clarified with the following two examples.

Example: 8.5 Verify that a second degree polynomial, and a first degree polynomial ob-
tained by least square smoothing of the another, take the same values at the points
of the Gauss quadrature of order 2.

Solution

Let us consider the second degree polynomial (n = 2)

f (x) = 1 + x + x2

We will now obtain a first degree polynomial (n = 1) which approximates f (x) in the
least square sense; i.e. find a polynomial

g(x) = a + bx

240
Fig. 8.9: Least square approximation of a quadratic polynomial by a linear one.
Intersection of both polynomials at the points of the Gauss quadrature of order 2

such that it minimizes the functional

Z +1 h i2 Z +1 h i2
M= f (x) − g(x) dx = (1 − a) + (1 − b)x + x2 dx
−1 −1

be a minimum.
The parameters a and b are obtained by making
Z +1 h i
∂M
= 0 =⇒ −2 (1 − a) + (1 − b)x + x2 dx = 0
∂a −1
Z +1 h i
∂M
= 0 =⇒ −2x (1 − a) + (1 − b)x + x2 dx = 0
∂b −1

4
which gives a = 3 y b = 1.
Figure 8.9 shows the two polynomials f (x) and g(x). Note that both polynomials
take the same values at the points of the Gauss quadrature of order 2.

Example: 8.6 Verify that a cubic polynomial and a quadratic one obtained by least square
smoothing of the former, take same values at the points of the 3rd order Gauss
quadrature.

Solution

Let us consider the third degree polynomial (n = 3)

f (x) = 1 + x + x2 + x3

We will obtain the second degree polynomial g(x) = a + bx + cx2 , such that
Z +1 h i2 Z +1 h i2
M= f (x) − g(x) dx = (1 − a) + (1 − b)x + (1 − c)x2 + x3 dx
−1 −1

241
Fig. 8.10: Least square interpolation of a cubic polynomial by a quadratic one.
Intersection of both polynomials at the points of the Gauss quadrature of order 3

is a minimum.
The three constants a, b, and c are obtained by solving the following system

∂M 1−c
= 0 =⇒ 1 − a + =0
∂a 3
∂M 1−b 1
= 0 =⇒ + =0
∂b 3 5
∂M 1−a 1−c
= 0 =⇒ + =0
∂c 3 5

which gives
8
a=1 ; b= y c=1
5

Figure 8.10 shows that the exact and interpolating polynomials take the same
p values
at the three points of the 3rd order Gauss quadrature (i,e. ξ = 0 and ξ = ± 3/3).

The following conclusions can be drawn from what is explained:

1. If the exact distribution of the gradient g (and the flux q) field is a

polynomial of nth degree and the approximate finite element solution is
a polynomial of n − 1th degree, the computation of q (or g) at the points
of the Gauss quadrature of nth order gives the exact values.

2. The evaluation of q or g at the Gauss quadrature points chosen for the

integration of K(e) yields a solution of one approximation order higher
than at any other point within the element.

242
Fig. 8.11: Optimal points for computation of the gradients and the fluxes in some
1D and 2D elements

The evaluation of gradients and fluxes at the so called “optimal” quadra-

ture points is therefore of higher accuracy than at any other element point.
The nodal stress values can subsequently be obtained from a local or global
smoothing of the Gauss point values.
The above concepts are rigorously true for 1D elements. For 2D and 3D
elements the sampling of the gradients and the fluxes at the “optimal” Gauss
quadrature points leads also to a substantial improvement in the results. Fi-
gure 8.11 shows the optimal points for computation of gradients and fluxes for
some 1D and 2D elements. Extrapolation to the 3D case is simple.
These concepts extend naturally for elasticity problems (Chapter 9) where
the Gauss point and optimal for the sampling of the strains ε and the stresses
σ . Figure 8.12 shows an example of the analysis of a cantilever beam analyzed
with 8-noded Serendipity rectangles. The accuracy of the shear stress sampled
at the 2 × 2 Gauss points is noticeable.

8.7 SELECTION OF THE QUADRATURE ORDER

The number of integration points is selected according to the degree of the
polynomials appearing in the element integrals. Isoparametric elements con-
tain rational terms within the integrals and exact integration is not longer
possible. The alternative is to choose a quadrature order which integrates ex-
actly the same expression for a rectangular or straight side triangular element.
This quadrature is termed in practice full integration or standard integration.
Remember that in these cases the Jacobian matrix is constant and the element
integrals have a simple analytical form.

243
Fig. 8.12: Exact and minimum quadrature rules for some rectangular and straight
side triangular elements

The minimum quadrature order for the stiffness matrix should preserve the
convergence rate of the element. This is achieved by choosing a quadrature
which integrates exactly all the complete polynomial terms contained in the
shape functions. For rectangular elements this quadrature is of lower order
than that required for the exact integration of the element stiffness matrix and,
thus, some economy is obtained. Figure 8.12 shows the exact and minimum
quadratures for some popular rectangular and triangular elements. Note that
both quadratures coincide for triangles.
Some authors associate the name “minimum quadrature” to that which
guarantees that the element can reproduce in the limit a constant gradient
(or strain) field [Z6,15]. This implies that the quadrature chosen should eval-
uate exactly the element area (or volume), which simply requires the exact
computation of the following integral
ZZ ZZ ¯ ¯
(e) ¯ (e) ¯
A = dA = ¯J ¯ dξ dη
A(e) A(e)

In rectangles and straight side triangles this condition is too weak as it

requires a single point quadrature which generally violates the minimum re-
quirement for preserving the element convergence as described above (with the
exception of 3-noded triangles).

244
Fig. 8.13: a) Propagable mechanisms in the 4-noded rectangle with a single inte-
gration point, b) Mechanisms in the 8-noded and 9-noded quadrilaterals induced by
reduced 2×2 quadrature. The mechanism in the 8-noded element are not propagable
in a mesh

Extreme care must also be taken so that a lower order quadrature does
not introduce internal mechanisms in the element. These mechanisms appear
when the interpolation field for the main variables field generates a gradient
(or strain) field which vanishes at the integration points, thus yielding a sin-
gular stiffness matrix. Sometimes these mechanisms are compatible between
adjacent elements and lead to the singularity of the global stiffness matrix and,
consequently, to an incorrent solution. A typical example of this situation is
the two mechanisms induced by the reduced one point quadrature in the four-
noded rectangle for 2D elasticity problems as shown in Figure 8.13a. This
mechanism invalidates this reduced quadrature for practical purposes unless
some stabilization techniques are used.
In some cases, the mechanisms induced by the reduced integration of the
stiffness matrix can not propagate in the mesh, and this preserves the cor-
rectness of the solution. This happens for the single mechanism originated by
the 2 × 2 reduced quadrature in the eight-noded rectangle for 2D elasticity
problems, as shown in Figure 8.13b. Unfortunately this is not the case for the
9-noded Lagrangian as the reduced integration introduces three mechanisms
(Figure 8.13) Two of which are propagable and can polute the solution and
hence it is not recommended practice.
Note that the minimum quadrature points coincide in most cases with the
optimum points for the computation of fluxes (or stresses). This can be easily
verified by comparing the minimum and optimum quadratures shown in Fig-

245
ures 8.9 and 8.10, respectively. This important coincidence is shown clearly in
the analysis of a cantilever beam using 8-noded Serendipity rectangles. Figure
8.14 shows that the shear force distribution within each element is parabolic
and, therefore, far from the correct linear solution. However, the tangential
stresses at the 2 × 2 Gauss quadrature coincide with the exact values and the
simple linear interpolation of the stresses gives the exact shear force distribu-
tion.

Fig. 8.14: Cantilever beam analyzed with four 8-noded Serendipity rectangles. Lin-
ear extrapolation of the shear force values from the transverse sections corresponding
to the 2 × 2 Gauss quadrature

246
8.8 THE PATCH TEST FOR 2D ELEMENTS
Application of the patch test to triangular and quadrilateral elements follows
the general lines described in Section 3.9 for 1D elements. Here once again the
test is a necessary and sufficient condition for the convergence of the element.
The application of the patch test to a 2D element can take the following
three different modalities.

Patch test A. A known linear field up is prescribed at all nodes of a patch

of 2D elements, where Qp are prescribed temperatures in a heat conduction
problem, or prescribed displacements in an elasticity problems, etc. For each
internal node j in the patch we verify that (Figure 8.13)

Kij apj − fjp = 0 (8.50)

where apj is the nodal unknown vector corresponding to the known field and
fjp is a vector resulting from any external flux (or force) required to satisfy the
governing differential equations for the known solution. Generally, in problems
expressed in cartesian coordinates fjp = 0.

Patch test B. Only the values of up corresponding to the boundaries of the

patch are inserted only and ai is found as (Figure 8.13)
p
ai = K−1
ii (fi − Kij uj ) i 6= j (8.51)

and compared against the exact value.

Patch tests A and B also involve the computation of the fluxes (stresses)
within the elements and the comparison with the expected “exact” values
(Figure 8.15).
Satisfaction of patch tests A and B is a necessary condition for convergence
of the element.

Patch test C. The assembled matrix system of the whole patch is written as

K̄ā = f̄ p (8.52)

where the bar denotes the patch values and f̄ p represents prescribed bound-
ary forces corresponding to the known solution. The solution for ā is sought
after fixing the minimum number of degrees of freedom necessary to eliminate
the singularity of the stiffness equation (i.e. one temperature for the heat
conduction problem or the displacements for 2D elasticity problems) and it is
compared with the known solution.
Patch test C allows us to detect any singularity in the stiffness matrix. This
test is therefore an assessment of the stability of the finite element solution and
hence provides not only a necessary but a sufficient condition for convergence.

247
 i i

Test A Test B
a prescribed on all nodes a prescribed at edges of patch
Kij aj = fi verified at node i ai = Kii–1 (fi – Kij aj ) (j = i) solved

Natural boundary
conditions specified

Minimum esential
boundary conditions
x
(a) (b)

Fig. 8.15: Patch tests of form A, B and C

When the patch is reduced to just one element the C test is termed the
single-element test. This test is one requirement of a good finite element formu-
lation as, on occasion, a larger patch may not reveal the intrinsic instabilities
of a single element. A typical example is the 8-noded isoparametric element
for elasticity problems with reduced 2 × 2 Gauss quadrature. Here the singular
deformation mode of a single element dissappears when several elements are
assembled. The satisfaction of the single-element test is not a sufficient con-
dition for convergence and the test of at least one internal element boundary
needs to be tested to assess sufficiency.

8.9 3D ISOPARAMETRIC ELEMENTS

3D elements with arbitrary geometry, as those shown in Figure 8.16, can be
easily derived using an isoparametric formulation. The element geometry is
defined in terms of the nodal coordinates, whereas the element integrals are
performed over normalized cubes or tetrahedra using simple transformations.
Let us consider first the case of hexahedral elements. The coordinates of a

248
 Fig. 8.16: 3D isoparametric elements. Actual and normalized geometries

point within a n-noded element are expressed in isoparametric form as

   
x X n xi 
x= y = Ni yi = N x(e) (8.53)
   
z i=1 zi

249
with
 
Ni
N = [N1 , N2 , . . . , Nn ] ; Ni =  Ni  ; Ni = f (ξ, η, ζ) (8.54)
Ni

where Ni are the shape functions used for the displacement field.
Eq.(8.53) relates the cartesian and the natural coordinates. This relation-
ship is uniquely defined if the sign of the determinant of the Jacobian trans-
formation x, y, z → ξ, η, ζ is positive over the element. This is usually fulfilled
except for very distorted element shapes.
The cartesian derivatives of the shape functions are computed following
the same procedure as explained for 2D elements. The chain rule of derivation
gives
   
 ∂Ni  ∂x ∂y ∂z  ∂Ni 

 
 
 


 ∂ξ   ∂ξ ∂ξ ∂ξ  
 ∂x 


 
   


    
 ∂N    ∂x ∂y

∂z  ∂Ni 
  
∂Ni
i 
=  = J(e) (8.55)

 ∂η    ∂η ∂η ∂η   ∂y  ∂x

 
   


 
  





 
  ∂x ∂z  
 
 ∂N
 i 
∂y  ∂Ni 

∂ζ ∂ζ ∂ζ ∂ζ ∂z
where J(e) is the Jacobian matrix. Using Eq.(8.53) gives
 
∂Ni ∂Ni ∂Ni
 ∂ξ xi ∂ξ yi ∂ξ zi 
X
n
 ∂Ni ∂Ni ∂Ni 

J(e)  
=  ∂η xi ∂η yi ∂η zi  (8.56)
i=1  
 ∂Ni ∂Ni ∂Ni 
xi yi zi
∂ζ ∂ζ ∂ζ

The cartesian derivatives of Ni are computed as

   
 ∂N i  
 ∂N i 


 
 
 


 
∂x  · ¸ 
 ∂ξ 

  −1 
∂Ni 
∂Ni (e)
∇ Ni = = J (8.57)

 ∂y   
 ∂η  

 
 
 

 ∂N
 i  
 ∂N i 

 
∂z ∂ζ

The volume differential is expressed as

¯ ¯
¯ (e) ¯
dx dy dz = ¯J ¯¯ dξ dη dζ
¯ (8.58)

250
 The strain matrix of an isoparametric prismatic element can be expressed
in terms of natural coordinates using Eq.(8.57) as
 
 b̄i 
Bi (ξ, η, ζ) = ∇ Ni = c̄i (8.59)
 ¯
di
with  
(e)
  
 J 

 b̄i  X 3 
 1k   ∂N
(e) i
c̄i = J 2k (8.60)
 ¯ 
 
 ∂ξk
di k=1 
 (e)  
J 3k
(e)
where J ij is the term ij of the inverse Jacobian matrix [J(e) ]−1 , ξ1 = ξ, ξ2 = η
and ξ3 = ζ.
The general form of the stiffness matrix of a 3D isoparametric hexahedral
element is therefore
ZZZ
(e)
Kij = BTi D Bj dV =
V (e)
Z +1 Z +1 Z +1 ¯ ¯
¯ ¯
= BTi (ξ, η, ζ)D Bj (ξ, η, ζ)¯J(e) ¯ dξ dη dζ =
−1 −1 −1
Z +1 Z +1 Z +1
= Gij (ξ, η, ζ) dξ dη dζ (8.61a)
−1 −1 −1

with ¯ ¯
¯ (e) ¯
Gij = BTi D Bj ¯ J ¯ (8.61b)

where b̄ij = b̄i b̄j , c̄ij = c̄i c̄j and d¯ij = d¯i d¯j , with b̄i , c̄i , d¯i as given in Eq.(7.84)
and dij are the terms of the constitutive matrix D of Eq.(6.7). Matrix G
will typically contain rational polynomials arising from the Jacobian inverse
contributions. Numerical integration is therefore the recommended option in
these cases.
Isoparametric tetrahedral elements follow a similar procedure. The geome-
try interpolation is once again defined by Eq.(8.53) with Ni expressed in terms
of volume or natural coordinates. The computation of the cartesian derivatives
of Ni for straight side tetrahedra is immediate from Eqs.(8.57). The element
integrals are directly expressed in terms of volume or natural coordinates. Nat-
ural coordinates are more convenient for curved tetrahedra. The derivation of
the stiffness matrix follows the same steps explained for hexahedral elements
simply substituting the coordinates ξ, η, ζ for α, β, γ, respectively. The result-
ing stiffness matrix has the form
Z 1 Z 1−α Z 1−α−β
(e)
Kij = Gij (α, β, γ) dα dβ dγ (8.62)
0 0 0

where G(α, β, γ) is deduced from Eq.(7.61b).

251
Fig. 8.17: Gauss quadratures of 1×1×1 and 2×2×2 points in hexahedral elements

8.10 NUMERICAL INTEGRATION IN THREE DIMENSIONS

8.10.1 Hexahedral elements
Let us consider the integration of a function f (x, y, z) over a hexahedral
isoparametric element. The following transformations are required
ZZZ Z 1 Z 1 Z 1 ¯ ¯
¯ ¯
f (x, y, z) dx dy dz = f (ξ, η, ζ) ¯J(e) ¯ dξ dη dζ =
V (e) −1 −1 −1

Z +1 Z +1 Z +1
= g(ξ, η, ζ) dξ dη dζ (8.63)
−1 −1 −1

The use of a Gauss quadrature leads to

Z +1 Z +1 Z +1 Z +1 Z np
+1 X
g(ξ, η, ζ) dξ dη dζ = Wp g(ξp , η, ζ) dη dζ =
−1 −1 −1 −1 −1 p=1
Z nq np
+1 X X
= Wp Wq g(ξp , ηq , ζ) dζ =
−1 q=1 p=1
nr X X nq np
X
= Wp Wq Wr g(ξp , ηq , ζr ) (8.64)
r=1 q=1 p=1

where np , nq and nr are the integration points via the ξ, η, ζ directions, re-
spectively, ξp , ηq , ζr are the coordinates of the integration point (p, q, r) and
Wp , Wq , Wr are the weights for each natural direction.
The local coordinates and weights for each quadrature are deduced from
Table 3.3 for the 1D case. We recall that a qth order quadrature integrates
exactly a 1D polynomial of degree 2q − 1. This rule helps us to identify the
number of integration point in each natural directions. Figure 8.17 shows the
sampling points for the 1 × 1 × 1 and 2 × 2 × 2 quadratures.

252
8.10.2 Tetrahedral elements
A Gauss quadrature for tetrahedral elements formulated in terms of volume
coordinates is written as
Z 1Z 1−L1Z 1−L1 −L2 np
X
f (L1 , L2 , L3 , L4 )dL1 dL2 dL3 = f (L1i , L2i , L3i , L4i )Wi
0 0 0 i=1
(8.65)
Figure 8.18 shows the position of the integration points and the correspond-
ing weights for the linear, quadratic and cubic quadratures.
Note that the weights in Figure 8.18 have been normalized so that their
sum is 1/6. In this manner, the element volume is computed exactly. Thus
Z Z Z Z +1 Z +1 Z +1
(e)
V = dV = |J(e) |dξdηdζ =
V (e) −1 −1 −1
Z +1 Z +1 Z +1
= |J(e) | dξdηdζ =
−1 −1 −1
np np
X X
(e) (e)
= |J | Wi = 6V Wi = V(e) (8.66)
i=1 i=1

8.11 NUMERICAL INTEGRATION OF THE ELEMENT

MATRICES
8.11.1 Isoparametric hexahedral elements
Combining Eqs.(8.61) and (8.64) yields the stiffness matrix of an isoparametric
hexahedral element as
ZZZ Z +1 Z +1 Z +1 ¯ ¯
(e) T ¯ ¯
Kij = Bi DBj dx dy dz = BTi DBj ¯J(e) ¯ dξ dη dζ =
V (e) −1 −1 −1
nq nr ·
np ¯ ¯¸
XX X ¯ (e) ¯
T
= Bi D Bj ¯J ¯ Wp Wq Wr =
p=1 q=1 r=1 p,q,r
np nq nr
X XX
= [Gij ]p,q,r Wp Wq Wr (8.67)
p=1 q=1 r=1

where Gij was given in Eq.(8.61b).

The computation of the equivalent nodal force vectors involving volume
integrals follows an identical procedure. For the nodal flux vector
Z Z Z Z +1 Z +1 Z +1 ¯ ¯
(e) ¯ (e) ¯
fi = Ni Q dxdydz = Ni Q ¯J ¯dξ dη dζ =
V (e) −1 −1 −1
np
XXX nr ·
nq ¯ ¯¸
¯ (e) ¯
= Ni Q¯J ¯ Wp Wq Wr (8.68)
p=1 q=1 r=1 p,q,r

253
 Figure np Precision Points L1 L2 L3 L4 wi
(a) 1 Linear a 1/4 1/4 1/4 1/4 1/6
a α β β β 1/24
b β α β β 1/24
(b) 4 Quadratic c β β α β 1/24
d β β β α 1/24
a 1/4 1/4 1/4 1/4 γ
b 1/3 1/6 1/6 1/6 δ
(c) 5 Cubic c 1/6 1/3 1/6 1/6 δ
d 1/6 1/6 1/3 1/6 δ
e 1/3 1/6 1/6 1/3 δ

2 3
α = 0.58541020 ; β = 0.13819660 ; γ = − ; δ=
15 40

Fig. 8.18: Coordinates and weights for the Gauss quadrature in tetrahedra

The case of surface fluxes is somewhat more complicated. To explain the

process let us asssume that a distributed flux tn acts orthogonally to the ele-
ment face corresponding to ζ = +1 defined by nodes 5 to 8 (Figure 8.19). The
computation of the equivalent nodal vector for this case requires evaluating the
term t dA where t contains the global components of the surface flux acting on
the element face and dA is the area differential. Thus, if n is the unit normal
to the face we have

t = tn n with n = [nx , ny , nz ]T (8.69)

Vector n is obtained by the cross product of vectors V~1 and V~2 tangent to
the lines η = constant and ξ = constant over the face, respectively. Thus
³ ∂x ´
V~1 = ~i + ∂y~j + ∂z ~k dξ
∂ξ ∂ξ ∂ξ ζ=+1
(8.70)
³ ∂x ∂y ∂z ´
V~2 = ~i + ~j + ~k dη
∂η ∂η ∂η ζ=+1

Eq.(8.70) shows that the components of V~1 and V~2 coincide with the terms
in the first and second row of the Jacobian matrix of Eq.(8.56).

254
 Fig. 8.19: Normal flux acting on a face of a hexahedral element

The unit normal vector is

V~1 × V~2
n= (8.71)
|V~1 × V~2 |

Noting that dA = |V~1 × V~2 | gives

 (e)

 J12 J23 − J22 J13 

1   1 (e)
n= J21 J13 − J11 J23 dξ dη = j dξ dη (8.72)
dA  
 dA
 
J11 J32 − J21 J12 ζ=+1

(e)
where the Jij terms are deduced from Eq.(8.56).
The final expression of the equivalent nodal force vector is
ZZ Z +1 Z +1
(e)
fti = Ni tn dA = Ni tn j(e) dξ dη =
A(e) −1 −1
np nq
XX
= [Ni j(e) tn ]p,q Wp Wq (8.73)
p=1 q=1

where j(e) is deduced from Eq.(8.72). Also note that in Eq.(8.73) Ni =

Ni (ξ, η, ζ = +1).

8.11.2 Isoparametric tetrahedral elements

The element stiffness matrix and the equivalent nodal force vector for isopara-
metric tetrahedra are computed by
np h i
(e)
X
Kij = Gij (α, β, γ) Wp (8.74a)
p
p=1

255
 np h i
(e)
X
fi = Ni Q|J(e) | Wp (8.74b)
p
p=1

where
Gij = BTi (α, β, γ)DBj (α, β, γ)|J(e) |
For the case of surface fluxes we have
np h i
(e)
X
f ti = Ni j(e) tn Wp (8.75)
p
p=1

where the different terms have the same meaning as for hexahedra.

8.11.3 Selection of the quadrature order

The selection of the quadrature order for hexahedral and tetrahedral elements
follows the same rules given for 2D solid elements. The full (exact) quadrature
for linear and quadratic prisms is 2 × 2 × 2 and 3 × 3 × 3, respectively. These
quadratures are also recommended for distorted shapes.
The reduced 1 × 1 × 1 and 2 × 2 × 2 quadratures induce spurious modes
in the 8-noded and 27-noded elements, respectively. The performance of the
20-noded Serendipity elements generally improves with the 2 × 2 × 2 reduced
quadrature. However, this quadrature must be used with extreme care as it
can lead to spurious mechanisms in certain problems, in a similar way as for
2D quadrilaterals.
For linear and quadratic tetrahedra the 1 and 4 points quadratures yield
exact integration. Cubic tetrahedra require a quadrature of quartic precission
for the exact integration of all the stiffness matrix terms.
The subroutines for computing K(e) and f (e) follow precisely the same steps
as for 2D elements.

8.12 PERFORMANCE OF 3D SOLID ELEMENTS

3D solid elements behave in general terms very similarly to their analogous 2D
elements. Hexahedral elements are more accurate than tetrahedra of the same
order. Low order elements like the 8-noded hexahedron or 4-noded tetrahedron
require fine meshes for problems with high gradients and higher order elements
give more satisfactory results in these cases.
The 27-noded Lagrange element performs better than the 20-noded Serendip-
ity one for representing a quadratic function on trilinearly distorted shapes.
The reasons are similar to those given to explain the better performance of the
9-noded element when linearly distorted. The performance of the 20-noded
Serendipity element in those cases can be improved by using 2 × 2 × 2 re-
duced integration. However, great care should be taken when using a reduced
quadrature for the quadratic Serendipity element.

256
 Despite its highest cost the 27-noded Lagrange element is therefore gener-
ally preferable for modelling smooth solutions on distorted geometries.
We emphasize that mesh generation is one of the crucial problems for prac-
tical 3D analysis. Here, tetrahedral elements are by far the more versatile
option for the discretization of complex 3D geometries. Much research on
the development of efficient mesh generators for tetrahedral and hexahedral
elements has been carried out in recent years. This issue is of even greater
importance if adaptive refinement strategies are used.

257
Chapter 9

2D SOLIDS. LINEAR TRIANGULAR AND

RECTANGULAR ELEMENTS

9.1 INTRODUCTION
This chapter initiates the application of the FEM to structures which satisfy
the assumptions of two-dimensional (2D) elasticity (i.e. plane stress or plane
strain). Many of the concepts in this chapter will be useful when dealing with
other structural problems in subsequent chapters. Therefore, this chapter is
introductory to the application of the FEM to continuous 2D and 3D struc-
tures.
There are a wide number of structures of practical interest which can be
analyzed following the assumptions of 2D elasticity. All these structures have
a sort of prismatic geometry. Depending on the relative dimensions of the
prism and the loading type, the following two categories can be distinguished:
Plane stress problems. A prismatic structure is under plane stress if one of
its dimensions (thickness) is much smaller than the other two and all the loads
are contained in the middle plane of the structure (Figure 9.1). Amongst the
structural problems that can be included in the plane stress category we find
the analysis of deep beams, plates and walls under in-plane loading, buttress
dams, etc.
Plane strain problems. A prismatic structure is under plane strain if one
of its dimensions (length) is larger than the other two and all the loads are
uniformly distributed along its length and they act orthogonally to the longi-
tudinal axis (Figure 9.2). Amongst the structures which follow the plane strain
assumption we can list containing walls, gravity dams, pressurised pipes, and
also many problems of geotechnical engineering (tunnels, foundations, etc.).
One of the main advantages of 2D elasticity theory is that it allows the
study of plane stress and plane strain problems in a unified manner. We
should recall however that each problem conceptually represents a class of
very different structural types.
2D elasticity theory provides a mathematical model by which the behaviour

259
 Fig. 9.1: Examples of plane stress problems

of a real 3D structure is represented by that of a 2D solid. The finite element

method allows us to obtain an approximation to the “exact” solution of the
2D elasticity equations using 2D solid elements. The accuracy of the finite
element solution will depend on the element type and the quality of the mesh
chosen.
The chapter starts with a brief description of the basic concepts of 2D elas-
ticity theory. Then the finite element solution using simple 3-noded triangles
and 4-noded quadrilaterals is presented. Most of the finite element expressions
are completely general and applicable to any other 2D solid element types. The
general derivation of the element shape functions and the formulation of higher
order triangular and quadrilateral elements and of isoparametric elements are
studied in the next chapter.

9.2 TWO DIMENSIONAL ELASTICITY THEORY

Next, we present the concepts of 2D elasticity theory needed for the application
of the FEM.

9.2.1 Displacement field

Both the plane stress and plane strain assumptions imply that the transverse
sections to the longitudinal axis z deform in the same manner and also that the
longitudinal displacement is negligible. Therefore, only a generic 2D transverse

260
 Fig. 9.2: Examples of plane strain problems

section defined in the plane x − y needs to considered for the analysis (Figures
9.1 and 9.2). The displacement field of the generic section is defined by the
displacements in the x and y direction of all its points. The displacement
vector is defined as ½ ¾
u(x, y)
u(x, y) = (9.1)
v(x, y)
where u(x, y) and v(x, y) are the displacements of the point in directions x and
y, respectively.

9.2.2 Strain field

The displacement field (9.1) allows the corresponding strains to be derived
from standard elasticity theory [T6]. This gives
∂u
εx =
∂x
∂v
εy =
∂y (9.2)
∂u ∂v
γxy = +
∂y ∂x
γxz = γyz = 0

The longitudinal strain εz is assumed to be zero in the plane strain case.

Conversely, εz is not zero in plane stress situations, although the conjugate
stress σz is assumed to be zero. Therefore, εz need not be considered for either

261
 Fig. 9.3: Deformation of an infinitesimal 2D domain and definition of strains

plane stress or plane strain problems as the work performed by the longitudinal
strains (i.e. σz εz ) is always zero. Consequently, the strain vector is defined in
cases simply as
ε = [εx , εy , γxy ]T (9.3)

The graphical meaning of the strains for 2D problems is shown in Figure

9.3.

9.2.3 Stress field

It is deduced from Eq.(9.2) that the shear stresses τxz and τyz are zero. Also,
for the same reasons as explained above, the longitudinal stress σz does not
contribute to the internal work and the stress vector is defined for both plane
stress and plane strain cases as (Figure 9.3)

σ = [σx , σy , τxy ]T (9.4)

9.2.4 Stress-strain relationship

The relationship between stresses and strains is derived from 3D elasticity
theory [T6] using the assumptions stated above (i.e. σz = 0 for plane stress,
εz = 0 for plane strain, and γxz = γyz = 0 in both cases). After same simple
algebra (see Example 4.1) the following matrix relationship can be obtained

σ=Dε (9.5)

262
where D is the elastic material matrix (or constitutive matrix)
 
d11 d12 0

D =  d21 d22 0  (9.6)
0 0 d33
It can be proved from the Maxwell-Betti theorem that D is always sym-
metrical [T6] and d12 = d21 . For isotropic elasticity we have

Plane stress Plane strain

E E(1 − ν)
d11 = d22 = d11 = d22 =
1 − ν2 (1 + ν)(1 − 2ν)
ν (9.7)
d12 = d21 = νd11 d12 = d21 = d11
1−ν
E E
d33 = =G d33 = =G
2(1 + ν) 2(1 + ν)
where E is the Young modulus and ν the Poisson’s ratio.
For an orthotropic material with principal orthotropy directions along the
1, 2, 3 axes, matrix D has the following expression [T6]:

Plane-stress
 
E1 ν21 E1 0
1  
D=  ν12 E2 E2 0  (9.8a)
1 − νxy νyx
0 0 (1 − ν12 ν21 )G12
Plane strain
 
aE1 bE1 0
1  
D=  cE2 dE2 0  (9.8b)
ad − bc
0 0 (ad − bc)G12
where
1 1 + ν21 1 + ν12
' + (9.9a)
G12 E1 E2
and
a = 1 − ν23 ν32 ; b = ν12 + ν32 ν13
(9.9b)
c = ν21 + νyz ν31 ; d = 1 − ν13 ν31
The symmetry of D requires
E2 ν12 E2 b
= (plane stress) and = (plane strain) (9.10)
E1 ν21 E1 c
If the orthotropy directions 1, 2 are inclined at an angle φ with respect to
the global axes of the structure 1, 2 (Figure 9.4) the constitutive relationship
is derived as follows. First, the strains in local axes 1, 2 are expressed in terms
of the global strains by
ε0 = Tε (9.11)

263
 Fig. 9.4: Orthotropic material with principal orthotropy directions 1, 2

with [C15]
 
cos2 φ sen2 φ sen φ cos φ

T= 2
sen φ cos2 φ −sen φ cos φ  (9.12)
2 2
−2senφ cos φ 2sen φ cos φ cos φ − sen φ

The global stresses are related to those in the local axes by [T6]

σ = TT σ 0 (9.13)

The stress-strain relationship in the local axes is written as

σ 0 = D 0 ε0 (9.14)

where D0 is given by Eq.(9.8).

Finally, from Eqs.(9.1), (9.13) and (9.14) we obtain

σ = TT D0 Tε = D ε (9.15)

with
D = TT D0 T (9.16)
It is easy to check that matrix D is also symmetrical.
The dij coefficients for anisotropic elasticity can be found in references [H2]
and [L1].
If the solid is subjected to initial strains such as thermal strains, the con-
stitutive relationship (9.5) must be modified. The total strain ε is now equal
to the sum of the elastic (εe ) and the initial (ε0 ) strains, whereas in Eq.(9.5)
all the strains were considered to be elastic. Since the stresses are proportional
to the elastic strains, the constitutive equation is now written as

σ = D εe = D(ε − ε0 ) (9.17)

For the case of initial strains due to thermal effects and isotropic material,
vector ε0 has the following expressions

264
 Plane stress Plane strain
   
 α∆T 
   α∆T 
 
ε0 = α∆T ε0 = (1 + ν) α∆T (9.18)

   
0  
0 
where α is the thermal expansion coefficient and ∆T is the temperature incre-
ment at each point.
The difference between the values of ε0 for plane stress and plane strain is
due to the different assumptions for σz and εz in each case (see Examples 4.2
and 4.3).
For anisotropic materials, the initial strains due to thermal effects are con-
sidered first in the principal directions of the material and then are transformed
to global axes to find the global components of ε◦ . In these cases the tangential
◦
strain τxy is not longer zero [H2], [Z6].
The solid can also be initially subjected to stresses defined by a vector
σ 0 . These initial stresses can have different sources. For instance, if a part of
the material is removed from a deformed structure under a set of loads, then
automatically a new deformation is originated due to the existance of initial
stresses. The total stresses in the new equilibrium configuration are obtained
by the sum of the initial ones and those originated in the deformation process.
For the more general case
σ = D(ε − ε0 ) + σ 0 (9.19)
where
0 T
σ 0 = [σx0 , σy0 , τxy ] (9.20)
is the initial stress vector. A practical example of initial stresses is the analysis
of a tunnel in geotechnical engineering, where the equilibrium of the excavated
zone depends on the initial stresses in the zone before the excavation. Initial
stresses are also very common in welded mechanical parts and here they are
usually termed “residual” stresses.

Example: 9.1 Find the constitutive equation for an isotropic elastic material under plane
stress and plane strain conditions.

-- Solution

The starting point is the constitutive equation for 3D isotropic elasticity [T6]

(σx − νσy − νσz ) (σy − νσx − νσz ) (σz − νσx − νσy )

εx = ; εy = ; εz =
E E E
2(1 + ν) 2(1 + ν) 2(1 + ν)
γxy = τxy ; γxz = τxz ; γyz = τyz
E E E
These equations will be now simplified using the plane stress and plane strain as-
sumptions.

265
Plane stress: σz = 0; γxz = γyz = 0

Substituting the plane stress conditions into the above equations we have

1 1 2(1 + ν)
εx = (σx − νσy ), εy = (σy − νσx ), γxy = τxy
E E E
ν
εz = − (σx + σy ) ; τxz = τyz = 0
E

These equations yield the relationship between σx , σy , τxy and the corresponding
strains as

E E E
σx = (εx + νεy ); σy = (εy + νεx ); τxy = γxy
1 − ν2 1 − ν2 (1 + ν)

from which the coefficients of D in Eq.(9.6) can be deduced.

Substituting the expressions of σx and σy for εz we find that

εz = −ν(εx + εy )

Therefore, the longitudinal strain εz can be obtained “a posteriori” in terms of εx

and εy .

Plane strain: εz = 0; γxz = γyz = 0

From the general equations relating strains and stresses we have

1 1
εx = (σx − νσy − νσz ); εy = (σy − νσx − νσz ) = 0
E E
1 2(1 + ν)
εz = 0 = (σz − νσx − νσy ) ; γxy = τxy ; τxz = τyz = 0
E E

From the condition εz = 0 we find σz = ν(σx + σy ). Substituting this value into the
other equations we find

E(1 − ν) ³ ν ´
σx = εx + εy
(1 + ν)(1 − 2ν) 1−ν
E(1 − ν) ³ ν ´
σy = εy + εx
(1 + ν)(1 − 2ν) 1−ν
E
τxy = γxy
2(1 + ν)

from which the expression (9.6) for D can be obtained.

The same procedure can be used for orthotropic or anisotropic materials taking as a
starting point the corresponding expressions of 3D elasticity [T6].

266
Example: 9.2 Find the initial strain vectors due to thermal effects for 2D isotropic elas-
ticity.

-- Solution

The main assumption is that the total strains are the sum of the elastic and the
thermal ones. Also, it is assumed that a thermal expansion (or contraction) originates
axial strains of value α∆T , where α is the thermal expansion coefficient and ∆T the
temperature increment. With these assumptions the total strains for 3D isotropic
elasticity can be written as (see first equation of Example 4.1).
1
εx = εex + ε0x = (σx − νσy − νσz ) + α∆T
E
1
εy = εey + ε0y = (σy − νσz − νσx ) + α∆T
E
e 1
εz = εz + εz = (σz − νσx − νσy ) + α∆T
0

E
2(1 + ν) 2(1 + ν) 2(1 + ν)
γxy = τxy ; γxz = τxz ; γyz = τyz
E E E

Plane stress σz = γxz = γyz = 0

Substituting these conditions into above equations we have

1 1
εx = (σx − νσy ) + α∆T ; εy = (σy − νσx ) + α∆T
E E
ν 2(1 + ν)
εz = − (σx + σy ) + α∆T ; γxy = τxy ; τxz = τyz = 0
E E
Solving for σx , σy and τxy gives
E h i
σx = (ε x − ε 0
x ) + ν(ε y − ε 0
y )
1 − ν2 h i
E
σy = (ε y − ε 0
y ) + ν(εx − ε0
x )
1 − ν2
E
τxy = γxy
2(1 + ν)

which can be written in the form σ = D (ε − ε0 ), with ε0 = α∆T [1, 1, 0]T being the
initial strain vector and D the matrix given in (9.6) and (9.7).

Plane strain εz = γxz = γyz = 0

From the general expressions we find

1
εx = (σx − νσy − νσz ) + α∆T
E
1
εy = (σy − νσx − νσz ) + α∆T
E
0 = E1 (σz − νσx − νσy ) + α∆T
2(1 + ν)
γxy = τxy ; τxz = τyz = 0
E

267
 From the third equation we find that

σz = ν(σx + σy ) − E α ∆T

Substituting this value into the first two equations yields

1h i
εx = (1 − ν 2 )σx − ν(1 + ν)σy + (1 + ν)α∆T
Eh i
1
εy = (1 − ν 2 )σy − ν(1 + ν)σx + (1 + ν)α∆T
E
2(1 + ν)
γxy = τxy
E

Solving for σx , σy and τxy gives

E(1 − ν) h ν i
σx = (εx − (1 + ν)α∆T ) + (εy − (1 + ν)α∆T )
(1 + ν)(1 − 2ν) 1−ν
E(1 − ν) h ν i
σy = (εy − (1 + ν)α∆T ) + (εx − (1 + ν)α∆T )
(1 + ν)(1 − 1ν) 1−ν
E
τxy = γxy
2(1 + ν)

which can be written in matrix form as σ = D(ε − ε0 ), where

ε0 = (1 + ν)α∆T [1, 1, 0]T

is the initial strain vector and D the matrix given in Eqs.(9.6) and (9.7).

Example: 9.3 Explain the meaning of the initial strains for the bar in Figure 9.5 subjected
to a uniform temperature increase.

-- Solution

Let us assume first that the bar is clamped at one end and free at the other end
(Figure 9.5a). Under a uniform temperature increment the bar will increase in length
by the amount
∆l = α ∆T l
and the corresponding “initial” strain is

∆l
ε0x = = α ∆T
l

Since the bar is free to move horizontally, the total elongation is equal to that pro-
duced by the thermal increment and, therefore, the elastic strain is equal to zero,
i.e.
εex = εx − ε0x = α ∆T − α ∆T = 0

Thus, from Eq.(9.17) it is deduced that the stresses in the deformed bar are zero.

268
Fig. 9.5: Interpretation of initial thermal strains, a) Clamped/free bar, b) Fully
clamped bar, c) 2D solid

Let us consider now the fully clamped bar of Figure 9.5b. To compute the ini-
tial strains let us assume that the bar points are free to move horizontally. Under
these conditions the “initial” elongation of the bar will coincide with that of the
clamped/free bar of Figure 9.5a, i.e. ε0x = α∆T . However, since the bar points have
the horizontal displacement restrained (due to the two clamped ends), the “elastic
strain”is now
εex = εx − ε0x = 0 − α∆T = −α∆T
and by using Eq.(9.17) it is deduced that the bar is subjected to a uniform axial force
of value N = −αEA∆T .
Therefore, the initial thermal strains can be interpreted as the strains induced in
the constraint-free body by a temperature increment occuring in some points (Figure
9.5c). Satisfaction of the kinematic (displacement) boundary conditions provides the
values of the actual (total) strains. The difference between total and initial strains
yields the “elastic” strains responsible for the stresses in the body.
It is also deduced from this example that a thermal increment produces no stresses
in a body which can move freely in space.

269
9.2.5 Virtual work expression
PVW is written for 2D elasticity problems as
ZZ ZZ
(δεx σx + δεy σy + δγxy τxy )t dA = (δubx + δvby )t dA +
A I A
X
+ (δutx + δvty )t ds + (δui Ui + δvi Vi ) (9.21)
l i

The second term in Eq.(9.21) represents the virtual work of the forces
per unit area bx , by ; the surface tractions tx , ty ; and the point loads Ui , Vi ,
respectively. The first term represents the work performed by the stresses
σx , σy , τxy over the virtual strains δεx , δεy y δγxy . A and l denote respectively
the area and the boundary of the transverse section of the solid and t its
thickness. For plane stress problems t is the actual thickness of the solid. For
plane strain situations the analysis domain is a unit slice and it is usual to
take t equal to one.
Eq.(9.21) can be written in matrix form as
ZZ ZZ I X
T T
δε σt dA = δu bt dA + δuT tt ds + δuTi qi (9.22)
A A l i

where
h iT h iT h iT
δε = δεx , δεy , δγxy ; δu = δu, δv ; b = bx , by
h iT h iT h iT
t = tx , t y ; δui = δui , δvi ; qi = Ui , Vi

(9.23)

The above equations show that the PVW integrals involve up to first deriva-
tives of the displacements only. Hence, C o continuous elements can be used.
This requirement holds for all elasticity elements (i.e. 2D/3D solids and ax-
isymmetric solids).
Eq.(9.22) is the starting point to derive the discretized finite element equa-
tions as described in the next section.

9.3 FINITE ELEMENT FORMULATION. THREE-NODED

SOLID TRIANGULAR ELEMENT
We will study first the simple 3-noded solid triangular element. This is con-
sidered to be the first element used for the analysis of structural problems.
Prior to the finite element era, Courant successfully used linear polynomical
approximations over triangular regions to solve differential equations in 2D
domains [C6]. Some years later Turner et al. [T12] in their classic paper pro-
posed the discretization of 2D solid domains into simple triangles as a way to

270
analyze solids using matrix structural techniques. This explains why the 3-
noded triangle is sometimes known as the Turner element. This element soon
became very popular among engineers and it was widely used in the analysis
of many structures in aeronautical and civil engineering [A10,11], [C18]. We
note the impact of this element in the study of gravity dams and tunnels for
practical civil engineering applications [Z12]. The key to the success of the
3-noded triangle is its simplicity which allows the assimilation of the FEM
and the standard matrix method for bar structures known to most structural
engineers. Conversely, it has limited accuracy due to the linear displacement
approximation yielding constant strain and stress fields. Hence, fine meshes
are required to capture accurate solutions in zones of high displacement gra-
dients. This is however not a serious problem due to its versatile geometry,
which is also very adequate for adaptive mesh refinement, as shown in Chap-
ter 15. In summary, the 3-noded triangular element has the ideal features to
introduce the application of the FEM to the analysis of 2D solids.

9.3.1 Discretization of the displacement field

Figure 9.6 shows the transverse section of a solid analized under the assump-
tions of plane elasticity. As usual the first step is the discretization of the
analysis domain as a mesh of finite elements. Figure 9.6 shows the mesh of
3-noded triangles chosen. The accuracy of the finite element solution can ob-
viously be improved by using a finer mesh.
A typical 3-noded solid triangular element is characterized by the num-
bering of its nodes and their coordinates x, y. The three nodes have a global
numbering i, j, k which corresponds to the local numbers 1, 2, 3 respectively.
It is convenient to use the local numbering to compute the element matrices
and vectors and to use the correspondance between local and global number-
ing (Figure 9.5) for the assembly process, as in the matrix analysis of bar
structures [L3].
Let us consider an individual triangle like that shown in Figure 9.6. The
two cartesian displacements of an arbitrary point within the element can be
expressed in terms of the nodal displacements as

u = N1 u 1 + N 2 u 2 + N3 u 3
(9.24)
v = N1 v1 + N2 v2 + N3 v3

where (ui , vi ) and Ni are the horizontal and vertical displacements and the
shape function of node i, respectively. There is not a fundamental reason to
choose the same approximation for the vertical and horizontal displacements.
However, the same interpolation for both displacements is typically used in
practice.

271
 Fig. 9.6: Discretization of a structure in 3-noded solid triangular elements

Eq.(9.24) is written in matrix form as

 

 u1 


 


 v 
1
½ ¾ · 
¸  

u N1 0 N2 0 N3 0  u2 
u= = (9.25)
v 0 N1 0 N2 0 N3 
 v2 



 



u
 3 

 
v3
or
u = N a(e) (9.26)
where ½ ¾
u
u= (9.27)
v
is the displacement vector of a point,
· ¸
Ni 0
N = [N1 , N2 , N3 ] ; Ni = (9.28)
0 Ni
are the shape function matrices of the element and the ith node, respectively,
and  (e) 
 a1 
  ½ ¾
ui
(e)
a = a(e)
(e)
2 with a i = (9.29)
 (e) 
  vi
a3

272
are the nodal displacement vectors of the element and of ith node, respectively.
(e)
Note that N and a(e) contain as many matrices Ni and vectors ai as
element nodes. This is a general rule which applies in all cases, as will be seen
throughout this book.
The shape functions for the 3-noded triangular element can be found as
follows.
The three nodes define a linear displacement field which can be written as
u = α1 + α2 x + α3 y
(9.30)
v = α4 + α5 x + α6 y
Since we have assumed the same interpolation for u and v, it is sufficient to
derive the shape function for one of the two displacements. Thus, for instance,
the nodal horizontal displacement values are deduced from Eq.(9.30) as
u1 = α1 + α2 x1 + α3 y1
u2 = α1 + α2 x2 + α3 y2 (9.31)
u3 = α1 + α2 x3 + α3 y3
Solving for α1 , α2 and α3 and substituting into Eq.(9.30) yields
1 h i
u= (a1 + b1 x + c1 y)u 1 + (a2 + b2 x + c2 y)u 2 + (a3 + b3 x + c3 y)u 3
2A(e)
(9.32a)
where A(e) is the element area and
ai = xj yk −xk yj , bi = yj −yk , ci = xk −xj ; i, j, k = 1, 2, 3 (9.32b)

Comparing Eqs.(9.32) and (9.24) the expression for the shape functions is
found to be as
1
Ni = (ai + bi x + ci y) , i = 1, 2, 3 (9.33)
2A(e)
The form of the linear shape functions is shown in Figure 9.7. It can be
checked that the shape function Ni takes the value one at node i and zero at
the other two nodes.

9.3.2 Discretization of the strain field

Substituting Eq.(9.24) into (9.2) gives the three characteristic strains as
∂u ∂N1 ∂N2 ∂N3
εx = = u1 + u2 + u3
∂x ∂x ∂x ∂x
∂v ∂N1 ∂N2 ∂N3
εy = = v1 + v2 + v3 (9.34)
∂y ∂y ∂y ∂y
∂u ∂v ∂N1 ∂N1 ∂N2 ∂N2 ∂N3 ∂N3
γxy = + = u1 + v1 + u2 + v2 + u3 + v3
∂y ∂x ∂y ∂x ∂y ∂x ∂y ∂x

273
 Fig. 9.7: Shape functions for the 3-noded solid triangular element

In matrix form
 
   .. .. 
 u1 
∂u ∂N1 ∂N2 ∂N3 
 






 ∂x
 0 . ∂x 0 . ∂x 0 

 v1



∂x

 

 
u  
∂v  ∂N1 .. ∂N2 .. ∂N3  2
ε= ∂y =  0 . 0 . 0 

 
  ∂y ∂y ∂y  v2 
 
   .. ..  
 ∂u + ∂v 
  ∂N1 ∂N1 ∂N2 ∂N2 ∂N3 ∂N3 

 u



∂y ∂x ∂y ∂x . ∂y ∂x . ∂y ∂x 
 3 

 
v3

(9.35)
or
ε = Ba(e) (9.36)
where
B = [B1 , B2 , B3 ] (9.37)
is the element strain matrix, and
 
∂Ni 0
 ∂x 
 ∂Ni 
 0
Bi =   (9.38)
 ∂y 

∂Ni ∂Ni
∂y ∂x

is the strain matrix of node i.

274
 Note that B contains as many Bi matrices as element nodes. This is also a
general property. Particularizing Eqs.(9.37) and (9.38) for the 3-noded triangle
we obtain (using Eq.(9.33))
 .. .. 
b 0 . b2 0 . b3 0
1  1 .. .. 

B=
2A(e) 0 c1 . 0 c2 . 0 c3 
 (9.39)
.. ..
c1 b1 . c2 b2 . c3 b3

and, therefore  
bi 0
1 
Bi = 0 ci 
 (9.40)
2A(e)
ci bi
We note that the expression of Bi in Eq.(9.38) is completely general and
applicable to any 2D solid element.

9.3.3 Discretization of the stress field

The discretized expression of the stress field within the element is obtained by
substituting Eq.(9.36) into (9.5) as

σ = Dε = DBa(e) (9.41)

If initial strains and stresses are considered we deduce from Eq.(9.17)

σ = D(ε − ε0 ) + σ 0 = DB a(e) − Dε0 + σ 0 (9.42)

Note that the strain matrix for the 3-noded triangle is constant (Eq.(9.40)).
This implies that both the strain and stress fields are constant within the ele-
ment. This is a consequence of the linear displacement interpolation chosen
which, naturally, has constant first derivatives. Therefore, a finer mesh will
be needed in zones where stress gradients are higher, so that the stress field is
accurately approximated.

9.3.4 Discretized equilibrium equations

The discretized equilibrium equations for the 3-noded triangle will be derived
by applying the PVW to an individual element. We note that the expressions
derived hereafter are completly general and aplicable to any 2D solid element.
Let us assume that the following external forces act on the element (Figure
9.8): a) distributed forces b acting per unit area (body forces), and b) dis-
tributed forces t acting along the element sides belonging to a boundary line
(surface tractions).
Note that the surface tractions due to the interaction of adjacent elements
are excluded “a priori” they cancel themselves out during the assembly process.

275
Fig. 9.8: Forces acting on a 3-noded solid triangle. The sides 13 and 23 belong to
the external boundary

As usual in the FEM, the equilibrium of the forces acting on the element
is enforced point wise at the nodes only. We must therefore define nodal point
loads which will balance the external forces and the internal forces due to
the element deformation. These “equilibrating nodal forces” are obtained by
applying the PVW to an individual element as
ZZ ZZ I
δεT σt dA = δuT bt dA + δuT tt ds+
A(e) A(e) l(e)
3
X 3
X (9.43)
+ δui Ui + δvi Vi
i=1 i=1

where δui and δvi are the nodal virtual displacements and Ui and Vi the cor-
responding equilibrating nodal forces. The virtual work performed by these
forces can be obtained from Eq.(9.43) as
ZZ ZZ I
T T T
δε σt dA − δu bt dA − δuT tt ds = [δa(e) ] q(e) (9.44)
A(e) A(e) l(e)

For the 3-noded triangular element

δa(e) = [δu1 , δv1 , δu2 , δv2 , δu3 , δv3 ]T

(9.45)
q(e) = [U1 , V1 , U2 , V2 , U3 , V3 ]T

Next we interpolate the virtual displacements in terms of the nodal values.

Following the same procedure as for deriving Eqs.(9.26) and (9.36) we obtain

δu = Nδa(e) ; δε = Bδa(e) (9.46a)

and
T T
δuT = [δa(e) ] NT ; δεT = [δa(e) ] BT (9.46b)

276
 Substituting the last equations into Eq.(9.44) gives
ZZ ZZ I
Th i T
[δa(e) ] BT σt dA − NT bt dA − NT tt dS = [δa(e) ] q(e)
A(e) A(e) l(e)
(9.47)
Since the virtual displacements are arbitrary it is finally deduced that
ZZ ZZ I
T T
B σt dA − N bt dA − NT tt ds = q(e) (9.48)
A(e) A(e) l(e)

Eq.(9.48) yields the equilibrating nodal forces q(e) in terms of the nodal
forces due to the element deformation (first integral), the body forces (second
integral) and the surface tractions (third integral). Substituting the stresses
in terms of the nodal displacements from Eq.(9.42) gives
ZZ ZZ I
BT (DBa(e) − Dε0 + σ 0 )t dA − NT bt dA − NT tt ds = q(e)
A(e) A(e) l(e)
(9.49)
and
hZ Z i ZZ
T (e)
B DBt dA a − BT Dε0 t dA +
Z(e)
ZA ZZ A(e) I (9.50)
T T
+ B σ t dA −
0
N bt dA − NT tt dS = q(e)
A(e) A(e) l(e)

or
K(e) a(e) − f (e) = q(e) (9.51)
where ZZ
(e)
K = BT D Bt dA (9.52)
A(e)

is the element stiffness matrix, and

(e) (e)
f (e) = fε(e) + fσ(e) + fb + ft (9.53)

is the equivalent nodal force vector for the element where

ZZ
fε(e) = BT Dε0 t dA (9.54)
A(e)
ZZ
fσ(e) =− BT σ 0 t dA (9.55)
A(e)
ZZ
(e)
fb = NT bt dA (9.56)
A(e)
I
(e)
ft = NT tt ds (9.57)
l(e)
are the equivalent nodal force vectors due to initial strains, initial stresses,
body forces and surface tractions, respectively.
The expressions for the element stiffness matrix and the equivalent nodal
force vectors given by Eqs.(9.52) - (9.57) are completely general and, therefore,

277
they are applicable to any 2D solid element. The particularization of these
expressions for the 3-noded triangular element is given in the next section.
The global equilibrium equations for the whole mesh are obtained by es-
tablishing that the nodes are in equilibrium, similarly as for 1D problems; i.e.
the sum of all the equilibrating nodal forces at a node balance the external
point loads and
X (e)
q i = pj (9.58)
e

where pj represents the vector of external point loads acting at node j and
the sum refers to all elements sharing the node. Eq.(9.58) is identical to that
studied in Chapter 1 for bar structures. The equilibrium equations for the
whole mesh can thus be obtained following identical procedures as explained
for bar structures. The global equilibrium equation is written in matrix form
as

Ka = f (9.59)

where K and f are the global stiffness matrix and the equivalent nodal force
vector for the whole mesh. Both K and f are assembled from the element
contributions in the standard manner.
We note once more that the equilibrating nodal forces due to the surface
traction along the element interface cancel themselves out during the assembly
process. Therefore, only the surface tractions acting on the external boundaries
of the structure must be considered in the analysis.

9.3.5 Stiffness matrix and equivalent nodal force vectors for the 3-
noded solid triangular element
Stiffness matrix
Eq.(9.52) can be written for the 3-noded triangle using (9.37) as
 
T
ZZ  B1 
 
K(e) = 2BT D [B1 , B2 , B3 ]t dA =
 T

A(e)
B3 
 T 
ZZ
B1 DB1 BT1 DB2 BT1 DB3
 .. 
=  . BT2 DB2 BT2 DB3 
  t dA (9.60)
A(e) ..
Sym. . BT3 DB3

(e)
A typical element stiffness submatrix, Kij , linking nodes i and j of the
element can be obtained as
ZZ
(e)
Kij = BTi DBj t dA (9.61)
A(e)

278
Substituting Eqs.(9.6) and (9.40) into Eq.(9.61) gives
   
ZZ · ¸ d11 d12 0 bj 0
(e) 1 bj 0 cj   1  
Kij =  d21 d22 0  (e)  0 cj  t dA (9.62)
A(e) 2A (e) 0 cj bj 2A
0 0 d33 cj bj

Since the integrand of Eq.(9.62) is constant we can easily obtain

³ t ´(e) · b b d + c c d bi cj d12 + bj ci d33
¸
(e) i j 11 i j 33
Kij = (9.63)
4A ci bj d21 + bi cj d33 bi bj d33 + ci cj d22
(e)
The form of Kij for plane stress and plane strain situations is simply ob-
tained by introducing the adequate values of the coefficients dij from Eq.(9.7).
(e)
Note that Kij is always symmetrical as d12 = d21 .

Equivalent nodal force vectors

a) Body forces
 
T
ZZ ZZ  N1 b 
 
(e)
fb = NT bt dA = 
NT2 b  t dA (9.64)

A(e) A(e)
NT3 b 
(e)
The nodal contribution of vector fb is
ZZ
(e)
fbi = NTi b t dA (9.65)
A(e)

If the body forces b are uniformly distributed over the element we obtain
using Eq.(9.33)
½ ¾
(At)(e) bx
fbi = (9.66)
3 by
i.e. the total force acting over the element is split into equal parts between the
three nodes of the triangle, as expected.
A particular case of body force is self weight with gravity acting in the
direction of the y-axis. In this case bx = 0 and by = −ρg where ρ and g are
the material density and the value of the gravity constant, respectively.

b) Surface loads
I
(e)
ft = NT tt ds (9.67)
l(e)

For a node i belonging to a loaded external boundary we have

I I ½ ¾
(e) Ni tx
fti = NTi tt ds = t ds (9.68)
l(e) l(e) Ni ty

279
 We note that the shape function of a node not belonging to the loaded
boundary takes a zero value. Thus, if the element side 1-2 is loaded with
(e)
uniformly distributed tractions tx and ty , vector ft is simply
 

 t 
 x
 


 ty 




(e)  

(e) (l12 t) tx
ft = (9.69)
2 

 ty 



 




0 

  
0

(e)
where l12 is the side lentgth. Eq.(9.69) shows that the traction force acting
along the element side is distributed into equal parts between the two side
(e)
nodes. The expressions of ft for loaded sides 1-3 and 2-3 are
   

 tx  
 0 

   


 ty 





 0 


(e) 
  (e)  
(e) (l13 t) 0 (e) (l23 t)  tx 
ft = ; ft = (9.70)
2 

 0 
 2 

 ty 


 
 
 




t x 
 
 t x 

   
  
ty ty

c) Forces due to initial strains

Substituting Eq.(9.37) into (9.54) gives

 
T
 B1 Dε 
0
ZZ ZZ  
fε(e) = BT Dε0 t dA = 
BT2 Dε0  t dA (9.71)

A(e) A(e)
BT3 Dε0 

and the equivalent nodal force of node i due to the initial strains is
ZZ
fε(e)
i
= BTi D ε0 t dA (9.72)
A(e)

If ε0 is constant over the element we obtain using Eqs.(9.6) and (9.40)

  
ZZ · ¸ d11 d12 0   εx 
0

1 bi 0 ci  
fε(e) =  d21 d22 0   ε0y  t dA =
i
2A (e) 0 ci bi
A(e)
0 0 0 
d33  γxy
½ ¾
t(e) bi (d11 ε0x + d12 ε0y ) + ci d33 γxy
0
= (9.73)
2 ci (d21 εx + d22 εy ) + bi d33 γxy
0 0 0

For initial thermal strains, the expressions (9.18) for ε0 should be used.

280
d) Forces due to initial stresses

Substituting Eq.(9.37) into (9.55) gives

 
T
 B1 σ 
0
ZZ ZZ  
fσ(e) = − BT σ 0 t dA = − BT2 σ 0 t dA (9.74)

 
A(e) A(e)
BT3 σ 0 

and the equivalent nodal force of node i due to the initial stresses is
ZZ
fσ(e)
i
=− BTi σ 0 t dA (9.75)
A(e)

For σ 0 being constant over the element, we obtain using Eqs.(9.20) and
(9.40)
 
ZZ · ¸ 
 σx0 
 ½ 0 ¾
1 bi 0 ci t(e) bi σx0 + ci τxy
fσ(e) =− σy t dA = −
0
i
A(e) 2A (e) 0 ci bi   0   2 ci σy0 + bi τxy0
τxy
(9.76)

The above expressions allow us to compute the matrices and vectors for
the simple 3-noded triangle for 2D elasticity applications. Some comments and
examples showing the behaviour of this element are given in Sections 9.5 and
9.6.
An example that illustrates the assembly and solution process is presented
below.

Example: 9.4 Analyze the plane structure of the figure below under self weight.

-- Solution

Mesh topology

Element Nodal connections

1 1 4 5
2 1 5 2
3 2 6 3
4 2 6 3
Plane strain situation
u1 = u4 = 0

281
 The assembly process is similar to that for matrix analysis of bar structures. The
global system of equations has the following form

1 2 3 4 5 6
 (1)

    r 1 + f1 + 

     (2) 

1 (1)
(K11 + K11 )
(2) (2)
K13 0 K12
(1) (1)
(K13 + K12 )
(2)
0 
 
 

f1 

a1
    

 (2) (3) (4) (2) (3) (3) (4) 
(K13 + K12 ) 

 
 (3) 
 (K33 + K11 + K13 0 (K32 + K12 )  
  
(f
(2)
+ f +

2 (4)
+K11 ) 
a 
 

3 1


 
 2  
  +f
(4)
) 

 
 
 
 1 

 K33
(4)
0 0
(4)
K32 
  
  

3 a3  (4) 
 (1) (1)  f3
 K22 K23 0  =
    (1) 
4 Symm. 
a4  
 r4 + f2 

 











 (1)
(K33 + K22 +
(2) (3)
K23 
 
 
 

 
a5  
(f3 + f2 +
(1) (2)

5 +K22 )
(3) 
 
 
 

  
    (3) 

   +f2 ) 
(K22 + K33 )    
(4) (3) a
 
 6
   

6 
 
(4) 
 f3 + f2 
(3)

(e) (e)
where Kij is obtained from Eq.(9.63) and fi from Eq.(9.64) with bx = 0 and
by = −ρg. In both cases t = 1 should be taken.

The above system can be solved in the usual way by elliminating the rows and columns
corresponding to the prescribed displacements a1 and a4 . Once the nodal displace-
ments have been obtained the corresponding reactions r1 and r4 can be computed.

Finally, the constant strains and stresses within each element can be found “a poste-
riori” from the known nodal displacements by Eqs.(9.36) and (9.41), respectively.

The reader is encouraged to repeat this exercise by him/herself.

9.4 THE FOUR NODED SOLID RECTANGULAR ELEMENT

9.4.1 Basic formulation
The 4-noded solid rectangle is the simplest of the quadrilateral solid element
family. This element was developed soon after the 3-noded solid triangle of
the previous section. However, its irregular behaviour has motivated much
research which we will summarize here.
Figure 9.9 shows a deep beam discretized in a mesh of 4-noded rectan-
gles. Let us consider an isolated element with the local coordinate system r
and s shown in Figure 9.9. The four nodal displacements define a four-term
polynomial interpolation for the displacement field. The simplest interpolation
satisfying the condition of interelement compatibility and geometric invariance
is
u(x, y) = α1 + α2 r + α3 s + α4 rs
(9.77)
v(x, y) = α5 + α6 r + α7 s + α8 rs

282
Fig. 9.9: Discretization of a deep beam with 4-noded solid rectangular elements.
Definition of the local axes r and s for an element

Eq.(9.77) implies a linear distribution of u and v along each element side,

thus guaranteeing continuity of the displacement field between adjacent ele-
ments. Note that the displacements vary as an incomplete quadratic polyno-
mial within the element. The four constants αi for each displacement compo-
nent are obtained from the following conditions expressed in the local system
r, s.

u = u1 and v = v1 for r = −a , s = −b
u = u2 and v = v2 for r = −a , s = −b
(9.78)
u = u3 and v = v3 for r=a , s=b
u = u4 and v = v4 for v=a , s=b

Substituting these conditions into Eq.(9.77) and solving for the αi param-
eter, Eq.(9.77) can be rewritten as follows (note that only the αi parameters
for one of the two displacements are needed, as the same interpolation is used
for u and v )

4
X 4
X
u= Ni u i ; v= Ni vi (9.79)
i=1 i=1

The shape functions Ni are

1³ r´ ³ s´ 1³ r ³ s´
N1 = 1− 1− ; N2 = 1+ ) 1−
4 a b 4 a b (9.80)
1 ³ r´ ³ s ´ 1 ³ r´ ³ s´
N3 = 1 + 1+ ; N4 = 1− 1+
4 a b 4 a b

283
 Eqs.(9.79) can be rewritten in matrix form as
 



u 1 



 


 v 1

 

 
 u 2


½ ¾ .. .. .. 
 

u N1 0 . N2 0 . N3 0 . N4 
0  v2 
u= = .. .. .. = N a(e)
v 0 N1 . 0 N2 . 0 N3 . 0

N4 
 u 3


 




v 3 



 




u 4 


 
v4
(9.81)
where · ¸
Ni 0
N = [N1 , N2 , N3 , N4 ] ; Ni =
 (e)  0 Ni
 a1(e) 

 
 
a   ½
ui
¾ (9.82)
(e) 2 (e)
a = (e) ; ai =


 a 3


 vi

 (e) 

a4
are the shape function matrix and the displacement vector for the element and
the node i, respectively.
The element strain matrix is obtained from Eqs.(9.2) and (9.79) as
 (e) 


 a1 
 (e) 
 
4
X (e) a 
ε= Bi ai = [B1 , B2 , B3 , B4 ] 2
(e) = Ba(e) (9.83)

 a3  
i=1 
 
 (e) 
a4
where Bi is given by precisely the same expression (9.40) derived for the 3-
noded triangle. For the computation of Bi note that
∂Ni ∂Ni ∂Ni ∂Ni
= and = (9.84)
∂x ∂r ∂y ∂s
The expression of the B matrix is shown in Box 4.1.
The stiffness matrix and the equivalent nodal force vectors for the element
are obtained via the PVW as previously explained for the linear triangle. The
element stiffness matrix is
ZZ
(e)
K = BT D Bt dr ds =
A(e)
 
BT1 DB1 BT1 DB2 BT1 DB3 BT1 DB4
ZZ  .. 

 . BT2 DB2 BT2 DB3 BT2 DB4 

=  ...  t. dr ds (9.85)
A(e)  BT3 DB3 BT3 DB4 
Sym. BT4 DB4
Box 4.1 shows that the strain matrix contains linear terms in r and s.
Therefore, the integrand of Eq.(9.85) contains quadratic terms. However, the

284
  
−b2 0 | b2 0 | b1 0 | −b1 0
 
B= 0 −a2 | 0 −a1 | 0 a1 | 0 a2 
−a2 b2 | −a1 b2 | a1 b1 | a2 b1

1 r 1 r
a1 = (1 + ) , a2 = (1 − )
4b a 4b a
1 1 1 1
b1 = (1 + ) , b2 = (1 − )
4a b 4a b
 
2a11 a36 c41 b36 −a14 −a36 c14 b63

 2a35 b63 c25 −a36 −a25 b36 c52 

 
 2a14 −a36 c14 b36 −a14 a63 
 
 2a25 b63 c52 a36 −a52 
 
K(e) =  

 2a14 a36 c41 b63 

 Symmetric 2a25 b63 c25 
 
 
 2a11 −b36 
2a25

aij = ai + aj , bij = ai − aj , cij = ai − 2aj

tbd11 tad22 td12 tad33 tbd33 td33
a1 = , a2 = , a3 = , a4 = , a5 = , a6 =
6a 6b 4 6b 6a 4

Box 9.1 Strain and stiffness matrices for a 4-noded solid rectangular element
of dimensions 2a × 2b

simplicity of the element geometry allows an explicit integration of all terms.

The resulting expression for K(e) is also shown in Figure 9.9.
In the same way, the equivalent nodal force vectors for the element are
obtained by Eqs.(9.54)-(9.57) using the new expressions for Ni and Bi . It is
interesting that the nodal contributions of a uniformly distributed load over
the element (see Eq.(9.65)) are

½ ¾
(e) (tA)(e) bx
fbi = (9.86)
4 by

i.e. the total force is distributed in equal parts between the four nodes, like
for the 3-noded triangle.
Similarly, a uniformly distributed traction acting over a side is distributed
in equal parts between the two side nodes.

285
Fig. 9.10: 4-noded solid rectangle subjected to pure bending, a) Initial geometry,
b) Element distorsion, c) Correct deformation of a beam segment in pure bending

9.4.2 Some remarks on the behaviour of the 4-noded solid rectangle

Both the 3-noded solid triangle and the 4-noded solid rectangle perform excel-
lently in problems where traction (or compression) is important. Conversely,
the accuracy of both elements strongly deteriorates in situations where bending
movements are involved, and very fine meshes are needed to obtain accurate
solutions in these cases (see Section 4.7).
The fact that the 4-noded solid rectangle cannot be used to model bending
dominated fields has a very instructive explanation. Let us consider the be-
haviour of an isolated element subjected to pure bending (Figure 9.10). The
exact solution from beam theory is [T6]

M
u(r, s) = rs
EI (9.87)
M a2 ³ r 2 ´ M b2 ³ s2 ´
v(r, s) = 1− 2 + 1− 2
2EI a 2EI b
Since the element sides are straight, the 4-noded rectangle can only repre-
sent the following bending mode (Figure 9.10)

u = ūrs ; v=0 (9.88)

It is obvious from the above that the element cannot correctly reproduce
the quadratic distribution of vertical displacements for the pure bending case.
This leads to excessive stiffness, which is a natural consequence of the inability
of the element sides to be curved.
Additionaly it is deduced from Eq.(9.87) that

∂u ∂v
γxy = + =0
∂y ∂x

i.e. the “exact” shear strain vanishes and only normal strains (and stresses)
exist.
The shear strain field for the element is obtained from Eq.(9.88) as

γxy = ūr (9.89)

286
i.e. the element has an “excess”of shear strain. This introduces an undesirable
additional stiffness which contributes to the poor ability of the element to
reproduce bending modes. Similar results are obtained for bending moments
acting on the horizontal sides simply by changing the coordinate r for s in
Eq.(9.89).
The deficiencies of the 4-noded solid rectangle also show for more irregular
quadrilateral shapes. These drawbacks are usually overcome in practice by
using very fine meshes. Other alternatives are possible, however, and some are
presented in the following sections.

Reduced integration of the shear stiffness terms

Eq.(9.89) clearly shows that the shear strain is zero at the element center only.
Therefore, the excess of shear strain can be eliminated by sampling the shear
strain at the element center (r = s = 0). This is simply achieved by using a
reduced one point Gauss quadrature for the shear terms in the stiffness matrix.
For this purpose the element stiffness matrix is split into two parts as

K(e) = K(e) (e)

a + Ks (9.90)

where K(e) (e)

a and Ks include the “axial” and “shear” contributions, respectively
given by
ZZ ZZ
K(e)
aij = BTai Da Baj t dA ; K(e)
sij = BTsi Dt Bsj t dA (9.91)
A(e) A(e)

with  
∂Ni · ¸
 0  ∂Ni ∂Ni
Bai = ∂x ; Bs i = ,
0 ∂Ni ∂y ∂x
∂y (9.92)
· ¸
d11 d12
Da = ; Ds = [ d33 ]
d12 d22
Matrix K(e)
a is integrated exactly, either analytically or via a 2×2 Gauss
quadrature, whereas a single integration point is used for K(e)
s . This “selected
integration” technique also improves the behaviour of 4-noded quadrilaterals
of arbitrary shape.
The reduced integration of K(e)
s can also be interpreted as a simple pro-
cedure to mitigate the excessive influence of the shear terms in the element
stiffness matrix. A disadvantage of reduced integration is that it produces a
quadrilateral element that is not geometric-invariant (Section 3.10.4), although
it passes the patch test and, therefore, it converges to the exact solution [3.3,
Cook]. In Chapter 7 we will study the application of reduced integration to
alleviate the influence of the transverse shear stiffness in Timoshenko beam
elements. However, reduced integration of the stiffness matrix terms should
always be looked upon with extreme caution, as they can lead to internal

287
Fig. 9.11: Four node rectangle, a) Addition of a central node, b) Shape functions
for the incompatible modes

mechanisms and to the violation of the patch test in some cases. Reduced in-
tegration procedures will be further considered in Chapters 9-12 when dealing
with plates and shells.

Addition of internal modes

A method to enhance the flexibility of the 4-noded rectangle is to add to the
original interpolation internal displacement modes which vanish at the element
boundaries. The simplest mode is a “bubble”function associated with an extra
central node (Figure 9.11a). The displacement field is expressed as
5
X 5
X
u= N i ui ; v= N5 v i (9.93)
i=1 i=1

where N1 , N2 , N3 , N4 are the linear functions of (9.80) and

h ³ r ´2 ih ³ s ´2 i
N5 = 1 − 1− (9.94)
a b
The internal d.o.f u5 and v5 (also called hierarchical d.o.f [C23]) can be
eliminated after the element stiffness matrix is obtained. Note that u5 and v5
are not absolute displacements and they represent the difference between the
total displacements of the central node and the bilinear field defined by the
four corner displacements. For instance, the horizontal displacement of the
central node is given by
à 4
!
X
u5 = u(0, 0) = Ni u i + u5 (9.95)
i=1 0,0

The behaviour of the modified 4-noded solid element can be improved by

using a reduced single point quadrature for the shear terms as described in the
previous section.

Addition of incompatible modes

The 4-noded rectangle can also be improved
³ ´2
by adding to
³ ´2
the original displace-
r s
ment field the displacement modes 1 − a and 1 − b which are needed

288
to reproduce the exact solution (9.87) (Figure 9.11b). The new displacement
field is
X4 h ³ r ´2 i h ³ s ´2 i
u= Ni u i + 1 − u5 + 1 − u6
i=1 a b
4 h ³ r ´2 i h ³ s ´2 i (9.96)
X
v= Ni vi + 1 − v5 + 1 − v6
i=1 a b
The additional generalized d.o.f. u5 , v5 , u6 , v6 (also called “nodells” d.o.f.)
are internal to each element and can be eliminated by static condensation.
However, the displacements along the interelemental boundaries are discon-
tinuous and the element is incompatible. Incompatible 4-noded quadrilaterals
formulated in this way fail to pass the patch test under constant stress (or
constant strain) states unless they are rectangular.
Fortunately, the element satisfies the patch test for arbitrary quadrilateral
shapes if the shear stiffness terms are evaluated using a reduced single point
Gauss quadrature, whereas the rest of the stiffness terms can be exactly inte-
grated [C15,T3]. The resulting element is geometrically invariant and does not
have spurious mechanisms. Box 9.2 shows the stiffness matrix for an homo-
geneous and isotropic element with reduced integration after eliminating the
internal incompatible d.o.f. by static condensation [C15], [F11].
The incompatible modes technique can also be successfully applied to 4-
noded quadrilaterals of arbitrary shape. More information can be found in
[C15] and [Z6].

Use of an assumed strain field

Another procedure to enhance the performance of the 4-noded quadrilateral
is to impose over the element an assumed strain field compatible with the
condition γxy = 0 for the pure bending case.
Dvorkin and Vassolo [D13] have proposed the following assumed strain field
εx = α1 + α2 x + α3 y ; εy = α4 + α5 x + α6 y ; γxy = α7 (9.97)
The αi parameters are expressed in terms of the nodal displacements by
sampling the assumed strains at a number of element points and equaling their
values with those given by the strains deduced from the original displacement
field. This leads to a substitute strain matrix from which the element stiffness
matrix can be directly obtained. More information on this element can be
found in [D13].

9.5 PERFORMANCE OF THE 3-NODED SOLID TRIANGLE

AND THE 4-NODED SOLID RECTANGLE
The 3-noded triangle and the 4-noded rectangle perform reasonably well under
pure tension or compression dominated situations. In general the 4-noded

289
 ½ ¾ ½ ¾ " # ½ ¾ " #
u 4
X ui ³ r ´2 u5 ³ s ´2 ½ u ¾
6
= Ni + 1− + 1−
v i=1
vi a v5 a v6

 
C1 C5 C2 −C6 C4 −C5 C3 C6

 C7 C6 C9 −C5 C10 −C6 C8 

 
 C1 −C5 C3 −C6 C4 C5 
 
Dt  C7 C6 C8 C5 C10 
(e)  
K =  
12(1 − α2 ) 
 C1 C5 C2 −C6 
 Symmetric C7 C6 C9 
 
 
 C1 −C5 
C7

Plane stress : D=E ; α=ν

E ν
Plane strain : D= ; α=
1 − ν2 1−ν

a a
C1 = (−m2 − 1.5m + 5.5) , C4 = (−m2 + 1.5m − 3.5)
b b
a
C2 = (m2 − 1.5m − 2.5) , C5 = 1.5 (1 + m)
b
a 2
C3 = (m + 1.5m − 0.5m) , C6 = 1.5 (1 + 3m)
b
C7 − C10 are obtained from C1 − C4 interchanging a by b

Box 9.2 Stiffness matrix for an homogeneous and isotropic four node rectan-
gular element of dimensions 2a×2b with incompatible modes

element is more accurate than the 3-noded triangle for the same number of
d.o.f. in these cases (Section 9.6 and Figures 9.12 and 9.13). However, the
behaviour of both elements deteriorates in bending situations. Still, the 4-
noded element has a superior performance for such problems. Indeed the
accuracy of the 4-noded rectangle increases substantially by adding the two
incompatible modes as described in Section 9.4.2. Good results for bending
dominated problems can be obtained by using higher order triangular and
rectangular elements. A comparison of different triangular and rectangular
element in a bending problem is presented in Section 9.6.
The poorer performance of the 3-noded triangle is compensated by its ver-
satility to discretize complex geometries using automatic mesh generators, and
so it remains very popular.
The accuracy of both elements increases by using a higher order approxi-

290
Fig. 9.12: Square plate under parabolic traction. Analysis with 3- and 6-noded
triangles and 4 and 9-noded rectangles

mation for the displacement field. This also allows curved sided elements to be
derived using an isoparametric formulation as described in the next chapter.

9.6 GENERAL PERFORMANCE OF TRIANGULAR AND

RECTANGULAR ELEMENTS
Figures 9.12 and 9.13 show two examples which lead us to draw some inter-
esting conclusions on the behaviour of rectangular and triangular elements.
The first example is the analysis of a square plate under a parabolic traction
acting symmetricaly on two opposite sides. Different meshes of 3 and 6-noded
triangles and 4 and 9-noded rectangles are used for the analysis. The numer-
ical results for the horizontal displacement of the central point on the loaded
side show that the 3-noded triangle is the less accurate of all elements stud-
ied. Nevertheless 1% error with respect to the “exact” analytical solution is
obtained using a fine mesh (Figure 9.12).
The accuracy increases notably for the same number of d.o.f. when 6-
noded triangles are used and, even more, when either the 4- or the 8-noded

291
Fig. 9.13: Cantilever deep beam under parabolic edge load (ν = 0.2). Analysis
with 3- and 6-noded triangles, 4- and 9-noded rectangles and the 4-noded rectangle
with two incompatible modes

rectangles are used, as shown in Figure 9.12 [G3], [Y1]. Similarly good results
are obtained with the 8-noded rectangle.
The second example is a cantilever deep beam under a parabolic edge load
(Figure 9.13). The analysis is performed using the same elements as in the
previous example and, in addition, the 4-noded rectangle is enhanced with two
incompatible modes studied in Section 9.4.2. Results plotted in Figure 9.13
show clearly the poor accuracy of the 3-noded triangle for bending dominated
problems. The solution improves slightly for the quadratic triangular element.
The 4-noded rectangle has an overstiff behaviour, as expected from its inability
to reproduce pure bending situations. The solution improves however when
finer meshes are used. Note the excellent performance of the incompatible
4-noded rectangle and also of the 9-noded Lagrange rectangle. Similar good
results are obtained using the 8-noded quadratic Serendipity rectangle.
These results can be generalized to other situations. Typically, rectangles
are more accurate than triangles for the same number of degrees of freedom.
However, triangular elements are more versatile due to their better ability to

292
model complex geometries with unstructured meshes using automatic mesh
generators.
As a rule, low order elements are simpler to use, although finer meshes are
needed in zones where high stress gradients exist. Higher order elements are
more competitive in these regions.

9.7 CONCLUDING REMARKS

This chapter has presented the basic concepts for the analysis of 2D solids with
the FEM. The steps followed in the formulation of the kinematic variables,
the strain and stress fields, the equilibrium expression via the PVW and the
discretization process are completely general and will repeatedly appear when
considering the finite element analysis of any other structure. The study of
this chapter is therefore essential as a general introduction to the analysis of
continuous structures with the FEM.
The procedure for deriving the element stiffness matrix and the equivalent
nodal force vector from the PVW has been detailed. The expressions of the
different matrices and vectors have general applicability to any element type.
The particular form of these matrices for the 3-noded linear triangle and the
4-noded rectangle has been given. The linear triangle has limited accuracy for
coarse meshes although its simplicity and versality make it probably the most
popular element for practical analysis of 2D solids with arbitrary geometry.
The basic 4-noded rectangle has some limitations when it comes to capturing
pure bending modes. These defficiencies can be overcome by “ad hoc” proce-
dures such as reduced integration, the addition of internal nodes and the use
of an assumed strain field.
The derivation of higher order triangular and quadrilateral elements of ar-
bitrary shape follows the concept studied in previous chapters for the Poisson
equation. The basic ingredients are: a systematic procedure to obtain the shape
functions, the use of an isoparametric formulation and numerical integration.
These topics will be studied in the course on computational structural mecha-
nics.

293
294
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