Finite Element Analysis

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Preface

v

2HAB=?A Finite Element Analysis was developed as a numerical method of stress analysis, but now it has been extended as a general method of solution to many complex engineering and physical science problems. As it involves lot of calculations, its growth is closely linked with the developments in computer technology. Now-a-days a number of finite element analysis packages are available commercially and number of users is increasing. A user without a basic course on finite element analysis may produce dangerous results. Hence now-a-days in many M.Tech. programmes finite element analysis is a core subject and in undergraduate programmes many universities offer it as an elective subject. The experience of the author in teaching this course to M.Tech (Geotechnical Engineering) and M.Tech. (Industrial Structures) students at National Institute of Technology, Karnataka, Surathkal (formerly, K.R.E.C. Surathkal) and to undergraduate students at SDM College of Eingineering and Technology, Dharwad inspired him to write this book. This is intended as a text book to students and as an introductory course to all users of finite element packages. The author has developed the finite element concept, element properties and stiffness equations in first nine chapters. In chapter X the various points to be remembered in discritization for producing best results is presented. Isoparametric concept is developed and applications to simple structures like bars, trusses, beams and rigid frames is explained thoroughly taking small problems for hand calculations. Application of this method to complex problems like plates, shells and nonlinear analysis is introduced. Finally a list of commercially available packages is given and the desirable features of such packages is presented. The author hopes that the students and teachers will find it as a useful text book. The suggestions for improvements are most welcome. DR S.S. BHAVIKATTI

Acknowledgements The author sincerely acknowledges Dr C.V. Ramakrishnan, Professor, Department of Applied Mechanics, IIT Delhi for introducing him the subject finite element analysis as his Ph.D. guide. The author thanks the authorities of Karnataka Regional Engineering College, Surathkal (presently National Institute of Technology, Karnataka, Surathkal) for giving him opportunity to teach this subject to M.Tech. (Industrial Structures and Geotechnical Engineering) students for several years. He thanks SDM College of Engineering and Technology, Dharwad for the opportunity given to him for teaching the course on FEA to VII semester BE (Civil) students. The author wishes to thank his M.Tech. Students Madhusudan (1987), Gowdaiah N.G. (1987), Parameshwarappa P.C. (1988), Kuriakose Mathew (1991), Vageesh S.M. (1991), Vageesh S.V. (1992), Manjunath M.B. (1992), Siddamal T.V. (1993), Venkateshan Y. (1994), Nagaraj B.N. (1995), Devalla Lakshmi Satish (1996) and Ajith Shenoy M. (1996) for carrying out their M.Tech thesis work under his guidance. Thanks are also due to clerical assistance he got from Mrs. Renuka Deshpande, Sri. R.M. Kanakapur and Sri. Rayappa Kurabagatti of Department of Civil Engineering of SDM College of Engineering & Technology, Dharwad in preparing the manuscript. He acknowledges the help rendered by Sri R.J.Fernandes, Sri Satish and Sri Chandrahas of SDM College of Engineering & Technology, Dharwad in preparing the drawings.

Contents

vii

Contents Preface

v

Acknowledgements

vi

1. Introduction 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8

General 1 General Description of the Method 1 Brief Explanation of FEA for a Stress Analysis Problem Finite Element Method vs Classical Method 4 FEM vs FDM 5 A Brief History of FEM 6 Need for Studying FEM 6 Warning to FEA Package Users 7 Questions 7 References 7

1

2

2. Basic Equations in Elasticity 2.1 2.2 2.3 2.4 2.5 2.6

9

Introduction 9 Stresses in a Typical Element 9 Equations of Equilibrium 12 Strains 14 Strain Displacement Equations 14 Linear Constitutive Law 15 Questions 20

3. Matrix Displacement Formulation 3.1 3.2 3.3 3.4

Introduction 21 Matrix Displacement Equations 21 Solution of Matrix Displacement Equations 28 Techniques of Saving Computer Memory Requirements Questions 32

21

30

viii

Contents

4. Element Shapes, Nodes, Nodal Unknowns and Coordinate Systems 4.1 4.2 4.3 4.4 4.5

Introduction 33 Element Shapes 33 Nodes 38 Nodal Unknowns 39 Coordinate Systems 40 Questions 53

5. Shape Functions 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8

55

Introduction 55 Polynomial Shape Functions 56 Convergence Requirements of Shape Functions 59 Derivation of Shape Functions Using Polynomials 61 Finding Shape Functions Using Lagrange Polynomials 82 Shape Functions for Serendipity Family Elements 89 Hermite Polynomials as Shape Functions 95 Construction of Shape Functions by Degrading Technique 98 Questions 102

6. Strain Displacement Matrix 6.1 6.2 6.3 6.4

33

104

Introduction 104 Strain—Displacement Matrix for Bar Element 104 Strain Displacement Matrix for CST Element 105 Strain Displacement Relation for Beam Element 107 Questions 108

7. Assembling Stiffness Equation—Direct Approach

110

7.1 Introduction 110 7.2 Element Stiffness Matrix for CST Element by Direct Approach 110 7.3 Nodal Loads by Direct Approach 114 Questions 117

8. Assembling Stiffness Equation—Galerkin’s Method, Virtual Work Method 8.1 Introduction 118 8.2 Galerkin’s Method 118 8.3 Galerkin’s Method Applied to Elasticity Problems Questions 127

119

118

Contents

9. Assembling Stiffness Equation—Variational Method 9.1 9.2 9.3 9.4 9.5 9.6

128

Introduction 128 General Variational Method in Elasticity Problems 128 Potential Energy in Elastic Bodies 134 Principles of Minimum Potential Energy 136 Rayleigh—Ritz Method 140 Variational Formulation in Finite Element Analysis 150 Questions 153

10. Discritization of a Structure 10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8

154

Introduction 154 Nodes as Discontinuities 154 Refining Mesh 156 Use of Symmetry 157 Finite Representation of Infinite Bodies 157 Element Aspect Ratio 158 Higher Order Element vs Mesh Refinement 159 Numbering System to Reduce Band Width 159 Questions 160

11. Finite Element Analysis—Bars and Trusses 11.1 11.2 11.3 11.4

161

Introduction 161 Tension Bars/Columns 161 Two Dimensional Trusses (Plane Trusses) 180 Three Dimensional Trusses (Space Trusses) 197 Questions 201

12. Finite Element Analysis—Plane Stress and Plane Strain Problems 12.1 Introduction 204 12.2 General Procedure when CST Elements are Used 12.3 Use of Higher Order Elements 216 Questions 217

204

204

13. Isoparametric Formulation 13.1 13.2 13.3 13.4 13.5

ix

Introduction 219 Coordinate Transformation 221 Basic Theorems of Isoparametric Concept 222 Uniqueness of Mapping 223 Isoparametric, Superparametric and Subparametric Elements

219

224

x

Contents 13.6 Assembling Stiffness Matrix 225 13.7 Numerical Integration 230 13.8 Numerical Examples 232 Questions 240 References 241

14. Analysis of Beams and Rigid Frames 14.1 14.2 14.3 14.4 14.5

242

Introduction 242 Beam Analysis Using two Noded Elements 242 Analysis of Rigid Plane Frame Using 2 Noded Beam Elements 259 A Three Dimensional Rigid Frame Element 266 Timoshenko Beam Element 269 Questions 278 References 279

15. Bending of Thin Plates 15.1 15.2 15.3 15.4 15.5 15.6

Introduction 280 Basic Relations in Thin Plate Theory 281 Displacement Models for Plate Analysis 282 Rectangular Plate Element with 12 Degrees of Freedom Rectangular Plate Element with 16 Degrees of Freedom Mindlin’s Plate Element 292 Questions 299 References 299

16. Analysis of Shells 16.1 16.2 16.3 16.4

284 289

301

Introduction 301 Force on Shell Element 301 Finite Element for Shell Analysis 302 Finite Element Formulation Using Four Noded Degenerated Quadrilateral Shell Element 307 Questions 317 References 317

17. Nonlinear Analysis 17.1 17.2 17.3 17.4

280

Introduction 318 Nonlinear Problems 318 Analysis of Material Nonlinear Problems 320 Analysis of Geometric Nonlinear Problems 325

318

Contents 17.5 Analysis of Both Material and Geometric Nonlinear Problems Questions 328 References 328

18. Standard Packages and Their Features 18.1 18.2 18.3 18.4 18.5

Introduction 329 Commercially Available Standard Packages 329 Structure of a Finite Element Analysis Program 330 Pre and Post Processors 331 Desirable Features of FEA Packages 333 Questions 333

xi

328

329

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Introduction

1

1 Introduction 1.1 GENERAL The finite element analysis is a numerical technique. In this method all the complexities of the problems, like varying shape, boundary conditions and loads are maintained as they are but the solutions obtained are approximate. Because of its diversity and flexibility as an analysis tool, it is receiving much attention in engineering. The fast improvements in computer hardware technology and slashing of cost of computers have boosted this method, since the computer is the basic need for the application of this method. A number of popular brand of finite element analysis packages are now available commercially. Some of the popular packages are STAAD-PRO, GT-STRUDEL, NASTRAN, NISA and ANSYS. Using these packages one can analyse several complex structures. The finite element analysis originated as a method of stress analysis in the design of aircrafts. It started as an extension of matrix method of structural analysis. Today this method is used not only for the analysis in solid mechanics, but even in the analysis of fluid flow, heat transfer, electric and magnetic fields and many others. Civil engineers use this method extensively for the analysis of beams, space frames, plates, shells, folded plates, foundations, rock mechanics problems and seepage analysis of fluid through porous media. Both static and dynamic problems can be handled by finite element analysis. This method is used extensively for the analysis and design of ships, aircrafts, space crafts, electric motors and heat engines.

1.2 GENERAL DESCRIPTION OF THE METHOD In engineering problems there are some basic unknowns. If they are found, the behaviour of the entire structure can be predicted. The basic unknowns or the Field variables which are encountered in the engineering problems are displacements in solid mechanics, velocities in fluid mechanics, electric and magnetic potentials in electrical engineering and temperatures in heat flow problems. In a continuum, these unknowns are infinite. The finite element procedure reduces such unknowns to a finite number by dividing the solution region into small parts called elements and by expressing the unknown field variables in terms of assumed approximating functions (Interpolating functions/Shape functions) within each element. The approximating functions are defined in terms of field variables of specified points called nodes or nodal points. Thus in the finite element analysis the unknowns are the field variables of the nodal points. Once these are found the field variables at any point can be found by using interpolation functions. After selecting elements and nodal unknowns next step in finite element analysis is to assemble element properties for each element. For example, in solid mechanics, we have to find the force-displacement i.e. stiffness characteristics of each individual element. Mathematically this relationship is of the form

2

Finite Element Analysis

[k ]e {δ }e = {F}e where [k]e is element stiffness matrix, {δ }e is nodal displacement vector of the element and {F}e is nodal force vector. The element of stiffness matrix kij represent the force in coordinate direction ‘i’ due to a unit displacement in coordinate direction ‘j’. Four methods are available for formulating these element properties viz. direct approach, variational approach, weighted residual approach and energy balance approach. Any one of these methods can be used for assembling element properties. In solid mechanics variational approach is commonly employed to assemble stiffness matrix and nodal force vector (consistant loads). Element properties are used to assemble global properties/structure properties to get system equations [ k ] {δ } = {F}. Then the boundary conditions are imposed. The solution of these simultaneous equations give the nodal unknowns. Using these nodal values additional calculations are made to get the required values e.g. stresses, strains, moments, etc. in solid mechanics problems. Thus the various steps involved in the finite element analysis are: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii)

Select suitable field variables and the elements. Discritise the continua. Select interpolation functions. Find the element properties. Assemble element properties to get global properties. Impose the boundary conditions. Solve the system equations to get the nodal unknowns. Make the additional calculations to get the required values.

1.3 A BRIEF EXPLANATION OF FEA FOR A STRESS ANALYSIS PROBLEM The steps involved in finite element analysis are clarified by taking the stress analysis of a tension strip with fillets (refer Fig.1.1). In this problem stress concentration is to be studies in the fillet zone. Since the problem is having symmetry about both x and y axes, only one quarter of the tension strip may be considered as shown in Fig.1.2. About the symmetric axes, transverse displacements of all nodes are to be made zero. The various steps involved in the finite element analysis of this problem are discussed below: Step 1: Four noded isoparametric element (refer Fig 1.3) is selected for the analysis (However note that 8 noded isoparametric element is ideal for this analysis). The four noded isoparametric element can take quadrilateral shape also as required for elements 12, 15, 18, etc. As there is no bending of strip, only displacement continuity is to be ensured but not the slope continuity. Hence displacements of nodes in x and y directions are taken as basic unknowns in the problem. Fillet

t

P b2

C

A B

Fig. 1.1

Typical tension flat

D

b1

Introduction 5

A 1

9

13 17 21 24 29 33

37

1

4

7

10 13 16 19 22

25

2

5

8

11 14 17 20 23

26

3

6

9

41

3

45

28

C

31

2 3

B 4

8

29

12 15 18 21 24 27 16 20 24 32 28 36 40

12

Fig. 1.2

32

30

P

33

D

44

48

Discretisation of quarter of tension flat n 10

6

3

4

xP 5

11

7

2

1

(a) Element no. 5

(b) Typical element

Fig. 1.3

Step 2: The portion to be analysed is to be discretised. Fig. 1.2 shows discretised portion. For this 33 elements have been used. There are 48 nodes. At each node unknowns are x and y components of displacements. Hence in this problem total unknowns (displacements) to be determined are 48 × 2 = 96. Step 3: The displacement of any point inside the element is approximated by suitable functions in terms of the nodal displacements of the element. For the typical element (Fig. 1.3 b), displacements at P are

∑N u = N u + N u v = ∑N v = N v + N v

u= and

i i

1 1

2 2

+ N 3 u3 + N 4 u 4

i i

1 1

2 2

+ N 3v3 + N 4 v4

…(1.2)

The approximating functions Ni are called shape functions or interpolation functions. Usually they are derived using polynomials. The methods of deriving these functions for various elements are discussed in this text in latter chapters. Step 4: Now the stiffness characters and consistant loads are to be found for each element. There are four nodes and at each node degree of freedom is 2. Hence degree of freedom in each element is 4 × 2 = 8. The relationship between the nodal displacements and nodal forces is called element stiffness characteristics. It is of the form

[k ]e {δ }e = {F}e , as explained earlier. For the element under consideration, ke is 8 × 8 matrix and δ e and Fe are vectors of 8 values. In solid mechanics element stiffness matrix is assembled using variational approach i.e. by minimizing potential energy. If the load is acting in the body of element or on the surface of element, its equivalent at nodal points are to be found using variational approach, so that right hand side of the above expression is assembled. This process is called finding consistant loads.

4

Finite Element Analysis

Step 5: The structure is having 48 × 2 = 96 displacement and load vector components to be determined. Hence global stiffness equation is of the form [k]

{δ }

= {F}

96 × 96 96 × 1 96 × 1 Each element stiffness matrix is to be placed in the global stiffness matrix appropriately. This process is called assembling global stiffness matrix. In this problem force vector F is zero at all nodes except at nodes 45, 46, 47 and 48 in x direction. For the given loading nodal equivalent forces are found and the force vector F is assembled. Step 6: In this problem, due to symmetry transverse displacements along AB and BC are zero. The system equation [ k ] {δ } = {F} is modified to see that the solution for {δ } comes out with the above values. This modification of system equation is called imposing the boundary conditions. Step 7: The above 96 simultaneous equations are solved using the standard numerical procedures like Gausselimination or Choleski’s decomposition techniques to get the 96 nodal displacements. Step 8: Now the interest of the analyst is to study the stresses at various points. In solid mechanics the relationship between the displacements and stresses are well established. The stresses at various points of interest may be found by using shape functions and the nodal displacements and then stresses calculated. The stress concentrations may be studies by comparing the values obtained at various points in the fillet zone with the values at uniform zone, far away from the fillet (which is equal to P/b2t).

1.4 FINITE ELEMENT METHOD VS CLASSICAL METHODS 1. In classical methods exact equations are formed and exact solutions are obtained where as in finite element analysis exact equations are formed but approximate solutions are obtained. 2. Solutions have been obtained for few standard cases by classical methods, where as solutions can be obtained for all problems by finite element analysis. 3. Whenever the following complexities are faced, classical method makes the drastic assumptions’ and looks for the solutions: (a) Shape (b) Boundary conditions (c) Loading Fig. 1.4 shows such cases in the analysis of slabs (plates). To get the solution in the above cases, rectangular shapes, same boundary condition along a side and regular equivalent loads are to be assumed. In FEM no such assumptions are made. The problem is treated as it is. 4. When material property is not isotropic, solutions for the problems become very difficult in classical method. Only few simple cases have been tried successfully by researchers. FEM can handle structures with anisotropic properties also without any difficulty. 5. If structure consists of more than one material, it is difficult to use classical method, but finite element can be used without any difficulty. 6. Problems with material and geometric non-linearities can not be handled by classical methods. There is no difficulty in FEM.

Introduction

5

Hence FEM is superior to the classical methods only for the problems involving a number of complexities which cannot be handled by classical methods without making drastic assumptions. For all regular problems, the solutions by classical methods are the best solutions. Infact, to check the validity of the FEM programs developed, the FEM solutions are compared with the solutions by classical methods for standard problems.

(a) Irregular shaper

(b) Irregular boundary condition

(c) Irregular loading

Fig. 1.4

1.5 FEM VS FINITE DIFFERENCE METHOD (FDM) 1. FDM makes pointwise approximation to the governing equations i.e. it ensures continuity only at the node points. Continuity along the sides of grid lines are not ensured. FEM make piecewise approximation i.e. it ensures the continuity at node points as well as along the sides of the element. 2. FDM do not give the values at any point except at node points. It do not give any approximating function to evaluate the basic values (deflections, in case of solid mechanics) using the nodal values. FEM can give the values at any point. However the values obtained at points other than nodes are by using suitable interpolation formulae. 3. FDM makes stair type approximation to sloping and curved boundaries as shown in Fig. 1.5. FEM can consider the sloping boundaries exactly. If curved elements are used, even the curved boundaries can be handled exactly. 4. FDM needs larger number of nodes to get good results while FEM needs fewer nodes. 5. With FDM fairly complicated problems can be handled where as FEM can handle all complicated problems.

6

Finite Element Analysis

Fig. 1.5

FDM approximation of shape

1.6 A BRIEF HISTORY OF FEM Engineers, physicists and mathematicians have developed finite element method independently. In 1943 Courant [1] made an effort to use piecewise continuous functions defined over triangular domain. After that it took nearly a decade to use this distribution idea. In fifties renewed interest in this field was shown by Polya [2], Hersh [3] and Weinberger [4]. Argyris and Kelsey [5] introduced the concept of applying energy principles to the formation of structural analysis problems in 1960. In the same year Clough [6] introduced the word ‘Finite Element Method’. In sixties convergence aspect of the finite element method was pursued more rigorously. One such study by Melesh [7] led to the formulation of the finite element method based on the principles of minimum potential energy. Soon after that de Veubeke [8] introduced equilibrium elements based on the principles of minimum potential energy, Pion [9] introduced the concept of hybrid element using the duel principle of minimum potential energy and minimum complementary energy. In Late 1960’s and 1970’s, considerable progress was made in the field of finite element analysis. The improvements in the speed and memory capacity of computers largely contributed to the progress and success of this method. In the field of solid mechanics from the initial attention focused on the elastic analysis of plane stress and plane strain problems, the method has been successfully extended to the cases of the analysis of three dimensional problems, stability and vibration problems, non-linear analysis. A number of books [10 – 20] have appeared and made this field interesting.

1.7 NEED FOR STUDYING FEM Now, a number of users friendly packages are available in the market. Hence one may ask the question ‘What is the need to study FEA?’. The above argument is not sound. The finite element knowledge makes a good engineer better while just user without the knowledge of FEA may produce more dangerous results. To use the FEA packages properly, the user must know the following points clearly: 1. Which elements are to be used for solving the problem in hand. 2. How to discritise to get good results. 3. How to introduce boundary conditions properly.

Introduction

7

4. How the element properties are developed and what are their limitations. 5. How the displays are developed in pre and post processor to understand their limitations. 6. To understand the difficulties involved in the development of FEA programs and hence the need for checking the commercially available packages with the results of standard cases. Unless user has the background of FEA, he may produce worst results and may go with overconfidence. Hence it is necessary that the users of FEA package should have sound knowledge of FEA.

1.8 WARNING TO FEA PACKAGE USERS When hand calculations are made, the designer always gets the feel of the structure and get rough idea about the expected results. This aspect cannot be ignored by any designer, whatever be the reliability of the program, a complex problem may be simplified with drastic assumptions and FEA results obtained. Check whether expected trend of the result is obtained. Then avoid drastic assumptions and get more refined results with FEA package. User must remember that structural behaviour is not dictated by the computer programs. Hence the designer should develop feel of the structure and make use of the programs to get numerical results which are close to structural behaviour.

QUESTIONS 1. Explain the concept of FEM briefly and outline the procedure. 2. Discuss the advantages and disadvantages of FEM over (i) Classical method (ii) Finite difference method. 3. Clearly point out the situations in which FEM is preferred over other methods. 4. When there are several FEM packages are available is there need to study this method? Discuss.

References 1. R. Courant, “Variational Methods for the Solutions of Problems of Equilibrium and Vibrations”, Bulletin of American Mathematical Society, Vol. 49, 1943. 2. G. Polya, Estimates for Eigen Values, Studies presented to Richard Von Mises, Academic Press, New York, 1954. 3. J. Hersch, “Equations Differentielles et Functions de cellules”, C.R. Acad. Science, Vol. 240, 1955. 4. H.F. Weinberger, “Upper and Lower Bounds for Eigen Values by Finite Difference Method”, Pure Applied Mathematics, Vol. 9, 1956. 5. J.H. Argyris and S. Kelsey, “Energy Theorems and Structural Analysis”, Aircraft Engineering, Vol. 27, 1955. 6. R.W. Clough, “The Finite Element Method in Plane Stress Analysis”, Proceeding of 2nd ASCE Conference on Electronic Computation, Pittsburg, PA, September, 1960. 7. R.J. Melosh, “Basis for the Derivation for the Direct Stiffness Method”, AIAA Journal, Vol. 1, 1963. 8. B. Fraeijs de Veubeke, “Upper and Lower Bounds in Matrix Structural Analysis”, AGARD ograph 72, B.F. de Veubeke (ed). Pergaman Press, New York, 1964. 9. T.H.H. Pian, “Derivation of Element Stiffness Matrices”, AIAA Journal, Vol. 2, 1964. pp. 556–57.

8

Finite Element Analysis 10. 11. 12. 13.

O.C. Zienkiewicz, The Finite Element Method in Engineering Science, McGraw-Hill, London 1971. K.H. Huebner, The Finite Element Methods for Engineers, John Wiley and Sons, 1971. Desai and Abel, Introduction to the Finite Element Method, CBS Publishers & Distributors, 1972. H.C. Martin and G. F. Carey, Introduction to Finite Element Analysis- Theory and Applications, Tata McGraw-Hill Publishing Company Ltd., New Delhi, 1975. 14. K.L. Bathe and E.L. Wilson, Finite Element Methods, Prentice Hall, 1976. 15. Y.K. Cheuny and M.F. Yeo, A Practical Introduction to Finite Element Analysis, Pitman Publishers, 1979. 16. R.D. Cook, D.S. Makus and M.F. Plesha, Concept and Applications of Finite Element Analysis, John Wiley and Sons, 1981. 17. J.N Reddy, An introduction to the Finite Element Method, McGraw-Hill International Edition, 1984. 18. C.S. Krishnamoorthy, Finite Element Analysis, Theory and Programming, Tata McGraw-Hill Publishing Company Ltd., New Delhi, 1987. 19. T.R. Chandrapatla and A.D. Belegundu, Introduction to Finite Elements in Engineering, Prentice Hall, 1991. 20. S. Rajasekharan, Finite Element Analysis in Engineering Design, Wheeler Publisher, 1993.

Basic Equations in Elasticity

9

2 Basic Equations in Elasticity 2.1 INTRODUCTION This chapter summarizes the results from theory of elasticity which are useful in solving the problems in structural and continuum mechanics by the finite element method.

2.2 STRESSES IN A TYPICAL ELEMENT In theory of elasticity, usually right hand rule is used for selecting the coordinate system. Fig. 2.1 shows various orientations of right hand rule of the coordinate systems. Equations derived for any one such orientation hold good for all other orientations of z z

y

x

y

x

(b)

y x

z

(a)

(c)

Fig. 2.1

coordinate system with right hand rule. In this Chapter orientation shown in Fig. 2.1(a) is used for the explanation. Fig. 2.2 shows a typical three dimensional element of size dx × dy × dz. Face abcd may be called as negative face of x and the face efgh as the positive face of x since the x value for face abcd is less than that for the face efgh. Similarly the face aehd is negative face of y and bfgc is positive face of y. Negative and positive faces of z are dhgc and aefb. The direct stresses σ and shearing stresses τ acting on the negative faces are shown in the Fig. 2.3 with suitable subscript. It may be noted that the first subscript of shearing stress is the plane and the second subscript is the direction. Thus the τ xy means shearing stress on the plane where x value is constant and y is the direction.

10

Finite Element Analysis z

z

dy e

a



dx dz 

f

b



d

xy



y

y

h

 yx





yz



x

xz

y zx

zy



z

c g x

x

Fig. 2.2

Fig. 2.3

In a stressed body, the values of stresses change from face to face of an element. Hence on positive face the various stresses acting are shown in Fig. 2.4 with superscript ‘+’. All these forces are listed in table 2.1 Note the sign convention: A stress is positive when it is on positive face in positive direction or on negative face in negative direction. In other words the stress is + ve when it is as shown in Figs 2.3 and 2.4. z





z 













zx

yz 

y

y 





 xz





zy



xy

x

x

Fig. 2.4



zx

Basic Equations in Elasticity Table 2.1

11

Stresses on a typical element Face

Stress on –ve Face

x

σx

σ +x = σ x +

τ xy

τ +xy

τ xz

τ +xz = τ xz +

σy

σ +y = σ y +

τ yx

τ +yx = τ yx +

τ yz

τ +yz = τ yz +

σz

σ +z = σ z +

τ zx

τ +zx

τ zy

τ +zy

y

z

Stresses on +ve Face

∂σ x dx ∂x ∂τ xy dx = τ xy + ∂x ∂τ xz dx ∂x ∂σ y ∂y ∂τ yx

∂y ∂τ yz ∂y

dy

dy dy

∂σ z dz ∂z ∂τ zx dz = τ zx + ∂z ∂τ zy dz = τ zy + ∂z

Note that stress on positive face is equal to the stress on negative face plus rate of change of that stress multiplied by the distance between the faces. z

Z dy

dx

Y dz

X

x

Fig. 2.5

y

12

Finite Element Analysis

Let the intensity of body forces acting on the element in x, y, z directions be X, Y and Z respectively as shown in Fig 2.5. The intensity of body forces are uniform over entire body. Hence the total body force in x, y, z direction on the element shown are given by (i) X dx dy dz in x – direction (ii) Y dx dy dz in y – direction and (iii) Z dx dy dz in z – direction

2.3 EQUATIONS OF EQUILIBRIUM Considering all the forces acting, we can write equations of equilibrium for the element.

∑F

=0

x

σ +x dy dz − σ x dy dz + τ +yx dx dz − τ yx dx dz + τ +zx dx dy − τ zx dx dy + X dx dy dz = 0

FGσ H

i.e.

x

+

F GH

IJ K

I JK

∂τ yx ∂σ x dx dy dz − σ x dy dz + τ yx + dy dx dz − τ yx dx dz ∂x ∂y

FG H

+ τ zx +

IJ K

∂τ zx dz dy dx − τ zx dx dy + X dx dy dz = 0 ∂z

…(i)

Simplifying and then dividing throughout by dx dy dz, we get

∂σ x ∂τ yx ∂τ zx + + + X=0 ∂x ∂y ∂z Similarly ΣFy = 0 and ΣFz = 0 equilibrium conditions give,

∂τ xy

and

∂σ y

…(ii)

∂τ yz ∂σ z ∂τ xz + + +Z =0 ∂x ∂y ∂z

…(iii)

∂y

+

∂τ zy

+Y =0

∂x

+

∂z

Now, Σ moment about x-axis = 0 through the centroid of the element gives

τ +yz dx dz

i.e.

Fτ GH

yz

+

∂τ yz ∂y

I JK

dy dx dy

LM N dz LF − MG τ dx dy 2 MNH

OP Q

dy dy dy dy + + τ yz dx dz − τ zy + τ zy dx dz =0 dx dz 2 2 2 2 dz + τ yz 2

zy

+

∂τ yz ∂z

I JK

dz dx dy

OP PQ

dz dz + τ zy dx dz =0 2 2

Neglecting the small quantity of higher (4th) order and dividing throughout by dx dy dz, we get

τ yz = τ zy Similarly the moment equilibrium conditions about y-axis and z-axis result into τ xz = τ zx

…(iv) …(v)

Basic Equations in Elasticity

τ xy = τ yx

and

13 …(vi)

Thus the stress vector is

σ

τ

= σ x σ y σ z τ xy τ yz τ xz

…(2.1)

and the equations of equilibrium are

∂τ xy ∂τ xz ∂σ x + + + X=0 ∂x ∂y ∂z

∂τ xy ∂x

+

∂σ y ∂y

∂τ yz

+

∂z

+Y =0

∂τ xz ∂τ yz ∂σ z + + +Z=0 ∂x ∂y ∂z

…(2.2)

τ xy = τ yx , τ yz = τ zy and τ xz = τ zx

…(2.3)

and and note that

2.4 STRAINS Corresponding to the six stress components given in equation 2.1, the state of strain at a point may be divided into six strain components as shown below:

lεq

T

= εx εy εz γ

xy

γ

yz

γ

…(2.4)

yx

2.5 STRAIN DISPLACEMENT EQUATIONS Taking displacement components in x, y, z directions as u, v and w respectively, the relations among components of strains and components of displacements are

LMF ∂u I MNGH ∂x JK ∂v 1 LF ∂u I + MG J ε = ∂y 2 MH ∂y K N ∂w 1 LF ∂u I + MG J ε = ∂z 2 MH ∂z K N εx =

∂u 1 + ∂x 2

2

2

y

z

γ xy =

FG ∂v IJ + FG ∂w IJ OP H ∂x K H ∂x K QP F ∂v I F ∂w I O +G J +G J P H ∂y K H ∂y K PQ F ∂ v I F ∂w I O +G J +G J P H ∂z K H ∂z K PQ 2

2

2

2

2

2

+

2

∂v ∂u ∂u ∂u ∂v ∂v ∂w ∂w + + ⋅ + ⋅ + ⋅ ∂x ∂y ∂x ∂y ∂x ∂y ∂x ∂y

14

Finite Element Analysis

and

γ yz =

∂w ∂v ∂u ∂u ∂v ∂v ∂w ∂w + + ⋅ + ⋅ + ⋅ ∂y ∂z ∂y ∂z ∂y ∂z ∂y ∂z

γ

∂u ∂w ∂u ∂u ∂v ∂v ∂w ∂w + + ⋅ + ⋅ + ⋅ ∂z ∂x ∂x ∂z ∂x ∂z ∂x ∂z

xz

=

…(2.5)

In equation 2.5, strains are expressed upto the accuracy of second order (quadratic) changes in displacements. These equations may be simplified to the first (linear) order accuracy only by dropping the second order changes terms. Then linear strain – displacement relation is given by:

εx =

∂u ∂x

γ xy =

∂u ∂v + ∂x ∂y

εy =

∂v ∂y

γ

yz

=

∂w ∂v + ∂y ∂z

εz =

∂w ∂z

γ

xz

=

∂w ∂u + ∂x ∂z

…(2.6)

Equations 2.6 are used in small deflection theories and equations 2.5 in large deflection theories.

2.6 LINEAR CONSTITUTIVE EQUATIONS The constitutive law expresses the relationship among stresses and strains. In theory of elasticity, usually it is considered as linear. In one dimensional stress analysis, the linear constitutive law is stress is proportional to strain and the constant of proportionality is called Young’s modulus. It is very well known as Hooke’s law. The similar relation is expressed among the six components of stresses and strains and is called ‘Generalized Hookes Law”. This may be stated as:

R|σ ||σ S|τσ ||τ |Tτ

x y z

xy yz xz

or in matrix form

U| R| D || ||D V| = S|DD || || D |W |TD

11 21 31 41 51 61

D12 D22 D32 D42 D52 D62

D13 D23 D33 D43 D53 D63

D14 D24 D34 D44 D54 D64

D15 D25 D35 D45 D55 D65

D16 D26 D36 D46 D56 D66

U| || V| || |W

R| ε || ε S|γε ||γ |Tγ

x y z xy yz xz

U| || V| || |W

…(2.7)

lσ q = D lεq,

where D is 6 × 6 matrix of constants of elasticity to be determined by experimental investigations for each material. As D is symmetric matrix [Dij = Dji], there are 21 material properties for linear elastic Anisotropic Materials. Certain materials exhibit symmetry with respect to planes within the body. Such materials are called Ortho tropic materials. Hence for orthotropic materials, the number of material constants reduce to 9 as shown below:

Basic Equations in Elasticity

R|σ ||σ S|τσ ||τ |Tτ

x y z

xy yz xz

U| R|D || || V| = S| || || |W |T

D12 D22

11

0 0 0 D44

D13 D23 D33

Sym

0 0 0 0 D55

0 0 0 0 0 D66

U| || V| || |W

R| ε || ε S|γε ||γ |Tγ

x y z xy yz xz

U| || V| || |W

15

…(2.8)

Using the Young’s Modulii and Poisons ratio terms the above relation may be expressed as:

εx =

σy σx σ − µ yx − µ zx z Ex Ey Ez

ε y = − µ xy

σy σ σx + − µ zy z Ex Ey Ez

σy σ σx − µ yz + z Ex Ey Ez τ xy τ yz τ , γ yz = , γ zx = zx γ xy = G xy G yz G zx ε z = − µ xz

…(2.9)

Note that there are 12 material properties in equations 2.9. However only nine of these are independent because the following relations exist

Ey Ey E Ex Ez E = , = z , = x µ xy µ yx µ yz µ zy µ zx µ xz

…(2.10)

For Isotropic Materials the above set of equations are further simplified. An isotropic material is the one that has same material property in all directions. In other word for isotropic materials, Ex = Ey = Ez say E and

µ xy = µ yx = µ yz = µ zy = µ xz = µ zx say µ

…(2.11)

Hence for a three dimensional problem, the strain stress relation for isotropic material is,

R| ε || ε || |S ε ||γ || ||γ |Tγ

x

y

z

xy

yz xz

U| R 1 || || E || || |V = ||S || || || || || || |W T

µ E 1 E

−

µ E µ − E 1 E

−

0

0

0

0

0

0

1− µ 2

0 1− µ 2

U| | 0 | || 0 | V 0 | || 0 | 1 − µ| | 2 W 0

R|σ ||σ || |Sσ ||τ || ||τ |Tτ

x

y

z

xy

yz xz

U| || || |V || || || |W

…(2.12)

16

Finite Element Analysis

Since G =

R|σ ||σ ||σ S|τ ||τ || Tτ

x y z

xy yz

xz

E and stress – strain relation is 2(1 − µ )

U| R|1 − µ || || || | V| = b1 + µg Eb1 − 2µ g |S| || || || || T W

µ 1− µ

µ µ 1− µ

0 0 0 1 − 2µ 2

0 0 0 0 1 − 2µ 2

U| || | 0 | V| 0 | | 1 − 2µ | 2 | W 0 0 0

R| ε || ε || ε S|γ ||γ || Tγ

x y z xy yz xz

U| || || V| …(2.13) || || W

In case of two dimensional elasticity, the above relations get further simplified. There are two types of two dimensional elastic problems, namely plane stress and plane strain problems.

Plane Stress Problems The thin plates subject to forces in their plane only, fall under this category of the problems. Fig. 2.6 shows a typical plane stress problem. In this, there is y y

x

o

o

Fig. 2.6

no force in the z-direction and no variation of any forces in z-direction. Hence

σ z = τ xz = τ yz = 0 The conditions τ xz = τ yz = 0 give γ

xz

=γ

yz

= 0 and the condition σ z = 0 gives,

b

g

σ z = µε x + µε y + 1 − µ ε z = 0

z

Basic Equations in Elasticity

εz = −

i.e.

d

µ εx + εy 1− µ

17

i

If this is substituted in equation 2.13 the constitutive law reduces to

R|σ S|σ |Tτ

x y xy

U| V| = 1 −Eµ |W

2

LM 1 MMµ MM0 N

µ 1

0 0 1− µ 2

0

OP Rε PP |Sε PP ||Tγ Q

x y xy

U| V| |W

…(2.14)

Plane Strain Problems A long body subject to significant lateral forces but very little longitudinal forces falls under this category of problems. Examples of such problems are pipes, long strip footings, retaining walls, gravity dams, tunnels, etc. (refer Fig. 2.7). In these problems, except for a small distance at the ends, state of stress is represented by any small longitudinal strip. The displacement in longitudinal direction (z-direction) is zero in typical strip. Hence the strain components,

y y

x x z z (a)

(b)

y y

x x z

z (c)

(d)

Fig. 2.7

18

Finite Element Analysis

y

X z (e)

Fig. 2.7 (contd)

ε z = γ xz = γ yz = 0 γ xz = γ yz = 0 means τ xz and τ yz are zero. ε z = 0 means εz =

σz E

−µ

(σ x + σ y ) E

=0

σ z = µ (σ x + σ y )

i.e.

Hence equation 2.13 when applied to plane strains problems reduces to

R|σ S|σ |Tτ

x y xy

F1 − µ U| G V| = (1 + µ )(E1 − 2µ) GG µ GH 0 |W

µ 1− µ 0

0 0 1 − 2µ 2

I Rε JJ | JJ S|γε K |T

x y xy

U| V| |W

…(2.15)

Axi-Symmetric Problems Axi-symmetric structures are those which can be generated by rotating a line or curve about an axis. Cylinders (refer Fig. 2.8) are the common examples of axisymmetric structures. If such structures are subjected to axisymmetric loadings like uniform internal or external pressures, uniform self weight or live load uniform over the surface, there exist symmetry about any axis. The advantage of symmetry may be made use to simplify the analysis. In these problems cylindrical coordinates can be used advantageously. Because of symmetry, the stress components are independent of the angular ( θ ) coordinate. Hence all derivatives with respect to θ vanish i.e. in these cases.

v = γ rθ = γ θ z = τ rθ = τ θ z = 0

Basic Equations in Elasticity

19

z, w

z, w

ru ,

ru ,

Q

Q ru ,

ru ,

(a)

(b)

Fig. 2.8

Hence there are only four nonzero components. The strain displacement relations for these components are

εr =

∂u , ∂r

εθ = γ rz =

u , r

εz =

∂w and ∂z

∂u ∂w + ∂z ∂r

…(2.16)

In these cases stress-strain relation is

R|σ |Sσ ||σ Tτ

r z

θ

rz

F1 − µ U| GG E |V = || (1 + µ )(1 − 2µ ) GG GH W

µ 1− µ

µ µ 1− µ

I Rε JJ ||ε 0 J Sε 1 − 2µ J | J |γ 2 K T 0 0

r z

θ rz

U| |V || W

…(2.17)

QUESTIONS 1. Draw a typical three dimensional element and indicate state of stress in their positive senses. 2. Derive the equations of equilibrium in case of a three dimensional stress system.

20

Finite Element Analysis 3. State and explain generalized Hooke’s law. 4. Give strain displacement relations in case of a three dimensional elasticity problem upto (i) accuracy of linear terms only (ii) accuracy of quadratic terms. 5. Explain the terms, ‘Anisotropic’, ‘Orthotropic’ and ‘Isotropic’ as applied to material properties. 6. Give constitutive laws for three dimensional problems of (i) orthotropic materials (ii) isotropic materials. 7. Explain the terms ‘Plane stress’ and ‘Plane strain’ problems. Give constitutive laws for these cases. 8. Explain the term ‘Axi-symmetric problems’ and give constitutive law for such problems.

Matrix Displacement Formulation

21

3 Matrix Displacement Formulation 3.1 INTRODUCTION Though mathematicians, physicists and stress analysts worked independently in the field of FEM, it is the matrix displacement formulation of the stress analysts which lead to fast development of FEM. Infact till the word FEM became popular, stress analyst worked in this field in the name of matrix displacement method. In matrix displacement method stiffness matrix of an element is assembled by direct approach while in FEM though direct stiffness matrix may be treated as an approach for assembling element properties (stiffness matrix as far as stress analysis is concerned), it is the energy approached which has revolutionized entire FEM. Hence in this chapter, a brief explanation of matrix displacement method is presented and solution techniques for simultaneous equations are discussed briefly.

3.2 MATRIX DISPLACEMENT EQUATIONS The standard form of matrix displacement equation is,

[k ] {δ } = {F} where

[k] is stiffness matrix

{δ } is displacement vector and {F} is force vector in the coordinate directions The element kij of stiffness matrix maybe defined as the force at coordinate i due to unit displacement in coordinate direction j. The direct method of assembling stiffness matrix for few standard cases is briefly given in this article.

1. Bar Element Common problems in this category are the bars and columns with varying cross section subjected to axial forces as shown in Fig. 3.1. For such bar with cross section A, Young’s Modulus E and length L (Fig. 3.2 (a)) extension/shortening δ is given by

δ =

PL EA

22

Finite Element Analysis P1

L1 P2

A3

P3

A2

P2

A1

x

L2

P1 L3

L2

L1

P3

(a)

L3

(b)

Fig. 3.1 A, E 1

2

P

P

1

L

(b)

(a)

1

L (c)

Fig. 3.2

∴ P=

EA δ L

EA L By giving unit displacement in coordinate direction 1, the forces development in the coordinate direction 1 and 2 can be found (Fig. 3.2 (b)). Hence from the definition of stiffness matrix, ∴ If δ = 1, P =

EA EA and k21 = − L L Similarly giving unit displacement in coordinate direction 2 (refer Fig. 3.2 (c)), we get k11 =

Matrix Displacement Formulation

k12 = − k =

Thus,

23

EA EA and k22 = L L

LM N

OP Q

EA 1 −1 L −1 1

…(3.5)

2. Truss Element Members of the trusses are subjected to axial forces only, but their orientation in the plane may be at any angle to the coordinate directions selected. Figure 3.3 shows a typical case in a plane truss. Figure 3.4 (a) shows a typical member of the truss with Young’s Modulus E, cross sectional area A, length L and at angle θ to x-axis y 



3



2

1

x

Fig. 3.3

(i) Unit displacement of end 1 in x-direction. Due to this, displacement along the axis is 1 × cos θ as shown in Fig. 3.4 (b). Hence forces development at the ends are as shown in figure.

EA cos θ L From the definition of elements of stiffness matrix, we get P=

k11 = P cos θ = k21 = P sin θ =

EA cos 2 θ L

EA cos θ sin θ L

k31 = − P cos θ = −

EA cos 2 θ L

k41 = − P sin θ = −

EA cos θ sin θ L

24

Finite Element Analysis 4

P

P

3 E,

2 

A,

L 1×

s Co





1

 1 (b)

P

(a)

1

s co



P 1×

P

(c)

P 1

1

P

 sin

1

P

(d)

sin



(e)

Fig. 3.4

(ii) Unit displacement in coordinate direction 2; This case is shown in Fig. 3.4 (c). In this case axial deformation is 1 × sin θ and the forces developed at each end are as shown in the figure. ∴P =

EA sin θ L

k12 = P cos θ = k22 = P sin θ =

EA sin θ cos θ L EA 2 sin θ L

k32 = − P cos θ = −

EA sin θ cos θ L

k42 = − P sin θ = −

EA 2 sin θ L

(iii) Unit displacement in coordinate direction 3, Extension along the axis is 1 × sin θ and hence the forces developed are as shown in the Fig. 3.4 (d) ∴P =

EA cos θ L

k13 = − P cos θ = −

EA cos 2 θ L

Matrix Displacement Formulation

k23 = − P sin θ = −

25

EA cos θ sin θ L

k33 = P cos θ =

EA cos 2 θ L

k43 = P sin θ =

EA cos θ sin θ L

(vi) Due to unit displacement in coordinate direction 4, Extension of the bar is equal to 1 × sin θ , and hence the forces developed are as shown in Fig. 3.4 (e). ∴P =

EA sin θ L

k14 = − P cos θ = −

EA sin θ cos θ L

k24 = − P sin θ = −

EA 2 sin θ L

k34 = P cos θ =

EA sin θ cos θ L

k44 = P sin θ =

EA 2 sin θ L

∴ The stiffness matrix is

LM cos θ EA M cos θ sin θ = L M –cos θ MM−cosθ sinθ N 2

k

2

cosθ sin θ sin 2 θ −cos θ sin θ –sin 2 θ

LM l EA M lm = L M −l MM−lm N 2

2

lm m2 −lm −m 2

−cos 2 θ −cosθ sin θ cos2 θ cos θ sin θ −l 2 −lm −lm − m 2 l2 lm lm m2

OP PP PP Q

−cos θ sin θ –sin 2 θ cosθ sin θ sin 2 θ

OP PP PP Q …(3.6)

Where l and m are the direction cosines of the member i.e. l = cos θ and m = cos (90 – θ ) = sin θ . (v) Beam Element In the analysis of continuous beams normally axial deformation is negligible (small deflection theory) and hence only two unknowns may be taken at each end of a element (Fig. 3.5). Typical element and the coordinates of displacements selected are shown in Fig. 3.5 (b). The end forces

26

Finite Element Analysis developed due to unit displacement in all the four coordinate directions are shown in Fig. 3.6 (a, b, c, d). y

1

3

E1, I1 x

2

5

7

E2, I2 4

L1

E3, I3 6

L2

9

E4, I4 8

L3

10

L4

z (a)

1

3

E,I, L 2

4

(b)

Fig. 3.5 6 EI L2

6 EI

12EI L3

1 12 EI L3

(a)

6EI L2

4EI L 6 EI L2

12 EI

6 EI

L3

L2

L2

 =1

2EI L

(b)

6 EI

2EI L

L2  =1 4EI L

1 12 EI

6 EI L2

6 EI

L3

(c)

L2

(d)

Fig. 3.6

From the definition of stiffness matrix and looking at positive senses indicated, we can write (a) Due to unit displacement in coordinate direction 1, k11 =

12 EI 3

L

k21 =

6EI 2

L

k31 = −

12 EI 3

L

k41 =

6EI

k42 =

2 EI L

L2

(b) Due to unit displacement in coordinate direction 2, 6 EI 4 EI k32 = − 2 k22 = L L2 L (c) Due to unit displacement in coordinate direction 3, k12 =

6EI

k13 = −

12 EI L3

k23 = −

6 EI L2

k33 =

12 EI L3

k43 = −

6 EI L2

Matrix Displacement Formulation

27

(d) Due to unit displacement in coordinate direction 4, k14 =

∴ k

6EI

k24 =

2

L

LM 12 FI M 6 L = L M−12 MN 6L

2 EI L

k34 = −

6 L −12 6 L 4L2 −6 L 2L2 −6 L 12 −6 L 2L2 −6 L 4 L2

3

6EI

k44 =

2

L

OP PP PQ

4 EI L

…(3.7)

If axial deformations in the beam elements are to be considered as in case of columns of frames, etc. (Fig. 3.7), it may be observed that axial force do not affect values of bending moment and shear force and vice versa is also true. Hence stiffness matrix for the element shown in Fig. 3.8 is obtained by combining the stiffness matrices of bar element and beam element and arranging in proper locations. For this case

k

LM EA MM L MM 0 M 0 =M MM− EA MM 0L MM MN 0

0

0

12EI

6 EI

L3 6EI L2

L2 4 EI L

0

0

−

12EI

−

L3 6EI

6 EI

L2 2EI L

L2

−

EA L 0

−

0 EA L

0

0

0

12EI

L3 6 EI − 2 L

0

OP 6 EI P P L P 2EI P L PP 0 P P 6 EI P − P L P 4 EI P L PQ

0

2

12EI −

L3 6 EI

…(3.8)

2

L2

(b)

(a)

Fig. 3.7 2

5 1

4 6

3

Fig. 3.8

28

Finite Element Analysis The following special features of matrix displacement equations are worth noting: (i) The matrix is having diagonal dominance and is positive definite. Hence in the solution process there is no need to rearrange the equations to get diagonal dominance. (ii) The matrix is symmetric. It is obvious from Maxwell’s reciprocal theorem. Hence only upper or lower triangular elements may be formed and others obtained using symmetry. (iii) The matrix is having banded nature i.e. the nonzero elements of stiffness matrix are concentrated near the diagonal of the matrix. The elements away from the diagonal are zero. Considerable saving is effected in storage requirement of stiffness matrix in the memory of computers by avoiding storage of zero values of stiffness matrices. The banded nature of matrix is shown in Fig. 3.9. B I

I II

II III

III IV

IV

Q P V

V

N

VI

VI

R VII

VII

VIII

VIII

IX

IX

Fig. 3.9

In this case instead of storing N ×N size matrix only N × B size matrix can be stored.

3.3 SOLUTION OF MATRIX DISPLACEMENT EQUATIONS The matrix displacement equations are linear simultaneous equations. These equations can be solved using Gaussian elimination method. Let the equations to be solved be

Matrix Displacement Formulation

LMa MMa MMaM MM M MNa

21

a12 a22

a13 ... a1k ... a1n a23 ... a2 k ... a2n

k1

ak 2

ak 3 ... akk ... akn

n1

an2

an3 ... ank ... ann

11

OP PP PP PP PQ

R| x U| R|b U| ||x || ||b || S|xM V| = S|bM V| || M || || M || |Tx |W |Tb |W 1

1

2

2

k

k

n

n

29

…(3.2)

i.e. [A] {x} = {b} The Gauss elimination method consists in reducing A matrix to upper triangular matrix and then finding the variables xn, xn–1 …, xk…., x2, x1 by back substitution Step 1: To eliminate x1 in the lower equations: (i) First equation is maintained as it is (ii) For equations below 1,

aij(1) = aij − and bij(1) = bi −

ai1 aij a11 ai1 b1 a11

At the end of this, the equations will be

LMa MM 0 MM 0M MM M MN 0

11

a12 (1) a22

a13 ... a1k ... a1n (1) ... a2(1k) ... a2(1n) a23

ak(12)

(1) (1) ... akn ak(13) ... akk

an(12)

(1) (1) ... ann an(13) ... ank

OP PP PP PP PQ

R| x U| R| b U| ||x || ||b || S|xM V| = S|bM V| || M || || M || |Tx |W |Tb |W 1 2

1 (1) 2

k

(1) k

n

(1) n

The above process is called pivotal operation on a11. For pivotal operation on akk, no changes are made in kth row but for the rows below kth, aij( k ) = aij( k – 1) −

(k) and bi =

bk( k – 1) ( k – 1) akk

aik( k – 1) ( k – 1) akk

bk(

k – 1)

akj( k – 1) for i, j = k + 1, …, n.

for i = k + 1, …, n.

After n – 1 pivotal operations, matrix equation is of the form

30

Finite Element Analysis

LMa MM 0 MM 0M MM 0 MM M N0

11

a12 (1) a22 0

a13 ... (1) ... a23 ( 2) a33 ...

0

0 ...

0

0 ...

a1k ... a2(1k) ... a3(2k) ...

a1n a2(1n) a3( 2n)

( k – 1) ( k – 1) ... akn akk

0 ...

( n – 1) ann

OP PP PP PP PP Q

R| x U| R| b ||x || || b |SxM |V = |S bM ||x || ||b || M || || M Tx W Tb 1

3

1 (1) 2 (2 ) 3

k

( k – 1) k

n

( k – 1) n

2

U| || |V || || W

…(3.3)

From the last equation,

xn =

bn ann

and then, n

bi − xi =

∑a

ij

xj

j =i +1

aij

, i = n – 1, n – 2 ...1

…(3.4)

Thus the required solution is obtained.

3.4 TECHNIQUES OF SAVING COMPUTER MEMORY REQUIREMENTS In FEM size of stiffness matrix of size 1000 × 1000 or even more is not uncommon. Hence memory requirement for storing stiffness matrix is very high. If user tries to implement Gaussian elimination straight way as described above, ends up with the problem of shortage of memory. The following techniques are used to reduce memory requirement for storing the stiffness matrices: (i) Use of symmetry and banded nature (ii) Partitioning of matrix (Frontal solution). (iii) Skyline storage.

(i) Use of Symmetry and Banded Nature Since the stiffness matrix is always symmetric and banded in nature, techniques have been developed to store only semiband width of non-zero elements and get the solution. If B is the semiband width of N × N matrix we need to store only N × B elements as indicated in Fig. 3.9(b). The diagonal of the given matrix is stored as the first column of the modified matrix. The computer coding is modified to use modified matrix for the solution of the given problem. The modification required is,

aij = a′i, ( j − i + 1)

(ii) Partitioning of the Matrix For larger systems, even this method of storage may be inadequate. In such cases the partitioning of the matrix is made as shown in Fig. 3.10.

Matrix Displacement Formulation

31

B I

I II

II III

III IV

IV

Q P V

V

N

VI

VI

R VII

VII

VIII

VIII

IX

IX

Fig. 3.10

Then only few of the triangular sub-matrices need to be stored in the computer core at a given time, while the remaining portions are kept in peripheral shortage like hard disk. It may be noted that the elimination performed using one row affects only the triangle of element within the band below that row. For example, in Fig. 3.10 reduction involving row PQ modifies only the triangle PQR. This permits us to carry out the elimination with only few of the sub-matrices of Fig. 3.10 in core. Frontal Solution Technique is developed on this scheme.

(iii) Skyline Storage Further saving in memory requirement is by making use of skyline storage technique. In this system of storage, if there are zeros at the top of a column, only the elements starting from non-zero value need to be stored. The line separating the top zeros from the first non-zero element is called the skyline. For the matrix given below the skyline is indicated.

MM MM MM MM MM MN

a11

a 12 a 22

0 a23 a33

0

a 14 0

a 25

a 34 a 44

0 0 a 55

0 0

0 0

a 36 a 46 a56 a 66

0 0 a57 a 67 a 77

PP 0 P P a P 0 P P a P a P P a PQ 0 0

48

68

78

88

Sky line

32

Finite Element Analysis

QUESTIONS 1. Define stiffness matrix and explains its special features. 2. By direct stiffness matrix approach, determine stiffness matrix for (a) Bar Element (b) Truss Element (c) Beam element neglecting axial deformation (d) Beam element (Frame Element), considering axial deformation also. 3. Briefly explain various attempts made to reduce memory requirement in storing stiffness matrix. 4. Explain the term “Skyline Storage Technique”.

Element Shapes, Nodes, Nodal Unknowns and Coordinate Systems

33

4 Element Shapes, Nodes, Nodal Unknowns and Coordinate Systems 4.1 INTRODUCTION In this chapter, element shapes, types of nodes, order of the element, types of nodal unknowns, are disscussed. Global – Local coordinate systems and natural coordinate systems are explained. Before taking up mathematical aspect of finite element analysis, these preliminaries are to be understood.

4.2 ELEMENT SHAPES Based on the shapes elements can be classified as (i) One dimensional elements (ii) Two dimensional elements (iii) Axi-symmetric elements and (iv) Three dimensional elements.

One Dimensional Elements These elements are suitable for the analysis of one dimensional problem and may be called as line elements also. Figure 4.1 shows different types of one dimensional elements. 2

1

3

1

3

1

1

x

3

Fig. 4.1

x

2

4

n–1

x

2

4

n

One dimensional elements

2

x

34

Finite Element Analysis

Two Dimensional Elements We need two dimensional elements to solve two dimensional problems. Common two dimensional problems in stress analysis are plane stress, plane strain and plate problems. Two dimensional elements often used is three noded triangular element shown in Fig. 4.2. It has the distinction of being the first and most used element. These elements are known as Constant Strain Triangles (CST) or Linear Displacement Triangles. 3

2

1

Fig. 4.2

Constant strain triangle

Six noded and ten noded triangular elements (Fig. 4.3) are also used by the analysts. Six noded triangular element is known as Linear Strain Triangle (LST) or as Quadratic Displacement Triangle. Ten noded 3

3

8 6

7

5 9

4

1

(a)

Fig. 4.3

2

1

10

4

6

5

2

(b)

(a) Linear strain triangle (b) Quadratic strain triangle

triangular elements are known as Quadratic Strain Triangles (QST) or Cubic Displacement Triangles. One can think of trying the use of still higher order triangular elements like Cubic Strain Triangles and Quartic Strain Triangles. A simple but less used two dimensional element is the four noded rectangular element whose sides are parallel to the global coordinate systems (Fig. 4.5). This systems is easy to construct automatically but it is not well suited to approximate inclined boundaries.

Element Shapes, Nodes, Nodal Unknowns and Coordinate Systems 3

35

3

10

12

9

11 16

13

13

11

10

8

12 15

14 1

4

7

6

5

15

2

19

17

14 18 4

1

20

8

21 6

5

7

2

(b)

(a)

Fig. 4.4

9

(a) Cubic strain triangle (15 noded) (b) Quartic strain triangle (21 noded) y 4

3

1

2

x

o

Fig. 4.5

4 noded rectangular element

Rectangular elements of higher order also can be used. Figure 4.6 shows a family of Lagrange rectangle in which nodes are in the form of grid points. Figure 4.7 shows the family of Serendipity rectangles which are having nodes only along the external boundaries.

Fig. 4.6

Lagrange family rectangular elements

36

Finite Element Analysis

Fig. 4.7 Serendipity family rectangular elements

Quadrilateral Elements are also used in finite element analysis (Fig. 4.8). Initially quadrilateral elements were developed by combining triangular elements (Fig. 4.9). But it has taken back stage after isoparametric concept was developed. Isoparametric concept is based on using same functions for defining geometries and nodal unknowns. Even higher order triangular elements may be used to generate quadrilateral elements. y 3

4

2

1

x

o

Fig. 4.8 y

Quardilateral element

y

y

3 4

2 1

x

Fig. 4.9

x

Quardilateral elements generated using triangular elements

x

Element Shapes, Nodes, Nodal Unknowns and Coordinate Systems

37

Using isoparametric concept even curved elements are developed to take care of boundaries with curved shapes (Fig. 4.10). y

y

x y

x y

x

Fig. 4.10

x

Curved two dimensional elements

Axi-symmetric Elements These are also known as ring type elements. These elements are useful for the analysis of axi-symmetric problems such as analysis of cylindrical storage tanks, shafts, rocket nozzles. Axi-symmetric elements can be constructed from one or two dimensional elements. One dimensional axi-symmetric element is a conical frustum and a two dimensional axi-symmetric element is a ring with a triangular or quadrilateral cross section. Two such elements are shown in Fig. 4.11.

Fig. 4.11

Axis-symmetric elements

38

Finite Element Analysis

Three Dimensional Elements Similar to the triangle for two dimensional problems tetrahedron is the basic element for three dimensional problems (Fig. 4.12). Tetrahedron is having four nodes, one at each corner. Three dimensional elements with eight nodes are either in the form of a general hexahedron or a rectangular prism, which is a particular case of a hexahedron. The rectangular prism element is many times called as a brick element also. In these elements also one can think of using higher order elements. (Fig. 4.12). 8

4

7

4

3

3

5 6 2

1

1

2

(a)

(b) 8

4 7

3

5 6 1

(d) 2 (c)

Fig. 4.12 (a) Tetrahedron element (b) Rectangular prism (brick) element (c) Arbitrary hexahedron element (d) Three dimensional quadratic element

4.3 NODES Nodes are the selected finite points at which basic unknowns (displacements in elasticity problems) are to be determined in the finite element analysis. The basic unknowns at any point inside the element are determined

Element Shapes, Nodes, Nodal Unknowns and Coordinate Systems

39

by using approximating/interpolation/shape functions in terms of the nodal values of the element. There are two types of nodes viz. external nodes and internal nodes. External nodes are those which occur on the edges/ surface of the elements and they may be common to two or more elements. In Fig. 4.13, nodes, 1 and 2 in one dimensional element, nodes 1 to 9 in 10 noded triangular element and nodes 1 to 8 in 9 noded lagrangian element are external nodes. These nodes may be further classified as (i) Primary nodes and (ii) Secondary nodes. 3

1

3

4

2

1, 2, 3, 4 – Primary nodes 5, 6, 7, 8 – Secondary nodes 9 – Internal node 7

1, 2, 3 – Primary nodes 4, 5, 6, 7, 8, 9 – Secondary nodes 10 – Internal node 4

8

3

7

9

1

1, 2 – Primary nodes 3, 4 – Internal nodes

10

4

8

6

5

2

1

(a)

Fig. 4.13

9

6

5 (b)

2

(a) 10 noded triangular element (b) 9 noded Lagrange element

Primary nodes occur at the ends of one dimensional elements or at the corners in the two or three dimensional elements. Secondary nodes occur along the side of an element but not at corners. Figure 4.13 shows such nodes. Internal nodes are the one which occur inside an element. They are specific to the element selected i.e. there will not be any other element connecting to this node. Such nodes are selected to satisfy the requirement of geometric isotropy while choosing interpolation functions. Figure 4.13 shows such nodes for few typical cases.

4.4 NODAL UNKNOWNS Basic unknowns may be displacements for stress analysis, temperatures for heat flow problems and the potentials for fluid flow or in the magneticfield problems. In the problems like truss analysis, plane stress and plane strain, it is enough if the continuity of only displacements are satisfied, since there is no change in the slopes at any nodal point. Such problems are classified as ‘zeroth’ continuity problems and are indicated as C0continuity problem. In case of beams and plates, not only the continuity of displacements, but the slope continuity also should be ensured. Since the slope is the first derivative of displacement, this type of problems are classified as ‘First order continuity problems and are denoted as C1 – continuity problems. In exact

F∂ wI GH ∂x ∂y JK 2

plate bending analysis even second order

continuity should be ensured. Hence the actual nodal

40

Finite Element Analysis

unknowns in these problems are w,

∂w ∂w ∂ 2 w where w is displacement. Such problems are classified as , , ∂x ∂y ∂x ∂y

C2– continuity problems. In general Cr continuity problems are those in which nodal unknowns are to be basic unknowns and up to rth derivatives of the basic unknowns.

4.5 COORDINATE SYSTEMS The following terms are commonly referred in FEM (i) Global coordinates (ii) Local coordinates and (iii) Natural coordinates. However there is another term ‘generalized coordinates’ used for defining a polynomial form of interpolation function. This has nothing to do with the ‘coordinates’ term used here to define the location of points in the element.

Global Coordinates The coordinate system used to define the points in the entire structure is called global coordinate system. Figure 4.14 shows the cartesian global coordinate system used for some of the typical cases. 1

2

(1)

x1

3

(2)

x

x2

x3

(a)

y 2

(6)

(5)

(1) (3) (2) 1

(8)

3

(7)

(4)

Fig. 4.14

(12) (9)

(11)

(10) 7 (b)

6

4

(13) 8

5

x

Global coordinate system

Local Coordinates For the convenience of deriving element properties, in FEM many times for each element a separate coordinate system is used. For example, for typical elements shown in Fig. 4.14, the local coordinates may be as shown in Fig. 4.15. However the final equations are to be formed in the common coordinate system i.e. global coordinate system only.

Element Shapes, Nodes, Nodal Unknowns and Coordinate Systems

41

1

x

y 2

y1 (1) (2) 1 1

x

x1 6

1

2 1

y (3)

y1

(4)

6

Fig. 4.15

x

1

7

6

Local coordinate system

Natural Coordinates A natural coordinate system is a coordinate system which permits the specification of a point within the element by a set of dimensionless numbers, whose magnitude never exceeds unity. It is obtained by assigning weightages to the nodal coordinates in defining the coordinate of any point inside the element. Hence such system has the property that ith coordinate has unit value at node i of the element and zero value at all other nodes. The use of natural coordinate system is advantages in assembling element properties (stiffness matrices), since closed form integrations formulae are available when the expressions are in natural coordinate systems. Natural coordinate systems for one dimensional, two dimensional and three dimensional elements are discussed below:

Natural Coordinates in One Dimension Consider the two noded line element shown in Fig. 4.16. let the natural coordinate of point P be (L1, L2) and the Cartesian coordinate be x. Node 1 and node 2 have the Cartesian coordinates x1 and x2. 1

P ( L1, L2)

x1

x

×

x – x1

x2 x2 – x

L

Fig. 4.16

2

x

42

Finite Element Analysis

Since natural coordinates are nothing but weightage to the nodal coordinates, total weightage at any point is unity i.e., L1 + L2=1 …(4.1) and also L1 x1 + L2 x2 = x …(4.2) i.e. In matrix form

LM 1 Nx

1

1 x2

OP RS L UV = RS1UV Q TL W TxW R L U L 1 1 OP = RS1UV ∴S V = M TL W Nx x Q TxW RSL UV = 1 LMx − x OP RS1UV TL W x − x N−1 1 Q TxW LM x −1OP RS1UV = 1 = 1 Q Tx W x x − x N− x 1

2

−1

1

2

1

2

τ

1 2

2

2

1

1

2

2

1

1

RS T

x2 − x 1 2 − x1 − x1 + x

UV W

Noting that x2 – x1 is the length of the element say, l, we can write

R|L U| R| x l− x U| S| V| = S| x − x V| TL W T l W 1

2

…(4.3)

1

2

The variation of L1 and L2 is shown in Fig. 4.17. L1 is 1 at node 1 and is zero at node 2 where as L2 is zero when referred to node 1 and is one when referred to node 2. The variation is linear. 1

2

1 (a) Variation of L1

1 (b) Variation of L2

Fig. 4.17

Variation of natural coordinates L1, L2

Element Shapes, Nodes, Nodal Unknowns and Coordinate Systems

43

The standard closed form integration over entire length is

z

x2

L1P Lq2 dx =

x1

b

p! q! l p + q + 1!

g

…(4.4)

Example. 4.1: Integrate the following over the entire length l of the element:

z l

(i)

z l

L21

dx

(ii)

o

L31 L2 dx

o

z l

Solution: (i)

L21 dx

o

z l

Using the standard formula,

L1P L2 dx = q

o

b

p! q! p + q + 1!

g

We note, p = 2, q = 0.

z l

Hence

b

L21 dx =

o

=

z l

(ii)

L31 L2 dx =

o

=

2! 0! l 2 + 0 + 1!

g

2 1 l= 2×3 3

Answer

3! 1! l 3 + 1 + 1!

b

g

3× 2 ×1 l l= 5×4×3×2 20

Answer

Natural Coordinate ξ In one dimensional problem, the following type of natural coordinate is also used. The natural coordinator ξ for any point in the element shown in Fig. 4.18 is defined as ξ =

FG H

PC x2 − x1 2

IJ K

where P is the point referred and

C is the centre point of nodes 1 and 2. The relationship between natural coordinate ξ and Cartesian coordinate x can be easily expressed as shown next page:

44

Finite Element Analysis 1

C

P

×

×

x1

x1  x 2 2

x





1

0



2

x

x2 

1

Natural coordinate ξ

Fig. 4.18

1 1 0

2

–1

Variation of ξ

Fig. 4.19

PC

x − xc

FG 1 IJ = 1 , where l is length of the element = x – x H 2K 2 x + x I 2 F = Gx − J 2 K l H x − x + 2x I 2 F = Gx − JK 2 l H l + 2x I 2 F = Gx − J l H 2 K lO 2 L = Mx − x − P l N 2Q

ξ=

2

2

1

2

1

1

1

1

1

l l ξ = x − x1 − 2 2 or

b

g

l 1 + ξ = x − x1 2

…(4.5)

It may be noted that, at node 1 where x = x1,

b

g

l 1 + ξ = x − x1 = 0 2 ∴ξ = −1

Element Shapes, Nodes, Nodal Unknowns and Coordinate Systems

45

and at node 2, where x = x2, we get

b

g

l 1 + ξ = x 2 − x1 = l 2

ξ =1

or

The variation of local coordinate ξ is as shown in Fig 4.19. The integration formula for integration over entire length is

z

x2

z 1

ξ pdx =

ξp

−1

x1

=

l 2 dξ since dx = dξ (refer equation 4.5) 2 l

l 1 ξp+1 2 p+1

1 −1

= 0, if p is odd

=

…(4.6a)

bg

l 1 l 2 = , if p is even 2 p+1 p+1

…(4.6b)

Natural Coordinates in Two Dimensions Natural coordinates for triangular and rectangular elements are discussed below: 1. Natural Coordinates for Triangular Elements: Consider the typical 3 noded triangular element shown in Fig. 4.20. Since there are three nodes, for any point there are three coordinates, say L1, L2 and L3. From the definition of natural coordinates, we have y C 3

P (L1, L2, L3)

B 2

A 1 X

Fig. 4.20

L1 + L2 + L3 = 1 L1 x1 + L2 x2 + L3 x3 = x L1 y1 + L2 y2 + L3 y3 = y

Typical 3 noded triangular element

…(4.7a) …(4.7b) …(4.7c)

46

Finite Element Analysis Expressing the above equations in matrix form,

LM 1 MMx Ny

1

1

1 x2 y2

1 x3 y3

R| L U| LM 1 S|L V| = Mx TL W MNy 1

2

1

3

1

R|L U| R|1U| S|L V| = S|xV| TL W T y W

OP PP Q

1 2

3

1 x2 y2

1 x3 y3

1 x2 y2

1 x3 y3

OP PP Q

−1

R|1U| S|xV| T yW

It can be shown that the determinant,

1 x1 y1

is equal to twice the area of triangle with corners (x1, y1), (x2, y2) and (x3, y3) Proof: Now,

1 Det = x1 y1

1 x2 y2

1 x 3 = x 2 y3 − x 3 y2 − x1 y3 − x3 y1 + x1 y2 − x2 y1 y3

b

g b

g b

g

Consider the triangle ABC shown in Fig. 4.21. Drop perpendiculars AD, BE and CF on to x-axis. y C 3 ( x3, y3) B (x , y ) 2 2 2

A (x1, y1) 1

x D

E

F

Fig. 4.21

Now, Area of triangle ABC =Area ADEC + Area CEFB – Area ADFB =

b

g

b

g

b

g

1 1 1 AD + CE DE + CE + BF EF − AD + BF DF 2 2 2

Element Shapes, Nodes, Nodal Unknowns and Coordinate Systems

b

g bx

g

b

g

b

g

=

1 y1 x3 − y1 x1 + y3 x 3 − y3 x1 + y3 x2 − y3 x 3 + y2 x 2 − y2 x 3 − y1 x 2 + y1 x1 − y2 x 2 + y2 x1 2

=

1 y1 x3 − y3 x1 + y3 x 2 − y2 x 3 − y1 x 2 + y2 x1 2

=

1 2

=

1 Det 2

2

− x3 −

1 y1 + y2 2

g bx

1 y1 + y3 2

− x1 +

1 y3 + y2 2

g bx

=

3

2

− x1

47

bx y − x y g − bx y − x y g + bx y − x y g 2 3

3 2

1 3

3 1

1 2

2 1

∴ Det = 2 Area of triangle ABC = 2A

R|L U| 1 LMx y − x y ∴ SL V = y −y |TL |W 2 A MMN x − x 1

2 3

2

2

3

3

3

2

LM MM N

3 2

x2 y3 − x3 y 2 1 x 3 y1 − x1 y3 = 2A x1 y 2 − x2 y1

…(4.8)

x 3 y1 − x1 y3 y3 − y1 x1 − x 3 y2 − y3 y3 − y1 y1 − y 2

x1 y2 − x2 y1 y1 − y2 x 2 − x1

x 3 − x2 x1 − x3 x2 − x1

OP PP Q

T

R|1U| S|xV| TyW

OP R|1U| 1 LMa PP S| xV| = 2 A MMa Q T y W Na

1 2 3

b1 b2 b3

c1 c2 c3

OP R|1U| PP S|xV| Q TyW

where

a1 = x2y3 – x3y2 a2 = x3y1 – x1y3 a3 = x1y2 – x2y1 b1 = y 2 – y 3 b2 = y 3 – y 1 b3 = y 1 – y 2 c1 = x3 – x2 c2 = x1 – x3 c3 = x2 – x1 [Note the cyclic order of subscript and absence of subscript of left hand term in right hand terms] Thus

R| L U| R| a + b x + c y U| | L | = || a + b2xA + c y || S| V| S| 2 A V| || L || || a + b x + c y || T W T 2A W 1

1

1

2

2

2

3

3

3

1

2

3

Referring to Fig. 4.22 and applying equation 4.8, we get Area of subtraingle CPB

LM1 = 2 A = Mx MNy 1

1 x2 y2

1 x3 y3

OP PP Q

…(4.9)

48

Finite Element Analysis y

y

C 3

C 3 Q1 P

A2

B 2

A1

B 2

A2 A3

A3

A 1

A 1

x

x

Fig. 4.22

Area coordinates for a triangle

i.e.,

2A1 = x2y3 – x3y2 – (xy3 – x3y) + xy2 – x2y = x2y3 – x3y2 + x(y2 – y3) + y(x3 – x2) = a1 + b1x + c1y Thirdly 2A2 = a2 + b2x + c2y and 2A3 = a3 + b3x + c3y Equation 4.9 reduces to ∴

R| L U| R| 22 AA U| R| A U| | L | = || 2 A || = 1 | A | S| V| S| 2 A V| A S| V| || L || || 2 A || || A || T W T 2A W T W 1

1

1

2

2

3

2

3

…(4.10)

3

where A1, A2 and A3 are the areas of sub-triangles PCB, PAC and PAB, which are opposite to nodes 1, 2 and 3 respectively. Hence the natural coordinates in triangles are also known as area coordinates. Note the following: To get the natural coordinate of A, P is to be moved to A. Then we find area A1 = A, A2 = A3 = 0 ∴ A (1, 0, 0) Similarly natural coordinates of B and C are A (0, 1, 0), C (0, 0,1) For the point Q1, natural coordinates are to by found by shifting P to Q, (refer Fig. 4.22(b)). In this case A1 = 0; A2 = ½ CQ1 h A3 = ½ BQ1 h Where h is perpendicular distance of A from BC.

Element Shapes, Nodes, Nodal Unknowns and Coordinate Systems

49

Then A = ½ BC h ∴ L1 =

A1 =0 A

1 CQ1h CQ A2 1 = 2 = L2 = 1 A BC BC h 2 1 BQ1h A3 BQ1 2 = = L3 = 1 A BC BC h 2 Thus natural coordinate representation of Q1 is

FG CQ , BQ IJ H BC BC K F CQ , 0, AQ IJ Q =G H AC BC K F BQ , AQ , 0IJ Q =G H AB AB K Q1 = 0,

1

1

2

Thirdly,

2

2

3

and

3

3

z

The closed form integration for the function

=

Example 4.2: Determine the values of (i)

z z z

L1 L2 L3 dA

A

(ii)

L31 L2 dA

A

(iii)

A

Solution: (i)

L21 L22 L33 dA

z

L1 L2 L3 dA

A

Now, p = 1, q = 1,

r=1

L1p Lq2 Lr3 dA p! q! r!

b p + q + r + 2g! 2 A

…(4.11)

50

Finite Element Analysis

z

∴ L1 L2 L3 dA =

(ii)

z

A

1! 1! 1! 1×1×1 A 2A = 2A = 1+1+1+ 2! 5×4×3×2×1 60

b

g

L31 L2 dA

A

In this case, p = 3,

(iii)

z

z

q = 1,

L31 L2 dA =

A

r=0

3 × 2 × 1× 1 × 1

3! 1! 0!

b3 + 1 + 0 + 2g! 2 A = 6 × 5 × 4 × 3 × 2 × 1 2A = 60 A

L21 L22 L33 dA

A

In this case p = 2, q = 2, r = 3

z

∴ L21 L22 L33 dA = A

=

b

2! 2! 3! 2A 2 +2 + 3+ 2!

g

2 × 1 × 2 × 1× 3 × 2 × 1 A 2A = 9×8×7×6×5×4×3×2×1 7960

Natural Coordinates for Rectangular Elements Natural coordinates for rectangular elements are as shown in Fig. 4.23. In these cases the centroid of the area is the origin. The relationships between the local coordinates and the Cartesian coordinates are based on isparametric concept, which is taken up in the latter chapter. It may be noted here that the coordinates ξ and η vary from –1 to 1. The relationship between global coordinates and the natural coordinates are x = Σ Li xi and y = Σ Li yi . The derivation of Li are discussed in the chapter ‘isoparametric elements’. When the expressions are formed in these coordinate systems, instead of seeking integrations in the closed form expressions, numerical technique is usually employed. 4 (–1, 1)

n

3 (1, 1)

P  ,n 

O

1 (–1, –1)

Fig. 4.23

2 (1, –1)

Local coordinate system for a rectangular element

Element Shapes, Nodes, Nodal Unknowns and Coordinate Systems

51

Natural Coordinates in Three Dimensions Natural coordinates for a 4 noded tetrahedron may be derived and it results into volume coordinates. Consider the typical tetrahedron shown in Fig. 4.24. 4, D

3 C

P

2

y x

1 A

2 B

Fig. 4.24

Tetrahedron coordinates

The natural coordinates are related to the Cartesian coordinates as follows:

R|1U| R| 1 |Sx|V = |Sx ||y|| || y Tz W T z

1

1

1

1

1

x2 y2 z2

x3 y3 z3

U| x | V y | z | W 1

4

4

4

R| L U| |SL |V ||L || TL W 1 2

3

…(4.12)

4

The above equation may be solved by inverting the 4 × 4 matrix. It gives Li =

1 x where 6V = 1 y1 z1

and

1 x2 y2 z2

1 x3 y3 z3

1 (ai + bi x + ci y + di z ), for i = 1, 2, 3 and 4 6V

1 x4 = 6 × volume of tetrahedron defined by nodes 1, 2, 3 and 4 y4 z4 x2 a1 = y2

x3 y3

x4 y4

1

1

1

b1 = y2

y3

y4

z2

z3

z4

z2

z3

z4

…(4.13)

52

Finite Element Analysis

1

1

1

1

1

1

c1 = x2 z2

x3 z3

x4 z4

and d1 = x 2

x3 y3

x4 y4

y2

The other constants are obtained by cyclic changes in the subscripts. It may be noted that the above equations are valid only when the nodes are numbered so that nodes 1, 2 and 3 are ordered counter clockwise when viewed from node 4. It is also necessary that for coordinates system of right hand rule is strictly adhered to. If Vi is the volume of the smaller tetrahedron which has vertices P and the three nodes other than the node i, then the tetrahedron coordinates can be considered as volume coordinates, defined as Li =

Vi for i=1, 2, 3 and 4 V

…(4.14)

The closed form integration formula for the volume coordinates is

z

L1p Lq2 Lr3 Ls4 dV =

v

b

p! q! r! s! 6V p + q + r + s + 3!

g

Example 4.3: Find the values of the following: (i)

z v

Solution: (i)

L1 L2 L3 L4 dV

z

(ii)

z

L21 L2 L4 dV

v

L1 L2 L3 L4 dV

v

In this case p = 1, q = 1, r = 1, s = 1 Hence using equation

z z

b

L1 L2 L3 L4 dV =

b1 + 1 + 1 + 1 + 3g! 6V = 7 × 6 × 5 × 4 × 3 × 2 × 1 × 6V

v

we get,

v

(ii)

z

p! q! r! s! × 6V p+q+r+s+3

L1p Lq2 Lr3 Ls4 dV =

=

g

1! 1! 1! 1!

V Answer 840

L21 L2 L4 dV

v

In this case p = 2, q = 1, r = 0 and s = 1

z

∴ L21 L2 L4 dV = ν

2! 1! 0! 1!

b2 + 1 + 0 + 1 + 3g! 6V

1×1×1×1

…(4.15)

Element Shapes, Nodes, Nodal Unknowns and Coordinate Systems

=

53

V Answer 2 ×1×1×1 6V = 420 7 × 6 ×5× 4 × 3× 2 ×1

Natural Coordinates for Hexahedron Figure 4.25 shows a typical hexahedron with natural coordinates. It has origin at centroid of the hexahedron. It may be noted that natural coordinates vary from –1 to +1. The natural coordinates are related to global coordinates as



n

z 

y x

Fig. 4.25

x = Σ Li xi y = Σ Li yi and z = Σ Li zi

…(4.16)

The derivation of Li is discussed under the chapter ‘Isoparametric Elements’. Later for integration numerical technique is preferred. Hence no discussion is taken up here about the closed form integrations for such cases.

QUESTIONS 1. Explain the following terms clearly (i) Nodes, primary nodes, secondary nodes and internal nodes (ii) Local coordinates, global coordinates, natural coordinates and area coordinates. (iii) Higher order elements and lower order elements. 2. Explain the terms (i) Constant strain triangle (CST) (ii) Linear strain triangle(LST) and (iii) Quadratic strain triangles (QST). 3. Explain the term Cr-continuity.

54

Finite Element Analysis 4. Derive the expressions for natural coordinates for a two noded element (i) In terms of L1 and L2, when range is 0 to 1 (ii) In terms of ξ , when range is –1 to 1. 5. Derive expressions for natural coordinates in a CST element. Show that they are nothing but area coordinates.

Shape Functions

55

5 Shape Functions 5.1 INTRODUCTION In the finite element analysis aim is to find the field variables at nodal points by rigorous analysis, assuming at any point inside the element basic variable is a function of values at nodal points of the element. This function which relates the field variable at any point within the element to the field variables of nodal points is called shape function. This is also called as interpolation function and approximating function. In two dimensional stress analysis in which basic field variable is displacement, u = Σ N i ui , v = Σ N i vi

…(5.1)

where summation is over the number of nodes of the element. For example for three noded triangular element, displacement at P (x, y) is

u = Σ N i ui = N 1 u1 + N 2 u2 + N 3 u3 v = Σ N i vi = N 1 v1 + N 2 v 2 + N 3 v3

R|u U| |v || 0 O |u S V N PQ |v | ||u || |Tv |W 1

i.e.,

RSuUV = LM N TvW N 0

1

1

0 N1

0 N2

N2 0

N3 0

2

2

3

3

3

R|u U| |u || RSuUV = LM N 0 OP |Su V = LM N TvW N 0 N Q |v | N 0 ||v || |Tv |W lδq =

2

2

or

1

3

1

N2 0

N3 0

0 N2

3

3

1

3

3

2×1

0 N1

2

2

or

R|u U| |u || 0 O |u S V N PQ | v | ||v || |Tv |W 1

1

N

lδq

e 2× 6 6×1

…(5.2a)

56

Finite Element Analysis

where q is displacement at any point in the element [N] shape function

lδ q

e

is vector of nodal displacements

Similarly in case of 6 noded triangular element

lδq = 2×1

In case of 4 noded rectangular element or

lδ q = 2×1

N

lδ q

…(5.2b)

N

lδ q

…(5.2c)

e 2 × 12 12 × 1

e 2 ×8 8×1

5.2 POLYNOMIAL SHAPE FUNCTIONS Polynomials are commonly used as shape functions. There are two reasons for using them: (i) They are easy to handle mathematically i.e. differentiation and integration of polynomials is easy. (ii) Using polynomial any function can be approximated reasonably well. If a function is highly nonlinear we may have to approximate with higher order polynomial. Fig. 5.1 shows approximation of a nonlinear one dimensional function by polynomials of different order. u u

u   1 

2x

u

u   1 

2 x

3x

u

1

(a) Constant

(b) Linear

Fig. 5.1

(c) Quadratic

Approximation with polynomials

One Dimensional Polynomial Shape Function A general one dimensional polynomial shape function of nth Order is given by,

bg

u x = α 1 + α 2 x + α 3 x 2 + ... α n +1 x n

In matrix form u = G where and

lαq

…(5.3) …(5.4)

G = 1, x , x 2 ... x n

lαq

T

= α 1 α 2 α 3 ... α n + 1

Thus in one dimensional nth order complete polynomial there are m = n + 1 terms.

…(5.5)

Shape Functions

57

Two Dimensional Polynomial Shape Function A general form of two dimensional polynomial model is

b g v b x , yg = α x+α y + ... + α +α lδ q = R|S|uv bb xx ,, yyggU|V| = G lα q = LMNG0 G0 OPQ lαq T W

u x , y = α 1 + α 2 x + α 3 y + α 4 x 2 + α 5 xy + α 6 y 2 + α 7 x 3 ... + α m y n m +1

m+2

m+3

2m

…(5.6)

yn

1

or

…(5.7)

1

G1 = [1 x y x2 xy y2 x3 ... yn]

where

lαq

T

= α 1 α 2 α 3 α 4 ... α 2 m

It may be observed that in two dimensional problem, total number of terms m in a complete nth degree polynomial is

bn + 1gbn + 2g

m= For first order complete polynomial n = 1,

…(5.8)

2

b1 + 1gb1 + 2g = 3

∴m =

2

The first three terms are α 1 + α 2 x + α 3 y

b2 + 1gb2 + 2g = 6

Similarly for n = 2, m =

2

and we know the first six terms are,

α 1 + α 2 x + α 3 y + α 4 x 2 + α 5 xy + α 6 y 2 Another convenient way to remember complete two dimensional polynomial is in the form of Pascal Triangle shown in Fig. 5.2 1

x x x x4 x5 x6

2 2

3 2

x y

xy

3

4

y

xy

xy2

x y

x y x 5y

y

3

2

xy xy4

2

xy3

3

2

y

2

xy

3

3

y 4

xy2

3

xy

xy2

4

xy 5

Fig. 5.2

4

y

5

y6

Pascal triangle

Constant

1

Linear

3

Quadratic

6

Cubic

10

Quartic

15

Quintic

21

Hexadic

28

58

Finite Element Analysis

Three Dimensional Polynomial Shape Function A general three dimensional shape function of nth order complete polynomial is given by

b g v b x , y , zg = α w b x , y , zg = α

u x , y , z = α 1 + α 2 x + α 3 y + α 4 z + α 5 x 2 + ... + α m x n −1 z m +1 +

α m + 2 x + α m + 3 y + α m + 4 z + α m + 5 x 2 + ... + α 2 m x n − 1 z

2m + 1 +

α 2 m + 2 x + α 2 m + 3 y + α 2 m + 4 z + ... + α 3m x

R|u b x , y , zg U| LMG δ b x , y , z g = Sv b x , y , z g V = G lα q = M 0 ||Tw b x , y , zg||W MN 0

1

or

0 G1 0

0 0 G1

n −1

…(5.9)

z

OP PP lαq Q

…(5.10)

Where G1 = [1 x y z x2 xy y2 yz z2 zx …zn zn-1x … zxn-1]

lα q

and

T

= α 1 α 2 α 3 ... α 3m

It may be observed that a complete nth order polynomial in three dimensional case is having number of terms m given by the expression

m= Thus when n = 1, m =

bn + 1gbn + 2gbn + 3g 6

b1 + 1gb1 + 2gb1 + 3g = 4 6

α1 + α2 x + α 3 y + α 4 z

i.e. For n = 2, m =

b2 + 1gb2 + 2gb2 + 3g = 10

6 Thus second degree complete polynomial is

α 1 + α 2 x + α 3 y + α 4 z + α 5 x 2 + α 6 xy + α 7 y 2 + α 8 yz + α 9 z 2 + α 10 zx Complete polynomial in three dimensions may be expressed conveniently by a tetrahedron as shown in Fig. 5.3.

5.3 CONVERGENCE REQUIREMENTS OF SHAPE FUNCTIONS Numerical solutions are approximate solutions. Stiffness coefficients for a displacements model have higher magnitudes compared to those for the exact solutions. In other words the displacements obtained by finite element analysis are lesser than the exact values. Thus the FEM gives lower bound values. Hence it is desirable that as the finite element analysis mesh is refined, the solution approaches the exact values. This requirement is shown graphically in Fig. 5.4. In order to ensure this convergence criteria, the shape functions should satisfy the following requirement:

Shape Functions 1

x

z

Constant

1

Linear

4

Quadratic

10

Cubic

20

Quartic

35

59

y x

2

zx

z2 yz

xy y x

2

3

xz2

xz

z3

2

yz 2

2

xy x

xyz xy 2

4

y

3

xz 3

3

yz2

z4

2 2

xz

xz

3

2 xyz

xy

yz 3 xyz 2

2

xy

2

yz2

2

xyz2 3

yz

xy 3 y4

x 2z 2

x 3z

x4

x 2y 2

x 3y

xz 3

z4

xyz 2

x 2y 2

yz 3

y 2z 2

xy 2z

15 terms

xy

3

3

y z

y4

Fig. 5.3

Terms at base of quartic tetrahedron of polynomial

1. The displacement models must be continuous within the elements and the displacements must be compatible between the adjacent elements. The second part implies that the adjacent elements must deform without causing openings, overlaps or discontinuities between the elements. This requirement is called ‘compatibility requirement’ .

Finite Element Analysis

5

0

Exact solution

No. of elements

% Error

60

–5 FEM solution –10

Fig. 5.4

Convergence of FEM solution

According to Felippa and Clough this requirement is satisfied, if the displacement and its partial derivatives upto one order less than the highest order derivative appearing in strain energy function is continuous. Hence in plane stress and plane strain problems, it is enough if continuity of displacement is satisfied, since strain energy function includes only first order derivatives of the displacement (SE = ½ stress × strain). It implies, it is enough if C0 continuity is ensured in plane stress and plane strain problems. In case of flexure problems (beams, plates, shells) the strain

F GH

energy terms include second derivatives of displacements like

I JK

d2 w 1 M2 where M = − EI . 2 EI dx 2

Hence to satisfy compatibility requirement, not only displacement continuity but slope continuity ( C1 -continuity) should be satisfied. Hence in flexure problems displacements and their first derivatives are selected as nodal field variables. 2. The displacement model should include the rigid body displacements of the element. It means in displacement model there should be a term which permit all points on the element to experience the same displacement. It is obvious, if such term do not exists, shifting of the origin of the coordinate system will cause additional stresses and strains, which should not occur. In the displacement model, u = α1 + α 2 x + α 3 y

the term α 1 provides for the rigid body displacement. Hence to satisfy the requirement of rigid body displacement, there should be constant term in the shape function selected. 3. The displacement models must include the constant strain state of the element. This means, there should exist combination of values of polynomial terms that cause all points in the element to experience the same strain. One such combination should occur for each possible strain. The necessity of this requirement is understood physically, if we imagine the refinement of the mesh. As these elements approach infinitesimal size, the strains within the element approach constant values. Unless the shape function term includes these constant strain terms, we cannot hope to converge to a correct solution. In the displacement model,

u = α 1 + α 2 x + α 3 y + α 4 x 2 + ... + α m y n

Shape Functions

61

v = α m + 1 + α m + 2 x + α m + 3 y + α m + 4 x 2 + ... + α 2 m y n

α 2 and α m+2 provide for uniform strain ε x , α 3 and α m+3 provide for uniform strain ε y An additional consideration in the selection of polynomial shape function for the displacement model is that the pattern should be independent of the orientation of the local coordinate system. This property is known as Geometric Isotropy, Spatial Isotropy or Geometric Invariance. There are two simple guidelines to construct polynomial series with the desired property of isotropy: 1. Polynomial of order n that are complete, have geometric isotropy. 2. Polynomial of order n that are not complete, yet contain appropriate terms to preserve ‘symmetry’ have geometric isotropy. The simple test for this property is to interchange x and y in two dimensional problems or x, y, z in cyclic order in three dimensional problems and see that the total expression do not change. However the arbitrary constants may change. For example, we wish to construct a cubic polynomial expression for an element that has eight nodal values assigned to it. In this situation, we have to drop two terms from the complete cubic polynomial which contains 10 terms. To maintain geometric isotropy drop only terms that occur in symmetric pairs i.e. x3, y3 or x2y, xy2. Thus the acceptable eight term cubic polynomials shape function exhibiting geometric isotropy are

α 1 + α 2 x + α 3 y + α 4 x 2 + α 5 xy + α 6 y 2 + α 7 x 2 y + α 8 xy 2 and

α 1 + α 2 x + α 3 y + α 4 x 2 + α 5 xy + α 6 y 2 + α 7 x 3 + α 8 y 3

In finite element analysis, the safest approach to reach correct solution is to pick the shape functions that satisfy all the requirements. For some problems, however, choosing shape functions that meet all the requirements may be difficult and may involve excessive numerical computations. For this reason some investigators have ventured to formulate shape functions for the elements that do not meet compatibility requirements. In some cases acceptable convergence has been obtained. Such elements are called ‘non-conforming elements’. The main disadvantage of using non-conforming elements is that we no longer know in advance that correct solution is reached.

5.4 DERIVATION OF SHAPE FUNCTIONS USING POLYNOMIALS Initially shape functions were derived interms of Cartesian coordinates. Polynomial function were used for this. After natural coordinates were identified and its advantage was noticed researchers started deriving shape functions in terms of natural coordinates. By this approach more elements could be developed. Advantages of using Lagrangian and Hermetian functions were discovered latter. To handle few special cases, degeneration technique was also developed. In this artiae these methods are illustrated for various elements.

Polynomial Functions In Terms of Cartesian Coordinates In this approach polynomials with number of constants exactly equal to nodal degrees of freedom of the element are selected. Care is taken to see that geometric isotropy is not lost. Using nodal values number of equations equal to number of constants in the polynomials are formed and then the constants found. Then the shape functions are identified. This procedure may also be called as generalized coordinate approach, since

62

Finite Element Analysis

the constants in the polynomial are called as generalized coordinates. This procedure is illustrated with few cases below. Example 5.4: Using generalized coordinate approach, find shape functions for two noded bar/truss element. Solution: Figure 5.5 shows the typical truss element. In this case nodal unknowns are displacements u1 and u2 along x-axis. For this element we have to select polynomial with only two constants to represent displacement at any point in the elements. Hence we select

u = α1 + α 2 x

…(5.11)

where α 1 and α 2 are generalized coordinates. This polynomial satisfies compatibility and completeness requirement. Writing equation 5.11 in the matrix form we have, P

1

2

×

u1

1

x

2

u2

x1

x2 1

l

N1 1

N2

Fig. 5.5 Bar/Truss element with two nodes

u= 1 x

RSα UV Tα W 1

2

since u = u1 at node 1 and equal to u2 at node 2, we have

lδ q = RSTuu UVW = LMN11 xx OPQ RSTαα UVW RSα UV = LM1 x OP RSu UV = 1 LM x −1OP RSu UV = 1 LMx Tα W N1 x Q Tu W x − x N− x 1Q Tu W l N−1 Rα U 1 L x − x O Ru U ∴ u= 1 x S V = 1 x M PS V l N−1 1 Q Tu W Tα W Ru U L x − x x − x O RSu UV 1 x − x −x +x S V=M = l Tu W N l l PQ Tu W Ru U = N N S V= N u + N u Tu W 1

1

1

2

2

2

−1

T

1

1

1

2

2

2

2

1

2

1

1

1

2

2

1

1

2

2

1

2

2

1

2

2

2

1 1

1

1

2

1

1

2

2

2

OP RSu UV 1 Q Tu W

− x1

1

2

Shape Functions

N1 =

where

63

x2 − x x − x1 and N 2 = l l

Thus the shape function [N] is N = N1 N 2 =

LM x N

2

− x l

x − x1 l

OP Q

Answer

...5.12.

Variation of shape function N1 and N2 is shown in Fig. 5.5. (b). Example 5.5: Using polynomial functions (generalized coordinates) determine shape functions for a two noded beam element. 1

2

x1 = 0

x2 = l

u1,

u2,

1

2

l

1

N1

1

N2

1

N3

1

N4

Fig. 5.6

Variation of shape functions in a beam element

Solution: The typical beam element is shown in Fig. 5.6. (a). In this case C1 – continuing is to be satisfied, since strain energy expression involves second differentiation term unknowns are the displacement and slope. i.e.,

d 2w . Hence in this case at each node, dx 2

64

Finite Element Analysis

R|w U| lδ q = |Sθw |V || || Tθ W 1

1

2

2

θ1 =

where

and

∂w1 ∂x

∂w2 ∂x Since there are four nodal values, we select polynomial with four constants. Thus θ2 =

w = α 1 + α 2 x + α 3 x2 + α 4 x3

…(5.13)

Equation 5.13 satisfies compatibility and completeness requirement. Now,

∂w = α 2 + 2α 3 x + 3α 4 x 2 dx For convenience we select local coordinate system. i.e., x1 = 0 x2 = l θ =

∴ w1 = α 1

θ1 = α 2 w2 = α 1 + α 2 l + α 3 l 2 + α 4 l 3

θ 2 = α 2 + 2α 3l + 3α 4 l 2

R|w U| LM1 lδ q = |Sθw |V = MM01 || || M Tθ W N0 1

1

i.e.,

2

2

R|α |α ∴ S ||α Tα

1 2 3 4

U| LM1 |V = M0 || MM1 W N0

0

0

1

0

l

2

l

1 2l

OP 0 P l P P 3l Q 0

3

2

−1

0 1

0 0

l l2 1 2l

R|w U| |Sθ |V = ||w || 3l Tθ W 2

4

1 4

− 2l

2 3

2

4

1

1

3

1

2

OP R|α PP |Sα l P |Tαα 3l Q | LMl 0 MM 0 l MM 00 00 N 0 0

4

4

U| |V || W −3l 2 −2l 3 3l 2 −l 3

2l l2 −2l l2

OP Rw U PP ||Sθ ||V PP ||Tθw || Q W 1

1

2

2

Shape Functions

LM 1 MM −03 = MM l2 MN l 2

3

0 1 −2 l 1 l2

0 0 3 l2 −2 l3

0 0 −1 l 1 l2

65

OP Rw U PP ||θ || PP S|w V| PQ |Tθ |W 1

1

2

2

∴ w = α1 + α 2 x + α 3 x2 + α 4 x3

R|α |Sα ||α Tα

= 1 x x2 x3

L 3x = M1 − MN l

1 2 3 4

U| |V = 1 x x || W

2

x3

LM 1 MM −03 MM l2 MN l 2

3

0 1 −2 l 1 l2

0 0 3 l2 −2 l3

0 0 −1 l 1 l2

OP Rw U PP ||θ || PP S|w V| PQ |Tθ |W 1

1

2

2

Rw U OP ||Sθ ||V PQ |w | |Tθ |W 1

2

2

+

2x l

3

x−

3

2x 2 x −x x 3x x + 2 − 3 + 2 2 l l l l l l 2

3

2

3

2

3

1

2

2

lδ q

= N1 N 2 N 3 N 4 where and

e

= N

lδ q

e

N = N1 N 2 N 3 N 4 N1 = 1 − 3x 2

3x 2 2x3 + l2 l3 2x3

N2 = x −

2x2 x3 + 2 l l

…(5.13)

x2 x3 + 2 l l l l Variation of these function is shown in Fig. 5.6 (b) (Note that at node 1, N1=1, N3 =

2

−

3

N2= N3= N4=0, and

N4 = −

∂ N2 ∂ N1 ∂ N 3 ∂ N 4 = 1, = = = 0 similarly at node 2, ∂x ∂x ∂x ∂x

N1= N2 = N4 = 0, N3 =1 and

∂ N1 ∂ N 2 ∂ N 3 ∂ N4 = = = 0 and = 1, ∂x ∂x ∂x ∂x

Example 5.6: Determine the shape functions for the Constant Strain Triangle (CST). Use polynomial functions. Solution: Figure 5.7 shows a typical CST element. Let the nodal variables be u1, u2, u3, v1, v2 and v3 i.e.,

66

Finite Element Analysis y 3 (x3, y3) 2 (x2, y2)

1 (x1, y1)

x

Fig. 5.7

lδ q

T

= u1 u2 u3 v1 v 2 v 3

From the consideration of compatibility and completeness the following displacement model is selected u = α1 + α 2 x + α 3 y v = α 4 + α5 x + α 6 y

…(5.14)

∴ u1 = α 1 + α 2 x1 + α 3 y1 u2 = α 1 + α 2 x 2 + α 3 y 2 u3 = α 1 + α 2 x 3 + α 3 y 3

R|u U| LM1 S|u V| = M1 Tu W MN1 1

i.e.,

2

3

U| V| W

R| S| T

x1 x2 x3

LM MM N

1 x1 α1 ∴ α 2 = 1 x2 1 x3 α3 1 x1 1 x2 1 x3

Now

y1 1 y 2 = x1 y3 y1

1 x2 y2

OP R|α U| PP S|α V| Q Tα W y O Ru U P| | y P Su V y PQ | Tu |W

y1 y2 y3

1 2

3

1

1

2

2

3

3

1 x3 = 2 A y3

Where A is the area of triangle with vertices at (x1, y1), (x2, y2) and (x3, y3) i.e., the area of the element.

R|α U| 1 LMx y ∴ Sα V = x y |Tα |W 2 A MMN x y 1

2

3

2

3

1

3

1

2

− x 3 y2 − x1 y3 − x 2 y1

y2 − y 3 y3 − y1 y1 − y2

x3 − x 2 x1 − x3 x 2 − x1

OP PP Q

T

R|u U| S|u V| Tu W 1

2

3

Shape Functions

La 1 M = a 2A M MNa

b1 b2 b3

1

2 3

c1 c2 c3

OP PP Q

T

R|u U| 1 LMa S|u V| = 2 A Mb MNc Tu W 1

1

2

1

3

1

where

a1 = x2 y3 – x3 y2 a2 = x3 y1 – x1 y3 b1 = y 2 – y 3 b2 = y 3 – y 1 c1 = x3 – x2 c2 = x1 – x3 same as used in deriving natural coordinates.

a2 b2 c2

a3 b3 c3

67

OP R|u U| PP S|u V| Q Tu W 1

2

3

a3 = x1 y2 –x2 y1 b3 = y 1 – y 2 c3 = x2 – x1,

∴ u = α1 + α 2 x + α 3 y

= 1 x y

R|α S|α Tα

1 2 3

U| V| = 1 W

La + b x + c y =M N 2A 1

1

LM MM N

a1 a2 1 x y b1 b2 2A c1 c2

OP R|u U| PP S|u V| Q Tu W 1

2 3

a2 + b2 x + c2 y a3 + b3 x + c3 y 2A 2A

1

OP R|Suu U|V Q |Tu |W 1

2 3

R|u U| S|u V| = Tu W 1

= N1 N 2 N 3

a3 b3 c3

2

N

lδ q

…(5.15)

e

3

where

N = N1 N 2 N 3

and

N1 =

a1 + b1 x + c1 y a + b2 x + c2 y a3 + b3 x + c3 y N2 = 2 and N 3 = 2A 2A 2A

R|v U| S|v V| Tv W 1

Similarly

v = N1 N 2 N 3

2 3

R|u U| |u | 0 O |u | L N 0 O lδ q S V= N PQ |v | MN 0 N PQ ||v || |Tv |W 1

R|u b x , ygU| = L N ∴ u b x , yg = S V T|v b x , yg W| MN 0

1

2

N2 0

N3 0

0 N1

0 N2

3

3

1

2 3

e

…(5.16)

68

Finite Element Analysis

Shape Functions In Terms of Natural Coordinate Systems Using polynomial functions and natural coordinate systems, shape functions can be derived easily. This approach makes it possible to find shape functions for more elements. This approach is illustrated with few standard cases below: Example 5.7: For a two noded bar element, determine the shape functions. Use natural coordinate system. Solution: As there are only two nodal values in this case, only linear function in natural coordinates are to be taken. Fig. 5.8 shown the typical element. Thus P (L1, L2)

1

×

L1 = 1 L2 = 0

2

L1 = 0 L2 = 1

Fig. 5.8

u = α 1 L1 + α 2 L2 = L1 L2

LMα OP Nα Q 1

…(5.17)

2

RSu UV = LM1 Tu W N0 Rα U L1 ∴ S V=M Tα W N0

OP RSα UV [Since L = 1 and L = 0 at node 1 and L = 0 and L = 1 at node 2] Q Tα W 0O Ru U 1 L1 0O Ru U S V= M PS V 1PQ Tu W 1 N0 1Q Tu W Rα U ∴u= L L S V Tα W L1 0OP RSu UV = L L Ru U = L L M STu VW N0 1Q Tu W Ru U Since u = N N S V by definition of shape function Tu W ∴

0 1

1

2

1

1

2

1

2

2

−1

1

1

1

2

2

2

1

1

2

2

1

1

2

2

1

1

2

2

1

1

2

2

∴

N1 = L1 and N2 = L2

Example 5.8: Derive the expression for shape function for a two noded bar element taking natural coordinate ξ as varying from -1 to 1. Solution: The typical bar element in the natural coordinate ξ varying from –1 to 1 is shown in Fig. 5.9 (a). Since there are only two nodal values, a linear polynomial is to be selected. Let displacement at any point

bg

P ξ be

Shape Functions P  

1 ×

2

=0

= –1

1

×

69

2

=1 1

N1 1

N2

Fig. 5.9

Variation of shape function in bar element

u = α1 + α 2 ξ

RSα UV Tα W RSu UV = LM1 −1OP RSα UV Since ξ = − 1 at node 1 and ξ = 1 at node 2 Tu W N1 1 Q Tα W Rα U L1 −1OP RSu UV ∴ S V=M Tα W N1 1 Q Tu W 1 L 1 −1O Ru U 1 L 1 1O Ru U = S V= M PS V 1 + 1 MN +1 1 PQ Tu W 2 N−1 1Q Tu W Rα U 1 L 1 1O Ru U ∴ u= 1 ξ S V= 1 ξ M PS V 2 N−1 1Q Tu W Tα W Ru U L1 − ξ 1 + ξ O RSu UV = N N RSu UV 1 = 1−ξ 1+ξ S V = M 2 Tu W N 2 2 PQ Tu W Tu W = 1 ξ

1

…(5.18)

2

1

1

2

2

−1

1

1

2

2

−T

1

1

2

2

1

1

2

2

1

1

1

1

2

1− ξ 1+ξ and N 2 = 2 2 Variation of shape functions is shown in Fig. 5.9 (b).

where

N1 =

2

2

2

…(5.19)

Example 5.9: Determine the shape functions for a three noded bar element with natural coordinate system as shown in Fig. 5.10

70

Finite Element Analysis

=0

= –1

=1

×

1

1

×

3

3

2

×

P  

2

N1

N2

N3

Fig. 5.10

Variation of shape functions in 3 noded bar element

Solution: In this case there are three nodal unknowns. Hence a polynomial with 3 generalized coordinates as shown below is selected

u = α 1 + α 2 ξ + α 3ξ 2

R|u U| LM1 ∴ lδ q = Su V = M1 |Tu |W MN1 1

2

e

3

−1 1 1 1 0 0

…(5.20)

OP R|α PP S|α Q Tα

1 2 3

U| V| W

Since ξ 1 = − 1 , ξ 2 = 1 and ξ 3 = 0

R|α S|α Tα

1 2 3

U| LM1 V| = M1 W MN1

OP PP Q

−1 1 1 1 0 0

R|u U| S|u V| Tu W

−1

1

2 3

LM 0 1 = 0 0 + 1b −1g + 1b −1g M MN−2 LM MM N

OP PP lδ q Q

0 −1 1 1 = 0 1 1 2 2 0 −2

T

e

OP PP lδ q Q

1 −1 −1 −1 0 2

LM MM N

T

0 0 2 1 = −1 1 0 2 1 1 −2

e

OP PP lδ q Q

e

Shape Functions

u = α1 + α 2 ξ + α 3 ξ = 1 ξ ξ 2

Now,

2

OP L −ξ + ξ PP lδ q = MMN 2 Q ξ L1 O = M ξ bξ − 1g bξ + 1g 1 − ξ P lδ q 2 2 N Q N N N lδ q 1 1 = ξ bξ − 1g , N = ξ bξ + 1g , N = 1 − ξ 2 2 = 1 ξ ξ2

1 2

LM 0 MM−1 N1

0 2 1 0 1 −2

R|α S|α Tα

1 2 3

U| V| W

ξ + ξ2 1 − ξ2 2

2

e

71

OP lδ q PQ

e

2

e

1

where N1

2

3

…(5.21)

e

2

2

3

The variation of the shape functions are as shown in Fig. 5.10 (b). Example 5.10: Determine the shape functions for a Constant Strain Triangular (CST) element in terms of natural coordinate systems. Solution: Let the natural coordinates of nodes 1,2,3 be L1, L2, L3 and shape functions be N1, N2, N3. The typical CST element is shown in Fig. 5.11 (a). Since the CST element is a linear displacement model, let the displacement function be y 3

3

2

2

1 1

1

x (a) Typical CST element

(b) Variation of N1

Fig. 5.11

u = α 1 L1 + α 2 L2 + α 3 L3

= L1 L2 L3

R|α S|α Tα

1 2 3

U| V| W

…(5.22)

72

Finite Element Analysis

OP R|α PP S|α Q Tα

R|u U| LM1 ∴ Su V = M0 |Tu |W MN0

0 0 1 0 0 1

1

2 3

1 2 3

U| V| W

Since the natural coordinates of node 1,2,3 are (1,0,0), (0,1,0) and (0,0,1) respectively.

LM1 lαq = M0 MN0 u = L1 L2 L3

Hence

OP PP Q

0 0 1 0 0 1

R|α S|α Tα

1 2 3

−1

R|u U| LM1 S|u V| = M0 Tu W MN0

OP R|u U| PP S|u V| Q Tu W

0 0 1 0 0 1

1

2 3

U| V| = W

R|u U| S|u V| = Tu W

L1 L2 L3

1

= L1 L2 L3

2

LM1 0 MM0 1 N0 0 R|u U| S|u V| Tu W

1

2 3

OP R|u U| PP S|u V| Q Tu W

0 0 1

1

2 3

1

N1 N 2 N 3

3

2 3

Where N1 = L1, N2 = L2 and N3 = L3 Similarly we can show,

v = L1v1 + L2 v 2 + L3v3 = N 1v1 + N 2 v2 + N 3 v3

R|v U| S|v V| Tv W 1

= N1 N 2 N 3

2

…(5.23)

3

∴

RSuUV = LM L Tv W N 0

1

L2 0

L3 0

0

0

0

N1

N2

N3

OP lδ q = LMN 0 OP lδ q Q N 0 NQ e

e

where N1 = L1, N2 = L2 and N3 = L3, variation of N1 is shown in Fig. 5.11 (b). Example 5.11: Determine the shape function for Linear Strain Triangular (LST) element. Use natural coordinate system. Solution: Figure 5.12 (a) shows the typical LST element. As there are three nodes along any side, it can be easily seen that displacement varies in the quadratic form (one order higher than the variation of strain). As there are six nodal values we have to pick a polynomial with six constants. Taking all the quadratic terms in natural coordinate system, we can select shape function as:

u = α 1 L21 + α 2 L22 + α 3 L23 + α 4 L1 L2 + α 5 L2 L3 + α 6 L3 L1

…(5.24)

Shape Functions

73

y 3

3

3

5

5

5 2

2 6

2

6

6

4

1

4

4

1 1

1

1

x

(b) Variation of N1

(a) Typical LST element

(c) Variation of N 4

Fig. 5.12

L1 R|u U| MM 0 ||u || MM 01 u ∴ S V = M4 ||u || MM 0 ||Tuu ||W MM 1 MN 4 1

2 3 4

5

6

0 1 0 1 4 1 4 0

0 0 1

0 0 0 1 4

0 1 4 1 4

OP PP R|αα || P 0 P Sα P |α 0 P |α P ||α 1P T 4 PQ

0 0 0

0 0 0

0

0

1 4

0

0

1 2 3 4 5 6

U| || V| || |W

since natural coordinates at nodes 1 to 6 are (1,0,0,), (0,1,0), (0,0,1), ( ½, ½, 0), (0, ½, ½) and (½, 0, ½) respectively. Then, we have

luq

e

lq

= A α

where {u}e is the vector of nodal displacements in x directions, [A] is the matrix shown in the above equation

lq

and α is the vector of generalized coordinates (constants in polynomial)

A

−1

LM 1 MM 0 0 =M MM−1 MM−01 N

lαq =

A

−1

0 0 1 0 0 1 −1 0 −1 −1 0 −1

luq

0 0 0 4 0 0

0 0 0 0 4 0

0 0 0 0 0 4

e

and we have started with

u = α 1 L21 + α 2 L22 + α 3 L23 + α 4 L1 L2 + α 5 L2 L3 + α 6 L3 L1

OP PP PP PP PQ

74

Finite Element Analysis

= L21 L22 L23 L1 L2 L2 L3 L3 L1

1 2 3 4 5 6

A

= L21 L22 L23 L1 L2 L2 L3 L3 L1

LM 1 MM 0 MM−01 MM 0 MN−1

eL

2 1

je

U| || V| || |W

−1

= L21 L22 L23 L1 L2 L2 L3 L3 L1

=

luq

e

0 0 1 0 0 1 −1 0 −1 −1 0 −1

0 0 0 4 0 0

0 0 0 0 4 0

0 0 0 0 0 4

je

OP PP PP luq PP PQ

e

j

− L1 L2 − L3 L1 , L22 − L1 L2 − L2 L3 , L23 − L2 L3 − L3 L1 , 4 L1 L2 , 4 L2 L3 , 4 L3 L1

= N1 N 2 N 3 N 4 N 5 N 6 where

R|α ||α S|αα ||α |Tα

luq

luq

e

e

N1 = L12 − L1 L2 − L3 L1 , N 2 = L22 − L1 L2 − L2 L3 , N 3 = L23 − L2 L3 − L3 L1 N4 = 4L1L2, N5 = 4L2L3, and N6 = 4L3L1

Now, N1 = L12 − L1 L2 − L3 L1 , = L1 (L1 – L2 – L3) = L1 [L1 – (1 – L1)] since L1 + L2 + L3 = 1 = L1 (2L1 – 1)

b

g

Similarly N 2 = L2 2 L2 − 1

b

g

and N 3 = L3 2 L3 − 1

Similarly it can be proved that

v = N1 N 2 N 3 N 4 N 5 N 6

RSuUV = LM N 0 OP lδ q Tv W N 0 N Q lδ q = u u u u u

∴ where

lvq

e

…(5.24)

e

T e

1

2

3

4

5

u6 v1 v 2 v3 v4 v5 v6

Shape Functions

N = N1 N 2 N 3 N 4 N5 N 6

and

lδ q

75

e

N1 = L1 (2L1 – 1); N2 = L2 (2L2 – 1); N3 = L3 (2L3 – 1); N4 = 4L1L2; N5 = 4L2L3, and N6 = 4L3L1 Variation of N1 and N4 are shown in Fig. 5.12 (b). Example 5.12: Determine the shape functions for 4 noded rectangular elements. Use natural coordinate system. Solution: The typical 4 noded rectangular element is shown in Fig. 5.13. y 

4 (–1, 1)

3 (1, 1)

b

0

b

1 (–1, –1)

2 (1, –1)

a

a x

Fig. 5.13

Typical 4 noded rectangular element

Taking the centroid of the rectangle as origin and ξ and η as natural coordinates, we have

x − xc y − yc and η = a b where 2a × 2b is the size of the element as shown in Fig. 5.13 and xc, yc are the coordinates of the origin.

ξ=

We need a polynomial in two dimension with 4 constants. Such polynomial is obtained by dropping ξ 2 and η 2 terms in second degree polynomial. Such polynomial maintains geometric isotropy also. Thus, u = α1 + α 2 ξ + α 3 η + α 4 ξ η

R|u U| LM1 luq = |Suu |V = MM11 || || M Tu W N1 3 4

−1

2

e

OP R|α PP |Sα ||α P −1Q Tα

−1 −1 1 1 −1 −1 1 1 1

1

1

1 2 3 4

…(5.25)

U| |V || W

76

Finite Element Analysis

R|α |α ∴ S ||α Tα

1 2 3 4

U| LM1 |V = M1 || MM1 W N1

−1 −1 1

−1

1

1

−1

1

OP −1P 1P P −1Q 1

−1

R|u U| |Su |V = ||u || Tu W 1

2

A

−1

3

luq

e

4

It can be shown that

A

−1

LM 1 MM−41 =M 4 MM− 1 MM 14 N4

1 4 1 4 1 − 4 1 − 4

1 4 1 4 1 4 1 4

1 4 1 − 4 1 4 1 − 4

OP PP PP PP PQ

∴ u = α1 + α 2 ξ + α 3 η + α 4 ξη

= 1 ξ η ξη

R|α |Sα ||α Tα

1 2 3 4

U| |V = 1 ξ || W

η ξη

LM 1 MM−41 MM 41 MM− 4 MN 14

1 4 1 4 1 − 4 1 − 4

1 4 1 4 1 4 1 4

1 4 1 − 4 1 4 1 − 4

OP PP PP luq PP PQ

e

LMb1 − ξ gb1 − ηg b1 + ξgb1 − ηg b1 − ξgb1 + ηg b1 − ξgb1 + ηg OP luq 4 4 4 MN 4 PQ = N N N N luq = N luq =

1

2

3

4

e

e

e

where

N1 =

b1 − ξgb1 − ηg 4

N2 =

b1 + ξgb1 − ηg 4

N3 =

b1 + ξgb1 + ηg 4

N4 =

b1 − ξgb1 + ηg 4

In short N1 to N4 may be written as

Ni =

b

gb

g

1 1 + ξξ i 1 + ηη i for i = 1, 2, 3 and 4 4

…(5.26)

Shape Functions

R|v U| |Sv |V ||v || Tv W

77

1

v = N1 N 2 N 3 N 4

Similarly

2 3 4

∴

RSuUV = LM N Tv W N 0

N2 0

1

N3 0

N4 0

0

0

0

N1

N2

N3

Note: Ni = 1 at node i and is zero at all other nodes.

OP lδ q N Q 0

4

e

Example 5.13: Determine the shape function for quadratic rectangular element shown in Fig. 5.14 y  4 (–1, 1)

3 (1, 1)

7, (0, 1)

b 6 (1, 0) 8 (–1, 0)

b

5 (0, –1)

1 (–1, –1)

a

2 (1, –1)

a x

0

Fig. 5.14

Typical quadratic rectangular element

Solution: There are 8 nodal values for u and 8 for v. Hence the displacement function is to be selected with only 8 constant. The polynomial has to maintain geometric isotropy also. This may be obtained by dropping

ξ 3 and η 3 terms from complete 3rd order polynomial. Thus, u = α 1 + α 2 ξ + α 3 η + α 4 ξ 2 + α 5 ξη + α 6 η 2 + α 7 ξ 2 η + α 8 ξη 2

R|u U| LM1 ||u || MM1 ||uu || MM11 luq = Su V = M1 || || M ||uu || MM11 |Tu |W MMN1 1

2 3 4

e

5

6 7

8

−1 −1 1 −1 1 1 −1 1 0 −1 1 0 0 1 −1 0

1 1 1 −1 1 1 1 −1 0 0 1 0 0 0 1 0

OP R|α PP ||α PP ||αα PP S|α PP ||α PP ||αα QT

1 −1 −1 1 −1 1 1 1 1 1 1 −1 1 0 0 0 0 0 1 0 0 0 0 0

1 2 3 4 5 6 7 8

U| || || V| = A lαq || || W

…(5.27)

78

Finite Element Analysis

lα q =

or

A −1

luq

e

It may be shown that,

A

−1

LM−1 MM 0 0 M 1 1 = M 4 M1 MM 1 MM−1 MN−1

But, u = 1 ξ η ξ 2 ξη η 2 ξ 2 η ξη 2

2 2 2 −1 −1 −1 2 0 0 0 0 2 0 −2 0 0 0 −2 0 2 0 1 1 1 −2 0 −2 0 −1 1 −1 0 0 0 0 1 1 1 0 −2 0 −2 0 −2 0 −1 1 1 2 1 1 −1 0 −2 0 2

OP PP PP PP PP PP Q

lαq

luq 1 L1 = M b1 − ξ gb1 − ηgb −ξ − ηgb −ξ − η − 1g , b1 + ξ gb1 − ηgbξ − η − 1g , 4 N4 = 1 ξ η ξ 2 ξη η 2 ξ 2η ξη 2

b

A

−1

e

gb gb g 41 b1 − ξ gb1 + ηgb−ξ + η − 1g , b gb gb g , 12 b1 + ξ gb1 + ηgb1 − ηg,

1 1+ξ 1+η ξ +η −1 , 4 1 1+ξ 1−ξ η −1 2

b

gb

gb

g b luq

gb

gb

1 1 1+ ξ 1− ξ η +1 , 1− ξ 1+ η 1− η 2 2 = N1 N 2 N 3 N 4 N 5 N 6 N 7 N 8

gOPQ luq

e

…(5.28)

e

Where N1 …N8 are as defined above. In other words, for corner nodes (i = 1, 2, 3, 4)

Ni =

b

gb

gb

g

1 1 + ξξ i 1 + ηη i ξξ i + ηη i − 1 4

…(5.29)

For mid side nodes,

e

jb

g

…(5.30)

e

jb

g

…(5.31)

If ξ i = 0 then N i =

1 1 − ξ 2 1 + ηη i (i.e., for 5, 7) 2

If η i = 0 then N i =

1 1 − η 2 1 + ξξ i (i.e., for 6, 8) 2

Shape Functions

79

Thus,

RSuUV = LM N Tv W MN 0

OP Ru U V NPS Q Tv W 0

1× 8

1× 8

e

1× 8

1× 8

e

[Note: Ni =1 at node i and is zero at all other nodes, variation is quadratic] Example 5.14: Determine the shape functions for a tetrahedron element. Solution: The typical tetrahedron is as shown in Fig. 5.15. 4 3

P

2

1

Fig. 5.15

Tetrahedron element

In Chapter 4, we have seen that the natural coordinates for such element are volume coordinates

Li =

Vi V

Where Li = Natural coordinates Vi = Volume of sub tetrahedron formed by the point and the nodes except ith and V = Volume of the tetrahedron ∴ Natural coordinates for the node point 1, 2, 3 and 4 are 1(1,0,0,0), 2(0,1,0,0), 3(0,0,1,0) and 4(0,0,0,1) Let the displacement in x direction at any point P in the element be

u = α 1 L1 + α 2 L2 + α 3 L3 + α 4 L4

LM1 0 ∴ luq = M MM0 N0 R|α U| LM1 |α | 0 ∴ S V=M ||α || MM0 Tα W N0 e

1

2 3 4

0 1 0 0

0 0 1 0

0 1 0 0

0 0 1 0

OP R|α U| PP |Sα |V PQ ||Tαα ||W 0O 0PP luq 0P P 1Q

0 0 0 1

…(5.32)

1

2 3 4

−1

e

LM1 0 =M MM0 N0

0 1 0 0

0 0 1 0

OP PP luq PQ

0 0 0 1

e

80

Finite Element Analysis

∴ u = α 1 L1 + α 2 L2 + α 3 L3 + α 4 L4

R|α U| |Sα |V = L ||α || Tα W luq = N 1

= L1 L2 L3 L4

2

L2 L3 L4

1

3 4

= L1 L2 L3 L4

N2 N3 N4

1

e

LM1 0 0 0OP MM0 1 0 0PP luq MN00 00 10 01PQ luq = N luq

where N1 = L1, N2 = L2, N3 = L3 and N4 = L4 Similarly for displacement v and w, we get v = [N] {v}e and w = [N] {w}e

where

R|u U| LMM N 0 S|v V| = M 0 N TwW MMN 0 0 lδ q = u u u T e

OP 0 P lδ q P NP PQ

1× 4

1× 4

1× 4

1× 4

1× 4

1× 4

1× 4

1× 4

2

3

e

…(5.34)

0

1× 4

1

e

e

e

u4 v1 v2 v3 v4 w1 w2 w3 w4

Note: Ni = 1 at node i and is zero at all other nodes. Variation is linear. Example 5.15: Explain the method of finding shape function for a hexahedral element. Solution: The typical element selected is shown in Fig. 5.16. The natural coordinates of various nodal points are 8

7 

4

3 5

 6

1

2

Fig. 5.16

Typical hexahedron element

1 (1, –1, –1) 2 (1, 1, –1) 3(1, 1, 1) 4 (1, –1, 1) 5 (–1, –1, –1) 6 (–1, 1, –1) 7 (–1, 1, 1) 8 (–1, –1, 1) There are only eight nodal values for defining displacement inside the element. Hence polynomial with 8 constants is to be selected for shape function. Keeping in view that geometric isotropy is to be maintained the following polynomial is selected.

Shape Functions

u = α 1 + α 2ξ + α 3η + α 4ζ + α 5ξη + α 6 ηζ + α 7ζξ + α 8ξηζ

81

…(5.35)

R|α U| S| M V| Tα W 1

= 1 ξ ηζ ξη ηζ ζξ ξηζ

i.e.,

8

LM1 MM1 MM11 ∴ luq = MM1 MM1 MM11 N e

lq

∴ α = A

−1

luq

1 1 1 1 −1 −1 −1 −1

−1 1 1 −1 −1 1 1 −1

−1 −1 −1 1 1 1 1 −1 −1 1 −1 −1 −1 1 1 1

1 −1 1 −1 −1 1 −1 1

1 2 3 4 5 6 7 8

U| || || V| = A lαq || || W

e

[A]-1 can be found. Then

u = 1 ξ η ζ ξη ηζ ζξ ξηζ

where

OP R|α PP ||α PP ||αα PP S|α PP ||α PP ||αα QT

1 −1 −1 −1 1 1 −1 1 1 1 −1 1 −1 1 −1 −1

lαq

= 1 ξ η ζ ξη ηζ ζξ ξηζ

A

N = 1 ξ η ζ ξη ηζ ζη ξηζ

A

−1

luq

e

= N

luq

e

−1

= N1 N 2 N 3 N 4 N 5 N 6 N 7 N 8 It will be found that

Ni =

where

b

gb

gb

1 1 + ξξ i 1 + ηη i 1 + ζζ i 8

R|u U| LMM N 0 0 OPP S|v V| = M 0 N 0 P lδ q TwW MMN 0 0 N PPQ lδ q = u ... u , v ... v , w T e

1× 8

1× 8

1× 8

1× 8

1× 8

1× 8

1× 8

1× 8

1× 8

1

8

Note: Ni = 1 at node i and is zero at all other nodes.

1

e

8

1

... w8

g

…(5.36)

82

Finite Element Analysis

5.5 FINDING SHAPE FUNCTIONS USING LAGRANGE POLYNOMIALS If only continuity of basic unknown (displacement) is to be satisfied, Lagrange polynomials can be used to derive shape functions. Lagrange polynomial in one dimension is defined by

b g ∏ xx −− xx n

Lk x =

m

m=1 m≠ k

k

…(5.37)

m

Thus, if n = 5 and k = 3,

b g b xb x−−xxgbgbxx −− xx gbgb xx −−xx gbgbxx− −x xg g

L3 x =

1

3

1

2

3

4

2

3

5

4

3

5

Obviously equation 5.37 takes the value equal to zero at all points except at point k. At point k its value is unity. This is exactly the property required for the interpolation functions. Hence Lagrange Polynomial can be straight way used as shape functions for one dimensional problems. The following example illustrates it. Example 5.16: Using Lagrange polynomial find shape functions for (i) Two noded bar element (ii) Three noded bar element and (iii) Five noded bar element Plot the variation of shape functions. Solution: (i) Two noded bar element The typical two noded bar elements is shown in Fig. 5.17 (a) General Lagrange Polynomial is Typical element

Variation of N1

Variation of N2 (a)

Fig. 5.17

Lk =

(a) Two noded bar element

b x − x gb x − x g ... d x − x idx − x i ... b x − x g b x − x gb x − x g ... d x − x id x − x i ... b x − x g 1

k

Now n = 2. Hence when k = 1

1

k −1

2

k

2

k

k −1

k +1

k

k +1

n

k

n

Shape Functions

83

x − x2 x −x N = 2 same as found earlier , x1 − x2 x 2 − x1 1× 8

∴ N1 = L1 = When k = 2,

N 2 = L2 =

x − x1 , same as found earlier variation of N1 and N2 is also shown in Fig. 5.17 (a). x2 − x1

(ii) For Three Noded Element n = 3. Hence when k = 1,

b x − x gb x − x g b x − x gb x − x g

N1 = L1 =

2

1

2

3

1

3

When k = 2

N 2 = L2 =

b x − x gb x − x g b x − x gb x − x g 1

2

and

N 3 = L3 =

b x − x gb x − x g b x − x gb x − x g 1

3

1

1

3

2

3

2

3

2

1

3

2

Typical 3 noded bar element

Variation of N1

Variation of N2

Variation of N3

Fig. 5.17 (contd)

(b) Three noded bar element

The typical element and the variation of its shape functions are shown in Fig. 5.17 (b). (iii) For Five Noded Element n=5

N1 = L1 =

b x − x gb x − x gb x − x gb x − x g b gb gb gb g 2

3

4

5

x1 − x2 x1 − x 3 x1 − x4 x1 − x5

84

Finite Element Analysis

b x − x gb x − x gb x − x gb x − x g b x − x gb x − x gb x − x gb x − x g b x − x gb x − x gb x − x gb x − x g = b x − x gb x − x gb x − x gb x − x g b x − x gb x − x gb x − x gb x − x g = b x − x gb x − x gb x − x gb x − x g b x − x gb x − x gb x − x gb x − x g = b x − x gb x − x gb x − x gb x − x g

N 2 = L2 = N 3 = L3 N 4 = L4 N 5 = L5

1

2

1

3

1

4

1

5

1

3

2

3

3

2

4

2

1

4

2

1

2

2

4

3

5

5

3

4

3

5

5

5

3

4

2

5

4

4

3

2

1

5

2

5

4

3

5

4

The five noded bar element and variation of its shape functions are shown in Fig. 5.17 (c) 1

3

4 (a) Typical

5

2

noded element

1

Variation of N1 1 Variation of N3 1 Variation of N4 1 Variation of N5 1

Variation of N2

Fig. 5.17 (contd)

(c) Typical 5 noded bar element

Lagrange Polynomial Approach for Two Dimensional Elements Although Lagrangian interpolation functions are for only one dimension, we may extend the concept to two and three dimensions by forming the product of the functions which hold good for the individual one dimensional coordinate directions i.e.,

Shape Functions

bg bg

N1 = L1 ξ L1 η

85

…(5.38)

Thus for 4 noded rectangular element shown in Fig. 5.18,

bg bg

ξ − ξ 2 η − η4 ξ 1 − ξ 2 η1 − η 4

N1 = L1 ξ L1 η =

=

bξ − 1g

b

gb

g

b

gb

η −1 1 1 = ξ −1 η −1 = 1−ξ 1−η −1 − 1 −1 − 1 4 4 

4 (–1, 1)

g

3 (1, 1)

2 (1, –1)

1 (–1, –1)

Fig. 5. 18

Typical 4 noded rectangular element

bg bg

N 2 = L2 ξ L2 η =

=

cξ − b−1gh bη − 1g = bξ + 1gbη − 1g = 1 b1 + ξgb1 − ηg 4 1 − b −1g −1 − 1 b−4g

bg bg

N 3 = L3 ξ L3 η = =

ξ − ξ1 η − η3 ξ 2 − ξ1 η2 − η3

ξ − ξ 4 η − η2 ξ 3 − ξ4 η3 − η 2

b

gb

ξ +1 η+1 1+ξ 1+η 1 − −1 1 − −1 = 4

b g b g

bg bg

N 4 = L4 ξ L4 η =

g

ξ − ξ 3 η − η1 ξ 4 − ξ 3 η 4 − η1

bξ − 1g η + 1 = bξ − 1gbη + 1g = b1 − ξgb1 + ηg −1 − b1g 1 − b −1g −4 4 b1 + ξξ gb1 + ηη g = =

Thus

Ni

i

i

4 Example 5.17: Using Lagrange functions write shape functions for the nine noded rectangular element shown in Fig. 5.19.

86

Finite Element Analysis  8 (0, 1)

7 (–1, 1)

4 (–1, 0)

9 (1, 1)

6 (1, 0)

5 (0, 0)

1 (–1, –1)

3 (1, –1)

2 (0, –1)

Fig. 5.19

Typical nine noded rectangular element

Solution: The natural coordinates of various nodes are as shown in the figure. For the Co continuity element in two dimensions,

bg bg

N i = Li ξ Li η

where Li refers to Langrangian function at node i. In this case there are 3 nodes in each direction. Hence n = 3 in Lagrange function

b

gb gb

g b gb g b gb gb g bξ − 0gbξ − 1g bη − 0gbη − 1g = ξ bξ − 1g η bη − 1g = b−1 − 0gb−1 − 1g b−1 − 0gb−1 − 1g 4 bξ − ξ gbξ − ξ g bη − η gbη − η g N = bξ − ξ gbξ − ξ g bη − η gbη − η g bξ + 1gbξ − 1g bη − 0gbη − 1g = bξ + 1gbξ − 1gη bη − 1g = b0 + 1gb0 − 1g b−1 − 0gb−1 − 1g b−2g bξ − ξ gbξ − ξ g bη − η gbη − η g N = bξ − ξ gbξ − ξ g bη − η gbη − η g bξ + 1gbξ − 0g bη − 0gbη − 1g = bξ + 1g ξ η bη − 1g = b1 + 1gb1 − 0g b−1 − 0gb−1 − 1g 4 bξ − ξ gbξ − ξ g bη − η gbη − η g N = bξ − ξ gbξ − ξ g bη − η gbη − η g ξ − ξ2 ξ − ξ3 η − η4 η − η7 ξ 1 − ξ 2 ξ 1 − ξ 3 η1 − η 4 η1 − η 7

N1 =

1

3

5

8

2

2

1

2

1

3

2

2

5

2

6

8

9

3

3

1

3

2

5

3

6

6

3

9

1

7

4

4

5

4

6

4

1

4

7

Shape Functions

bξ − 0gbξ − 1g bη + 1gbη − 1g = ξ bξ − 1gbη + 1gbη − 1g b gb g b0 + 1gb0 − 1g −2 bξ − ξ gbξ − ξ g bη − η gbη − η g = bξ − ξ gbξ − ξ g bη − η gbη − η g bξ + 1gbξ − 1g bη + 1gbη − 1g = bξ + 1gbξ − 1gbη + 1gbη − 1g = b0 + 1gb0 − 1g b0 + 1gb0 − 1g 1 bξ − ξ gbξ − ξ g bη − η gbη − η g = bξ − ξ gbξ − ξ g bη − η gbη − η g bξ + 1gbξ − 0g bη + 1gbη − 1g = bξ + 1g ξ bη + 1gbη − 1g = b1 + 1gb1 − 0g b0 + 1gb0 − 1g −2 bξ − ξ gbξ − ξ g bη − η gbη − η g = ξ b − ξ gbξ − ξ g bη − η gbη − η g bξ − 0gbξ − 1g bη + 0gbη − 0g = ξ bξ − 1gbη + 1g η = b−1 − 0gb −1 − 1g b1 + 1gb1 − 0g 4 bξ − ξ gbξ − ξ g bη − η gbη − η g = bξ − ξ gbξ − ξ g bη − η gbη − η g bξ + 1gbξ − 1g bη + 1gbη − 0g = bξ + 1gbξ − 1gbη + 1gbηg = b0 + 1gb0 − 1g b1 + 1gb1 − 0g −2 bξ − ξ gbξ − ξ g bη − η gbη − η g = bξ − ξ gbξ − ξ g bη − η gbη − η g bξ + 1gbξ − 0g bη + 1gbη − 0g = bξ + 1g ξ bη + 1gbηg = b1 + 1gb1 − 0g b1 + 1gb1 − 0g 4 =

N5

−1 − 0 −1 − 1 4

5

N6

4

6

5

8

5

7

6

9

8

7

9

8

3

6

9

4

1

7

4

5

2

8

3

9

8

9

2

8

7

5

1

9

7

8

2

3

9

7

9

2

6

5

7

8

N9

5

8

7

N8

6

4

4

6

N7

87

5

6

9

8

3

9

Ni =

1 ξη ξ + ξ i η + η i 4

6

Thus in this case, for corner nodes,

b

gb

g

…(5.41)

For nodes 2 and 8 where ξ = 0 ;

Ni =

bξ + 1gbξ − 1g η bη + η g i

−2

88

Finite Element Analysis For nodes 4 and 6, where η i = 0

Ni = and for central node,

b

ξ ξ + ξi

g bη − 1gbη + 1g −2

bξ + 1gbξ − 1g bη + 1gbη − 1g

N5 =

4

Example 5.18: Using Langrange functions, derive shape function for hexahedron (brick) element. Solution: Typical hexahedron element is shown in Fig. 5.20. The coordinates for various nodal points are 1 (1, –1, –1), 2 (1, 1, –1), 3 (1, 1, 1), 4 (1, –1, 1), 5 (–1, –1, –1), 6 (–1, 1, –1), 7 (–1, 1, 1), 8 (–1, –1, 1) In general, shape function by Lagrange function for three dimensional case is given by

bg bg bg

N i = Li ξ Li η Li ζ 8



7

4 3



5

6

z 1 2

y x

Fig. 5.20

∴ N1 =

bξ − ξ g 5

ξ1 − ξ 5

Typical hexahedron element

b gb b gb

bg bg bg bξ − ξ g bη − η g bζ − ζ g = bξ − ξ g bη − η g bζ − ζ g

N 2 = L2 ξ L2 η L2 ζ 6

2

6

gb gb

g b g

gb

gb

g

ξ +1 η −1 ζ −1 ξ +1 η−1 ζ −1 η − η2 ζ − ζ 4 = = 1 + 1 −1 − 1 −1 − 1 8 η1 − η 2 ζ 1 − ζ 4

1

2

1

3

2

3

Shape Functions

89

bξ + 1gbη + 1gbζ − 1g = bξ + 1gbη + 1gbζ − 1g −8 b1 + 1gb1 + 1gb−1 − 1g = L bξ g L bηg L bζ g bξ − ξ g η − η ζ − ζ = =

N3

3

3

7

3

4

2

ξ 3 − ξ 7 η3 − η4 ζ 3 − ζ 2

=

b

gb

bg bg bg

N 4 = L4 ξ L4 η L4 ζ = =

gb

g

ξ +1 η+1 ζ +1 ξ + 1η + 1ζ + 1 = 1+1 1+1 1+1 8

b

ξ − ξ 8 η − η3 ζ − ζ 1 ξ 4 − ξ8 η4 − η3 ζ 4 − ζ 1

gb

gb

g

ξ +1 η −1 ζ +1 ξ +1 η −1 ζ +1 = −8 1 + 1 −1 − 1 1 + 1

bg bg bg bξ − ξ g η − η ζ − ζ =

N 5 = L5 ξ L5 η L5 ζ 1

6

8

ξ 5 − ξ1 η5 − η6 ζ 5 − ζ 8

=

bξ − 1g η − 1 ζ − 1 = bξ − 1gbη − 1gbζ − 1g −1 − 1 −1 − 1 −1 − 1

bg bg bg

−8

N 6 = L6 ξ L6 η L6 ζ =

b

bg bg bg

N 7 = L7 ξ L7 η L7 ζ =

gb

gb

gb

gb

g

ξ −1 η +1 ζ −1 ξ − ξ 2 η − η5 ζ − ζ 7 ξ −1 η+1 ζ −1 = = ξ 6 − ξ 2 η 6 − η5 ζ 6 − ζ 7 −1 − 1 1 + 1 −1 − 1 8

b

g

ξ −1 η +1 ζ +1 ξ − ξ 3 η − η8 ζ − ζ 6 ξ − 1 η + 1ζ + 1 = = ξ 7 − ξ 3 η 7 − η8 ζ 8 − ζ 6 −1 − 1 1 + 1 1 + 1 −8

bg bg bg

N 8 = L8 ξ L8 η L8 ζ =

b

gb

g

b

gb

gb

g

ξ −1 η −1 ζ +1 ξ −1 η−1 ζ +1 ξ − ξ 4 η − η7 ζ − ζ 5 = = ξ 8 − ξ 4 η8 − η 7 ζ 8 − ζ 5 −1 − 1 −1 − 1 1 + 1 8

In general it may be noted that,

Ni =

b

1 1 + ξξ i 8

g b1 + ηη g b1 + ζζ g i

i

…(5.42)

5.6 SHAPE FUNCTIONS FOR SERENDIPITY FAMILY ELEMENTS Figure 5.21 shows Serendipity family elements. These elements may be called as boundary node family elements also. In these elements nodes are only on the boundaries. Zienkiewcz called them as ‘Serendip family’ elements by referring to the famous princess of Serendip noted for chance discoveries. The terms linear, quadratic, cubic and quartic are used since the variation of shape functions about a boundary is of that order. The shape functions are found from the consideration that Ni = 1 for ith node and is zero when referred

90

Finite Element Analysis

to any other node. Discovery of these elements clubbed with isoparametric concept (explained in ch.13) has made major break through in the finite element analysis. In this article derivation of shape functions for this family of elements is presented, through examples.

(a) Linear

(b) Quadratic

(c) Cubic

(d) Quartic

Fig. 5.21

Serendipity family elements

Example 5.19: Using ‘serendipity concept’ derive shape functions for 4 noded rectangular element. Solution: Figure 5.22 shows the typical element in natural coordinate system. 

4 (–1, 1)

 =1

3 (1, 1)



1

0

2 (1, –1)

1 (–1, –1)

Fig. 5.22

Typical serendipity linear element

N1 has to satisfy the conditions (a) along ξ = 1, N1 = 0. (b) along η = 1, N1 = 0 and

(c) at ξ = − 1,

η = − 1 N1 = 1.

Shape Functions Hence let

b

gb

g

N1 = C 1 − ξ 1 − η , where C is arbitrary constant. Conditions (a) and (b) are satisfied. Condition (c) gives,

b gb g b1 − ξgb1 − ηg =

1 = C 1 + 1 1 + 1 or C = ∴ N1

1 4

4

On the same lines we can get,

and

N2 =

b1 + ξ gb1 − ηg

N3 =

b1 + ξgb1 + ηg

N4 =

b1 − ξgb1 − ηg

4

4

4

These are same as given in equations 5.26 and 5.40, which were derived from different approaches. Example 5.20: Using serendipity concept find shape functions for quadratic serendipity family element. Solution: Figure 5.23 shows a typical element of this type.  4 (–1, 1)

 

1

7 (0, 1)

0 3 (1, 1)   

8 ( –1, 0)

0

6 (1.0, 0)   

0

 1

0

2 (1, –1)

1 (–1, –1)

Fig. 5.23

 1

5 (0, –1)

 

1

0

Typical serendipity quadratic element

The conditions to be satisfied by N1 are, (a) along line ξ = 1,

N1 = 0

(b) along line η = 1,

N1 = 0

(c) along line 5–8,

N1 = 0

91

92 i.e.,

Finite Element Analysis

ξ + η + 1 = 0. (d) At ξ = − 1 and η = −1 , N1 = 1 Hence let

b

gb

gb

N1 = C 1 − ξ 1 − η 1 + ξ + η

g

It satisfies the requirements a, b, c. In other words it ensues N1 = 0 at nodes 2, 3, … 8. From condition (d),

b gb gb

g

1= C 1+ 1 1+ 1 1− 1−1 ∴C = − ∴ For the corner node

N1 = −

b1 − ξgb1 − ηgb1 + ξ + ηg

N2 = −

b1 + ξgb1 − ηgb1 − ξ + ηg

N3 = −

b1 + ξ gb1 + ηgb1 − ξ − ηg

N4 = −

b1 − ξgb1 + ηgb1 + ξ − ηg

1 4

4

Similarly we can show that

and

4

4

4

For mid-side node 5, the conditions to be satisfied are (a) Along ξ = 1 N5 = 0 (b) Along η = 1 N5 = 0 (c) Along ξ = − 1 N5 = 0 (d) At node 5 where ξ = 0 , η = − 1 , N5 = 1

b

gb

gb

∴ Let N 5 = C 1 − ξ 1 − η 1 + ξ

g

This form satisfies N1 = 0 at all nodes other than node 5. From the condition ‘d’ we get, 1 = C (1 – 0) (1 + 1) (1 + 0) 1 ∴C= 2

∴ N5 =

g e

jb

1 − ξ2 1 − η 1 1− ξ 1−η 1+ ξ = 2 2

b

gb

gb

g

Similarly it may be shown that

b1 + ξg e1 − η j 2

N6 =

2

…(5.44a)

Shape Functions

e1 − ξ j b1 + ηg

93

2

N7 =

…(5.44b)

2

b1 − ξg e1 − η j = 2

N8

and

…(5.44c)

2

Example 5.21: Determine the shape functions for cubic serendipity family element. Solution: Typical element is shown in Fig. 5.24. Shape functions for corner nodes: N1 = 0 for all nodes except 1 and is 1 for node 1. N1 = 0 is satisfied for nodes 2, 7, 8, 3 if 1 − ξ = 0 N1 = 0 is satisfied for nodes 3, 9, 10, 4 if 1 − η = 0  4

10

9

3

A

11

8

0

12

7

1

5

Fig. 5.24

6

2

Cubic serendipity family element

The points 5, 6, 7, 8, 9, 10, 11, 12 lie on the circle shown in Figure. The radius of this circle = OA

=

FG 1IJ H 3K

2

+ 1 since A

FG 1 , 1IJ and O = b0 , 0g = H3 K

∴ The equation of the circle is

ξ 2 + η2 =

10 3

10 3

94

Finite Element Analysis

10 =0 3

ξ 2 + η2 − Satisfies N1 = 0 for nodes 5 to 12.

b

gb

g FGH

IJ K

10 Satisfies N1 = 0 for all nodes except for node 1. 9

N1 = C 1 − ξ 1 − η ξ 2 + η 2 − For node 1 N1 = 1.

b

g FGH

gb

∴ 1 = 1 − ξ 1 1 − η1 ξ 12 + η12 − But we know ξ 1 = η 1 = − 1

FG H

∴1= 2 × 2 × 1+ 1− C=

or

9 32

10 9

IJ K

IJ K

10 32 C= C 9 9

b

gb

g FGH

b

gb

g{ e

9 10 1 − ξ 1 − η ξ 2 + η2 − 32 9

∴ N1 = =

IJ K

j

}

1 1 − ξ 1 − η 9 ξ 2 + η 2 − 10 32

Similarly it may be shown that

b

gb

g{ e

j }

b

gb

g{ e

j

b

gb

g{ e

j }

N2 =

1 1 + ξ 1 − η 9 ξ 2 + η 2 − 10 32

N3 =

1 1 + ξ 1 + η 9 ξ 2 + η 2 − 10 32

N4 =

1 1 − ξ 1 + η 9 ξ 2 + η 2 − 10 32

For mid side node 5, 1 − ξ = 0 ensures N5 = 0 at nodes 2, 7, 8, 3

1 − η = 0 ensures N5 = 0 at nodes 3, 9, 10, 4 1 + ξ = 0 ensures N5 = 0 at nodes 4, 11, 12, 1. 1 − 3ξ = 0 ensures N5 = 0 at node 6.

b

gb

gb

gb

∴ Let N 5 = C 1 − ξ 1 − η 1 + ξ 1 − 3ξ 1 At node 5, ξ = − , η = − 1 and N 5 = 1 3

g

}

Shape Functions

∴1=C × C=

i.e.,

∴ N5 =

Similarly,

95

2 4 ×2× ×2 3 3

9 32

b

gb

gb

gb

9 1 − ξ 1 − η 1 + ξ 1 − 3ξ 32

e

jb

gb

g

e

jb

gb

g

e

jb

gb

g

e

jb

gb

g

e

jb

gb

g

e

jb

gb

g

e

jb

gb

g

e

jb

gb

g

=

9 1 − ξ 2 1 − η 1 − 3ξ 32

N6 =

9 1 − ξ 2 1 − η 1 + 3ξ 32

N7 =

9 1 − η 2 1 + ξ 1 − 3η 32

N8 =

9 1 − η 2 1 + ξ 1 + 3η 32

N9 =

9 1 − ξ 2 1 + η 1 + 3ξ 32

N10 =

9 1 − ξ 2 1 + η 1 − 3ξ 32

N11 =

9 1 − η 2 1 − ξ 1 + 3η 32

N12 =

9 1 − η 2 1 − ξ 1 − 3η 32

g

5.7 HERMITE POLYNOMIALS AS SHAPE FUNCTIONS Similar to Lagrangian functions for Co continuity elements, there are Hermite polynomials for cn continuity elements. Hermite polynomial in one dimension is denoted as Hn(x). It is a polynomial of order 2n + 1. Thus H1(x) is a first order polynomial which is cubic in x. H2(x) is a second order polynomial and is of 5th order. The speciality of Hermite polynomials is their values and the values of their derivatives upto order n are unity or zero at the end points of the interval 0 to 1. The elements of a set of Hermitian polynomials representing

bg

n these properties may be written as Hmi x where m is the order of derivatives, i the node number and n is the

order of the Hermitian function. To illustrate it, let us take a two noded beam element (refer Fig. 5.25) in which nodal unknowns are the displacement w and the slope

∂w . Since the element has four degrees of freedom, we have to select the ∂x

polynomial with only 4 constants i.e. the first order (cubic) Hermitian polynomials as shape function.

96

Finite Element Analysis

x=0

P

1

x1, w 1,

2

×

x2, w 2,

1

S=0

x 2

S=1 S l

Fig. 5.25

Beam element with natural coordinated varying from zero to one

These are given as

bg bsg = ls bs − 1g b sg = s b3 − 2sg b sg = ls bs − 1g

1 H01 s = 1 − 3s 2 + 2 s 3 1 H11 1 H02 1 H12

Where s =

2

2

2

x − x1 x − x1 and l = x2 − x1 = x 2 − x1 l

The displacement model for the beam element is

w= N

lδ q

e

R|w U| |Sθ |V ||w || Tθ W 1

w = N1 N 2 N 3 N 4

i.e.,

1

…(5.47)

2

2

where

Then

θ =

∂w ∂x

LM N

R|w U| ∂ N O |θ | PS V ∂ x Q |w | |Tθ |W 1

∂w ∂ N1 ∂ N 2 ∂ N 3 = ∂x ∂x ∂ x ∂x

4

1

2

2

We select Hermitian first order polynomial functions as shape functions i.e.,

bg bsg = ls bs − 1g

1 N1 = H01 s = 1 − 3s 2 + 2 s 3

1 N 2 = H11

2

…(5.48)

Shape Functions

97

bg b g b sg = ls b s − 1g

1 N 3 = H02 s = s2 s − 2s 1 N 4 = H12

2

Now it may be observed that, at node 1, N1 = 1, N2 = N3 = N4 = 0

∂ N1 ∂ N2 ∂N ∂ N4 = 0, =1 3 = =0 ∂s ∂s ∂s ∂s

and Similarly at node 2,

N1 = 0, N2 = 0, N3 = 1, N4 = 0

∂ N1 ∂ N 2 ∂ N 3 ∂ N4 = = 0 and =1 ∂s ∂s ∂s ∂s This is exactly the requirement of the shape functions. Hence first order Hermite functions are suitable for beam element in which C1 continuity is to be satisfied. If we select natural coordinate as shown in Fig. 5.26, it may be observed that

x=0

Fig. 5.26

= –1 1 A

=0

S=0



=1

x

2

×

×

C

P

B

S = 0.5

S

S=1

Beam element with natural coordinates ξ varying from –1 to 1

ξ=

b

g b

g

PC 2 = AP − AC = 2 s − 0.5 = 2s – 1 l l 2

Substituting this in equation 5.47 we get, Hermite polynomial of first order as, 1 = N1 = H01

2 − 3ξ + ξ 3 4

e

1 = 1 − ξ − ξ2 + ξ3 N 2 = H11 1 N 3 = H02 =

e

1 2 + 3ξ − ξ 3 4

e

j 8l

j

…(5.49)

j

l −1 − ξ + ξ 2 + ξ 3 8 Hermitian shape functions in two dimensions may be constructed as multiplication of Hermitian functions in x and y directions. Thus, 1 = N 4 = H12

bg b xg

bg b yg

bg b xg

bg b yg

1 N1 = H01 x

1 H 01 y

1 N 2 = H11 x

1 H11 y

1 N 3 = H02

1 H02

1 N 4 = H12

1 H12

98

Finite Element Analysis

5.8 CONSTRUCTION OF SHAPE FUNCTIONS BY DEGRADING TECHNIQUE In the problems in which stress concentration is quite high in some portion and its variation is low in some other portion, a stress analyst prefers to use higher order elements in the area of stress concentration and lower order elements in the other areas. Fig. 5.27 shows a case in which this has been done using LST elements and CST elements. In this problem left side portion is having stress variation fast. Hence LST elements are used while on the right hand side CST elements are used. When this is done we come across few elements which do not belong entirely to CST or LST category. In element No.5, we find there are five nodes while in element No.7, there are four nodes. For these odd noded elements, we can find shape function by degrading higher order (in this case LST) elements. This degrading technique is illustrated in this article.

1

13

9

5 2

6

Area of stress concentration

10

11

7

3

Fig. 5.27

12

8

4

CST and LST elements used in a problem

(i) Degrading Six Noded Triangular Element to Five Noded Triangular Element Let the typical 6 noded and 5 noded triangular elements be as shown in Fig. 5.28. Five noded element is to be obtained by dropping node 6 from LST element. Now for LST element, 3

3

5

5 6 ×

6 2

2

4 1

1

Fig. 5.28

u=

4

∑N u

i i

Typical LST and 5-noded triangular elements

= N1u1 + N 2 u2 + N 3 u3 + N 4 u4 + N 5u5 + N 6 u6

where N1 = L1 (2L1 – 1)

N2 = L2 (2L2 – 1)

N3 = L3 (2L3 – 1)

…(5.50)

Shape Functions

99

N5 = 4L2L3 N6 = 4L3L1 N4 = 4L1L2 In this case variation of displacement is quadratic along all the three sides of the triangle. In case of five noded triangle, the variation along line 1–3 should be linear, since there are only two nodes along this line. Hence,

u6′ =

u1 + u3 2

Replacing u6 by u6′ in equation 5.46, we get

u = N 1u1 + N 2 u2 + N 3u3 + N 4 u4 + N 5u5 + N 6

FG H

= N1 +

FG H

IJ K

IJ K

FG u H

1

+ u3 2

IJ K

N6 N u1 + N 2 u2 + N 3 + 6 u3 + N 4 u4 + N 5u5 2 2

…(5.52a)

5

=

∑ N u , for five noded triangular element. i i

i =1

= N 1′u1 + N 2′ u2 + N 3′u3 + N 4′ u4 + N 5′u5

…(5.52b)

comparing equation 5.52a and 5.52b, we conclude, N 1′ = N 1 +

b

g

b

N 3′ = N 3 = L3

g

g since L + L + L = 1 = L 1 − 2 L = L b2 L − 1g 4L L N + = L b 2 L − 1g + = L 2L − 1 + 2L 2 2 −1 + 2 b1 − L g = L 1 − 2 L

= L1 −1 + 2 1 − L2

N 2′ = N 2

b

4 L3 L1 N6 = L1 2 L1 − 1 + = L1 2 L1 − 1 + 2 L3 2 2

2

1

2

1

3

2

2

6

3 1

3

3

2

3

3

3

1

2

N 4′ = N 4 = 4 L1 L2 N 5′ = N 5 = 4 L2 L3 (ii) Degrading LST element to 4-Noded Triangular Element Let the typical LST and 4-Noded triangular elements be as shown in Fig. 5.29. In this case it is possible to get shape functions for 4-noded triangular element by degrading LST element from which nodes 5 and 6 are to be eliminated. If nodes 5 and 6 do not exist, the variation of displacements along lines 2-3 and 1-3 should be linear. Thus in 4 noded triangle u6′ =

u1 + u3 u2 + u3 and u5′ = 2 2

100

Finite Element Analysis 3

3

× 5

5

6

2

6 ×

2

4

4

1

1

Fig. 5.29

Typical LST and 4 noded triangular element

Substituting these in equation 5.45, we get

u = N 1u1 + N 2 u2 + N 3 u3 + N 4 u4 + N 5

FG H

= N1 +

FG H

IJ K

IJ K

FG H

FG u H

2

IJ K

FG H N I + Ju 2 K

+ u3 u + u3 + N6 1 2 2

N6 N N u1 + N 2 + 5 u2 + N 3 + 5 2 2 2

6

3

IJ K

+ N 4 u4

…(5.34a)

But for 4 noded element,

u = N1′u1 + N 2′ u2 + N 3′ u3 + N 4′ u4

…(5.34b)

Comparing 5.52a and 5.52b, we get, N 1′ = N 1 +

b

g

4 L3 L1 N6 = L1 2 L1 − 1 + = L1 2 L1 − 1 + 2 L3 2 2

b

g

= L1 −1 + 2 1 − L2 , since L1 + L2 + L3 = 1 = L1 1 − 2 L2 N 2′ = N 2 +

b

g

4 L2 L3 N5 = L2 2 L2 − 1 + = L2 2 L2 − 1 + 2 L3 2 2

b

g

= L2 −1 + 2 1 − L1 , since L1 + L2 + L3 = 1 = L2 1 − 2 L1 N 3′ = N 3 +

b

g

4 L2 L3 4 L3 L1 N5 N + 6 = L3 2 L3 − 1 + + 2 2 2 2

= L3 2 L3 − 1 + 2 L2 + 2 L1 = L3 −1 + 2 since L1 + L2 + L3 = 1 = L3 Example 5.22: By degrading technique develop shape function for the seven noded rectangular element shown in Fig. 5.30 (b). Given that for eight noded element shown in Fig. 5.30 (a)

Ni =

b

gb

gb

g

1 1 + ξξ i 1 + ηη i ξξ i + ηη i − 1 for nodes 1, 2, 3 and 4. 4

Shape Functions

e

jb

g

e

jb

g

Ni =

1 1 − ξ 2 1 + ηη i for nodes 5 and 7 2

Ni =

1 1 − η 2 1 + ξξ i for nodes 6 and 8 2 

4

 3

7

4

6 8

1

5

Fig. 5.30

7

3

6

8

2

1

5

2

(a) Eight noded element (b) Seven noded element

For eight nodded element,

b

gb

gb

g

b

gb

gb

g

b

gb

gb

g

b

gb

gb

e

jb

g

b

ge

j

e

jb

g

b

ge

j

N1 =

1 1 − ξ 1 − η −ξ − η − 1 4

N2 =

1 1+ξ 1−η ξ −η −1 4

N3 =

1 1+ ξ 1+η ξ + η −1 4

N4 =

1 1 − ξ 1 + η −ξ + η − 1 4

N5 =

1 1 − ξ2 1 − η 2

N6 =

1 1 + ξ 1 − η2 2

N7 =

1 1 − ξ2 1 + η 2

g

1 1 − ξ 1 − η2 2 For seven noded element, variation of displacement along the edge 1–4 is linear. Hence N8 =

u8 =

u1 + u4 2

101

102

Finite Element Analysis

∴ u = N1u1 + N 2 u2 + N 3u3 + N 4 u4 + N 5u5 + N 6 u6 + N 7 u7 + N 8

FG H

= N1 +

FG H

IJ K

IJ K

FG u + u IJ H 2 K 1

4

N8 N u1 + N 2 u2 + N 3u3 + N 4 + 8 u4 + N 5 u5 + N 6 u6 + N 7 u7 2 2

...(5.54a)

If the shape function for 7 noded element is N ′ , then,

u = N 1′u1 + N 2′ u2 + N 3′ u3 + N 4′ u4 + N 5′u5 + N 6′ u6 + N 7′ u7

…(5.54b)

comparing equation 5.54a and 5.54b, we conclude N 1′ = N 1 +

=

b

b

gb

gb

g

b

ge

N8 1 1 = 1 − ξ 1 − η −ξ − η − 1 + 1 − ξ 1 − η 2 2 4 4

gb

gb

g

b

gb

j

gb g

1 1 1 − ξ 1 − η −ξ − η − 1 + 1 + η = 1 − ξ 1 − η −ξ 4 4

N 2′ = N 2 ; N 3′ = N 3 N 4′ = N 4 +

b

gb

gb

g

b

ge

N8 1 1 = 1 − ξ 1 + η −ξ + η − 1 + 1 − ξ 1 − η 2 2 4 4

b

gb

gb

b

gb

gb g

=

1 1 − ξ 1 + η −ξ + η − 1 + 1 − η 4

=

1 1 − ξ 1 + η −ξ 4

g

j …(5.55)

N 5′ = N 5 , N 6′ = N 6 , N 7′ = N 7 .

QUESTIONS 1. Explain the term ‘Shape Functions’. Why polynomial terms are preferred for shape functions in finite element method ? 2. State and explain the convergence requirements of polynomial shape functions. 3. Explain the term ‘geometric isotropy / geometric Invariance’. Why polynomial shape functions should satisfy these requirement ? How do you check a polynomial for this requirement ? 4. Determine the shape function for a two noded bar element using (i) Cartesian coordinate system (ii) Local coordinate system ranging from 1 to zero. (iii) Local coordinate system with range -1 to 1. 5. Determine the shape function for a three noded bar element (i) using polynomial form in local coordinates (ii) using Lagrangian functions. Plot their shapes.

Shape Functions

103

6. Using generalized coordinate approach, determine shape functions for a two noded beam element and apply necessary checks. 7. Determine the shape functions for a CST element. Show that they are nothing but area coordinate 8. Explain the method of finding shape functions for LST element in terms of local coordinates L1, L2, and L3, 9. Derive shape functions for a rectangular element for plane stress / plane strain analysis starting with (i) Polynomial form (ii) Lagrange functions Apply the checks 10. Explain the method of finding shape function for 8 noded rectangular element to be used for plane stress/plane strain problems. 11. Using Lagrange functions determine the shape function for 9 noded rectangular element. Plot the variation of shape function of a typical corner node, a typical mid side node and the central node. 12. Explain the situations where you need 4 noded and 5 noded triangular elements. By degradation technique derive the shape functions for them. Apply necessary checks. 13. Write short note on Hermite polynomial. 14. Write short note on Lagrange functions. 15. Using Lagrange functions derive the shape functions for a hexahedron element.

104

Finite Element Analysis

6 Strain Displacement Matrix 6.1 INTRODUCTION In the previous chapter we saw the shape function [N] which established the relationship between displacement at any point in the element with nodal displacements of the element. In this chapter we establish the relationship between strain at any point in the element with nodal displacement. We define strain displacement relation as

lεq =

B

e

lδ q

…(6.1)

e

lq bδ g is displacement vector of nodal values of the element

Where ε is strain at any point in the element. e

[B]e is strain displacement matrix In this article strain displacement matrix will be found for few standard cases.

6.2 STRAIN-DISPLACEMENT MATRIX FOR BAR ELEMENTS There is only one strain component in such element i.e. ε x . Thus for bar elements [refer Fig. 6.1]

lεq = ε

1

x1  = –1

x

=

∂u ∂x

RS UV T W

=

u ∂u N 1 u2 ∂x

=

∂u N1 ∂x

N2

RSu UV Tu W 1

×

2

2

×

xc  =0

x 

l

Fig. 6.1



x2 =1

Strain Displacement Matrix

But N1 =

x2 − x l

N2 =

x − x1 l

105

(from equation 5.12) ∴

∂N1 ∂N 2 1 1 = − and = l l ∂x ∂x

LM N

lq

∴ ε = εx = −

1 l

1 l

OP LMu OP Q Nu Q 1

2

= [B] {q} B =

Thus

1 −1 1 l

…(6.2)

The above relationship may be derived using shape function in terms of natural coordinate ' ξ ' also

lεq = ε

x

=

where

∴

∂u ∂ξ ∂u ∂ξ ∂ N1 = = ∂x ∂x ∂ξ ∂x ∂ξ 2 x − x2 = x − x2 l l 2

N1 =

1−ξ 1+ξ and N 2 = 2 2

LM N

lq

2 1 − 2 l

B =

1 2

1

(from equation 4.5)

∂N1 1 =− 2 ∂ξ

2 ∂ξ =− l ∂x

RSu UV Tu W 2

ξ=

∴ ε = εx =

where

N2

and

(from equation 5.19)

∂N 2 1 = 2 ∂ξ

OP LMu OP = 1 −1 1 lδ q Q Nu Q l 1 2

e

lq

= B δ

e

1 −1 1 l

6.3 STRAIN DISPLACEMENT MATRIX FOR CST ELEMENT At any point in the constant strain triangle element [refer Fig. 6.2] used for plane stress or plane strain problems, there are three strain components i.e. ε x , ε y ,γ

xy .

R| ε lεq = S ε |Tγ

They are given by

x y xy

R| ∂u U| U| || ∂∂xv || V| = S| ∂y V| W | ∂u + ∂v | |T ∂y ∂x |W

106

Finite Element Analysis y

c1

c2

3

Cyclic

Side 1

b1 b2

2

Side 2

b3

Side 3 1

Cyclic

b1  y 2

y3

c1  x 3

x2

b 2  y 3

y1

c 2  x 1

x3

b 3  y1

y2

c 3  x 2

x1

c3 x

Fig. 6.2

The displacement vector is

R|u U| |u || 0 O |u PS V N Q |v | ||v || |Tv |W 1

luq = RSTuvUVW = LMN N0

1

2

N2 0

0 N1

N3 0

0 N2

3

3

1

2

3

R| ∂u U| LM ∂N || ∂∂xv || MM ∂x ∴lε q = S = 0 ||∂u ∂y ∂u V|| MM ∂N |T ∂y + ∂x |W MMN ∂y

1

1

∂N 2 ∂x

∂N 3 ∂x

0

0

∂N 2 ∂y

∂N 3 ∂y

0

0

∂N 1 ∂y ∂N 1 ∂x

∂N 2 ∂y ∂N 2 ∂x

But in these elements (from equation 5.15 and 5.16) N1 =

a1 + b1 x + c1 y 2A

N2 =

LM MM N

a2 + b2 x + c2 y a3 + b3 x + c3 y and N 3 = 2A 2A

b1 b2 1 ∴ε = 0 0 2A c1 c2

lq

b3 0 c3

0 0 c1 c2 b1 b2

0 c3 b3

OP PP lδ q Q

e

lq

= B δ

e

OP P ∂N P ∂y P ∂N P P ∂x PQ 0

3

3

R|u U| ||u || S|uv V| ||v || |Tv |W 1 2

3

1

2

3

Strain Displacement Matrix

107

where

LM MM N

b1 b2 1 ∴ B = 0 0 2A c1 c2

b3 0 c3

0 0 c1 c2 b1 b2

0 c3 b3

OP PP Q

…(6.3)

c1 = x3 – x2 b1 = y 2 – y 3 b2 = y 3 – y 1 c2 = x1 – x3 b3 = y 1 – y 2 c3 = x2 – x1 Note b1, b2, b3 and c1, c2, c3 are in cyclic order and they are constant for the triangle. Hence the strains are constant within the element. Therefore the element is referred as constant strain triangle.

where

6.4 STRAIN DISPLACEMENT MATRIX FOR A BEAM ELEMENT We know from constitutive law

lσ q =

lεq

D

From definition of strain displacement matrix

lεq = ∴ lσ q =

B

lδ q

D

B

e

…(6.4)

lδ q

e

In case of beam, stress resultant is the moment, which is given by the expression, M = EI

∂ 2v

…(6.5)

∂x 2

where v-displacement in y-direction. Treating the stress resultant M similar to the stress σ and curvature similar to strain, [D] and [ε ] matrix are [D] = EI

…(6.6)

lεq = ∂∂x v 2

and

…(6.7)

2

From equation 5.43, we have

v= N

lδ q

e

R| v U| |Sθ |V || v || Tθ W 1

= N1

N2

N3

N4

1

2

2

for the typical beam element shown in Fig. 6.3 and from equation 5.44, N1 = 1 – 3s2 + 2s3 N2 = ls(s – 1)2

108

Finite Element Analysis N3 = s2 (3 – 2s) N4 = ls2 (s – 1) v1,

v2,

P

1

2

×

x

x1

x2 S=

S=0

S

S=1

x – x1 l

l

Fig. 6.3

x − x1 l

s=

where

lq

∴ ε =

=

∴ B =

∂2 ∂x

L l q MMN ∂∂xN δ

N

2

LM∂ N MN ∂x 2

1

e

2

=

1

2

∂ 2 N3

∂ 2 N4

∂x

∂x

∂x

2

2

∂x 2

∂ 2 N2 ∂x 2

∂ 2 N2

2

∂2v

2

∂ 2 N3 ∂x 3

∂ 2 N4 ∂x 2

OP PQ

OP lδ q PQ

e

lq

= B δ

e

∂Ni ∂s ∂Ni 1 ∂Ni = = ∂x ∂x ∂s l ∂s

Now

∴

FG H

1 ∂ ∂ 2 Ni = l ∂s ∂x 2 ∴ B =

1 l2

=

1 l2

IJ FG 1 K Hl

LM ∂ N MN ∂s 2

2

1

∂N i ∂s

IJ = 1 K l

∂ 2 N2 ∂s 2

b

2

∂ 2 Ni ∂s 2

∂ 2 N3 ∂s 2

−6 + 12s l 6s − 4

g

∂ 2 N4 ∂s 2

b

OP PQ

6 − 12s l 6s − 2

g

…(6.8)

QUESTIONS 1. 2. 3. 4. 5.

Selecting interpolation function for a bar element in its natural coordinates, find the strain matrix. Selecting interpolation function for a CST element in its natural coordinates, find the strain matrix. Determine the strain matrix for a beam element. Use shape functions in natural coordinates form. Determine the natural coordinates (area coordinates) for a CST element. Assemble strain vector. For a CST element shown in Fig. 6.4. Obtain the strain–displacement matrix. Assume Possion’s ratio is zero and Young's modulus is constant.

Strain Displacement Matrix y

(2, 5)

(4, 3)

(1, 1)

x

Fig. 6.4

109

110

Finite Element Analysis

7 Assembling Stiffness Equations—Direct Approach 7.1 INTRODUCTION In assembling the element stiffness equation, [k ] {δ } = {F}, the first step is to derive the expression for element stiffness property and nodal force vector. The first method developed in this field was direct approach. Latter on variational approach and Galerkin’s weighted residual approaches developed. Direct approach is the extension of matrix displacement approach. For one dimensional elements this is exactly same as explained in Chapter 3. In fact no distinction was seen between matrix and finite element method. The analysts tried to extend the matrix method to two dimensional problems also. The only element in which it could be done successfully was the three noded triangular element (CST–element). This concept is explained in this chapter. Lumped mass concept of assembling the nodal force vector is also presented.

7.2 ELEMENT STIFFNESS MATRIX FOR CST ELEMENT BY DIRECT APPROACH Turner was first to suggest it and that was real starting point of FEM. Consider the typical element shown in Fig. 7.1. It is subjected to constant stresses along its all the three edges. Let the constant stresses be

σ x , σ y , τ xy = τ yx . Assembling stiffness matrix means finding nodal equivalent set of forces which are statically equivalent to the constant stress field acting at the edges of the elements. The equivalent nodal forces to be found are F1, F2, F3 … F6 as shown in Fig. 7.2. We have six unknown nodal forces, but only three equations of equilibrium. Hence it is not possible to determine F1, F2 … F6 in terms of σ x , σ y , τ xy mathematically. Turner resolved the uniform stress distribution into an equivalent force system at midsides as shown in Fig. 7.3. Note side i is the side opposite to node i. With this notation,

b

g

b

g

b

g

b

g

Fm1x = σ x y3 − y2 t + τ xy x 2 − x 3 t where t is the thickness of the element.

Fm1y = σ y x 2 − x3 t + τ xy y3 − y2 t

b

g

b

g

Fm2 x = − σ x y3 − y1 t + τ xy x3 − x1 t

b

g

b

g

Fm2 y = σ y x3 − x1 t − τ xy y3 − y1 t

...(7.1)

Assembling Stiffness Equations—Direct Approach

111

y  

yx



xy

y

3





x

x

2

1  

xy



y

yx

x

Fig. 7.1

Stress field in CST element

y F6 3

F3

F5

1

F2

2

F4

F1 x

0

Fig. 7.2

Equivalent nodal forces

b

g

b

g

Fm3 x = σ x y2 − y1 t − τ xy x 2 − x1 t

b

g

b

g

Fm3 y = − σ y x2 − x1 t + τ xy y2 − y1 t After this Turner transferred half of mid side forces to nodes at the end of sides to get equivalent nodal forces. Thus he got

112

Finite Element Analysis y

Fm 1 y Fm 2 y Fm 1 x Fm 2 x Fm 3 y

2

Fm 3 x 1

x

0

Fig. 7.3

F1 =

b

1 Fm 2 x + Fm 3 x 2

b

Equivalent mid-side forces

g

g

b

g

g

b

b

g

b

g

=

1 −σ x y3 − y1 t + τ xy x 3 − x1 t + σ x y2 − y1 t − τ xy x 2 − x1 t 2

=

t σ x − y3 + y1 + y2 − y1 + τ xy x 3 − x1 − x2 + x1 2

=

t σ x y2 − y3 + τ xy x 3 − x 2 2

F2 =

b b

g

b

1 Fm1 x + Fm 3 x 2

b

g

g

g

b

g

b

g

b

g

b

g

b

g

b

g

b

g

b

g

=

t σ x y3 − y2 + τ xy x2 − x 3 + σ x y2 − y1 − τ xy x 2 − x1 2

=

t σ x y3 − y1 + τ xy x1 − x 3 2

F3 =

b

1 Fm1x + Fm 2 x 2

g

b

g

b

g

b

g

b

g

=

t σ x y3 − y2 + τ xy x 2 − x 3 − σ x y3 − y1 + τ xy x 3 − x1 2

=

t σ x y1 − y2 + τ xy x 2 − x1 2

Assembling Stiffness Equations—Direct Approach

F4 =

d

1 Fm2 y + Fm 3 y 2

i

b

g

b

g

b

g

b

g

b

g

b

g

b

g

b

g

b

g

b

g

=

t σ y x3 − x1 − τ xy y3 − y1 − σ y x 2 − x1 + τ xy y2 − y1 2

=

t σ y x 3 − x 2 + τ xy y2 − y3 2

F5 =

d

1 Fm1y + Fm 3 y 2

i

b

g

b

g

b

g

b

g

=

t σ y x 2 − x 3 + τ xy y3 − y2 − σ y x2 − x1 + τ xy y2 − y1 2

=

t σ y x1 − x3 + τ xy y3 − y1 2

F6 = =

d

1 Fm1y + Fm 2 y 2

i

b

g

b

g

b

g

b

g

t σ y x 2 − x3 + τ xy y3 − y2 + σ y x3 − x1 − τ xy y3 − y1 2

t σ y x 2 − x1 + τ xy y1 − y2 2 Thus the force vector as derived by Turner is =

113

LMb MMb t b = M 2 M0 MM 0 MN 0

1 2

F

3

0 0 0 c1 c2 c3

c1 c2 c3 b1 b2 b3

OP PP PP PP PQ

R|σ S|σ Tτ

x y

xy

U| V| W

…(7.2)

where b1, b2, b3, c1, c2 and c3 have the expressions as used while deriving shape function expressions. i.e. b1 = y 2 – y 3 b2 = y 3 – y 1 b3 = y 1 – y 2 c1 = x3 – x2 c2 = x1 – x3 c3 = x2 – x1

LMb MMb MMb0 MM 0 MN 0

1

2

But,

3

0 0 0 c1 c2 c3

c1 c2 c3 b1 b2 b3

OP PP PP PP PQ

= 2A B

T

…(7.3)

114

Finite Element Analysis

lσ q = D lε q =

and

lq

D B δ

e

We have got

lFq = 2t 2 A = B where V is the volume

T

B

T

lq

D B δ

= B

T

lq

D B t A δ

…(7.4)

e

lq

D B V δ

l q lδ q lkq = B D

∴ F = k

e

e

T

where

e

B V =

since [B] and [D] are constants.

z

B

T

…(7.5)

D B dV ,

7.3 NODAL LOADS BY DIRECT APPROACH In the stiffness equation [k ] {δ } = {F}, the right hand side term {F} refers to the nodal forces. Generally, while subdividing a structure, nodal locations are selected so as to coincide with the external forces applied. This can be easily done in case of concentrated loads acting on the structure. But in case of distributed loads like self weight, uniformly distributed load, uniformly varying load, we need a technique of transferring the load as nodal loads. There are two procedures for it, namely direct procedure and variational approach. In this chapter the direct procedure is dealt. This procedure was first to be used in the finite element method. In this procedure classical structural analysis background is utilized or a portion is assigned to each node and load on that region as the nodal load. The latter method is called as lumped load method. Consider the self weight of the uniform bar element shown in Fig. 7.4. Half the bar length may be

1

 Al 2

1

l

2  Al 2

 Al 2

l (b) 2  Al 2 (a)

Fig. 7.4

Assembling Stiffness Equations—Direct Approach

115

assumed to contribute to each node. Hence at each node vertical downward load is =

ρAl 2

Where ρ is unit weight of the material, A the cross sectional area and l is the length of the element. In case of beam element subject to uniformly distributed load (refer Fig. 7.5(a); (i) Lumped load procedure: half the region is assigned to each node as shown in Fig. 7.5(b). Its equivalent

wl is taken at the centre of gravity of the element as shown in Fig. 7.5(c). Then at the 2

node, the forces are

wl 2 wl . Hence Lumped load vector is and 8 2

lFq = LMM wl2 N

wl 2 8

T

wl 2

−

wl 2 8

OP PQ

W/unit length

l wl 2

wl 2

l 4

l 4 l 2

l 2 wl 2

wl 2

wl2 8

wl2 8

Fig. 7.5

(ii) Classical Structural Analysis Approach: In case of beam the end reactions, for a fixed beam [shown in Fig. 7.6.(a)] are

wl 2 wl and 12 2 Hence the equivalent nodal loads are

LM− wl MN 2

wl 2 12

−

wl 2

−

wl 2 12

OP PQ

116

Finite Element Analysis y

W/unit length

x l (a)

wl2 12

wl2 12 wl 2

wl 2 (b)

wl2 12

wl2 12 wl 2

wl 2 (c)

Fig. 7.6

(iii) In case of a CST element

force is .

1 area may be assigned to each node and hence equivalent nodal 3 rd

1 the self weight as shown in Fig. 7.7 3 rd y 3

F1 = F2 = F3 =

1  At 3

2

F6 F5

1

F4 x

Fig. 7.7

For complex loading and elements this method may not be of much use. The distribution obtained in lumped load approach may be one of the possible distribution. We cannot say confidently that the distribution considered is exact.

Assembling Stiffness Equations—Direct Approach

117

Because of difficulties and uncertainties associated with direct approach not much progress could be made with direct approach. The real break through was found in finite element analysis only when variational approach was discovered.

QUESTIONS 1. Derive stiffness matrix for a CST element by direct approach. 2. Differentiate between the terms ‘lumped loads’ and consistent loads.

118

Finite Element Analysis

8 Assembling Stiffness Equations—Galerkin's Method, Virtual Work Method 8.1 INTRODUCTION Finite Element Analysis is a versatile method which is used for solving a set of differential equations specified over a region, the solution satisfying specified boundary conditions. In the solid mechanics we try to get the displacements in a structure by solving the equations of equilibrium specified over the structure and the displacements obtained are such that the specified support conditions and the values of loads are satisfied. Galerkin has given a method for solving such differential equations which can be used by the analysts of solid mechanics, fluid mechanics, heat flow, electrical engineering. Galerkin’s general method is briefly explained first and then its application in elasticity is presented. It will be found that in elasticity problems this method turns out to be principle of virtual work.

8.2 GALERKIN’S METHOD Let the governing differential equation on a specified region V be L (u) = P …(8.1) Where L is a differential operator on a basic unknown u. The value of u to be found has to satisfy specified values on the boundary of the region. If we take u as approximate solution, then we may get error ε ( x ) at a point x and

bg bg

ε x =Lu −P

…(8.2)

ε ( x ) is called residual at point x. The solution sought revolve around setting the residual, relative to a weighting function wi to zero i.e. to get

z

b

g

wi Lu − P dV = 0 for i = 1 to n

…(8.3)

v

Depending upon the selection of weighting function, there can be different approaches. In the Galerkin’s method the equation 8.3 is taken as,

z v

b

g

ψ Lu − P = 0

…(8.4)

Assembling Stiffness Equations—Galerkin's Method, Virtual Work Method

119

where ψ is also chosen from the basis function used for constructing approximate solution function u . Let n

u =

∑Q G i

…(8.5)

i

i −1

where Qi is the basic unknown vector and Gi are basis functions. Gi are usually polynomial in space coordinates x, y, z. Then in Galerkin’s method the weighting function φ is taken as n

φ=

∑ψ

i

…(8.6)

Gi

i −1

In the above equation φ i are arbitrary, except at the points where boundary conditions are satisfied. Since ψ is constructed similar to that as u , Galerkin’s method leads to simplified method. Thus in Galerkin’s method we choose basis function Gi and determine φ i in u =

n

∑Q G i

i −1

i

to satisfy

zb

g

Lu − PdV = 0 where

v

coefficient φ i are arbitrary except at specified boundaries.

8.3 GALERKIN’S METHOD APPLIED TO ELASTICITY PROBLEMS In case of three dimensional problems in elasticity the equations of equilibrium are:

∂σ x ∂τ xy ∂τ xz + + + Xb = 0 ∂x ∂y ∂z ∂τ xy ∂x and

+

∂σ y ∂y

+

∂τ yz ∂z

…(8.7) (Refer 2.2)

+ Yb = 0

∂τ xz ∂τ yz ∂σ z + + + Zb = 0 ∂x ∂y ∂z

The stresses {σ } are in terms of displacements {u}. The displacements are arbitrary in the entire region of structure except at specified boundaries and the solution has to see that specified load values are obtained, wherever loading is existing. For simplicity we take two dimensional problem in elasticity and then extend the results to three dimensional problems. In two dimensional elasticity, the equations of equilibrium is

∂σ x ∂τ xy + + Xb = 0 ∂x ∂y and

∂τ xy ∂x

+

∂σ y ∂y

+ Yb = 0

…(8.8)

If Xs and Ys are the surface forces and α is the angle made by normal to surface with x-axis (refer Fig. 8.1), then from the equilibrium of the element we find,

120

Finite Element Analysis A ys

Normal to surface

xs

x

90 – 

B

C

xy



Fig. 8.1

y

Two dimensionally stressed element

Xs AC t = σ x AB t + τ xy BC t AB BC + τ xy AC AC

Xs = σ x

i.e.

= σ x cos α + τ xy sin α = σ x l + τ xy m where l and m are the direction cosines of normal to the surface. Similarly, considering the equilibrium of the forces on the element in y direction, we get

Ys = τ xy l + σ y m Thus the surfaces forces on the element are

Xs = σ x l + σ xy m Ys = τ xy l + σ y m

and Now consider the integral

zz

LMF ∂σ MNGH ∂x

x

+

∂τ xy ∂y

I JK

+ Xb δu +

zz

F ∂τ GH ∂x

xy

…(8.9)

+

∂σ y ∂y

I JK

OP PQ

+ Yb δv dx dy

where δu and δv are the elemental displacements in x and y directions. The above bounded integral is zero, since

∂τ xy ∂σ y ∂σ x ∂τ xy + + X b = 0 and + + Yb = 0 (from equation 8.8). Thus, ∂x ∂y ∂x ∂y

zz LMMNFGH

I JK

F GH

I OP JK PQ

∂τ xy ∂σ y ∂σ x ∂τ xy + + X b δu + + + Yb δv dx dy = 0 ∂x ∂y ∂x ∂y

…(8.10)

Now we can expand the integral 8.10 using Green’s theorem. According to Green’s theorem, if φ ( x , y ) and ψ ( x, y) are continuous functions then their first and second partial derivatives also continuous,

Assembling Stiffness Equations—Galerkin's Method, Virtual Work Method

zz

LM∂φ ∂ψ + ∂φ ∂ψ OP dx dy = − N ∂x ∂x ∂y ∂y Q

zz zz

φ

z

LM∂ ψ + ∂ ψ OP dx dy + φ L∂ψ l + ∂ψ mO ds MN ∂x ∂y PQ MN ∂x ∂y PQ 2

2

2

2

121

…(8.11)

Now let us consider the integration of the first term in equation 8.11

∂ψ ∂ψ = 0, from Greens function we get = ∂u and ∂y ∂x

Assuming φ = σ x

Similarly

zz

zz

∂σ y ∂y

∂σ x ∂u dx dy ∂x

∂σ x δu dx dy = − ∂x

zz

zz zz zz zz LMN

∂σ y

δu dx dy = −

∂y

on the same lines, we can get,

and

b g dx dy +

∂ δu ∂x

z

σ x l δu ds

δu dx dy can be found by taking φ = σ y,

It leads to

σx

∂τ xy ∂x

∂φ ∂φ =0 = δv , ∂x ∂y

zz zz zz

σy

z

b g

∂ δu dx dy + σ y m δv ds ∂y

b g

z

δv dx dy = −

τ xy

∂ δv dx dy + τ xy l δv ds ∂x

δu dx dy = −

τ xy

∂ δu + τ xy m δu ds ∂y

∂τ xy ∂y

b g

z

Hence equation 8.10 is equal to

−

σx

b g

b g

b gOPQ dx dy

b g

∂ ∂ ∂ ∂ δu + σ y δv + τ xy δv + τ xy δu ∂y ∂x ∂y ∂x

z

+ σ x l δu + σ y m δv + τ xy l δv + τ xy m δu ds + Regrouping the terms, we get,

− +

zz b

zz RST

σx

zz

Xb δu dx dy +

b g

zz

Yb δv dx dy = 0

b g

b g

b

gUVW dx dy

∂ ∂ ∂ ∂ δu + σ y δv + τ xy δv + δu ∂x ∂y ∂x ∂y

g

X b δu + Yb δv dx dy +

zd

i

σ x l + τ xy m δu ds +

zd

i

τ xy l + σ y m δv ds = 0

…(8.12)

122

Finite Element Analysis Now consider the term

zz b

g

X b δu + Yb δv dx dy in equation 8.12

Xb is the body force in x direction and δu is arbitrary (virtual) displacement in x direction. Similarly Yb and δv are the body force and virtual displacements in y directions. Hence the above term represents virtual work done by the body forces. Now consider the term,

zd

i

d

i

σ x l + τ xy m δu + τ xy l + σ y m δv ds

from equation 8.9, we have X s = σ x l + τ xy m and

Ys = τ xy l + σ y m

zd

i

d

i

σ x l + τ xy m δu + τ xy l + σ y m δv ds

Hence,

z

=

Xs δu + Ys δv ds

…(8.13)

Thus the above expression represents the virtual work done by surface forces. Barring the first term the other terms in equation 8.12, represents the virtual work done by external forces due to virtual displacements

δu and δv. Now let us try to attach physical meaning to first term in equation 8.12. Consider the element of unit thickness shown in Fig. 8.2. Let a virtual displacement δu be given to the element. Dotted position shows the element with virtual displacement. Hence work done by σ x stresses are

LM N

= σ x dy δu +

b g OPQ

b g

∂ ∂ δu dx − σ x dy δu = σ x δu dx dy ∂x ∂x

y



u

u



d

dx

b gd u

x

dx

0

Fig. 8.2

x

Element with virtual displacement in n-direction

…(8.14)

Assembling Stiffness Equations—Galerkin's Method, Virtual Work Method

123

It may be noted that due to virtual displacement δu the change in strain δε x is given by

δε x =

LMδu + ∂ bδug dxOP − δu N ∂x Q = ∂ bδug ∂x dx

Substituting it in equation 8.14, we get virtual work done by

σ x stresses = σ x δε x dx dy Similarly it can be shown that the terms

zz

σy

i.e.

b g

∂ δv dx dy and ∂y

zz

zz

τ xy

σ y δε y dx dy and

represent the work done by σ y and τ xy stresses.

∴ First term in equation 8.12 is

δU =

…(8.15)

RS ∂ bδvg + ∂ bδugUV dx dy ∂x T ∂y W

zz

τ xy δγ xy dx dy ,

zz d

i

σ x δε x + σ y δε y + τ xy δγ xy dx dy

…(8.16)

Thus the first term represents the work done by internal forces with negative sign. Therefore equation 8.12 may be looked as,

−δU + δ We = 0 δU = δ We

…(8.17)

where U is internal work done and We external work done. Thus in elasticity problems Galerkin’s method turns out to be the principle of virtual work which may be stated as a deformable body is in equilibrium when the total work done by external forces is equal to the total work done by internal forces. The work done considered in the above derivation is called virtual, since the forces and deformations considered are not related. The displacements ' δu' and ' δv' are arbitrary. It may also be noted that the principle of virtual work is independent of the material properties.

Extention to Three Dimensional Problems The principal of virtual work holds good for the three dimensional problems also. In this case

δ We =

zzz b

g

X b δu + Yb δv + Zb δw dV +

v

zz b zz l q l q

g

X s δu + Ys δv + Z s δw ds

…(8.18)

s

where s is the surface on which forces are acting. The above expression in the matrix form is,

δ We =

where

zzz l q l q T

δu

lδuq

T

X dV +

δu

s

= δu δv δw

T

p ds

…(8.19)

124

Finite Element Analysis

lXq l pq

and

T

= X b Yb

Zb

T

= X s Ys

Zs

In three dimensional case, internal work expression is,

δU =

zzz d zzz l q l q

i

σ x δε x + σ y δε y + σ z δε z + τ xy δγ xy + τ yz δγ yz + τ xz δγ xz dV

v

δU =

or

δε

T

σ

…(8.20)

v

where and

lδεq lσ q

T

T

= δε x

δε y

δε z

γ

= σx

σy

σ z τ xy

xy

γ

yz

γ

xz

τ yz τ zx

From principal of virtual work,

δU = δ We Example 8.1: Using virtual displacement principle, determine the forces developed in the three bar truss shown in Fig. 8.3 (a).

A

A

A

1

B

30°

2 2m

60°

30°

D (,u) v

B

D

60° u

D

B

v

20 kN

3

C

C (a)

C (b)

Fig. 8.3

Solution: Fig. 8.3 (a) shows the given truss. The length of various members are

(c)

Assembling Stiffness Equations—Galerkin's Method, Virtual Work Method

l1 =

125

2 2 = 4m = 2.3094m l2 = 2m l3 = Cos 60 Cos 30

Total displacement of point D is u in x direction and v in y direction. The displacements u and v are separated and shown in Figs 8.2 (b) and 8.3 (c) respectively. From it we can see that total strains in the various members are

ε1 =

u cos 30 + v sin 30 = 0.375u + 0.2165 v 2.3094

ε2 =

u = 0.5 u 2

ε3 =

u cos 60 – v sin 30 = 0.125 u − 0.2165 v 4

…(a)

Let a unit virtual displacement be given in x direction as shown in Fig. 8.4. Then due to δu = 1, the strains introduced are, A

B

D

u=1

C

Fig. 8.4

δε 1′ =

1 × cos 30 = 0.375 2.3094

δε ′2 =

1 = 0.5 2

δε 3′ =

1 × cos 60 = 0.125 4

126

Finite Element Analysis

∴ Internal work done in a member when unit virtual displacement is introduced in x direction is

δU =

zzz

σ x δε ′x dV

= EAL ε x δε ′x since σ x = Eε x and V = AL

Internal work done in the truss = δU1 + δU 2 + δU 3

= EA [2.3094 (0.375u + 0.2165v) 0.375 + 2 × 0.54 × 0.5 + 4 (0.125u – 0.2165v) 0.125] = EA (0.88726u – 0.07924v) work done by load δ We = 0, since no displacement in y direction. Equating internal work to external work, we get 0.88726u = 0.07924v …(b) ∴ u = 0.0893v Now consider virtual work principle when unit virtual displacement is given to point D in y direction

δv = 1. Referring to Fig. 8.5, we can note virtual strains in the members as, A

D

u=1

B

C

Fig. 8.5

δε 1′ =

1 × sin 30 = 0.2165 2 .3094

δε ′2 = 0 δε ′3 = −

1 × sin 60 = − 01265 . 4

Assembling Stiffness Equations—Galerkin's Method, Virtual Work Method

127

∴ Work done by internal forces δU = δU1 + δU 2 + δU 3 = EA [2.3094 (0.375u + 0.2165v) 0.2165 + 0 + 4 (0.125u – 0.2165v) (–0.2165)] = EA (0.0792u + 0.2957v) From equation (b) u= 0.0893v

∴ δU = EA 0.3028v work done by external forces

δWe = 20 × 1 = 20 Equating internal work to external work, we get 0.3028v EA = 20

v=

or Thus v =

20 66.048 = 0.3028EA EA

66.048 5.898 and u = 0.0893 v = EA EA ∴ ε 1 = 0.375u + 0.2165v =

ε 2 = 0.5u = 0.5 ×

26.045 EA

5898 2.949 . = EA EA

ε 3 = 0125 . u − 0.2165v = −

13562 . EA

∴ F1 = Aσ x = AEε 1 = 26.045kN

F2 = AEε 2 = 2.949kN

Answer

F3 = –13.562kN

Comments on Gelerkin’s Method This is a method which can be applied to any problem involving solution of a set of equations subject to specified boundary values. In mechanics of solids it turns out to be virtual work method.

QUESTIONS 1. Write short note on Galerkin’s method. 2. Show that in elasticity problems Galerkin’s method turns out to be the principle of virtual work. 3. State and explain the principle of virtual work.

128

Finite Element Analysis

9 Assembling Stiffness Equations—Variational Method 9.1 INTRODUCTION This method of assembling stiffness equations is widely used in elasticity problem. In calculus we know a function has extreme value when its first derivative with respect to variables is zero. The function is maximum, if the second derivative is negative and is minimum, if its second derivative is positive. The first derivative of function of a function is called first variance. The function of a function is termed a functional and the statement that the first variance of functional is zero is termed as first variance attains a stationary value. In many engineering problems there are such functional, the first variance of which attain stationary values. In elasticity problems potential energy of the body of the structure is such functional. In this chapter we will first study general mathematical method of variational approach, then assemble the expression for potential energy in a deformable solid and derive principle of minimum potential energy. Few simple problems are solved to explain the procedure. Then Raileigh–Ritz method is explained, which is useful for complex structural problems, encountered in finite element analysis. The general procedure of assembling stiffness matrix and load vector using the principle of minimum potential energy after expressing potential energy as a function with arbitrary constant using Raileigh – Ritz method is presented. It may be noted that variational approach is possible only if a suitable functional is available, otherwise the Galerkin’s method of weighted residual is to be used.

9.2 GENERAL VARIATIONAL METHOD IN ELASTICITY PROBLEMS

b

g

Let y be a function of x. Then, F y , y ′ , y ′′ is a functional. Say our concern is to find y = y (x), such that the first variance of

zb

x2

I =

g

F y , y ′ , y ′′ dx

…(9.1)

x1

is made stationary satisfying the boundary conditions

b g y bx g = y

y x1 = y1

and

…(9.2) 2 2 Figure 9.1 shows a typical function y = y (x). The continues line shows the exact function and dotted line shows an approximate function. Since to start with we don’t know exact solution y = y (x), we have approximate solution y = y x .

bg

Assembling Stiffness Equations—Variational Method

129

y ()x  y

y

0

x1

x

X

x2

Fig. 9.1

bg bg

δr = y x − y x .

Then,

…(9.3)

We are interested in finding the solution with

δI =0

z FGH

x2

i.e.

x1

…(9.4)

IJ K

∂F ∂F ∂F δy ′′ dx = 0 δy + δ y′ + ∂y ∂ y′ ∂ y ′′

…(9.5)

Let us now find integration of second and third terms by parts.

z

x2

x1

z

x2

and

x1

LM N

∂F ∂F δy ′ dx = ∂y ∂ y′ ∂ y′

LM N

∂F ∂F δy ′′ dx = ∂ y′ ∂ y ′′ ∂ y ′′

OP Q

L ∂ F ∂ y ′OP =M N∂ y ′′ Q

− x1 x2

x1

z

x2

x2

x1

OP Q

z

x2

x2

− x1

x1

FG IJ H K

d ∂F δy dx dx ∂ y ′

…(9.6a)

FG IJ H K

d ∂F δ y ′ dx dx ∂ y ′′

L d F ∂ F I ∂ yOP −M G MN dx H ∂ y ′′ JK PQ

x2

+ x1

z

x2

x1

FG IJ H K

d 2 ∂F δ y dx d x 2 ∂ y ′′

…(9.6b)

Hence substituting equation 9.6 in equation 9.5, we get

z

x2

x1

F ∂ F − d F ∂ F I + d F ∂ F I I δy dx + LMR|S ∂ F − d F ∂ F I U|V δyOP + L ∂ F ∂y ′O GH ∂ y dx GH ∂ y′ JK dx GH ∂ y ′′ JK JK MN|T ∂ y′ dx GH ∂ y ′′ JK |W PQ MN ∂ y′′ PQ 2

x2

x2

x1

x1

= 0 …(9.7)

2

130

Finite Element Analysis Since δ y is arbitrary, all the three terms in equation 9.7 should be zero. Thus we have

F I GH JK

F ∂F I = 0 GH ∂ y ′′ JK LMR ∂F − d F ∂F I UδyOP MNS|T∂ y′ dx GH ∂ y′′ JK V|W PQ LM ∂ F ∂ y ′OP = 0 N ∂ y ′′ Q

∂F d ∂F d2 + 2 − ∂ y dx ∂ y ′ dx

…(9.8a)

x2

…(9.8b)

x1

x2

and

…(9.8c)

x1

Equation 9.8 (a) is known as Euler–Lagrange equation. Equations 9.8 (b) and 9.8 (c) are known as boundary conditions. To satisfy equation 9.8 (b).

FG IJ H K

b g

∂F d ∂F | at x1 = 0 − ∂ y ′ dx ∂ y ′′

b g

∂F d ∂F − | at x2 = 0 ∂ y ′ dx ∂ y ′′

δ y x1 = 0 or Similarly at boundary x = x2,

δ y x2 = 0 or

FG IJ H K

To satisfy equation 9.8 (c),

b g

∂F | at x1 = 0 ∂ y ′′

b g

∂F | at x2 = 0 ∂ y ′′

δ y ′ x1 = 0 or and at x = x2,

δ y ′ x2 = 0 or The conditions like

b g

b g δ y ′ b x g = 0 , and δ y ′ b x g = 0

δ y x1 = 0 , δ y x2 = 0

1

2

…(9.9)

are known as kinematic boundary conditions. In solid mechanics they specify displacement requirement at support points. The conditions like

FG IJ H K ∂F d F ∂F I − G J| ∂ y ′ dx H ∂ y ′′ K

∂F d ∂F | at x1 = 0 − ∂ y ′ dx ∂ y ′′

at x2

=0

Assembling Stiffness Equations—Variational Method

131

∂F ∂F | x = x1 = 0 and | at x2 = 0 ∂ y ′′ ∂ y ′′ are known as natural boundary conditions. In solid mechanics, they specify force conditions like moment and shear force conditions at supports. Thus the first variance of the functional for stationary value yields Euler–Lagrange equation [equation 9.8(a)], kinematics boundary conditions [equation. 9.9(a)] and natural boundary conditions [equation. 9.9(b)]. A finite element analyst who is interested in solving a set of equations subject to a set of boundary values aim at first identifying the functional which satisfy Euler–Lagrange equation and satisfies boundary values specified. Then tries to solve Euler–Lagrange equation. In solid mechanics it has been identified that total potential energy is suitable functional, the first variance of which yields equation of equilibrium satisfying the boundary conditions. This statement is verified with the following simple problems in solid mechanics. Example 9.1: Show that the condition that first variance of total potential energy is stationary, is equivalent to satisfying equilibrium equation and boundary conditions in case of (i) Simply supported beam subjected to udl. (ii) Cantilever beam subjected to udl. Solution: Figs 9.2 (a) and 9.2 (b) show the typical beams considered. p/unit length

p/unit length

l

l

x1 = 0

x2 = l

x1 = 0

x2 = l

(a)

(b)

Fig. 9.2

In beams, strain energy due to bending

z l

=

0

z z l

=

0 l

=

0

M2 dx 2 EI

F EI d y I GH dx JK 2

2

2

dx

2EI

F I GH JK

EI d 2 y 2 dx 2

2

dx

132

Finite Element Analysis Potential energy due to load

z l

=−

py dx

0

∴ Total energy is given by

z LM z MN z l

Π=

0 l

=

F I GH JK

EI d 2 y 2 dx 2

0

2

z l

−

py dx

0

F I GH JK

EI d 2 y 2 dx 2

OP PQ

2

− py dx

l

=

F dx

0

F I GH JK

EI d 2 y F= 2 dx 2

where

2

− py

From equation 9.8, we know that the first variance of

FG IJ H K

z

F dx is stationary, means

FG ∂ F IJ = 0 H ∂ y ′′ K LM ∂ F − d F ∂ F I OP δy = 0 MN ∂ y ′ dx GH ∂ y ′′ JK PQ LM ∂ F δ y ′OP = 0 N ∂ y ′′ Q

∂F d ∂F d2 + 2 − ∂ y dx ∂ y ′ dx

(a)

x2

(b)

x1

x2

and

x1

In this problem we are trying EI F = 2

∴

since

Fd yI GH dx JK

∂F =−p ∂y y ′′ =

∂2y ∂x2

2

2

2

− py

∂F ∂F ∂2y = 0 and = EI 2 ∂ y′ ∂ y ′′ ∂x

(c)

Assembling Stiffness Equations—Variational Method

133

Hence Euler–Lagrange equation (equation a) for the beams is,

−p − 0 +

F GH

I JK

∂2 ∂2y EI 2 = 0 2 ∂x ∂x

i.e.,

EI

∂4y − p=0 ∂x4

This is the well known equation of equilibrium in structural mechanics. Thus in case of beams Euler– Lagrange equation yields equation of equilibrium. Now let us see whether boundary conditions are satisfied or not. Noting that y is exact solution and δy is the difference between exact and approximate solution, we find at supports y = 0 means δ y = 0 y ′ = 0 means δ y ′ = 0

and In case of simply supported beam, At x = x1

b g b g

δy x1 = y x1 = 0 hence equation (b) is satisfied. From equation (c), we have EI d 2 y dF = = M, where M is moment. dy ′′ dx 2

At x = x1 we know moment is zero. Hence equation (c) is also satisfied. At x = x2,

b g b g ∂F = M bx g = 0 | ∂ y ′′ δy x 2 = y x 2 = 0

and

x = x2

2

Thus the boundary conditions at x = x2 are also satisfied. Hence the condition that first variance of total potential energy to be stationary, satisfies Euler–Lagrange equation as well as the boundary conditions. Consider the cantilever beam [refer Fig. 9.2(b)]. At x = x1,

b g b g δy ′ b x g = y ′ b x g = 0

Equation (b) is satisfied since δy x1 = y x1 = 0 and equation (c) is satisfied, since 1

At x = x2,

1

FG IJ H K d F d yI EI =0 0− dx GH dx JK

∂F d ∂F =0 − ∂ y dx ∂ y ′′ 2

yields,

2

134

Finite Element Analysis

d 3y =0 dx 3

EI =

i.e.,

i.e., shear force = 0 We know at free end, shear force is zero. Hence equation (c) is satisfied. Thus, if potential energy is selected as a suitable function, its first variance is stationary, satisfies equilibrium equation as well as the boundary conditions.

9.3 POTENTIAL ENERGY IN ELASTIC BODIES Potential energy is the capacity to do the work by the forces acting on deformable bodies. The forces acting on a body may be classified as external forces and internal forces. External forces are the applied loads while internal forces are the stresses developed in the body. Hence the total potential energy is the sum of internal and external potential energies. We will derive first the potential energy in a spring which is uniaxially stressed member then we will derive the expression for potential energy in a three dimensionally stressed body.

u

P

0 + ku ku was also zero. Hence the average force during the extension is . = 2 2 Hence the energy stored in the spring due to straining, = Average force × Extension

Fig. 9.3

1 1 kuu = ku 2 2 2 ∴ Total Potential Energy in the Spring =

Π=

L

Stiffness k

(a) Potential Energy in a Spring: Figure 9.3 shows a typical spring. Let its stiffness (load per unit extension) be k and length L. Due to a force P let it extend by u. The load P moves down by distance u. Hence it looses its capacity to do work by Pu. Hence the external potential energy in this case Wp = –Pu …(9.10) When the load has reached equilibrium position after extension of spring by u, the force in spring = ku. But when extension was zero the resisting force

…(9.11)

1 2 ku − pu 2

…(9.12)

(b) Potential Energy in a There Dimensional Body: Consider a body of volume V subjected to (i) body forces Xb, Yb, Z b, in x, y, z directions. (ii) surface forces Xs, Ys, Zs, on surface S1 Let u, v and w be the displacement components. Then the potential energy of the external forces

zzz b

Wp = −

v

g

X b u + Yb v + Zb w dV −

zz b s1

g

X s u + Ys v + Z s w ds

…(9.13a)

Assembling Stiffness Equations—Variational Method

=−

zzz l q l u

T

q

X b dV −

v

where

zz l q l u

T

q

X s ds

135

…(9.13b)

s1

luq = u v w lX q = X Y Z lX q = X Y Z T

T

b

b

b

b

s

s

s

T

s

To find the internal energy due to straining, let us consider one by one stress component. Figure 9.4 shows a typical element of size dx × dy × dz subjected to σ x stresses and displacement in x-directions. The work done by σ x stresses in the element y







u

u



 

u





x

du

x

x

dy dx x

0

Fig.9.4

F ∂ bδug dxI − σ dy dz δu GH ∂ x JK d bδ ug dx dy dz , since du = ε

= σ x dydz δu +

= σx

x

∂x

dx

x

∴ The work done by σ x stresses in the element is = σ x dε x dx dy dz As the strain increases from zero to the final value ε x , the work stored as strain energy is

z

εx

=

0

σ x dε x du

lq

In the three dimensional element, there are six stress components σ Hence the total strain energy in the element

T

= σ x σ y σ z τ xy τ yz τ zx .

136

Finite Element Analysis

L =M MM N

z

z z

εx

εγ

εz

σ x ∂ε x + σ γt ε γ + σ z dε z +

0

0

z

γ xy

0

z

τ xy γ xy +

0

z

OP PP Q

γ zx

γ yz 2

τ yz dγ γ z +

0

τ zx dγ zx dV

0

The integration is from initial to the final state of stress. Let Uo be the strain energy per unit volume. Then from the above equation, we get

dU 0 = σ x dε x + σ y dε y + σ z dε z + τ xy dγ

xy

+ τ yz dγ

yz

+ τ zx dγ zx

(a)

But from chain rule of differentiation, we know

∂ U0 =

∂ U0 ∂ U0 ∂ U0 ∂ U0 dε x + dε y + dε z + dγ ∂ε x ∂ε y ∂ε z ∂γ xy

xy

+

∂ U0 dγ ∂γ yz

yz

+

∂ U0 dγ zx ∂ε zx

(b)

Comparing equations (a) and (b) we get,

∂ U0 ∂ U0 ∂ U0 =σx, =σy, = σz ∂ε x ∂ε y ∂ε z ∂ U0 ∂ U0 ∂ U0 = τ xy , = τ yz and = τ zx ∂γ xy ∂γ yz ∂γ zx

…(9.14)

Thus we find that the first derivate of strain energy per unit volume with respect to a strain component is the corresponding stress component i.e.,

RS dU UV = lσ q T dε W = D lε q 0

…(9.15)

Integration equation 9.15 we get,

lq

lq lq l q

1 1 T T ε ε σ D ε = 2 2 To get total strain energy of the solid, the term U0 be integrated over the entire volume of the solid. Thus, U0 =

U =

zzz

U 0 dV =

v

1 2

zzz l q l q ε

T

σ dV =

v

1 2

zzz l q ε

T

lq

D ε dV

v

∴ Total potential energy of the solid

Π = U + Wp i.e.,

Π=

1 2

zzz l q ε

v

T

lq

D ε dV −

zzz l q l u

T

q

X b dV −

v

zz l q l u

T

…(9.16a)

q

X s ds

…(9.16b)

s1

9.4 PRINCIPLE OF MINIMUM POTENTIAL ENERGY From the expression for total potential energy (9.16.a) we know

Π = U + Wp ∴ δ Π = δ U + δ Wp .

…(9.17)

Assembling Stiffness Equations—Variational Method

137

Comparing equation 9.13 (a) with equation 8.18, we find potential energy of external forces Wp is equal but opposite to total virtual work done by external forces. Thus,

Wp = − We ∴ δ Wp = − δ We ∴ From Equation 9.17, we conclude δ Π = δ U − δ We . But from principle of virtual work (equation 8.17) we know, δ U = δ We ∴ δΠ = 0

…(9.18)

Hence we can conclude that a deformable body is in equilibrium when the total potential energy is having stationary value. By taking second variance of potential energy, it has been proved by researchers that the value is positive definite. And hence it is concluded that the condition that value of total potential energy is stationary correspond to minimum value. Hence we have principle of minimum potential energy in solid mechanics, which. may be stated as “of all the possible displacement configurations a body can assume which satisfy compatibility and boundary conditions, the configuration satisfying equilibrium makes the potential energy assume a minimum value”. This is the variation principle in solid mechanics. Example 9.2: Assemble equations of equilibrium for the spring system shown in Fig. 9.5 by direct approach. Show that minimization of potential energy also yields same result. F1

F2

1

k1

2

k2 u1

F3 3

k3 u2

u3

Fig. 9.5

Solution: Consider the free body diagram of nodes 1, 2 and 3 shown in Fig. 9.6. Let the displacement of nodes be u1, u2, u3. Then the extensions of spring 1, 2 and 3 are

δ 1 = u1

δ 2 = u2 − u1 and δ 3 = u3 − u2

...(1)

F1

F2 F3

k1 δ1

k2 δ2 (a)

k2 δ2

k3 δ3 (b)

k3 δ3 (c)

Fig. 9.6

Equations of equilibrium are, − k1 δ 1 + k 2 δ 2 + F1 = 0

...(2a)

− k2 δ 2 + k3 δ 3 + F2 = 0

...(2b)

− k 3 δ 3 + F3 = 0

...(2c)

138

Finite Element Analysis From equations 1 and 2, we get

b g − k bu − u g + k bu − u g + F = 0 − k bu − u g + F = 0 bk + k g u − k u = F −k u + b K + k g u − k u = F − k1 u1 + k 2 u2 − u1 + F1 = 0

2

i.e.,

2

1

3

3

2

2

3

3

2

3

2

1

2

2

1

...(3a)

3

2

3 3

2

...(3b)

− k 3 u2 + k 3 u3 = F3

...(3c)

OP R|u U| R|F U| PP S|u V| = S|F V| Q Tu W TF W

...(4)

1

2 1

2

and The above equations in the matrix form are

LMbk + k g MM − k N 0 1

−k2

0

k2 + k3

− k3

−k3

k3

2

2

1

1

2

2

3

3

Now, let us see the potential energy approach. Total potential energy it the system is,

Π=

1 1 1 k1 δ 12 + k 2 δ 22 + k 3 δ 23 − F1u1 − F2 u2 − F3u3 2 2 2

=

∴

b

1 1 k1 u12 + k2 u2 − u1 2 2

g

2

+

b

1 k3 u3 − u2 2

g

2

− F1u1 − F2u2 − F3u3

∂Π = 0, gives ∂u1

b

gb g k u − k bu − u g − F = 0 bk + k g u − k u = F

k1u1 + k 2 u2 − u1 −1 − F1 = 0 i.e.,

1 1

i.e.,

2

1

2

1

2

1

...(a)

1

2

2

1

∂Π = 0 gives ∂u2

b

g b gb g u + bk + k g u − k u = F

k 2 u2 − u1 + k 3 u3 − u2 −1 = F2 − k2

1

2

3

2

3 3

2

∂Π = 0 gives, ∂u3

and

b

g

k 3 u3 − u2 − F3 = 0 or

...(b)

− k 3 u2 + k 3 u3 = F3

...(c)

Assembling Stiffness Equations—Variational Method

139

In the matrix form,

LMbk + k g MM − k N 0 1

−k2

2

2

k2 + k3 −k3

OP R|u U| R|F U| − k P Su V = S F V k PQ | Tu |W |TF |W 0

3

3

1

1

2

2

3

3

...(5)

Equations 4 and 5 are exactly same. Thus the condition that potential energy should have extreme value (minimum) leads to equations of equilibrium. Example 9.3: Determine the displacements of nodes 1 and 2 in the spring system shown in Fig. 9.7. Use minimum of potential energy principle to assemble equations of equilibrium. k1 = 60 N/m ( 1)

k3 = 100 N/m 100 N

1

2

( 3)

80 N

k2 = 75 N/m ( 2)

Fig. 9.7

Solution: Let u1 and u2 be the displacements of nodes 1 and 2. Then the extensions of springs are

δ 1 = u1

Π= = ∴ i.e.,

bk + k 1

δ 2 = u1

δ 3 = u2 − u1

1 1 1 k1 δ 12 + k 2 δ 22 + k 3 δ 23 − 100u1 − 80u2 2 2 2

b

1 1 1 k1 u12 + k 2 u12 + k 3 u2 − u1 2 2 2

b

g

2

− 100u1 − 80u2

gb g

∂Π = 0 → k1u1 + k 2 u1 + k 3 u2 − u1 −1 − 100 = 0 . ∂u1

g

+ k 3 u1 − k 3 u2 = 100

2

∴

b

g

∂Π = 0 → k 3 u2 − u1 − 80 = 0 . ∂u2

...(b)

− k 3u1 + k 3 u2 = 80

LMk + k + k N −k 1

...(a)

2

3

3

−k3 k3

OP RSu UV = RS100UV Q Tu W T80 W 1

2

Substituting the values of k1,k2 and k3, we get.

LM 235 N−100

OP RSu UV = RS100UV Q Tu W T80 W

−100

1

100

2

140

Finite Element Analysis

∴

RSu UV = LM100 1 u 23500 10000 − N100 T W

OP RS100UV = RS1333 . U 80 Q T W T2.133VW

100 235

1

2

Answer

Example 9.4: Solve the three bar truss problem given in Example 8.1 by minimum potential energy principle. Solution: As given in example 8.1, l1 = 2.309 m l2 = 2 m and l3 = 4.0 m.

ε 1 = 0.375u + 0.2165v ε 2 = 0.5u . u − 0.2165v ε 3 = 0125

and Strain energy of a bar

=

1 stress × strain × volume 2

=

1 Eε × ε × AL 2

=

1 EAl ε 2 2

∴ Potential energy of the structure is Π=Σ

1 EAli ε i2 − 20v 2

b

1 EA 2.3094 0.375u + 0.2165v 2 Which is exactly same as in example 8.1. =

g

2

b g

2

b

+ 2 0.5u + 4 0125 . u − 0.2165v

g

2

− 20v

∴ u = 0.0893u

b

g

b

gb

g

∂Π 1 = 0 → EA 2.3094 × 2 0.375u + 0.2165v × 0.2165 + 4 0125 . u − 0.2165v −0.2165 − 20 = 0 ∂v 2 This equation is also exactly same as in example 8.1 for vertical virtual displacement. Hence here also

v=

66.048 5898 . and u = EA EA

and member forces are

and

F1=26.045 kN F2=2.949 kN F3= - 13.562 kN

Answer

9.5 THE RAYLEIGH–RITZ METHOD The Rayleigh–Ritz method of expressing field variables by approximate method clubbed with minimization of potential energy has made a big break through in finite element analysis. In this article Rayleigh – Ritz method is explained with simple problems.

Assembling Stiffness Equations—Variational Method

141

In 1870 Rayleigh used an approximating field with single degree of freedom for studies on vibration problems. In 1909 he used approximating field with several functions, each function satisfying boundary conditions and associating with separate degree of freedom. Ritz applied this technique to static equilibrium and Eigenvalue problems. The procedure for static equilibrium problem is given below: Consider an elastic solid subject to a set of loads. The displacements and stresses are to be determined. Let u, v and w be the displacements in x, y and z coordinate directions. Then for each of displacement component an approximate solution is taken as

u=

∑ a φ b x , y , zg for i = 1 to m i

i

1

b g w = ∑ a φ b x , y , z g for k = m + 1 to m

v=

∑a

φ j x , y , z for j = m1 + 1 to m2

j

k

k

…(9.12)

2

The function φ i are usually taken as polynomials satisfying the boundary conditions. ‘a’ are the amplitudes of the functions. Thus in equation 9.21 there are n number of unknown ‘a’ values. Substituting these expressions for displacement in strain displacements and stress strain relations, potential energy expression 9.16 can be assembled. Then the total potential energy

d

Π = Π a1 , a 2 . .. a m1 , a m1 +1 ... a m2 a m2 +1 ... a m

i

From the principle of minimum potential energy,

dΠ = 0 for i = 1 to m . dai

…(9.22)

From the solution of m equation of 9.22, we get the values of all ‘a’ . With these values of ‘ai’s and φ i ’s satisfying boundary conditions, the displacements are obtained. Then the strains and stresses can be assembled. The Rayleigh – Ritz procedure is illustrated with small problems below: Example 9.5: Using Ragleigh–Ritz method determine the expressions for deflection and bending moments in a simply supported beam subjected to uniformly distributed load over entire span. Find the deflection and moment at midspan and compare with exact solutions. Solution: Figure 9.8 shows the typical beam. The Fourier series y =

α

∑

ai sin

m = 1, 3

for simply supported beams since y = 0 and M = EI

d2y

∂x2

m πx is the ideal function l

= 0 at x = 0 and x = l are satisfied. For the simplicity

w/unit length

A

B l

y x=0

x=l

Fig. 9.8

x

142

Finite Element Analysis

let us consider only two terms in the series i.e. let

y = a1 sin

z l

Π=

0

3π x πx + a2 sin l l

F I GH JK

EI d 2 y 2 dx 2

Substituting y in equation (b) we get

z l

=

0

=

=

LM MN

3π x π2 π x 9π 2 EI − 2 a1 sin − 2 a2 sin 2 l l l l

EI π 4 2 l4 EI π 4 2 l4

z z z l

0 l

0

l

Noting that

0

OP PQ

2

z

...(a)

l

dx −

wy dx

z FGH

2

l

dx − w a1 sin 0

IJ K

3π x πx + a2 sin dx l l

FG a sin π x + 9a sin 3π x IJ dx − w LM−a l cos π x − a l cos 3π x OP H l l K N π l 3π l Q FG a sin π x + 18a a sin π x sin 3π x + 81a sin 3π x IJ dx − wl LM2a + 2a OP H π N 3π Q l l l l K l

2

1

2

1

2

0

2 1

2

πx sin dx = l 2

z 0

z

FG H

l

0

sin

2

1

IJ K

3π x πx sin dx = l l

z l

y=

EI π 4 2 l4

=

EI π 4 4 l3

zFGH z FGH l

cos

0

IJ K

2π x 4π x − cos dx = 0 l l

IJ K FG a l + 81 l a IJ − 2 wl FG a + a IJ H 2 2 K π H 3K ea + 81a j − 2πwl FGH a + a3 IJK

sin 2

0

Π to be minimum,

2

1 2π x 1 1 − cos dx = 2 2 l

and

we get,

2 2

1 2

l

3π x = l

l

0

1 6π x 1 1 – cos dx = 2 2 l

2 1

2 1

2 2

2

1

2 2

2

1

∂Π ∂Π = 0 and = 0. ∂a1 ∂a2 i.e.,

or

and

...(b)

0

2 wl EIπ 4 2a1 − =0 3 π 4l a1 =

4 wl 4 EI π 5

2 wl EIπ 4 81 × 2a2 − =0 3π 4l 3

Assembling Stiffness Equations—Variational Method

a2 =

or

∴ y= ∴ Max. deflection which occurs at x = y max =

143

4 wl 4 243EI π 5 4 wl 4 4 wl 4 3π x πx + sin sin 5 5 l l EIπ 243EIπ

l is 2

4wl 4 EIπ 5

−

4wl 4 wl 4 = 243EIπ 5 76.82EI

we know the exact solution is

y max =

5 wl 4 wl 4 = 384 EI 76.8EI

Thus the deflection is almost exact.

M x = EI

Now,

LM MN

= EI −

LM MN

M centre = −

we know the exact value is

F GH

9π 2 3π x π2 πx d2y sin sin = − − EI a a 1 2 2 2 2 l l dx l l 4wl 2 EIπ

3

sin

3π x πx 4 wl 2 x 9 − sin 3 l l 243EIπ

OP PQ

OP PQ

I JK

4wl 2 4 wl 2 x 9 wl 2 + 3 3 = EIπ 243EIπ 8.05

wl 2 . 8

By taking more terms in Furier series more accurate results can be obtained. Example 9.6: Using Rayleigh-Ritz method, determine the expressions for displacement and stress in a fixed bar subject to axial force P as shows in Fig. 9.9. Draw the displacement and stress variation diagram. Take 3 terms in displacement function. P ×

E – Youngs modulus

1

n A – Cross sectional area

l/2 x=0

l/2 x = /2 l

x=l

Fig. 9.9

144

Finite Element Analysis

Solution: Let the displacement at load point be u1. Then the strain energy of the bar

z

FG IJ H K

l

1 du U = EA 2 dx 0

2

dx

and potential energy due to external forces = –Pu1 1 ∴Π= 2

z l

EA

0

FG du IJ H dx K

2

dx − Pu1

Let the displacement at any point be given by, u = a1 + a2 x + a3 x2 This function has to satisfy the boundary conditions (i) at x = 0, u = 0 (ii) at x = l, u = 0 From Boundary condition (i), we get 0 = a1 From Boundary condition (ii), we get, 0 = a1 + a2l + a3 l2 ∴ From equations (1) and (2) we get 0 = a2 l + a3 l2 or a2 = –a3 l

(1) (2)

(3)

∴ u = –a3 l x + a3 x = a 3 − lx + x 2 2

At

x=

l 2

LM MN

u = u1 = a3 −l

u1 = −

i.e., Now

l l2 + 2 4

OP PQ

a 3l 2 4

...(4)

b

du = a3 −l + 2 x dx

z l

g

b

1 ∴Π= EA a32 −l + 2 x 2 0

1 = EA a 32 2

z l

el

2

g

2

dx + Pa3

l2 4

j

− 4lx + 4 x 2 dx + Pa 3

0

LM MN

1 4x3 = EA a32 l 2 x − 2lx 2 + 2 3

OP PQ

l

+ Pa3 0

l2 4

l2 4

Assembling Stiffness Equations—Variational Method

= ∴

F I GH JK

1 l3 l2 + Pa3 EA a32 2 3 4

dΠ l3 l2 = 0 → EA a 3 + P =0 3 4 da3 3p 4 EAl

a3 = − ∴u=−

3p −lx + x 2 4 EAl

∴ u1 = −

3 pl a3 l 2 = 4 16

b

du = Ea3 −l + 2 x dx

σ =E =E

g

−3 p 3p −l + 2 x = l − 2x 4 EAl 4 Al

∴ σ 0 = σ x=0 =

3p 4A

σ1 = σ x= l = 0 2

3p 4A The variation of displacement and stresses are shown in Fig. 9.10.

σ 2 = σ x =l = −

3Pl 16 × Parabolic variation

u X



(a)

+

–

(b)

Fig. 9.10

(a) Variation of u (b) Variation of stress



3P 4A

145

146

Finite Element Analysis

Example 9.7: Determine the displacement and stress in a bar of uniform cross section due to self weight only when held as shown in Fig. 9.11. Use (i) two terms (ii) three terms, for approximating polynomial. Verify the expression for total extension with the exact value. 0

y

l

x

Fig. 9.11

Solution: Let ‘ ρ ’ be unit weight and E Young’s modulus of the material of the bar. If A is the cross section of the bar then,

U=

zzz l q l q z FGH IJK LMN OPQ z FGH IJK zzz l q l q z

1 T ∈ σ dV 2

v

l

=

0

=

and

1 du 2 dx l

1 2

EA

0

We = −

u

T

E

du dx T

du A dx dx

...(1)

2

dx

X b dV

v

l

= − u ρ A dx

...(2)

0

Polynomial function for displacement may be taken as

u = a0 + a1 x + a2 x 2 + a3 x 3 + ... + an xnn The boundary condition to be satisfied is At x = 0, u = 0

...(3)

Assembling Stiffness Equations—Variational Method

147

From this we get 0 = a0

∴ u = a1 x + a2 x 2 + a3 x 3 + ... + an x n (i) When only two terms of polynomial equations are used,

u = a1 x

∴U =

1 2

z z z

∴

l

EA

0

FG du IJ H dx K

du = a1 dx 2

dx =

1 2

z l

0

EA a12 dx = 1 EA a12 l 2

l

Wp = − u ρ A dx 0

l

=−

0

Lx O a x ρ A dx = − ρ A a M P MN 2 PQ 2

1

= − ρA a1

l

1

0

l2 2

1 l2 EA a12 l − ρ Aa1 2 2 From minimization condition, we get ∴Π=

dΠ l2 = 0 i.e., 0 = EAa1l − ρ A 2 da1 or

a1 = ∴ u=

and

ρl 2E

...(4)

ρl x 2E

...(5)

du ρl = dx 2 The displacement and stress variations are shown in Fig. 9.12. Extension of the bar = ul – u0

σ =E

=

ρl ρl 2 l−0= 2E 2E

...(6)

...(7)

148

Finite Element Analysis

Three terms

u

Pl2 2E

Two terms

x (a)

Three terms

u

Two terms

x (b)

Fig. 9.12

(a) Variation of displacement (b) Variation of stress

(ii) When three terms are considered for displacement in equation 3:

u = a1 x + a2 x 2

du = a1 + 2a2 x dx

∴

Π = U + Wp =

=

1 2

z l

EA

0

1 EA 2

z l

FG du IJ H dx K

dx −

u ρ A dx

0

ba + 2a xg 1

z l

2

ze l

2

2

0

dx − ρ A

0

LM MN 1 L = EAMa l + 2a a l 2 N

Fa x + a x I − ρ AG H 2 3 JK 4 O F l + a l IJ + a l P − ρA G a 3 Q H 2 3K

x2 x3 1 = EA a12 x + 4a1a2 + 4a22 2 2 3 2 1

∴

j

a1 x + a2 x 2 dx

1 2

OP PQ

l

2

3

1

0

2

2

2 3 2

dΠ 1 l2 = 0 → EA 2a1l + 2a2 l 2 − ρ A = 0 2 2 da1

1

l

2

3

2

0

Assembling Stiffness Equations—Variational Method

i.e.,

a1 + a2 l =

ρl 2E

149

...(8)

LM N

OP Q

dΠ 1 8 l3 = 0 → EA 2a1l 2 + a2 l 3 − ρA = 0 2 3 3 da2 i.e.,

4 ρl a2 l = 3 3E From equation 8 and 9 we get, a1 +

...(9)

ρl ρl ρl 1 a2 l = − =− 3 3E 2 E 6E ∴ a2 = − Substituting it in equation 8, we get

a1 =

FG H

ρ 2E

IJ K

ρl ρ ρl − − l = 2E 2E E

∴ u = a1 x + a2 x 2 =

LM MN

ρl ρ 2 ρ x2 x− x = lx − E 2E E 2

OP PQ

du =ρ l− x dx The variations of displacement and stress in this case also are shown in Fig. 9.12.

σ =E

Extension of the bar u1 − u0 =

LM MN

OP PQ

ρ 2 l2 ρl 2 = l − 2 2E E

Actual extension of the bar [refer Fig. 9.13]

Px l

dx

Px

x

Fig. 9.13

150

Finite Element Analysis

z z l

=

0

Px dx = AE

l

0

LM OP MN PQ

ρAxdx ρ x 2 = AE E 2

l

= 0

ρl 2 2E

Thus total extension of the bar obtained is exact in both the cases.

Comments on Rayleigh–Ritz Method In this method the approximating functions must satisfy the boundary conditions and should be easy to use. Polynomials are normally used. Some times sine-cosine terms are also used. Results can be obtained for complex problems. But for complex problems it is difficult to say whether the results obtained are accurate enough to use. The doubt will arise due to the following two reasons (i) Whether this is the only function which can be used (ii) How many terms in the function are to be used. The best way to ensure the accuracy is to get result using a certain number of terms and then use additional terms to get the results. If the difference is negligible, we can conclude that the satisfactory result is obtained. However it may be noted that the lowest terms in the series should not be omitted in the approximating functions.

9.6 VARIATIONAL FORMULATION IN FINITE ELEMENT ANALYSIS From variational principle, we have concluded that a body is in equilibrium when potential energy is minimum. From Rayleigh-Ritz method we have found approximating functions satisfying the boundary conditions can be used to get the solutions. In finite element analysis we use approximating functions for the elements but not for entire body and use the principle of minimizing potential energy to get the solutions for complex structures. The similarity and the differences in Rayleigh-Ritz method and finite element method are as listed below: (a) Similarity: (i) Both methods use approximating functions as trial solution (ii) Both methods take linear combinations of trial functions. (iii) In both methods completeness condition of the function should be satisfied (iv) In both methods solution is sought by making a functional stationary. Difference (i) Rayleigh-Ritz method assumes trial functions over entire structure, while finite element method uses trial functions only over an element. (ii) The assumed functions in Rayleigh-Ritz method have to satisfy boundary conditions over entire structure while in finite element analysis, they have to satisfy continuity conditions at nodes and some times along the boundaries of the element. However completeness condition should be satisfied in both methods. Now let us see the variational method in finite element analysis in detail. The potential energy of a structure is

Π = Internal potential energy – External potential energy = Strain energy – Work done by external forces =U–W These expressions involve integration terms.

Assembling Stiffness Equations—Variational Method

151

The above expression refers to entire structure. Since the integration of the summation is the same as the sum of the individual integrals, we can apply the principle to each element separately. Thus n

∑

Π=

n

e =1

∴

dΠ =0= dδ

n

∑

Πe =

Ue −

e =1

∑ We e =1

∑ d lδ q − ∑ d lδ q dUe

dWe

e

…(9.23)

e

l q is vector of nodal displacement in the structure. And lδ q is the vector of nodal displacements in the element.

Where δ e

Now, Ue = Strain energy of the element

zzz l q

=

1 ε 2

v

T

σ dV

lεq = B lδ q δ = D lε q = D B lδ q 1 ∴ Ue = d B lδ q i D B lδ q dV 2

But and

zzz zzz l q

T

e

e

v

1 2

=

If

lXq = lX T

b

Yb Z b

q and l X q = lX

δ

equation 9.13 as

Wp = −

s

B

luq = lu T

…(9.24)

e

T

zz l q l

q

X b dV −

u

T

q

X s ds

…(9.25)

s1

q

v w

lq

u= N δ

But

D B δ

q

v

where

l q dV ,

T

Ys Z s , then the work done by these forces is given by

zzz l q l u

e

v

T

s

T

e

where δ e is the nodal displacement vector of the element.

∴W =−

zzz d l q i l q zz d zzz l q l q zz l q N δ

T

e

X b dV −

v

=−

T

s

e

s1

δ

v

l q i l X q ds

N δ

T e

N

T

δ

X b dV −

s1

T e

N

T

l X q ds s

152

Finite Element Analysis Hence,

Πe = Ue + Wp =

1 2

zzz l q

T

δ

B

e

lq

T

D B δ

e

dV −

v

zzz l q δ

T e

N T Xb dV −

v

zz l q

zzz zzz v

i.e.,

1 2 B 2

B

lq

T

D B δ

lq

T

D B δ

e

dV −

dV =

e

v

zzz

zzz

N

l X q dV −

T

b

v

N

N

e

T

lX q ds . s

s1

From principle of minimum potential energy (equation 9.23), 0=

T

δ

T

l X q dV + b

v

zz

zz

N

T

l X q ds s

s1

N

T

l X q ds s

s1

Thus element equilibrium equation is

k

e

lδ q = b F g e

k

where

e

=

zzz zzz

e

B

T

D B dV

v

lF q

and

The matrix k

e

=

zzz

e

=

v

B

T

N

T

lX q dV + b

…(9.26)

zz

N

T

lX q ds s

s1

lq

D B dV is called stiffness matrix of the element and the load vector F

e

v

is called consistent load. Thus

k

e

= B

T

lq

lq=

D B is stiffness matrix of the element and F

e

zzz v

N

T

l X q dV − b

zz

N

T

l X q ds s

s1

is called consistent load vector. The above equation of equilibrium is to be assembled for entire structure and boundary conditions are to be introduced. Then the solution of equilibrium equations result into nodal displacements of all the nodal points. Once these basic unknowns are known, then displacement at any point may be obtained by the relation

l q . The strains are assembled using the relation lεq = B lδ q lσ q = D lεq . u= N δ

e

e

and then stresses also can be found

e

Comments on Variational Method It is a versatile method. Using this method stiffness matrices and consistent load vectors can be assembled easily. This method has made finite element analysis a versatile method. All complex problems can be solved. This is universally used method in solid mechanics.

Assembling Stiffness Equations—Variational Method

153

QUESTIONS 1. State and explain the principle of minimum potential energy. 2. Derive the general equation for determining the stiffness of an element with usual notations in the form

k (i) (ii) (iii) (iv)

e

=

z

B

T

D B dV .

Explain the principle of Rayleigh–Ritz method. Write short note on variational principles. Derive Euler–Lagrange equation for an integral function using variational principle. Using principle of minimum potential energy derive the expressions for consistent loads for body forces and surface forces.

154

Finite Element Analysis

10 Discretization of Structures 10.1 INTRODUCTION The process of modeling a structure using suitable number, shape and size of the elements is called discretization. The modeling should be good enough to get the results as close to actual behavior of the structure as possible. In this chapter various aspects of discretization of structures are discussed.

10.2 NODES AT DISCONTINUITIES In a structure we come across the following types of discontinuities: (a) Geometric (b) Load (c) Boundary conditions (d) Material.

(a) Geometric Discontinuities Wherever there is sudden change in shape and size of the structure there should be a node or line of nodes. Figure 10.1 shows some of such situations. Node

(a)

Line of nodes

t1

t2 (b)

Fig. 10.1

(a) Bar subject to axial forces (b) Plate with varying

Discretization of Structures

155

Node

Node

Node

(c)

(d)

Fig. 10.1 (contd)

(c) Slab (d) Dam with a crack

(b) Discontinuity of Loads Concentrated loads and sudden change in the intensity of uniformly distributed loads are the sources of discontinuity of loads. A node or a line of nodes should be there to model the structure. Some of these situations are shown in Fig. 10.2. Node

Node

P1

P2

(a)

Line of nodes

(b)

Fig. 10.2

(a) FEM model (b) Slab with different udls

156

Finite Element Analysis

(c) Discontinuity of Boundary conditions If the boundary condition for a structure suddenly change we have to discretize such that there is node or a line of nodes. This type of situations are shown in Fig. 10.3. Line of nodes

Node Column

Wall

Node Column

Fig. 10.3

Slab with intermediate wall and columns

(d) Material Discontinuity Node or node lines should appear at the places where material discontinuity is seen. Material 1

Material 2

Fig. 10.4

Material discontinuity

18

13 9 7 8 2 4 1

Fig. 10.5

3

11

10 6 5

12 10

7 8

9

Refined mesh near curved boundary of a dam

10.3 REFINING MESH To get better results the finite element mesh should be refined in the following situations

Discretization of Structures

157

(a) To approximate curved boundary of the structure (b) At the places of high stress gradients. Such situations are shown in Fig. 10.5 and Fig. 1.2.

10.4 USE OF SYMMETRY Wherever there is symmetry in the problem it should be made use. By doing so lot of memory requirement is reduced or in other words we can use more elements (refined mesh) for the same capacity of computer memory. When symmetry is to be used, it is to be noted that at right angles to the line of symmetry displacement is zero. In the tension bar example shown in Fig. 1.2, biaxial symmetry of the problem is utilized and only quarter of the bar is taken for the analysis.

10.5 FINITE REPRESENTATION OF INFINITE BODIES Soil is a typical example of infinite body. Wherever settlement of soil is to be studied or study is required about soil structure interaction, the mass of soil is to be modeled. It is well known fact that the soil mass away from footing is not affected. However the question remains how much mass of soil is to be considered. Best way to handle this type of problems is to consider a certain mass of soil and determine the settlement under the load. Then increase or decease the mass of soil, analyses and again compare the results. This type of preliminary study helps in identifying the mass of soil to be considered. Figure 10.6 shows the finite element idealization of one such problem. There are research reports that for homogeneous soil mass, H should be 4 to 6 times and V should be 10 to 12 times the footing width or diameter. B or D

V = 10 to 12 Bor D

H = 4 to 6 Bor D

Fig. 10.6

158

Finite Element Analysis

10.6 ELEMENT ASPECT RATIO The shape of the element also affects the accuracy of analysis. Defining the aspect ratio as ratio of largest to smallest size in an element, the conclusion of many researchers is aspect ratio should be as close to unity as possible. For a two dimensional rectangular element, the aspect ratio is conveniently defined as length to breadth ratio. To study the effect of aspect ratio on the accuracy of results, Desai and Abel analyzed a beam with 12 elements of different aspect ratios as shown in Figs. 10.7 and 10.8 shows the plot of inaccuracy of the displacement verses the aspect ratio. From this it can be concluded that the aspect ratio closer to unity yields better results. 12 units

12 units

1 8 units 1

3

2

4

5

6

8

7

2

1

5

9

2

6

10

Aspect ratio:

4

5

6

9 10 11 12

8 units

7

Aspect ratio:

3

8

8 =8 1

3

7

11

9

10

Aspect ratio:

4

11

4 =2 2

1

2

3

4

5

6

7

8

9

10

11

12

8

12

12 / 4 = 1.125 8/3

Aspect ratio:

1

2

3

4

5

6

7

8

9

10

11

12

Aspect ratio:

12 / 2 = 4.5 8/6

Fig. 10.7

12 / 3 =2 8/4

12

% error in displacement

Discretization of Structures

1

–5

2

3

4

5

6

7

159

8 Aspect ratio Exact solution

×

–10 × –15

Fig. 10.8

Effect of aspect ratio on accuracy of result

10.7 HIGHER ORDER ELEMENTS VS REFINED MESH Accuracy of calculation increases if higher order elements are used. Accuracy can also be increased by using more number of elements. Limitation on use of number of elements comes from the total degrees of freedom the computer can handle. The limitation may be due to cost of computation time also. Hence to use higher order elements we have to use less number of such elements. The question arises whether to use less number of higher order elements or more number of lower order elements for the same total degree of freedom. There are some studies in this matter keeping degree of accuracy per unit cost as the selection criteria. However the cost of calculation is coming down so much that such studies are not relevant today. Accuracy alone should be selection criteria which may be carried out initially on the simplified problem and based on it element may be selection for detailed study.

10.8 NUMBERING SYSTEM TO REDUCE BAND WIDTH Storing global stiffness matrix in the computer memory imposes a serious limitation on the number of elements/ degrees of freedom to be used. In elasticity problem the stiffness matrix is symmetric and banded. In a problem with 1000 degree of freedom the size of stiffness matrix is 1000 × 1000. If it has semi band width of 28, we can store only 1000 × 28 elements and handle the solution using suitable programming technique. The size of semi band width of stiffness matrix depends upon the numbering system adopted for nodes. The semi band width B is given by the expression B = (D + 1) f …(10.1) where D is maximum difference in node number in an element after considering all elements f-degrees of freedom per node. The semi band widths for various types of numbering for a two dimensional problems are shown in Fig. 10.9. It may be concluded that the semi band width is minimum if numbering is in shorter direction and restarted from the initial end after reaching other end. In many standard packages numbering is done automatically to keep the semi band width least.

160

Finite Element Analysis 1

14 15

28

29

2

3

4

5

6

7

13

12

11

10

9

16

17

18

19

20

27

26

25

24

23

30

31

32

33

34

1

8

8

21

15

22

22

35

29

2

3

2 3

4

5

10

11

20

21

30

12

19

22

29

8

13

18

23

28

7

14

17

24

15

16

25

6

7

10

11

12

13

16

17

18

19

20

23

24

25

26

27

30

31

32

33

14

21

28

34

35

26

31

B = (8 + 1) × 2 = 18

9

6

5

9

B = (13 + 1) × 2 = 28 1

4

27

26

31

1

32

2

33

3

34

4

35

5

B = (9 + 1) × 2 = 20

6

11

16

21

7

12

17

22

8

13

18

23

28

9

14

19

24

29

10

15

20

25

27

30

32

33

34

35

B = (6 + 1) × 2 = 14

Fig. 10.9

QUESTIONS 1. Write short notes on: (a) Effect of element aspect ratio on accuracy (b) Numbering nodes for band width minimization (c) Mesh refinement vs higher order elements. 2. Discuss the various points to be considered while descretizing a structure for finite element analysis. 3. Briefly explain how problem involving infinite bodies are handled in finite element analysis.

Finite Element Analysis—Bars and Trusses

161

11 Finite Element Analysis—Bars and Trusses 11.1 INTRODUCTION Under ‘bars’ we consider the analysis of members subject to axial forces only. These members are having one dimension (length) considerably large compared to cross sectional dimensions. Tension bars and columns fall under this category. In case of pin connected frames (trusses), members can be assumed to have only axial forces. In this chapter the analysis of the following three types of members is explained: (i) Tension bars/columns (ii) Two dimensional trusses (plane trusses) (iii) Three dimensional trusses (space trusses) Various steps involved in finite element analysis is thoroughly presented and few simple problems are solved with hand calculations.

11.2 TENSION BARS/COLUMNS The typical member considered for explaining the procedure is shown in Fig.11.1. In this problem we see cross section varies in 3 steps A1, A2 and A3. There are three point loads P1, P2 and P3. The surface forces are xs1, xs2, and xs3 and Xb is the body force. The surface forces may be due to frictional forces, viscous drag or surface shear. The body force is due to self weight. The material of the bar is same throughout. Step 1: Selecting suitable field variables and elements: In all stress analysis problems, displacements are selected as field variables. In the tension bar or columns at any point there is only one component of displacement to be considered, i.e., the displacement in x direction. Since there is only one degree of freedom and it needs only Co continuity, we select bar element shown in Fig. 11.2. In this case there are only two nodes. Step 2: Discritise the continua In this problem there are geometric discontinuities at x = 200 mm, 500 mm and 650 mm. There is additional point of discontinuity at x = 350 mm, where concentrated load P1 is acting. Hence we discritise the continua as shown in Fig. 11.3 using four bar elements.

162

Finite Element Analysis

x5

200 mm

xb

150 mm

xs2 300 mm

xb

P1

P2

x5

150 mm

xb P3 1

Fig. 11.1 

 1 1

2

2

x xc

x1  = –1

x2

x

 =0





=1

Fig. 11.2

Hence nodal displacement vector is

lδ q = RSTδδ UVW 1

2

In finite element analysis the nodes may be numbered in any fashion, but to keep the band width minimum we number the nodes continuously. In this problem there are five nodes and in all such problem there is definite relationship between number of nodes and number of element i.e. Number of node = Number of elements + 1. There is only one degree of freedom at each node. Hence total degree of freedom in the problem is = Number of nodes × degree of freedom at each node =5×1=5 i.e.

lδ q

T

= δ1 δ2 δ 3 δ 4 δ5

...(11.2)

Finite Element Analysis—Bars and Trusses

163

1

200 mm

1

2 2

150 mm 3

3

P1

150 mm

4

P2

4 150 mm

5 P3

x

x

(a)

(b)

Fig. 11.3

For any element local node number is 1 and 2 only, but global coordinate numbers for each element are different. For example, local coordinate numbers 1 and 2 for element 3 refers to global numbering system 3 and 4 respectively. The relation between the local and global node number is called connectivity details. In this problem the connectivity detail is as shown in Fig. 11.4. From this Figure it can be seen that the connectivity detail can be easily generated also. Thus For element (i), Local node number 1 = i Local node number 2 = i + 1

Element

Nodes 1

2

1

1

2

2

2

3

3

3

4

4

4

5

Fig. 11.4

Local numbers

Global numbers

164

Finite Element Analysis

Step 3: Select Interpolation Functions In chapter 5 we have seen interpolation functions [N] is given by

luq = N lδ q

...(11.3)

e

and for bar elements [N] = [N1 N2], where

and

N1 =

x2 − x 1 − ξ = 2 le

N2 =

x − x1 1 + ξ = 2 le

...(11.4)

Step 4: Element Properties In this step we assemble element stiffness matrix and nodal force vector of the element. At any point in the element,

luq = u lεq = ε and lσ q = σ , all in x direction, which is the only direction for these elements.

From strain displacement relations,

lεq = ε = dudx = dxd =

RSδ UV = L dN Tδ W MN dx 1

N1 N 2

1

2

RS UV T W

dN 2 dx

OP RSδ UV Q Tδ W 1

...(11.5)

2

δ1 1 −1 1 le δ2

RSδ UV , where Tδ W lσ q = σ = D lεq = B

1

B =

2

1 −1 1 le

= E ε , since D = E Element stiffness matrix

k e=

...(11.6)

zzz z LMN OPQ B

T

D B dV

v

l

=

0

=

1 −1 1 EA −1 1 A dx = 2 E le 1 le le

LM N

OP Q

EA 1 −1 x le2 −1 1

le 0

=

LM N

z LMN OP Q

EA 1 −1 le −1 1

OP Q

1 −1 dx −1 1

...(11.7)

Finite Element Analysis—Bars and Trusses

165

Consistant Load Equivalent nodal loads are to be calculated for each type of load acting on the element (i) Body Force: Xb is the only body force in this case. In case of self weight X b = ρ where ρ is unit weight of the material. From equation 9.26 the consistant load due to this body force is given by

lF q = e

zzz

N

v

ξ=

since

z

l X q dV = RSNN UV ρ T W 2 = bx − x g l le

T

b

1

x − xc le 2

b

A dx

2

0

c

e

l 2 dx or dx = e dξ 2 le and limits of integration will be from –1 to 1 dξ =

we get

R|1 − ξ U| lFq = S1 +2 ξ V ρ A l2 dξ || || T 2 W L ξ OP = l A ρ l l A dξ = AX Mξ − 2 4 MN 2 PQ 2

z 1

e

e

b

−1

Now

Similarly

z z

1−ξ ρb 2

1

2

e

e

e

b

b

−1

1+ξ 1 l ρ b A e dξ = Ale ρ b 2 2 2

l q = Al 2ρ LMN11OPQ

∴ F

e

b

...(11.8)

e

Noting that Ale is volume of the element, we find that half the self weight goes to each node. (ii) Surface Load: If Xs is the intensity of surface load, T = Xs × perimeter is the load per unit length of the element. Then consistant load corresponding to it is

zz z

lFq = lN q X ds RN U RN U l = S V T dx = S V T dξ TN W TN W 2 R|1 − ξ U| l 2 = S T dξ Tl R1U V ||1 + ξ || 2 = 2 ST1VW T 2 W e

T

s

z

l

0

z

l

1

2

−1

1

e

s

2

l

−1

e

s

e

...(11.9)

166

Finite Element Analysis Thus the consistant load for such surface traction is also half the total load at each node.

(iii) Point Load: Point loads can be directly added to nodal force vector.

l q = RSTFF UVW can be

After finding consistant load due to all types of loads, element nodal force vector F assembled.

e1

e

e2

...(11.10)

Step 5: Global Properties From step 3, we have

LM N

OP Q

1

2

EA1 1 −1 1 ke = 1 le1 −1 1 2

k

e3

LM N

k

OP Q

3

e2

LM N

2

OP Q

LM N

4

4

EA3 1 −1 3 = le3 −1 1 4

k

OP Q

3

EA2 1 −1 2 = le2 −1 1 3 5

EA4 1 −1 4 = le 4 −1 1 5

e4

For each element their position corresponding to global rows and columns are indicated above. Now global stiffness matrix {k} of size 5 × 5 is to be assembled. First this is made a null matrix and then one by element stiffness matrix is added to corresponding element in global matrix. After first element stiffness matrix is placed in global stiffness matrix, it looks as-

LM A MM l A − E MM l MM 0 MM 00 N

A1 le1 A1 le1 0 0 0

−

1

e1

1

e1

OP PP 0P P 0P 0P P 0QP

0 0 0 0 0 0 0 0 0 0 0

After second element stiffness is placed in global stiffness matrix, it looks as

LM A MM l A MM− l E MM 0 MM 0 MN 0 1

e1

1

e1

A1 le1 A1 A + 2 le1 le1 A − 2 le2 0 0 −

0 A2 le2 A2 le2 0 0

−

OP PP 0P PP 0P P 0P 0PQ

0 0 0 0 0 0

Finite Element Analysis—Bars and Trusses

167

Final stiffness matrix in global system

LM A MM l A MM− l M K= EM 0 MM MM 0 MM 0 N

A1 le1 A1 A + 2 le1 le2 A − 2 le 2 −

1

e1

1

e1

−

0

0

A2 le 2

0

A2 A + 3 le 2 le 3 A − 3 le 3

0 0

A3 le 3 A3 A + 4 le3 le4 A4 − le 4 −

0

OP PP 0 P PP 0 P P A P − P l P A P l PQ 0

...(11.11)

4

e4 4

e4

Thus we find the stiffness matrix is a symmetric matrix and its half the band width is equal to maximum difference in nodes of any element multiplied by degrees of freedom at each node plus 1, that is 2 in this problem Load Vector {F} Load vector {F}T = [F1 F2 F3 F4 F5] Let the element load vectors be

lFq = RSTFF UVW; lFq = RSTFF lFq = RSTFF UVW ; lFq = RSTFF 11

e1

21

e2

12

22

31

e3

41

e4

32

42

Then global load vector {F}is given by

R| F ||F + F lF q = S F + F ||F + F T| F 11

12

21

22

31

32

41

42

U| || V| || W

UV W UV W

...(11.12)

Thus we can assemble global / structure stiffness equation as

k 5 x5

lδ q = lFq 5 x1

5 x1

...(11.13)

Step 6: Boundary Conditions In this problem there is only one boundary condition i.e. δ 1 = 0 or it may have specified value. There are two methods of imposing the boundary conditions: (i) Elimination Approach (ii) Penalty Approach

168

Finite Element Analysis

(i) Elimination Approach In this method the known displacement is removed from the list of unknowns and the equations are reduced. If δ 1 is known displacement , then k11 k 21

LM MM MMkk MNk

LMk MMk MNkk

31 41 51

22 32 42 52

k12 k 22 k 32 k 42 k52

k13 k 23 k 33 k 43 k53

k14 k 24 k 34 k 44 k 54

k 23 k 33

k 24 k 34

k 25 k 35

k 43 k 53

k 44 k 54

k 45 k 55

OP R|δ U| R| F U| PP ||δ || ||F || PP S|δδ V| = S|FF V| PQ ||Tδ ||W ||T F ||W U| LMF − k δ OP |V = M F − k δ P || MMF − k δ PP W NF − k δ Q

k15 k 25 k 35 k 45 k 55

OP R|δ PP |Sδ PQ ||Tδδ

2 3 4 5

1

1

2

2

3

3

4

4

5

5

2

21 1

3

31 1

4

41 1

5

51 1

...(11.14)

Note elements corresponding to row and column of k11 are eliminated. If δ 1 is zero F

T

= F2 F3 F4 F5 .

This method is useful when hand calculations are made. If computers are used, computer coding becomes too lengthy.

(ii) Penalty Approach The round off errors involved in computations are advantageously used in imposing boundary conditions. It involve adding a very large number to the diagonal element and right hand side vector corresponding to the displacement on which boundary condition is specified. Thus to impose δ 1 = a1 , the modified equation will be

LMk + C MM k MM ⋅⋅ MN k 11

21

51

k12 k 22 ⋅ ⋅ k 52

k13 k 23 ⋅ ⋅ k 53

k14 k 24 ⋅ ⋅ k 54

k15 k 25 ⋅ ⋅ k 55

OP R|δ U| R|F + Ca U| PP ||δ || || F || PP S| ⋅⋅ V| = S| ⋅⋅ V| PQ ||Tδ ||W ||T F ||W 1 2

5

1

1

2

...(11.15)

5

Because of the modified equation 1, δ 1 results into δ 1 = a1

Other values are obtained as usual. Thus the required result is achieved without much changes in computer coding. The value of C selected should be much larger than k11, not less than 108 times k11 so that with round off errors δ 1 comes out to be a1. The author used C = 1 × 1020 to 1 × 1030 and got satisfactory results. Step 7: Solution of Simultaneous Equations After imposing the boundary conditions, the simultaneous equations 11.13 are to be solved. Any method of solving simultaneous equations can be employed. Gauss elimination is commonly employed. In many programs to save the memory in storing stiffness matrix k, half the band width of the matrix is stored and Choleski’s decomposition method employed. The solution gives the unknown nodal values. Step 8: Additional Calculations The additional calculations required may be to find strains and stresses at various points. These calculations are carried out element by element. From the list of global nodal values δ , for each element nodal values δ 1 and δ 2 of the element under consideration is picked up. Then displacement within the element.

Finite Element Analysis—Bars and Trusses

lq

u= N δ

e

= N1 N1

169

RSδ UV Tδ W 1

2

since ‘ ξ ’ coordinate of the point under consideration is known ‘u’ can be found. Then

lεq = ε = B lδ q lσ q = σ = D lεq = Eε = E B lδ q e

and

...(11.16)

e

...(11.17)

e

Calculation of Reactions Another important stress resultant required in the stress analysis is the reactions at support. This can be found from the equilibrium conditions of the support. For example, in this problem support is at node 1 and at this point displacement δ 1 is zero. Hence if R1 is the reaction of the support in direction 1,then k11δ 1 + k12δ 2 + k13δ 3 + k14δ 4 + k15δ 5 = F1 + R1

or

R1 = k11δ 1 + k12δ 2 + k13δ 3 + k14δ 4 + k15δ 5 − F1 In general R1 = k11δ 1 + k12δ 2 + ... + k1 N δ N − F1 Where N is total number of nodal displacements

...(11.18)

Example 11.1: The thin plate of uniform thickness 20 mm, is as shown in Fig. 11.5(a). In addition to the self weight, the plate is subjected to a point load of 400N at mid-depth. The Young's modulus E = 2 × 105 N/mm2 and unit weight ρ = 0.8 × 10−4 N/mm2. Analyse the plate after modeling it with two elements and find the stresses in each element. Determine the support reactions also. 125 mm

A1

250 mm

xb

P = 400 N xb

xb

x 1

1

2 (b)

400N 250 mm

xb

400N

A2

100 mm

x (a)

Fig. 11.5

2

3

170

Finite Element Analysis

Solution: A1 = 125 × 20 = 2500 mm2 A2 = 100 × 20 = 2000 mm2 The plate is modeled with two elements as shown in Fig. 11.5 (b)

=

k

e1

k

= e2

∴ k

LM N LM 1 N−1

OP LM OP LM OP Q N Q N Q −1O 2 × 10 × 2000 L 1 −1O L 8 −8OP = = 2 × 10 M P M P 1Q 250 N−1 1 Q N−8 8 Q −10 0 O L 10 −10 0 O P M P 10 + 8 −8P = M−10 18 −8P −8 −8 8 QP 8 QP NM 0

10 −10 2 × 105 × 2500 1 −1 EA1 1 −1 = = 2 × 105 −1 1 −10 10 250 le1 −1 1 EA2 le 2

LM 10 = M−10 NM

5

5

Consistant Loads: Due to body force only

lFq = RSTFF lFq = RSTFF lFq = RSTFF

e1

e

e2 11

e1

12 21

e2

22

UV = X Al RS1UV 2 T1W W UV = 0.8 × 10 × 2500 × 250 RS1UV = RS25UV 2 T1W T25W W UV = 0.8 × 10 × 2000 × 250 RS1UV = RS20UV 2 T1W T20W W e

b

−4

−4

Apart form these there is a 400N concentrated load at node 2. Hence,

LM 25 OP R| 25 U| lFq = M25 + 20 + 400P = S445V MN 20 PQ |T 20 |W

Hence the stiffness equation is,

2 × 10

5

LM 10 MM−10 N0

−10 0 18 −8 −8 8

OP R|δ PP S|δ Q Tδ

1 2 3

U| R| 25 U| V| = S|445V| W T 20 W

The boundary condition is δ 1 = 0 . Hence the reduced equation is,

LM18 −8OP RSδ UV = LM445 − 10 × 0OP = RS445UV N−8 8 Q Tδ W N 20 − 0 × 0 Q T 20 W L18 −8 OP Rδ U LM 445 OP 2 × 10 M MN 0 8 − 188 × 8PQ STδ VW = MN20 + 188 × 445PQ

2 × 105

2 3

5

2

3

Finite Element Analysis—Bars and Trusses

2 × 105

i.e.

LM18 N0

OP RS UV LM QT W N

OP Q

−8 δ2 445 = 4.444 δ 3 217.778

217.778

∴ δ3=

171

= 2.45 × 10 −4 mm

4.444 × 2 × 10 5

from equation 1, we have

2 × 105 18δ 2 − 8δ 3 = 445 2 × 105 18δ 2 − 8 × 2.45 × 10 −4 = 445

. × 10 −3 = 2.225 × 10 −3 18δ 2 − 196

δ 2 = 2.325 × 10 −4 mm from the relation

lq

σ =E B δ σ 1 = 2 × 105

e

we get,

RS T

UV W

0 1 −1 1 = 0.186 N/mm2 2.325 × 10 −4 250

σ 2 = 2 × 105

1 −1 1 250

R|2.325 × 10 S|2.45 × 10 T

−4

−4

U| V| = 0.01 N/mm W

2

Reaction at Support:

R1 = k11 k12 k13

R|δ S|δ Tδ

1 2 3

U| V| − F = 2 × 10 W 1

5

10 − 10

R| 0 0 S2.325 × 10 |T 2.45 × 10

−4

−4

U| V| − 25 W

∴ R1 = 490N

[Obviously in this simple problem reaction = total load]. Example 11.2: Assemble the element properties for a bar with uniformly varying area subject to self weight only as shown in Fig. 11.6(a). Neglect the possibility of buckling. Treat it as a single element.

ξ=

x − xc l 2 , dξ = dx or dx = e dξ 2 le / 2 le

luq = u = LMN1 −2 ξ εx =

1+ξ 2

du du dξ = ⋅ dx dξ dx

OP RSδ UV Q Tδ W 1

2

172

Finite Element Analysis b1

xb 



1

1

t - constant

2

2

x xc

x1  = –1

x2

x

 =0





=1

(b)

b2

2

x (a)

Fig. 11.6

LM N

= − B =

where

OP RS UV = 1 Q T W l

1 1 2 δ1 2 2 le δ 2

RSδ UV = B lδq Tδ W

−1 1

1

2

e

e

1 −1 1 le

Area which varies linearly also can be represented in natural coordinate system as

A= N

lAq = LMN1 −2 ξ

1+ξ 2

[Check: when ξ = − 1 , A = A1 ; when ξ = + 1 , A = A2 ] Element stiffness:

k

e

=

zzz z z RST UVW B

T

l

B

−1 l

=

−1

T

1

2

l

D B dV =

v

k =

z

OP RS A UV Q TA W

D B A

B

T

D B A dx

0

le dξ 2

LM N

1 −1 1 1−ξ 1+ξ −1 1 E 2 2 le 1 le

OP RS A UV l dξ Q TA W 2 1

2

e

Finite Element Analysis—Bars and Trusses

z LMN

OP 1 − ξ 1 + ξ RS A UV dξ −1 1 Q TA W E L 1 −1O L ξ ξ O RA U = ξ− ξ+ P S V M P M 2 2 PQ T A W 4l N−1 1 Q MN 1 −1 A E L 1 −1O = M P 2 2 RST A UVW = 4El LMN−1 1 OPQ 2 lA + A q 4l N−1 1 Q A + A L 1 −1O E A L 1 −1O =E MN−1 1 PQ = l MN−1 1 PQ 2l =

E 4le

1

−1

1

173

1

2

−1

2

2

e

1

1

2

−1

1

1

2

e

1

2

e

where A = Average area =

2

e

...(11.19)

e

A1 + A2 2

Consistant Load Only body force is acting and it is in the x-direction

l X q = X = ρ A dx = ρ A l2 dξ R|1 − ξ U| lFq = N l X q dV = S1 +2 ξ V ρ A l2 dξ || || T 2 W R|1 − ξ U| l LM1 − ξ 1 + ξ OP RS A UV dξ = ρl 2 =ρ S V 2 |1 + ξ | N 2 2 Q TA W 8 |T 2 |W e

b

e

zzz

z 1

T

b

e

−1

v

z 1

e

1

e

2

−1

z 1

Now,

b1 − ξ g

ze

dξ =

=

z

e1 − ξ j dξ = LMNξ − ξ33 OPQ

−1

=1− −1

ξ3 + 3

2

FG H

OP PQ

2

1

−1

1 1 − −1 − 1 − 3 3

FG H

IJ K

1 1 4 − −1 + = 3 3 3

1

2

j

8 3

1

2

2

2

−1

=1−1+

1

2

1 − 2ξ + ξ 2 dξ

L = Mξ − ξ MN

=

−1

LMb1 − ξ g 1 − ξ OP R A U dξ MN 1 − ξ b1 + ξ g PQ ST A VW

1

2

−1

Similarly

z 1

IJ K

174

Finite Element Analysis

zb 1

1+ξ

and

−1

g

2

dξ =

8 3

LM 4 OP R A U l q MM 38 PP |S |V MN 3 PQ |T A |W ρl L2 1O R A U = M PS V 6 N1 2 Q T A W ρl R2 A + A U = S V 6 TA + 2A W Hence the element equilibrium equation is E A L 1 −1O Rδ U ρl R 2 A + A U MN−1 1 PQ STδ VW = 6 ST A + 2 A VW l 8 ρl 3 ∴ F e= 8 4 3

1

2

1

2

1

2

1

1

1

2

e

...(11.20)

2

1

2

Answer

...(11.21)

2

Example 11.3: Determine the extension of the bar shown in Fig. 11.7 due to self weight and a concentrated load of 400N applied at its end. Given b1 = 150 mm b2 = 75 mm t = 20 mm

ρ = 0.8 × 10−4 N / mm3

E = 2 × 105 N/mm2 b1

xb 

t - 20 mm 600 mm



1

1

2

2

x  = –1

 =0 (b)

b2 400 N (a)

Fig. 11.7





=1

Finite Element Analysis—Bars and Trusses

175

Solution: A1 = 150 × 20 = 3000 mm2; ∴ Average area A =

∴ k =

A2 = 75 × 20 = 1500 mm2;

3000 + 1500 = 2250 mm 2 2

LM N

OP Q

LM N

OP Q

E A 1 −1 2 × 105 × 2250 1 −1 = −1 1 le −1 1 600

since it is the only element, above expression is global stiffness matrix also. Due to self weight

lFq = ρ6l RST2AA++2 AA UVW 1

e

e

1

=

2

2

RS T

UV RS UV W T W

60 0.8 × 10−4 × 600 2 × 3000 + 1500 = 3000 + 2 × 1500 48 6

Due to concentrated load

0 U lFq = RST400 VW R 60 U ∴ lF q = S V T448W 60 U lFq = lFq = RST448 VW e

e

since there is only one element,

e

The equation is

=

LM N

OP RSδ UV = RS 60 UV Q Tδ W T448W

2 × 105 × 2250 1 −1 −1 1 600

1

2

The boundary condition is δ 1 = 0 . Hence the equation reduces to

=

2 × 105 × 2250 = δ 2 = 448 600

δ 2 = 5.973 × 10−4 mm

or

∴ Extension of the bar = δ 2 − δ 1 = 5.973 × 10–4 mm

Temperature Stresses Change in the temperature in a member causes stresses, if its free expansion is prevented. Let ∆ T be the rise in temperature and α be the coefficient of thermal expansion. To find the stresses developed due to change in temperature we can use any one of the following two methods: (i) Direct Approach (ii) Variational Approach

176

Finite Element Analysis

(i) Direct Approach The free expansion of the element shown in Fig. 11.8(a) will be le α e ∆ T . This is as good 1

Ele e T

2

Ele e T

lc le e T

Fig. 11.8

as applying tensile forces E e Ae α e ∆T at the ends of the element as shown in Fig. 11.8(b). Hence the nodal load vector due to rise in temperature in the element is

lF q = E Aα ∆T RST−11UVW

...(11.23)

eT

This may be added to nodal load vector {F} due to the other loads. If u is the final displacement of any point in the element, displacement due to elastic strain = Total displacement – Free expansion = u − le α e ∆T

lq

∴ Elastic Strain

=ε= B δ

lq

= B δ Elastic Stress σ = Ee ε

e

e

−

le α e ∆T le

− α e ∆T

lq

= Ee B δ

e

...(11.24)

− E e α e ∆T

...(11.25)

[Note: Free expansion will not cause stresses in a member]

(ii) Variation Approach Strain due to change in temperature may be treated as initial strain as shown in Fig. 11.9 

E 1

 



0

Fig. 11.9

Finite Element Analysis—Bars and Trusses

ε 0 = α ∆T

Thus, Stress strain relation is

177

...(11.25)

b

σ = E ε − ε0

g

...(11.26)

If our interest is to find the stresses due to temperature rise only, then equation 9.16(b) for total potential energy reduces to

Π=

=

1 2

zzz

lq

v

e

∑

1 E e Ae le 4

∑

1 E e Ae le 4

e

∑

D ε dV =

∑

e

=

T

1 E e Ae le 4

e

=

lεq

zb g b ze l q z el q

1 2

zl q ε

T

− ε0

j e B lδ q

T

lq

e

T

−1 1

T e

B

e e

dξ

g

1

δ

Ee ε

ε − ε 0 dξ

−1

B δ

T

−1

1

ε − ε0

l q A2l

1

B δ

−1

e

e

j

− ε 0 dξ

lq

−2 δ

T

T

B

j

ε 0 + ε 02 dξ

dΠ = 0 , gives d δ

lq

Minimization of potential energy

1 = E e Ae le 4

ze

lq

1

2 B

T

B δ

−1

e

−2 B

T

j

ε 0 + 0 dξ = 0

The first term correspond to element stiffness matrix as found earlier. The second term corresponds to load vector due to temperature changes. Thus the load vector due to temperature effect is given by

lF q eT

substituting B = and

E A l 1 = e e e 2 le

z 1

B

T

ε 0 dξ

−1

1 −1 1 le ε 0 = α ∆T , we get

lF q = E

e

eT

= Ee

RS UV T W R−1U A α ∆T S V T1 W

lq

Ae le 1 −1 α ∆T ε 2 le 1

e

1 −1

...(11.27)

178

Finite Element Analysis This load vector may be added to load vector due to body forces, surface forces or the applied load and

l q are found,

the problem solved to get displacements due to all the causes. After nodal displacements δ member stress may be found by

b

g

σ = Ee ε − ε 0 = Ee ε − Ee ε 0

lq

= Ee B δ

e

− E e α e ∆T

...(11.28)

Example 11.4: Determine the nodal displacements at node 2, stresses in each material and support reactions in the bar shown in Fig. 11.10, due to applied force P = 400 × 103N and temperature rise of 30oC. Given: A2 = 1200 mm2 A1 = 2400 mm2 l1 = 300 mm l 2 = 400 mm 5 E2 = 2 × 105 N/mm2 E1 = 0.7 × 10 N/mm2 and

α 1 = 22 × 10−6 /C°

α 2 = 12 × 10 −6 /C°

2

1

P

3 Steel

Aluminium 300 mm

400 mm

(A1, l1, E1, 1)

(A2, l2, E2, 2)

Fig. 11.10

Solution:

k e=

LM N

OP Q

E1 A1 1 −1 −1 1 l1

LM N

1

LM N

OP Q

2 ← Global

OP Q

B

560 −560 1 0.7 × 105 × 2400 1 −1 = 103 = −560 560 2 −1 1 300 2

∴ k

B

L 600 −600OP 2 2 × 10 × 1200 L 1 −1O = 10 M = M P 400 N−1 1 Q N−600 600 Q 3 LM 560 −560 0 OP LM 560 −560 = 10 M−560 560 + 600 −600P = 10 M−560 1160 MN MN 0 −600 −600 600 PQ 5

k

3 ← Global

3

2

3

3

0 −600 600

OP PP Q

Finite Element Analysis—Bars and Trusses

Nodal Force Vector Due to temperature changes

lF q = 0.7 × 10 × 2400 × 22 × 10 5

eT 1

−6

R−1U R−110880UV × 30 S V = S T 1 W T 110880 W

–86400U lF q = 2 × 10 × 1200 × 12 × 10 × 30 RST−11UVW = RST86400 VW R| −110880 U| R|−110880U| ∴ lF q = S110880 − 86400V = S 24480 V |T 86400 |W |T 86400 |W −6

5

eT 2

B

Global

1 2

Global

2 3

eT

Due to applied forces

R| 0 U| lFq = S400000V |T 0 |W ∴ Load vector due to applied loads and temperature effect is

. U R| −110880 + 0 U| R|−11088 lFq = S24480 + 400000V = 10 S 424.48 |V |T 86400 + 0 |W |T 86.40 |W 3

The equilibrium equation is

10

3

LM 560 MM−560 N0

OP R|δ PP S|δ Q Tδ

−560 0 1160 −600 −600 600

1

. U U| R|−11088 V| = 10 S| 424.48 |V| W T 86.40 W 3

2 3

The boundary conditions are δ 1 = δ 3 = 0 ∴ The equation reduces to

1160 δ 2 = 424.48 i.e.

δ 2 = 0.36593

Answer

lq −E α

σ = Ee B δ

e

∴ σ 1 = 0.7 × 105 × = 39.18 N/mm2

e

e

∆T

RS T

UV W

0 1 −1 1 − 0.7 × 105 × 22 × 10−6 × 30 0.36593 300 Answer

179

180

Finite Element Analysis

σ 2 = 2 × 105 ×

1 −1 1 400

= 50.965 N/mm2

R1 = k11 k12 k13

= 103 560 −560

5

−6

× 30

Answer

U| V| − F W R| 0 U| 0 S0.36593V + 110080 = –94041N |T 0 |W

R|δ S|δ Tδ

1

2

1

3

R3 = 103 0 −600 600

[Check:

RS0.36593UV − 2 × 10 × 12 × 10 T 0 W

R| 0 U| S|0.36593V| − 86400 = –305959N T 0 W

Answer

Answer

∑ H = 0 → − 94040 + 400000 − 305959 = 0 ].

11.3 TWO DIMENSIONAL TRUSSES (PLANE TRUSSES) Fig. 11.11 shows a typical plane truss. The truss may be statically determinate or indeterminate. In the analysis all joints are assumed pin connected and all loads act at joints only. These assumptions result into no bending of any member. All members are subjected to only direct stresses–tensile or compressive. Now we are interested to see the finite element analysis procedure for such trusses. y P8

P7

P6

P5

P1

P2

P3

P4

P9

x

Fig. 11.11

A typical plane truss

Step 1: Field Variables and Elements Joint displacements are selected as basic field variables. Since there is no bending of the members, we have to ensure only displacement continuity (Co-continuity) and there is no need to worry about slope continuity (C1continuity). Hence we select two noded bar elements for the analysis of trusses. Since the members are subjected to only axial forces, the displacements are only in the axial directions of the members. Therefore the

Finite Element Analysis—Bars and Trusses

181

nodal variable vector for the typical bar element shown in Fig. 11.12 is

lδ ′q = RSTδδ ′′ UVW 1

...(11.29)

2

where δ 1′ and δ 2′ are in the axial directions of the element. But the axial direction is not same for all members. If we select x-y as global coordinate system, there are two displacement components at every node. Hence the nodal variable vector for a typical element is,

lδ q

T

= δ1 δ 2 δ 3 δ 4

...(11.30) x

2 

1

 

2



1

3



4

1

1 

y 1

 1

1

 



1

2

1

x

Fig. 11.12

Typical element and its nodal displacements

as shown in Fig. 11.12 From the Figure it is clear that

δ 1′ = δ 1 cosθ + δ 2 sin θ δ 2′ = δ 3 cosθ + δ 4 sin θ If l and m are the direction cosines,

l = cosθ , m = sin θ , ∴ δ 1′ = l δ 1 + m δ 2 δ 2′ = l δ 3 + m δ 4

i.e.

i.e.

l q RST UVW LMN

R|δ OP |Sδ Q |δ |Tδ

δ′ l m 0 0 δ′ = 1 = 0 0 l m δ ′2

lδ ′q = L lδ q

1 2 3 4

U| |V || W ...(11.31)

182

Finite Element Analysis

L =

where

LM l N0

OP Q

m 0 0 0 l m

and [L] is called transformation (or rotation) matrix. If the coordinates (x1, y1) and (x2, y2) of node 1 and 2 of the elements are known, we can find

l= le =

where

x2 − x1 y − y1 ,m = 2 le le

bx

2

− x1

g + by 2

2

− y1

g

2

...(11.32)

Step 2: Discritising A member may be taken as an element conveniently. Hence in the typical truss considered. There are (a) (b) (c) (d)

4 – top chord members 4 – bottom chord members 5 – vertical members and 8 – diagonal members

∴ Total elements selected are –21 There are 10 nodal points and they are numbered as shown in Fig. 11.13. 2

5

4

6

15

6

17

9 14 3

Fig. 11.13

8

12 18

2

5

10

21

11 16

1

8

19

10

1

7

13 20

3

7

4

9

Numbering nodes and members

The numbering is such that the band width is minimum. In this case maximum difference in the node numbers of an element is in diagonal members and is equal to 3. The degree of freedom of each node is 2, one in x-direction and another in y-direction. Hence the maximum band width = (3 + 1) × 2 = 8 Total degrees of freedom is = Total number of nodes × degree of freedom of each node = 10 × 2 = 20

lq

∴ δ

T

= δ 1 δ 2 δ 3 δ 4 ... δ 19 δ 20

Finite Element Analysis—Bars and Trusses

183

The nodal connectivity details is as given below: Element No.

Element Node 1

Element Node 2

1 2 3 : : 20 21

1 3 5 : : 7 8

3 5 7 : : 10 9

Global numbers

Step 3: Interpolation Functions Since bar element is used,

luq = N lδ ′q

LM x ′ − x ′ N l

N = N1 N 2 =

where

2

OP LM Q N

x ′ − x1′ = 1 − ξ ′ 1 + ξ ′ 2 2 l

OP Q

Step 4: Element Properties (a) Stiffness Matrix: In the analysis of bars and columns, we have seen the element stiffness matrix is

k

e

=

LM N

OP Q

EA 1 −1 l −1 1

when viewed in local coordinate system, the truss is also a one dimensional two noded bar element. Hence the element stiffness matrix of truss element in local coordinate system, k ′

LM OP N Q 1 ∴ U = lδ ′q k ′ lδ ′q 2 lδ ′q = L lδ q 1 ∴ U = c L lδ q h k ′ L lδ q 2 1 1 = lδ q L k ′ L lδ q = lδ q 2 2 k′ e =

e

is given by

EA 1 −1 le −1 1 T

e

T

e

T

k e= L

where

T

T

T

lq

ke δ

k′ L

...(11.33a)

and it may be called as element stiffness matrix in global coordinate system.

∴ k

e

LM l m =M MM 0 N0

OP PP PQ

0 0 E e Ae 1 −1 −1 1 l le m

LM N

OP LMl Q N0

OP Q

m 0 0 0 l m

184

Finite Element Analysis

E A = e e le

LM l MMm MN 00

OP PP LM l l N− l P mQ 0 0

LM l MM lm MM−−lml N 2

OP Q

m −l −m E A = e e −m l m le

2

lm m2 − lm −m 2

−l 2 −lm − lm −m2 l2 lm lm m 2

OP PP PP Q

...(11.33b)

[Note: The above expression is same as equation 3.6, which was obtained by direct approach] (b) Consistant Loads: Since all loads are acting at joints which are node points also, the load vector can be assembled straight way. Normally load vector {F} of size equal to degrees of freedom is generated as null vector and then load vector values are supplied. In the problem shown in Fig. 11.11, F6 = –P1, F10 = –P2, F14 = –P3, F3 = –P9, F8 = –P7, F12 = –P6, F16 = –P5 and F20 = –P4 Step 5: Global Properties It may be noted that for member i j.

δ 1 = δ 2i − 1 , δ 2 = δ 2i , δ 3 = δ 2 j −1 and δ 4 = δ 2 j − 1 Hence the position of elements of [k]e in global stiffness matrix for member i j are as shown below:

LM l MM lm MM−−lml N

2i − 1 2

k

e

=

E e Ae le

2

2i

lm m2 −lm − m2

2 j −1 2

2j

OP PP PP Q

−l −lm 2i − 1 2i −lm − m2 2 l lm 2 j − 1 2j lm m2

The global stiffness matrix of required size is initially developed as null matrix. Then for element number 1, element matrix is generated and they are added to the existing values of elements of global matrix at approximate places. Then next element stiffness matrix is generated and placed in appropriate positions in global matrix. The process is continued till all elements are handled. Assembling of load vector is already explained in step 4. Step 6: Boundary Conditions If hand calculations are made usually elimination approach is used and if computers are used penalty approach is used for imposing boundary conditions. The method is exactly same as explained in the analysis of columns and tension members. Step 7: Solution of Simultaneous Equations This step is also same as explained in the analysis of tension bars and columns. Step 8: Additional Calculations Analysts are interested in finding stresses and forces in the members of the truss. We know

σ = Ee ε But

ε=

Change in length δ 2′ − δ 1′′ = Original length le

Finite Element Analysis—Bars and Trusses

∴ σ = Ee = =

185

δ ′2 − δ 1′ le

RS UV T W

δ 1′ Ee −1 1 δ ′2 le

R|δ U| lδ ′q = L lδ q = LMN0l m0 0l m0 OPQ |Sδδ |V || || Tδ W Ll m 0 0 OP lδ q E ∴σ = −1 1 M l N0 0 l mQ 1

2

But

3 4

e

e

lq

Ee −l − m l m δ ...(11.34) le Using equation 11.34 stresses are calculated in all the elements / members. If forces are required the stresses may be multiplied by the cross sectional areas. Positive value indicates tension and the negative compression. The reactions at supports may be calculated on the lines explained in the analysis of tension bars and columns. To make the analysis procedure clear, a small problem is solved below with hand calculations. =

Example 11.4: For the three – bar truss shown in Fig. 11.14, determine the nodal displacements and the stress in each member. Find the support reactions also. Take modulus of elasticity as 200 GPa. Solution: Element numbers, node numbers and displacement numbers are as shown in Fig. 11.15. Taking node 1 as the origin, the coordinates of various nodes are 1 (0,0), 2 (800, 0), 3 (400, 400).

∴ lel = le2 = le3 =

b800 − 0g + b0 − 0g = 800 mm b400 − 800g + b400 − 0g = 400 2 mm b400 − 0g + b400 − 0g = 400 2 mm 2

2

2

2

2

2

2

150 kN C

m

00 20

20 0

2

m m

0

m

400 mm

1500 mm 2

B

A 400 mm

400 mm

Fig. 11.14

186

Finite Element Analysis 

y

6

3 

5

2

3 



2

1 

2

1

1

4



2 3

Fig. 11.15

Element No.

Node 1

Node 2

Element

1 2 3

1 2 1

2 3 3

Global

θ1 = 0

θ 2 = 135°

θ 3 = 45°

l1 = 1.0 l2 = – 0.707 m2 = 0.707 m1 = 0 E1 = E2 = E3 = 200GPa = 200 kN/mm2

LM l MM l m MM−−l lm N 2 1

k

e1

=

E1 A1 le1

l1m1 m12 − l1m1

1 1 2 1

−m12

1 1

LM 375 0 −375 0 0 0 =M MM−375 0 375 N0 0 0 LM 0.5 −0.5 200 × 2000 M−0.5 0.5 = 400 2 M−0.5 0.5 MN 0.5 −0.5 1

2

3

k

e2

3

4

4

OP PP PQ

l3 = 0.707 m3 = 0.707

− l12 −l1m1

−l1m1 − m12

l12 l1m1

l1m1 m12

OP LM 1 PP = 200 × 1500 M 0 PP 800 MM−01 Q N

Global Numbers

0 0 0 0

1 2 3 4 5

6

OP PP PQ

−0.5 0.5 0.5 −0.5 0.5 −0.5 −0.5 0.5

Global numbers

3 4 5 6

0 −1 0 0 0 1 0 0

OP PP PQ

0 0 0 0

Finite Element Analysis—Bars and Trusses

. LM 35355 −35355 . =M MM−35355 . 35355 . N 3

4

5

6

OP PP PQ

−35355 −35355 . . 35355 . −35355 35355 . 35355 . . −35355 35355 . 35355 . . −35355 −35355 . . 35355 .

Global numbers

3 4 5 6

LM 0.5 0.5 −0.5 −0.5OP 1 2 200 × 2000 M 0.5 0.5 −0.5 −0.5P k = M P 0 5 0 5 0 5 0 5 − − . . . . 400 2 MN−0.5 −0.5 0.5 0.5 PQ 56 −35355 −35355 . 35355 . . . O 1 LM 35355 P −35355 −35355 35355 . 35355 . . . P 2 =M MM−35355 −35355 . . 35355 . 35355 . P 5 P −35355 . . 35355 . 35355 . Q 6 N−35355 −375.0 + 0+ 0+ 0+ .0 + LM 375 −353.55 35355 353.55 0 0 . MM 0+ 0+ 0+ 0+ 0+ MM 35355 −353.55 353.55 0 0 . MM 0+ 0+ .0 + MM−3750 .0 + 00+ 375 −353.55 −353.55 353.55 k =M MM 0 + 0+ 0+ 0+ 0+ −35355 0 0 . 353.55 35355 . MM MM −353 −353.55 + 35355 0+ 0+ . + 353.55 + .55 −353.55 0 0 35355 . MM 0+ 353.55 + −353.55 + −353.55 + MM 0 + 0 35355 . N −353.55 −353.55 0 1

2

5

6

Global numbers

e3

1

2

5

6

Global numbers

OP PP 0+ P −353.55 P P 0 + PP 353.55 P P 0+ P −353.55 P PP −353.55 + P 353.55 P PP 353.55 + P 353.55 PQ 0+ −353.55

F1 = F2 = F3 = F4 = F5 = 0 and F6 = –150

LM 728.55 . MM 35355 −3750 . ∴ M MM 0 . MM−−35355 . N 35355

OP R|δ PP ||δ PP Sδδ PP ||δ PQ ||Tδ

−375.0 −35355 −35355 35355 . 0 . . 35355 . 0 0 . . −35355 −35355 0 728.55 −353.75 −35355 . 35355 . −35355 −35355 0 . 35355 . 35355 . . −35355 −35355 . . 35355 . 7071 . 0 −35355 −35355 . 35355 . . 0 707.1

1 2 3 4 5 6

U| R| 0 U| || || 0 || V| = S| 00 V| || || 0 || |W |T−150|W

187

188

Finite Element Analysis The boundary conditions are,

δ1 = δ 2 = δ 4 = 0 Hence the equation reduces to

LM 728.55 . MM−35355 35355 . N LM728.55 MM 0 N 0 LM728.55 ∴M 0 NM 0

OP R|δ U| R| 0 U| PP S|δ V| = S| 0 V| Q Tδ W T−150W 35355 . O Rδ U R 0 U P| | | | 17157 . P Sδ V = S 0 V 53553 . PQ | Tδ |W |T−150|W 35355 . O Rδ U R 0 U P| | | | 17157 . P Sδ V = S 0 V 480.56QP | Tδ |W |T−150|W

−35355 . 35355 . 70710 . 0 0 707.1 −35355 . 53553 . 17157 . −35355 . 556.98 0

3

5

6

1

2 3

3

5

6

from equation 3; 480.56δ 6 = − 150 i.e. δ 6 = − 0.312 mm

b

g

−0.312 = 0 . from equation 2; 556.98δ 5 + 17157 δ 5 = 0.1 mm

b g

b

g

from equation 1; 728.55δ 3 − 35355 . 01 . + 35355 . −0.312 = 0 ∴ δ3 =

353.55 × 0.1 + 353.55 × 0.312 = 0.2 mm 728.55

R|δ U| E |δ | = −l − m l m S V l ||δ || Tδ W R| 0 U| 200 |0| = −1 0 1 0 S V 800 ||0.2|| = –0.05 T0W 1

∴ σ1

2

1

1

1

1

1

1

3 4

∴ P1 = σ 1 A1 = − 0.05 × 1500 = − 75 kN

σ2 =

200 400 2

0.707 − 0.707 − 0.707

Answer

Answer

R| 0.2 U| | 0.0 |V 0.707 S || 01. || T−0.312W

Finite Element Analysis—Bars and Trusses

1

=

2 2

b

g

0.707 0.2 − 0 − 01 . − 0.312 = 0.053

∴ P2 = σ 2 A2 = − 0.053 × 2000 = 106 kN

σ3 =

200 400 2

=

1 2 2

−0.707 − 0.707 0.707

b

Answer

Answer

R| 0.2 U| | 0.0 |V 0.707 S || 0.1 || T−0.312W g

0.707 −0 − 0 + 01 . − 0.312 = 0.053 kN/mm2 Answer

∴ P3 = σ 3 A3 = 0.053 × 2000 = 106 kN

Answer

35355 −375.0 0 −35355 R1 + 0 = 72855 . . .

∴ R1 = 0

189

R| 0.0 U| || 0.0 || 0.2 −35355 . S || 0.0 V|| . ||W ||T−001.312

Answer

35355 0 0 −35355 R2 + 0 = 35355 . . .

R| 0.0 U| || 00..20 || −35355 . S || 0.0 V|| . ||W ||T−001.312

∴ R2 = 75 kN Answer

Similarly R3 = 75 kN Answer Example 11.5: If the support B of truss shown in Fig. 11.15 yields by 0.1 mm, determine the member forces due to applied load and yielding of the support. Solution: The equation of equilibrium remains as in the previous problem. Only boundary condition changes. In this case δ 4 = 0.1 mm . The other two boundary conditions are same as in the previous case i.e., δ 1 = δ 2 = 0 .

190

Finite Element Analysis

Imposing the boundary conditions δ 1 = δ 2 = 0 , the equation reduces to

OP R|δ PP |Sδ PQ ||Tδδ

. LM 72855 . MM−35355 35355 . MN−35355 .

−35355 . −35355 . 35355 . 35355 . 35355 . −35355 . 35355 . 707.1 0 −35355 . 0 707.1

3 4 5 6

U| R| 0 U| |V = |S 0 |V || || 0 || W T−150W

Introducing δ 4 = − 0.1 we get

i.e.,

. LM 72855 . MM−35355 . N 35355 . LM 72855 . MM−35355 . N 35355 LM728.55 MM 0 N 0 ∴

LM728.55 MM 0 N 0

. × 01 . U OP R|δ U| R| 0 − 35355 | . × 01 . V PP S|δ V| = S| 0 + 35355 707.1 Q Tδ W T−150 − 35355 . × 01 .| W 35355 . O Rδ U R −35355 . U| | | | P 0 P Sδ V = S +35355 . V| | | | P 707.1 Q Tδ W T−185.355W 353.55O Rδ U R 35.355 U P | | | . |V 17157 . P Sδ V = S −18198 . | 535.53PQ | W Tδ |W |T−168198

−35355 . 35355 . 7071 . 0 0

3

5

6

−35355 . 707.1 0 −35355 . 535.53 . 17157

3 5 6

3 5 6

OP R|δ PP S|δ Q Tδ

−35355 . 35355 . 53553 . 17157 . 0 480.56

δ6 = −

3 5 6

U| R| 35.355 U| . V| = S| −18198 V| − 111957 . W T W

111318 . = − 0.2318 mm 480.56

b

g

53553 . δ 5 + 17157 . × 0.2318 = − 18198 .

b

g

b

∴ δ 5 = 0.04028 mm

g

728.55δ 3 − 35355 . 0.04028 + 35355 . −0.2318 = 35.355

P1 =

=

E1 A1 −l1 le1

200 × 1500 800

δ 3 = 0.18056 mm

−m1 l1 m1

R|δ |Sδ ||δ Tδ

1 2 3 4

U| |V || W

R| 0 U| | 0 |V = 67.71 kN −1 0 1 0 S . ||018056 || − 01 . T W

Finite Element Analysis—Bars and Trusses

191

. R|018056 U| 01 . − 200 × 2000 | |V = 4.24 kN 0.707 − 0.707 − 0.707 0.707 S P = 0 04028 . 400 2 || | T−0.2318|W R| 0 U| 200 × 2000 | 0 |V = –95.76 k P = −0.707 − 0.707 0.707 0.707 S 400 2 ||0.04028|| T−0.2318W R| 67.71 U| lPq = S 4.24 V Answer |T−95.76|W 2

3

Temperature Stresses (i) Direct Approach: When viewed in local coordinate system, this element is one dimensional and for such element thermal forces due to ∆T rise in temperature are E e Ae α e ∆T as shown in Fig. 11.16 FeT    sin FeT 2

 cos FeT

y

FeT  sin   FeT

E e A e e

1

x

T

FeT  cos

Fig. 11.16

∴ FeT′ = E e Ae α e ∆T

RS−1UV T 1W

l q is load vector due to temperature effect in local coordinate system.

where FeT′

Ee A e e

T

192

Finite Element Analysis In global x-y system, it is seen easily that the load vector is,

lF ′ q = E eT

Ae α e

e

R|−1 × cosα U| |−1 × sin α |V = E ∆T S || 1 × cosα || T 1 × sin α W

Ae α e

e

R| −l U| |−m| ∆T S V || l || TmW

(ii) Variational Approach: If we represent the load vector due to temperature effect by

lF ′ q in local eT

coordinate system, then from equation 11.27, we get,

RS−1UV T 1W

∴ FeT′ = E e Ae α e ∆T

l q be the corresponding load vector in global system. Then potential energy due to this load is

Let FeT′

obviously same whether expressed in local or in global coordinates system. Thus,

lδ ′q lF ′ q = lδ q lF q T e

T e

eT

but from equation 11.31,

eT

lδ ′q = L lδ q lF ′ q = lδ q lF q lF q = L lF ′ q e

lq

∴ δ

T e

L

T

e

T e

eT

eT

T

i.e.,

eT

LM l m =M MM 0 N0

OP PP PQ

eT

R| |S || T

U| |V || W

0 −l −1 0 −m E e Ae α e ∆T = E e Ae α e ∆T l 1 l m m

RS UV T W

...(11.35)

The stress and force in the element can be found as was done in case of tension bars /columns i.e.,

b

=

and

g d B lδ q − l α ∆T i = El

σ = E ε − ε0 Ee le

P = Aσ =

e

e

e

l q − E α ∆T

−l − m l m δ

e

lq −E

E e Ae −l − m l m δ le

e

e

Ae α ∆T

e

e

...(11.36)

...(11.37)

Finite Element Analysis—Bars and Trusses

193

Stresses Due to Lack of Fit At the time of fabricating the statically indeterminate frames, if a member is found to be slightly shorter or longer (lacks in Fit), the member is forced in position. This is possible by introducing initial forces in the member. If the member is longer by δ l , the initial force applied on joint in outward direction is,

P0 = i.e. initial stress = E e or initial strain ε 0 =

E e Ae δl le

δl le

δl le

This problem of initial stress may be handled on the lines similar to stresses due to temperature effect. The term α ∆ T in case of temperature effect is to be replaced by

δl in this case. le

Example 11.5: Fig. 11.17 shows an indeterminate pin connected plane stress with cross sectional area of diagonal members equal to 2000 mm2 and all other members with cross sectional area of 1000 mm2. If Youngs modulus E = 200kN/mm2 (i) Assemble global stiffness matrix (ii) Determine load vector if temperature of member 1–3 increases by 25oC. Given α = 12 × 10−6 / °C (iii) Determine load vector if member 1–3 is longer by 0.2 mm. (iv) Introduce Boundary Conditions 3

4

3

6 4

2 5

1

2 1 4000 mm

Fig. 11.17

Solution: Joints and members are numbered as shown in Fig. 11.17.

3000 mm

194

Finite Element Analysis Member No.

End 1

End 2

L

m

le in mm

Ae in mm2

1 2 3 4 5 6

1 2 3 1 1 2

2 3 4 4 3 4

1.0 0.0 –1.0 0.0 0.8 –0.8

0.0 1.0 0.0 1.0 0.6 0.6

4000 3000 4000 3000 5000.0 5000.0

1000 1000 1000 1000 2000 2000

We know

LM l MM lm MM−−lml N 2

ke =

E e Ae le

2

LM 1 200 × 1000 M 0 = MM−1 4000 N0 1

k

1

3

k

2

4

LM 50 0 =M MM−50 N0

O 0PP 0P P 0Q

LM0 0 =M MM0 N0

6

5

B

0

0

6

0

50

0

0

0

0

66.67

0

0

0

LM 0.64 200 × 2000 M 0.48 = MM−0.64 5000 N−0.48

Global

1 2 3 4

B

3 4 5 6

1

2

LM 50 0 = M MM−50 N0 LM0 0 =M MM0 N0

3

←Global

0

1

2

5

0

0

66.67

0

0

0

−66.67 0

2 7 8 5

6

0 0 0 0

4

B

OP PP PQ

0.48 −0.64 −0.48 0.36 −0.48 −0.36 −0.48 0.64 0.48 −0.36 0.48 0.36

← Global

B

1 2 5 6

B

OP PP PQ

0 −50 0 0 0 50 0 0

8 8

4 ← Global

3

7

OP −66.67 P 0 P P 66.67 Q

−66.67 0

OP PP PQ

OP PP PP Q

← Global

0 0 0 −1 0 0 0 1 5

7

OP PP PQ

0 0 0 0 6

0 −50 0

2

4

8 ←Global

7

1

k5

3

0 −1 0 0 0 1 0 0

0 1 0 −1

1

k4

2

LM0 200 × 1000 M0 = MM0 3000 N0 5

k3

−l 2 −lm −lm − m2 l2 lm lm m2

lm m2 −lm − m2

1 2 3 4

OP −66.67 P 0 P P 66.67 Q 6

0

← Global

B

3 4 5 6

Finite Element Analysis—Bars and Trusses 1

LM 51.2 38.4 =M MM−512. N−38.4 3

k6

LM 51.2 −38.4 =M MM−512. N 38.4

2

5

38.4

−51.2

28.8

−38.4

−38.4

512 .

−28.8

38.4

4

7

−38.4

−512 .

28.8

38.4

38.4

51.2

−28.8 −38.4

Global Stiffness Matrix k 1

2

3

4

50.0 0 51.2

0 0 38.4

–50.0

0

0 0 38.4

0 66.67 28.87

0

–50

0

50.0 0 51.2

0

–51.2

0

0

OP −28.8 P 38.4 P P 28.8 Q

−38.4

B

1 2 5 6

8

←Global

38.4

3

O −28.8 PP −38.4 P P 28.8 Q

B

4 7 8

6

0 51.2

0 –38.4

–38.4

–28.8

0 0 –38.4

0

0

0 0 –38.4

0 66.67 28.80

0

0

0

–66.67

–28.8

8

0

0 1

–51.2

38.4

2

3

4 38.4

–28.8

0 0 38.4

–50

0

5

0 0 38.4

66.67 0 28.8

0

0

6

–50

0

50 0 51.2

0 0 –38.4

7

0 0 –38.4

0 66.67 28.80

8

0 –28.8

–66.67

0 50 51.2

38.4

–66.67

0

–66.67

0

38.4

7

0

–38.4

–51.2 0

← Global

5

0 –38.4

6

0

195

196

Finite Element Analysis

LM1012. MM 38.4 MM −050 k = MM−512. MM−38.4 MM 00 N

38.4 0 −50 95.47 0 0 0 1012 . −38.4 0 −38.4 95.47 0 0 −38.40 0 −28.8 −66.67 0 . 38.4 −513 −66.67 38.4 −28.8

. 0 0 −512 −38.4 0 −38.4 −28.8 −66.67 0 0 . 38.4 −512 0 −66.67 38.4 −28.8 1012 . 38.4 0 −50 38.4 95.47 0 0 0 1012 . −50.0 −38.4 0 0 −38.4 95.47

OP PP PP PP PP PP Q

Answer

(ii) Load vector for temperature forces:

F5T = E 5 A5 α 5

R| −l U| |−m| ∆T S V || l || TmW

= 200 × 2000 × 12 × 10−6

lF q T

T

R|−0.8U| R|−96.0U| |−0.6|V = |S−72.0|V × 25 S || 0.8 || || 96.0 || T 0.6 W T 72.0 W

B

Global

1 2 5 6

= −96.0 − 72.0 0 0 96.0 72.0

(ii) Load vector if the member 5(1–3) is longer by 0.2mm

lF q = E 5

5

R| −l U| R|−0.8U| R|−128.0U| δ l |−m| 0.2 |−0.6| | −96.0 | A = 200 × 2000 × S V S V=S V l | l | 500 | 0.8 | | 128.0 | |T m |W |T 0.6 |W |T 96.0 |W 5

e

B

Global

1 2 5 6

Equations after introducing boundary conditions: Since it is too big problem for hand calculations penalty method may be used and solution may be obtained using standard programs. Now ∆ 1 = ∆ 2 = ∆ 4 = 0

Finite Element Analysis—Bars and Trusses

197

For temperature forces the final equation is:

LM1012. + C MM 38.4 MM −500 .0 k = MM −512. MM −38.4 MM 00 N

38.4 0 −50 95.47 + C 0 0 . 0 1012 −38.4 . +C 0 −38.4 9547 0 0 −38.40 0 −28.8 −66.67 . 0 38.4 −513 38.4 −66.67 −28.8

OP R|δ PP ||δ PP ||δδ PP S|δ PP ||δ PP ||δδ QT

. 0 0 −512 −38.4 0 −38.4 −28.8 −66.67 . 0 0 38.4 −512 0 −66.67 38.4 −28.8 . 1012 38.4 0 −50 38.4 95.47 0 0 . 0 1012 −50.0 −38.4 0 0 −38.4 95.47

1 2 3 4 5 6 7 8

U| R| 0 U| || || 0 || || || 00 || V| = S|96.0V| || ||72.0|| || || 0 || W T0W

In case of stresses due to lack of fit only, only right hand side changes,

lF q

T

= 0 0 0 0 128.0 96.0 0 0

11.4 THREE DIMENSIONAL TRUSSES (SPACE TRUSSES) Typical two noded truss element is shown in Fig. 11.18(a). In this x ′ is the local coordinate system. While x, y, z are the global coordinate system. Fig. 11.18(b) shows the 2



x

2

2

x

2

lc 

y

1

y

1

1

1

x

x

z

z (a)

(b)

Fig. 11.18

displaced position of the element. In local system, the displacement vector =

RSδ ′ UV Tδ ′ W 1

2

In global system, the displacement vector is

lδ q

T

= δ1 δ2 δ 3 δ 4 δ5 δ 6

198

Finite Element Analysis If l, m and n are the direction cosines, we know,

x 2 − x1 y − y1 z − z1 and n = 2 ,m = 2 le le le

l=

bx

The length of the member le =

2

− x1

g + by 2

2

− y1

g + bz 2

2

− z1

g

2

. We know,

δ 1′ = lδ 1 + mδ 2 + nδ 3 δ ′2 = lδ 4 + mδ 5 + nδ 6

and In matrix form,

RSδ ′ UV = LMl Tδ ′ W N0 1

2

lδ ′q = L lδ q Ll m n L=M N0 0 0

where

Rδ U OP ||Sδ ||V Q|M | |Tδ |W 1

m n 0 0 0 0 0 l m n

2

6

...(11.38)

0 0 0 l m n

OP Q

is called transformation matrix. From equation 11.33(a), we know the relationship between global and local stiffness matrices is,

k

k

e

= L

T

L

e

LM l 0 OP MMm 0 PP E A n 0 Ll = M PM l M 0 l P N− l MM 0 mPP MN 0 n PQ LM l lm M lm m E A M ln mn = M l M −l −lm MM−lm −m NM− ln −mn e

e

lk ′q

LM l MMm n =M MM 0 MM 00 N

e

e

2

2

e

e

e

2

2

OP PP PP PP PQ

0 0 0 E e Ae 1 −1 −1 1 l le m n

LM N

OP LM l Q N0

m n −l −m −n −m −n l m n

m n 0 0 0 0 0 l m n

OP Q

OP Q

OP PP PP PP PQ

−l 2 −lm − ln ln mn −lm −m2 − mn − ln −mn −n 2 n2 − ln l 2 ln lm 2 − mn lm m mn 2 −n ln mn n2

...(11.39)

Finite Element Analysis—Bars and Trusses

199

Example 11.6: The tripod shown in Fig. 11.19 carries a vertically downward load of 10kN at joint 4. If Young's modulus of the material of tripod stand is 200kN/mm2 and the cross sectional area of each leg is 2000mm2, determine the forces developed in the legs of the tripod. 10KN 4

5m

2m

y

1

3 2m

x

3m

z

2m 2m

2

Fig. 11.19

Solution: Each member is taken as a bar element in space. The coordinates of various joints are 1(–3, 0, 0); 2(2, 0, 2), 3(2, 0, –2) and 4(0, 5, 0) Length of each element is given by

bx

le =

2

− x1

g + by 2

2

− y1

g + bz 2

2

− z1

g

2

and the direction cosines are given by

x 2 − x1 y − y1 z − z1 and n = 2 ,m = 2 le le le

l=

The details of the three elements are given below in the tabular form: Element No.

Node 1

Node 2

le in mm

l

m

n

1 2 3

1 2 3

4 4 4

5831 5745 5745

0.514 –0.348 –0.348

0.857 0.870 0.870

0 –0.348 –0.348

From equation 11.39, 1

k1

. LM 1811 . MM 30184 0 = M . MM −1811 . MM−30184 N 0

2

30184 . 50.350 0 −30184 . −50.350 0

3

10

11

0 −1811 −30184 . . 0 −30184 −50.350 . 0 0 0 0 1811 30184 . . 0 30184 50.350 . 0 0 0

12

OP PP PP PP PQ

0 0 0 0 0 0

← Global

B

1 2 3 10 11 12

200

Finite Element Analysis

4

k2

5

LM 8.425 . MM−21097 8.425 =M MM −8.425 . MM −21097 N 8.425

10

11

12

OP PP PP PP PQ

8.425 −21097 −8.425 21097 −8.425 . . 52.707 −21037 2109 −52.707 21097 . . . 8.425 −21097 −8.425 21097 −8.425 . . 21097 8.425 −21097 8.425 −8.425 . . 52.707 −21097 −52.707 21097 −21097 . . . 21097 8.425 −21097 8.425 −8.425 . .

8

7

k3

6

LM 8.425 . MM−21097 −8.425 =M MM −8.425 . MM 21097 N 8.245

10

9

11

12

OP PP PP PP PQ

8.425 −21097 −8.425 −8.425 21097 . . 52.707 21097 21097 −52.707 −21097 . . . 21097 8.425 8.425 −21097 −8.425 . . 21097 8.425 8.425 −21097 −8.425 . . 52.707 21097 −52.707 −21097 −21097 . . . 8.425 −21097 −8.425 −8.425 21097 . .

← Global

B

4 5 6 10 11 12

← Global

B

7 8 9 10 11 12

Global matrix of size 12 × 12 can be assembled. The load vector is F

T

= 0 0 0 0 0 0 0 0 0 0 − 10 0

Hence the equilibrium equation can be assembled. Noting that δ 1 = δ 2 = ... = δ 9 = 0 , the reduced matrix equation will be of size 3 × 3, the elements being from rows and columns of 10, 11, 12. The reduced equation of equilibrium is i.e.,

18110 . 30148 . 0 8.425 −21097 . 8.425 −8.425 8.425 −21097 . 30148 . 50.35 0 −21097 . 52.707 −21097 . −21097 . 52.707 21097 . 0 8.425 −8.425

i.e.

LM 34.96 MM−12.046 N 0

0 −21097 . 21097 .

R|δ || |Sδ || || Tδ

0 8.425 8.425

OP R|δ PP S|δ Q Tδ

−12.046 0 155764 . 0 0 16.950

10

11

12

10 11 12

U| R| 0 U| || || || |V = |S−10|V || || || || || || W T0 W

U| R| 0 U| V| = S|−10V| W T0W

Finite Element Analysis—Bars and Trusses

LM34.96 MM 0 N 0

i.e.

OP R|δ PP S|δ Q Tδ

−12.046 0 151589 . 0 0 16.950

10 11 12

201

U| R| 0 U| V| = S|−10V| W T0W

∴ ∆ 12 = 0 From equation 2, 151584 δ 12 = − 10 or δ 11 = − 0.065697 .

b

g

34.96 δ 10 − 12.046 −0.06597 = 0 ∴ δ 10 = − 0.02273

F1 =

200 × 2000 −0.514 − 0.857 0 0.514 0.857 5831

= –4.680 kN

R| 0 U| || 0 || 0 0 S V| 0 02273 − . || || ||T 0.06597 |W 0

F2 =

b

g

b

g

F3 =

200 × 2000 0 − 0 + 0 − 0.348 −0.02273 + 0.87 −0.06597 + 0 5745

b

g

b

g

200 × 2000 0 − 0 + 0 − 0.348 −0.02273 + 0.87 −0.06597 + 0 5745 = –3.445 kN

= – 3.445 kN

lF q

T

= −4.680 − 3.445 − 3.445

Answer

Note: As the degrees of freedom is high three dimensional problems are not suited for hand calculations. The computer program with penalty method of imposing boundary conditions is ideally suited for such problems.

QUESTIONS 1. Differentiate between a bar element and a truss element 2. Using variational approach derive element stiffness matrix of (a) bar element (b) plane truss element 3. Derive the expressions for nodal load vector in a two noded bar element due to (a) Body force (b) Surface load Use variational approach

202

Finite Element Analysis 4. Explain the elimination method and penalty method of imposing boundary conditions. Comment on the two methods. 5. In axially loaded cases, how do you find the support reactions after getting required displacements? Explain. 6. Using variational approach determine the expression for consistant load, due to rise in temperature ∆ T in an element. 7. Determine the nodal displacement, element stresses and support reactions of the axially loaded bar as shown in Fig. 11.20. Take E = 200 GPa and P = 30 kN 400 mm 2

250 mm2 1

2

150 mm

3

P

4

150 mm

300 mm

Fig. 11.20

[Answer: ∆ 2 = 0.062307mm ∆ 3 = 0.034615mm

σ 1 = 83.08 N/mm2 σ 2 = 36.92 N/mm2 σ 3 = 23.08 N/mm2 R1 = –20.77 kN R4 = –9.23 kN] 8. Obtain the forces in the plane truss shown in Fig. 11.21 and determine the support reactions also. Use finite element method. Take E = 200 GPa and A = 2000 mm2. 20 KN 15 KN 3

3000 mm

1

2 4000 mm

Fig. 11.21

[Answer: {F}T = [ 0 –37.25 18.75] in kN R1 = –11.25 kN R2 = –15 kN R4 = 31.25 kN]

Finite Element Analysis—Bars and Trusses

203

9. Determine the forces in the members of the truss shown in Fig. 11.22 Take E = 200GPa, A = 2000 mm2. [Answer: {F}T = [ 20 –26.25 4.961 0 –6.2752] in kN] 30 KN 3

4

3

4

5

2

2

1 1 4m

Fig. 11.22

3m

204

Finite Element Analysis

12 Finite Element Analysis—Plane Stress and Plane Strain Problems 12.1 INTRODUCTION In this chapter, first we see the general method for the analysis of plane stress/plane strain problems using CST elements. Then a small problem is taken up and hand calculations are made. The idea of such approach is to explain the steps involved is computer programming and make it clear by giving physical feel of the calculations involved. The lengthy calculations involved are pointed out when higher order elements are used.

12.2 GENERAL PROCEDURE WHEN CST ELEMENTS ARE USED The general procedure is explained referring to tension bar problem shown in Fig. 1.2 and the dam section analysis problem shown in Fig. 10.5. Step 1: Field Variable and Element: Since plane stress and plane strain problems are two dimensional problems, we need two dimensional elements. Any one from the family of triangular elements (CST/LST/QST) are ideally suited for these problems. Any one element from the family of two dimensional isoparametric elements (to be explained in next chapter) also may be used. In these elements there are two degree of freedom at each node i.e. the displacement in xdirection and displacement in y-direction. Hence total degree of freedom in (i) each element = 2 × No. of nodes per element (ii) structure = 2 × No. of nodes in entire structure. For a CST element shown in Fig. 12.1, the displacement vector may be taken as

lδ q = lδ = lu T e

1

δ2 δ3 δ4 δ5 δ6

1

u2

u3

v1 v 2

v3

q

q

…(12.1a)

e

or as

lδq = lu T

1

v1 u2

v2

u3

v3

q

…(12.1b)

In most of the programs the order shown in equation 12.1 (b) is selected. Hence in this chapter the

lq

displacement vector δ is used in the form of equation 12.1 (b). Then the x and y displacements of the node in global system are referred as 2n – 1th and 2nth displacements.

Finite Element Analysis—Plane Stress and Plane Strain Problems y

y 



 

1 

5

3 

5

2

4

6

3

3





6



 2

1

x

0

4

2

2

1 

205



3

1

x

0

(a)

(b)

Fig. 12.1

Nodal variables in CST elements

Step 2: Discritization Discritization of the structure should be made keeping in mind all the points listed in Chapter 10. For all nodes x and y coordinates are to be supplied/generated. Then nodal connectivity detail is to be supplied. For the dam analysis problem shown in Fig. 10.5, the nodal connectivity detail is of the form shown in Table 12.1. Table 12.1

Nodal connectivity Element No.

1

2

3

1 2 : 7 8 : 10 :

1 2

2 7

7 8

4 4

11 5

10 11

6

12

11

Local Numbers Global Numbers

Step 3: Shape/Interpolation Functions As shown in equation 5.15, the shape function terms are

N1 = where a1 = x2 y3 – x3 y2 b1 = y 2 – y 3 c1 = x3 – x2

and

a1 + b1 x + c1 y a + b2 x + c2 y a + b3 x + c3 y , N2 = 2 and N 3 = 3 2A 2A 2A a2 = x 3 y 1 – x 1 y 3 a3 = x 1 y 2 – x 2 y 1 b2 = y 3 – y 1 b3 = y 1 – y 2 c2 = x1 – x3 c3 = x2 – x1 1 x1 2 A = 1 x2 1 x3

y1 y2 y3

When we select nodal displacement vector as shown in Fig. 12.1 (b),

206

Finite Element Analysis

b g |RS|uv bbxx,, yygg|UV| = LMNN0 T W

u x, y =

N2

0 N1

1

0 N2

0

N3

0 N3

0

OP lδ q Q

e

…(12.3)

Step 4: Element Properties Since strain vector

R|ε lεq = Sε |Tε

x y 2

R| ∂u U| U| || ∂∂xv || V| = S| ∂y V| W | ∂u + ∂v | |T ∂y ∂x |W

and nodal displacement vector is in the form 12.3, the strain displacement vector ({ε } = [B]{δ }), [B] is given by

LM MM N

b1 0 b2 1 B = 0 c1 0 2A c1 b1 c2

0 c2 b2

b3

0 c3 b3

0 c3

OP PP Q

…(12.4)

According to variational principal (equation 9.26)

k e=

zzz b g B

T

D B dV

v

Since [B], [D] are constant matrices we get …(12.5) [k]e = [B]T [D][B]V where V = A t This is exactly same as equation 7.4 which was obtained by Turner by the direct approach. In equation 12.5, [D] is the elasticity matrix, which is as presented in Chapter 2 (equation 2.14 and 2.15). In case of isotropic materials, for plane stress case,

D = and for plane strain case,

E 1 − µ2

LM 1 MMµ MM 0 N

µ 1 0

LM1 − µ E M D = µ 1 + b gb1 − 2µ g MM µ0 MN

0 0 1− µ 2

µ 1− µ 0

Using equation 12.5, the element stiffness matrix can be found.

OP PP PP Q

…(12.6)

0 0 1 − 2µ 2

OP PP PP Q

…(12.7)

Finite Element Analysis—Plane Stress and Plane Strain Problems

207

Consistent Loads

z

Consistent loads can be derived using the equation

lFq = e

N

T

lX q dV + b

z

N

T

lT qds

(equation 9.26)

If there are nodal forces, they are to be added directly to the vector {F}e. Step 5: Global Properties Using nodal connectivity details the exact position of every term of stiffness matrix and nodal vector must be identified and placed in global stiffness matrix. Step 6: Boundary Conditions Since in most of the problems in plane stress and plane strain degree of freedom is quite high, the computers are to be used. These problems are not suitable for hand calculations. When computer programs are to be developed, imposition of boundary condition is conveniently done by penalty method. Step 7: Solution of Simultaneous Equations Gauss elimination method or Cholesky’s decompositions method may be used. In elasticity problems, there exits symmetry and banded nature of stiffness matrix. Hence the programs are developed to store only half the band width of stiffness matrix and solve simultaneous equations using Choleski’s decomposition method. Step 8: Additional Calculations After getting nodal displacements stresses and strains in each element is assembled using the relations

lεq = B lδ q lσ q = D B lδ q e

and

e

The calculated value of stress for an element is constant. It is assumed to represent the value at the centroid of the element. As a designer is normally interested in the principal stresses, for each element these values also may be calculated. Example 12.1: Find the nodal displacements and element stresses in the propped beam shown in Fig. 12.2. Idealize the beam into two CST elements as shown in the figure. Assume plane stress condition. Take µ = 0.25, E = 2 × 105 N/mm2, Thickness = 15mm. Y 3

4

50 KN

1 500 mm 2

2

1 750 mm

Fig. 12.2

x

208

Finite Element Analysis

Solution: For element (1), global nodal numbers are 1, 3, 4. Local numbers 1, 2, 3 selected are indicated in Fig. 12.3. Selecting node 4 as the origin of global coordinate system. v3

v4 4

u4

3

(3)

u3

(2)

1

v1 u1

1 (1)

Fig. 12.3

1(0, 0), 2(750, 500) and 3(0, 500)

1 0 0 2 A = 1 750 500 = 750 × 500 − 0 = 750 × 500 1 0 500

1 ∴ B = 750 × 750

1 = 750

LM 0 MM 0 N−15.

LM 0 MM 0 N−750

OP PP Q

0 500 0 −500 0 −750 0 0 0 750 0 0 500 750 −500

OP PP Q

0 1 0 −1 0 −15 . 0 0 0 15 . 0 0 1 15 . −1

LM1 − µ µ 0 OP E M P D = b1 + µgb1 − 2µ g MM µ0 1 −0 µ 1 −02µ PP MN 2 PQ L0.75 0.25 0 OP LM3 2 × 10 M = 0.25 0.75 0 P = 0.2 × 10 M1 . × 0.5 M 125 MN 0 0 0.25PQ MN0 5

5

OP PP Q

1 0 3 0 0 1

Finite Element Analysis—Plane Stress and Plane Strain Problems

OP LM 0 PP MM 0 Q N−15.

LM MM N

3 1 0 1 5 D B = × 0.2 × 10 1 3 0 750 0 0 1 0.2 × 105 = 750

LM 0 MM 0 N−15.

[k]1 = t A [B]T [D][B]

LM 0 MM 0 15 × 750 × 500 1 1 = × M 2 750 M 0 MM−1 MN 0 u v u LM 2.25 0 0 MM 0 6.75 −15. 0 3.0 −15 . = 1000