322892315 notes-dixit-pdf

FINITE ELEMENT METHODS
IN ENGINEERING
(Lecture Notes)
UDAY S. DIXIT
E-mail: uday@iitg.ernet.in
Department of Mechanical Engineering
Indian Institute of Technology Guwahati
Guwahati – 781039, Assam, India
May 2007

PREFACE
Finite Element Method (FEM) initially gained popularity as a method of stress analysis owing to
its origin in solving the problems of structural mechanics. At many places, FEM could actually
come as a replacement of experimental techniques. It was sooner realized that FEM can be very
helpful in solving the problems of heat transfer, fluid mechanics, electromagnetism, and
manufacturing process modeling, to speak of only a few areas. Precisely speaking, FEM is a
numerical method for solving the integral and differential equations. Therefore, this booklet
attempts to present FEM as a tool of solving governing equations of the physical systems. The
application examples and exercise problems have been chosen from a number of different fields
of engineering.
At present, due to its increasing importance in industries and academic research, each
year increasing number of students are learning FEM as a part of their curriculum at post-
graduate or final year undergraduate level. However, there is a lack of good textbooks suited for a
one semester course, although there are a good number of reference books on the subject. Many
colleges in India lack teachers specialized to teach this course. My own learning of FEM was
through the excellent lectures of some of the professors of Indian Institute of Technology Kanpur
in early 1990s. I supplemented the lectures with a number of books and sometimes journal
papers. I started teaching this course at Indian Institute of Technology Guwahati since 1998.
Soon, I felt the need of providing some handouts to students for compensating their ability to
learn this subject from available books. Meantime, Quality Improvement Programme (QIP)
became very active in this Institute, which provided me with an opportunity to bring out the
collection of my lectures of a one-semester course in the form of this booklet. I am thankful to
QIP of Indian Institute of Technology Guwahati for sponsoring the publication of my lecture
notes. My personal thanks also go to QIP coordinator Prof. Rajeev Tiwari, who was always
available for providing the valuable suggestions.
The organization of the subject matter and content in this booklet has evolved as a result
of the experience gained in classroom teaching. For example, initially after just giving a brief
introduction to FEM, I used to teach calculus of variation, which appeared somewhat frightening
to some engineering students, whose closeness with mathematics had diminished over the years.
Considering this, I first introduced to them direct FEM formulation and made them solve a
number of one-dimensional problems in order to get confidence and interest in the subject. This
took initial 3 lectures (Chapter 1 of this booklet) after which I went in sequence to calculus of
variation, classical methods for solving boundary value problems, Galerkin and Ritz FEM
methods applied to one-dimensional problems, and finally 2-D & 3-D FEM problems. The
lectures put emphasis mainly on clear understanding of fundamental concepts. It is assumed that
the readers have sufficient knowledge of using computers. At the end of each chapter, I have
included a number of exercise problems including the problems requiring the use of a computer.
Solutions of some of these problems may be provided on demand.
I am thankful to my students for using these notes, providing valuable feedback and
discussing with me the exercise problems. I hope that these notes will be useful to students,
teachers and practicing engineers interested in learning FEM and after going through these
lecture-notes the readers will face no difficulty in referring to advanced topics from the books and
journals. I shall welcome any constructive feedback on these notes and will be grateful for
pointing out errors if any in these notes. The readers may send me an e-mail at uday@iitg.ernet.in
or usd1008@yahoo.com.in.
Uday S. Dixit
May 2007

Contents
1 Finite Element Method: A Quick Introduction 1
1.1 Introduction 1
1.2 Direct FEM Formulation of Axial Rod Problem 2
1.2.1 Pre-processing 2
1.2.2 Elemental Stiffness Matrix and Load Vector 3
1.2.3 Assembly Procedure 5
1.2.4 Application of Boundary Conditions and Solution 6
1.2.5 Post-processing 6
1.3 Direct FEM Formulation of Beam Problem 7
1.3.1 Pre-processing 7
1.3.2 Elemental Stiffness Equations 8
1.3.3 Assembly Procedure 10
1.3.4 Boundary Conditions and Solutions 11
1.3.5 Post-processing 12
1.4 Conclusions 12
References 12
Exercise 1 13
2 Introduction to Calculus of Variation 19
2.1 Introduction 19
2.2 Functional 19
2.3 Extremization of A Functional 20
2.4 Obtaining the Variational Form from A Differential Equation 28
2.5 Principle of Virtual Work 32
2.6 Principle of Minimum Potential Energy 33
2.7 Conclusions 33
References 34
Exercise 2 34
3 Some Classical Function Approximation Methods for Solving
Differential Equations
39
3.1 Introduction 39
3.2 Ritz Method 39
3.3 Galerkin Method 42
3.4 The Least Square Method 44
3.5 Collocation Method 45
3.6 Sub-Domain Method 45
3.7 Conclusions 46

vi
References 46
Exercise 3 47
4 Ritz and Galerkin FEM Formulation 51
4.1 Introduction 51
4.2 Completeness and Compatibility 51
4.3 Concepts of shape Functions 52
4.4 Developing the Elemental Equations by Ritz method 55
4.5 Developing the Elemental Equation by Galerkin method 59
4.6 Conclusions 60
Exercise 4 61
5 Some One-Dimensional C0
Continuity FEM Formulation 63
5.1 Introduction 63
5.2 Steady-State Heat Conduction 63
5.3 Longitudinal Deformation of A Rod 68
5.4 Fluid Flow Problem 71
5.5 Conclusion 71
Exercise 5 72
6 Finite Element Formulation for Bending of Beams 75
6.1 Introduction 75
6.2 Galerkin FEM Formulation 75
6.2.1 Weak Form 76
6.2.2 Choosing Suitable Approximating Function 76
6.2.3 Hermitian Shape Function 77
6.2.4 Elemental Equations 78
6.2.5 Assembly Boundary Condition and Solution 78
6.3 Ritz Formulation 79
6.4 Summary 80
Exercise 6 81
7 Finite Element Formulation for Trusses and Frames 85
7.1 Introduction 85
7.2 Formulation for A Truss 86
7.3 An Example 89
7.4 FEM Formulation for The Frames 92
7.5 Summary 93
Exercise 7 93

vii
8 Introduction to 2-D and 3-D FEM 97
8.1 Introduction 97
8.2 Triangular Elements 97
8.3 Tetrahedral Elements 102
8.4 Rectangular Elements 103
8.5 Brick Elements 104
8.6 Governing Differential Equation for 2-D Heat Conduction 105
8.7 Weak Form and FEM Formulation 105
8.8 A Note on The Assembly in Two Dimensions 108
8.9 Poisson Equation In 3-D 109
8.10 Fluid Flow Problem 109
8.11 Torsion of Circular and Noncircular Cross-Section 110
8.12 Summary 110
References 110
Exercise 8 111
9 Numerical Integration 113
9.1 Introduction 113
9.2 One Dimensional Integration Formulae 113
9.2.1 Newton-Cotes Quadrature 114
9.2.2 Gauss Quadrature 116
9.3 Two Dimensional Integration Formulae 117
9.3.1 Integration over Square Region 117
9.3.2 Integration over Triangular Region 118
9.4 Conclusions 119
References 120
Exercise 9 122
10 Further Details on 2-D FEM 125
10.2 Natural Coordinates and Iso-Parametric, Sub-Parametric and Super-
Parametric Elements
125
10.3 Four-Noded Quadrilateral Elements 127
10.4 Serendipity Elements 128
10.5 Eight-Noded Curvilinear Elements 130
10.6 Conclusions 131
References 131
Exercise 10 131
11 FEM Formulation for Plane Stress and Plane Strain Problems 137

viii
11.2 Basic Equations 138
11.3 Boundary Conditions 139
11.4 FEM Formulation 139
11.5 Shape Functions 142
11.6 Numerical Evaluation of Elements Matrices and Vectors 143
11.7 Assembly of Element Matrices 145
11.8 Boundary Conditions and Solutions 145
11.9 Gradient Estimates 145
11.10 An Example 146
11.11 Summary 146
Exercise 11 148
12 Free Vibration Problems 151
12.2 Vibration of A Rod 151
12.3 Vibration of A Beam 152
Exercise 12 157
13 Finite Element Formulation of Time Dependent Problems 161
13.2 Classification of Partial Differential Equations 161
13.3 Time Response of A parabolic Equation 162
13.4 Forced Vibration Problems 164
References 165
Exercise 13 165
14 FEM Formulation of Plate Problem 167
14.2 Thin Plate Formulation 167
14.3 Various Thin Plate Elements 169
14.3.1 Rectangular Element with Corner Nodes 169
14.3.2 Triangular Element with Corner Nodes 170
14.3.3 Quadrilateral and Parallelogram Elements 171
14.3.4 16 Noded Rectangular Shape Function 171
14.4 Thick Plate Formulation 171
14.5 Locking Phenomenon 174
14.6 An Example 174

ix
14.7 Summary and Conclusion 176
References 176
Exercise 14 177
15 Finite Element Formulation of 2-D Flow problems 179
15.2 Discretization of The Strip 180
15.3 Governing Equations and Boundary Conditions 181
15.4 Weak Formulation 182
15.5 Finite Element Approximation 184
15.6 Finite Element Equations 186
15.7 Application of Boundary Conditions 190
15.8 Post-Processing 192
15.9 Conclusion 192
Exercise 15 193
16 Error Analysis in Finite Element Method 197
16.2 Error Measures 197
16.3 Types of Error Estimates 200
16.3.1 A Priori Error Estimates 200
16.3.1.1 h-Convergence 200
16.3.1.2 p- Convergence 201
16.3.1.3 hp- Convergence 201
16.3.2 Posteriori Error Estimates 201
16.3.2.1 ZZ Error Estimate 201
16.3.2.2 Residual Method 202
16.3.2.3 Superconvergent Patch Recovery (SPR) Technique 204
16.3.2.4 Higher Order Approximation of Primary Variables
(HOAPV)
207
16.4 Error Estimates by Recovery 210
Further Readings 211
Exercise 16 212
17 Miscellaneous 213
17.2 Difference Between FEM and FDM 213
17.3 Finite Element Solutions Versus Exact Solution 214
17.4 Accuracy of Derivatives of The Primary Variables 215

x
17.5 Essential and Natural Boundary Conditions 215
17.6 Mesh Refinement 215
17.7 Effect of The Geometry of A Particular Element 216
17.8 Solving The Problems of Fracture Mechanics using FEM 216
17.9 Infinite Elements 217
17.10 Ill-Conditioned System 218
17.11 Patch Test 219
Exercise 17 220
Bibliography 221

Chapter 1
FINITE ELEMENT METHOD: A QUICK INTRODUCTION
(Lectures 1-3)
1.1 INTRODUCTION
The finite element method (FEM) is a numerical method to solve differential and
integral equations. Since the behavior of physical systems can be represented by differential
equations, finite element method can be used to analyze a number of physical problems. Method
originated as a technique to analyze complex structural systems. The discovery of method is
often attributed to Courant (1943). The use of method in structural (aircraft) analysis was first
reported by Turner et al. (1956). The method received its name from Clough (1960). In finite
element method, region of interest is divided into a number of elements. Differential equations
are reduced to algebraic equations by using appropriate approximations for the variables over the
elements. Boundary conditions of any complexity can be applied very easily. Complicated
geometries and variations of material properties pose not much problem. Hence, the method has
emerged as a versatile and powerful tool of computational engineering.
Aim of this chapter is to introduce the reader to the finite element methodology. For this
purpose, two 1-dimensional problems have been considered- axial rod problem and beam
problem. In axial rod problem, one is usually interested to find out the axial displacement of
each point of the rod under the action of prescribed load, whereas in the beam problem, at each
point, vertical deflection and its slope need to be found out. Thus, the axial rod problem is a one-
degree freedom per node problem and the beam problem is a two degrees freedom per node
problem. However, it will be seen that the finite element procedure is similar for both the
problems. In fact, it is similar for any problem irrespective of its dimension and degrees of
freedom. The finite element method follows the following steps:
• Pre-processing: In this step, the geometry is discretized into a number of small
elements. The elements can be of different shapes. Each element is characterized by
number of points called ‘nodes’ present in the element. Complete system of elements is
called mesh and the process of generating the elements is called mesh generation.
• Obtaining elemental equations: In this step, algebraic equations are obtained for each
element. A number of methods can be used for this purpose. In this article, they are
derived using direct FEM formulation, in which algebraic equations are obtained
directly from the physics of the problem.
1

• Assembly: In this step, the elemental stiffness equations are assembled to yield a global
system of equations.
• Application of boundary conditions: In this step, the assembled system of equations is
modified by inserting prescribed boundary conditions.
• Solution: In this step, modified global system of equations is solved to obtain solution
in the form of values of primary variables at nodes, such as nodal displacements in axial
rod problem and nodal deflections and slopes in beam problem.
• Post-processing: In this step, various secondary quantities are computed from the
obtained solution. For example, stresses and strains are computed from the obtained
nodal displacements in axial rod problem.
1.2 DIRECT FEM FORMULATION OF AXIAL ROD PROBLEM
Consider an axial rod loaded with a force P at the end (Fig. 1.1). In general, the rod may
be of variable cross-section, non-homogeneous material and may be loaded with concentrated
forces at different points as well as distributed forces at different segment of the rod. However,
to introduce the finite element method a trivial problem of uniform axial rod loaded with force P
at the end is chosen. It is desired to find out deflections, strains and stresses at different points of
the rod.
A governing differential equation of the problem with axial deflection u as the
independent variable and point coordinate x as the dependent variable can be obtained. In the
finite element method, the differential equation is converted into algebraic equation. However,
for this particular problem, the algebraic equation can directly be obtained from the physics of
the problem. Hence, the methodology described here is called direct finite element formulation.
Figure 1.1: Axial rod loaded at one end
1.2.1 Pre-processing
First step in the finite element is to discretize the rod into a number of small segments,
each one being called an element. For example, in Fig. 1.2, the rod has been divided into three
elements. The end points of each element are called nodes. Thus in this problem, there are total
2

three elements and four nodes. Each element is designated by its two nodes and coordinates of
each node are stored. This step is called pre-processing or mesh generation.
Figure 1.2: (a) Discretization of the rod (b) A typical element
1.2.2 Elemental stiffness matrix and load vector
In order to obtain governing algebraic equations, deflection in each element is assumed
to be linear. This will be indeed so, if the element is composed of homogeneous material
following Hook’s law, has uniform cross-sectional area and loads are only point loads acting at
the ends. Fig. 1.2(b) shows a general element, with end nodes designated by i and j. The tensile
strain in the element is given by,
h
uu ij
t
−
=ε (1.1)
where ui and uj are axial deflections at nodes i and j respectively and h is the element length
(equal to L/3 in this case). Corresponding tensile stress is
h
uu
E
ij
t
−
=σ (1.2)
where E is the Young’s modulus of elasticity. The force Fj applied at the j th
node is stress times
the cross-sectional area, A. Hence,
j
ij
F
h
uu
AE =
−
(1.3)
Thus, a relationship between force Fj and nodal deflections is obtained. Similar relation between
force Fi and nodal deflections can be obtained in the following. The compressive strain, in the
element is
h
uu ji
c
−
=ε (1.4)
and the corresponding compressive stress is
h
uu
E ji
c
−
=σ (1.5)
3

Hence,
i
ji
F
h
uu
AE −=
−
(1.6)
Note the negative sign at the right hand side of the above equation. This is because force Fi is
assumed tensile in Fig. 1.2(b). Note also that equation (1.3) and (1.6) suggest that Fi= Fj. These
indeed are the condition for the rod to be in equilibrium and equations (1.3) and (1.6) are same.
However, we retain two equations at this stage and write them in the matrix form as,
⎭
⎬
⎫
⎩
⎨
⎧−
=
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
−
−
j
i
j
i
F
F
u
u
h
AE
11
11
(1.7)
In the compact form, this can be written as
[ ]{ } { }eee
Fuk = (1.8)
where
[ ] ⎥
⎦
⎤
⎢
⎣
⎡
−
−
=
11
11
h
AE
ke
(1.9)
{ } (1.10)
⎭
⎬
⎫
⎩
⎨
⎧
=
j
ie
u
u
u
and
{ } (1.11)
⎭
⎬
⎫
⎩
⎨
⎧−
=
j
ie
F
F
F
Compare equation (1.8) with equation for a spring loaded with force F:
kx =F (1.12)
In analogy with this, matrix [ke
] is called elemental stiffness matrix and its elements have units
N/m in SI system, { }e
u is elemental displacement or primary variable vector and {Fe
} is the
elemental load vector.
Let us observe the elemental system of equations given by equation (1.7). This system
cannot be solved to yield the values of ui and uj, because of the following reasons:
1. In general, Fi and Fj are internal forces, which are unknown.
2. Even if the values of Fi and Fj are known, the elemental stiffness matrix cannot be
inverted to yield solution, because this matrix is singular and its rank is 1. Physically this
means that just by prescribing the values of two end forces, one cannot predict the
displacement, because infinite numbers of rigid body modes are possible.
4

In order to overcome the first difficulty i.e. to get rid of internal forces, the elemental stiffness
are assembled. Second difficulty is overcome by prescribing one geometric boundary conditions
(i.e. prescribing axial deflection at one node). Following subsection illustrates the assembly
procedure and the next subsection illustrates the application of boundary conditions.
1.2.3 Assembly procedure
For the given problem, let us write the elemental equations for three elements. These
are:
⎭
⎬
⎫
⎩
⎨
⎧−
=
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
−
−
2
1
2
1
11
11
F
F
u
u
h
AE (1.13)
⎭
⎬
⎫
⎩
⎨
⎧−
=
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
−
−
3
2
3
2
11
11
F
F
u
u
h
AE (1.14)
⎭
⎬
⎫
⎩
⎨
⎧−
=
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
−
−
4
3
4
3
11
11
F
F
u
u
h
AE (1.15)
These elemental stiffness equations can be assembled to yield global stiffness equations, having
u1, u2, u3 and u4 as unknowns. In the assembled system of equations, internal forces will be
eliminated. There are various ways to understand assembly operation. We follow a simple
approach, in which elemental system of equations for each element is written in global form and
then they are algebraically added. Thus, the equation (1.13-1.15) are written as,
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧−
=
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
−
−
0
0
0000
0000
0011
0011
2
1
4
3
2
1
F
F
u
u
u
u
h
AE (1.16)
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
−
=
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
−
−
0
0
0000
0110
0110
0000
3
2
4
3
2
1
F
F
u
u
u
u
h
AE (1.17)
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
−
=
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
−
−
4
3
4
3
2
1
0
0
1100
1100
0000
0000
F
F
u
u
u
u
h
AE (1.18)
Additions of these, yields
5

1 1
2
3
44
1 0 0 1 0 0 0 0 0 0 0 0
1 0 0 1 1 0 0 1 0 0 0 0 0
0 0 0 0 1 0 0 1 1 0 0 1 0
0 0 0 0 0 0 0 0 1 0 0 1
u F
uAE
uh
Fu
−+ + − + + + + + + ⎧ ⎫ ⎧ ⎫⎡ ⎤
⎪ ⎪ ⎪ ⎪⎢ ⎥− + + + + − + + + ⎪ ⎪ ⎪ ⎪⎢ ⎥ =⎨ ⎬ ⎨ ⎬
⎢ ⎥+ + − + + + + − ⎪ ⎪ ⎪ ⎪
⎢ ⎥ ⎪ ⎪ ⎪ ⎪+ + + + + − + +⎣ ⎦ ⎩ ⎭⎩ ⎭
(1.19)
Note that in system of equations (1.19), internal forces F2 and F3 got eliminated. However, F1
and F4 still remain. They can be eliminated only by putting boundary conditions. Also note that
the assembled global stiffness matrix is singular, with rank 3. Thus one nodal displacements
need to be prescribed.
1.2.4 Application of boundary conditions and solution
For the present problem, F4 is equal to the externally applied load P. This is called force
or natural boundary condition. However, F1 is unknown. On the node at which F1 acts, u1=0.
This is called essential or geometric boundary condition. There are various ways to apply this
boundary condition. A simple way is to replace the first equation by u1=0. Thus, assembled
system of equations, after the application of boundary conditions, becomes,
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
=
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
−
−−
−−
Pu
u
u
u
h
AE
0
0
0
1100
1210
0121
0001
4
3
2
1
(1.20)
There are various methods to solve this linear system of equations. Solution yields,
u1=0, u2=
AE
PL
3
, u3=
AE
PL
3
2
, u4=
AE
PL
(1.21)
Notice that these are exact displacements, obtainable from elementary strength of materials.
This is no surprise, as the exact displacement function is linear and a linear displacement field
(via constant strain) was assumed in each element.
1.2.5 Post-processing
After the nodal displacements have been obtained, strains and stresses in the elements
can be computed. This is a part of post-processing. In this example, strain in the element 2 is
( ) AE
P
L
uu
h
uu
=
−
=
−
=
3/
2323)2(
ε (1.22)
and the stress is given by
A
P
E == )2()2(
εσ (1.23)
6

In the same way, stresses in other elements may be computed. The displacement at any point
inside the element can be found by linear interpolation.
1.3 DIRECT FEM FORMULATION OF BEAM PROBLEM
Consider a beam rigidly fixed at the ends and loaded in the center by a load P (Fig. 1.3).
In general, beam can be of any arbitrary cross-section loaded with any complex loading function.
For the sake of simplicity, only a beam of uniform cross-sectional area is considered and
deflection due to only bending is considered. Deflection due to shear is not taken into
consideration.
Figure 1.3: Fixed-fixed beam with a central load
1.3.1 Pre-processing
We divide the entire beam into two 2-noded equal elements (Fig. 1.4(a)). Element 1 is
composed of nodes 1 and 2 and element 2 is composed of nodes 2 and 3. We introduce here the
concept of connectivity matrix, which we have not done in Section 2 in order to avoid loading
lot of information in one go. Connectivity matrix is a simple representation of element-node
relations, in which row indicates element number, column indicates local (elemental) node
number and element of the matrix denotes global node number. Thus, the connectivity matrix for
the present mesh is:
(1.24)⎥
⎦
⎤
⎢
⎣
⎡
32
21
Given connectivity matrix and coordinates of the node, the mesh can be easily constructed.
7

Figure 1.4: (a) Discretization of the beam (b) A typical element
1.3.2 Elemental stiffness equations
From elementary mechanics of materials, it is known that deformation of axial rod is
characterized by axial displacement of each point, where as in beam problem, at each point
vertical displacement as well as slope needs to be prescribed. Thus, a typical node in the element
has two degrees of freedom, vertical deflection and slope. Fig. 1.4(b) shows a typical two
nodded element. On two nodes i and j, forces Fi and Fj and moments Mi and Mj are acting. In
general, Fi depends on the elastic property of the element and displacements at the two nodes.
Hence
Fi = k11 vi+ k12 θi+ k13 vj+ k14 θj (1.25)
where k’s are coefficients dependent on the geometry and material of the element. Similar
equation can be written for Mi, Fj and Mj. Thus, the elemental equations become
(1.26)
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
=
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
j
j
i
i
j
j
i
i
M
F
M
F
v
v
kkkk
kkkk
kkkk
kkkk
θ
θ
44434241
34333231
24232221
14131211
In order to derive the values of coefficients, we proceed as follows. Let us assume that vj and θj =
0 in Fig. 1.4(b). First two equations of system of equation given by (1.26) reduce to
(1.27)
⎭
⎬
⎫
⎩
⎨
⎧
=
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
i
i
i
i
M
Fv
kk
kk
θ2221
1211
Element then becomes a cantilever beam loaded by a vertical force Fi and Moment Mi at one
end. The deflection and slope of that end can be obtained from elementary strength of materials
using the following equations:
i
ii
v
EI
hM
EI
hF
=−
23
23
(1.28)
8

i
ii
EI
hM
EI
hF
θ=+−
2
2
(1.29)
where h is the element length equal to L/2. In the matrix form, the equations can be written as,
⎭
⎬
⎫
⎩
⎨
⎧
=
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
−
−
i
i
i
i v
M
F
EI
h
EI
h
EI
h
EI
h
θ
2
23
2
23
(1.30)
Inverting the above equations, we obtain,
⎭
⎬
⎫
⎩
⎨
⎧
=
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
i
i
i
i
M
Fv
hh
h
h
EI
θ23
46
612
(1.31)
Comparing this with (1.27):
h
EI
k
h
EI
kk
h
EI
k
4612
2222112311 ==== (1.32)
In order to derive other terms of first two columns of (1.26), we make use of following equations
of equilibrium:
Fi + Fj = 0 (1.33)
Mi + Mj – Fi h =0 (1.34)
Third equation of (1.26) gives:
k31vi + k32 θi = Fj = -Fi = -(k11vi+k12θi) (1.35)
Hence,
k31 = - k11 k32 = - k12 (1.36)
From fourth equation we get
k41vi+k42 θi = Mj = -Mi +Fi h = -( k21vi + k22 θi) + ( k11vi + k12 θi )h (1.37)
Solving this we get
k41=6EI/h2
and k42 = 2EI/h2
(1.38)
To obtain other elements of the matrix, node i is fixed, then the third and fourth equations of
(1.26) reduce to
9

(1.39)
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧
=
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
j
j
j
j
M
Fv
kk
kk
θ4443
3433
Element then becomes a cantilever beam loaded by a vertical force Fj and moment Mj at one
end. The deflection and slope of that end can be obtained from elementary strength of materials.
They are given by the following equations:
j
jj
v
EI
hM
EI
hF
=+
23
23
(1.40)
j
jj
EI
hM
EI
hF
θ=+
2
2
(1.41)
In the matrix form, the equations can be written as,
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧
=
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
j
j
j
j v
M
F
EI
h
EI
h
EI
h
EI
h
θ
2
23
2
23
(1.42)
Inverting above equation, we obtain,
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧
=
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
−
−
j
j
j
j
M
Fv
hh
h
h
EI
θ23
46
612
(1.43)
Comparing this with (1.39):
h
EI
k
h
EI
kk
h
EI
k
4612
4424334333 =−=== (1.44)
Similarly, from equilibrium consideration, we can obtain
k13 = -12EI/ h3
k14 = - k23 = 6EI/h2
k24= 2EI/h2
(1.45)
Thus, the elemental stiffness matrix is given by,
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−−−
−
−
22
22
3
4626
612612
2646
612612
hhhh
hh
hhhh
hh
h
EI
(1.46)
The resulting stiffness matrix is exact, not approximate, for the given problem.
1.3.3 Assembly procedure
10

In order to perform the assembly, elemental equations can be written in global form.
First elemental equation in global coordinate system is,
( )
( )
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
=
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−−−
−
−
0
0
000000
000000
004626
00612612
002646
00612612
1
2
1
2
1
1
3
3
2
2
1
1
22
22
3
M
F
M
F
v
v
v
hhhh
hh
hhhh
hh
h
EI
θ
θ
θ
(1.47)
Here, superscript (1) on forces and moments indicate contribution from element 1. Second
elemental equation in global coordinates is
( )
( )
( )
( )
⎪
⎪
⎪
⎪
⎭
⎪
⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎨
⎧
=
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−−−
−
−
1
3
1
3
2
2
2
2
3
3
2
2
1
1
22
223
0
0
462600
61261200
264600
61261200
000000
000000
M
F
M
F
v
v
v
hhhh
hh
hhhh
hh
h
EI
θ
θ
θ
(1.48)
Adding this system of equations, following global system of equations is obtained:
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
=
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−−−
−++−
−+−+−−
−
−
3
3
1
1
3
3
2
2
1
1
22
2222
22
3 0
462600
61261200
26446626
612661212612
002646
00612612
M
F
P
M
F
v
v
v
hhhh
hh
hhhhhhhh
hhhh
hhhh
hh
h
EI
θ
θ
θ
(1.49)
Note that,
F2
(1)
+ F2
(2)
= P and M2
(1)
+ M2
(2)
= 0 (1.50)
1.3.4 Boundary conditions and solutions
It can be verified that the rank of assembled global stiffness matrix is 4. Hence,
minimum two essential boundary conditions are required. However, in this case, we have four
essential (geometric) boundary conditions:
v1=θ1= v3= θ3=0 (1.51)
11

Hence unknowns for the problem are v2 and θ2 and we need only two equations. We choose third
and fourth equations of equation (1.49) as the right hand side of these equations is known to us.
After substituting the values of prescribed degrees of freedom, these equations reduce to,
⎭
⎬
⎫
⎩
⎨
⎧
=
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
080
024
2
2
23
Pv
hh
EI
θ
(1.52)
Solving this, we get
( )
EI
PL
EI
LP
EI
Ph
v
19224
2/
24
333
2 === and θ2=0 (1.53)
Reader can verify that same values are obtained from elementary strength of materials.
1.3.5 Post-processing
By finite element analysis, we get nodal deflections and slope. The task of post-
processing is to find out the slopes and deflection at any point of the beam and shear force and
bending moment. Knowing the shear force and bending moment at any section of the beam, the
stresses can be calculated. In Section 1.2.5, it was suggested that the displacement at any point
inside the element can be found by linear interpolation of the nodal displacements. Many a
times, students do the mistake of linearly interpolating the nodal deflections in a beam problem
too. If you do this, you are not making use of the information of nodal slopes. With slopes and
defections known at the nodes, the displacement can be expressed as a cubic polynomial in an
element. The deflection at a point can be found by evaluating the value of the cubic polynomial
at that point. The slope at a point can be found by finding out the value of the first derivative of
the cubic polynomial. For bending moment calculation, second derivative and for shear force the
third derivative of the cubic polynomial is to be calculated.
1.4 CONCLUSIONS
In this chapter, finite element method has been introduced by taking the one-
dimensional problems as examples. For two and three-dimensional problems, methodology is
similar. As the equations are developed element by element and then assembled, incorporation
of non-homogeneous material properties becomes quite easy. The objective of the present
chapter is to expose the reader with the FEM and many details have been omitted.
We are trying to understand FEM as a tool to solve differential equations. Thus, the
FEM can be applied to number of engineering problems. Although it originated as a technique of
12

solving elastic structure problem, of late it has been applied to plastic deformation problems
also. It has been widely used for solving heat transfer, fluid mechanics and electromagnetism
problems. In manufacturing area, FEM has been used to model metal forming, metal cutting and
non-traditional machining processes. The problems of dynamics and vibrations are also
successfully solved using finite element method.
REFERENCES
1. Clough, R. W., “The Finite Element in Plane Stress Analysis”, Proc. 2nd
A. S. C. E. conf.
on Electronic Computation, Pittsburgh, Pa., Sept. 1960.
2. Courant, R., “Variational Methods for the Solution of Problems of Equilibrium and
Vibrations,” Bulletin of the American Mathematical Society, Vol. 49, 1943, pp. 1-23.
3. Turner, M. J., Clough R. W., Martin, H. C. and Topp, L. J., “Stiffness and Deflection
Analysis of Complex Structures”, J. Aero. Sci., Vol. 23, 1956, pp. 805-823.
EXERCISE 1
Q.1: A short rod of length l is rigidly supported at both ends and an axial load P is applied at the
mid-length. Taking 2 equal finite elements, find out the displacement at the point of application
of load. Also find out the support reactions. The cross-sectional area of the rod is A and Young’s
modulus of elasticity E.
Figure: Q1
Q.2: The short rod (cross-sectional area A, Young’s modulus of elasticity E) shown in figure is
fixed at one end, the other end being held by a spring of spring constant k. A load P is applied at
the mid length. Using direct finite element formulation, find out the spring compression. (Solve
by two methods. In the first method, take 2 elements in the rod and put spring force as the
natural boundary condition. In the second method, taking 2 elements in the rod and treating
spring as the third element apply essential boundary conditions at the both ends.
13

Figure: Q2
Q.3: A cantilever beam of length l, second moment of inertia I and Young’s modulus of
elasticity E is loaded by a load P. Take only one finite element and find out the deflection and
slope at the free end. Compare it with the solution obtained using strength of material’s
approach. Using the fact that deflection of any point of this beam is a cubic polynomial in x (the
distance of the point from the fixed end), find out the deflection at a distance of l/2 from the
fixed end.
Figure: Q3
Q.4: Fourier’s law of heat conduction in a rod gives:
d
d
T
q kA
x
= −
where k is the thermal conductivity, A is the area of the rod and T is the temperature. Using
direct FEM approach, obtain the elemental stiffness and right hand side (load) vector for solving
1-dimesional heat conduction problem. For the rod shown below, find out the temperature at
node 2 by taking 2 elements. The rod is made of steel having the thermal conductivity 50 W/m-
K.
14

Figure: Q4
Q.5: One end of a steel rod is fixed and other end is pulled by an unknown force F. It is known
that due to application of the load the mid-length point of the rod moves by a distance of 0.2
mm. Using FEM with 2 equal elements, find out the value of unknown force. The length of the
rod is 1 m, area 1cm2
and Young’s modulus of elasticity 200 GN/m2
. Note that in this problem
you know only one boundary condition and in lieu of the other boundary condition you have the
information about mid-point displacement.
Figure: Q5
Q.6: Using direct FEM approach, find out the currents in all resistors of the circuit shown below.
Treat each resistor as one element and potentials at the two ends as primary variables. Elemental
equations can be derived using Ohm’s law and assembly can be carried out using Kirchhoff’s
current law.
Figure: Q6
15

Q.7: Two beams of length l each are joined by a pin joint and then combined beam is subjected
to a load P. Can you find the deflection at the load using FEM? Do you have to make any special
comment about this problem?
Figure: Q7
Q.8: A cantilever beam of length is supported on a spring of spring constant k at its free end.
Using FEM, find out the deflection of the beam if a load P is applied at the mid-length of the
beam.
Figure: Q8
Problems requiring the use of computer:
Q.9: The cross-sectional area of a rod varies as
0( ) 2
x
A x A
l
⎛ ⎞
= −⎜ ⎟
⎝ ⎠
where A0 is the area at the fixed end, x is the longitudinal displacement from the fixed end and l
is the length of the rod. A load P is applied at the free end and you have to find the displacement
at free end using FEM. Solve this problem 10 times by discretizing the rod in 1 to 10 elements.
In each element average area of the element should be taken. Plot the obtained displacement
versus number of elements and comment on convergence. Take suitable numerical values for
plotting the graph. Otherwise express the displacement in some non-dimensional form.
16

Figure: Q9
Q.10: One dimensional seepage through a porous medium is governed by Darcy’s law, which
gives the flow in terms of the gradient of the total potentialφ. The law is similar to Fourier’s law
of heat conduction, i.e.,
q kA
x
φ∂
= −
∂
where q is the flow, k is the permeability coefficient and A is the cross-section area of the porous
medium. In the problem shown in figure, potentials on the two sides the porous medium is h1
and h2. The thickness of the porous medium is t, and permeability coefficient on left and right
sides is kl and kr, variation being linear across thickness. Solve this problem using FEM. Study
the convergence by taking various numbers of elements.
Figure: Q10
17

Chapter 2
INTRODUCTION TO CALCULUS OF VARIATION
(Lectures 4-5)
2.1 INTRODUCTION
In the previous chapter, we introduced finite element method as a method to solve
differential equations. More often, the behavior of a physical system is described by governing
differential equations. However, sometimes, it is convenient to derive an integral expression
(called variational form), minimization or maximization of which leads to same solution as
obtained by solving the governing differential equations. Given a variational form, one can
obtain the governing differential equations and solve differential equations by suitable numerical
method including FEM. Differential equations can also be transformed into a variational form.
The branch of mathematics, which deals with transforming a variational form to differential
equation form and vice versa is called calculus of variation. We will study the necessary
techniques of calculus of variation in this chapter. Similar to calculus, where we are often
interested in finding out a point at which a function attains minimum/maximum value, in
calculus of variation, we find out the function that provides minimum/maximum value of the
variational form. This chapter will provide a brief introduction to calculus of variation.
2.2 FUNCTIONAL
In calculus, we come across functions. A function provides a dependent variable, whose
value depends on one or many independent variables. For example, y = f (x) = x3
is a function, in
which for each value of independent variable x, there is a scalar value of dependent variable y.
Similarly, in function z = x2
+ y2
, for each value of x and y, there is a value of z.
Now consider the definite integral
( )
5 5
0 0
dI ydx f x x= =∫ ∫ (2.1)
Here, for each particular function, there is a scalar value of I. For example, when f(x) = 1, the
value of I is 5. When f(x) = x, the value of I becomes 2.5. In literature, I is called a functional
(function of function), whose value depends on independent function f (x). Following integral
expression is also a functional:
( ) ( ), , d
b
a
I y F x y y x′=∫ (2.2)
19

where F depends on x, y ≡ y(x) and y′ ≡
d
d
y
x
. Mathematically, a functional is an operator I,
mapping y into a scalar value.
A functional l(y) is said to be linear in y, if and only if, it satisfies the following relation:
( ) ( ) ( )l y z l y l zα β α β+ = + (2.3)
for any scalars, α and β and independent functions y and z. A functional B(y, z) is said to be
bilinear, if it is linear in each of its argument y and z, i.e.,
( ) ( ) (1 2 1 2, ,B y y z B y z B y zα β α β+ = + ), (2.4)
( ) ( ) ( )1 2 1, ,B y z z B y z B y zα β α β+ = + 2, (2.5)
The functional is symmetric if
( ) ( ),B y z B z y= , (2.6)
An example of a linear functional is
( ) 0
d
L
I u u f= ∫ x (2.7)
where f is a constant function and u is the independent variable function. An example of a
symmetric bilinear functional is
( ) 0
d d
,
d d
L u v
dI u v E A x
x x
= ∫ (2.8)
where E and A are constant functions and u and v are independent variable function.
2.3 EXTREMIZATION OF A FUNCTIONAL
Let
d
, , d
d
b
a
y
I F x y
x
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
∫ x be some functional. The relation between y and x is not known and
the problem consists of finding this relation so that I is a maximum or a minimum. Assume that
y0(x) is some known relation between y and x, which extremizes I. Let another function in the
neighborhood of y0(x) is denoted by y0(x) + εη(x), where η(x) is an arbitrary (but sufficiently
differentiable) function of x and ε is an arbitrary small quantity. The function η(x) does not
violate the geometric boundary conditions of the problem. Hence, wherever y is prescribed η is
zero. The function εη(x) is called δy, the variation of y at a given x, δ being a variational
operator. Variational operator δ is in many ways similar to differential operator d and has similar
type of mathematical properties. However, they are conceptually different. Differential of a
function dy is a first order approximation to the change in function along a particular curve,
20

while the variational of a function δy is a first order approximation to the change from curve to
curve.
Figure 2.1: Variation of a function
Figure 2.1 shows the plot of function y0 with solid line. Assume that the value of
function at one end point is prescribed, then a general function y0(x)+εη(x) is shown by a dotted
curve. If we put the general function in the functional, I will be obtained as a function of ε. The
condition for extremum of this function is,
d
0
d
I
ε
= (2.9)
However, if y0 itself is extremum, then above condition should hold good at ε = 0, i.e.,
0
d
0
d
I
εε =
= (2.10)
Replacing y by y0(x) + εη(x) in functional I,
( )0 0, ,
b
'
a
dI F x y y xεη εη′= + +∫ (2.11)
where a dash in the superscript indicates differentiation with respect to x. At a fixed value of x,
one can write using Taylor’s theorem,
( )
( )
( )
( )'
0 0
2 2
2 2 2
2 2
, , ( , , ) ...
2! 2!
F F F F F
F x y y F x y y
y y y y y y'
εη εη
εη εη εηεη
⎛ ⎞
⎜ ⎟
⎝ ⎠
′∂ ∂ ∂ ∂ ∂
′ ′ ′= + + + + +
′ ′∂ ∂ ∂ ∂ ∂ ∂
+
(2.12)
where all the derivatives are evaluated at y0 and 0
'
y . Note that while expanding F by Taylor
series, we treat x as fixed and y and as two independent variables. Once x is fixed, y and
become variables instead of function and expression (2.12) is possible. Integrating the above
expression between a and b, and taking the derivative with respect to ε,
y'
y'
21

2 2 2
2
2 2
d
0 2 ... d
d
b
a
I F F F F F
x
y y y y y y
η η ε η ε ηη ε
ε
⎧ ⎫⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂ ∂⎪ ⎪
′ ′= + + + + + +⎨ ⎬⎜ ⎟⎜ ⎟′ ′∂ ∂ ∂ ∂ ∂ ∂⎪ ⎪⎝ ⎠ ⎝ ⎠⎩ ⎭
∫ ′∫ (2.13)(2.13)
Applying the condition for extremum, we getApplying the condition for extremum, we get
0
0
b
a
I F Fd
d
d
x
y yε
η η
=
⎛ ⎞∂ ∂
ε
′= = +⎜ ⎟′∂ ∂⎝ ⎠
∫ (2.14)
The
0
d
d
I
εε =
is also called the first variation of I, δI. Thus, the condition for extremizing a
functional is δI=0. is Similarly,
2
2
0
d
d
I
ε
ε =
is called the second variation of I, δ2
I, which can tell if
the function is minimized, maximized or neither minimized nor maximized. Integrating the right
hand side of the above equation by parts, so as to reduce the order of derivative of η, Eq. (2.14)
gets transformed to
d
d
d
b
b
a
a
F F F
x
y x y y
η η
⎧ ⎫⎛ ⎞∂ ∂ ∂
− +⎨ ⎬⎜ ⎟′ ′∂ ∂ ∂⎝ ⎠⎩ ⎭
∫ (2.15)
Thus,
d
d
d
b
a
b a
F F F F
I x
y x y y y
δ η η
⎧ ⎫⎛ ⎞∂ ∂ ∂ ∂
0η= − + −⎨ ⎬⎜ ⎟′ ′ ′∂ ∂ ∂ ∂⎝ ⎠⎩ ⎭
∫ = (2.16)
If the value of y is prescribed at a boundary, η at that boundary will be zero as there is no
variation at that point. At other places η is arbitrary. We can also put in Eq. (2.16), an arbitrary
η, which is 0 at both the boundaries. In that case,
2
1
d
d 0
d
x
x
F F
x
y x y
η
⎛ ⎞⎛ ⎞∂ ∂
−⎜ ⎟⎜ ⎟′∂ ∂⎝ ⎠⎝ ⎠
∫ = (2.17)
In view of the η being arbitrary, Eq. (2.17) implies that
d
0
d
F F
y x y
⎛ ⎞∂ ∂
− =⎜ ⎟′∂ ∂⎝ ⎠
(2.18)
Thus, extremization of the functional I requires the satisfaction of the above differential
equation. Substituting Eq. (2.18), in Eq. (2.16), we have
0
b a
F F
y y
η η
∂ ∂
− =
′ ′∂ ∂
(2.19)
At a particular boundary, say at point a, either η is zero or arbitrary and can be made zero. Thus,
0
b
F
y
η
∂
=
′∂
(2.20)
22

In the same way, it can be shown that
0
a
F
y
η
∂
=
′∂
(2.21)
Eqs. (2.20-2.21) are the two boundary conditions, which must be satisfied along with the
differential equation given by Eq. (2.18) for the extremization of the functional. Boundary
conditions imply that at the boundaries either η should be 0 i.e. the value of y should be
prescribed or F
y
∂
′∂
should be zero. The first type of boundary condition is called geometric or
essential boundary condition, whilst the second type of boundary condition is called natural
boundary condition. The Eq. (2.18) is called Euler-Lagrangian equation.
Example 2.1: Find out the Euler-Lagrangian equation, which extremizes the following
functional:
( )
1
2 2
0
2 dI y y xy′= + +∫ x (2.22)
Solution: Using Eq. (2.18), we have
( )
d
2 2 2 0 . .,
d
y x y i e y x y 0
x
′ ′′+ − = + − = (2.21)
As
2
F
y
y
∂
′=
′∂
(2.22)
The boundary conditions are:
Either = 0 or y is prescribed at x = 0 and 1 (2.23)y'
Note that the variational form of the differential equation (2.21) does not depend on the
prescribed value of y at the boundaries.
If F depends on several dependent variable, i.e.
where each yi = yi (x), the analysis proceeds as before, leading to n separate but simultaneous
equations for the yi(x),
1 1 2 2( , , , ,......., , , )' ' '
n nF F y y y y y y x=
d
0, 1,......, .
d '
i i
F F
i
y x y
⎛ ⎞∂ ∂
− = =⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠
n (2.24)
with corresponding boundary conditions. With n independent variables, we need to extremise
multiple integrals of the form
1 1
1
..... ( , ,......, , ,......., )d .....dn
n
y y
nI F y x x x x
x x
∂ ∂
= ∫ ∫
∂ ∂
(2.25)
23

Using the same kind of analysis as before, we find that the extremising function y = y(x1,
………..,xn) must satisfy
1
0
i
n
i i x
F F
y x y=
⎛ ⎞∂ ∂ ∂
− =⎜ ⎟∑
⎜ ⎟∂ ∂ ∂⎝ ⎠
(2.26)
where ixy stands for
i
y
x
∂
∂
.
We will now derive the necessary differential equation for extremizing
( , , , )d
b
a
I F x y y' y'' x= ∫ . We can apply the procedure adopted for extremizing
d
, , d
d
b
a
y
I F x y
x
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
∫ x , however, we will now follow a short but not so rigorous approach. We will
now make use of the variational operator δ, and the fact that this operator behaves in the same
way as a differential operator. We also make use of the following two properties:
Property 1: Differentiation and variation commute i.e., ( )y ' = y'δ δ .
Proof: We can write,
( ) ( )
d d
( )
d d
y ' = y
d
dx x x
η
δ δ εη ε= = (2.27)
and
d d d d d d d
( ) ( )
d d d d d d d
y y y
y' y
y
x x x x x x x
η η
δ δ εη ε ε
⎛ ⎞
= = + − = + − =⎜ ⎟
⎝ ⎠
(2.28)
Both expressions are equal, hence proved.
Property 2: Integration and variation commute
i.e. ( ) ( )d db b
a ay x x y xδ δ=∫ ∫
Proof:
( ) ( ) ( )d db b b
a a ay x x y x y xδ εη= + −∫ ∫ ∫ d
d( )d db b b
a a ay x x y xεη= + −∫ ∫ ∫
( )d db b
a ax y xεη δ= =∫ ∫
24

Now, using second property, we can write
( , , , db
a )I F x y y y xδ δ ′ ′′= ∫ (2.29)
The first variation of F is given by
( ) ( ) (, , , , , , , , ,F x y y y F x y y y n F x y y yδ εη εη ε′ ′′ ′ ′ ′′ ′′ ′ ′′= + + + − ) (2.30)
where ε is a small quantity. The expansion using Taylor series leads to
( )2F F F
F y y y O
y y y
δ δ δ δ
∂ ∂ ∂
′ ′′= + + +
′ ′′∂ ∂ ∂
ε (2.31)
Noting that yδ εη≡ etc. As 0ε → , we can write
F F F
F y y
y y y
yδ δ δ δ
∂ ∂ ∂
′ ′′= + +
′ ′′∂ ∂ ∂
(2.32)
Thus,
'
'
b
a
F F F
dI y y y
y y y
δ δ δ δ
⎛ ⎞∂ ∂ ∂
x′′= + +∫ ⎜ ′′∂ ∂ ∂⎝ ⎠
⎟ (2.33)
Making use of the property 1, we can write
2
2
d d
( ) ( ) d
' d '' d
b
a
F F F
I y y y
y y x y x
δ δ δ δ
⎛ ⎞∂ ∂ ∂
= + +⎜ ⎟∫ ⎜ ⎟∂ ∂ ∂⎝ ⎠
x
(2.34)
Integrating the second and third terms by parts,
d d d d
( ) ( ) ( ) d ( )
d ' d '' d ' d
b bb
a a a
F F F F F
I y y y x y
y x y x y x y y'' x
δ δ δ δ δ δ
⎛ ⎞∂ ∂ ∂ ∂ ∂
= − − + +∫ ⎜ ⎟
∂ ∂ ∂ ∂ ∂⎝ ⎠
y (2.35)
Integrating the third term by parts,
2
2
d d
d
d ''
y x
d
b
a
F F F
I y y
y x y' yx
δ δ δ δ
⎧ ⎫⎛ ⎞ ⎛ ⎞∂ ∂ ∂⎪ ⎪
= − +∫ ⎨ ⎬⎜ ⎟ ⎜ ⎟
∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎪ ⎪⎩ ⎭
+
d
' d ''
b b bF F F
y y' y
a a ay y'' x y
δ δ δ
∂ ∂ ∂
+ −
∂ ∂ ∂
(2.36)
At this point, note the point that if you had adopted the procedure of putting
0 ( ) ( )y y x xεη= + in the functional, then the condition for extremum would have been
0
d
0
d
b
a
I F F F
dx
y y yε
η η η
ε =
⎛ ⎞∂ ∂ ∂
′ ′′= = + +⎜ ′ ′′∂ ∂ ∂⎝ ⎠
∫ ⎟ (2.37)
25

Multiplying the integral in equation (2.37) by ε, we can see that this is same as equation (2.33).
Thus it is seen that the necessary condition for extremization is 0Iδ = . We can argue it in a
different way also. Assume that 0Iδ ≠ . In that case it is possible to increase and decrease the
functional by giving a small variation. The extremum is reached when it is not possible to
increase or decrease the functional by giving a small variation. Note also that if a function can be
increased by giving a small variation, it can be decreased also by giving a variation in opposite
direction. Thus, the first variation must vanish for extremization.
Applying the necessary condition for extremization, we get
2
2
d d d
( ) ( ) d ( ) 0
d ' '' ' dd
b
a
b bF F F F F F
I y x y
a ay x y y y x y'' y''x
δ δ δ
⎡ ⎤ ⎧ ⎫⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂
= − + + − + =⎢ ⎥∫ ⎨ ⎬⎜ ⎟
∂ ∂ ∂ ∂ ∂ ∂⎢ ⎥ ⎝ ⎠⎩ ⎭⎣ ⎦
y'δ (2.38)
Thus, the differential equation is
2
2
d d
0
d ' ''d
F F F
y x y yx
⎛ ⎞ ⎛ ⎞∂ ∂ ∂
− +⎜ ⎟ ⎜ ⎟
∂ ∂ ∂⎝ ⎠ ⎝ ⎠
= (2.39)
The boundary conditions are
d
( )
' d ''
F F
y
y x y
δ
⎧ ⎫∂ ∂
0− =⎨ ⎬
∂ ∂⎩ ⎭
at x = a and at x = b (2.40)
and
' 0
''
F
y
y
δ
∂
=
∂
at x = a and at x = b (2.41)
The equation (2.40) suggests that at a boundary point, we can have either δ y = 0 or the quantity
in curly bracket equal to 0. The former is called the essential (or geometric) boundary, whilst the
latter is called the natural boundary condition. Similarly in equation (2.41), the boundary
condition ' 0yδ = is called essential boundary condition and /F y′′∂ ∂ = 0 is natural boundary
condition. By convention, the boundary conditions associated with the variational operator are
always called essential boundary condition and other boundary conditions are called natural
boundary conditions. Note that ' 0yδ = does not mean that the slope is zero. It means that the
slope has been prescribed.
Example 2.2: Total potential energy of a beam of length l is loaded by a load of intensity q(x) is
given by:
26

2
0
1
'' d
2
l
I EIw qw x
⎛
= −∫ ⎜
⎝ ⎠
⎞
⎟ (2.42)
where E and I are the Young’s modulus of elasticity and second moment of area about the
perpendicular to the plane of bending respectively and are a known function of x, and w is the
unknown beam deflection function. Find out the governing differential equation and boundary
condition.
Solution: Minimization of the total potential energy will lead to finding out the beam deflection
under load intensity q.
Eq. (2.39 ) gives
2
2
d
0 ( '') 0
d
q EIw
x
− − + =
or
2
2
d
( '')
d
EIw q
x
= (2.43)
which is the well known Euler-Bernoulli beam equation.
The boundary conditions are
[
d
0 ( '')]
d
EIw w 0
x
δ− = at x=0, l (2.44)
and
'' ' 0EIw wδ = at x = 0, l (2.45)
The term
d
( '')
d
EIw
x
represents the shear force and the bending moment. We will not
talk about the sign convention at this stage. Thus, at both the ends of the beam, two sets of the
boundary conditions need to be satisfied:
EIw''
Set1: either the shear force is zero or the differential is prescribed.
Set2: either the bending moment is zero or the slope is prescribed.
27

The boundary conditions concerning the slope and deflection are called ‘geometric’ or
‘essential’ boundary conditions. Boundary conditions involving shear and bending moment are
called ‘natural’ or ‘force’ boundary conditions. Boundary conditions for three differently
supported beams are shown in Fig. 2.2.
( )
0
d
0
d
''
''
EIw
EIw
x
=
= ( )
0
d
0
d
''
''
EIw
EIw
x
=
=
Free-free beam
0
0''
w
EIw
=
=
0
0''
w
EIw
=
=
Simply-supported beam
0
0'
w
w
=
=
d
( )
d
0
''
''
EIw
x
EIw
=
=
0
Cantilever beam
Figure 2.2: Boundary conditions for 3 different types of beams
2.4 OBTAINING THE VARIATIONAL FORM FROM A DIFFERENTIAL
EQUATION
Let us learn the procedure of converting a differential equation into a variational form.
Consider the differential equation
( ) 0L fφ − = (2.46)
where L is linear or non linear differential operator, φ is a scalar function defined over the
domain D and f is a known scalar function. Multiplying the equation by a variation of φ and
integrating it over the domain
28

[ ( ) ] d 0
D
L f Dφ φ− δ =∫ (2.47)
We keep on manipulating equation (2.47) using integration by parts till we are able to put it in
the variational form
[ * ( ) ]d 0
D
L f Dδ φ φ− =∫ (2.48)
Then, we say that the variational form of the given differential equation is
( ) [ * ( ) ]d
D
I L f Dφ φ φ= −∫ (2.49)
In the process, the order of derivatives gets reduced. Generally, in a particular term, we attempt
to keep the order of derivative on δφ and on the assocaited expression same. The procedure will
be clear after seeing the Examples (2.3-2.5).
Example 2.3: The governing differential equation for a rod loaded with axial force is
d d
0
d d
u
EA q
x x
⎛ ⎞
+ =⎜ ⎟
⎝ ⎠
(2.50)
where E is Young’s modulus of elasticity, A is the cross-sectional area, q is the load intensity
(load per unit length in axial direction) and u is the axial displacement as a function of axial
coordinate x. Obtain the variational form of this equation. Assume that the boundary conditions
are
d
At =0, 0; at = ,
d
u
x u x l EA P
x
= = (2.51)
where l is the length of the rod.
Solution: As a first step, multiply the above governing differential equation by δu and integrate
between 0 to l. Thus,
0
d d
d 0
d d
l u
EA q u x
x x
δ
⎡ ⎤⎛ ⎞
+ =⎜ ⎟⎢ ⎥
⎝ ⎠⎣ ⎦
∫ (2.52)
Integrating equation (2.51) by parts,
( )
0 0 0
d d d
d
d d d
l l l
u u
u EA u q x u EA x
x x
δ δ δ d 0
x
+ −∫ ∫ = (2.53)
As the variational operator behaves like a differential operator and δu at x=0 is 0, we can write
29

2
0 0
1 d d
d ( )d
2 d d
l l
l
u u
EA x qu x EA u
x x
δ δ δ
⎡ ⎤⎛ ⎞ ⎛ ⎞
0− −⎢ ⎥⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
∫ ∫ = (2.54)
Therefore,
2
0
1 d
d ( )
2 d
l u
EA qu x P u l
x
δ
⎡ ⎤⎡ ⎤⎛ ⎞
0− −⎢ ⎥⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦
∫ = (2.55)
Hence, the variational form is given by
2
0
1 d
d
2 d
l
u
( )I EA qu x Pu l
x
⎡ ⎤⎛ ⎞
= −⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦
∫ − (2.56)
The reader may observe that this is the total potential energy of the rod. Thus, the displacement
function of the rod will be one which extremizes the total potential energy amongst the functions
that satisfy the essential boundary condition i.e., u=0 at x=0.
Example 2.4: The heat conduction in a rod without heat generation is governed by
2
2
d
0
d
T
k
x
= (2.57)
where k is the thermal conductivity and T is the temperature. Assume that the temperatures at the
two ends of the rod are prescribed. Obtaining the variational form of the problem.
Solution: Multiplying the differential equation by δT and integrating between 0 and l (length of
the rod),
2
20
d
d = 0
d
l T
k T x
x
δ∫ (2.58)
Integrating by parts
( )0
0
d d d
d = 0
d d d
l
lT
k T T k x
x x x
δ − δ∫
T
(2.59)
Making use of the fact that variation of the temperature at the ends is 0 and the properties of the
variational operator, we can write
0
1 d d
d
2 d d
l
T T
k
x x
⎛ ⎞
0xδ =⎜ ⎟
⎝ ⎠
∫ (2.60)
or
30

0
1 d d
d
2 d d
l
T T
k x
x x
⎛ ⎞
0δ =⎜ ⎟
⎝ ⎠
∫ (2.61)
Hence, the variational form is given by
2
0
1 d
d
2 d
l
T
I k
x
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
∫ x (2.62)
Thus, for finding out the temperature distribution, extremize equation (2.62) while satisfying the
essential boundary condition at the ends. Unlike in the case of rod subjected to axial load, we
cannot assign any name to the functional I in equation (2.62).
Example 2.5: Steady-state heat conduction in a isotropic and homogeneous plate, in which the
temperature across thickness direction remains constant is governed by the following differential
equation:
2 2
2 2
0
T T
x y
∂ ∂
+ =
∂ ∂
(2.63)
Assume that the temperatures at the edges have been prescribed. Obtain the variational form of
the problem.
Solution: We first multiply the differential equation by δT and integrate it over the domain.
Thus,
2 2
2 2
d
A
T T
T A
x y
δ
⎛ ⎞∂ ∂
0+ =⎜ ⎟
∂ ∂⎝ ⎠
∫ (2.64)
Now, we can integrate this equation by parts to reduce the order of derivative; however, it is
better to make use of divergence theorem. Thus, the equation (2.64) can be written as
( ) ( )2
d . d . d
A A A
I T T A T T A T T Aδ δ δ δ= ∇ = ∇ ∇ − ∇ ∇ =∫ ∫ ∫ 0 (2.65)
Applying divergence theorem,
( ) ( )ˆ. d . d
A
I T n T T T A
Γ
δ δ Γ δ= ∇ − ∇ ∇ =∫ ∫ 0 (2.66)
where Γ is the boundary of the domain. As the temperature is prescribed in the boundary, δT
becomes 0 there. Thus, equation (2.66) becomes
( ). d d
A A
T T T T
I T T A A
x x y y
δ δ
δ δ
⎛ ⎞∂ ∂ ∂ ∂
= − ∇ ∇ =− + =⎜
∂ ∂ ∂ ∂⎝ ⎠
∫ ∫ 0⎟ (2.67)
Using the properties of the variational operator, we can write
31

22
1
d
2A
T T
I
x y
δ δ
⎧ ⎫⎛ ⎞∂ ∂⎪ ⎪⎛ ⎞
=− + =⎨ ⎬⎜ ⎟⎜ ⎟
∂ ∂⎝ ⎠ ⎝ ⎠⎪ ⎪⎩ ⎭
∫ 0A (2.68)
Thus, the variational form is
22
1
d
2A
T T
I A
x y
⎧ ⎫⎛ ⎞∂ ∂⎪ ⎪⎛ ⎞
=− +⎨ ⎜ ⎟⎜ ⎟
∂ ∂⎝ ⎠ ⎝ ⎠⎪ ⎪⎩ ⎭
∫ ⎬ (2.69)
2.5 PRINCIPLE OF VIRTUAL WORK
We obtained the variational form of the rod subjected to axial loading by converting the
differential equation into the variational form. We could have obtained it using the principle of
virtual work. According to it, when we subject a loaded body in equilibrium to small compatible
virtual displacements (which do not violate the essential boundary conditions), the total internal
virtual work is equal to the total external virtual work. The internal virtual work per unit volume
will be equal to the product of real stresses and virtual strain, which can be integrated to yield
the total internal virtual work. Note that virtual displacement are imaginary small displacements,
which when applied do not cause any change in the forces are stresses. During the application of
virtual displacements, the stresses can be considered constant. That is why we are able to take
internal virtual work per unit volume equal to the product of real stresses and virtual strain. The
virtual strain are found by taking the derivatives of the virtual displacement function. Therefore,
the virtual displacement function must be differentiable. The external work will be the
summation of works done by all forces acting on the body when subjected to virtual
displacements.
If the rod of Example (2.3) is subjected to virtual displacement δu, the internal work will be
( )int 0
d d
d
d d
l u
W u EA x
x x
δ= ∫ (2.63)
and the external work will be
0
d
l
extW u q x P uδ δ= +∫ ( )l (2.64)
Applying the principle of virtual work
( )int
0
0
d d
d d ( )
d d
l
l
ext
u
W W u EA x u q x P u l
x x
δ δ δ− = − − =∫ ∫ 0 (2.64)
The equation (2.64) is called integral form of the problem. If we treat δ as the variational
operator, we can easily obtain equation (2.56) i.e., the variational form of the problem.
32

2.6 PRINCIPLE OF MINIMUM POTENTIAL ENERGY
In the process of obtaining the variational form of the rod problem, we have obtained a
functional whose extremization alongwith the satisfaction of essential boundary condition will
provide us the solution. We indicated that this functional is actually the total potential energy of
the rod composed of (a) the strain energy of elastic distortion, and (b) the potential possessed by
applied loads. We could have written this functional using the principle of minimum potential
energy for the stable conservative mechanical system. For a conservative mechanical system,
one can express the energy content of the system in terms of its configuration, without reference
to whatever deformation history or path may have led to the configuration.
The statement of principle of minimum potential energy is as follows. Among all
possible configurations of a conservative system satisfying internal compatibility and essential
boundary conditions, those that keep the body in stable equilibrium make the potential energy
minimum with respect to small admissible variations of displacement. This principle is
applicable evenif the material behavior is non-linear.
The reader may have a doubt how one can know that the functional in equation (2.56) is
to be minimized and not maximized, because the functional was obtained by putting the
necessary condition for extremization. For ascertaining that the functional has to be minimized,
one needs to calculate second variation δ2
I , which we have not done in this chapter. However,
often the form of the functional and the physics of the problem can provide the answer. For
example, we can see that equation (2.56) is unbounded for maximization problem. We can
choose the function u=-M, where M is a large positive number and can make the functional as
large as we wish. Thus, there is no physically realistic solution for a maximization problem.
Therefore, physically realistic solution should correspond to minimization problem.
2.7 CONCLUSIONS
In this chapter, we have briefly introduced calculus of variation. We can model a physical
problem in the form of differential equations or in the form of integral. The differential form is
called strong form, because it contains the higher order derivatives, whilst the integral form
containing lower order derivatives is called the weak form. In most of the physical problems, it
is possible to convert one form into the other. The integral form can also be obtained using
principle of virtual work or principle of minimum potential energy. These principles have been
33

developed for mechanical systems also. Thus, knowledge of calculus of variation will enable
you to obtain the variational form for other physical problems too.
REFERENCES (for further reading, not cited in the text)
1. J.N. Reddy, An Introduction to the Finite Element Method, McGraw-Hill, New York,
1993.
2. R.D. Cook, D.S. Malkus and M.E. Plesha, Concepts and Applications of Finite Element
Analysis, 3rd
ed., John Wiley, New York 1989.
3. K.J. Bathe, Finite Element Procedures in Engineering Analysis, Prentice-Hall, Englewood
Cliffs, NJ 1982.
4. K.F. Riley, M.P. Hobson and S.J. Bence, Mathematical Methods for Physics and
Engineering, Cambridge University Press, Cambridge, 1998.
EXERCISE 2
Q.1: The functional governing static buckling of the column in the figure shown below is
2 22
2
2
0 0
1 1
dx dx +
2 2
L L
2
L
d w P dw
EI kw
dxdx
⎛ ⎞ ⎛ ⎞
∏ = −⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
∫ ∫ (a)
where 0
0
0, 0x
x
dw
w
dx=
=
= = (b)
Invoke the stationary conditions 0δ ∏ = to derive the problem-governing differential
equation and the natural boundary conditions.
Figure: Q1
Q.2: Obtain the variational form of heat transfer problem in a rod.
34

Assume A, k, r are constants.
Figure: Q2
Differential equation:
2
2
d
0
d
T r
x kA
+ =
Boundary conditions:
1. Essential boundary conditions: T = 0 at x = 0
2. Natural boundary conditions 1d
d 2
TT
x
= at x = 1
Q.3: A certain physical problem has the functional
2
,
0
1
50 dx
2
L
xφ φ
⎛ ⎞
∏ = −⎜ ⎟
⎝ ⎠
∫
Essential boundary conditions are 0φ = at x = 0 and x = L. Express φ as a function of x, which
extremizes the functional.
Q.4: The potential energy of an isotropic plate that carries lateral pressure q is
2 2
, , , , ,( ) 2(1 )( ) 2 d
2
xx yy xx yy xyp
D q
w w w w w x
D
ν
⎧ ⎫
= + − − − −⎨ ⎬
⎩ ⎭
∏ ∫∫ dy
where D is a constant called flexural rigidity. Show that the governing differential equation
is , where is the bi-harmonic operator.4
/q D∇ = 4
∇
Q.5: Steady state heat conduction without heat generation is governed by the following
differential equation (for domain with constant thermal conductivity):
35

2 2
2 2
0
T T
x y
∂ ∂
+ =
∂ ∂
If the temperatures at the boundary are specified, obtain the variational form of the above
equation.
Q.6: Consider the following boundary value problem
2
2
d
1 0
d
u
u x 1
x
+ = ≤ ≤
with u(0) = 0 and
d
d
u
x
= 0 at x = 1. Convert this problem into variational form.
Q.7: The bending of a beam is governed by the following differential equation
2 2
2 2
dd d
( )
d d d
z
zz y
mv
EI p
x x x
0+ − =
with essential boundary conditions at x = 0 are v = v*
and *d
d
z
v
x
θ= and natural boundary
conditions at x = l are
2
*
2
d
d
zz z
v
EI M
x
= and
2
*
2
d d
( )
d d
zz z y
v
EI m V
x x
+ = − . Where, is the
distributed moment per unit length about z-axis and
zm
yp is the distributed force per unit length in
y-direction. Obtain the variational functional.
Q.8: Using calculus of variations, show that the shortest curve joining two points is a straight
line.
Q.9: frictionless wire in a vertical plane connects two points A and B, A being higher than B.
Let the position of A be fixed at the origin of an xy-coordinate system, but allow the B to lie
anywhere on the vertical line x = x0. Find the shape of the wire such that a bead of mass m placed
on it at A will slide under gravity to B in the shortest possible time.
Q.10: Consider a mechanical system whose configuration can be uniquely defined by a number
of coordinates (usually distances and angles) together with time t, and which only experiences
forces derivable from a potential. Hamilton’s principle states that in moving from one
configuration at time t1 the motion of such a system is such as to make stationary.
iq
36

1
0
1 1L ( ,..., , ,..., , )d
t
n nt
L q q q q t= ∫ t
The Lagrangian L is defined in terms of the kinetic energy T and the potential energy V (with
respect to some reference situation) by L = T-V. Here V is a function of the only, not of the
. Applying the EL equation to L, we obtain Lagrange’s equations,
iq
iq
d
, 1,..., .
di i
L L
i n
q t q
⎛ ⎞∂ ∂
= =⎜ ⎟
∂ ∂⎝ ⎠
Using Hamilton’s principle, derive the wave equation for small transverse oscillations of a taut
string.
37

Chapter 3
SOME CLASSICAL FUNCTION APPROXIMATION METHODS
FOR SOLVING DIFFERENTIAL EQUATIONS
(Lectures 6 and 7)
3.1 INTRODUCTION
In this chapter, we will study some methods invented before the advent of FEM for
solving the differential equations. Methods discussed in this chapter, approximate the function
globally as a weighted summation of several linearly independent functions. The weights of
various functions are found such that in some sense, the error is minimized.
3.2 RITZ METHOD
The method was introduced by a German physicist and mathematician, W. Ritz in 1908 [1].
It operates directly on the variational form of the differential equation. Therefore, it is also called
direct method in variational problems. If the variational form is given, start solving without
bothering about Euler-Lagarange equation. If differential equation has been provided, you have to
convert it to a variational form. Some persons attach the name of Lord Rayleigh with the method
and call it “Rayleigh-Ritz” method. However, based on the study of history, Leissa has observed
that Rayleigh method is quite different from the Ritz method and therefore Rayleigh’s name
should not be attached with the method [2]. The method works as follows:
• Approximate the function as a weighted sum of linearly independent functions. The
approximated function must satisfy the essential boundary conditions of the problem. The
natural boundary conditions need not be satisfied. The chosen functions must be
complete, in the sense that starting from the constant term, the successively higher degree
terms should be taken (without missing the terms in between) from a series of complete
functions. In the context of polynomials, the series 1, x, x2
, x3
is complete but 1, x2
, x3
, x4
are incomplete. To give an example, if u is a function of x, it can be represented as
1
n
i i
i
u a φ
=
= ∑ (3.1)
where ϕi is the ith
basis function, and ai is the corresponding weight or coefficient.
• In Eq. (3.1), either the functions must be chosen in such a way that they satisfy the
essential boundary conditions, else put the boundary conditions in Eq. (3.1), which will
39

provide linear equations corresponding to each essential boundary conditions. Express
the coefficients equal to number of these equations in the form of other coefficients and
modify Eq. (3.1).
• Put the modified equation that satisfies the essential boundary conditions into variational
form. The variational form will now be a function of the coefficients. Extremize the
variational form. For this purpose take partial derivatives of the variational form with
respect to the coefficients and make them 0. this will give the equations equal to the
number of coefficients, which can be solved to yield the coefficients. Having known the
coefficients, approximate solution can be constructed.
Example 3.1: Consider the following boundary value problem:
2
2
d
1
d
u
u
x
+ = 0 1x≤ ≤ with ( )0u 0= and
d
0
d
u
x
= at 1x = (3.2)
Solve this equation using Ritz’s method.
Solution: First we have to convert it into variational form. For this, let
2
1
0 2
d
1 d
d
u
I u u
x
δ
⎛ ⎞
= + −⎜ ⎟∫ ⎜ ⎟
⎝ ⎠
xδ (3.3)
We write the above expression again, after integrating the first term by parts. Thus,
( )1 1 1
0 00
d d d
d d
d d d
u u 1
0 dI u u x u u x u x
x x x
δ δ δ δ δ
⎛ ⎞
= − + −∫ ∫⎜ ⎟
⎝ ⎠
∫ (3.4)
In view of the boundary conditions and fact that variational and differential operators are
commutative, the expression becomes:
1 1
0 0
d d
d d
d d
u u 1
0 dI x u u x u x
x x
δ δ δ δ
⎛ ⎞
= − + −∫ ∫⎜ ⎟
⎝ ⎠
∫ (3.5)
As the variational operator behaves like a differential operator,
( )
2
1 1 2
0 0
1 d 1
d d
2 d 2
u 1
0 dI x u x
x
δ δ δ δ
⎛ ⎞
= − + −∫ ∫⎜ ⎟
⎝ ⎠
u x∫ (3.6)
or
40

( )
2
1 1 2
0 0
1 d 1
d d
2 d 2
u 1
0 dI x u x u
x
δ δ
⎡ ⎤⎛ ⎞
= − + −⎢ ∫ ∫⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦
x⎥∫ (3.7)
Therefore,
2
1 2
0
1 d 1
d
2 d 2
u
I u u
x
⎧ ⎫⎪ ⎪⎛ ⎞
= − + −∫ ⎨ ⎜ ⎟
⎝ ⎠⎪ ⎪⎩ ⎭
x⎬ (3.8)
Now let us solve this problem using Ritz’s method. If we approximate u by , then in
view of the first essential boundary conditions i.e.
u a bx= +
( )0u 0= , a = 0. Hence approximate u is bx
and
d
d
u
b
x
= . Putting the value of u and
d
d
u
x
in the variational form,
1 2 2 2
0
1 1
d
2 2
I b b x bx
⎛
= − + −∫ ⎜
⎝ ⎠
x
⎞
⎟ (3.9)
Extremize this with respect to b, i.e.
( )1 2
0
d
d
d
I
b bx x x
b
= − + −∫
1
0
3 2
b
b
⎡ ⎤
= − + − =⎢⎣ ⎦⎥ (3.10)
This gives
3
4
b = − . Thus,
3
4
u x= − .
Exact solution of this differential equation is
sin cos 1u A x B x= + + (3.11)
where constants A and B are to be determined from the boundary conditions. Since ( )0 0u = ,
1B = − . Also,
d
cos sin
d
u
A x B x
x
= − (3.12)
At 1x = ,
d
0
d
u
x
= . Hence ( ) ( )cos 1 sin 1A B= , or ( ) ( )tan 1 tan 1A B= = − .Thus,
( )tan 1 sin cos 1u x= − − +x
8
(3.13)
The value of exact u at and 1 are0.5x =
( )0.5 0.6242u = − (3.14)( )1 0.850u = −
The approximate solutions at these points are
( )0.5 0.375u = − ( )1 0.7u = − 5 (3.15)
Now let us add one more term in the approximating function and take .2
u a bx cx= + +
41

Essential boundary condition i.e. ( )0u 0= gives a = 0. Thus,
2
u bx cx= + (3.16)
d
2
d
u
b cx
x
= + (3.17)
( ) ( ) ( )
221 2
0
1 1
2
2 2
2
dI b cx bx cx bx cx x
⎧ ⎫
= − + + + − +∫ ⎨ ⎬
⎩ ⎭
(3.18)
Now, we have to extremize this with respect to b and c. Thus,
{
1
2
0
( 2 ) ( ) d
I
b cx bx cx x x x
b
∂
= − + + + − =∫
∂
} 0 (3.19)
and
{
1
2 2 2
0
( 2 )2 ( ) d
I
b cx x bx cx x x x
c
∂
= − + + + − =∫
∂
} 0
x
(3.20)
After integration and simplification, Eq. (3.18) and (3.19) provide respectively,
8b+9c=-6 (3.21)
45b + 68c = -20 (3.22)
Solving them, we get, b= -1.6402 and c=0.7913. Thus, the approximate solution is
(3.23)2
1.6402 0.7913u x= − +
It gives u(0.5)=-0.6223 and u(1)=-0.8489, very near to the exact solution (see Eq. (3.14)).
Increasing the terms in the approximation polynomial will keep on increasing the accuracy. Now,
let us see how the approximate solutions satisfy the natural boundary conditions. When we take
linear approximation, du/dx is constant and equal to –0.75 everywhere. This obviously gives
large error in the natural boundary conditions. However, when we take quadratic approximation,
d
1.6402 1.5826
d
u
x
x
= − + (3.24)
giving the value –0.0576 at x=1, much nearer to exact value of 0. Thus, we see that with Ritz’s
procedure, the natural boundary conditions also get satisfied.
3.3 GALERKIN METHOD
Boundary value problems may be solved without converting into variational form by the method
proposed by Soviet Scientist B.G. Galerkin (1871-1945). Galerkin published the method in 1915
[3]. Before this, Bubnov [4] had applied this method to some specific problems, but did not give
the method in general form. To recognize the work of Bubnov, some researchers call the method
42

as Bubnov-Galerkin method. Galerkin’s method works directly on the differential equation,
which is called the strong form. The variational form is the weak form, because the highest order
of derivative gets reduced, thus weakening the continuity requirement on primary variables.
Consider a differential equation
L(u)+ f = 0 (3.25)
where L is the differential operator and f is the know function. The boundary conditions may be
on 1,2,..........i i iB u q S i= = (3.26)
The method works as follows:
• Approximate the function as a weighted sum of linearly independent functions as in
Ritz’s method. However, here the approximated function must satisfy the essential as
well as natural boundary conditions of the problem. The chosen functions must be
complete. To give an example, if u is a function of x satisfying all the boundary
conditions, it can be represented as
0
1
n
i i
i
u aψ ψ
=
= + ∑ (3.27)
where Ψi is the ith
basis function, and ai is the corresponding coefficient
• Substitute the approximating function in Eq. (3.25) and obtain the residue as follows:
0
1
(
n
i i
i
)R L a fψ ψ
=
= +∑ + (3.28)
If you are very lucky, the residue R will be zero and accidentally (?) you have found the
exact solution. In general, it will be non-zero and you have to minimize it by adjusting the
coefficients ai.
• In Galerkin’s method coefficients are found by making the weighted integrals zero. The n
linearly independent basis functions Ψi s act as weights. Thus,
d 0 1,2,....,i
D
R D i nψ = =∫ (3.29)
where D is the domain and integral is the definite integral over the domain. Eq. (3.29)
provides us n simultaneous equations and we can determine n unknown coefficients, thus
obtaining the approximating function.
Example 3.2: Solve the differential equation of Example (3.1) using Galerkin method.
43

Solution: Let us take a quadratic approximation i.e.,
(3.30)2
u a bx cx= + +
In view of the essential boundary conditions, a becomes 0. The natural boundary condition
gives
2b c 0+ = (3.31)
Thus, the approximating function becomes
2
( 2u c x x= − )
1
(3.32)
The residue is given by
2
2 ( 2 )R c c x x= + − − (3.33)
The coefficient c is found by minimizing the weighted integral of R, the weight being (x2
-
2x). Thus,
1
2 2
0
( 2 )[2 ( 2 ) 1]dx x c c x x x− + − −∫ = 0 (3.34)
giving c= 5/6. Hence, the approximate solution is
25
( 2
6
u x= − )x (3.35)
This solution satisfies both the boundary conditions. Its values at x=0.5 and x=1 are -0.625
and –0.8333. Let us compare the values at these points with the values obtained by Ritz
method and exact solution. Table 3.1 shows this comparison. At one point, Galerkin method is
closer to the exact method, while at other place Ritz method is closer.
Table 3.1 The values of u at two points obtain by Galerkin, Ritz and exact method
The value of uPoint
Galerkin method Ritz method Exact method
x=0.5 -0.6250 -0.6223 -0.6242
x=1.0 -0.8333 -0.8489 -0.8508
3.4 THE LEAST SQUARE METHOD
In Galerkin’s method, the error is minimized by taking the weighted integral of residue. The
residue can also be minimized in a least square sense giving rise to the least square method. The
approximating function should satisfy both the natural and essential boundary conditions. When
44

the residue given by equation (3.28) is minimized in a least square sense, we get n simultaneous
equations as follows:
2
d 0 1,2,.........,
Di
R D i
a
∂
= =∫
∂
n (3.36)
Example 3.3: Solve the differential equation of Example (3.1) by using the least square method.
Solution: Let us take the approximating function of Example (3.2). The residue is given by
equation (3.33). There is only one unknown c. Therefore,
(
1 2
2
0
2 ( 2 ) 1 dc c x x x
c
∂
+ − − =
∂ ∫ ) 0
0
(3.37)
or
( )( )
1
2 2
0
2 2 ( 2 ) 2 ( 2 ) 1 dx x c c x x x+ − + − − =∫ (3.38)
Solving it we get c= 5/7. Therefore, the solution is
25
( 2
7
u x= − )x (3.39)
3.5 COLLOCATION METHOD
In this method, the residue is made equal to 0 at n points called collocation points. This gives n
simultaneous equations, which can be solved for n unknown coefficients. The method is very
simple. However, here the solution depends on the chosen collocation points. For example, if the
residue in equation (3.33) is made equal to zero at x=0.5, we get c=4/5, giving a solution
( 24
2
5
u x= − )x (3.40)
If the residue is made equal to zero at x=1/3, we get c=9/13 giving
( 29
2
13
u x= − )x (3.41)
Of course if we take many collocation points and approximate the function higher degree of
polynomial, we may expect to get better solution even by this method.
3.6 SUB-DOMAIN METHOD
In this method, the domain is divided into n sub-domains and the integration of the
residue is made zero over each sub-domain to generate n equations. Supposing the
approximating function is given by equation (3.27). To find out the unknown coefficients, we can
45

divide the whole domain into n sub-domain, integrate the residue over each subdomain and force
it to zero for each subdomain. Thus, we make the residue zero in an average sense over each sub-
domain.
If we take the approximating function of Example (3.2) and solved it by sub-domain
method, we
shall get the solution as
( 23
2
4
u x= − )x (3.42)
You can verify it easily. Ofcourse, as there is only one unknown in the approximating function,
the entire domain is taken as subdomain.
3.7 CONCLUSIONS
In this chapter, a number of classical methods have been introduced for solving the differential
equations. These methods can form the basis of finite element, each method giving rise to one
type of formulation. The main difference between these methods and finite element is that here a
continuous function is approximated for the whole domain, whereas in the finite element method
a number of locally continuous functions are chosen.
REFERENCES
1. Ritz, W., Variationsproblem der
mathematischen Physik, Journal
Uber eine neue Methode zur Losung gewisser
fur Reine und Angewande Mathematik, Vol. 135,
1908, pp. 1-61.
2. Leissa, A. W., The historical bases of the Rayleigh and Ritz methods, Journal of
Sound and Vibration, Vol. 287, 2005, pp. 961-978.
3. Galerkin, B.G., Rods and Plates. Series occurring in various questions concerning
the elastic equilibrium of rods and plates, Engineers Bulletin (Vestnik Inzhenerov),
Vol. 19, 1915, pp. 897-908 (in Russian).
4. Bubnov, I.G., Report on the works of Professor Timoshenko which were awarded the
Zhuranski Prize. Symposium of the Institute of Communication Engineers, No. 81,
All union Special Planing office (SPB), 1913 (in Russian).
46

EXERCISE 3
Q.1: A certain problem of one dimensional heat transfer is governed by the equation
2
2
d
1 0
dx
φ
φ+ + = and boundary conditions 1φ = at 0x = and
d
1
dx
φ
= at 1x = . Solve this
problem by using Galerkin method.
Q.2: Given a differential equation:
22
20
02 2
dd
0
d d
zz
v
EI v
x x
ρω
⎛ ⎞
− =⎜ ⎟
⎝ ⎠
;
with boundary conditions:
0
0
d
0
d
v
v
x
= = at and l (both essential).0x =
EIzz is a function of x and ω is a constant unknown angular velocity.
a) Derive the variational functional associated with this problem. State explicitly the
conditions to be satisfied by the vδ as well as the properties of the δ operator used
in the derivation.
b) Using the approximation
( ai are unknown constants) and the Ritz method, derive
the algebraic equations satisfied by the unknown constants in the following form:
( )0
1
n
i i
i
v aφ
=
= ∑ x
=( )2
1
0
n
ij ij j
j
k M aω
=
−∑ for i =1,2,….n .
State explicitly the condition to be satisfied by iφ .
Q.3: The variational functional of the problem shown in Fig. Q3 is
2
/ 20
1 d
d ( )
2 d
l
x l
u
I AE pu x Pu
x =
⎛ ⎞⎛ ⎞
= − + +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
∫
a) Using the approximation
( )
1 2
x l x
u a
l
−
= (a1 is unknown constant), express I in
terms of a1 (Note: A, E, p are constants.).
b) Find a1 using Ritz method.
47

p (distributed force)
P (point force)
l/2
l
Figure: Q3
Q.4: Solve the for the following differential equation by Galerkin method:
2
2
2
d
0
d
u
cu x
x
− − + = for 0 1x≤ ≤ , the boundary conditions being
( ) ( )
d d
0 1, 1
d d
u u
x x
4
3x x
= = = = .
Q.5: Solve the following problem by Galerkin method:
Differential Equation:
d d
0
d d
u
u f
x x
⎛ ⎞
− =⎜ ⎟
⎝ ⎠
for 0 < x < 1 ;
Boundary conditions: (1) Essential: 2 at =0u x=
(2) Natural :
d
0
d
u
x
= at x=1
Take f to be a linear function of x : 1 2b b x+ (b1, b2 are constants).
Take
2
1 2 1 22 ( , constants)u a x a x a a= + + = as the approximation for
the Galerkin method.
Q.6: A certain problem of one dimensional heat transfer is governed by the equation
2
2
d
1 0
dx
φ
φ+ + = and boundary conditions 1φ = at 0x = and
d
1
dx
φ
= at 1x = .
Use Ritz method to solve this problem. Approximate the function by a quadratic polynomial and
compare with the exact solution.
48

Q.7: Solve the following differential equation by Ritz method:
2
2
2
d
0
d
u
cu x
x
− − + = for 0 1x≤ ≤ for the boundary conditions: ( ) ( )
d d
0 1, 1
d d
u u
x x
4
3x x
= = = = .
Solve it by all other methods which you have studied in this chapter.
49

Chapter 4
RITZ AND GALERKIN FEM FORMULATION
(Lectures 8-10)
4.1 INTRODUCTION
In Ritz FEM, the finite element equations are obtained by following a procedure similar
to classical Ritz method. The difference is that in FEM, piecewise continuous functions are
chosen instead of choosing a globally continuous function for the whole domain. The elemental
equations can be obtained by writing the variational expression for an element, putting a
continuous interpolation function in that expression and obtaining the coefficients of the
interpolation function by extremizing the variational expression. The interpolation function,
which approximates the actual solution should be complete and should satisfy the compatibility
conditions. The following subsection explains the concept of completeness and compatibility.
Galerkin FEM follows procedure similar to classical Galerkin method. Here, also the piecewise
continuous approximation is employed. However, before applying the method, equations are
converted in weak form.
4.2 COMPLETENESS AND COMPATIBILITY
In the context of classical Ritz method, completeness means that basis functions of the
approximating functions are such that if enough terms are taken, the primary variables and their
derivatives can be approximated as accurately as we wish. In the context of finite elements, the
set of basis functions are said to be complete if they can approximate the primary variables and
their derivatives appearing in the variational form as accurately as we wish by reducing the size
of element. Thus, leaving aside the computational difficulties, if the size of the elements approach
0, the exact values of the primary variables and their derivatives (upto the order appearing in
variational form) should be obtained.
A polynomial series is complete if it is of high enough degree and no terms are omitted.
Fourier series are also complete. By high enough degree we mean that the highest order
derivative appearing in the variational form can be represented. For example, if
2
0
d
d
d
l u
I u x
x
⎧ ⎫⎪ ⎪⎛ ⎞
= −∫ ⎨ ⎬⎜ ⎟
⎝ ⎠⎪ ⎪⎩ ⎭
(4.1)
then is a complete approximation (or interpolation) function, as it can represent du/dx
by b. However, u=a is not complete because it makes du/dx zero. The interpolation
u ax b= +
51

function is complete. It represents2
u a bx cx= + +
d
d
u
x
by 2b cx+ . However, is not
complete. Why? After all it can represent
2
u a cx= +
d
d
u
x
by . The reason is that it cannot approximate a
constant derivative term. The approximate
2cx
d
d
u
x
will always be zero at x = 0. In practice, one may
have a non-zero du/dx at x=0.
Now, consider the following variational expression:
22 3
20
d d
d d
l
3
x l
u u
I qu dx
x x =
⎛ ⎞⎛ ⎞ ⎛ ⎞
⎜ ⎟= − −⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
∫ (4.2)
For this, a complete approximate function will be . We have taken cubic
polynomial because we have to represent
2
u a bx cx dx= + + + 3
3
3
d
d
y
x
at x l= , though inside the integrand highest order
of derivative is 2 only. Thus, for finding out the completeness, the degree of approximating
polynomial should be equal to the highest order of derivative in the whole variational expression.
We should also ensure that whatever be the approximation, the I should not become
infinite. This means the approximating function should have continuity equal to at least one order
less than the highest order derivatives in the integral. In Eq, (4.1) approximate should be at least
continuous through the domain. This will ensure that0
C
d
d
u
x
is always finite inside the integral.
The piece-wise continuous polynomial may be different in two neighboring elements. However,
the values of primary variables should come same from both the polynomials at the common
node. Eq. (4.1) requires continuity i.e. u and1
C
d
d
u
x
should be same for both the elements at
common nodes. Thus, for compatibility, one should check the highest order of the derivative
inside the integral expression. If the highest order of derivative is m, the function should be
continuous upto order (m-1) throughout the domain i.e. the function should be continuous.
Note that for compatibility, we check the highest order of the derivative in the integral expression
of the full variational form.
1m
C −
4.3 CONCEPT OF SHAPE FUNCTIONS
Consider the problem of minimizing I in Eq. (4.1). For this problem, the lowest degree of
incomplete polynomial is u a . At the same time, there should be atleast continuity. Ifbx= + 0
C
52

we take a 2 noded element of length h, we can express the constants a and b in terms of the nodal
values of the primary variables. This will automatically ensure continuity. It the local
coordinate of the node 1 is 0 and that of node 2 is h and the values of the primary variable at these
nodes are u1 and u2 respectively, then
0
C
( )1 0u a b a= + = (4.3)
and
2 1u a bh u bh= + = + (4.4)
Substituting the value of a from Eq. (4.1) into Eq. (4.3), we get
2 1u u
b
h
−
= (4.5)
Hence, by expressing the constants a and b in terms of the nodal values of the primary variable,
the piece-wise continuous polynomial can be expressed as,
( )2 1
1
u u
u u x
h
−
= + 1 21
x x
u u
h h
⎛ ⎞
= − +⎜ ⎟
⎝ ⎠
1 1 2 2N u N u= + (4.6)
where N1 and N2 are called shape functions, because seeing them we can know the shape of
piecewise continuous approximating polynomial. In this case, it is linear.
We can also have 3-noded elements. In that case, a quadratic interpolation function,
can be taken. Let the coordinates of 3 nodes be x1, x2, and x3 respectively. Then,2
u a bx cx= + +
(4.7)2
1 1u a bx cx= + + 1
2
3
(4.8)2
2 2u a bx cx= + +
(4.9)2
3 3u a bx cx= + +
Solving this, we express the coefficients a, b and c in terms of nodal values of the primary
variables and arrange the expression to get the following form:
(4.10)1 1 2 2 3 3u N u N u N u= + +
where N1, N2 and N3 are the shape functions corresponding to 3 nodes respectively. The same
procedure can be extended to n-noded element. There one has to solve n simultaneous equation
for obtaining the coefficients of polynomial interpolation function in terms of the nodal values of
the primary variables. These coefficients are substituted back in the polynomial expression and a
rearrangement provides the following form:
(4.11)1 1 2 2
1
............
n
n n i i
i
u N u N u N u N u
=
= + + + = ∑
53

where Ni is the ith
shape function and ui is the value of primary variable at ith
node. However, this
procedure of obtaining the shape functions is tedious. A somewhat simpler way is to obtain the
shape function on the basis of the three properties of the shape functions.
One-dimensional C0
polynomial shape functions will satisfy the following 3 properties:
Property 1: All shape functions of an n-noded element are polynomial of (n-1) degree. This is
because for an n-noded element, interpolation function will be of (n-1) degree. It should be
possible to represent a polynomial function of (n-1) degree such that its value is zero at all nodes
except one node. That one node can be any one out of the n nodes. Hence, the shape functions
associated with all nodes should be of the same degree.
Property 2: For any shape function
( )i j ijN x δ= (4.12)
where Ni (xj) is the value of ith
shape function at jth
node and δij is the Kronecker delta. This can be
justified as follows. Let us assume a variable field such that it is non-zero at ith
node and all other
nodal values of the variable are zero. In that case,
i iu N u= (4.13)
Now, at x= xi, u=ui. Hence, we have from Eq. (41.3)
(4.14)( ) or ( ) 1i i i i i iu N x u N x= =
=
i =
At x=xj, j≠i, the value of the u is 0. Hence, from Eq. (4.13)
(4.15)0 ( ) or ( ) 0i j i i jN x u N x=
Hence, we have proved the property 2.
Property 3: The shape functions sum to unity. This can be proved as follows. Assume that u is
constant and equal to c throughout. Thus, the nodal values of the variable will also be c. From
Eq. (4.11), we can write
(4.16)
1 1
or 1
n n
i
i i
c cN N
= =
= ∑ ∑
Lagrange’s interpolation functions satisfy these properties. For n-noded element,
Lagrange shape functions are given by
54

2 3 4
1
2 1 3 1 4 1 1
1 3 4
2
1 2 3 2 4 2 2
1 3 4 1
1 3
( )( )( ).........( )
( )( )( ).........( )
( )( )( ).........( )
( )( )( ).........( )
.
.
.
( )( )( ).........( )
( )(
n
n
n
n
n
n
n n
x x x x x x x x
N
x x x x x x x x
x x x x x x x x
N
x x x x x x x x
x x x x x x x x
N
x x x x
−
− − − −
=
− − − −
− − − −
=
− − − −
− − − −
=
− − 4 1)( ).........( )n n nx x x −− − x
1
n
(4.17)
It can be easily seen that for these shape functions, first two properties are satisfied. It is also easy
to show that at all nodes, in view of the property 2. However, we have to prove that
the sum of shape functions is 1 everywhere. We argue it as follows. It is clear that can be
of at most (n-1) degree polynomial. Thus,
1
1
n
i
i
N
=
=∑
1
n
i
i
N
=
∑
2
0 1 2
1
..........
n
n
i
i
N a a x a x a x −
=
= + + + +∑ (4.18)
Its value at all the nodes is 1. Hence,
2 1
0 1 1 2 1 1
2
0 1 2 2 2 2
2 1
0 1 2
1 ..........
1 ..........
.
.
.
1 ..........
n
n
n
n
n
n n n n
a a x a x a x
a a x a x a x
a a x a x a x
1
−
−
−
= + + + +
= + + + +
= + + + +
(4.19)
This is a system of n equations in n unknowns and should give a unique solution. One solution is
a0=1 and all other coefficients 0 (by inspection). Hence, from Eq. (4.18),
1
1
n
i
i
N
=
=∑ (4.20)
throughout the element.
4.4 DEVELOPING THE ELEMENTAL EQUATIONS BY RITZ METHOD
Consider the problem
2
2
d
1 0
d
u
u
x
+ − = (4.21)
55

with boundary conditions
and(0) 0u =
d
0 at 1
d
u
x
x
= = (4.22)
Let us develop the element equations for this problem. Consider an element whose length is h.
Adopting the local coordinate system, the coordinate of the first node is 0 and that of second is h.
For obtaining the variational form of the differential equation:
2
20
d
1 d 0
d
h u
I u u
x
δ
⎡ ⎤
= + −⎢ ⎥
⎣ ⎦
∫ xδ = (4.23)
Integrating by parts and using the properties of the variational operator
( )
2
2
0 0 0
d 1 d 1
d d
d 2 d 2
h h hh
o
u u
I u x u x u x
x x
δ δ δ δ δ
⎛ ⎞ ⎛ ⎞
= − + −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
∫ ∫ ∫ d 0= (4.24)
Hence,
2
2
00 0 0
1 d 1 d
dx+ d d
2 d 2 d
h h h hu u
I u x u x u
x x
⎛ ⎞ ⎛ ⎞
= − − +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
∫ ∫ ∫ (4.25)
We observe that completeness upto first degree polynomial and continuity throughout the
domain is enough. Thus, 2-noded elements are enough. However, we can also take higher order
elements for better accuracy. Let be the approximation inside an n-noded element
0
C
0
C e
u
= (4.26){ }
1
2
1 2 ....... .
.
e
n
n
u
u
u N N N N u
u
⎧ ⎫
⎪ ⎪
⎪ ⎪⎪ ⎪
= ne
⎢ ⎥ ⎢⎨ ⎬⎣ ⎦ ⎣
⎪ ⎪
⎪ ⎪
⎪ ⎪⎩ ⎭
{ }ne
u N⎢ ⎥⎣ ⎦= ⎥⎦
Putting this value in I,
{ } { } { } { } { }
0
, , 1 2
d
d
0
01 1
d d d ...
.2 2
.
d
d
h h h
ne ne ne ne ne
x x no o o
h
u
x
I u N N u x u N N u x u N x u u u
u
x
⎧ ⎫⎛ ⎞
−⎜ ⎟⎪ ⎪
⎝ ⎠⎪ ⎪
⎪
⎪
⎪ ⎪
⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥= − + − +⎢ ⎥ ⎢ ⎥ ⎨ ⎬⎣ ⎦ ⎣ ⎦⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎛ ⎞⎪ ⎪
⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭
∫ ∫ ∫
⎪
⎪ (4.27)
which can also be written as
56

{ } { } { } { } { }
0
, ,
d
d
0
01 1
d d d
.2 2
.
d
d
h h h
ne ne ne ne ne ne
x xo o o
h
u
x
I u N N u x u N N u x u N x u
u
x
⎧ ⎫⎛ ⎞
−⎜ ⎟⎪ ⎪
⎝ ⎠⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥= − + − +⎢ ⎥ ⎨⎣ ⎦⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎛ ⎞⎪ ⎪
⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭
∫ ∫ ∫
⎪
⎬
(4.28)
For extremizing it,
0ne
I
u
∂
=
⎢ ⎥∂ ⎣ ⎦
(4.29)
{ } { } { } { } { }
0
, ,0 0 0
d
d
0
0
d d d
.
.
d
d
h h h
ne ne
x x
h
u
x
N N u x N N u x N x
u
x
⎧ ⎫⎛ ⎞
−⎜ ⎟⎪ ⎪
⎝ ⎠⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎢ ⎥− + − +⎢ ⎥ ⎨ ⎬⎣ ⎦⎣ ⎦
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎛ ⎞⎪ ⎪
⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭
∫ ∫ ∫ 0=
(4.30)
Hence,
{ } { } { } { }
0
, ,0 0 0
d
d
0
0
d d d
.
.
d
d
h h h
ne
x x
h
u
x
N N x N N x u N x
u
x
⎧ ⎫⎛ ⎞
−⎜ ⎟⎪ ⎪
⎝ ⎠⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪⎡ ⎤⎢ ⎥ − = − +⎢ ⎥ ⎨ ⎬⎣ ⎦⎣ ⎦⎢ ⎥⎣ ⎦ ⎪ ⎪
⎪ ⎪
⎪ ⎪
⎛ ⎞⎪ ⎪
⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭
∫ ∫ ∫ (4.31)
This is the element equation.
For 2-noded elements, we have for a typical element,
01
2
d
d1 1 2 11 2
1 1 1 26 d
2 d h
uh
u xh
u hh u
x
⎧ ⎫⎛ ⎞⎧ ⎫ −− ⎜ ⎟⎪ ⎪⎪ ⎪⎡ − ⎤ ⎧ ⎫⎡ ⎤ ⎡ ⎤ ⎝ ⎠⎪ ⎪ ⎪ ⎪
− = +⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥⎢ ⎥ ⎢ ⎥− ⎛ ⎞⎣ ⎦ ⎣ ⎦ ⎩ ⎭⎣ ⎦ ⎪ ⎪ ⎪ ⎪− ⎜ ⎟⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎝ ⎠⎩ ⎭
(4.32)
57

Reader should verify equation (4.32) by substituting the expressions for the shape functions and
their derivatives for the 2-noded elements and integrating from 0 to h. Taking 2-elements and
assembling the elemental equations and putting the boundary condition that the first derivative of
u vanishes at x=1, we get
1
2
3
d
(0)
1.833 2.083 0 0.25 d
2.083 3.667 2.083 0.5 0
0 2.083 1.833 0.25 0
u
u x
u
u
⎧ ⎫
−⎪ ⎪− −⎧ ⎫⎡ ⎤ ⎧ ⎫
⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥− − = − +⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥
⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥− −⎣ ⎦ ⎩ ⎭⎩ ⎭ ⎪ ⎪
⎩ ⎭
(4.33)
(Note: At this stage, learn the faster way of assembly. You need not write the elemental
equations in global form. First, make a format of final global equations with empty entries. Then,
keep on putting the entries corresponding to elemental equations in corresponding places of
global system of equations. If 2 are more entries are kept at the same place, they simply get
added.)
As u1 =0, the first row and first column of Eq. (4.33) get eliminated. Solving the remaining 2-by-
2 matrix, we get
(4.34)2 30.6035 and 0.8222u u= − = −
Thus, ( )1 0.8222u = −
If we take 3 elements we get and for 6 elements( )1 0.8377u = − ( )1 0.847u = − 5 . Verify it.
The exact solution is
( )tan 1 sin cos 1u x= − − +x (4.35)
The value of exact u at and 1 are0.5x =
and( )0.5 0.6242u = − ( )1 0.8508u = − (4.36)
We have illustrated the Ritz-FEM by a simple one-dimensional problem, but the procedure
is same for all problems. For any element, we substitute the approximating function in terms of
the unknown nodal displacement and extremize the function with respect to the displacement
vector. When you develop the elemental equations for a differential equation containing the
independent variable x, do not forget to express this in the form of local variable.
58

4.5 DEVELOPING THE ELEMENTAL EQUATION BY GALERKIN METHOD
For obtaining the finite element equations by Galerkin method, one has to obtain the
weak form of the differential equation. Galerkin method is a special form of weighted residual
method. In the weighted residual method, we integrate the weighted residual over the domain and
make it 0. Let w be the weight function. Then, for the example of Section 4.4, we have
2
20
d
1 d 0
d
h u
w u x
x
⎡ ⎤
+ − =⎢ ⎥
⎣ ⎦
∫ (4.37)
Integrating by parts,
0 0 0
0
d d d
d d d
d d d
h
h h hu w u
w x wu x
x x x
− + −∫ ∫ ∫ 0w x = (4.38)
Seeing it, we know that taking
1
1 2
2
u
u N N
u
⎧ ⎫
= ⎢ ⎥ ⎨ ⎬⎣ ⎦
⎩ ⎭
(4.39)
is enough. Writing in a condensed form,
{ }ne
u N u= ⎢ ⎥⎣ ⎦ (4.40)
In Galerkin method, w is approximated in the same way as the shape function, i.e.,
{ }ne
w w N⎢ ⎥= ⎣ ⎦
(4.41)
Putting equations (4.40-4.41) in equation (4.38),
{ }
0
0 0 0
.
. , , { }d { } { }d
.
h h h
ne ne ne ne ne ne
h
du
dx
w w N x N x u x w N N u x w N x
du
dx
⎧ ⎫⎛ ⎞
−⎜ ⎟⎪ ⎪
⎝ ⎠⎪ ⎪
⎪ ⎪
⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− + −⎢ ⎥ ⎢ ⎥∫ ∫ ∫⎨ ⎬ ⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
⎪ ⎪
⎪ ⎪
⎪ ⎪⎛ ⎞
⎜ ⎟⎪ ⎪
⎝ ⎠⎩ ⎭
{ }d 0= (4.42)
As the nodal values of the weights are arbitrary, we get
59

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