FINITE ELEMENT METHODS
IN ENGINEERING
(Lecture Notes)
UDAY S. DIXIT
E-mail: [email protected]
Department of Mechanical Engineering
Indian Institute of Technology Guwahati
Guwahati – 781039, Assam, India
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 underpost-graduate 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 [email protected]
Uday S. Dixit
May 2007Contents
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
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 C
0Continuity 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
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
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
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
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
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 one-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.
• 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
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
u
u
j i 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
u
u
E
j i t−
=
σ
(1.2) where E is the Young’s modulus of elasticity. The force Fj applied at the j th node is stress timesthe cross-sectional area, A. Hence,
j i j
F
h
u
u
AE
−
=
(1.3) Thus, a relationship between force Fj and nodal deflections is obtained. Similar relation betweenforce Fi and nodal deflections can be obtained in the following. The compressive strain, in the
element is
h
u
u
i j c−
=
ε
(1.4) and the corresponding compressive stress is
h
u
u
E
i j c−
=
σ
(1.5)Hence, i j
F
ih
u
u
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 iF
F
u
u
h
AE
1
1
1
1
(1.7) In the compact form, this can be written as
[ ]
k
e{ } { }
u
e=
F
e (1.8) where[ ]
⎥
⎦
⎤
⎢
⎣
⎡
−
−
=
1
1
1
1
h
AE
k
e (1.9){ }
(1.10)⎭
⎬
⎫
⎩
⎨
⎧
=
j i eu
u
u
and{ }
(1.11) ⎭ ⎬ ⎫ ⎩ ⎨ ⎧− = j i e F F FCompare 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,
{ }
ue is elemental displacement or primary variable vector and {Fe} is theelemental 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.
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 1 1 1 1 F F u u h AE (1.13) ⎭ ⎬ ⎫ ⎩ ⎨ ⎧− = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − 3 2 3 2 1 1 1 1 F F u u h AE (1.14) ⎭ ⎬ ⎫ ⎩ ⎨ ⎧− = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − 4 3 4 3 1 1 1 1 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 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 2 1 4 3 2 1 F F u u u u h AE (1.16) ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ − = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 3 2 4 3 2 1 F F u u u u h AE (1.17) ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ − = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − 4 3 4 3 2 1 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 F F u u u u h AE (1.18)
1 1 2 3 4 4 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 u AE u h F u − + + − + + + + + + ⎧ ⎫ ⎧ ⎫ ⎡ ⎤ ⎪ ⎪ ⎪ ⎪ ⎢− + + + + − + + + ⎥ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎨ ⎬ ⎨= ⎬ ⎢ + + − + + + + − ⎥ ⎪ ⎪ ⎪ ⎪ ⎢ + + + + + − + + ⎥ ⎪ ⎪ ⎪ ⎪ ⎣ ⎦ ⎩ ⎭ ⎩ ⎭ (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,
⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − − − − P u u u u h AE 0 0 0 1 1 0 0 1 2 1 0 0 1 2 1 0 0 0 1 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 u u h u u = − = − = 3 / 2 3 2 3 ) 2 (ε
(1.22)and the stress is given by
A P E = = (2) ) 2 (
ε
σ
(1.23)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)⎦
⎤
⎢
⎣
⎡
3
2
2
1
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 = k11vi+ k12θi+ k13vj+ k14θj (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 k k k k k k k k k k k k k k k k θ θ 44 43 42 41 34 33 32 31 24 23 22 21 14 13 12 11
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 F v k k k k θ 22 21 12 11
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 i
v
iEI
h
M
EI
h
F
=
−
2
3
2 3 (1.28)i i i
EI
h
M
EI
h
F
+
=
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 2 3 2 2 3 (1.30)
Inverting the above equations, we obtain,
⎭
⎬
⎫
⎩
⎨
⎧
=
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
i i i iM
F
v
h
h
h
h
EI
θ
2 36
4
6
12
(1.31)Comparing this with (1.27):
h EI k h EI k k h EI k11 =123 12 = 21 =6 2 22 =4 (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
(1.39) ⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ = ⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ j j j j M F v k k k k
θ
44 43 34 33Element 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 j j v EI h M EI h F = + 2 3 2 3 (1.40) j j j EI h M EI h F
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 2 3 2 2 3 (1.42)
Inverting above equation, we obtain, ⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ = ⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − j j j j M F v h h h h EI
θ
2 3 6 4 6 12 (1.43)Comparing this with (1.39):
h EI k h EI k k h EI k33 =123 34 = 43 =−6 2 44 = 4 (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,
⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − − − − − 2 2 2 2 3 4 6 2 6 6 12 6 12 2 6 4 6 6 12 6 12 h h h h h h h h h h h h h EI (1.46)
The resulting stiffness matrix is exact, not approximate, for the given problem.
In order to perform the assembly, elemental equations can be written in global form. First elemental equation in global coordinate system is,
( ) ( ) ⎪ ⎪ ⎪ ⎪ ⎭ ⎪⎪ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪⎪ ⎪ ⎪ ⎨ ⎧ = ⎪ ⎪ ⎪ ⎪ ⎭ ⎪⎪ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪⎪ ⎪ ⎪ ⎨ ⎧ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − − − − − 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 6 2 6 0 0 6 12 6 12 0 0 2 6 4 6 0 0 6 12 6 12 1 2 1 2 1 1 3 3 2 2 1 1 2 2 2 2 3 M F M F v v v h h h h h h h h h h h h 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 2 2 2 2 30
0
4
6
2
6
0
0
6
12
6
12
0
0
2
6
4
6
0
0
6
12
6
12
0
0
0
0
0
0
0
0
0
0
0
0
0
0
M
F
M
F
v
v
v
h
h
h
h
h
h
h
h
h
h
h
h
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 2 2 2 2 2 2 2 2 3 0 4 6 2 6 0 0 6 12 6 12 0 0 2 6 4 4 6 6 2 6 6 12 6 6 12 12 6 12 0 0 2 6 4 6 0 0 6 12 6 12 M F P M F v v v h h h h h h h h h h h h h h h h h h h h h h h h 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:
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,
⎭ ⎬ ⎫ ⎩ ⎨ ⎧ = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 0 8 0 0 24 2 2 2 3 P v h h EI
θ
(1.52)Solving this, we get
(
)
EI
PL
EI
L
P
EI
Ph
v
192
24
2
/
24
3 3 3 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-one-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
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.
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.
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.
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
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.
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
d
I
=
∫
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:( )
b(
, ,
)
d
awhere 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
α
y
+
β
z
=
α
l y
+
β
l z
(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,
)
(
2B
α
y
+
β
y z
=
α
B y z
+
β
B y z
,
)
(2.4)(
,
1 2)
(
,
1)
(
)
B y z
α
+
β
z
=
α
B y z
+
β
B y z
,
2 (2.5) The functional is symmetric if( ) ( )
,
B y z
=
B z y
,
(2.6) An example of a linear functional is( )
0Ld
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 d I 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 andthe 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,
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 ' ad
I
=
∫
F x y
+
εη
y
+
εη
′
x
(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
y
0' . Note that while expanding F by Taylorseries, 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'
2 2 2 2 2 2 d 0 2 ... d d b a I F F F F F x yη yη ε y η ε y yηη ε y ε ⎧ ⎛∂ ∂ ⎞ ⎛ ∂ ∂ ∂ ⎞ ⎫ ⎪ ′ ′ ⎪ = ⎨ +⎜∂ +∂ ′ ⎟+⎜ ∂ + ∂ ∂′ + ∂ ⎟+ ⎬ ⎪ ⎝ ⎠ ⎝ ⎠ ⎪ ⎩ ⎭
∫
′∫
(2.13) (2.13)Applying the condition for extremum, we get Applying the condition for extremum, we get
0 0 b a I F F d d d ε= yη y η x ⎛∂ ∂ ⎞ 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, δ
2I, 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
η
yη
⎧∂ − ⎛∂ ⎞⎫ +∂ ⎨∂ ⎜∂ ′⎟⎬ ∂ ′ ⎝ ⎠ ⎩ ⎭∫
(2.15) Thus,d
d
d
b a b aF
F
F
F
I
x
y
x
y
y
y
δ
=
⎧
⎨
∂
−
⎜
⎛
∂
⎟
⎞
⎫
⎬
η
+
∂
η
−
∂
η
0
′
′
′
∂
⎝
∂
⎠
∂
∂
⎩
⎭
∫
=
(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 xF
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
η
yη
∂ −∂ = ′ ′ ∂ ∂ (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)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
d
I
=
∫
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) As2
F
y
y
∂
=
′
′
∂
(2.22)The boundary conditions are:
Either y'= 0 or y is prescribed at x = 0 and 1 (2.23) 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,
'n, )
F
=
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 n I F y x x x x x x ∂ ∂ = ∫ ∫ ∂ ∂ (2.25)
Using the same kind of analysis as before, we find that the extremising function y = y(x1, ………..,xn) must satisfy 1 i
0
n i i xF
F
y
=x
y
⎛
⎞
∂
∂
∂
−
∑
⎜
⎜
⎟
⎟
=
∂
∂
⎝
∂
⎠
(2.26) where i xy
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 extremizingd , , d d b a y I F x y x ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠
∫
x, however, we will now follow a short but not so rigorous approach. We willnow 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.,
( )
δ
y ' = y'
δ
.Proof: We can write,
( )
d( )
d ( ) d d y ' = y d d x x xη
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 xη
η
δ
=δ
⎛⎜ ⎟⎞= +εη
− = +ε
− =ε
⎝ ⎠ (2.28)Both expressions are equal, hence proved.
Property 2: Integration and variation commute
i.e.
