The document describes the finite element method (FEM) for analyzing physical systems modeled by differential equations. It provides an introduction to FEM and presents examples of applying FEM to one-dimensional axial rod and beam problems. The key steps of FEM include pre-processing (discretization), obtaining elemental equations, assembly, applying boundary conditions, solution, and post-processing. Direct FEM formulation is used to directly obtain algebraic equations from the physics for the axial rod and beam problems in 1D.

1. 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

2. 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

3. 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

4. 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

5. 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
10.1 Introduction 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

6. viii
11.1 Introduction 137
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
12.1 Introduction 151
12.2 Vibration of A Rod 151
12.3 Vibration of A Beam 152
12.4 Conclusions 156
Exercise 12 157
13 Finite Element Formulation of Time Dependent Problems 161
13.1 Introduction 161
13.2 Classification of Partial Differential Equations 161
13.3 Time Response of A parabolic Equation 162
13.4 Forced Vibration Problems 164
13.5 Conclusions 165
References 165
Exercise 13 165
14 FEM Formulation of Plate Problem 167
14.1 Introduction 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

7. ix
14.7 Summary and Conclusion 176
References 176
Exercise 14 177
15 Finite Element Formulation of 2-D Flow problems 179
15.1 Introduction 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
16.1 Introduction 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
16.5 Conclusions 211
Further Readings 211
Exercise 16 212
17 Miscellaneous 213
17.1 Introduction 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

8. 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
17.12 Conclusions 220
Exercise 17 220
Bibliography 221

9. 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

10. • 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

11. 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

12. 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

13. 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

14. 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

15. 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

16. 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

17. 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

18. (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

19. 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

20. 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

21. 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

22. 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

23. 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

24. 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

25. 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

26. 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

27. 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

28. 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

29. 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

30. 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

31. 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

32. 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

33. 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

34. 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

35. 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

36. [ ( ) ] 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

37. 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

38. 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

39. 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

40. 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

41. 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

42. 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

43. 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

44. 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

45. 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

46. 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

47. ( )
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

48. 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

49. 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

50. 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

51. 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

52. 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

53. 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

54. 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

55. 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

56. 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

57. 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

58. 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

59. 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

60. 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

61. 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

62. { } { } { } { } { }
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

63. 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

64. 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

65. { } { } { } { }
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
⎧ ⎫⎛ ⎞
−⎜ ⎟⎪ ⎪
⎝ ⎠⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪⎡ ⎤⎢ ⎥ − = − +⎢ ⎥ ⎨ ⎬⎣ ⎦⎣ ⎦⎢ ⎥⎣ ⎦ ⎪ ⎪
⎪ ⎪
⎪ ⎪
⎛ ⎞⎪ ⎪
⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭
∫ ∫ ∫ (4.43)
This is the elemental equation, which is same as that obtained by Ritz method. Rest of the
procedure is similar to that described in the previous section.
If the variational form of a differential equation exists, the both methods provide the
same results for the same approximation. However, for many differential equations, the variational
form does not exist, for example, the differential equation
2
2
d d
5 7
d d
y y
y
x x
+ + = 0 (4.44)
does not have a variational form. For solving this equation by FEM, we can use Galerkin procedure
but not the Ritz procedure.
4.6 CONCLUSIONS
In this chapter two approaches of finite element formulation have been described- Ritz
FEM and Galerkin FEM. If the varaitional form is given, then it is convenient to apply Ritz FEM. If
the differential equation has been provided, then it is convenient to apply Galerkin FEM. One can
convert the problem from a variational form to differential form and vice versa and use any method.
However, for certain differential equations, the variational form does not exist at all.
60

66. EXERCISE 4
Q.1: Solve the following boundary value problem by Ritz and Galerkin FEM:
Differential Equation:
2
2
d
5 7
d
y
xy
x
+ =
Boundary conditions are at x=0, y=0 and at x=5, y=6.
Solve first by taking 2 elements and then 3 elements and compare the results.
Q.2: Solve the following boundary value problem by Galerkin FEM:
2
2
d d
5 7
d d
y y
y
x x
+ + = 0
The boundary conditions are at x=0, y=0 and at x=1, dy/dx=0. Solve by taking 2 or 3 elements.
Q.3: To solve the following problem, use Ritz FEM:
2
1
0
d
Minimize 5 d 10 (1)
d
u
I x u
x
⎛ ⎞
= −⎜ ⎟
⎝ ⎠
∫ ; u(0)=0
Start by taking one element and keep on increasing the elements till the convergence is achieved.
Q.4: To solve the following problem, use Ritz FEM:
2
1
0
d
Minimize 5 d
d
u
I u x
x
⎛ ⎞⎛ ⎞
= −⎜⎜ ⎟⎜⎝ ⎠⎝ ⎠
∫ ⎟⎟
; u(0)=0
Start by taking one element and keep on increasing the elements till the convergence is achieved.
Q.5: At a point, the values of the shape functions of a 4-noded C0
continuity element are:
N1=0.1; N 2=N3= 0.2. Find our N4.
Q.6: Solve the problem in Q.4 by taking one 3-noded element.
Q.7: Solve the following problem by Ritz and Galerkin FEM:
2
2
d
Differential equation: 5 ( 0.5) 0 0 1
d
d
Boundary conditions: At =0, =0 and at =1, 0.
d
y
x x
x
y
x y x
x
δ+ − = ≤ ≤
=
Take 2 2-noded elements.
61

67. Chapter 5
SOME ONE-DIMENSIONAL C0
CONTINUITY FEM
FORMULATIONS
(Lectures 11-12)
5.1 INTRODUCTION
In the previous chapter, you have learned the techniques of FEM formulation using Ritz and
Galerkin methods. In this chapter, we shall take up a number of one-dimensional problems of
interest to engineers and obtain the FEM formulations for them. Before starting the FEM
procedure, the governing equations for various problems have been developed.
5.2 STEADY-STATE HEAT CONDUCTION
Suppose we have to find out the temperature distribution of the rod shown in Fig. 5.1. If
the cross-sectional area of the rod is small compared to its length, then the temperature across a
particular cross-section can be considered constant. Thus, the temperature is a function of
longitudinal coordinate x. The cross-sectional area and thermal conductivity of the rod also may
be considered as the function x.
The governing law for heat conduction problems is Fourier heat conduction equation
given by
T
q k
x
∂
= −
∂
(5.1)
where q is the heat flow per unit area (heat flux) in direction x, T is the temperature and k is the
thermal conductivity. The boundary conditions are at x=0, T=T0 and at x=xL, q=qR. For obtaining
the governing differential equation for this problem, we take an infinitesimal small element of
length ‘dx’ as shown in Fig. 5.2 and obtain the heat balance. In steady-state the heat generated
will be equal to net heat coming out of the rod. If is heat generated per unit volume, then the
heat generation in the element is , where A is the average cross-sectional area of the rod.
At the left hand side, the heat entering the rod is Aq. We can say that the heat coming out of the
rod through the left hand side cross-section of the element is –Aq. At the right hand side cross-
section, the heat coming out of the element is Aq+d(Aq), where d(Aq) denotes the differential
increase in the heat flow. The convective heat transfer from the surface is hc(T-Tf)pdx, where hc is
Q
dQA x
63

68. the convective heat transfer coefficient, Tf is the temperature of the surrounding fluid and p is the
average perimeter of the cross-section of the infinitesimal element.
Fig. 5.1: Finite element discretization of a rod
Convective heat loss
dx
Aq Aq +d(Aq)
Fig. 5.2 Heat transfer from an infinitesimal element
Thus,
+ d( ) - - ( ) d = dc fAq Aq Aq h T -T p x QA x (5.2)
The above equation provides us
( )
d
( )
dx
c fAq QA h T T p 0− − − = (5.3)
Using Eq. (5.1), we can write
64

69. d d
( )
dx dx
c f
T
kA QA h T T p
⎛ ⎞
0− − − − =⎜ ⎟
⎝ ⎠
(5.4)
Thus, the governing differential equation is
( )
d
, ( )
dx
c fAkT x QA h T T p 0+ + − = (5.5)
where ,x denotes the differentiation with respect to x. Remember qR is the heat flowing from left
to right.
From here we can proceed towards Galerkin FEM formulation. However, we shall show
Ritz FEM formulation for it. For this, we first obtain the variational form. The variation form of
this problem in an element of length h is given by
(5.6)( )2 2
,0
0
1 1 d
d
2 2 d
h
h
x c c f
T
AkT h pT QAT h pT x kA T
x
⎧ ⎫
∏ = + − + −⎨ ⎬
⎩ ⎭
∫
Considering the completeness and compatibility, a 2-noded element with Lagrange shape
functions is good enough. Putting
[ ]{ }eT N T= (5.7)
[ ], , { eT x N x T= } (5.8)
in the variational form, we get
[ ] [ ] [ ] [ ]
[ ] [ ] [ ]
0 0
T T
1 20 0
1 1
{ } , , dx{ } { } dx{ }
2 2
d
dx
{ } dx { } dx
d
dx
T Th h
T T
e e e e c e
h h
T T
e e c f
T N x Ak N x T T N h p N T
T
kA
T QA N T h pT N T T
T
kA
∏ = +
⎧
⎪⎪ ⎪
− − + ⎨ ⎬
⎪ ⎪−
⎪ ⎪⎩ ⎭
∫ ∫
∫ ∫
⎫
⎪
(5.9)
Minimizing with respect to {Te}T
, we get elemental equations:
65

70. [ ] [ ] [ ] [ ]
[ ] [ ]
0 0
T T 0
0 0
0 , , dx{ } dx{
d
d
dx dx
d
dx
T Th h
e c
h h
c f
h
N x Ak N x T N h p N T
T
kA
x
QA N h pT N
T
kA
= + }e
⎧ ⎫
⎪ ⎪
⎪ ⎪
− − + ⎨ ⎬
⎪ ⎪−
⎪ ⎪⎩ ⎭
∫ ∫
∫ ∫
(5.10)
These elemental equations can be assembled and solved after applying the boundary conditions.
Fig. 5.3: A hollow cylinder
We can solve the heat conduction problem expressed in polar coordinates also. In polar
coordinates, the steady-state heat conduction without heat generation is
d d
d d
T
kr
r r
⎛ ⎞
⎜
⎝ ⎠
⎟=0 (5.11)
We solve this problem using Galerkin FEM. For this purpose, we write the residual as the
weighted integral of the jth
element and make it to 0. Thus,
( )
0
d d
d 0
d d
h
j
j
T
R w R r k r
r r
⎛ ⎞
= +∫ ⎜ ⎟
⎝ ⎠
=
where Rj is the radius of jth
node and r is the local coordinate. The integration by part yields
( ) ( )
00
d d d
d 0
d d d
h h
j j
T w T
w R r k R r k r
r r r
⎛ ⎞
+ − +∫⎜ ⎟
⎝ ⎠
=
66

71. Observing the weak form, we assess that Completeness in and continuity of T is adequate.
Thus, we can approximate the temperature as
'
,T T
1 ,
j
k
Tr r
T a br
h h T
⎧ ⎫⎪ ⎪⎡ ⎤
= + = − ⎨ ⎬⎢ ⎥⎣ ⎦ ⎪ ⎪⎩ ⎭
Therefore,
d 1 1
,
d
j
k
TT
r h h T
⎧ ⎫⎪ ⎪⎡ ⎤
= − ⎨ ⎬⎢ ⎥⎣ ⎦ ⎪ ⎪⎩ ⎭
Putting this approximation in the weak form,
( ) { } { }
1
1' '
0
2
2
d
d
d
d
d
h
ne
j
T
R k
r
k R r N N r T
T
R k
r
⎧ ⎫
⎪ ⎪
⎪ ⎪⎡ ⎤+ =∫ ⎨ ⎬⎣ ⎦
⎪ ⎪
⎪ ⎪⎩ ⎭
or
'2 2
1 1
'
2 22 2
1 1
1 1 2
j
j
k
T R k Thh h
k R h
T R k T
h h
⎡ ⎤− ⎧ ⎫⎧ ⎫ −⎪ ⎪ ⎪ ⎪⎛ ⎞⎢ ⎥ + =⎨ ⎬ ⎨ ⎬⎜ ⎟⎢ ⎥ ⎝ ⎠ ⎪ ⎪− ⎩ ⎭ ⎪ ⎪⎩ ⎭⎢ ⎥⎣ ⎦
or
1 11 1
1 12 2
j jj
k k
T HR
k
h T Hπ
⎧ ⎫ ⎧ ⎫−⎛ ⎞ ⎡ ⎤ ⎪ ⎪ ⎪ ⎪
+ =⎨ ⎬ ⎨ ⎬⎜ ⎟ ⎢ ⎥− −⎪ ⎪ ⎪ ⎪⎣ ⎦⎝ ⎠ ⎩ ⎭ ⎩ ⎭
where Hj and Hk denote the heat flow rate, if the length of the cylinder is taken unity. Taking two
elements, the assembled matrix is
1 1
2
33
3 3 0
2 2
13 3 5 5 0
2 2 2 2 2
5 50
2 2
T H
T
HT
π
⎡ ⎤−
⎧ ⎫ ⎧ ⎫⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥− + − =⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎪ ⎪ ⎪ ⎪−⎢ ⎥ ⎩ ⎭⎩ ⎭−⎢ ⎥⎣ ⎦
Let us impose the boundary conditions:
*
1 aT T= *
3 bT T=
Applying boundary condition, we get
* *
2 0
3 8 5
0
2 2 2
aT T T− + − =
67

72. ( ) * *
2 00.375 0.625
FEM
aT T= + T
The exact solution is given by
( )
* *
*
ln
ln
b a
aexact
ab
a
T T RT T
RR
R
− ⎛ ⎞= + ⎜ ⎟
⎝ ⎠⎛ ⎞
⎜ ⎟
⎝ ⎠
Thus,
( ) * *
2 0.415 0.585a bT Exact T T= +
If =2000
C The FEM solution is T2 141.5 0
C*
1 aT T=
Exact solution is T2 136.90 0
C.* 0
0 100 CT =
5.3 LONGITUDINAL DEFORMATION OF A ROD
L
q
Figure 5.4: A rod subjected to axial loads
Let us consider a rod loaded with axial force of intensity q. Note that q is the force per unit
length and has unit of N/m in SI system. However, q may be a function of x. The cross-
sectional area A and material properties are also function of A. (Do not infer from the figure
that we are describing a rod of uniform cross-sectional area. The formulation is quite general,
although Fig. 5.4 looks like a uniform rod. You consider it a bar where although the height of
the cross-section is same everywhere, the width of the cross-section varies with x.) The
governing differential equation is given by
d d
0
d d
u
EA q
x x
⎛ ⎞
+ =⎜ ⎟
⎝ ⎠
(5.12)
Boundary conditions are:
(i) At 0, 0x u= = (5.13)
68

73. (ii) At
d
,
d
u
x L
x
0= = (5.14)
You may easily derive this differential equation by writing the expression of the total
potential energy and applying Euler-Lagrange equations.
Let e
y be the approximate solution of u for an element e whose end coordinates are ex
and 1ex + . Taking weighted residual with respect to the weight function w and obtaining the
weak form, we get
11 1
d d d d d
d d
d d d d d
ee e
ee e
xx xe e
xx x
y w y y
w EA q x EA qw x wEA
x x x x x
++ +
⎛ ⎞⎛ ⎞ ⎡ ⎤
⎜ + ⎟ = − + +⎜ ⎟ ⎢ ⎥⎜ ⎟⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦⎝ ⎠
∫ ∫ 0
e
=
(5.15)
The highest order of derivative appearing in the weak form is 1, thus a complete function of
the form a+bx is enough. Since the highest order of derivative inside the integral is 1, the
function need to be C0
continuous everywhere in the domain (0<x<1). Thus, the value of y at
any node should come same from both the elements of which the node is the part. A suitable
linear approximation for the element is
1
1 2
2
e y
y N N
y
⎧ ⎫
= ⎢ ⎥ ⎨ ⎬⎣ ⎦
⎩ ⎭
(5.16)
where N1 and N2 are called shape functions and y1 and y2 are the nodal values of the
displacement u. Their values can be found from the following expression:
1
1 2
1 1
( ) ( )
;
( ) ( )
e
e e e e
x x x x
N N
x x x x
+
+ +
− −
= =
− + −
e
(5.17)
Differentiating Eq. (5.16) with respect to x,
1 ' '1 2
1 2
2 2
d dd
d d d
e yN Ny
N N
yx x x
1y
y
⎧ ⎫⎢ ⎥ ⎢ ⎥= =
⎧ ⎫
⎨ ⎬⎢ ⎥ ⎣ ⎦⎣ ⎦ ⎩ ⎭ ⎩ ⎭
⎨ ⎬ (5.18)
Here, ( ′ ) indicates differentiation with respect to x.
Now, a linear weight function is given by,
1
1 2
2
N
w w w
N
⎧ ⎫
= ⎢ ⎥ ⎨ ⎬⎣ ⎦
⎩ ⎭
(5.19)
where w1 and w2 are the weights at the nodes. With these approximations, the weak form of
equation (5.14) is approximated as,
1 1'
11 1 1 1 1' '
1 2 1 2 1 2 1 2'
2 22 2 2 2
( )'
d d
( )'
e e
e
e
x x e
e
x x
NN y E A y
w w EA N N x q w w x w w
y NN E
+ +
⎧ ⎫ ⎧ ⎫−⎧ ⎫⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥− + +⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦
⎩ ⎭ ⎩ ⎭⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭
∫ ∫ 0
A y
=
(5.20)
69

74. Without the loss of generality, we can take xe
=0 and xe+1
=he
, where he
is the length of the element
e. Then, Eq. (5.20) becomes
'
1 1 1 1 1' '
1 2 1 2 1 2 1 2'
22 2 2 20 0
( )'
d d
( )'
e e
h h
N y N E A
w w EA N N x w w q x w w
NN y E A
⎧ ⎫ ⎧ ⎫ ⎧ ⎫−⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ − =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦
⎩ ⎭⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭ ⎩ ⎭
∫ ∫
1
2
y
y
1 1
2 2
y
y
(5.21)
As the weights are arbitrary, the above equation becomes
'
1 1 1 1' '
1 2'
22 2 20 0
( )'
d d
( )'
e e
h h
N y N E A
EA N N x q x
NN y E A
⎧ ⎫ ⎧ ⎫ ⎧ ⎫−⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ = +⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎣ ⎦
⎩ ⎭⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭ ⎩ ⎭
∫ ∫ (5.22)
This is the finite element elemental equation. For Uniform load intensity in an element, the
equation becomes,
1 1
2 2 2
( )'1 1 2
1 1 ( )'
2
e
e e
qh
y E AEA
h y E Aqh
⎧ ⎫
⎪ ⎪⎧ ⎫ ⎧ ⎫−−⎡ ⎤ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪
= +⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥−⎣ ⎦ ⎪ ⎪ ⎪ ⎪⎪ ⎪⎩ ⎭ ⎩ ⎭
⎪ ⎪⎩ ⎭
1 1
2
y
y
d
(5.23)
In the above equation, vectors on the right hand side are load vectors and internal load vectors
respectively.
If there is a concentrated load Q at the second node, the load vector becomes,
1
20
( )
e
h
e N
Q x h x
N
δ
⎧ ⎫
− ⎨ ⎬
⎩ ⎭
∫ (5.24)
where the concentrated load has been converted into distributed load by using Dirac delta
function. The value of the above integral is the value of integrand at the second node. Thus,
the load vector is
0 0
1
Q
Q
⎧ ⎫ ⎧ ⎫
=⎨ ⎬ ⎨ ⎬
⎩ ⎭ ⎩ ⎭
(5.25)
This means the concentrated loads can be accounted for by adding these loads at the position
of node.
70

75. 5.4 FLUID FLOW PROBLEM
Consider the flow of a viscous and incompressible fluid flowing between two parallel
plate. Assume that the flow is fully developed. Because of the symmetry, we can consider
only one half the portion. The governing differential equation for this problem is
2
2
d d
d d
u P
y x
μ = (5.26)
where u is the velocity, μ is the viscosity and P is the pressure. If at a particular cross-section,
the pressure gradient dP/dx is specified, we can find out the velocity distribution along the
radial direction.
The boundary conditions are:
No-slip condition: At the plate surface, u = 0 (5.27)
Symmetry condition:
d
d
u
y
μ = 0 (5.28)
Symmetry condition comes because of the fact that at the line of symmetry there cannot be
any shear force.
The formulation of this problem is similar to the heat conduction or rod loaded with axial
loads problems. You may solve this problem and compare it with the exact solution. The
exact solution is quadratic in y. Therefore, if you take 3-noded element, you will get the
perfect matching with the exact solution with one element only. If you take 2-noded element,
you will never get zero error. Ofcourse, with more number of elements, the error will be very
small.
What is the point in solving this equation by using FEM? We already know the exact
solution. The importance of FEM will come when the viscosity will vary as a function of y. In
that case, it may be difficult to find out the exact solution. However, FEM can provide
approximate solution easily.
5.5 CONCLUSION
In this chapter, we have studied the problems of heat conduction, solid mechanics and
fluid flow problems. The problems from three different areas have been chosen to emphasize
the applicability of FEM as a general tool. By now you might have got the feeling that
although FEM was started as a tool of structural mechanics, it has the potential of application
in other areas too.
71

76. EXERCISE 5
Q.1: The elastic rod is divided into 2 elements of equal length with 3 nodes per element (2 end
points and the mid point) and 1 dof (i.e. u) per node .the corresponding expressions for the
elemental coefficient matrix and the right side vector are. Prove that the elemental stiffness and
load vector are:
[ ]
7 8 1
8 16 8
3
1 8 7
e
e
AE
k
l
−⎡ ⎤
⎢= − −⎢
⎢ ⎥−⎣ ⎦
⎥
⎥ [ ]
e
1
4
6
1
e pl
f
⎡ ⎤
⎢ ⎥= ⎢ ⎥
⎢ ⎥⎣ ⎦
where le
is the length of the element. Find the global coefficient matrix and the right side vector.
Apply the essential boundary conditions. Assuming 2u u4= , determine the nodal displacement
and .2u 3u
Prove that the exact solution is:
( )
2
px l x
u
AE
−
=
Compare the exact solution at the nodes 2 and 3 with their FEM values.
Figure: Q1
Q 2: In a one-dimensional stress analysis of an axial rod one end is fixed and the other end
just touches a wall at room temperature. The rod is heated by 100°C . Model the rod by two
linear elements. Give all the finite element matrix equations to evaluate the reactions at the
ends. Also find the thermal stresses at point P at the center of element (2).
(Make use of initial strain approach)
72

77. Figure: Q2
A1 = 2 cm2
A2 = 1 cm2
l1 = 50 cm l2 = 50 cm
E = 200 GPa ΔT = 100°C α = 17.3 × 10-6
Q.3: Consider the following problem:
D.E:
d d
= 0;
dx dx
T
kA r
⎛ ⎞
+⎜ ⎟
⎝ ⎠
B.C.: (i) Essential: T = T *
at x = 0,
(ii) Natural (convection boundary condition): ( )
d
= - -
dx
0
T
kA hA T T
⎛ ⎞
⎜ ⎟
⎝ ⎠
at x = l.
( )
d
= - -
dx
0
T
kA hA T T
⎛ ⎞
⎜ ⎟
⎝ ⎠
x
Origin
T = Temperature difference
A = area of cross section
r = heat generated per unit
length
k = Thermal conductivity
h = heat transfer coefficient
T0 = Ambient temperature
T *
= Specified temperature at
x = 0.
l
Fig.Q3
Obtain FEM formulation using Ritz FEM and solve by taking 2 elements.
73

78. Q.4: The equation for one-dimensional steady-state heat conduction in absence of heat generation
is:
d d
( )
d d
T
kA
x x
= 0
Let us take a rod with end-point coordinates of x=1 and x=3 respectively. Assume that k=1
throughout the domain and non-uniform cross-sectional area is given by A = x. Boundary
conditions are:
T(at x=1)=300K, T (at x=3)=400K.
You know that the stiffness matrix of an element for this problem is given by
2
0
1 1
d
1 1
h kA
x
h
−⎡ ⎤
∫ ⎢ ⎥−⎣ ⎦
where h is the length of the element.
Now two approaches can be used for computing the stiffness matrix. In one A is treated as
constant and equal to average area of the element and in the other A is treated as varying. Solve
this problem by both the approaches and compare them by studying their performance with
different number of elements. (You can easily find out the exact solution also.) Submitted a short
paper (limited to maximum 3 pages) on this. You may use MATLAB or write a program in
C/C++.
74

79. Chapter 6
FINITE ELEMENT FORMULATION FOR BENDING OF BEAMS
(Lecture 13)
6.1 INTRODUCTION
By now you are familiar with finite element procedure for 1-dimensional problems. We have
taken several examples to illustrate the Galerkin and Ritz FEM formulations. However, so far we
have encountered the problems requiring only C0
continuity for the approximating functions. In
this chapter, we will learn the formulation of the problems requiring C1
continuity i.e., the value
of the function as well as its first derivative should be continuous. To understand the finite
element formulation of such problems, we have chosen the bending of an Euler-Bernoulli beam.
The governing differential equation of this problem is a fourth order differential equation,
requiring C1
continuity. The procedure developed in this chapter will be useful for any fourth
order ordinary differential equation.
6.2 GALERKIN FEM FORMULATION
In strength of materials course, we derive the following differential equation:
4
4
d
d
v
EI q
x
= (6.1)
where EI is called the flexural rigidly, which is the product of Young’s modulus of elasticity and
second moment of the cross-section with respect to centroidal axis and perpendicular to the plane
of bending, v is the vertical defection as a function of longitudinal coordinate x and q is the load
intensity (load per unit length) function. To completely solve this equation, we need 4 boundary
conditions. However, out of these four boundary conditions, at least one boundary condition
should be in the form of prescribed deflection and in addition one other boundary condition
should be prescribed deflection or slope. It is not essential to prescribe the second or third
derivative of the deflection if in place of these the slope and deflections are prescribed. Thus, the
boundary conditions on slope and deflection are called essential boundary conditions, whilst the
other boundary conditions are called natural boundary condition. In the following subsections, we
explain the steps of Galerkin FEM formulation.
75

80. 6.2.1 WEAK FORM
The first step in Galerkin FEM is to obtain the weak form of the differential equation. For
this purpose, we multiply the residual of this differential equation by a weight function w and
integrate it by part so as to evenly distribute the order of differentiation on v and w. In the weak
form, both v and w will be having the derivatives upto the second order only. Carrying out the
integration by parts two times, we obtain
4 3 2 2 2
4 3 2 2 20 0 0
0 0
d d d d d d
d d
d d d d d d
l l
l l lv v v w w v
EI q w x EI w EI EI x qw x
x x x x x x
⎛ ⎞
− = − + −⎜ ⎟
⎝ ⎠
∫ ∫ d 0=∫ (6.2)
Here, l is the length of the beam. We have skipped certain steps in writing equation (6.2). You are
supposed to verify its correctness by doing it yourself.
In the weak form, if we put w equal to δv, we can get the variational form. Then, seeing
the terms on the boundary, we can recognize, that at x=0 and l:
3
3
d
Either 0 or =0
d
v
EI v
x
δ= (6.3)
and
2
2
d d
Either 0 or =0
d d
v v
EI
x x
δ= (6.4)
The first boundary condition in each of the equations (6.3-6.4) is called the natural boundary
condition, whilst the boundary conditions of having 0 variation in the deflection v and slope
dv/dx are called the essential (geometric) boundary condition. From the strength of materials,
recall that EI d2
v/dx2
is the bending moment and EI d3
v/dx3
is the negative of the shear force. (The
sign convention for the bending moment and shear force may differ from book to book.)
6.2.2 CHOOSING SUITABLE APPROXIMATING FUNCTIONS
After obtaining the weak form, we have to choose the suitable approximating functions
within a element. Observing the weak form in equation (6.2), we see that highest order of the
derivative on v is 3, therefore, the approximating function should be thrice differentiable. A third
degree polynomial is that type function. Thus, one may take approximate v within an element as
2
v a bx cx dx= + + + 3
(6.5)
Inside the integral, the highest order of derivative is 2, therefore, the overall approximation
should be C1
continuous. If we take a 4-noded Lagrange element, it will not guarantee that the
slope at the end points will be same from two adjacent elements. However, if we obtain the
constants a, b, c and d in equation (6.5) by expressing them in terms of the slopes and
76

81. deflections at the ends of the element, the continuity of slope is ensured. In Galerkin FEM, w is
approximated in the same way as v . In the next subsection, we shall express the approximating
functions in terms of nodal values of slope and deflection.
6.2.3 HERMITIAN SHAPE FUNCTION
We denote the end points (nodes) of a beam element by 1 and 2 and use them as
subscript for specifying the value at the point. The coordinate of point 1 is 0 and that of 2 is h.
Then,
1
1
2
2
2
2
d
d
d
2 3
d
v a
v
b
x
v a bh ch dh
v
b ch dh
x
=
⎛ ⎞
=⎜ ⎟
⎝ ⎠
= + + +
⎛ ⎞
= + +⎜ ⎟
⎝ ⎠
3
'
(6.6)
With the help of these set of equations, expressing the constants in terms of the nodal values of
slopes and deflection, putting them in equation (6.5) and rearranging, we get
'
1 1 2 1 3 2 4 2v N v N v N v N v= + + + (6.7)
where N1 , N2, N3 and N4 are called Hermitian shape function. Their values are given by
2 3
1 1 3 2
x x
N
h h
⎛ ⎞ ⎛ ⎞
= − +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
(6.8)
2
2 1
x
N x
h
⎛ ⎞
= −⎜ ⎟
⎝ ⎠
(6.9)
2 3
3 3 2
x x
N
h h
⎛ ⎞ ⎛ ⎞
= −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
(6.10)
2
4
x x
N x
h h
⎡ ⎤⎛ ⎞
= −⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦
(6.11)
If the coordinates of the first node is not 0, but is xi, the shape functions are given by
2 3
1 1 3 2i ix x x x
N
h h
− −⎛ ⎞ ⎛
= − +⎜ ⎟ ⎜
⎝ ⎠ ⎝
⎞
⎟
⎠
(6.12)
( )
2
2 1 i
i
x x
N x x
h
−⎛
= − −⎜
⎝ ⎠
⎞
⎟ (6.13)
77

82. 2 3
3 3 2i ix x x x
N
h h
− −⎛ ⎞ ⎛
= −⎜ ⎟ ⎜
⎝ ⎠ ⎝
⎞
⎟
⎠
(6.14)
( )
2
4
i
i
ix x x x
N x x
h h
⎡ ⎤− −⎛ ⎞
⎢ ⎥= − −⎜ ⎟
⎢ ⎥⎝ ⎠⎣ ⎦
(6.15)
6.2.4 ELEMENTAL EQUATIONS
For obtaining the elemental stiffness matrix for a beam element of length h, we change l to
h in equation (6.2) and substitute the approximations
1
'
1
1 2 3 4
2
'
2
v
v
v N N N N
v
v
⎧ ⎫
⎪ ⎪
⎪ ⎪
= ⎢ ⎥ ⎨ ⎬⎣ ⎦
⎪ ⎪
⎪ ⎪
⎩ ⎭
(6.16)
and
1 1
2' '
1 1 2 2
3 3
4 4
ne
N N
N N
w w w w w w
N N
N N
2
⎧ ⎫ ⎧ ⎫
⎪ ⎪ ⎪ ⎪
⎪ ⎪ ⎪
⎢ ⎥ ⎢ ⎥= =
⎪
⎨ ⎬ ⎨⎣ ⎦ ⎣ ⎦ ⎬
⎪ ⎪ ⎪ ⎪
⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭⎪
1
1
2
2
V
M
V
M
1
1
2
2
V
M
V
M
(6.17)
Thus,
1 1 1
2 1 2
1 2 3 4
3 2 30 0
4 2 4
d d
h h
ne ne ne
N v N
N v N
w EI N N N N x w q x w
N v N
N v N
′′ −⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ⎧ ⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪′′ ′ −⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥′′ ′′ ′′ ′′ = +⎡ ⎤⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦′′⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪′′ ′⎩ ⎭ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭
∫ ∫ (6.18)
As the nodal weights are arbitrary, equation (6.18) gives the following elemental equation:
1 1 1
2 1 2
1 2 3 4
3 2 30 0
4 2 4
d d
h h
N v N
N v N
EI N N N N x q x
N v N
N v N
′′ −⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ⎧ ⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪′′ ′ −⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪
′′ ′′ ′′ ′′ = +⎡ ⎤⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎣ ⎦′′⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪′′ ′⎩ ⎭ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭
∫ ∫ (6.19)
Here, the first term on the left hand side of the equality sign is called the stiffness term, the last
term on the right hand side is called internal load vector and the middle term is called the load
vector.
6.2.5 ASSEMBLY BOUNDARY CONDITION AND SOLUTION
After obtaining the elemental equations, we assemble them as usual. In the process internal
loads viz. shear force and bending moments get cancelled except at boundary. These are tackled
by means of the boundary conditions. Either their values are specified or in lieu of that the slope
and/or deflections are prescribed. We can then solve the equations.
78

83. 6.3 RITZ FEM FORMULATION
Many a times, it is possible to find a functional, minimization of which along with the
satisfaction of geometric boundary conditions means solution of governing differential equation
along with geometric and force boundary condition. Ritz method can be used in that situation.
The method is illustrated with the help of the example of a Beam. In a typical beam element of
length h, the total potential energy may be written as
2
2
1 1 2 2 1 1 2 22
0 0
1
d d
2
h h
v
EI x qv x V v V v M v M v
x
⎛ ⎞∂
′ ′∏ = − − + − +⎜ ⎟⎜ ⎟∂⎝ ⎠
∫ ∫
(6.20)
where the first term is the strain energy and all other terms are work potential. Here, v is the beam
deflection, V the shear force and M the moment. The subscript 1 and 2 indicate nodes. The
highest order of derivative in this expression is 3, since the shear force V contains the third
derivative of v. Hence, v should be of the form . Since the highest order of
derivative inside the integral is 2, the approximating function should be C1
continuous. One
suitable approximation is
2
a bx cx dx+ + + 3
1
1
1 2 3 4
2
2
e
v
v
v N N N N
v
v
⎧ ⎫
⎪ ⎪′⎪ ⎪
= ⎡ ⎤⎨ ⎬⎣ ⎦
⎪ ⎪
⎪ ⎪′⎩ ⎭ (5.27)
Here, N1, N2, N3 and N4 are called Hermitian shape functions. Then,
[ ]
[ ] [ ]
1 1
2 1
1 1 2 2 1 2 3 4
3 20
4 2
1 1
2 1
1 1 2 2 1 1 2 2
3 20
4
1
d
2
d
′′
2
⎧ ⎫ ⎧ ⎫
⎪ ⎪ ⎪′′ ′ ⎪
⎪ ⎪ ⎪
′ ′ ′′ ′′ ′′ ′′Π = ⎡ ⎤
⎪
⎨ ⎬ ⎨⎣ ⎦′′
⎬
⎪ ⎪ ⎪ ⎪
⎪ ⎪ ⎪′′ ′⎩ ⎭ ⎩ ⎭⎪
⎧ ⎫ ⎧ ⎫
⎪ ⎪ ⎪ ⎪
⎪ ⎪ ⎪
′ ′ ′ ′− −
⎪
⎨ ⎬ ⎨
−
⎬
⎪ ⎪ ⎪ ⎪
⎪ ⎪ ⎪−⎩ ⎭ ⎩ ⎭
∫
∫
h
h
N v
N v
v v v v EI N N N N x
N v
N v
N V
N M
q v v v v x v v v v
N V
N ⎪M
]
(5.32)
For minimizing this expression, we differentiate the expression with respect to [v1 v′1 v2 v′2]
and set equal to zero. Thus,
[ 1 1 2 2
0
∂ ∏
=
′ ′∂ v v v v
(5.33)
79

84. 1 1 1
2 1 2
1 2 3 4
3 2 30 0
4 2 4
d d
h h
N v N
N v N
EI N N N N x q x
N v N
N v N
′′ −⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ⎧ ⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪′′ ′ −⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪
′′ ′′ ′′ ′′ = +⎡ ⎤⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎣ ⎦′′⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪′′ ′⎩ ⎭ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭
∫ ∫
1
1
2
2
V
M
V
M
(5.34)
This is the elemental formulation. The technique of assembly is carried as usual and followed
by application of boundary conditions to determine the solution.
6.4 SUMMARY
In this chapter FEM formulation of beam problem is carried out by using Galerkin and Ritz
FEM formulation. Both the formulations yield the same results. The procedure can be
employed for any fourth order differential equation.
80

85. EXERCISE 6
Q.1. A beam is modeled by one element. From finite element equation evaluate the
deflection at free end due to concentrated load, P.
If the beam is acted upon by a uniformly distributed load of intensity q per unit length
find the deflection at the free end. Here, first take one element and then keep on
increasing the number of elements. Plot the error in end point deflection versus number
of elements.
Give reasons why the deflection result in the first case is exact and not so in the second
case.
load q / unit length
1 2
P
1 2
Figure: Q1
Given A, I, E and l are the cross-sectional area, second moment of inertia, Young’s
modulus of elasticity and length respectively.
Q.2. A load P is applied at the end of the cantilever which is supported on a spring of
stiffness k. The length of the cantilever is L, flexural rigidity is EI.
(a) Write down the essential and natural boundary conditions of the problem.
81

86. Figure: Q2
(b) Take one element to solve this problem by FEM. The stiffness matrix for that
element is given by
2 2
2
12 6 12 6
6
4 2
12 6
4
L L L L
EI
L
L
SYM
L L
−⎡ ⎤
⎢ ⎥
⎢ ⎥
−⎢ ⎥
⎢ ⎥
⎢ ⎥−⎢ ⎥
⎢ ⎥
⎢ ⎥
⎣ ⎦
You have to generate load matrix, apply boundary conditions and obtain the slope and
deflection at the ends
Q.3. The functional governing static buckling of the column in fig. is
2 22
2
20 0
1 d d 1
d d
2 d 2 d 2
L L
L
w P w
EI x x
x x
⎛ ⎞ ⎛ ⎞
= −⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
∏ ∫ ∫ kw+ (a)
where L x L
w w =
= and the essential boundary conditions are
0
0
0, 0x
x
dw
w
dx=
=
= = (b)
Invoke the stationary condition 0δ ∏ = to derive the problem-governing differential
equation and the natural boundary conditions.
82

87. For the case, when load P is zero, carry out the FEM formulation by Ritz FEM.
Figure: Q3
Q.4. The distribution of bending moment M in a beam subjected to a loading by a
distributed load w(x) per unit length satisfies the equation ( )2 2
d d .M x w x= A beam of
unit length is simply supported (i.e. , M = 0) at both ends and carries a load ( ) sinw x xπ=
per unit length.
Carry out FEM formulation for obtaining the bending moment distribution. Obtain the
bending moment distributions by taking 2 and 3 elements respectively.
83

88. Chapter 7
FINITE ELEMENT FORMULATION FOR TRUSSES AND FRAMES
(Lectures 14-15)
7.1 INTRODUCTION
A truss is a structure to support the load. When a load is applied on the truss, it is
supposed to have no displacement of the load or any part of the structure. However, in practice,
the structure undergoes elastic deformation. The members of the truss are straight slender bars or
rod and in ideal case, are connected with each other by pin joints. In actual case, the members
may be riveted, bolted or welded at the end. However, as long as they are slender and cannot
carry bending load, they may be assumed to be connected through pin joints. The loads in a truss
are applied only at the joints. In cases, where the weight of the member is not assumed to be
negligible, the assumption is made that half of the weight of each member acts as an applied force
at the joints at each end of the member.
There are two types of trusses, plane trusses and space trusses. In plane trusses, both the
truss structure and the applied loads lie in the same plane. In space trusses either the structure or
the applied loads or both lie in different planes. Each member of the truss can be considered as a
rod/bar subjected to an axial load. Therefore, its stiffness can be easily calculated by the method
introduced in Chapter 1. However, one has to transform, the stiffness matrix into a global form so
that the assembly can be easily done.
Frames are the structures in which at least one member carries the transverse load. The
members of the frames resist axial as well bending loads. Here, also the local stiffness matrix can
be easily written as the combination of the stiffness matrices of a beam and a rod subjected to
axial load. The local stiffness matrix has to be converted to global form for assembly purpose.
85

89. 7.2 FORMULATION FOR A TRUSS
Figure 7.1: Rotated coordinate system
Figure 7.1 shows two Cartesian coordinate systems, which have same origin but are
rotated with respect to each other. A vector having the components in the coordinate
system
'
, and'
u v w′
' '
x y z′− − has components u, v and w in the x-y-z system. These components are related
as
[ ]
u u
v R v
w w
′⎧ ⎫ ⎧ ⎫
⎪ ⎪ ⎪ ⎪
′ =⎨ ⎬ ⎨ ⎬
⎪ ⎪ ⎪ ⎪′⎩ ⎭ ⎩ ⎭
(7.1)
where
[ ]
1 1 1
2 2 2
3 3 3
l m n
R l m n
l m n
⎡ ⎤
⎢ ⎥=⎢ ⎥
⎢ ⎥⎣ ⎦
(7.2)
is called rotation matrix. It is orthogonal i.e. its inverse is equal to its transpose. Thus
[ ]
T
u u
v R v
w w
′⎧ ⎫ ⎧ ⎫
⎪ ⎪ ⎪ ⎪
′=⎨ ⎬ ⎨ ⎬
⎪ ⎪ ⎪ ⎪′⎩ ⎭ ⎩ ⎭
(7.3)
If { } [ ]
T
d u v w= is displacement vector, then
86

90. { } [ ]{ }d R d′ = and { } [ ] { }
T
d R d′= (7.4)
If is a force vector, then{ }
T
x y zr F F F⎡= ⎣ ⎤⎦
{ } [ ]{ }r R r′ = and { } [ ] { }
T
r R r′= (7.5)
If { } [ ]{ }d T d′ = where is not necessarily orthogonal and may not even be square, then[ ]T
{ } [ ] { }
T
r T r′= (7.6)
This can be proved as follows.
From work equality
{ } { } { } { }
T T
d r d rδ δ ′ ′= or { } { } { } [ ] { }
TT T
d r d T rδ δ ′=
from which we can write, { } { } [ ] { } 0
TT
d r T rδ ⎡ ⎤′− =
⎣ ⎦
Therefore { } [ ] { }
T
r T r′=
Now consider the elemental equations in the global x-y-z and the local ' '
x y z′− − system,
[ ]{ } [ ]{ } { }ork d r k d r′ ′= ′= (7.7)
Now if { } [ ]{ }d T d′ =
then { } [ ] { }
T
r T r′=
As [ ]{ } { } [ ] { }
T
k d r T r′= =
[ ] [ ]{ } [ ] [ ][ ]{ }
T T
T k d T k T d′ ′ ′= =
We get
[ ] [ ] [ ][ ]
T
k T k T′= (7.8)
This is the equation for transforming the local stiffness equation into the global stiffness
equation.
For a plane truss, the local stiffness matrix is given as
87

91. [ ]
1 1
1 1
AE
k
L
−⎡ ⎤
′ = ⎢ ⎥−⎣ ⎦
(7.9)
The local degrees of freedom are related to the global degrees of freedom as
[ ]
1
1
u
u v
T
u
1
2
2
2
u
v
⎧ ⎫
⎪ ⎪⎧ ⎫′⎪ ⎪ ⎪
=
⎪
⎨ ⎬ ⎨
′
⎬
⎪ ⎪ ⎪ ⎪⎩ ⎭
⎪ ⎪⎩ ⎭
(7.10)
where
[ ]2 4
cos sin 0 0
0 0 cos sin
T
θ θ
θ θ×
⎡ ⎤
= ⎢ ⎥
⎣ ⎦
(7.11)
Therefore, the formula for transforming the local stiffness matrix into the global one is
[ ] [ ] [ ]'
4 2 2 42 2
T
k T k T× ××
⎡ ⎤= ⎣ ⎦
From this, we get
[ ]
2 2
2 2
2
2
cos cos sin cos cos sin
sin cos sin sin
cos cos sin
sin
AE
k
L
θ θ θ θ θ θ
θ θ θ θ
θ θ θ
θ
⎡ ⎤− −
⎢ ⎥
− −⎢ ⎥=
⎢ ⎥
⎢ ⎥
⎣ ⎦
(7.12)
For space truss,
[ ] 11 1
2 6
1 1 1
0 0 0
00 0
nl m
T
l m n×
⎡ ⎤
= ⎢ ⎥
⎣ ⎦
Rest of the procedure is similar to that for plane truss.
88

92. 7.3 AN EXAMPLE
Figure 7.2 shows a plane truss. The point 1 is pinned and point 3 is supported on a roller support.
The roller can roll in a plane inclined at 450
from the horizontal. Let us solve this problem using
FEM.
Figure 7.2: A typical plane truss
Let for element 1-2, AE = 3√2, for 2-3, AE = 3, for 3-1, AE = 3
For element 1-2, θ = 1350
( )
( )
( )
( )
1
1
1 1
11
1
2
2
1
2
2
1 1 1 1
2 2 2 2
1 1 1
3 2 2 2 2
1 13 2
2 2
1
2
x
y
x
y
F
U
FV
U F
V
F
⎡ ⎤
− −⎢ ⎥ ⎧ ⎫
⎢ ⎥ ⎧ ⎫ ⎪ ⎪
⎢ ⎥− ⎪ ⎪ ⎪ ⎪
⎢ ⎥ ⎪ ⎪ ⎪ ⎪
=⎨ ⎬ ⎨ ⎬⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥−
⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎩ ⎭
⎪ ⎪⎢ ⎥ ⎩ ⎭
⎢ ⎥
⎣ ⎦
For element 2-3, θ = -900
89

93. ( )
( )
( )
( )
2
2
2 2
22
2
3
3
2
3
3
0 0 0 0
1 0 13
0 03
1
x
y
x
y
F
U
FV
U F
V
F
⎧ ⎫
⎧ ⎫⎡ ⎤ ⎪ ⎪
⎪ ⎪⎢ ⎥ ⎪ ⎪− ⎪ ⎪ ⎪ ⎪⎢ ⎥ =⎨ ⎬ ⎨ ⎬
⎢ ⎥ ⎪ ⎪ ⎪ ⎪
⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎣ ⎦ ⎩ ⎭
⎪ ⎪⎩ ⎭
For element 1-3, θ = 1800
( )
( )
( )
( )
3
1
1 3
11
3
3
3
3
3
3
1 0 1 0
0 0 03
1 03
0
x
y
x
y
F
U
FV
U F
V
F
⎧ ⎫
− ⎧ ⎫⎡ ⎤ ⎪ ⎪
⎪ ⎪⎢ ⎥ ⎪ ⎪
⎪ ⎪ ⎪ ⎪⎢ ⎥ =⎨ ⎬ ⎨ ⎬
⎢ ⎥ ⎪ ⎪ ⎪ ⎪
⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎣ ⎦ ⎩ ⎭
⎪ ⎪⎩ ⎭
Assembling,
( )
( )
( )
( )
1
1
1 1
2
2
3 3
3
3
0.5 1 0.5 0 0.5 0.5 1 0
0.5 0 0.5 0 0.5 0.5 0 0
100.5 0.5 0.5 0 0.5 0 0 0
00.5 0.5 0.5 0 0.5 1 0 1
1 0 0 0 1 0
0 0 0 1 0 0 1 0
x
y
x
y
F
U
FV
U
V
U F
V F
⎧ ⎫
+ − + − − ⎧ ⎫⎡ ⎤ ⎪ ⎪
⎪ ⎪⎢ ⎥ ⎪ ⎪− + + − ⎪ ⎪⎢ ⎥ ⎪ ⎪
⎪ ⎪⎢ ⎥− + − + ⎪ ⎪ ⎪ ⎪
=⎨ ⎬ ⎨ ⎬⎢ ⎥
− − + + − ⎪ ⎪ ⎪ ⎪⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥−
⎪ ⎪ ⎪ ⎪⎢ ⎥
− + + ⎪ ⎪⎣ ⎦ ⎪ ⎪⎩ ⎭
⎩ ⎭
As U1 = 0, V1 = 0, we can eliminate first 2 rows and columns to get,
( )
( )
2
2
33
3
3
100.5 0.5 0 0
00.5 1.5 0 1
0 0 1 0
0 1 0 1
x
y
U
V
FU
V F
⎧ ⎫− ⎧ ⎫⎡ ⎤
⎪ ⎪⎪ ⎪⎢ ⎥− − ⎪ ⎪⎪ ⎪⎢ ⎥ =⎨ ⎬ ⎨ ⎬
⎢ ⎥ ⎪ ⎪ ⎪ ⎪
⎢ ⎥ ⎪ ⎪ ⎪ ⎪−⎣ ⎦ ⎩ ⎭ ⎩ ⎭
At node 3, force along the inclined plane is zero. Hence,
( ) ( )0 0
3 3
cos45 sin 45 0x yF F− =
or,
90

94. ( ) ( )3 3
0x yF F− =
To enforce it, subtract fourth row from third, to get
( )
2
2
3
3 3
100.5 0.5 0 0
00.5 1.5 0 1
00 1 1 1
0 1 0 1 y
U
V
U
FV
⎧ ⎫− ⎧ ⎫⎡ ⎤
⎪ ⎪⎪ ⎪⎢ ⎥− − ⎪ ⎪ ⎪ ⎪⎢ ⎥ =⎨ ⎬ ⎨ ⎬
⎢ ⎥− ⎪ ⎪ ⎪ ⎪
⎢ ⎥ ⎪ ⎪ ⎪ ⎪−⎣ ⎦ ⎩ ⎭ ⎩ ⎭
As the normal displacement is zero at node 3.
0 0
3 3cos45 cos 45 0U V+ =
i.e. . Replace fourth equation by this.3 3 0U V+ =
2
2
3
3
0.5 0.5 0 0 10
0.5 1.5 0 1 0
0 1 1 1 0
0 0 1 1 0
U
V
U
V
− ⎧ ⎫⎡ ⎤
⎪ ⎪⎢ ⎥− − ⎪ ⎪ ⎪ ⎪⎢ ⎥ =⎨ ⎬ ⎨ ⎬
⎢ ⎥− ⎪ ⎪ ⎪ ⎪
⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎣ ⎦ ⎩ ⎭
⎧ ⎫
⎪ ⎪
⎩ ⎭
Solving the above equation, we get
2 2 3 340, 20, 10, 10U V U V= = = − =
91

95. 7.4 FEM FORMULATION FOR THE FRAMES
Elemental equation for the frame is
3 2 3 2 1
1
2 2
1
2
2
2
3 2 3 2
2 2
0 0 0 0
0 12 6 0 12 6
0 6 4 0 6 2
0 0 0 0
0 12 6 0 12 6
0 6 2 0 6 4
AE AE
L L
EI EI EI EI
u
L L L L
vEI EI EI EI
LL L L
AE AE
L L v
EI EI EI EI
L L L L
EI EI EI EI
L LL L
u
θ
θ
⎡ ⎤
−⎢ ⎥
⎢ ⎥
⎢ ⎥− ′⎧ ⎫⎢ ⎥
⎪ ⎪⎢ ⎥ ′
⎪ ⎪⎢ ⎥−
⎪ ⎪′⎢ ⎥ ⎪ ⎪
⎨ ⎬⎢ ⎥ ′⎪ ⎪⎢ ⎥−
⎪ ⎪⎢ ⎥ ′
⎪ ⎪⎢ ⎥
′⎪ ⎪− − − ⎩ ⎭⎢ ⎥
⎢ ⎥
⎢ ⎥
−⎢ ⎥
⎣ ⎦
=generalizes force vector (7.13)
or
[ ]{ } { }'
k d' R= (7.14)
We denote the displacement vector in rotated coordinate system by
{ }
T
d u v u vθ θ⎡ ⎤′ =
⎣ ⎦1 1 1 2 2 2
' ' ' ' ' '
(7.15)
Note that slash (/) as a superscript here does not represent differentiation. It refers rotated
coordinate. In the global coordinate system, the displacement vector is
[ T
1 1 1 2 2 2{ }d u v u vθ= ]θ (7.16)
Both displacement vectors are related in the following manner:
1 1
1 1
1 1
2 2
2 2
2 2
cos sin 0 0 0 0
sin cos 0 0 0 0
0 0 1 0 0 0
0 0 0 cos sin 0
0 0 0 sin cos 0
0 0 0 0 0 1
u u
v v
u u
v v
φ φ
φ φ
θ θ
φ φ
φ φ
θ θ
′⎧ ⎫ ⎧ ⎫⎡ ⎤
⎪ ⎪ ⎪ ⎪⎢ ⎥′ −⎪ ⎪ ⎪ ⎪⎢ ⎥
⎪ ⎪ ⎪ ⎪′ ⎢ ⎥⎪ ⎪ ⎪ ⎪
=⎨ ⎬ ⎨ ⎬⎢ ⎥
′⎪ ⎪ ⎪ ⎪⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥′ −
⎪ ⎪ ⎪ ⎪⎢ ⎥
′⎪ ⎪ ⎪ ⎪⎣ ⎦⎩ ⎭ ⎩ ⎭
(7.17)
92

96. where φ is the angle which the frame element makes from horizontal. This can be written as
{ } [ ]{ }d' T d= (7.18)
Then the stiffness matrix in global system is obtained as
[ ] [ ] [ ][ ]T
k T k T′= (7.19)
In the frame problems, if the load acts other than joints, they can be transferred to joint by
calculating the load vectors as is done in beams and rods.
7.5 SUMMARY
In this chapter FEM formulation for trusses and frames is carried out. We have solved one
example for the truss problem. For the frames, we have not solved any problem. As the size of the
assembled stiffness matrix in frame problems is usually large, these problems are suitable for
solving with the help of a computer.
EXERCISE 7
Q.1: A truss problem was solved in this chapter. Solve it again by yourself and find out the
axial forces in all the elements. Find out the support reactions also.
Q.2: Shown below is a five member truss. It is made of steel. The cross-sectional area each
member is 6cm2
. It is loaded by a vertical force of 100N at the rightmost top point.
Find out the defection of that point.
Figure: Q2
Q.3: Frame shown below has all the members of equal length. You assume the length, A
93

97. and EI as unity. If a unit load is applied as shown, find out the deflection at the point of
application of the load and maximum stress in the frame.
Figure: Q3
Q.4: In the figure a cantilever beam is supported on a stepped rod. Take one beam element and
two rod elements to solve it. The stiffness matrix for beam element is
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−−−
−
−
22
22
3
4626
612612
2646
612612
hhhh
hh
hhhh
hh
h
EI
.
For the rod element you already remember (else derive it). Solve this problem to find out the
deflections and stresses.
Figure: Q4
1 m
10 mm dia, Steel 3 mm dia, Al.
6 mm dia Al 0.2 mm
0.2 mm
1 kN
94

98. Q.5: For the truss shown below, find out the displacements at 2, 3 and 4 when a vertical load P is
applied at 4. Find out the stresses in the all elements.
Figure: Q5
95

99. Chapter 8
INTRODUCTION TO 2-D and 3-D FEM
(Lectures 16-19)
8.1 INTRODUCTION
So far we have discussed the problems involving the solutions of one-dimensional
differential equations. Now we move towards solving the physical problems involving two-
dimensional and three-dimensional differential equations. In this chapter, we shall study the FEM
formulation for Laplace and Poisson equations. We shall start this chapter with a discussion on
different types of elements and then develop the finite element equations. The detailed derivation
of two-dimensional equation will be provided. The reader can easily extend the procedure to
obtain FEM equations for 3-dimensional equations.
8.2 TRIANGULAR ELEMENTS
In one-dimension, a finite element can be described by two nodes. The line passing
through these nodes forms a line element. In two dimensions, a minimum of three nodes are
required to make an element. These nodes should be able to form a triangle.
Figure 8.1: A triangular element
Fig. 8.1 shows a 3-noded triangular element in 2-dimension. Let us say we have to
interpolate temperature T in it. We need to express the constants in terms of nodal temperatures.
As there are three nodes, we can determine three constants. So what type of function should be
taken? For better visualization, one can construct a triangle similar to Pascal’s triangle, as shown
in Fig. 8.2.
97

100. 1
x y
x2
xy y2
x3
x2
y xy2
y3
x4
x3
y x2
y2
xy3
y4
x5
x4
y x3
y2
x2
y3
xy4
y5
Figure 8.2: The triangle for helping in choosing the appropriate interpolation function
In Fig. 8.2, the first row contains only one term i.e.1, in which sum of powers of x and y
is 0. The second row contains two terms i.e. x and y, in which the sum of powers of x and y is 1.
The third row contains, the terms in which the powers of x and y sum to 1 and so on. For the
three noded element, we choose first two rows of the triangle. Thus,
T a bx cy= + + (8.1)
This linear equation must be satisfied at the nodes. Therefore,
1 1 1 2 2 2 3 3, ,T a bx cy T a bx cy T a bx cy= + + = + + = + + 3 (8.2)
or
1 1 1
2 2 2
3 3 3
1
1
1
x y Ta
x y b T
cx y T
⎡ ⎤ ⎧⎧ ⎫ ⎫
⎪ ⎪ ⎪ ⎪⎢ ⎥ =⎨ ⎬ ⎨ ⎬⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎩ ⎭⎣ ⎦ ⎩ ⎭
(8.3)
From Eq. (8.1),
[ ] [ ]
1
1 1 1 1
2 2 2 1 2 3 2
3 3 3 3
1
[1 ] 1 1
1
x y T Ta
T x y b x y x y T N N N T
c x y T T
−
⎡ ⎤ ⎧ ⎫ ⎧ ⎫⎧ ⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥= = =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥
⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥⎩ ⎭ ⎣ ⎦ ⎩ ⎭ ⎩ ⎭
(8.4)
where N1, N2 and N3 are shape functions given by,
2 2 2 3 3 2 3 3
1
1 2 2 1 2 3 3 2 3 1 3 1
1 1 3 3 3 1 3 1
2
1 2 2 1 2 3 3 2 3 1 3 1
1 2 2 1 2 2
3
( ) ( ) ( ) area( 23)
( ) ( ) ( ) area(123)
( ) ( ) ( ) area( 31)
( ) ( ) ( ) area(123)
( ) ( )
xy x y x y x y x y y x P
N
x y x y x y x y x y y x
x y xy xy x y x y y x P
N
x y x y x y x y x y y x
x y x y x y xy
N
− + − + −
= =
− + − + −
− + − + −
= =
− + − + −
− + −
= 1 1
1 2 2 1 2 3 3 2 3 1 3 1
( ) area( 12)
( ) ( ) ( ) area(123)
xy yx P
x y x y x y x y x y y x
+ −
=
− + − + −
(8.5)
98

101. where P is the point having coordinates (x, y). It is clearly seen that the sum of shape functions is
1. We can define the natural coordinates of a point (x, y) as
2 2 2 3 3 2 3 3
1
1 2 2 1 2 3 3 2 3 1 3 1
1 1 3 3 3 1 3 1
2
1 2 2 1 2 3 3 2 3 1 3 1
1 2 2 1 2 2
3
( ) ( ) ( ) area( 23)
( ) ( ) ( ) area(123)
( ) ( ) ( ) area( 31)
( ) ( ) ( ) area(123)
( ) ( )
xy x y x y x y x y y x P
x y x y x y x y x y y x
x y xy xy x y x y y x P
x y x y x y x y x y y x
x y x y x y xy
ξ
ξ
ξ
− + − + −
= =
− + − + −
− + − + −
= =
− + − + −
− + −
= 1 1
1 2 2 1 2 3 3 2 3 1 3 1
( ) area( 12)
( ) ( ) ( ) area(123)
xy yx P
x y x y x y x y x y y x
+ −
=
− + − + −
(8.6)
The natural coordinates of every point will be bounded between 0 and 1 and
1 2 3 1ξ ξ ξ+ + = (8.7)
Thus, there are only 2 independent natural coordinates. For a 3-noded triangular element
1 1 2 2 3; ;N N N 3ξ ξ ξ= = =
2
3
(8.8)
The 3-noded linear triangular element is also called constant strain triangle (CST) element,
because if we approximate the displacement components using this element, strains will be
constant.
1 4 2
3
56
Figure 8.3: A six noded triangular element
A six noded triangular element is shown with 3 nodes at the corner and 3 mid-side nodes.
This element can carry out quadratic interpolation. Taking first three rows from the triangle
shown in Fig. 8.2, T can be approximated as
2
T a bx cy dx exy fy= + + + + + (8.9)
We can find out the shape functions, the way we have done for 3-noded element. However, here
we derive the shape functions in a different way using natural coordinates. In natural coordinates,
the shape functions corresponding ith
node are:
2 2 2
1 2 3 1 2 2 3 1i i i i i i iN a b c d e fξ ξ ξ ξ ξ ξ ξ ξ ξ= + + + + + (8.10)
99

102. Note that in the above expression, the sum of the powers of natural coordinates is 2 in each term.
One may wonder why a constant term or term containing natural coordinates raised to degree one
are not present. The answer is that even if a constant term is present, it can be multiplied by
2
1 2 3( )ξ ξ ξ+ + , which equal to 1 only. Similarly, a term containing a single degree natural
coordinate can be multiplied by 1 2 3( )ξ ξ ξ+ + . Thus, ultimately, we can obtain the form given in
Eq. (8.10).
Let us first derive the expression for N4, which can be written as
2 2 2
4 4 1 4 2 4 3 4 1 2 4 2 3 4 1N a b c d e f 3ξ ξ ξ ξ ξ ξ ξ ξ ξ= + + + + + (8.11)
It is zero at all other nodes and 1 at the node 4. At node 1, ξ1=1, ξ2=0 and ξ3=0. Thus,
4 0a = (8.12)
Similarly, it can be shown that
4 4 0b c= = (8.13)
At node 5, 2 3
1
2
ξ ξ= = , thus
4 4
1
0 or 0
4
e e= = (8.14)
At node 6, 1 3
1
2
ξ ξ= = , thus
4 4
1
0 or 0
4
f f= =
Finally, at node 4, 1 2
1
2
ξ ξ= = , thus
4 4
1
1 or 4
4
d d= = .
Therefore,
4 14N 2ξ ξ= (8.15)
Similarly, it can be shown (rather it can be directly written by noting the similarity) that
5 2 3 6 14 and 4N N 3ξ ξ ξ ξ= = (8.16)
Let us now derive the expression for N1, which can be written as
2 2 2
1 1 1 1 2 1 3 1 1 2 1 2 3 1 1 3N a b c d e fξ ξ ξ ξ ξ ξ ξ ξ= + + + + + ξ (8.17)
This shape function is 0 at all nodes expect at node 1, where its value is 1.
100

103. At node 2, ξ1=ξ3=0 and ξ2=1, giving
1 0b = (8.18)
At node 3, ξ1=ξ2=0 and ξ3=1, giving
1 0c = (8.19)
At node 5, ξ1=0, ξ2= ξ3=1/2, giving
1 0e = (8.20)
At node 1, ξ2=ξ3=0 and ξ1=1, giving
1 1a = (8.21)
At node 4, ξ1=ξ2=1/2 and ξ3=0, giving
1 1
10 or 1
4 4
a d
d+ = = − (8.22)
Lastly, at node 6, ξ1=ξ3=1/2 and ξ2=0, giving
1 1f = − (8.23)
Therefore,
2
1 1 1 2 1N 3ξ ξ ξ ξ ξ= − − (8.24)
The above expression can be simplified as
2 2
1 1 1 2 3 1 1 1 1 1( ) (1 ) (2N ξ ξ ξ ξ ξ ξ ξ ξ ξ= − + = − − = −1) (8.25)
By analogy, we can write,
2 2 2 3 3 3(2 1) and (2 1)N Nξ ξ= − = −ξ ξ (8.26)
Figure 8.4: A triangular element interpolating a M degree polynomial
101

104. We have derived the shape functions for a linear (degree 1) and a quadratic (degree 2)
element. Now, let us derive the expression for degree M polynomial. The similar procedure can
be followed. However, Argyris et al. [1] and some others [2, 3] have developed the simple
procedure. In this procedure, each node i is represented by triplet (I, J, K). For example, the
corner nodes in the baseline are (M, 0, 0) and (0, M, 0). On the baseline first coordinate keeps
decreasing by one from left to right as we move from one node to other and second coordinate
keeps increasing by 1. For any node,
(8.27)I J K M+ + =
A simple way to obtain triplet corresponding to a node is to assume that nodes are equi-spaced on
sides (they need not be actually), and multiplying the natural coordinates of the nodes by M. For a
node having designation (I, J, K), the shape function is given by,
1 2( ) ( ) ( )I J K
i I J KN l l l 3ξ ξ ξ= (8.28)
where
0 1 1
0 1
( )( )........( )
( )
( )( )........( )
n n
n
n n n n
x x x x x x
l x
x x x x x x
−
1−
− − −
=
− − −
(8.29)
where starting from x0=0, xi s are the 1-d natural coordinates. (You may divide a line with end
coordinates 0 and 1 into n equal parts to obtain the coordinates x0, x1, …….,xn. The coordinate of xn
will be 1.) Take . Using this method, for cubic triangle,0
0 1l =
1 1 1 1 4 1 2 1 10 1
1 9
(3 1)(3 2) , (3 1), 27
2 2
N N N 2 3ξ ξ ξ ξ ξ ξ ξ ξ= − − = − = ξ (8.30)
where N1 is the corner node, N4 is the mid-side node adjacent to node 1 on the line connecting
nodes 1 & 2 and N10 is the internal node.
8.3 TETRAHEDRAL ELEMENTS
These are 3-D elements. Formulae may be derived in the same manner. Linear element
has 4 nodes, quadratic 10 nodes and cubic 20 nodes. Figure 8.5 shows tetrahedral elements with 4
nodes and 10 nodes, respectively. We can define the coordinates of an in inside point by
1
VolumeP234
Volume1234
ξ = etc. (8.31)
Then shape functions for linear elements are
iN iξ= (8.32)
For a quadratic tetrahedron, we have the following relation for the typical nodes:
102

105. For corner node
1 1(2 1)N 1ξ ξ= − (8.33)
For mid-side node
5 14N 2ξ ξ= (8.34)
You can easily write the shape functions for the other nodes.
Figure 8.5: Tetrahedral elements (a) 4 noded (b) 10 noded
8.4 RECTANGULAR ELEMENTS
Figure 8.6 shows 4 and 9 noded rectangular elements. With the help of a four noded
element, a variable, say temperature T, may be interpolated as
( )(T a bx cy dxy A Bx C Dy= + + + = + + ) (8.35)
Thus, the interpolation function is the product of two linear functions in x and y . We can obtain
the constants a, b, c and d (or A, B, C and D) in terms of the nodal values T1, T2 , T3 and T4 and
express the T in the following form:
1 1 2 2 3 3 4 4T N T N T N T N T= + + + (8.36)
103

106. However, the shape functions can be easily obtained if we consider the shape function at a node
to be the product of one-dimensional shape functions along x and y. Thus, if x1=x4=p, x2=x4=q and
y1=y4=r, y2=y4=s, then
1 2
3 4
( ) ( ) ( ) ( )
;
( ) ( ) ( ) (
( ) ( ) ( ) ( )
;
( ) ( ) ( ) (
)
)
x q y s x p y s
N N
p q r s q p r s
x p y r x q y r
N N
q p s r p q s r
− − − −
= =
− − − −
− − − −
= =
− − − −
(8.37)
These shape function satisfy the three properties mentioned in Chapter 4. These shape functions
are called Lagrangian shape functions and the corresponding elements shown in Fig. 8.6 are
called Lagrangian elements. You can easily derive the shape function for 9 noded elements,
which will be products of 1-dimensional quadratic shape functions.
Figure 8.6: Rectangular elements (a) 4 noded (b) 9 noded
8.5 BRICK ELEMENTS
Extending the rectangular elements to 3-dimensions, we obtain brick elements. Figure 8.7
shows 8-noded and 20-noded brick elements. The element shown in Fig. 8.7 (a) is called
Lagrangian elements. The corresponding quadratic element will be a 27 noded element. The 20-
noded element is actually a Serendipity element, which we shall discuss later. The shape
functions for 8-noded elements can be obtained as the product of shape functions along x, y and z.
However, the shape functions for 20 noded-element cannot be obtained by simple multiplication
of 1-dimensional shape functions.
104

107. Figure 8.7: Brick elements (a) 8 noded (b) 20 noded
8.6 GOVERNING DIFFERENTIAL EQUATION FOR 2-D HEAT CONDUCTION
A FE model is developed and implemented for solving a 2-D steady state heat conduction
problem for an orthotropic, non-homogeneous material. The governing equation is given by
0x y
T T
k k
x x y y
⎛ ⎞∂ ∂ ∂ ∂⎛ ⎞
Q+ + =⎜ ⎟⎜ ⎟
∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠
(8.38)
where T is the temperature field, kx is the thermal Conductivity in x direction, ky is the thermal
conductivity in y direction and Q is the rate of heat generation per unit volume.
As the material is non-homogenous, both kx and ky are functions of x and y. The boundary
conditions are
i. Essential: T = T *(x, y), prescribed temperature field on the boundary Г1 .
ii. Natural: Boundary heat flux conditions on boundary Г2.
qn = q*( x, y) is the prescribed boundary heat flux, or
qn = β ( T - T∞) is the convective heat flux.
where
ˆ ˆn x x y
T T
q k n k ˆyn
x x
∂ ∂
= − −
∂ ∂
(8.39)
is the normal heat flux on the boundary Г2 , β is the convective heat transfer coefficient and T∞ is
the ambient temperature. We shall carry out FEM formulation using Galerkin procedure.
8.7 WEAK FORM AND FEM FORMULATION
Once the discretization of the domain is done, the weak form is developed on the
arbitrary typical element. The element is denoted by e
Ω with boundary .e
Γ Let be the finitee
T
105

108. element temperature field within an element. As it is not the exact temperature field, on
substituting in the left hand side of equation (8.38), in general, a non-zero residue, e
R will be
obtained, i.e.
e e
e
x y
T T
R k k
x x y y
⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂
= +⎜ ⎟ ⎜ ⎟
∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠
Q+
0
(8.40)
The weighted integral statement of the governing equation is then given by,
(8.41)d
e
e
wR
Ω
Ω =∫∫
where w is the weighting function.
Thus,
d d 0
e
e e
x y
T T
w k k Q x y
x x y yΩ
⎡ ⎤⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂
− − −⎢ ⎜ ⎟ ⎜ ⎟
∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦
∫ =⎥ (8.42)
Using the component form of gradient theorem, i.e.,
ˆd
e e
x
F
w w
x xΩ Γ
∂ ∂⎛ ⎞
Ω =⎜ ⎟
∂ ∂⎝ ⎠
∫ ∫ dFn s (8.43)
and
(
F w
w F )wF
x x x
∂ ∂ ∂
− = −
∂ ∂ ∂
(8.44)
Equation (8.42) becomes
d d d 0
e
e
x y x x y y
w T w T T T
k k wQ x y w k n k n s
x x y y x xΓΩ
⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞
+ − − +⎜ ⎟ ⎜
∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠
∫ ∫ =⎟ (8.45)
or
ˆd d d d d 0
e e e
e e
x x n
w T w T
k k x y wQ x y wq s
x x y yΩ Ω
⎛ ⎞∂ ∂ ∂ ∂
+ = −⎜ ⎟
∂ ∂ ∂ ∂⎝ ⎠
∫ ∫
Γ
=∫ (8.46)
where is the heat flux in the direction of unit normal n.ˆnq
For a convective boundary, the natural boundary condition is a balance of energy transfer
across the boundary due to conduction and/or convection (i.e. Newton’s law of cooling):
(x x y y
T T
k n k n T T )
x x
β ∞
∂ ∂
+ = − −
∂ ∂
(8.47)
106

109. where β is the convective conductance ( or convective heat transfer coefficient), T is the
ambient temperature of the surrounding fluid medium, and qn is the specified heat flux, if any. It
is the presence of the term
∞
( )T Tβ ∞− that requires some modification of equation (8.46).
In the weak form of (8.38) the boundary integral is modified to account for the convective heat
transfer term in (8.45):
d d d 0
e
e
x y x x y y
w T w T T T
k k wQ x y w k n k n s
x x y y x xΓΩ
⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞
+ − − +⎜ ⎟ ⎜ ⎟
∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠
∫ ∫ = (8.48)
or, ( )
1 2
d d d d 0
e
e e
x y n
w T w T
k k wQ x y wq s w T T s
x x y y
β ∞
Γ ΓΩ
⎛ ⎞∂ ∂ ∂ ∂
+ − − + −⎜ ⎟
∂ ∂ ∂ ∂⎝ ⎠
∫ ∫ ∫ = (8.49)
or, ( ) ( )B , l 0w T w− =
where w is the weight function, and B(.,.) and l(.) are the bilinear forms
( )
e
B , d d d
e
x y
w T w T
w T k k x y wT s
x x y y
β
Ω Γ
⎛ ⎞∂ ∂ ∂ ∂
= + +⎜ ⎟
∂ ∂ ∂ ∂⎝ ⎠
∫ ∫
ds
(8.50)
( )l d d d
e e e
xw wQ x y wT s wqβ ∞
Ω Γ Γ
= + +∫ ∫ ∫ (8.51)
The finite element model is then obtained by substituting the finite element approximation of the
form
(
1
,
n
e e
j j
j
T T N x
=
= ∑ )y
)y
e
y
ds
(8.52)
and
(
1
,
n
e e
j j
j
w w N x
=
= ∑ (8.53)
for T and w respectively, into (8.49).
This leads to the following equations:
( )
1
n
e e e e
ij ij j i i
j
K H T F P
=
+ = +∑ (8.54)
where,
( ), , , , d d
e
e
ij x i x j x y i y j yK k N N k N N x
Ω
= +∫
e
d ,
e
e e e e
ij i j i iH N N s P N Tβ β ∞
Γ Γ
= =∫ ∫
107

110. d d de e
i i x i iF QN x y q N s f Q= + ≡∫ ∫
e e
i+
Once the elemental matrices are obtained, they are assembled to form the set of linear
simultaneous equations, the solution of which yields the temperature field. The assembly is based
on the principle of maintaining the continuity of primary variables i.e. fluxes.
The variation form of the differential equation can be obtained easily by putting δT in place of w
in the weak form. It is given by,
2 2 21 1 1
ˆ( , ) ( , ) d ( )d
2 2 2
e
e
e
x y nk T x k T y QT q T hT hTT s
Γ
∞Ω
⎡ ⎤
∏ = + − Ω + − + −∫ ∫⎢ ⎥⎣ ⎦
(8.55)
Then, Ritz FEM can be employed to obtain the FEM equations.
8.8 A NOTE ON THE ASSEMBLY IN TWO DIMENSIONS
The concept of connectivity matrix was introduced in Chapter 1 itself. In two
dimensional problems, it gains major importance. The connectivity matrix is the matrix in which
a row denotes the element number and a column denotes the local node number. An element of
the connectivity matrix indicates global node number. For Figure 8.8 (a) the connectivity matrix
is given by
1 2 6 5
2 3 7 6
3 4 8 7
5 6 10 9
6 7 11 10
7 8 12 11
9 10 14 13
10 11 15 14
11 12 16 15
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎣ ⎦
Figure 8.8: Mesh of (a) rectangular elements (b) triangular elements
108

111. For Fig. 8.8 (b), the connectivity matrix is given by
1 2 3
3 2 4
3 4 7
3 7 6
3 6 5
1 3 5
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎣ ⎦
The task of assembly is to identify the global location of each element in the local matrix and load
vector and put the element in that place.
8.9 POISSON EQUATION IN 3-D
The steady state heat conduction equation in 3-dimension is given by
0x y z
T T T
k k k
x x y y z z
⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞
Q+ + +⎜ ⎟⎜ ⎟ ⎜ ⎟
∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠
= (8.56)
The FEM formulation for this problem will provide the stiffness equation as
{ } { } { }( ), , , , , , d d d
e
e
x x x y y y z z zK k N N k N N k N N x y z
Ω
⎢ ⎥⎢ ⎥ ⎢ ⎥= + +⎣ ⎦ ⎣ ⎦⎣ ⎦∫ (8.57)
The reader can derive this as well as the load vector.
8.10 FLUID FLOW PROBLEM
For an incompressible and irrotational two-dimensional flow, the vorticity is 0. Therefore,
0
u v
y x
∂ ∂
− =
∂ ∂
(8.58)
Where u and y are the fluid velocities in the x and y directions, respectively. The continuity
condition is given by
0
u v
x y
∂ ∂
+ =
∂ ∂
(8.59)
For solving it, define a potential function φ(x, y) such that
; andu v
x y
φ φ∂ ∂
= =
∂ ∂
(8.60)
Equation (8.58) is identically satisfied. Substitution of equation (8.60) into equation (8.59)
provides
109

112. 2 2
2 2
0
x y
φ φ∂ ∂
+ =
∂ ∂
(8.61)
This is a Laplace equation and FEM formulation can be obtained similar to the case of steady-
state heat conduction. At any boundary,
nv
n
φ∂
=
∂
(8.62)
where
n
φ∂
∂
denotes the derivative of velocity potential in the direction of unit normal vector to the
boundary. This is a natural boundary condition. In addition, at one point, one need to provide an
arbitrary value of velocity potential, which forms an essential boundary condition.
8.11 TORSION OF CIRCULAR AND NONCIRCULAR CROSS-SECTION
The torsional behavior of a shaft of a circular and non-circular cross-section is governed by
2 2
2 2
2G
x y
φ φ
θ
∂ ∂
0+ + =
∂ ∂
(8.63)
where θ is the angle of twist per unit length, G is the shear modulus of the shaft material and φ is
the stress function. The shear stress components at any point can be calculated as
;zx zy
y x
φ φ
τ τ
∂ ∂
= = −
∂ ∂
(8.64)
The boundary condition on φ is that it is zero on the boundary of the shaft. Thus, the problem is
similar to heat conduction problem with heat generation and prescribed temperature on the
boundary.
8.12 SUMMARY
In this chapter an introduction to 2-D and 3-D FEM is provided. We have limited our
discussion to Laplace and Poisson equations in this chapter. In this chapter, we have used simple
types of elements. More details of FEM formulation with isoparametric elements will be provided
in the next chapter.
REFERENCES
1. J.H. Argyris, I. Fried, and D.W. Scharpf, ‘The TET 20 and the TEA 8 elements for the matrix
displacement method,’ Aero. J., 72, 618-25, 1968.
110

113. 2. P. Silvester, ‘Higher order polynomial triangular elements for potential problems’, Int. J. Eng.
Sci., 7, 849-61, 1969.
3. R.L. Taylor, ‘On completeness of shape functions for finite element analysis’ Int. J. Num.
Meth. Eng., 4, 17-22, 1972.
EXERCISE 8
Q.1: Derive all the shape functions for a cubic triangular element using simplified formula
described in this chapter.
Q.2: Sketch a 15 noded triangular element. Find out the expression for the shape functions of this
element.
Q.3. Figure shows the mesh consisting of linear triangular elements. Write down the connectivity
matrix of the mesh.
Figure: Q.3
Q.4: In the triangle shown in the figure a point has natural coordinates (1/3, 1/3, 1/3). Find out its
x and y coordinates.
Figure: Q.4
Q.5: A square element has a point heat source of 100 W at the center. Find out its contribution at
all the corner nodes.
Q.6: Develop the expression for the stiffness matrix for a triangular element for solving steady-
sate heat conduction problem. Treat the material as orthotropic.
Q.7: Develop the expression for the stiffness matrix for a rectangular element of sides a and b for
solving steady-sate heat conduction problem. Treat the material as orthotropic.
111

114. Chapter 9
NUMERICAL INTEGRATION
(Lectures 20-21)
9.1 INTRODUCTION
In finite element method, often the integrations arising in the computation of stiffness
matrix and load vector are carried out numerically. The word ‘quadrature’ is also used to mean
numerical integration. In this chapter, we shall describe the commonly used quadrature formulae.
The accuracy of numerical integration affects the solution accuracy to a great extent. Therefore, a
complete chapter has been devoted on numerical integration.
9.2 ONE DIMENSIONAL INTEGRATION FORMULAE
Integration is basically summation. Figure 1 shows the function f(x) graphically in the
form of the curve 1-2-3-4 The end points of the curve are 1 and 4. Let us consider the sum
1 2 3( ) ( ) ( )S f x x f x x f x x= Δ + Δ + Δ (9.1)
This sum is basically the hatched area in the figure. It need not be equal to the area under the
curve f(x) on the first quadrant. However, if we divide the curve in many more parts, say n parts,
instead of just 3 parts as in Fig.1, then the summation
1
( )
n
i
i
S f x
=
x= Δ∑ (9.2)
will be very close to the area under the curve on the first quadrant. As n and , the
summation S in Eq. (9.2) tends to the area under the curve f(x) on positive quadrant. The
summation in Eq. (9.2) when n and
→ ∞ 0xΔ →
→ ∞ 0xΔ → is denoted as ( )df x x∫ . The symbol ‘∫’ is
distorted form of ‘S’ which implies summation. As integration is basically summation, all
numerical integration formula involve summation. We cannot ofcourse take n=∞, when summing
equation (9.2) manually or even using a fastest computer. Therefore, in numerical integration
procedure, the approximation is involved. There are always finite terms in the summation.
However, depending on the function and the type of quadrature formula, sometimes the error can
be zero also. In this section, we describe two commonly used numerical integration procedure.
113

115. Figure 9.1: The plot of function f(x) and illustration of the concept of integration
9.2.1 Newton-Cotes quadrature
Observe equation (9.2), if we treat Δx as the weight, the integral ( )df x x∫ can be
considered approximately as weighted sum of function values at discrete points. The weights (or
interval sizes Δx) need not be equal. Similarly, the discrete points need not be equally spaced.
However, in Newton-Cotes quadrature, the points are usually equally spaced and are always
decided before deciding the weights. It means that the points at which the function is to be
evaluated are determined a priori, usually at equal intervals. A polynomial is passed through
these points and exactly integrated to obtain a close form expression. The coefficients associated
with the function values at taken as weights in quadrature formula.
As ‘n’ values of the function define a polynomial of degree (n-1), the errors will be of
the order O(∆n
) in a n-point formula, where ∆ is the point spacing. For a general n point Newton-
Cotes ‘quadrature’ formula, the integral can be written as
( ) ( )
1
1
1
d =
n
i i
i
I f H fξ ξ
−
=
= ∑∫ ξ (9.3)
where Hi are weights. We have taken the limits from -1 to +1. If the limits are different you can
always transform the limits from -1 to +1, by linear transformation of the variable. In the
following paragraph, we develop 2-point Newton-Cotes formula.
For n = 2, a straight line can be easily fitted. Thus, the given function is approximated as
( )f a bξ ξ= + (9.4)
At 1,ξ = −
( 1)f a b− = − (9.5)
At 1,ξ = +
(1)f a b= + (9.6)
114

116. From Eq. (9.5) and (9.6), we obtain
( ) ( )
1
1 1
2
a f f= − +⎡ ⎤⎣ ⎦ (9.7)
( ) ( )
1
1
2
b f f= + − −1⎡ ⎤⎣ ⎦ (9.8)
Thus,
( ) ( ) ( ) ( ) ( )
1 1
1 1 1 1
2 2
f f f f fξ ξ= − + + + − −⎡ ⎤ ⎡⎣ ⎦ ⎣ ⎤⎦
1
(9.9)
Therefore,
( ) ( ) ( )
1
1
d 1I f fξ ξ
−
= = − +∫ f (9.10)
The above equation is called trapezoidal rule.
Now, we develop a three point Newton-Cotes quadrature formula. For n = 3,
( ) 2
f a b cξ ξ ξ= + + (9.11)
Let us choose the points, 1, 0, 1ξ = − +
( )
1
1
2
d 2
3
f aξ ξ
+
−
= +∫ c (9.12)
Now,
( )1f a b c− = − +
( )0f a=
( )1f a b c= + +
Thus,
( ) ( )1 0c b f f− = − −
( ) ( )1 0c b f f+ = −
Solving these, we get
( ) ( ) ( )1 1 2
2
f f f
c
− + −
=
0
Now putting the value of a and c in Eq. (9.12)
115

117. ( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
1
1
1 1 2
2 0 1 1 0
3 3 3
1
1 4 0 1
3
f d f f f f
f f f
ξ ξ
+
−
= + − + −
= − + +⎡ ⎤⎣ ⎦
∫
(9.13)
This is called Simpson’s ‘one third’ rule.
In the same way, we can obtain 4-point formula. For n = 4,
( ) ( ) ( )
1
1
1 1 1
1 3 3 1
4 3 3
f d f f f fξ ξ
+
−
⎡ ⎤⎛ ⎞ ⎛ ⎞
= − + − + +⎜ ⎟ ⎜ ⎟⎢
⎝ ⎠ ⎝ ⎠⎣ ⎦
∫ ⎥
f
(9.14)
9.2.2 Gauss quadrature
In this method,
( ) ( )
1
1
1
d
n
i i
i
f wξ ξ
+
−
=
= ∑∫ ξ (9.15)
where iξ is the sampling point and wi is the associated weight. Here, the sampling points and
associated weights are optimally chosen. Therefore, the accuracy gets improved. We develop one
and two Gauss-point formula.
One Gauss-point formula:
Let ( )f a bξ ξ= + can be evaluated exactly.
( ) ( ) (
1
1 1 1 11
da b w f w a b )ξ ξ ξ
+
−
+ = = +∫ ξ
1
(9.16)
or
1 12a w a w bξ= + (9.17)
This is true for any arbitrary a and b.
Hence, w1 = 2 and 1 0.ξ =
Thus for one gauss point formula, evaluate the function at 0ξ = and multiply by 2.
Two Gauss point- formula:
Let ( ) 2 3
f a b c dξ ξ ξ ξ= + + + can be evaluated exactly.
( ) ( ) ( )
1
1 1 2 21
df w f w fξ ξ ξ
+
−
= +∫ ξ (9.18)
or
( ) ( ) ( )
1
2 3 2 3 2
1 1 1 1 2 2 21
a b c d d w a b c d w a b c d 3
2ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ
+
−
+ + + = + + + + + + +∫
116

118. or
( ) ( ) ( ) (2 2 3 3
1 2 1 1 2 2 1 1 2 2 1 1 2 2
2
2
3
c
a w w a w w b w w c w wξ ξ ξ ξ ξ ξ+ = + + + + + + + )d
As a, b, c, d are arbitrary, equating the coefficients on both the sides,
1 2 2w w+ = (9.19)
1 1 2 2 0w wξ ξ+ = (9.20)
2 2
1 1 2 2
2
3
w wξ ξ+ = (9.21)
3 3
1 1 2 2 0w wξ ξ+ = (9.22)
Solution is 1 2
1 1
,
3 3
ξ ξ= − =
w1 = 1, w2 = 2
For three Gauss-point formula
1 1
2 2
3 3
5
0.77459667,
9
8
0,
9
5
0.77459667,
9
w
w
w
ξ
ξ
ξ
=− =
= =
=+ =
Following important points are to be noted in 1-D Gauss-quadrature:
(1) As any Gauss - quadrature formula must evaluate a constant function accurately
1
1
1
d 2
n
i
i
a a wξ
+
−
=
= = a∑∫ (9.23)
Thus, 2iw =∑
(2) With n- Gauss point formula, (2n-1) degree polynomial can be evaluated exactly.
9.3 TWO DIMENSIONAL INTEGRATION FORMULAE
One dimensional Gauss-quadrature can be extended to 2 or 3-dimensions. We describe how
the integration can be carried out in square and triangular domains.
9.3.1 Integration over square region
Figure 9.2 shows a square element. The numerical integration can be carried out as follows:
117

119. ( )
1 1
1 1
, d dI f ξ η ξ η
−
− −
= ∫ ∫
( )
1 1
1
1
, d
n
i i
i
w f ξ η ξ
+
−
=
= ∑ ∫
where n1 is the number of Gauss-points in η direction. Application of 1-dimensional Gauss-
quadrature again provides:
I ( )
1 2
1 1
,
n n
i j j
i j
w w f iξ η
= =
= ∑ ∑
( )
1 2
1 1
,
n n
i j j i
i j
w w f ξ η
= =
= ∑∑
where n2 is the number of Gauss-points in ξ direction.
For applying this formula, four corners of the region must be (-1, -1), (1,-1), (1,1) and (-1,1).
If a rectangular region is provided, it can be transformed to square region in natural coordinates
ξ-η by linear transformation. Evenif, there is a quadrilateral region; it can be transformed to
square region in natural coordinates. We shall discuss this in next Chapter.
Figure 9.2: A square domain
9.3.2 Integration over triangular region
Figure 9.3 shows a triangular element in natural coordinates. The coordinates of the
corners of the triangle are (0, 0), (1, 0) and (0, 1). If a f is function of three natural coordinates, it
can be integrated over the triangular domain as
( )
11 1
1 2 3 2 10 0
, , d dI f
ξ
ξ ξ ξ ξ ξ
−
= ∫ ∫ (9.24)
118

120. Note that the sum of three natural coordinates is 1. Therefore, ξ3 can be eliminated to express f as
another function F of first two natural coordinates.
Figure 9.3: A triangular element
In terms of first two natural coordinates, the integration is
( )
21 1
1 2 1 20 0
, d dI F
ξ
ξ ξ ξ ξ
−
= ∫ ∫ (9.25)
This can be integrated by applying the Gauss-point formula two times, first along ξ1 with limit 0
to 1-ξ2 and then along ξ2 with limits 0 to 1. Note that the limits have to be transformed to “from -1
to +1”. The calculations are tedious. Simplified expressions have been developed. One formula is
given by
( ) (21 1
1 2 1 2 1 20 0
1
, d d ,
Gn
i i
i
i
F w
ξ
)Fξ ξ ξ ξ ξ ξ
−
=
= ∑∫ ∫ (9.26)
The weights wi and the coordinates ( )1 2,i i
ξ ξ of the Gauss points corresponding to various values
of nG are provided in Table 9.1. For further details, reader can see the reference for numerical
integration [1-4].
9.4 CONCLUSIONS
In this Chapter, numerical integration formulae have been discussed. First, the numerical
integration in one-dimension using Newton-Cotes and Gauss quadrature has been described. The
advantage of Gauss-quadrature scheme has been highlighted. Then the Gauss-quadrature formula
119

121. has been applied extended to two dimensions. The reader can also extend them to three
dimensions.
REFERENCES
1. Carnahan, B., Luther, H.A. and Wilkes, J.O., 1969, Applied Numerical Methods, John
Wiley, New York.
2. Froberg, C.E., 1969, Introduction to Numerical Analysis, Addison-Wesley, Reading,
MA.
3. Hammer, P.C., Marlowe, O.J. and Stroud, A.H., 1956, “Numerical Integration over
Simplexes and Cones,” Mathematical Tables and other Aids to Computations, Vol. 10,
pp. 130-137, The National Research Council, Washington, DC, 1956.
4. Cowper, G.R., 1973, Gaussian Quadrature Formulas for Triangles, International Journal
for Numerical Methods in Engineering, Vol. 7, pp. 405-408.
120

122. Table 9.1: Gauss points and weights for integration on a triangular domain
nG Coordinates ( )1 2,i i
ξ ξ of the Gauss points Weights wi
1
A = (1/3,1/3)
wA = ½.
Formula exact for complete 1st
degree polynomial.
3
A = (1/2,1/2)
B = (0,1/2)
C = (1/2,0)
wA = wB = wC = 1/6
Formula exact for complete 2nd
degree polynomial.
7
A = (1,0),
B = (0,1),
C = (0,0),
D = (1/2,1/2)
E = (0,1/2),
F = (1/2,0),
G = (1/3,1/3)
wA = wB = wC = 3/120
wD = wE = wF = 8/120
wG = 27/120.
Formula exact for complete 3rd
degree polynomial.
7
( )1 1, ,A α β= ( )2 2, ,D α α=
( )1 1, ,B β α= ( )2 2, ,E β α=
( )1 1, ,C β β= ( )2 2,F ,α β=
G = (1/3,1/3)
where
1 2
1 2
9 2 15 6 15
,
21 21
6 15 9 2 15
,
21 21
α α
β β
+
= =
− −
= =
+
wA = wB = wC =
155 15
2400
−
wD = wE = wF =
155 15
2400
+
wG = 9/80.
Formula exact for complete 5th
degree polynomial.
121

123. EXERCISE 9
Q.1: Integrate
2 3 4 5
3 4 5 6
( ) 10 20
10 100 1000 10000
x x x x
f x x= + − + − +
between 8 and 12 using 3 point Gauss formulation. The three points and corresponding weights
are
Points Weights
0 8/9
0.6± 5/9
Q.2: Consider the following element used to solve heat conduction problem. Sides are of 4 mm
size.
Figure: Q2
The uniform heat generation in the element is
Watt/m3
(2.0 | | )(2.0 | | )Q x= − − y
where x and y are in mm. Obtain the load vector due to heat generation using appropriate
Gauss point formula. Use the following table
122

124. Number of Gauss points Location Weight
1 0.0 2.0
2 ± 0.5773502692 1.0
3
± 0.7745966692
0.0
0.55555555556
0.88888888889
Q.3: Integrate 2 2
x y+ over the triangle shown in the Figure.
Figure: Q3
Q.4: Consider the evaluation of
1 1
1 1
( , )d df x y x y
+ +
− −
∫ ∫ . How many total Gauss points should be used
for the evaluation of it, if the f(x, y) is
(a)
2 2
x y+ (b)
2 3 4
3x y y+ (c)
x y
x y
+
−
123

125. Chapter 10
FURTHER DETAILS ON 2D-FEM
(Lecture 22)
10.1 INTRODUCTION
In this Chapter, we shall provide more details on 2D-FEM. First, it will be emphasized that
use of natural coordinates is very convenient in FEM formulation. They allow us to write
formulae for the shape functions of a particular type of element. When the functions are
expressed in the form of natural coordinates, it becomes very convenient to carry out numerical
integration. Type of mapping between physical and natural coordinates influences the behavior of
the element. In this context, a description of iso-parametric, sub-parametric and super-parametric
elements have been provided. Certain other types of element have been introduced.
10.2 NATURAL COORDINATES AND ISO-PARAMETRIC, SUB-PARAMETRIC
AND SUPER-PARAMETRIC ELEMENTS
You have already become familiar with natural coordinates and have seen that how
convenient it becomes to perform numerical integration in natural coordinates. This is not the
only advantage of using natural coordinates, as will be clear after studying this Chapter. First, let
us learn dealing with the natural coordinates in 1-dimensional elements.
Let us consider that there is a 2-noded rod element, the coordinates of the two nodes
being x1 and x2. You have to transform the coordinates from physical to natural coordinate
system, such that x1 transforms to –1 and x2 transforms to +1. A linear transformation of the
form x a bξ= + is appropriate for this purpose you can easily find out the constants a and b. We
already have carried out linear interpolation of the primary variable of the problem using C0
continuity shape functions. The physical variable can be interpolated in the same way. Thus,
1 1 2 2x N x N x= + (10.1)
where N1 = (1-ξ)/2 and N2 = (1+ξ)/2.
The elements in which the field variable and physical variables are approximated in the
same way are called iso-parametric element. Instead of two nodes, if we use 3 nodes in an
element and approximate a primary variable such as temperature in the following way:
1 1 2 2 3 3T N T N T N T= + + (10.2)
and transform the physical coordinate by Eq. (10.1), it will be called sub-parametric element and
the corresponding formulation as sub-parametric formulation. In sub-parametric formulation, the
125

126. primary variable is approximated by a higher degree of polynomial than geometry. In super-
parametric formulation, reverse is the case.
Let us consider that the stiffness matrix is
[
2
1
1
1 2
2
d / d
[ ] d / d d / d d
d / d
x
e
x
N x
K N x N
N x
⎧ ⎫
= ∫ ⎨ ⎬
⎩ ⎭
]x x (10.3)
In the above expression, we have to transform everything to natural coordinates. From Eq.(10.1),
we have
2 1d d
2
x x
dx Jξ ξ
−
= = (10.4)
where J is called Jacobian and it is the ratio of the sizes of the element in physical and natural
coordinate system. (Basically, in one-dimension, the Jacobian is the ratio of infinitesimal length
in the physical coordinates to the corresponding mapped length in natural coordinates. If the
transformation is linear, the value of the Jacobian is same everywhere. In two dimension, the
determinant of the Jacobian matrix is the ratio of infinitesimal area in physical coordinate system
to that in natural coordinate system. In three dimension, the determinant of the Jacobian matrix is
the ratio of infinitesimal volume in physical coordinate system to that in natural coordinate
system.
For finding out the derivative of shape function with respect to x, we employ chain rule.
Thus,
11 1d d dd
d d d d
N N N
J
x x
1ξ
ξ ξ
−
= = (10.5)
Similarly,
12 2d d dd
d d d d
N N N
J
x x
2ξ
ξ ξ
−
= = (10.6)
By substituting Eqs (10.4-10.6) in (10.3) and integrating, you can verify that
1 1
[ ]
1 1
e
K
−⎡ ⎤
= ⎢ ⎥−⎣ ⎦
(10.7)
For beam element field variable (deflection) is approximated as a cubic polynomial, but the
geometry is approximated by linear functions only. Therefore, it is a subparmetric formulation.
The stiffness matrix for a beam element is
126

127. { }
2
1
[ ]d
x
e ''
x
K EI N N⎡ ⎤ = ∫⎣ ⎦
''
x (10.8)
The dx can be transformed as before. It can be easily seen that
2 2
1 2
2 2
d d
( )
d d
i iN N
J
x ξ
−
= (10.9)
10.3 FOUR-NODED QUADRILATERAL ELEMENT
In quadrilateral element, transform geometry in the following way:
4
1
i i
i
x N x
=
= ∑ ,
(10.10)
4
1
i i
i
y N
=
= ∑ y
a a c a y a cy
Cook et al. [1] has provided an interesting exercise problem. The problem is as follows:
The quadrilateral element shown might be used in finite element model. Imagine that its x-
direction displacement field is , where the ai are constants. How does
u vary with x or y along each side? Do you think the element will be compatible with its
neighbors?
1 2 3 4u a a x a y a xy= + + +
y
4
2
3
450
x1
Figure 10.1: A quadrilateral element
Let us solve this problem. Along line 1-2, y=0, i.e. u=a1+a2x. Thus, the displacement field
is linear. Along line 2-3, x=c (a constant), u 1 2 3 4= + + + , which is linear in y.
Similarly, it can be shown that the approximated displacement field is linear along 3-4. Along
line 1-4, y=x. Therefore, . Thus, the displacement varies in a quadratic
manner along 1-4. A quadratic function cannot be fitted just by the nodal values of 1 and 4, the
2
1 2 3 4u a a y a y a y= + + +
127

128. other nodes will influence it. Therefore, this element is not compatible. If there is another element
whose one edge coincides with 1-4, then approximate displacement field need not be the same for
both the elements. The reason is that only the nodes 1 and 4 are common between the two
elements, not the other nodes.
Now, if the element is transformed into an square element using Eq. (10.10), then the
displacement field is approximated as,
1 1 2 2 3 3 4 4u N u N u N u N u= + + + (10.11)
where Ni are the Lagrangian shape functions. Now along line 1-4, N3 and N2 are zero. Also, the
natural coordinate ξ=-1 along line 1-4. Therefore,
1
(1 ) (1 )
2 2
u u 4u
η η− +
= + (10.12)
This is a linear equation. The displacement field along the line 1-4 depends the nodal values of
the nodes 1 and 4. Same thing will be true for the element that interfaces with this element along
1-4. Therefore, this element will be compatible.
10.4 SERENDIPITY ELEMENTS
The dictionary meaning of the world ‘serendipity’ is ‘the faculty of making fortunate
discoveries by accident’. The English author Horace Walpole coined this word in 1754 based on a
fairy tale of a country named Serendip (now Sri Lanka). In the tale, three Princes of Serendip
keep making accidental discoveries during their travel. Serendipity elements of FEM are
something similar to that. The elements probably have been discovered as a result of numerical
experimentation and found to be better. You will see the derivation of the shape functions is
carried out in some what unusual way. In this section, the derivation of shape function for 8-
noded rectangular serendipity element will be carried out to make you understand the concepts.
Figure 10.2 shows an 8-noded rectangular serendipity element. It is similar to a 9-noded
element, but there is no internal node. Let us derive the shape functions for this element.
Figure 10.2: A 8-noded serendipity element
128

129. For C0
element, the shape functions should be such that the value of a shape function
should be 1 at the corresponding node and 0 at the other nodes. Also, the sum of the shape
functions should be 1. Moreover, each shape function should be a quadratic function. We shall
derive the expressions in natural coordinates, in which node is point (-1, 1) and node 2 (1, 1). Let
us see what happens if N1 is chosen corresponding to a 4-noded square element, i.e.
1
1
(1 )(1 )
4
N ξ η= − − (10.13)
This shape function is 1 at node number 1 and 0 at the nodes 2, 3, 4, 6 and 7. Fine! But what
about its value at nodes 5 and 8. At node 5, ξ=0 and η=-1 and therefore, N1= ½. Similarly, at node
8, ξ=-1 and η=0. Here, also N1= ½. How can we modify the shape function in Eq. (10.13) so that
it is 0 at 5 and 8 without affecting its value at other nodes? A little thought will show that the
following shape function can achieve this purpose:
1
1 1
(1 )(1 )
4 2
N ξ η= − − − −5 8
1
2
N N (10.14)
At node 5, N5 is 1 and therefore the term 5
1
2
N− is equal to -1/2, thus making the overall
expression 0. The last term does the same thing at node 8. At other nodes neither the last term nor
the second last term has any effect, as these terms are 0 by the very characteristics of the shape
functions.
In the similar manner, we can write
2
1 1
(1 )(1 )
4 2
N ξ η= + − − −5 6
1
2
N N (10.15)
3
1 1
(1 )(1 )
4 2
N ξ η= + + − −7 6
1
2
N N (10.16)
4
1 1
(1 )(1 )
4 2
N ξ η= − + − −7 8
1
2
N N (10.17)
Note that
8
5 8 5
1
6 7 8
5 6 7 8
1 1
(1 )(1 ) (1 )(1 )
4 2 2 4 2
1 1
(1 )(1 ) (1 )(1 )
4 2 2 4 2
i
i
N N N N
N
N N N N
N N N N
ξ η ξ η
ξ η ξ η
=
= − − − − + + − − −
+ + + − − + − + − −
+ + + +
∑ 6
7
2
2
(10.18)
This simplifies to
129

130. 8
1
1 1 1 1
(1 )(1 ) (1 )(1 ) (1 )(1 ) (1 )(1 )
4 4 4 4
i
i
N ξ η ξ η ξ η ξ
=
= − − + + − + + + + − +∑ η
(10.19)
The above expression is the sum of the Lagrangian functions of a 4-noded element, which is 1.
Thus, the shape functions in Eqs. (10.14-10.17) can solve our purpose. But wait, we have yet to
find out the expressions for N5, N6, N7 and N8. If we form N5 as the product of a quadratic shape
function in ξ and a linear shape function in η, our purpose will be solved, for then the shape
function will be 0 at ξ=-1 and +1 and η=1. It will be non-zero and equal to 1, only at ξ =0 and
η=-1. That’s what we need. Thus,
2
5
1
(1 )( 1)
2
N η η ξ= − − (10.20)
Similarly,
2
6
1
(1 )( 1)
2
N ξ ξ η= + − + (10.21)
2
7
1
(1 )( 1)
2
N η η ξ= + − + (10.22)
2
8
1
(1 )( 1)
2
N ξ ξ η= − − (10.23)
Thus, we have derived all the shape functions of a 8-noded Serendipity element.
10.5 EIGHT-NODED CURVILINEAR ELEMENT
The 8-noded curvilinear element can be transformed to square element by
8 8
1 1
,i i i i
i i
x N x y N y
= =
= =∑ ∑ (10.24)
where the shape functions correspond to 8-noded square serendipity element and they are
functions of the natural coordinates. For this element,
i i iN N Nx y
x yξ ξ ξ
∂ ∂ ∂∂ ∂
= +
∂ ∂ ∂ ∂ ∂
(10.25)
and
i i iN N Nx y
x yη η η
∂ ∂ ∂∂ ∂
= +
∂ ∂ ∂ ∂ ∂
(10.26)
Transforming the equations in the matrix form,
130

131. i i i
i ii
N x y N N
x x
J
NN x y
y y
ξ ξ ξ
η ηη
∂ ∂ ∂⎧ ⎫ ∂ ∂⎡ ⎤
N
⎧ ⎫ ⎧
⎪ ⎪ ⎢ ⎥
⎫
⎪ ⎪ ⎪∂ ∂ ∂ ⎪∂ ∂⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥= =⎨ ⎬ ⎨ ⎬ ⎨∂ ∂∂ ∂ ∂⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪
⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪
⎪
⎬
∂ ∂∂ ∂∂ ⎩ ⎭ ⎩ ⎭⎣ ⎦⎩ ⎭
(10.27)
Here, J is the Jacobian and its determinant is equal to (dx dy)/ (dξ dη).
From Eq. (10.27), it is seen that
1
ii
i i
NN
x
J
N N
y
ξ
η
−
∂⎧ ⎫∂⎧ ⎫
⎪ ⎪⎪ ⎪ ∂∂⎪ ⎪ ⎪
=
⎪
⎨ ⎬ ⎨∂ ∂
⎬
⎪ ⎪ ⎪ ⎪
⎪ ⎪ ⎪∂ ∂⎩ ⎭ ⎩ ⎭⎪
(10.28)
The elements of Jacobian matrix can be computed as
8
1
i
i
i
Nx
x
ξ ξ=
∂∂
= ∑
∂ ∂
and so on (10.29)
10.6 CONCLUSIONS
In this chapter, the importance of natural coordinates has been highlighted. It has been
explained that how the derivatives in physical coordinates can be obtained from the known
derivatives in the natural coordinates. Four noded quadrilateral and 8 noded serendipity elements
have been introduced. As you have now understand the concept of numerical integration and
Jacobian, you are ready to write your own FEM codes for solving two-dimensional problems. In
exercise some problems have been included.
REFERENCES
1. R.D. Cook, D.S. Malkus and M.E. Plesha, Concepts and applications of finite element
analysis, 3rd
ed., John Wiley, New York 1989
EXERCISE 10 (Computer Assignment)
You may use ANSYS, IDEA-S or codes written in C++/C/FORTRAN.
Q.1: This problem involves essential boundary conditions only. Consider a rectangular domain
with temperatures specified on each side as shown in Fig. Q1(a). An analytical solution to this
problem is given by
( ) ( )
( )
1
1 2 1
1
sinh
1 12
, sin
sinh
n
n
n y
n x L
T x y T T T
n Wn L
L
π
π
ππ
+
∞
=
⎡ ⎤⎛ ⎞
⎜ ⎟⎢ ⎥− + ⎛ ⎞ ⎝ ⎠⎢ ⎥= + − ⎜ ⎟
⎛ ⎞⎝ ⎠⎢ ⎥
⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
∑ .
131

132. Solve the problem to obtain the temperature field and plot the temperature contour. (Take
L=W=1000 m. and k =1 with T1 = 500 K and T2 = 0K.). Use two types of meshes as shown in
figures and compare the temperature values at the nodes with the exact solution.
Figure: Q1(a)
Figure: Q.1(b)
Q2: This problem involves Essential and Flux boundary conditions. Consider the rectangular
domain with two adjacent surfaces insulated and the temperature values specified on the other
two surfaces as shown in figure. ( b =100) .
132

133. Figure: Q2 (a)
Analytical solution to this problem is given by,
( )
cosh
6
, 100 cos
6
cosh
3
y
xb
T x y
b
b
π
π
π
⎛ ⎞
⎜ ⎟
⎛ ⎞⎝ ⎠= ⎜ ⎟
⎛ ⎞ ⎝ ⎠
⎜ ⎟
⎝ ⎠
.
Obtain the temperature field and plot temperature contours, using two types of meshes shown
below
Figure: Q 2(b)
Q3: Convective and flux boundary conditions only. Obtain the temperature field and plot the
temperature contours for the following problem. Use the mesh as shown in Fig. Q3(b)
133

134. Figure: Q3(a)
Figure: Q3(b)
Q4: Heat generation and essential boundary conditions only. Consider an infinite plate in the y-z
plane with thickness 2L. Heat is generated throughout the volume of the plate at a rate of Q
W/m3
. Both the surfaces of the plate are maintained at a temperature of Ts. As the problem is
essentially one-dimensional it can be solved by insulating the top and bottom surfaces as shown
in Fig. Q4(b).
(b)
(a)
Figure: Q4
134

135. Take Q=100 W/m2
, L = 1 m and k = 1W/mK. Use the grids as shown in Fig. Q1(b). Obtain the
temperature field and plot the temperature contours. Also, find out the temperature gradients.
Q.5: This problem involves essential and convective boundary conditions only. Consider a
square plate of side 1 m. The boundary conditions are as shown in Fig. Q5. Obtain the
temperature field and plot temperature contours. Use the grids as shown in Fig. Q1(b). Also plot
the temperature gradients contour.
Figure: Q5
Q.6: This problem involves non-homogenous material with heat flux and essential boundary
conditions only. Consider a square plate of non-homogeneous material as shown in Fig. Q6(a).
Use the grid shown in Fig.Q6 (b) to obtain the temperature field. Plot the temperature and
temperature gradients contours also.
135

136. Figure: Q6 (a)
Figure: Q6(b)
136

137. Chapter 11
FEM FORMULATION FOR PLANE STRESS AND PLANE STRAIN
PROBLEMS
(Lecture 23-24)
11.1 INTRODUCTION
Consider a linear elastic solid of domain Ω and having uniform thickness bounded by two
parallel planes on any closed boundary Γ as shown in Fig. 11.1. If the thickness in z direction is
small compared with the size Ω of the domain, the problem may be approximated as a plane
stress problem. The following assumptions are made. The body forces, if any exist, cannot vary in
the thickness direction and cannot have components in the z direction; the applied boundary
forces must be uniformly distributed across the thickness (i.e. constant in the z direction); and no
loads can be applied on the parallel planes bounding to the bottom surfaces. The assumption that
the forces are zero on the parallel planes implies that for plane stress problems the stresses in the
z direction are negligibly small i.e.,
σxz=0 σyz=0 σz=0 (11.1)
Figure11.1: A solid of domain Ω and support conditions
Plane strain is defined as a deformation state in which there is no deformation in z-
direction and deformations in other directions are functions of x and y but not of z. Thus, stain
components .0=== zxyzz εεε In plane strain problems non-zero stress components are
zxyyx σσσσ and,, . However, zσ is not an independent component and can be obtained if
yx σσ and are known. This makes the FEM formulation for plane stress and plane strain
137

138. problems similar. Only difference is in the constitutive matrices for both problems. In this article
FEM formulation for plane stress and plane strain problems will be discussed.
11.2 BASIC EQUATIONS
The governing equations for the plane elasticity problems are given by
2
2
t
u
f
yx
x
xyx
∂
∂
=+
∂
∂
+
∂
∂
ρ
σσ
(11.2)
2
2
t
v
f
yx
y
yxy
∂
∂
=+
∂
∂
+
∂
∂
ρ
σσ
(11.3)
where ƒx and ƒy denote the body forces per unit volume along the x and y directions, respectively
and ρ is the density of the material. σx , σy are the normal stresses and u, v are the displacements
in x and y directions respectively, σxy is the shear stress on the xz and yz planes. Strain-
displacement relations are given by
x
u
x
∂
∂
=ε , y
v
y
ε
∂
=
∂
,
x
v
y
u
xy
∂
∂
+
∂
∂
=ε2 (11.4)
For plane stress problems, stress and strain are related by the constitutive matrix D, in the
following manner:
(11.5)
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
xy
y
x
xy
y
x
d
dd
dd
ε
ε
ε
σ
σ
σ
200
0
0
33
2221
1211
where dij (dij = dji) are the elasticity (material) constants for an orthotropic material with the
material principal directions coinciding with the co-ordinate axes (x,y) used to describe the
problem. For an isotropic material of plane stress dij are given by
22211
1 v
E
dd
−
== , 22112
1 v
Ev
dd
−
== ,
)1(2
33
v
E
d
+
=
(11.6)
where E is Young's modulus of the material and v is Poisson's ratio. For plane strain problems:
,
)21)(1(
)1(
2211
νν
ν
−+
−
==
E
dd ,
)21)(1(
2112
νν
ν
−+
==
E
dd
)1(2
33
v
E
d
+
= (11.7)
138

139. 11.3 BOUNDARY CONDITIONS
For the given problem, essential or geometric boundary conditions are
, on Γu (11.8)
−
= uu
−
= vv
and natural boundary conditions are
(11.9)xyxyxxx tnnt
−
=+= σσ
on Γt (11.10)yyyxxyy tnnt
−
=+= σσ
where , are the components of the unit normal vector n on the boundary Γ. Γu and Γt are
portions of the boundary Γ (Γ = Γu U Γt). , are specified boundary stresses or tractions, and
, are specified displacements. Only one element of each pair, ( u, tx) and (v, ty) may be
specified at a boundary point.
xn yn
xt
−
yt
−
−
u
−
v
11.4 FEM FORMULATION
There are a number of techniques of FEM formulation, the most popular being Rayleigh-Ritz
technique and Galerkin technique. Some other methods are collocation method, sub-domain
method, least square method, over-determined collocation method. In this article, FEM
formulation will be obtained by Rayleigh-Ritz method.
In the Rayleigh-Ritz FEM method, the variables whose values are to be determined are
approximated by piecewise continuous polynomials. The coefficients of these polynomials are
obtained by minimizing the total potential energy of the system. In FEM, usually, these
coefficients are expressed in terms of unknown values of primary variables. Thus, if an element
has got 4 nodes, the displacement field u can be approximated as
∑=
=
4
1i
iiuNu (11.11)
where ui are the nodal displacements in x-direction and Ni are the shape functions, which are
functions of coordinates.
For plane elastic body, the total potential energy of an element is given by,
1
2e e e
e
iij ij i i i
v v
dv f u dv t u ds
Γ
Π σ ε= − −∫ ∫ ∫ (11.12)
139

140. where, ve
denotes the volume of element e, Γt is the boundary of domain Ωe
, ijσ and ijε are the
components of stress and strain tensors, respectively and ƒi and are the components of body
force and boundary stress vectors, respectively.
it
−
yxyx σσσσσσ === 221211 ,, (11.13)
yxyx ttttffff ==== 2121 ,,, (11.14)
The first term in equation (11.12) corresponds to strain energy stored in the element, the second
represents the work potential of the body force, and the third represent the work potential of
surface forces. For plane stress problems with thickness he
, it is assumed that all quantities are
independent of the thickness co-ordinates z. Hence,
1
( 2
2e
e e
x x y y xy xyh σ ε σ ε σ ε
Ω
Π = + +∫ )d dx y
)ds( )d d (
e e
e e
x y x yh f u f v x y h t u t v
Ω Γ
− + − +∫ ∫ (11.15)
where ƒx and ƒy are the body forces per unit area, and tx and ty are boundary forces per unit
length. Equation (11.15) can be rewritten as
∫
⎭
⎬
⎫
⎩
⎨
⎧
⎭
⎬
⎫
⎩
⎨
⎧
−
⎭
⎬
⎫
⎩
⎨
⎧
∫
⎭
⎬
⎫
⎩
⎨
⎧
−∫
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎝
⎛
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
=Π
ΓΩΩ eee
ds
t
t
v
u
hdxdy
f
f
v
u
hdxdyh
y
x
T
e
y
x
T
e
xy
y
x
T
xy
y
x
ee
σ
σ
σ
ε
ε
ε
2
2
1
(11.16)
The finite element model of the plane elasticity equations is developed using the matrix
form in (11.16). The displacements u and v are approximated by the Lagrange family of
interpolation functions (shape functions). Let u and v are approximated over Ωe
by the finite
element interpolations
, (11.17)∑≈
=
n
i
e
i
e
i yxNuu
1
),( ∑≈
=
n
i
e
i
e
i yxNvv
1
),(
where is the number of nodes representing the element e, are the displacement shape
functions, and are the nodal displacements in x- and y- directions respectively. The
displacements and strains over element e are given by
n e
iN
e
iu e
iv
140

141. ⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧
∑
∑
=
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧
=
=
n
i
e
i
e
i
n
i
e
i
e
i
e
e
Nv
Nu
v
u
1
1
= (11.18)
⎪
⎪
⎪
⎪
⎪
⎭
⎪
⎪
⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
n
n
n
n
v
v
v
u
u
u
NNN
NNN
.
.
...0.....00
0..00...
2
1
2
1
21
21
= (11.19)[ ]{ }ee
n
n
n
n
N
v
u
v
u
v
u
NNN
NNN
Δ≡
⎪
⎪
⎪
⎪
⎪
⎭
⎪
⎪
⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
..
..
.
.0..0.0
0...00 2
2
1
1
21
21
and
(11.20)}]{][[}{},]{[}{ eeeeeee
BDB Δ=Δ= σε
where is called Gradient matrix and [Te
] is the matrix of differential operators.
Substituting these expressions for the displacements and strains into (16)
]][[][ eee
NTB =
(11.21)
}){}{}{]([}{][}{
][}{})]{][[]([}{
eeeTeTe
y
xTeTee
y
xTeTeeeeeTeTeee
Qfkds
t
t
Nh
dxdy
f
f
NhdxdyBDBh
e
ee
−−ΔΔ=
⎭
⎬
⎫
⎩
⎨
⎧
∫ Δ−
⎭
⎬
⎫
⎩
⎨
⎧
∫ Δ−Δ∫ Δ=Π
Γ
ΩΩ
Minimizing this i.e. differentiating the above expression with respect to { }e
Δ , we get
(11.22)}{}{}{][ eeee
Qfk +=Δ
where ,dxdyBDBhk eeTeee
e
]][[][][ ∫=
Ω
, (11.23)dxdy
f
f
Nhf
y
xTeee
e
⎭
⎬
⎫
⎩
⎨
⎧
∫=
Ω
][}{ ds
t
t
NhQ
y
xTeee
e
⎭
⎬
⎫
⎩
⎨
⎧
∫=
Γ
][}{
The element stiffness matrix [ke
] is of order 2n x 2n and the elemental load vector
141

142. (11.24)}{}{][ eee
QfF +=
is of order 2n x 1 where n is the number of nodes of the element.
11.5 SHAPE FUNCTIONS
Shape functions or interpolation functions Ni are used in the finite element analysis to
interpolate the nodal displacements of any element to any point within each element. The
interpolation functions for the four nodded quadrilateral elements shown in Fig. 11.2 are
4
)1)(1(
1
ηξ −−
=N ,
4
)1)(1(
2
ηξ −+
=N ,
4
)1)(1(
3
ηξ ++
=N ,
4
)1)(1(
4
ηξ +−
=N (11.25)
where ξ and η are the natural co-ordinates for the physical co-ordinates x and y, respectively. In
natural coordinate system, the coordinates of four nodes are (-1,-1), (1,-1), (1,1) and (-1,1).
One of the earliest finite elements is a three nodded triangular element shown in Fig. 11.3.
An arbitrarily located point P divides a triangle 1-2-3 into three sub-areas A1, A2, and A3. Then,
the natural coordinates of the point P are defined as ratios of areas:
A
A
A
A
A
A 3
3
2
2
1
1 === ξξξ (11.26)
where A is the area of triangle 1-2-3. Since A=A1 + A2 + A3, the ξi are not independent. They
satisfy the constraint equation
1321 =++ ξξξ (11.27)
For this triangle, shape functions in terms of natural coordinates are given as
332211 ξξξ === NNN (11.28)
It can be shown that displacement field obtained using these shape functions provides constant
strain inside the triangular element. This element is, therefore, called constant strain triangle
(CST).
Figure 11.2: A quadrilateral element
Figure 11.2 A quadrilateral element
142

143. Side 1
Side 2 A2 P A1
A3 2
1 Side 3
Figure 11.3: A triangular element
11.6 NUMERICAL EVALUTION OF ELEMENT MATRICES AND VECTORS
The evaluation of the element matrices in equation (11.22) is done by using numerical
integration techniques. For all area and line integrals Gauss-Quadrature rule is used. All physical
domain integration is transformed to the (ξ,η) plane as shown in Fig. 11.4.
As a result,
∫ ∫=∫
− −Ω
1
1
1
1
),(),( ηξηξ ddJfdxdyyxf (11.29)
where |J| is the determinant of the Jacobian matrix of the transformation and is given by
==
43
21
JJ
JJ
J
∑∑
∑∑
==
==
n
i i
e
i
n
i i
e
i
n
i i
e
i
n
i i
e
i
yNxN
yNxN
1 ,1 ,
1 ,1 ,
ηη
ξξ
(11.30)
η
(-1,1) (1,1)
4 3
ξ
(0,0)
1 2
(-1,-1) (1,-1)
Figure 11.4: Natural co-ordinate system.
143

144. where n=number of nodes of the element,
=shape function corresponding to node i,),( ηξe
iN
(xi,yi)=physical co-ordinates of nodes i.
In writing equation (11.26), the isoparametic formulation has been used, i.e. the interpolation
functions used for the geometry variables (x,y) and field variables (u,v) are same. Also, the spatial
derivatives are transformed to the (ξ,η) plane using
(11.32)
⎭
⎬
⎫
⎩
⎨
⎧
=
⎭
⎬
⎫
⎩
⎨
⎧ −
η
ξ
,
,1
,
,
][
i
i
yi
xi
N
N
J
N
N
where,
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
−
−
=−
J
J
J
J
J
J
J
J
J
13
24
1
][
Finally, the Gauss-Quadrature scheme gives
∫ ∫ = ∑ ∑ (11.28)
− −
1
1
1
1
),( ηξηξ ddf
= =
1
1
2
1
, )(
n
r
n
s
srsr fww ηξ
where, n1=number of Gauss points in ξ direction,
n2=number of Gauss points in ηdirection,
wr, ws=weights of corresponding Gauss points.
If the displacement field within each element is assumed to be bilinear then, 2 x 2 Gauss
quadrature exactly integrates all terms of the elemental stiffness matrix.
Now, considering the evolution of boundary integral of the type
(11.29)∫=
Γe
dssNqQ i
e
n
e
i )(
Where is a known function (here boundary stress) of the distance s along the boundary Γe
. It is
not necessary to compute such integrals when a portion of Γ does not coincide with the boundary
Γ of the total Ω. This is because for any interior boundary the stresses from adjacent elements
cancel each other. The 2-D line integral in (x, y) plane is transformed to 1-D line integral in the
natural co-ordinate plane by using the fact that along any element side one of the natural co-
ordinates is constant as shown in Fig. 11.3. Thus,
e
nq
∫=∫ (11.30)
−
1
10
)(),( ξξ dJfdsyxf b
s
144

145. where the boundary Jacobian
22
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
∂
∂
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
∂
∂
=
ξξ
yx
Jb .
11.7 ASSEMBLY OF ELEMENT MATRICES
Once the element matrices are obtained, they are assembled to form the set of linear
simultaneous equations, the solution of which yields the displacement field. The assembly is
based on the principle of maintaining the continuity of the primary variable, in this case
displacement and the equilibrium of the secondary variables, here forces and tractions.
11.8 BOUNDARY CONDITIONS AND SOLUTIONS
There are two types of boundary conditions,
1. Essential or geometric boundary conditions which are imposed on the primary variable like
displacements, and
2. Natural or force boundary conditions which are imposed on the secondary variable like forces
and tractions.
The force boundary conditions are imposed during the evaluation of the element matrices
itself while the prescribed displacement boundary conditions are imposed after the assembly of
the element matrices. Then the global system of linear equations are solved by any numerical
technique to get the displacements at global nodes.
11.9 GRADIENT ESTIMATES
In finite element calculations, one often have a need for accurate estimates of the derivatives
of the primary variable. For example, in plane stress or plane strain analysis, the primary
unknowns to be computed are the displacement components of the nodes. However in many cases
the strains and stresses are the prime importance, which are computed from the derivatives of the
displacements. As the finite element solutions is only an interpolate solutions, it was exact at the
nodes and approximate elsewhere. Such accuracy is rare but, in general, one finds that the
computed values of the primary variables are most accurate in the nodes points. Thus, for the sake
of simplicity it is assumed that the element’s nodal values are exact. It was observed that
derivatives estimates are least accurate at the nodes, generally and most accurate at the Gauss
points. These points are also called as Barlow points or optimal points. Thus, the center of the
linear element is taken as the optimal position for sampling the first derivative.
145

146. 11.10 AN EXAMPLE
As an example of finding out the stress in whole domain, consider the problem of a plate
with the hole loaded by uniformly distributed tensile load (Fig. 11.5). This problem was solved
on ANSYS, commercial FEM software. Because of the symmetry of the problem, only a quarter
plate needs to be analyzed. On the line of symmetries, tractions and normal displacement
components will be zero. Fig.11.6 shows the finite element mesh and boundary conditions.
Here, quadrilateral elements have been used. Figure 11.7 shows the contours of longitudinal
stresses. Applied tensile stress was 100 MPa. Note that in the domain far away from the hole
stresses are close to this value. There is a region of high stress concentration near the highest
vertical point of the circle, the stresses there being around 336 MPa. One can obtain the contours
of other stress components and equivalent stresses also.
11.11 SUMMARY
In this article finite element formulation of plane stress and plane strain problems is
discussed. Difference between plane strain and plane stress formulations is only that both have
different constitutive matrices. After obtaining the finite element solution in the form of nodal
displacements, stresses and strains can be found by post-processing the results. One example
problem was solved in ANSYS and has been presented here to demonstrate the effectiveness of
FEM for finding the stresses in the whole domain.
Figure 11.5: A plate with a hole loaded by uniformly distributed tensile load on both sides
146

147. Figure 11.6: Finite element mesh for solving quarter plate problem
Figure 11.7: Contours of longitudinal stresses
147

148. Exercise 11
Q.1: Prove that in plane stress analysis using a rectangular element of size 2a × 2b, the [B]
matrix is given by
( ) 0 ( ) 0 ( ) 0 ( ) 0
1
0 ( ) 0 ( ) 0 ( ) 0 (
4
( ) ( ) ( ) ( ) ( ) ( ) ( ) (
b y b y b y b y
a x a x a x a x
ab
a x b y a x b y a x b y a x b y
− − − + − +⎡ ⎤
⎢ ⎥− − − + + −
⎢ ⎥
− − − − − + − + + − − +⎢ ⎥⎣ ⎦
)
)
Prove that this element possess the rigid body motion and constant strain capability.
Q.2: Transform the following quadrilateral element into a square element in natural
coordinates. Find out the Jacobian and its determinant.
(10,15)
Figure: Q2
Q.3: Write down the expression for all the nine shape functions of a 9-noded Lagrangian element.
The physical coordinates of the element are shown in figure. Find out the physical coordinate at
ξ=0.5 and η=0.5. Natural coordinates of the nodes are (-1,-1), (-1,1), (1,1), (-1,1), (0,-1), (1,0),
(0,1), (-1,0), (0,0).
Figure: Q3
(0,0)
(0,10)
3(10,10)
(10,0)
)
x
y
1(0,0)
8(0,5)
4(0,10)
5(5,0)
6(10,5)
7(5,10)
9 (5,5)
2(10,0)
148

149. Q.4: A plate of thickness 1 mm is loaded at one surface by a force of 10N/mm as shown in the
figure. The length of the plate is 100 cm and width 20 cm. If a mesh as shown in Figure Q4 is
used, then
(A) Find out the contribution of load to various nodes.
(B) Solve this problem first by fixing the nodes 1, 2 and 3 in both along the length and width
directions. After that solve it by fully fixing the node 2, but allowing nodes 1 and 3 to move along
the width direction. Calculate displacements and stresses. You may use professional FEM
package to solve it. Comment on the results obtained.
1
2
3
Figure: Q4
Q.5: In a plane stress square element of 1 m2
area, the u-v deflections (in mm) at 4 nodes are (0,0),
(0,0), (0,1), (0,1). Thickness of the element is 1 cm. Take E as 200GPa and Poisson’s ratio as zero.
Find out the strain energy contained in the element.
149

150. Chapter 12
FREE VIBRATION PROBLEMS
(Lecture 25)
12.1 INTRODUCTION
After discussing the FEM formulation for some two dimensional problems, we now
discuss time dependent problems. In solid mechanics, two types of time dependent problems are
solved. In one type of problems, the dynamic response i.e. change in the displacement of the
particle with time is studied. In the second type, we find out the natural frequencies of vibrations.
In this chapter, we shall find out the natural frequencies of vibration for one-dimensional
problems. One-dimensional problems have been chosen to make the treatment simple. After
reading this chapter, the reader will be able to carry out the finite element formulation for 2-
dimensional and 3-dimensional problems.
12.2 VIBRATION OF A ROD
The governing differential equation for the vibration of a rod is
2
2
0
u u
EA A
x x t
ρ
∂ ∂ ∂⎛ ⎞
− =⎜ ⎟
∂ ∂ ∂⎝ ⎠
(12.1)
where u is the axial displacement function, E is Young’s modulus of elasticity of the rod, A is the
cross-sectional area of the rod and ρ is the density of the rod. If the parameters E, A and ρ are not
the function of x, Eq. (12.1) can be solved analytically. However, if E, A and ρ vary along the rod,
the numerical methods become necessary. Let us carry out FEM formulation for equation (12.1).
Let the approximating function in an element be ue
. In that case, the residual is given by
2
2
e e
e u u
R EA A
x x t
ρ
⎛ ⎞∂ ∂ ∂
= −⎜ ⎟
∂ ∂ ∂⎝ ⎠
(12.2)
Making the weighted integral of the residual equal to 0, we get
0
d
h
e
wR x 0=∫ (12.3)
where we assume that x is a local coordinate and the differential equation (12.1) has been
expressed in local coordinates. Thus,
2
20
d
e e
h u u
w EA A x
x x t
ρ
⎛ ⎞⎛ ⎞∂ ∂ ∂
0− =⎜ ⎜ ⎟⎜ ∂ ∂ ∂⎝ ⎠⎝ ⎠
∫ ⎟⎟ (12.4)
Integrating this equation by parts to reduce the order of the derivative with respect to x, we get
151

151. 2
20 0
0
d
he e
h hu w u u
wEA EA x Aw x
x x x t
ρ
∂ ∂ ∂ ∂
− −
∂ ∂ ∂ ∂∫ ∫ d 0
e
=
}
(12.5)
The approximating function ue
is a function of x and t. It can be expressed in terms of the nodal
displacements as in the case of static problem, where the nodal displacements will be functions of
t. Considering the completeness and compatibility, 2-noded C0
continuity element is good
enough. Thus,
{1
1 2
2
e
ne
u
u N N N u
u
⎧ ⎫
= =⎢ ⎥ ⎢ ⎥⎨ ⎬⎣ ⎦ ⎣ ⎦
⎩ ⎭
(12.6)
where double dots indicate second derivative with respect to time. You already know the
expression for the shape functions, i.e.,
1 1 ; 2
x x
N N
h h
= − = (12.7)
In Galerkin FEM, the weight functions are approximated in the same way as the primary
variable. Thus,
1
1 2
2
N
w w w
N
⎧ ⎫
= ⎢ ⎥ ⎨ ⎬⎣ ⎦
⎩ ⎭
(12.8)
Substituting the expressions ue
and w in the weak form, we get
'
1 1' '
2 1 1 2 1 2'0
220
1 1
1 2 1 20
2 2
d
d 0
e e
h
h
h
N uu u
w EA w EA w w EA N N x
ux x N
N u
w w A N N x
N u
ρ
⎧ ⎫ ⎧ ⎫∂ ∂ ⎪ ⎪
⎢ ⎥− − ⎢ ⎥ ⎨ ⎬ ⎨⎣ ⎦ ⎣ ⎦∂ ∂ ⎪ ⎪ ⎩ ⎭⎩ ⎭
⎧ ⎫ ⎧ ⎫
− =⎢ ⎥ ⎢ ⎥⎨ ⎬ ⎨ ⎬⎣ ⎦ ⎣ ⎦
⎩ ⎭ ⎩ ⎭
∫
∫
⎬
(12.9)
where a dash on superscript indicates the differentiation with respect to x. Noting that node 1
corresponds to x=0 and node 2 corresponds to x=h, and taking 1 2w w⎢ ⎥⎣ ⎦ common, Eq. (12.9)
can be written as
'
2 1 1 1 1' '
1 2 1 2 1 2'0 0
2 2 22
1
d d
h h
u
EA
N u Nx
w w EA N N x A N N x
u N uu N
EA
x
ρ
⎡ ⎤⎧ ⎫∂
−⎢ ⎥⎪ ⎪ ⎧ ⎫∂ ⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥⎢ ⎥− −⎢ ⎥ ⎢ ⎥⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎣ ⎦ ⎣ ⎦⎣ ⎦⎢ ⎥∂ ⎪ ⎪ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭⎪ ⎪ ⎩ ⎭⎢ ⎥
⎪ ⎪∂⎢ ⎥⎩ ⎭⎣ ⎦
∫ ∫ 0
u
= (12.10)
As the nodal weights are arbitrary, we get the following elemental equations:
152

152. '
1 1 1 1' '
1 2 1 2'0 0
2 2 22
1
d d
h h
u
EA
N u N u 2x
EA N N x A N N x
u N u uN
EA
x
ρ
⎧ ⎫∂
−⎪ ⎪⎧ ⎫ ∂⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪
⎢ ⎥ + =⎢ ⎥⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨⎣ ⎦⎣ ⎦ ∂⎪ ⎪ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭
⎪
⎬
⎪ ⎪⎩ ⎭
⎪ ⎪∂⎩ ⎭
∫ ∫ (12.11)
If ρ, E and A are constant in an element, the above system of equations may be written as
1 1
2 2
2
2 1 1 1
1 2 1 16
u
EA
u u 1xAh EA
u uh u
EA
x
ρ
⎧ ⎫∂
−⎪ ⎪∂−⎧ ⎫ ⎧ ⎫⎡ ⎤ ⎡ ⎤ ⎪ ⎪
+ =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥− ∂⎣ ⎦ ⎣ ⎦⎩ ⎭ ⎩ ⎭ ⎪ ⎪
⎪ ⎪∂⎩ ⎭
(12.12)
We observe that the elemental element equations are not algebraic equation, but are ordinary
differential equations in time. Writing equation (12.12) in notational form:
{ } { }[ ] [ ] {e ne e ne e
im u k u F+ = } (12.13)
where [me
] is the elemental mass matrix, [ke
] is the stiffness matrix and right hand side is the
internal load vector.
Let us observe the elemental mass matrix. If we sum all the elements of the mass matrix,
we get ρAh, which is the total mass of the element. Thus, as a result of FEM formulation, the total
mass of the elements has been distributed in the elements of the stiffness matrix. The mass matrix
obtained this way is called consistent mass matrix, because its derivation is consistent with the
derivation of stiffness matrix. A simpler way, would have been to distributed total mass at two
nodes. Then, the mass matrix would be
1 0
0 12
e Ah
m
ρ ⎡ ⎤
⎡ ⎤ = ⎢ ⎥⎣ ⎦
⎣ ⎦
(12.14)
This type of mass matrix is called lumped mass matrix and has the advantage of being strongly
diagonally dominant.
Assembly and application of boundary condition provides, the following equations:
[ ]{ } { }[ ] 0M U K U+ = (12.15)
where [M] is the global mass matrix, [K] is the global stiffness matrix and U is the global
displacement vector. In the above equation, rows and columns corresponding to the fixed
boundary conditions have been eliminated. To solve it, assume
{ } { }sinU A tω= (12.16)
Then, the system of equations become
[ ]{ } { }2
[ ] 0M A K Aω− + = (12.17)
153

153. This is an eigen value problem, ω2
corresponding to eigen value. In general, the frequencies
predicted by FEM are expected to be higher than the actual frequencies, because FEM makes the
structure stiffer by constraining the motion. The static deflections predicted by FEM are expected
to be lower than the actual deflection. Refining the mesh improves the accuracy.
12.3 VIBRATION OF A BEAM
In the absence of damping, the governing differential equation of the motion of the free
vibration of a beam is
2 2 2
2 2 2
0
v v
EI m
x x t
⎛ ⎞∂ ∂ ∂
+ =⎜ ⎟⎜ ⎟∂ ∂ ∂⎝ ⎠
(12.18)
where m is the mass per unit length of the beam and it may be a non-constant function of x. Here,
the vertical defection of the beam is a function both x and t. Assuming that, like in the static beam
problem, the vertical deflection in an element can be expressed as
1
1
1 2 3 4
2
2
( )
( )
( )
( )
e
v t
v t
v N N N N
v t
v t
⎧ ⎫
⎪ ⎪′⎪ ⎪
= ⎡ ⎤⎨ ⎬⎣ ⎦
⎪ ⎪
⎪ ⎪′⎩ ⎭
(12.19)
The above equation is same as equation (5.27) except that nodal deflections/slopes are now
function of time. For assume deflection ve
, the residual is obtained as
2 2 2
2 2
e e
e v
R EI m 2
v
x x t
⎛ ⎞∂ ∂ ∂
= +⎜ ⎟⎜ ⎟∂ ∂ ∂⎝ ⎠
(12.20)
We make the weighted residual inside the element equal to zero, i.e.,
2 2 2
2 2 2
0 0
d
e eh h
e v v
wR x w EI m x
x x t
⎛ ⎞⎛ ⎞∂ ∂ ∂
= +⎜ ⎜ ⎟∫ ∫ ⎜ ⎟⎜ ∂ ∂ ∂⎝ ⎠⎝ ⎠
d 0=⎟
⎟
(12.21)
The first part of the expression is integrated by twice, so that the order of differentiation of ve
becomes two. The second part is left as it is. Equation (12.21) then becomes
2 2 2 2 2
2 2 2 2 2
0 0
0 0
d d
h h
e e eh hv w v w v v
w EI EI EI x m x
x xx x x x t
⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂
− + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟∫ ∫⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠
0
e
= (12.22)
In Galerkin FEM, the basis functions for approximating the deflections and weight
functions are same i.e.,
154

154. [
T
1
T2
1 2 3 4
3
4
w
w
w N N N N
w
w
⎧ ⎫
⎪ ⎪
⎪ ⎪
= ⎨ ⎬
⎪ ⎪
⎪ ⎪⎩ ⎭
] (12.23)
Putting the above approximation in equation (12.22),
[ ] [ ]
[ ]
''
1 1 1
'' ' '
2 21 1'' '' '' ''
1 2 3 4 1 2 3 4 1 2 3 4''
0 0 32 23
' ''' 42 24
2
2
1 2 3 4
d d
h h
N v vN
N Nv v
w w w w EI N N N N x m N N N N x
Nv vN
Nv vN
v
EI
x x
w w w w
1
⎡ ⎤⎧ ⎫ ⎧ ⎫ ⎧⎧ ⎫ ⎫
⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪
⎢ ⎥⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎡ ⎤ +⎢ ⎥∫ ∫⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎣ ⎦
⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪
⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭⎩ ⎭ ⎩ ⎭⎪ ⎪⎢ ⎥⎩ ⎭⎣ ⎦
⎛∂ ∂
−
∂ ∂
=
0
2
2
0
2
2
2
2
h
h
v
EI
x
v
EI
x x
v
EI
x
⎧ ⎫⎛ ⎞⎞
⎪ ⎪⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎪ ⎪⎝ ⎠⎝ ⎠
⎪ ⎪
⎛ ⎞⎪ ⎪∂
+⎜ ⎟⎪ ⎪⎜ ⎟∂⎪⎝ ⎠ ⎪
⎨ ⎬
⎛ ⎞⎛ ⎞⎪ ⎪∂ ∂
+⎜ ⎟⎜ ⎟⎪ ⎪⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠⎪ ⎪
⎪ ⎪
⎛ ⎞∂⎪ ⎪−⎜ ⎟⎪ ⎪⎜ ⎟∂⎝ ⎠⎩ ⎭ (12.24)
where (..) denotes double partial derivative with respect to time.
Since the nodal weights w1, w2 are arbitrary, the elemental finite element equations
become
[ ]
''
1 1 11
'' ' '
2 21 1'' '' '' ''
1 2 3 4 1 2 3 4''
0 0 32 23
' ''' 42 24
d d
h h
N v vN
N Nv v
EI N N N N x m N N N N x
Nv vN
Nv vN
⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎡ ⎤ +∫ ∫⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨⎣ ⎦
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭⎩ ⎭ ⎩ ⎭⎪ ⎪⎩ ⎭
⎪
⎬
⎪
⎪
= Internal force vector (12.25)
The first integral on the left hand side is same as obtained in the left hand side of equation (6.19)
and is called elemental stiffness matrix, whereas the second integral is called elemental mass
matrix. The elemental stiffness matrix for a beam is given by equation (1.46). The elemental mass
matrix for the beam element is
155

155. 2 2
2 2
156 22 54 13
22 4 13 3
54 13 156 22420
13 3 22 4
h h
h h h hm
h h
h h h h
−⎡ ⎤
⎢ ⎥
−⎢ ⎥
⎢ ⎥−
⎢ ⎥
⎢ ⎥− − −⎣ ⎦
(12.26)
The elemental matrices can be assembled in the usual manner. The internal force vectors will
cancel each other, except at the nodes where essential boundary conditions are prescribed.
However, the rows and column corresponding to essential boundary conditions may be
eliminated from the global matrices. In view of this, the global finite element equations are
obtained as,
[ ]{ } [ ]{ } { }+ =K D M D 0 (12.27)
where is the global stiffness matrix and[ ]K [ ]M is the global mass matrix. In free vibrations,
the structure undergoes the harmonic motion. Thus,
{ } { }sin tω=D A and { } { }2
sin tω ω= −D A (12.28)
where {A} is the vector containing the amplitudes and ω is the circular frequency. Putting the
equation (12.28) in equation (12.27), the following eigen value problem for free vibration is
obtained:
2
[ ] [ ] { } { }ω⎡ ⎤−
⎣ ⎦
K M A 0= (12.29)
From the above equation, the natural frequencies and mode shapes may be found out. The
solution accuracy increases with finer discretization.
12.4 CONCLUSIONS
In this chapter, finite element formulations of the free vibration of rod and beam have been
carried out. The formulations provide the eigen value problems. Usually, in these problems, the
accuracy of the fundamental frequency is more than the accuracy of other natural frequency. The
accuracy keeps on decreasing for higher modes. The discretization should be carried out
depending on how many modes are important for the problem.
156

156. EXERCISE 12
Q.1:
Figure: Q1
Fig. Q1(a) is a schematic diagram of single-link flexible manipulator with length L,
Young’s modulus E, transverse area moment of inertia I, mass per unit length m, hub inertia
Jh subjected to a torque T. In this figure, XOY and POQ represent the fixed and rotating co-
ordinate frames respectively. The manipulator is assumed to vibrate dominantly in the XOY
plane only. Considering the manipulator long and slender, transverse shear and rotary inertia
effects are neglected and hence the manipulator is modeled by Euler-Bernoulli beam theory.
For small values of angular displacement θ , angular velocity θ and deflection w, the
governing linear partial differential equations for undamped motion are obtained as:
2 2 2 2
2 2 2 2
d
d
w w
EI m m x q
x x t t
θ⎛ ⎞∂ ∂ ∂
+ +⎜ ⎟⎜ ⎟∂ ∂ ∂⎝ ⎠
= (1)
2 2 2
2 2 2
0
d d
d
d d
L
h
w
J mx x x T
t t t
θ θ⎛ ⎞∂
+ +⎜ ⎟⎜ ⎟∂⎝ ⎠
∫ = (2)
157

157. Note that m and I are functions of x, the distance of the point on the manipulator from the
fixed point O and transverse load q is a function of both x and time t. Tip mass Mp can be
tackled by using Dirac-delta function i.e. by adding a term Mpδ(x-L) in m. The boundary
conditions at the torque end are:
(0, ) 0w t = (3)
0
0
x
w
x =
∂
=
∂
(4)
The natural boundary conditions at the free end are:
2
2
0
x L
w
x =
∂
=
∂
(5)
2
2
0
x L
w
EI
x x
=
⎛ ⎞∂ ∂
=⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠
(6)
For FEM formulation, the manipulator is divided into n elements. A typical
element having five degree of freedom is shown in Fig. Q1(b). Here, are the
transverse deflections and slopes at the first and second nodes of the element.
2211 ,,, wwww ′′
Now, ansewr the following questions:
(A) Prove that the finite element formulation by Galerkin’s approach yields the following
elemental mass matrix for the ith
element:
11 12 13 14 15
21
2
31
41
2 2
51
156 22 54 13
22 4 13 3
420
54 13 156 22
13 3 22 4
i
m m m m m
m h
m h
m h h h h
m h
m h h h h
⎡ ⎤
⎢ ⎥−⎢ ⎥
⎢=
⎢
−⎢ ⎥
⎢ ⎥
− − −⎢ ⎥⎣ ⎦
e
iM 2
h
h
⎥−
⎥
m h i i
(7)
where mi is the mass per unit length of the element obtained by multiplying the average
area of the element with mass density, h is the element length and
;( )2 2
11 140 3 3 1= − + ( )12 21 21 10 7= = −m m h i ; ( )2
13 31 7 5 3= = −m m h i
;14 41 21 (10 3)m m h i= = − ( )2
15 51 7 5 2= = − −m m h i .
(B) Prove that to take into account, the hub inertia and tip mass, following terms are added in
the assembled mass matrix of size (2n+3) × (2n+3):
(i) Jh + MpL2
in the first element of the principal diagonal
158

158. (ii)MpL at the element (1, 2n+2) and (2n+2, 1)
(iii) Mp at the element (2n+2, 2n+2).
(C) Prove that the element stiffness matrix is obtained as
2
3
2 2
0 0 0 0 0
0 12 6 12 6
0 6 4 6 2
0 12 6 12 6
0 6 2 6 4
e
iK
⎡ ⎤
⎢ ⎥−⎢ ⎥
⎢= −
⎢
− − −⎢ ⎥
⎢ ⎥
−⎣ ⎦
i
h h
EI
h h h h
h
h h
h h h h
2 ⎥
⎥
(8)
where Ii is the average moment of inertia of the ith
element. The size of an element
being small, taking the average value of cross-sectional area and area moment of inertia
yields sufficiently accurate results.
Q.2: Carry out the FEM formulation of a rod vibrating freely with damping. For a uniform rod,
taking 2 elements, find out the fundamental frequencies with and without damping.
Q.3: Find out the fundamentally frequency and corresponding mode shape for a uniform rod
subjected to the following boundary conditions:
(A) free-free (B) fixed-free (C) fixed-fixed
In each case, first find out the exact solution, then study the convergence of FEM solution starting
from the 1 element model to 10 element model.
Q.4: A rod is pushed by a force F on a smooth surface. The rigid body displacement of the rod is
denoted by a, which is the displacement of the node 1. The displacement of the particles relative
to node 1 is denoted by u, which is a function of x. The governing equations of motion are:
2 2
2 2
2
20
( )
0
( )
d
l
u a u
E A m
x t
a u
F m x
t
∫
∂ ∂ +
− =
∂ ∂
∂ +
=
∂
where E is the Young modulus of elasticity, A is the cross-sectional area and m is the mass per
unit length.
F
l
Figure: Q4
159

159. The one-element finite element formulation of this problem leads to:
[ ] 1 1
1
2 2
d
[ ]
d
0
F
a a
u
M u K u EA
x
u u
⎧ ⎫
⎧ ⎫ ⎧ ⎫ ⎪ ⎪
⎪ ⎪ ⎪ ⎪ ⎪ ⎪
+ = −⎨ ⎬ ⎨ ⎬ ⎨ ⎬
⎪ ⎪ ⎪ ⎪ ⎪ ⎪
⎩ ⎭ ⎩ ⎭ ⎪ ⎪⎩ ⎭
Find out the expression for mass matrix [M] and stiffness matrix [K].
160

160. Chapter 13
FINITE ELEMENT FORMULATION OF TIME DEPENDENT
PROBLEMS
(Lecture 26-28)
13.1 INTRODUCTION
In this Chapter we will study the problems in which we are interested in finding out the
time response. By time response, we mean the value of the variable with respect to time. In
previous chapter, we found the frequency of free vibrations and mode shape. If the mode shape,
frequency and initial conditions are known, then the periodic motion is known completely. In the
forced vibration, we have a forcing function that may depend on time. Therefore, the motion at
each time has to be found incrementally. Usually, a finite-difference approximation is made for
the time derivative. Thus, these type of problems involve FEM and finite difference schemes in
combination.
13.2 CLASSIFICATION OF PARTIAL DIFFERENTIAL EQUATIONS
Let the differential equation be
(
2 2 2
2 2
2 2 2
z z z z z
a h b f g cz f x
x x y y x y
∂ ∂ ∂ ∂ ∂
+ + + + + =
∂ ∂ ∂ ∂ ∂ ∂
), y (13.1)
We say that Eq. (13.1) is of
(1) Elliptic type, when 2
0ab h− >
(2) Parabolic type, when 2
0ab h− =
(3) Hyperbolic type, when 2
0ab h− <
Consider the Laplace’s equation given by
2 2
2 2
0
z z
x y
∂ ∂
+ =
∂ ∂
(13.2)
Here, . Therefore, ab-h2
>0. Thus, the Laplace equation is of
elliptic type. Consider one dimensional transient heat conduction equation given by
1, 1, 0a b h f g c= = = = = =
2
2
1
0
T T
tx α
∂ ∂
− =
∂∂
(13.3)
where t is the time and α is the thermal diffusivity. Here, 1, 0, 0a b h= = = . Therefore, ab-
h2
=0. Hence, this equation is of parabolic type. For two-dimensional transient heat conduction
161

161. 2 2
2 2
1
0
T T T
tx y α
∂ ∂ ∂
+ −
∂∂ ∂
= (13.4)
Here, a=1, b=1 and h=0. Therefore, ab-h2
>0. Hence, this equation is of elliptic type. Consider the
forced vibration of a rod problem with damping. The governing equation is given by
2 2
2 2
( ) 0
u u
EA A q t
x t
ρ
∂ ∂
+ + =
∂ ∂
(13.5)
where E is the Young’s modulus, A is the cross-sectional area, ρ is the density and q is the
forcing load intensity. Here, a= EA, b= ρA and h=0. Therefore, ab-h2
>0. Thus, this equation is of
elliptic type.
13.3 TIME RESPONSE OF A PARABOLIC EQUATION
You have carried out the finite element formulation of free vibration problems. You can
easily do the finite element formulation for the transient problems. In that, the stiffness and mass
matrices are obtained in a similar manner. The load vector is similar to load vector in static
problem, the applied load is distributed to the nodes in the same manner. However, the nodal
forces will now be a function of time.
As a result of the finite element formulation, one gets the equation of the following form:
[ ]{ } [ ]{ } { }( )M u k u F t+ = (13.6)
Subjected to the initial conditions
{ } { }00
u u= (13.7)
We have to carry out the finite difference approximation of the time derivative. Therefore, let
( ){ } { }
{ } { }1
1
1
1 fs s
s s
s
u u
u u
t
α α α+
+
+
−
− + = ≤
Δ
or 0 1≤ (13.8)
where the subscript s indicates the time step number and Δts+1=ts+1-ts. Usually, time interval
between 2 successive increments kept same, although not necessary. For the time being, we will
assume that time interval between two successive intervals is Δt. Thus,
{ } { } ( ){ } { }1 1
1s s s s
u u t u uα α α+ +
⎡ ⎤= + Δ − + ≤ ≤⎣ ⎦ for 0 1 (13.9)
Let us consider the different cases,
Case1: 0,α = is called the forward difference (or Euler) scheme. It is conditionally stable and
the order of accuracy is ( )O tΔ . It is called explicit scheme, because the value of u at time
increment s+1 can be obtained from the known information at time increment s.
162

162. Case 2: 1,α = is called the backward difference scheme. It is stable and the order of accuracy
is ( )O tΔ .
Case 3:
1
,
2
α = is called Crank-Nicholson scheme. It is also stable and the order of accuracy
is ( )( )2
O tΔ .
Case 4:
2
,
3
α = is called Galerkin method, It is also stable and the order of accuracy is
( )( )2
O tΔ .
Case 2, 3 and 4 are called the implicit methods. In these methods, in order to find u at time
increment s+1, one needs the information at time increment s as well as s+1.
The finite element equations at two time intervals are given by
{ } [ ] { } { }ss ss
M u k u F+ = (13.10)
{ } [ ] { } { }11 1 s 1s ss
M u k u F++ +
+ = +
(13.11)
We can see that
[ ]{ } ( )[ ]{ } [ ] { } { }( )1 11 1
1s ss s s
t M u t M u M u uα α+ ++ +
Δ + Δ − = s
− (13.12)
Substituting equations (13.10-13.11) in it,
{ } [ ] { }( ) ( ) { } [ ] { }( ) [ ] { } { }( )1 11 1 11
1s ss s s ss s
t F k u t F k u M u uα α+ ++ + ++
Δ − + Δ − − = −s s
(13.13)
Rearranging,
[ ] [ ] { } ( ) [ ] { } { } ( ){ }1 1 11 11 1
1 1s s ss s ss s
M t k u M t k u t F Fα α α+ + ++ ++ +
⎡ ⎤ ⎡ ⎤ s
α⎡ ⎤+ Δ = − − Δ + Δ + −⎣ ⎦⎣ ⎦ ⎣ ⎦
(13.14)
Note that for 0α = we obtain left hand side as [ ]{ } 1
.s
M u +
when the mass matrix is diagonal, the
equation becomes explicit and we can solve for { } 1s
u +
directly without inverting.
For all numerical schemes, in which 1/ 2α < , the α - family of approximation is stable
only if the time step satisfied the following (stability) condition.
( )
2
1 2
crt t
α λ
Δ < Δ ≡
−
(13.15)
where λ is the largest eigen value of the finite element equations.
163

163. 13.4 FORCED VIBRATION PROBLEMS
If the frequency of excitations applied to a structure is less than roughly one-third of the
structure’s lowest natural frequencies of vibration, then the problem is quasistatic. Two types of
dynamic problems are wave propagations and structural dynamics problems. Time history can be
obtained by model methods and direct integration method. The finite element formulation
provides the following equations:
t t t t t t t t
M U C U k U R+Δ +Δ +Δ +Δ
+ + = (13.16)
The Newmark method is an extension of the linear acceleration method. The following
assumptions are used:
( )1t t t t t t
U Uδ δ+Δ +Δ
U t⎡ ⎤= + − + Δ⎣ ⎦U (13.17)
21
2
t t t t t t t
U U U t U U tα α+Δ +Δ⎡ ⎤⎛ ⎞
= + Δ + − + Δ⎜ ⎟⎢ ⎥
⎝ ⎠⎣ ⎦
(13.18)
where α and δ are parameters that can be determined to obtain integration accuracy and
stability. When 1/ 2, 1/ 6,δ α= = it corresponds to linear acceleration method. For constant
average acceleration, 1/ 2, 1/ 4δ α= = . The average acceleration is ( )1
2
t t t
U U+Δ
+ .
Now we present Step-by-step solution using Newmark integration method. The algorithm
has been taken from the book of Bathe [1].
A. Initial conditions:
1. Form stiffness matrix K, mass matrix M, and damping matrix C.
2. Initialize 0 0 0
U, U, and U.
3. Select time step and parameterstΔ α and δ and calculate integration constants.
( )
2
0.50; 0.25 0.5δ α δ≥ ≥ +
( )
0 1 2 32
4 5 6 7
1 1 1
; ;
2
1; 2 ; 1 ;
2
a a a a
t t t
t
a a a t a
δ
α α α α
δ δ
1;
tδ δ
α α
= = = = −
Δ Δ Δ
Δ ⎛ ⎞
= − = − = Δ − = Δ⎜ ⎟
⎝ ⎠
4. Form effective stiffness matrix 0 1
ˆ ˆK : K K M+ C.a a= +
5. Triangularize ˆ ˆK : K LDL .T
=
B. For each time step:
164

164. 1. Calculate effective loads at timet t+ Δ :
0 2 3 1 4 5
ˆR R M( U U U) ( U Ut t t t t t t t t t
a a a C a a a+Δ +Δ
= + + + + + + U)
2. Solve for displacements at time t t+ Δ :
t+ t t+ t ˆLDL U= R.T Δ Δ
3. Calculate accelerations and velocities at timet t+ Δ :
( )0 2U U U Ut t t t t t t
a a+Δ +Δ
= − − - 3 Ua
6 7U U+ U+ Ut t t t t t
a a+Δ +Δ
=
13.5 CONCLUSIONS
In this chapter, we have discussed the time dependent problems. The procedure uses
finite difference method along with finite element. Only some representative methods have been
discussed.
REFERENCE
1. Bathe, K.J., 1982, Finite Element Procedures, Prentice-Hall of India, New Delhi.
EXERCISE 13
Q.1: A rod is pushed by a force F on a smooth surface. The rigid body displacement of the rod is
denoted by a, which is the displacement of the node 1. The displacement of the particles relative
to node 1 is denoted by u, which is a function of x. The governing equations of motion are:
2 2
2 2
2
20
( )
0
( )
d
l
u a u
E A m
x t
a u
F m x
t
∫
∂ ∂ +
− =
∂ ∂
∂ +
=
∂
where E is the Young modulus of elasticity, A is the cross-sectional area and m is the mass per
unit length.
l
F
Figure: Q1
165

165. The one-element finite element formulation of this problem leads to:
[ ] 1 1
1
2 2
d
[ ]
d
0
F
a a
u
M u K u EA
x
u u
⎧ ⎫
⎧ ⎫ ⎧ ⎫ ⎪ ⎪
⎪ ⎪ ⎪ ⎪ ⎪
+ = −
⎪
⎨ ⎬ ⎨ ⎬ ⎨ ⎬
⎪ ⎪ ⎪ ⎪ ⎪
⎩ ⎭ ⎩ ⎭
⎪
⎪ ⎪⎩ ⎭
Solve this problem on computer by giving the suitable values to various parameters.
Q.2: Show that an explicit problem is only conditionally stable.
Q.3: A copper cylinder of diameter 10 cm and length 5 m is heated at one end by a flame of
7000
C temperature. The ambient temperature is 200
C. Find out the temperature-time history using
FEM. Take the data from standard books on heat transfer.
Q.4: For all numerical schemes, in which 1/ 2α < , the α - family of approximation is stable
only if the time step satisfied the following (stability) condition.
( )
2
1 2
crt t
α λ
Δ < Δ ≡
−
Prove it.
166

166. Chapter 14
FEM FORMULATION OF PLATE PROBLEM
(Lecture 29-31)
14.1 INTRODUCTION
A flat plate supports transverse load by bending action. In the classical theory of plates
certain assumptions are made initially to simplify the problems to two dimensions. Such
assumptions concern the linear variation of strains and stresses on lines normal to plane of
plate. The state of deformation of plate can described entirely by one quantity. This is the
lateral displacement w of the ‘middle plane’ of the plate.
In classical Kirchhoff thin plate theory it is assumed that the normal to the mid plane
remains normal to it even after deformation. This means that cross sections do not deform.
However, in thick plate theory the deformation of cross section is taken into account.
FEM formulation of thin plate problems involves solving a fourth order differential
equation. The potential energy functional will contain second derivatives of the unknown
functions. Because of this, continuity conditions, between elements have to be imposed not
only on the lateral displacement but also on its derivatives. This is to ensure that plate
remains continuous and does not ‘kink’. If ‘kinking’ occurs, the second derivative or
curvature becomes infinite and certain infinite terms occur in the energy expression.
Determination of suitable shape functions is much more complex.
The potential energy functional of thick plate problem contains only first derivatives.
Here, rotations of the cross-section about two axes cannot be determined from the slopes. At
each node one has to find the lateral displacement w and rotations θx and θy. Only these three
quantities need to be continuous; their slopes can be discontinuous. Thus, in thick plate
problems, same shape functions as used in the plane stress and plane strain problems can be
used. However, solving a thin plate problem using thick plate formulation poses numerical
difficulties. In this article, both the formulations are discussed.
14.2 THIN PLATE FORMULATION
The thin formulation is based on the classical Kirchhoff’s theory. Here, the main
assumptions are that the points on the mid-surface z=0 move only in the z direction as the
plate deforms in bending and a line that is straight and normal to the mid-surface before
loading is assumed to remain straight and normal to the mid-surface after loading (see
167

167. line OP in Fig. 14.1). Thus transverse shear deformation is assumed to be zero. A point
not on the mid-surface has displacement components u and v in the x and y directions,
respectively. Since w,x and w,y are small, from Fig.14.1 one can see,
yx zwvzwu ,, −=−= (14.1)
Hence, the strain-displacement relations are given by,
xyxyxy
yyyy
xxxx
zwvu
zwv
zwu
,2,,
,,
,,
−=+=
−==
−==
γ
ε
ε
(14.2)
It can be easily shown [Cook et al.,1989] that for a plate with zero initial strain,
{ } }]{[ κkDM −= (14.3)
where
T
xyyyx
T
xyyx wwwMMMM ],2,,[}{and][}{ x== κ (14.4)
zzzzMzzM
t
t
xy
t
t
yy
t
t
xx ∫∫=∫=
−−−
2/
2/
2/
2/
2/
2/
ddd σσσ (14.5)
For isotropic material,
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
=
2/)1(00
0
0
][
D
DD
DD
Dk
ν
ν
ν
where
)1(12 2
3
ν−
=
Et
D (14.6)
D is called flexural rigidity of the plate.
The strain energy expression is given by
{ } { } ADU k
T
A
d][
2
1
κκ∫= (14.7)
Now, }]{[ dNw = (14.8)
where {d}=array containing nodal degrees of freedom in the element
Then,
{ } }]{[ dB=κ (14.9)
where [B] is gradient matrix obtaining by operating on [N]
Following Ritz FEM procedure, the following stiffness matrix is obtained
168

168. [ ] ∫=
A
k
T
ABDBk d]][[][ (14.10)
The load vectors can be found in the similar manner.
w,x u=-zw x , Midsurface
z, w w
169
dx
z
P
t/2 z w,x
x, w w 0
x,u
t/2 w,x
p
o
(a) (b)
Figure 14.1: (a) Differential element of thin plate before loading (b) after loading
14.3VARIOUS THIN PLATE ELEMENTS
As discussed earlier, thin plate formulation requires both w and its normal slope across an
interface to be continuous. This condition increases the mathematical and computational
difficulties. It is, however, relatively simple to obtain shape functions which, while preserving
continuity of w, may violate its slope continuity between elements, even though they satisfy the
slope continuity at the node where such continuity is imposed. If such chosen functions can
represent ‘constant strain’ state and in addition pass the Patch test (Zienkiwicz, 1993), then
convergence can still be found. These type of shape functions are called ‘non-conforming’ shape
functions and associated elements are called ‘non-conforming’ elements. The shape functions
satisfying all continuity requirements are called ‘conforming’ shape functions and associated
elements are called ‘conforming’ elements. It has been found that on many occasions,
‘conforming’ elements yield an inferior accuracy compared to ‘non-conforming’ elements.
Following subsections describe the commonly used thin plate elements.
14.3.1 RECTANGUAR ELEMENT WITH CORNER NODES
Consider the 12 d.o.f rectangular element of Fig. 14.2. Degree of freedom w,x3 and w,y3
are slopes of the plate mid-surface at node 3. Lateral displacement w of this element has the form
[Gallagher, 1975]
[ ] { }axyyxyxyyxxyxyxyxw 33322322
1= (14.11)

169. where vector {a} contains 12 coefficients, which must be exchanged for the twelve nodal
degrees of freedom {d} by the usual process. If the d. o. f. at each nodes are w, xw ∂∂ / and
, then shape functions at Ith
nodes are given byyw ∂∂ /
[ ])1)(1()1(,)1()1)(1(),2)(1)(1(
2
1
00
2
00
2
00
22
0000 −++−++−−++++ ξηξξηηξηηξηξηξ ii ab (14.12)
with
ii
cc byyaxx
ηηηξξξ
ηξ
..
/)(,/)(
00 ==
−=−=
It can be shown that this element is non-conforming. In fact it is impossible to specify simple
polynomial expressions for shape functions ensuring full compatibility when only w and its
slopes are prescribed at nodes [Irons and Draper, 1965]. However, this element does permit a
state of constant strain (curvature) to exist.
Z,w
2 4
w3 a
y
a
3 w,x3
b b 3
w,y3
x
Figure 14.2: Rectangular element with corner nodes
14.3.2 TRIANGULAR ELEMENT WITH CORNER NODES
Fig. 14.3 shows a triangular node with nine degrees of freedom per node. A nine-term field
for w is appropriate to this element. However, a complete cubic contains 10 terms. Possible nine
term fields include
[ ] { }ayxyyxxyxyxw 322322
1= (14.13)
[ ] { }ayxyyxxyxyxyxw 322322
1 += (14.14)
Equation (14.13) does not contain term xy. The resulting element is therefore incapable of
passing a constant-twist patch test, and is considered unacceptable. Equation (14) leads to an
element that lacks geometric isotropy and has poor convergence properties. For certain
orientations of triangle sides, the shape functions cannot be obtained as the corresponding
170

170. transformation matrix becomes singular. An alternative is to add a central node to the formulation
and eliminate it by static condensation. This would allow a complete cubic to be used but it was
found that an element derived on this basis does not converge.
w3
3 w,y3
y
w,x3
w2
w,x1 2 w,x2
1 w,y1 w,y2
w1
x
Figure 14.3: A triangular element with corner nodes
14.3.3 QUADRILATERAL AND PARALLEOGRAM ELEMENTS
Henshall et al. (1972) studied the performance of a quadrilateral element and concluded
that reasonable accuracy is attainable. Their paper gives all the details of transformations required
for an isoparametric mapping and the resulting need for numerical integration. Only for the case
of a parallelogram is it possible to achieve constant curvature using natural coordinates. Stiffness
matrices of these elements have been worked out by Dawe (1966).
14.3.4 16 NODED RECTANGULAR SHAPE FUNCTION
With a rectangular element of Fig. 14.2, the specification of ∂ as a nodal
parameter is always possible as it does not involve ‘excessive continuity’. It is easy to show that
for such an elemental polynomial shape functions giving compatibility can be easily determined.
A polynomial expansion involving sixteen constants (equal to the number of nodal parameters)
could for instance be written retaining terms which do not produce a higher order variation of w
or its normal slope than cubic along the sides.
yxw ∂∂/2
14.4 THICK PLATE FORMULATION
Thick plate formulation is based on Mindlin plate theory. The main assumptions of the
Mindlin plate theory are:
1. Stress component along the normal to the mid-plane of the plate is negligible. (Same as in
Kirchhoff plate theory.)
2. Transverse displacement does not vary in the thickness direction. (Same as in Kirchhoff
plate theory).
171

171. 3. The normal to the mid-plane remains straight but not necessarily normal to it after
deformation.
In view of the Mindlin plate assumptions, for small deformation, we have the following
displacement and strain components:
where and θy are the counterclockwise rotations about the y and negative x axis respectively.θx
(14.15)
y x
y x
x y
yx
xy
yz x
xz y
u z v z
z z
x y
z
x y
w
y
w
x
θ θ
θ θ
ε ε
θθ
γ
γ θ
γ θ
= =
∂ ∂
= =
∂ ∂
∂⎛ ⎞∂
= +⎜ ⎟
∂ ∂⎝ ⎠
∂
= +
∂
∂
= +
∂
Since σz=0, the stress-strain relations for the isotropic linearly elastic material become,
)(
1 2 yxx
E
νεεσ +=
ν−
)(
1 2 xyy
E
νεε
ν
σ +
−
=
=xyτ xy
E
γ
ν)1(2 +
(14.16)
yzsyz
E
γ
ν
ατ
)1(2 +
=
zxszx
E
γ
ν
ατ
)1(2 +
=
where E is the Young modulus, ν is the Poisson ratio and sα is the shear correction
in energy V is given by,
factor.
The stra
2
1
=V
/ 2
(
t
/ 2
)d dx x y y z z xy xy yz yz xz xzA t
z Aσ ε τ γ τ γ τ γ
−
+ + + (14.17)
where t is the thickness of the plate and A is the area of the plate. Substituting
σ ε σ ε+ +∫ ∫
the
expressions of stresses and strains in the above equations, the expression for V becomes:
172

172. dA
y
t
xy
t
x
tEt
V
yxyx
A
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
∂
∂
+
∂
∂
∂
∂
+⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∫
−
=
2
2222
2
1261212
1 θθθνθ
ν
dA
x
y
y
t
y
yt
y
tEt xx
A
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
∂
∂
∂
∂
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
∂
∂
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
∂
∂
∫
+
+
θθθθ
ν 61212)1(22
1 22222
dA
x
w
x
w
y
w
y
wEt
+ν
xxyyA s
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
∂
∂
+⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
++
∂
∂
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
∂
∂
+∫+ θθθθα 22
)1(22
1
2
2
2
2
(14.18)
When the total potential energy term is minimized, the strain energy functional provides
the stiffness matrix.
The Mindlin plate element requires C0
continuity. The types of shape functions used
for plane stress and plane strain problems can be used for plates also. For n nodded
element, the primary variables are approximated as,
∑=
=
=
ni
i
ii wNw
1
(14.19)
(14.20)∑=
=
=
ni
i
xiix N
1
θθ
∑=
=
=
ni
i
yiiy N
1
θθ (14.21)
where Ni is Lagrange shape function for ith
node.
Following the standard procedure, the elemental stiffness matrix can be obtained
as,
[ ] [ ] [ ][ ] [ ] [ ][ ]dABDBdABDBk sjs
T
A
sibjb
T
A
biij
ee
∫+∫= (14.22)
where
[ ]
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
000
0000
0000
,,
,
,
xiyi
yi
xi
bi
NN
N
N
B (14.23)
[ ]
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
=
000
000
,
,
iyi
ixi
si
NN
NN
B (14.24)
173

173. [ ]
( ) ( )
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
=
2
1
00
01
01
112 2
3
ν
ν
ν
ν
Eh
Db (14.25)
[ ]
( ) ⎥
⎦
⎤
⎢
⎣
⎡
+
=
10
01
12 ν
α
Eh
D ss (14.26)
In the right hand side of expression (14.22), first term indicates stiffness due to bending,
while the second term indicates stiffness due to shear.
14.5 LOCKING PHENOMENON
As discussed in the previous section, the stiffness matrix is composed of two parts-matrix
due to bending and that due to shear. In a very thin plate, the stiffness matrix due to shear
dominates. Hence, the solution gives spurious zero values of bending deformation. This is
called the locking of the mesh. Thus, solving thin plate problem by a thick plate formulation
may not provide the correct results.
One way to reduce locking is to use selective integration scheme. In this scheme, the
bending stiffness matrix is computed using full Gauss-Quadrature integration, but the shear
stiffness matrix is computed using reduced Gauss-quadrature integration. Thus, using a 4-
noded rectangular element, full (2 by 2) integration is used for bending stiffness matrix and 1
Gauss-Quadrature point is used for shear stiffness matrix.
14.6 AN EXAMPLE
To illustrate the application of plate elements in stress analysis, an example of a plate
clamped at four edges and loaded by uniform pressure is taken (Fig. 14.4). Thin plate
rectangular elements are used. Mesh is shown in Fig. 14.5. Fig.14.6 and Fig. 14.7 show the
contours of xσ and yσ respectively. It is seen that stresses at edges are more compared to
the stresses at center. This example is a simple one for which even the closed form solution is
available. FEM can be applied to cases involving complicated geometries and non-
homogeneous as well as anisotropic materials.
174

174. Figure 14.4: A plate clamped at four edges and loaded with uniform pressure
Figure 14.5: Finite element mesh
Figure 14.6: Contours of σx
175

175. Figure 14.7: Contours of σy
14.7 SUMMARY AND CONCLUSION
In this article, FEM formulation of thin and thick plates has been discussed. The thin
plate formulation is complicated because of the fourth order governing differential involved.
The thick plate formulation requires only C0
continuity element. However, it does not provide
good results for thin plate problem due to phenomenon of locking. Locking can be avoided
by following a selective integration procedure and it is possible to write a robust code for
solving both thick and thin plate problem. It is to be mentioned that for stress analysis
purpose, even 3-dimensional elements can be used. However, solving the plate problem using
3-dimensional stress analysis will require more memory. Moreover, the system of equation
becomes ill-conditioned and poses numerical difficulties.
REFERENCES:
1.Cook,R.D.,Malkus.D.S.,Plesha.M.E., “Concepts and Applications of Finite Element
Analysis”, Jhon Wiley & Sons Inc., 1989
2.Dawe.D.J., “Parallelogram element in the solution of rhombic cantilever plate problems”,
J.of Stain analysis,Vol.3,1966
3.Gallagher,R.H.,”Finite element analysis: Fundamentals”, Prentice-Hall, Englewood cliffs,
NJ,1975.
4.Henshell.R.D.,Walters.D., and Warburton.G.B., “A new family of curvilinear plate bending
elements for vibration and stability”, J.Sound and Vibration,Vol.20,1972,pp.327-343.
5.Irons.B.M and Draper.J.K., “Inadequacy of nodal connections in a stiffness solution for
plate bending”,J.A.I.A.A.,Vol.3,1965
6.Zienkiewicz.O.C., “The Finite Element Method”, Tata McGraw-Hill Publishing Company
Limited, New Delhi.,1993.
176

176. EXERCISE 14
Q.1: The governing equation for thin plate bending is given by
4 4 4
4 2 2 4
2
w w w
x x y y
∂ ∂ ∂ q
D
+ + =
∂ ∂ ∂ ∂
Obtain the variational form for this. Obtain FEM equations.
Q.2: Flow of an incompressible viscous fluid with negligible inertia is governed by the
following differential equation
4 4 4
4 2 2 4
2 0
x x y y
ψ ψ ψ∂ ∂ ∂
+ + =
∂ ∂ ∂ ∂
with boundary conditions:
Essential: u=u*
, v=v*
on Γu
Natural:
* *
andx x yt t t t= = y
where
xx xyx x
y yxy yy
t n
t n
σ σ
σ σ
⎡ ⎤⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪
= ⎢ ⎥⎨ ⎬ ⎨ ⎬
⎪ ⎪ ⎪ ⎪⎢ ⎥⎩ ⎭ ⎩ ⎭⎣ ⎦
2ij ij ijpσ δ μ= − + ε
, ,y xu vψ ψ= = −
where ψ is the stream function. This is similar to thin plate equation. Find out the variational
form of the equation and obtain FEM formulation.
Q.3: For thin plate FEM analysis, if we choose a 4 noded rectangular element with degrees of
freedom at the nodes as , / and /w w x w y∂ ∂ ∂ ∂ , then this element will be incompatible.
Mathematically, prove this statement. Also, prove that if we add a degree of freedom
2
w
x y
∂
∂ ∂
, then
this element will be compatible.
Q.4: Formulate the problem of vibration of thick plate.
Q.5: Consider a simply supported thick square plate loaded with a central concentrated load.
Using a FEM package obtain the maximum deflection of the plate. Next, solve the same problem
by fixing the plate on one edge and simply supporting on the other three edges. Next, solve the
problem by fixing two edges and simply supporting the other 2 edges. Similarly solve the
177

177. problem by fixing 3 and simply supporting 1 edge. Lastly, solve the problem by fixing all 4
edges. Compare the maximum deflection and maximum stresses in all cases.
178

178. Chapter 15
FINITE ELEMENT FORMULATION OF 2-D FLOW PROBLEMS
(Lecture 32-35)
15.1 INTRODUCTION
Till now we have carried out finite element formulation of linear problem. In this chapter,
we shall study non-linear problems. To illustrate the concept, problem of strip drawing has been
formulated in details. The steady-state strip drawing process can be considered as the flow of a
viscous fluid, where viscosity is dependent on the flow stress and strain-rate. After reading this
chapter, you will be able to carry out the finite element formulation of Navier-Stokes equations.
Navier-Stokes equations for 2-dimensional flow of viscous, incompressible fluids;
2 2
2 2
1u u p u u
u v
x y x x
υ
ρ
⎛ ⎞∂ ∂ ∂ ∂ ∂
+ = − + +⎜
∂ ∂ ∂ ∂ ∂⎝ ⎠y
⎟ (15.1)
2 2
2 2
1v v p v v
u v
x y y x
υ
ρ
⎛ ⎞∂ ∂ ∂ ∂ ∂
+ = − + +⎜
∂ ∂ ∂ ∂ ∂⎝ ⎠y
⎟ (15.2)
0
u v
x y
∂ ∂
+ =
∂ ∂
(15.3)
Here, u and v are the velocities along x and y-axis, p is the pressure, ρ is the density and υ is the
kinematic viscosity. The boundary condition can be in the form of prescribed velocity (essential
boundary condition) and prescribed traction (natural boundary condition). There are two popular
ways of solving this problem in a constant volume. One is to carry out FEM formulation with u, v
and p as the primary variable which is called as mixed (p-v) formulation. The second way is to
eliminate p using penalty method. In this, the Eq.(15.3) is written as
0
u v p
x y λ
∂ ∂
+ + =
∂ ∂
(15.4)
where λ is a very large number. Ideally for incompressible flow, this number should be infinite.
However, in practice, this number should be chosen depending upon the precision of computer.
From this equation,
u v
p
x y
λ
⎛ ⎞∂ ∂
= − +⎜ ⎟
∂ ∂⎝ ⎠
(15.5)
179

179. This value can be substituted in the Eq. (15.1). In this chapter we will be solving Navier-Stokes
equation, considering following features:
1) A mixed pressure-velocity formulation will be adopted.
2) We shall consider that inertia force is not large, therefore convective acceleration terms
(the terms on the left hand side of inequality constraints) will not be considered.
3) The viscosity will not be considered constant. It will depend on deformation. Thus
despite the elimination of inertial forces, the problem remains non-linear.
Now we will model plane-strain steady-state process of strip drawing. Index notation has been
used in the formulation of the model.
15.2 DISCRETIZATION OF THE STRIP
In FE modeling of drawing process where a curved die has been chosen, mesh
generation can be done as shown in the Fig. 15.1. The mesh selected for the current problem
consists of 56 elements. More elements have been selected in the deformation zone. This is the
preprocessing stage of the problem. For velocity approximation nine noded element is chosen and
for pressure four noded element is selected. It will be seen from the weak-formulation that four
noded pressure and nine noded velocity are sufficient to formulate the process.
Figure 15.1: A typical 56 element mesh and nodal structure for pressure and velocity
180

180. 15.3 GOVERNING EQUATIONS AND BOUNDARY CONDITIONS
The non-dimensionalzed governing equations for FE modeling of the plain-strain drawing
problem will be,
11 22 0ε ε+ = (15.6)
11 12
1 2
21 22
1 2
0
0
x x
x x
σ σ
σ σ
∂ ∂
+ =
∂ ∂
∂ ∂
+ =
∂ ∂
(15.7)
where 11 22,ε ε are strain rate tensors in x1 and x2 directions respectively and
11 12 21 22, , ,σ σ σ σ are corresponding stresses.
The boundary conditions specified on any boundary can be classified as essential boundary
condition i.e., the specification of the primary variable vector (such as velocity vector vi) and the
natural boundary condition i.e., the specification of the secondary variable (such as traction
vector). The Fig.15.1 shows the strip has been distinguished with different boundaries.
1. Entry and exit boundaries (AF and DE)
On the surface AF and DE
v1 = U1, v2 = 0 on AF (15.8)
v1 = U2, v2 = 0 on DE (15.9)
where U1 and U2 are respectively the inlet and exit velocities.
2. The top free surfaces (AB and CD)
t1 = 0, v2 = 0 on AB and CD (except two nodes near entry to die gap), (15.10)
t1 = 0, t2 = 0 on the two nodes near entry to die gap. (15.11)
3. The axis of symmetry (FE)
t1 = 0, v2 = 0 on FE (15.12)
4. The die- strip interface (BC)
v2+ v1tanφ = 0 on BC (15.13)
181

181. where φ is the angular position of the point on the surface.
The second boundary condition will be;
| ts | = μ | tn | (15.14)
This has been shown in Fig.15.2, where ts and tn are the tangential and normal components of the
stress vector t which is defined by i j j in tσ = and μ is the coefficient of friction in the case of
Coulomb’s friction model.
Figure15.2: Schematic diagram of die-strip interface BC
15.4 WEAK FORMULATION
Let v1, v2, p be the functions that satisfy all essential boundary conditions exactly, thus
the weighted residual will become;
( ) 11 12 21 22
11 22 1 2 1 2
1 2 1 2
d d 0p
A
w w w
x x x x
σ σ σ σ
ε ε
⎡ ⎤⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂
− + + + + + =⎢ ⎥⎜ ⎟ ⎜ ⎟
∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦
∫ x x (15.15)
where A denotes domain of a typical area element.
The above expression can be written in compact form as;
, 1 2d d 0ii p i j j i
A
w w x xε σ⎡ ⎤− + =⎣ ⎦∫ (15.16)
The first part of the integral contains the first order derivatives, which cannot be weakened
further, but the second part contains second order derivatives, so it has to be reduced in weak
form. The second part can be written as
182

182. ( ) ( ), ,d di j j i i j i i j i jj
A A
w A w w Aσ σ σ ,
⎡ ⎤⎡ ⎤ = −⎣ ⎦ ⎣ ⎦∫ ∫ (15.17)
The Eq. (15.17) becomes,
( ) ( ), ,d di i p i j i i j i jj
A A
w A w w Aε σ σ⎡ ⎤− + −
⎣ ⎦
0=
∫ ∫ (15.18)
In the second part of the Eq. (15.18), the term ( ),i j i j
wσ can be written as ( )i j i jw nσ
But according to Cauchy’s equation i j j in tσ =
Thus Eq. (15.18) can be expressed as,
( ),d d di i p i i i j i j
A Ai
w A t w w Aε σ
Γ
− + Γ −
∫ ∫ ∫ 0= (15.19)
The last term in Eq. (15.19) can be expressed as,
( ) ( ) ( ), , , ,
1
d d
2
i j i j i j i j j i i j j i
A A
w A w w w wσ σ , A⎡ ⎤= + + −
⎣ ⎦∫ ∫ (15.20)
which reduces to
( ), ,
1
d d
2
i j i j i j i j j i
A A
w A w wσ σ , A⎡ ⎤= +⎣ ⎦∫ ∫ (15.21)
The Eq. (15.21) can be written as
( ) ( ), di j i j i j i j
A A
w A wσ σ ε=
∫ ∫ dA
d 0Γ =
(15.22)
Thus, the weighted residual becomes
( )d di i p i j i j i i
A A i
w A w A t wε σ ε
Γ
+ −
∫ ∫ ∫ (15.23)
183

183. Equation (15.23) has first two terms in first order derivatives of velocity component, thus the
weak form has been obtained for further FE modeling. Further the equation is simplified by
substituting the following terms,
i j i j i jp Sσ δ= − + and 2ij ijS με= (15.24)
Using the relationship in Eq. (15.24), we can write
( ) ( ) ( )
( ) ( )2
i j i j i j i j i j
i i i j i j
w p S w
p w w
σ ε δ ε
ε με ε
= − +
= − +
(15.25)
The above equation can be further expanded as
( ) ( ) ( ) ( ) ( ) ( )11 22 11 11 12 12 22 222 2ij ij w p w w w w wσ ε ε ε μ ε ε ε ε ε ε= − + + + +⎡ ⎤ ⎡⎣ ⎦ ⎣ ⎤⎦
0=
(15.26)
Thus the entire weak form of weighted residual is
1 2 3 4
1 2
d d d d
A A
I A I A I I
Γ Γ
+ − Γ − Γ
∫ ∫ ∫ ∫ (15.27)
where
[ ]
( ) ( ) ( ) ( ) ( )
1 11 22
2 11 22 11 11 12 12 22 22
3 1 1
4 2 2
2 2
pI w
I p w w w w w
I t w
I t w
ε ε
ε ε μ ε ε ε ε ε ε
= +
= − + + + +⎡ ⎤ ⎡⎣ ⎦ ⎣
=
=
⎤⎦
(15.28)
Γ1 and Γ2 are respectively those parts of the boundary where tractions t1 and t2 are specified.
15.5 FINITE ELEMENT APPROXIMATION
In the weak form of the governing equation, as velocity is of first order derivative and
pressure is of zeroth order, C0
continuity is sufficient. But to avoid numerical difficulties
approximations for velocity has to be chosen one degree higher than of pressure variable. Thus
there will be nine shape functions for both components of velocity and four shape functions for
pressure approximation. The approximation for v1 and v2 is
184

184. ( )
( )
( )
( )
{ }
1 1
2 1
1 2 3 91
2 1 2 3 9
1 9
2 9
0 0 0 ........ 0 |
|0 0 0 .......0
e
e
e
v
e
e
v
v
N N N Nv
N v
v N N N N
v
v
⎧ ⎫
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪⎢ ⎥⎧ ⎫ ⎪ ⎪
= = ⎢ ⎥⎢ ⎥⎨ ⎬ ⎨ ⎬ ⎣ ⎦
⎢ ⎥⎩ ⎭ ⎪ ⎪⎣ ⎦
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪⎩ ⎭
(15.29)
In Galerkin formulation, weight functions for velocity and pressure are approximated using same
shape functions as that of velocity and pressure respectively.
{ } [ ]{1
2
e
v v
w
w N
w
⎧ ⎫
= =⎨ ⎬
⎩ ⎭
}vw (15.30)
The approximation for pressure is
{ } { }
1
2
1 2 3 4
3
4
e
e
Tp p p p
pe
e
p
p e
p N N N N N p
p
p
⎧ ⎫
⎪ ⎪
⎪ ⎪
⎢ ⎥= =⎨ ⎬⎣ ⎦
⎪ ⎪
⎪ ⎪
⎩ ⎭
(15.31)
The weight function is
{ } { }e
p p pw N w⎢ ⎥= ⎣ ⎦ (15.32)
In order to model curved boundary, geometry is approximated by 9-noded shape functions. Thus
for geometry approximation, same shape functions can be considered as that of the velocity
variable. Therefore,
{ } { }1 1
e
x N x= ⎢ ⎥⎣ ⎦ and { } { }2
e
2x N x= ⎢ ⎥⎣ ⎦ (15.33)
As the boundary of the element consists of three nodes, evaluation of the integrals over the
boundaries Γi will be,
185

185. {
1
1
2
1 1 2 3 11
3
1
v
b b b b
v b
v
w
w N N N w N w
w
⎧ ⎫
⎪ ⎪
⎪ ⎪⎢ ⎥= = ⎢ ⎥⎨ ⎬ ⎣ ⎦⎣ ⎦
⎪ ⎪
⎪ ⎪⎩ ⎭
}
}
(15.34)
and
{
1
2
2
2 1 2 3 22
3
2
v
b b b b
v b
v
w
w N N N w N w
w
⎧ ⎫
⎪ ⎪
⎪ ⎪⎢ ⎥= = ⎢ ⎥⎨ ⎬ ⎣ ⎦⎣ ⎦
⎪ ⎪
⎪ ⎪⎩ ⎭
(15.35)
Further t1 and t2 are approximated as
{ }
1
1
2
1 1 2 3 11
3
1
b
b b b b
b b
b
t
t N N N t N t
t
⎧ ⎫
⎪ ⎪
⎪ ⎪⎢ ⎥= = ⎢ ⎥⎨ ⎬ ⎣ ⎦⎣ ⎦
⎪ ⎪
⎪ ⎪⎩ ⎭
(15.36)
{ }
1
2
2
2 1 2 3 22
3
2
b
b b b b
b b
b
t
t N N N t N t
t
⎧ ⎫
⎪ ⎪
⎪ ⎪⎢ ⎥= = ⎢ ⎥⎨ ⎬ ⎣ ⎦⎣ ⎦
⎪ ⎪
⎪ ⎪⎩ ⎭
(15.37)
where t1 and t2 are the vectors of the nodal value of the traction.
( ) ( ) ( )2 2
1 2 3
1 1 1
, ,
2 2 2
b b b
N N N 2
1ζ ζ ζ ζ ζ= − = + = − (15.38)
where ζ is the natural coordinate on the boundary.
15.6 FINITE ELEMENT EQUATIONS
Equation (15.28) has to be expressed in matrix form to obtain the FEM equations. For that
purpose,
186

186. { }
1
1
11
2
22
2
12
1 2
2 1
2
1
2
v
x
v
x
v v
x x
ε
ε ε
ε
⎧ ⎫∂
⎪ ⎪
∂⎪ ⎪⎧ ⎫
⎪ ⎪⎪ ⎪ ∂⎪ ⎪
= =⎨ ⎬ ⎨
∂⎪ ⎪ ⎪
⎩ ⎭
⎬
⎪
⎪ ⎪⎛ ⎞∂ ∂
⎪ ⎪+⎜ ⎟
∂ ∂⎪ ⎪⎝ ⎠⎩ ⎭
(15.39)
( ){ }
( )
( )
( )
1
1
11
2
22
2
12
1 2
2 1
2
1
2
w
xw
w
w w
x
w
w w
x x
ε
ε ε
ε
⎧ ⎫∂
⎪ ⎪
∂⎪ ⎪⎧ ⎫
⎪ ⎪⎪ ⎪ ∂⎪ ⎪ ⎪
= =⎨ ⎬ ⎨
∂⎪ ⎪ ⎪
⎪ ⎪
⎪
⎬
⎪
⎪ ⎪⎩ ⎭ ⎛ ⎞∂ ∂
⎪ ⎪+⎜ ⎟
∂ ∂⎪ ⎪⎝ ⎠⎩ ⎭
(15.40)
The strain rate vector { }ε and ( ){ }wε becomes
{ } [ ]{ }e
B vε = and ( ){ } [ ]{ }e
vw B wε = (15.41)
where
[ ]
91 2
1 1 1
91 2
1 1
9 91 1 2 2
2 1 2 1 2
0 0
0 0 0
1 1 1 1 1 1
2 2 2 2 2 2
NN N
x x x
NN N
B
x x
N NN N N N
1
1
0
x
x x x x x x
⎡ ⎤∂∂ ∂
− − − − −⎢ ⎥
∂ ∂ ∂⎢ ⎥
⎢ ⎥∂∂ ∂
= − − − − −⎢ ⎥
∂ ∂⎢ ⎥
⎢ ⎥∂ ∂∂ ∂ ∂ ∂
− − − −⎢ ⎥
∂ ∂ ∂ ∂ ∂ ∂⎣ ⎦
∂
(15.42)
Further,
{ }{ } { } [ ]{ }11 22 1 1 0
T e
m B vε ε ε+ = = (15.43)
where
187

187. { }
1
1
0
m
⎧ ⎫
⎪ ⎪
= ⎨ ⎬
⎪ ⎪
⎩ ⎭
(15.44)
Similarly
( ) ( ) { } [ ]{ }11 22
T e
vw w m Bε ε+ = w (15.45)
Substituting Eqs. (15.39-15.45) in Eq. (15.28),
{ } { }{ } [ ]{ }
{ } { }{ } { }
[ ] { }
1 2
1 2
1 2 1 1 2 2
1 2
d d
[ ] d d
2 [ ] d d d
T Te p e
p
A
T T
e T p e
v
A
Te e
v
A
w N m B v x x
w B m N p x x
w B B v x x t w t wμ
Γ Γ
−
+ −
⎡ ⎤+ =⎣ ⎦
∫
∫
∫ ∫ dΓ + Γ
∫
(15.46)
Using the notations,
{ }{ } [ ]
{ }{ } { }
[ ]
1 2
1 2
1 2
d d
[ ] d d
2 [ ] d d
Te p
pv
A
T T
e T p
vp pv
A
e T
vv
A
K N m B x x
K B m N x x K
K B B x xμ
⎡ ⎤ = −⎣ ⎦
⎡ ⎤ = − =⎣ ⎦
⎡ ⎤ =⎣ ⎦
∫
∫
∫
e
T
b
(15.47)
Thus the finite element equation in local variable form will be,
{ } { } { } { } { } { }
1 2
1 1 2 2
1 1 1
nb nbne
T T
e e e b b b
e b b
w K w f w fδ
= = =
⎡ ⎤ = +⎣ ⎦∑ ∑ ∑ (15.48)
where
{ } { }{ } { }
{ } { }{ } { }
1 1
1
2 2
2
d
d
Tb b
b b
b
Tb b
b b
b
f N N t
f N N t
Γ
Γ
= Γ
= Γ
∫
∫
(15.49)
188

188. Here ne is the no. of area elements, nb1, nb2 are the number of boundary elements on Γ1 and Γ2.
Similarly,
{ }
{ }
{ }
e
p
e
e
v
w
w
w
⎧ ⎫
⎪ ⎪
= ⎨ ⎬
⎪ ⎪
⎩ ⎭
, { }
e
e
e
p
v
δ
⎧ ⎫⎧ ⎫⎪⎪ ⎪⎪
= ⎨⎨ ⎬⎬
⎪ ⎪⎪ ⎪⎩ ⎭⎩ ⎭
and
[ ]0 e
pv
e
e e
vp vv
K
K
K K
⎡ ⎤⎡ ⎤
⎣ ⎦⎢ ⎥⎡ ⎤ =⎣ ⎦ ⎢ ⎥⎡ ⎤ ⎡ ⎤
⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦
(15.50)
For the purpose of numerical evaluation, the variables of the area integrals in Eq. (15.47) are
transformed to the natural coordinates (ξ, η) using the following transformation;
( ) ( )
1 1
1 2
1 1
........ d d ....... | | d d
eA
x x J ξ η
+ +
− −
=
∫ ∫ ∫ (15.51)
1 1
2 2
x x
J
x x
ξ η
ξ η
∂ ∂
∂ ∂
=
∂ ∂
∂ ∂
(15.51)
where |J| is the elemental Jacobian matrix. Similarly the boundary integrals are transformed by
the relation
( ) ( )
1
1
........ d .......... | | dbJ ζ
+
−
Γ
Γ =
∫ ∫ (15.52)
where | Jb | is the Jacobian for boundary element and is given by
2 2
1 2
| |b
x x
J
ζ ζ
⎛ ⎞ ⎛ ⎞∂ ∂
= +⎜ ⎟ ⎜ ⎟
∂ ∂⎝ ⎠ ⎝ ⎠
(15.53)
Along the boundary, the coordinates (x1, x2) are approximated using 1-D quadratic shape
functions. All elemental matrices are evaluated using 3×3 Gauss quadrature. Similarly the
elemental vectors are evaluated using 3 point gauss quadrature.
The assembled finite element can be written as
{ } [ ]{ } { } { }T
W K W FΔ =
T
(15.54)
189

189. where { } [ ] { }, ,W K Δ are the global vector of nodal values of weight function, global
coefficient matrix and global vector of nodal values of pressure and velocity. { is the global
right hand side vector. Since weight functions are arbitrary, final FEM expression will be
}F
[ ]{ } { }K Δ = F (15.55)
15.7 APPLICATION OF BOUNDARY CONDITIONS
The friction conditions at a typical node on the tool work interface (Fig. 15.3) is
given by Eq. (15.14).
Figure 15.3: Shear and normal components of traction at work-tool interface if the strip
Using the Eq. (15.14), we can express ts and tn in the form of t1 and t2 as follows;
1 2
1 2
cos sin
| sin cos
s
n
t t t
t t t
φ φ
φ φ
= − +
= − −
(15.56)
Thus putting the decomposed form ts and tn into Eq. (15.14) we get,
1 2 1 2( cos sin ) ( sin cost t t t )φ φ μ φ φ− + = − − (15.57)
1 2(cos sin ) (sin cos ) 0t tφ μ φ φ μ φ∴ − − + + = (15.58)
Thus the above Eq. (15.58) becomes,
1 2(1 tan ) (tan ) 0t tμ φ φ μ+ − + = (15.59)
The above expression for an element for FEM can be written as
190

190. ( )
( )
( )
( )
( )
( )
1 1
1 2
2 2
1 2
3 3
1 2
(1 tan ) (tan ) 0
b b
b b
b b
t t
t t
t t
μ φ φ μ
⎧ ⎫ ⎧ ⎫
⎪ ⎪ ⎪ ⎪
⎪ ⎪ ⎪ ⎪
+ − +⎨ ⎬ ⎨ ⎬
⎪ ⎪ ⎪ ⎪
⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭
= (15.60)
After multiplying the above equation by
{ }{ }
1
1
| |
T
b b bN N J dζ
+
−
∫ (15.61)
it becomes
{ } { } { }1 2(1 tan ) (tan ) 0b b
f fμ φ φ μ+ − + = (15.62)
This above equation holds good at all nodes of the element. At the middle node say ‘k’ (global
node number), there is no contribution from the neighboring elements and therefore, in terms of
the global right hand side of the vector, Eq. (15.62) can be expressed as
{ } { } { }( 2 1) ( 2 )
(1 tan ) (tan ) 0
p pd k d k
F Fμ φ φ μ+ − +
+ − + = (15.63)
where dp is the total number of pressure variables. This Eq. (15.63) holds good at the end node
also.
Procedure for applying the condition (15.14) at the node ‘k’ is as follows:
• Replace
th
row of global coefficient matrix [K] by the following linear
combination:
th
− row of
( 2 1)pd k+ −
(1 tan ) times ( 2 1)pd kμ φ+ +
[K] – (tanφ + μ) times row of [K].
th
( 2 )pd k+
• Make the
th
row of global right hand vector {F} zero.( 2 1)pd k+ −
• The velocity boundary condition at the node is applied by replacing
th
row of
[K] by Eq. (15.13).
( 2 )pd k+
• Make
th
row of [K] zero.( 2 )pd k+
• Set
th
element( 2 , 2 1)p pd k d k+ + −
191

191. of [K] to element of [K] to 1.
th
tan ( 2 , 2 )p pd k d kφ + +
• Make
th
row of {F} to zero.( 2 )pd k+
The essential boundary conditions at the other boundaries are applied as follows:
In the stiffness matrix, all the elements of row and columns corresponding to the
specified degree of freedom excepting the diagonal term are made equal to zero.
The diagonal term is replaced by unity.
The right hand side vector element corresponding to the specified degree of freedom is
replaced by the specified value and other elements are modified by subtracting from them
the products of corresponding coefficient element and specified value.
After imposing the boundary condition as discussed above section, the resulting non-linear
algebraic equations (15.55) are solved iteratively by the Householder method, because the
resulting matrix of the mixed formulation is ill-conditioned. The solution is obtained in the form
of primary variables for nodal pressure and velocity.
15.8 POST-PROCESSING
The evaluation of the secondary variables like drawing force, die-pressure, separation force
and strain, strain-rate contours are termed as post-processing of the finite element method. There
are certain precautions to be followed in post-processing step. Drawing force can be calculated by
integrating the stresses across the thickness or dividing the total power by the velocity. The latter
method provides more accurate solution, because errors get subdued due to integration. Stresses
should preferably be calculated at Gauss-points corresponding to 2 Gauss-point formula.
Interfacial tangential stress should be calculated by multiplying the coefficient of friction to
interfacial normal stress.
15.9. CONCLUSION
The Flow formulation has been extensively used for the analysis of metal forming
process considering the material as rigid-plastic. The steady-state drawing process has been
solved using mixed pressure-velocity FE formulation. In the mixed pressure-velocity finite
element formulation, no pressure boundary conditions are employed and therefore the pressure
field is computed in this method needs further refinement. FEM is preferred to other methods in
the analysis of drawing process, as it can easily incorporate non-homogeneity of deformation,
process dependent material properties and different friction models.
192

192. EXERCISE 15
Q1: In the rolling process, the governing equations are given as
11 22 0ε ε+ =
11 12
1 2
21 22
1 2
0
0
x x
x x
σ σ
σ σ
∂ ∂
+ =
∂ ∂
∂ ∂
+ =
∂ ∂
Figure: Q1
The essential boundary conditions are shown in the figure. The two natural boundary conditions
at the roll-strip interface are given as
v2+ v1tanφ = 0 and | ts | = μ | tn |
where φ is the angular position of the point on the surface, μ is the Coloumb’s coefficient of
friction and where ts and tn are the tangential and normal components of the stress vector t. Due
to the existence of neutral point, friction conditions will change. Modify these boundary
conditions accordingly in the FEM modeling of rolling process.
Q2: The governing equations for the steady, constant volume flow in axisymmetric problem is
given as
193

193. ( )
( )
0
1
0
1
0
rr zz
rrr r rz
r z
rzz z zz
r z
rv v
p v v
r z r r z r
rv v
p v v
r z r r z
θθ
θθ
ε ε ε
σ σσ
σ σ
+ + =
∂⎛ ⎞∂ ∂ ∂⎛ ⎞
+ − + −⎜ ⎟⎜ ⎟
∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠
∂⎛ ⎞∂ ∂ ∂⎛ ⎞
+ − + =⎜ ⎟⎜ ⎟
∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠
=
The boundary conditions for the domain of the present problem are shown in the figure.
Figure: Q2
Carry out the FEM modeling of this problem using these governing equations and boundary
conditions.[Ref. Dixit and Dixit, 1995, An analysis of the steady–state wire drawing of the strain
hardening materials, J. Mater. Process. Technol. 47, pp.201-229].
Q3: The governing equation of a spring is given as
( )0F k k x= + x
Figure: Q3
where , , F = 2000 N0 1000 N/mk = 2
100 N/mk =
194

194. The equations can be written as
1
2
1 1
1 1 0
u F
k
u
− −⎧ ⎫⎡ ⎤ ⎧
=⎨ ⎬ ⎨ ⎬⎢ ⎥−⎣ ⎦ ⎩⎩ ⎭
⎫
⎭
Solve the problem by finite element method.
Q.4: The governing equation of a large deflection bending of elastic beam is
2
22 2
2 2
d 1
0
d 2
d d 1
0
d d 2
u w
EA f
x x x
w u w
EI EA q
x x x x x
⎧ ⎫⎛ ⎞∂ ∂⎪ ⎪⎛ ⎞
− + − =⎜ ⎟⎨ ⎬⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎪ ⎪⎝ ⎠⎩ ⎭
⎧ ⎫⎛ ⎞⎧ ⎫∂ ∂ ∂⎪ ⎪⎛ ⎞
− − +⎜ ⎟⎨ ⎬ ⎨ ⎬⎜ ⎟⎜ ⎟∂ ∂ ∂⎝ ⎠⎩ ⎭ ⎪ ⎪⎝ ⎠⎩ ⎭
− =
where u is the longitudinal displacement, w is the transverse deflection, E is modulus of elasticity,
A is the cross-sectional area, f and q are the axial distributed and transverse loading respectively.
Carry out the FEM formulation to solve this problem.
195

195. Chapter 16
ERROR ANALYSIS IN FINITE ELEMENT METHOD
(Lecture 36-38)
16.1 INTRODUCTION
Finite element method (FEM) provides an approximate solution of differential equations.
There is a need to calculate the error, which is the difference between the exact solution and the
approximate solution (i.e, FEM). Error in finite element solutions are divided mainly into three
categories:
1. Domain approximation error, which is due to the approximation of the domain.
2. Quadrature and finite arithmetic errors, which are due to the numerical evaluation of
integrals and the numerical computation on a computer.
3. Approximation error, which is due to the approximation of the solution.
In the formulation of the finite element method, usually the displacement or primary variable
field is approximated by polynomials. This approximation is the main source of error in the
solution. As this error is inherent in the method, the amount of error in the solution must be
determined in order to judge the quality of the results obtained. This error information can be
used to improve the results, by improving the primary variable approximation, and to monitor the
convergence of the solution. The differences between the exact and approximate solution, i.e.
errors decrease as the size of the subdivision ‘h’ gets smaller or as ‘p’, the order of the
polynomial in the trial function used, increases.
In this chapter, various error measures are described. Two common estimates of error are a
priori and a posteriori error estimates. A discussion regarding these is presented. Recovery
method of estimating error is discussed in detail.
16.2 ERROR MEASURES
The subject of error estimation for finite element solutions and a consequent adaptive
analysis, in which the approximation is successively refined to reach predetermined standards of
accuracy, is central to the effective use of finite element codes for practical engineering, analysis.
The main problem in the error estimation is the cost of computations and implementing such
computations into an existing code structure and hence a fully adaptive finite element structure
must be obtained. Various error measures are explained with the help of an example.
Consider the second order differential equilibrium equation
197

196. (16.1)0q-DSuSq-Lu T
==
in a domain , with prescribed displacementΩ
uu = on the boundary (16.2)uΓ
and prescribed derivatives
tSu = on the boundary (16.3a)tΓ
with
tu ΓΓΓ ∪= (16.3b)
where L is linear second-order differential operator, S is first order differential operator, q is a
constant vector, D is some matrix and u is primary variable.
In a finite element approximation, we obtain the approximating equations by a standard
Galerkin process (or equivalently by minimizing the potential energy) to obtain
0f-Ku = (16.4)
where is the elemental stiffness matrix and( ) Ω= ∫Ω
dD(SN)SNK
T
∫ ∫Ω Γ
+Ω= ΓtNqNf TT
dd
t
is the load vector. Here, N is the shape function (interpolation
function) matrix. The approximate solution u for an element is related to nodal solution une
in
the following manner:
ˆ
uˆ =N une
(16.5)
The derivatives are calculated as
(16.6)ˆ = ne
(SN)uσ
The approximate solutions u,ˆ ˆσ differs from the exact values u, σ and the difference is the
error. Thus, primary variable and derivative errors are
u-ue = (16.7a)
ˆσ = −e σ σ (16.7b)
The specification of local error in the above equations is generally not convenient and
occasionally misleading. For this reason various ‘norms’ representing some integral quantity are
198

197. often introduced to measure the error. The most common measures are the ‘energy norm’ and
‘L2 norm’. The energy norm for general problems is
∫Ω
Ω= 2
1
)( dLeee T
(16.8)
where .uue ˆ−=
A more direct measure is the L2 norm, which can be associated with the errors in any
quantity. Thus for the displacement u, the L2 norm of the error e is
( )2
1
2
Ω= ∫Ω
dL
eee T
(16.9)
and for derivatives,
2
1
))((
2
Ω= ∫Ω
dL σ
T
σ () eeeσ (16.10)
The ‘root mean square’ (RMS) error for whole domain Ω is given by
2
1
2
2
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜
⎝
⎛
=
Ω
e
Δu
L
(16.11)
Similarly for derivatives
2
1
2 2
L
⎛ ⎞
⎜=
⎜ ⎟
⎝ ⎠
Δσ
Ω
se
⎟ (16.12)
The error for the whole domain is given by summing the element contributions. Thus,
∑=
=
m
i
i
1
22
ee (16.13)
where i represent an element contribution and m is the total number of elements. For an optimal
mesh, it is considered that the contributions to the square of the norm is equal for all elements.
The relative percentage error η can be given by
%100×=
u
e
η (16.14)
199

198. 16.3 TYPES OF ERROR ESTIMATES
The error estimators for finite element analysis that exist today can be divided into two
main categories: a priori error estimates and a posteriori estimates. Finding out the error before
the solution is called a priori and after the solution is called a posteriori.
16.3.1 a priori error estimates
In a priori error estimation, the effect of the proposed improved solution is predicted with
out actually finding the solution. A priori error estimates provide only a qualitative description on
the rate of convergence of the finite element solution. Hence, it is difficult to directly make use of
this type of error estimates in a mesh refinement process of adaptive mesh generation, which
usually requires a quantitative description of the error distribution as input information. However,
this type of error estimates provides an excellent tool for predicting the convergence rate during
the adaptive refinement process and also for the estimation of exact energy norms.
16.3.1.1 h – convergence
For h version refinement, the polynomial degree of the interpolation function, p, is kept
constant. If the mesh is refined uniformly and the size of the element, h ,approaches zero, the
error estimate is given in the form
),min( λp
u Che ≤ (16.15)
For 2-D problems, h is approximately proportional to the inverse of the square root of the total
degree of freedom of the mesh. Hence,
min( , )
2
p
ue CN
λ−
≤ (16.16)
where N is the total degree of freedom of the mesh, λ is the strength of singularities and C is a
constant dependent on the problem but independent of p and λ . The mesh is called optimal if the
sequence of mesh is designed in such a way that the error is equally distributed in each element
and the influence of the singularities is eliminated and
2
p
p
ue Ch CN
−
≤ ≈ (16.17)
200

199. 16.3.1.2 p-CONVERGENCE
For p-version refinement the mesh size is fixed and p is increased uniformly. The error
estimate is given by
β−
≤ CNeu (16.18)
where β is a positive constant dependent on smoothness of the exact solution and quality of the
mesh. Hence the rate of convergence will depend very much on the design of the mesh. If a
properly designed mesh is used, an exponential rate of convergence can be obtained. If the mesh
is not well designed, the performance will be affected especially when singularities are present. In
the case of p-refinement on a uniform mesh in the presence of singularities we have λβ = and
the rate is double than that of the uniform h-refinement. The convergence rate of the p-refinement
is always better than that of the uniform h-refinement but care should be taken for the element
size near singularities as oscillation of derivatives of solution may occur.
16. 3.1.3 hp-convergence
This simply means that the mesh size, h, is refined simultaneously with the increase of
value of p. The error estimate is
)( ϑ
αN
u Cee −
≤ (16.19)
where α and ϑ are positive constants dependent on the smoothness of the solution. In practice,
there are substantial difficulties in the implementation of the hp-version algorithm as the optimal
mesh for h-refinement depends on p.
16.3.2 Posteriori error estimates
In posteriori error estimation, the error norms are based on already determined solution
and hence the effect of increasing polynomial order or decreasing element size is known and not
estimated as is done in a priori error estimation. In addition, because the error is found on a local
basis and then summed globally, a posteriori error analyses allow for element level refinement,
which is advantageous. They are relatively inexpensive and simple to calculate. Due to the above
reasons, a posteriori error estimator has become more popular and was shown to be effective and
convergent in many classes of application.
16.3.2.1 ZZ error estimate
The main advantages of using the Zienkiewicz-Zhu (ZZ) error estimator over the other
types of error estimators is the simplicity of its implementation and its cost effectiveness. This is
201

200. due to the fact that in practical finite element computations some smoothing procedure, which
may or may not be superconvergent, will always be employed at the post-processing stage of the
computing process to recover the derivatives of the finite element solutions in order to achieve
more acceptable approximations. Using such recovered derivatives, the ZZ error estimator can be
calculated at a fraction of the total cost of the computation. However, the quality and the
reliability of the error estimator is obviously dependent on the accuracy of the recovered solutions
and therefore on the smoothing procedures.
To obtain acceptable results for stress, resort is generally made to a nodal averaging or projection
process in which it is assumed that the recovered derivative is interpolated by the same
function as the displacement, i.e.
*
σ
= N*
σ *
σ (16.20)
where *
σ is improved nodal derivative and
ˆ( )d
Ω
0− Ω =∫
T
N *
σ σ (16.21)
where is elemental FEM derivative.ˆσ
It is intuitively obvious that is in fact a better approximation than*
σ σˆ and we shall use it to
estimate the error i.e.σe
*
ˆσ ≈ −e σ σ (16.22)
to evaluate various error norms.
16.3.2.2 Residual method
This method is very much useful in p-refinement schemes. In this method, we temporarily
introduce a single degree of freedom associated with a shape function of order p+1 into the
previous refinement degree of freedom system, where p is the highest order of the shape function
present on the element edge and/or face considered. Let us consider that the displacement field of
an element is given by,
{ } [ ]{ }ii dNu = (16.23)
202

201. Consider that, at the refinement step m, the element displacement fields of an element of order p,
is given by,
{ } [ ]{ }nn dNu = (16.24)
where [Nn] contains all existing nodal and hierarchical shape functions {dn} is the vector of
known element displacements. The corresponding finite element equation is written as,
[ ]{ } { }=nnK nd fn (16.25)
If all new hierarchical shape functions of order p+1, [Nh], are temporarily added into the element,
the above equations are upgraded to,
{ } [ ]{ } [ ]{ }hhnn dNdNu +=' (16.26)
and
(16.27)
⎭
⎬
⎫
⎩
⎨
⎧
=
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
h
n
h
n
hhnh
nhnn
f
f
d
d
KK
KK
respectively, where {dh} are the temporarily added element displacements, which correspond to
the shape function [Nh],[Khh] are the sub-stiffness matrices that correspond to the interaction
between {dh} and {dn}, and where {fh} is the vector of element loads on the added degree of
freedom. The approximate values of {dh} can be obtained from the second equation of equation
(16.27):
{ } [ ] { } [ ] { }( )1−
= −hh nhK K
T
h hd f nd (`16.28)
To obtain the approximate value of each component of {dh}, equation (16.28) is simply written
as,
203

202. ( )
{ } { }( )
( )
j
j
j
−
=
nh
hh
K
K
T
h n
h
f d
d (16.29)
where j is the index number of {dh} and (Khh) is the diagonal term of [Khh} at the particular j.
The difference between the two fields (before and after adding the higher order shape functions)
is defined as the estimated displacement error. The error in displacement fields can be expressed
as,
{ } { } { }uu'eu −= (16.30)
16.3.2.3 Superconvergent Patch Recovery (SPR) technique
Implementation of this recovery method is very simple and cost effective. This can be
widely used for linear, quadratic and cubic elements for both one and two-dimensional problems.
The gradients at nodes and element boundaries which are obtained by extrapolation of the
gradients at Gauss points from the finite element analysis do not possess the inter element
continuity and are of low accuracy. To obtain a better gradient field, the Superconvergent Patch
Recovery (SPR) technique proposed by Zienkiewiz and Zhu is being used here. Some typical 1-
D and 2-D patches are shown in Fig 16.1. We denote the gradient by q. In this recovery process,
it is assumed that the accurate nodal values
*
pq belong to a polynomial expansion of the same
complete order ‘p’ as that present in the basis function N and which is valid over an element
patch surrounding the particular assembly node considered. Such a ‘patch’ represents a union of
elements containing this vertex node. This polynomial expansion will be used for each
component of and one can get
*
pq
*
pq
Pa=*
pq (16.31)
where P contains the appropriate polynomial terms and a is a set of unknown parameters. For one
dimensional elements ‘P’one can write,
[ ]p2
xxx,1,P ,........,= (16.32)
204

203. Figure 16.1: Superconvergent points for some 1-D and 2-D patches
and
[ ]T
a 1,321 ,.........., += paaaa (16.33)
Thus for two dimensions (triangular) and linear expansion we have
[ ]yx,,1=P (16.34)
and for quadratic
[ ]22
,,,,,1 yxyxyx=P (16.35)
For a bilinear quadrilateral
[ ]xyyx ,,,1=P (16.36)
and similar forms for higher order expansions can be used.
The determination of the unknown parameters a of the expansion given in equation
(16.36) is best made by ensuring a least square fit of this to the set of superconvergent or a least
high accuracy sampling points existing in the patch considered if such points are available. To
obtain this one can minimize
( ) ( )( )
2
1
,
*
)( ∑=
−=
n
i
iiii yxyx ph qqaF
205

204. (16.37)( ) ( )(
2
1
,∑=
−=
n
i
iiii yxyx aPqh )
where are the global co –ordinates of sampling points,)( ii yx mkn = is the total number of
sampling points and k is the number of the sampling points on each element
mj of the element patch. The minimization condition of F (a) implies that a
satisfies
).......3,2,1( mmj =
(16.38)( ) ( ) ( ) ( iiii
n
i
iiii
n
i
yxyxyxyx ,,,
1
,
1
h
TT
qPaPP ∑∑ ==
= )
) )
This can be solved in matrix form as
(16.39)bAa 1−
=
where
and (16.40)( ) ( iiii
n
i
yxyx ,,
1
PPA T
∑=
= ( ) ( iiii
n
i
yxyx ,,
1
h
T
qPb ∑=
=
Once the parameters a are determined, the recovered nodal values are calculated by insertion of
appropriate co-ordinates into the expression for .*
hq
It will be observed that, element patches will overlap for internal misdside nodes and nodes in the
element interior. This means that such recovered nodal values are frequently evaluated from two
or more patches. In this case aggregate values of this patches is used for such nodes. A more
difficult situation arises at the domain boundary where a local patch, such as shown in Fig.16.2,
may involve only one or two elements. For one such element situation (corner node) the size of
the patch is insufficient for determination of the parameters a and the corner node values are
determined from an interior patch shown.
206

205. Figure 16.2: Boundary nodal recovery
16.3.2.4 Higher Order Approximation of Primary Variables (HOAPV)
From the above discussion, it is clear that a higher degree fitted polynomial is
expected to give more accurate results. In some cases, when the matrix P in equation (16.40)
becomes singular, SPR technique fails to give accurate results. To circumvent the difficulty, here
the HOAPV method is introduced [3]. It can be shown that the finite element solution obtained by
the linear interpolation functions can be improved by using HOAPV. In this, the recovered
solution for the primary variable itself is obtained by fitting a polynomial one order higher than
the finite element polynomial and their blended functions are used. For example, consider the
one-dimensional second order differential equation. If one has nodal solution data, then fitting a
second-degree polynomial will ensure to provide finite residue at each point of the domain.
A typical node inside the mesh is surrounded by two nodes on each side as shown in
Fig. 16.3a. A quadratic polynomial can be fitted on that node passing through three points. At
node j, the quadratic polynomial is of the form
2
xcxbaT jjjj
++= (16.41)
Similarly, the quadratic polynomial on the (j+1) node is
21111
xcxbaT jjjj ++++
++= (16.42)
Now, the temperature function in between the nodes j and (j+1) can be written as
1
21
+
+= jj
TNTNT (16.43)
where N1 and N2 are standard linear shape functions.
207

206. Figure 16.3: 1-D & 2-D elements illustrating the higher order blending function
The above scheme is based on curve-fitting using parabolic bending scheme. It can be
easily shown that the above approximation yields highly accurate results compared to standard
post-processing procedure. To derive the expression for error, take the coordinates of the nodes
j-1, j, j+1and j+2 as –1, 0, 1 and 2 respectively. Assume that the exact polynomial Texact is given
by the following equation
Texact = a + bx + cx2
+ dx3
(16.44)
Since the fitted polynomial given by equation (16.44) is exact at the nodes j-1, j, j+1, it is equal to
the exact value of primary variables at these three points. Hence, the following set of equations is
obtained:
dcbacba
aa
dcbacba
jjj
j
jjj
+++=++
=
−+−=+−
(16.45)
Solution of the above equation yields
ccdbbaa jjj
=+== ,, (16.46)
Therefore, the approximation function Tj
is given by
dxcxbxacxxdbaT j
+++=+++= 22
)( (16.47)
208

207. In the same way, Tj+1
given by Eqn.(16.42) is exact at the nodes j (x=0), j+1 (x=1), j+2 (x=2),
and so equal to the exact value of primary variables at these three points. Therefore, the
coefficients in Eqn.(16.42) are given by
dccanddbbaa jjj
32, 111
+=−== +++
(16.48)
Then, the approximating function Tj+1
is given by
dxxcxbxaT j
)23( 221
−+++=+
(16.49)
Now, in the element j – (j+1), N1 is (1-x) and N2 is x. Hence,
dxxxcxbxaT )33( 322
+−+++= (16.50)
Subtracting the equation for exact solution from the above equation, error in the primary variable
becomes
dxxxeT )32( 23
+−= (16.51)
The roots of the polynomial inside the brackets from the above equation are 0, 1 and 0.5,
indicating that blended function gives exact value not only at the nodes, but also at the middle
point. One can also find the L2 norm
( )[ ]
2
1
1
0
223
32
2 ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+−= ∫ dxdxxxe LT (16.52)
The first derivative will be exact at points
32
1
2
1
± and the second derivative at the middle of
the element. There can be various ways to find out error and the mesh refinement strategy. The
simplest way to find out the error is to calculate residue r by putting back the expression for T in
the differential equation. Zienkiewicz and Morgan [4] have proposed an error norm as
∫Ω
Ω−= erde E
2
(16.53)
where the error e = T – Texact. The exact value of T is not known. One way is to treat T of next
refinement level as exact. The refinement criterion may also be based on the absolute maximum
value of residue in various elements.
A procedure has been developed for two-dimensional problems using Laplace or
Poisson equation. A 2-D finite element mesh using four nodded elements is shown in Fig.
16.3(b). A typical node I inside the mesh is usually a member of four elements. There are a total
of nine nodes in these four elements, hence a quadratic at the ith
node may be constructed using
nine points. The quadratic is of the form given below:
222
8
2
7
2
6
2
5432 yxyxxyyxxyyxfi ααααααααα ++++++++= (16.54)
209

208. Approximation of the primary variable inside an element is carried out in the following manner:
44332211 fNfNfNfNf e
+++= (16.55)
In which N1, N2………are the linear shape functions and fi is quadratic function given by Eqn
(16.54) defined for the ith
node. Note that the ith
node is at the center surrounded by other eight
nodes, when a polynomial fi is constructed using nine nodes. In the case of heat transfer problems,
the primary variables (temperature) T can be written as
44332211 TNTNTNTNT e
+++= (16.56)
These primary variables themselves are the improved ones and hence gradients found using these
values are also improved ones.
There may be situations where all the four nodes of the element may not be at the
center of the patch of the elements, This situation is tackled in the following manner. If only one
node is available over which the polynomial can be defined, then at all the other nodes, the same
function is taken. If the functions can be defined in two adjacent nodes, then the same functions
are used on the corresponding opposite nodes. If the polynomial is defined on the diagonal nodes,
then in the other two diagonal nodes, average value of the function is taken. If the polynomial can
be defined on three nodes, then in the fourth node, average of two adjacent nodes is taken for the
coefficients of the polynomial.
16.4. ERROR ESTIMATES BY RECOVERY
One of the most important applications of the recovery method is its use in the
computation of the a posteriori error estimators. With the residual solutions available, evaluate
errors simply by replacing the exact values of quantities such as u, σ , etc., which are in general
known, in equations (16.7a, 16.7b), by the recovered values which are much more accurate than
the direct finite element solution. Now, error estimators in various norms such as
u-uee *
ˆˆ =≈ (16.57)
222
ˆˆ
LLL
u-uee *
=≈ (16.58)
σσσ u-uee *
ˆˆ =≈ (16.59)
Error estimators formulated by replacing the exact solution with the recovered solution are
sometimes called recovery based error estimators.
210

209. The accuracy or the quality of the error estimator is measured by the effectivity index
θ, which is defined as
e
eˆ
=θ (16.60)
Assuming that the true error convergences as
h
hC h− =u u p
(16.61)
and the error of the post-processing solution
*
p
C h α+
− =u u*
(16.62)
for some super convergent solution with 1≥α . It is shown that for all estimators based on
recovery can establish the following bounds for the effectivity index:
e
e
e
e **
+≤≤− 11 θ (16.63)
Where e is the actual error and e*
is the error of the recovered solution, e.g.
**
u-ue = (16.64)
Effectivity index approaches to unity as h approaches 0.
1≥α indicates whether the recovered solution has a higher rate of convergence than FE solution.
Two important conclusions follow:
1. any recovery process which results in reduced error will give a reasonable error estimator
and more importantly,
2. if the recovered solution converges at a higher rate, then the finite element solution shall
always have asymptotically exact estimation.
16.5CONCLUSIONS
In this chapter, a brief introduction of error analysis has been provided. The error analysis is
very important for adaptive refinement. It also provides an idea about the quality of the solution.
FURTHER READINGS
1. Zienkiewicz, O.C. and Zhu, J.G., 1987, A simple error estimator and adaptive procedure
for practical engineering analysis, Int. J. Num. Meth. Eng., 24, 337-357.
211

210. 2. Zienkiewicz, O.C. and Zhu, J.Z., 1992, The super convergent patch recovery and a
posteriori error estimation in the finite element method, Part I & Part II, Int. J. Num.
Meth. Eng., 33,1331-1364.
3. Sk.Karimulla, S.K Dwivedy and U.S. Dixit, “An efficient post-processing strategy for
finite element analysis of heat transfer problems”, CD Proceedings Recent Advances in
Heat and Mass Transfer, Jan. 6-8 (2002), IIT Guwahati (India)
4. Zienkiewicz, O.C. and Taylor, R.L., 1989, The Finite Element Method (Vol.1) 4th
Edition, McGraw-Hill, New York.
EXERCISE 16
Q.1: Prove that in solving the problem of rod subjected to axial load using 3 noded Lagrangian
element, longitudinal stress is expected to be accurate corresponding to the points of 2 point
Gauss-quadrature formula.
Q.2: A square plate is subjected to the temperature of 500K at three sides and 0K at the fourth
side. Solve this problem by FEM by discretizing the domain into 16 equal elements. Find out the
recovered solution using Superconvergent Patch Recovery technique. Using the recovered
solution, find out the error of each element and overall error of the solution. You are already
familiar with the exact solution. Find out the error using the exact solution. Compare it with the
estimated error.
Q.3: Solve the problem of a plate with hole subjected to tensile load in one direction and find out
the error with coarse and fine mesh.
Q.4: The governing equation for certain heat conduction problem is
2
2
d
0
d
T
k q
x
+ =
where T is a temperature as a function of x, k is thermal conductivity and q is the rate of heat
generation per unit volume. For formulation purpose, an element of length h is transformed to
natural coordinates ξ having the coordinate of the first node as –1 and that of second as +1. The
interpolation function for temperature is given as
2
1 1 2 2 (1 )T N T N T aξ= + + −
where T1 and T2 are temperatures at the 2 nodes and a is some constant. Find out the elemental
stiffness matrix and elemental load vector for this problem. Also, carry out the error analysis.
212

211. Chapter 17
MISCELLANEOUS
(Lecture 39-40)
17.1 INTRODUCTION
In the last sixteen chapters, we have covered various topics of finite element method. This
material is sufficient for one semester introductory course on FEM. One can refer other textbooks
and papers for further knowledge. One can also extend the knowledge gained in the previous
chapter to many unexplored areas. Finite element method is still an important research area and
newer developments are taking place. In this chapter, we shall touch upon miscellaneous topics in
the FEM and conclude the first course on Finite Element Methods in Engineering. In this chapter,
there are some topics which have not been described at all in the previous chapters, whilst some
other topics are revisited.
17.2 DIFFERENCE BETWEEN FEM AND FDM
The basic concept of Finite Element Method (FEM) is as follows:
• The solution is approximated in the form of piecewise continuous functions.
These functions may be algebraic, trigonometric or any other type.
• Coefficients of these functions are obtained in a manner so that the error is
minimized.
Different formulation methods differ in terms of the form of functions, definition of error and
procedure for minimizing the error.
In the Finite Difference Method (FDM), derivatives are replaced by discrete analog of
derivatives. For example,
h
xuhxu
x
u )()(
d
d −+
≈ (17.1)
where h is a small finite step length. Thus, the system of differential equations gets approximated
to system of linear simultaneous algebraic equations, like in FEM. However, the basic concept
here is different from FEM. Here, we do not approximate the variable by any piecewise
continuous function.
In certain cases, the finite element and finite difference may provide the same solution.
For example, in axial rod problem, if the axial displacement in a element is approximated as,
211 u
h
x
u
h
x
bxau +⎟
⎠
⎞
⎜
⎝
⎛
−=+= (17.2)
213

212. where u1 and u2 are axial displacements at the 2 end nodes. Then,
h
uu
x
u 12
d
d −
= (17.3)
which is the same as finite difference approximation. On the hand, if axial displacement is
approximated by some function like
xbau sin+= (17.4)
Then derivative approximation will be different from the derivative approximation in FDM.
It is not convenient to apply FDM method to a problem involving a boundary of very
complex geometry and involving non-homogeneous material. If the proper choice of piecewise
continuous polynomials is chosen, the finite element procedure may take much less time than
FDM. In certain problems, FDM gives better results.
17.3 FINITE ELEMENT SOLUTIONS VERSUS EXACT SOLUTION
In general, the finite element procedure tends to make the structure stiffer than the actual
structure. This is because; we are imposing constraints on the structure by way of forcing the
deflection field to follow certain pattern. Hence, deflections are expected to be lesser than the
actual deflections. Similarly, natural frequencies are expected to be more than the actual
frequencies.
This can be understood mathematical. Consider the finite element formulation, in which
[K] is the stiffness matrix, {u} is the nodal deflection vector and {P} is the load vector. The
system of equation to be solved is given by
[K]{u}={P} (17.5)
Total potential energy is given by
Π=
2
1
{u}T
[K]{u} – {u}T
{P} (17.6)
Substituting equation (17.5) in equation (17.6), the total potential energy expression becomes,
Π=
2
1
{u}T
{P} – {u}T
{P} = -
2
1
{u}T
{P} (17.7)
The potential energy given by this expression will be more or equal to the actual potential energy,
because in the actual system potential energy will be minimized. Hence,
{u}T
{P}<{uactual}T
{P} (17.8)
214

213. 17.4 ACCURACY OF DERIVATIVES OF THE PRIMARY VARIABLES
Deflections are expected to be most accurate at the nodes. Stresses are expected to be
most accurate at certain points called Barlow points, which often coincide with the Gauss-
Quadrature points. In a two nodded one dimensional elements stresses are expected to be accurate
at the center. Similarly, in a 3-noded 1-dimensional element with end point coordinates of –1 and
1, stresses are expected to be accurate at coordinates ±1/√3. Similar thing is applicable in higher
dimension elements. In 4-noded rectangular element, the stresses are expected to be most
accurate at the center. In 8 and 9 noded rectangular elements, the 4 Barlow points are given by
(±1/√3, ±1/√3). However, this is not a very strict rule.
17.5 ESSENTIAL AND NATURAL BOUNDARY CONDITIONS
In structural mechanics, the essential boundary conditions correspond to prescribed
displacement and rotations. If in the potential energy functional, the highest derivative of a state
variable (with respect to a space coordinate) is m, the order of derivative in the essential boundary
conditions is at most (m-1). The second class of boundary conditions, namely the natural
boundary conditions are also called force boundary conditions, because in structural mechanics,
the natural boundary conditions correspond to prescribed boundary forces and moments. The
highest derivatives in these boundary conditions are of order m to 2 m.
17.6 MESH REFINEMENT
The solution accuracy is very much dependent on the type of mesh. Modifying the mesh with
a view to obtain a more accurate solution is called mesh refinement. There are 4 methods of
carrying mesh refinement:
(1) h-refinement: In this refinement method, the size of the element is reduced.
Consequently, the number of elements in a domain increases. This refinement method
is most common. However, the mesh has to be generated again.
(2) p-refinement: In this refinement, element sizes remain the same, but the degree of
polynomial approximation in an element increases. The advantages of this method
include faster convergence and avoidance of new mesh generation at each refinement
stage.
(3) hp-refinement: This is the combination of both the above methods. It is a very effective
method for most of the problems.
215

214. (4) r-method: In this method, nodes are relocated and a new mesh is constructed. Size of
the elements and degrees of approximation remain the same. This is not a very
common method.
Refinement should be carried out where the primary variables is expected to be a high
degree polynomial function. For example, if a plate is having a hole, then near the hole, the mesh
should be refined. If a cantilever beam problem is solved using constant strain triangular (CST)
elements, then there should be enough elements in the vertical directions also, because the strain
is linear in vertical direction and CST gives only constant strain. So many elements are needed in
the vertical direction in order to properly represent the strain.
17.7 EFFECT OF THE GEOMETRY OF A PARTICULAR ELEMENT
If the shape of the element is not proper, computational difficulties are introduced. For
example, while using CST elements, three angles of the triangle should be close to 600
. Similarly,
in a rectangular element the ratio of higher side to lower side should not be more than 10. A
square element is better. As the aspect ratio (ratio of higher side to lower side) increases,
contribution due to parasitic shear increases and locking phenomenon gets activated. Parasitic
shear is the presence of spurious shear when a rectangular element is subjected to pure bending.
In a quadrilateral element, angle between two sides should not be much far away from 900
. In
eight nodded, rectangular element, the middle node should be in between 1/3 to 2/3 of length of
the side from the end nods.
17.8 SOLVING THE PROBLEMS OF FRACTURE MECHANICS USING FEM
For solving the fracture mechanics problem, Quarter-Point Elements (QPE) are employed.
Fig. 1 is a three-node quarter-point bar element. In this the middle node is exactly at a distance of
(L/4) from one end-node, where L is the length of the bar. In this element, the stresses vary
as .Thus, there is a singularity at x=0, where the crack tip may be located. The six-node
plane triangle can display the r-1/2
in the strain field if its side nodes are moved at quarter points
near the crack tip as shown in Fig. 17.2. In the similar way rectangular QPE’s can be obtained.
1/2
x−
216

215. ξ =-1 ξ =0 ξ =+1
1 2 3
x,u
L/4 3L/4
Figure 17.1: Three-node quarter-point bar element
y,v
θ
r
C2 B2 x,u
C1 B1
l/4
l
Figure 17.2: Mesh of quarter plate elements around the crack tip
17.9 INFINITE ELEMENTS
Many stress analysis problems deal with an unbounded medium for example, problem of
a load supported on the ground. In this problem, a finite element model must be terminated
somewhere short of infinity. The analysis then becomes expensive and many elements have to be
taken. In order to solve such types of problems economically, a special type of element called
infinite element can be used. The shape functions of this element will cause the stresses to reach
to zero at infinity.
Two types of shape functions can be used. The first type is the decay shape functions that
approach 0 as coordinate approaches infinity. The second type is the growth shape function which
grow to infinity at one particular node. The first type of shape function are applied to stresses,
whilst the normal shape functions are used for boundaries. The second type of shape functions are
used to approximate geometry, whilst the stresses are approximated by the normal shape
functions.
217

216. 17.10 ILL-CONDITIONED SYSTEM
Consider the set of equations
⎭
⎬
⎫
⎩
⎨
⎧
−
=
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
−
−
00.2
00.4
02.100.1
00.100.1
y
x
for which (17.9)
⎭
⎬
⎫
⎩
⎨
⎧
=
⎭
⎬
⎫
⎩
⎨
⎧
100
104
y
x
and a very similar set of equations
⎭
⎬
⎫
⎩
⎨
⎧
−
=
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
−
−
00.2
00.4
01.100.1
00.100.1
y
x
for which (17.10)
⎭
⎬
⎫
⎩
⎨
⎧
=
⎭
⎬
⎫
⎩
⎨
⎧
200
204
y
x
A 1% change in one coefficient has changed the results by a factor of two. These equation sets
are both ill- conditioned, which means that their solutions are sensitive to small changes in the
coefficient matrix or the vector of constants.
Ill-conditioned system is produced because of the physics of the problem. For example,
consider the two d.o.f. structure with linear springs of stiffness k1 and k2 (Fig. 17.3). FEM
formulation provides,
⎭
⎬
⎫
⎩
⎨
⎧
=⎥
⎦
⎤
⎢
⎣
⎡
+−
−
0211
11 P
kkk
kk
(17.11)
The rows of stiffness matrix are almost linearly dependent if k1>>k2 and hence the stiffness
matrix becomes ill-conditioned. This is not case if k2>>k1.
u2
,x,uP 1 k1 2 k2
u1
Figure 17.3: An example that can yield ill-conditioned system if k1>>k2
In structural analysis, the major cause of ill-conditioning is a large difference in stiffnesses, with
the stiffer region being supported by the more flexible region. Physically, the stiffer region has
one or more displacement states that are almost rigid-body motions within a more flexible
218

217. supporting structure. The limiting case is a structure without any supports: it has only rigid-body
motion in static analysis, and its stiffness matrix is singular.
Following measures should be adopted for avoiding the ill-conditioning:
(1) It is recommended that double-precision arithmetic be used for all phases of a finite
element analysis.
(2) Attempt should be to develop a model without large discrepancies in stiffness. A
comparatively stiff region may be modeled as perfectly rigid by use of a constraint
transformation. On the other hand, perhaps a stiff region can be made more flexible.
(3) One possibility in modeling is to use relative motions, rather than absolute motions, for
troublesome d. o. f. For example, in Fig. 3, one could introduce the d. o. f. ur = u1-u2.
Instead of equation (11) we now have
⎭
⎬
⎫
⎩
⎨
⎧
=
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
P
P
u
u
k
k r
22
1
0
0
(17.12)
This system is well-conditioned for all values of k1 and k2.
17.11 PATCH TEST
In some cases considerable difficulty is experienced in finding displacement functions for an
element which will automatically be continuous along the whole interface between adjacent
elements. The discontinuity of displacement will cause infinite strains at the interfaces. However,
if, in the limit, as the size of the subdivision decreases continuity is restored, then formulation
will tend to the correct answer. This condition is always reached if
(1) a constant strain condition automatically ensures displacement continuity.
(2) the constant strain criterion is satisfied i.e., if nodal displacements are compatible with a
constant strain condition such constant strain will in fact be obtainable.
To test whether a non-conforming element will provide satisfactory result, a patch test is
conducted. In this test, nodal displacements corresponding to any state of constant strain are
imposed on an arbitrary patch of elements. If nodal equilibrium is simultaneously achieved
without the imposition of external, nodal, forces and if a state of constant stress is obtained, then
clearly no external work has been lost through inter-element discontinuity. Elements that pass
such a patch test will converge.
219

218. 17.12 CONCLUSIONS
FEM has been applied to a number of areas. If one can find the governing differential
equations of the problem, FEM can be applied. It has been applied to stress-analysis of elastic as
well as plastic deformation problems. Specific applications include metal forming, metal cutting
and non-traditional machining.
EXERCISE 17
Q.1: Mathematically show that a Quarter-point element can be used for solving the problems
when the derivative of the primary variable becomes infinite at a singular point.
Q.2: In solving an axial rod problem using FEM with 3 noded element, the middle node need not
be exactly at the middle. Find out the zone in which the middle node can be placed.
Q.3: Take a 3-noded one dimensional element 1-2-3, in which the third node is at infinity. Show
that the geometry can be approximated as
1 1 2 2x N x N x= +
where
1 2
2 1
;
1 1
N N
ξ ξ
ξ ξ
+
= − =
− −
Approximate the primary variable u by 3-noded Lagrangian shape function and obtain the
expression for du/dx.
Q.4: Show that in a CST element, if one angle is very small, the resulting matrix will be ill-
conditioned.
220

219. BIBLIOGRAPHY
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. T.J.T. Hughes, The finite element method, Prentice-Hall, Englewood Cliffs, NJ,
1986.
5. O.C. Zeinkiewicz and R.L. Taylor, The finite element method, 3rd
ed., McGraw-Hill,
1989.
6. T.R. Chandrupatla, A,D. Belegundu, Introduction to Finite Element in Engineering,
2nd Edition, Prentice-Hall International, Inc, Englewood Cliffs, NJ, 1997.
221