FINITE ELEMENT METHOD FOR TWO-DIMENSIONAL ELASTICITY
PROBLEM HANIMAH OTHMAN A dissertation submitted in partial fulfilment of the requirements for the award of the degree of Master of Science (Mathematics) Faculty of Science Universiti Teknologi Malaysia APRIL 2010

To Mum and Dad To Prof Shah To My Husband my infinite element in gratitude for encouraging me to persist in the completion of this thesis and for loving companionship and support through my whole life iii

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ACKNOWLEDGEMENT
Bismillah, I would like to thank God Almighty for giving me the will and strength to complete this research. I would express my sincere thanks to Prof. Dr. Shaharuddin bin Salleh for sharing his knowledge and experience with me. His constant patient and tolerant in guiding me during the process of completing this project is so much appreciated. A special thank dedicated to the librarians at Perpustakaan Sultanah Zanariah for their assistance in guiding and supplying me with reading sources and relevant literature studies. I am also grateful to all beloved persons in my life that inspire and motivate me a lot throughout all the hard works while carrying out this research. Last but not least, thanks to all my friends for sharing ideas and encouraging me to finish this thesis. For me, your support was of immense help to me. Thank you again.

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ABSTRACT
The purpose of this study is to develop a system which can be used to solve two-dimensional elasticity problem. A manual solution has been made to solve this two-dimensional elasticity problem based on the heat problem. In the first step, the main component is solved to get the stiffness matrix. Next, the initial strain and load vector, and the boundary vector are computed to complete the solution. The solution is computed and verified using MATLAB.

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ABSTRAK
Kajian ini bertujuan untuk membangunkan satu sistem yang boleh digunakan bagi menyelesaikan masalah kekenyalan dua dimensi. Satu penyelesaian secara manual untuk menyelesaikan masalah kekenyalan dua dimensi ini telah dibentuk dalam disertasi ini berdasarkan masalah haba. Dalam langkah pertama, komponen-komponen utama yang diselesaikan bagi mendapatkan matrik stiff. Seterusnya, nilai awal kekenyalan dan vektor limpahan serta vektor sempadan diselesaikan bagi menyempurnakan penyelesaian. Setiap penyelesaian dikira dan ditentusahkan menggunakan perisian MATLAB.

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TABLE OF CONTENTS   FINITE ELEMENT METHOD FOR TWO-DIMENSIONAL ELASTICITY PROBLEM i
ACKNOWLEDGEMENT iv
ABSTRACT v
ABSTRAK vi
TABLE OF CONTENTS LIST OF FIGURES LIST OF SYMBOLS vii x xi
GLOSSARY OF TERMS LIST OF APPENDICES 1
RESEARCH FRAMEWORK xii xiii 1
1.1
Introduction 1
2
1.2
1.3
1.4
1.5
Background of the problem Statement of the problem Objectives of the study Scope of the study 2   2   3   3   1.6
Outline of the research 4
LITERATURE REVIEW
2.1
5
Introduction 5     2.2
Elasticity 5   2.3
Types of Elasticity 2.4
Linear Elasticity 7 8     2.4.1
Initial strain 9

3
4
2.4.2
Plane stress 2.4.3
Plane strain 2.5
Two-Dimensional Elasticity 2.5.1
Weak Form of Equilibrium of Two-Dimensional Elasticity 15   2.5.2
Formulation of Two-Dimensional Elasticity 17   2.6
Conclusion 19
COMPUTATIONAL MODELS
3.1
20
Introduction 20     viii 10   13   14   3.2
Exact and Approximation Methods 3.2.1
Finite Element Method 3.2.2
Finite Difference Method 3.2.3
Boundary Element Method 20   21   22   22   3.3
Manual Calculation for One-Dimensional Problem 3.3.1
Example on One-Dimensional Elasticity 3.3.2
Strains 25   3.3.3
Stress-Strains Relationship 23   23 26     3.4
Manual Calculation for Torsion Problem 3.5
Some Method from the Journal 26 30     3.5.1
Introduction to Finite Element Techniques in Stress Analysis 30   3.5.2
What makes problems in theoretical stress analysis difficult? 31   3.5.3
3.5.4
The Finite Element Method Example from the journal 32   33   3.6
Conclusion 37
CASE STUDY IN TWO-DIMENSIONAL PROBLEM
4.1
38
Introduction 38

5
4.2
4.3
Case Study: Two-Dimensional Elasticity Problem Finite Element Formulation of the problem 4.4
Manual calculation 4.4.1
4.4.2
Element 1 Element 2 45   51   4.4.3
Element 3 55   4.5
Conclusion 64
CONCLUSION AND RECOMMENDATION
5.1
65
Conclusion 65     ix 38   39   45   5.2
Recommendation 65
REFERENCES 67
A
WATERFALL MODEL 69
B
C
D
GANTT CHART DETAILS IN CALCULATIONS MATLAB PROGRAMMING ALGORITHM 70 71 76

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LIST OF FIGURES FIGURE NO. TITLE
Figure 2.1 (a) Direct or normal stress Figure 2.1 (b) Tangential or shear stress Figure 2.1 (c) Change in volume with no change in shape Figure 3.1 Finite Element Discretisations of a machine component Figure 3.2 Finite Difference Discretisations of a machine component Figure 3.3 Boundary Element Discretisations of a machine component Figure 3.4 The illustration 2-noded bar element Figure 3.5 The iilustration for torsion problem Figure 3.6 The illustration for torsion problem for element 1 Figure 3.7 Two-fold symmetry of geometry Figure 3.8 Less mesh elements Figure 3.9 More mesh elements Figure 3.10 Mesh No 1 Figure 3.11 Mesh No 2 Figure 4.1 An illustration for case study Figure 4.2 An element from the problem Figure 4.3 Element 1 Figure 4.4 Element 2 Figure 4.5 Element 3
PAGE
6   6 6   21   22   23 23 26 27 34 35 35 36 36 38 39 45 51 55

LIST OF SYMBOLS
∆ - Matrix differential operator for shape function - Thermal expansion coefficient xi

GLOSSARY OF TERMS
FEM FDM BEM dof - - - - Finite Element Method Finite Difference Method Boundary Element Method Degree of freedom xii

xiii
LIST OF APPENDICES
APPENDIX TITLE A
B
C
D
PAGE
Waterfall Model 69 Gantt Chart 70 Details in calculations 71 Matlab programming algorithm 76

CHAPTER 1 1 RESEARCH FRAMEWORK 1.1 Introduction
Historically, Kevin and his friends such as Lame, Boussinesq, Cerruti and others are one of the first experimentally observed about the theory of elasticity. They mentioned that the rigorous to some boundary value problem of elasticity theory for the regions bounded by the surfaces prescribed by a single parameter have been obtained. However, Saint-Venant was reported his observation in the books of “On torsion of prisms” and “On bending of prisms” in which “the semi-inverse method” and Saint-Venant’s principle were suggested. It should be proclaimed as the origin of elasticity theory as an applied discipline. Thus the elasticity problems represent, not a flux and temperature, but the traction vector and the displacement vector need to be found. The problem of finding a flux and temperature at certain nodes in heat problem is such a way in giving ideas about elasticity problem. Some problems in elasticity field such as torsion need detail calculations and a system must be used if it comes to a big number. It contains some crucial parts and need to spend more times when doing manually. The system developed here is just for certain problem using certain techniques in finite element method (FEM). By using this finite element method, all the calculations contain such as the differentiation and integration parts in most calculation and solving a matrix system

2 using Gauss elimination system. These entire things are the main parts in solving this type of problem. Elasticity is just one of the applications in finite element method. The methods were in action when some forces are given to a solid that make changes in their shape. When this occurs, the shape for that solid are not permanently in their new shape but still have a chance to back normal but it depends on the force that given on it.
1.2 Background of the problem
The elasticity problem is common problem that many engineers faced it. They give the problem to model and solve it numerically. This problem will be solved using finite element method for grid rectangle shape function. It depends on the value given such as the area, coordinate of the model and the flux at beginning. For instance, it does not give directly, but need to find based on the model graphically. However, the process are the same as heat flow problem except only for certain part such as the value of load vector, boundary vector and followed by initial strain vector. Each such solution has different shape function based on the number of element. Recent comparisons with every element show the different flux and temperature for every nodal point with surprising accuracy. The stress and strain value needs to be found by doing some other calculation before go through to the real problem. This success suggests that more complicated of three-dimensional problem might provide accurate physical models of more complex elasticity phenomena or even though with other shape function.
1.3 Statement of the problem
Given a two-dimensional domain for an axially loaded elastic bar with a constant traction vector along the boundary by assuming a rectangular disk. The forces and the traction vector are given. In the same time, the mass density and the acceleration due to gravity are provided. The problem is to determine the finite

3 element formulation of elasticity for one element by getting the value of the boundary vector, the load vector and the initial strain vector. The result shows that one-fourth of this total force is distributed to each nodal point. The same results are obtained when considering the other elements and this implies that the uniform traction is equivalent to the nodal forces.
1.4 Objectives of the study
The main objectives of this research are to: 1. Study the characteristics and problems of two-dimensional elasticity from previous. 2. Derive finite element formula for two-dimensional elasticity for grid rectangle problem. 3. Determine the solvability of the formulated finite element method. 4. Provide a finite element technique of two-dimensional elasticity for grid rectangle problem using software MATLAB.
1.5 Scope of the study
This research will only consider the study on two-dimensional elasticity problem which is the extension from finite element method. A two-dimensional elasticity problem will be solved using finite element method and Gauss elimination is a technique to solve the matrix part. Apart from that, several requirements will only get from the problem itself or will be given directly. The solution of two dimensional elasticity problem will only focus on the problem which is using a grid rectangle as a method of solution in order to get their shape function.

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1.6 Outline of the research
This dissertation has five chapters including the preceding introductory part that have been discussed in the earlier sections. Chapter 2 will provide with some information on the literature studies related to this research. This chapter starts with the history of elasticity problem followed by some introductions on finite element method. Then, there will be a section that discusses on two-dimensional elasticity problem and finally some brief introduction on finite element method. In Chapter 3 will focus on the computational methods. This chapter starts with the process in order to complete the programming parts that have been used in some research by the other researchers. Some journal about using the elasticity or some parts in elasticity will be referred. All solutions are obtained by using MATLAB computer programming. Chapter 4 is mainly about the elasticity problems in details and some explanations about the problem. The manual solutions are used to compare the solutions using programming system. Most of the problems are nearly same as heat problem because it has a strong connection between both problems. There are conditions for elasticity to occur in heat problem such as for one-dimensional and two-dimensional problem. Manual solutions will be produced in this section. Finally, the last chapter will be the conclusion part that summarized the whole research. There are also some suggestions regarding related research area will be proposed in the last chapter, as well as the interesting applications of two dimensional elasticity problem in daily physical problems. Attachments have been provided in order to complete this research. Appendix A is a research methodology based on waterfall model. Appendix B is a Gantt chart that shows the timeline and milestones of the project that reflects with the Appendix A. While, the Appendix C shows the detail for calculations inside and the Appendix D shows the algorithm for the calculations.

CHAPTER 2 2 LITERATURE REVIEW Introduction 2.1
This chapter gives the historical background and development of the elasticity problem and related topic inside. It is divided into six main sections and contains the other small sections. Section 2.2 includes the historical background of elasticity. The literature regarding the elasticity and its types are given respectively in Section 2.3 while Section 2.4 gives an explanation on its components in terms of calculations and Section.2.5 will discusses the weak form of the formula which leads to the approximations. Finally, Section 2.6 gives the conclusion of this chapter.
2.2 Elasticity
Flexibility is one of the conditions or properties of being elastic. In Physics, the properties of returning to an initial form or states following deformation while the degree to which this property is exhibited. The theory of elasticity is one of the vital branches of mechanics of deformable bodies (Mohammed Ameen, 2005). A body is said to be elastic if it recovers its original shape when the forces causing deformations are removed.

6 Elasticity refers to a solid that responds to stress or the application of force on a given unit area which helps to illuminate the properties of objects for example steel bars, rubber bands and human bones. A solid does not respond to outside forces through flowing or compressing (Robert, W. L., 1973). The English physicist Robert Hooke (1635-1703), proposed the Hooke's law, that relates strain to stress. Hooke's law is depicted simply as "the strain is proportional to the stress," which is expressed as, F = ks, where F is the applied force and s, the resulting change in dimension, and k, a constant whose value is related to the nature and size of the object under stress. The value of k is directly proportional to the object's cross-sectional area or thickness. When the material becomes harder, it makes higher value of k. Plastic deformation is application involving elasticity. Plastic deformation describes a permanent change in shape or size as a result of stress. By contrast, elastic deformation is only a temporary change in dimension. The other one is a rubber band. It can be deformed to a length many times its original size, but after release, it returns to its original shape. Below are the examples of elasticity. (a) (b) (c)
A’ F B’ C F C’ D D’ A A B A’ B F A B’ D C B D’ C’ F
Figure 2-1: (a) Direct or normal stress (b) Tangential or shear stress (c) Change in volume with no change in shape (All deformations exaggerated)

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2.3 Types of Elasticity
There are three types of elasticity which are linear elasticity or one dimensional elasticity, two-dimensional elasticity and three-dimensional elasticity. In general, for one-dimensional problem, it contains the matrix D that must be constant and symmetric, initial strains, plane stress and plane strain. However, for two-dimensional problem, it looks as well as three-dimensional problem based on their components. But, the different is only on their boundary of integrands. It is independent from z-coordinate for two-dimensional problem while it depends on z-coordinate for three-dimensional problem. So, the boundary of the integrands only depends on their area. At the same time, the three-dimensional problem depends on their volume for its integrands. For further knowledge, there are some explanations about one-dimensional and two-dimensional problem below. Below are the finite element equations for one element for three-dimensional problem. dV. (2.1) dS dS. (2.2) dV. (2.3) dV. (2.4) Where K is the stiffness matrix, is the boundary vector, is the load vector and is the initial strain vector, V is volume, h and t is the length for certain boundary, S is the boundary, b is the body force and is the initial strain.

8
2.4 Linear Elasticity
The concept of linear elasticity was first proposed by Robert Hooke in 1678 when he wrote, “Ut tension sic vis,“ or “the power of any springy body is in the same proportion as the extension (Robert, W. L., 1973). The elasticity field equations related to the kinematics of small deformation theory and the equilibrium of the associated internal stress field. Based on these physical concepts, three strain displacement relations, six compatibility equations and three equilibrium equations were developed for the general three-dimensional case. Our interest here is limited to a two-dimensional case only. Our discussion on elasticity state with one-dimensional case, followed by two and three-dimensional cases. One-dimensional elasticity is called linear elasticity, and it is one example of constitutive relation between strains and stresses. Other examples are plasticity, viscoelasticity, viscoplasticity and creep. Actually, linear elasticity is the simplest constitutive theory. In one dimension, linear elasticity is expressed by Hooke’s law in 1676; . (2.5) Where E is Young’s modulus. In order to maintains the concept is given by . (2.6) Where ; … … … … … … ; . (2.7)

9 In general, D is positive definite since singular, det 0 which for : 0 for all 0. D is non where C is the compliance or flexibility matrix. From here, it appears that zero stresses imply zero strains.
2.4.1 Initial strain
From the general case, stress is given by and the strain by . There exists a concept of linearity between both of them given by to become where C = where D is symmetric. . While, it can be inverted The concept of linearity leads to the conclusion that zero stress implies zero strains. At the same time, there exists a situation of non-zero strains for zero stresses. This happens when we introduce the , the initial strain, are the strains for zero stress. As mentioned before, where C = , . (2.8) For example, some materials are free to expand when there is a difference in temperature. This means, 0 and the thermal strain is given by ∆ 1 1 1 0 0 0 . (2.9) In the equation, is the thermal expansion coefficient 1/ and ∆ is the change in temperature. The D matrix will change if another coordinate system is chosen. If every plane is a symmetry plane, if the D matrix takes the same form irrespective of the coordinate system, and the coefficients are Young’s modulus and Poisson’s ratio,

D = 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 2 1 0 0 0 0 0 2 10 1 0 0 0 0 0 2 . (2.10) From (2.10), (2.9) and (2.8), ∆ 1 1 1 0 0 0 . (2.11)
2.4.2 Plane stress
Plane stress is defined as stress involving non-zero , . Here, the stress tensor S becomes 0 0 0 0 0 . (2.12) when the boundary conditions are . (2.13) . (2.14) 0. (2.15) It is important that plane stress conditions require that the component 0. Then,

11 ; . (2.16) Finally, the equilibrium conditions become 0. (2.17) 0. (2.18) 0. (2.19) Defining the matrices 0 0 ; ; . (2.20) And it can be written as 0. (2.21) After some observation, for plane stress requires that and equal zero. Both of them, the traction vector t along the boundary of the body and the body force b must be lie in the xy-plane. We assumed that nothing happened or depends on z coordinate from the assumption above. For a plane, the only non-zero stresses are , . From this conditions, and assuming isotropy and that , will obtain 1 0 1 0 2 1 0 0 ∆ 1 1 0 . (2.22)

12 It follows by ∆ . (2.23) = = 0. (2.24) Here, 1 0 1 0 2 1 0 0 ; Equation (2.22) can be written into the form of (2.8), where 1 1 0 0 1 0 0 ; = ∆ 1 1 0 . (2.25) . This equation leads to . (2.26) And ∆ 1 1 0 . (2.27) This situation is the same as plane strain. This suggests in-plane stresses will direct determine the in-plane strains, and vice versa. This also means, when the stress components are known. The remaining strain components can be determined.

13
2.4.3 Plane strain
The plane strain exists from non-zero strains in , and . As before, nothing depends on z-coordinates. So, plane strain constitutes the two-dimensional relationships, as follows: , ; , ; 0. (2.28) 0 ; ; 0 . (2.29) By using this conditions and assuming isotropy will obtain 1 0 1 0 1 0 0 2 Also, it follows by ∆ 1 1 0 . (2.30) ∆ . (2.31) = = 0. (2.32) Then, it is possible to write (2.31) as ∆ . (2.33) This situation is the same as plane stress. We consider that in-plane strains will directly determine the in-plane stresses, and vice versa. It means, when the strains components are known, the remaining stress components can also be determined.

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2.5 Two-Dimensional Elasticity
The two-dimensional formulation elasticity follows from the three dimensional elasticity. We introduce some assumption where the constitutive relation and the displacement vectors are already known. However, for two-dimensional problems, the thickness t may vary, but in order to approximate the requirements for the existence of plane stress or plane strain, such a thickness variation must be small (Ottosen & Petersson, 1992). The finite element formulation for one element with evidence notation and the result is for stiffness matrix, the boundary vector, the load vector and the initial strain vector. t dA. (2.34) t d t d . (2.35) ε t dA. (2.36) t dA. (2.37) Where . (2.38) And finally = . (2.39)

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2.5.1 Weak Form of Equilibrium of Two-Dimensional Elasticity
In order to derive the weak form of two-dimensional elasticity problem, need to consider the weak form of three-dimensional elasticity problem, and this is given by d . (2.40) The weight vector v is arbitrary. It is allowable to choose its components according to , ; , ; 0. (2.41) This implies that . (2.42) From the following redefinition of matrices, 0 0 ; ; . (2.43) It follows that (2.42) is fulfilled. We note that this definition of the operator occurs also for a plane stress as well as for plane strain. We also introduce the definitions ; . (2.44) With the definitions (2.43) and (2.44), the weak form (2.40) becomes d d d . (2.45)

16 For two-dimensional problem, only consider the coordinates in x and y directions. From here, it will come out with a region in xy-plane denoted by A and the boundary for this region denoted by . The thickness t is assumed to be negligible two-dimensional problems. It follows that, d dz d d . (2.46) d dz d d . (2.47) Now, consider the surface integral in (2.45). By using this equation, assume that the rigid body have two different surface, which are the upper surface , and the lower surface . Then we get that d d d d d . (2.48) The given t for traction vector are include the components and along . Referring to (2.44), we have 0, and gives and . (2.49) For further steps in plane stress, the d 0 and reduces (2.48) becomes d . (2.50) In other hand, for plane strain, the , and , . It may be different from zero. Along the upper surface surface , having , we have = 1 and along the lower = -1. This condition implies that = and v is independent of z-coordinate where (2.48) will reduce to (2.50).

17 Therefore, by inserting (2.46), (2.47) and (2.50) in (2.45) yields d d d . (2.51) However, there are different between t for thickness and t for traction vector. The thickness may vary but in order for existence of plane stress or plane strain, it must be small. If the thickness t is constant, then we will get d d d . (2.52)
2.5.2 Formulation of Two-Dimensional Elasticity
The formulation of two-dimensional elasticity is closely same as three dimensional elasticity problem. Need to make some assumption before continuing the procedure. Now, the displacement vector is already known and given by . (2.53) And need to approximate by arbitrary weight vector v is the equation before this, will get . By using Galerkin method, choose the . Since v is arbitrary, then it follows by c. From where B = . Now, by inserting the weak form in here gives d d d 0. (2.54) And as the c matrix is arbitrary, will get d d d . (2.55)

18 By using the kinematic relation, the strains are derived from the displacements in ( and ). It doesn’t matter whether plane stress or plane strain is assumed. . (2.56) From , and , will get . (2.57) By using equation (2.8), then it becomes Combining (2.58) and (2.55), the results is .
. (2.58) d d d d . (2.59) Along the xy-plane, there exist the unit vector n normal to the boundary and the traction vector t along the boundary. This traction vector can be written as t = Sn . (2.60) The traction vector t is the natural boundary condition along the boundary and the displacement vector u is the essential boundary condition along the boundary . The total boundary consist of and . The boundary condition are given by t = Sn = h on . (2.61) u = g on . (2.62)

19 By using the boundary condition (2.61) and (2.62) above, we now reformulate the equation in (2.59), d d d d d . (2.63) For easy to write and understand, the formulation in the compact form for each element are same in equation (2.34), (2.35), (2.36), (2.37), (2.38) and (2.39).
2.6 Conclusion
The reviewing of the literature regarding the elasticity of two-dimensional problem including its formulation of the formulae and their weak form are included in this chapter. The explanation of the historical background and the development for each part in elasticity formulae which comes directly based on three-dimensional problem. The discussion will consider only for two-dimensional problem and not interested in three-dimensional case.

CHAPTER 3 3 COMPUTATIONAL MODELS 3.1 Introduction
Most of the engineering problems require finding the exact closed form solutions which may be rarely available. Some of the difficulties in finding exact solutions stem from the presence of irregularity shape boundaries, odd-shape holes and stiffeners, complex boundary conditions, nonlinearities due to the material constitutive relations and strain displacement relation ceasing to obey simple linear relationship. The need to reduce the complexity of the problems to a convenient degree is new sans. Exact solutions to the problems may be far from the one sought. The other alternative is to make use of approximation methods available and solve the problem without introducing too many assumptions.
3.2 Exact and Approximation Methods
Exact methods refer to the analytical method that makes use of the mathematical theories of differential and integral equations for solving problems. Exact methods can be used to solve only trivial problems in a familiar fact. On the

21 other hand, approximations using numerical methods are suitable for obtaining highly accurate solutions of practical engineering problems.
3.2.1 Finite Element Method
Finite element method is a numerical approach for solving boundary value problems involving differential equations to produce approximate solutions. The concerning of finite element method formulation like heat conduction, diffusion, torsion of elastic shafts, and the elastic problems are such diverse problems. Finite element method is the most popular numerical method available to engineers. Figure 3.1 Finite Element Discretisations of a machine component Certain characteristic feature of FEM that the main body, is divided into smaller parts or sub domains called elements. We assumed that the variables are known at certain points over the elements. The solution is then approximated over each element and is quantified in terms of values at certain locations within the element called nodal points. Every nodal point has its own boundary element. Then, each element need to establish of stiffness relations. Material properties and equilibrium conditions for each element are used in this establishment. In finite element method, the element size, shape, and approximating scheme can be set to suit the problem. The method can accurately simulate solutions to problems of complex geometry and loading (Sadd, M. H., 2005). For two dimensional problems, the displacements, strains, stresses, tractions and body forces

22 do not depend on the z-coordinate. The weak form of two-dimensional case is according to equation (2.41), (2.42), (2.43), (2.44) and (2.45) as mention before.
3.2.2 Finite Difference Method
The finite difference method is one such method has been use widely in continuum mechanics such as fluid flow and heat conduction problem. The differential equations of the problem are approximated by their finite difference equivalents in a light manner, leading to a system of algebraic equations where the solution represents an approximate solution of the problem. Finite difference method has been used to solve problems of moderate complexity. However, existence of abnormality boundary geometry or unusual boundary conditions is difficult to handle by this method. Figure 3.2 Finite Difference Discretisations of a machine component
3.2.3 Boundary Element Method
The boundary element method or sometimes called as domain methods, is a method based on the boundary integral equation formulation of the problem. In contrast to the domain methods, need only the boundary discretisation. As a result, improved numerical accuracy together with a lesser equipment on computational resources is possible using this method.

23 Figure 3.3 Boundary Element Discretisations of a machine component
3.3 Manual Calculation for One-Dimensional Problem 3.3.1 Example on One-Dimensional Elasticity
Below are given an example on one-dimensional elasticity problem which is consisting of stress and strain calculation and also its relationship between both of them. For a two-nodes bar element; 1 2 x Figure 3.4 The illustration for two-noded bar element From here, we assume for coordinate (x), deformation (u) and force (F) at each node becoming as x , u and F at nodal point one and x , u and F at nodal point two. The degree of freedom (dof) per node is equal to one are given. From this example, knowing that the nodes per element are equal to two and the degree of freedom per element is equal to two since two multiply by one is equal to two. Then, x . (3.1) By using the given condition, the two similar equations which are

24 . (3.2) . (3.3) From the above equation, it gives solutions as . (3.4) Hence . (3.5) Equation (3.5) defines the deformation at any point ‘x’ within the element in terms of and i.e nodal values. In matrix form 1 . (3.6) Now, having the two nodal points will produce the matrix in form of 1 1 . (3.7) In the compact form, A . (3.8) After some operations to get this form, Then it returns to this form, . (3.9)

25 . (3.10)
3.3.2 Strains
Strains are defined as a change in distance between two neighbouring points or change of angle between two intersecting lines as a result of deformation. Since it is a one-dimensional problem, therefore only one component of strain exists. For example are as follow. . (3.11) Then, it similar as . (3.12) Or in matrix form . (3.13) . (3.14) . (3.15) Where B is called as a strain displacement matrix.

26
3.3.3 Stress-Strains Relationship
The stresses occurring in the element are now related to strains. In this step, the elastic properties of the element have to be taken into consideration. From equation (2.6) since where D is termed the elasticity matrix and contains the elastic properties, i.e. quantities such as modulus of elasticity and Poisson’s ratio. . (3.16) . (3.17) C . (3.18) Where C is called stress displacement matrix.
3.4
10 0
Manual Calculation for Torsion Problem
y 0 8 x 1 10 2 2 1 Melosh element Figure 3.5 The illustration for torsion problem 7 2 6 2 Serendipity element 3 5 4

27 From this problem, the value of , and need to be found. The value of x 10 ⁄ and the value of 10 ⁄ are given. The problem needs a solution from each element. It means that, every element will give either different or same product based on their behaviour. Finding the equations for its nodal point and also its Jacobian value are as given below. When dealing with more than one element and have a same fashion, the notation must be different rather than the usual one. It’s because, the identifying for the value of each element since it’s have a same value are ne. = 0 ( 0, 10 2) (-1,1) (10 (-1,1) 4 , 10 4) = 0 = 0 (-1,-1) (0,0)
1
Melosh element (1,-1) ( 10 4 ,0) = 0 Figure 3.6 The illustration for torsion problem for element 1 Here are the equations for each nodal point, 1 4 1 1 1 1 . 1 . 1 . 1 1 . (3.19)

28 Now, the and value are 1 Then their inverse, 1 1 1 . (3.20) 1 1 1 1 . (3.21) And then, its derivative for x and y to and are as follows, 10 8 . 0. 0. 10 4 (3.22) And then, its derivative for N to and , 4 1 . 4 1 . 4 1 . 4 1 . ∂ 4 1 . 4 1 4 1 4 1 . (3.23) And now the Jacobian matrix, 0 0 (3.24)

29 32 10 0 0 (3.25) After that, the B matrix is given by 100 2 1 2 2 1 2 2 1 2 2 2 1 (3.26) Insert all the requirements in the form of stiffness matrix, load vector and boundary vector to solve the problem. 1 | | (3.27) 2 | | (3.28) 5 4 10 10 8 1 4 1 4 1 1 0 0 10 2 10 2 1 1 (3.29) As a conclusion for this problem, continuing the same step for the second element and find the total value for the stiffness matrix and the load vector are needed. But, it differs for the boundary vector because we only get the value for the whole boundary and did not depend on the element.

30
3.5 Some Method from the Journal 3.5.1 Introduction to Finite Element Techniques in Stress Analysis
In FEM, the techniques of theoretical and numerical stress analysis have been developed with advanced stage but it is still true that cannot solve all the real engineering system. It means that the engineers still need to carry out some idealization from the real situation if the system is to be made a reasonable time and at a reasonable cost when it’s applied in design practice. In real problem, there might be having a various levels of precision. The purpose is to get the precision rather than using the sophisticated modelling technique while the complexity in formulation and solution is not necessary. The responsibility of the designer is to implement decision in stress analysis even though the method such as the FEM is already available. The formulation and solution of problems in stress analysis include three main points such as conditions of equilibrium, conditions of compatibility or geometric fit and some stress-strain characteristics for the elements of the system. The system may carry on along one of two paths. First the three main points may be manipulated in an explicit approach in the usual vector. Second, the energy principles of mechanics may be used in which the main points are handled in an implicit method. The value of modelling for stress analysis depends to a significant amount on the experience and opinion of the practitioner. The engineers need to be confidence in formulating the theories and bases for numerical solution. In order to interpret the results depends very much on a thorough understanding of the fundamental assumptions and implications of the principles of solid mechanics. In this paper, the engineers are concerned with introducing the finite element technique to obtain an accurate answer. On the other hand, approximate solutions to difficult stressing problems are also considered. The objective is to understand the nature of the approximations involved in the solution. So that, user can creates the problem in proper ways, understand the meaning of the data and make a decision based on the value of data obtained.

31
3.5.2 What makes problems in theoretical stress analysis difficult?
In stress analysis, it concern only with determining deformations (displacements) and distributions of stresses (intensity of internal forces) in deformable solids. The properties of stiffness and mass are distributed in a permanent approach. Means that, a body can deform in an infinite number of ways (depending on the load) and it has an infinite number of degrees of freedom. The mathematical description of the mechanical behaviour of such bodies is called as continuum mechanics. Some excellent details of molecular structure of the material were ignored. It means that, the stresses and strains are differing in a continuous method because they are continuous ‘function of position’ and they are described by differential equations. The engineers are still trying to solve those equations to satisfy given boundary conditions which are too difficult in most real situations. Possibilities open to us are: (i) Try to develop an approximate theory appropriate to a particular problem, in term of mechanics of materials. For design purposes, we idealize the real situation to be able in this theory. For many purposes, this is sufficient. (ii) Approach the problem by the supposed inverse or semi-inverse methods. In the previous method, some solution to the differential equations is accepted and the related problem situation is examined to see if it has any practical significance. Since the equations are linear, a superposition of several solutions could be examined. In the semi-inversed method, some assumptions were made about the distribution of displacements or stresses or both. However, the boundary conditions such that the formulae for stresses or displacements cannot be obtained and the solutions can only be obtained numerically. (iii) As a beginning of simplifying assumptions, an approximate theory for some specified boundary value problem may be obtained. As advantageous applications to the theory of plates and shells are the uses of

32 variation principles. This approach is at a higher level of idealization than in (i) above but the equations may be such that direction solution to give in formulae for stresses and displacements may not be possible. (iv) The formulated boundary value problem was solved approximately. Since the finite difference, collocation and other techniques are available, but concerning with the finite element method only.
3.5.3 The Finite Element Method
It is not an easy task to formulate a displacement model for the entire region of a solid. Ones need to divide the region into a number of finite sized sub-regions or elements. The expressions for strain energy stored and work done by external forces element by element and sum to get the total energy involved can be determined. This process equal to the infinitely flexible continuum. In order to determine the amplitudes of the various displacements, it will be parameters that relate with the displacement model. Depending on the problem and the degree of precision required in the solution, the finite element shows that the component need to be analyzed. It will include assemblages of lines; plane figures or solid such as will be used in the examples that follow. In each case, for linear elastic analysis, the modelling leads to governing matrix equations of equilibrium of the form (3.30) which is precisely the same form as ( ). The q’s are the basic unknowns displacements at selected points in the discretisation, the Q’s are corresponding generalized forces whilst is a stiffness matrix for the elements. After getting the value of q from equation (3.36), the strains and stresses are calculated at chosen points in the individual elements and the analysis is then complete. As consideration moves from one situation to the next, it is only the detailed contents of equation (3.36) which change.

33 In a real application, it is necessary to form a model with a large number of degrees of freedom. But, at the same time, at least tens of unknowns will be involved even for the trivial problem. As if for the modest one, more than hundred unknowns will be involved. All this calculation requires a sophisticated programming to get the answer. It is not necessary to know all the rules given such as the choice of admissible displacement models, how integrations are carried out numerically or how various items of data are manipulated in the computer etc. The most important thing is that the finite element process be understood; otherwise it is not possible to understand the significance of various items of input (design of the mesh etc.). Below are a few finite element examples similar to the example in the previous section.
3.5.4 Example from the journal
The components of practical of the non-uniformity of stress and displacement fields may result from: (a) A geometric abnormality such as a hole or groove in an otherwise uniform geometry and simple loading. (b) An abnormality in the loading on a simple shaped component. (c) A combination of (a) and (b). The finite element method becomes valuable since it is suspect that an exact theoretical solution exists so that we cannot turn. On the other hand, choose an example of class (a) for the analytic solution exists: the case of a uniform plate of infinite size supporting a uniform tension and containing a single circular hole.

34 12” y x Figure 3.7 Two-fold symmetry of geometry The modelling of this problem is as follows. The arrangement consist two fold symmetry of geometry and load pattern so that only the quadrant of Figure 3.7 need to be analyzed. A six node isoperimetric element has been chosen since it allows the curved boundary of the hole to be corresponding more accurate compared to simple triangles. The stress field is non-uniform in the neighbourhood of the hole but we know that this non-uniformity evens out away from the hole. This fact was exploited in the graded nature of the two meshes shown in Figure 3.8 and Figure 3.9 many degrees of freedom area incorporated only where they are needed so that the total amount of computation is only as much as is judged necessary only for the purpose. One mesh is more refined than the other with a view to improving the quality of the approximation.

35 Figure 3.8 Less mesh elements Figure 3.9 More mesh elements In this example, the main interest is in showing how the stress distribution along the Ox axis peaks at the edge of the hole. Because the displacement model built into the particular element is quadratic, equilibrium is only satisfied in some

36 average manner lumped at the nodes, the computed stresses have a discontinuous distribution which, for the meshes used, shows up quite noticeably where the stress gradient is high, but is not important in regions away from the hole. See Figure 3.10. STRESS dy/p 3.0 2.0 1.0 MESH No 1 - CURVED SIDED TRIANGLES x- STRAIGHT SIDED TRIANGLES 0 1 2 3 4 5 DISTANCE x, (inc) Figure 3.10 Mesh No 1 STRESS dy/p 3.0 MESH No 2 0 2.0 1.0 1 2 3 4 5 DISTANCE x, (inc) Figure 3.11 Mesh No 2 Due to the symmetry, there should be no shearing stresses along the Ox and Oy axes; however, with the meshes shown, small shears were generated there and the magnitudes also provide a useful measure of the approximation in the solution. In

37 some finite element packages, the output might comprise principal stresses and their directions, then for axes Ox and Oy, we would find that the principal directions are inclined to them instead of being at 0
°
and 90
°
. This provides an alternative measure of the accuracy of the discretisation and the user of the package would have to exercise some judgement to decide if it was adequate for his purpose.
3.6 Conclusion
In brief, having taken an advantage on the example given in this chapter is needed. The one-dimensional problem given is the basic knowledge in order to solve the case study in the next chapter. This situations same as for torsion problem which is a part of elasticity problem. Other than that, we also find some ideas from the journal taken. But, most of them were used for engineering problem with a large element inside.

CHAPTER 4 4 CASE STUDY IN TWO-DIMENSIONAL PROBLEM Introduction 4.1
In many cases especially in engineering problem, solving some problems with a manual calculation will take a big number of times. In real time situation, it is necessary to use some systems so that it will give more accurate and convenient solution when dealing with a large number. In this case, the problem stated here is just a simple problem but still can be expanded with a user friendly system.
4.2 Case Study: Two-Dimensional Elasticity Problem
(0,4) (0,2) (0,0) 4 3 (2,4) 0.5
2 4 E1 3 4 0.5
1 3 (4,2) E2 E3 1 2 (4,0) 0.5
0 (2,0) Figure 4.1 An illustration from case study

39 The problem above with its local and global nodal points is given. No value for z-axis because of two-dimensional problem. However, the value for thickness t as one are given, Poisson ratio v as half, Young modulus E as twelve and as 3 5 .
4.3 Finite Element Formulation of the problem
The finite element formulation for this problem consists of two-dimensional solid mechanics is defined according to plane stress or plane strain conditions. This formulation looks like as follows: Consider a typical four-nodded element, 4 1 3 2 Figure 4.2 An element from the problem Assuming an isoparametric element i.e. degree of approximation used to describe coordinate transformation is equal to the degree of approximation used to represent the dependent variable, . (4.1) . (4.2) . (4.3) . (4.4) where
′
are the shape functions as indicated previously. These shape functions are the functions of natural coordinates. One can write

40 0 0 0 0 0 0 0 0 . (4.5) Or Similarly , . (4.6) 0 0 0 0 0 0 0 0 . (4.7) Or . (4.8) two direct strains, For plane stress and plane strain problem, the strain vector and and one shearing strain, . consist of , . (4.9) From the theory of electricity, the following relationship exists between strains and displacements. . 4.10

41 Substituting for ‘u’ and ‘v’. . (4.11) . (4.12) . (4.13) In form of matrix 0 0 0 0 0 0 0 0 . (4.14) Or Where , . (4.15) is strain displacement matrix of dimension (3X8). 0 0 0 0 0 0 0 0 . (4.16) For a plane elasticity problem, the state of stress at any point may be represented by these stress components , . (4.17)

42 The stress and strain components are related by the problem, matrix. For plane elasticity is a three by three matrix, the elements of which depend upon whether the problem is plane stress or plain strain. For plane stress problem , 1 0 1 0 2 1 0 0 . (4.18) Or , 1 1 0 0 0 0 . (4.19) Hence 1 1 0 0 0 0 . (4.20) For plane strain , 1 0 1 0 0 0 . (4.21) Hence 1 0 1 0 Now element stiffness matrix can be determined using 0 0 . (4.22)

43 . (4.23) The complication in the calculation of is due to the numerical evaluation of integral. Note that a master element was defined in natural coordinate system, while the finite element mesh was defined in rectangular coordinate system. In other words, the shape function is expressed in terms of and while in matrix its derivatives are sought with respect to ‘x’ and ‘y’. Hence a coordinate transformation is required such as the transformation of a finite element mesh to the master element. Consider an i th shape function, , . The derivative of with respect to ‘x’ and ‘y’ can be obtained by using chain rule of differentiation. . + . . (4.24) . + . . (4.25) Or . (4.26) Which gives the relation between the derivatives of with respect to global (x,y) and local coordinates. The above matrix is called the Jacobian matrix. . (4.27) Inversion of Jacobian matrix gives . (4.28)

44 Therefore, . (4.29) To calculate Jacobian inverse, need to calculate , , , and | | using equation (4.6). . (4.30) . (4.31) . (4.32) . (4.33) Or . (4.33) To calculate , determinant of must be greater than zero at every point in element domain. Therefore the functions , and , must be continuous differentiable and invertible. . (4.34) With replacement of dA = dxdy , . (4.35)

45 Where | | , . (4.36) The elemental area in the element is transformed to where | | refers to its determinant. , . (4.37) , . (4.38)
4.4 Manual calculation 4.4.1 Element 1
A (0,4) 4 3 H (2,4) B (0,2) 1 2 G (2,2) Figure 4.3 Element 1 Here is the illustration for element one from the whole problem. The value of Young modulus and Poisson ratio were given. From these two requirements, will getting the value for D either for plane strain or plane stress, vice versa. But, for this time, we use the formula for plane stress only. The value for E is equal to twelve and v is equal to half. After some calculation, the value for D is 1 1 0 0 0 0 4 4 2 0 2 4 0 0 0 1 . (4.39)

46 Here are the equations for each nodal point, 1 1 1 1 1 . 1 . 1 . 1 . (4.40) Now, the value for and is, 1 1 1 1 1 1 1 1 . (4.41) 1 1 1 1 . (4.42) And then, its derivative for x and y to and are as follows 1. 0. 0. 1. (4.43) Hence its derivative for N to and are, . . . . . . . . (4.44)

47 And now the Jacobian matrix, . (4.45) Then their inverse, . (4.46) After that, the B matrix is, 1 4 1 4 1 4 4 1 1 4 1 4 – 1 4 1 4 (4.47) 1 4 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 (4.48) Insert all the requirements in the form of stiffness matrix, load vector and boundary vector to solve the problem. Here are the formula to get the stiffness matrix, load vector and boundary vector. Because of these three same shape rectangles, need to use and as a notation. Notice that dA = dxdy = | | d d . 1 1 4 0 1 1 0 1 0 1 0 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 4 4 2 0 2 4 0 0 0 1

1 4 0 1 1 0 1 1 0 1 1 0 1 1 1 0 1 0 1 1 0 1 1 48 0 1 1 1 (4.49) 3 3 1 1 3 3 1 1 1 3 1 3 1 3 3 1 3 1 3 1 1 3 3 1 1 3 1 3 1 3 1 (4.50) 3

49 4 3 1 10 3 3 0 0 0 0 0 0 0 0 14 3 1 20 3 3 1 4 3 1 14 3 3 10 0 0 0 0 0 0 3 0 0 3 20 3 20 3 3 14 3 1 0 0 0 0 0 0 0 0 10 3 4 3 3 1 3 20 3 4 3 0 0 0 0 0 0 0 0 1 3 10 3 1 14 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 14 1 20 3 3 0 0 0 0 0 0 0 0 4 3 1 3 3 10 3 1 4 3 3 20 3 0 0 0 0 0 0 0 0 1 14 3 3 10 3 (4.51) As a conclusion for this problem, continuing the same step for the second element and find the total value for the stiffness matrix and the load vector are necessary. But, it differs for the boundary vector because the value is for the whole boundary and did not depend on the number of element. Now, the calculation for the load vector are proceed to produce. Nevertheless, using b for body force. It is same when using as a notation for body force. After 10 3 0 0 0 0 0 0 0 0 20 3 4 3 3 1 3 14 3 1 3 20 3 3 10 3 1 14 0 0 0 0 0 0 3 0 0 4 3 1 some integration and make use the value of to get component: . (4.52)

And now, by using the value above, the load vector will be determined. 50 (4.53) And this is their expanded element, 0 0 0 0 0 0 0 0 (4.54) Other than that, it requires us to find the strains and stresses for each of the element consist in the problem. The strains are denoted by while the stresses are denoted by . (4.55) 1 1 0 (4.56)

51 From constitutive relations, (4.57) 24 24 0 (4.58)
4.4.2 Element 2
B (0,2) 4 3 G (2,2) C (0,0) 1 2 D (2,0) Figure 4.4 Element 2 Here is the illustration for element two from the whole problem. The method and techniques are same as before. However, the different is only on their coordinate of each nodal point. Since only concerning on a grid rectangle which is having a same size, calculation is needed for the first element only. It is because, the other elements also have the same value based on its same size. Here are the equations for each nodal point, 1 4 1 4 1 4 1 4 1 1 1 1 1 1 1 1 (4.59)

52 Now, their and value, 1 1 1 1 (4.60) 1 1 1 And then, its derivative for x and y to and are as follows 1 (4.61) 1 0 0 1 (4.62) Hence its derivative for N to and , (4.63) And now the Jacobian matrix, (4.64) Then their inverse, (4.65)

53 After that, we get the B matrix, 1 4 1 4 1 4 4 1 1 4 1 4 – 1 4 1 4 (4.66) 1 4 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 (4.67) Since having a same value for each component in stiffness matrix, now just take the value at the end of the process. 3 1 3 1 3 1 3 1 1 3 1 3 1 3 1 3 3 1 3 1 (4.68) 3 1 3 1 1 3 1 3 1 3 1 3

54 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 20 3 4 3 3 1 10 3 3 0 0 0 0 14 3 0 0 1 0 0 4 3 1 20 3 3 14 3 1 0 0 0 0 10 3 0 0 3 0 0 3 20 3 1 4 3 0 0 1 14 3 3 10 0 3 0 0 0 0 0 10 3 3 14 3 0 0 0 0 4 3 1 20 3 3 1 0 0 3 20 3 4 3 0 0 0 0 1 3 10 0 3 0 0 0 1 14 3 1 4 3 3 20 3 0 0 0 0 1 14 0 3 0 0 0 3 10 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 14 3 1 10 3 3 0 0 0 0 0 0 20 3 4 3 3 1 0 0 4 3 1 3 20 3 0 0 3 10 3 1 14 0 3 0 0 0 (4.69) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Then, for its load vector yields, (4.70)

55 And this is their expanded element, 0 0 0 0 0 0 (4.71) 0 0 Now, the strains and stresses value is, 1 1 0 (4.72) From constitutive relations: 24 24 0 (4.73)
4.4.3 Element 3
G (2,2) 4 3 F (4,2) D (2,0) 1 2 E (4,0) Figure 4.5 Element 3

56 Here are the equations for each nodal point, 1 4 1 4 1 4 1 4 1 1 1 1 1 . 1 . 1 . 1 . (4.74) Now, the value for and is, 1 2 1 1 1 1 2 1 1 1 1 2 1 1 1 1 . And then, its derivative for x and y to and are as follows 1. 0. 0. 1. 1 2 1 1 . (4.75) (4.76) (4.77) Hence the derivative for N to and are, . . . .

57 . . . . (4.78) And now the Jacobian matrix, . (4.79) Then their inverse, 1 0 0 1 . (4.80) After that, we get the B matrix, 1 4 0 1 1 0 1 1 1 0 1 4 4 1 1 0 1 1 0 1 4 4 1 1 4 1 4 – 1 4 1 4 (4.81) 1 1 0 1 1 0 1 1 0 1 1 (4.82)

58 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 1 3 0 0 1 0 0 0 0 0 0 3 1 0 0 1 3 3 0 0 0 0 0 0 0 0 1 1 3 0 0 0 0 0 0 1 3 0 0 3 1 0 0 0 0 0 0 3 1 0 0 3 1 0 0 0 0 0 0 3 1 3 0 0 1 0 0 0 0 0 0 1 3 1 0 0 3 0 0 0 0 0 0 1 3 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 (4.83) Then, for its load vector yields, (4.84)

59 And this is their expanded element, 0 0 0 0 0 0 (4.85) 0 0 Now, its strains and stresses value are, 1 1 0 (4.86) From constitutive relations: 24 24 0 (4.87) After having found the stiffness matrix and load vector for each element, need to expand it based on the number of nodal points. Now, the boundary vector is produced by using the expanded elements just now. After that, need to solve some linear algebra which is consisting thirty two unknown to be found.

3 3 60 2 2 2 2 (4.88)

61 0 0 0 14 3 1 20 3 3 0 0 0 0 0 4 3 1 10 3 3 0 0 0 0 1 4 3 3 20 3 1 14 3 3 10 3 0 20 3 3 2 2 4/3 0 0 0 1 10 3 3 8 4 3 4 1 0 0 0 0 0 0 4 3 1 20 3 3 14 3 1 0 0 0 0 0 0 10 3 3 0 0 2 2 1 14 3 3 20 3 4 8 1 14 3 3 10 3 0 0 0 1 10 3 8 3 3 0 0 10 3 3 14 3 1 40 3 0 14 3 0 0 0 0 0 3 10 3 0 0 0 1 14 3 3 20 3 4 3 1 0 0 0 0 3 10 3 3 10 3 0 28 3 0 1 4 3 0 40 3 4 3 1 0 0 0 0 0 0 0 14 3 1 20 3 4 3 3 1 10 3 3 0 0 1 14 3 3 10 3 0 1 4 3 3 20 3 0 0 0 0 0 0 0 0 0 0 0 0 0 10 3 4 3 3 1 20 3 3 14 3 1 0 0 3 20 3 4 3 1 0 3 10 3 1 14 3 0 0 0 0 0 0 0 8 3 3 2 10 3 0 10 3 3 14 3 1 44 3 1 10 3 3 14 3 1 2 1 4 3 4 8 1 4 3 1 26 3 3 10 3 3 10 3 0 28 3 3 10 3 0 0 0 0 0 0 0 (4.89) 0 20 3 3 14 3 1 10 3 4 3 3 1 3 10 3 1 14 3 0 0 0 0 0 0 0 0 3 20 3 4 3 1 Then, the boundary vector becomes, . (4.90)

62 Then the final is, 15 2 25 2 15 2 25 2 (4.91)

63 Here is the final process consist a big number of unknown. 4 3 1 10 3 3 0 0 0 0 0 0 0 0 14 3 1 20 3 3 0 0 0 0 1 4 3 3 20 3 1 14 3 3 10 3 0 20 3 3 2 0 0 0 4 3 2 1 10 3 3 0 0 0 0 8 4 3 4 1 3 20 3 4 8 1 14 3 3 10 3 0 0 0 0 2 2 1 14 3 0 0 4 3 1 20 3 3 14 3 1 0 0 0 0 10 3 3 0 0 0 0 4 3 1 0 1 14 3 3 20 3 0 0 0 0 3 10 3 0 0 8 3 3 0 0 10 3 3 14 3 1 40 3 0 14 3 1 10 3 0 0 0 0 3 10 3 0 28 3 0 1 4 3 3 10 3 0 40 3 4 3 1 0 0 0 0 0 0 0 14 3 1 20 3 4 3 3 1 10 3 3 0 0 0 0 0 0 0 0 1 4 3 3 20 3 1 14 3 3 10 3 0 0 0 0 0 0 0 0 10 3 4 3 3 1 20 3 3 14 3 1 0 0 0 0 3 10 3 1 14 3 3 20 3 4 3 1 0 0 0 0 0 0 2 10 3 1 44 3 1 10 3 3 8 3 3 0 10 3 3 14 3 14 3 1 2 3 10 3 0 28 3 3 10 3 1 4 3 1 26 3 3 10 3 1 4 3 4 8 0 0 0 20 3 3 14 3 1 4 3 1 10 3 3 0 0 0 0 0 0 0 0 3 20 3 4 3 1 3 10 3 1 14 3 0 0 0 0 0 2 0 1 0 0 0.5
0 0 0.5
1 0.5
2 2 2 2 2 3 3 15 2 25 2 15 2 25 2 (4.92)

64 Then it becomes as 14 3 20 3 3 4 3 3 10 3 4 3 3 20 3 14 3 10 3 3 3 0 0 8 10 8 10 10 3 4 3 20 3 3 14 3 10 3 3 14 3 3 20 3 4 3 11 38 3 1 3 2 3 4 3 17 74 3 3 3 2 2 2 2 15 2 25 2 15 2 25 2 8 10 (4.93)
4.5 Conclusion
From the previous calculations, apart from so many problems in elasticity field has been solved. Many of the problems were from the engineers. Consider the calculation and method as well as the formulation for that case only. Because of the limitation, not all the requirements needed could be found such as the initial strain vector. Since not dealing with heat conduction, the increment in terms of its temperature is assumed to become zero.

CHAPTER 5 5 CONCLUSION AND RECOMMENDATION Conclusion 5.1
In this dissertation, the discussion is mainly about the numerical solution of the two-dimensional elasticity problem for grid rectangle only. It differs from manual calculations for solving the two-dimensional elasticity problem which requires the user to find all the value for each part which is including all the mesh calculations in order to find the final result. Furthermore, the presented programming system has the advantage that it needs less time to solve the problem and is easier for the user to manage the system.
5.2 Recommendation
Here are some several possibilities for future research. In this dissertation, the scope of the study had been limited to the two-dimensional problem for grid rectangle only. The extension of the programming system can be done for three dimensional elasticity problem. Further work on improving the numerical techniques used for computing the two-dimensional elasticity problem can be done for the other shape function such as for un-grid rectangle or grid triangle needs to be carried out. Note that these

66 techniques weaken but do not entirely remove the precision. It is possible to continue the process further in a systematic way, by combining such shapes. For example, by combining the grid rectangle with the un-grid rectangle in the same problem but it must be same for the dimensionality of the problem. The numerical experiments in this dissertation were based on solving the two dimensional elasticity problem which is merely same as heat problem. The integral equation was discretised by some method with the trapezoidal rule. The matrices equation is then solved and verified using MATLAB that makes use of Gauss elimination. However, the numerical experiment cannot proceed because of the result of matrix has many unknown. It is because the grid rectangle in the case study has many nodal points. In addition, the numerical experiment has been calculated using MATLAB and the result still not available. Therefore, further work on improving the numerical techniques used in this dissertation needs to be carried out. It is so interesting when combining some shape function. But, it is not easy to do either manually or developing a system. It requires higher thinking of knowledge.

REFERENCES
1. Advance Computing Kuala Lumpur (1999), Course FEM and Modeling, Universiti Teknologi Malaysia, Kuala Lumpur 2. A. I Lurie (2005), Theory of Elasticity, Springer-Verlag Berlin Heidelberg, New York 3. Crandall, S. H., Dahl, N. C. and Lardner, T. J. (1972), ‘An introduction to the mechanics of solids’. 2 nd Edition, McGraw Hill 4. Desai, C. S. and Abel, J. F. (1972), ‘Introduction to the finite element method’. Van Nostrand Reinhold 5. Doyle, J. F. and Phillips, J. W. (1989), Manual on Experimental Stress Analysis, Society for Experimental Mechanics 6. Dym, C. L. and Shames, I. H. (1973), ‘Solid mechanics: a variational approach’, McGraw Hill 7. Entwistle, K. M. (1999), Basic Principles of the Finite Element Method, Maney Publishing 8. Fung, Y. C. (1965), ‘Foundation of solid mechanics’, Prentice Hall 9. Hunter, S. C. (1976), ‘Mechanics of continuous media’. Ellis Horwood Publishers (John Wiley) 10. Lanczos, C. (1970), ‘The variational principles of mechanics’, 4 th Edition, Toronto University Press 11. Martin, H. C. and Carey, G. F. (1973), ‘Introduction to finite element analysis: theory and application’, McGraw Hill 12. Mohammed Ameen (2005), Computational Elasticity, Alpha Science International Ltd, India 13. Niels Ottosen and Hans Petersson (1992), Introduction to the Finite Element Method, Prentice Hall International,United Kingdom

68 14. Richards, T. H. (1977), ‘Energy methods in stress analysis’, Ellis Horwood Publishers (John Wiley) 15. Roark, R. J. and Young, W. C. (1975), ‘Formulas for stress and strain’, 5 th Edition, McGraw Hill 16. Robert, W. L. (1973), Elasticity, Prentice Hall, Englewood Cliffs, New Jersey 17. Sadd, M. H. (2005), Elasticity: Theory , Applications, and Numerics, Elsevier Inc, United States of America 18. Timoshenko, S. (1955), ‘Strength of Materials’, Vols. 1 and 2, 3 rd edition, Van Nostrand 19. Zienkiewicz, O. C. (1977), ‘The finite element method’, 3 rd Edition, McGraw Hill 20. Answer.com, URL :
http://www.answers.com/topic/elasticity
, (retrieved 8 th April 2010)

A APPENDIX A WATERFALL MODEL

B APPENDIX B GANTT CHART
70

71
C APPENDIX C DETAILS IN CALCULATIONS
[1] More calculation for load vector given here. (4.90) 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 72

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 15 2 25 2 0 0 0 0 2 15 2 25 73

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 74

Then the final is, 0 0 0 0 0 0 0 0 0 0 0 0 15 2 25 2 15 2 25 2 75

76
D APPENDIX D MATLAB PROGRAMMING ALGORITHM
1. Specify Read Input File (.txt) and define Output File (.txt) 2. Read all Parameters in the Input File 3. Create Plot in a Figure according to these parameters: a. Coordinates (coords) b. No. Spatial Coordinates (2D) c. No. of Nodes d. List of Nodes e. No. of Elements 4. Specify Read Input File (.txt) and define Output File (.txt) 5. Read all Parameters in the Input File 6. Create Plot in a Figure according to these parameters: a. Coordinates (coords) b. No. Spatial Coordinates (2D) c. No. of Nodes d. List of Nodes e. No. of Elements 7. Define Material Stiffness a. Computes elasticity tensor C_{ijkl} = shear modulus and Poissons ratio 8. Define Material Stress a. Computes stress sigma_{ij} given strain epsilon_{ij} 9. Defines the number of integration points:be used for each element type a. 2D element – Rectangular element 10. Defines positions of integration points 11. Defines integration weights w_i a. 2D element – Rectangular element 12. Calculates shape functions for various element types 13. Assemble the element stiffness a. Set up integration points and weights b. Loop over the integration points c. Compute shape functions and derivatives wrt local coords

77 d. Compute the Jacobian matrix and its determinant e. Convert shape function derivatives:derivatives wrt global coords f. Compute the material tangent stiffness (d stress/d strain) ds/de is just C_ijkl for linear elasticity - this notation is used to allow extension to nonlinear problems g. Compute the element stiffness 13. Element distributed load vector a. Compute the jacobian matrix and its determinant 14. Assemble the global stiffness matrix a. Loop over all the elements b. Extract coords of nodes, DOF for the current element c. Add the current element stiffness : the global stiffness 15. Assemble the global traction vector a. Extract the coords of the nodes on the appropriate element face b. Compute the element load vector c. Assemble the element load vector into global vector d. Print nodal displacements, element strains and stresses:a file e. Loop over all the elements f. Extract coords of nodes, DOF for the current element g. Set up integration points h. Loop over the integration points i. Compute shape functions & derivatives wrt local coords j. Compute the coords of the integration point k. Compute the jacobian matrix and its determinant a. Convert shape function derivatives:derivatives wrt global coords b. Compute the (infinitesimal) strain by differentiating displacements 16. End of File