FEM

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

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 postgraduate 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

Contents 1

Finite Element Method: A Quick Introduction 1.1 Introduction 1.2 Direct FEM Formulation of Axial Rod Problem 1.2.1 Pre-processing 1.2.2 Elemental Stiffness Matrix and Load Vector 1.2.3 Assembly Procedure 1.2.4 Application of Boundary Conditions and Solution 1.2.5 Post-processing 1.3 Direct FEM Formulation of Beam Problem 1.3.1 Pre-processing 1.3.2 Elemental Stiffness Equations 1.3.3 Assembly Procedure 1.3.4 Boundary Conditions and Solutions 1.3.5 Post-processing 1.4 Conclusions References Exercise 1

1 1 2 2 3 5 6 6 7 7 8 10 11 12 12 12 13

2

Introduction to Calculus of Variation 2.1 Introduction 2.2 Functional 2.3 Extremization of A Functional 2.4 Obtaining the Variational Form from A Differential Equation 2.5 Principle of Virtual Work 2.6 Principle of Minimum Potential Energy 2.7 Conclusions References Exercise 2

19 19 19 20 28 32 33 33 34 34

3

Some Classical Function Approximation Methods for Solving Differential Equations 3.1 Introduction 3.2 Ritz Method 3.3 Galerkin Method 3.4 The Least Square Method 3.5 Collocation Method 3.6 Sub-Domain Method 3.7 Conclusions

39 39 39 42 44 45 45 46

vi References Exercise 3

46 47

4

Ritz and Galerkin FEM Formulation 4.1 Introduction 4.2 Completeness and Compatibility 4.3 Concepts of shape Functions 4.4 Developing the Elemental Equations by Ritz method 4.5 Developing the Elemental Equation by Galerkin method 4.6 Conclusions Exercise 4

51 51 51 52 55 59 60 61

5

Some One-Dimensional C0 Continuity FEM Formulation 5.1 Introduction 5.2 Steady-State Heat Conduction 5.3 Longitudinal Deformation of A Rod 5.4 Fluid Flow Problem 5.5 Conclusion Exercise 5

63 63 63 68 71 71 72

6

Finite Element Formulation for Bending of Beams 6.1 Introduction 6.2 Galerkin FEM Formulation 6.2.1 Weak Form 6.2.2 Choosing Suitable Approximating Function 6.2.3 Hermitian Shape Function 6.2.4 Elemental Equations 6.2.5 Assembly Boundary Condition and Solution 6.3 Ritz Formulation 6.4 Summary Exercise 6

75 75 75 76 76 77 78 78 79 80 81

7

Finite Element Formulation for Trusses and Frames 7.1 Introduction 7.2 Formulation for A Truss 7.3 An Example 7.4 FEM Formulation for The Frames 7.5 Summary Exercise 7

85 85 86 89 92 93 93

vii 8

Introduction to 2-D and 3-D FEM 8.1 Introduction 8.2 Triangular Elements 8.3 Tetrahedral Elements 8.4 Rectangular Elements 8.5 Brick Elements 8.6 Governing Differential Equation for 2-D Heat Conduction 8.7 Weak Form and FEM Formulation 8.8 A Note on The Assembly in Two Dimensions 8.9 Poisson Equation In 3-D 8.10 Fluid Flow Problem 8.11 Torsion of Circular and Noncircular Cross-Section 8.12 Summary References Exercise 8

97 97 97 102 103 104 105 105 108 109 109 110 110 110 111

9

Numerical Integration 9.1 Introduction 9.2 One Dimensional Integration Formulae 9.2.1 Newton-Cotes Quadrature 9.2.2 Gauss Quadrature 9.3 Two Dimensional Integration Formulae 9.3.1 Integration over Square Region 9.3.2 Integration over Triangular Region 9.4 Conclusions References Exercise 9

113 113 113 114 116 117 117 118 119 120 122

10

Further Details on 2-D FEM 10.1 Introduction 10.2 Natural Coordinates and Iso-Parametric, Sub-Parametric and SuperParametric Elements 10.3 Four-Noded Quadrilateral Elements 10.4 Serendipity Elements 10.5 Eight-Noded Curvilinear Elements 10.6 Conclusions References Exercise 10

125 125 125

FEM Formulation for Plane Stress and Plane Strain Problems

137

11

127 128 130 131 131 131

viii 11.1 Introduction 11.2 Basic Equations 11.3 Boundary Conditions 11.4 FEM Formulation 11.5 Shape Functions 11.6 Numerical Evaluation of Elements Matrices and Vectors 11.7 Assembly of Element Matrices 11.8 Boundary Conditions and Solutions 11.9 Gradient Estimates 11.10 An Example 11.11 Summary Exercise 11

137 138 139 139 142 143 145 145 145 146 146 148

12

Free Vibration Problems 12.1 Introduction 12.2 Vibration of A Rod 12.3 Vibration of A Beam 12.4 Conclusions Exercise 12

151 151 151 152 156 157

13

Finite Element Formulation of Time Dependent Problems 13.1 Introduction 13.2 Classification of Partial Differential Equations 13.3 Time Response of A parabolic Equation 13.4 Forced Vibration Problems 13.5 Conclusions References Exercise 13

161 161 161 162 164 165 165 165

14

FEM Formulation of Plate Problem 14.1 Introduction 14.2 Thin Plate Formulation 14.3 Various Thin Plate Elements 14.3.1 Rectangular Element with Corner Nodes 14.3.2 Triangular Element with Corner Nodes 14.3.3 Quadrilateral and Parallelogram Elements 14.3.4 16 Noded Rectangular Shape Function 14.4 Thick Plate Formulation 14.5 Locking Phenomenon 14.6 An Example

167 167 167 169 169 170 171 171 171 174 174

ix 14.7 Summary and Conclusion References Exercise 14

176 176 177

15

Finite Element Formulation of 2-D Flow problems 15.1 Introduction 15.2 Discretization of The Strip 15.3 Governing Equations and Boundary Conditions 15.4 Weak Formulation 15.5 Finite Element Approximation 15.6 Finite Element Equations 15.7 Application of Boundary Conditions 15.8 Post-Processing 15.9 Conclusion Exercise 15

179 179 180 181 182 184 186 190 192 192 193

16

Error Analysis in Finite Element Method 16.1 Introduction 16.2 Error Measures 16.3 Types of Error Estimates 16.3.1 A Priori Error Estimates 16.3.1.1 h-Convergence 16.3.1.2 p- Convergence 16.3.1.3 hp- Convergence 16.3.2 Posteriori Error Estimates 16.3.2.1 ZZ Error Estimate 16.3.2.2 Residual Method 16.3.2.3 Superconvergent Patch Recovery (SPR) Technique 16.3.2.4 Higher Order Approximation of Primary Variables (HOAPV) 16.4 Error Estimates by Recovery 16.5 Conclusions Further Readings Exercise 16

197 197 197 200 200 200 201 201 201 201 202 204 207

Miscellaneous 17.1 Introduction 17.2 Difference Between FEM and FDM 17.3 Finite Element Solutions Versus Exact Solution 17.4 Accuracy of Derivatives of The Primary Variables

213 213 213 214 215

17

210 211 211 212

x 17.5 Essential and Natural Boundary Conditions 17.6 Mesh Refinement 17.7 Effect of The Geometry of A Particular Element 17.8 Solving The Problems of Fracture Mechanics using FEM 17.9 Infinite Elements 17.10 Ill-Conditioned System 17.11 Patch Test 17.12 Conclusions Exercise 17 Bibliography

215 215 216 216 217 218 219 220 220 221

Chapter 1

FINITE ELEMENT METHOD: A QUICK INTRODUCTION (Lectures 1-3)

1.1

INTRODUCTION The finite element method (FEM) is a numerical method to solve differential and

integral equations. Since the behavior of physical systems can be represented by differential equations, finite element method can be used to analyze a number of physical problems. Method originated as a technique to analyze complex structural systems. The discovery of method is often attributed to Courant (1943). The use of method in structural (aircraft) analysis was first reported by Turner et al. (1956). The method received its name from Clough (1960). In finite element method, region of interest is divided into a number of elements. Differential equations are reduced to algebraic equations by using appropriate approximations for the variables over the elements. Boundary conditions of any complexity can be applied very easily. Complicated geometries and variations of material properties pose not much problem. Hence, the method has emerged as a versatile and powerful tool of computational engineering. Aim of this chapter is to introduce the reader to the finite element methodology. For this purpose, two 1-dimensional problems have been considered- axial rod problem and beam problem. In axial rod problem, one is usually interested to find out the axial displacement of each point of the rod under the action of prescribed load, whereas in the beam problem, at each point, vertical deflection and its slope need to be found out. Thus, the axial rod problem is a onedegree freedom per node problem and the beam problem is a two degrees freedom per node problem. However, it will be seen that the finite element procedure is similar for both the problems. In fact, it is similar for any problem irrespective of its dimension and degrees of freedom. The finite element method follows the following steps: •

Pre-processing: In this step, the geometry is discretized into a number of small elements. The elements can be of different shapes. Each element is characterized by number of points called ‘nodes’ present in the element. Complete system of elements is called mesh and the process of generating the elements is called mesh generation.

•

Obtaining elemental equations: In this step, algebraic equations are obtained for each element. A number of methods can be used for this purpose. In this article, they are derived using direct FEM formulation, in which algebraic equations are obtained directly from the physics of the problem.

1

•

Assembly: In this step, the elemental stiffness equations are assembled to yield a global system of equations.

•

Application of boundary conditions: In this step, the assembled system of equations is modified by inserting prescribed boundary conditions.

•

Solution: In this step, modified global system of equations is solved to obtain solution in the form of values of primary variables at nodes, such as nodal displacements in axial rod problem and nodal deflections and slopes in beam problem.

•

Post-processing: In this step, various secondary quantities are computed from the obtained solution. For example, stresses and strains are computed from the obtained nodal displacements in axial rod problem.

1.2

DIRECT FEM FORMULATION OF AXIAL ROD PROBLEM Consider an axial rod loaded with a force P at the end (Fig. 1.1). In general, the rod may

be of variable cross-section, non-homogeneous material and may be loaded with concentrated forces at different points as well as distributed forces at different segment of the rod. However, to introduce the finite element method a trivial problem of uniform axial rod loaded with force P at the end is chosen. It is desired to find out deflections, strains and stresses at different points of the rod. A governing differential equation of the problem with axial deflection u as the independent variable and point coordinate x as the dependent variable can be obtained. In the finite element method, the differential equation is converted into algebraic equation. However, for this particular problem, the algebraic equation can directly be obtained from the physics of the problem. Hence, the methodology described here is called direct finite element formulation.

Figure 1.1: Axial rod loaded at one end

1.2.1

Pre-processing First step in the finite element is to discretize the rod into a number of small segments,

each one being called an element. For example, in Fig. 1.2, the rod has been divided into three elements. The end points of each element are called nodes. Thus in this problem, there are total

2

three elements and four nodes. Each element is designated by its two nodes and coordinates of each node are stored. This step is called pre-processing or mesh generation.

Figure 1.2: (a) Discretization of the rod (b) A typical element

1.2.2

Elemental stiffness matrix and load vector In order to obtain governing algebraic equations, deflection in each element is assumed

to be linear. This will be indeed so, if the element is composed of homogeneous material following Hook’s law, has uniform cross-sectional area and loads are only point loads acting at the ends. Fig. 1.2(b) shows a general element, with end nodes designated by i and j. The tensile strain in the element is given by,

εt =

u

j

− ui

(1.1)

h

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

σt = E

u j − ui

(1.2)

h

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,

AE

u j − ui h

= Fj

(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

εc =

ui − u

j

h

(1.4)

and the corresponding compressive stress is

σc = E

ui − u j h

(1.5)

3

Hence,

AE

ui − u j h

= − Fi

(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,

AE ⎡1 − 1⎤ ⎧u i ⎫ ⎨ ⎬= h ⎢⎣− 1 1⎥⎦ ⎩u j ⎭

⎧− Fi ⎫ ⎬ ⎨ ⎩ Fj ⎭

(1.7)

In the compact form, this can be written as

[k ] {u } = {F }

(1.8)

− 1⎤ 1⎥⎦

(1.9)

e

e

e

where

⎡ 1 [k ] = AE h ⎢− 1 e

⎣

ui ⎫ ⎬ ⎩ j⎭

{u } = ⎧⎨ u e

(1.10)

and − Fi ⎫ ⎬ ⎩ Fj ⎭

{F } = ⎧⎨ e

Compare equation (1.8) with equation for a spring loaded with force F: kx =F

(1.11)

(1.12)

In analogy with this, matrix [ke] is called elemental stiffness matrix and its elements have units N/m in SI system, {u e } is elemental displacement or primary variable vector and {Fe} is the elemental load vector. Let us observe the elemental system of equations given by equation (1.7). This system cannot be solved to yield the values of ui and uj, because of the following reasons: 1. In general, Fi and Fj are internal forces, which are unknown. 2. Even if the values of Fi and Fj are known, the elemental stiffness matrix cannot be inverted to yield solution, because this matrix is singular and its rank is 1. Physically this means that just by prescribing the values of two end forces, one cannot predict the displacement, because infinite numbers of rigid body modes are possible.

4

In order to overcome the first difficulty i.e. to get rid of internal forces, the elemental stiffness are assembled. Second difficulty is overcome by prescribing one geometric boundary conditions (i.e. prescribing axial deflection at one node). Following subsection illustrates the assembly procedure and the next subsection illustrates the application of boundary conditions.

1.2.3

Assembly procedure For the given problem, let us write the elemental equations for three elements. These

are:

AE ⎡1 − 1⎤ ⎧u1 ⎫ ⎧− F1 ⎫ ⎨ ⎬=⎨ ⎬ h ⎢⎣− 1 1⎥⎦ ⎩u 2 ⎭ ⎩ F2 ⎭

(1.13)

AE ⎡1 − 1⎤ ⎧u 2 ⎫ ⎧− F2 ⎫ ⎨ ⎬=⎨ ⎬ h ⎢⎣− 1 1⎥⎦ ⎩u 3 ⎭ ⎩ F3 ⎭

(1.14)

AE ⎡1 − 1⎤ ⎧u3 ⎫ ⎧− F3 ⎫ ⎨ ⎬=⎨ ⎬ h ⎢⎣− 1 1⎥⎦ ⎩u 4 ⎭ ⎩ F4 ⎭

(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,

⎡1 − 1 0 0⎤ ⎧u1 ⎫ ⎧− F1 ⎫ ⎪ ⎢ ⎥⎪ ⎪ ⎪ AE ⎢− 1 1 0 0⎥ ⎪u 2 ⎪ ⎪ F2 ⎪ = ⎨ ⎬ ⎨ ⎬ 0 0 0 ⎥ ⎪u 3 ⎪ ⎪ 0 ⎪ h ⎢0 ⎢ ⎥ 0 0 0 ⎦ ⎪⎩u 4 ⎪⎭ ⎪⎩ 0 ⎪⎭ ⎣0 ⎡0 ⎢ AE ⎢0 h ⎢0 ⎢ ⎣0

0 0 0 ⎤ ⎧u1 ⎫ 1 − 1 0⎥ ⎪⎪u 2 ⎪⎪ ⎥⎨ ⎬ = − 1 1 0 ⎥ ⎪u 3 ⎪ ⎥ 0 0 0 ⎦ ⎪⎩u 4 ⎪⎭

⎡0 ⎢ AE ⎢0 h ⎢0 ⎢ ⎣0

⎧ 0 ⎫ ⎪− F ⎪ ⎪ 2⎪ ⎬ ⎨ ⎪ F3 ⎪ ⎪⎩ 0 ⎪⎭

0 0 0 ⎤ ⎧u1 ⎫ 0 0 0 ⎥ ⎪⎪u 2 ⎪⎪ ⎥⎨ ⎬ = 0 1 − 1⎥ ⎪u3 ⎪ ⎥ 0 − 1 1 ⎦ ⎪⎩u 4 ⎪⎭

⎧ 0 ⎫ ⎪ 0 ⎪ ⎪ ⎪ ⎨ ⎬ F − ⎪ 3⎪ ⎪⎩ F4 ⎪⎭

(1.16)

(1.17)

(1.18)

Additions of these, yields

5

⎡1 + 0 + 0 − 1 + 0 + 0 ⎢ AE ⎢ −1 + 0 + 0 1 + 1 + 0 0 −1+ 0 h ⎢0 + 0 + 0 ⎢ 0+0+0 ⎣0 + 0 + 0

0 + 0 + 0 0 + 0 + 0⎤ 0 − 1 + 0 0 + 0 + 0 ⎥⎥ 0 +1 +1 0 + 0 −1 ⎥ ⎥ 0 + 0 −1 0 + 0 +1 ⎦

⎧u1 ⎫ ⎪u ⎪ ⎪ 2⎪ ⎨ ⎬= ⎪u3 ⎪ ⎪⎩u4 ⎭⎪

⎧− F1 ⎫ ⎪0 ⎪ ⎪ ⎪ ⎨ ⎬ ⎪0 ⎪ ⎪⎩ F4 ⎪⎭

(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, 0 0 0 ⎤ ⎧u1 ⎫ ⎧ 0 ⎫ ⎡1 ⎢ 2 − 1 0⎥⎥ ⎪⎪u 2 ⎪⎪ ⎪⎪ 0 ⎪⎪ AE ⎢− 1 ⎨ ⎬=⎨ ⎬ 2 − 1 ⎥ ⎪u3 ⎪ ⎪ 0 ⎪ h ⎢0 − 1 ⎢ ⎥ 0 − 1 1 ⎦ ⎪⎩u 4 ⎪⎭ ⎪⎩ P ⎪⎭ ⎣0

(1.20)

There are various methods to solve this linear system of equations. Solution yields, u1=0,

u2=

PL 2 PL PL , u3= , u4= 3 AE AE 3 AE

(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

ε ( 2) =

u3 − u 2 u3 − u 2 P = = (L / 3) AE h

(1.22)

P A

(1.23)

and the stress is given by

σ ( 2) = Eε ( 2) =

6

In the same way, stresses in other elements may be computed. The displacement at any point inside the element can be found by linear interpolation.

1.3

DIRECT FEM FORMULATION OF BEAM PROBLEM Consider a beam rigidly fixed at the ends and loaded in the center by a load P (Fig. 1.3).

In general, beam can be of any arbitrary cross-section loaded with any complex loading function. For the sake of simplicity, only a beam of uniform cross-sectional area is considered and deflection due to only bending is considered. Deflection due to shear is not taken into consideration.

Figure 1.3: Fixed-fixed beam with a central load

1.3.1

Pre-processing We divide the entire beam into two 2-noded equal elements (Fig. 1.4(a)). Element 1 is

composed of nodes 1 and 2 and element 2 is composed of nodes 2 and 3. We introduce here the concept of connectivity matrix, which we have not done in Section 2 in order to avoid loading lot of information in one go. Connectivity matrix is a simple representation of element-node relations, in which row indicates element number, column indicates local (elemental) node number and element of the matrix denotes global node number. Thus, the connectivity matrix for the present mesh is:

⎡1 ⎢2 ⎣

2⎤ 3 ⎥⎦

(1.24)

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

7

Figure 1.4: (a) Discretization of the beam (b) A typical element

1.3.2

Elemental stiffness equations From elementary mechanics of materials, it is known that deformation of axial rod is

characterized by axial displacement of each point, where as in beam problem, at each point vertical displacement as well as slope needs to be prescribed. Thus, a typical node in the element has two degrees of freedom, vertical deflection and slope. Fig. 1.4(b) shows a typical two nodded element. On two nodes i and j, forces Fi and Fj and moments Mi and Mj are acting. In general, Fi depends on the elastic property of the element and displacements at the two nodes. Hence Fi = k11 vi+ k12 θi+ k13 vj+ k14 θj

(1.25)

where k’s are coefficients dependent on the geometry and material of the element. Similar equation can be written for Mi, Fj and Mj. Thus, the elemental equations become ⎡k11 ⎢k ⎢ 21 ⎢k31 ⎢ ⎣k 41

k12 k13 k 22 k 23 k 32 k 33 k 42 k 43

k14 ⎤ k 24 ⎥⎥ k 34 ⎥ ⎥ k 44 ⎦

⎧ vi ⎫ ⎧ Fi ⎫ ⎪θ ⎪ ⎪M ⎪ ⎪ i ⎪ ⎪ i⎪ ⎨v ⎬ = ⎨ F ⎬ ⎪ j⎪ ⎪ j ⎪ ⎪θ j ⎪ ⎪M j ⎪ ⎩ ⎭ ⎩ ⎭

(1.26)

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 ⎧ Fi ⎫ ⎡k11 k12 ⎤ ⎧ vi ⎫ ⎢k k ⎥ ⎨θ ⎬ = ⎨ M ⎬ ⎣ 21 22 ⎦ ⎩ i ⎭ ⎩ i⎭

(1.27)

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:

Fi h 3 M i h 2 − = vi 3EI 2 EI

(1.28)

8

Fi h 2 M i h + = θi EI 2 EI

−

(1.29)

where h is the element length equal to L/2. In the matrix form, the equations can be written as, ⎡ h3 ⎢ ⎢ 3 EI ⎢− h2 ⎢⎣ 2 EI

− h2 2 EI h EI

⎤ ⎥ ⎧ F ⎫ ⎧v ⎫ ⎥ ⎨ i ⎬ = ⎨ i⎬ ⎥ ⎩ M i ⎭ ⎩θ i ⎭ ⎥⎦

(1.30)

Inverting the above equations, we obtain,

⎡12 6h ⎤ ⎧vi ⎫ ⎧ Fi ⎫ =⎨ ⎬ ⎢ 2 ⎥⎨ ⎬ ⎣6h 4h ⎦ ⎩θ i ⎭ ⎩M i ⎭

EI h3

(1.31)

Comparing this with (1.27):

k11 =

12 EI h3

k12 = k 21 =

6 EI h2

k 22 =

4 EI h

(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)

k31vi + k32 θi = Fj = -Fi = -(k11vi+k12θi)

(1.35)

Third equation of (1.26) gives: Hence, k31 = - k11

k32 = - k12

(1.36)

From fourth equation we get k41vi+k42 θi = Mj = -Mi +Fi h = -( k21vi + k22 θi) + ( k11vi + k12 θi )h

(1.37)

Solving this we get k41=6EI/h2 and k42 = 2EI/h2

(1.38)

To obtain other elements of the matrix, node i is fixed, then the third and fourth equations of (1.26) reduce to

9

⎡k 33 k 34 ⎤ ⎪⎧v j ⎪⎫ ⎧⎪ F j ⎫⎪ ⎢k ⎥⎨ ⎬ = ⎨ ⎬ ⎣ 43 k 44 ⎦ ⎪⎩θ j ⎪⎭ ⎪⎩M j ⎪⎭

(1.39)

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:

Fj h3 3EI Fj h2 2 EI

+

+

M j h2 2 EI M jh EI

=vj

(1.40)

=θ j

(1.41)

In the matrix form, the equations can be written as, ⎡ h3 ⎢ ⎢ 3 EI ⎢ h2 ⎢⎣ 2 EI

h2 2 EI h EI

⎤ ⎥ ⎥ ⎥ ⎥⎦

(1.42)

⎧⎪ F j ⎫⎪ ⎧⎪ v j ⎫⎪ ⎨ ⎬ = ⎨ ⎬ ⎪⎩ M j ⎪⎭ ⎪⎩θ j ⎪⎭

Inverting above equation, we obtain,

EI h3

⎡12 − 6h ⎤ ⎧⎪v j ⎫⎪ ⎧⎪ F j ⎫⎪ ⎬=⎨ ⎬ ⎢ 2 ⎥⎨ θ h h 6 4 − ⎪ ⎣ ⎦ ⎩ j ⎪⎭ ⎪⎩M j ⎪⎭

(1.43)

Comparing this with (1.39):

k 33 =

12 EI h3

k 34 = k 43 = −

6 EI h2

k 44 =

4 EI h

(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, 6h ⎡12 ⎢ 4h 2 EI ⎢6h h 3 ⎢⎢− 12 − 6h ⎢⎣6h 2h 2

− 12 − 6h 12 − 6h

6h ⎤ ⎥ 2h 2 ⎥ − 6h ⎥ ⎥ 4h 2 ⎥⎦

(1.46)

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

1.3.3

Assembly procedure

10

In order to perform the assembly, elemental equations can be written in global form. First elemental equation in global coordinate system is,

⎡12 6h − 12 6h ⎢ 2 2 ⎢6 h 4 h − 6 h 2 h EI ⎢− 12 − 6h 12 − 6h ⎢ h 3 ⎢6 h 2h 2 − 6h 4h 2 ⎢0 0 0 0 ⎢ ⎢⎣0 0 0 0

0 0 0 0 0 0

0⎤ ⎥ 0⎥ 0⎥ ⎥ 0⎥ 0 ⎥ ⎥ 0 ⎦⎥

⎧ F1 ⎫ ⎧v1 ⎫ ⎪M ⎪ ⎪θ ⎪ 1 ⎪ 1 ⎪ ⎪ ⎪ ⎪⎪ F2(1) ⎪⎪ ⎪⎪v 2 ⎪⎪ ⎨ ⎬ = ⎨ (1) ⎬ ⎪θ 2 ⎪ ⎪M 2 ⎪ ⎪v3 ⎪ ⎪0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩θ 3 ⎪⎭ ⎩⎪0 ⎭⎪

(1.47)

Here, superscript (1) on forces and moments indicate contribution from element 1. Second elemental equation in global coordinates is

⎡0 ⎢0 ⎢ EI ⎢0 ⎢ h 3 ⎢0 ⎢0 ⎢ ⎢⎣0

0 0

0 0

0

12

0

6h

0 − 12 0

6h

⎤ ⎥ ⎥ 6h − 12 6h ⎥ ⎥ 4h 2 − 6h 2h 2 ⎥ − 6h 12 − 6h ⎥ ⎥ 2h 2 − 6h 4h 2 ⎥⎦ 0 0

0 0

0 0

⎧0 ⎫ ⎧v1 ⎫ ⎪0 ⎪ ⎪θ ⎪ ⎪ ⎪ ⎪ 1⎪ ⎪ F2(2 ) ⎪ ⎪⎪v2 ⎪⎪ ⎪ ⎪ ⎨ ⎬ = ⎨ (2 ) ⎬ ⎪θ 2 ⎪ ⎪M 2 ⎪ ⎪v3 ⎪ ⎪ F (1) ⎪ ⎪ 3 ⎪ ⎪ ⎪ ⎪⎩θ 3 ⎪⎭ ⎪⎩M 3(1) ⎪⎭

(1.48)

Adding this system of equations, following global system of equations is obtained:

6h ⎡12 ⎢ 4h 2 ⎢6 h EI ⎢− 12 − 6h ⎢ h 3 ⎢6 h 2h 2 ⎢0 0 ⎢ ⎢⎣0 0 Note that,

1.3.4

− 12 − 6h 12 + 12

6h

0

2h 2 − 6 h + 6h

0 − 12

− 6h + 6h 4 h 2 + 4h 2 − 12 − 6h

− 6h

6h

2h

2

12 − 6h

0 ⎤ ⎥ 0 ⎥ 6h ⎥ ⎥ 2h 2 ⎥ − 6h ⎥ ⎥ 4h 2 ⎥⎦

⎧ F1 ⎫ ⎧v1 ⎫ ⎪M ⎪ ⎪θ ⎪ ⎪ 1⎪ ⎪ 1⎪ ⎪⎪ P ⎪⎪ ⎪⎪v 2 ⎪⎪ ⎨ ⎬ = ⎨0 ⎬ ⎪θ 2 ⎪ ⎪ ⎪ ⎪v3 ⎪ ⎪ F3 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩θ 3 ⎪⎭ ⎩⎪M 3 ⎭⎪

F2(1) + F2(2) = P and M2(1) + M2(2) = 0

(1.49)

(1.50)

Boundary conditions and solutions It can be verified that the rank of assembled global stiffness matrix is 4. Hence,

minimum two essential boundary conditions are required. However, in this case, we have four essential (geometric) boundary conditions: v1=θ1= v3= θ3=0

(1.51)

11

Hence unknowns for the problem are v2 and θ2 and we need only two equations. We choose third and fourth equations of equation (1.49) as the right hand side of these equations is known to us. After substituting the values of prescribed degrees of freedom, these equations reduce to,

EI h3

⎡ 24 ⎢ ⎣0

0 ⎤ ⎧v 2 ⎫ ⎧P⎫ = ⎨ ⎬ ⎨ ⎬ ⎥ 8 h 2 ⎦ ⎩θ 2 ⎭ ⎩0 ⎭

(1.52)

Solving this, we get

Ph 3 P(L / 2 ) PL3 and θ =0 2 = = v2 = 24 EI 24 EI 192 EI 3

(1.53)

Reader can verify that same values are obtained from elementary strength of materials.

1.3.5

Post-processing By finite element analysis, we get nodal deflections and slope. The task of post-

processing is to find out the slopes and deflection at any point of the beam and shear force and bending moment. Knowing the shear force and bending moment at any section of the beam, the stresses can be calculated. In Section 1.2.5, it was suggested that the displacement at any point inside the element can be found by linear interpolation of the nodal displacements. Many a times, students do the mistake of linearly interpolating the nodal deflections in a beam problem too. If you do this, you are not making use of the information of nodal slopes. With slopes and defections known at the nodes, the displacement can be expressed as a cubic polynomial in an element. The deflection at a point can be found by evaluating the value of the cubic polynomial at that point. The slope at a point can be found by finding out the value of the first derivative of the cubic polynomial. For bending moment calculation, second derivative and for shear force the third derivative of the cubic polynomial is to be calculated.

1.4

CONCLUSIONS In this chapter, finite element method has been introduced by taking the one-

dimensional problems as examples. For two and three-dimensional problems, methodology is similar. As the equations are developed element by element and then assembled, incorporation of non-homogeneous material properties becomes quite easy. The objective of the present chapter is to expose the reader with the FEM and many details have been omitted. We are trying to understand FEM as a tool to solve differential equations. Thus, the FEM can be applied to number of engineering problems. Although it originated as a technique of

12

solving elastic structure problem, of late it has been applied to plastic deformation problems also. It has been widely used for solving heat transfer, fluid mechanics and electromagnetism problems. In manufacturing area, FEM has been used to model metal forming, metal cutting and non-traditional machining processes. The problems of dynamics and vibrations are also successfully solved using finite element method.

REFERENCES 1. Clough, R. W., “The Finite Element in Plane Stress Analysis”, Proc. 2nd A. S. C. E. conf. on Electronic Computation, Pittsburgh, Pa., Sept. 1960. 2. Courant, R., “Variational Methods for the Solution of Problems of Equilibrium and Vibrations,” Bulletin of the American Mathematical Society, Vol. 49, 1943, pp. 1-23. 3.

Turner, M. J., Clough R. W., Martin, H. C. and Topp, L. J., “Stiffness and Deflection Analysis of Complex Structures”, J. Aero. Sci., Vol. 23, 1956, pp. 805-823.

EXERCISE 1 Q.1: A short rod of length l is rigidly supported at both ends and an axial load P is applied at the mid-length. Taking 2 equal finite elements, find out the displacement at the point of application of load. Also find out the support reactions. The cross-sectional area of the rod is A and Young’s modulus of elasticity E.

Figure: Q1 Q.2: The short rod (cross-sectional area A, Young’s modulus of elasticity E) shown in figure is fixed at one end, the other end being held by a spring of spring constant k. A load P is applied at the mid length. Using direct finite element formulation, find out the spring compression. (Solve by two methods. In the first method, take 2 elements in the rod and put spring force as the natural boundary condition. In the second method, taking 2 elements in the rod and treating spring as the third element apply essential boundary conditions at the both ends.

13

Figure: Q2 Q.3: A cantilever beam of length l, second moment of inertia I and Young’s modulus of elasticity E is loaded by a load P. Take only one finite element and find out the deflection and slope at the free end. Compare it with the solution obtained using strength of material’s approach. Using the fact that deflection of any point of this beam is a cubic polynomial in x (the distance of the point from the fixed end), find out the deflection at a distance of l/2 from the fixed end.

Figure: Q3 Q.4: Fourier’s law of heat conduction in a rod gives:

q = − kA

dT dx

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/mK.

14

Figure: Q4 Q.5: One end of a steel rod is fixed and other end is pulled by an unknown force F. It is known that due to application of the load the mid-length point of the rod moves by a distance of 0.2 mm. Using FEM with 2 equal elements, find out the value of unknown force. The length of the rod is 1 m, area 1cm2 and Young’s modulus of elasticity 200 GN/m2. Note that in this problem you know only one boundary condition and in lieu of the other boundary condition you have the information about mid-point displacement.

Figure: Q5 Q.6: Using direct FEM approach, find out the currents in all resistors of the circuit shown below. Treat each resistor as one element and potentials at the two ends as primary variables. Elemental equations can be derived using Ohm’s law and assembly can be carried out using Kirchhoff’s current law.

Figure: Q6

15

Q.7: Two beams of length l each are joined by a pin joint and then combined beam is subjected to a load P. Can you find the deflection at the load using FEM? Do you have to make any special comment about this problem?

Figure: Q7 Q.8: A cantilever beam of length is supported on a spring of spring constant k at its free end. Using FEM, find out the deflection of the beam if a load P is applied at the mid-length of the beam.

Figure: Q8

Problems requiring the use of computer: Q.9: The cross-sectional area of a rod varies as

⎛ A( x) = A0 ⎜ 2 − ⎝

x⎞ ⎟ l⎠

where A0 is the area at the fixed end, x is the longitudinal displacement from the fixed end and l is the length of the rod. A load P is applied at the free end and you have to find the displacement at free end using FEM. Solve this problem 10 times by discretizing the rod in 1 to 10 elements. In each element average area of the element should be taken. Plot the obtained displacement versus number of elements and comment on convergence. Take suitable numerical values for plotting the graph. Otherwise express the displacement in some non-dimensional form.

16

Figure: Q9 Q.10: One dimensional seepage through a porous medium is governed by Darcy’s law, which gives the flow in terms of the gradient of the total potentialφ. The law is similar to Fourier’s law of heat conduction, i.e.,

q = −kA

∂φ ∂x

where q is the flow, k is the permeability coefficient and A is the cross-section area of the porous medium. In the problem shown in figure, potentials on the two sides the porous medium is h1 and h2. The thickness of the porous medium is t, and permeability coefficient on left and right sides is kl and kr, variation being linear across thickness. Solve this problem using FEM. Study the convergence by taking various numbers of elements.

Figure: Q10

17

Chapter 2

INTRODUCTION TO CALCULUS OF VARIATION (Lectures 4-5)

2.1 INTRODUCTION In the previous chapter, we introduced finite element method as a method to solve differential equations. More often, the behavior of a physical system is described by governing differential equations. However, sometimes, it is convenient to derive an integral expression (called variational form), minimization or maximization of which leads to same solution as obtained by solving the governing differential equations. Given a variational form, one can obtain the governing differential equations and solve differential equations by suitable numerical method including FEM. Differential equations can also be transformed into a variational form. The branch of mathematics, which deals with transforming a variational form to differential equation form and vice versa is called calculus of variation. We will study the necessary techniques of calculus of variation in this chapter. Similar to calculus, where we are often interested in finding out a point at which a function attains minimum/maximum value, in calculus of variation, we find out the function that provides minimum/maximum value of the variational form. This chapter will provide a brief introduction to calculus of variation.

2.2 FUNCTIONAL In calculus, we come across functions. A function provides a dependent variable, whose value depends on one or many independent variables. For example, y = f (x) = x3 is a function, in which for each value of independent variable x, there is a scalar value of dependent variable y. Similarly, in function z = x2 + y2, for each value of x and y, there is a value of z. Now consider the definite integral 5

5

0

0

I = ∫ ydx = ∫ f ( x ) dx

(2.1)

Here, for each particular function, there is a scalar value of I. For example, when f(x) = 1, the value of I is 5. When f(x) = x, the value of I becomes 2.5. In literature, I is called a functional (function of function), whose value depends on independent function f (x). Following integral expression is also a functional: b

I ( y ) = ∫ F ( x, y, y′ ) dx a

(2.2)

19

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

dy . Mathematically, a functional is an operator I, dx

mapping y into a scalar value. A functional l(y) is said to be linear in y, if and only if, it satisfies the following relation:

l (α y + β z ) = α l ( y ) + β l ( z )

(2.3)

for any scalars, α and β and independent functions y and z. A functional B(y, z) is said to be bilinear, if it is linear in each of its argument y and z, i.e.,

B (α y1 + β y 2 , z ) = α B ( y1 , z ) + β B ( y 2 , z )

(2.4)

B ( y, α z1 + β z 2 ) = α B ( y, z1 ) + β B ( y, z 2 )

(2.5)

The functional is symmetric if

B ( y, z ) = B ( z , y )

(2.6)

An example of a linear functional is L

I ( u ) = ∫ u f dx 0

(2.7)

where f is a constant function and u is the independent variable function. An example of a symmetric bilinear functional is L

I ( u, v ) = ∫ E A 0

d u dv dx dx dx

(2.8)

where E and A are constant functions and u and v are independent variable function.

2.3 EXTREMIZATION OF A FUNCTIONAL ⎛ ⎝

Let I = ∫ab F ⎜ x, y,

dy ⎞ ⎟ dx be some functional. The relation between y and x is not known and dx ⎠

the problem consists of finding this relation so that I is a maximum or a minimum. Assume that y0(x) is some known relation between y and x, which extremizes I. Let another function in the neighborhood of y0(x) is denoted by y0(x) + εη(x), where η(x) is an arbitrary (but sufficiently differentiable) function of x and ε is an arbitrary small quantity. The function η(x) does not violate the geometric boundary conditions of the problem. Hence, wherever y is prescribed η is zero. The function εη(x) is called δy, the variation of y at a given x, δ being a variational operator. Variational operator δ is in many ways similar to differential operator d and has similar type of mathematical properties. However, they are conceptually different. Differential of a function dy is a first order approximation to the change in function along a particular curve,

20

while the variational of a function δy is a first order approximation to the change from curve to curve.

Figure 2.1: Variation of a function Figure 2.1 shows the plot of function y0 with solid line. Assume that the value of function at one end point is prescribed, then a general function y0(x)+εη(x) is shown by a dotted curve. If we put the general function in the functional, I will be obtained as a function of ε. The condition for extremum of this function is,

dI =0 dε

(2.9)

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

dI dε

=0

(2.10)

ε =0

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

I = ∫ F ( x, y0 + εη , y0' + εη ′ ) dx b

(2.11)

a

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, ⎛ ∂F

F ( x, y, y′ ) = F ( x, y 0 , y 0' ) + ⎜

⎝ ∂y

⎞ ∂ 2 F (εη ) ∂F ∂ 2F ∂ 2 F (εη ′ ) + + ... εη ′ ⎟ + 2 εηεη ′ ) + 2 ( 2! 2! ∂y′ ∂y∂y′ ∂y' ⎠ ∂y 2

εη +

2

(2.12)

where all the derivatives are evaluated at y0 and y'0 . Note that while expanding F by Taylor series, we treat x as fixed and y and y' as two independent variables. Once x is fixed, y and

y' 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,

21

⎫⎪ b⎧ dI ∂F ⎞ ⎛ ∂ 2 F 2 ∂ 2F ∂2F ⎞ ⎪ ⎛ ∂F η ′ ⎟ + ⎜ ε 2 η + 2ε ηη ′ + ε = ∫ ⎨0 + ⎜ η + + ...⎬ dx 2 ⎟ a ∂y′ ⎠ ⎝ ∂y ∂y′∂y ∂y′ ⎠ dε ⎩⎪ ⎝ ∂y ⎭⎪

(2.13)

Applying the condition for extremum, we get dI dε

The dI dε

b ⎛ ∂F ∂F ⎞ =0=∫ ⎜ η + η ′ ⎟ dx a y ∂ ∂y ′ ⎠ ⎝ ε =0

(2.14)

is also called the first variation of I, δI. Thus, the condition for extremizing a ε =0

2 functional is δI=0. is Similarly, d I dε 2

is called the second variation of I, δ2I, which can tell if ε =0

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

⎧ ∂F d ⎛ ∂F ⎞ ⎫ ∂F ∫a ⎩⎨ ∂y − dx ⎝⎜ ∂y′ ⎠⎟ ⎭⎬η dx + ∂y′ η b

b

(2.15) a

Thus, b

⎧ ∂F

δI = ∫ ⎨ a

⎩ ∂y

−

∂F ∂F d ⎛ ∂F ⎞ ⎫ ⎜ ′ ⎟ ⎬η dx + ′ η − ′ η = 0 dx ⎝ ∂y ⎠ ⎭ ∂y b ∂y a

(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, x2 ⎛

∫x

1

∂F d ⎛ ∂F ⎞ ⎞ − ⎜ ⎜ ⎟ ⎟η dx = 0 ⎝ ∂y dx ⎝ ∂y′ ⎠ ⎠

(2.17)

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

∂F d ⎛ ∂F ⎞ − ⎜ ⎟=0 ∂y dx ⎝ ∂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

∂F ∂F η − η =0 ∂y ′ b ∂y′ a

(2.19)

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

∂F η =0 ∂y′ b

(2.20)

22

In the same way, it can be shown that

∂F η =0 ∂y′ a

(2.21)

Eqs. (2.20-2.21) are the two boundary conditions, which must be satisfied along with the differential equation given by Eq. (2.18) for the extremization of the functional. Boundary conditions imply that at the boundaries either η should be 0 i.e. the value of y should be prescribed or ∂F should be zero. The first type of boundary condition is called geometric or ∂y ′

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:

I = ∫ ( y′ 2 + y 2 + 2 xy ) dx 1

0

(2.22)

Solution: Using Eq. (2.18), we have

2 y + 2x −

d ( 2 y′ ) = 0 i.e., y + x − y′′ = 0 dx

(2.21)

As

∂F = 2 y′ ∂y′

(2.22)

The boundary conditions are: Either y' = 0 or y is prescribed at x = 0 and 1

(2.23)

Note that the variational form of the differential equation (2.21) does not depend on the prescribed value of y at the boundaries. If F depends on several dependent variable, i.e. F = F ( y1 , y1' , y2 , y'2 ,......., yn , y'n , x) where each yi = yi (x), the analysis proceeds as before, leading to n separate but simultaneous equations for the yi(x),

∂F d ⎛ ∂F − ⎜ ∂yi dx ⎜⎝ ∂y'i

⎞ ⎟⎟ = 0, i = 1,......, n. ⎠

(2.24)

with corresponding boundary conditions. With n independent variables, we need to extremise multiple integrals of the form

I = ∫ .....∫ F ( y,

∂y ∂y ,...... , , x1 ,......., xn ) dx1.....dxn ∂x1 ∂xn

(2.25)

23

Using the same kind of analysis as before, we find that the extremising function y = y(x1, ………..,xn) must satisfy

∂F n ∂ −∑ ∂y i =1 ∂xi

will

⎞ ⎟=0 ⎟ ⎠

(2.26)

∂y . ∂xi

where y xi stands for We

⎛ ∂F ⎜ ⎜ ∂y x ⎝ i

now

b

I = ∫ F ( x, y, y' , y'' ) dx .

derive We

the can

necessary apply

the

differential

equation

for

extremizing

procedure

adopted

for

extremizing

a

dy ⎞ b ⎛ I = ∫a F ⎜ x, y, ⎟ dx , however, we will now follow a short but not so rigorous approach. We will dx ⎠ ⎝

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., (δ y )' = δ y' .

Proof: We can write,

(δ y )' =

d d dη (δ y ) = (εη ) = ε dx dx dx

(2.27)

and

dy dy dη d y dη ⎛ dy ⎞ d = +ε − =ε ⎟ = ( y + εη ) − dx dx d x dx dx ⎝ dx ⎠ dx

δ ( y' ) = δ ⎜

(2.28)

Both expressions are equal, hence proved. Property 2: Integration and variation commute i.e. δ ∫ab y ( x )dx = ∫ab (δ y )dx Proof:

δ ∫ab y ( x ) dx = ∫ab ( y + εη )dx − ∫ab ( y ) dx = ∫ab ydx + ∫ab ( εη )dx − ∫ab ydx

= ∫ab ( εη )dx = ∫ab δ ydx

24

Now, using second property, we can write

δ I = ∫ab δ F ( x, y, y′, y′′ ) dx

(2.29)

δ F ( x, y, y′, y′′ ) = F ( x, y + εη , y′ + εη ′, y′′ + ε n′′ ) − F ( x, y, y′, y′′ )

(2.30)

The first variation of F is given by

where ε is a small quantity. The expansion using Taylor series leads to

δF =

( )

∂F ∂F ∂F δy+ δ y′ + δ y′′ + O ε 2 ∂y ∂y′ ∂y′′

(2.31)

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

δF =

∂F ∂F ∂F δy+ δ y′ + δ y′′ ∂y ∂y′ ∂y′′

(2.32)

Thus, b ⎛ ∂F

∂F

∂F

⎞

δI = ∫⎜ δ y+ δ y '+ δ y′′ ⎟ dx ∂y ' ∂y′′ a ⎝ ∂y ⎠

(2.33)

Making use of the property 1, we can write b ⎛ ∂F

⎞ ∂F d ∂F d 2 (δ y ) + ( ) δ I = ∫ ⎜⎜ δ y + δ y ⎟⎟ dx ∂y ' dx ∂y '' dx 2 a ⎝ ∂y ⎠ (2.34) Integrating the second and third terms by parts, b

b ⎛ ∂F

b

⎞ d ∂F d ∂F d ∂F ∂F d (δ y ) δ I = ∫ ⎜ δ y − ( )δ y − ( ) (δ y ) ⎟ dx + δy + ∂y ' ∂y'' dx dx ∂y ' dx ∂y '' dx a ⎝ ∂y ⎠ a a

(2.35)

Integrating the third term by parts, b⎪ ⎧ ∂F

δI = ∫⎨

a⎩ ⎪ ∂y

δy−

b ∂F b d ∂F b ∂F d ⎛ ∂F ⎞ d 2 ⎛ ∂F ⎞ ⎫⎪ + δ y + δ y' − δ y (2.36) y y d x + δ δ ⎬ ⎜ ⎟ ⎜ ⎟ a ∂y'' a dx ∂y '' a ∂y ' dx ⎝ ∂y' ⎠ dx 2 ⎝ ∂y '' ⎠ ⎭⎪

At this point, note the point that if you had adopted the procedure of putting

y = y0 ( x) + εη ( x) in the functional, then the condition for extremum would have been dI dε

ε =0

b ⎛ ∂F ∂F ∂F ⎞ =0=∫ ⎜ η+ η′ + η ′′ ⎟dx a ′ ′′ ∂ ∂ ∂ y y y ⎝ ⎠

(2.37)

25

Multiplying the integral in equation (2.37) by ε, we can see that this is same as equation (2.33). Thus it is seen that the necessary condition for extremization is δ I = 0 . We can argue it in a different way also. Assume that δ I ≠ 0 . 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 b ⎡ ∂F

δI = ∫⎢

a⎢ ⎣

∂y

−

⎧ ∂F d ⎛ ∂F ⎞ ⎫ b ∂F b d ∂F d 2 ∂F ⎤ ( )+ 2 ( ) ⎥ δ y dx + ⎨ δ y' = 0 − ⎜ ⎟ ⎬ (δ y ) + a ∂y'' a dx ∂y ' dx ∂y '' ⎥⎦ ⎩ ∂y ' dx ⎝ ∂y'' ⎠ ⎭

(2.38)

Thus, the differential equation is

∂F d ⎛ ∂F ⎞ d 2 ⎛ ∂F ⎞ − ⎜ ⎟+ ⎜ ⎟=0 ∂y dx ⎝ ∂y ' ⎠ dx 2 ⎝ ∂y '' ⎠

(2.39)

The boundary conditions are

⎧ ∂F d ∂F ⎫ − ( )⎬δ y = 0 ⎨ ⎩ ∂y ' dx ∂y '' ⎭

at x = a and at x = b

(2.40)

and

∂F δ y'= 0 ∂y ''

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 δ y ' = 0 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 δ y ' = 0 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

l 1 ⎛ ⎞ I = ∫ ⎜ EIw ''2 − qw ⎟ dx ⎠ 0⎝ 2

(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

d2

−q − 0 +

dx 2

( EIw '') = 0

or

d2 dx 2

( EIw '') = q

(2.43)

which is the well known Euler-Bernoulli beam equation. The boundary conditions are [0−

d ( EIw '')]δ w = 0 at x=0, l dx

(2.44)

and

EIw '' δ w ' = 0 at x = 0, l

The term

(2.45)

d ( EIw '') represents the shear force and EIw'' the bending moment. We will not dx

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: Set1: either the shear force is zero or the differential is prescribed. Set2: either the bending moment is zero or the slope is prescribed.

27

The boundary conditions concerning the slope and deflection are called ‘geometric’ or ‘essential’ boundary conditions. Boundary conditions involving shear and bending moment are called ‘natural’ or ‘force’ boundary conditions. Boundary conditions for three differently supported beams are shown in Fig. 2.2.

EIw'' = 0

(

EIw'' = 0

)

(

d EIw'' = 0 dx

)

d EIw'' = 0 dx

Free-free beam

w=0

w=0

EIw'' = 0

EIw'' = 0

Simply-supported beam d ( EIw'' ) = 0 dx

w=0 w' = 0

EIw'' = 0

Cantilever beam

Figure 2.2: Boundary conditions for 3 different types of beams

2.4 OBTAINING THE VARIATIONAL FORM FROM A DIFFERENTIAL EQUATION Let us learn the procedure of converting a differential equation into a variational form. Consider the differential equation

L (φ ) − f = 0

(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

∫ [ L (φ ) −

f ]δφ d D = 0

(2.47)

D

We keep on manipulating equation (2.47) using integration by parts till we are able to put it in the variational form

δ ∫ [ L * (φ ) − f φ ]d D = 0

(2.48)

D

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

I (φ ) =

∫ [ L * (φ ) −

f φ ]d D

(2.49)

D

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 ⎛ du ⎞ ⎜ EA ⎟+q = 0 dx ⎝ dx ⎠

(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

At x =0, u = 0; at x =l , EA

du =P dx

(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,

∫

l 0

⎡ d ⎛ du ⎞ ⎤ ⎢ dx ⎜ EA dx ⎟ + q ⎥ δ u dx = 0 ⎝ ⎠ ⎣ ⎦

(2.52)

Integrating equation (2.51) by parts,

δ u EA

du dx

l

l

+ 0

∫ δ u qdx − 0

l

d

du

∫ d x (δ u ) E A d x d x = 0

(2.53)

0

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

29

2 l ⎡1 du ⎞ ⎛ du ⎞ ⎤ ⎛ δ − E A d x ⎢ ⎥ ∫0 ⎢ 2 ⎜⎝ d x ⎟⎠ ⎥ ∫0 δ ( q u )d x − δ ⎜⎝ E A d x u ⎟⎠ l = 0 ⎣ ⎦ l

(2.54)

Therefore, 2 ⎡ l⎡1 ⎤ ⎤ ⎛ du ⎞ q u d x P u ( l ) δ ⎢∫ ⎢ EA ⎜ − − =0 ⎥ ⎥ ⎟ 0 ⎝ dx ⎠ ⎥⎦ ⎣⎢ ⎣⎢ 2 ⎦⎥

(2.55)

Hence, the variational form is given by l

I =

∫ 0

2 ⎡1 ⎤ ⎛ du ⎞ ⎢ EA ⎜ ⎟ − q u ⎥ d x − P u (l ) ⎝ dx ⎠ ⎥⎦ ⎣⎢ 2

(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

k

d 2T =0 dx 2

(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),

∫

l 0

d 2T δ T dx = 0 dx 2

k

(2.58)

Integrating by parts

k

dT δT dx

l

− 0

∫

l 0

d dT dx = 0 (δT ) k dx dx

(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 l

⎛1

∫ δ ⎜⎝ 2 k 0

dT dT ⎞ ⎟ dx = 0 dx dx ⎠

(2.60)

or

30

l

⎛ 1 dT dT ⎞ δ∫ ⎜ k ⎟ dx = 0 2 dx dx ⎠ 0 ⎝

(2.61)

Hence, the variational form is given by 2

l

1 ⎛ dT ⎞ I = ∫ k⎜ ⎟ dx 2 ⎝ dx ⎠ 0

(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:

∂ 2T ∂ 2T + =0 ∂x 2 ∂y 2

(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,

⎛ ∂ 2T ∂ 2T ∫A ⎜⎝ ∂x2 + ∂y 2

⎞ ⎟ δ T dA = 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

δ I = ∫ ∇ 2T δ T dA = ∫ ∇. ( ∇T δ T ) dA − ∫ ∇T .∇ (δ T ) dA = 0 A

A

(2.65)

A

Applying divergence theorem,

δ I = ∫ ( ∇T .nˆ ) δ TdΓ − ∫ ∇T .∇ (δ T ) dA = 0 Γ

(2.66)

A

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

⎛ ∂T ∂δ T ∂T ∂δ T ⎞ + ⎟ dA = 0 ∂x ∂x ∂y ∂y ⎠ A⎝

δ I = − ∫ ∇T .∇ (δ T ) dA = − ∫ ⎜ A

(2.67)

Using the properties of the variational operator, we can write

31

2 2 1 ⎧⎪⎛ ∂T ⎞ ⎛ ∂T ⎞ ⎫⎪ δ I = − δ ∫ ⎨⎜ ⎟ + ⎜ ⎟ ⎬ dA = 0 2 ⎪⎝ ∂x ⎠ ⎝ ∂y ⎠ ⎪ A ⎩ ⎭

(2.68)

Thus, the variational form is 2 2 1 ⎧⎪⎛ ∂T ⎞ ⎛ ∂T ⎞ ⎫⎪ I = − ∫ ⎨⎜ ⎟ ⎬ dA ⎟ +⎜ 2 ⎪⎝ ∂x ⎠ ⎝ ∂y ⎠ ⎪ A ⎩ ⎭

(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

Wint =

∫

l

0

d du (δ u ) EA dx dx dx

(2.63)

and the external work will be l

W ext =

∫ δ u q d x + P δ u (l )

(2.64)

0

Applying the principle of virtual work

Wint − Wext = ∫

l

0

l

d du (δ u ) EA dx − ∫ δ u qdx − P δ u (l ) = 0 dx dx 0

(2.64)

The equation (2.64) is called integral form of the problem. If we treat δ as the variational operator, we can easily obtain equation (2.56) i.e., the variational form of the problem.

32

2.6 PRINCIPLE OF MINIMUM POTENTIAL ENERGY In the process of obtaining the variational form of the rod problem, we have obtained a functional whose extremization alongwith the satisfaction of essential boundary condition will provide us the solution. We indicated that this functional is actually the total potential energy of the rod composed of (a) the strain energy of elastic distortion, and (b) the potential possessed by applied loads. We could have written this functional using the principle of minimum potential energy for the stable conservative mechanical system. For a conservative mechanical system, one can express the energy content of the system in terms of its configuration, without reference to whatever deformation history or path may have led to the configuration. The statement of principle of minimum potential energy is as follows. Among all possible configurations of a conservative system satisfying internal compatibility and essential boundary conditions, those that keep the body in stable equilibrium make the potential energy minimum with respect to small admissible variations of displacement.

This principle is

applicable evenif the material behavior is non-linear. The reader may have a doubt how one can know that the functional in equation (2.56) is to be minimized and not maximized, because the functional was obtained by putting the necessary condition for extremization. For ascertaining that the functional has to be minimized, one needs to calculate second variation 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

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

1 ∏ = 2

L

∫ 0

⎛ d 2w ⎞ EI ⎜ 2 ⎟ ⎝ dx ⎠

where w

2

P dx − 2

x=0

= 0,

2

L

1 ⎛ dw ⎞ 2 ∫0 ⎜⎝ d x ⎟⎠ d x + 2 kw L

dw dx

= 0

(a)

(b)

x=0

Invoke the stationary conditions δ ∏ = 0 to derive the problem-governing differential equation and the natural boundary conditions.

Figure: Q1 Q.2: Obtain the variational form of heat transfer problem in a rod.

34

Assume A, k, r are constants.

Figure: Q2 Differential equation:

d 2T r + = 0 2 dx kA Boundary conditions: 1. Essential boundary conditions: T = 0 at x = 0 2. Natural boundary conditions

T dT = 1 at x = 1 dx 2

Q.3: A certain physical problem has the functional L

∏ =

⎛1

∫ ⎜⎝ 2 φ 0

2 ,x

⎞ − 5 0 φ ⎟d x ⎠

Essential boundary conditions are φ = 0 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

∏

p

=

D ⎧ q⎫ 2 2 ⎨( w, xx + w, yy ) − 2(1 − ν )( w, xx w, yy − w, xy ) − 2 ⎬ dx dy ∫∫ D⎭ 2 ⎩

where D is a constant called flexural rigidity. Show that the governing differential equation is ∇ 4 = q / D , where ∇ 4 is the bi-harmonic operator.

Q.5: Steady state heat conduction without heat generation is governed by the following differential equation (for domain with constant thermal conductivity):

35

∂ 2T ∂ 2T + =0 ∂x 2 ∂y 2 If the temperatures at the boundary are specified, obtain the variational form of the above equation. Q.6: Consider the following boundary value problem

d2u +u =1 dx 2 with u(0) = 0 and

0 ≤ x ≤1

du = 0 at x = 1. Convert this problem into variational form. dx

Q.7: The bending of a beam is governed by the following differential equation

d2 d 2 v dmz ( EI )+ − py = 0 zz dx 2 dx 2 dx with essential boundary conditions at x = 0 are v = v* and conditions at x = l are EI zz

dv = θ z* and natural boundary dx

d2v d d2v * = M and ( EI ) + mz = −Vy* . Where, mz is the z zz dx dx 2 dx 2

distributed moment per unit length about z-axis and p y 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 qi (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.

36

t1

L = ∫ L(q1 ,..., qn , q1 ,..., q n , t ) dt t0

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 qi only, not of the

qi . Applying the EL equation to L, we obtain Lagrange’s equations,

∂L d ⎛ ∂L = ⎜ ∂qi dt ⎝ ∂qi

⎞ ⎟, ⎠

i = 1,..., n.

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

37

Chapter 3

SOME CLASSICAL FUNCTION APPROXIMATION METHODS FOR SOLVING DIFFERENTIAL EQUATIONS (Lectures 6 and 7)

3.1 INTRODUCTION In this chapter, we will study some methods invented before the advent of FEM for solving the differential equations. Methods discussed in this chapter, approximate the function globally as a weighted summation of several linearly independent functions. The weights of various functions are found such that in some sense, the error is minimized.

3.2 RITZ METHOD The method was introduced by a German physicist and mathematician, W. Ritz in 1908 [1]. It operates directly on the variational form of the differential equation. Therefore, it is also called direct method in variational problems. If the variational form is given, start solving without bothering about Euler-Lagarange equation. If differential equation has been provided, you have to convert it to a variational form. Some persons attach the name of Lord Rayleigh with the method and call it “Rayleigh-Ritz” method. However, based on the study of history, Leissa has observed that Rayleigh method is quite different from the Ritz method and therefore Rayleigh’s name should not be attached with the method [2]. The method works as follows: •

Approximate the function as a weighted sum of linearly independent functions. The approximated function must satisfy the essential boundary conditions of the problem. The natural boundary conditions need not be satisfied. The chosen functions must be complete, in the sense that starting from the constant term, the successively higher degree terms should be taken (without missing the terms in between) from a series of complete functions. In the context of polynomials, the series 1, x, x2, x3 is complete but 1, x2, x3, x4 are incomplete. To give an example, if u is a function of x, it can be represented as n

u = ∑ aiφi i =1

(3.1)

where ϕi is the ith basis function, and ai is the corresponding weight or coefficient. •

In Eq. (3.1), either the functions must be chosen in such a way that they satisfy the essential boundary conditions, else put the boundary conditions in Eq. (3.1), which will

39

provide linear equations corresponding to each essential boundary conditions. Express the coefficients equal to number of these equations in the form of other coefficients and modify Eq. (3.1). •

Put the modified equation that satisfies the essential boundary conditions into variational form. The variational form will now be a function of the coefficients. Extremize the variational form. For this purpose take partial derivatives of the variational form with respect to the coefficients and make them 0. this will give the equations equal to the number of coefficients, which can be solved to yield the coefficients. Having known the coefficients, approximate solution can be constructed.

Example 3.1: Consider the following boundary value problem:

d 2u dx 2

+ u = 1 0 ≤ x ≤ 1 with u ( 0 ) = 0 and

du = 0 at x = 1 dx

(3.2)

Solve this equation using Ritz’s method. Solution: First we have to convert it into variational form. For this, let

⎛ d 2u

⎞

δ I = ∫01 ⎜⎜ 2 + u − 1⎟⎟ δ u dx ⎝ dx ⎠

(3.3)

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

δI =

du 1 1 ⎛ du ⎞ d δ u 0 − ∫0 ⎜ ⎟ (δ u ) dx + ∫01 uδ udx − ∫01 δ udx dx ⎝ dx ⎠ dx

(3.4)

In view of the boundary conditions and fact that variational and differential operators are commutative, the expression becomes:

δ I = − ∫01

du ⎛ d u ⎞ δ ⎜ ⎟ dx + ∫01 uδ udx − ∫01 δ udx dx ⎝ dx ⎠

(3.5)

As the variational operator behaves like a differential operator,

δI =

2

( )

⎛ du ⎞ 11 1 2 ⎜ ⎟ dx + ∫0 δ u dx − ∫0 δ udx 2 ⎝ dx ⎠ 2

1 − ∫01 δ

(3.6)

or

40

⎡ 1 1 ⎛ du ⎞2 ⎤ 1 δ I = δ ⎢ − ∫0 ⎜ ⎟ dx + ∫01 u 2 dx − ∫01 udx ⎥ 2 ⎢⎣ 2 ⎝ dx ⎠ ⎥⎦

( )

(3.7)

Therefore,

⎧⎪ 1 ⎛ du ⎞2 1 ⎫⎪ I = ∫01 ⎨− ⎜ ⎟ + u 2 − u ⎬dx ⎩⎪ 2 ⎝ dx ⎠ 2 ⎭⎪

(3.8)

Now let us solve this problem using Ritz’s method. If we approximate u by u = a + bx , then in view of the first essential boundary conditions i.e. u ( 0 ) = 0 , a = 0. Hence approximate u is bx and

du du in the variational form, = b . Putting the value of u and dx dx 1 ⎛ 1 ⎞ I = ∫01 ⎜ − b 2 + b 2 x 2 − bx ⎟dx 2 ⎝ 2 ⎠

(3.9)

Extremize this with respect to b, i.e.

(

)

b 1⎤ dI ⎡ = ∫01 −b + bx 2 − x dx = ⎢ −b + − ⎥ = 0 3 2⎦ db ⎣ This gives b = −

(3.10)

3 3 . Thus, u = − x . 4 4

Exact solution of this differential equation is

u = A sin x + B cos x + 1

(3.11)

where constants A and B are to be determined from the boundary conditions. Since u ( 0 ) = 0 ,

B = −1 . Also,

du = A cos x − B sin x dx At x = 1 ,

(3.12)

du = 0 . Hence A cos (1) = B sin (1) , or A = B tan (1) = − tan (1) .Thus, dx

u = − tan (1)sin x − cos x + 1

(3.13)

The value of exact u at x = 0.5 and 1 are

u ( 0.5) = −0.6242

u (1) = −0.8508

(3.14)

The approximate solutions at these points are

u ( 0.5) = −0.375

u (1) = −0.75

(3.15)

Now let us add one more term in the approximating function and take u = a + bx + cx 2 .

41

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

u = bx + cx 2

(3.16)

du = b + 2cx dx

(3.17)

(

(3.18)

2 1 ⎧ 1 I = ∫01 ⎨ − ( b + 2cx ) + bx + cx 2 2 ⎩ 2

) − (bx + cx2 )⎫⎬⎭dx 2

Now, we have to extremize this with respect to b and c. Thus,

{

}

(3.19)

}

(3.20)

∂I 1 = ∫ −(b + 2cx) + (bx + cx 2 ) x − x dx = 0 ∂b 0 and

{

∂I 1 = ∫ −(b + 2cx)2 x + (bx + cx 2 ) x 2 − x 2 dx = 0 ∂c 0 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

u = −1.6402 x + 0.7913 x 2

(3.23)

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,

du = −1.6402 + 1.5826 x dx

(3.24)

giving the value –0.0576 at x=1, much nearer to exact value of 0. Thus, we see that with Ritz’s procedure, the natural boundary conditions also get satisfied. 3.3 GALERKIN METHOD Boundary value problems may be solved without converting into variational form by the method proposed by Soviet Scientist B.G. Galerkin (1871-1945). Galerkin published the method in 1915 [3]. Before this, Bubnov [4] had applied this method to some specific problems, but did not give the method in general form. To recognize the work of Bubnov, some researchers call the method

42

as Bubnov-Galerkin method. Galerkin’s method works directly on the differential equation, which is called the strong form. The variational form is the weak form, because the highest order of derivative gets reduced, thus weakening the continuity requirement on primary variables. Consider a differential equation L(u)+ f = 0

(3.25)

where L is the differential operator and f is the know function. The boundary conditions may be

Bi u = qi on Si

i = 1, 2,..........

(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 n

u = ψ 0 + ∑ aiψ i i =1

(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: n

R = L( ∑ aiψ i + ψ 0 ) + f i =1

(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,

∫ Rψ i dD = 0

i = 1, 2,...., n

(3.29)

D

where D is the domain and integral is the definite integral over the domain. Eq. (3.29) provides us n simultaneous equations and we can determine n unknown coefficients, thus obtaining the approximating function. Example 3.2: Solve the differential equation of Example (3.1) using Galerkin method.

43

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

u = a + bx + cx 2

(3.30)

In view of the essential boundary conditions, a becomes 0. The natural boundary condition gives

b + 2c = 0

(3.31)

Thus, the approximating function becomes

u = c( x 2 − 2 x)

(3.32)

The residue is given by

R = 2c + c( x 2 − 2 x) − 1

(3.33)

The coefficient c is found by minimizing the weighted integral of R, the weight being (x22x). Thus, 1

2 2 ∫ ( x − 2 x)[2c + c( x − 2 x) − 1]dx

=

0

(3.34)

0

giving c= 5/6. Hence, the approximate solution is

5 u = ( x 2 − 2 x) 6

(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 Point

The value of u Galerkin method

Ritz method

Exact method

x=0.5

-0.6250

-0.6223

-0.6242

x=1.0

-0.8333

-0.8489

-0.8508

3.4 THE LEAST SQUARE METHOD In Galerkin’s method, the error is minimized by taking the weighted integral of residue. The residue can also be minimized in a least square sense giving rise to the least square method. The approximating function should satisfy both the natural and essential boundary conditions. When

44

the residue given by equation (3.28) is minimized in a least square sense, we get n simultaneous equations as follows:

∂ 2 ∫ R dD = 0 ∂ai D

i = 1, 2,........., 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, 2 ∂ 1 2 c + c ( x 2 − 2 x ) − 1 dx = 0 ∫ ∂c 0

(

)

(3.37)

or

∫ 2(2 + (x 1

0

2

)(

)

− 2 x) 2c + c( x 2 − 2 x) − 1 dx = 0

(3.38)

Solving it we get c= 5/7. Therefore, the solution is

u=

5 2 ( x − 2 x) 7

(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

u=

4 2 x − 2x 5

(

)

(3.40)

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

u=

9 2 x − 2x 13

(

)

(3.41)

Of course if we take many collocation points and approximate the function higher degree of polynomial, we may expect to get better solution even by this method.

3.6 SUB-DOMAIN METHOD In this method, the domain is divided into n sub-domains and the integration of the residue is made zero over each sub-domain to generate n equations.

Supposing

the

approximating function is given by equation (3.27). To find out the unknown coefficients, we can

45

divide the whole domain into n sub-domain, integrate the residue over each subdomain and force it to zero for each subdomain. Thus, we make the residue zero in an average sense over each subdomain. If we take the approximating function of Example (3.2) and solved it by sub-domain method, we shall get the solution as

u=

3 2 x − 2x 4

(

)

(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., Uber eine neue Methode zur Losung gewisser Variationsproblem der  Reine und Angewande Mathematik, Vol. 135, mathematischen Physik, Journal fur 1908, pp. 1-61. 2. Leissa, A. W., The historical bases of the Rayleigh and Ritz methods, Journal of Sound and Vibration, Vol. 287, 2005, pp. 961-978. 3.

Galerkin, B.G., Rods and Plates. Series occurring in various questions concerning the elastic equilibrium of rods and plates, Engineers Bulletin (Vestnik Inzhenerov), Vol. 19, 1915, pp. 897-908 (in Russian).

4. Bubnov, I.G., Report on the works of Professor Timoshenko which were awarded the Zhuranski Prize. Symposium of the Institute of Communication Engineers, No. 81, All union Special Planing office (SPB), 1913 (in Russian).

46

EXERCISE 3 Q.1: A certain problem of one dimensional heat transfer is governed by the equation

dφ d 2φ = 1 at x = 1 . Solve this + φ + 1 = 0 and boundary conditions φ = 1 at x = 0 and 2 dx dx problem by using Galerkin method. Q.2: Given a differential equation:

d 2 v0 ⎞ d2 ⎛ EI − ρ ω 2v0 = 0 ; zz 2 ⎜ 2 ⎟ dx ⎝ dx ⎠ with boundary conditions:

v0 =

dv0 = 0 at x = 0 and l (both essential). dx

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 δ v as well as the properties of the δ operator used in the derivation. b) Using the approximation n

v0 = ∑ aiφi ( x ) ( ai are unknown constants) and the Ritz method, derive i =1

the algebraic equations satisfied by the unknown constants in the following form: n

∑(k j =1

ij

− ω 2 M ij ) a j = 0 for i =1,2,….n .

State explicitly the condition to be satisfied by φi . Q.3:

The variational functional of the problem shown in Fig. Q3 is 2 ⎛ 1 ⎞ ⎛ du ⎞ I = ∫ ⎜ − AE ⎜ ⎟ + pu ⎟dx + ( Pu ) x =l / 2 0⎜ ⎟ ⎝ dx ⎠ ⎝ 2 ⎠ l

a)

Using the approximation u = a1

x (l − x ) l2

(a1 is unknown constant), express I in

terms of a1 (Note: A, E, p are constants.). b)

Find a1 using Ritz method.

47

p (distributed force)

P (point force)

l/2 l Figure: Q3 Q.4: Solve the for the following differential equation by Galerkin method:

−

d 2u − cu + x 2 = 0 dx 2

for

0 ≤ x ≤ 1,

the

boundary

conditions

being

4 du du ( x = 0 ) = 1, ( x = 1) = . dx dx 3 Q.5: Solve the following problem by Galerkin method: Differential Equation:

d ⎛ du ⎞ ⎜ u ⎟ − f = 0 for 0 < x < 1 ; dx ⎝ dx ⎠ Boundary conditions: (1) Essential: u = 2 at x =0 (2) Natural :

du =0 dx

at x=1

Take f to be a linear function of x : b1 + b2 x (b1, b2 are constants). 2 Take u = 2 + a1 x + a2 x

(a1 , a2 = constants) as the approximation for

the Galerkin method. Q.6: A certain problem of one dimensional heat transfer is governed by the equation

dφ d 2φ = 1 at x = 1 . + φ + 1 = 0 and boundary conditions φ = 1 at x = 0 and 2 dx dx Use Ritz method to solve this problem. Approximate the function by a quadratic polynomial and compare with the exact solution.

48

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

−

d 2u 4 du du − cu + x 2 = 0 for 0 ≤ x ≤ 1 for the boundary conditions: ( x = 0 ) = 1, ( x = 1) = . 2 dx dx 3 dx

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

49

Chapter 4

RITZ AND GALERKIN FEM FORMULATION (Lectures 8-10) 4.1 INTRODUCTION In Ritz FEM, the finite element equations are obtained by following a procedure similar to classical Ritz method. The difference is that in FEM, piecewise continuous functions are chosen instead of choosing a globally continuous function for the whole domain. The elemental equations can be obtained by writing the variational expression for an element, putting a continuous interpolation function in that expression and obtaining the coefficients of the interpolation function by extremizing the variational expression. The interpolation function, which approximates the actual solution should be complete and should satisfy the compatibility conditions. The following subsection explains the concept of completeness and compatibility. Galerkin FEM follows procedure similar to classical Galerkin method. Here, also the piecewise continuous approximation is employed. However, before applying the method, equations are converted in weak form.

4.2 COMPLETENESS AND COMPATIBILITY In the context of classical Ritz method, completeness means that basis functions of the approximating functions are such that if enough terms are taken, the primary variables and their derivatives can be approximated as accurately as we wish. In the context of finite elements, the set of basis functions are said to be complete if they can approximate the primary variables and their derivatives appearing in the variational form as accurately as we wish by reducing the size of element. Thus, leaving aside the computational difficulties, if the size of the elements approach 0, the exact values of the primary variables and their derivatives (upto the order appearing in variational form) should be obtained. A polynomial series is complete if it is of high enough degree and no terms are omitted. Fourier series are also complete. By high enough degree we mean that the highest order derivative appearing in the variational form can be represented. For example, if

I=

⎧

⎫⎪ − u ⎬dx ⎩⎪⎝ dx ⎠ ⎭⎪

l ⎪⎛ du ⎞ ∫0 ⎨⎜ ⎟

2

(4.1)

then u = ax + b 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

51

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

du by b + 2cx . However, u = a + cx 2 is not dx

complete. Why? After all it can represent

du by 2cx . The reason is that it cannot approximate a dx

constant derivative term. The approximate

du will always be zero at x = 0. In practice, one may dx

have a non-zero du/dx at x=0. Now, consider the following variational expression:

⎛ ⎛ d 2u ⎞ 2 ⎞ ⎛ d 3u ⎞ I = ∫ ⎜ ⎜ 2 ⎟ − qu ⎟dx − ⎜ 3 ⎟ 0 ⎜ dx ⎟ ⎠ ⎝ dx ⎠ x = l ⎝⎝ ⎠ l

(4.2)

For this, a complete approximate function will be u = a + bx + cx 2 + dx3 . We have taken cubic polynomial because we have to represent

d3 y at x = l , though inside the integrand highest order dx 3

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

C 0 continuous through the domain. This will ensure that

du is always finite inside the integral. dx

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 C 1 continuity i.e. u and

du should be same for both the elements at dx

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 C m −1 continuous. Note that for compatibility, we check the highest order of the derivative in the integral expression of the full variational form.

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 + bx . At the same time, there should be atleast C 0 continuity. If

52

we take a 2 noded element of length h, we can express the constants a and b in terms of the nodal values of the primary variables. This will automatically ensure C 0 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

u1 = a + b ( 0 ) = a

(4.3)

u2 = a + bh = u1 + bh

(4.4)

and

Substituting the value of a from Eq. (4.1) into Eq. (4.3), we get

b=

u2 − u1 h

(4.5)

Hence, by expressing the constants a and b in terms of the nodal values of the primary variable, the piece-wise continuous polynomial can be expressed as,

u = u1 +

( u2 − u1 ) x h

x ⎛ x⎞ = ⎜1 − ⎟ u1 + u2 h ⎝ h⎠

= N1u1 + N 2u2

(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, u = a + bx + cx 2 can be taken. Let the coordinates of 3 nodes be x1, x2, and x3 respectively. Then,

u1 = a + bx1 + cx12

(4.7)

u2 = a + bx2 + cx22

(4.8)

u3 = a + bx3 + cx32

(4.9)

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:

u = N1u1 + N 2u2 + N3u3

(4.10)

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: n

u = N1u1 + N 2u2 + ............ + N nun = ∑ N i ui i =1

(4.11)

53

where Ni is the ith shape function and ui is the value of primary variable at ith node. However, this procedure of obtaining the shape functions is tedious. A somewhat simpler way is to obtain the shape function on the basis of the three properties of the shape functions. One-dimensional C0 polynomial shape functions will satisfy the following 3 properties: Property 1: All shape functions of an n-noded element are polynomial of (n-1) degree. This is because for an n-noded element, interpolation function will be of (n-1) degree. It should be possible to represent a polynomial function of (n-1) degree such that its value is zero at all nodes except one node. That one node can be any one out of the n nodes. Hence, the shape functions associated with all nodes should be of the same degree. Property 2: For any shape function

Ni ( x j ) = δ ij

(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,

u = Ni ui

(4.13)

Now, at x= xi, u=ui. Hence, we have from Eq. (41.3)

ui = Ni ( xi )ui or Ni ( xi ) = 1

(4.14)

At x=xj, j≠i, the value of the u is 0. Hence, from Eq. (4.13)

0 = Ni ( x j )ui or Ni ( x j ) = 0

(4.15)

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 n

c = ∑ cNi i =1

or

n

∑ Ni = 1

i =1

(4.16)

Lagrange’s interpolation functions satisfy these properties. For n-noded element, Lagrange shape functions are given by

54

N1 =

( x2 − x)( x3 − x)( x4 − x).........( xn − x) ( x2 − x1 )( x3 − x1 )( x4 − x1 ).........( xn − x1 )

N2 =

( x1 − x)( x3 − x)( x4 − x).........( xn − x) ( x1 − x2 )( x3 − x2 )( x4 − x2 ).........( xn − x2 )

. .

(4.17)

. Nn =

( x1 − x)( x3 − x)( x4 − x).........( xn −1 − x) ( x1 − xn )( x3 − xn )( x4 − xn ).........( xn −1 − xn )

It can be easily seen that for these shape functions, first two properties are satisfied. It is also easy n

to show that ∑ N i = 1 at all nodes, in view of the property 2. However, we have to prove that i =1

n

the sum of shape functions is 1 everywhere. We argue it as follows. It is clear that ∑ N i can be i =1

of at most (n-1) degree polynomial. Thus, n

2 n −1 ∑ N i = a0 + a1 x + a2 x + .......... + an x

i =1

(4.18)

Its value at all the nodes is 1. Hence, 1 =a0 + a1 x1 + a2 x12 + .......... + an x1n −1 1 =a0 + a1 x2 + a2 x2 2 + .......... + an x2n−1

(4.19)

. . . 1 =a0 + a1 xn + a2 xn 2 + .......... + an xn n−1

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), n

∑ Ni = 1

i =1

(4.20)

throughout the element.

4.4 DEVELOPING THE ELEMENTAL EQUATIONS BY RITZ METHOD Consider the problem

d 2u + u −1 = 0 dx 2

(4.21)

55

with boundary conditions

u (0) = 0 and

du = 0 at x = 1 dx

(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: h

⎡ d 2u

⎤

δ I = ∫ ⎢ 2 + u − 1⎥δ udx = 0 0 dx ⎣ ⎦

(4.23)

Integrating by parts and using the properties of the variational operator

⎛ du ⎞

1 ⎛ du ⎞

2

1

δ I = ⎜ ⎟ δ u o − ∫ δ ⎜ ⎟ dx + ∫ δ ( u 2 )dx − ∫ δ udx = 0 0 2 0 2 0 ⎝ dx ⎠ ⎝ dx ⎠ h

h

h

h

(4.24)

Hence,

I = −∫

h

0

2

h1 h 1 ⎛ du ⎞ ⎛ du ⎞ h 2 u x − udx + ⎜ ⎟ u 0 dx+ d ⎜ ⎟ ∫ ∫ 0 0 2 ⎝ dx ⎠ 2 ⎝ dx ⎠

(4.25)

We observe that completeness upto first degree polynomial and C 0 continuity throughout the domain is enough. Thus, 2-noded elements are enough. However, we can also take higher order

C 0 elements for better accuracy. Let u e be the approximation inside an n-noded element

⎧u1 ⎫ ⎪u ⎪ ⎪⎪ 2 ⎪⎪ e u = ⎢⎣ N1 N 2 ....... N n ⎥⎦ ⎨. ⎬ = ⎢⎣ N ⎥⎦ {u ne } = ⎢⎣u ne ⎥⎦ {N } ⎪. ⎪ ⎪ ⎪ ⎪⎩un ⎪⎭

(4.26)

Putting this value in I, ⎧ ⎛ du ⎞ ⎫ ⎪− ⎜ dx ⎟ ⎪ ⎪ ⎝ ⎠0 ⎪ ⎪ 0 ⎪ ⎪ ⎪ h1 h1 h 0 ⎪ ⎪ I = − ∫ ⎣⎢u ne ⎦⎥ { N , x } ⎣⎢ N , x ⎦⎥ {u ne } dx + ∫ ⎣⎢u ne ⎦⎥ { N } ⎢⎣ N ⎥⎦ {u ne } dx − ∫ ⎣⎢u ne ⎦⎥ { N } dx + ⎢⎣u1 u2 ... un ⎥⎦ ⎨ ⎬ o 2 o 2 o . ⎪ ⎪ ⎪ . ⎪ ⎪ ⎪ ⎪⎛ du ⎞ ⎪ ⎪⎜⎝ dx ⎟⎠ ⎪ h ⎭ ⎩

(4.27)

which can also be written as

56

⎧ ⎛ du ⎞ ⎫ ⎪− ⎜ dx ⎟ ⎪ ⎪ ⎝ ⎠0 ⎪ ⎪ 0 ⎪ ⎪ ⎪ h1 h1 h 0 ⎪ (4.28) ne ne ne ne ne ne ⎪ I = −∫ ⎢⎣u ⎥⎦ { N,x } ⎢⎣ N, x ⎥⎦ {u } dx + ∫ ⎢⎣u ⎥⎦ { N} ⎢⎣ N ⎥⎦ {u } dx − ∫ ⎢⎣u ⎥⎦ { N} dx + ⎢⎣u ⎥⎦ ⎨ ⎬ o 2 o 2 o ⎪ . ⎪ ⎪ . ⎪ ⎪ ⎪ ⎪⎛ du ⎞ ⎪ ⎪⎜⎝ dx ⎟⎠ ⎪ h ⎭ ⎩ For extremizing it,

∂I =0 ∂ ⎢⎣u ne ⎥⎦

(4.29)

⎧ ⎛ du ⎞ ⎫ ⎪ − ⎜ dx ⎟ ⎪ ⎪ ⎝ ⎠0 ⎪ ⎪ 0 ⎪ ⎪ ⎪ h h h ⎪ 0 ⎪ ne ne ⎢ ⎥ − ∫ { N, x } ⎣ N , x ⎦ {u } dx + ∫ { N } ⎣⎢ N ⎦⎥ {u } dx − ∫ { N }dx + ⎨ ⎬=0 0 0 0 . ⎪ ⎪ ⎪ . ⎪ ⎪ ⎪ ⎪⎛ du ⎞ ⎪ ⎪⎜⎝ dx ⎟⎠ ⎪ h ⎭ ⎩

(4.30)

Hence,

⎧ ⎛ du ⎞ ⎫ ⎪− ⎜ dx ⎟ ⎪ ⎪ ⎝ ⎠0 ⎪ ⎪ 0 ⎪ ⎪ ⎪ h ⎪ ⎡{ N } ⎢ N ⎥ dx − h { N } ⎢ N ⎥ dx ⎤ {u ne } = − h { N }dx + ⎪ 0 ⎨ ⎬ ∫0 ⎣⎢ , x ⎣ , x ⎦ ∫0 ⎣ ⎦ ⎦⎥ ∫0 ⎪ . ⎪ ⎪ . ⎪ ⎪ ⎪ ⎪⎛ du ⎞ ⎪ ⎪⎜⎝ dx ⎟⎠ ⎪ h ⎭ ⎩

(4.31)

This is the element equation. For 2-noded elements, we have for a typical element,

⎧ h⎫ − ⎪ u ⎡ 1 ⎡ 1 −1⎤ h ⎡ 2 1 ⎤ ⎤ ⎧ 1 ⎫ ⎪ 2 ⎪⎪ ⎢ ⎢ ⎥− ⎢ ⎥⎥ ⎨ ⎬ = ⎨ ⎬ + ⎣ h ⎣ −1 1 ⎦ 6 ⎣1 2 ⎦ ⎦ ⎩u2 ⎭ ⎪− h ⎪ ⎩⎪ 2 ⎭⎪

⎧ ⎛ du ⎞ ⎫ ⎪− ⎜ ⎟ ⎪ ⎪ ⎝ dx ⎠ 0 ⎪ ⎨ ⎬ ⎪ ⎛ du ⎞ ⎪ ⎪⎩ ⎝⎜ dx ⎠⎟h ⎭⎪

(4.32)

57

Reader should verify equation (4.32) by substituting the expressions for the shape functions and their derivatives for the 2-noded elements and integrating from 0 to h. Taking 2-elements and assembling the elemental equations and putting the boundary condition that the first derivative of u vanishes at x=1, we get

⎧ du ⎫ − (0) 0 ⎤ ⎧u1 ⎫ ⎧−0.25⎫ ⎪ dx ⎪ ⎡ 1.833 −2.083 ⎪ ⎢ −2.083 3.667 −2.083⎥ ⎪u ⎪ = ⎪−0.5 ⎪ + ⎪0 ⎨ ⎬ ⎨ ⎬ ⎨ ⎬ 2 ⎢ ⎥ ⎪ ⎢⎣ 0 −2.083 1.833 ⎥⎦ ⎩⎪u3 ⎭⎪ ⎪⎩−0.25⎪⎭ ⎪0 ⎪ ⎪ ⎩ ⎭

(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-by2 matrix, we get

u2 = −0.6035 and u3 = −0.8222

(4.34)

Thus, u (1) = −0.8222 If we take 3 elements we get u (1) = −0.8377 and for 6 elements u (1) = −0.8475 . Verify it. The exact solution is

u = − tan (1)sin x − cos x + 1

(4.35)

The value of exact u at x = 0.5 and 1 are

u ( 0.5) = −0.6242

and

u (1) = −0.8508

(4.36)

We have illustrated the Ritz-FEM by a simple one-dimensional problem, but the procedure is same for all problems. For any element, we substitute the approximating function in terms of the unknown nodal displacement and extremize the function with respect to the displacement vector. When you develop the elemental equations for a differential equation containing the independent variable x, do not forget to express this in the form of local variable.

58

4.5 DEVELOPING THE ELEMENTAL EQUATION BY GALERKIN METHOD For obtaining the finite element equations by Galerkin method, one has to obtain the weak form of the differential equation. Galerkin method is a special form of weighted residual method. In the weighted residual method, we integrate the weighted residual over the domain and make it 0. Let w be the weight function. Then, for the example of Section 4.4, we have

∫

h

0

⎡ d 2u ⎤ w ⎢ 2 + u − 1⎥dx = 0 ⎣ dx ⎦

(4.37)

Integrating by parts, h

h dw du h h du w −∫ dx + ∫ wudx − ∫ wdx = 0 0 0 dx 0 0 dx dx

(4.38)

Seeing it, we know that taking

⎧u ⎫ u = ⎢⎣ N1 N 2 ⎦⎥ ⎨ 1 ⎬ ⎩u2 ⎭

(4.39)

is enough. Writing in a condensed form,

u = ⎣⎢ N ⎦⎥ {u ne }

(4.40)

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

w = ⎢⎣ w ne ⎥⎦ { N } (4.41) Putting equations (4.40-4.41) in equation (4.38),

⎧ ⎛ du ⎞ ⎫ ⎪− ⎜ dx ⎟ ⎪ ⎪ ⎝ ⎠0 ⎪ ⎪. ⎪ h h ⎪ ⎪ h ⎢ ne ⎥ ⎢ wne ⎥ ⎨. − ∫ w { N , x} ⎢⎣ N , x ⎥⎦ {u ne }dx + ∫ ⎢ wne ⎥{N} ⎢⎣ N ⎥⎦ {u ne }dx − ∫ ⎢ wne ⎥{N}dx = 0 (4.42) ⎬ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 0 0 ⎪. ⎪ 0 ⎪ ⎪ ⎪⎛ du ⎞ ⎪ ⎪⎜ dx ⎟ ⎪ ⎩⎝ ⎠h ⎭ As the nodal values of the weights are arbitrary, we get

59

⎧ ⎛ du ⎞ ⎫ ⎪ − ⎜ dx ⎟ ⎪ ⎠0 ⎪ ⎪ ⎝ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎡ h { N } ⎢ N ⎥ d x − h { N } ⎢ N ⎥ d x ⎤ {u ne } = − h { N }d x + ⎪ 0 ⎨ ⎬ ,x ⎣ ,x ⎦ ∫0 ⎣ ⎦ ⎦⎥ ∫0 ⎣⎢ ∫0 ⎪ . ⎪ ⎪ . ⎪ ⎪ ⎪ ⎪⎛ du ⎞ ⎪ ⎪⎜⎝ d x ⎟⎠ ⎪ h ⎩ ⎭

(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

d2 y dy +5 + 7y = 0 2 dx dx

(4.44)

does not have a variational form. For solving this equation by FEM, we can use Galerkin procedure but not the Ritz procedure. 4.6 CONCLUSIONS In this chapter two approaches of finite element formulation have been described- Ritz FEM and Galerkin FEM. If the varaitional form is given, then it is convenient to apply Ritz FEM. If the differential equation has been provided, then it is convenient to apply Galerkin FEM. One can convert the problem from a variational form to differential form and vice versa and use any method. However, for certain differential equations, the variational form does not exist at all.

60

EXERCISE 4 Q.1: Solve the following boundary value problem by Ritz and Galerkin FEM: Differential Equation:

d2 y + 5 xy = 7 dx 2

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:

d2 y dy +5 + 7y = 0 2 dx dx 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

⎛ du ⎞ Minimize I = ∫ 5 ⎜ ⎟ dx − 10u (1) ; 0 ⎝ dx ⎠ 1

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:

⎛ ⎛ du ⎞2 ⎞ Minimize I = ∫ ⎜ ⎜ ⎟ − 5u ⎟ dx ; 0 ⎜ dx ⎟ ⎝⎝ ⎠ ⎠ 1

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:

d2 y Differential equation: 2 + 5 δ ( x − 0.5) = 0 dx

0 ≤ x ≤1

Boundary conditions: At x =0, y =0 and at x =1,

dy = 0. dx

Take 2 2-noded elements.

61

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

q = −k

∂T ∂x

(5.1)

where q is the heat flow per unit area (heat flux) in direction x, T is the temperature and k is the thermal conductivity. The boundary conditions are at x=0, T=T0 and at x=xL, q=qR. For obtaining the governing differential equation for this problem, we take an infinitesimal small element of length ‘dx’ as shown in Fig. 5.2 and obtain the heat balance. In steady-state the heat generated will be equal to net heat coming out of the rod. If Q is heat generated per unit volume, then the

 dx , where A is the average cross-sectional area of the rod. heat generation in the element is QA 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 crosssection, 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

63

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

Aq +d(Aq)

Aq

dx Fig. 5.2 Heat transfer from an infinitesimal element

Thus,

Aq + d( Aq ) - Aq - hc (T f - T ) pdx = Q Adx

(5.2)

The above equation provides us

d  − h (T − T ) p = 0 ( Aq ) − QA c f dx

(5.3)

Using Eq. (5.1), we can write

64

d ⎛ dT ⎞  ⎜ −kA ⎟ − QA − hc (T f − T ) p = 0 dx ⎝ dx ⎠

(5.4)

Thus, the governing differential equation is

d  + h (T − T ) p = 0 ( AkT , x ) + QA c f dx

(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

h ⎧1 1 dT ⎫  ∏ = ∫ ⎨ AkT, x2 + hc pT 2 − QAT + hc pT f ⎬dx − kA T 0 2 dx ⎩2 ⎭

(

)

h

0

(5.6)

Considering the completeness and compatibility, a 2-noded element with Lagrange shape functions is good enough. Putting

T = [ N ]{Te } T , x = [ N , x ]{Te }

(5.7) (5.8)

in the variational form, we get

∏e =

T T h h 1 1 {Te }T ∫ [ N , x ] Ak [ N , x ] dx{Te } + {Te }T ∫ [ N ] hc p [ N ] dx{Te } 0 0 2 2 dT ⎫ ⎧ kA ⎪ h dx ⎪⎪  [ N ]T dx − {T }T h h pT [ N ]T dx + [T T ] ⎪⎨ − {Te }T ∫ QA ⎬ 1 2 e ∫0 c f 0 ⎪ − kA d T ⎪ dx ⎭⎪ ⎩⎪

(5.9)

Minimizing with respect to {Te}T, we get elemental equations:

65

0=

T

∫0 [ N , x ] h

Ak [ N , x ] dx{Te } + ∫

T

[N ] 0 h

hc p [ N ] dx{Te }

dT ⎧ kA ⎪ h dx  [ N ]T dx − h h pT [ N ]T dx + ⎪⎨ − ∫ QA c f ∫ 0 0 ⎪ − kA d T ⎪⎩ dx

⎫ ⎪ 0 ⎪ ⎬ ⎪ ⎪ h⎭

(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 ⎛ dT ⎜ kr dr ⎝ dr

⎞ ⎟ =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, h

Rj = ∫w 0

d ⎛ dT ⎜ Rj + r k dr ⎝ dr

(

)

⎞ ⎟ dr = 0 ⎠

where Rj is the radius of jth node and r is the local coordinate. The integration by part yields

⎛ dT ⎞ w Rj + r ⎜ k ⎟ ⎝ dr ⎠

(

)

h

h dw

−∫ 0

0

dT Rj + r)k ( dr dr

dr = 0

66

Observing the weak form, we assess that Completeness in T , T ' and continuity of T is adequate. Thus, we can approximate the temperature as

⎡ r r⎤ T = a + br = ⎢1 − , ⎥ ⎣ h h⎦

⎧⎪T j ⎫⎪ ⎨ ⎬ ⎪⎩Tk ⎪⎭

Therefore,

dT ⎡ 1 1 ⎤ = − , dr ⎢⎣ h h ⎥⎦

⎪⎧T j ⎪⎫ ⎨ ⎬ ⎩⎪Tk ⎭⎪

Putting this approximation in the weak form, h

(

∫ k Rj + r

0

) {N '} ⎡⎣ N ' ⎤⎦ dr {T }ne

⎧ ⎪ R1k ⎪ =⎨ ⎪R k ⎪⎩ 2

⎫ ⎪ 1 ⎪ ⎬ dT ⎪ dr 2 ⎪⎭

dT dr

or

⎡1 2 h k ⎢⎢ 1 ⎢⎣ − h 2

⎤ h ⎞ ⎧⎪T j ⎫⎪ h2 ⎥ ⎛ ⎜ Rj + ⎟h ⎨ ⎬ = ⎥ 1 2 ⎠ ⎪⎩Tk ⎭⎪ ⎝ h 2 ⎥⎦

− 1

' ⎪⎧− R1k T1 ⎪⎫ ⎨ ⎬ ' ⎪⎩ R2 k T2 ⎪⎭

or

⎛ R j 1 ⎞ ⎡1 + ⎟ ⎢ k⎜ ⎝ h 2 ⎠ ⎣ −1

− 1⎤ ⎧⎪T j ⎫⎪ 1 ⎧⎪ H j ⎫⎪ ⎨ ⎬= ⎨ ⎬ 1 ⎥⎦ ⎩⎪Tk ⎭⎪ 2π ⎩⎪− H k ⎭⎪

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

⎡3 −3 2 2 ⎢ ⎢− 3 3 +5 2 2 ⎢ 2 ⎢0 −5 ⎢⎣ 2

0 ⎤ ⎥ −5 ⎥ 2⎥ 5 ⎥ 2 ⎦⎥

⎧T1 ⎫ ⎪ ⎪ 1 ⎨T2 ⎬ = ⎪T ⎪ 2π ⎩ 3⎭

⎧ H1 ⎫ ⎪ ⎪ ⎨0 ⎬ ⎪− H ⎪ ⎩ 3⎭

Let us impose the boundary conditions:

T1 = Ta*

T3 = Tb*

Applying boundary condition, we get

3 8 5 − Ta* + T2 − T0* = 0 2 2 2

67

T2(

FEM )

= 0.375 Ta* + 0.625 T0*

The exact solution is given by

T( exact ) = Ta* +

Tb* − Ta* ln ⎛⎜ R ⎞⎟ Ra ⎠ R ln ⎛⎜ b ⎞⎟ ⎝ ⎝ Ra ⎠

Thus,

T2 ( Exact ) = 0.415 Ta* + 0.585 Tb* If T1 = Ta* =2000C

The FEM solution is T2 141.5 0C

T0* =100 0 C

Exact solution is T2 136.90 0C.

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 crosssectional 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 ⎛ du ⎞ EA ⎟ + q = 0 dx ⎜⎝ dx ⎠

(5.12)

Boundary conditions are: (i) At x = 0, u = 0

(5.13)

68

du =0 dx

(ii) At x = L,

(5.14)

You may easily derive this differential equation by writing the expression of the total potential energy and applying Euler-Lagrange equations. Let y e be the approximate solution of u for an element e whose end coordinates are xe and xe +1 . Taking weighted residual with respect to the weight function w and obtaining the weak form, we get xe +1

∫

xe

⎛ d ⎛ dy e w ⎜ ⎜ EA ⎜ ⎜ dx ⎝ dx ⎝

⎞ ⎞ ⎟⎟ + q ⎟ dx = ⎟ ⎠ ⎠

xe +1

∫

xe

⎡ dw ⎤ dy e dy e EA + qw⎥ dx + w EA ⎢− dx dx ⎢⎣ dx ⎥⎦

xe +1

=0 xe

(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 (00. Thus, the Laplace equation is of elliptic type. Consider one dimensional transient heat conduction equation given by

∂ 2T 1 ∂T − =0 ∂x 2 α ∂t

(13.3)

where t is the time and α is the thermal diffusivity. Here, a = 1, b = 0, h = 0 . Therefore, abh2=0. Hence, this equation is of parabolic type. For two-dimensional transient heat conduction

161

∂ 2T ∂ 2T 1 ∂T + − =0 ∂x 2 ∂y 2 α ∂t

(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

EA

∂2u ∂2u + ρ A + q(t ) = 0 ∂x 2 ∂t 2

(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

{u}0 = {u0 }

(13.7)

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

(1 − α ){u}s + α {u}s +1 =

{u}s +1 − {u}s Δts +1

for 0 ≤ α ≤ 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,

{u}s +1 = {u}s + Δt ⎡⎣(1 − α ){u}s + α {u}s +1 ⎤⎦

for 0 ≤ α ≤ 1

(13.9)

Let us consider the different cases, Case1: α = 0, is called the forward difference (or Euler) scheme. It is conditionally stable and the order of accuracy is O ( Δt ) . It is called explicit scheme, because the value of u at time increment s+1 can be obtained from the known information at time increment s.

162

Case 2: α = 1, is called the backward difference scheme. It is stable and the order of accuracy is O ( Δt ) . Case 3: α =

(

is O ( Δt )

2

). 2 3

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

Case 4:

(

1 , is called Crank-Nicholson scheme. It is also stable and the order of accuracy 2

)

O ( Δt ) . 2

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

M {u}s + [ k ]s {us } = { F }s

(13.10)

M {u}s +1 + [ k ]s +1 {us +1} = { F }s +1

(13.11)

We can see that

Δt s +1α [ M ]{u}s +1 + Δt s +1 (1 − α ) [ M ]{u}s = [ M ] ({u}s +1 − {u}s )

(13.12)

Substituting equations (13.10-13.11) in it,

Δts +1α

({ F }

s +1

− [ k ]s +1 {u}s +1 + Δts +1 (1 − α ) { F }s − [ k ]s {u}s = [ M ] ({u}s +1 − {u}s ) (13.13)

)

(

)

Rearranging, ⎡⎣[ M ] + α Δt s +1 [ k ]s +1 ⎤⎦ {u}s +1 = ⎡⎣ M − (1 − α ) Δt s +1 [ k ]s +1 ⎤⎦ {u}s + Δt s +1 ⎡⎣α { F }s +1 + (1 − α ){ F }s ⎤⎦

(13.14)

Note that for α = 0 we obtain left hand side as [ M ]{u}s +1 . when the mass matrix is diagonal, the equation becomes explicit and we can solve for {u}s +1 directly without inverting. For all numerical schemes, in which α < 1/ 2 , the α - family of approximation is stable only if the time step satisfied the following (stability) condition.

Δt < Δtcr ≡

2 (1 − 2α ) λ

(13.15)

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

163

13.4 FORCED VIBRATION PROBLEMS If the frequency of excitations applied to a structure is less than roughly one-third of the structure’s lowest natural frequencies of vibration, then the problem is quasistatic. Two types of dynamic problems are wave propagations and structural dynamics problems. Time history can be obtained by model methods and direct integration method. The finite element formulation provides the following equations:

M

t +Δt

U + C

t +Δt

U + k

t +Δt

U =

t +Δt

R

(13.16)

The Newmark method is an extension of the linear acceleration method. The following assumptions are used:

U = t U + ⎡⎣(1 − δ ) tU + δ

t +Δt

U ⎤⎦ Δt

t +Δt

(13.17)

⎡⎛ 1 ⎤ ⎞ U = tU + tU Δt + ⎢⎜ − α ⎟ tU + α t +ΔtU ⎥ Δt 2 ⎠ ⎣⎝ 2 ⎦

t +Δt

(13.18)

where α and δ are parameters that can be determined to obtain integration accuracy and stability. When δ = 1/ 2, α = 1/ 6, it corresponds to linear acceleration method. For constant average acceleration, δ = 1/ 2, α = 1/ 4 . The average acceleration is

1 t  t +Δt  U+ U . 2

(

)

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.

 and 0 U.  2. Initialize 0 U, 0 U, 3. Select time step Δt and parameters α and δ and calculate integration constants.

δ ≥ 0.50; a0 =

1 δ ; a1 = 2 α Δt α Δt

a4 =

δ − 1; α

a5 =

α ≥ 0.25 ( 0.5 + δ ) a2 =

2

1 1 ; a3 = − 1; 2α α Δt

Δt ⎛ δ ⎞ ⎜ − 2 ⎟ ; a6 = Δt (1 − δ ) ; a7 = δ Δt 2 ⎝α ⎠

ˆ: K ˆ = K + a M+a C. 4. Form effective stiffness matrix K 0 1 ˆ: K ˆ = LDLT . 5. Triangularize K B. For each time step:

164

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

Rˆ =

t +Δt

 + a t U)  + C (a t U + a t U  + a t U)  R + M(a0 t U + a2 t U 3 1 4 5

2. Solve for displacements at time t + Δ t : t+Δt

LDLT

ˆ U= t+Δt R.

3. Calculate accelerations and velocities at time t + Δ t : t +Δt

 = a ( t +Δt U − t U ) − a t U  -a t U  U 0 2 3 t +Δt

 a t U+  a t +Δt U  U = t U+ 6 7

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:

∂ u ∂ (a + u ) − m = 0 ∂x ∂t ∂ (a + u ) F = ∫m dx ∂t 2

EA

2

2

2

2

l

0

2

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: Q1

165

The one-element finite element formulation of this problem leads to: ⎧a ⎫ ⎧a ⎪ ⎪ [ M ] ⎨u ⎬ + [ K ] ⎪⎨u ⎪ ⎪ ⎪ ⎩u ⎭ ⎩u 1

1

2

2

⎧F ⎫ ⎪ du ⎪ ⎪ ⎬ = ⎨− EA dx ⎪ ⎪ ⎭ ⎪0 ⎩

1

⎫ ⎪ ⎪ ⎬ ⎪ ⎪⎭

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 α < 1/ 2 , the α - family of approximation is stable

only if the time step satisfied the following (stability) condition.

Δt < Δtcr ≡

2 (1 − 2α ) λ

Prove it.

166

Chapter 14

FEM FORMULATION OF PLATE PROBLEM (Lecture 29-31)

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

14.2 THIN PLATE FORMULATION The thin formulation is based on the classical Kirchhoff’s theory. Here, the main assumptions are that the points on the mid-surface z=0 move only in the z direction as the plate deforms in bending and a line that is straight and normal to the mid-surface before loading is assumed to remain straight and normal to the mid-surface after loading (see

167

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,

u = − zw, x

v = − zw, y

(14.1)

Hence, the strain-displacement relations are given by,

ε x = u, x = − zw, xx ε y = v, y = − zw, yy

(14.2)

γ xy = u , y +v, x = −2 zw, xy It can be easily shown [Cook et al.,1989] that for a plate with zero initial strain,

{M } = −[ Dk ]{κ }

(14.3)

where

{M } = [M x

M xy ]T

My

t/2

M x = ∫ σ x z dz

and t/2

t/2

−t / 2

−t / 2

M y = ∫ σ y z dz

−t / 2

{κ } = [w, xx

w, yy

2w, xy ]T

∫ σ xy z dz

(14.4) (14.5)

For isotropic material,

⎡D [ Dk ] = ⎢⎢νD ⎢⎣0

νD D 0

⎤ Et 3 ⎥ where D = ⎥ 12(1 − ν 2 ) (1 − ν ) D / 2⎥⎦ 0 0

(14.6)

D is called flexural rigidity of the plate. The strain energy expression is given by

1 T {κ } [ Dk ]{κ } dA A2

Now,

U=∫

(14.7)

w = [ N ]{d }

(14.8)

where {d}=array containing nodal degrees of freedom in the element Then,

{κ } = [ B]{d}

(14.9)

where [B] is gradient matrix obtaining by operating on [N] Following Ritz FEM procedure, the following stiffness matrix is obtained

168

[k ] = ∫ [ B]T [ Dk ][ B]

dA

(14.10)

A

The load vectors can be found in the similar manner. u=-zw x , Midsurface

w,x z, w

w dx z t/2

p o

P z x, w

w

w,x

0

x,u

t/2

w,x (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]

[

w = 1 x y x 2 xy y 2 x 3 x 2 y xy 2 y 3 x 3 y xy 3

] {a}

(14.11)

169

where vector {a} contains 12 coefficients, which must be exchanged for the twelve nodal degrees of freedom {d} by the usual process. If the d. o. f. at each nodes are w, ∂w / ∂x and

∂w / ∂y , then shape functions at Ith nodes are given by

[

]

1 (ξ0 +1)(η0 +1)(2 + ξ0 +η0 −ξ 2 −η 2 ), bηi (ξ0 +1)(η0 +1)2 (1−η0 ) , aξi (ξ0 +1)2 (η0 +1)(ξ0 −1) (14.12) 2 with

ξ = ( x − x c ) / a, η = ( y − y c ) / b ξ 0 = ξ .ξ i η 0 = η.η i 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 a

w3

3 b x

b

3

y a w,x3

w,y3

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

[ w = [1 x

w = 1 x y x2

y 2 x 3 x 2 y xy 2 y 3

y x 2 xy y 2 x 3 x 2 y + xy 2

] {a} y ] {a} 3

(14.13) (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

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. 3

w3

y

w,y3

w,x3 w,x1 1 w1

w2 2 w,x2 w,y2

w,y1 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 ∂ 2 w / ∂x∂y 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.

14.4 THICK PLATE FORMULATION Thick plate formulation is based on Mindlin plate theory. The main assumptions of the Mindlin plate theory are: 1. Stress component along the normal to the mid-plane of the plate is negligible. (Same as in Kirchhoff plate theory.) 2. Transverse displacement does not vary in the thickness direction. (Same as in Kirchhoff plate theory).

171

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: u = zθ y

εx = z

v = zθ x

∂θ y ∂x

εy = z

∂θ x ∂y

∂θ y ⎞ ⎛ ∂θ x + ⎟ ∂y ⎠ ⎝ ∂x ∂w = + θx ∂y ∂w = +θy ∂x

γ xy = z ⎜ γ

yz

γ xz

(1 4 .1 5 )

where θx and θy are the counterclockwise rotations about the y and negative x axis respectively. Since σz=0, the stress-strain relations for the isotropic linearly elastic material become,

σx =

E (ε x + νε y ) 1 −ν 2

σy =

E (ε y + νε x ) 1 −ν 2

τ xy =

E γ xy 2(1 + ν )

τ yz = α s

E γ yz 2(1 + ν )

τ zx = α s

E γ zx 2(1 + ν )

(14.16)

where E is the Young modulus, ν is the Poisson ratio and α s is the shear correction factor. The strain energy V is given by,

V =

1 2

∫∫

t/2

A −t / 2

(σ x ε x + σ y ε y + σ z ε z + τ xy γ xy + τ yz γ yz + τ xz γ xz )dzdA

(14.17)

where t is the thickness of the plate and A is the area of the plate. Substituting the expressions of stresses and strains in the above equations, the expression for V becomes:

172

Et 1 V = ∫A 2 1 −ν 2

⎡ t 2 ⎛ ∂θ ⎞ 2 νt 2 ∂θ y ∂θ t 2 ⎛ ∂θ y x x ⎜⎜ ⎢ ⎜ + ⎟ + x y x 12 6 12 ∂ ∂ ∂ ⎢⎣ ⎝ ⎠ ⎝ ∂y ⎡ t 2 ⎛ ∂θ ⎢ ⎜⎜ x ⎢⎣12 ⎝ ∂y

Et 1 + ∫A 2 2(1 + ν )

⎞ ⎟⎟ ⎠

2

⎤ ⎥ dA ⎥⎦

2 2 ⎞ t 2 ⎛ ∂θy ⎞ t 2 ∂θ x ∂θy ⎤ ⎟⎟ + ⎜⎜ ⎟⎟ + ⎥ dA 6 ∂y ∂x ⎥ ⎠ 12 ⎝ ∂y ⎠ ⎦

2 ⎡ 2 ⎛ ∂w ⎞ 2 ∂w ∂w ⎤ 1 Et ⎛ ∂w ⎞ 2 + ∫A + θ x + ⎜ ⎟ + 2θ x α s ⎢θ y + ⎜⎜ ⎟⎟ + 2θ y ⎥ dA ∂y ∂x ⎥ 2 2(1 + ν ) ⎢ ⎝ ∂x ⎠ ⎝ ∂y ⎠ ⎣ ⎦

(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, i =n

w = ∑ N i wi

(14.19)

i =1

i =n

θ x = ∑ N iθ xi

(14.20)

i =1

i =n

θ y = ∑ N iθ yi

(14.21)

i =1

where Ni is Lagrange shape function for ith node. Following the standard procedure, the elemental stiffness matrix can be obtained as,

[k ] = ∫ [B ] [D ][B ]dA + ∫ [B ] [D ][B ]dA T

ij

Ae

bi

T

b

bj

Ae

si

s

sj

(14.22)

where

⎡0 N i , x 0 0 0 ⎤ ⎢ ⎥ [Bbi ]= ⎢0 0 N i , y 0 0 ⎥ ⎢0 N N 0 0 ⎥ i, y i,x ⎣ ⎦ ⎡ N i,x N i

0

⎣⎢ N i , y 0

Ni

[Bsi ]= ⎢

0⎤ ⎥ 0 0 ⎦⎥

0

(14.23)

(14.24)

173

[Db ]=

Eh 3 12 1 − ν 2

[Ds ]=

αs

(

)

⎡ ⎢1 ν ⎢ ⎢ν 1 ⎢ ⎢0 0 ⎣

0 0 (1 − ν ) 2

Eh ⎡ 1 2(1 + ν ) ⎢⎣0

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

0 ⎤ 1 ⎥⎦

(14.25)

(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 4noded 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 σ x and σ y 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

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

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

EXERCISE 14 Q.1: The governing equation for thin plate bending is given by

∂w ∂w ∂w q +2 + = ∂x ∂x ∂y ∂y D 4

4

4

4

2

2

4

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

∂ψ ∂ψ ∂ψ +2 + =0 ∂x ∂x ∂y ∂y 4

4

4

2

4

2

4

with boundary conditions: Essential: u=u*, v=v* on Γu * Natural: t x = t x

⎧⎪t x ⎫⎪ ⎪⎩t y ⎪⎭

and

t y = t *y

⎡σ xx

σ xy ⎤ ⎧⎪nx ⎫⎪

⎣⎢σ xy

σ yy ⎦⎥ ⎪⎩n y ⎪⎭

where ⎨ ⎬ = ⎢

⎥⎨

⎬

σ ij = − pδ ij + 2μεij u = ψ , y v = −ψ , x 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 w, ∂w / ∂x and ∂w / ∂y , then this element will be incompatible.

∂2 w , then Mathematically, prove this statement. Also, prove that if we add a degree of freedom ∂x∂y 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

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

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;

u

⎛ ∂ 2u ∂ 2u ⎞ 1 ∂p ∂u ∂u +v =− +υ ⎜ 2 + ⎟ ∂x ∂y ∂y 2 ⎠ ρ ∂x ⎝ ∂x

u

1 ∂p ∂v ∂v +v =− +υ ∂x ∂y ρ ∂y

⎛ ∂ 2v ∂ 2v ⎞ ⎜ 2 + ⎟ ∂y 2 ⎠ ⎝ ∂x

∂u ∂v + =0 ∂x ∂y

(15.1)

(15.2)

(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

∂u ∂v p + + =0 ∂x ∂y λ

(15.4)

where λ is a very large number. Ideally for incompressible flow, this number should be infinite. However, in practice, this number should be chosen depending upon the precision of computer. From this equation,

⎛ ∂u ∂v ⎞ p = −λ ⎜ + ⎟ ⎝ ∂x ∂y ⎠

(15.5)

179

This value can be substituted in the Eq. (15.1). In this chapter we will be solving Navier-Stokes equation, considering following features: 1) A mixed pressure-velocity formulation will be adopted. 2) We shall consider that inertia force is not large, therefore convective acceleration terms (the terms on the left hand side of inequality constraints) will not be considered. 3) The viscosity will not be considered constant. It will depend on deformation. Thus despite the elimination of inertial forces, the problem remains non-linear. Now we will model plane-strain steady-state process of strip drawing. Index notation has been used in the formulation of the model.

15.2 DISCRETIZATION OF THE STRIP In FE modeling of drawing process where a curved die has been chosen, mesh generation can be done as shown in the Fig. 15.1. The mesh selected for the current problem consists of 56 elements. More elements have been selected in the deformation zone. This is the preprocessing stage of the problem. For velocity approximation nine noded element is chosen and for pressure four noded element is selected. It will be seen from the weak-formulation that four noded pressure and nine noded velocity are sufficient to formulate the process.

Figure 15.1: A typical 56 element mesh and nodal structure for pressure and velocity

180

15.3 GOVERNING EQUATIONS AND BOUNDARY CONDITIONS The non-dimensionalzed governing equations for FE modeling of the plain-strain drawing problem will be,

ε11 + ε 2 2 = 0 ∂ σ 11 ∂ σ 12 + =0 ∂ x1 ∂x2 ∂ σ 21 ∂ σ 22 + =0 ∂ x1 ∂x2 where

(15.6)

(15.7)

ε1 1 , ε 2 2 are strain rate tensors in x1 and x2 directions respectively and

σ 11 , σ 12 , σ 21 , σ 22 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)

v2+ v1tanφ = 0 on BC

(15.13)

4. The die- strip interface (BC)

181

where φ is the angular position of the point on the surface. The second boundary condition will be; | ts | = μ | tn |

(15.14)

This has been shown in Fig.15.2, where ts and tn are the tangential and normal components of the stress vector t which is defined by σ i j n j = t i 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;

⎡ ⎤ ⎛ ∂ σ 11 ∂ σ 12 ⎞ ⎛ ∂ σ 21 ∂ σ 2 2 ⎞   − + + + + + ε ε w w w ( ) ⎢ ⎥ d x1 d x 2 = 0 (15.15) ⎜ ⎟ ⎜ ⎟ p 11 22 1 2 ∫A ∂x2 ⎠ ∂x2 ⎠ ⎝ ∂ x1 ⎝ ∂ x1 ⎣ ⎦ where A denotes domain of a typical area element. The above expression can be written in compact form as;

∫ ⎡⎣ − ε A

ii

w p + σ i j , j w i ⎤⎦ d x1 d x 2 = 0

(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

∫ ⎡⎣σ

⎤⎦d A =

i j , j wi

A

∫ ⎡⎣ (σ

i j wi

) , j − (σ i j wi , j )⎤⎦d A

(15.17)

A

The Eq. (15.17) becomes,

∫ − ε

i i w p dA

+

A

∫ ⎡⎣ (σ

i j wi

) , j − (σ i j w i , j )⎤⎦d A = 0

(15.18)

A

(

In the second part of the Eq. (15.18), the term σ i j wi

) , j can be written as (σ i j wi n j )

But according to Cauchy’s equation σ i j n j = ti

Thus Eq. (15.18) can be expressed as,

∫ − ε

i i w p dA

+

∫ t w d Γ − ∫ (σ i

i

Γi

A

i j wi , j

) dA = 0

(15.19)

A

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

1 σ i j w i , j ) d A = σ i j ⎡⎣ ( w i , j + w j , i ) + ( w i , j − w j , i ) ⎤⎦ d A ( ∫ ∫ 2 A

(15.20)

A

which reduces to

1 σ i j w i , j ) d A = σ i j ⎡⎣ w i , j + w j , i ⎤⎦ d A ( ∫ ∫ 2 A

(15.21)

A

The Eq. (15.21) can be written as

∫ (σ

) d A = ∫ σ i j εi j ( w ) d A

i j wi , j

A

(15.22)

A

Thus, the weighted residual becomes

∫ ε A

iiwp

dA +

∫σ A

i j εi j

( w ) d A − ∫ ti wi d Γ

=0

(15.23)

Γi

183

Equation (15.23) has first two terms in first order derivatives of velocity component, thus the weak form has been obtained for further FE modeling. Further the equation is simplified by substituting the following terms,

σ i j = − pδ i j + S i j and S ij = 2 μεij

(15.24)

Using the relationship in Eq. (15.24), we can write

σ i j εi j ( w ) = ( − pδ i j + S i j ) εi j ( w ) = − p εi i ( w ) + 2 μεi j εi

j (w )

(15.25)

The above equation can be further expanded as

σ ij εij ( w ) = − p ⎡⎣ ε11 ( w ) + ε 2 2 ( w ) ⎤⎦ + 2 μ ⎡⎣ ε1 1ε1 1 ( w ) + 2 ε1 2 ε1 2 ( w ) + ε 2 2 ε 2 2 ( w ) ⎤⎦ (15.26) Thus the entire weak form of weighted residual is

∫I A

1

dA +

∫I A

2

dA −

∫ I dΓ − ∫ I dΓ = 0 3

Γ1

4

(15.27)

Γ2

where

I1 = I2 =

[ε11 + ε22 ] w p − p ⎡⎣ ε11 ( w ) + ε 22 ( w ) ⎤⎦ + 2 μ ⎡⎣ ε11ε11 ( w ) + 2 ε12 ε12 ( w ) + ε 22 ε 22 ( w ) ⎤⎦

I 3 = t1 w1 I 4 = t2 w2 (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

⎧ v1 ⎫ ⎢ N 1 ⎨ ⎬=⎢ ⎩ v 2 ⎭ ⎢⎣ 0

0

N2

0

N3

0 ........ N 9

N1

0

N2

0

N 3 .......0

⎧ ( v )e ⎫ ⎪ 1 1 ⎪ ⎪ ( v )e ⎪ ⎪ 2 1⎪ ⎪⎪ | ⎪⎪ e ⎨ ⎬ = ⎢⎣ N v ⎥⎦ v | ⎪ ⎪ ⎪ v e ⎪ ⎪ ( 1 )9 ⎪ ⎪ e⎪ ⎩⎪ ( v 2 )9 ⎭⎪

0⎥ ⎥ N 9 ⎥⎦

{ }

(15.29) In Galerkin formulation, weight functions for velocity and pressure are approximated using same shape functions as that of velocity and pressure respectively.

⎧ w1 ⎫ ⎬= w ⎩ 2⎭

{w v } = ⎨

[ N v ]{ w v e }

(15.30)

The approximation for pressure is

p = ⎢⎣ N 1p

N 2p

N 3p

⎧ p1e ⎫ ⎪ e⎪ ⎪ p2 ⎪ N 4p ⎥⎦ ⎨ ⎬ = N p e ⎪ p3 ⎪ ⎪ e⎪ ⎩ p4 ⎭

{ } {pe} T

(15.31)

The weight function is

{w p } =

{

⎢⎣ N p ⎥⎦ w p e

}

(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,

{ x1 } =

⎢⎣ N ⎥⎦

{ x } and 1

e

{x2 } =

⎢⎣ N ⎥⎦

{x } 2

e

(15.33)

As the boundary of the element consists of three nodes, evaluation of the integrals over the boundaries Γi will be,

185

w1 = ⎢⎣ N 1b N 2b

⎧ w 1v ⎫ ⎪ 1⎪ b⎥⎪ 2 ⎪ N 3 ⎦ ⎨ w v1 ⎬ = ⎢⎣ N b ⎥⎦ w1b ⎪ 3 ⎪ ⎪⎩ w v1 ⎪⎭

w 2 = ⎢⎣ N 1b N 2b

⎧ w 1v ⎪ 2 b⎥⎪ 2 N 3 ⎦ ⎨ w v2 ⎪ 3 ⎪⎩ w v 2

{ }

(15.34)

⎫ ⎪ ⎪ b ⎬ = ⎢⎣ N b ⎥⎦ w 2 ⎪ ⎪⎭

(15.35)

and

{

}

Further t1 and t2 are approximated as

t1 = ⎣⎢ N 1b N 2b

⎧ t b1 ⎫ ⎪ 1⎪ b⎥⎪ 2 ⎪ N 3 ⎦ ⎨ t b1 ⎬ = ⎢⎣ N b ⎦⎥ t1b ⎪ 3 ⎪ ⎪⎩ t b1 ⎪⎭

{ }

(15.36)

t 2 = ⎢⎣ N 1b N 2b

⎧ t b1 ⎪ 2 b⎥⎪ 2 N 3 ⎦ ⎨ t b2 ⎪ 3 ⎪⎩ t b2

⎫ ⎪ ⎪ b ⎬ = ⎢⎣ N b ⎥⎦ t 2 ⎪ ⎪⎭

(15.37)

{ }

where t1 and t2 are the vectors of the nodal value of the traction.

N 1b =

(

1 ζ 2

2

)

− ζ , N 2b =

(

1 ζ 2

2

+ζ

),

N 3b =

(

1 1−ζ 2

2

)

(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

⎫ ⎪ ⎪ ⎪⎪ ⎬ ⎪ ⎞⎪ ⎟⎪ ⎠ ⎭⎪

(15.39)

⎧ ⎫ ∂ w1 ⎪ ⎪ x ∂ ⎪ ⎪ 1 ⎫ ε11 ( w ) ⎪⎪ ⎪⎪ ∂w2 ⎪⎪ ε22 ( w ) ⎬ = ⎨ ⎬ ∂x2 ⎪ ⎪ ⎪ 2 ε12 ( w ) ⎭⎪ ⎪ 1 ⎛ ∂ w1 ∂ w 2 ⎞ ⎪ ⎪ + ⎜ ⎟⎪ ∂ x1 ⎠ ⎭⎪ ⎪⎩ 2 ⎝ ∂ x 2

(15.40)

⎧ ⎪ {ε} = ⎨ ⎪ ⎩

⎧ ⎪ {ε ( w )} = ⎪⎨ ⎪ ⎩⎪

{

⎧ ∂ v1 ⎪ ∂ x1 ε11 ⎫ ⎪⎪ ⎪ ⎪ ∂v2 ε 22 ⎬ = ⎨ ∂x2 ⎪ ⎪ 2 ε12 ⎭ ⎪ 1 ⎛ ∂ v1 ∂ v 2 ⎪ + ⎜ ⎪⎩ 2 ⎝ ∂ x 2 ∂ x1

}

The strain rate vector {ε} and ε ( w ) becomes

{ε} = [ B ]{v e }

and

{ε ( w )} = [ B ]{w ve }

(15.41)

where

⎡ ∂N1 ⎢ ∂ x1 ⎢ ⎢ 0 [B ] = ⎢ ⎢ ⎢ 1 ∂N1 ⎢ ⎣ 2 ∂x2

0

∂N 2 ∂ x1

∂N1 ∂ x1

0

∂N 2 −−−−− ∂ x1

1 ∂N1 2 ∂ x1

1 ∂N 2 2 ∂x2

1 ∂N 2 −− −− 2 ∂ x1

0

−−−−−

∂N 9 ∂ x1 0 1 ∂N 9 2 ∂x2

⎤ ⎥ ⎥ ⎥ ∂N 9 ⎥ ∂ x1 ⎥ 1 ∂N 9 ⎥ ⎥ 2 ∂ x1 ⎦ 0

(15.42)

Further,

ε1 1 + ε 22 = {1 1 0}{ε } = {m }

T

[ B ]{v e }

(15.43)

where

187

⎧1 ⎫ {m } = ⎪⎨1 ⎪⎬ ⎪0⎪ ⎩ ⎭

(15.44)

Similarly

ε1 1 ( w ) + ε 2 2 ( w ) = {m }

T

[ B ]{w ve }

(15.45)

Substituting Eqs. (15.39-15.45) in Eq. (15.28),

∫ { } { N } {m } − w ep

T

T

p

[ B ]{v e } d x1d x 2

A

+

∫ { } − w ve

T

{ } { p }dx dx

[ B ]T {m } N

A

+

∫ 2 μ ⎡⎣ w

e v

T

p

e

1

{ }

⎤ [ B ]T [ B ] v e d x1 d x 2 = ⎦

(15.46)

2

∫ t w dΓ + ∫ t w dΓ 1

1

2

Γ1

A

2

Γ2

Using the notations,

∫ − { N } {m }

⎡ K epv ⎤ = ⎣ ⎦

T

p

[ B ] d x1d x 2

A

e ⎤ ⎡ K vp ⎣ ⎦=

∫

{ }

− [ B ]T {m } N

{

T

p

d x1 d x 2 = K epv

}

T

(15.47)

A

e ⎤ ⎡ K vv ⎣ ⎦ =

∫ 2μ[B]

T

[ B ] d x1d x 2

A

Thus the finite element equation in local variable form will be, ne

∑ {w } e

e =1

T

nb1

nb2

{ } = ∑ { } { } + ∑ {w } { f }

⎡K ⎤ δ ⎣ ⎦ e

e

w1b

T

f1b

b =1

b 2

T

b 2

(15.48)

b =1

where

{ f } = ∫ {N b 1

b

}{ N b }T {t1b } d Γ

b

}{ N b }

Γ1b

{ } = ∫ {N f 2b

T

{ }dΓ t 2b

(15.49)

Γ b2

188

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

⎫ ⎧⎪ ⎪⎧ p } { ⎪ = δ , w ⎬ { } ⎨⎨ { } { } ⎪⎭ ⎩⎪ ⎩⎪ v e

⎧ w ep ⎪ =⎨ e ⎪ wv ⎩

e

e

e

⎡ [0 ] ⎪⎫ ⎪⎫ e ⎬ ⎬ and ⎡⎣ K ⎤⎦ = ⎢⎢ ⎡K e ⎤ ⎭⎪ ⎭⎪ ⎣⎢ ⎣ vp ⎦

⎡ K epv ⎤ ⎤ ⎣ ⎦ ⎥ e ⎤⎥ ⎡ K vv ⎣ ⎦ ⎦⎥

(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;

∫

(........ ) d x1d x 2

=

Ae

+1

∫ ∫ −1

∂ x1 ∂ξ J = ∂ x2 ∂ξ

+1 −1

(....... ) | J

∂ x1 ∂η ∂x2 ∂η

| dξ dη

(15.51)

(15.51)

where |J| is the elemental Jacobian matrix. Similarly the boundary integrals are transformed by the relation +1

∫ (........) d Γ = ∫ (..........) | J b | d ζ −1

Γ

(15.52)

where | Jb | is the Jacobian for boundary element and is given by 2

| J b |=

⎛ ∂ x1 ⎞ ⎛ ∂x2 ⎞ ⎜ ⎟ +⎜ ⎟ ⎝ ∂ζ ⎠ ⎝ ∂ζ ⎠

2

(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

{W }T [ K ]{Δ } = {W }T {F }

(15.54)

189

where

{W } , [ K ] , {Δ } are

the global vector of nodal values of weight function, global

coefficient matrix and global vector of nodal values of pressure and velocity.

{F }

is the global

right hand side vector. Since weight functions are arbitrary, final FEM expression will be

[ 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;

t s = − t1 cos φ + t 2 sin φ t n | = − t1 sin φ − t 2 cos φ

(15.56)

Thus putting the decomposed form ts and tn into Eq. (15.14) we get,

( − t1 cos φ + t 2 sin φ ) = μ ( − t1 sin φ − t 2 cos φ )

(15.57)

∴ − t1 (cos φ − μ sin φ ) + t 2 (sin φ + μ cos φ ) = 0

(15.58)

Thus the above Eq. (15.58) becomes,

t1 (1 + μ tan φ ) − t 2 (tan φ + μ ) = 0

(15.59)

The above expression for an element for FEM can be written as

190

⎧ ( t b1 ) ⎫ 1 ⎪ ⎪ ⎪ 2 ⎪ ⎨ ( t b )1 ⎬ (1 + μ tan φ ) − ⎪ 3 ⎪ ⎪⎩ ( t b )1 ⎪⎭

⎧ ( t b1 ) ⎫ 2 ⎪ ⎪ ⎪ 2 ⎪ ⎨ ( t b )2 ⎬ (tan φ + μ ) = 0 ⎪ 3 ⎪ ⎪⎩ ( t b )2 ⎪⎭

(15.60)

After multiplying the above equation by +1

∫ {N

b

}{ N b }T

| Jb | dζ

(15.61)

−1

it becomes

{ f } (1 + μ tan φ ) − { f } (tan φ + μ ) = {0} b

b 2

1

(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

{ F }( d

p

+ 2 k − 1)

(1 + μ tan φ ) − { F }( d

p

+2k )

(tan φ + μ ) = {0}

(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: •

th Replace ( d p + 2 k − 1) row of global coefficient matrix [K] by the following linear th combination: (1 + μ tan φ ) tim es ( d p + 2 k − 1) row of th [K] – (tanφ + μ) times ( d p + 2 k ) row of [K].

•

th Make the ( d p + 2 k − 1) row of global right hand vector {F} zero.

•

th The velocity boundary condition at the node is applied by replacing ( d p + 2 k ) row of

[K] by Eq. (15.13). •

th Make ( d p + 2 k ) row of [K] zero.

•

th Set ( d p + 2 k , d p + 2 k − 1) element

191

th of [K] to tan φ ( d p + 2 k , d p + 2 k ) element of [K] to 1.

•

th Make ( d p + 2 k ) row of {F} to zero.

The essential boundary conditions at the other boundaries are applied as follows: ƒ

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.

ƒ

The diagonal term is replaced by unity.

ƒ

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

EXERCISE 15 Q1: In the rolling process, the governing equations are given as

ε11 + ε 2 2 = 0 ∂ σ 11 ∂ σ 12 + =0 ∂ x1 ∂x2 ∂ σ 21 ∂ σ 22 + =0 ∂ x1 ∂x2

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

ε rr + εθθ + ε zz = 0 ∂ v r ⎞ ⎛ 1 ∂ ( r σ rr ) ⎛ ∂vr + vz + p ⎜ vr ⎟−⎜ ∂r ∂z ⎠ ⎝ r ∂r ⎝ ∂ v z ⎞ ⎛ 1 ∂ ( r σ rz ) ⎛ ∂v z + vz + p ⎜ vr ⎟−⎜ ∂r ∂z ⎠ ⎝ r ∂r ⎝

∂ σ rz σ θθ ⎞ − ⎟=0 ∂z r ⎠ ∂ σ zz ⎞ ⎟=0 ∂z ⎠

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

F = (k0 + k x ) x

Figure: Q3 where k 0 = 1000 N /m , k = 100 N /m 2 , F = 2000 N

194

The equations can be written as

⎡ 1 − 1⎤ ⎧ u1 ⎫ ⎧ − F ⎫ k⎢ ⎬ ⎥⎨ ⎬ = ⎨ ⎣ −1 1 ⎦ ⎩u 2 ⎭ ⎩ 0 ⎭ Solve the problem by finite element method. Q.4: The governing equation of a large deflection bending of elastic beam is

⎧⎪ ⎛ ∂ u 1 ⎛ ∂ w ⎞ 2 ⎞ ⎫⎪ + ⎜ ⎨ E A ⎜⎜ ⎟ ⎟⎟ ⎬ − f = 0 ⎪⎩ ⎝ ∂ x 2 ⎝ ∂ x ⎠ ⎠ ⎪⎭ ⎛ ∂ u 1 ⎛ ∂ w ⎞ 2 ⎞ ⎫⎪ d2 ⎧ ∂ 2 w ⎫ d ⎧⎪ − + ⎜ ⎨ EI ⎬− ⎨ E A ⎜⎜ ⎟ ⎟⎟ ⎬ − q = 0 dx 2 ⎩ ∂ x 2 ⎭ d x ⎪⎩ ⎝ ∂ x 2 ⎝ ∂ x ⎠ ⎠ ⎪⎭

d − dx

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

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

Lu - q = S T DSu - q = 0

(16.1)

in a domain Ω , with prescribed displacement

u = u on the boundary Γ u

(16.2)

Su = t on the boundary Γ t

(16.3a)

and prescribed derivatives

with

Γ = Γu ∪ Γ t

(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

Ku - f = 0 where

K=

∫ (SN )

T

Ω

D(SN) d Ω

is

the

elemental

(16.4) stiffness

matrix

and

f = ∫ N T q dΩ + ∫ N T t dΓ is the load vector. Here, N is the shape function (interpolation Ω

Γt

ˆ for an element is related to nodal solution une in function) matrix. The approximate solution u the following manner:

uˆ =N une

(16.5)

The derivatives are calculated as

σˆ = (SN)une

(16.6)

ˆ , σˆ differs from the exact values u , σ and the difference is the The approximate solutions u error. Thus, primary variable and derivative errors are

e = u- u eσ = σ − σˆ

(16.7a) (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

often introduced to measure the error. The most common measures are the ‘energy norm’ and ‘L2 norm’. The energy norm for general problems is 1

e = ( ∫ e T Le dΩ) 2

(16.8)

Ω

e = u − uˆ .

where

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

e

L2

=

(∫ e edΩ)

1 2

T

(16.9)

Ω

and for derivatives,

eσ

= ( ∫ ( e σ ) ( e σ ) dΩ ) T

L2

1 2

Ω

(16.10)

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

⎛ e2 ⎜ L2 Δu = ⎜ ⎜ Ω ⎝

1

⎞2 ⎟ ⎟⎟ ⎠

(16.11)

Similarly for derivatives

⎛ es 2 L2 Δσ = ⎜ ⎜ Ω ⎝

1

⎞2 ⎟ ⎟ ⎠

(16.12)

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

2

m

=∑ e i =1

2 i

(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

η=

e u

× 100%

(16.14)

199

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

eu ≤ Ch min( p ,λ )

(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,

eu ≤ CN

− min( p ,λ ) 2

(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

eu ≤ Ch ≈ CN p

−p 2

(16.17)

200

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

eu ≤ CN − β

(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 β = λ 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

eu ≤ Ce ( −αN

ϑ

)

(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

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 σ * is interpolated by the same function as the displacement, i.e.

σ * = Nσ *

(16.20)

where σ * is improved nodal derivative and

∫

Ω

NT (σ* − σˆ )d Ω = 0

(16.21)

where σˆ is elemental FEM derivative. It is intuitively obvious that σ * is in fact a better approximation than σˆ and we shall use it to estimate the error eσ i.e.

eσ ≈ σ * − σˆ

(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,

{u} = [Ni ]{di }

(16.23)

202

Consider that, at the refinement step m, the element displacement fields of an element of order p, is given by,

{u} = [Nn ]{dn }

(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,

[K nn ]{dn } = {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,

{u'} = [Nn ]{dn } + [Nh ]{dh }

(16.26)

and

⎡K nn K nh ⎤ ⎧dn ⎫ ⎧fn ⎫ ⎢K ⎥⎨ ⎬ = ⎨ ⎬ ⎣ nh K hh ⎦ ⎩dh ⎭ ⎩fh ⎭

(16.27)

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):

{dh } = [K hh ] ({fh } − [K nh ] {dn }) −1

T

(`16.28)

To obtain the approximate value of each component of {dh}, equation (16.28) is simply written as,

203

( dh ) j

=

(f

h

− {K nh }

T

{dn }) j

(16.29)

( K hh ) j

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,

{e u } = {u'} − {u}

(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 1D 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 q p belong to a polynomial expansion q p 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 q p and one can get

q *p = Pa

(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,

[

P = 1, x, x 2 ,........, x p

]

(16.32)

204

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

[

a = a1 , a 2 a3, .........., a p +1

]

T

(16.33)

Thus for two dimensions (triangular) and linear expansion we have

P = [1, x, y ] and for quadratic

(16.34)

[

P = 1, x, y, x 2 , xy, y 2

]

(16.35)

For a bilinear quadrilateral

P = [1, x, y, xy ]

(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

F(a) = ∑ (qh (xi y i ) − q (xi , y i )) n

i =1

* p

205

2

n

= ∑ (qh ( xi y i ) − P(xi , y i )a )

(16.37)

i =1

where ( xi y i ) are the global co –ordinates of sampling points, n = mk is the total number of sampling points and k is the number of the sampling points on each element mj ( m j = 1,2,3.......m) of the element patch. The minimization condition of F (a) implies that a satisfies n

n

∑ P (x y )P(x , y )a = ∑ P (x , y )q (x , y ) T

i =1

i,

i

i

i

T

i

i =1

i

h

i

i

(16.38)

This can be solved in matrix form as

a = A −1b

(16.39)

where n

n

i =1

i =1

A = ∑ P T (xi , y i )P(xi , y i ) and b = ∑ P T ( xi , y i )qh ( xi , y i )

(16.40)

Once the parameters a are determined, the recovered nodal values are calculated by insertion of appropriate co-ordinates into the expression for q*h . 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

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

T j = a j + b j x + c j x2

(16.41)

Similarly, the quadratic polynomial on the (j+1) node is

T

j +1

= a j +1 + b j +1 x + c j +1 x 2

(16.42)

Now, the temperature function in between the nodes j and (j+1) can be written as

T = N 1T j + N 2T

j +1

(16.43)

where N1 and N2 are standard linear shape functions.

207

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:

aj −bj +cj = a−b+c−d aj =a

(16.45)

a +b +c = a+b+c+d j

j

j

Solution of the above equation yields

a j = a, b j = b + d , c j = c

(16.46)

Therefore, the approximation function Tj is given by

T j = a + (b + d ) x + cx 2 = a + bx + cx 2 + dx

(16.47)

208

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

a j +1 = a, b j +1 = b − 2d

c j +1 = c + 3d

and

(16.48)

Then, the approximating function Tj+1 is given by

T

j +1

= a + bx + cx 2 + (3 x 2 − 2 x)d

(16.49)

Now, in the element j – (j+1), N1 is (1-x) and N2 is x. Hence,

T = a + bx + cx 2 + ( x − 3 x 2 + 3 x 3 )d

(16.50)

Subtracting the equation for exact solution from the above equation, error in the primary variable becomes

eT = (2 x 3 − 3 x 2 + x)d

(16.51)

The roots of the polynomial inside the brackets from the above equation are 0, 1 and 0.5, indicating that blended function gives exact value not only at the nodes, but also at the middle point. One can also find the L2 norm

⎛ ⎞ 2 = ⎜⎜ ∫ 2 x 3 − 3x 2 + x d dx ⎟⎟ ⎝0 ⎠ 1

eT

L2

The first derivative will be exact at points

[(

)]

1 2

(16.52)

1 1 ± and the second derivative at the middle of 2 2 3

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

e

2 E

= − ∫ erdΩ

(16.53)

Ω

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:

f i = α + α 2 x + α 3 y + α 4 xy + α 5 x 2 + α 6 y 2 + α 7 xy 2 + α 8 x 2 y + αx 2 y 2 (16.54)

209

Approximation of the primary variable inside an element is carried out in the following manner:

f e = N1 f1 + N 2 f 2 + N 3 f 3 + N 4 f 4

(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

T e = N 1T1 + N 2T2 + N 3T3 + N 4T4

(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

e ≈ eˆ = u * - uˆ e

L2

≈ eˆ

L2

(16.57)

= u * - uˆ

e σ ≈ eˆ σ = u * - uˆ

σ

L2

(16.58) (16.59)

Error estimators formulated by replacing the exact solution with the recovered solution are sometimes called recovery based error estimators.

210

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

θ, which is defined as

eˆ

θ=

(16.60)

e

Assuming that the true error convergences as

u − u h = Ch h p

(16.61)

u − u* = C* h p +α

(16.62)

and the error of the post-processing solution

for some super convergent solution with α ≥ 1 . It is shown that for all estimators based on recovery can establish the following bounds for the effectivity index:

1−

e* e

≤ θ ≤ 1+

e* e

(16.63)

Where e is the actual error and e* is the error of the recovered solution, e.g.

e * = u - u*

(16.64)

Effectivity index approaches to unity as h approaches 0.

α ≥ 1 indicates whether the recovered solution has a higher rate of convergence than FE solution. Two important conclusions follow: 1. any recovery process which results in reduced error will give a reasonable error estimator and more importantly, 2. if the recovered solution converges at a higher rate, then the finite element solution shall always have asymptotically exact estimation.

16.5CONCLUSIONS In this chapter, a brief introduction of error analysis has been provided. The error analysis is very important for adaptive refinement. It also provides an idea about the quality of the solution.

FURTHER READINGS 1.

Zienkiewicz, O.C. and Zhu, J.G., 1987, A simple error estimator and adaptive procedure for practical engineering analysis, Int. J. Num. Meth. Eng., 24, 337-357.

211

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

k

d 2T +q =0 dx 2

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

T = N1T1 + N 2T2 + (1 − ξ 2 ) a where T1 and T2 are temperatures at the 2 nodes and a is some constant. Find out the elemental stiffness matrix and elemental load vector for this problem. Also, carry out the error analysis.

212

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,

du u ( x + h) − u ( x) ≈ dx h

(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,

x⎞ x ⎛ u = a + bx = ⎜1 − ⎟u1 + u 2 h ⎝ h⎠

(17.2)

213

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

du u 2 − u1 = dx h

(17.3)

which is the same as finite difference approximation. On the hand, if axial displacement is approximated by some function like

u = a + b sin x

(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=

1 {u}T[K]{u} – {u}T{P} 2

(17.6)

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

1 {u}T {P} – {u}T{P} 2

=-

1 {u}T {P} 2

(17.7)

The potential energy given by this expression will be more or equal to the actual potential energy, because in the actual system potential energy will be minimized. Hence, {u}T{P}>k2 and hence the stiffness matrix becomes ill-conditioned. This is not case if k2>>k1.

u2

u1

P

1

k1

2

k2

,x,u

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

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

⎡ k1 ⎢0 ⎣

0 ⎤ ⎧u r ⎫ ⎧ P ⎫ ⎨ ⎬=⎨ ⎬ k 2 ⎥⎦ ⎩u 2 ⎭ ⎩ P ⎭

(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

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

x = N1 x1 + N 2 x2 where

N1 = −

2ξ ; 1−ξ

N2 =

1+ ξ 1−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 illconditioned.

220

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.

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