Notes Dixit

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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 May 2007

Transcript of Notes Dixit

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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 May 2007

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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 [email protected] or [email protected]. Uday S. Dixit May 2007

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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

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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

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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

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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

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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

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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

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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.

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• 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

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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 ijt

−=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

huu

E ijt

−=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,

jij F

huu

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 jic

−=ε (1.4)

and the corresponding compressive stress is

h

uuE ji

c

−=σ (1.5)

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Hence,

iji F

huu

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

FF

uu

hAE

1111

(1.7)

In the compact form, this can be written as

[ ] eee Fuk = (1.8)

where

[ ] ⎥⎦

⎤⎢⎣

⎡−

−=

1111

hAEk e (1.9)

(1.10) ⎭⎬⎫

⎩⎨⎧

=j

ie

uu

u

and

(1.11) ⎭⎬⎫

⎩⎨⎧−

=j

ie

FF

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, eu 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.

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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

1111

FF

uu

hAE (1.13)

⎭⎬⎫

⎩⎨⎧−

=⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡−

3

2

3

2

1111

FF

uu

hAE (1.14)

⎭⎬⎫

⎩⎨⎧−

=⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡−

4

3

4

3

1111

FF

uu

hAE (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,

⎪⎪⎭

⎪⎪⎬

⎪⎪⎩

⎪⎪⎨

⎧−

=

⎪⎪⎭

⎪⎪⎬

⎪⎪⎩

⎪⎪⎨

⎥⎥⎥⎥

⎢⎢⎢⎢

⎡−

00

0000000000110011

2

1

4

3

2

1

FF

uuuu

hAE (1.16)

⎪⎪⎭

⎪⎪⎬

⎪⎪⎩

⎪⎪⎨

⎧−

=

⎪⎪⎭

⎪⎪⎬

⎪⎪⎩

⎪⎪⎨

⎥⎥⎥⎥

⎢⎢⎢⎢

−−

0

0

0000011001100000

3

2

4

3

2

1

FF

uuuu

hAE (1.17)

⎪⎪⎭

⎪⎪⎬

⎪⎪⎩

⎪⎪⎨

−=

⎪⎪⎭

⎪⎪⎬

⎪⎪⎩

⎪⎪⎨

⎥⎥⎥⎥

⎢⎢⎢⎢

−−

4

3

4

3

2

1

00

11001100

00000000

FF

uuuu

hAE (1.18)

Additions of these, yields

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1 1

2

3

44

1 0 0 1 0 0 0 0 0 0 0 01 0 0 1 1 0 0 1 0 0 0 0 0

0 0 0 0 1 0 0 1 1 0 0 1 00 0 0 0 0 0 0 0 1 0 0 1

u FuAEuh

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,

⎪⎪⎭

⎪⎪⎬

⎪⎪⎩

⎪⎪⎨

=

⎪⎪⎭

⎪⎪⎬

⎪⎪⎩

⎪⎪⎨

⎥⎥⎥⎥

⎢⎢⎢⎢

−−−

−−

Puuuu

hAE

000

110012100121

0001

4

3

2

1 (1.20)

There are various methods to solve this linear system of equations. Solution yields,

u1=0, u2=AEPL

3 , u3=

AEPL

32

, u4= AEPL

(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

( ) AEP

Luu

huu

=−

=−

=3/

2323)2(ε (1.22)

and the stress is given by

APE == )2()2( εσ (1.23)

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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) ⎥⎦

⎤⎢⎣

⎡3221

Given connectivity matrix and coordinates of the node, the mesh can be easily constructed.

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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

FMF

v

v

kkkkkkkkkkkkkkkk

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

MFv

kkkk

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:

iii vEIhM

EIhF

=−23

23

(1.28)

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iii

EIhM

EIhF 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 vMF

EIh

EIh

EIh

EIh

θ2

232

23

(1.30)

Inverting the above equations, we obtain,

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

i

i

i

i

MFv

hhh

hEI

θ23 46612

(1.31)

Comparing this with (1.27):

hEIk

hEIkk

hEIk 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

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(1.39) ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧⎥⎦

⎤⎢⎣

j

j

j

j

M

Fv

kkkk

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:

jjj vEIhM

EIhF

=+23

23

(1.40)

jjj

EIhM

EIhF

θ=+2

2

(1.41)

In the matrix form, the equations can be written as,

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎥⎥⎥

⎢⎢⎢⎢

j

j

j

j v

M

F

EIh

EIh

EIh

EIh

θ2

232

23

(1.42)

Inverting above equation, we obtain,

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧⎥⎦

⎤⎢⎣

j

j

j

j

M

Fv

hhh

hEI

θ23 46612

(1.43)

Comparing this with (1.39):

hEIk

hEIkk

hEIk 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

6126122646

612612

hhhh

hhhhhh

hh

hEI (1.46)

The resulting stiffness matrix is exact, not approximate, for the given problem.

1.3.3 Assembly procedure

10

Page 19: Notes Dixit

In order to perform the assembly, elemental equations can be written in global form.

First elemental equation in global coordinate system is,

( )

( )

⎪⎪⎪⎪

⎪⎪⎪⎪

⎪⎪⎪⎪

⎪⎪⎪⎪

=

⎪⎪⎪⎪

⎪⎪⎪⎪

⎪⎪⎪⎪

⎪⎪⎪⎪

⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢

−−−−

00

0000000000000046260061261200264600612612

12

12

1

1

3

3

2

2

1

1

22

22

3 M

F

MF

v

v

v

hhhhhh

hhhhhh

hEI

θ

θ

θ

(1.47)

Here, superscript (1) on forces and moments indicate contribution from element 1. Second

elemental equation in global coordinates is

( )

( )

( )

( ) ⎪⎪⎪⎪

⎪⎪⎪⎪

⎪⎪⎪⎪

⎪⎪⎪⎪

=

⎪⎪⎪⎪

⎪⎪⎪⎪

⎪⎪⎪⎪

⎪⎪⎪⎪

⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢

−−−−

13

13

22

22

3

3

2

2

1

1

22

223

00

4626006126120026460061261200

000000000000

M

F

M

F

v

v

v

hhhhhhhhhhhh

hEI

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

46260061261200

2644662661266121261200264600612612

M

F

PMF

v

v

v

hhhhhh

hhhhhhhhhhhh

hhhhhh

hEI

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

Page 20: Notes Dixit

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

223

Pvhh

EIθ

(1.52)

Solving this, we get

( )EI

PLEI

LPEI

Phv19224

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

Page 21: Notes Dixit

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

Page 22: Notes Dixit

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:

ddTq kAx

= −

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

Page 23: Notes Dixit

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

Page 24: Notes Dixit

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 xA x Al

⎛ ⎞= −⎜ ⎟⎝ ⎠

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

Page 25: Notes Dixit

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 kAxφ∂

= −∂

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

Page 26: Notes Dixit

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 0dI 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:

( ) ( ), , db

aI y F x y y x′= ∫ (2.2)

19

Page 27: Notes Dixit

where F depends on x, y ≡ y(x) and y′ ≡ ddyx

. 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

dL

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 xx 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, , dd

ba

yI F x yx

⎛ ⎞= ⎜ ⎟⎝ ⎠

∫ 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

Page 28: Notes Dixit

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 0d

Iε= (2.9)

However, if y0 itself is extremum, then above condition should hold good at ε = 0, i.e.,

0

d 0d

I

εε =

= (2.10)

Replacing y by y0(x) + εη(x) in functional I,

( )0 0, , b '

adI 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 22 2 2

2 2, , ( , , ) ...2! 2!

F F F F FF x y y F x y yy y y y y y'

εη εηεη 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

Page 29: Notes Dixit

2 2 22

2 2

d 0 2 ... dd

b

a

I F F F F F xy y y y y yη η ε η ε ηη ε

ε⎧ ⎫⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂ ∂⎪ ⎪′ ′= + + + + + +⎨ ⎬⎜ ⎟⎜ ⎟′ ′∂ ∂ ∂ ∂ ∂ ∂⎪ ⎪⎝ ⎠ ⎝ ⎠⎩ ⎭

∫ ′

∫ (2.13) (2.13)

Applying the condition for extremum, we get Applying the condition for extremum, we get

00

b

a

I F Fd dd

xy yεη η

=

⎛ ⎞∂ ∂ε

′= = +⎜ ⎟′∂ ∂⎝ ⎠∫ (2.14)

The 0

dd

I

εε =

is also called the first variation of I, δI. Thus, the condition for extremizing a

functional is δI=0. is Similarly, 2

20

dd

I

εε =

is called the second variation of I, δ2I, 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 dd

bb

aa

F F Fxy x y y

η η⎧ ⎫⎛ ⎞∂ ∂ ∂

− +⎨ ⎬⎜ ⎟′ ′∂ ∂ ∂⎝ ⎠⎩ ⎭∫ (2.15)

Thus,

d dd

b

ab a

F F F FI xy 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 0d

x

x

F F xy x y

η⎛ ⎞⎛ ⎞∂ ∂

−⎜ ⎟⎜ ⎟′∂ ∂⎝ ⎠⎝ ⎠∫ = (2.17)

In view of the η being arbitrary, Eq. (2.17) implies that

d 0d

F Fy 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

0b a

F Fy yη η∂ ∂

− =′ ′∂ ∂

(2.19)

At a particular boundary, say at point a, either η is zero or arbitrary and can be made zero. Thus,

0b

Fyη∂

=′∂

(2.20)

22

Page 30: Notes Dixit

In the same way, it can be shown that

0a

Fyη∂

=′∂

(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 Fy∂′∂ 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

02 dI y y xy′= + +∫ x (2.22)

Solution: Using Eq. (2.18), we have

( )d2 2 2 0 . .,d

y x y i e y x y 0x

′ ′′+ − = + − = (2.21)

As

2F yy∂ ′=′∂

(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 iy x y

⎛ ⎞∂ ∂− = =⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠

n (2.24)

with corresponding boundary conditions. With n independent variables, we need to extremise

multiple integrals of the form

1 11

..... ( , ,...... , , ,......., ) d .....dnn

y ynI F y x x x x

x x∂ ∂

= ∫ ∫∂ ∂

(2.25)

23

Page 31: Notes Dixit

Using the same kind of analysis as before, we find that the extremising function y = y(x1,

………..,xn) must satisfy

10

i

n

i i x

F Fy x y=

⎛ ⎞∂ ∂ ∂− =⎜ ⎟∑ ⎜ ⎟∂ ∂ ∂⎝ ⎠

(2.26)

where ixy stands for

i

yx∂∂

.

We will now derive the necessary differential equation for extremizing

( , , , ) db

aI F x y y' y'' x= ∫ . We can apply the procedure adopted for extremizing

d, , dd

ba

yI F x yx

⎛ ⎞= ⎜ ⎟⎝ ⎠

∫ 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 ddx x xηδ δ εη ε= = (2.27)

and

d d d d d d d( ) ( )d d d d d d dy y yy' y yx x x x x x x

η ηδ δ εη ε ε⎛ ⎞= = + − = + − =⎜ ⎟⎝ ⎠

(2.28)

Both expressions are equal, hence proved.

Property 2: Integration and variation commute

i.e. ( ) ( )d db ba ay x x y xδ δ=∫ ∫

Proof:

( ) ( ) ( )d db b ba a ay x x y x y xδ εη= + −∫ ∫ ∫ d

d ( )d db b ba a ay x x y xεη= + −∫ ∫ ∫

( )d db ba ax y xεη δ= =∫ ∫

24

Page 32: Notes Dixit

Now, using second property, we can write

( , , , dba )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 FF y y y Oy y y

δ δ δ δ∂ ∂ ∂′ ′′= + + +′ ′′∂ ∂ ∂

ε (2.31)

Noting that yδ εη≡ etc. As 0ε → , we can write

F F FF y yy y y

yδ δ δ δ∂ ∂ ∂′ ′′= + +′ ′′∂ ∂ ∂

(2.32)

Thus,

''

b

a

F F F dI y y yy y y

δ δ δ δ⎛ ⎞∂ ∂ ∂ x′′= + +∫ ⎜ ′′∂ ∂ ∂⎝ ⎠

⎟ (2.33)

Making use of the property 1, we can write

2

2d d( ) ( ) d

' d '' d

b

a

F F FI y y yy 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 FI y y y x yy x y x y x y y'' x

δ δ δ δ δ δ⎛ ⎞∂ ∂ ∂ ∂ ∂

= − − + +∫ ⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠y (2.35)

Integrating the third term by parts,

2

2d d dd ''

y xd

b

a

F F FI y yy x y' yx

δ δ δ δ⎧ ⎫⎛ ⎞ ⎛ ⎞∂ ∂ ∂⎪ ⎪= − +∫ ⎨ ⎬⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎪ ⎪⎩ ⎭

+d

' d ''b b bF F Fy y' ya 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 0d

b

a

I F F F dxy y yε

η η ηε =

⎛ ⎞∂ ∂ ∂′ ′′= = + +⎜ ′ ′′∂ ∂ ∂⎝ ⎠∫ ⎟ (2.37)

25

Page 33: Notes Dixit

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

2d d d( ) ( ) d ( ) 0d ' '' ' dd

b

a

b bF F F F F FI y x ya ay x y y y x y'' y''x

δ δ δ⎡ ⎤ ⎧ ⎫⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂

= − + + − + =⎢ ⎥∫ ⎨ ⎬⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎢ ⎥ ⎝ ⎠⎩ ⎭⎣ ⎦y'δ (2.38)

Thus, the differential equation is 2

2d d 0d ' ''d

F F Fy x y yx

⎛ ⎞ ⎛ ⎞∂ ∂ ∂− +⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠

= (2.39)

The boundary conditions are

d ( )' d ''

F F yy x y

δ⎧ ⎫∂ ∂ 0− =⎨ ⎬∂ ∂⎩ ⎭

at x = a and at x = b (2.40)

and

' 0''

F yyδ∂

=∂

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

Page 34: Notes Dixit

2

0

1 '' d2

lI 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

2d0 ( '') 0d

q EIwx

− − + =

or

2

2d ( '')d

EIw qx

= (2.43)

which is the well known Euler-Bernoulli beam equation.

The boundary conditions are

[d0 ( '')]d

EIw w 0x

δ− = at x=0, l (2.44)

and

'' ' 0EIw wδ = at x = 0, l (2.45)

The term d ( '')d

EIwx

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

Page 35: Notes Dixit

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 0d

''

''

EIw

EIwx

=

= ( )

0d 0d

''

''

EIw

EIwx

=

=

Free-free beam

0

0''

w

EIw

=

=

0

0''

w

EIw

=

=

Simply-supported beam

0

0'

w

w

=

=

d ( )d

0

''

''

EIwxEIw

=

=

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

Page 36: Notes Dixit

[ ( ) ] d 0D

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 0D

L f Dδ φ φ− =∫ (2.48)

Then, we say that the variational form of the given differential equation is

( ) [ * ( ) ]dD

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 0d d

uE A qx 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

dAt =0, 0; at = ,dux u x l EA Px

= = (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 0d d

l uE A q u xx x

δ⎡ ⎤⎛ ⎞ + =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦∫ (2.52)

Integrating equation (2.51) by parts,

( )0 0 0

d d ddd d d

l l lu uu E A u q x u E A xx x

δ δ δ d 0x

+ −∫ ∫ = (2.53)

As the variational operator behaves like a differential operator and δu at x=0 is 0, we can write

29

Page 37: Notes Dixit

2

0 0

1 d dd ( )d2 d d

l l

l

u uE A x qu x E A ux x

δ δ δ⎡ ⎤⎛ ⎞ ⎛ ⎞ 0− −⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦∫ ∫ = (2.54)

Therefore, 2

0

1 d d ( )2 d

l uE A qu x P u lx

δ⎡ ⎤⎡ ⎤⎛ ⎞ 0− −⎢ ⎥⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦∫ = (2.55)

Hence, the variational form is given by 2

0

1 d d2 d

l u ( )I E A qu x P u lx

⎡ ⎤⎛ ⎞= −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

∫ − (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 0d

Tkx

= (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 = 0d

l Tk T xx

δ∫ (2.58)

Integrating by parts

( )0

0

d d d d = 0d d d

llTk T T k x

x x xδ − 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 d2 d d

l T Tkx x

⎛ ⎞ 0xδ =⎜ ⎟⎝ ⎠∫ (2.60)

or

30

Page 38: Notes Dixit

0

1 d d d2 d d

l T Tk xx x

⎛ ⎞ 0δ =⎜ ⎟⎝ ⎠∫ (2.61)

Hence, the variational form is given by 2

0

1 d d2 d

l TI kx

⎛ ⎞= ⎜ ⎟⎝ ⎠∫ 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 0T Tx 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 dA

T T T Ax 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 . dA A A

I T T A T T A T T Aδ δ δ δ= ∇ = ∇ ∇ − ∇ ∇ =∫ ∫ ∫ 0 (2.65)

Applying divergence theorem,

( ) ( )ˆ. d . dA

I T n T T T AΓ

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 dA A

T T T TI T T A Ax x y y

δ δδ δ⎛ ⎞∂ ∂ ∂ ∂

= − ∇ ∇ =− + =⎜ ∂ ∂ ∂ ∂⎝ ⎠∫ ∫ 0⎟ (2.67)

Using the properties of the variational operator, we can write

31

Page 39: Notes Dixit

221 d2A

T TIx y

δ δ⎧ ⎫⎛ ⎞∂ ∂⎪ ⎪⎛ ⎞=− + =⎨ ⎬⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠⎪ ⎪⎩ ⎭

∫ 0A (2.68)

Thus, the variational form is 221 d

2A

T TI Ax 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 dd d

l uW u EA xx xδ= ∫ (2.63)

and the external work will be

0

dl

extW u q x P uδ δ= +∫ ( )l (2.64)

Applying the principle of virtual work

( )int 00

d d d d ( )d d

ll

extuW W u EA x u q x P u l

x xδ 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

Page 40: Notes Dixit

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 δ2I , 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

Page 41: Notes Dixit

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

22

0 0

1 1dx dx +2 2

L L

2 Ld w P dwE I kw

dxdx⎛ ⎞ ⎛ ⎞∏ = −⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠∫ ∫ (a)

where 0

0

0 , 0x

x

dwwdx=

=

= = (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

Page 42: Notes Dixit

Assume A, k, r are constants.

Figure: Q2

Differential equation:

2

2

d 0d

T rx kA

+ =

Boundary conditions:

1. Essential boundary conditions: T = 0 at x = 0

2. Natural boundary conditions 1dd 2

TTx

= at x = 1

Q.3: A certain physical problem has the functional

2,

0

1 50 dx2

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 qw w w w w xD

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

Page 43: Notes Dixit

2 2

2 2 0T Tx 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 0d

u u x 1x

+ = ≤ ≤

with u(0) = 0 and ddux

= 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

zzz y

mvEI px x x

0+ − =

with essential boundary conditions at x = 0 are v = v* and *dd z

vx

θ= and natural boundary

conditions at x = l are 2

*2

ddzz z

vEI Mx

= and 2

*2

d d( )d dzz z y

vEI m Vx 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

Page 44: Notes Dixit

1

01 1L ( ,..., , ,..., , ) d

t

n ntL 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 nq t q

⎛ ⎞∂ ∂= =⎜ ⎟∂ ∂⎝ ⎠

Using Hamilton’s principle, derive the wave equation for small transverse oscillations of a taut

string.

37

Page 45: Notes Dixit

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

ni i

iu 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

Page 46: Notes Dixit

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

2d 1d

u ux

+ = 0 1x≤ ≤ with ( )0u 0= andd 0dux

= at 1x = (3.2)

Solve this equation using Ritz’s method.

Solution: First we have to convert it into variational form. For this, let

210 2

d 1 dd

uI u ux

δ⎛ ⎞

= + −⎜ ⎟∫ ⎜ ⎟⎝ ⎠

xδ (3.3)

We write the above expression again, after integrating the first term by parts. Thus,

( )1 1 10 00

d d d d dd d du u 1

0 dI u u x u u x u xx 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 10 0

d d d dd du u 1

0 dI x u u x u xx x

δ δ δ δ⎛ ⎞= − + −∫ ∫⎜ ⎟⎝ ⎠

∫ (3.5)

As the variational operator behaves like a differential operator,

( )2

1 1 20 0

1 d 1d d2 d 2

u 10 dI x u x

xδ δ δ δ⎛ ⎞= − + −∫ ∫⎜ ⎟

⎝ ⎠u x∫ (3.6)

or

40

Page 47: Notes Dixit

( )2

1 1 20 0

1 d 1d d2 d 2

u 10 dI x u x u

xδ δ

⎡ ⎤⎛ ⎞= − + −⎢ ∫ ∫⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

x⎥∫ (3.7)

Therefore,

21 20

1 d 1 d2 d 2

uI u ux

⎧ ⎫⎪ ⎪⎛ ⎞= − + −∫ ⎨ ⎜ ⎟⎝ ⎠⎪ ⎪⎩ ⎭

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 ddu bx

= . Putting the value of u and ddux

in the variational form,

1 2 2 20

1 1 d2 2

I b b x bx⎛= − + −∫ ⎜⎝ ⎠

x⎞⎟ (3.9)

Extremize this with respect to b, i.e.

( )1 20

d dd

I b bx x xb

= − + −∫ 1 0

3 2bb⎡ ⎤= − + − =⎢⎣ ⎦⎥

(3.10)

This gives 34

b = − . Thus, 34

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 sindu A x B xx

= − (3.12)

At 1x = , d 0dux

= . 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 are 0.5x =

( )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 . 2u a bx cx= + +

41

Page 48: Notes Dixit

Essential boundary condition i.e. ( )0u 0= gives a = 0. Thus,

2u bx cx= + (3.16)

d 2du b cxx

= + (3.17)

( ) ( ) ( )221 20

1 122 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 ) ( ) dI b cx bx cx x x x

b∂

= − + + + − =∫∂ 0 (3.19)

and

1 2 2 2

0( 2 )2 ( ) dI 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) 21.6402 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 1.5826du xx

= − + (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

Page 49: Notes Dixit

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

01

ni i

iu 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:

01

(n

i ii

)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,....,iD

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

Page 50: Notes Dixit

Solution: Let us take a quadratic approximation i.e.,

(3.30) 2u 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

22 ( 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 ( 26

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 u Point

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

Page 51: Notes Dixit

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 ia∂

= =∫∂

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 22

02 ( 2 ) 1 dc c x x x

c∂

+ − − =∂ ∫ ) 0

0

(3.37)

or

( )( )1 2 2

02 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 ( 27

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 25

u x= − )x (3.40)

If the residue is made equal to zero at x=1/3, we get c=9/13 giving

( 29 213

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

Page 52: Notes Dixit

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 24

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

Page 53: Notes Dixit

EXERCISE 3

Q.1: A certain problem of one dimensional heat transfer is governed by the equation 2

2

d 1 0dx

φ φ+ + = and boundary conditions 1φ = at 0x = and d 1dxφ

= at 1x = . Solve this

problem by using Galerkin method.

Q.2: Given a differential equation:

22

2002 2

dd 0d dzz

vEI vx x

ρ ω⎛ ⎞

− =⎜ ⎟⎝ ⎠

;

with boundary conditions:

00

d 0dvvx

= = 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:

( )01

n

i ii

v a φ=

= ∑ x

=( )2

1

0n

ij ij jj

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

uI AE pu x Pux =

⎛ ⎞⎛ ⎞= − + +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠∫

a) Using the approximation ( )

1 2

x l xu 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

Page 54: Notes Dixit

p (distributed force)

P (point force)

l/2

l

Figure: Q3

Q.4: Solve the for the following differential equation by Galerkin method: 2

22

d 0d

u cu xx

− − + = for 0 1x≤ ≤ , the boundary conditions being

( ) ( )d d0 1, 1d du ux x 4

3x x= = = = .

Q.5: Solve the following problem by Galerkin method:

Differential Equation:

d d 0d d

uu fx x

⎛ ⎞ − =⎜ ⎟⎝ ⎠

for 0 < x < 1 ;

Boundary conditions: (1) Essential: 2 at =0u x=

(2) Natural : d 0dux

= at x=1

Take f to be a linear function of x : 1 2b b x+ (b1, b2 are constants).

Take 21 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 0dx

φ φ+ + = and boundary conditions 1φ = at 0x = and d 1dxφ

= at 1x = .

Use Ritz method to solve this problem. Approximate the function by a quadratic polynomial and

compare with the exact solution.

48

Page 55: Notes Dixit

Q.7: Solve the following differential equation by Ritz method: 2

22

d 0d

u cu xx

− − + = for 0 1x≤ ≤ for the boundary conditions: ( ) ( )d d0 1, 1d du ux x 4

3x x= = = = .

Solve it by all other methods which you have studied in this chapter.

49

Page 56: Notes Dixit

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

0d dd

l uI u xx

⎧ ⎫⎪ ⎪⎛ ⎞= −∫ ⎨ ⎬⎜ ⎟⎝ ⎠⎪ ⎪⎩ ⎭

(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

Page 57: Notes Dixit

function is complete. It represents 2u a bx cx= + +ddux

by 2b cx+ . However, is not

complete. Why? After all it can represent

2u a cx= +

ddux

by . The reason is that it cannot approximate a

constant derivative term. The approximate

2cx

ddux

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 dd d

l

3x l

u uI qu dxx x

=

⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟= − −⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠∫ (4.2)

For this, a complete approximate function will be . We have taken cubic

polynomial because we have to represent

2u a bx cx dx= + + + 3

3

3

dd

yx

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 that 0C ddux

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 and 1C ddux

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.

1mC −

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. If bx= + 0C

52

Page 58: Notes Dixit

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

0C

( )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 ubh−

= (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 uu u x

h−

= + 1 21 x xu uh 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, 2u a bx cx= + +

(4.7) 21 1u a bx cx= + + 1

2

3

(4.8) 22 2u a bx cx= + +

(4.9) 23 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 21

............n

n n i ii

u N u N u N u N u=

= + + + = ∑

53

Page 59: Notes Dixit

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 1n n

ii i

c cN N= =

= ∑ ∑

Lagrange’s interpolation functions satisfy these properties. For n-noded element,

Lagrange shape functions are given by

54

Page 60: Notes Dixit

2 3 41

2 1 3 1 4 1 1

1 3 42

1 2 3 2 4 2 2

1 3 4 1

1 3

( )( )( ).........( )( )( )( ).........( )

( )( )( ).........( )( )( )( ).........( )

.

.

.( )( )( ).........( )

( )(

n

n

n

n

nn

n n

x x x x x x x xNx x x x x x x x

x x x x x x x xNx x x x x x x x

x x x x x x x xNx x x x

− − − −=

− − − −

− − − −=

− − − −

− − − −=

− − 4 1)( ).........( )n n nx x x −− − x

1n

(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,

11

ni

iN

==∑

1

ni

iN

=∑

20 1 2

1..........

n ni

iN a a x a x a x −

== + + + +∑ (4.18)

Its value at all the nodes is 1. Hence,

2 10 1 1 2 1 1

20 1 2 2 2 2

2 10 1 2

1 ..........

1 .............

1 ..........

nn

nn

nn 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

1n

ii

N=

=∑ (4.20)

throughout the element.

4.4 DEVELOPING THE ELEMENTAL EQUATIONS BY RITZ METHOD

Consider the problem

2

2

d 1 0d

u ux

+ − = (4.21)

55

Page 61: Notes Dixit

with boundary conditions

and (0) 0u =d 0 at 1du xx= = (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 0d

h uI u ux

δ⎡ ⎤

= + −⎢ ⎥⎣ ⎦

∫ xδ = (4.23)

Integrating by parts and using the properties of the variational operator

( )2

2

0 0 0

d 1 d 1d dd 2 d 2

h h hh

o

u uI u x u x u xx x

δ δ δ δ δ⎛ ⎞ ⎛ ⎞= − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ ∫ ∫ d 0= (4.24)

Hence,

2

200 0 0

1 d 1 ddx+ d d2 d 2 d

h h h hu uI u x u x ux 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

0C

0C eu

= (4.26)

1

2

1 2 ....... ..

en

n

uu

u N N N N u

u

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪= ne⎢ ⎥ ⎢⎨ ⎬⎣ ⎦ ⎣⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

neu N⎢ ⎥⎣ ⎦= ⎥⎦

Putting this value in I,

0

, , 1 2

dd001 1d d d ....2 2.

dd

h h hne ne ne ne nex x no o o

h

ux

I u N N u x u N N u x u N x u u u

ux

⎧ ⎫⎛ ⎞−⎜ ⎟⎪ ⎪⎝ ⎠⎪ ⎪⎪⎪⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥= − + − +⎢ ⎥ ⎢ ⎥ ⎨ ⎬⎣ ⎦ ⎣ ⎦⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎪ ⎪⎪ ⎪⎪ ⎪⎛ ⎞⎪ ⎪⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭

∫ ∫ ∫

⎪⎪ (4.27)

which can also be written as

56

Page 62: Notes Dixit

0

, ,

dd001 1d d d.2 2.

dd

h h hne ne ne ne ne nex xo o o

h

ux

I u N N u x u N N u x u N x u

ux

⎧ ⎫⎛ ⎞−⎜ ⎟⎪ ⎪⎝ ⎠⎪ ⎪⎪ ⎪⎪ ⎪⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥= − + − +⎢ ⎥ ⎨⎣ ⎦⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎪ ⎪⎪ ⎪⎪ ⎪⎛ ⎞⎪ ⎪⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭

∫ ∫ ∫⎪⎬

(4.28)

For extremizing it,

0ne

Iu∂

=⎢ ⎥∂ ⎣ ⎦

(4.29)

0

, ,0 0 0

dd00

d d d..

dd

h h hne nex x

h

ux

N N u x N N u x N x

ux

⎧ ⎫⎛ ⎞−⎜ ⎟⎪ ⎪⎝ ⎠⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎢ ⎥− + − +⎢ ⎥ ⎨ ⎬⎣ ⎦⎣ ⎦⎪ ⎪⎪ ⎪⎪ ⎪⎛ ⎞⎪ ⎪⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭

∫ ∫ ∫ 0= (4.30)

Hence,

0

, ,0 0 0

dd00

d d d..

dd

h h hnex x

h

ux

N N x N N x u N x

ux

⎧ ⎫⎛ ⎞−⎜ ⎟⎪ ⎪⎝ ⎠⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎡ ⎤⎢ ⎥ − = − +⎢ ⎥ ⎨ ⎬⎣ ⎦⎣ ⎦⎢ ⎥⎣ ⎦ ⎪ ⎪⎪ ⎪⎪ ⎪⎛ ⎞⎪ ⎪⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭

∫ ∫ ∫ (4.31)

This is the element equation.

For 2-noded elements, we have for a typical element,

01

2

dd1 1 2 11 2

1 1 1 26 d2 d h

uhu xhu hh u

x

⎧ ⎫⎛ ⎞⎧ ⎫ −− ⎜ ⎟⎪ ⎪⎪ ⎪⎡ − ⎤ ⎧ ⎫⎡ ⎤ ⎡ ⎤ ⎝ ⎠⎪ ⎪ ⎪ ⎪− = +⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥⎢ ⎥ ⎢ ⎥− ⎛ ⎞⎣ ⎦ ⎣ ⎦ ⎩ ⎭⎣ ⎦ ⎪ ⎪ ⎪ ⎪− ⎜ ⎟⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎝ ⎠⎩ ⎭

(4.32)

57

Page 63: Notes Dixit

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 d2.083 3.667 2.083 0.5 0

0 2.083 1.833 0.25 0

uu xuu

⎧ ⎫−⎪ ⎪− −⎧ ⎫⎡ ⎤ ⎧ ⎫⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥− − = − +⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥

⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥− −⎣ ⎦ ⎩ ⎭⎩ ⎭ ⎪ ⎪⎩ ⎭

(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 are 0.5x =

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.

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Page 64: Notes Dixit

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 0d

h uw u xx

⎡ ⎤+ − =⎢ ⎥

⎣ ⎦∫ (4.37)

Integrating by parts,

0 0 0

0

d d d d d dd d d

hh h hu w uw x wu x

x x x− + −∫ ∫ ∫ 0w x = (4.38)

Seeing it, we know that taking

11 2

2

uu N N

u⎧ ⎫

= ⎢ ⎥ ⎨ ⎬⎣ ⎦⎩ ⎭

(4.39)

is enough. Writing in a condensed form,

neu N u= ⎢ ⎥⎣ ⎦ (4.40)

In Galerkin method, w is approximated in the same way as the shape function, i.e.,

new w N⎢ ⎥= ⎣ ⎦

(4.41)

Putting equations (4.40-4.41) in equation (4.38),

0

0 0 0

.

. , , d d

.

h h hne ne ne ne ne ne

h

dudx

w w N x N x u x w N N u x w N x

dudx

⎧ ⎫⎛ ⎞−⎜ ⎟⎪ ⎪⎝ ⎠⎪ ⎪⎪ ⎪⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− + −⎢ ⎥ ⎢ ⎥∫ ∫ ∫⎨ ⎬ ⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎪ ⎪⎪ ⎪⎪ ⎪⎛ ⎞⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭

d 0= (4.42)

As the nodal values of the weights are arbitrary, we get

59

Page 65: Notes Dixit

0

, ,0 0 0

dd00

d d d..

dd

h h hnex x

h

ux

N N x N N x u N x

ux

⎧ ⎫⎛ ⎞− ⎜ ⎟⎪ ⎪⎝ ⎠⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎡ ⎤⎢ ⎥ − = − +⎢ ⎥ ⎨ ⎬⎣ ⎦⎣ ⎦⎢ ⎥⎣ ⎦ ⎪ ⎪⎪ ⎪⎪ ⎪⎛ ⎞⎪ ⎪⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭

∫ ∫ ∫ (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 d5 7d d

y y yx 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.

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Page 66: Notes Dixit

EXERCISE 4

Q.1: Solve the following boundary value problem by Ritz and Galerkin FEM:

Differential Equation: 2

2

d 5 7d

y xyx

+ =

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 d5 7d d

y y yx 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

dMinimize 5 d 10 (1)duI x ux

⎛ ⎞= −⎜ ⎟⎝ ⎠∫ ; 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

dMinimize 5 dduI u xx

⎛ ⎞⎛ ⎞= −⎜⎜ ⎟⎜⎝ ⎠⎝ ⎠∫ ⎟⎟ ; 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

dDifferential equation: 5 ( 0.5) 0 0 1d

dBoundary conditions: At =0, =0 and at =1, 0.d

y x xx

yx y xx

δ+ − = ≤ ≤

=

Take 2 2-noded elements.

61

Page 67: Notes Dixit

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

Tq kx

∂= −

∂ (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

Page 68: Notes Dixit

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

Page 69: Notes Dixit

d d ( )dx dx c f

TkA 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 dd2 2 d

hh

x c c fTAkT h pT QAT h pT x kA Tx

⎧ ⎫∏ = + − + −⎨ ⎬⎩ ⎭∫

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 T1 20 0

1 1 , , dx dx 2 2

ddx dx dxddx

T Th hT Te e e e c e

h hT Te e c f

T N x Ak N x T T N h p N T

TkAT QA N T h pT N T T

TkA

∏ = +

⎧⎪⎪ ⎪− − + ⎨ ⎬⎪ ⎪−⎪ ⎪⎩ ⎭

∫ ∫

∫ ∫⎫⎪ (5.9)

Minimizing with respect to TeT, we get elemental equations:

65

Page 70: Notes Dixit

[ ] [ ] [ ] [ ]

[ ] [ ]

0 0

T T 0

0 0

0 , , dx dx

dd

dx dxddx

T Th h

e c

h h

c f

h

N x Ak N x T N h p N T

TkAx

QA N h pT NTkA

= + 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 dd d

Tkrr 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 0d d

hjj

TR w R r k rr 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 0d d d

h hj j

T w Tw R r k R r k rr r r

⎛ ⎞+ − +∫⎜ ⎟⎝ ⎠

=

66

Page 71: Notes Dixit

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 rT a brh h T

⎧ ⎫⎪ ⎪⎡ ⎤= + = − ⎨ ⎬⎢ ⎥⎣ ⎦ ⎪ ⎪⎩ ⎭

Therefore,

d 1 1,d

j

k

TTr h h T

⎧ ⎫⎪ ⎪⎡ ⎤= − ⎨ ⎬⎢ ⎥⎣ ⎦ ⎪ ⎪⎩ ⎭

Putting this approximation in the weak form,

( ) 1

1' '

02

2

dd

ddd

h nej

TR kr

k R r N N r TTR kr

⎧ ⎫⎪ ⎪⎪ ⎪⎡ ⎤+ =∫ ⎨ ⎬⎣ ⎦ ⎪ ⎪⎪ ⎪⎩ ⎭

or

'2 2 1 1'

2 22 2

1 1

1 1 2j

jk

T R k Thh hk R hT R k T

h h

⎡ ⎤− ⎧ ⎫⎧ ⎫ −⎪ ⎪ ⎪ ⎪⎛ ⎞⎢ ⎥ + =⎨ ⎬ ⎨ ⎬⎜ ⎟⎢ ⎥ ⎝ ⎠ ⎪ ⎪− ⎩ ⎭ ⎪ ⎪⎩ ⎭⎢ ⎥⎣ ⎦

or

1 11 11 12 2

j jj

k k

T HRk

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 02 213 3 5 5 02 2 2 2 2

5 50 2 2

T HT

HTπ

⎡ ⎤− ⎧ ⎫ ⎧ ⎫⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥− + − =⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎪ ⎪ ⎪ ⎪−⎢ ⎥ ⎩ ⎭⎩ ⎭−⎢ ⎥⎣ ⎦

Let us impose the boundary conditions:

*1 aT T= *

3 bT T=

Applying boundary condition, we get

* *2 0

3 8 5 02 2 2aT T T− + − =

67

Page 72: Notes Dixit

( ) * *2 00.375 0.625FEM

aT T= + T

The exact solution is given by

( )* *

* lnln

b aaexact ab

a

T T RT T RRR

− ⎛ ⎞= + ⎜ ⎟⎝ ⎠⎛ ⎞⎜ ⎟

⎝ ⎠

Thus,

( ) * *2 0.415 0.585a bT Exact T T= +

If =2000C The FEM solution is T2 141.5 0C *1 aT T=

Exact solution is T2 136.90 0C. * 00 100 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 0d d

uEA qx x⎛ ⎞ + =⎜ ⎟⎝ ⎠

(5.12)

Boundary conditions are:

(i) At 0, 0x u= = (5.13)

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Page 73: Notes Dixit

(ii) At d,dux Lx

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 ey 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 dd dd d d d d

ee e

ee e

xx xe e

xx x

y w y yw EA q x EA qw x w EAx x x x x

++ +⎛ ⎞⎛ ⎞ ⎡ ⎤⎜ + ⎟ = − + +⎜ ⎟ ⎢ ⎥⎜ ⎟⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦⎝ ⎠

∫ ∫ 0e

=

(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

11 2

2

e yy 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:

11 2

1 1

( ) ( );

( ) ( )e

e e e e

x x x xN N

x x x x+

+ +

− −= =

− + −e (5.17)

Differentiating Eq. (5.16) with respect to x,

1 ' '1 21 2

2 2

d ddd d d

e yN Ny N Nyx x x

1yy

⎧ ⎫⎢ ⎥ ⎢ ⎥= =⎧ ⎫

⎨ ⎬⎢ ⎥ ⎣ ⎦⎣ ⎦ ⎩ ⎭ ⎩ ⎭⎨ ⎬ (5.18)

Here, ( ′ ) indicates differentiation with respect to x.

Now, a linear weight function is given by,

11 2

2

Nw 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

ee

x x e

ex x

NN y E A yw w EA N N x q w w x w w

y NN E

+ +⎧ ⎫ ⎧ ⎫−⎧ ⎫⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥− + +⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎩ ⎭ ⎩ ⎭⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭∫ ∫ 0

A y=

(5.20)

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Page 74: Notes Dixit

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 eh hN y N E Aw 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 eh hN y N E AEA 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 21 1 ( ) '

2

e

e e

qhy 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

( )eh

e NQ x h x

⎧ ⎫− ⎨ ⎬

⎩ ⎭∫ (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 01

QQ

⎧ ⎫ ⎧ ⎫=⎨ ⎬ ⎨ ⎬

⎩ ⎭ ⎩ ⎭ (5.25)

This means the concentrated loads can be accounted for by adding these loads at the position

of node.

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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 dd d

u Py 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: dduy

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.

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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 18 16 8

31 8 7

ee

AEkl

−⎡ ⎤⎢= − −⎢⎢ ⎥−⎣ ⎦

⎥⎥ [ ]

e14

61

e plf⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

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:

( )

2px l x

uAE−

=

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)

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Page 77: Notes Dixit

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 dxTkA r⎛ ⎞ +⎜ ⎟

⎝ ⎠

B.C.: (i) Essential: T = T * at x = 0,

(ii) Natural (convection boundary condition): ( )d = - - dx 0TkA hA T T⎛ ⎞

⎜ ⎟⎝ ⎠

at x = l.

( )d = - - dx 0TkA 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.

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Q.4: The equation for one-dimensional steady-state heat conduction in absence of heat generation

is:

d d( )d d

TkAx 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

20

1 1d

1 1

h kA xh

−⎡ ⎤∫ ⎢ ⎥−⎣ ⎦

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++.

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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

dd

vEI qx

= (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.

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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 00 0

d d d d d dd dd d d d d d

l ll l lv v v w w vEI 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

dEither 0 or =0d

vEI vx

δ= (6.3)

and 2

2

d dEither 0 or =0d d

v vEIx 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 d2v/dx2 is the bending moment and EI d3v/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 2v 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

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Page 81: Notes Dixit

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

12

2

2

2

dd

d 2 3d

v av bx

v a bh ch dhv b ch dhx

=

⎛ ⎞ =⎜ ⎟⎝ ⎠= + + +

⎛ ⎞ = + +⎜ ⎟⎝ ⎠

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 2x xNh h

⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(6.8)

2

2 1 xN xh

⎛ ⎞= −⎜ ⎟⎝ ⎠

(6.9)

2 3

3 3 2x xNh h

⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(6.10)

2

4x xN xh 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 xN

h h− −⎛ ⎞ ⎛= − +⎜ ⎟ ⎜

⎝ ⎠ ⎝

⎞⎟⎠

(6.12)

( )2

2 1 ii

x xN x x

h−⎛= − −⎜

⎝ ⎠

⎞⎟ (6.13)

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Page 82: Notes Dixit

2 3

3 3 2i ix x x xN

h h− −⎛ ⎞ ⎛= −⎜ ⎟ ⎜

⎝ ⎠ ⎝

⎞⎟⎠

(6.14)

( )2

4i

iix x x x

N x xh 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 42'2

v

vv N N N N

v

v

⎧ ⎫⎪ ⎪⎪ ⎪= ⎢ ⎥ ⎨ ⎬⎣ ⎦⎪ ⎪⎪ ⎪⎩ ⎭

(6.16)

and

1 1

2' '1 1 2 2

3 3

4 4

ne

N NN N

w w w w w wN NN N

2

⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥= = ⎪⎨ ⎬ ⎨⎣ ⎦ ⎣ ⎦ ⎬⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭⎪

1

1

2

2

VMVM

1

1

2

2

VMVM

(6.17)

Thus,

1 1 1

2 1 21 2 3 4

3 2 30 0

4 2 4

d dh h

ne ne ne

N v NN v N

w EI N N N N x w q x wN v NN v N

′′ −⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪′′ ′ −⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥′′ ′′ ′′ ′′ = +⎡ ⎤⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦′′⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪′′ ′⎩ ⎭ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭

∫ ∫ (6.18)

As the nodal weights are arbitrary, equation (6.18) gives the following elemental equation:

1 1 1

2 1 21 2 3 4

3 2 30 0

4 2 4

d dh h

N v NN v N

EI N N N N x q xN v NN 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.

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Page 83: Notes Dixit

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

22

1 1 2 2 1 1 2 220 0

1 d d2

h hvEI 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

2a bx cx dx+ + + 3

1

11 2 3 4

2

2

e

vv

v N N N Nvv

⎧ ⎫⎪ ⎪′⎪ ⎪= ⎡ ⎤⎨ ⎬⎣ ⎦⎪ ⎪⎪ ⎪′⎩ ⎭ (5.27)

Here, N1, N2, N3 and N4 are called Hermitian shape functions. Then,

[ ]

[ ] [ ]

1 1

2 11 1 2 2 1 2 3 4

3 20

4 2

1 1

2 11 1 2 2 1 1 2 2

3 20

4

1 d2

d

′′

2

⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪′′ ′ ⎪⎪ ⎪ ⎪′ ′ ′′ ′′ ′′ ′′Π = ⎡ ⎤

⎪⎨ ⎬ ⎨⎣ ⎦′′ ⎬⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪′′ ′⎩ ⎭ ⎩ ⎭⎪

⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪′ ′ ′ ′− − ⎪⎨ ⎬ ⎨ − ⎬⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪−⎩ ⎭ ⎩ ⎭

h

h

N vN v

v v v v EI N N N N xN vN v

N VN M

q v v v v x v v v vN VN ⎪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 20∂∏

=′ ′∂ v v v v (5.33)

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Page 84: Notes Dixit

1 1 1

2 1 21 2 3 4

3 2 30 0

4 2 4

d dh h

N v NN v N

EI N N N N x q xN v NN v N

′′ −⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪′′ ′ −⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪′′ ′′ ′′ ′′ = +⎡ ⎤⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎣ ⎦′′⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪′′ ′⎩ ⎭ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭

∫ ∫1

1

2

2

VMVM (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.

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Page 85: Notes Dixit

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.

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Page 86: Notes Dixit

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

64 2

12 6

4

L L L L

EIL

LSYM

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

220 0

1 d d 1d d2 d 2 d 2

L L

Lw P wEI x x

x x⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠∏ ∫ ∫ kw+ (a)

where L x Lw w

== and the essential boundary conditions are

00

0, 0x

x

dwwdx=

=

= = (b)

Invoke the stationary condition 0δ∏= to derive the problem-governing differential

equation and the natural boundary conditions.

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Page 87: Notes Dixit

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 2d 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.

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Page 88: Notes Dixit

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.

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Page 89: Notes Dixit

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 uv R vw w

′⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪′ =⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪′⎩ ⎭ ⎩ ⎭

(7.1)

where

[ ]1 1 1

2 2 2

3 3 3

l m nR 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

[ ]Tu uv R vw w

′⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪′=⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪′⎩ ⎭ ⎩ ⎭

(7.3)

If [ ]Td u v w= is displacement vector, then

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Page 90: Notes Dixit

[ ] d R d′ = and [ ] Td R d ′= (7.4)

If is a force vector, then T

x y zr F F F⎡= ⎣ ⎤⎦

[ ] r R r′ = and [ ] Tr R r′= (7.5)

If [ ] d T d′ = where is not necessarily orthogonal and may not even be square, then [ ]T

[ ] Tr T r′= (7.6)

This can be proved as follows.

From work equality

T Td r d rδ δ ′ ′= or [ ] TT Td r d T rδ δ ′=

from which we can write, [ ] 0TTd r T rδ ⎡ ⎤′− =⎣ ⎦

Therefore [ ] Tr 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 [ ] Tr T r′=

As [ ] [ ] Tk d r T r′= =

[ ] [ ] [ ] [ ][ ] T TT k d T k T d′ ′ ′= = We get

[ ] [ ] [ ][ ]Tk 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

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Page 91: Notes Dixit

[ ] 1 11 1

AEkL

−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦ (7.9)

The local degrees of freedom are related to the global degrees of freedom as

[ ]1

1

uu v

Tu

1

22

2

uv

⎧ ⎫⎪ ⎪⎧ ⎫′⎪ ⎪ ⎪= ⎪

⎨ ⎬ ⎨′

⎬⎪ ⎪ ⎪ ⎪⎩ ⎭

⎪ ⎪⎩ ⎭

(7.10)

where

[ ]2 4

cos sin 0 00 0 cos sin

Tθ θ

θ θ×

⎡ ⎤= ⎢ ⎥⎣ ⎦

(7.11)

Therefore, the formula for transforming the local stiffness matrix into the global one is

[ ] [ ] [ ]'4 2 2 42 2

Tk T k T× ××

⎡ ⎤= ⎣ ⎦

From this, we get

[ ]

2 2

2 2

2

2

cos cos sin cos cos sinsin cos sin sin

cos cos sinsin

AEkL

θ θ θ θ θ θθ θ θ θ

θ θ θθ

⎡ ⎤− −⎢ ⎥− −⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦

(7.12)

For space truss,

[ ] 11 12 6

1 1 1

0 0 000 0nl m

Tl m n×

⎡ ⎤= ⎢ ⎥⎣ ⎦

Rest of the procedure is similar to that for plane truss.

88

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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

11 1

111

22122

1 1 1 12 2 2 2

1 1 13 2 2 2 2

1 13 22 2

12

x

y

x

y

FUFV

U FV

F

⎡ ⎤− −⎢ ⎥ ⎧ ⎫⎢ ⎥ ⎧ ⎫ ⎪ ⎪⎢ ⎥− ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥−⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎩ ⎭ ⎪ ⎪⎢ ⎥ ⎩ ⎭

⎢ ⎥⎣ ⎦

For element 2-3, θ = -900

89

Page 93: Notes Dixit

( )( )( )( )

2

22 2

222

33233

0 0 0 01 0 13

0 031

x

y

x

y

FUFV

U FV

F

⎧ ⎫⎧ ⎫⎡ ⎤ ⎪ ⎪⎪ ⎪⎢ ⎥ ⎪ ⎪− ⎪ ⎪ ⎪ ⎪⎢ ⎥ =⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎣ ⎦ ⎩ ⎭ ⎪ ⎪⎩ ⎭

For element 1-3, θ = 1800

( )( )( )( )

3

11 3

113

33333

1 0 1 00 0 03

1 030

x

y

x

y

FUFV

U FV

F

⎧ ⎫− ⎧ ⎫⎡ ⎤ ⎪ ⎪

⎪ ⎪⎢ ⎥ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥ =⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎣ ⎦ ⎩ ⎭ ⎪ ⎪⎩ ⎭ Assembling,

( )( )

( )( )

11

1 1

2

2

3 3

33

0.5 1 0.5 0 0.5 0.5 1 00.5 0 0.5 0 0.5 0.5 0 0

100.5 0.5 0.5 0 0.5 0 0 000.5 0.5 0.5 0 0.5 1 0 1

1 0 0 0 1 00 0 0 1 0 0 1 0

x

y

x

y

FUFV

UVU FV 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 0 000.5 1.5 0 1

0 0 1 00 1 0 1

x

y

UV

FUV F

⎧ ⎫− ⎧ ⎫⎡ ⎤ ⎪ ⎪⎪ ⎪⎢ ⎥− − ⎪ ⎪⎪ ⎪⎢ ⎥ =⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪−⎣ ⎦ ⎩ ⎭ ⎩ ⎭ At node 3, force along the inclined plane is zero. Hence,

( ) ( )0 03 3cos 45 sin 45 0x yF F− =

or,

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Page 94: Notes Dixit

( ) ( )3 30x yF F− =

To enforce it, subtract fourth row from third, to get

( )

2

2

3

3 3

100.5 0.5 0 000.5 1.5 0 100 1 1 1

0 1 0 1 y

UVU

FV

⎧ ⎫− ⎧ ⎫⎡ ⎤⎪ ⎪⎪ ⎪⎢ ⎥− − ⎪ ⎪ ⎪ ⎪⎢ ⎥ =⎨ ⎬ ⎨ ⎬⎢ ⎥− ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪−⎣ ⎦ ⎩ ⎭ ⎩ ⎭

As the normal displacement is zero at node 3. 0 0

3 3cos 45 cos 45 0U V+ =

i.e. . Replace fourth equation by this. 3 3 0U V+ =

2

2

3

3

0.5 0.5 0 0 100.5 1.5 0 1 00 1 1 1 00 0 1 1 0

UVUV

− ⎧ ⎫⎡ ⎤⎪ ⎪⎢ ⎥− − ⎪ ⎪ ⎪ ⎪⎢ ⎥ =⎨ ⎬ ⎨ ⎬⎢ ⎥− ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎣ ⎦ ⎩ ⎭

⎧ ⎫⎪ ⎪

⎩ ⎭

Solving the above equation, we get

2 2 3 340, 20, 10, 10U V U V= = = − =

91

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7.4 FEM FORMULATION FOR THE FRAMES

Elemental equation for the frame is

3 2 3 2 1

12 2 1

2

2

23 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 AEL L

EI EI EI EIuL L L LvEI EI EI EI

LL L LAE AEL L v

EI EI EI EIL L L L

EI EI EI EIL LL L

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

[ T1 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 0sin cos 0 0 0 0

0 0 1 0 0 00 0 0 cos sin 00 0 0 sin cos 00 0 0 0 0 1

u uv v

u uv v

φ φφ φ

θ θφ φφ φ

θ θ

′⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥′ −⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪′ ⎢ ⎥⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬⎢ ⎥′⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥′ −⎪ ⎪ ⎪ ⎪⎢ ⎥′⎪ ⎪ ⎪ ⎪⎣ ⎦⎩ ⎭ ⎩ ⎭

(7.17)

92

Page 96: Notes Dixit

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

[ ] [ ] [ ][ ]Tk 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

Page 97: Notes Dixit

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

4626612612

2646612612

hhhhhh

hhhhhh

hEI

.

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

Page 98: Notes Dixit

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

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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.

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Page 100: Notes Dixit

1

x y

x2 xy y2

x3 x2y xy2 y3

x4 x3y x2y2 xy3 y4

x5 x4y x3y2 x2y3 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

111

x y Tax y b T

cx y T

⎡ ⎤ ⎧⎧ ⎫ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥ =⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎩ ⎭⎣ ⎦ ⎩ ⎭

(8.3)

From Eq. (8.1),

[ ] [ ]

11 1 1 1

2 2 2 1 2 3 2

3 3 3 3

1[1 ] 1 1

1

x y T TaT 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 31

1 2 2 1 2 3 3 2 3 1 3 1

1 1 3 3 3 1 3 12

1 2 2 1 2 3 3 2 3 1 3 1

1 2 2 1 2 23

( ) ( ) ( ) area( 23)( ) ( ) ( ) area(123)

( ) ( ) ( ) area( 31)( ) ( ) ( ) area(123)

( ) ( )

xy x y x y x y x y y x PNx y x y x y x y x y y x

x y xy xy x y x y y x PNx y x y x y x y x y y x

x y x y x y xyN

− + − + −= =

− + − + −

− + − + −= =

− + − + −

− + −= 1 1

1 2 2 1 2 3 3 2 3 1 3 1

( ) area( 12)( ) ( ) ( ) area(123)

xy yx Px y x y x y x y x y y x

+ −=

− + − + −

(8.5)

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Page 101: Notes Dixit

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 31

1 2 2 1 2 3 3 2 3 1 3 1

1 1 3 3 3 1 3 12

1 2 2 1 2 3 3 2 3 1 3 1

1 2 2 1 2 23

( ) ( ) ( ) area( 23)( ) ( ) ( ) area(123)

( ) ( ) ( ) area( 31)( ) ( ) ( ) area(123)

( ) ( )

xy x y x y x y x y y x Px y x y x y x y x y y x

x y xy xy x y x y y x Px 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 Px 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

2T 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 21 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)

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Page 102: Notes Dixit

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

21 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 24 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, 2 312

ξ ξ= = , thus

4 41 0 or 04

e e= = (8.14)

At node 6, 1 312

ξ ξ= = , thus

4 41 0 or 04

f f= =

Finally, at node 4, 1 212

ξ ξ= = , thus

4 41 1 or 44

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 21 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.

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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

1 110 or 1

4 4a d d+ = = − (8.22)

Lastly, at node 6, ξ1=ξ3=1/2 and ξ2=0, giving

1 1f = − (8.23)

Therefore,

21 1 1 2 1N 3ξ ξ ξ ξ ξ= − − (8.24)

The above expression can be simplified as

2 21 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

Page 104: Notes Dixit

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 Ki I J KN l l l 3ξ ξ ξ= (8.28)

where

0 1 1

0 1

( )( )........( )( )( )( )........( )

n nn

n n n n

x x x x x xl xx 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, 00 1l =

1 1 1 1 4 1 2 1 10 11 9(3 1)(3 2) , (3 1), 272 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

1Volume P234Volume 1234

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

Page 105: Notes Dixit

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)

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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 sN Np q r s q p r sx p y r x q y rN Nq 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.

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Page 107: Notes Dixit

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 yT Tk 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 yT Tq k n k ˆynx 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 eΩ with boundary .eΓ Let be the finite eT

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Page 108: Notes Dixit

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, eR will be

obtained, i.e. e e

ex y

T TR k kx x y y

⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂= +⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

Q+

0

(8.40)

The weighted integral statement of the governing equation is then given by,

(8.41) de

ewRΩ

Ω =∫∫

where w is the weighting function.

Thus,

d d 0e

e e

x yT Tw k k Q x y

x x y yΩ

⎡ ⎤⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂− − −⎢ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦

∫ =⎥ (8.42)

Using the component form of gradient theorem, i.e.,

ˆde e

xFw w

x xΩ Γ

∂ ∂⎛ ⎞ Ω =⎜ ⎟∂ ∂⎝ ⎠∫ ∫ dFn s (8.43)

and

(F ww F )wFx x x

∂ ∂ ∂− = −

∂ ∂ ∂ (8.44)

Equation (8.42) becomes

d d d 0e

e

x y x x y yw T w T T Tk k wQ x y w k n k n sx x y y x xΓΩ

⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞+ − − +⎜ ⎟ ⎜∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠∫ ∫ =⎟ (8.45)

or

ˆd d d d d 0e e e

e e

x x nw T w Tk k x y wQ x y wq sx x y y

Ω 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 yT Tk n k n T T )x x

β ∞

∂ ∂+ = − −

∂ ∂ (8.47)

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Page 109: Notes Dixit

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 0e

e

x y x x y yw T w T T Tk k wQ x y w k n k n sx x y y x xΓΩ

⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞+ − − +⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠∫ ∫ = (8.48)

or, ( )1 2

d d d d 0e

e e

x y nw T w Tk k wQ x y wq s w T T sx 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 de

x yw T w Tw T k k x y wT sx x y y

βΩ Γ

⎛ ⎞∂ ∂ ∂ ∂= + +⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠

∫ ∫

ds

(8.50)

( )l d d de 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 ej j

j

T T N x=

= ∑ )y

)y

e

y

ds

(8.52)

and

(1

,n

e ej j

j

w w N x=

= ∑ (8.53)

for T and w respectively, into (8.49).

This leads to the following equations:

( )1

ne e e eij ij j i i

j

K H T F P=

+ = +∑ (8.54)

where,

( ), , , , d de

eij x i x j x y i y j yK k N N k N N x

Ω

= +∫

e

d ,e

e e e eij i j i iH N N s P N Tβ β ∞

Γ Γ

= =∫ ∫

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Page 110: Notes Dixit

d d de ei 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 1ˆ( , ) ( , ) d ( )d2 2 2

ee

ex 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 52 3 7 63 4 8 75 6 10 96 7 11 107 8 12 119 10 14 1310 11 15 1411 12 16 15

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

Figure 8.8: Mesh of (a) rectangular elements (b) triangular elements

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Page 111: Notes Dixit

For Fig. 8.8 (b), the connectivity matrix is given by

1 2 33 2 43 4 73 7 63 6 51 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 zT T Tk 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 de

ex 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,

0u vy x

∂ ∂− =

∂ ∂ (8.58)

Where u and y are the fluid velocities in the x and y directions, respectively. The continuity

condition is given by

0u vx y

∂ ∂+ =

∂ ∂ (8.59)

For solving it, define a potential function φ(x, y) such that

; and u vx yφ φ∂ ∂

= =∂ ∂

(8.60)

Equation (8.58) is identically satisfied. Substitution of equation (8.60) into equation (8.59)

provides

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Page 112: Notes Dixit

2 2

2 2 0x yφ 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,

nvnφ∂

=∂

(8.62)

where nφ∂

∂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 2Gx yφ 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 zyy xφ φτ 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.

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Page 113: Notes Dixit

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.

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Page 114: Notes Dixit

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

ii

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.

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Page 115: Notes Dixit

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

11

d =n

i ii

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)

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Page 116: Notes Dixit

From Eq. (9.5) and (9.6), we obtain

( ) ( )1 1 12

a f f= − +⎡ ⎤⎣ ⎦ (9.7)

( ) ( )1 12

b f f= + − −1⎡ ⎤⎣ ⎦ (9.8)

Thus,

( ) ( ) ( ) ( ) ( )1 11 1 1 12 2

f f f f fξ ξ= − + + + − −⎡ ⎤ ⎡⎣ ⎦ ⎣ ⎤⎦

1

(9.9)

Therefore,

( ) ( ) ( )1

1d 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,

( ) 2f a b cξ ξ ξ= + + (9.11)

Let us choose the points, 1, 0, 1ξ = − +

( )1

1

2d 23

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 22

f f fc

− + −=

0

Now putting the value of a and c in Eq. (9.12)

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Page 117: Notes Dixit

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

1

1

1 1 22 0 1 1 03 3 3

1 1 4 0 13

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 11 3 3 14 3 3

f d f f f fξ ξ+

⎡ ⎤⎛ ⎞ ⎛ ⎞= − + − + +⎜ ⎟ ⎜ ⎟⎢ ⎝ ⎠ ⎝ ⎠⎣ ⎦∫ ⎥

f

(9.14)

9.2.2 Gauss quadrature

In this method,

( ) ( )1

11

dn

i ii

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 11da 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 3f a b c dξ ξ ξ ξ= + + + can be evaluated exactly.

( ) ( ) ( )1

1 1 2 21df w f w fξ ξ ξ

+

−= +∫ ξ (9.18)

or

( ) ( ) ( )1 2 3 2 3 21 1 1 1 2 2 21

a b c d d w a b c d w a b c d 32ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ

+

−+ + + = + + + + + + +∫

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Page 118: Notes Dixit

or

( ) ( ) ( ) (2 2 3 31 2 1 1 2 2 1 1 2 2 1 1 2 2

223ca 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 21 1 2 2

23

w wξ ξ+ = (9.21)

3 31 1 2 2 0w wξ ξ+ = (9.22)

Solution is 1 21 1,3 3

ξ ξ= − =

w1 = 1, w2 = 2

For three Gauss-point formula

1 1

2 2

3 3

50.77459667,9

80,9

50.77459667,9

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

11

d 2n

ii

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:

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Page 119: Notes Dixit

( )1 1

1 1, d dI f ξ η ξ η

− −= ∫ ∫

( )1 1

11

, dn

i ii

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 ji j

w w f iξ η= =

=∑ ∑

( )1 2

1 1

,n n

i j j ii 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

ξξ ξ ξ ξ ξ

−= ∫ ∫ (9.24)

118

Page 120: Notes Dixit

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

ξξ 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 01

, d d ,Gn

i ii

i

F wξ )Fξ ξ ξ ξ ξ ξ−

=

= ∑∫ ∫ (9.26)

The weights wi and the coordinates ( )1 2,i iξ 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

Page 121: Notes Dixit

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.

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Table 9.1: Gauss points and weights for integration on a triangular domain

nG Coordinates ( )1 2,i iξ 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.

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Page 123: Notes Dixit

EXERCISE 9

Q.1: Integrate

2 3 4 53 4 5 6( ) 10 20

10 100 1000 10000x x x xf 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

Page 124: Notes Dixit

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 2x 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 2x y+ (b) 2 3 43 x y y+ (c)x yx y+−

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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

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Page 126: Notes Dixit

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

11 2

2

d / d[ ] d / d d / d d

d / d

xe

x

N xK 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 d2

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 ddd d d dN N NJx x

1ξξ ξ

−= = (10.5)

Similarly,

12 2d d ddd d d dN N NJx x

2ξξ ξ

−= = (10.6)

By substituting Eqs (10.4-10.6) in (10.3) and integrating, you can verify that

1 1[ ]

1 1eK

−⎡ ⎤= ⎢ ⎥−⎣ ⎦

(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

Page 127: Notes Dixit

2

1

[ ]dx

e ''

xK EI N N⎡ ⎤ = ∫⎣ ⎦

'' x (10.8)

The dx can be transformed as before. It can be easily seen that 2 2

1 22 2

d d( )d d

i iN NJx ξ

−= (10.9)

10.3 FOUR-NODED QUADRILATERAL ELEMENT

In quadrilateral element, transform geometry in the following way: 4

1i i

ix N x

== ∑ , (10.10)

4

1i i

iy 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

21 2 3 4u a a y a y a y= + + +

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Page 128: Notes Dixit

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 2u u 4uη 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

Page 129: Notes Dixit

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.

11 (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:

11 1(1 )(1 )4 2

N ξ η= − − − −5 812

N N (10.14)

At node 5, N5 is 1 and therefore the term 512

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

21 1(1 )(1 )4 2

N ξ η= + − − −5 612

N N (10.15)

31 1(1 )(1 )4 2

N ξ η= + + − −7 612

N N (10.16)

41 1(1 )(1 )4 2

N ξ η= − + − −7 812

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 21 1(1 )(1 ) (1 )(1 )4 2 2 4 2

ii

N N N NN

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

Page 130: Notes Dixit

8

1

1 1 1 1(1 )(1 ) (1 )(1 ) (1 )(1 ) (1 )(1 )4 4 4 4i

iN 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,

25

1 (1 )( 1)2

N η η ξ= − − (10.20)

Similarly,

26

1 (1 )( 1)2

N ξ ξ η= + − + (10.21)

27

1 (1 )( 1)2

N η η ξ= + − + (10.22)

28

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 ix 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 yx yξ ξ ξ

∂ ∂ ∂∂ ∂= +

∂ ∂ ∂ ∂ ∂ (10.25)

and

i i iN N Nx yx yη η η

∂ ∂ ∂∂ ∂= +

∂ ∂ ∂ ∂ ∂ (10.26)

Transforming the equations in the matrix form,

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Page 131: Notes Dixit

i i i

i ii

N x y N Nx xJ

NN x yy 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

NNx J

N Ny

ξ

η

∂⎧ ⎫∂⎧ ⎫⎪ ⎪⎪ ⎪ ∂∂⎪ ⎪ ⎪= ⎪

⎨ ⎬ ⎨∂ ∂ ⎬⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪∂ ∂⎩ ⎭ ⎩ ⎭⎪

(10.28)

The elements of Jacobian matrix can be computed as

8

1

ii

i

Nx xξ 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 11

sinh1 12, sinsinh

n

n

n yn x LT x y T T T

n Wn LL

ππ

ππ

+∞

=

⎡ ⎤⎛ ⎞⎜ ⎟⎢ ⎥− + ⎛ ⎞ ⎝ ⎠⎢ ⎥= + − ⎜ ⎟ ⎛ ⎞⎝ ⎠⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

∑ .

131

Page 132: Notes Dixit

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

Page 133: Notes Dixit

Figure: Q2 (a)

Analytical solution to this problem is given by,

( )cosh

6, 100 cos6cosh

3

yxbT x yb

b

ππ

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

Page 134: Notes Dixit

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

Page 135: Notes Dixit

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

Page 136: Notes Dixit

Figure: Q6 (a)

Figure: Q6(b)

136

Page 137: Notes Dixit

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

Page 138: Notes Dixit

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

tuf

yx xxyx

∂∂

=+∂

∂+

∂∂

ρσσ (11.2)

2

2

tvf

yx yyxy

∂∂

=+∂

∂+

∂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

xu

x ∂∂

=ε , yvy

ε ∂=∂

, xv

yu

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

ddddd

εεε

σ

σσ

20000

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 vEdd−

== , 22112 1 vEvdd−

== , )1(233 v

Ed+

=

(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−+

−==

Edd ,)21)(1(2112 νν

ν−+

==Edd

)1(233 vEd+

= (11.7)

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Page 139: Notes Dixit

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

1iiiuNu (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,

12e e e

eiij ij i i i

v vdv f u dv t u ds

ΓΠ σ ε= − −∫ ∫ ∫ (11.12)

139

Page 140: Notes Dixit

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 ( 22e

e ex x y y xy xyh σ ε σ ε σ ε

Ω

Π = + +∫ ) d dx y

)ds

( )d d (e e

e ex 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

∫⎭⎬⎫

⎩⎨⎧

⎭⎬⎫

⎩⎨⎧

−⎭⎬⎫

⎩⎨⎧

∫⎭⎬⎫

⎩⎨⎧

−∫⎟⎟⎟⎟

⎜⎜⎜⎜

⎪⎭

⎪⎬

⎪⎩

⎪⎨

⎪⎭

⎪⎬

⎪⎩

⎪⎨

=ΠΓΩΩ eee

dstt

vu

hdxdyff

vu

hdxdyhy

xT

e

y

xT

e

xy

y

x

T

xy

y

xee

σ

σσ

ε

εε

221

(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

ei

ei yxNuu

1),( ∑≈

=

n

i

ei

ei 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 eiN

eiu e

iv

140

Page 141: Notes Dixit

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

∑=

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

=

=

ni

ei

ei

ni

ei

ei

e

e

Nv

Nu

vu

1

1 = (11.18)

⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎪

⎥⎦

⎤⎢⎣

n

n

n

n

v

vvu

uu

NNNNNN

.

.

...0.....000..00...

2

1

2

1

21

21

= (11.19) [ ] ee

n

n

n

n N

vu

vuvu

NNNNNN

Δ≡

⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎪

⎥⎦

⎤⎢⎣

..

..

.

.0..0.00...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

Qfkdstt

Nh

dxdyff

NhdxdyBDBh

e

ee

−−ΔΔ=⎭⎬⎫

⎩⎨⎧

∫ Δ−

⎭⎬⎫

⎩⎨⎧

∫ Δ−Δ∫ Δ=Π

Γ

ΩΩ

Minimizing this i.e. differentiating the above expression with respect to eΔ , we get

(11.22) ][ eeee Qfk +=Δ

where , dxdyBDBhk eeTeee

e]][[][][ ∫=

Ω

, (11.23) dxdyff

Nhfy

xTeee

e ⎭⎬⎫

⎩⎨⎧

∫=Ω

][ dstt

NhQy

xTeee

e ⎭⎬⎫

⎩⎨⎧

∫=Γ

][

The element stiffness matrix [ke] is of order 2n x 2n and the elemental load vector

141

Page 142: Notes Dixit

(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(

1ηξ −−

=N , 4

)1)(1(2

ηξ −+=N ,

4)1)(1(

3ηξ ++

=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:

AA

AA

AA 3

32

21

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

Page 143: Notes Dixit

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

JJJJ

J ∑∑∑∑

==

==ni i

ei

ni i

ei

ni i

ei

ni i

ei

yNxNyNxN

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

Page 144: Notes Dixit

where n=number of nodes of the element,

=shape function corresponding to node i, ),( ηξeiN

(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

NN

JNN

where,

⎥⎥⎥⎥

⎢⎢⎢⎢

=−

JJ

JJ

JJ

JJ

J13

24

1][

Finally, the Gauss-Quadrature scheme gives

∫ ∫ = ∑∑ (11.28) − −

1

1

1

1),( ηξηξ ddf

= =

1

1

2

1, )(

n

r

n

ssrsr 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 ien

ei )(

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,

enq

∫=∫ (11.30) −

1

10)(),( ξξ dJfdsyxf b

s

144

Page 145: Notes Dixit

where the boundary Jacobian 22

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=ξξyxJb .

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.

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Page 146: Notes Dixit

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

Page 147: Notes Dixit

Figure 11.6: Finite element mesh for solving quarter plate problem

Figure 11.7: Contours of longitudinal stresses

147

Page 148: Notes Dixit

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 ( ) 01 0 ( ) 0 ( ) 0 ( ) 0 (

4( ) ( ) ( ) ( ) ( ) ( ) ( ) (

b y b y b y b ya x a x a x a x

aba 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

Page 149: Notes Dixit

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

Page 150: Notes Dixit

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 0u uEA Ax 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 ee u uR EA A

x x tρ

⎛ ⎞∂ ∂ ∂= −⎜ ⎟∂ ∂ ∂⎝ ⎠

(12.2)

Making the weighted integral of the residual equal to 0, we get

0d

h ewR 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

20d

e eh u uw EA A xx 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

Page 151: Notes Dixit

2

20 00

dhe eh hu w u uwEA EA x Aw x

x x x tρ∂ ∂ ∂ ∂

− −∂ ∂ ∂ ∂∫ ∫ 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,

11 2

2

ene

uu 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 ; 2x xN Nh h

= − = (12.7)

In Galerkin FEM, the weight functions are approximated in the same way as the primary

variable. Thus,

11 2

2

Nw w w

N⎧ ⎫

= ⎢ ⎥ ⎨ ⎬⎣ ⎦⎩ ⎭

(12.8)

Substituting the expressions ue and w in the weak form, we get

'1 1' '

2 1 1 2 1 2'0220

1 11 2 1 20

2 2

d

d 0

e e h

h

h

N uu uw EA w EA w w EA N N xux x N

N uw w A N N x

N uρ

⎧ ⎫ ⎧ ⎫∂ ∂ ⎪ ⎪ ⎢ ⎥− − ⎢ ⎥ ⎨ ⎬ ⎨⎣ ⎦ ⎣ ⎦∂ ∂ ⎪ ⎪ ⎩ ⎭⎩ ⎭⎧ ⎫ ⎧ ⎫

− =⎢ ⎥ ⎢ ⎥⎨ ⎬ ⎨ ⎬⎣ ⎦ ⎣ ⎦⎩ ⎭ ⎩ ⎭

⎬ (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 02 2 22

1

d dh h

uEAN u Nx

w w EA N N x A N N xu N uu NEA

x

ρ

⎡ ⎤⎧ ⎫∂−⎢ ⎥⎪ ⎪ ⎧ ⎫∂ ⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥⎢ ⎥− −⎢ ⎥ ⎢ ⎥⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎣ ⎦ ⎣ ⎦⎣ ⎦⎢ ⎥∂ ⎪ ⎪ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭⎪ ⎪ ⎩ ⎭⎢ ⎥⎪ ⎪∂⎢ ⎥⎩ ⎭⎣ ⎦

∫ ∫ 0u

= (12.10)

As the nodal weights are arbitrary, we get the following elemental equations:

152

Page 152: Notes Dixit

'1 1 1 1' '

1 2 1 2'0 02 2 22

1

d dh h

uEAN u N u 2x

EA N N x A N N xu 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 11 2 1 16

uEAu u 1xAh EAu uh uEA

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 eim 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 00 12

e Ahm 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)

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Page 153: Notes Dixit

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 0v vEI mx 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

11 2 3 4

2

2

( )( )( )( )

e

v tv t

v N N N Nv tv 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 ee vR EI m 2

vx x t

⎛ ⎞∂ ∂ ∂= +⎜ ⎟⎜ ⎟∂ ∂ ∂⎝ ⎠

(12.20)

We make the weighted residual inside the element equal to zero, i.e.,

2 2 2

2 2 20 0

de eh he v vwR 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 20 00 0

d dh he e eh hv w v w v vw EI EI EI x m x

x xx x x x t

⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂− + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟∫ ∫⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

0e

= (12.22)

In Galerkin FEM, the basis functions for approximating the deflections and weight

functions are same i.e.,

154

Page 154: Notes Dixit

[

T1

T21 2 3 4

3

4

ww

w N N N Nww

⎧ ⎫⎪ ⎪⎪ ⎪= ⎨ ⎬⎪ ⎪⎪ ⎪⎩ ⎭

] (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 dh h

N v vNN Nv v

w w w w EI N N N N x m N N N N xNv vNNv vN

vEIx x

w w w w

1⎡ ⎤⎧ ⎫ ⎧ ⎫ ⎧⎧ ⎫ ⎫⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪⎢ ⎥⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎡ ⎤ +⎢ ⎥∫ ∫⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎣ ⎦⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭⎩ ⎭ ⎩ ⎭⎪ ⎪⎢ ⎥⎩ ⎭⎣ ⎦

⎛∂ ∂−∂ ∂

=

02

20

2

2

2

2

h

h

vEIx

vEIx x

vEIx

⎧ ⎫⎛ ⎞⎞⎪ ⎪⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎪ ⎪⎝ ⎠⎝ ⎠⎪ ⎪⎛ ⎞⎪ ⎪∂+⎜ ⎟⎪ ⎪⎜ ⎟∂⎪⎝ ⎠ ⎪

⎨ ⎬⎛ ⎞⎛ ⎞⎪ ⎪∂ ∂+⎜ ⎟⎜ ⎟⎪ ⎪⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠⎪ ⎪

⎪ ⎪⎛ ⎞∂⎪ ⎪−⎜ ⎟⎪ ⎪⎜ ⎟∂⎝ ⎠⎩ ⎭ (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 dh h

N v vNN Nv v

EI N N N N x m N N N N xNv vNNv 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

Page 155: Notes Dixit

2 2

2 2

156 22 54 13

22 4 13 354 13 156 22420

13 3 22 4

h h

h h h hmh 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.

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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 2dd

w wEI m m x qx x t t

θ⎛ ⎞∂ ∂ ∂+ +⎜ ⎟⎜ ⎟∂ ∂ ∂⎝ ⎠

= (1)

2 2 2

2 2 20

d d dd d

L

hwJ mx x x T

t t tθ θ⎛ ⎞∂+ +⎜ ⎟⎜ ⎟∂⎝ ⎠∫ = (2)

157

Page 157: Notes Dixit

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)

00

x

wx =

∂=

∂ (4)

The natural boundary conditions at the free end are: 2

2 0x L

wx =

∂=

∂ (5)

2

2 0x L

wEIx 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

212

31

412 2

51

156 22 54 13

22 4 13 3420

54 13 156 22

13 3 22 4

i

m m m m mm h

m hm h h h hm h

m h h h h

⎡ ⎤⎢ ⎥−⎢ ⎥⎢=⎢

−⎢ ⎥⎢ ⎥

− − −⎢ ⎥⎣ ⎦

eiM 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

; ( )2 211 140 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= = − ( )215 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

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Page 158: Notes Dixit

(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

23

2 2

0 0 0 0 00 12 6 12 6

0 6 4 6 20 12 6 12 6

0 6 2 6 4

eiK

⎡ ⎤⎢ ⎥−⎢ ⎥⎢= −⎢

− − −⎢ ⎥⎢ ⎥

−⎣ ⎦

i

h hEI

h h h hh 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

( ) dl

u a uE A mx t

a uF m xt

∂ ∂ +− =

∂ ∂∂ +

=∂

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

Page 159: Notes Dixit

The one-element finite element formulation of this problem leads to:

[ ] 1 1

1

2 2

d[ ]d

0

Fa a

uM u K u EAx

u u

⎧ ⎫⎧ ⎫ ⎧ ⎫ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪+ = −⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭ ⎪ ⎪⎩ ⎭

Find out the expression for mass matrix [M] and stiffness matrix [K].

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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 22 2 2z z z z za h b f g cz f xx 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 0z zx 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 0T Ttx 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

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Page 161: Notes Dixit

2 2

2 2

1 0T T Ttx 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 ( ) 0u uEA A q tx 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

00u u= (13.7)

We have to carry out the finite difference approximation of the time derivative. Therefore, let

( ) 11

1

1 fs ss s

s

u uu u

tα α α+

++

−− + = ≤

Δor 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 11

s s s su 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.

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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 ( )( )2O tΔ .

Case 4: 2 ,3

α = is called Galerkin method, It is also stable and the order of accuracy is

( )( )2O 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 ssM u k u F+ = (13.10)

[ ] 11 1 s 1s ssM u k u F++ +

+ =+

(13.11)

We can see that

[ ] ( )[ ] [ ] ( )1 11 11s 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 111s 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 11 1s s ss s ss s

M t k u M t k u t F Fα α α+ + ++ ++ +⎡ ⎤ ⎡ ⎤

sα⎡ ⎤+ Δ = − − Δ + Δ + −⎣ ⎦⎣ ⎦ ⎣ ⎦ (13.14)

Note that for 0α = we obtain left hand side as [ ] 1.

sM u

+ when the mass matrix is diagonal, the

equation becomes explicit and we can solve for 1su

+ 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 2crt tα λ

Δ < Δ ≡−

(13.15)

where λ is the largest eigen value of the finite element equations.

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Page 163: Notes Dixit

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 tM 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 tU Uδ δ+Δ +ΔU t⎡ ⎤= + − + Δ⎣ ⎦U (13.17)

212

t t t t t t tU 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 ( )12

t t tU 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 0U, U, and U.

3. Select time step and parameters tΔ α and δ and calculate integration constants.

( )20.50; 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 at t t

ta a a t a

δα α α αδ δ

1;

tδ δα α

= = = = −Δ Δ Δ

Δ ⎛ ⎞= − = − = Δ − = Δ⎜ ⎟⎝ ⎠

4. Form effective stiffness matrix 0 1ˆ ˆK : K K M+ C.a a= +

5. Triangularize ˆ ˆK : K LDL .T=

B. For each time step:

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Page 164: Notes Dixit

1. Calculate effective loads at time t t+Δ :

0 2 3 1 4 5R R M( U U U) ( U Ut t t t t t t t t ta 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 time t t+Δ :

( )0 2U U U Ut t t t t t ta a+Δ +Δ= − − - 3 Ua

6 7U U+ U+ Ut t t t t ta 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

( ) dl

u a uE A mx t

a uF m xt

∂ ∂ +− =

∂ ∂∂ +

=∂

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

Page 165: Notes Dixit

The one-element finite element formulation of this problem leads to:

[ ] 1 1

1

2 2

d[ ]d

0

Fa a

uM u K u EAx

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

7000C temperature. The ambient temperature is 200C. 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 2crt tα λ

Δ < Δ ≡−

Prove it.

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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

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Page 167: Notes Dixit

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

zwvzwu

,2,,

,,,,

−=+=

−==−==

γ

εε

(14.2)

It can be easily shown [Cook et al.,1989] that for a plate with zero initial strain,

][ κkDM −= (14.3)

where T

xyyyxT

xyyx wwwMMMM ],2,,[and][ x== κ (14.4)

zzzzMzzMt

txy

t

tyy

t

txx ∫∫=∫=

−−−

2/

2/

2/

2/

2/

2/ddd σσσ (14.5)

For isotropic material,

⎥⎥⎥

⎢⎢⎢

−=

2/)1(0000

][D

DDDD

Dk

νν

ν where

)1(12 2

3

ν−=

EtD (14.6)

D is called flexural rigidity of the plate.

The strain energy expression is given by

ADU kT

Ad][

21 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

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Page 168: Notes Dixit

[ ] ∫=A

kT 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 333223221= (14.11)

Page 169: Notes Dixit

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 by yw ∂∂ /

[ ])1)(1()1(,)1()1)(1(),2)(1)(1(21

002

002

0022

0000 −++−++−−++++ ξηξξηηξηηξηξηξ ii ab (14.12)

with

ii

cc byyaxxηηηξξξ

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 3223221= (14.13)

[ ] ayxyyxxyxyxyxw 3223221 += (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

Page 170: Notes Dixit

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).

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Page 171: Notes Dixit

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

(1 4 .1 5)

y x

y xx y

yxxy

yz x

xz y

u z v z

z zx y

zx y

wywx

θ θ

θ θε ε

θθγ

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 νεεν

σ +−

=

=xyτ xyE γν )1(2 +

(14.16)

yzsyzE γν

ατ)1(2 +

=

zxszxE γν

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

21

=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

Page 172: Notes Dixit

dAy

txy

tx

tEtV yxyxA

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂

∂+

∂∂

∂+⎟

⎠⎞

⎜⎝⎛∂∂

∫−

=22222

2 12612121 θθθνθ

ν

dAxy

yt

yyt

ytEt xx

A⎥⎥⎦

⎢⎢⎣

∂∂

∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∫+

+θθθθ

ν 61212)1(221 22222

dAxw

xw

yw

ywEt

+ν xxyyA s⎥⎥⎦

⎢⎢⎣

∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

++∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∫+ θθθθα 22)1(22

1 22

22 (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

iii wNw

1 (14.19)

(14.20) ∑==

=

ni

ixiix N

1θθ

∑==

=

ni

iyiiy N

1θθ (14.21)

where Ni is Lagrange shape function for ith node.

Following the standard procedure, the elemental stiffness matrix can be obtained

as,

[ ] [ ] [ ][ ] [ ] [ ][ ]dABDBdABDBk sjsT

Asibjb

T

Abiij

ee

∫+∫= (14.22)

where

[ ]⎥⎥⎥

⎢⎢⎢

=

000

00000000

,,

,

,

xiyi

yi

xi

bi

NN

NN

B (14.23)

[ ]⎥⎥⎦

⎢⎢⎣

⎡=

000000

,

,

iyi

ixisi NN

NNB (14.24)

173

Page 173: Notes Dixit

[ ] ( ) ( )⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢

−−

=

2100

0101

112 2

3

νν

ν

νEhDb (14.25)

[ ] ( ) ⎥⎦

⎤⎢⎣

⎡+

=1001

12 να EhD 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

Page 174: Notes Dixit

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

Page 175: Notes Dixit

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

Page 176: Notes Dixit

EXERCISE 14

Q.1: The governing equation for thin plate bending is given by 4 4 4

4 2 2 42w 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 42 0

x x y yψ 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 nt 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 2wx 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

Page 177: Notes Dixit

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

Page 178: Notes Dixit

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 uu vx y x x

υρ

⎛ ⎞∂ ∂ ∂ ∂ ∂+ = − + +⎜∂ ∂ ∂ ∂ ∂⎝ ⎠y ⎟ (15.1)

2 2

2 2

1v v p v vu vx y y x

υρ

⎛ ⎞∂ ∂ ∂ ∂ ∂+ = − + +⎜∂ ∂ ∂ ∂ ∂⎝ ⎠y ⎟ (15.2)

0u vx 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

0u v px 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 vpx y

λ⎛ ⎞∂ ∂

= − +⎜ ⎟∂ ∂⎝ ⎠ (15.5)

179

Page 179: Notes Dixit

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

Page 180: Notes Dixit

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

Page 181: Notes Dixit

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 2211 22 1 2 1 2

1 2 1 2

d d 0pA

w w wx x x xσ σ σ σε 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 iA

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

Page 182: Notes Dixit

( ) ( ), ,d di j j i i j i i j i jjA 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 jjA 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 jwσ 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 σΓ

− + Γ −∫ ∫ ∫ 0= (15.19)

The last term in Eq. (15.19) can be expressed as,

( ) ( ) ( ), , , ,1d d2i 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

( ), ,1d d2i 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 εΓ

+ −∫ ∫ ∫ (15.23)

183

Page 183: Notes Dixit

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 dA A

I A I A I IΓ 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 2pI w

I p w w w w w

I t wI 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

Page 184: Notes Dixit

( )

( )

( )

( )

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

ev

e

e

v

vN N N Nv

N vv 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

ev v

ww N

w⎧ ⎫

= =⎨ ⎬⎩ ⎭

vw (15.30)

The approximation for pressure is

1

21 2 3 4

3

4

e

eTp p p p

pe

e

p

p ep N N N N N pp

p

⎧ ⎫⎪ ⎪⎪ ⎪⎢ ⎥= =⎨ ⎬⎣ ⎦⎪ ⎪⎪ ⎪⎩ ⎭

(15.31)

The weight function is

ep 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 1ex N x= ⎢ ⎥⎣ ⎦ and 2

e2x 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

Page 185: Notes Dixit

11

21 1 2 3 11

31

v

b b b bv b

v

w

w N N N w N w

w

⎧ ⎫⎪ ⎪⎪ ⎪⎢ ⎥= = ⎢ ⎥⎨ ⎬ ⎣ ⎦⎣ ⎦⎪ ⎪⎪ ⎪⎩ ⎭

(15.34)

and

1

22

2 1 2 3 2232

v

b b b bv b

v

w

w N N N w N w

w

⎧ ⎫⎪ ⎪⎪ ⎪⎢ ⎥= = ⎢ ⎥⎨ ⎬ ⎣ ⎦⎣ ⎦⎪ ⎪⎪ ⎪⎩ ⎭

(15.35)

Further t1 and t2 are approximated as

112

1 1 2 3 1131

b

b b b bb b

b

t

t N N N t N t

t

⎧ ⎫⎪ ⎪⎪ ⎪⎢ ⎥= = ⎢ ⎥⎨ ⎬ ⎣ ⎦⎣ ⎦⎪ ⎪⎪ ⎪⎩ ⎭

(15.36)

12

22 1 2 3 22

32

b

b b b bb 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 21 2 3

1 1 1, ,2 2 2

b b bN N N 21ζ 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

Page 186: Notes Dixit

1

111

222

212

1 2

2 1

212

vxvx

v vx x

εε ε

ε

⎧ ⎫∂⎪ ⎪∂⎪ ⎪⎧ ⎫

⎪ ⎪⎪ ⎪ ∂⎪ ⎪= =⎨ ⎬ ⎨ ∂⎪ ⎪ ⎪⎩ ⎭

⎬⎪

⎪ ⎪⎛ ⎞∂ ∂⎪ ⎪+⎜ ⎟∂ ∂⎪ ⎪⎝ ⎠⎩ ⎭

(15.39)

( ) ( )( )( )

1

111

222

212

1 2

2 1

212

wxw

ww wx

ww wx x

ε

ε ε

ε

⎧ ⎫∂⎪ ⎪∂⎪ ⎪⎧ ⎫

⎪ ⎪⎪ ⎪ ∂⎪ ⎪ ⎪= =⎨ ⎬ ⎨ ∂⎪ ⎪ ⎪⎪ ⎪

⎪⎬⎪

⎪ ⎪⎩ ⎭ ⎛ ⎞∂ ∂⎪ ⎪+⎜ ⎟∂ ∂⎪ ⎪⎝ ⎠⎩ ⎭

(15.40)

The strain rate vector ε and ( ) wε becomes

[ ] eB vε = and ( ) [ ] evw 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 12 2 2 2 2 2

NN Nx x x

NN NBx x

N NN N N N1

1

0

x

x x x x x x

⎡ ⎤∂∂ ∂− − − − −⎢ ⎥∂ ∂ ∂⎢ ⎥

⎢ ⎥∂∂ ∂= − − − − −⎢ ⎥∂ ∂⎢ ⎥⎢ ⎥∂ ∂∂ ∂ ∂ ∂

− − − −⎢ ⎥∂ ∂ ∂ ∂ ∂ ∂⎣ ⎦

(15.42)

Further,

[ ] 11 22 1 1 0 T em B vε ε ε+ = = (15.43)

where

187

Page 187: Notes Dixit

110

m⎧ ⎫⎪ ⎪= ⎨ ⎬⎪ ⎪⎩ ⎭

(15.44)

Similarly

( ) ( ) [ ] 11 22T 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 ep

AT Te T p e

v

A

Te ev

A

w N m B v x x

w B m N p x x

w B B v x x t w t wμΓ Γ

+ −

⎡ ⎤+ =⎣ ⎦

∫ ∫ dΓ + Γ∫

(15.46)

Using the notations,

[ ]

[ ]

1 2

1 2

1 2

d d

[ ] d d

2 [ ] d d

Te ppv

AT Te T p

vp pv

A

e Tvv

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 21 1 1

nb nbne T Te 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 bb b

b

Tb bb b

b

f N N t

f N N t

Γ

Γ

= Γ

= Γ

∫ (15.49)

188

Page 188: Notes Dixit

Here ne is the no. of area elements, nb1, nb2 are the number of boundary elements on Γ1 and Γ2.

Similarly,

epeev

ww

w

⎧ ⎫⎪ ⎪= ⎨ ⎬⎪ ⎪⎩ ⎭

, e

ee

p

⎧ ⎫⎧ ⎫⎪⎪ ⎪⎪= ⎨⎨ ⎬⎬⎪ ⎪⎪ ⎪⎩ ⎭⎩ ⎭

and[ ]0 e

pvee evp vv

KK

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 21 1

........ d d ....... | | d deA

x x J ξ η+ +

− −=∫ ∫ ∫ (15.51)

1 1

2 2

x x

Jx xξ 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 21 2| |b

x xJζ 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

[ ] TW K W FΔ = T (15.54)

189

Page 189: Notes Dixit

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 tt t t

φ φφ 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

Page 190: Notes Dixit

( )( )( )

( )( )( )

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

| |Tb b bN N J dζ

+

−∫ (15.61)

it becomes

1 2(1 tan ) (tan ) 0b bf 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 kF 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

Page 191: Notes Dixit

of [K] to element of [K] to 1. thtan ( 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

Page 192: Notes Dixit

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

Page 193: Notes Dixit

( )

( )

0

1 0

1 0

rr zz

rrr r rzr z

rzz z zzr z

rv vp v vr z r r z r

rv vp v vr 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 N 0 1000 N /mk = 2100 N/mk =

194

Page 194: Notes Dixit

The equations can be written as

1

2

1 11 1 0

u Fk

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 0d 2

d d 1 0d d 2

u wEA fx x x

w u wEI EA qx 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

Page 195: Notes Dixit

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

Page 196: Notes Dixit

(16.1) 0q-DSuSq-Lu T ==

in a domain , with prescribed displacement Ω

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

u=N une (16.5)

The derivatives are calculated as

(16.6) ˆ = ne(SN)uσ

The approximate solutions u , ˆ σ 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

Page 197: Notes Dixit

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= 21

)( 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

( )212

Ω= ∫Ω dL

eee T (16.9)

and for derivatives,

21

))((2

Ω= ∫Ω dL σ

Tσ () eeeσ (16.10)

The ‘root mean square’ (RMS) error for whole domain Ω is given by

21

2

2

⎟⎟⎟

⎜⎜⎜

⎛=

Ωe

Δu L (16.11)

Similarly for derivatives

2

12 2L

⎛ ⎞⎜=⎜ ⎟⎝ ⎠

ΔσΩse ⎟ (16.12)

The error for the whole domain is given by summing the element contributions. Thus,

∑=

=m

ii

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×=ue

η (16.14)

199

Page 198: Notes Dixit

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( λpu 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( , )

2p

ue CNλ−

≤ (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

2p

pue Ch CN

≤ ≈ (16.17)

200

Page 199: Notes Dixit

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

)( ϑαNu 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

Page 200: Notes Dixit

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.

= N*σ *σ (16.20)

where *σ is improved nodal derivative and

ˆ( )dΩ

0− Ω =∫ TN *σ σ (16.21)

where is elemental FEM derivative. σ

It is intuitively obvious that is in fact a better approximation than *σ σ 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

Page 201: Notes Dixit

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

ff

dd

KKKK

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

Page 202: Notes Dixit

( ) ( )( )

jj

j

−=

nh

hh

K

K

Th n

h

f dd (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

Page 203: Notes Dixit

Figure 16.1: Superconvergent points for some 1-D and 2-D patches

and

[ ]Ta 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

iiiii yxyx ph qqaF

205

Page 204: Notes Dixit

(16.37) ( ) ( )(2

1,∑

=

−=n

iiiii 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( mm j =

(16.38) ( ) ( ) ( ) ( iiii

n

iiiii

n

i

yxyxyxyx ,,,1

,1

hTT qPaPP ∑∑

==

= )

) )

This can be solved in matrix form as

(16.39) bAa 1−=

where

and (16.40) ( ) ( iiii

n

iyxyx ,,

1PPA T∑

=

= ( ) ( iiii

n

iyxyx ,,

1h

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

Page 205: Notes Dixit

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 2xcxbaT 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

Page 206: Notes Dixit

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:

dcbacbaaa

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)

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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

( )[ ] 21

1

0

223 322 ⎟⎟

⎞⎜⎜⎝

⎛+−= ∫ dxdxxxe

LT (16.52)

The first derivative will be exact at points 32

121± 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−= erdeE

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

82

72

62

5432 yxyxxyyxxyyxfi ααααααααα ++++++++= (16.54)

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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)

222ˆˆ

LLLu-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.

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The accuracy or the quality of the error estimator is measured by the effectivity index

θ, which is defined as

ee

=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

*pC 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:

ee

ee **

+≤≤− 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.

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Page 210: Notes Dixit

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 0d

Tk qx

+ =

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.

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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,

hxuhxu

xu )()(

dd −+

≈ (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 uhxu

hxbxau +⎟⎠⎞

⎜⎝⎛ −=+= (17.2)

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Page 212: Notes Dixit

where u1 and u2 are axial displacements at the 2 end nodes. Then,

huu

xu 12

dd −

= (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= 21

uT[K]u – uTP (17.6)

Substituting equation (17.5) in equation (17.6), the total potential energy expression becomes,

Π= 21

uT P – uTP = - 21

uT 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,

uTP<uactualTP (17.8)

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Page 213: Notes Dixit

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.

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Page 214: Notes Dixit

(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/ 2x−

216

Page 215: Notes Dixit

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

Page 216: Notes Dixit

17.10 ILL-CONDITIONED SYSTEM

Consider the set of equations

⎭⎬⎫

⎩⎨⎧−

=⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡−

−00.2

00.402.100.100.100.1

yx

for which (17.9) ⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

100104

yx

and a very similar set of equations

⎭⎬⎫

⎩⎨⎧−

=⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡−

−00.2

00.401.100.100.100.1

yx

for which (17.10) ⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

200204

yx

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 Pkkkkk

(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,u P 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

Page 217: Notes Dixit

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

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡PP

uu

kk r

22

1

00

(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

Page 218: Notes Dixit

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 22 1;

1 1N Nξ 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

Page 219: Notes Dixit

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