Content Module 1: Introduction to Finite Element Analysis (04) Lecture 1: Introduction (Page: 1-6)

1.1.2 Background of Finite Element Analysis 1.1.3 Numerical Methods 1.1.4 Concepts of Elements and Nodes 1.1.5 Degrees of Freedom

Content Module: 2 Finite Element Formulation Techniques (04) Lecture 1: Virtual Work and Variational Principle (Page: 1-4)

2.1.1 Introduction 2.1.2 Principle of Virtual Work 2.1.3 Variational Principle 2.1.4 Weighted Residual Method

Content Module:3 Element Properties (09)

Lecture 1: Natural Coordinates (Page:1-8) 3.1.2 Two Dimensional Triangular Elements

3.1.3 Shape Function using Area Coordinates

Lecture 2: Triangular Elements (Page:9-17) 3.2.1 Shape function using Cartesian coordinates

3.2.2 Higher Order Triangular Elements

3.2.2.1 Shape function for six node element

3.2.3 Construction of Shape Function by Degrading Technique

3.2.3.1 Five node triangular element

Lecture 3: Rectangular Elements (Page:18-24) 3.3.1 Shape Function for Four Node Element

3.3.1.1 Shape function using Cartesian coordinates

3.3.1.2 Shape Function using natural coordinates

3.3.2 Shape Function for Eight Node Element

Lecture 4: Lagrange and Serendipity Elements (Page:25-33) 3.4.1 Lagrange Interpolation Function

3.4.1.1 Shape function for two node bar element

3.4.1.2 Shape function for three node bar element

3.4.1.3 Shape function for two dimensional elements

3.4.2 Serendipity Elements

Lecture 5: Solid Elements (Page:34-39) 3.5.1 Tetrahedral Elements

3.5.2 Brick Elements

Lecture 6: Isoparametric Formulation (Page:40-47) 3.6.1 Necessity of Isoparametric Formulation

3.6.2 Coordinate Transformation

3.6.3 Concept of Jacobian Matrix

Lecture 7: Stiffness Matrix of Isoparametric Elements(Page:48-55) 3.7.1 Evaluation of Stiffness Matrix of 2-D Isoparametric Elements

3.7.2 Evaluation of Stiffness Matrix of 3-D Isoparametric Elements

Lecture 8: Numerical Integration: One Dimensional(Page:56-60) 3.8.1 Gauss Quadrature for One-Dimensional Integrals

3.8.2 One- Point Formula

3.8.3 Two-Point Formula

Lecture 9: Numerical Integration: Two and Three

Dimensional(Page:61-67)

3.9.1 Gauss Quadrature for Two-Dimensional Integrals

3.9.2 Gauss Quadrature for Three-Dimensional Integrals

3.9.3 Numerical Integration of Element Stiffness Matrix

3.9.4 Gauss Quadrature for Triangular Elements

3.9.5 Gauss Quadrature for Tetrahedron

Worked out Examples (Page:68-74) Example 3.1 Calculation of displacement using area coordinates

Example 3.2 Derivation of shape function of four node triangular element

Example 3.3 Numerical integration for three dimensional problems

Content Module 4: Analysis of Frame Structures (06) Lecture 1: Stiffness of Truss Members(Page: 1-6)

4.1.1 Introduction

4.1.2 Element Stiffness of a Truss Member

4.1.3 Member Stiffness with Varying Cross Section

4.1.4 Generalized Stiffness Matrix of a Plane Truss Member

Lecture 2: Analysis of Truss(Page: 7-14) 4.2.1 Element Stiffness of a 3 Node Truss Member

4.2.2 Worked Out Example

Lecture 3: Stiffness of Beam Members(Page: 15-21) 4.3.1 Introduction

4.3.2 Derivation of Shape Function

4.3.3 Derivation of Element Stiffness Matrix

4.3.4 Generalized Stiffness Matrix of a Beam Member

Lecture 4: Finite Element Analysis of Continuous Beam(Page: 22-30) 4.4.1 Equivalent Loading on Beam Member

4.4.1.1 Varying Load

4.4.1.2 Concentrated Load

4.4.2 Worked Out Example

Lecture 5: Plane Frame Analysis(Page: 31-38) 4.5.1 Introduction

4.5.2 Member Stiffness Matrix

4.5.3 Generalized Stiffness Matrix

4.5.4 Worked Out Example

Lecture 6 Analysis of Grid and Space Frame(Page: 38-48) 4.6.1 Introduction

4.6.2 Member Stiffness Matrix

4.6.3 Generalized Stiffness Matrix

4.6.4 Worked Out Example

4.6.5 Analysis of Space Frame

Content Module 5: FEM for Two and Three Dimensional Solids (08) Lecture 1: Constant Strain Triangle(Page:1-6)

5.1.1 Element Stiffness Matrix for CST

5.1.2 Nodal Load Vector for CST

Lecture 2: Linear Strain Triangle(Page:7-12) 5.2.1 Element Stiffness Matrix for LST

5.2.2 Nodal Load Vector for LST

5.2.3 Numerical Example using CST

Lecture 3: Rectangular Elements(Page:13-19) 5.3.1 Computation of Element Stiffness

5.3.2 Computation of Nodal Loads

Lecture 4: Numerical Evaluation of Element Stiffness(Page:20-30) 5.4.1 Numerical Example

5.4.2 Evaluation of Stiffness using One Point Gauss Quadrature

5.4.3 Evaluation of Stiffness using Two Point Gauss Quadrature

Lecture 5: Computation of Stresses, Geometric Nonlinearity and Static

Condensation(Page:31-38) 5.5.1 Computation of Stresses

5.5.2 Geometric Nonlinearity

5.5.2.1 Steps to include effect of geometrical nonlinearity

5.5.3 Static Condensation

Lecture 6: Axisymmetric Element(Page:39-45) 5.6.1 Introduction

5.6.2 Relation between Strain and Displacement

5.6.3 Relation between Stress and Strain

5.6.4 Axisymmetric Shell Element

Lecture 7: Finite Element Formulation of Axisymmetric

Element(Page:45-52) 5.7.1 Stiffness Matrix of a Triangular Element

5.7.2 Stiffness Matrix of a Quadrilateral Element

Lecture 8: Finite Element Formulation for 3 Dimensional

Elements(Page:53-59) 5.8.1 Introduction

5.8.2 Strain Displacement Relation 5.8.3 Element Stiffness Matrix

5.8.4 Element Load Vector

5.8.4.1 Gravity load

5.8.4.2 Surface pressure

5.8.5 Stress Computation

Worked out Examples(Page:60-61) Example 5.1 Calculation of nodal loads on a triangular element

Content Module 6 FEM for Plates and Shells (06) Lecture 1: Introduction to Plate Bending Problems(Page: 1-10)

6.1.1 Introduction 6.1.2 Notations and Sign Conventions 6.1.3 Thin Plate Theory

6.1.3.1 Basic relationships 6.1.3.2 Constitutive equations 6.1.3.3 Calculation of moments and shear forces

6.1.4 Thick Plate Theory 6.1.5 Boundary Conditions

Lecture 2: Finite Element Analysis of Thin Plate(Page: 11-17)

6.2.1 Triangular Plate Bending Element 6.2.2 Rectangular Plate Bending Element

Lecture 3: Finite Element Analysis of Thick Plate (Page: 18-24)

6.3.1 Introduction 6.3.2 Strain Displacement Relation 6.3.3 Element Stiffness Matrix 6.3.4 Nodal Load Vector

Lecture 4: Finite Element Analysis of Skew Plate (Page: 25-31)

6.4.1 Introduction 6.4.2 Finite Element Analysis of Skew Plate Element

Lecture 5: Introduction to Finite Strip Method(Page: 32-37)

6.5.1 Introduction 6.5.2 Finite Strip Method 6.5.2.1 Boundary conditions 6.5.3 Finite Element Formulation

Lecture 6: Finite Element Analysis of Shell(Page: 38-45)

6.6.1 Introduction 6.6.2 Classification of Shells 6.6.3 Assumptions for Thin Shell Theory 6.6.4 Overview of Shell Finite Elements 6.6.5 Finite Element Formulation of a Degenerated Shell

6.6.5.1 Jacobian matrix 6.6.5.2 Strain displacement matrix 6.6.5.3 Stress strain relation 6.6.5.4 Element stiffness matrix

Content Module 7: Additional Applications of FEM (03)

Lecture 1: Finite Elements for Elastic Stability(Page: 1-8)

7.1.1 Introduction 7.1.2 Buckling of Truss Members 7.1.3 Buckling of Beam-Column Members 7.1.4 Buckling of Plate Bending Elements

Lecture 2: Finite Elements in Fluid Mechanics(Page: 9-15)

7.2.1 Governing Fluid Equations 7.2.2 Finite Element Formulation

7.2.2.1 Displacement based finite element formulation 7.2.2.2 Pressure based finite element formulation

7.2.3 Finite Element Formulation of Infinite Reservoir

Lecture 3: Dynamic Analysis(Page: 16-23)

7.3.1 Hamilton’s Principle 7.3.2 Finite Element Form in MDOF System 7.3.3 Solid Body with Distributed Mass

7.3.3.1 Mass matrix for truss element 7.3.3.2 Mass matrix for beam element 7.3.3.3 Mass matrix for plane frame element 7.3.3.4 Mass matrix for space frame element

7.3.4 Time History Analysis

Content Module 1: Introduction to Finite Element Analysis (04) Lecture 1: Introduction (Page: 1-6)

1.1.2 Background of Finite Element Analysis 1.1.3 Numerical Methods 1.1.4 Concepts of Elements and Nodes 1.1.5 Degrees of Freedom

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Lecture 1: Introduction 1.1.1 Introduction The Finite Element Method (FEM) is a numerical technique to find approximate solutions of partial differential equations. It was originated from the need of solving complex elasticity and structural analysis problems in Civil, Mechanical and Aerospace engineering. In a structural simulation, FEM helps in producing stiffness and strength visualizations. It also helps to minimize material weight and its cost of the structures. FEM allows for detailed visualization and indicates the distribution of stresses and strains inside the body of a structure. Many of FE software are powerful yet complex tool meant for professional engineers with the training and education necessary to properly interpret the results.

Several modern FEM packages include specific components such as fluid, thermal, electromagnetic and structural working environments. FEM allows entire designs to be constructed, refined and optimized before the design is manufactured. This powerful design tool has significantly improved both the standard of engineering designs and the methodology of the design process in many industrial applications. The use of FEM has significantly decreased the time to take products from concept to the production line. One must take the advantage of the advent of faster generation of personal computers for the analysis and design of engineering product with precision level of accuracy. 1.1.2 Background of Finite Element Analysis The finite element analysis can be traced back to the work by Alexander Hrennikoff (1941) and Richard Courant (1942). Hrenikoff introduced the framework method, in which a plane elastic medium was represented as collections of bars and beams. These pioneers share one essential characteristic: mesh discretization of a continuous domain into a set of discrete sub-domains, usually called elements.

• In 1950s, solution of large number of simultaneous equations became possible because of the digital computer.

• In 1960, Ray W. Clough first published a paper using term “Finite Element Method”. • In 1965, First conference on “finite elements” was held. • In 1967, the first book on the “Finite Element Method” was published by Zienkiewicz and

Chung. • In the late 1960s and early 1970s, the FEM was applied to a wide variety of engineering

problems.

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• In the 1970s, most commercial FEM software packages (ABAQUS, NASTRAN, ANSYS, etc.) originated. Interactive FE programs on supercomputer lead to rapid growth of CAD systems.

• In the 1980s, algorithm on electromagnetic applications, fluid flow and thermal analysis were developed with the use of FE program.

• Engineers can evaluate ways to control the vibrations and extend the use of flexible, deployable structures in space using FE and other methods in the 1990s. Trends to solve fully coupled solution of fluid flows with structural interactions, bio-mechanics related problems with a higher level of accuracy were observed in this decade.

With the development of finite element method, together with tremendous increases in computing power and convenience, today it is possible to understand structural behavior with levels of accuracy. This was in fact the beyond of imagination before the computer age. 1.1.3 Numerical Methods The formulation for structural analysis is generally based on the three fundamental relations: equilibrium, constitutive and compatibility. There are two major approaches to the analysis: Analytical and Numerical. Analytical approach which leads to closed-form solutions is effective in case of simple geometry, boundary conditions, loadings and material properties. However, in reality, such simple cases may not arise. As a result, various numerical methods are evolved for solving such problems which are complex in nature. For numerical approach, the solutions will be approximate when any of these relations are only approximately satisfied. The numerical method depends heavily on the processing power of computers and is more applicable to structures of arbitrary size and complexity. It is common practice to use approximate solutions of differential equations as the basis for structural analysis. This is usually done using numerical approximation techniques. Few numerical methods which are commonly used to solve solid and fluid mechanics problems are given below.

• Finite Difference Method • Finite Volume Method • Finite Element Method • Boundary Element Method • Meshless Method

The application of finite difference method for engineering problems involves replacing the governing differential equations and the boundary condition by suitable algebraic equations. For

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example in the analysis of beam bending problem the differential equation is reduced to be solution of algebraic equations written at every nodal point within the beam member. For example, the beam equation can be expressed as:

EIq

dxwd=4

4

(1.1.1)

To explain the concept of finite difference method let us consider a displacement function variable namely ( )w f x=

Fig. 1.1.1 Displacement Function

Now, w f (x x) f (x)

So, Lt Lt i+1 iΔx 0 Δx 0

dwΔw f(x+ Δx)- f(x) 1= = = w - wdxΔx Δx h

(1.1.2)

Thus,

2

i+1 i i+2 i+1 i+1 i i+2 i+1 i2 2 2

d w d 1 1 1= w - w = w - w - w + w = w - 2 w + wdx dx h h h

(1.1.3)

3

i+3 i+2 i+2 i+1 i+1 i3 3

i+3 i+2 i+1 i3

d w 1= w - w - 2w +2w + w - wdx h

1= w - 3w +3w - wh

(1.1.4)

4

4

i+4 i+3 i+3 i+2 i+2 i+1 i+1 i4 4

i+4 i+3 i+2 i+1 i4

i+2 i+1 i i-1 i-24

d w 1= w - w - 3w +3w +3w - 3w - w + wdx h

1= w - 4w +6w - 4w + wh1= w - 4w +6w - 4w + wh

(1.1.5)

Thus, eq. (1.1.1) can be expressed with the help of eq. (1.1.5) and can be written in finite difference form as:

42112 )464( h

EIqwwwww iiiii =+−+− ++−− (1.1.6)

Fig. 1.1.2 Finite difference equation at node i

Thus, the displacement at node i of the beam member corresponds to uniformly distributed load can be obtained from eq. (1.1.6) with the help of boundary conditions. It may be interesting to note that, the concept of node is used in the finite difference method. Basically, this method has an array of grid points and is a point wise approximation, whereas, finite element method has an array of small interconnecting sub-regions and is a piece wise approximation.

Each method has noteworthy advantages as well as limitations. However it is possible to solve various problems by finite element method, even with highly complex geometry and loading conditions, with the restriction that there is always some numerical errors. Therefore, effective and reliable use of this method requires a solid understanding of its limitations.

1.1.4 Concepts of Elements and Nodes Any continuum/domain can be divided into a number of pieces with very small dimensions. These small pieces of finite dimension are called ‘Finite Elements’ (Fig. 1.1.3). A field quantity in each element is allowed to have a simple spatial variation which can be described by polynomial terms. Thus the original domain is considered as an assemblage of number of such small elements. These elements are connected through number of joints which are called ‘Nodes’. While discretizing the structural system, it is assumed that the elements are attached to the adjacent elements only at the nodal points. Each element contains the material and geometrical properties. The material properties inside an element are assumed to be constant. The elements may be 1D elements, 2D elements or 3D elements. The physical object can be modeled by choosing appropriate element such as frame

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element, plate element, shell element, solid element, etc. All elements are then assembled to obtain the solution of the entire domain/structure under certain loading conditions. Nodes are assigned at a certain density throughout the continuum depending on the anticipated stress levels of a particular domain. Regions which will receive large amounts of stress variation usually have a higher node density than those which experience little or no stress.

Fig. 1.1.3 Finite element discretization of a domain

1.1.5 Degrees of Freedom A structure can have infinite number of displacements. Approximation with a reasonable level of accuracy can be achieved by assuming a limited number of displacements. This finite number of displacements is the number of degrees of freedom of the structure. For example, the truss member will undergo only axial deformation. Therefore, the degrees of freedom of a truss member with respect to its own coordinate system will be one at each node. If a two dimension structure is modeled by truss elements, then the deformation with respect to structural coordinate system will be two and therefore degrees of freedom will also become two. The degrees of freedom for various

types of element are shown in Fig. 1.1.4 for easy understanding. Here ( ), ,u v w and ( ), ,x y zθ θ θ

represent displacement and rotation respectively.

6

Fig. 1.1.4 Degrees of Freedom for Various Elements

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Lecture 2: Basic Concepts of Finite Element Analysis 1.2.1 Idealization of a Continuum A continuum may be discretized in different ways depending upon the geometrical configuration of

the domain. Fig. 1.2.1 shows the various ways of idealizing a continuum based on the geometry.

Fig. 1.2.1 Various ways of Idealization of a Continuum 1.2.2 Discretization of Technique The need of finite element analysis arises when the structural system in terms of its either geometry, material properties, boundary conditions or loadings is complex in nature. For such case, the whole structure needs to be subdivided into smaller elements. The whole structure is then analyzed by the assemblage of all elements representing the complete structure including its all properties.

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The subdivision process is an important task in finite element analysis and requires some skill and knowledge. In this procedure, first, the number, shape, size and configuration of elements have to be decided in such a manner that the real structure is simulated as closely as possible. The discretization is to be in such that the results converge to the true solution. However, too fine mesh will lead to extra computational effort. Fig. 1.2.2 shows a finite element mesh of a continuum using triangular and quadrilateral elements. The assemblage of triangular elements in this case shows better representation of the continuum. The discretization process also shows that the more accurate representation is possible if the body is further subdivided into some finer mesh.

Fig. 1.2.2 Discretization of a continuum

1.2.3 Concepts of Finite Element Analysis FEA consists of a computer model of a continuum that is stressed and analyzed for specific results. A continuum has infinite particles with continuous variation of material properties. Therefore, it needs to simplify to a finite size and is made up of an assemblage of substructures, components and members. Discretization process is necessary to convert whole structure to an assemblage of members/elements for determining its responses. Fig. 1.2.3 shows the process of idealization of actual structure to a finite element form to obtain the response results. The assumptions are required to be made by the experienced engineer with finite element background for getting appropriate response results. On the basis of assumptions, the appropriate constitutive model can be constructed. For the linear-elastic-static analysis of structures, the final form of equation will be made in the form of F=Kd where F, K and d are the nodal loads, global stiffness and nodal displacements respectively.

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Fig. 1.2.3From classical to FE solution

Varieties of engineering problem like solid and fluid mechanics, heat transfer can easily be solved by the concept of finite element technique. The basic form of the equation will become as follows where action, property and response parameter will vary for case to case as outlined in Table 1.2.1.

Classical Actual Structure

0yx z Fx y z

∂σ∂σ ∂σ∂ ∂ ∂ Ω+ + + =

(Partial Differential Equations)

Assumptions Equilibrium Equations

Stress-Strain Law Compatibility Conditions

T T Td d F d d F d

(Principles of Virtual Work)

FEM Structural Model F = K d

(Algebraic Equations)

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[ ] [ ] 1F K d OR d K F−= =

Table 1.2.1 Response parameter for different cases Property Action Response

Solid Stiffness Load Displacement Fluid Viscosity Body force Pressure/Velocity

Thermal Conductivity Heat Temperature

1.2.4 Advantages of FEA 1. The physical properties, which are intractable and complex for any closed bound solution,

can be analyzed by this method. 2. It can take care of any geometry (may be regular or irregular). 3. It can take care of any boundary conditions. 4. Material anisotropy and non-homogeneity can be catered without much difficulty. 5. It can take care of any type of loading conditions. 6. This method is superior to other approximate methods like Galerkine and Rayleigh-Ritz

methods. 7. In this method approximations are confined to small sub domains. 8. In this method, the admissible functions are valid over the simple domain and have nothing

to do with boundary, however simple or complex it may be. 9. Enable to computer programming.

1.2.5 Disadvantages of FEA 1. Computational time involved in the solution of the problem is high. 2. For fluid dynamics problems some other methods of analysis may prove efficient than the

FEM. 1.2.6 Limitations of FEA

1. Proper engineering judgment is to be exercised to interpret results. 2. It requires large computer memory and computational time to obtain intend results. 3. There are certain categories of problems where other methods are more effective, e.g., fluid

problems having boundaries at infinity are better treated by the boundary element method.

Property Action Response

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4. For some problems, there may be a considerable amount of input data. Errors may creep up in their preparation and the results thus obtained may also appear to be acceptable which indicates deceptive state of affairs. It is always desirable to make a visual check of the input data.

5. In the FEM, many problems lead to round-off errors. Computer works with a limited number of digits and solving the problem with restricted number of digits may not yield the desired degree of accuracy or it may give total erroneous results in some cases. For many problems the increase in the number of digits for the purpose of calculation improves the accuracy.

1.2.7 Errors and Accuracy in FEA Every physical problem is formulated by simplifying certain assumptions. Solution to the problem, classical or numerical, is to be viewed within the constraints imposed by these simplifications. The material may be assumed to be homogeneous and isotropic; its behavior may be considered as linearly elastic; the prediction of the exact load in any type of structure is next to impossible. As such the true behavior of the structure is to be viewed with in these constraints and obvious errors creep in engineering calculations.

1. The results will be erroneous if any mistake occurs in the input data. As such, preparation of the input data should be made with great care.

2. When a continuum is discretised, an infinite degrees of freedom system is converted into a model having finite number of degrees of freedom. In a continuum, functions which are continuous are now replaced by ones which are piece-wise continuous within individual elements. Thus the actual continuum is represented by a set of approximations.

3. The accuracy depends to a great extent on the mesh grading of the continuum. In regions of high strain gradient, higher mesh grading is needed whereas in the regions of lower strain, the mesh chosen may be coarser. As the element size decreases, the discretisation error reduces.

4. Improper selection of shape of the element will lead to a considerable error in the solution. Triangle elements in the shape of an equilateral or rectangular element in the shape of a square will always perform better than those having unequal lengths of the sides. For very long shapes, the attainment of convergence is extremely slow.

5. In the finite element analysis, the boundary conditions are imposed at the nodes of the element whereas in an actual continuum, they are defined at the boundaries. Between the nodes, the actual boundary conditions will depend on the shape functions of the element forming the boundary.

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6. Simplification of the boundary is another source of error. The domain may be reduced to the shape of polygon. If the mesh is refined, then the error involved in the discretised boundary may be reduced.

7. During arithmetic operations, the numbers would be constantly round-off to some fixed working length. These round–off errors may go on accumulating and then resulting accuracy of the solution may be greatly impaired.

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Lecture 3: Introduction to Elasticity 1.3.1 Stresses and Equilibrium Let consider an infinitesimal element of sides dx, dy and dz as shown in Fig. 1.3.1. The stresses are acting on the elemental volume dV because of external and or body forces. These stresses can be represented by six independent components as given below.

, , , , ,T

x y z xy yz zxσ σ σ σ τ τ τ =

(1.3.1)

Here, ,x y zandσ σ σ are normal stresses and ,xy yz zxandτ τ τ are shear stresses. Applying the conditions of static equilibrium for forces along the direction of X axis (i.e., 0xF =∑ ), following expression will be obtained.

0yxx zxxdxdydz dxdydz dxdydz F dxdydz

x y zτσ τ

Ω

∂∂ ∂+ + + =

∂ ∂ ∂ (1.3.2)

Fig. 1.3.1Stresses on an infinitesimal element

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Where, FΩx is the component of body force along x direction. Now, dividing dxdydz on the above expression, following equilibrium condition is obtained.

yxx zxxf

x y zτσ τ∂∂ ∂

+ + = −∂ ∂ ∂

(1.3.3)

Similarly, applying equilibrium condition along Y and Z directions, one can find the following relations.

xy y zyyf

x y zτ σ τ∂ ∂ ∂

+ + = −∂ ∂ ∂

(1.3.4)

yzxz zzf

x y zττ σ∂∂ ∂

+ + = −∂ ∂ ∂

(1.3.5)

Here, FΩy and FΩz are the component of body forces along Y and Z directions respectively. Satisfying moment equations (i.e., 0; 0 0;x y zM M and M= = =∑ ∑ ∑ ), one can obtain the following relations.

; andxy yx yz zy xz zxτ τ τ τ τ τ= = = (1.3.6)

Using eq. (1.3.6), the equilibrium equations (1.3.3 to 1.3.5), can be rewritten in the following form. xyx zx

x

xy y yzy

yzzx zz

Fx y z

Fx y z

Fx y z

τσ τ

τ σ τ

ττ σ

Ω

Ω

Ω

∂∂ ∂+ + = −

∂ ∂ ∂∂ ∂ ∂

+ + = −∂ ∂ ∂

∂∂ ∂+ + = −

∂ ∂ ∂

(1.3.7)

Eq. (1.3.7) is known as equation of equilibrium. Let assume an element of area ∆Γ on the surface of the solid in equilibrium (Fig.1.3.2) and FΓx, FΓy and FΓz are the components of external forces per unit area and are acting on the surface. Consideration of equilibrium along the three axes directions gives the following relations.

x xy zx x

xy y yz y

zx yz z z

l m n Fl m n Fl m n F

σ τ τ

τ σ τ

τ τ σ

Γ

Γ

Γ

+ + =

+ + =

+ + =

(1.3.8)

Here, l, m and n are the direction cosines of the normal to the boundary surface. Eq. (1.3.8) is known as static boundary condition.

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Fig. 1.3.2 Forces acting on an elemnt on the boundary

1.3.2 Strain-Displacement Relations The displacement at any point of a deformable body may be expressed by the components of u, v and w parallel to the Cartesian coordinate’s axes. The components of the displacements can be described as functions of x, y and z. Displacements basically the change of position during deformation. If point P (x,y,z) is displaced to P’ (x’,y’,z’), then the displacement along X, Y and Z direction (Fig. 1.3.3) will become

x’ = x + u or u = x’ – x y’ = y + v or v = y’ – y z’ = z + w or w = z’ – z

Therefore, the normal strain can be written as:

0x x

u uLtx x

ε∆ →

∆ ∂= =

∆ ∂(As L

Lε ∆= for uniform strain in axial member)

Similarly, andy zv wy z

ε ε∂ ∂= =∂ ∂

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Fig. 1.3.3 Deformation of an elastic body

Let consider points P, Q and R are before deformation and points P’,Q’ and R’ are after deformation as shown in Fig. 1.3.4 below. Now for small deformation, rotation of PQ will become

1 0limx

v vx x

θ∆ →

∆ ∂= =

∆ ∂

Similarly, rotation of PR due to deformation will be: 2 0limy

u uy y

θ∆ →

∆ ∂= =

∆ ∂

Thus, the total change of angle between PQ and PR after deformation is as follows which is defined as shear strain in X-Y plane.

1 2xy yxv ux y

γ γ θ θ ∂ ∂= = + = +

∂ ∂

Fig. 1.3.4 Derivation of shear strain

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Similarly, shear strains in Y-Z and X-Z plane will become

;yz zy xz zxw v w uy z x z

γ γ γ γ∂ ∂ ∂ ∂= = + = = +

∂ ∂ ∂ ∂

The strain can be expressed as partial derivatives of the displacements u, v and w. The above expressions for strain-displacement relationship are true only for small amplitude of deformation. However, the strain-displacement relations are expressed by the following equations for large magnitude of deformation.

∂∂

+

∂∂

+

∂∂

+∂∂

=222

21

xw

xv

xu

xu

xε (1.3.9)

∂∂

+

∂∂

+

∂∂

+∂∂

=222

21

yw

yv

yu

yv

yε (1.3.10)

∂∂

+

∂∂

+

∂∂

+∂∂

=222

21

zw

zv

zu

zw

zε (1.3.11)

∂∂

∂∂

+∂∂

∂∂

+∂∂

∂∂

+∂∂

+∂∂

=yw

xw

yv

xv

yu

xu

yu

xv

xyγ (1.3.12)

∂∂

∂∂

+∂∂

∂∂

+∂∂

∂∂

+∂∂

+∂∂

=zw

yw

zv

yv

zu

yu

zv

yw

yzγ (1.3.13)

∂∂

∂∂

+∂∂

∂∂

+∂∂

∂∂

+∂∂

+∂∂

=xw

zw

xv

zv

xu

zu

xw

zu

zxγ (1.3.14)

The eqs.(1.3.9 to 1.3.14) are known as Green-Lagrange strain displacement equation. The components of the strain εx, εy, εz, γxy, γyz and γzx define the state of strains in the deformed body, and can be written in a matrix form as

Tx y z xy yz zx

(1.3.15)

The relations given in eqs.(1.3.9 to 1.3.14) are non-linear partial differential equations in the unknown component of the displacements. In case of small deformations, the products and squares of the first derivatives are assumed to be negligible compared with the derivatives themselves in many problems of stress analysis. Thus the strain-displacement relations in eqs. (1.3.9 to 1.3.14 ) reduce to linear relations as follows.

18

x

y

z

xy

yz

zx

uxvywzv ux yw vy zu wz x

ε

ε

ε

γ

γ

γ

∂=∂∂

=∂∂

=∂∂ ∂

= +∂ ∂∂ ∂

= +∂ ∂∂ ∂

= +∂ ∂

(1.3.16)

Eq. (1.3.16) is known as Von-Karman strain displacement equation. The above equation can be expressed in a matrix form as given below.

∂∂

∂∂

∂∂

∂∂∂∂

∂∂

∂∂

∂∂

∂∂

=

wvu

xz

yz

xy

z

y

x

zx

yz

xy

z

y

x

0

0

0

00

00

00

γγγεεε

(1.3.17)

The above assumption will be incorrect in case of large deformation problems. In these cases, geometric nonlinearity has to be considered. 1.3.3 Linear Constitutive Relations Hooke’s law states that the six component of stress may be described as linear function of six components of strain. The relation for a linear elastic, anisotropic and homogeneous material are expressed as follows.

19

=

zx

yz

xy

z

y

x

zx

yz

xy

z

y

x

CCC

CCCCCC

γγγεεε

τττσσσ ε

666261

262221

161211

.........

...

...

(1.3.18)

or [ ] εσ C= (1.3.19)

Where [C] is constitutive matrix. If the material has three orthogonal planes of symmetry, it is said to be orthotropic. In this case only nine constants are required for describing constitutive relations as given below.

=

zx

yz

xy

z

y

x

zx

yz

xy

z

y

x

CCSymmetry

CCCCCCC

γγγεεε

τττσσσ

τ

66

55

44

33

2322

131211

000000000000

(1.3.20)

The inverse relation for strains and stresses may be expressed as

1C D (1.3.21)

An isotropic is one for which every plane is a plane of symmetry of material behavior and only two constants (Young Modulus, E and Poisson ratio µ) are required to describe the constitutive relation. The following equation includes the effect due to temperature changes as may be necessary in certain cases of stress analysis.

or

x x

y

z

xy

yz

zx zx

1 0 0 0 11 0 0 0 . 1

1 0 0 0 . 11 T2 1 0 0 . 0E

2 1 0 . 02 1 0

(1.3.22)

20

T and α in eq. (1.3.22) denote the difference of temperature and coefficient of thermal expansion respectively. The inverse relation of stresses in terms of strain components can be expressed as

( )( )

( )

−−

−

−

−−

−−

=

000111

21

221

0221

00221

000100010001

µα

γγγεεε

µ

µ

µµ

µµµµµ

τττσσσ

TEE

zx

yz

xy

z

y

x

zx

yz

xy

z

y

x

(1.3.23)

where ( )( )µµ 211 −+=

EE

1.3.4 Two-Dimensional Stress Distribution The problems of solid mechanics may be formulated as three-dimensional problems and finite element technique may be used to solve them. In many practical situations, the geometry and loading will be such that the problems may be formulated to two-dimensional or one-dimensional problems without much loss of accuracy. The relation between strain and displacement for two dimensional problems can be simplified from eq. (1.3.16) and can be written as follows.

x

y

xy

uxvyv ux y

ε

ε

γ

∂=∂∂

=∂∂ ∂

= +∂ ∂

(1.3.24)

The above expression can be written in a combined form: 2 22

2 2y xyx

y x x yε γε ∂ ∂∂

+ =∂ ∂ ∂ ∂

(1.3.25)

Eq. (1.3.25) is the compatibility equation since it states the geometric requirements. This condition will ensure adjacent elements to remain free from discontinuities such as gaps and overlaps. 1.3.4.1 Plane stress problem The plane stress problem is characterized by very small dimensions in one of the normal directions. Some typical examples are shown in Fig. 1.3.5. In these cases, it is assumed that no stress

21

component varies across the thickness and the stress components σz, τxz and τyz are zero. The state of stress is specified by σx, σy and τxy only and is called plane stress.

Fig. 1.3.5 Plane stress example: Thin plate with in-plane loading

The stress components may be expressed in terms of strain, which is as follows.

x x

y y2

xy xy

1 0 1E E T1 0 1

1 11 00 0

2

(1.3.26)

The strain components can also be expressed in terms of the stress, which is given below.

x x

y y

xy xy

1 0 11 1 0 T 1E

0 0 2 1 0

(1.3.27)

It can also be shown that

( ) Tyxz αµµεε

µµε

−+

++−−

=11

1and 0== zxyz γγ (1.3.28)

1.3.4.2 Plane strain problem Problems involving a long body whose geometry and loading do not vary significantly in the longitudinal direction are referred to as plane strain problems. Some typical examples are given in Fig. 1.3.6. In these cases, a constant longitudinal displacement corresponding to a rigid body

22

translation and displacements linear in z corresponding to rigid body rotation do not result in strain. As a result, the following relations arise.

0=== zxyzz γγε (1.3.29)

The constitutive relation for elastic isotropic material for this case may be given by,

( )( )

−−

−−

−

−+=

011

21

22100

0101

211 µα

γεε

µµµ

µµ

µµτσσ

TEE

xy

y

x

xy

y

x

(1.3.30)

Also TEyxz ασσµσ −+= )( and 0== zxyz ττ (1.3.31)

The strain components can be expressed in terms of the stress as follows.

x x

y y

xy xy

1 0 11

1 0 1 T 1E

0 0 2 0

(1.3.32)

(a) Retaining wall (b) Dam

Fig. 1.3.6 Plane strain examples

23

1.3.4.3Axisymmetric Problem Many problems in stress analysis which are of practical interest involve solids of revolution subject to axially symmetric loading. A circular cylinder loaded by a uniform internal or external pressure, circular footing resting on soil mass, pressure vessels, rotating wheels, flywheels etc. The strain-displacement relations in these type of problems are given by

x

y

xy

ux

uxvyu vy x

θ

ε

ε

ε

γ

∂=∂

=

∂=∂∂ ∂

= +∂ ∂

(1.3.33)

The two components of displacements in any plane section of the body along its axis of symmetry define completely the state of strain and therefore the state of stress. The constitutive relations are given below for such types of problems.

( )( )

( )

−−

−−

−+=

xy

y

x

xy

y

x

E

γεεε

µµµµ

µµµµµµ

µµτσσσ

θθ

221000

010101

211 (1.3.34)

24

Lecture 4: Steps in Finite Element Analysis 1.4.1 Loading Conditions There are multiple loading conditions which may be applied to a system. The load may be internal and/or external in nature. Internal stresses/forces and strains/deformations are developed due to the action of loads. Most loads are basically “Volume Loads” generated due to mass contained in a volume. Loads may arise from fluid-structure interaction effects such as hydrodynamic pressure of reservoir on dam, waves on offshore structures, wind load on buildings, pressure distribution on aircraft etc. Again, loads may be static, dynamic or quasi-static in nature. All types of static loads can be represented as:

• Point loads • Line loads • Area loads • Volume loads

The loads which are not acting on the nodal points need to be transferred to the nodes properly using finite element techniques. 1.4.2 Support Conditions In finite element analysis, support conditions need to be taken care in the stiffness matrix of the structure. For fixed support, the displacement and rotation in all the directions will be restrained and accordingly, the global stiffness matrix has to modify. If the support prevents translation only in one direction, it can be modeled as ‘roller’ or ‘link supports’. Such link supports are commonly used in finite element software to represent the actual structural state. Sometimes, the support itself undergoes translation under loadings. Such supports are called as ‘elastic support’ and are modeled with ‘spring’. Such situation arises if the structures are resting on soil. The supports may be represented in finite element modeling as:

• Point support • Line support • Area support • Volume support

1.4.3 Type of Engineering Analysis Finite element analysis consists of linear and non-linear models. On the basis of the structural system and its loadings, the appropriate type of analysis is chosen. The type of analysis to be carried out depends on the following criteria:

25

• Type of excitation (loads) • Type of structure (material and geometry) • Type of response

Considering above aspects, types of engineering analysis are decided. FEA is capable of using multiple materials within the structure such as:

• Isotropic (i.e., identical throughout) • Orthotropic (i.e., identical at 900) • General anisotropic (i.e., different throughout)

The Equilibrium Equations for different cases are as follows: 1. Linear-Static: Ku F= (1.4.1) 2. Linear-Dynamic ( ) ( ) ( ) ( )Mu t Cu t Ku t F t+ + = (1.4.2)

3. Nonlinear - Static NLKu F F+ = (1.4.3)

1. Nonlinear-Dynamic ( ) ( ) ( ) ( ) ( )NLMu t Cu t Ku t F t F t+ + + = (1.4.4)

Here, M, C, K, F and U are mass, damping, stiffness, force and displacement of the structure respectively. Table 1.4.1 shows various types of analysis which can be performed according to engineering judgment.

Table 1.4.1 Type of analysis

Excitation

Structure

Response

Basic analysis type

Static Elastic Linear Linear-Elastic-Static Analysis Static Elastic Nonlinear Nonlinear-Elastic-Static Analysis Static Inelastic Linear Linear-Inelastic-Static Analysis Static Inelastic Nonlinear Nonlinear-Inelastic-Static Analysis Dynamic Elastic Linear Linear-Elastic-Dynamic Analysis Dynamic Elastic Nonlinear Nonlinear-Elastic-Dynamic Analysis Dynamic Inelastic Linear Linear-Inelastic-Dynamic Analysis Dynamic Inelastic Nonlinear Nonlinear-Inelastic-Dynamic Analysis

26

1.4.4 Basic Steps in Finite Element Analysis The following steps are performed for finite element analysis.

1. Discretisation of the continuum: The continuum is divided into a number of elements by imaginary lines or surfaces. The interconnected elements may have different sizes and shapes.

2. Identification of variables: The elements are assumed to be connected at their intersecting points referred to as nodal points. At each node, unknown displacements are to be prescribed.

3. Choice of approximating functions: Displacement function is the starting point of the mathematical analysis. This represents the variation of the displacement within the element. The displacement function may be approximated in the form a linear function or a higher-order function. A convenient way to express it is by polynomial expressions. The shape or geometry of the element may also be approximated.

4. Formation of the element stiffness matrix: After continuum is discretised with desired element shapes, the individual element stiffness matrix is formulated. Basically it is a minimization procedure whatever may be the approach adopted. For certain elements, the form involves a great deal of sophistication. The geometry of the element is defined in reference to the global frame. Coordinate transformation must be done for elements where it is necessary.

5. Formation of overall stiffness matrix: After the element stiffness matrices in global coordinates are formed, they are assembled to form the overall stiffness matrix. The assembly is done through the nodes which are common to adjacent elements. The overall stiffness matrix is symmetric and banded.

6. Formation of the element loading matrix: The loading forms an essential parameter in any structural engineering problem. The loading inside an element is transferred at the nodal points and consistent element matrix is formed.

7. Formation of the overall loading matrix: Like the overall stiffness matrix, the element loading matrices are assembled to form the overall loading matrix. This matrix has one column per loading case and it is either a column vector or a rectangular matrix depending on the number of loading cases.

8. Incorporation of boundary conditions: The boundary restraint conditions are to be imposed in the stiffness matrix. There are various techniques available to satisfy the boundary conditions. One is the size of the stiffness matrix may be reduced or condensed in its final form. To ease computer programming aspect and to elegantly incorporate the boundary conditions, the size of overall matrix is kept the same.

9. Solution of simultaneous equations: The unknown nodal displacements are calculated by the multiplication of force vector with the inverse of stiffness matrix.

27

10. Calculation of stresses or stress-resultants: Nodal displacements are utilized for the calculation of stresses or stress-resultants. This may be done for all elements of the continuum or it may be limited to some predetermined elements. Results may also be obtained by graphical means. It may desirable to plot the contours of the deformed shape of the continuum.

The basic steps for finite element analysis are shown in the form of flow chart below:

Fig. 1.4.1 Flowchart for steps in FEA

28

1.4.5 Element Library in FEA Software A real structure can be modeled with various ways with appropriate assumptions. The structure may be divided into following categories:

• Cable or tension structures • Skeletal or framed structures • Surface or spatial structures • Solid structures • Mixed structures

The configuration of structural elements depends upon the geometry of the structural system and the number of independent space coordinates (i.e., x, y and z) required to describe the problem. Thus, the element can be categorized as one, two or three dimensional element. One dimensional element can be represented by a straight line whose ends will be nodal points. The skeletal structures are generally modeled by this type of elements. The pin jointed bar or truss element is the simplest structural element. This element undergoes only axial deformation. The beam element is another type of element which undergoes in-plane transverse displacements and rotations. The frame element is the combination of truss and beam element. Thus, the frame element has axial and in-plane transverse displacements and rotations. This element is generally used to model 1D, 2D and 3D skeletal structural systems. Two-dimensional elements are generally used to model 2D and 3D continuum. These elements are of constant thickness and material properties. The shapes of these elements are triangular or rectangular and it consists of 3 to 9 or even more nodes. These elements are used to solve many problems in solid mechanics such as plane stress, plane strain, plate bending. Three-dimensional element is the most cumbersome which is generally used to model the 3-D continuum. The elements have 6 to 27 numbers of nodes or more. Because of large degrees of freedom, the analysis is time consuming using 3-D elements and difficult to interpret its results. However, for accurate analysis of the irregular continuum, 3-D elements are useful. To analyze any real structure, appropriate elements are to be assigned for the finite element analysis. In standard FEA software, following types of element library are used to discretize the domain.

• Truss element • Beam element • Frame element • Membrane/ Plate/Shell element • Solid element • Composite element • Shear panel • Spring element

29

• Rigid/Link element • Viscous damping element

The different types of elements available in standard finite element software are shown in Fig. 1.4.2.

1D Elements (Truss, beam, grid and frame)

2D Elements (Plane stress, Plane strain, Axisymmetric, Plate and Shell)

3D Elements

Fig. 1.4.2 Various types of elements for computer modeling

Content Module: 2 Finite Element Formulation Techniques (04) Lecture 1: Virtual Work and Variational Principle (Page: 1-4)

2.1.1 Introduction 2.1.2 Principle of Virtual Work 2.1.3 Variational Principle 2.1.4 Weighted Residual Method

1

Lecture 1: Virtual Work and Variational Principle 2.1.1 Introduction Finite element formulation can be constructed from governing differential equations over a domain. This can be formulated by various ways like Virtual work method, Variational method, Weighted Residual Method etc. 2.1.2 Principle of Virtual Work The principle of virtual work is a very useful approach for solving varieties of structural mechanics problem. When the force and displacement are unrelated to the cause and effect relation, the work is called virtual work. Therefore, the virtual work may be caused by true force moving through imaginary displacements or vice versa. Thus, the principle of virtual work can be divided into two categories: (a) principle of virtual forces and (b) principle of virtual displacements. The principle of virtual forces establishes the compatibility conditions. The principle of virtual displacements establishes the conditions of equilibrium and is used in the displacement model of the finite element technique.

The external virtual work is the work done by real load moving through imaginary displacements in a structure. These loads include both the load distributed over the entire surface and volume. Thus, the virtual work done by the external force is:

x x

E y y

z z

F FW u v w F d u v w F d

F F

(2.1.1)

Where, δu, δv and δw are the components of the virtual displacements in x, y and z direction respectively. FΓx, FΓy and FΓz are the surface forces and FΩx, FΩy and FΩz are the body forces in x, y and z direction respectively. In the above equation, the integration is carried out over the entire surface in the first term and over the entire volume in the second term. The above expression can be rewritten as:

T TEW d F d d F d

(2.1.2)

Here, Td u v w . For the three dimensional stress-strain condition, there are six

components of stresses ( x y z xy yz zx, , , , , ) and six components of strains in virtual

displacement fields ( x y z xy yz zx, , , , , ). Therefore, the virtual internal work can be

expressed as follows:

2

x

y

zx y z xy yz zx

xy

yz

zx

U d

(2.1.3)

Or

TU d

(2.1.4)

According to principle of virtual work, the work done by external forces due to the virtual displacement of a structure in equilibrium is equal to the work done by the internal forces for the virtual internal displacement. Therefore, EW U Thus eqs. (2.1.2) and (2.1.4) can be made equal and can be related as follows:

T T Td F d d F d d

(2.1.5)

2.1.3 Variational Principle Variational formulation is the generalized method of formulating the element stiffness matrix and load vector using the variational principle of solid mechanics. The strain energy in a structural body is given by the relation

12

TU dε σΩ

= Ω∫∫∫ (2.1.6)

For a 3D structural problem, stress has six components: Tx y z xy yz zx, , , , , .

Similarly, there are six components of strains: Tx y z xy yz zx, , , , , . Now the strain-

displacement relationship can be expressed as [ ] B dε = , where d is the displacement vector in

x, y and z directions and [B] is called as the strain displacement relationship matrix. Again, the stress can be represented in terms of its constitutive relationship matrix: [ ] Dσ ε= . Here [ ]D

is called as the constituent relationship matrix. Using the above relationship in the strain energy equation one can arrive

[ ] [ ] 12

TU B d D B d d

Ω

= Ω ∫∫∫ (2.1.7)

Applying the variational principle one can express

3

[ ] [ ][ ] TUF B D B d dd Ω

∂= = Ω∂ ∫∫∫

(2.1.8)

Now, from the relationship of [ ] F K d= , one can arrive at the element stiffness matrix as:

[ ] [ ] [ ][ ]TK B D B dΩ

= Ω∫∫∫ (2.1.9)

Thus, by the use of variational principle, the stiffness matrix of a structural element can be obtained as expressed in the above equation. 2.1.4 Weighted Residual Method Virtual work and Variational method are applicable and adequate for most of the problems. However, in some cases functional analogous to potential energy cannot be written because of not having clear physical meaning. For some applications, such as in fluid mechanics problem, functional needed for a variational approach cannot be expressed. For some types of fluid flow problems, only differential equations and boundary conditions are available. For Such problems weighted residual method can be used for obtaining the solutions. Approximate solutions of differential equation satisfy only part of conditions of the problem. For example a differential equation may be satisfied only at few points, rather than at each. The strategy used in weighted residual method is to first take an approximate solution and then its validity is assessed. The different methods in weighted Residual Method are

• Collocation method • Least square method • Method of moment • Galerkin method

The mathematical statement of a physical problem can be defined as: In domain Ω,

Du f 0 (2.1.10) Where,

D is the differential operator u = u(x) = dependent variables such as displacement, pressure, velocity,

potential function x = independent variables such as coordinates of a point f = a function of x which may be constant or zero

If u is an approximate solution then residual in domain Ω,

R Du f (2.1.11) According to the weighted residual method, the weak form of above equation will become

4

i

i

w R d 0 for i 1,2,3,...,n

or

w Du f d 0

(2.1.12)

Where weighting function wi = wi(x) is chosen from the approximate basis function used for constructing approximated solution u .

5

Lecture 2: Galerkin Method 2.2.1 Introduction Galerkin method is the most widely used among the various weighted residual methods. Galerkin method incorporates differential equations in their weak form, i.e., before starting integration by parts it is in strong form and after by parts it will be in weak form, so that they are satisfied over a domain in an integral. Thus, in case of Galerkin method, the equations are satisfied over a domain in an integral or average sense, rather than at every point. The solution of the equations must satisfy the boundary conditions. There are two types of boundary conditions:

• Essential or kinematic boundary condition • Non essential or natural boundary condition

For example, in case of a beam problem (4

4

yEI q 0x

) differential equation is of forth order.

As a result, displacement and slope will be essential boundary condition where as moment and shear will be non-essential boundary condition. 2.2.2 Galerkin Method for 2D Elasticity Problem For a two dimensional elasticity problem, equation of equilibrium can be expressed as

xyxxF 0

x y

(2.2.1)

xy yyF 0

x y

(2.2.2)

Where, x yF and F are the body forces in X and Y direction respectively. Let assume,

x yand are surface forces in X and Y direction and as angle made by normal to surface

with X– axis (Fig. 2.2.1). Therefore, force equilibrium of element can be written as:

x x xy

x x xy x xy x xy

F PQ t OP t OQ t

OP OQF cos sin cos Cos 90PQ PQ

x x xyThus, F m (2.2.3)

Where, ℓ and m are direction cosines of normal to the surface. Similarly,

y xy yF m (2.2.4)

6

Fig. 2.2.1 Elemental stress in 2D

Adopting Galerkin’s approach

xy xy yxx xF u F v dxdy 0

x y x y

(2.2.5)

Where u and v are weighting functions i.e elemental displacements in X and Y directions respectively. Now one can expand above equation by using Green’s Theorem.

Green Theorem states that if x, y and x, y are continuous functions then their first and

second partial derivatives are also continuous. Therefore, 2 2

2 2dxdy dxdy m dsx x y y x y x y

(2.2.6)

Assuming, x ; u; 0x y

one can rewrite with the use of above relation as

xx x

uu dx dy dx dy u ds

x x

(2.2.7)

Similarly, assuming y; 0 and vx y

yy y

vv dx dy dx dy m v ds

y y

(2.2.8)

7

Again, assuming x y; v; 0x y

x yx y x y

vv dx dy dx dy v ds

y x

(2.2.9)

And assuming, x y; 0; ux y

x yx y x y

uu dx dy dx dy m u ds

y y

Putting values of eqs. (2.2.7), (2.2.8) and (2.2.9), in eq. (2.2.5), one can get the following relation:

x y xy xy

x y xy xy x y

u v v u dx dyx y x y

u m v v m u ds F u dx dy F v dx dy 0

(2.2.10) Rearranging the terms of above expression, the following relations are obtained.

x y xy xy x yu v v u dx dy F u F v dx dyx y x y

x xy xy ym uds m vds 0 (2.2.11)

Here, x yF and F are the body forces and u & v are virtual displacements in X and Y directions

respectively. Considering first term of eq. (2.2.11), virtual displacement u is given to the element of unit thickness. Dotted position in Fig. 2.2.2 shows the virtual displacement. Thus, work done by x :

x x xdy u u dx dy u u dxdyx x

(2.2.12)

Similarly, considering second term of eq. (2.2.11), virtual work done by body forces is

x yF u F v dx dy

8

Putting eqs. (2.2.3) & (2.2.4) in third term of eq. (2.2.11) we get the virtual work done by surface forces as:

x yF uds F vds

Fig. 2.2.2 Element subjected to stresses

Due to virtual displacement u , change in strain x is given by:

x

u u dx ux u

dx x

(2.2.13)

The virtual work done by x x xis . .dxdy . Similarly all the individual term in the first term of eq. (2.2.11) can be derived from eq. (2.2.13) which will be as follows:

x x xu dxdy dxdyx

y y yv dxdy dxdyy (2.2.14)

xy xy xyv u dxdyx y

Now, the work done by internal forces will be

9

x x y y xy xyU dxdy (2.2.15)

If external work done is represented by WE and U is the internal work done then,

E EU w 0 or U w (2.2.16)

Thus in elasticity problems, Galerkin’s method turns out to be the principle of virtual work, which can be stated that “A Deformable body is said to be in equilibrium, if the total work done by external forces is equal to the total work done by internal forces.” The work done above is virtual as either forces or deformations are also virtual. Thus, Galerkin’s approach can be followed in all problems involving solution of a set of equations subjected to specified boundary values. 2.2.3 Galerkin Method for 2D Fluid Flow Problem Let consider the two dimensional incompressible fluid equation which can be expressed by pressure variable only as follows.

2p 0 (2.2.17) Where p is the pressure inside the fluid domain. The above equation can be expressed in 2D form as:

2 2

2 2

ii

p p 0x y

orp, 0

(2.2.18)

Applying weighted residual method, the weak form of the above equation will become

i iiw p, d 0

(2.2.19)

Integrating by parts of the above expression, the following relation can be obtained.

i i i,i iw p, d w p, d 0

i,i i i ior w p, d w p, d

(2.2.20)

If the nodal pressure and interpolation functions are denoted by p and N respectively, then the pressure at any point inside the fluid domain can be expressed as

p N p

Similarly, the weighted function can also be written with the help of interpolation function as

w N w

10

Thus, i,ip L p L N p B p , where, Lx y

= differential operator.

Similarly, i,iw L W L N w B w

T Ti,i iThus, w p, d w B B p d

(2.2.21)

T Ti i

pw p, d = w N dn

(2.2.22)

Here, Γdenotes the surface of the fluid domain and n represents the direction normal to the surface. Thus, from eq. (2.2.20), one can write the expression as:

T T T T pThus, w B B p d w N dn

Or, G p S (2.2.23)

Where,

T T T

T

G B B d N N N N dx x y y

pand S N dn

(2.2.27)

Here, n is the direction normal to the surface. Thus, solving the above equation with the prescribed boundary conditions, one can find out the pressure distribution inside the fluid domain by the use of finite element technique.

11

Lecture 3: Finite Element Method: Displacement Approach 2.3.1 Choice of Displacement Function Displacement function is the beginning point for the structural analysis by finite element method. This function represents the variation of the displacement within the element. On the basis of the problem to be solved, the displacement function needs to be approximated in the form of either linear or higher-order function. A convenient way to express it is by the use of polynomial expressions. 2.3.1.1 Convergence criteria The convergence of the finite element solution can be achieved if the following three conditions are fulfilled by the assumed displacement function.

a. The displacement function must be continuous within the elements. This can be ensured by choosing a suitable polynomial. For example, for an n degrees of polynomial, displacement function in I dimensional problem can be chosen as:

2 3 40 1 2 3 4 ..... n

nu x x x x xα α α α α α= + + + + + + (2.3.1) b. The displacement function must be capable of rigid body displacements of the element. The

constant terms used in the polynomial (α0 to αn) ensure this condition. c. The displacement function must include the constant strains states of the element. As

element becomes infinitely small, strain should be constant in the element. Hence, the displacement function should include terms for representing constant strain states.

2.3.1.2 Compatibility Displacement should be compatible between adjacent elements. There should not be any discontinuity or overlapping while deformed. The adjacent elements must deform without causing openings, overlaps or discontinuous between the elements.

Elements which satisfy all the three convergence requirements and compatibility condition are called Compatible or Conforming elements. 2.3.1.3 Geometric invariance Displacement shape should not change with a change in local coordinate system. This can be achieved if polynomial is balanced in case all terms cannot be completed. This ‘balanced’ representation can be achieved with the help of Pascal triangle in case of two-dimensional polynomial. For example, for a polynomial having four terms, the invariance can be obtained if the following expression is selected from the Pascal triangle.

0 1 2 3u x y xyα α α α= + + + (2.3.2)

12

The geometric invariance can be ensured by the selection of the corresponding order of terms on either side of the axis of symmetry.

1

x y

x2 xy y2

x3 x2y xy2 y3

x4 x3y x2y2 xy3 y4

Fig. 2.3.1 Pascal’s Triangle

2.3.2 Shape Function In finite element analysis, the variations of displacement within an element are expressed by its nodal displacement ( i iu N u=∑ ) with the help of interpolation function since the true variation of

displacement inside the element is not known. Here, u is the displacement at any point inside the element and ui are the nodal displacements. This interpolating function is generally a polynomial with n degree which automatically provides a single-valued and continuous field. In finite element literature, this interpolation function (Ni) is referred to “Shape function” as well. For linear interpolation, n will be 1 and for quadratic interpolation n will become 2 and so on. There are two types of interpolation functions namely (i) Lagrange interpolation and (ii) Hermitian interpolation. Lagrange interpolation function is widely used in practice. Here the assumed function takes on the same values as the given function at specified points. In case of Hermitian interpolation function, the slopes of the function also take the same values as the given function at specified points. The derivation of shape function for varieties of elements will be discussed in subsequent lectures. 2.3.3 Degree of Continuity Let consider φ as an interpolation function in a piecewise fashion over finite element mesh. While such interpolation function φ can be ensured to vary smoothly within the element, the transition between adjacent elements may not be smooth. The term Cm is considered to define the continuity of a piecewise displacement. A function Cm is continuous if its derivative up to and including degree m are inter-element continuous. For example, for one dimensional problem, φ= φ(x) is C0 continuous if φ is continuous, but φ,x is not. Similarly, φ= φ(x) is C1 continuous if φ and φ,x are continuous, but φ,xx is not. In general, C0 element is used to model plane and solid body and C1 element is used to model beam, plate and shell like structure, where inter-element continuity of slope is necessary to ensure. Let assume a linear function for bar like element:

1 0 1xφ α α= + This function is C0 continuous as

13

φ1,x is discontinuous. If the interpolation function is considered as 22 0 1 2x xφ α α α= + + then

2, 1 22x xφ α α= + is also continuous but 2, 22xxφ α= is discontinuous. As a result, this function φ2 will

become C1 continuous. 2.3.4 Isoparametric Elements If the shape functions (Ni) used to represent the variation of geometry of the element are the same as the shape functions (N´i) used to represent the variation of the displacement then the elements are called isoparametric elements. For example, the coordinates (x,y) inside the element are defined by the shape functions (Ni) and displacement (u,v) inside the element are defined by the shape functions (N´i) as below.

i i i i

i i i i

x N x u N u

y N y v N v

(2.3.3)

If Ni = N´i, then the element is called isroparametric. Fig. 2.3.2(a) shows the two dimensional 8 node isoparametric element. If the geometry of element is defined by shape functions of order higher than that for representing the variation of displacements, then the elements are called superparametric (Fig. 2.3.2(b)). If the geometry of element is defined by shape functions of order lower than that for representing the variation of displacements then the elements are called subparametric (Fig. 2.3.2(c)).

Fig. 2.3.2 Shape functions for geometry and displacements

14

2.3.5 Various Elements Selection of the order of the polynomial depends on the type of elements. For example, in case of one dimensional element having single degrees of freedom with two nodes, the displacement function can be chosen as 0 1u xα α= + . However, if the same has two degrees of freedom at each

node, then the chosen displacement function should be 2 30 1 2 3u x x xα α α α= + + + . Various types of

elements used in finite element analysis are given below: 1. One dimensional elements.

(a) Two node element (b) Three node element

Fig. 2.3.3 One dimensional elements

2. Two dimensional elements (a) Triangular element (b) Rectangular element (c) Quadrilateral element (d) Quadrilateral formed by two triangles (e) Quadrilateral formed by four triangles Few of the elements with number of nodes are shown in Fig. 2.3.4.

15

Fig. 2.3.4 Two dimensional elements

3. Three dimensional elements. (a) Tetrahedron

(b) Rectangular brick element Few of the three dimensional solid elements are shown in Fig. 2.3.5.

16

Fig. 2.3.5 Three dimensional elements

17

Lecture 4: Stiffness Matrix and Boundary Conditions 2.4.1 Element Stiffness Matrix The stiffness matrix of a structural system can be derived by various methods like variational principle, Galerkin method etc. The derivation of an element stiffness matrix has already been discussed in earlier lecture. The stiffness matrix is an inherent property of the structure. Element stiffness is obtained with respect to its axes and then transformed this stiffness to structure axes. The properties of stiffness matrix are as follows:

• Stiffness matrix is symmetric and square. • In stiffness matrix, all diagonal elements are positive. • Stiffness matrix is be positive definite

2.4.2 Global Stiffness Matrix A structural system is an assemblage of number of elements. These elements are interconnected together to form the whole structure. Therefore, the element stiffness of all the elements are first need to be calculated and then assembled together in systematic manner. It may be noted that the stiffness at a joint is obtained by adding the stiffness of all elements meeting at that joint. To start with, the degrees of freedom of the structure are numbered first. This numbering will start from 1 to n where n is the total degrees of freedom. These numberings are referred to as degrees of freedom corresponding to global degrees of freedom. The element stiffness matrix of each element should be placed in their proper position in the overall stiffness matrix. The following steps may be performed to calculate the global stiffness matrix of the whole structure.

a. Initialize global stiffness matrix [ ]K as zero. The size of global stiffness matrix will be equal to the total degrees of freedom of the structure.

b. Compute individual element properties and calculate local stiffness matrix [ ]k of that

element. c. Add local stiffness matrix[ ]k to global stiffness matrix [ ]K using proper locations

d. Repeat the Step b. and c. till all local stiffness matrices are placed globally. The steps to be followed in the computer program are shown in the form of flow chart in Fig. 2.4.1 for assembling the local stiffness matrix to global stiffness matrix.

18

Fig. 2.4.1 Assemble of stiffness matrix from local to global

19

2.4.3 Boundary Conditions Under this section, procedure to include the effect of boundary condition in the stiffness matrix for the finite element analysis will be discussed. The solution cannot be obtained unless support conditions are included in the stiffness matrix. This is because, if all the nodes of the structure are included in displacement vector, the stiffness matrix becomes singular and cannot be solved if the structure is not supported amply, and it cannot resist the applied loads. A solution cannot be achieved until the boundary conditions i.e., the known displacements are introduced.

In finite element analysis, the partitioning of the global matrix is carried out in a systematic way for the hand calculation as well as for the development of computer codes. In partitioning, normally the equilibrium equations can be partitioned by rearranging corresponding rows and columns, so that prescribed displacements are grouped together. For example, let consider the equation of equilibrium is expressed in compact form as:

[ ] F K d= (2.4.1)

Where, [K] is the global stiffness matrix, d is the displacement vector consisting of global degrees of freedom, and F is the load vector corresponding to degrees of freedom. By the method of partitioning the above equation can be partitioned in the following manner.

[ ]

K KF d

F dK Kαα α βα α

β ββα ββ

=

(2.4.2)

Where, subscripts α refers to the displacements free to move and β refers to the prescribed support displacements. As the prescribed displacements dβ are known, eq. (2.4.2) may be written in expanded form as: [ ] F = K d + K dα αα α αβ β (2.4.3)

Thus it is possible to obtain the free displacement of the structure dα as

[ ] -1d = K F - K dα αα α αβ β (2.4.4)

If the displacements at supports dβ are zero, then the above equation can be simplified to the following expression.

[ ] -1d = K Fα αα α (2.4.5)

Thus, by rearranging assembled matrix, the portion corresponding to the unknown displacements in eq. (2.4.4) can be taken out for the solution purpose. This is possible as the known displacements dβ are restrained, i.e., displacements are zero. If the support has some known displacements, then eq. (2.4.4) can be used to find the solution. If the few supports of the structures yield, then the above method may be modified by partitioning the stiffness matrix into three parts as shown below:

20

K K KF d

F K K K d

F dK K K

(2.4.6)

Here, α refers to unknown displacement; β refers to known displacement (≠0) and γ refers to zero displacement. Thus, the above equation can be separated and solved independently to find required unknown results as shown below.

1

F K d K d K d

or, K d F K d as d 0

Thus, d K F K d

(2.4.7)

For computer programming, several techniques are available for handling boundary conditions. One of the approaches is to make the diagonal element of stiffness matrix corresponding to zero displacement as unity and corresponding all off-diagonal elements as zero. For example, let consider a 3×3 stiffness matrix with following force-displacement relationship.

1 11 12 13 1

2 21 22 23 2

3 31 32 33 3

F k k k dF k k k dF k k k d

= (2.4.8)

Now, if the third node has zero displacement (i.e., d3= 0) then the matrix will be modified as follows to incorporate the boundary condition.

1 11 12 1

2 21 22 2

3

00

0 0 0 1

F k k dF k k d

d

= (2.4.9)

Thus, while inverting whole matrix, d3 will become zero automatically.

To incorporate known support displacement in computer programming following procedure may be adopted. Considering the displacement d2 has known value of δ, 1st row of eq. (2.4.8) can be written as:

1 11 1 12 2 13 3F k d k d k d= × + × + × (2.4.10)

Or

1 12 11 1 13 3F k k d k dδ− × = × + × (2.4.11)

Now the 2nd row of eq. (2.4.8) has to become:

21

2dδ = (2.4.12)

Similarly 3rd row will be:

3 32 31 1 33 3F k k d k dδ− × = × + × (2.4.13) Thus above three equations can be written in a combined form as

1 12 11 13 1

2

32 31 33 3

00 1 0

0

F k k k dd

F k k d

δδδ

− = − (2.4.14)

Another approach may also be followed to take care the known restrained displacements by assigning a higher value δ (say δ =1020) in the diagonal element corresponding to that displacement.

1 11 12 13 120 20

22 21 22 23 2

3 31 32 33 3

10 10F k k k d

k k k k dF k k k d

δ × × = × (2.4.15)

20 2022 21 1 22 2 23 310 k k d k 10 d k d

As d3 is corresponding to zero displacement, the above equation can be simplified to the following. 20 20

22 21 1 22 220 20

22 22 2

2

10 k k d k 10 d

or 10 k k 10 dd known displacement is ensured

If the overall stiffness matrix is to be formed in half band form then the numbering of nodes should be such that the bandwidth is minimum. For this the labels are put in a systematic manner irrespective of whether the joint displacements are unknowns or restraints. However, if the unknown displacements are labeled first then the matrix operations can be restricted up to unknown displacement labels and beyond that the overall stiffness matrix may be ignored.

Content Module:3 Element Properties (09)

Lecture 1: Natural Coordinates (Page:1-8) 3.1.2 Two Dimensional Triangular Elements

3.1.3 Shape Function using Area Coordinates

Lecture 2: Triangular Elements (Page:9-17) 3.2.1 Shape function using Cartesian coordinates

3.2.2 Higher Order Triangular Elements

3.2.2.1 Shape function for six node element

3.2.3 Construction of Shape Function by Degrading Technique

3.2.3.1 Five node triangular element

Lecture 3: Rectangular Elements (Page:18-24) 3.3.1 Shape Function for Four Node Element

3.3.1.1 Shape function using Cartesian coordinates

3.3.1.2 Shape Function using natural coordinates

3.3.2 Shape Function for Eight Node Element

Lecture 4: Lagrange and Serendipity Elements (Page:25-33) 3.4.1 Lagrange Interpolation Function

3.4.1.1 Shape function for two node bar element

3.4.1.2 Shape function for three node bar element

3.4.1.3 Shape function for two dimensional elements

3.4.2 Serendipity Elements

Lecture 5: Solid Elements (Page:34-39) 3.5.1 Tetrahedral Elements

3.5.2 Brick Elements

Lecture 6: Isoparametric Formulation (Page:40-47) 3.6.1 Necessity of Isoparametric Formulation

3.6.2 Coordinate Transformation

3.6.3 Concept of Jacobian Matrix

Lecture 7: Stiffness Matrix of Isoparametric Elements(Page:48-55) 3.7.1 Evaluation of Stiffness Matrix of 2-D Isoparametric Elements

3.7.2 Evaluation of Stiffness Matrix of 3-D Isoparametric Elements

Lecture 8: Numerical Integration: One Dimensional(Page:56-60) 3.8.1 Gauss Quadrature for One-Dimensional Integrals

3.8.2 One- Point Formula

3.8.3 Two-Point Formula

Lecture 9: Numerical Integration: Two and Three

Dimensional(Page:61-67)

3.9.1 Gauss Quadrature for Two-Dimensional Integrals

3.9.2 Gauss Quadrature for Three-Dimensional Integrals

3.9.3 Numerical Integration of Element Stiffness Matrix

3.9.4 Gauss Quadrature for Triangular Elements

3.9.5 Gauss Quadrature for Tetrahedron

Worked out Examples (Page:68-74) Example 3.1 Calculation of displacement using area coordinates

Example 3.2 Derivation of shape function of four node triangular element

Example 3.3 Numerical integration for three dimensional problems

1

Lecture 1: Natural Coordinates Natural coordinate system is basically a local coordinate system which allows the specification of a point within the element by a set of dimensionless numbers whose magnitude never exceeds unity. This coordinate system is found to be very effective in formulating the element properties in finite element formulation. This system is defined in such that the magnitude at nodal points will have unity or zero or a convenient set of fractions. It also facilitates the integration to calculate element stiffness. 3.1.1 One Dimensional Line Elements The line elements are used to represent spring, truss, beam like members for the finite element analysis purpose. Such elements are quite useful in analyzing truss, cable and frame structures. Such structures tend to be well defined in terms of the number and type of elements used. For example, to represent a truss member, a two node linear element is sufficient to get accurate results. However, three node line elements will be more suitable in case of analysis of cable structure to capture the nonlinear effects. The natural coordinate system for one dimensional line element with two nodes is shown in Fig. 3.1.1. Here, the natural coordinates of any point P can be defined as follows.

1 21 x xN and Nl l

(3.1.1)

Where, x is represented in Cartesian coordinate system. Similarly, x/l can be represented as ξ in natural coordinate system. Thus the above expression can be rewritten in the form of natural coordinate system as given below.

1 21N and Nx x (3.1.2)

Now, the relationship between natural and Cartesian coordinates can be expressed from eq. (3.1.1) as

1

2

1 1 10

NNx l

(3.1.3)

Here, N1 and N2 is termed as shape function as well. The variation of the magnitude of two linear shape functions (N1 and N2) over the length of bar element are shown in Fig. 3.1.2. This example displays the simplest form of interpolation function. The linear interpolation used for field variable φ can be written as

1 1 2 2N Nf x f f (3.1.4)

2

Fig. 3.1.1 Two node line element

Fig. 3.1.2 Linear interpolation function for two node line element Similarly, for three node line element, the shape function can be derived with the help of natural coordinate system which may be expressed as follows:

3

2

22

1 22

2 22

23

2

3 211 3 2

4 4 4 422

x xl lN

x xN Nl l

Nx xl l

x x x x x x

(3.1.5)

The detailed derivation of the interpolation function will be discussed in subsequent lecture. The variation of the shape functions over the length of the three node element are shown in Fig. 3.1.3

Fig. 3.1.3 Variation of interpolation function for three node line element

Now, if φ is considered to be a function of L1 and L2, the differentiation of φ with respect to x for two node line element can be expressed by the chain rule formula as

1 2

1 2

d L L. .dx L x L xf f f

(3.1.6)

Thus, eq.(3.1.4) can be written as

1 21 1L Landx l x l

(3.1.7)

Therefore,

4

2 1

1ddx l L L

(3.1.7)

The integration over the length l in natural coordinate system can be expressed by

1 2! !

1 !p q

l

p qL L dl lp q

(3.1.9)

Here, p! is the factorial product p(p-1)(p-2)….(1) and 0! is defined as equal to unity. 3.1.2 Two Dimensional Triangular Elements The natural coordinate system for a triangular element is generally called as triangular coordinate system. The coordinate of any point P inside the triangle is x,y in Cartesian coordinate system. Here, three coordinates, L1, L2 and L3 can be used to define the location of the point in terms of natural coordinate system. The point P can be defined by the following set of area coordinates:

𝐿1 = 𝐴1𝐴

; 𝐿2 = 𝐴2𝐴

; 𝐿3 = 𝐴3𝐴

(3.1.10)

Where, 𝐴1= Area of the triangle P23 𝐴2= Area of the triangle P13 𝐴3= Area of the triangle P12

A=Area of the triangle 123 Thus,

𝐴 = 𝐴1 + 𝐴2 + 𝐴3 and

𝐿1 + 𝐿2 + 𝐿3 = 1 (3.1.11) Therefore, the natural coordinate of three nodes will be: node 1 (1,0,0); node 2 (0,1,0); and node 3 (0,0,1).

5

Fig. 3.1.4 Triangular coordinate system

The area of the triangles can be written using Cartesian coordinates considering x, y as coordinates of an arbitrary point P inside or on the boundaries of the element:

A = 121 𝑥1 𝑦11 𝑥2 𝑦21 𝑥3 𝑦3

A1 = 12

1 𝑥 𝑦1 𝑥2 𝑦21 𝑥3 𝑦3

A2 = 12

1 𝑥 𝑦1 𝑥3 𝑦31 𝑥1 𝑦1

A3 = 12

1 𝑥 𝑦1 𝑥1 𝑦11 𝑥2 𝑦2

The relation between two coordinate systems to define point P can be established by their nodal coordinates as

1𝑥𝑦 =

1 1 1𝑥1 𝑥2 𝑥3𝑦1 𝑦2 𝑦3

𝐿1𝐿2𝐿3 (3.1.12)

Where, 𝑥 = 𝐿1𝑥1 + 𝐿2𝑥2 + 𝐿3𝑥3

6

𝑦 = 𝐿1𝑦1 + 𝐿2𝑦2 + 𝐿3𝑦3 The inverse between natural and Cartesian coordinates from eq.(3.1.12) may be expressed as

𝐿1𝐿2𝐿3 = 1

2𝐴𝑥2𝑦3−𝑥3𝑦2 𝑦2 − 𝑦3 𝑥3 − 𝑥2𝑥3𝑦1−𝑥1𝑦3 𝑦3 − 𝑦1 𝑥1 − 𝑥3𝑥1𝑦2−𝑥2𝑦1 𝑦1 − 𝑦2 𝑥2 − 𝑥1

1𝑥𝑦 (3.1.13)

The derivatives with respect to global coordinates are necessary to determine the properties of an element. The relationship between two coordinate systems may be computed by using the chain rule of partial differentiation as

1 2 3

1 2 3

1 2 3

1 2 3

3i

i 1 i

L L L. . .x L x L x L x

b b b. . .2A L 2A L 2A L

b .2A L

(3.1.14)

Where, b1 = y2 – y3; b2 = y3 – y1 and b3 = y1 – y2. Similarly, following relation can be obtained. 3

i

i 1 i

c .y 2A L

(3.1.15)

Where, c1 = x3 – x2; c2 = x1 – x3 and c3 = x2 – x1. The above expressions are looked cumbersome. However, the main advantage is the ease with which polynomial terms can be integrated using following area integral expression.

p q r

1 2 3A

p!q!r!L L L dA 2Ap q r 2 !

(3.1.16)

Where 0! is defined as unity. 3.1.3 Shape Function using Area Coordinates The interpolation functions for the triangular element are algebraically complex if expressed in Cartesian coordinates. Moreover, the integration required to obtain the element stiffness matrix is quite cumbersome. This will be discussed in details in next lecture. The interpolation function and subsequently the required integration can be obtained in a simplified manner by the concept of area coordinates. Considering a linear displacement variation of a triangular element as shown in Fig. 3.1.5, the displacement at any point can be written in terms of its area coordinates.

7

1 1 2 2 3 3u L L L

Or, Tu f (3.1.17)

where, T T1 2 3 1 2 3L L L andf

And 𝐿1 = 𝐴1𝐴

; 𝐿2 = 𝐴2𝐴

; 𝐿3 = 𝐴3𝐴

(3.1.18)

Here, A is the total area of the triangle. It is important to note that the area coordinates are dependent as 𝐿1 + 𝐿2 + 𝐿3 = 1 . It may be seen from figure that at node 1, L1 = 1 while L2 = L3 = 0. Similarly for other two nodes: at node 2, L2 = 1 while L1 = L3 = 0, and L3 = 1 while L2 = L1 = 0. Now, substituting the area coordinates for node 1, 2 and 3, the displacement components at nodes can be written as

1

i 2

3

u 1 0 0u u 0 1 0

u 0 0 1

(3. 1.19)

Thus, from the above expression, one can obtain the unknown coefficient :

1

2

3

1 0 0 u0 1 0 u0 0 1 u

(3. 1.20)

Fig. 3.1.5 Area coordinates for triangular element

Now, eq.(3.1.17) can be written as:

8

1

T T2 i

3

1 0 0 u 1 0 0u 0 1 0 u 0 1 0 u

0 0 1 u 0 0 1

f f

(3. 1.21)

The above expression can be written in terms of interpolation function as Tiu N u

Where,

T1 2 3 1 2 3

1 0 0N L L L 0 1 0 L L L

0 0 1

(3. 1.22)

Similarly, the displacement variation v in Y direction can be expressed as follows.

Tiv N v (3.1.23)

Thus, for two displacement components u and v of any point inside the element can be written as:

T Ti

T Ti

N 0 uud = =

vv 0 N

(3.1.24)

Thus, the shape function of the element will become

1 2 3

1 2 3

L L L 0 0 0N

0 0 0 L L L

(3.1.25)

It is important to note that the shape function Ni become unity at node i and zero at other nodes of the element. The displacement at any point of the element can be expressed in terms of its nodal displacement and the interpolation function as given below.

1 1 2 2 3 3

1 1 2 2 3 3

u = N u + N u + N uv = N v + N v + N v

(3.1.26)

9

Lecture 2: Triangular Elements The triangular element can be used to represent the arbitrary geometry much easily. On the other hand, rectangular elements, in general, are of limited use as they are not well suited for representing curved boundaries. However, an assemblage of rectangular and triangular element with triangular elements near the boundary can be very effective (Fig. 3.2.1). Triangular elements may also be used in 3-dimensional axi-symmetric problems, plates and shell structures. The shape function for triangular elements (linear, quadratic and cubic) with various nodes (Fig. 3.2.2) can be formulated. An internal node will exist for cubic element as seen in Fig. 3.2.2(c).

Fig. 3.2.1 Finite element mesh consisting of triangular and rectangular element

Fig. 3.2.2 Triangular elements

In displacement formulation, it is very important to approximate the variation of displacement in the element by suitable function. The interpolation function can be derived either using the Cartesian coordinate system or by the area coordinates.

10

3.2.1 Shape function using Cartesian coordinates Polynomials are easiest way of mathematical operation for expressing variation of displacement. For example, the displacement variation within the element can be represented by the following function in case of two dimensional plane stress/strain problems.

u = α0 + α1x + α2y (3.2.1)

v = α3 + α4x + α5y (3.2.2) where α0, α1, α2 ….. are unknown coefficients. Thus the displacement vectors at any point P, in the element (Fig.3.2.3) can be expressed with the following relation.

𝑑 = 𝑢𝑣 = 1 𝑥 𝑦 0 0 00 0 0 1 𝑥 𝑦

⎩⎪⎨

⎪⎧𝛼0𝛼1𝛼2𝛼3𝛼4𝛼5⎭⎪⎬

⎪⎫

(3.2.3)

Or, d=[ 𝜙] α (3.2.4)

Fig. 3.2.3 Triangular element in Cartesian Coordinates

Similarly, for “m” node element having three degrees of freedom at each node, the displacement function can be expressed as 𝑢 = 𝛼0 + 𝛼1𝑥 + 𝛼2𝑦 + 𝛼3𝑥2 + 𝛼4𝑥𝑦 + 𝛼5𝑦2+ . . … . . . +𝛼𝑚−1𝑦𝑛 𝑣 = 𝛼𝑚 + 𝛼𝑚+1𝑥 + 𝛼𝑚+2𝑦 + 𝛼𝑚+3𝑥2 + 𝛼𝑚+4𝑥𝑦+. . . . . +𝛼2𝑚−1𝑦𝑛 (3.2.5) 𝑤 = 𝛼2𝑚 + 𝛼2𝑚+1𝑥 + 𝛼2𝑚+2𝑦 + 𝛼2𝑚+3𝑥2 + 𝛼2𝑚+4𝑥𝑦+. . . . . +𝛼3𝑚−1𝑦𝑛

11

Hence, in such case,

𝑑 = 𝑢𝑣𝑤 =

ϕ𝑇 0 00 ϕ𝑇 00 0 ϕ𝑇

𝛼 (3.2.6)

Where, 𝛼𝑇 = [𝛼0 𝛼1 . . .𝛼3𝑚−1] and, [𝜙]𝑇 = [1 𝑥 𝑦 𝑥2 𝑥𝑦 . . . . 𝑦𝑛] Now, for a linear triangular element with 2 degrees of freedom, eq. (3.2.3) can be written in terms of the nodal displacements as follows.

𝑑 =

⎩⎪⎨

⎪⎧𝑢1𝑢2𝑢3𝑣1𝑣2𝑣3⎭⎪⎬

⎪⎫

=

⎣⎢⎢⎢⎢⎡1 𝑥1 𝑦1 0 0 01 𝑥2 𝑦2 0 0 01 𝑥3 𝑦3 0 0 00 0 0 1 𝑥1 𝑦10 0 0 1 𝑥2 𝑦20 0 0 1 𝑥3 𝑦3⎦

⎥⎥⎥⎥⎤

⎩⎪⎨

⎪⎧𝛼0𝛼1𝛼2𝛼3𝛼4𝛼5⎭⎪⎬

⎪⎫

(3.2.7)

Where, d is the nodal displacements. To simplify the above expression for finding out the shape function, the displacements in X direction can be separated out which will be as follows:

𝑢𝑖=𝑢1𝑢2𝑢3 =

1 𝑥1 𝑦11 𝑥2 𝑦21 𝑥3 𝑦3

∝0∝1∝2

(3.2.8)

To obtain the polynomial coefficients, α the matrix of the above equation are to be inverted. Thus, 1

1 10 1 12 3 3 2 3 1 1 3 1 2 2 1

2 21 2 2 3 3 1 21 2

3 3 3 2 1 3 2 12 3 3

x y u u1 x y x y x y x y x y x y1x y1 u y y y y uy y

2Ax y1 x x x x x xu u

131 2

31 2 2

31 2 3

uaa a1 b b b u

2Acc c u

(3.2.9)

Where, A is the area of the triangle and can be obtained as follows.

A = 121 𝑥1 𝑦11 𝑥2 𝑦21 𝑥3 𝑦3

(3.2.10)

Now, eq. (3.2.1) can be written from the above polynomial coefficients.

12

2 3 3 2 2 3 3 2 1

3 1 1 3 3 1 1 3 2

1 2 2 1 1 2 2 1 2

1u x y x y y y x x x y u2A

1 x y x y y y x x x y u2A1 x y x y y y x x x y u

2A

(3.2.11)

Thus, the interpolation function can be obtained from the above as:

2 3 3 2 2 3 3 2

1

2 3 1 1 3 3 1 1 3

3

1 2 2 1 1 2 2 1

1 x y x y y y x x x y2AN

1N N x y x y y y x x x y2A

N 1 x y x y y y x x x y2A

(3.2.12)

Such three node triangular element is commonly known as constant strain triangle (CST) as its strain is assumed to be constant inside the element. This property may be derived from eq. (3.2.1) and eq.(3.2.2). For example, in case of 2-D plane stress/strain problem, one can express the strain inside the triangle with the help of eq.(3.2.1) and eq.(3.2.2):

0 1 2x 1

3 4 5y 5

xy 2 2

u ( x y)x xv ( x y)y yv ux y

(3.2.13)

CST is the simplest element to develop mathematically. As there is no variation of strain inside the element, the mesh size of the triangular element should be small enough to get correct results. This element produces constant temperature gradients ensuring constant heat flow within the element for heat transfer problems. 3.2.2 Higher Order Triangular Elements Higher order elements are useful if the boundary of the geometry is curve in nature. For curved case, higher order triangular element can be suited effectively while generating the finite element mesh. Moreover, in case of flexural action in the member, higher order elements can produce more accurate results compare to those using linear elements. Various types of higher order triangular

13

elements are used in practice. However, most commonly used triangular element is the six node element for which development of shape functions are explained below. 3.2.2.1 Shape function for six node element Fig. 3.2.4 shows a triangular element with six nodes. The additional three nodes (4, 5, and 6) are situated at the midpoints of the sides of the element. A complete polynomial representation of the field variable can be expressed with the help of Pascal triangle:

( ) 2 20 1 2 3 4 5,x y x y x xy yφ α α α α α α= + + + + + (3.2.14)

Fig. 3.2.4 (a) Six node triangular element (b) Lines of constant values of the area coordinates Using the above field variable function, one can reach the following expression using interpolation function and the nodal values.

6

i ii 1

x, y N x, y

f f (3.2.15)

Here, the every shape function must be such that its value will be unity if evaluated at its related node and zero if evaluated at any of the other five nodes. Moreover, as the field variable representation is quadratic, each interpolation function will also become quadratic. Fig. 3.2.4(a) shows the six node element with node numbering convention along with the area coordinates at three corners. The six node element with lines of constant values of the area coordinates passing through the nodes is shown in Fig. 3.2.4(b). Now the interpolation functions can be constructed with the help of area coordinates from the above diagram. For example, the interpolation function N1 should be unity at node 1 and zero at all other five nodes. According to the above diagram, the value of L1 is 1

14

at node 1 and ½ at node 4 and 6. Again, L1 will be 0 at nodes 2, 3 and 5. To satisfy all these conditions, one can propose following expression:

1 1 1 2 3 1 11N x, y N L ,L ,L L L2

(3.2.16)

Evaluating the above expression, the value of N1 is becoming ½ at node 1 though it must become unity. Therefore, the above expression is slightly modified satisfying all the conditions and will be as follows:

1 1 1 1 11N 2L L L 2L 12

(3.2.17)

Eq. (3.2.17) assures the required conditions at all the six nodes and is a quadratic function, as L1 is a linear function of x and y. The remaining five interpolation functions can also be obtained in similar fashion applying the required nodal conditions. Thus, the shape function for the six node triangle element can be written as given below.

1 1 1

2 2 2

3 3 3

4 1 2

5 2 3

6 3 1

N L 2L 1

N L 2L 1

N L 2L 1N 4L LN 4L LN 4L L

(3.2.18)

Such six node triangular element is commonly known as linear strain triangle (LST) as its strain is assumed to vary linearly inside the element. In case of 2-D plane stress/strain problem, the element displacement field for such quadratic triangle may be expressed as

( )( )

2 20 1 2 3 4 5

2 26 7 8 9 10 11

,

,

u x y x y x xy y

v x y x y x xy y

α α α α α α

α α α α α α

= + + + + +

= + + + + + (3.2.19)

So the element strain can be derived from the above displacement field as follows.

x 1 3 4

y 8 10 11

xy 2 4 5 7 9 10

u 2 x yxv x 2 yyv u x 2 y 2 x yx y

(3.2.20)

The above expression shows that the strain components are linearly varying inside the element. Therefore, this six node element is called linear strain triangle. The main advantage of this element is that it can capture the variation of strains and therefore stresses of the element.

15

3.2.3 Construction of Shape Function by Degrading Technique Sometimes, the geometry of the structure or its loading and boundary conditions are such that the stresses developed in few locations are quite high. On the other hand, variations of stresses are less in some areas and as a result, refinement of finite element mesh is not necessary. It would be economical in terms of computation if higher order elements are chosen where stress concentration is high and lower order elements at area away from the critical area. Fig. 3.2.5 shows graphical representations where various order of triangular elements are used for generating a finite element mesh.

Fig. 3.2.5 Triangular elements with different number of nodes Fig. 3.2.5 contains four types of element. Type 1 has only three nodes, type 2 element has five nodes, type 3 has four nodes and type 4 has six nodes. The shape function for 3-node and 6-node triangular elements has already been derived. The shape functions of 6-node element can suitably be degraded to derive shape functions of other two types of triangular elements. 3.2.3.1 Five node triangular element Let consider a six node triangular element as shown in Fig. 3.2.6(a) whose shape functions and nodal displacements are (N1, N2, N3, N4, N5, N6) and (u1, u2, u3, u4, u5, u6) respectively. Similarly, for a five node triangular element as shown in Fig. 3.2.6(b), the shape functions and nodal displacements are considered as (N’1, N’2, N’3, N’4, N’5) and (u’1, u’2, u’3, u’4, u’5) respectively. Thus, the displacement at any point in a six node triangular element will become

16

1 1 2 2 3 3 4 4 5 5 6 6u N u N u N u N u N u N u (3.2.21)

Where, N1, N2, …, N6 are the shape functions and is given in eq.(3.2.18). If there is no node between 2 and 3, the displacement along line 2-3 is considered to vary linearly. Thus the displacement at an assumed node 5´ may be written as

2 35

u uu '2 (3.2.22)

Substituting, the value of u’5 for u5 in eq.(3.2.21) the following expression will be obtained.

2 31 1 2 2 3 3 4 4 5 6 5

u uu N u N u N u N u N N u2

(3.2.23)

Fig. 3.2.6 Degrading for five node element

Thus, the displacement function can be expressed by five nodal displacements as:

5 51 1 2 2 3 3 4 4 6 5

N Nu N u N u N u N u N u2 2

(3.2.24)

However, the displacement function for the five node triangular element can be expressed as

1 1 2 2 3 3 4 4 5 5u N u N u N u N u N u (3.2.25)

Comparing eq.(3.2.24) and eq.(3.2.25) and observing node 6 of six node triangle corresponds to node 5 of five node triangle, we can write

5 51 1, 2 2 3 3 4 4 5 5

N NN ' N N N , N N , N N an dN N2 2

(3.2.26)

Hence, the shape function of a five node triangular element will be

17

1 1 1

2 32 2 2 2 1

5 2 33 3 3 3 3 1

4 1 2

5 3 1

N L 2L 14L LN L 2L 1 L 1 2L

2N 4L LN N L 2L 1 L 1 2L2 2

N 4L L

N 4L L

(3.2.27)

Thus, for a five node triangular element, the above shape function can be used for finite element analysis.

18

Lecture 3: Rectangular Elements Rectangular elements are suitable for modelling regular geometries. Sometimes, it is used along with triangular elements to represent an arbitrary geometry. The simplest element in the rectangular family is the four node rectangle with sides parallel to x and y axis. Fig. 3.3.1 shows rectangular elements with varying nodes representing linear, quadratic and cubic variation of function.

Fig. 3.3.1 Rectangular elements

3.3.1 Shape Function for Four Node Element Shape functions of a rectangular element can be derived using both Cartesian and natural coordinate systems. A four term polynomial expression for the field variable will be required for a rectangular element with four nodes having four degrees of freedom. Since there is no complete four term polynomial in two dimensions, the incomplete, symmetric expression from the Pascal’s triangle may be chosen to ensure geometric isotropy. 3.3.1.1 Shape function using Cartesian coordinates For the derivation of interpolation function, the sides of the rectangular element (Fig. 3.3.2) are assumed to be parallel to the global Cartesian axes. From the Pascal’s triangle, a linear variation may be assumed to define filed variable to ensure inter-element continuity.

0 1 2 3,x y x y xy f (3.3.1)

19

Fig. 3.3.2 Rectangular element in Cartesian coordinate

Applying nodal conditions, the above expression may be written in matrix form as

01 1 1 1 1

12 2 2 2 2

23 3 3 3 3

34 4 4 4 4

1111

x y x yx y x yx y x yx y x y

ffff

(3.3.2)

The unknown polynomial coefficients may be obtained from the above equation with the use of nodal field variables.

10 1 1 1 1 1

1 2 2 2 2 2

2 3 3 3 3 3

3 4 4 4 4 4

1111

x y x yx y x yx y x yx y x y

f f f f

(3.3.3)

Thus, the field variable at any point inside the element can be described in terms of nodal values as

10 1 1 1 1 1

1 2 2 2 2 2

2 3 3 3 3 3

3 4 4 4 4 4

1

21 2 3 4

3

4

11

, 1 111

x y x yx y x y

x y x y xy x y xyx y x yx y x y

N N N N

f f

f f f

ffff

(3.3.4)

From the above expression, the shape function Ni can be derived and will be as follows.

20

2 41

1 2 1 4

312

2 1 2 3

4 23

3 4 3 2

3 14

4 3 4 1

x x y yNx x y y

y yx xNx x y y

x x y yNx x y y

x x y yNx x y y

− −= − −

−−= − − − −

= − − − −

= − −

(3.3.5)

Now, substituting the nodal coordinates in terms of (x1, y1) as (–a, –b) at node 1; (x2, y2) as (a, –b) at node 2; (x3, y3) as (a, b) at node 3 and (x4, y4) as (–a, b) at node 4 the above expression can be re-written as:

( )( )

( )( )

( )( )

( )( )

1

2

3

4

14

141

41

4

N x a y bab

N x a y bab

N x a y bab

N x a y bab

= − −

= + −

= + +

= − +

(3.3.6)

Thus, the shape function N can be found from the above expression in Cartesian coordinate system. 3.3.1.2 Shape function using natural coordinates The derivation of interpolation function in terms of Cartesian coordinate system is algebraically complex as seen from earlier section. However, the complexity can be reduced by the use of natural coordinate system, where the natural coordinates will vary from -1 to +1 in place of –a to +a or –b to +b. The transformation of Cartesian coordinates to Natural coordinates are shown in Fig. 3.3.3.

21

Fig. 3.3.3 Four node rectangular element

From the figure, the relation between two coordinate systems can be expressed as

x x y yanda b

ξ η− −= = (3.3.7)

Here, 2a and 2b are the width and height of the rectangle. The coordinate of the center of the rectangle can be written as follows:

1 2 1 4

2 2x x y yx and y+ +

= = (3.3.8)

Thus, from eq. (3.3.7) and eq.(3.3.8), the nodal values in natural coordinate systems can be derived which is shown in Fig. 3.3.4(b). With the above relations variations of &x h will be from -1 to +1.

Now the interpolation function can be derived in a similar fashion as done in section 3.3.1.1. The filed variable can be written in natural coordinate system ensuring inter-element continuity as:

0 1 2 3,f x h x h xh (3.3.9)

The coordinates of four nodes of the element in two different systems are shown in Table 3.3.1 for ready reference for the derivation purpose. Applying the nodal values in the above expression one can get

01

12

23

34

1 1 1 11 1 1 11 1 1 11 1 1 1

ffff

(3.3.10)

22

Table 3.3.1 Cartesian and natural coordinates for four node element

Thus, the unknown polynomial coefficients can be found as

10 1 1

1 2 2

2 3 3

3 4 4

1 1 1 1 1 1 1 11 1 1 1 1 1 1 111 1 1 1 1 1 1 141 1 1 1 1 1 1 1

f f f f f f f f

(3.3.11)

The field variable can be written as follows using eq.(3.3.9) and eq.(3.3.11).

0 1

1 2

2 3

3 4

1

21 2 3 4

3

4

1 1 1 11 1 1 11, 1 11 1 1 14

1 1 1 1

N N N N

f f

f x h x h xh x h xh f f

ffff

(3.3.12)

Where, Ni are the interpolation function of the element in natural coordinate system and can be found as:

1

2

3

4

1 14

1 14

1 14

1 14

i

NN

NNN

x h

x h

x h

x h

(3.3.13)

Node Cartesian Coordinate Natural Coordinate x y ξ η

1 x1 y1 -1 -1 2 x2 y2 1 -1 3 x3 y3 1 1 4 x4 y4 -1 1

23

3.3.2 Shape Function for Eight Node Element The shape function of eight node rectangular element can be derived in similar fashion as done in case of four node element. The only difference will be on choosing of polynomial as this element is of quadratic in nature. The derivation will be algebraically complex in case of using Cartesian coordinate system. However, use of the natural coordinate system will make the process simpler as the natural coordinates vary from -1 to +1 in the element. The variation of filed variable ϕ can be expressed in natural coordinate system by the following polynomial.

2 2 2 20 1 2 3 4 5 6 7,f x h x h x xh h x h xh (3.3.14)

It may be noted that the cubic terms ξ3 and η3 are omitted and geometric invariance is ensured by choosing the above expression. Fig. 3.3.4 shows the natural nodal coordinates of the eight node rectangle element in natural coordinate system. The nodal field variables can be obtained from the above expression after putting the coordinates at nodes.

01

12

23

34

5 4

6 5

7

8

1 1 1 1 1 1 1 11 1 1 1 1 1 1 11 1 1 1 1 1 1 11 1 1 1 1 1 1 11 0 1 0 0 1 0 01 1 0 1 0 0 0 01 0 1 0 0 1 0 01 1 0 1 0 0 0 0

i

ffff

ff f ff

6

7

iA

(3.3.15)

Fig. 3.3.4 Natural coordinates of eight node rectangular element

Replacing the unknown coefficient αi in eq.(3.3.14) from eq.(3.3.15), the following relations will be obtained.

24

12 2 2 2, 1 iA f x h x h x xh h x h xh f

1

2

3

42 2 2 2

5

6

7

8

1 1 1 1 2 2 2 20 0 0 0 0 2 0 20 0 0 0 2 0 2 01 1 1 1 2 0 2 0111 1 1 1 0 0 0 041 1 1 1 0 2 0 21 1 1 1 2 0 2 01 1 1 1 0 2 0 2

ffff

x h x xh h x h xhffff

1

2

3

41 2 3 4 5 6 7 8

5

6

7

8

N N N N N N N N

ffffffff

(3.3.16)

Thus, the interpolation function will become

1 2

3 4

5 6

7 8

1 1 1 1 1 1; ;

4 41 1 1 1 1 1

; ;4 4

1 1 1 1 1 1; ;

2 21 1 1 1 1 1

; 2 2

N N

N N

N N

N N

x h x h x h x h

x h x h x h x h

x x h x h h

x x h x h h

(3.3.17)

The shape functions of rectangular elements with higher nodes can be derived in similar manner using appropriate polynomial satisfying all necessary criteria. However, difficulty arises due to the inversion of large size of the matrix because of higher degree of polynomial chosen. In next lecture, the shape functions of rectangular element with higher nodes will be derived in a much simpler way.

25

Lecture 4: Lagrange and Serendipity Elements In last lecture note, the interpolation functions are derived on the basis of assumed polynomial from Pascal’s triangle for the filed variable. As seen, the inverse of the large matrix is quite cumbersome if the element is of higher order. 3.4.1 Lagrange Interpolation Function An alternate and simpler way to derive shape functions is to use Lagrange interpolation polynomials. This method is suitable to derive shape function for elements having higher order of nodes. The Lagrange interpolation function at node i is defined by

nj 1 2 i 1 i 1 n

ij 1 i 1 i 2 i i 1 i i 1 i ni jj i

.... ....f ( )

.... ....

xx xx xx xx xx xxx

x x x x x x x x x xx x (3.4.1)

The function fi (ξ) produces the Lagrange interpolation function for ith node, and ξj denotes ξ coordinate of jth node in the element. In the above equation if we put ξ = ξj, and j ≠ i, the value of the function fi(ξ) will be equal to zero. Similarly, putting ξ = ξi, the numerator will be equal to denominator and hence fi(ξ) will have a value of unity. Since, Lagrange interpolation function for ith node includes product of all terms except jth term; for an element with n nodes, fi(ξ) will have n-1 degrees of freedom. Thus, for one-dimensional elements with n-nodes we can define shape function

as i iN ( ) f ( )x x .

3.4.1.1 Shape function for two node bar element Consider the two node bar element discussed as in section 3.1.1. Let us consider the natural coordinate of the center of the element as 0, and the natural coordinate of the nodes 1 and 2 are -1 and +1 respectively. Therefore, the natural coordinate ξ at any point x can be represented by,

12 x x1

l

x (3.4.2)

Fig. 3.4.1 Natural coordinates of bar element

26

The shape function for two node bar element as shown in Fig. 3.4.1 can be derived from eq.(3.4.1) as follows:

21 1

1 2

12 1

2 1

1 1N f 11 (1) 2

1 1N f 11 (1) 2

xx x x x

x x

xx x x x

x x (3.4.3)

Graphically, these shape functions are represented in Fig.3.4.2.

Fig. 3.4.2 Shape functions for two node bar element

3.4.1.2 Shape function for three node bar element For a three node bar element as shown in Fig. 3.4.3, the shape function will be quadratic in nature. These can be derived in the similar fashion using eq.(3.4.1) which will be as follows:

2 31 1

1 2 1 3

21 32 2

2 1 2 3

1 23 3

3 1 3 2

( ) 1 1N f 11 ( 2) 2

1 1N f 1

1 1

1 ( ) 1N f 12 (1) 2

xx xx x xx x x x

x x x x

xx xx x xx x x

x x x x

xx xx x xx x x x

x x x x

(3.4.4)

27

Fig. 3.4.3 Quadratic shape functions for three node bar element

3.4.1.3 Shape function for two dimensional elements We can derive the Lagrange interpolation function for two or three dimensional elements from one dimensional element as discussed above. Those elements whose shape functions are derived from the products of one dimensional Lagrange interpolation functions are called Lagrange elements. The Lagrange interpolation function for a rectangular element can be obtained from the product of appropriate interpolation functions in the ξ direction [fi(ξ )] and η direction [fi(η)]. Thus,

i i iN , f fx h x h Where, i = 1,2,3, …., n-node (3.4.5)

The procedure is described in details in following examples. Four node rectangular element The shape functions for the four node rectangular element as shown in the Fig. 3.4.4 can be derived by applying eq.(3.4.3) eq.(3.4.5) which will be as follows.

`

2 21 1 1

1 2 1 2

N , f f

1 1 1 1 11 (1) 1 (1) 4

xx hhx h x h

x x h h

x h x h

(3.4.6)

Similarly, other interpolation functions can be derived which are given below.

`

2 2 1

3 2 2

4 1 2

1N , f f 1 141N , f f 1 141N , f f 1 14

x h x h x h

x h x h x h

x h x h x h

(3.4.7)

These shape functions are exactly same as eq.(3.3.13) which was derived earlier by choosing polynomials.

28

Fig. 3.4.4 Four node rectangular element

Nine node rectangular element In a similar way, to the derivation of four node rectangular element, we can derive the shape functions for a nine node rectangular element. In this case, the shape functions can be derived using eq.(3.4.4) and eq.(3.4.5).

1 1 11 1 1N , f f 1 1 1 12 2 4

x h x h x x h h xh x h (3.4.8)

In a similar way, all the other shape functions of the element can be derived. The shape functions of nine node rectangular element will be:

1 2

3 4

2 25 6

2 27 8

2 29

1 1N 1 1 , N 1 14 41 1N 1 1 , N 1 14 41 1N 1 1 , N 1 12 21 1N 1 1 , N 1 12 2

N 1 1

xh x h xh x h

xh x h xh x h

h x h x x h

h x h x x h

x h

(3.4.9)

29

Fig. 3.4.5 Nine node rectangular element

Thus, it is observed that the two dimensional Lagrange element contains internal nodes (Fig. 3.4.6) which are not connected to other nodes.

Fig. 3.4.6 Two dimensional Lagrange elements and Pascal triangle 3.4.2 Serendipity Elements Higher order Lagrange elements contains internal nodes, which do not contribute to the inter-element connectivity. However, these can be eliminated by condensation procedure which needs extra computation. The elimination of these internal nodes results in reduction in size of the element matrices. Alternatively, one can develop shape functions of two dimensional elements which contain nodes only on the boundaries. These elements are called serendipity elements (Fig. 3.4.7) and their interpolation functions can be derived by inspection or the procedure described in previous lecture

30

(Module 3, lecture 3). The interpolation function can be derived by inspection in terms of natural coordinate system as follows:

(a) Linear element

i i i1N , 1 14

x h xx hh (3.4.10)

(b) Quadratic element (i) For nodes at 1, 1x h

i i i i i1N , 1 1 14

x h xx hh xx hh (3.4.11a)

(ii) For nodes at 1, 0x h

2i i

1N , 1 12

x h xx h (3.4.11b)

(iii) For nodes at 0, 1x h

2i i

1N , 1 12

x h x hh (3.4.11c)

(c) Cubic element (i) For nodes at 1, 1x h

2 2i i i

1N , 1 1 9 1032

x h xx hh x h (3.4.12a)

(ii) For nodes at 11,3

x h

2i i i

9N , 1 1 1 932

x h xx h hh (3.4.12b)

And so on for other nodes at the boundaries.

31

Fig. 3.4.7 Two dimensional serendipity elements and Pascal triangle Thus, the nodal conditions must be satisfied by each interpolation function to obtain the functions serendipitously. For example, let us consider an eight node element as shown in Fig. 3.4.8 to derive its shape function. The interpolation function N1 must become zero at all nodes except node 1, where its value must be unity. Similarly, at nodes 2, 3, and 6, ξ = 1, so including the term ξ − 1 satisfies the zero condition at those nodes. Similarly, at nodes 3, 4 and 7, η = 1 so the term η − 1 ensures the zero condition at these nodes.

Fig. 3.4.8 Two dimensional eight node rectangular element Again, at node 5, (ξ, η) = (0, −1), and at node 8, (ξ, η) = (−1, 0). Hence, at nodes 5 and 8, the term (ξ + η + 1) is zero. Using this reasoning, the equation of lines are expressed in Fig. 3.4.9. Thus, the

32

interpolation function associated with node 1 is to be of the form 1 1 1 1 1N y h x x h

where, ψ1 is unknown constant. As the value of N1 is 1 at node 1, the magnitude unknown constant ψ1 will become -1/4. Therefore, the shape function for node 1 will become

11 1 1 14

N h x x h .

Similarly, ψ2 will become -1/4 considering the value of N2 at node 2 as unity and the shape function

for node 2 will be 2 211 1 1 1 1 14

N y h x x h x h x h . In a similar

fashion one can find out other interpolation functions from Fig. 3.4.9 by putting the respective values at various nodes. Thus, the shape function for 8-node rectangular element is given below.

Fig. 3.4.9 Equations of lines for two dimensional eight node element

21 5

22 6

23 7

24 8

1 1N 1 1 1 , N 1 1 ,4 21 1N 1 1 1 , N 1 1 ,4 21 1N 1 1 1 , N 1 1 ,4 21 1N 1 1 1 and N 1 14 2

x h xh x h

x h xh x h

x h xh x h

x h xh x h

(3.4.13)

33

It may be observed that the Lagrange elements have a better degree of completeness in polynomial function compare to serendipity elements. Therefore, Lagrange elements produce comparatively faster and better accuracy.

34

Lecture 5: Solid Elements There are two basic families of three-dimensional elements similar to two-dimensional case. Extension of triangular elements will produce tetrahedrons in three dimensions. Similarly, rectangular parallelepipeds are generated on the extension of rectangular elements. Fig. 3.5.1 shows few commonly used solid elements for finite element analysis.

Fig. 3.5.1 Three-dimensional solid elements Derivation of shape functions for such three dimensional elements in Cartesian coordinates are algebraically quite cumbersome. This is observed while developing shape functions in two dimensions. Therefore, the shape functions for the two basic elements of the tetrahedral and parallelepipeds families will be derived using natural coordinates. The polynomial expression of the field variable in three dimensions must be complete or incomplete but symmetric to satisfy the geometric isotropy requirements. Completeness and symmetry can be

35

ensured using the Pascal pyramid which is shown in Fig. 3.5.2. It is important to note that each independent variable must be of equal strength in the polynomial.

Fig. 3.5.2 Pascal pyramid in three dimensions The following 3-D quadratic polynomial with complete terms can be applied to an element having 10 nodes.

2 2 20 1 2 3 4 5 6 7 8 9, ,f x h z x h z x h z xh hz zx (3.5.1)

However, the geometric isotropy is not an absolute requirement for field variable representation to derive the shape functions. 3.5.1 Tetrahedral Elements The simplest element of the tetrahedral family is a four node tetrahedron as shown in Fig. 3.5.3. The node numbering has been followed in sequential manner, i.e, in this case anti-clockwise direction. Similar to the area coordinates, the concept of volume coordinates has been introduced here. The coordinates of the nodes are defined both in Cartesian and volume coordinates. Point P(x, y, and z) as shown in Fig. 3.5.2 is an arbitrary point in the tetrahedron.

36

Fig. 3.5.3 Four node tetrahedron element The linear shape function for this element can be expressed as,

[ ]1 2 3 4TN L L L L= (3.5.2)

Here, 4321 ,,, LLLL are the set of natural coordinates inside the tetrahedron and are defined as follows

VVL i

i = (3.5.3)

Where Vi is the volume of the sub element which is bound by point P and face i and V is the total volume of the element. For example L1 may be interpreted as the ratio of the volume of the sub element P234 to the total volume of the element 1234. The volume of the element V is given by the determinant of the nodal coordinates as follows:

4321

4321

4321

1111

61

zzzzyyyyxxxx

V =

(3.5.4)

The relationship between the Cartesian and natural coordinates of point P may be expressed as

=

4

3

2

1

4321

4321

4321

11111

LLLL

zzzzyyyyxxxx

zyx

(3.5.5)

It may be noted that the identity included in the first row ensure the matrix invertible.

37

14321 =+++ LLLL (3.5.6)

The inverse relation is given by

=

zyx

cbaVcbaVcbaVcbaV

VLLLL 1

61

4444

3333

2222

1111

4

3

2

1

(3.5.7)

Here, iV is the volume subtended from face i and terms ,, ii ba and ic represent the projected area of

face i on the zyx ,, coordinate planes respectively and are given as follows:

)()()(

)()()(

)()()(

ljjlkllkjkkji

ljjlkllkjkkji

ljjlkllkjkkji

xyxyxyxyxyxycxzxzxzxzxzxzbyzyzyzyzyzyza

−+−+−=

−+−+−=

−+−+−=

(3.5.8)

lkji ,,, will be in cyclic order (i.e., 1 2 3 4 1). The volume coordinates fulfil all nodal

conditions for interpolation functions. Therefore, the field variable can be expressed in terms of nodal values as

1 1 2 2 3 3 4 4, ,x y z L L L Lf f f f f (3.5.9)

Though the shape functions (i.e., the volume coordinates) in terms of global coordinates is algebraically complex but they are straightforward. The partial derivatives of the natural coordinates with respect to the Cartesian coordinates are given by

,6Va

xL ii =∂∂

,6Vb

yL ii =∂∂

Vc

zL ii

6=

∂∂

(3.5.10)

Similar to area integral, the general integral taken over the volume of the element is given by,

Vsrqp

srqpdVLLLL s

v

rqp 6.)!3(

!!!!4321 ++++

=∫

(3.5.11)

The four node tetrahedral element is a linear function of the Cartesian coordinates. Hence, all the first partial derivatives of the field variable will be constant. The tetrahedral element is a constant strain element as the element exhibits constant gradients of the field variable in the coordinate directions. Higher order elements of the tetrahedral family are shown in Fig. 3.5.1. The shape functions for such higher order three dimensional elements can readily be derived in volume coordinates, as for higher-order two-dimensional triangular elements. The second element of this family has 10 nodes and a cubic form for the field variable and interpolation functions. +++++++++++ 3.5.2 Brick Elements

38

Various orders of elements of the parallelepiped family are shown in Fig. 3.5.1. Fig. 3.5.4 shows the eight-node brick element with reference to a global Cartesian coordinate system and then with reference to natural coordinate system. The natural coordinates for the brick element can be relate Cartesian coordinate system by

,x x y y z zanda b c

ξ η ζ− − −= = = (3.5.12)

Here, 2a, 2b and 2c are the length, height and width of the element. The coordinate of the center of the brick element can be written as follows:

1 51 2 1 4,2 2 2

z zx x y yx y and z ++ += = = (3.5.13)

Thus, from eq.(3.5.12) and eq.(3.5.13), the nodal values in natural coordinate systems can be derived which is shown in Fig. 3.5.4(b). With the above relations variations of , &x h z will be from -1 to +1.

Now the interpolation function can be derived in several procedures as done in case of two dimensional rectangular elements. For example, the interpolation function can be derived by inspection in terms of natural coordinate system as follows:

i i i i1N , , 1 1 18

x h z xx hh zz (3.5.14)

Fig. 3.5.4 Eight node brick element By using field variable the following terms of the polynomial may be used for deriving the shape function for eight-node brick element.

0 1 2 3 4 5 6 7, ,f x h z x h z xh hz zx zhx (3.5.14) The above equation is incomplete but symmetric. However, such representations are quite often used and solution convergence is achieved in the finite element analysis. Again, the shape functions for

39

three dimensional 8-node or 27-node brick elements can be derived using Lagrange interpolation function. For this we need to introduce interpolation function in the ζ-direction. Thus, for example, the Lagrange interpolation function for a three dimensional 8 node brick element can be obtained from the product of appropriate interpolation functions in the ξ, η and ζ directions. Therefore, the shape function will become

i i i iN , , f f fx h z x h z Where, i = 1,2,3, …., n-node (3.5.15)

Thus using the Lagrange interpolation function the shape function at node 1 can be expressed as

2 2 21 1 1 1

1 2 1 2 1 2

N , , f f f

1 1 1 1 1 1 11 (1) 1 (1) 1 (1) 4

xx hh zzx h z x h z

x x h h z z

x h z x h z

(3.5.16)

Using any of the above concepts, the interpolation function for 8-node brick element can be found as follows:

1 1

3 4

5 6

7 8

1 1N 1 1 1 , N 1 1 1 ,4 41 1N 1 1 1 , N 1 1 1 ,4 41 1N 1 1 1 , N 1 1 1 ,4 41 1N 1 1 1 , N 1 1 14 4

x h z x h z

x h z x h z

x h z x h z

x h z x h z

(3.5.17)

The shape functions of rectangular parallelepiped elements with higher nodes can be derived in similar manner satisfying all necessary criteria.

40

Lecture 6: Isoparametric Formulation 3.6.1 Necessity of Isoparametric Formulation The two or three dimensional elements discussed till now are of regular geometry (e.g. triangular and rectangular element) having straight edge. Hence, for the analysis of any irregular geometry, it is difficult to use such elements directly. For example, the continuum having curve boundary as shown in the Fig. 3.6.1(a) has been discretized into a mesh of finite elements in three ways as shown.

(a) The Continuum to be discritized (b) Discritization using Triangular Elements (c)

Discritization using rectangular elements (d) Discritization using a combination of rectangular and quadrilateral elements

41

Fig 3.6.1 Discretization of a continuum using different elements

Figure 3.6.1(b) presents a possible mesh using triangular elements. Though, triangular elements can suitable approximate the circular boundary of the continuum, but the elements close to the center becomes slender and hence affect the accuracy of finite element solutions. One possible solution to the problem is to reduce the height of each row of elements as we approach to the center. But, unnecessary refining of the continuum generates relatively large number of elements and thus increases computation time. Alternatively, when meshing is done using rectangular elements as shown in Fig 3.6.1(c), the area of continuum excluded from the finite element model is significantly adequate to provide incorrect results. In order to improve the accuracy of the result one can generate mesh using very small elements. But, this will significantly increase the computation time. Another possible way is to use a combination of both rectangular and triangular elements as discussed in section 3.2. But such types of combination may not provide the best solution in terms of accuracy, since different order polynomials are used to represent the field variables for different types of elements. Also the triangular elements may be slender and thus can affect the accuracy. In Fig. 3.6.1(d), the same continuum is discritized with rectangular elements near center and with four-node quadrilateral elements near boundary. This four-node quadrilateral element can be derived from rectangular elements using the concept of mapping. Using the concept of mapping regular triangular, rectangular or solid elements in natural coordinate system (known as parent element) can be transformed into global Cartesian coordinate system having arbitrary shapes (with curved edge or surfaces). Fig. 3.6.2 shows the parent elements in natural coordinate system and the mapped elements in global Cartesian system.

42

43

(a) Natural Coordinate System (b) Global Coordinate System

Fig. 3.6.2 Mapping of isoparametric elements in global coordinate system

3.6.2 Coordinate Transformation The geometry of an element may be expressed in terms of the interpolation functions as follows.

1 1 2 21

1 1 2 21

1 1 2 21

...

...

...

n

n n i ii

n

n n i ii

n

n n i ii

x N x N x N x N x

y N y N y N y N y

z N z N z N z N z

=

=

=

= + + + =

= + + + =

= + + + =

∑

∑

∑

(3.6.1)

i

i i i

Where,n=No.of NodesN =Interpolation Functionsx ,y ,z =Coordinates of Nodal Points of the Element

One can also express the field variable variation in the element as

44

( ) ( )1

, , , ,n

i ii

Nφ ξ η ζ ξ η ζ φ=

=∑ (3.6.2)

As the same shape functions are used for both the field variable and description of element geometry, the method is known as isoparametric mapping. The element defined by such a method is known as an isoparametric element. This method can be used to transform the natural coordinates of a point to the Cartesian coordinate system and vice versa. Example 3.6.1 Determine the Cartesian coordinate of the point P (ξ= 0.8, η= 0.9) as shown in Fig. 3.6.3.

Fig. 3.6.3 Transformation of Coordinates

Solution: As described above, the relation between two coordinate systems can be represented through their interpolation functions. Therefore, the values of the interpolation function at point P will be

1

2

3

4

(1 )(1 ) (1 0.8)(1 0.9) 0.0054 4

(1 )(1 ) (1 0.8)(1 0.9) 0.0454 4

(1 )(1 ) (1 0.8)(1 0.9) 0.8554 4

(1 )(1 ) (1 0.8)(1 0.9) 0.0954 4

nN

nN

nN

nN

ξ

ξ

ξ

ξ

− − − −= = =

+ − + −= = =

+ + + += = =

− + − += = =

Thus the coordinate of point P in Cartesian coordinate system can be calculated as

45

4

14

1

0.005 1 0.045 3 0.855 3.5 0.095 1.5 3.275

0.005 1 0.045 1.5 0.855 4.0 0.095 2.5 3.73

i ii

i ii

x N x

y N y

=

=

= = × + × + × + × =

= = × + × + × + × =

∑

∑

Thus the coordinate of point P (ξ= 0.8, η= 0.9) in Cartesian coordinate system will be 3.275, 3.73. Solid isoparametric elements can easily be formulated by the extension of the procedure followed for 2-D elements. Regardless of the number of nodes or possible curvature of edges, the solid element is just like a plane element which is mapped into the space of natural co-ordinates, i.e,

1,1,1 ±=±=±= ζηξ .

3.6.3 Concept of Jacobian Matrix A variety of derivatives of the interpolation functions with respect to the global coordinates are necessary to formulate the element stiffness matrices. As the both element geometry and variation of the shape functions are represented in terms of the natural coordinates of the parent element, some additional mathematical obstacle arises. For example, in case of evaluation of the strain vector, the operator matrix is with respect to x and y, but the interpolation function is with andx h . Therefore,

the operator matrix is to be transformed for taking derivative with andx h . The relationship between

two coordinate systems may be computed by using the chain rule of partial differentiation as x y x yand

x y x yx x x h h h

(3.6.3)

The above equations can be expressed in matrix form as well.

x yx xJ

x yy y

x x x

h h h

(3.6.4)

The matrix [J] is denoted as Jacobian matrix which is:

x y

x yx x

h h

. As we know, 1

n

i ii

x N x=

=∑

where, n is the number of nodes in an element. Hence, 111

1

n

i i ni i

ii

N xNxJ x

ξ ξ ξ=

=

∂∂∂

= = =∂ ∂ ∂

∑∑

Similarly one can calculate the other terms J12, J21 and J22 of the Jacobian matrix. Hence,

46

1 1

1 1

n ni i

i ii in n

i ii i

i i

N Nx yJ

N Nx y

x x

h h

(3.6.5)

From eq. (3.6.4), one can write

1x J

y

x

x

(3.6.6)

Considering * *11 12* *21 22

J JJ J

are the elements of inverted [J] matrix, we may arise into the following

relations.

* *11 12

* *21 22

J Jx

J Jy

x h

x h

(3.6.7)

Similarly, for three dimensional case, the following relation exists between the derivative operators in the global and the natural coordinate system.

[ ]

x y zx x

x y z Jy y

x y zz z

ξ ξ ξ ξ

η η η η

ζ ζ ζ ζ

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = = ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

(3.6.8)

Where,

47

[ ]

x y z

x y zJ

x y z

ξ ξ ξ

η η η

ζ ζ ζ

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

= ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

(3.6.9)

[J] is known as the Jacobian Matrix for three dimensional case. Putting eq. (3.6.1) in eq. (3.6.9) and after simplifying one can get

[ ]1

i i ii i i

ni i i

i i ii

i i ii i i

N N Nx y z

N N NJ x y z

N N Nx y z

ξ ξ ξ

η η η

ζ ζ ζ

=

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

= ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

∑ (3.6.10)

From eq. (3.6.8), one can find the following expression.

[ ] 1

x

Jy

z

ξ

η

ζ

−

∂ ∂ ∂∂ ∂ ∂ = ∂ ∂

∂∂

∂∂

(3.6.11)

Considering [ ]* * *11 12 13

1 * * *21 22 23* * *31 32 33

J J JJ J J J

J J J

−

=

we can arrived at the following relations.

* * *11 12 13

* * *21 22 23

* * *31 32 33

J J Jx

J J Jy

J J Jz

x h z

x h z

x h z

(3.6.12)

48

Lecture 7: Stiffness Matrix of Isoparametric Elements 3.7.1 Evaluation of Stiffness Matrix of 2-D Isoparametric Elements For two dimensional plane stress/strain formulation, the strain vector can be represented as

* *11 12

* *21 22

* * * *11 12 21 22

x

y

xy

u uu J Jxv v vJ Jy

v u v v u uJ J J Jx y

x h

x h

x h x h

(3.7.1)

The above expression can be rewritten in matrix form

* *11 12

* *21 22

* * * *21 22 11 12

0 00 0

u

uJ JJ J

vJ J J J

v

x

h

x

h

(3.7.2)

For an n node element the displacement u can be represented as, 1

n

i ii

u N u

and similarly for v &

w. Thus,

1

1

1

11

1

0 0

0 0

0 0

0 0

n

n

n

n

n n

NNuu

NNuuvv NN

v NN v

x xx

h h h

x x x

h h h

(3.7.3)

As a result, eq. (3.7.2) can be written using eq. (3.7.3) which will be as follows.

49

11 n

* * 1 n11 12

* * n21 22

* * * * 1 n 121 22 11 12

1 n

n

uN N 0 0 ::N NJ J 0 00 0uJ J0 0

N N vJ J J J 0 0:

N N :0 0v

x x h h x x h h

(3.7.4)

Or,

B d (3.7.5)

Where d is the nodal displacement vector and [B] is known as strain displacement relationship matrix and can be obtained as

1 n

* * 1 n11 12

* *21 22

* * * * 1 n21 22 11 12

1 n

N N 0 0

N NJ J 0 00 0J JB 0 0

N NJ J J J 0 0

N N0 0

x x h h x x h h

(3.7.6)

It is necessary to transform integrals from Cartesian to the natural coordinates as well for calculation of the elemental stiffness matrix in isoparametric formulation. The differential area relationship can be established from advanced calculus and the elemental area in Cartesian coordinate can be represented in terms of area in natural coordinates as:

dA dx dy J d d x h (3.7.7)

Here J is the determinant of the Jacobian matrix. The stiffness matrix for a two dimensional

element may be expressed as

T

A

T t B D B dxdyk B D B d

(3.7.8)

Here, [B] is the strain-displacement relationship matrix and t is the thickness of the element. The above expression in Cartesian coordinate system can be changed to the natural coordinate system as follows to obtain the elemental stiffness matrix

50

1 1

T

1 1

k t B D B J d d

x h (3.7.9)

Though the isoparametric formulation is mathematically straightforward, the algebraic difficulty is significant. Example 3.7.1: Calculate the Jacobian matrix and the strain displacement matrix for four node two dimensional quadrilateral elements corresponding to the gauss point (0.57735, 0.57735) as shown in Fig. 3.6.4.

Fig. 3.7.1 Two dimensional quadrilateral element Solution: The Jacobian matrix for a four node element is given by,

1 1

1 1

n ni i

i ii in n

i ii i

i i

N Nx yJ

N Nx y

x x

h h

For the four node element one can find the following relations.

1 11

1 1 N N1 1N , , 4 4 4

x h h x x h

2 22

1 1 N N1 1N , , 4 4 4

x h h x x h

3 33

1 1 N N1 1N , , 4 4 4

x h h x x h

51

4 44

1 1 N N1 1N , , 4 4 4

x h h x x h

Now, for a four node quadrilateral element, the Jacobian matrix will become

1 131 2 4

2 2

3 331 2 4

4 4

1 1

2 2

3 3

4 4

x yNN N Nx y

Jx yNN N Nx y

x y1 1 1 1x y4 4 4 4 x y1 1 1 1

4 4 4 4 x y

x x x x h h h h h h h h x x x x

Putting the values of ξ & η as 0.57735 and 0.57735 respectively, one will obtain the following.

1 1

2 2

3 3

4 4

N N0.10566 0.10566

N N0.10566 0.39434

N N0.39434 0.39434

N N0.39434 0.10566

x h x h x h x h

Hence, 4

111

0.10566 1 0.10566 3 0.39434 3.5 0.39434 1.5 1.0ii

i

NJ xx

Similarly, J12 =0.64632, J21 =0.25462 and J22 =1.14962. Hence,

1.00000 0.646320.25462 1.14962

J

Thus, the inverse of the Jacobian matrix will become: * *

* 11 12* *21 22

1.1671 0.65610.2585 1.0152

J JJ

J J

Hence strain displacement matrix is given by,

52

1 n

* * 1 n11 12

* *21 22

* * * * 1 n21 22 11 12

1 n

N N 0 0

N NJ J 0 00 0J JB 0 0

N NJ J J J 0 0

N N0 0

x x h h x x h h

1.1671 0.6561 0 00 0 0.2585 1.0152

0.2585 1.0152 1.1671 0.6561

0.10566 0.10566 0.39434 0.39434 0 0 0 00.10566 0.39434 0.39434 0.10566 0 0 0 0

0 0 0 0 0.10566 0.10566 0.39434 0.394340 0 0 0 0.10566 0.39434 0.39434 0.10566

0.0540 0.3820 0.2015 0.5294 0 0 0 0

0 0 0 0 0.0800 0.4276 0.2984 0.20920.0800 0.4276 0.2984 0.2092 0.0540 0.3820 0.2015 0.5294

3.7.2 Evaluation of Stiffness Matrix of 3-D Isoparametric Elements Stiffness matrix of 3-D solid isoparametric elements can easily be formulated by the extension of the procedure followed for plane elements. For example, the eight node solid element is analogous to the four node plane element. The strain vector for solid element can be written in the following form.

53

1 0 0 0 0 0 0 0 00 0 0 0 1 0 0 0 00 0 0 0 0 0 0 0 10 1 0 1 0 0 0 0 00 0 0 0 0 1 0 1 00 0 1 0 0 0 1 0 0

x

y

z

xy

yz

zx

uxuyuzvxvyvzwxwywz

εεεγγγ

∂ ∂ ∂

∂ ∂

∂ ∂ ∂ ∂ = ∂

∂ ∂

∂ ∂ ∂

∂ ∂

∂ (3.7.10)

The above equation can be expressed as

* * *11 12 13

* * *21 22 23

* * *31 32 33

* * * * * *21 22 23 11 12 13

* * * * * *31 32 33 21 22 23

* * *31 32 33

0 0 0 0 0 00 0 0 0 0 00 0 0 0 0 0

0 0 00 0 0

uxv

J J JyJ J Jw

J J Jzu v J J J J J Jy x J J J J J Jv w J J Jz yx wz x

ε

∂ ∂

∂ ∂

∂ ∂ = = ∂ ∂ +∂ ∂

∂ ∂ + ∂ ∂ ∂ ∂ + ∂ ∂

* * *11 12 130 0 0

u

u

u

v

v

vJ J J

w

w

v

ξ

η

ζ

ξ

η

ζ

ξ

η

ζ

∂ ∂ ∂

∂ ∂

∂ ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂

(3.7.11)

54

For an 8 node brick element u can be represented as, 8

1i i

i

u N u

and similarly for v & w.

8

1

ii

i

Nu uξ ξ=

∂∂=

∂ ∂∑ , 8

1

ii

i

Nu uη η=

∂∂=

∂ ∂∑ & 8

1

ii

i

Nu uζ ζ=

∂∂=

∂ ∂∑

8

1

ii

i

Nv vξ ξ=

∂∂=

∂ ∂∑ , 8

1

ii

i

Nv vη η=

∂∂=

∂ ∂∑ & 8

1

ii

i

Nv vζ ζ=

∂∂=

∂ ∂∑ (3.7.12)

8

1

ii

i

Nw wξ ξ=

∂∂=

∂ ∂∑ , 8

1

ii

i

Nw wη η=

∂∂=

∂ ∂∑ & 8

1

ii

i

Nw wζ ζ=

∂∂=

∂ ∂∑

Hence eq. (3.7.11) can be rewritten as

* * *11 12 13

* * *21 22 23

* * *31 32 33

* * * * * *21 22 23 11 12 13

* * * * * *31 32 33 21 22 23

* * * * * *31 32 33 11 12 13

0 0

0 00 0 0 0 0 0

0 0 0 0 0 00 0

0 0 0 0 0 00 0 0 0

0 0 00 0 0 0

i

i

i

i i

i i

N

NJ J J

J J J NJ J J

N NJ J J J J JJ J J J J J

N NJ J J J J J

ξ

η

ζε

η ξ

ζ

∂∂

∂ ∂

∂ ∂

= × ∂ ∂

∂ ∂

∂ ∂ ∂ ∂

8

1

0

i

ii

i

i i

uvw

N Nη

ζ ξ

=

∂ ∂ ∂ ∂

∑ (3.7.13)

Thu, the strain-displacement relationship matrix [B] for 8 node brick element is

55

[ ]

* * *11 12 13

* * *21 22 23

* * *31 32 33

* * * * * *21 22 23 11 12 13

* * * * * *31 32 33 21 22 23

* * * * * *31 32 33 11 12 13

0 0

0 00 0 0 0 0 0

0 0 0 0 0 00 0

0 0 0 0 0 00 0 0 0

0 0 00 0 0 0

i

i

i

i i

i i

N

NJ J J

J J J NJ J J

BN NJ J J J J J

J J J J J JN NJ J J J J J

ξ

η

ζ

η ξ

ζ

∂∂

∂ ∂

∂ ∂

= × ∂ ∂

∂ ∂

∂ ∂ ∂ ∂

8

1

0

i

i iN Nη

ζ ξ

=

∂ ∂ ∂ ∂

∑ (3.7.14)

The stiffness matrix may be found by using the following expression in natural coordinate system.

1 1 1

T T T

V 1 1 1

k B D B d B D B dxdydz [B] [D][B]d d d J

x h z (3.7.15)

56

Lecture 8: Numerical Integration: One Dimensional The integrations, we generally encounter in finite element methods, are quite complicated and it is not possible to find a closed form solutions to those problems. Exact and explicit evaluation of the integral associated to the element matrices and the loading vector is not always possible because of the algebraic complexity of the coefficient of the different equation (i.e., the stiffness influence coefficients, elasticity matrix, loading functions etc.). In the finite element analysis, we face the problem of evaluating the following types of integrations in one, two and three dimensional cases respectively. These are necessary to compute element stiffness and element load vector.

d ; , d d ; , , d d d ;f x x f x h x h f x h z x h z (3.8.1)

Approximate solutions to such problems are possible using certain numerical techniques. Several numerical techniques are available, in mathematics for solving definite integration problems, including, mid-point rule, trapezoidal-rule, Simpson’s 1/3rd rule, Simpson’s 3/8th rule and Gauss Quadrature formula. Among these, Gauss Quadrature technique is most useful one for solving problems in finite element method and therefore will be discussed in details here. 3.8.1 Gauss Quadrature for One-Dimensional Integrals The concept of Gauss Quadrature is first illustrated in one dimension in the context of an integral in

the form of 2

1

1 x

1 xI d from f (x)dx

f x x . To transform from an arbitrary interval of x1≤ x ≤ x2

to an interval of -1 ≤ ξ ≤ 1, we need to change the integration function from f(x) to ϕ(ξ) accordingly. Thus, for a linear variation in one dimension, one can write the following relations.

2

1

1 2 1 1 2 2

1 2 1

2x 1

x 1

1 1 x x x N x N x2 2

1 1 1 1so for 1, x x x x2 2

1, x x

I f (x)dx d

x x

x

x

f x x

Numerical integration based on Gauss Quadrature assumes that the function ϕ(ξ) will be evaluated over an interval -1 ≤ ξ ≤ 1. Considering an one-dimensional integral, Gauss Quadrature represents the integral ϕ(ξ) in the form of

n1

i i 1 1 2 21 i 1

I d w w w .. w

h h

f x x f x f x f x f x (3.8.2)

57

Where, the ξ1, ξ2, ξ3, ..., ξn represents n numbers of points known as Gauss Points and the corresponding coefficients w1, w2, w3, …, wn are known as weights. The location and weight coefficients of Gauss points are calculated by Legendre polynomials. Hence this method is also sometimes referred as Gauss-Legendre Quadrature method. The summation of these values at n sampling points gives the exact solution of a polynomial integrand of an order up to 2n-1. For example, considering sampling at two Gauss points we can get exact solution for a polynomial of an order (2×2-1) or 3. The use of more number of Gauss points has no effect on accuracy of results but takes more computation time. 3.8.2 One- Point Formula Considering n = 1, eq.(3.8.2) can be written as

1

1 11( )d w ( )

f x x f x (3.8.3)

Since there are two parameters 1 1w and x , we need a first order polynomial for ϕ(ξ) to evaluate the

eq.(3.8.3) exactly. For example, considering, 0 1a af x x ,

1

0 1 1 11

0 1 0 1 1

Error a a d w 0

2a w a a 0

x x f x

x

0 1 1 1 1a 2 w w a 0 x (3.8.4)

Thus, the error will be zero if 1 1w 2 and 0 x . Putting these in eq.(3.8.3), for any general ϕ, we

have

1

1I d 2 0

f x x f (3.8.5)

This is exactly similar to the well known midpoint rule. 3.8.3 Two-Point Formula If we consider n = 2, then the eq.(3.8.2) can be written as

1

1 1 2 21d w w

f x x f x f x (3.8.6)

This means we have four parameters to evaluate. Hence we need a 3rd order polynomial for ϕ(ξ) to exactly evaluate eq.(3.8.6).

Considering, 2 30 1 2 3a a a af x x x x

1

2 30 1 2 3 1 1 2 21

Error a a a a d w w

x x x x f x f x

58

2 3 2 30 2 1 0 1 1 2 1 3 1 2 0 1 2 2 2 3 2

22 a a w a a a a w a a a a 03

x x x x x x

2 2 3 31 2 0 1 1 2 2 1 1 1 2 2 2 1 1 2 2 3

22 w w a w w a w w a w w a 03

x x x x x x

Fig 3.8.1 One-point Gauss Quadrature

Requiring zero error yields

1 2

1 1 2 2

2 21 1 2 2

3 31 1 2 2

w w 2w w 0

2w w3

w w 0

x x

x x

x x

(3.8.7)

These nonlinear equations have the unique solution as

1 2 1 2w w 1 1 3 0.5773502691 x x (3.8.8)

From this solution, we can conclude that n-point Gaussian Quadrature will provide an exact solution if ϕ(ξ) is a polynomial of order (2n-1) or less. Table 3.8.1 gives the values of 1 1w and x for Gauss

Quadrature formulas of orders n = 1 through n = 6. From the table it can be observed that the gauss

59

points are symmetrically placed with respect to origin and those symmetrical points have the same weights. For accuracy in the calculation maximum number digits for gauss point and gauss weights should be taken. The Location and weights given in the Table 3.8.1 must be used when the limits of integration ranges from -1 to 1. Integration limits other than [-1, 1], should be appropriately changed to [-1, 1] before applying these values. Table 3.8.1 Gauss points and corresponding weights Number of points, n

Gauss Point Location, 1x Weight, 1w

1 0.0 2.0 2 ±0.5773502692 (= 1 3 ) 1.0

3 0.0 0.8888888889 (=8/9)

±0.7745966692 (= 6 ) 0.5555555556 (=5/9)

4 ±0.3399810436 0.6521451549 ±0.861363116 0.3478548451

5 0.0 0.5688888889 ±0.5384693101 0.4786286705 ±0.9061798459 0.2369268851

6 ±0.2386191861 0.4679139346 ±0.6612093865 0.3607615730 ±0.9324695142 0.1713244924

Example 1:

Evaluate 1

x20

2xI e dxx 2

using one, two and three point gauss Quadrature.

Solution: Before applying the Gauss Quadrature formula, the existing limits of integration should be changed from [0, 1] to [-1, +1]. Assuming, a bxx , the upper and lower limit can be changed. i.e., at x =

0, ξ = -1 and at x = 1, ξ = +1. Thus, putting these conditions and solving for a & b, we get a = -1 and b = 2. The relation between two coordinate systems will become 2x 1x and d 2dxx .

Therefore the initial equation can be written as

60

11

221

122I e dx

1 22

x

x x

Or,

112

21

4 11I e d2 1 8

x

x x x

Using one point gauss Quadrature:

1 1w 2, 0 and

I 2 0 x

f

Or 0.51 4I 2 e 2.220152 7

Using two point gauss Quadrature:

1 2

1

2

w w 10.5773502692

0.5773502692

x x

Putting these values and calculating, I 2.39831

Using three point gauss Quadrature:

1 0.555555556w =

1 0.774596669ξ = −

2 0.888888889w =

2 0.000000000ξ =

3 0.555555556w =

3 0.774596669ξ =

and I 2.41024

This may be compared with the exact solution as exactI 2.41193

61

Lecture 9: Numerical Integration: Two and Three Dimensional Numerical integrations using Gauss Quadrature method can be extended to two and three dimensional cases in a similar fashion. Such integrations are necessary to perform for the analysis of plane stress/strain problem, plate and shell structures and for the three dimensional stress analysis. 3.9.1 Gauss Quadrature for Two-Dimensional Integrals For two dimensional integration problems the above mentioned method can be extended by first evaluating the inner integral, keeping η constant, and then evaluating the outer integral. Thus,

n n n1 1 1

i i j i i j1 1 1 i 1 i 1 i 1

I , d d w , d w w ,

f x h x h f x h h f x h

Or,

n n

i j i ji 1 j 1

I w w ,

f x h (3.9.1)

In a matrix form we can rewrite the above expression as

[ ]

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )

1 1 1 2 1 1

2 1 2 2 2 21 2

1 2

, , ,, , ,

, , ,

n

nn

n n n n n

ww

I w w w

w

φ ξ η φ ξ η φ ξ ηφ ξ η φ ξ η φ ξ η

φ ξ η φ ξ η φ ξ η

≈

(3.9.2)

Example 1:

Evaluate the integral:

y d 4 x b 3

2 2

y c 4 x a 2

I 1 x 2 y dxdy

Solution: Before applying the Gauss Quadrature formula, the above integral should be converted in terms of

and x h and the existing limits of y should be changed from [-4,4] to [-1, 1] and that of x is from

[2,3] to [-1,1].

62

b a b a 5 dx ; dx2 2 2 2

d c d cy 4 ; dy 4d

2 2

x x x

h h h

1 11 122

1 1 1 1

22 2 2

3I 2 2 4 d d , d d2

3where , 2 2 4 2 3 1 22

h hx x

h x h x

x h x h f x h x h

xf x h h x h

Fig. 3.9.1 Gauss points for two-dimensional integral

63

1 1 2 2

2 2

1 1

2

2

2 1

2

2

2 2

1

1 1 1 1; ; ;3 3 3 3

1 2, 2 3 1 54.498573 3

1343, 2 118.83018

2 3

1343, 2 0.61254

2 3

,

x h x h

f x h

f x h

f x h

f x

2

2

2

1343 2 0.28093

2 3

h

11 1 1 21 2

2 1 2 2 2

w, ,I w w

, , w

f x h f x h f x h f x h

54.49857 0.28093 1 1 1

118.83018 0.61254 1

= 174.22222 agrees with the exact value 174.22222

3.9.2 Gauss Quadrature for Three-Dimensional Integrals In a similar way one can extend the gauss Quadrature for three dimensional problems also and the integral can be expressed by.

n n n1 1 1

i j k i j k1 1 1 i 1 j 1 k 1

I , , d d d w w w , ,

f x h z x h z f x h z (3.9.3)

The above equation will produce exact value for a polynomial integrand if the sampling points are selected as described earlier sections.

64

3.9.3 Numerical Integration of Element Stiffness Matrix As discussed earlier notes, the element stiffness matrix for three dimensional analyses in natural coordinate system can be written as

1 1 1

T T T

V 1 1 1

k B D B d B D B dxdydz [B] [D][B]d d d J

x h z (3.9.4)

Here, [B] and [D] are the strain displacement relationship matrix and constitutive matrix respectively and integration is performed over the domain. As the element stiffness matrix will be calculated in natural coordinate system, the strain displacement matrix [B] and Jacobian matrix [J] are functions of , andx h z . In case of two dimensional isoparametric element, the stiffness matrix will be

simplified to

1 1

T

1 1

k t [B] [D][B]d d J

x h (3.9.5)

This is actually an 8×8 matrix containing the integrals of each element. We do not need to integrate elements below the main diagonal of the stiffness matrix as it is symmetric. Considering,

T, t[B] [D][B] Jf x h , the element stiffness matrix will become after numerical integration as

n n

i j i ji 1 j 1

k w w ,

f x h (3.9.6)

Using a 2×2 rule, we get

2 21 1 1 1 2 1 2 2 1 2 1 2 2 2k w , w w , w w , w , f x h f x h f x h f h h (3.9.7)

Where 1 2 1 1 2 2w w 1.0, 0.57735...., and 0.57735.... x h x h Here, wn is the weight

factor at integration point n. A suitable computer program can be written to calculate the element stiffness matrix through the numerical integration. The process of obtaining stiffness matrix using Gauss Quadrature integration will be demonstrated through a numerical example in module 5. 3.10.4 Gauss Quadrature for Triangular Elements The procedure described for the rectangular element will not be applicable directly. The Gauss Quadrature is extended to include triangular elements in terms of triangular area coordinates.

n

i i i1 2 3 i 1 2 2

i 1A

I L , L ,L dA w L ,L ,L

f f (3.9.8)

Where, L terms are the triangular area coordinates and the wi terms are the weights associated with those coordinates. The locations of integration points are shown in Fig. 3.9.2.

65

Fig. 3.9.2 Gauss points for triangles

The sampling points and their associated weights are described below: For sampling point =1 (Linear triangle)

1 1 11 1 2 3

1w 1 L L L3

(3.9.9)

For sampling points =3 (Quadratic triangle)

1 1 11 1 2 3

2 2 22 1 2 3

3 3 33 1 2 3

1 1w L L , L 03 21 1w L 0, L L3 21 1 1w L , L 0, L3 2 2

(3.9.10)

For sampling point = 7 (Cubic triangle)

1 1 11 1 2 3

2 2 22 1 2 3

3 3 23 1 2 3

4 4 44 1 3 2

5 5 55 1 2 3

6 6 66 1 3 2

7 7 77 1 2 3

27 1w L L L60 38 1w L L ,L 060 28 1w L 0,L L60 28 1w L L ,L 060 23w L 1,L L 0

603w L L 0,L 1

603w L L 0,L 1

60

(3.9.11)

66

3.10.5 Gauss Quadrature for Tetrahedron The Gauss Quadrature for triangles can be effectively extended to include tetrahedron elements in terms of tetrahedron volume coordinates.

n

i i i i1 2 3 4 i 1 2 3 4

i 1A

I L ,L ,L ,L dA w L , L ,L ,L

f f (3.9.12)

Where, L terms are the volume coordinates and the wi terms are the weights associated with those coordinates. The locations of Gauss points are shown in Fig. 3.9.3.

Fig. 3.9.3 Gauss points for tetrahedrons

The sampling points and their associated weights are described below: For sampling point = 1 (Linear tetrahedron)

1 1 1 11 1 2 3 4

1w 1 L L L L4

(3.9.13)

For sampling points = 4 (Quadratic tetrahedron)

1 1 1 11 1 2 3 4

2 2 2 22 2 1 3 4

3 3 3 33 3 1 2 4

4 4 4 44 4 1 2 3

1w L 0.5854102,L L L 0.138196641w L 0.5854102,L L L 0.138196641w L 0.5854102,L L L 0.138196641w L 0.5854102,L L L 0.13819664

(3.9.14)

For sampling points = 5 (Cubic tetrahedron)

67

1 1 1 11 1 2 3 4

2 2 2 22 1 2 3 4

3 3 3 33 2 1 3 4

4 4 4 44 3 1 2 4

5 5 5 45 4 1 2 3

4 1w L L L L5 4

9 1 1w L ,L L L20 3 69 1 1w L ,L L L20 3 69 1 1w L ,L L L20 3 69 1 1w L ,L L L20 3 6

(3.9.15)

68

Worked out Examples Example 3.1 Calculation of displacement using area coordinates The coordinates of a three node triangular element is given below. Calculate the displacement at point P if the displacements of nodes 1, 2 and 3 are 11 mm, 14mm and 17mm respectively using the concepts of area coordinates.

A = 121 𝑥1 𝑦11 𝑥2 𝑦21 𝑥3 𝑦3

= 12

1 2 31 5 41 3 6

= 12 [(30-12) - (12-9) + (8-15)] = 8

2 = 4

𝐴1 = 12

1 𝑥 𝑦1 𝑥2 𝑦21 𝑥3 𝑦3

= 12

1 3 41 5 41 3 6

= 12 [(30-12) - (18-12) + (12-20)] = 4

2 = 2

Fig. Ex.3.1 Nodal coordinates of a triangular element

𝐴2 = 12

1 𝑥 𝑦1 𝑥3 𝑦31 𝑥1 𝑦1

= 12

1 3 41 3 61 2 3

= 12 [(9-12) - (9-8) + (18-12)] = 2

2 = 1

69

𝐴3 = 12

1 𝑥 𝑦1 𝑥1 𝑦11 𝑥2 𝑦2

= 12

1 3 41 2 31 5 4

= 12 [(8-15) - (12-20) + (9-8)] = 2

2 = 1

𝑁1 = 𝐴1𝐴

= 24 = 0.5

𝑁2 = 𝐴2𝐴

= 14 = 0.25

𝑁3 = 𝐴3𝐴

= 14 = 0.25

u = 𝑁1𝑢1+𝑁2𝑢2+𝑁3𝑢3 = 0.5 x 11 + 0.25 x 14 + 0.25 x 17 = 13.25 mm

70

Example 3.2 Derivation of shape function of four node triangular element Derive the shape function of a four node triangular element.

Fig. Ex.3.2 Degrading for four node element

The procedure for four node triangular element is the same as five node triangular element to derive its interpolation functions. Here, node 5 and 6 are omitted and therefore displacements in these nodes can be expressed in terms of the displacements at their corner nodes. Hence,

2 3 1 3

5 6u u u uu ' an d u '

2 2

(3.11.1)

Substituting the values of u’5 and u’6 in eq.(3.3.8), the following relations can be obtained.

2 3 3 11 1 2 2 3 3 4 4 5 6

6 5 5 61 1 2 2 3 3 4 4

u u u uu N u N u N u N u N N

2 2N N N NN u N u N u N u2 2 2

(3.11.2)

Now, the displacement at any point inside the four node element can be expressed by its nodal displacement with help of shape function.

1 1 2 2 3 3 4 4u N u N u N u N u (3.11.3)

Comparing eq. (3.11.2) and eq. (3.11.3), one can find the following relations.

6 3 11 1 1 1 1 2

5 2 32 2 2 2 2 1

5 6 2 3 3 13 3 3 3 3

4 4 1 2

N 4L LN N L 2L 1 L 1 2L2 2N 4L LN N L 2L 1 L 1 2L2 2

N N 4L L 4L LN N L 2L 1 L2 2

N N 4L L

(3.11.4)

Thus, the shape functions for the four node triangular element are

71

1 1 2

2 2 1

3 3

4 1 2

N L 1 2L

N L 1 2L

N LN 4L L

(3.11.5)

Example 3.3 Numerical integration for two dimensional problems

Evaluate the integral: 3

2

2

11 32I x x dx

using one, two and three point gauss Quadrature.

Also, find the exact solution for comparison of accuracy. Solution: The existing limits of integration should be changed from [-2, +3] to [-1, +1]. Assuming, a+bxx ,

the upper and lower limit can be changed. i.e., at 1 12, 1x x and at 2 23, 1x x . Thus,

putting these limits and solving for a & b, we get a = -0.2 and b = 0.4. The relation between two coordinate systems will become:

0.2 0.4xx or 5 1x

2x and dx 2.5d x

Thus, the initial equation can be written as

3

2

2

11 32I x x dx

= 21

1

5 1 5 12.5 11 322 2

d

x x x

(i) Exact Solution:

3

2

233 2

2

11 32

11 323 2

99 89 96 22 642 3

37.5 83.33333 120.83333

I x x dx

x x x

Thus, Iexact = -120.83333

72

(ii) One Point Formula:

1

1 11

I d w

f x x f x

For one point formula in Gauss Quadrature integration, 1 12, 0w x . Thus,

2

15 0 1 5 0 12 2.5 11 32

2 2

1 11 5 32 131.254 2

I

Thus, % of error = (120.83333-131.25)×100/120.83333 = 8.62% (iii) Two Point Formula: Here, for two point formula in Gauss Quadrature integration,

1 2 1 21w w 1.0 and3

x x

. Thus,

2 1 1 2 2

2 25 5 5 51 1 1 13 3 3 31.0 2.5 11 32 1.0 2.5 11 322 2 2 2

I w w f x f x

0.88996 10.37713 32 2.5 3.77671 21.3771 32 2.548.3333 2.5

= 120.83325

Thus, % of error = (120.83333-120.83325)×100/120.83333 = 6.62×10-05

(iv) Three Point Formula: Here, for three point formula in Gauss Quadrature integration,

1 1

2 2

3 3

w 0.8889, 0.00.5556, 0.77460.5556, 0.7746

ww

x x x

Thus,

3 1 1 2 2 3 3I w w w f x f x f x

73

2

3

2

2

5 0 1 5 0 10.8889 2.5 11 322 2

5 0.7746 1 5 0.7746 10.5556 2.5 11 322 2

5 0.7746 1 5 0.7746 10.5556 2.5 11 322 2

I

3 0.8889 2.5 0.25 5.5 32

0.5556 2.5 5.9365 26.8015 32

0.5556 2.5 2.0635 15.8015 32

2.5 23.3336 0.4100 25.41202.5 48.3356 120.839

I

Thus, % of error = (120.83333-120.839)×100/120.83333 = 4.69×10-03. However, difference of results will approach to zero, if few more digits after decimal points are taken in calculation. Example 3.4 Numerical integration for three dimensional problems

Evaluate the integral:

1 1 1 2 2 2

1 1 1I 1 2 1 3 2 d d d

x h z x h z

Solution: Using two point gauss Quadrature formula for the evaluation of three dimensional integration, we have the following sampling points and weights.

1 2

1

2

1

2

1

2

w w 10.5773502692

0.57735026920.5773502692

0.57735026920.5773502692

0.5773502692

x x h h z z

Putting the above values, in 2 2 2, , 1 2 1 3 2f x h z x h z one can find the following values

in 8 (i.e., 2 × 2 × 2) sampling points.

74

1 1 1

1 1 2

1 2 1

1 2 2

2 1 1

2 1 2

2 2 1

2 2 2

, , 160.8886

, , 0.8293

, , 11.5513

, , 0.0595

, , 0.8293

, , 0.0043

, , 0.0595

, , 0.0003

f x h z

f x h z

f x h z

f x h z

f x h z

f x h z

f x h z

f x h z

Now, 2 2 2

i j k i j ki 1 j 1 k 1

I w w w , ,

f x h z

Thus, 1 1 1 1 1 1 1 1 2 1 1 2 2 2 2 2 2 2I w w w , , w w w , , w w w , , f x h z f x h z f x h z = 174.222, where as

Iexact = 174.222.

Content Module 4: Analysis of Frame Structures (06) Lecture 1: Stiffness of Truss Members(Page: 1-6)

4.1.1 Introduction

4.1.2 Element Stiffness of a Truss Member

4.1.3 Member Stiffness with Varying Cross Section

4.1.4 Generalized Stiffness Matrix of a Plane Truss Member

Lecture 2: Analysis of Truss(Page: 7-14) 4.2.1 Element Stiffness of a 3 Node Truss Member

4.2.2 Worked Out Example

Lecture 3: Stiffness of Beam Members(Page: 15-21) 4.3.1 Introduction

4.3.2 Derivation of Shape Function

4.3.3 Derivation of Element Stiffness Matrix

4.3.4 Generalized Stiffness Matrix of a Beam Member

Lecture 4: Finite Element Analysis of Continuous Beam(Page: 22-30) 4.4.1 Equivalent Loading on Beam Member

4.4.1.1 Varying Load

4.4.1.2 Concentrated Load

4.4.2 Worked Out Example

Lecture 5: Plane Frame Analysis(Page: 31-38) 4.5.1 Introduction

4.5.2 Member Stiffness Matrix

4.5.3 Generalized Stiffness Matrix

4.5.4 Worked Out Example

Lecture 6 Analysis of Grid and Space Frame(Page: 38-48) 4.6.1 Introduction

4.6.2 Member Stiffness Matrix

4.6.3 Generalized Stiffness Matrix

4.6.4 Worked Out Example

4.6.5 Analysis of Space Frame

1

Lecture 1: Stiffness of Truss Members 4.1.1 Introduction Analysis of frame structures can be carried out by the approach of stiffness method. However, such types of structures can also be analyzed by finite element method. A unified formulation will be demonstrated based on finite element concept in this module for the analysis of frame like structures. A truss structure is composed of slender members pin jointed together at their end points. Truss element can resist only axial forces (tension or compression) and can deform only in its axial direction. Therefore, in case of a planar truss, each node has components of displacements parallel to X and Y axis. Planar trusses lie in a single plane and are used to support roofs and bridges. Such members will not be able to carry transverse load or bending moment. The major benefits of use of truss structures are: lightweight, reconstructable, reconfigurable and mobile. Configuration of few standard truss structures are shown in Fig. 4.1.1.

Fig. 4.1.1 Configuration of various truss structures

4.1.2 Element Stiffness of a Truss Member Since, the truss is an axial force resisting member, the displacement along its axis only will be developed due to axial load. Therefore, using Pascal’s triangle, the displacement function of truss member for development of shape function can be expressed as:

( ) [ ] 00 1

1

1u x x xα

α αα

= + = (4.1.1)

2

Fig. 4.1.2 Axial force on the member along X axis Applying boundary conditions as shown in Fig. 4.1.2: At x= 0, u(0)= u 1 and at x=L, u(L) = u 2

Thus, 10 u=α and L

uu 121

−=α

. Therefore,

( ) [ ] uNuLxu

Lxxu =+

−= 211

(4.1.2)

Here, N is the shape function of the element and is expressed as:

[ ]

−=

Lx

LxN 1 (4.1.3)

So we get the element stiffness matrix as [ ] [ ] [ ][ ]Tk B D B d

Ω

= Ω∫∫∫ (4.1.4)

Where, [ ] [ ]

−==

LLdxNdB 11

So, the stiffness matrix will become:

[ ] [ ]

−

−=

−

−== ∫∫ 11

11111

1

00 LAEdx

LLL

LAEAdxBEBLL T

Thus, the stiffness matrix of the truss member along its member axis will be:

[ ]

−

−=

1111

LAEk (4.1.5)

4.1.3 Element Stiffness of Truss Member with Varying Cross Section

3

Now, let us find the stiffness matrix of a pin-jointed member of length L with respect to local axis, having cross sectional areas A1 and A2 at the two ends of the member as shown in the figure below.

Fig. 4.1.3 Member with varying cross section

From the above figure, the cross sectional area at a distance of x from left end can be expressed as:

xL

AAAAx12

1−

+= (4.1.6)

As it is a pin-jointed member, the displacement at any point may be expressed in terms of nodal displacement as 1 1 2 2u N u N u= + .

Similarly the cross sectional area at any point may be represented in terms of the cross sectional area of the two ends. Thus 2211 ANANAx +=

Where the shape functions are: 1 21 ;x xN NL L

= − =

Now, the strain may be written as:

2122

11 11 u

Lu

Lu

xNu

xN

xu

x +−=∂∂

+∂∂

=∂∂

=ε = [ ] [ ] uBu

u

L=

−2

1

111 (4.1.7)

As the stress is proportional to strain according to Hook’s law, the stress-strain relationship will be as follows:

[ ] [ ] uBEu

u

LEE xx =

−==2

1

11εσ

(4.1.8)

Now the strain energy may be expressed as

dxAEdvU xx

LT

xV

xT

x εεσε ∫∫ ==02

121 [ ] [ ] dxAuBEBu x

LTT∫=

021

(4.1.9)

Applying Castigliano’s theorem, the force will become:

4

[ ] [ ] dxAuBEBuUF x

LT∫=∂

∂=

0

[ ] [ ] [ ] dku

udxA

LE

x

TL

=

−−= ∫2

1

02 1111

(4.1.10)

Thus, the stiffness matrix will be:

[ ] ∫

−

+

−

−=

L

dxxL

AAALEk

0

1212

11

11

L

xL

AAxALE

0

21212 211

11

−

+

−

−=

−

+

−

−=

21111 12

1AAA

LE ( )

−

−+=

1111

2 21 AAL

E

(4.1.11)

4.1.4 Generalized Stiffness Matrix of a Plane Truss Member Let us consider a member making an angle ‘θ’ with X axis as shown in the figure below. By resolving the forces along local X and Y direction, the following relations are obtained.

1 1 1

2 2 2

1 1 1

2 2 2

cos sin

cos sin

sin cos

sin cos

x x y

x x y

y x y

y x y

F F F

F F F

F F F

F F F

θ θ

θ θ

θ θ

θ θ

= +

= +

= − +

= − +

(4.1.12)

Where, 1xF and 2xF are the axial forces along the member axis X . Similarly, 1yF and 2yF are

the forces perpendicular to the member axis X .

Fig. 4.1.4 Inclined truss member

5

The relationship expressed in eq. (4.1.12) can be rewritten in matrix form as follows:

11

11

22

22

cos 0 0cos 0 0

0 0 cos0 0 cos

xx

yy

xx

yy

FF sinFF sinFF sinFF sin

θ θθ θ

θ θθ θ

− = −

(4.1.13)

Now, the above equation can be expressed in short as:

[ ] FTF = (4.1.14)

Here, [T] is called transformation matrix. This relates between the global (𝑋,𝑌 axis) and member axis (𝑋,𝑌 axis). Similarly, the relations of nodal displacements between two coordinate systems may be written as: [ ] d T d= (4.1.15)

Again, the equation stated in (4.1.5) can be generalized and expressed with respect to the member axis including force and displacement vector as:

1 1

1 1

2 2

2 2

1 0 1 00 0 0 01 0 1 0

0 0 0 0

x

y

x

y

F uF vAEF uLF v

− = −

(4.1.16)

Where, the nodal forces in Y direction are zero. The above equation may also be expressed in short as:

[ ] dkF = (4.1.17) Where, the matrices in the above equation are written with respect to the member axis. Now, eq. (4.1.17) can be rewritten with the use of eq. (4.1.14) and (4.1.15) as given below.

[ ] [ ][ ] dTkFT = (4.1.18) Or,

[ ] [ ][ ] dTkTF 1−= (4.1.19)

6

Here, the transformation matrix [T] is orthogonal, i.e., [T]-1 is equal to [T]T. Therefore, from the above relationship, the generalized stiffness matrix can be expressed as:

[ ] [ ] [ ][ ]TkTk T= (4.1.20) Thus,

[ ]

cos sin 0 0 1 0 1 0 cos sin 0 0sin co s 0 0 0 0 0 0 sin co s 0 0

0 0 cos sin 1 0 1 0 0 0 cos sin0 0 sin co s 0 0 0 0 0 0 sin co s

AEkL

θ θ θ θθ θ θ θ

θ θ θ θθ θ θ θ

− − − = − − −

(4.1.21)

Or,

[ ]

−−

−−

−−

−−

=

θθθθθθ

θθθθθθ

θθθθθθ

θθθθθθ

22

22

22

22

sincossinsincossin

cossincoscossincos

sincossinsincossin

cossincoscossincos

LAEk (4.1.22)

The above stiffness matrix can be used for the analysis of two-dimensional truss problems.

7

Lecture 2: Analysis of Truss 4.2.1 Element Stiffness of a 3 Node Truss Member

Fig. 4.2.1 3-node truss member Here, the displacement function using Pascal’s triangle can be expressed as:

( ) 20 1 2u x x xα α α= + +

02

1

2

1 x xααα

=

(4.2.1)

Applying boundary conditions: At x= 0, u(0)= u 1 , x=L/2, u(L/2) = u

2 and at x=L, u(L) = u

3

And solving for 0α , 1α and 2α

10 u=α ,

1 2 31

3 4u u uL

α − + −=

and 3

3212

242L

uuu +−=α

Therefore,

( ) [ ] 2 2 2

1 2 32 2 2

3 2 4 4 21 x x x x x xu x u u u N uL L L L L L

= − + + − + − + =

(4.2.2)

Here, N is the shape function of the element and is expressed as:

[ ]2 2 2

2 2 2

3 2 4 4 21 x x x x x xNL L L L L L

= − + − − +

(4.2.3)

Now, the element stiffness matrix can be written as [ ] [ ] [ ][ ]Tk B D B d

Ω

= Ω∫∫∫

(4.2.4)

Where, [ ] [ ]2 2 2

3 4 4 8 1 4d N x x xBd x L L L L L L

= = − + − − +

So, the stiffness matrix will be:

[ ] [ ] [ ][ ]Tk B D B dΩ

= Ω∫∫∫ [ ] [ ]0

L TB E B Adx= ∫

8

= 𝐴𝐸

⎩⎪⎨

⎪⎧−

3𝐿

+4𝑥𝐿2

4𝐿−

8𝑥𝐿2

−1𝐿

+4𝑥𝐿2⎭⎪⎬

⎪⎫

𝐿

0× −

3𝐿

+4𝑥𝐿2

4𝐿−

8𝑥𝐿2

−1𝐿

+4𝑥𝐿2 𝑑𝑥

=𝐴𝐸𝐿2

⎣⎢⎢⎢⎢⎢⎡ 9 +

16𝑥2

𝐿2−

24𝑥𝐿

−12 +40𝑥𝐿

−32𝑥2

𝐿23 −

16𝑥𝐿

+16𝑥2

𝐿2

−12 +40𝑥𝐿

−32𝑥2

𝐿216 −

64𝑥𝐿

+64𝑥2

𝐿2−4 +

24𝑥𝐿

−32𝑥2

𝐿2

3 −16𝑥𝐿

+16𝑥2

𝐿2−4 +

24𝑥𝐿

−32𝑥2

𝐿21 −

8𝑥𝐿−

16𝑥2

𝐿2 ⎦⎥⎥⎥⎥⎥⎤

𝐿

0𝑑𝑥

(4.2.5)

After integrating the above equation, the stiffness matrix of the 3-node truss member will become:

[𝑘] = AE3L

7 −8 1−8 16 −81 −8 7

(4.2.6)

4.2.2 Worked Out Example Analyze the truss shown below by finite element method. Assume the cross sectional area of the inclined member as 1.5 times the area (A) of the horizontal and vertical members. Assume modulus of elasticity is constant for all the members and is E.

Fig. 4.2.2 Plane truss

9

Solution The analysis of truss starts with the numbering of members and joints as shown below:

Fig. 4.2.3 Numbering of members and nodes The member information for the truss is shown in Table 4.2.1. The member and node numbers, modulus of elasticity, cross sectional areas are the necessary input data. From the coordinate of the nodes of the respective members, the length of each member is computed. Here, the angle θ has been calculated considering anticlockwise direction. The signs of the direction cosines depend on the choice of numbering the nodal connectivity. Now, let assume the coordinate of node 1 as (0, 0). The coordinate and restraint joint information are given in Table 4.2.2. The integer 1 in the restraint list indicates the restraint exists and 0 indicates the restraint at that particular direction does not exist. Thus, in node no. 2, the integer 0 in x and y indicates that the joint is free in x and y directions.

Table 4.2.1 Member Information for Truss

Member No.

Starting Node

Ending Node

Value of θ

Area Modulus of Elasticity

1 1 2 90° A E 2 2 3 315° 1.5A E 3 3 1 180° A E

Table 4.2.2 Nodal Information for Plane Truss

Node No. Coordinates Restraint List x y x y

1 0 0 1 1 2 0 L 0 0 3 L 0 1 1

10

The stiffness matrices of each individual member can be found out from the stiffness matrix equation as shown below.

[ ]

−−

−−

−−

−−

=

θθθθθθ

θθθθθθ

θθθθθθ

θθθθθθ

22

22

22

22

sinsincossinsincos

sincoscossincoscos

sinsincossinsincos

sincoscossincoscos

LAEk

Thus the local stiffness matrices of each member are calculated based on their individual member properties and orientations and written below.

and

Global stiffness matrix can be formed by assembling the local stiffness matrices into globally. Thus the global stiffness matrix are calculated from the above relations and obtained as follows:

5 6 1 2 5

6

1

2

[ ]

−

−

−

=

0000

0101

1000

0101

3 LAEk

3 4 5 6 3

4

5

6

[ ]

−−

−−

−−

−−

=

1111

1111

1111

1111

243

2 LAEk

1 2 3 4 1

2

3

4

[ ]

−

−=

1010

0000

1010

0000

1 LAEk

1 2 3 4 5 6 1 2 3 4 5 6

11

[ ]

−−

−+−−

−+−−

−−

−−

=

243

243

243

24300

2431

243

243

24301

243

243

2431

24310

243

243

243

24300

001010010001

LAEK

The equivalent load vector for the given truss can be written as:

1

1

2

2

3

3

00

2

00

x

y

x

y

x

y

FFF P

FF PFF

= =

Let us assume that u and v are the horizontal and vertical displacements respectively at joints. Thus the displacement vector will be expressed as follows:

=

=

00

00

2

2

3

3

2

2

1

1

vu

vuvuvu

d

Therefore, the relationship between the force and the displacement will be:

12

1

1

2

2

3

3

1 0 0 0 1 00 1 0 1 0 0

03 3 3 30 0 04 2 4 2 4 2 4 22 3 3 3 30 1 1

4 2 4 2 4 2 4 23 3 3 3 01 0 1

4 2 4 2 4 2 4 2 03 3 3 30 0

4 2 4 2 4 2 4 2

x

y

x

y

FF

P uAEP vL

FF

− −

− −

= =− − + − − − + −

− −

From the above relation, the unknown displacements u2 and v2 can be found out through computer programming. However, as numbers of unknown displacements in this case are only two, the solution can be obtained by manual calculations. The above equation may be rearranged with respect to unknown and known displacements in the following form:

F k k dF k k dα αα αβ α

β βα ββ β

=

Thus the developed matrices for the truss problem can be rearranged as:

−−

−+−−

−−

−−+

−

−−

=

243

24300

243

243

2431

24301

243

243

001010010100

243

243

1024

31243

243

24300

243

243

LAE

. The above relation may be condensed into following

2

2

3 32 4 2 4 2

3 314 2 4 2

uP AEvP L

− = − +

2P

P

Fx1

Fy1

Fx3

Fy3

u2

v2

0 0

0

0

13

The unknown displacements can be derived from the relationships expressed in the above equation.

1

2

2

3 3 3 312 24 24 2 4 2 4 2 4 2

3 3 3 3314 2 4 2 4 2 4 2

u P PAE Lv P PL AE

−− + = = − +

Thus the unknown displacement at node 2 of the truss structure will become:

2

2

8 233

3

u PLv AE

+ =

Support Reactions: The support reactions Ps can be determined from the following relation:

s csP P K dβα α = − + Where, Pcs correspond to equivalent loadings at supports. Thus, the support reaction of the present truss structure will be:

0 00 0 1

8 20 33 330 4 2 4 2 3

0 3 34 2 4 2

sAE PLPL AE

− + − = − + −

−−

=

PPP

2230

Member End Actions: Now, the member end actions can be obtained from the corresponding member stiffness and the nodal displacements. The member end forces are derived as shown below. Member –1

1

1

2

2

00 0 0 0 0

00 1 0 1 3

8 20 0 0 0 0330 1 0 1 3

3

mx

my

mx

my

FF PAE PLF L AEF P

− − = = + −

14

Member – 2

2

2

3

3

8 21 1 1 1 2331 1 1 1 23 3

1 1 1 1 24 2 01 1 1 1 2

0

mx

my

mx

my

F PF PAE PLF PAELF P

− − + − − − = = − − − − −

Member –3

=

−

−

=

0000

0000

0000010100000101

1

1

3

3

AEPL

LAE

FFFF

my

mx

my

mx

Thus the member forces in all members of the truss will be:

( ) ( )2 2

3 3

2 2 2 200

m

P P

F P P P

= − + = −

The reaction forces at the supports of the truss structure will be:

032

2

R

PF

PP

− = −

Thus the member force diagram will be as shown in Fig. 4.2.4.

Fig. 4.2.4 Member Force Diagram

15

Lecture 3: Stiffness of Beam Members 4.3.1 Introduction A beam is a structural member which is capable of withstanding load primarily by resisting bending. The primary tool for analysis of beam is the Euler–Bernoulli beam equation. Other methods for determining the deflection of beams include "slope deflection method" and "method of virtual work". For calculation of internal forces of beam include "moment distribution method", force or flexibility method and stiffness method. However, all these methods have limitations if either of geometry, loading, material properties or boundary conditions becomes arbitrary in nature. Finite element techniques can well handle such cases and relieve the analyzer of making simplifications to arrive approximate solutions. 4.3.2 Derivation of Shape Function The degrees of freedom at each node for a beam member will be (i) vertical deflection and

(ii) rotation. For a beam member, the slope of the elastic curve θ is given by: dxdv

=θ , where

the variable v is the displacement function of the beam. As the beam has two degrees of freedom at each node, the variation of v will be cubic and can be expressed using Pascal’s triangle as:

( ) 33

2210 xxxxv αααα +++= [ ]

=

3

2

1

0

321

αααα

xxx (4.3.1)

and

[ ]

==

3

2

1

0

23210

αααα

θ xxdxdv (4.3.2)

Fig. 4.3.1 Beam element

16

Now, applying boundary conditions, the following expressions from the above relations can be obtained: At x=0:

[ ]0

11

2

3

1 0 0 0V

αααα

=

; [ ]0

11

2

3

0 1 0 0

αα

θαα

=

;

At x=L:

0

12 32

2

3

1V L L L

αααα

=

;

0

122

2

3

0 1 2 3L L

αα

θαα

=

Thus combining the above expressions one can write:

[ ]

01

112 3

222

32

1 0 0 00 1 0 010 1 2 3

V

AV L L L

L L

ααθ

αααθ

= =

(4.3.3)

So,

10 1

1 22 3

2 22

3 2

1 0 0 00 1 0 010 1 2 3

V

VL L LL L

αα θαα θ

− =

−

−−−=

2

2

1

1

2323

22

1212

132300100001

θ

θV

V

LLLL

LLLL (4.3.4)

Therefore,

( ) [ ]

−

−−−=

2

2

1

1

2323

2232

1212

132300100001

1

θ

θV

V

llll

llllxxxxv [ ]

=

2

2

1

1

4321

θ

θV

V

NNNN (4.3.5)

Where,

33

221

231 xL

xL

N +−= ; 2

32

22

Lxx

LxN +−= ;

3

3

2

2

323Lx

LxN −= and

2

32

4 Lx

LxN +−=

(4.3.6)

17

N is called shape function which interpolates the beam displacement in terms of its nodal displacements. 4.3.3 Derivation of Element Stiffness Matrix Now, the strain displacement relationship matrix [B] can be expressed from the following expressions with the help of eq. (4.3.1):

[ ] [ ] [ ][ ] dABBxdx

vd 1

4

2

1

0

2

2

6200 −==

== α

α

α

α

α

χ (4.3.7)

Where, [ ] [ ] [ ]

1

22 3

22

2

1 0 0 00 1 0 0

0 0 2 6 ; ;10 1 2 3

V

B x A dVL L L

L L

θ

θ

= = =

From the moment curvature relationship, we can write:

[ ][ ] dABEIdx

vdEIEIM 12

2−=== χ (4.3.8)

Strain energy,

[ ] [ ]dxMU TL

χ∫=0 2

1 = [ ] [ ] [ ][ ] ∫ −−L

TTT dxdABBAdEI

0

11

2 (4.3.9)

Thus,

[ ] [ ] [ ][ ] dxdABBAEIdUF T

LT 1

0

1 −−∫=∂∂

= (4.3.10)

So, the stiffness matrix will be:

[ ] [ ] [ ] [ ][ ]dxABBAEIk TL

T 1

0

1 −−∫= [ ] [ ] [ ] [ ] 1

0

1 −− ∫= AdxBBAEIL

TT (4.3.11)

Now, [ ][ ] [ ]dxx

x

dxBBLL

T 6200

6200

00∫∫

= ∫

=L

dx

xxx0

236120012400

00000000

=

32

2

12600

6400

0000

0000

LL

LL

(4.3.12)

18

So,

[ ] [ ] [ ] 1

32

2

1

12600

6400

0000

0000

−−

= A

LL

LLAEIk T

[ ]

−

−−−

−

−

−

−

=

2323

22

32

2

2

32

2

32

1212

13230010

0001

12600

6400

0000

0000

1100

2300

1210

2301

LLLL

LLLL

LL

LL

LL

LL

LL

LL

EIk

−

−−

−

−=

2323

22

1212

13230010

0001

6200

6000

0200

6000

LLLL

LLLL

l

EI

−

−−−

−−=

LLLL

LLLL

LLLL

LLLL

EI

4626

612612

6646

612612

22

2323

222

2323

Thus, the element stiffness of a beam member is:

[ ]2 2

3

2 2

12 6 12 66 4 6 212 6 12 6

6 2 6 4

L LL L L LEIk

L LLL L L L

− − = − − − −

(4.3.13)

4.3.4 Generalized Stiffness Matrix of a Beam Member Consider a beam member making an angle ‘θ’ with X axis as shown in Fig 4.3.2 below. By resolving the forces along local X and Y direction, the following relations are obtained.

19

1 1 1

2 2 2

1 1 1

2 2 2

1 1

2 2

cos sin

cos sin

sin cos

sin cos

x x y

x x y

y x y

y x y

F F F

F F F

F F F

F F F

M MM M

θ θ

θ θ

θ θ

θ θ

= +

= +

= − +

= − +

=

=

(4.3.14)

Where, 1xF and 2xF are the axial forces along the member axis X . Similarly, 1yF and 2yF are

the forces perpendicular to the member axis X . 1M and 2M are the moment about its axis at node 1 and 2 respectively.

Fig. 4.3.2 Inclined beam member The relationship expressed in eq. (4.3.14) can be rewritten in matrix form as follows:

11

11

11

22

22

22

cos 0 0 0 0cos 0 0 0 0

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

xx

yy

xx

yy

FsinFFsinFMMFsinFFsinFMM

θ θθ θ

θ θθ θ

− = −

(4.3.15)

Now, the above equation can be expressed in short as:

[ ] F T F= (4.3.16)

20

Similarly, the displacement vector in local coordinate system (𝑋,𝑌) may be transformed to global (𝑋,𝑌) coordinate system by the following relation.

[ ] d T d= (4.3.17)

The force-displacement relation in local coordinate system may be expressed as:

1 13 2 3 2

1 1

2 21 1

2 2

2 23 2 3 2

22

2 2

0 0 0 0 0 012 6 12 60 0

6 4 6 20 0

0 0 0 0 0 012 6 12 60 0

6 2 6 40 0

EI EI EI EIFx uL L L L

Fy vEI EI EI EIM L L L LFx u

EI EI EI EIFy vL L L LM EI EI EI EI

L L L L

θ

θ

− − = − − −

−

(4.3.18)

The matrices in the above equation are written with respect to the member axis. Now, the eq. (4.3.18) can be rewritten as follows with the use of eqs. (4.3.16) and (4.3.17).

[ ] [ ][ ] dTkFT = (4.3.19)

Or,

[ ] [ ] 1F T k T d− = (4.3.20)

Here, the transformation matrix [T] is orthogonal. Thus, from the above relationship, the generalized stiffness matrix can be expressed as:

[ ] [ ] [ ][ ][ ]TkTk T= (4.3.21) Considering cosλ θ= and sinµ θ= the above expression can be written as follows:

21

[ ]

3 2 3 2

2 2

3 2 3 2

2 2

0 0 0 0 0 012 6 12 60 00 0 0 0 0 0 0 0

0 0 0 0 0 0 0 06 4 6 20 00 0 1 0 0 0 0 0 1 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 12 6 12 6 0 0 0 00 00 0 0 0 0 1 0 0 0 0 0 1

6 2 6 40 0

L L L L

L L L Lk EI

L L L L

L L L L

λ µ λ µµ λ µ λ

λ µ λ µµ λ µ λ

−− − − = − − − − − −

(4.3.22) Thus, the generalized stiffness matrix of a beam member is derived as:

[ ]

2 2

3 3 2 3 3 2

2 2

3 3 2 3 3 2

2 2 2 2

2 2

3 3 2 3 3 2

2 2

3 3 2 3 3 2

2 2 2 2

12 12 6 12 12 6

12 12 6 12 12 6

6 6 4 6 6 2

12 12 6 12 12 6

12 12 6 12 12 6

6 6 2 6 6 4

L L L L L L

L L L L L L

L L L L L Lk EI

L L L L L L

L L L L L L

L L L L L L

µ µλ µ µ µλ µ

µλ λ λ µλ λ λ

µ λ µ λ

µ µλ µ µ µλ µ

µλ λ λ µλ λ λ

µ λ µ λ

− − − −

− − − −

= − −

− − − −

− −

(4.3.23)

22

Lecture 4: Analysis of Continuous Beam

4.4.1 Equivalent Loading on Beam Member In finite element analysis, the external loads are necessary to be acting at the joints, which does not happen always; as some forces may act on the member. The forces acting on the member should be replaced by equivalent forces acting at the joints. These joint forces obtained from the forces on the members are called equivalent joint loads. These joint loads are combined with the actual joint loads to provide the combined joint loads, which are then utilized in the analysis. 4.4.1.1 Varying Load Let a beam is loaded with a linearly varying load as shown in the figure below. The equivalent forces at nodes can be expressed using finite element technique. If w(x) is the function of load, then the nodal load can be expressed as follows.

[ ] ( )TQ N w x dx= ∫ (4.4.1)

The loading function for the present case can be written as:

( ) 2 11

w ww x w xL−

= + (4.4.2)

Fig. 4.4.1 Varying load on beam From eqs. (4.4.1) and (4.4.2), the equvalent nodal load will become

( )

( )

( )

( )

−−

+

+

+

=

−

+−

+−

+−

=

=

∫

∫

∫

∫

221

21

221

21

0

2

2

30

2

2

3

30

2

2

30

2

2

3

3

2

2

1

1

2030

207

203

3020

203

207

32

2

132

Lww

Lww

Lww

Lww

dxxwLx

Lx

dxxwLx

Lx

dxxwxLx

Lx

dxxwLx

Lx

M

F

M

F

Q

L

L

L

L

(4.4.3)

23

Now, if w1=w2=w, then the equivalent nodal force will be:

2

2

2

12

2

12

wL

wL

QwL

wL

= −

(4.4.4)

4.4.1.2 Concentrated Load Consider a force F is applied at a point is regarded as a limiting case of intense pressure over infinitesimal length, so that p(x)dx approaches F. Therefore,

[ ] ( ) * TTQ N p x dx N F = = ∫ (4.4.5)

Fig. 4.4.2 Concentrated load on beam Here, [N*] is obtained by evaluating [N] at point where the concentrated load F is applied. Thus,

[ ]

3 2 3 2

3 2 3 2

3 2 3 2

2 2

3 2 3 2

3 2 3 2

3 2 3 2

2 2

2 3 2 31 1

2 2

*2 3 2 3

x x a aL L L Lx x a ax aL L L LN at dista nce a

x x a aL L L Lx x a aL L L L

− + − +

− + − + = = − + − + − −

(4.4.6)

24

Therefore,

−

+−

+−

+−

=

=

La

La

La

La

aLa

La

La

La

M

F

M

F

Q

2

2

3

2

2

3

3

2

2

3

2

2

3

3

2

2

1

1

32

2

132

(4.4.7)

Now, if load F is acting at midspan (i.e., a=L/2), then equivalent nodal load will be

2

8

2

8

F

FL

QF

FL

= −

(4.4.8)

With the above approach, the equivalent nodal load can be found for various loading function acting on beam members. 4.4.2 Worked Out Example Analyze the beam shown below by the stiffness method. Assume the moment of inertia of member 2 as twice that of member 1. Find the bending moment and reactions at supports of the beam assuming the length of span, L as 4 m, concentrated load (P) as 15 kN and udl, w as 4 kN/m.

Fig. 4.4.3 Example of a continuous beam

25

Solution Step 1: Numbering of Nodes and Members The analysis of beam starts with the numbering of members and joints as shown below:

Fig. 4.4.4 Numbering of nodes and members The member AB and BC are designated as (1) and (2). The points A,B,C are designated by nodes 1, 2 and 4. The member information for beam is shown in tabulated form as shown in Table 4.4.1. The coordinate of node 1 is assumed as (0, 0). The coordinate and restraint joint information are shown in Table 4.4.2. The integer 1 in the restraint list indicates the restraint exists and 0 indicates the restraint at that particular direction does not exist. Thus, in node no. 2, the integer 0 in rotation indicates that the joint is free rotation.

Table 4.4.1Member Information for Beam

Member number

Starting node Ending node Rigidity modulus

1 1 2 EI 2 2 3 2EI

Table 4.4.2 Nodal Information for Beam

Node No. Coordinates Restraint List x y Vertical Rotation

1 0 0 1 1 2 L 0 1 0 3 2L 0 1 0

26

Step 2: Formation of member stiffness matrix: The local stiffness matrices of each member are given below based on their individual member properties and orientations. Thus the local stiffness matrix of member (1) is: 1 2 3 4

Similarly, the local stiffness matrix of member (2) is: 3 4 5 6

Step 3: Formation of global stiffness matrix: The global stiffness matrix is obtained by assembling the local stiffness matrix of members (1) and (2) as follows: 1 2 3 4 5 6

[ ]

−

−−−

−

−

=

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

k

4626

612612

2646

612612

22

2323

22

2323

1

1 2 3 4

[ ]

−

−−−

−

−

=

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

k

812412

12241224

412812

12241224

22

2323

22

2323

2

3 4 5 6

[ ]

−

−−−

−

−−−

−

−

=

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

K

81241200

1224122400

41212626

1224636612

002646

00612612

22

2323

222

23223

22

2323 1 2 3 4 5 6

27

Step 4: Boundary condition: The boundary conditions according to the support of the beam can be expressed in terms of the displacement vector. The displacement vector will be as follows

2

3

000

0

dθ

θ

=

Step 5: Load vector: The concentrated load on member (1) and the distributed load on member (2) are replaced by equivalent joint load. The equivalent joint load vector can be written as

Fig. 4.4.5 Equivalent Load

Step 6 : Determination of unknown displacements:. The unknown displacement can be obtained from the relationship as given below:

−

−

+−

−

−

=

12

2

128

22

8

2

2

2

wL

wL

wLPL

wLP

PL

P

F

28

[ ] [ ] 1

F K d

d K F−

=

=

−

−

+−

−

−

×

−

−−−

−

−−−

−

−

=

−

12

2

128

22

8

2

81241200

1224122400

41212626

1224636612

002646

00612612

0

000

2

2

1

22

2323

222

23223

22

2323

3

2

wL

wL

wLPL

wLP

PL

P

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

θ

θ

The above relation may be condensed into following

Step 7: Determination of member end actions: The member end actions can be obtained from the corresponding member stiffness and the nodal displacements. The member end actions for each member are derived as shown below. Member-(1)

+−

−=

38

4420 2

2

3

2

wLPL

wLPL

EIL

θθ

EIwL

EIPL

8080

32

2 −=θ

EIwL

EIPL

60160

3

3 +−=θ

1 2 2

22 2

3

12 42 18 12 8 12

4 8 1 32012 12

PL wL PL wLEI EILL L

EI EI EIwL wLL L

θθ

− − − − = × = −

29

−

×

−

−−−

−

−

=

44

000

4626

612612

2646

612612

20 2

22

2323

22

2323

2

2

1

1

wLPL

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

EIL

MFMF

−

+−

−

−

=

2020

403

403

4040

403

403

2

2

wLPL

wLP

wLPL

wLP

Member-(2)

+−

−×

−

−−−

−

−

=

38

044

0

812412

12241224

412812

12241224

2

2

22

22

3

3

2

2

wLPL

wLPL

LL

LLLL

LL

LLLL

LEI

MFMF

−−

−

+

=

12

403

20

30403

403

20

2

2

wL

PwL

wLPL

PwL

Actual member end actions: Member (1)

−

+

−

+−

−

−

=

8

2

8

2

2020

403

403

4040

403

403

2

2

2

2

1

1

PL

P

PL

P

wLPL

wLP

wLPL

wLP

MFMF

−−

+

−

−

=

20403

403

4017

40406

403

4023

2

2

wLPL

wLP

wLPL

wLP

Member (2)

−

+

−−

−

+

=

12

2

12

2

12

403

20

30403

403

20

2

2

2

2

3

3

2

2

wL

wL

wL

wL

wL

PwL

wLPL

PwL

MFMF

−

+

+

=

0403

209

20403

403

2011

2

PwL

wLPL

PwL

The support reactions at the supports A, B and C are

+

+

−

=

=

43

209

24025

403

4023

PwL

PwL

wLP

RRR

F

C

B

A

R

Putting the numerical values of L, P and w (P=15, L=4, w=4) the member actions and support reactions will be as follows:

30

Member end actions:

=

0

075.6

7.7

925.9

3

3

2

2

M

F

M

F

,

−

=

7.7

575.7

4.7

425.7

2

2

1

1

M

F

M

F

Support reactions:

=

=

075.65.17

425.7

C

B

A

R

RRR

F

31

Lecture 5: Plane Frame Analysis 4.5.1 Introduction The plane frame is a combination of plane truss and two dimensional beam. All the members lie in the same plane and are interconnected by rigid joints in case of plane frame. The internal stress resultants at a cross-section of a plane frame member consist of axial force, bending moment and shear force. 4.5.2 Member Stiffness Matrix In case of plane frame, the degrees of freedom at each node will be (i) axial deformation, (ii) vertical deformation and (iii) rotation. Thus the frame members have three degrees of freedom at each node as shown in Fig. 4.5.1 below.

Fig. 4.5.1 Plane frame element Therefore, the stiffness matrix of the frame in its local coordinate system will be the combination of 2-d truss and 2-d beam matrices:

[𝑘]=

⎣⎢⎢⎢⎢⎢⎢⎢⎢⎡𝐴𝐸𝐿

0 0 −𝐴𝐸𝐿

0 0

0 12𝐸𝐼𝐿3

6𝐸𝐼𝐿2

0 −12𝐸𝐼𝐿3

6𝐸𝐼𝐿2

0 6𝐸𝐼𝐿2

4𝐸𝐼𝐿

0 −6𝐸𝐼𝐿2

2𝐸𝐼𝐿

− 𝐴𝐸𝐿

0 0 𝐴𝐸𝐿

0 0

0 −12𝐸𝐼𝐿3

− 6𝐸𝐼𝐿2

0 12𝐸𝐼𝐿3

− 6𝐸𝐼𝐿2

0 6𝐸𝐼𝐿2

2𝐸𝐼𝐿

0 −6𝐸𝐼𝐿2

4𝐸𝐼𝐿 ⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎤

(4.5.1)

4.5.3 Generalized Stiffness Matrix In plane frame the members are oriented in different directions and hence it is necessary to transform stiffness matrix of individual members from local to global co-ordinate system

u1 v1 θ1 u2 v2 θ2

32

before formulating the global stiffness matrix by assembly. The generalized stiffness matrix of a frame member can be obtained by transferring the matrix of local coordinate system into its global coordinate system. The transformation matrix can be expressed as:

[T]=

⎣⎢⎢⎢⎢⎡

cos 𝜃 sin𝜃 0 0 0 0− sin 𝜃 cos 𝜃 0 0 0 0

0 0 1 0 0 00 0 0 cos 𝜃 sin𝜃 00 0 0 − sin𝜃 cos 𝜃 00 0 0 0 0 1⎦

⎥⎥⎥⎥⎤

(4.5.2)

Now, the generalized stiffness matrix of the member can be obtained from the relation of [𝐾] = [𝑇]𝑇[𝐾][𝑇] . Thus considering 𝜆 = cos𝜃 and 𝜇 = sin𝜃 the stiffness matrix in global coordinate system can be written as follows:

[K] = EI

⎣⎢⎢⎢⎢⎡λ −µ 0 0 0 0µ λ 0 0 0 00 0 1 0 0 00 0 0 λ −µ 00 0 0 µ λ 00 0 0 0 0 1⎦

⎥⎥⎥⎥⎤

×

⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡

AEL

0 0 −AEL

0 0

012EI

L36EIL2

0 −12EI

L36EIL2

06EIL2

4EIL

0 −6EIL2

2EIL

−AEL

0 0AEL

0 0

0 −12EI

L3−

6EIL2

012EI

L3−

6EIL2

06EIL2

2EIL

0 −6EIL2

4EIL ⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤

×

⎣⎢⎢⎢⎢⎡λ µ 0 0 0 0−µ λ 0 0 0 00 0 1 0 0 00 0 0 λ µ 00 0 0 −µ λ 00 0 0 0 0 1⎦

⎥⎥⎥⎥⎤

=

⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡

EAL

λ2 +12EI

L3μ2

EAL

λμ −12EI

L3λμ −

6EIL2

μ −EAL

λ2 −12EI

L3μ2 −

EAL

λμ +12EI

L3λμ −

6EIL2

μ

EAL

λμ −12EI

L3λμ

EAL

μ2 +12EI

L3λ2

6EIL2

λ −EAL

λμ +12EI

L3λμ −

EAL

μ2 −12EI

L3λ2

6EIL2

λ

−6EIL2

μ6EIL2

λ4EI

L6EIL2

μ −6EIL2

λ2EI

L

−EAL

λ2 −12EI

L3μ2 −

EAL

λμ +12EI

L3λμ

6EIL2

μ EAL

λ2 +12EI

L3μ2

EAL

λμ −12EI

L3λμ

6EIL2

μ

−EAL

λμ +12EI

L3λμ −

EAL

μ2 −12EI

L3λ2 −

6EIL2

λ EAL

λμ −12EI

L3λμ

EAL

μ2 +12EI

L3λ2 −

6EIL2

λ

−6EIL2

μ6EIL2

λ2EI

L6EIL2

μ −6EIL2

λ4EI

L ⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤

33

(4.5.3) 4.5.4 Worked Out Example Analyse the plane frame shown below by the stiffness method. Assume the modulus of elasticity of the horizontal member is 1.5 times that of the vertical member and length of the vertical member is 1.5 times that of horizontal member. Find the bending moment and reactions at support assuming the length, cross section area and modulus of elasticity of vertical member as 3.0 m, 0.4 x 0.4 m2 and 2 x 1011 N/mm2, respectively.

Fig. 4.5.2 Plane frame Solution Step 1: Numbering of Nodes and Members The numbering of members and joints of the plane frame are as shown below:

Fig. 4.5.3 Numbering of Nodes and Members The members AB and BC are designated as (1) and (2). The points A, B and C are designated by nodes 1, 2 and 3. The member information for the frame is shown in tabulated form as shown in Table 1(a). The coordinate of node 1 is assumed as (0,0). The coordinate and restraint joint information are shown in Table 1(b). The integer 1 in the restraint list indicates

34

the restraint exists and 0 indicates the restraint at that particular direction does not exist. Thus, in node no. 2, the integer 0 all the restraint type indicates that the joint is in free all the three directions.

Table 4.5.1 Member Information for Beam

Member number Starting node Ending node Rigidity modulus 1 1 2 EI 2 2 3 1.5EI

Table 4.5.2 Nodal Information for Beam Node no. Coordinates Restraint list

X Y Axial Vertical Rotation 1 0 0 1 1 1 2 0 1.5L 0 0 0 3 L 1.5L 1 1 1

Step 2: Formation of member stiffness matrix: The individual member stiffness matrices can be found out directly from eqn. shown above. Thus the stiffness matrices of each member in global coordinate system are given below based on their individual member properties and orientations. Thus the stiffness matrix of member (1) is:

[𝑘]1=

⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡

12EI(1.5L)3

0 − 6EI(1.5L)2

− 12EI(1.5L)3

0 − 6EI(1.5L)2

0 AE(1.5L)

0 0 − AE(1.5L)

0

− 6EI(1.5L)2

0 4EI(1.5L)

6EI(1.5L)2

0 2EI1.5L

− 12EI(1.5L)3

0 6EI(1.5L)2

12EI(1.5L)3

0 6EI(1.5L)2

0 − AE(1.5L)

0 0 AE(1.5L)

0

− 6EI(1.5L)2

0 2EI(1.5L)

6EI(1.5L)2

0 4EI(1.5L) ⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤

Similarly, the stiffness matrix of member (2) is :

1 2 3 4 5 6

1

2

3

4

5

6

4 5 6 7 8 9

35

[𝑘]2 =

⎣⎢⎢⎢⎢⎢⎢⎢⎢⎡A(1.5 E)

L0 0 −A(1.5 E)

L0 0

0 12(1.5 E)IL3

6(1.5 E)IL2

0 −12(1.5 E)IL3

6(1.5 E)IL2

0 6(1.5 E)IL2

4(1.5 E)IL

0 −6(1.5 E)IL2

2(1.5 E)IL

− A(1.5 E)L

0 0 A(1.5 E)L

0 0

0 −12(1.5 E)IL3

− 6(1.5 E)IL2

0 12(1.5 E)IL3

− 6(1.5 E)IL2

0 6(1.5 E)IL2

2(1.5 E)IL

0 −6(1.5 E)IL2

4(1.5 E)IL ⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎤

Step 3 : Formulation of global stiffness matrix: The global stiffness matrix is obtained by assembling by assembling the local stiffness matrix of member (1) and (2) as follows:

[K] =

⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡32EI9L3

0 − 8EI3L2

− 32EI9L3

0 − 8EI3L2

0 0 0

0 2AE3L

0 0 −2AE3L

0 0 0 08EI3L2

0 8EI3L

8EI3L2

0 4EI3L

0 0 0

−32EI9L3

0 8EI3L2

(32EI9L3

+ 1.5EAL

) 0 8EI3L2

− 1.5EAL

0 0

0 −2AE3L

0 0 (2AE3L

+ 18EIL3

) 9EIL2

0 −18EIL3

9EIL2

− 8EI3L2

0 4EI3L

8EI3L2

9EIL2

(8EI3L2

+ 6EIL

) 0 −9EIL2

3EIL

0 0 0 −1.5AEL

O O 1.5AEL

0 0

0 0 0 0 −18EIL3

− 9EIL2

0 18EIL3

− 9EIL2

0 0 0 0 9EIL2

3EIL

0 −9EIL2

6EIL ⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤

Step 4: Boundary conditions: The boundary conditions according to the support of the frame can be expressed in terms of the displacement vector. The displacement vector will be as follows:

𝑑 =

⎣⎢⎢⎢⎢⎢⎢⎢⎡

000𝛿𝑥𝐵𝛿𝑦𝐵𝜃𝐵000 ⎦⎥⎥⎥⎥⎥⎥⎥⎤

Here, 𝛿𝑥𝐵 , 𝛿𝑦𝐵 and 𝜃𝐵 indicate the displacement in X-direction, displacement in Y-direction and rotation at point B. Step 5: Load vector:

4

5

6

7

8

9

1 2 3 4 5 6 7 8 9

1

2

3

4

5

6

7

8

9

36

The distributed load on member (2) can be replaced by its equivalent joint load as shown in the figure below.

Fig. 4.5.4 Equivalent Joint Loads

Thus, the equivalent joint load vector can be written as

𝐹 =

⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡

0000

−𝑤𝐿2

−𝑤𝐿2

120

−𝑤𝐿2

𝑤𝐿2

12 ⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤

Step 6: Determination of unknown displacements: The unknown displacements can be obtained from the relationship of F = [K]d or d =[k]-1 F. Now eliminating the rows and columns in the stiffness matrix and force matrix, corresponding to zero elements in displacement matrix, the reduced matrix will be as follows.

𝛿𝑥𝐵𝛿𝑦𝐵𝜃𝐵

=

⎣⎢⎢⎢⎢⎡

32𝐸𝐼9𝐿3

+1.5𝐸𝐴𝐿

08𝐸𝐼3𝐿2

0 2𝐴𝐸3𝐿

+18𝐸𝐼𝐿3

9𝐸𝐼𝐿2

8𝐸𝐼3𝐿2

9𝐸𝐼𝐿2

8𝐸𝐼3𝐿

+6𝐸𝐼𝐿⎦⎥⎥⎥⎥⎤−1

⎣⎢⎢⎢⎡

0

−𝑤𝐿2

−𝑤𝐿2

12 ⎦⎥⎥⎥⎤

Thus, the unknown displacements will be:

𝛿𝑥𝐵𝛿𝑦𝐵𝜃𝐵

=1

1010

0.04327𝑤−1.7127𝑤−5.4978𝑤

Step 7: Determination of member end actions: The member end actions can be obtained from the corresponding member stiffness and the nodal displacements. The member end actions for each member are derived as shown below.

37

Member – (1) In case of member (1), the member forces will be:𝐹𝑚1 = [𝐾](1)𝑑(1)

⎣⎢⎢⎢⎢⎢⎡𝐹𝑥1𝐹𝑦1𝑀1𝐹𝑥2𝐹𝑦2𝑀2⎦

⎥⎥⎥⎥⎥⎤

= 106

⎣⎢⎢⎢⎢⎡

56.17 0 −126.4 −56.17 0 −126.4 0 7110 0 0 −7110 0

−126.4 0 379.2 126.4 0 189.6 −56.17 0 379.2 56.17 0 126.4

0 −7110 0 0 7110 0−126.4 0 189.6 126.4 0 379.2 ⎦

⎥⎥⎥⎥⎤

×

⎣⎢⎢⎢⎢⎡

000

4.327 𝑋 10−12𝑤−1.7127 𝑋 10−10𝑤−5.4978 𝑋 10−10𝑤⎦

⎥⎥⎥⎥⎤

=

⎣⎢⎢⎢⎢⎡

0.0697𝑤1.2177𝑤

−0.10479𝑤−0.06925𝑤−1.21661𝑤−0.20793𝑤⎦

⎥⎥⎥⎥⎤

It is to be noted that Fm are the end actions due to joint loads. Hence it must be added to the corresponding end actions in the restrained structure in order to obtain the end actions due to the loads. Therefore, Fmactual are the true member end actions due to actual loading system can be expressed as 𝐹𝑚𝑎𝑐𝑡𝑢𝑎𝑙 = Fm + Ffm Where, Ffm are the end actions in the restrained structure. Since there is no load acting on member (1), the actual end actions will be:

Fmactual =

⎣⎢⎢⎢⎢⎡

0.0697𝑤1.2177𝑤

−0.10479𝑤−0.06925𝑤−1.21661𝑤−0.20793𝑤⎦

⎥⎥⎥⎥⎤

+

⎣⎢⎢⎢⎢⎡000000⎦⎥⎥⎥⎥⎤

=

⎣⎢⎢⎢⎢⎡

0.0697𝑤1.2177𝑤

−0.10479𝑤−0.06925𝑤−1.21661𝑤−0.20793𝑤⎦

⎥⎥⎥⎥⎤

Member (2) In similar way, the member forces in member (2) will be Fm(2) = [K](2)d(2)

38

⎣⎢⎢⎢⎢⎢⎡𝐹𝑥2𝐹𝑦2𝑀2𝐹𝑥3𝐹𝑦3𝑀3⎦

⎥⎥⎥⎥⎥⎤

= 109

⎣⎢⎢⎢⎢⎡

16 0 0 −16 0 00 0.284 0.426 0 −0.284 0.4260 0.426 0.853 0 −0.426 0.426

−16 0 0 16 0 00 −0.284 −0.426 0 0.284 −0.4260 0.426 0.426 0 −0.426 0.853 ⎦

⎥⎥⎥⎥⎤

×

⎣⎢⎢⎢⎢⎡ 4.327 𝑋 10−12𝑤−1.7127 𝑋 10−10𝑤−5.4978 𝑋 10−10𝑤

000 ⎦

⎥⎥⎥⎥⎤

=

⎣⎢⎢⎢⎢⎡

0.069232𝑤−0.28325𝑤−0.54215𝑤−0.06923𝑤0.283245𝑤−0.3076𝑤 ⎦

⎥⎥⎥⎥⎤

The actual member forces in the member (2) will be:

Fmactual =

⎣⎢⎢⎢⎢⎡

0.069232𝑤−0.28325𝑤−0.54215𝑤−0.06923𝑤0.283245𝑤−0.3076𝑤 ⎦

⎥⎥⎥⎥⎤

+

⎣⎢⎢⎢⎢⎡

01.5𝑤

0.75𝑤0

1.5𝑤−0.75𝑤⎦

⎥⎥⎥⎥⎤

=

⎣⎢⎢⎢⎢⎡

0.0692𝑤1.2167𝑤0.2078𝑤−0.0692𝑤1.7832𝑤−1.0576𝑤⎦

⎥⎥⎥⎥⎤

Lecture 6 Analysis of Grid and Space Frame 4.6.1 Introduction The property of a grid member is basically a combination of 2-d beam with torsional effect. The plane frame is assumed to be loaded in its own plane where as loading in the grid is normal to its plane. As a result torsional effects are included in the grid analysis. Thus the grid member can withstand bending moment, shear force as well as torsional moment. 4.6.2 Element Stiffness Matrix for Grid Members The degrees of freedom at each node of the grid member will be (i) vertical deformation and (ii) rotation in two different directions.

39

Fig. 4.6.1 Degrees of freedom of grid element Therefore, the stiffness matrix of the grid in its local coordinate system will be:

[ ]

−

−

−

−−−

−

−

=

3232

22

3232

22

1260

1260

640

620

0000

1260

1260

620

640

0000

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LGI

LGI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LGI

LGI

k

yyyy

yyyy

xx

yyyy

yyyy

xx

(4.6.1)

Here, the G is the modulus of torsional rigidity. 4.6.3 Generalized Stiffness Matrix The generalized stiffness matrix of a grid member can be obtained by transferring the matrix of local coordinate system into its global coordinate system. The transformation matrix can be expressed as:

[T]=

⎣⎢⎢⎢⎢⎡

cos 𝜃 sin𝜃 0 0 0 0− sin 𝜃 cos 𝜃 0 0 0 0

0 0 1 0 0 00 0 0 cos 𝜃 sin𝜃 00 0 0 − sin𝜃 cos 𝜃 00 0 0 0 0 1⎦

⎥⎥⎥⎥⎤

θx1 θz1 v1 θx2 θy2 v2

Mx1

Mz1

Fy1

Mx2

Mz2

Fy2

40

Now, the generalized stiffness matrix of the member can be obtained from the relation of [𝑘] = [𝑇]𝑇𝑘[𝑇] . Thus considering 𝜆 = cos𝜃 and 𝜇 = sin𝜃 the stiffness matrix in global coordinate system can be written as follows:

[𝑘] =

⎣⎢⎢⎢⎢⎡𝜆 −𝜇 0 0 0 0𝜇 𝜆 0 0 0 00 0 1 0 0 00 0 0 𝜆 −𝜇 00 0 0 𝜇 𝜆 00 0 0 0 0 1⎦

⎥⎥⎥⎥⎤

×

⎣⎢⎢⎢⎢⎢⎢⎢⎢⎡GIxL

0 0 −GIxL

0 0

0 4EIyL

− 6EIyL2

0 2EIyL

6EIyL2

0 −6EIyL2

12EIyL3

0 −6EIyL2

− 12EIyL3

− GIxL

0 0 GIxL

0 0

0 2EIyL

− 6EIyL2

0 4EIyL

− 6EIyL2

0 6EIyL2

− 12EIyL3

0 6EIyL2

12EIyL3 ⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎤

×

⎣⎢⎢⎢⎢⎡𝜆 𝜇 0 0 0 0−𝜇 𝜆 0 0 0 00 0 1 0 0 00 0 0 𝜆 𝜇 00 0 0 −𝜇 𝜆 00 0 0 0 0 1⎦

⎥⎥⎥⎥⎤

−−−

+

−−+−

+

−

−+

+−+−

−−−

+−

+−−+

−

−

+−+−

−+

=

322322

222

222

222

222

322322

222

222

222

222

12661266

644622

644624

12661266

622644

624644

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LGI

LEI

LGI

LEI

LEI

LGI

LEI

LGI

LEI

LEI

LGI

LEI

LGI

LEI

LEI

LGI

LEI

LGI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LGI

LEI

LGI

LEI

LEI

LGI

LEI

LGI

LEI

LEI

LGI

LEI

LGI

LEI

LEI

LGI

LEI

LGI

yyyyyy

yyxyxyyxyx

yyxyxyyxyx

yyyyyy

yyxyxyyxyx

yyxyxyyxyx

λµλµ

λλµλµλλµλµ

µλµµλµλµµλ

λµλµ

λλµλµλλµλµ

µλµµλµλµµλ

(4.6.2)

4.6.4 Worked Out Example Analyze the grid shown below by the stiffness method. Draw the shear force and bending moment diagram assuming the cross sectional area and modulus of elasticity of each member as 0.3×0.3 m2 and 2×1011 N/m2 respectively. Assume EI = 3GJ. The length of member AB and BC is 4 m and 5 m respectively.

41

Fig. 4.6.2 Grid structure

Solution Step 1: Numbering of Nodes and Members The numbering of members and joints of the plane frame are as shown in the figure below:

Fig. 4.6.3 Numbering of nodes and members

The member AB and BC are designated as (1) and (2). The points A, B and C are designated by nodes 1, 2 and 3. The member information for the grid is shown in tabulated form as shown in Table 4.6.1. The coordinate of node 1 is assumed as (0, 0). The coordinate and restraint joint information are shown in Table 4.6.2. The integer 1 in the restraint list indicates the restraint exists and 0 indicates the restraint at that particular direction does not exist. Thus, in node no. 2, the integer 0 all the restraint type indicates that the joint is free in all the three directions.

Table 4.6.1 Member Information Member number Starting node Ending node

1 1 2 2 2 3

42

Table 4.6.2 Member Coordinates

ode No

coordinates Restraint list x z Vertical Rotation Rotation

1 0 0 1 1 1 2 4 0 0 0 0 3 4 5 1 1 1

Step 2: Formation of member stiffness matrix: The individual member stiffness matrices can be found out directly. Thus the stiffness matrices of each member in global coordinate system are given below based on their individual member properties and orientations. As the member AB is horizontal, i.e., θ = 0, the values of Cos θ = 1 and Sin θ = 0. Thus the stiffness matrix of member (1) is: 1 2 3 4 5 6

[ ]

−

−−−

−

−

−

−

=

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LGJ

LGJ

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LGJ

LGJ

k AB

460260

61206120

0000

260460

61206120

0000

22

2323

22

2323

Assuming EI=3GJ=3K, the above equation can be written as

1 2 3 4 5 6

[ ]

−

−−−

−

−

−

−

=

LK

LK

LK

LK

LK

LK

LK

LK

LK

LK

LK

LK

LK

LK

LK

LK

LK

LK

LK

LK

k AB

121806180

1836018360

0000

618012180

1836018360

0000

22

2323

22

2323

1 2 3 4 5 6

1

2

3

4

5

6

43

As the member BC member is also horizontal, the value of Cos θ = 1 and Sin θ = 0 and thus, the stiffness matrix will be: Step 3: Formation of global stiffness matrix: The global stiffness matrix can be obtained by assembling the local stiffness matrix of members (AB) and (BC). Now looking at the grid structure, the displacements at the fixed supports, are known and all are equal to zero. Only the displacement at co-ordinates 4, 5, 6 are unknown. So the global system stiffness matrix, corresponding to the displacement at co-ordinate 4, 5, 6 will be:

[𝐾] =

⎣⎢⎢⎢⎡𝐾𝐿𝐴𝐵

+ 12𝐾𝐿𝐵𝐶

18𝐾𝐿𝐵𝐶2

018𝐾𝐿𝐵𝐶2

36𝐾𝐿𝐴𝐵3

+ 36𝐾𝐿𝐵𝐶3

− 18𝐾𝐿𝐴𝐵2

0 − 18𝐾𝐿𝐴𝐵2

𝐾𝐿𝐵𝐶

+ 12𝐾𝐿𝐴𝐵⎦

⎥⎥⎥⎤

= K 2.65 0.72 00.72 0.8505 −1.125

0 −1.125 3.2

Step 4: Boundary condition: The boundary conditions according to the support of the grid structure can be expressed in terms of the displacement vector. The displacement vector will be as follows

6 5 4 7 8 9

[ ]

−

−−−

−

−

−

−

=

LK

LK

LK

LK

LK

LK

LK

LK

LK

LK

LK

LK

LK

LK

LK

LK

LK

LK

LK

LK

k BC

121806180

1836018360

0000

618012180

1836018360

0000

22

2323

22

2323

6 5 4 7 8 9

44

4

5

6

000

000

dd d

d

=

Here, d4, d5, and d6 indicate the displacement vectors at point B. Step 5: Load vector: The distributed load on member (1) can be replaced by its equivalent joint load as shown in the figure below.

Fig. 4.6.4 Equivalent load

Thus the equivalent load vector will be:

−

−

−

=

000

12

2

012

2

0

2

2

wL

wL

wL

wL

P

45

Step 6: Determination of unknown displacements:. The unknown displacements can be obtained from the relationship of [ ] F K d= or

[ ] 1d K F−= . Now, eliminating the rows in the force matrix, corresponding to zero

element in displacement matrix, the reduced matrix will be as follows.

𝑑4𝑑5𝑑6 = k−1

2.65 0.72 00.72 0.8505 −1.125

0 −1.125 3.2−1

⎣⎢⎢⎢⎡

0

−4𝑤2

16𝑤12 ⎦

⎥⎥⎥⎤

=1𝑘

0.662 −1.047 −0.368−1.047 3.856 1.3556−0.368 1.355 0.789

0−2𝑤43𝑤

Thus, the unknown displacements will be:

𝑑4𝑑5𝑑6 =

1𝐾

1.603𝑤−5.905𝑤— 1.658𝑤

Step 7: Determination of member end actions: The member end actions can be obtained from the corresponding member stiffness and the nodal displacements. The member end actions for each member are derived as shown below. Member - AB In case of member (AB), the member forces will be: Fm(AB) = [K](AB) d(AB)

⎣⎢⎢⎢⎢⎡𝑇1𝐹1𝑀1𝑇2𝐹2𝑀2⎦

⎥⎥⎥⎥⎤

= 𝐾

⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡

14

0 0−14

0 0

03643

1842

0 −3643

1842

01842

124

0 −1842

64

−14

0 014

0 0

0 −3643

−1842

03643

−1842

01842

64

0 −1842

124 ⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤

×1𝐾

⎣⎢⎢⎢⎢⎡

000

1.603𝑤−5.905𝑤−1.658𝑤⎦

⎥⎥⎥⎥⎤

Thus,

⎣⎢⎢⎢⎢⎡𝑇1𝐹1𝑀1𝑇2𝐹2𝑀2⎦

⎥⎥⎥⎥⎤

=

⎣⎢⎢⎢⎢⎡−0.4𝑤1.456𝑤4.156𝑤

0.4𝑤−1.456𝑤

1.70𝑤 ⎦⎥⎥⎥⎥⎤

46

It is to be noted that Fm are the end actions due to joint loads. Hence it must be added to the corresponding end actions in the restrained structure in order to obtain the end actions due to the loads. Therefore, Fm Actual are the true member end actions due to actual loading system and can be expressed as

Fm Actual = Fm + Ffm Where, Ffmare the end actions in the restrained structure. Since there is no load acting on member (1), the actual end action will be:

𝐹𝑚𝐴𝑐𝑡𝑢𝑎𝑙 =

⎣⎢⎢⎢⎢⎡−0.4𝑤1.456𝑤4.156𝑤

0.4𝑤−1.456𝑤

1.70𝑤 ⎦⎥⎥⎥⎥⎤

+

⎣⎢⎢⎢⎢⎢⎢⎢⎢⎡

04𝑤2

42𝑤120

4𝑤2

−42𝑤12 ⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎤

=

⎣⎢⎢⎢⎢⎡−0.4𝑤3.46𝑤5.49𝑤0.40𝑤0.54𝑤0.34𝑤⎦

⎥⎥⎥⎥⎤

Member - BC In similar way, the member forces in member (BC) will be: Fm(BC) = [K](BC) d(BC)

⎣⎢⎢⎢⎢⎡𝑇2𝐹2𝑀2𝑇3𝐹3𝑀3⎦

⎥⎥⎥⎥⎤

= 𝐾

⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡

15

0 0−14

0 0

03653

1852

0 −3653

1852

01852

125

0 −1852

65

−15

0 015

0 0

0 −3653

−1852

03653

−1852

01852

65

0 −1852

125 ⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤

×1𝐾

⎣⎢⎢⎢⎢⎡−1.658𝑤−5.905𝑤

1.603000 ⎦

⎥⎥⎥⎥⎤

Thus,

47

⎣⎢⎢⎢⎢⎡𝑇2𝐹2𝑀2𝑇3𝐹3𝑀3⎦⎥⎥⎥⎥⎤

=

⎣⎢⎢⎢⎢⎡−0.33𝑤−0.55𝑤−0.40𝑤0.33𝑤0.55𝑤−2.33𝑤⎦

⎥⎥⎥⎥⎤

Since there is no load acting on member (BC), the actual end action will be:

𝐹𝑚𝐴𝑐𝑡𝑢𝑎𝑙 =

⎣⎢⎢⎢⎢⎡−0.33𝑤−0.55𝑤−0.40𝑤0.33𝑤0.55𝑤−2.33𝑤⎦

⎥⎥⎥⎥⎤

+

⎣⎢⎢⎢⎢⎡000000⎦⎥⎥⎥⎥⎤

=

⎣⎢⎢⎢⎢⎡−0.33𝑤−0.55𝑤−0.40𝑤0.33𝑤0.55𝑤−2.33𝑤⎦

⎥⎥⎥⎥⎤

Thus, the reaction forces at the support and load at the joints will be:

⎣⎢⎢⎢⎢⎢⎢⎢⎡𝑇1𝐹1𝑀1𝑇2𝐹2𝑀2𝑇3𝐹3𝑀3⎦⎥⎥⎥⎥⎥⎥⎥⎤

=

⎣⎢⎢⎢⎢⎢⎢⎢⎡−0.4𝑤3.46𝑤5.49𝑤

000

0.33𝑤0.55𝑤−2.33𝑤⎦

⎥⎥⎥⎥⎥⎥⎥⎤

4.6.5 Analysis of Space Frame Space frames are an increasingly common architectural technique especially for large roof spans in commercial and industrial buildings. The rigid jointed frames such as building frames are usually three dimensional space structures. Thus in case of certain structures like multi-storeyed buildings, it is necessary consider 3-dimensional effects for analysis. The space frame constitutes the final step of increasing complexity. It consists of plane frame and grid actions. The displacement and rotation vector associated with each joint have three components in case of space frame structures. There are six equilibrium equations associated with each joint. The degrees of freedom at each node of the space frame member will be (i) displacement in three perpendicular directions and (ii) rotations in three different directions. Therefore, the degrees of freedom in each node of the member will be six as shown in the figure below. The stiffness matrix in local coordinate system considering all possible degrees of freedom will be as given in Table 4.6.1.

48

Fig. 4.6.5 Degrees of freedom for space frame member

Table 4.6.1 Stiffness matrix of space frame member

3 2 3 2

3 2 3 2

2 2

2 2

3

0 0 0 0 0 0 0 0 0 0

12 6 12 60 0 0 0 0 0 0 0

12 6 12 60 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

6 4 6 20 0 0 0 0 0 0 0

6 4 6 20 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

120

X X

Z Z Z Z

Y Y Y Y

X X

Y Y Y Y

Z Z Z Z

X X

Z

EA EAL L

EI EI EI EIL L L L

EI EI EI EIL L L L

GI GIL L

EI EI EI EIL L L L

EI EI EI EIL L L L

EA EAL L

EIL

−

−

− − −

−

−

−

−

− 2 3 2

3 2 3 2

2 2

2

6 12 60 0 0 0 0 0 0

12 6 12 60 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

6 2 6 40 0 0 0 0 0 0 0

6 2 6 40 0 0 0 0 0 0 0

Z Z Z

Y Y Y Y

X X

Y Y Y Y

Z Z Z Z

EI EI EIL L L

EI EI EI EIL L L L

GI GIL L

EI EI EI EIL L L L

EI EI EI EIL L L L

− −

−

− − −

The generalized stiffness matrix of a rigid jointed space frame member can be obtained by transferring the matrix of local coordinate system into its global coordinate system. The transformation matrix will become a square matrix of size 12×12 in this case as the degrees of freedom for each node/joint is six.

1

2

3

5

6

7

8

9

10

11

12

4

12

11

10

9 8 7 6 5 4 3 2 1

Content Module 5: FEM for Two and Three Dimensional Solids (08) Lecture 1: Constant Strain Triangle(Page:1-6)

5.1.1 Element Stiffness Matrix for CST

5.1.2 Nodal Load Vector for CST

Lecture 2: Linear Strain Triangle(Page:7-12) 5.2.1 Element Stiffness Matrix for LST

5.2.2 Nodal Load Vector for LST

5.2.3 Numerical Example using CST

Lecture 3: Rectangular Elements(Page:13-19) 5.3.1 Computation of Element Stiffness

5.3.2 Computation of Nodal Loads

Lecture 4: Numerical Evaluation of Element Stiffness(Page:20-30) 5.4.1 Numerical Example

5.4.2 Evaluation of Stiffness using One Point Gauss Quadrature

5.4.3 Evaluation of Stiffness using Two Point Gauss Quadrature

Lecture 5: Computation of Stresses, Geometric Nonlinearity and Static

Condensation(Page:31-38) 5.5.1 Computation of Stresses

5.5.2 Geometric Nonlinearity

5.5.2.1 Steps to include effect of geometrical nonlinearity

5.5.3 Static Condensation

Lecture 6: Axisymmetric Element(Page:39-45) 5.6.1 Introduction

5.6.2 Relation between Strain and Displacement

5.6.3 Relation between Stress and Strain

5.6.4 Axisymmetric Shell Element

Lecture 7: Finite Element Formulation of Axisymmetric

Element(Page:45-52) 5.7.1 Stiffness Matrix of a Triangular Element

5.7.2 Stiffness Matrix of a Quadrilateral Element

Lecture 8: Finite Element Formulation for 3 Dimensional

Elements(Page:53-59) 5.8.1 Introduction

5.8.2 Strain Displacement Relation 5.8.3 Element Stiffness Matrix

5.8.4 Element Load Vector

5.8.4.1 Gravity load

5.8.4.2 Surface pressure

5.8.5 Stress Computation

Worked out Examples(Page:60-61) Example 5.1 Calculation of nodal loads on a triangular element

Lecture 1: Constant Strain Triangle The triangular elements with different numbers of nodes are used for solving two dimensional solid members. The linear triangular element was the first type of element developed for the finite element analysis of 2D solids. However, it is observed that the linear triangular element is less accurate compared to linear quadrilateral elements. But the triangular element is still a very useful element for its adaptivity to complex geometry. These are used if the geometry of the 2D model is complex in nature. Constant strain triangle (CST) is the simplest element to develop mathematically. In CST, strain inside the element has no variation (Ref. module 3, lecture 2) and hence element size should be small enough to obtain accurate results. As indicated earlier, the displacement is expressed in two orthogonal directions in case of 2D solid elements. Thus the displacement field can be written as

ud

v (5.1.1)

Here, u and v are the displacements parallel to x and y directions respectively. 5.1.1 Element Stiffness Matrix for CST A typical triangular element assumed to represent a subdomain of a plane body under plane stress/strain condition is represented in Fig. 5.1.1. The displacement (u, v) of any point P is represented in terms of nodal displacements

1 1 2 2 3 3

1 1 2 2 3 3

u N u N u N uv N v N v N v= + += + +

(5.1.2)

Where, N1, N2, N3 are the shape functions as described in module 3, lecture 2.

Fig. 5.1.1 Linear triangular element for plane stress/strain

The strain-displacement relationship for two dimensional plane stress/strain problem can be simplified in the following form from three dimensional cases (eq.1.3.9 to1.3.14).

2 2

2 2

12

12

x

y

xy

u u vx x x

v u vy y y

v u u u v vx y x y x y

(5.1.3)

In case of small amplitude of displacement, one can ignore the nonlinear term of the above equation and will reach the following expression.

x

y

xy

uxvyv ux y

ε

ε

γ

∂=∂∂

=∂∂ ∂

= +∂ ∂

(5.1.4)

Hence the element strain components can be represented as,

31 21 2 3

31 21 2 3

3 31 2 1 21 2 3 1 2 3

x

y

xy

NN Nu u u ux x x x

NN Nv v v vy y y y

N NN N N Nu v u u u v v vy x y y y x x x

ε

ε ε

γ

∂∂ ∂∂= = + +∂ ∂ ∂ ∂

∂∂ ∂∂= = = + + ∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂∂ ∂

= + = + + + + +∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

Or,

131 2

2

331 2

1

23 31 2 1 2

3

0 0 0

0 0 0x

y

xy

uNN Nux x xuNN Nvy y yvN NN N N N

y y y x x x v

εε ε

γ

∂∂ ∂

∂ ∂ ∂ ∂∂ ∂ = = ∂ ∂ ∂

∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

(5.1.5)

Or, [ ] B dε = (5.1.6)

In the above equation [B] is called as strain displacement relationship matrix. The shape functions for the 3 node triangular element in Cartesian coordinate is represented as,

2 3 3 2 2 3 3 2

1

2 3 1 1 3 3 1 1 3

3

1 2 2 1 1 2 2 1

1 x y x y y y x x x y2AN

1N x y x y y y x x x y2A

N 1 x y x y y y x x x y2A

Or,

1 1 1

1

2 2 2 2

3

3 3 3

1 x y2AN1N x y

2AN 1 x y

2A

(5.1.7)

Where,

1 2 3 3 2x y x y , 2 3 1 1 3x y x y , 3 1 2 2 1x y x y ,

1 2 3y y , 2 3 1y y , 3 1 2y y , (5.1.8)

1 3 2x x , 2 2 1x x , 3 2 1x x ,

Hence the required partial derivatives of shape functions are,

1 1 ,2

Nx A

β∂=

∂ 2 2 ,

2Nx A

β∂=

∂ 3 3 ,

2Nx A

β∂=

∂

1 1 ,2

Ny A

γ∂=

∂ 2 2 ,

2Nx A

γ∂=

∂ 3 3 ,

2Nx A

γ∂=

∂

Hence the value of [B] becomes:

[ ]

31 2

31 2

3 31 2 1 2

0 0 0

0 0 0

NN Nx x x

NN NBy y y

N NN N N Ny y y x x x

∂∂ ∂ ∂ ∂ ∂

∂∂ ∂= ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

Or, [ ]1 2 3

1 2 3

1 2 3 1 2 3

0 0 01 0 0 0

2B

A

β β βγ γ γ

γ γ γ β β β

=

(5.1.9)

According to Variational principle described in module 2, lecture 1, the stiffness matrix is represented as,

[ ] [ ] [ ][ ]Tk B D B dΩ

= Ω∫∫∫ (5.1.10)

Since, [B] and [D] are constant matrices; the above expression can be expressed as

[ ] [ ] [ ][ ] [ ] [ ][ ]T T

V

k B D B d V B D B V= =∫∫∫ (5.1.11)

For a constant thickness (t), the volume of the element will become A.t . Hence the above equation becomes,

[ ] [ ] [ ][ ]Tk B D B At= (5.1.12)

For plane stress condition, [D] matrix will become:

[ ] 2

1 01 0

110 0

2

EDµ

µµ

µ

= − −

(5.1.13)

Therefore, for a plane stress problem, the element stiffness matrix becomes,

[ ] ( )

1 1

2 21 2 3

3 31 2 32

1 11 2 3 1 2 3

2 2

3 3

00

1 0 0 0 00

1 0 0 0 004 1 10 00

20

EtkA

β γβ γ

µ β β ββ γ

µ γ γ γγ βµ

µ γ γ γ β β βγ βγ β

= − −

(5.1.14)

Or,

[ ] ( )

( )

( )

( )

2 21 1 1 2 1 2 1 3 1 3 1 1 1 2 2 1 1 3 3 1

2 22 2 2 3 2 3 2 1 1 2 2 2 2 3 3 2

2 223 3 3 1 1 3 3 2 2 3 3 3

2 21 1 1 2 1 2 1 3 1 3

2 22 2 2 3

12

12

14 1 2

.

C C C C C

C C C CEtk

C C CA

C C CSym C C

µβ γ β β γ γ β β γ γ β γ µβ γ β γ µβ γ β γ

µβ γ β β γ γ µβ γ β γ β γ µβ γ β γ

µβ γ µβ γ β γ µβ γ β γ β γµ

γ β γ γ β β γ γ β βγ β γ γ

++ + + + +

++ + + +

+=+ + +−

+ + ++ + 2 3

2 23 3C

β βγ β

+

(5.1.15)

Where, ( )12

Cµ−

=

Similarly for plane strain condition, [D] matrix is equal to,

[ ] ( )( )

( )( )

1 01 0

1 1 21 20 0

2

EDµ µ

µ µµ µ

µ

−

= − + − −

(5.1.16)

Hence the element stiffness matrix will become:

[ ] ( )

( )( )

2 21 1 1 2 1 2 1 3 1 3 1 1 1 2 2 1 1 3 3 1

2 22 2 2 3 2 3 2 1 1 2 2 2 2 3 3 2

2 23 3 3 1 1 3 3 2 2 3 3 3

2 21 1 1 2 1 2 1 3 1 3

2 22 2 2 3 2 3

3

( 1)1

12 1

.

M M MM M

M CEtkA M M M

Sym M MM

β γ β β γ γ β β γ γ µ β γ µβ γ β γ µβ γ β γβ γ β β γ γ µβ γ β γ µ β γ µβ γ β γ

β γ µβ γ β γ µβ γ β γ µ β γµ γ β γ γ β β γ γ β β

γ β γ γ β βγ

+ + + + + ++ + + + +

+ + + +=

+ + + ++ +

2 23β

+

(5.1.17)

Where ( )1M µ= −

5.1.2 Nodal Load Vector for CST From the principle of virtual work,

T T Td u F d u F d

(5.1.18)

Where, FΓ, and FΩ are the surface and body forces respectively. Using the relationship between stress-stain and strain displacement, one can derive the following expressions:

[ ][ ] [ ] [ ] andD B d , B d u N dσ δ ε δ δ δ= = =

(5.1.19) Hence eq. (5.1.18) can be rewritten as,

TT T T T TSd B D B d d d N F d d N F d

(5.1.20)

Or, TT TSB D B d d N F d N F d

(5.1.21)

Here, [Ns] is the shape function along the boundary where forces are prescribed. Eq.(5.1.21) is equivalent to k d F , and thus, the nodal load vector becomes

T TSF N F d N F d

(5.1.22)

For a constant thickness of the triangular element eq.(5.1.22) can be rewritten as

T TS

S A

F t N F ds t N F dA

(5.1.23)

For the a three node triangular two dimensional element, one can represent Fand F as,

x

y

FF

F

and x

y

FF

F

For example, in case of gravity load on CST element, x

y

F 0F

F g

For this case, the shape functions in terms of area coordinates are:

1 2 3

1 2 3

L L L 0 0 0N

0 0 0 L L L

(5.1.24)

As a result, the force vector on the element considering only gravity load, will become,

1

2

3

1 1 1A A A

2 2 2

3 3 3

L 0 0 0

L 0 0 0

L 0 0 00F t dA t dA dA

0 L L g Lg

0 L L g L

0 L L g L

gt

(5.1.25)

The integration in terms of area coordinate is given by,

1 2 3 22

p q r

A

p! q! r !L L L dA A( p q r )!

=+ + +∫ (5.1.26)

Thus, the nodal load vector will finally become

F

0 00 00 0

1!0!0! gAt2Agt (1 0 0 2)! 3

0!1!0! gAt2A(0 1 0 2)! 3

gAt0!0!1! 2A3(0 0 1 2)!

000gAt1311

(5.1.27)

Lecture 2: Linear Strain Triangle 5.2.1 Element Stiffness Matrix for LST In case of CST, it is observed that the strain within the element remains constant. Though, these elements are able to provide enough information about displacement pattern of the element, but it is unable to provide adequate information about stress inside an element. This limitation will be significant enough in regions of high strain gradients. The use of a higher order triangular element called Linear Strain Triangle (LST) significantly improves the results at these areas as the strin inside the element is varying. The LST element has six nodes (Fig. 5.2.1) and hence, twelve degrees of freedom. Thus the displacement function can be chosen as follows.

2 20 1 2 3 4 5

2 26 7 8 9 10 11

u x y x xy yv a a x a y a x a xy a y

α α α α α α= + + + + +

= + + + + + (5.2.1)

Fig. 5.2.1 Linear strain triangle element

Therefore, the element strain matrix is obtained as

x 1 3 4

y 8 10 11

xy 2 7 4 9 5 10

u 2 x yxv x 2 yyv u ( ) ( 2 )x (2 )yx y

(5.2.2)

In the area coordinate system as discussed in module 3, lecture 3 we can write the shape function for the six node triangular element as

1 1 1 2 2 2 3 3 3

4 1 2 5 2 3 6 3 1

N L 2L 1 N L 2L 1 N L 2L 1N 4L L N 4L L N 4L L

(5.2.3)

The displacement (u,v) of any point within the element can be represented in terms of their nodal displacements with the use of interpolation function.

6

16

1

i ii

i ii

u N u

v N v

=

=

=

=

∑

∑ (5.2.4)

Using eq.(5.2.4) we can rewrite eq.(5.2.2) as,

1

2

3

43 5 61 2 4

5

63 5 61 2 4

1

23 5 6 3 5 61 2 4 1 2 4

3

4

5

6

0 0 0 0 0 0

0 0 0 0 0 0

uuuuN N NN N Nux x x x x xuN N NN N Nvy y y y y yvN N N N N NN N N N N N

y y y y y y x x x x x x vvvv

ε

∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ = ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

Or, [ ] B dε = (5.2.5)

Where,

[ ]

3 5 61 2 4

3 5 61 2 4

3 5 6 3 5 61 2 4 1 2 4

0 0 0 0 0 0

0 0 0 0 0 0

N N NN N Nx x x x x x

N N NN N NBy y y y y y

N N N N N NN N N N N Ny y y y y y x x x x x x

∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂∂ ∂ ∂= ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

(5.2.6)

Using Chain rule,

31 1 1 1 2 1

1 2 3

. . . LN N L N L Nx L x L x L x

∂∂ ∂ ∂ ∂ ∂ ∂= + +

∂ ∂ ∂ ∂ ∂ ∂ ∂

As discussed in module 3, lecture 1, we can write the above expression as,

31 1 1 2 1 1

1 2 3

. . .2 2 2

bN b N b N Nx A L A L A L

∂ ∂ ∂ ∂= + +

∂ ∂ ∂ ∂

( )1 11. 4 1

2N b Lx A

∂= −

∂

Similarly we can evaluate expressions for other terms and can be written as,

( ) ( ) ( )

( ) ( ) ( )

3 31 1 2 21 2 3

5 642 1 1 2 3 2 2 3 1 3 3 1

. 4 1 . 4 1 . 4 12 2 2

4 4 4

N bN b N bL L Lx A x A x A

N NN L b L b L b L b L b L bx x x

∂∂ ∂= − = − = −

∂ ∂ ∂∂ ∂∂

= + = + = +∂ ∂ ∂

And,

( ) ( ) ( )

( ) ( ) ( )

3 31 1 2 21 2 3

5 642 1 1 2 3 2 2 3 1 3 3 1

. 4 1 . 4 1 . 4 12 2 2

4 4 4

N aN a N aL L Ly A y A y A

N NN L a L a L a L a L a L ay y y

∂∂ ∂= − = − = −

∂ ∂ ∂∂ ∂∂

= + = + = +∂ ∂ ∂

Where,

1 2 3 2 3 1 3 1 2

1 2 3 2 3 1 3 1 2

a x x a x x a x xb y y b y y b y y= − = − = −= − = − = −

The stiffness matrix of the element is represented by,

[ ] [ ] [ ][ ]Tk B D B dΩ

= Ω∫∫∫ (5.2.7)

The, [D] matrix is the constitutive matrix which will be taken according to plane stress or plane strain condition. The nodal strain and stress vectors are given by,

1 2 3 1 2 3 1 2 3

T

n x x x y y y xy xy xyε ε ε ε ε ε ε γ γ γ=

(5.2.9a)

1 2 3 1 2 3 1 2 3

T

n x x x y y y xy xy xyσ σ σ σ σ σ σ τ τ τ=

(5.2.9b)

[ ] [ ][ ] [ ][ ] [ ]

1

2

2 1

00n

n n

n n

BB d

B Bε

=

(5.2.10)

Referring to section 3.3.1, using proper values of area coordinates in [B] matrix, one can find

[ ]1 2 3 2 3

1 1 2 3 1 3

1 3 2 1

3 4 0 41 3 4 4 0

23 0 4 4

n

b b b b bB b b b b b

Ab b b b b

− − = − − − −

(5.2.11a)

And,

[ ]1 2 3 2 3

2 1 2 3 1 3

1 3 2 1

3 4 0 41 3 4 4 0

23 0 4 4

n

a a a a aB a a a a a

Aa a a a a

− − = − − − −

(5.2.11b)

Thus, the element stiffness can be evaluated by putting the values from eq. (5.2.11) in eq. (5.2.7). 5.2.2 Nodal Load Vector for LST Similar to 3-node triangular element, the load will be lumped at each node which can be computed using the earlier expression,

T TSF N F d N F d

(5.2.12)

And for element with constant thickness,

T TS

S A

F t N F ds t N F dA

(5.2.13)

5.2.3 Numerical Example using CST Determine the displacements at the nodes for the following 2D solid continuum considering a constant thickness of 25 mm, Poisson’s ratio, µ as 0.25 and modulus of elasticity E as 2 x 105 N/mm2. The continuum is discritized with two CST plane stress elements.

Fig. 5.2.2 Geometry and discretization of the continuum The element 1 is connected with node 1, 3 and 4 and let assume its Cartesian coordinates are (x1, y1), (x3, y3) and (x4, y4) respectively. If we consider nodes 1, 3 and 4 are similar to node 1, 2 and 3 in eq.(5.1.9) then the [B] can be written as

[ ]1 2 3

1 2 3

1 2 3 1 2 3

0 0 01 0 0 0

2B

A

β β βγ γ γ

γ γ γ β β β

=

By introducing values of β & γ discussed in previous lecture note, we can get value of [B] as

[ ]0 1 1 0 0 0

1 0 0 0 3 0 31500

3 0 3 0 1 1B

− = − − −

For plain stress problem, putting the values of E and µ one can find the following values.

[ ]4

2

1 0 16 4 04 101 0 4 16 0

1 31 0 0 60 0

2

EDµ

µµ

µ

× = = − −

Therefore the stiffness matrix for the element 1 will be

[ ] [ ] [ ][ ]1

Tk tA B D B=

Putting values of t, A, [B] & [D]we will get,

[ ] 41

750 0 -750 0 -250 -2500 222.2222 -222.2222 -166.6667 0 166.6667-750 -222.2222 972.2222 166.6667 250 -416.6667

4 100 -16

k = × ×6.6667 166.6667 2000 0 -2000

-250 0 250 0 83.3333 -83.3333250 166.6667 -416.6667 -2000 -83.3333 2083.3333

Similarly element 2 is connected with nodes 1, 2 and 3 and global coordinates of these nodes are (x1, y1), (x2, y2) and (x3, y3) respectively. For this element, by proceeding in a similar manner to element 1 we can calculate [B] matrix as,

[ ]1 1 0 0 0 0

1 0 0 0 0 3 31500

0 3 3 1 1 0B

− = − − −

Hence, the elemental stiffness matrix becomes,

[ ] 42

222.2222 -222.2222 0 0 166.6667 -166.6667-222.2222 972.2222 -750 250 -416.6667 166.66670 -750 750 -250 250 0

4 100 250 -250

k = × × 83.3333 -83.3333 0

166.6667 -416.6667 250 -83.3333 2083.3333 -2000-166.6667 166.6667 0 0 -2000 2000

By assembling the stiffness matrices into global stiffness matrix [K],

[ ] 4

972.2222 -222.2222 0 -750 0 166.6667 -416.6667 -250-222.2222 972.2222 -750 0 250 -416.6667 166.6667 00 -750 9

4 10K = × ×

72.2222 -222.2222 -416.6667 250 0 166.6667-750 0 -222.2222 972.2222 166.6667 0 250 -416.66670 250 -416.6667 166.6667 2083.3333 -83.3333 0 -2000166.6667 -416.6667 250 0 -83.3333 2083.3333 -2000 0-416.6667 166.6667 0 250 0 -2000 2083.3333 -83.3333250 0 166.6667 -416.6667 -2000 0 -83.3333 2083.3333

Now, applying equation [ ] [ ] F K d= , the following expression can be written.

1

2

3

4 4

1

2

3

4

972.2222 -222.2222 0 -750 0 166.6667 -416.6667 -250-222.2222 972.2222 -750 0 250 -416.

4 10

u

u

u

u

v

v

v

v

FFFFFFFF

= × ×

6667 166.6667 00 -750 972.2222 -222.2222 -416.6667 250 0 166.6667-750 0 -222.2222 972.2222 166.6667 0 250 -416.66670 250 -416.6667 166.6667 2083.3333 -83.3333 0 -2000166.6667 -416.6667 250 0 -83.3333 2083.3333 -2000 0-416.6667 166.6667 0 250 0 -2000 2083.3333 -83.3333250

1

2

3

4

1

2

3

4 0 166.6667 -416.6667 -2000 0 -83.3333 2083.3333

uuuuvvvv

Putting boundary conditions 1 1 2 4 4 0u v u u v= = = = = and adopting elimination technique for applying boundary condition we get expression,

24

3

3

0 972.2222 250 025000 4 10 250 2083.3333 2000

0 0 2000 2083.3333

vuv

= × ×

Solving the above expression, the unknown nodal displacements may be obtained as follows. 5

2 25.96 10v −= × mm, 53 10.02 10u −= − × mm and 5

3 96.92 10v −= × mm.

Lecture 3: Rectangular Elements The rectangular elements are widely used for solving two dimensional continuums. The main advantage of this type of element is the easy formulation and easy development of computer code. The element stiffness of such elements is derived here using the concept of isoparametric formulation. 5.3.1 Computation of Element Stiffness In case of a four node rectangular element, the geometry and displacement filed can be expressed in terms of their nodal values with the help of interpolation function. As the formulation will be isoparametric, the interpolation function will become same for expressing both the variables. Thus, coordinates and displacements at any point inside the element (Fig. 5.3.1) can be expressed as

1 1 2 2 3 3 4 4

1 1 2 2 3 3 4 4

x N x N x N x N xy N y N y N y N y (5.3.1)

And

1 1 2 2 3 3 4 4

1 1 2 2 3 3 4 4

u N u N u N u N uv N v N v N v N v (5.3.2)

The above equations can be written in matrix form as

1

1

2

1 2 3 4 2

1 2 3 4 3

3

4

4

0 0 0 00 0 0 0

xyx

N N N N yxN N N N xx

yxy

(5.3.3)

And

1

1

2

1 2 3 4 2

1 2 3 4 3

3

4

4

0 0 0 00 0 0 0

uvu

N N N N vuN N N N uv

vuv

(5.3.4)

The shape function four node rectangular element is derived and shown in module 3, lecture 4. However the shape functions are reproduced here for easy reference for the derivation of the stiffness matrix.

1

2

3

4

1 14

1 14

1 14

1 14

i

NN

NNN

x h

x h

x h

x h

(5.3.5)

Fig. 5.3.1 Four node rectangular element

The strain-displacement relationship for two dimensional plane stress/strain problem can be simplified in the following form from three dimensional cases (eq.1.3.9 to1.3.14).

2 2

2 2

12

12

x

y

xy

u u vx x x

v u vy y y

v u u u v vx y x y x y

(5.3.6)

In case of small amplitude of displacement, one can ignore the nonlinear term of the above equation and will reach the following expression.

x

y

xy

uxvyv ux y

ε

ε

γ

∂=∂∂

=∂∂ ∂

= +∂ ∂

(5.3.7)

Using the shape function the above expression can be written as

[ ]

1

131 2 4

2

231 2 4

3

33 31 1 2 2 4 4

4

4

0 0 0 0

0 0 0 0x

y

xy

uvNN N Nux x x xvNN N N B duy y y yvN NN N N N N N

y x y x y x y x uv

εεγ

∂∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ = = ∂ ∂ ∂ ∂

∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

(5.3.8)

Here, [B] is known as strain displacement relationship matrix. The derivatives of the shape functions are calculated using the chain rule.

. .

. .

i i i

i i i

N N Nx yx y

N N Nx yx y

ξ ξ ξ

η η η

∂ ∂ ∂∂ ∂= +

∂ ∂ ∂ ∂ ∂∂ ∂ ∂∂ ∂

= +∂ ∂ ∂ ∂ ∂

(5.3.9)

Here, i is referred to number of nodes in an element and will be 4 in this case. Converting above expression in matrix form

i i i

i ii

N x y N Nx xJ

N NN x yy y

x x x

h hh

(5.3.10)

The matrix [J] is referred to Jacobian matrix which is discussed in Lecture 7, module 3. Using eq. (5.3.1) one can write

31 2 41 2 3 4

NN N Nx x x x x x x x x x

(5.3.11)

Putting the values of the nodal coordinates and shape functions of the four node element in the above equation the following relations will be obtained.

1 1 1 1 1 1 1 10 0

4 4 4 4 2x aa a

h h h hx

(5.3.12a) Similarly,

1 1 1 1 1 1 1 10 0 0

4 4 4 4y b b

h h h hx

(5.3.12b) 1 1 1 1 1 1 1 1

0 0 04 4 4 4

x a a

x x x x

h (5.3.12c)

1 1 1 1 1 1 1 10 0

4 4 4 4 2y bb b

x x x xh

(5.3.12d) Substituting above values in Jacobian matrix the following relations will be obtained.

1

2 002

20 02

aaJ and J

bb

(5.3.13)

Thus, eq.(5.3.10) can be written as

1

22 .0

2 20 .

i i ii

i i i i

N N NNax aJ

N N N Ny b b

x x x

h h h

(5.3.14)

After derivation of the shape functions expressed in eq.(5.3.5), the following values will be obtained.

31 1 2 4

31 1 2 4

1 1 1 12 . ; ; ;2 2 2 2

1 1 1 12 . ; ; ;2 2 2 2

NN N N Nx a a x a x a x a

NN N N Ny b b y b y b y b

h h h hx

x x x xh

(5.3.15)

So, the strain displacement relationship matrix, [B] will become as follows.

[ ]

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

1 1 1 10 0 0 0

2 2 2 21 1 1 1

0 0 0 02 2 2 2

1 1 1 1 1 1 1 12 2 2 2 2 2 2 2

a a a a

Bb b b b

b a b a b a b a

η η η η

ξ ξ ξ ξ

ξ η ξ η ξ η ξ η

− − + +− − − + + −

= − − − − + − + + − +− − − −

(5.3.16)

The element stiffness matrix will become

[ ] [ ] [ ][ ] [ ] [ ][ ]T Tk t B D B dx dy t B D B J d dξ η= =∫∫ ∫∫ (5.3.17)

It is seen that the above is expressed in terms of ξ and η and hence can be numerically integrated by the Gauss Quadrature rule. The stiffness matrix for each element can be found which needs to be globally assembled for getting the global stiffness matrix to obtain the solution. The stiffness matrix of higher order rectangular element can be derived in a similar fashion. For example, in case of eight node rectangle element, the size of [B] matrix will become 16 × 3 which was 8 × 3 for four node element. Thus the size of element stiffness for eight node element will become 16 × 16. 5.3.2 Computation of Nodal Loads If a distributed load acts on a side of a four node rectangular element, the nodal load vector can be calculated the similar procedure as discussed in case of triangular element. If an element as shown below is subjected to a linearly varying intensities of load at its one side, then the magnitude of this at any point on the side can be expressed by its interpolation function as follows.

2

3

1 12 2

xx

x

qq

qη η − + =

(5.3.18)

Here, qx2 and qx3 are the force intensities per unit length at nodes 2 and 3 respectively. The load at nodes can be calculated from the following expression.

2 1S

x xF N q d= Γ∫ (5.3.19)

As ξ=1 along the side 2-3, the interpolation function will become

( )( )

( )( )

( )( )

( )( )

( )

( )2

1 104

1 1 14 2

1 1 14 2

01 14

SN

ξ η

ξ η η

ξ η η

ξ η

− − + − − = =

+ + +

− +

(5.3.20)

If the element thickness is t, then dΓ1 =t.dl. Thus the eq.(5.3.19) can be replaced as

( )

( )

12

31

01

1 122 21

20

xx

x

qF t dl

q

ηη η

η

+

−

−

− + = +

∫ (5.3.21)

Fig. 5.3.2 Varying load on a four node element After integrating the above expression, the nodal load vector along x direction will become as follows.

2 3

2 3

02

230

x xx

x x

q qtFq q

+ = +

(5.3.22)

Lecture 4: Numerical Evaluation of Element Stiffness Derivation of element stiffness for a four node rectangle element has been demonstrated in last lecture. The stiffness matrix of each element can be calculated easily by developing a suitable computer algorithm. To help students for developing their own computer code, a numerical example has been solved and demonstrated here. 5.4.1 Numerical Example Calculate the stiffness matrix for the given four node rectangular element by the Gauss Quadrature integration rule using one point and two point formula assuming plane stress formulation. Consider, the thickness of element = 20 cm, E=2 × 103 kN/cm2 and µ =0.

Fig. 5.4.1 Element Dimension

5.4.2 Evaluation of Stiffness using One Point Gauss Quadrature For the calculation of stiffness matrix, first, 1×1 Gauss Quadrature integration procedure has been carried out. Thus, the natural coordinate of the sampling point will become 0,0 and weight will become 2.0 which is shown in the figure below.

Fig. 5.4.2 Natural coordinates for one point Gauss Quadrature

For a four node quadrilateral element, the shape functions and their derivatives are as follows.

4)1)(1(N1

ηξ −−= ,

4)1)(1(N2

ηξ −+= and

4)1)(1(N3

ηξ ++= and

4)1)(1(N4

ηξ +−=

4)1(1 ξ

ξ−−

=∂∂N ;

4)1(2 ξ

ξ−

=∂∂N ;

4)1(3 ξ

ξ+

=∂∂N ;

4)1(4 ξ

ξ+−

=∂∂N

4)1(1 η

η−−

=∂∂N ;

4)1(2 η

η+−

=∂∂N ;

4)1(3 η

η+

=∂∂N ;

4)1(4 η

η−

=∂∂N

The Jacobian matrix can be found from the following relations.

1 1

2

1 131 2 4

2 2 2

3 331 2 4 3 3

4 4 4 4

(1 ) (1 ) (1 ) (1 )4 4 4 4| |

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

x yx yNN N Nx y x y

Jx yNN N N x yx y x y

ξ ξ ξ ξξ ξ ξ ξ

η η η ηη η η η

∂∂ ∂ ∂ − − + − + + − + ∂ ∂ ∂ ∂ = = ∂ − − − + + + + −∂ ∂ ∂ ∂ ∂ ∂ ∂

Considering the sampling point, (ξ=0 and η=0 ), the value of the Jacobian, [J] is

−−

−+−

=

500507007000

41

41

41

41

41

41

41

41

][J

Thus,

=

250035

][J and * *

1 11 12* *21 22

1 035[ ]

1025

J JJ

J J−

= =

; and |𝐽| = 875

Now, the strain vector for the element will become

∂∂∂∂∂∂∂∂

×

=

η

ξ

η

ξ

ε

v

v

u

u

JJJJJJ

JJ

*12

*11

*22

*21

*22

*21

*12

*11

0000

][

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

×

=

4

4

3

3

2

2

1

1

4321

4321

4321

4321

*12

*11

*22

*21

*22

*21

*12

*11

0000

0000

0000

0000

0000

][

vuvuvuvu

NNNN

NNNN

NNNN

NNNN

JJJJJJ

JJ

ηηηη

ξξξξ

ηηηη

ξξξξ

ε

[ ] ]['0000

][*12

*11

*22

*21

*22

*21

*12

*11

dBdBJJJJJJ

JJ=×

=ε

−−

−−

−−

−−

=

410

410

410

410

410

410

410

410

0410

410

410

41

0410

410

410

41

][ /B

and [ ]'

0351

2510

251000

000351

][ BB

=

7.143 0 7.143 0 7.143 0 7.143 0[B] 0 10 0 10 0 10 0 10

10 7.143 10 7.143 10 7.143 10 7.143

For plane stress condition

×=

−−=

2/100010001

102

2100

0101

1][ 3

2µ

µµ

µEC

×

×=

2/100010001

102]][[ 3BC

7.143 0 7.143 0 7.143 0 7.143 00 10 0 10 0 10 0 1010 7.143 10 7.143 10 7.143 10 7.143

Assume the values of gauss weight, w = 2, the stiffness matrix [k] at this sampling point is [ ] [ ] [ ] [ ] | |Tk tw B C B J= , Where t is thickness of the element. Thus,

[ ] 3

7.143 0 100 10 7.143

7.143 0 101 0 0

0 10 7.14320 2 875 2 10 10 0 1 0

7.143 0 100 0 1/ 2

0 10 7.1437.143 0 4

0 10 7.143

7.143 0 7.143 0 7.143 0 7.143 00 10 0 10 0 10 0 1010 7.143 10 7.143 10 7.

k

− − − − −

− = × × × × × ×

−

− − −

× − −− − − 143 10 7.143

−

3

7.071 2.5 0.714 2.5 7.071 2.5 0.714 2.52.5 8.785 2.5 5.214 2.5 8.785 2.5 5.2140.714 2.5 7.071 2.5 0.714 2.5 7.071 2.5

2.5 5.214 2.5 8.785 2.5 5.214 2.5 8.78510

7.071 2.5 0.714 2.5 7.071 2.5 0.714 2.52.5 8.785 2.5 5

− − − −− − −

− − − −− − − −

= ×− − − −− − − − .214 2.5 8.785 2.5 5.2140.714 2.5 7.071 2.5 0.714 2.5 7.071 2.5

2.5 5.214 2.5 8.785 2.5 5.214 2.5 8.785

− − − −

− − − −

5.4.3 Evaluation of Stiffness using Two Point Gauss Quadrature

In this case, 2×2 Gauss Quadrature integration procedure has been carried out to the calculate the stiffness matrix of the same element for a comparison. The natural coordinate of the sampling point is shown in the figure below.

Fig. 5.4.3 Natural Coordinates for Two Points Gauss Quadrature

The natural co-ordinates of the sampling points for 2×2 Gauss Quadrature integration are

1 +0.57735 +0.57735 2 -0.57735 +0.57735 3 -0.57735 -0.57735 4 +0.57735 -0.57735

For a four node quadrilateral element, the shape functions and their derivatives are as follows.

4)1)(1(N1

ηξ −−= ,

4)1)(1(N2

ηξ −+= ,

4)1)(1(N3

ηξ ++= and

4)1)(1(N4

ηξ +−=

4)1(1 ξ

ξ−−

=∂∂N ;

4)1(2 ξ

ξ−

=∂∂N ;

4)1(3 ξ

ξ+

=∂∂N ;

4)1(4 ξ

ξ+−

=∂∂N

4)1(1 η

η−−

=∂∂N ;

4)1(2 η

η+−

=∂∂N ;

4)1(3 η

η+

=∂∂N ;

4)1(4 η

η−

=∂∂N

The Jacobian matrix will be

1 1

2

1 131 2 4

2 2 2

3 331 2 4 3 3

4 4 4 4

(1 ) (1 ) (1 ) (1 )4 4 4 4| |

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

x yx yNN N Nx y x y

Jx yNN N N x yx y x y

ξ ξ ξ ξξ ξ ξ ξ

η η η ηη η η η

∂∂ ∂ ∂ − − + − + + − + ∂ ∂ ∂ ∂ = = ∂ − − − + + + + −∂ ∂ ∂ ∂ ∂ ∂ ∂

(a) At sampling point 1, (ξ=0.57735, η=0.57735) The value of the Jacobian, [J] at sampling point 1 will become

−++++−−−

+−++−+−−

=

500507007000

4)0.577351(

4)0.577351(

4)0.577351(

4)0.577351(

4)0.577351(

4)0.577351(

4)0.577351(

4)0.577351(

][J

Thus,

=

250035

][J and * *

1 11 12* *21 22

1 035[ ]

1025

J JJ

J J−

= =

; Thus |𝐽| = 875

∂∂∂∂∂∂∂∂

×

=

η

ξ

η

ξ

ε

v

v

u

u

JJJJJJ

JJ

*12

*11

*22

*21

*22

*21

*12

*11

0000

][

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

×

=

4

4

3

3

2

2

1

1

4321

4321

4321

4321

*12

*11

*22

*21

*22

*21

*12

*11

0000

0000

0000

0000

0000

][

vuvuvuvu

NNNN

NNNN

NNNN

NNNN

JJJJJJ

JJ

ηηηη

ξξξξ

ηηηη

ξξξξ

ε

[ ] ]['0000

][*12

*11

*22

*21

*22

*21

*12

*11

dBdBJJJJJJ

JJ=×

=ε

1 0.57735 1 0.57735 1 0.57735 1 0.577350 0 0 04 4 4 4

1 0.57735 1 0.57735 1 0.57735 1 0.577350 0 0 04 4 4 4[ ]

1 0.57735 1 0.57735 1 0.57735 1 0.577350 0 0 04 4 4 4

1 0.57735 1 0.57735 1 0.57735 1 0.577350 0 0 04 4 4 4

B

− − + +− −

− + + −− −′ = − − + + − −

− + + −

− −

0.1057 0 0.1057 0 0.3943 0 0.3943 00.1057 0 0.3943 0 0.3943 0 0.1057 0

[ ']0 0.1057 0 0.1057 0 0.3943 0 0.39430 0.1057 0 0.3943 0 0.3943 0 0.1057

B

− − − − = − − − −

Thus,

1 0 0 0 0.1057 0 0.1057 0 0.3943 0 0.3943 0350.1057 0 0.3943 0 0.3943 0 0.1057 01[ ] 0 0 0

0 0.1057 0 0.1057 0 0.3943 0 0.3943251 1 0 0.1057 0 0.3943 0 0.3943 0 0.10570 025 35

B

− − − − = × − − − −

−−−−−−

−−=

0113.00042.00113.00158.0003.00158.0003.00042.00042.000158.000158.000042.0000113.000113.00003.00003.0

][B

For plane stress condition

×=

−−=

2/100010001

102

2100

0101

1][ 3

2µ

µµ

µEC

×

×=

2/100010001

102]B][C[ 3

−−−−−−

−−

0113.00042.00113.00158.0003.00158.0003.00042.00042.000158.000158.000042.0000113.000113.00003.00003.0

The values of gauss weights are wi=wj=1.0. Therefore, the stiffness matrix [k] at this sampling

point is [ ] [ ] [ ] [ ] | |Ti j ij ij ijk tw w B C B J= , where t is thickness of the element. Thus at sampling point

1,

[ ] 31

0.003 0 0.00420 0.0042 0.003

0.003 0 0.01581 0 0

0 0.0158 0.00320 1 1 875 2 10 0 1 0

0.0113 0 0.01580 0 1/ 2

0 0.0158 0.01130.0113 0 0.0042

0 0.0042 0.0113

0.003 0 0.003 0 0.0113 0 0.

k

− − − − −

− = × × × × × × × ×

−

− − − 0113 0

0 0.0042 0 0.0158 0 0.0158 0 0.00420.0042 0.003 0.0158 0.003 0.0158 0.0113 0.0042 0.0113

− − − − − −

[ ] 41

0.0632 0.0223 0.0848 0.0223 0.2358 0.0834 0.0878 0.08340.0785 0.0834 0.2174 0.0834 0.2929 0.0223 0.0030

0.4672 0.0834 0.3162 0.3109 0.2358 0.31090.8866 0.0834 0.8111 0.0223 0.2929

100.8795 0.3109 0.3275 0.31

k

− − −− − − −

− − − −− −

= ×− − 09

1.0927 0.0834 0.01130.4755 0.0834

0.2847

sym

−

(b) At sampling point 2, (ξ=-0.57735, η=0.57735) The value of the Jacobian, [J] at sampling point 2 can be calculated in a similar way and finally the strain-displacement relationship matrix and then the stiffness matrix [k2] can be evaluated and is shown below.

[ ] 32

0.0113 0 0.00420 0.0042 0.0113

0.0113 0 0.01580 0.0158 0.0113

20 1 1 875 2 100.003 0 0.0158

0 0.0158 0.0030.003 0 0.0042

0 0.0042 0.0030

0.0113 0 0.0113 0 0.0030 0 0.0030 00 0.0042 0 0.015

k

− − − − − − = × × × × × × −

− − −

− − 8 0 0.0158 0 0.00420.0021 0.0056 0.0079 0.0056 0.0079 0.0015 0.0021 0.0015

− − − −

[ ] 42

0.4756 0.0833 0.3276 0.0833 0.2357 0.0223 0.0878 0.02230.2847 0.3110 0.0112 0.3110 0.2929 0.0833 0.0030

0.8797 0.3110 0.3164 0.0833 0.2357 0.08331.0930 0.3110 0.8113 0.0833 0.2929

100.4673 0.0833 0.0848 0.08

k

− − − −− − − −

− − − −− −

= ×− 33

0.8868 0.0223 0.21740.0632 0.0223

0.0785

sym

−

(c) At sampling point 3, (ξ=-0.57735, η=-0.57735) The value of the strain-displacement relationship matrix and then the stiffness matrix [k3] can be evaluated and is shown below.

[ ] 33

0.0113 0 0.01580 0.0158 0.0113

0.0113 0 0.00420 0.0042 0.0113

20 1 1 875 2 100.0030 0 0.0042

0 0.0042 0.00300.0030 0 0.0158

0 0.0158 0.0030

0.0030 0 0.0113 0 0.0030 0 0.0030 00 0.0158 0 0.

k

− − − − − − = × × × × × × −

− − −

− − 0042 0 0.0042 0 0.01580.0079 0.0056 0.0021 0.0056 0.0021 0.0015 0.0079 0.0015

− − − −

[ ] 43

0.8797 0.3110 0.3276 0.3110 0.2357 0.0833 0.3164 0.08331.0930 0.0833 0.0112 0.0833 0.2929 0.3110 0.8113

0.4756 0.0833 0.0878 0.0223 0.2357 0.02230.2847 0.0833 0.0030 0.3110 0.2929

100.0632 0.0223 0.0848 0.02

k

− − − − −− − − −

− − −− −

= ×− 23

0.0785 0.0833 0.21740.4673 0.0833

0.8868

sym

−

(d) At sampling point 4, (ξ=0.57735, η=-0.57735) The value of the strain-displacement relationship matrix and then the stiffness matrix [k3] can be evaluated and is shown below.

[ ] 34

0.0030 0 0.01580 0.0158 0.0030

0.0030 0 0.00420 0.0042 0.0030

20 1 1 875 2 100.0113 0 0.0042

0 0.0042 0.01130.0113 0 0.0158

0 0.0158 0.0113

0.0030 0 0.0030 0 0.0113 0 0.0113 00 0.0158 0 0.

k

− − − − − − = × × × × × × −

− − −

− − 0042 0 0.0042 0 0.01580.0079 0.0015 0.0021 0.0015 0.0021 0.0056 0.0079 0.0056

− − − −

[ ] 44

0.4673 0.0833 0.0848 0.0833 0.2357 0.3110 0.3164 0.31100.8868 0.0223 0.2174 0.0223 0.2929 0.0833 0.8113

0.0632 0.0223 0.0878 0.0833 0.2357 0.08330.0785 0.0223 0.0030 0.0833 0.2929

100.4756 0.0833 0.3276 0.08

k

− − − −− − − −

− − −− −

= ×− − 33

0.2847 0.3110 0.01120.8797 0.3110

1.0930

sym

−

The stiffness matrix of the element can be computed as the sum of the values at the four sampling points: 1 2 3 4[ ] [ ] [ ] [ ] [ ]k k k k k= + + + . Thus, the final value of the stiffness matrix will become

[ ] 4

1.8857 0.5000 0.4857 0.5000 0.9429 0.5000 0.4571 0.31102.3429 0.5000 0.4571 0.5000 1.1714 0.5000 1.6286

1.8857 0.5000 0.4571 0.5000 0.9429 0.50002.3429 0.5000 1.6286 0.5000 1.1714

101.8857 0.5000 0.4857 0.

k

− − − − −− − − −

− − − −− − −

= ×− − 5000

2.3429 0.5000 0.45711.8857 0.5000

2.3429

sym

−

Lecture 5: Computation of Stresses, Geometric Nonlinearity and Static Condensation 5.5.1 Computation of Stresses After solving the static equation of F = [K]d, the nodal displacement d can be obtained in global coordinate system. The element nodal displacement d can then be calculated from the

nodal connectivity of the element. Using strain-displacement relation and then stress-strain relation, the stress at the element level are derived.

σ= DεD B d (5.5.1)

Here, is the stress at the Gauss point of the element as the sampling points for the integration

has been considered as Gauss points. Here, [D] is the constitutive matrix, [B] is the strain displacement matrix of the element. As a result these stresses at Gauss points need to extrapolate to the corresponding nodes of the element. It is well established that 2 × 2 Gauss integration points are the optimal sampling points for two dimensional isoparametric elements. The ‘local stress smoothing’ is a technique that can be used to extrapolate stresses computed at Gauss points to nodal points. The stresses are computed at four Gauss points (I, II, III and IV) of an 8

node element as shown in Fig. 5.5.1. For example, at point III, r = s =1 an d ξ = η = 31 .

Therefore the factor of proportionality is 3 ; i.e.,

3ξ=r and 3η=s (5.5.2) Stresses at any point P in the element are found by the usual shape function as

∑ ′= idiP N σσ for i = 1,2,3,4 (3.5.3)

In the above equation, Pσ is yx σσ , and xyτ at point P. diN ′ are the bilinear shape functions

written in terms of r and s rather than ξ and η as

( )( )srNdi ±±=′ 1141 (5.5.4)

diN ′ are evaluated at r and s coordinates of point P. Let the point P coincides with the corner 1.

To calculate stress 1xσ at corner 1 from xσ values at the four Gauss points, substitution of r and s into the shape functions will give

xIVxIIIxIIxIxI σσσσσ 500.0134.0500.08666.1 −+−= (5.5.5)

Fig. 5.5.1 Natural coordinate systems used in extrapolation of stresses from Gauss points The resultant extrapolation matrix thus obtained may be written as

( ) ( )( ) ( )

( ) ( )( ) ( )

( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )

+−−+++−−−++−−−+++−−−−+−−−−+−−−−+

=

IV

III

II

I

σσσσ

σσσσσσσσ

4314314314314314314314314314314314314314314314312315.02315.0

5.02315.02312315.02315.0

5.02315.0231

8

7

6

5

4

3

2

1

(5.5.6)

Here, 1σ , 2σ …….. 8σ are the smoothened nodal values and Iσ … IVσ are the stresses at the Gauss points. Smoothened nodal stress values for four node rectangular element can be also be evaluated in a similar fashion. The relation between the stresses at Gauss points and nodal point for four nodel element will be

1 I

2 II

3 III

IV4

3 1 3 11 12 2 2 2

1 3 1 31 12 2 2 2

3 1 3 11 12 2 2 2

1 3 1 31 12 2 2 2

(5.5.7)

The stress at particular node joining with more than one element will have different magnitude as calculated from adjacent elements (Fig. 5.5.2(a)). The stress resultants are then modified by finding the average of resultants of all elements meeting at a common node. A typical stress distribution for adjacent elements is shown in Fig. 5.5.2(b) after stress smoothening.

Fig. 5.5.2 Stress smoothening at common node 5.5.2 Geometric Nonlinearity As discussed earlier, nonlinear analysis is mainly of two types: (i) Geometric nonlinearity and (ii) Material nonlinearity. For geometric nonlinearity consideration, the relation between strain and displacement is of utmost importance in the finite element formulation for stress analysis problems. In case of plane stress/strain problem, the nonlinear term of the strain expression are dropped for the sake of simplicity in the analysis. However, for large displacement problems, the nonlinear strain term plays a vital role to obtain accurate response. The generalized strain-displacement relations for the two-dimensional plane stress/strain problems are rewritten here to derive the nonlinear solution.

2 2

2 2

12

12

x

y

xy

u u vx x x

v u vy y y

v u u u v vx y x y x y

ε

ε

γ

∂ ∂ ∂ = + + ∂ ∂ ∂ ∂ ∂ ∂

= + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

= + + + ∂ ∂ ∂ ∂ ∂ ∂

(5.5.8)

The displacements at any point inside the node are expressed in terms of their nodla displacements. Thus,

[ ] 1

n

i i i ii

u N u N u=

= =∑ and [ ] 1

n

i i i ii

v N v N v=

= =∑ (5.5.9)

Therefore,

[ ] [ ]

[ ]

[ ]

[ ]

1

1

2

2

ii i

i

i

i

Nu u B ux xv B vxu B uyv B vy

∂∂= =

∂ ∂∂

=∂∂

=∂∂

=∂

(5.5.10)

Here, [B1] and [B2] are the derivative of the shape function [Ni] with respect to x and y respectively. The vectors ui and vi represent the nodal displacements vectors in x and y directions respectively. The vector of strains at any point inside an element, ε may be expressed in terms of nodal displacement as

[ ] B dε =

(5.5.11) where [B] is the strain displacement matrix. d is the nodal displacement vector and may be expressed as

i

i

ud

v =

(5.5.12) The matrix [B] may be expressed with two components as

[ ] [ ] [ ]l nlB B B= +

(5.5.13) where, [Bl] and [Bnl] are the linear and nonlinear part of the strain-displacement matrix respectively and are expressed as follows:

[ ][ ] [ ][ ] [ ][ ] [ ]

=

12

2

1

00

BBB

BBl

(5.5.14) and

[ ]

[ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]

=

1212

2222

1111

21

21

21

21

BBvBBu

BBvBBu

BBvBBu

BTTTT

TTTT

TTTT

nl

(5.5.15) 5.5.2.1 Steps to include effect of geometrical nonlinearity The nonlinear geometric effect of the structure at a particular instant of time can be obtained by performing the following steps.

1. Calculation of displacement d1 considering linear part of strain matrix [Bl]. 2. Evaluation of nonlinear part of the strain matrix [Bnl] (eq5.5.15) adopting d1 from

previous step. 3. Evaluation of total strain matrix [B] = [Bl] + [Bnl]. 4. Calculation of displacement d2 considering both linear and nonlinear part of strain

matrix [B]. 5. Repetition of steps 2 to 4 with d2, from which modified displacement, d3 are

obtained. 6. Step 5 is carried out until the displacements for two consecutive iteration converge

i.e.,

1j j

j

d d

dε−

−<

Where ε is any pre-assigned small value and j is the number of iterations. 5.5.3 Static Condensation The higher order Lagrangian elements (i.e., nine node, sixteen node rectangular element) contain number of internal nodes. This is necessary sometimes for the completeness of the desired polynomial used in displacement function for derivation of interpolation function. These internal nodes are not connected to the adjoining elements in the assemblage (Fig. 5.5.3).

Fig. 5.5.3 Internal nodes of Nine node elements Thus, the displacements of these nodes are not required to formulate overall equilibrium equations of the structure. This limits the usefulness of these elements. A technique known as “static condensation” can be used to suppress the degrees of freedom associated with the internal nodes in the final computation. The technique of static condensation is explained below. The equilibrium equation for a system are expressed in the finite element form as

F = K d (5.5.16)

Where, F, [K] and d are the load vector, stiffness matrix and displacement vector for the entire structure. The above equation can be rearranged by separating the relevant terms corresponding to internal and external nodes of the elements.

i ii ie i

e ei ee e

F K K d=

F K K d

(5.5.17)

Here, di and de are the displacement vectors corresponding to internal and external nodes respectively. Similarly, Fi and Fe are force vectors corresponding to internal and external nodes. Now, the above expression can be written in the following form separately.

i ii i ie eF K d K d (5.5.18)

e ei i ee eF K d K d (5.5.19)

The stiffness matrix and nodal load vector corresponds to the internal nodes can be separated out. For this, eq.(5.5.18) can be rewritten as

1 1i ii i ii ie ed K F K K d (5.5.20)

Substituting the value of di obtained from the above equation in eq.(5.5.19), the following expression will be obtained.

1 1e ei ii i ii ie e ee eF K K F K K d K d (5.5.21)

Here, the equations are reduced to a form involving only the external nodes of the elements. The above reduced substructure equations are assembled to achieve the overall equations involving only the boundary unknowns. Thus the above equation can be rewritten as

1 1e ei ii i ee ei ii ie e eF K K F K K K K d d (5.5.22)

Or

c c eF = K d (5.5.23)

Where, and-1 -1c e ei ii i c ee ei ii ie eF = F - K K F K = K - K K K d . Here, [Kc]

is called condensed or reduced stiffness matrix and Fc is the condensed or effective nodal load vector corresponding to external nodes of the elements. In this process, the size of the matrix for inversion will be comparatively small. The unknown displacements of the exterior nodes, de can be obtained by inverting the matrix [Kc] in eq.(5.5.23). Once, the values of de are obtained, the displacements of internal nodes di can be found from eq.(5.5.20).

Lecture 6: Axisymmetric Element 5.6.1 Introduction Many three-dimensional problems show symmetry about an axis of rotation. If the problem geometry is symmetric about an axis and the loading and boundary conditions are symmetric about the same axis, the problem is said to be axisymmetric. Such three-dimensional problems can be solved using two-dimensional finite elements. The axisymmetric problem are most conveniently defined by polar coordinate system with coordinates (r, θ, z) as shown in Fig. 5.6.1. Thus, for axisymmetric analysis, following conditions are to be satisfied.

1. The domain should have an axis of symmetry and is considered as z axis. 2. The loadings on the domain has to be symmetric about the axis of revolution, thus they

are independent of circumferential coordinate θ. 3. The boundary condition and material properties are symmetric about the same axis and

will be independent of circumferential coordinate.

Fig. 5.6.1 Cylindrical coordinates

Axisymmetric solids are of total symmetry about the axis of revolution (i.e., z-axis), the field variables, such as the stress and deformation is independent of rotational angle θ. Therefore, the field variables can be defined as a function of (r, z) and hence the problem becomes a two dimensional problem similar to those of plane stress/strain problems. Axisymmetric problems includes, circular cylinder loaded with uniform external or internal pressure, circular water tank, pressure vessels, chimney, boiler, circular footing resting on soil mass, etc. 5.6.2 Relation between Strain and Displacement An axisymmetric problem is readily described in cylindrical polar coordinate system: r, z and θ. Here, θ measures the angle between the plane containing the point and the axis of the coordinate

P( r,z, )θ•

system. At θ = 0, the radial and axial coordinates coincide with the global Cartesian X and Y coordinates. Fig. 5.6.2 shows a cylindrical coordinate system and the definition of the position vectors. Let ˆˆ ˆr, z and be unit vectors in the radial, axial, and circumferential directions at a point in the cylindrical coordinate system.

Fig. 5.6.2 Cylindrical Coordinate System If the loading consists of radial and axial components that are independent of θ and the material is either isotropic or orthotropic and the material properties are independent of θ, the displacement at any point will only have radial (𝑢𝑟) and axial (𝑢𝑧) components. The only stress components that will be nonzero are 𝜎𝑟𝑟 , 𝜎𝑧𝑧, 𝜎𝜃𝜃 𝑎𝑛𝑑 𝜏𝑟𝑧 .

(a) Element in r-z plane (b) Element in r-θ plane

Fig. 5.6.3 Deformation of the axisymmetric element

A differential element of the body in the r-z plane is shown in Fig. 5.6.3(a). The element undergoes deformation in the radial direction. Therefore, it initiates increase in circumference and associated circumferential strain. Let denote the radial displacement as u, the circumferential displacement as v, and the axial displacement as w. Dashed line represents the deformed positions of the body in Fig. 5.6.3(b). The radial strain can be calculated from the above diagram as

r1 u uε = u +×dr - u =dr r r

(5.6.1)

Since the rz plane is effectively the same as a rectangular coordinate system, the axial strain will become

z1 w wε = w+×dz - w =dz z z

(5.6.2)

Considering the original arc length versus the deformed arc length, the differential element undergoes an expansion in the circumferential direction. Before deformation, let the arc length is assumed as ds = rdθ. After deformation, the arc length will become ds = (r+u) dθ. Thus, the tangential strain will be

r +u d - rd uε =rd r

(5.6.3)

Similarly, the shear strain will be

rz

r z0 and 0

u wz r

(5.6.4)

Thus, there are four strain components present in this case and is given by

0

0

1 0

r

z

rz

ur rw

uz zwu

r ru wz r z r

θ

εε

εεγ

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂= = =

∂ ∂ ∂ ∂ +

∂ ∂ ∂ ∂

(5.6.5)

5.6.3 Relation between Stress and Strain The stress strain relation for axisymmetric case can be derived from the three dimensional constitutive relations. We know the stress-strain relation for a three-dimensional solid is

⎩⎪⎨

⎪⎧𝜎𝑥𝜎𝑦𝜎𝑧𝜏𝑥𝑦𝜏𝑦𝑧𝜏𝑧𝑥⎭

⎪⎬

⎪⎫

= 𝐸(1+𝜇)(1−2𝜇)

⎣⎢⎢⎢⎢⎢⎢⎢⎢⎡1 − 𝜇 𝜇 𝜇 0 0 0

𝜇 1 − 𝜇 𝜇 0 0 0

𝜇 𝜇 1 − 𝜇 0 0 0

0 0 0 1−2𝜇2

0 0

0 0 0 0 1−2𝜇2

0

0 0 0 0 0 1−2𝜇2 ⎦⎥⎥⎥⎥⎥⎥⎥⎥⎤

⎩⎪⎨

⎪⎧

ϵ𝑥ϵ𝑦ϵ𝑧𝜈𝑥𝑦𝜈𝑦𝑧𝜈𝑧𝑥⎭

⎪⎬

⎪⎫

(5.6.6) The stresses acting on a differential volume of an axisymmetric solid under axisymmetric loading is shown in Fig. 5.6.4.

Fig. 5.6.4 Stresses acting on a differential volume

Now, comparing the stress-strain components present in the axisymmetric case, the stress-strain relation can be expressed from the above expression as follows

⎩⎪⎨

⎪⎧𝜎𝑟

𝜎𝑧

𝜎𝜃

𝜏𝑟𝑧⎭⎪⎬

⎪⎫

= 𝐸(1+𝜇)(1−2𝜇)

⎣⎢⎢⎢⎢⎡1 − 𝜇 𝜇 𝜇 0

𝜇 1 − 𝜇 𝜇 0

𝜇 𝜇 1 − 𝜇 0

0 0 0 1−2𝜇2 ⎦⎥⎥⎥⎥⎤

⎩⎪⎨

⎪⎧𝜀𝑟

𝜀𝑧

𝜀𝜃

𝜈𝑟𝑧⎭⎪⎬

⎪⎫

(5.6.7)

Thus, the constitutive matrix [D] for the axisymmetric elastic solid will be

[D] = 𝐸(1+𝜇)(1−2𝜇)

⎣⎢⎢⎢⎢⎡1 − 𝜇 𝜇 𝜇 0

𝜇 1 − 𝜇 𝜇 0

𝜇 𝜇 1 − 𝜇 0

0 0 0 1−2𝜇2 ⎦⎥⎥⎥⎥⎤

(5.6.8)

5.6.4 Axisymmetric Shell Element A cylindrical liquid storage container like structures (Fig. 5.6.5) may be idealized using axisymmetric shell element for the finite element analysis. It may be noted that the liquid in the container may be idealized with two dimensional axisymmetric elements. Let us consider the radius, height and, thickness of the circular tank are R, H and h respectively.

Fig. 5.6.5 Thin wall cylindrical container

The strain energy of the axisymmetric shell element (Fig. 5.6.6) including the effect of both stretching and bending are expressed as

( )0

H

y yθ θ y y1U = Nε + N ε + M χ 2πRdy2 ∫ (5.6.9)

Here, Ny and Nθ are the membrane force resultants and My is the bending moment resultant. The shell is assumed to be linearly elastic, homogeneous and isotropic. Thus the force and moment resultants can be expressed in terms of the mid-surface change in curvature χy as follows.

Fig 5.6.6 Axisymmetric plate element Here, the strain-displacement relation is given by

[ ] σ ε D= (5.6.10)

In which,

y

yM

NNθσ

=

, y

y

θ

εε ε

χ

=

and [ ] 22

1 01 0

10 0

12

EhDh

µ

µµ

= −

(5.6.11)

The generalized strain vector can be expressed in terms of the displacement vectors as follows.

[ ] B dε = (5.6.12)

Where,

dv

u

=

and [ ]2

2

0

1 0

0

y

R

y

B

∂ ∂ =

∂ − ∂

(5.6.13)

Here, u and v are the displacement components in two perpendicular directions. With the use of stress and strain vectors, the potential energy expression are written in terms of displacement vectors as

[ ] [ ][ ] ( )0

TH

Td B D B d1U = 2πR dy2× ∫ (5.6.14)

Thus, the element stiffness are derived as

[ ] [ ] [ ][ ]0

2π H

Tk R B D B dy= ∫ (5.6.15)

Similarly, neglecting the rotary inertia, the kinetic energy can be expressed as

[ ] [ ] ( )0

THTd N m N1T = 2πR dyd

2× ∫ (5.6.16)

Where, m denotes the mass of the shell element per unit area and d represents the velocity

vector. Thus, the element mass matrix is given by

[ ] [ ] [ ]0

2π eL

TM Rm N N dy = ∫ (5.6.17)

Lecture 7: Finite Element Formulation of Axisymmetric Element Finite element formulation for the axisymmetric problem will be similar to that of the two dimensional solid elements. As the field variables, such as the stress and strain is independent of rotational angle θ, circumferential displacement will not appear. Thus, the displacement field variables are expressed as

n

i ii=1

n

i ii=1

u r,z = N r,z u

w r,z = N r,z w

(5.7.1)

Here, ui and wi represent radial and axial displacements respectively at nodes. Ni (r, z) are the shape functions. As the geometry and field variables are independent of rotational angle θ, the interpolation function Ni (r, z) can be expressed similar to 2-dimensional problems by replacing the x and y terms with r and z terms respectively.

5.7.1 Stiffness Matrix of a Triangular Element Fig. 5.7.1 shows the cylindrical coordinates of a three node triangular element. Hence the analysis of the axisymmetric element can be approached in a similar way as the CST element. Thus the field variables of such an element can be expressed as

0 1 2

3 4 5

u r zw r z

α α αα α α

= + += + +

(5.7.2)

Or,

[ ] d φ α= (5.7.3)

Where,

u

dw

=

, [ ] 1 0 0 00 0 0 1

r zr z

φ

=

and 0 1 2 3 4 5Tα α α α α α α=

Using end conditions,

1 0

2 1

3 2

1 3

2 4

3 5

1 0 0 01 0 0 01 0 0 00 0 0 10 0 0 10 0 0 1

i i

j j

k j

i i

j j

k k

u r zu r zu r zw r zw r zw r z

αααααα

=

(5.7.4)

Or,

[ ]

[ ] 1

d A

A d

α

α −

=

⇒ = (5.7.5)

Here d are the nodal displacement vectors.

Fig. 5.7.1 Axisymmetric three node triangle in cylindrical coordinates

Putting above values in eq.(5.7.3), the following relations will be obtained.

[ ][ ] 1 [ ]d A d N dφ −= = (5.7.6)

Or,

1

2

3

1

2

3

0 0 00 0 0

i j k

i j k

rr

N N N rud

N N N zwzz

= =

(5.7.7)

Using a similar approach as in case of CST elements, the three shape functions [ ]1 2 3, ,N N N can

be assumed as,

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

1 2 3 3 2 2 3 3 2

2 3 1 1 3 3 1 1 3

3 1 2 2 1 1 2 2 1

1,21,

21,

2

N r z r z r z z z r r r zA

N r z r z r z z z r r r zA

N r z r z r z z z r r r zA

= − + − + −

= − + − + −

= − + − + −

Or,

( ) ( )

( ) ( )

( ) ( )

1,21,

21,

2

i i i i

j j j j

k k k k

N r z r zA

N r z r zA

N r z r zA

α β γ

α β γ

α β γ

= + +

= + +

= + +

(5.7.8)

Where,

( )122

i j k k j j k i i k k i j j i

i j k j k i k i j

i k j i i k i j i

i j j k k i i k j i k j

r z r z r z r z r z r zz z z z z zr r r r r r

A r z r z r z r z r z r z

α α α

β β β

γ γ γ

= − = − = −

= − = − = −

= − = − = −

= + + − − −

(5.7.9)

Putting the value of u,w in eq. (5.7.7) from eq. (5.6.5),

[ ]

1

2

3

1

2

3

0 0 0

0 0 0

0 0 0

ji k

ji k

ji k

j ji k i k

NN Nrr r rrNN Nrr r r B dzNN Nzz z z

N NN N N N zz z z r r r

ε

∂ ∂ ∂ ∂ ∂ ∂ = = ∂∂ ∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

(5.7.10)

Thus, the strain displacement matrix can be expressed as,

[ ]

0 0 0

0 0 012 0 0 0

i j k

ji k

i j k

i j k i j k

NN NB r r r

A

β β β

γ γ γγ γ γ β β β

=

(5.7.11)

Where, 3

i j kr r rr

+ += . Thus the stiffness matrix will become

[ ] [ ] [ ][ ]Tk B D B d= Ω∫∫∫

Or, [ ] [ ] [ ][ ] [ ] [ ][ ]2

0

2T Tk B D B r d dA B D B r dr dzπ

θ π= =∫ ∫ ∫ ∫ ∫ (5.7.12)

Since, the term [B] is dependent of ‘r’ terms; the term [ ] [ ][ ]TB D B cannot be taken out of

integration. Yet, a reasonably accurate solution can be obtained by evaluating the [B] (denoted as [B]) matrix at the centroid.

Hence, [ ] [ ] [ ][ ]2 Tk r B D B dr dzπ= ∫ ∫

Or,

[ ] [ ] [ ][ ]2Tk B D B rAπ (5.7.13)

5.7.2 Stiffness Matrix of a Quadrilateral Element

50

The strain-displacement relation for axisymmetric problem derived earlier (eq.(5.6.5)) can be rewritten as

r

z

rz

ur

1 0 0 0 0 u0 0 0 1 0 z

w10 0 0 0 rrw0 1 1

urw

0 0z

zur

u wz r u

θ

εε

εεγ

∂∂ ∂ ∂ ∂ ∂ ∂ ∂= = = ∂

∂ ∂ ∂ ∂ + ∂

∂ ∂

(5.7.14)

Applying chain rule of differentiation equation we get,

* *11 12* *21 22

* *11 12* *21 22

uuξr

J J 0 0 0 uuηJ J 0 0 0z

w w0 0 J J 0rξ 0 0 J J 0w w0 0 0 0 1z η

u u

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ =∂ ∂ ∂ ∂

∂ ∂ ∂ ∂

(5.7.15)

Hence, the strain components are calculated as

* *11 12

r * *21 22

z * *11 12

θ * *21 22

rz

uξ

J J 0 0 0 u1 0 0 0 0εηJ J 0 0 00 0 0 1 0εw0 0 J J 01ε 0 0 0 0 ξ0 0 J J 0r

0 1 1 0 0 w0 0 0 0 1η

u

γ

∂ ∂

∂ ∂ = ∂ ∂ ∂

∂

Or,

51

* *11 12

r * *21 22

z

* * * *rz21 22 11 12

u

uJ J 0 0 00 0 J J 0

w10 0 0 0r

wJ J J J 0

u

x h x h

(5.7.16)

With the use of interpolation function and nodal displacements, , , ,u u w wx h x h

can be expressed

for a four node quadrilateral element as

31 2 4

31 2 4

31 2 4

31 2 4

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

uNN N Nu

NN N Nu

w NN N N

w NN N N

x x x xx

h h h h h

x x x x x

h h h h h

1

2

3

4

1

2

3

4

uuuwwww

(5.7.17)

Putting eq. (5.7.17) in eq. (5.7.16) we get,

31 2 4

* *31 2 411 12

r * *21 22

z31 2 4

* * * *rz31 2 421 22 11 12

NN N N 0 0 0 0

NN N NJ J 0 0 0 0 0 0 00 0 J J 0

NN N N1 0 0 0 00 0 0 0r

NN N NJ J J J 0 0 0 0 0

x x x x

h h h h x x x x h h h

1

2

3

4

1

2

3

41 2 3 4 1 2 3 4

uuuuwwwwN N N N N N N N

h

(5.7.18) Thus, the strain displacement relationship matrix [B] becomes

52

31 2 4

* *31 2 411 12

* *21 22

31 2 4

* * * *31 2 421 22 11 12

1 2 3 4 1 2 3 4

NN N N 0 0 0 0

NN N NJ J 0 0 0 0 0 0 00 0 J J 0

B NN N N1 0 0 0 00 0 0 0r

NN N NJ J J J 0 0 0 0 0

N N N N N N N N

x x x x h h h h x x x x h h h h

(5.7.19)

For a four node quadrilateral element, ( ) 1 1

1

1ξ 1 η N N1η 1 ξN and 4ξ 4 η 4( ) ( ) ( )− − ∂ ∂− −

= ⇒ = − = −∂ ∂

( ) 2 12

1ξ 1 η N N1η 1 ξN and 4ξ 4 η 4( ) ( ) ( )+ − ∂ ∂− +

= ⇒ = = −∂ ∂

(5.7.20)

( ) 2 13

1ξ 1 η N N1η 1 ξN and 4ξ 4 η 4( ) ( ) ( )+ + ∂ ∂+ +

= ⇒ = =∂ ∂

( ) 2 14

1ξ 1 η N N1η 1 ξN and 4ξ 4 η 4( ) ( ) ( )− + ∂ ∂+ −

= ⇒ = − =∂ ∂

Thus, the [B] matrix will become

* *11 12

* *21 22

* * * *21 22 11 12

J J 0 0 00 0 J J 0

B 10 0 0 0r

J J J J 0

1 1 1 10 0 0 0

4 4 4 41 1 1 1

0 0 0 04 4 4 4

1 1 1 10 0 0 0

4 4 4 41 1 1 1

0 0 0 04 4 4 4

1 1 1 1 1 1 1 1 1 1 1 1 14 4 4 4 4 4

h h h h

x x x x

h h h h

x x x x

x h x h x h x h x h x h 1 1 14 4

x h x h (5.7.21) The stiffness matrix for the axisymmetric element finally can be found from the following expression after numerical integration.

[ ] [ ] [ ][ ] [ ] [ ][ ]1 1

T

1 1

B D B .2πr. .dξdηTk B D JB d+ +

Ω − −

= Ω =∫ ∫ ∫ (5.7.22)

53

Lecture 8: Finite Element Formulation for 3 Dimensional Elements 5.8.1 Introduction Solid elements can easily be formulated by the extension of the procedure followed for two dimensional solid elements. A domain in 3D can be discritized using tetrahedral or hexahedral elements. For example, the eight node solid brick element is analogous to the four node rectangular element. Regardless of the possible curvature of edges or number of nodes, the solid element can be mapped into the space of natural co-ordinates, i.e the 1,1,1 ±=±=±= ζηξ just like a plane element.

Fig. 5.8.1Eight node brick element For three dimensional cases, each node has three degrees of freedom having u, v, and w as displacement field in three perpendicular directions (X, Y and Z). In this case, one additional dimension increases the computational expense manifolds. 5.8.2 Strain Displacement Relation The strain vector for three dimensional cases can be written in the following form

54

1 0 0 0 0 0 0 0 00 0 0 0 1 0 0 0 00 0 0 0 0 0 0 0 10 1 0 1 0 0 0 0 00 0 0 0 0 1 0 1 00 0 1 0 0 0 1 0 0

x

y

z

xy

yz

zx

uxuyuzvxvyvzwxwywz

εεεγγγ

∂ ∂ ∂

∂ ∂

∂ ∂ ∂ ∂ = ∂

∂ ∂

∂ ∂ ∂

∂ ∂

∂ (5.8.1)

The following relation exists between the derivative operators in the global co-ordinates and the natural co-ordinate system by the use of chain rule of partial differentiation.

∂∂∂∂∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

=

∂∂∂∂∂∂

z

y

x

zyx

zyx

zyx

ζζζ

ηηη

ξξξ

ζ

η

ξ

(5.8.2)

Where the Jacobian Matrix will be

[ ]

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

=

ζζζ

ηηη

ξξξ

zyx

zyx

zyx

J (5.8.3)

For an isoparamatric element the coordinates at a point inside the element can be expressed by its nodal coordinate.

1 1 1and

n n n

i i i i i ii i i

x N x ; y N y z N z= = =

= = =∑ ∑ ∑ (5.8.4)

55

Substituting the above equations into the Jacobian matrix for an eight node brick element, we get

[ ]

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

=∑=

ii

ii

ii

ii

ii

ii

ii

ii

ii

i

zNyNxN

zNyNxN

zNyNxN

J

ζζζ

ηηη

ξξξ8

1 (5.8.5)

The strain displacement relation is given by [ ] dB=ε , where,

=

wvu

d .

The displacements in the x, y and z direction are u, v, and w respectively. Let consider the inverse of Jacobian matrix as

[ ]

=−

*33

*32

*31

*23

*22

*21

*13

*12

*11

1

JJJJJJJJJ

J (5.8.6)

Thus, the relation between two coordinate systems can be rewritten as 1

11 12 13

21 22 23

31 32 33

* * *

* * *

* * *

x y zx J J J

x y z J J Jy

J J Jx y z

z

ξ ξ ξ ξ ξ

η η η η η

ζ ζ ζ ζ ζ

− ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂ = = ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂∂

∂ ∂ ∂ ∂ ∂∂

(5.8.7)

Thus, one can write the following relations

8 8

11 12 13 11 12 131 1

* * * * * *i i ii i i

i i

N N Nu u u uJ J J J J J u a ux ξ η ζ ξ η ζ= =

∂ ∂ ∂ ∂ ∂ ∂ ∂= + + = + + = ∂ ∂ ∂ ∂ ∂ ∂ ∂

∑ ∑

8 8

21 22 23 21 22 231 1

* * * * * *i i ii i i

i i

N N Nu u u uJ J J J J J u b uy ξ η ζ ξ η ζ= =

∂ ∂ ∂ ∂ ∂ ∂ ∂= + + = + + = ∂ ∂ ∂ ∂ ∂ ∂ ∂

∑ ∑

8 8

31 32 33 31 32 331 1

* * * * * *i i ii i i

i i

N N Nu u u uJ J J J J J u c uz ξ η ζ ξ η ζ= =

∂ ∂ ∂ ∂ ∂ ∂ ∂= + + = + + = ∂ ∂ ∂ ∂ ∂ ∂ ∂

∑ ∑

Similarly, 8 8 8 8 8 8

1 1 1 1 1 1andi i i i i i i i i i i i

i i i i i i

v v v w w wa v ; b v ; c v ; a w ; b w c wx y z x y z= = = = = =

∂ ∂ ∂ ∂ ∂ ∂= = = = = =

∂ ∂ ∂ ∂ ∂ ∂∑ ∑ ∑ ∑ ∑ ∑

Using above relations, the strain vector can be written as

56

8

1

0 00 00 0

00

0

x i

y ii

z ii

ixy i ii

yz i i

zx i i

uxv

aybw u

cz vu v b a

wy x c bv w c az yu wz x

εεε

εγγγ

=

∂ ∂

∂ ∂

∂ ∂ = = = ∂ ∂ + ∂ ∂ ∂ ∂ +

∂ ∂ ∂ ∂ + ∂ ∂

∑ (5.8.8)

Now, the strain displacement relationship matrix [B] can be identified from the above equation by

comparing it to [ ] dB=ε . 5.8.3 Element Stiffness Matrix The element stiffness matrix can be generated similar to two dimensional case using the following relations.

[ ]1 1 1

1 1 1

Tk [ B ] [ D ][ B ]d d d Jξ η ζ+ + +

− − −

= ∫ ∫ ∫ (5.8.9)

The size of the constitutive matrix [D] for solid element will be 6 × 6 and is already discussed in module 1, lectures 3. For eight node brick element, the size of stiffness matrix will become 24 × 24 as number of nodes in one element is 8 and the degrees of freedom at each node is 3. It is well established that 2 × 2 × 2 Gauss integration points are the optimal sampling points for eight node isoparametric brick elements.

5.8.4 Element Load Vector The forces on an element can be generated due to its self weight or externally applied force which may be concentrated or distributed in nature. The distributed load may be uniform or non-uniform. All these types of loads are to redistributed to the nodes using finite element formulation.

5.8.4.1 Gravity load The load vector due to body forces in general is given by

TQ [ N ] X dΩ

= Ω∫ (5.8.10)

57

where X is the body forces per unit volume. The nodal load vector at any node i may be expressed as

Ti iQ [ N ] X d

Ω

= Ω∫ (5.8.11)

In case of gravity load, the force will act in the global negative Z direction. Therefore,

=

i

i

i

i

NN

NN

000000

][ and 00X

gρ

= −

(5.8.12)

Here, the mass density of the material is ρ and the acceleration due to gravity is g. Thus, eq.(5.8.11) will become

00i

i

Q dN gρΩ

= Ω − ∫ (5.8.13)

For isoparametric element the, the above expression will become

ζηξρ

dddJN

Q

i

i ∫ ∫ ∫+

−

+

−

+

−

−=

1

1

1

1

1

100

(5.8.14)

Using Gauss Quadrature integration rule, the above expression may be evaluated as

1 1 1

00

i j k

n n n

i i j k i j ki j k

i ( , , )

Q w w w J ( , , )N g

ξ η ζ

ξ η ζρ= = =

= −

∑∑∑ (5.8.15)

Where, n is the number of nodes in an element. For eight node linear brick element the value of n will be 8 and the integration order suggested is 2×2×2. Similarly, for twenty node quadratic brick element, the value of n will be 20 and the integration order suggested is 3×3×3.

5.8.4.2 Surface pressure Let assume a uniform surface pressure of intensity q is acting normal to the element face. The load vector due to surface pressure is given by

∫∫= dApNQ Ts ][ (5.8.16)

The nodal load at any node i may be expressed as ∫∫= dApNQ Ts

ii ][ (5.8.17)

In case of surface load, the value of ][ siN in the above equation will become

58

=si

si

si

si

NN

NN

000000

][ (5.8.18)

Here siN is the interpolation function for the node i. For example, the value of s

iN can be obtained

by substituting ξ =1 in siN for face 1. Thus, the surface pressure is expressed as,

=

3

3

3

qnqmql

p (5.8.19)

Where, 333 ,, nml are the direction cosines. Thus, eq.(5.8.17) can be expressed using eq.(5.8.19) in the following form.

dAqnNqmNqlN

Qsi

si

si

i ∫ ∫+

−

+

−

=1

1

1

1

3

3

3

(5.8.20)

The value of dA is can be evaluated considering the cross product of vectors along the natural coordinates parallel to the loaded faces of the element. Thus,

ζηddeedA 32 ×= (5.8.21)

5.8.5 Stress Computation Using the relation of F = [K]d, the unknown nodal displacement vector d are calculated in global coordinate system. Once the nodal displacements are obtained, the strain components as each node can be computed using strain-displacement relations for each element. Similarly element stress can be calculated using stress-strain relation. These stresses at Gauss points are extrapolated to the corresponding nodes of the element to find the nodal stresses. In general, for three dimensional state of stress there are at least three planes, called principal planes. The corresponding stress vector is perpendicular to the plane and where there are no normal shear stresses. These three stresses which are normal to these principal planes are called principal stresses. The principal stresses σ1, σ2, and σ3 are computed from the roots of the cubic equation represented by the determinant of the flowing.

0x xy xz

xy y yz

xz yz z

σ σ τ ττ σ σ ττ τ σ σ

−− =

− (5.8.22)

The characteristic equation has three real roots σi, due to the symmetry of the stress tensor. The principal stresses are arranged so that σ1 > σ2 > σ3. The maximum shear stress can be computed from the following relations.

59

1 2 2 3 3 1max = largest of and

2 2 2,

σ σ σ σ σ στ

− − −

(5.8.23)

These three shear stress components will occur on planes oriented at 450 from the principal planes. The distortion energy theory suggests that the total strain energy can be divided into two components. They are (i) volumetric strain energy and (ii) distortion or shear strain energy. It is anticipated that yield develops if the distortion component exceeds that at the yield point for a simple tensile test. From the concept of distortion energy theory, the equivalent stress which is historically known as Von Mises stress are defined as

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

2e

σ σ σ σ σ σσ

− + − + −

(5.8.24)

The Von Mises stresses offer a measure of the shear or distortional stress in the material. In general, this type of stress tends to cause yielding in metals.

60

Worked out Examples Example 5.1 Calculation of nodal loads on a triangular element A CST element as shown in Fig. 5.I gets axial loading of (Fx1) 10 kN/m in X direction and (Fy1) 20 kN/m in Y direction. Compute the nodal loads in the element.

Fig. 5.I Distributed loading on a triangular element

From the above figure, the length of sides 1, 2 and 3 are calculated and will be 10, 8 and 6 cm respectively. First, let consider side 1:

[ ] 1

10

lTF N F dsΓ= ∫

Or,

1 1 1

2 2 2

13 3 31

10 0 0

2 2

3 3

0 0 0 0 00 00 0 10

100 0 0 0 0 00 0 00 0 0

l l lx

y

L L LFL L L

F ds ds dsF

L LL L

= = = ×

∫ ∫ ∫

Putting, ( )1 2

! !1 !

p q p qL L ds lp q

=+ +∫ , we will get,

61

11

0 0 01 1 0.51 1 0.50.110 100 0 02 20 0 00 0 0

lF

= × × = × × =

Similarly for side 2,

2 2

1 1

23 3 22

21 10 0

3 3

0 0 0 00 0 0 0 0 0

0 0 0 0 020

0 0 20 1 0.820 0 0 0 0 00 0 1 0.8

l lx

y

L L

FL L lF ds ds kNFL L

L L

= = = × × =

∫ ∫

Since no force is acting on side 3,

3

000000

F

=

Hence, the nodal load vector in all the nodes in x and y directions will become,

00.50.50.80

0.8

F

=

kN.

Content Module 6 FEM for Plates and Shells (06) Lecture 1: Introduction to Plate Bending Problems(Page: 1-10)

6.1.1 Introduction 6.1.2 Notations and Sign Conventions 6.1.3 Thin Plate Theory

6.1.3.1 Basic relationships 6.1.3.2 Constitutive equations 6.1.3.3 Calculation of moments and shear forces

6.1.4 Thick Plate Theory 6.1.5 Boundary Conditions

Lecture 2: Finite Element Analysis of Thin Plate(Page: 11-17)

6.2.1 Triangular Plate Bending Element 6.2.2 Rectangular Plate Bending Element

Lecture 3: Finite Element Analysis of Thick Plate (Page: 18-24)

6.3.1 Introduction 6.3.2 Strain Displacement Relation 6.3.3 Element Stiffness Matrix 6.3.4 Nodal Load Vector

Lecture 4: Finite Element Analysis of Skew Plate (Page: 25-31)

6.4.1 Introduction 6.4.2 Finite Element Analysis of Skew Plate Element

Lecture 5: Introduction to Finite Strip Method(Page: 32-37)

6.5.1 Introduction 6.5.2 Finite Strip Method 6.5.2.1 Boundary conditions 6.5.3 Finite Element Formulation

Lecture 6: Finite Element Analysis of Shell(Page: 38-45)

6.6.1 Introduction 6.6.2 Classification of Shells 6.6.3 Assumptions for Thin Shell Theory 6.6.4 Overview of Shell Finite Elements 6.6.5 Finite Element Formulation of a Degenerated Shell

6.6.5.1 Jacobian matrix 6.6.5.2 Strain displacement matrix 6.6.5.3 Stress strain relation 6.6.5.4 Element stiffness matrix

1

Lecture 1: Introduction to Plate Bending Problems 6.1.1 Introduction A plate is a planer structure with a very small thickness in comparison to the planer dimensions. The forces applied on a plate are perpendicular to the plane of the plate. Therefore, plate resists the applied load by means of bending in two directions and twisting moment. A plate theory takes advantage of this disparity in length scale to reduce the full three-dimensional solid mechanics problem to a two-dimensional problem. The aim of plate theory is to calculate the deformation and stresses in a plate subjected to loads. A flat plate, like a straight beam carries lateral load by bending. The analyses of plates are categorized into two types based on thickness to breadth ratio: thick plate and thin plate analysis. If the thickness to width ratio of the plate is less than 0.1 and the maximum deflection is less than one tenth of thickness, then the plate is classified as thin plate. The well known as Kirchhoff plate theory is used for the analysis of such thin plates. On the other hand, Mindlin plate theory is used for thick plate where the effect of shear deformation is included. 6.1.2 Notations and Sign Conventions Let consider plates to be placed in XY plane. Representation of plate surface slopes W,x, W,y by right hand rule produces arrows that point in negative Y and positive X directions respectively. Both surface slopes and rotations are required for plate elements. Signs and subscripts of rotations and slopes are reconciled by replacing θx by Ψy and θy by -Ψx

Fig. 6.1.1 Notations and sign conventions

6.1.3 Thin Plate Theory

2

Classical thin plate theory is based upon assumptions initiated for beams by Bernoulli but first applied to plates and shells by Love and Kirchhoff. This theory is known as Kirchhoff’s plate theory. Basically, three assumptions are used to reduce the equations of three dimensional theory of elasticity to two dimensions.

1. The line normal to the neutral axis before bending remains straight after bending. 2. The normal stress in thickness direction is neglected. i.e., 0zσ = . This assumption converts

the 3D problem into a 2D problem. 3. The transverse shearing strains are assumed to be zero. i.e., shear strains γxz and γyz will be

zero. Thus, thickness of the plate does not change during bending. The above assumptions are graphically shown in Fig. 6.1.2.

Fig. 6.1.2 Kirchhoff plate after bending

6.1.3.1 Basic relationships Let, a plate of thickness t has mid-surface at a distance 𝑡

2 from each lateral surface. For the analysis

purpose, X-Y plane is located in the plate mid-surface, therefore z=0 identifies the mid-surface. Let u, v, w be the displacements at any point (x, y, z).

3

Fig. 6.1.3 Thin plate element Then the variation of u and v across the thickness can be expressed in terms of displacement w as

u = -z 𝜕𝑤𝜕𝑥

v = -z 𝜕𝑤𝜕𝑦

(6.1.1)

Where, w is the deflection of the middle plane of the plate in the z direction. Further the relationship between, the strain and deflection is given by,

εx = 𝜕𝑢𝜕𝑥

= -z 𝜕2𝑤𝜕𝑥2

= z χx

εy = 𝜕𝑣𝜕𝑦

= -z 𝜕2𝑤𝜕𝑦2

= z χy (6.1.2)

γxy = 𝜕𝑢𝜕𝑦

+ 𝜕𝑣𝜕𝑥

= -2z 𝜕2𝑤

𝜕𝑥𝜕𝑦 = z χxy

where, ε corresponds to direct strain

γ corresponds to shear strain χ corresponds to curvature along respective directions.

Or in matrix form, the above expression can written as

𝜀𝑥𝜀𝑦γxy

= −𝑧

⎣⎢⎢⎢⎡𝜕2

𝜕𝑥2𝜕2

𝜕𝑦2

𝜕2

𝜕𝑥𝜕𝑦⎦⎥⎥⎥⎤

𝑤 (6.1.3)

Or, ε = −z∆𝑤 (6.1.4)

Where, ε is the vector of in-plane strains, and ∆ is the differential operator matrix. 6.1.3.2 Constitutive equations

4

From Hooke’s law,

[ ]Dσ ε= (6.1.5)

Where,

[ ] ( )2

1 01 0

1 10 02

EDυ

υυ

υ

= − −

(6.1.6)

Here, [D] is equal to the value defined for 2D solids in plane stress condition (i.e., 0zσ = ).

6.1.3.3 Calculation of moments and shear forces Let consider a plate element of 𝑑𝑥 × 𝑑𝑦 and with thickness t. The plate is subjected to external uniformly distributed load p. For a thin plate, body force of the plate can be converted to an equivalent load and therefore, consideration of separate body force is not necessary. By putting eq. (6.1.4) in eq. (6.1.5),

[ ]z D wσ = − ∆ (6.1.7)

It is observed from the above relation that the normal stresses are varying linearly along thickness of the plate (Fig. 6.1.4(a)). Hence the moments (Fig. 6.1.4(b)) on the cross section can be calculated by integration.

[ ] [ ]/ 2 / 2 3

2

/ 2 / 2 12

X t t

yt t

xy

MtM M zdt z dt D w D w

Mσ

− −

= = = − ∆ = − ∆

∫ ∫ (6.1.8)

5

(a) Stresses in plate

(b) Forces and moments in plate

Fig. 6.1.4 Forces on thin plate

6

On expansion of eq. (6.1.8) one can find the following expressions.

( ) ( )3 2 2

2 2212 1X P x yEt w wM D

x yυ χ υχ

υ ∂ ∂

= − + = + ∂ ∂−

( ) ( )3 2 2

2 2212 1y P y xEt w wM D

y xυ χ υχ

υ ∂ ∂

= − + = + ∂ ∂− (6.1.9)

( )( )3 2 1

12 1 2P

xy yx xy

DEt wM Mx y

υχ

υ−∂

= = = −+ ∂ ∂

Where, DP is known as flexural rigidity of the plate and is given by,

( )3

212 1PEtD

υ=

− (6.1.10)

Let consider the bending moments vary along the length and breadth of the plate as a function of x and

y. Thus, if Mx acts on one side of the element, ' xx x

MM M dxx

∂= +

∂ acts on the opposite side. Considering

equilibrium of the plate element, the equations for forces can be obtained as

0yx QQ px y

∂∂+ + =

∂ ∂ (6.1.11)

xyxx

MM Qx y

∂∂+ =

∂ ∂ (6.1.12)

xy yy

M MQ

x y∂ ∂

+ =∂ ∂

(6.1.13)

Using eq.6.1.9 in eqs.6.1.12 & 6.1.13, the following relations will be obtained. 2 2

2 2x Pw wQ D

x x y ∂ ∂ ∂

= − + ∂ ∂ ∂ (6.1.14)

2 2

2 2y Pw wQ D

y x y ∂ ∂ ∂

= − + ∂ ∂ ∂ (6.1.15)

Using eqs. (6.1.14) and (6.1.15) in eq. (6.1.11) following relations will be obtained. 4 4 4

4 2 2 42P

w w w px x y y D

∂ ∂ ∂+ + = −

∂ ∂ ∂ ∂ (6.1.16)

6.1.4 Thick Plate Theory Although Kirchhoff hypothesis provides comparatively simple analytical solutions for most of the cases, it also suffers from some limitations. For example, Kirchhoff plate element cannot rotate independently of the position of the mid-surface. As a result, problems occur at boundaries, where the undefined transverse shear stresses are necessary especially for thick plates. Also, the Kirchhoff theory is only

7

applicable for analysis of plates with smaller deformations, as higher order terms of strain-displacement relationship cannot be neglected for large deformations. Moreover, as plate deflects its transverse stiffness changes. Hence only for small deformations the transverse stiffness can be assumed to be constant. Contrary, Reissner–Mindlin plate theory (Fig. 6.1.5) is applied for analysis of thick plates, where the shear deformations are considered, rotation and lateral deflections are decoupled. It does not require the cross-sections to be perpendicular to the axial forces after deformation. It basically depends on following assumptions,

1. The deflections of the plate are small. 2. Normal to the plate mid-surface before deformation remains straight but is not necessarily

normal to it after deformation. 3. Stresses normal to the mid-surface are negligible.

Fig. 6.1.5 Bending of thick plate

Thus, according to Mindlin plate theory, the deformation parallel to the undeformed mid surface, u and v, at a distance z from the centroidal axis are expressed by,

8

𝑢 = 𝑧𝜃𝑦 (6.1.17) v = −zθ𝑥 (6.1.18)

Where θx and θy are the rotations of the line normal to the neutral axis of the plate with respect to the x and y axes respectively before deformation. The curvatures are expressed by

yx x

θχ

∂=

∂ (6.1.19)

xy y

θχ ∂= −

∂ (6.1.20)

Similarly the twist for the plate is given by,

y xxy y x

θ θχ∂ ∂

= − ∂ ∂ (6.1.21)

Using eqs.6.1.9-6.1.10, the bending stresses for the plate is given by

( )3

2

1 01 0

12 1 10 02

x x

y y

xy xy

MEtM

M

υ χυ χ

υυ χ

= − −

(6.1.22)

Or

[ ] M D χ= (6.1.23)

Further, the transverse shear strains are determined as

xz ywx

γ θ ∂= +

∂ (6.1.24)

yz xwy

γ θ ∂= − +

∂ (6.1.25)

The shear strain energy can be expressed as

Us = 12𝛼𝐺𝐴 ∬ (𝛾𝑥)2 + 𝛾𝑦

2 𝑑𝑥 𝑑𝑦

𝐴

= 12𝛼𝐺𝐴 ∬ 𝜃𝑦 + 𝜕𝑤

𝜕𝑥2

+ −𝜃𝑥 + 𝜕𝑤𝜕𝑦2 𝑑𝑥 𝑑𝑦

𝐴 (6.1.26)

Where, G = 𝐸2(1+𝜇)

. The shear stresses are

𝜏𝑥𝑧𝜏𝑦𝑧 = 𝐸

2(1+𝜇)1 00 1

𝛾𝑥𝛾𝑦 (6.1.27)

Hence the resultant shear stress is given by,

𝑄𝑥𝑄𝑦 = 𝐸𝑡𝛼

2(1+𝜇) 1 00 1

𝛾𝑥𝛾𝑦 (6.1.28)

Or, 𝑄 = [𝐷𝑠]𝛾 (6.1.29)

9

Here "𝛼" is the numerical correction factor used to characterize the restraint of cross section against warping. If there is no warping i.e., the section is having complete restraint against warping then α = 1 and if it is having no restraint against warping then α = 2/3. The value of α is usually taken to be π2/12 or 5/6. Now, the stress resultant can be combined as follows.

[ ][ ]

00 s

M DQ D

χγ

=

(6.1.30)

Or,

⎩⎪⎨

⎪⎧𝑀𝑥𝑀𝑦𝑀𝑥𝑦𝑄𝑥𝑄𝑦 ⎭

⎪⎬

⎪⎫

=

⎣⎢⎢⎢⎡ 𝐸𝑡3

12(1−𝜇2) 1 𝜇 0𝜇 1 00 0 1−𝜇

2

0 00 00 0

0 0 00 0 0

𝐸𝑡2(1+𝜇)

𝛼 00 𝛼⎦

⎥⎥⎥⎤

⎩⎪⎨

⎪⎧

χ𝑥χ𝑦χ𝑥𝑦𝛾𝑥𝛾𝑦 ⎭⎪⎬

⎪⎫

(6.1.31)

Or,

⎩⎪⎨

⎪⎧𝑀𝑥𝑀𝑦𝑀𝑥𝑦𝑄𝑥𝑄𝑦 ⎭

⎪⎬

⎪⎫

=

⎣⎢⎢⎢⎡ 𝐸𝑡3

12(1−𝜇2) 1 𝜇 0𝜇 1 00 0 1−𝜇

2

0 00 00 0

0 0 00 0 0

𝐸𝑡2(1+𝜇)

𝛼 00 𝛼⎦

⎥⎥⎥⎤

⎩⎪⎪⎪⎨

⎪⎪⎪⎧

𝜕𝜃𝑦𝜕𝑥

− 𝜕𝜃𝑥𝜕𝑦

𝜕𝜃𝑦𝜕𝑦

− 𝜕𝜃𝑥𝜕𝑥

𝜃𝑦 + 𝜕𝑤𝜕𝑥

−𝜃𝑥 + 𝜕𝑤𝜕𝑦⎭⎪⎪⎪⎬

⎪⎪⎪⎫

(6.1.32)

The above relation may be compared with usual stress-strain relation. Thus, the stress resultants and their corresponding curvature and shear deformations may be considered analogous to stresses and strains. 6.1.5 Boundary Conditions For different boundaries of the plate (Fig. 6.1.6), suitable conditions are to be incorporated in plate equation for solving the governing differential equations. For example, following conditions need to satisfy along y direction of the plate for various boundaries.

10

Fig. 6.1.6 Plate with four boundaries

1. Simply support edge (Along y direction)

( ) [ ], 0, 0 & 0xw x y M x const y b= = = ≤ ≤

2. Clamped Edge (Along y direction)

( ) ( ) [ ], 0, , 0 & 0ww x y x y x const y bx

∂= = = ≤ ≤

∂

3. Free Edge (Along y direction)

[ ]0, 0, & 0xyx x

MM Q x const y b

x∂

= + = = ≤ ≤∂

Similar to the above, the boundary conditions along x direction can also be obtained. Once the displacements w(x,y) of the plate at various positions are found, the strains, stresses and moments developed in the plate can be determined by using corresponding equations.

11

1

x y

x2 xy y2

x3 x2y xy2 y3

x4 x3y x2y2 xy3 y4

Lecture 2: Finite Element Analysis of Thin Plate 6.2.1 Triangular Plate Bending Element A simplest possible triangular bending element has three corner nodes and three degrees of freedom per nodes 𝑤,𝜃𝑥, 𝜃𝑦 as shown in Fig. 6.2.1.

Fig. 6.2.1 Triangular plate bending element As nine displacement degrees of freedom present in the element, we need a polynomial with nine independent terms for defining, w(x,y) . The displacement function is obtained from Pascal’s triangle by choosing terms from lower order polynomials and gradually moving towards next higher order and so on. Thus, considering Pascal triangle, and in order to maintain geometric isotropy, we may consider the displacement model in terms of the complete cubic polynomial as, ( ) ( )2 2 3 2 2 3

0 1 2 3 4 5 6 7 8,w x y x y x xy y x x y xy yα α α α α α α α α= + + + + + + + + + (6.2.1)

12

Corresponding values for 𝜃𝑥 ,𝜃𝑦are,

( )2 22 4 5 7 82 2 3x

w x y x xy yy

θ α α α α α∂= = + + + + +∂

(6.2.2)

( )2 21 3 4 6 72 3 2y

w x y x xy yx

θ α α α α α∂= − = − − − − − +

∂ (6.2.3)

In the matrix form,

( )( )( )

0

1

22 2 3 2 2 3

32 2

4

2 2 5

6

7

8

1

0 0 1 0 2 0 2 3

0 1 0 2 0 3 2 0

x

y

x y x xy y x x y xy ywx y x xy y

x y x xy y

αααα

θ αθ α

ααα

+ = +

− − − − − +

(6.2.4)

Putting the nodal displacements and rotations for the triangular plate element as shown in Fig. 6.2.1 in the above equation, one can express following relations.

( )( )

1

1

1 2 32 2 2

2 22 2

2 22 2

2 2 2 3 2 2 23 3 3 3 3 3 3 3 3 3 3 3

32 2

3 3 3 3 3 33

23 3 3 3

1 0 0 0 0 0 0 0 00 0 1 0 0 0 0 0 00 1 0 0 0 0 0 0 01 0 0 0 0 00 0 1 0 0 2 0 0 30 1 0 0 0 0 0

1

0 0 1 0 2 0 2 3

0 1 0 2 0 3

x

y

x

y

x

y

w

y y ywy y

y y

x y x x y y x x y x y ywx y x y x y

x y x y

θθ

θθ

θθ

− =

− − − + − + − − − − ( )

0

1

2

3

4

5

6

7

2 83 3 32 0x y

ααααααααα

+

(6.2.5)

Or,

[ ] 1idα −= Φ (6.2.6)

Further, the curvature of the plate element can be written as

𝜒𝑥𝜒𝑦𝜒𝑥𝑦

=

⎩⎪⎨

⎪⎧−

𝜕2𝑤𝜕𝑥2

− 𝜕2𝑤𝜕𝑦2

2𝜕2𝑤𝜕𝑥𝜕𝑦 ⎭

⎪⎬

⎪⎫

(6.2.7)

Again, from eq. (6.2.1) the following equations can be obtained.

13

( )

2

3 6 72

2

5 7 82

2

4 7

2 6 2

2 2 6

2

w x yxw x y

yw x y

x y

α α α

α α α

α α

∂= + +

∂∂

= + +∂

∂= + +

∂ ∂

(6.2.8)

The above equation is expressed in matrix form as

( )

0

1

2

3

4

5

6

7

8

0 0 0 2 0 0 6 2 00 0 0 0 0 2 0 2 60 0 0 0 2 0 0 4 0

x

y

xy

x yx y

x y

ααα

χ αχ αχ α

ααα

− − − = − − −

+

(6.2.9)

Or, [ ] Bχ α=

Thus,

[ ][ ] 1B dχ −= Φ (6.2.10)

Further for isotropic material,

( )3

2

1 01 0

12 1 10 02

x x

y y

xy xy

MEtM

M

υ χυ χ

υυ χ

= − −

(6.2.11)

Or

[ ] [ ][ ][ ] 1M D D B dχ −= = Φ (6.2.12)

Now the strain energy stored due to bending is

𝑈 = 12 ∫ ∫ [𝜒]𝑇𝑀𝑑𝑥𝑑𝑦𝑏

0𝑎0 = 1

2 ∫ ∫ 𝑑𝑇[𝜙−1]𝑇[𝐵]𝑇[𝐷][𝐵][𝜙−1]𝑑𝑑𝑥𝑑𝑦𝑏0

𝑎0 (6.2.13)

Hence the force vector is written as

𝐹 = 𝜕𝑈𝜕𝑑

= [𝜙−1]𝑇 ∫ ∫ [𝐵]𝑇[𝐷][𝐵]𝑑𝑥𝑑𝑦 [𝜙−1]𝑑𝑏0

𝑎0 = [𝑘]𝑑 (6.2.14)

Thus, [k] is the stiffness matrix of the plate element and is given by

14

[𝑘] = [𝜙−1]𝑇 ∫ ∫ [𝐵]𝑇[𝐷][𝐵]𝑑𝑥𝑑𝑦 [𝜙−1]𝑏0

𝑎0 (6.2.15)

For a triangular plate element with orientation as shown in Fig. 6.2.2, the stiffness matrix defined in local coordinate system [k] can be transformed into global coordinate system.

[𝐾] = [𝑇]𝑇[𝑘 ] [𝑇] (6.2.16) Where, [K] is the elemental stiffness matrix in global coordinate system and [T] is the transformation matrix given by

[ ]

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

x x

y y

x x

y y

x x

y y

l ml m

l mTl m

l ml m

=

(6.2.17)

Here, (lx , mx) and (ly, my) are the direction cosines for the lines OX and OY respectively as shown in Fig.

6.2.2.

Fig. 6.2.2 Local and global coordinate system 6.2.2 Rectangular Plate Bending Element A rectangular plate bending element is shown in Fig. 6.2.3. It has four corner nodes with three degrees of freedom 𝑤,𝜃𝑥 ,𝜃𝑦 at each node. Hence, a polynomial with 12 independent terms for defining w(x,y) is necessary.

15

Fig 6.2.3 Rectangular plate bending element Considering Pascal triangle, and in order to maintain geometric isotropy the following displacement function is chosen for finite element formulation.

2 2 3 20 1 2 3 4 5 6 7

2 3 3 38 9 10 11

w(x, y) x y x xy y x x y

xy y x y xy

(6.2.18)

Hence Corresponding values for 𝜃𝑥,𝜃𝑦are,

2 2 3 2x 2 4 5 7 8 9 10 11

w x 2 y x 2 xy 3 y x 3 xyy

(6.2.19)

2 2 2 3y 1 3 4 6 7 8 10 11

w 2 x y 3 x 2 xy y 3 x y yx

(6.2.20)

The above can be expressed in matrix form

02 2 3 2 2 3 3 3

12 2 3 2x

2 2 2 3y

11

w 1 x y x xy y x x y xy y x y xy0 0 1 0 x 2y 0 x 2xy 3y x 3xy0 1 0 2x y 0 3x 2xy y 0 3x y y

(6.2.21) In a similar procedure to three node plate bending element the values of α can be found from the following relatios.

16

1

x1

y12 3

22

x22 3

y2

3

x3

y3

4

x4

y4

w 1 0 0 0 0 0 0 0 0 0 0 00 0 1 0 0 0 0 0 0 0 0 00 1 0 0 0 0 0 0 0 0 0 0

w 1 0 b 0 0 b 0 0 0 b 0 00 0 1 0 0 2b 0 0 0 3b 0 00 1 0 0 b 0 0 0 b 0 0 b

w 1

w

0

1

2

3

4

52 2 3 2 2 3 3 3

62 2 3 3

72 2 2 3

82 3

92 3

102

11

a b a ab b a a b ab b a b ab0 0 1 0 a 2b 0 a 2ab 3b a 3ab0 1 0 2a b 0 3a 2ab b 0 3a b b1 a 0 a 0 0 a 0 0 0 0 00 0 1 0 a 0 0 a 0 0 a 00 1 0 2a 0 0 3a 0 0 0 0 0

(6.2.22) Thus,

[ ] 1 dα −= Φ (6.2.23)

Further,

𝜒𝑥𝜒𝑦𝜒𝑥𝑦

=

⎩⎪⎨

⎪⎧−

𝜕2𝑤𝜕𝑥2

− 𝜕2𝑤𝜕𝑦2

2𝜕2𝑤𝜕𝑥𝜕𝑦 ⎭

⎪⎬

⎪⎫

(6.2.24)

Where,

2

3 6 7 102

2

5 8 9 112

22 2

4 7 8 10 11

w 2 6 x 2 y 6 xyxw 2 2 x 6 y 6 xy

yw 2 x 2 y 3 x 3 y

x y

(6.2.25)

Thus, putting values of eq. (6.2.25) in eq. (6.2.24), the following relation is obtained.

0x

1y

22xy

11

2y 6xy0 0 0 0 0 6x 0 0 026y 6xy0 0 0 0 0 0 0 02 2x

4y 6y0 0 0 0 0 0 0 6x2 4x

(6.2.26)

Or,

-1B d (6.2.27)

17

Further in a similar method to triangular plate bending element we can estimate the stiffness matrix for rectangular plate bending element as,

( )

3

2

1 01 0

12 1 10 02

x x

y y

xy xy

MEtM

M

υ χυ χ

υυ χ

= − −

(6.2.28)

Or

[ ] [ ][ ][ ] 1M D D B dχ −= = Φ (6.2.29)

The bending strain energy stored is

𝑈 = 12 ∫ ∫ [𝜒]𝑇𝑀𝑑𝑥𝑑𝑦𝑏

0𝑎0 = 1

2 ∫ ∫ 𝑑𝑇[𝜙−1]𝑇[𝐵]𝑇[𝐷][𝐵][𝜙−1]𝑑𝑑𝑥𝑑𝑦𝑏0

𝑎0 (6.2.30)

Hence, the force vector will become

𝐹 = 𝜕𝑈𝜕𝑑

= [𝜙−1]𝑇 ∫ ∫ [𝐵]𝑇[𝐷][𝐵]𝑑𝑥𝑑𝑦 [𝜙−1]𝑑𝑏0

𝑎0 = [𝑘]𝑑 (6.2.31)

Where, [k] is the stiffness matrix given by

[𝑘] = [𝜙−1]𝑇 ∫ ∫ [𝐵]𝑇[𝐷][𝐵]𝑑𝑥𝑑𝑦 [𝜙−1]𝑏0

𝑎0 (6.2.30)

The stiffness matrix can be evaluated from the above expression. However, the stiffness matrix also can be formulated in terms of natural coordinate system using interpolation functions. In such case, the numerical integration needs to be carried out using Gauss Quadrature rule. Thus, after finding nodal displacement, the stresses will be obtained at the Gauss points which need to extrapolate to their corresponding nodes of the elements. By the use of stress smoothening technique, the various nodal stresses in the plate structure can be determined.

18

Lecture 3: Finite Element Analysis of Thick Plate 6.3.1 Introduction Finite element formulation of the thick plate will be similar to that of thin plate. The difference will be the additional inclusion of energy due to shear deformation. Therefore, the moment curvature relation derived in first lecture of this module for thick plate theory will be the basis of finite element formulation. The relation is rewritten in the below for easy reference to follow the finite element implementation.

⎩⎪⎨

⎪⎧𝑀𝑥𝑀𝑦𝑀𝑥𝑦𝑄𝑥𝑄𝑦 ⎭

⎪⎬

⎪⎫

=

⎣⎢⎢⎢⎡ 𝐸𝑡3

12(1−𝜇2) 1 𝜇 0𝜇 1 00 0 1−𝜇

2

0 00 00 0

0 0 00 0 0

𝐸𝑡2(1+𝜇)

𝛼 00 𝛼⎦

⎥⎥⎥⎤

⎩⎪⎨

⎪⎧

χ𝑥χ𝑦χ𝑥𝑦𝛾𝑥𝛾𝑦 ⎭⎪⎬

⎪⎫

(6.3.1)

Or,

[ ] [ ][ ] [ ]

00 s

M DQ D

χγ

=

(6.3.2)

The above relation is comparable to stress-strain relations.

[ ] P PPCσ ε= (6.3.3)

Where,

𝜀𝑃 =

⎩⎪⎨

⎪⎧

χ𝑥χ𝑦χ𝑥𝑦𝛾𝑥𝛾𝑦 ⎭⎪⎬

⎪⎫

= [𝐵]𝑑𝑖 (6.3.4)

Where [B] is the strain displacement matrix and di is the nodal displacement vector. Thus, combining eqs. (6.3.3) and (6.3.4), the following expression is obtained. [ ] [ ] iP P

C B dσ = (6.3.5)

6.3.2 Strain Displacement Relation Let consider a four node isoparametric element for the thick plate bending analysis purpose. The variation of displacement w and rotations, θx and θy within the element are expressed in the form of nodal values.

4

14

14

1

i ii

x i xii

y i yii

w N w

N

N

θ θ

θ θ

=

=

=

=

=

=

∑

∑

∑

(6.3.6)

19

Where, the shape function for the four node element is expressed as,

( )( )1 1 14i i iN ξ ξ ηη= + + (6.3.7)

Here, i iandξ η are the local coordinates andξ η of the ith node. Using eq. (6.3.6), eq. (6.3.4) can be rewritten as

4

14

1

4 4

1 1

4 4

1 14 4

1 1

ix yi

i

iy xi

i

i ixy yi xi

i i

ix i yi i

i i

iy i xi i

i i

NxNy

N Ny x

Nw NxNw Ny

χ θ

χ θ

χ θ θ

γ θ

γ θ

=

=

= =

= =

= =

∂=

∂∂

= −∂

∂ ∂= −

∂ ∂

∂= +

∂∂

= +∂

∑

∑

∑ ∑

∑ ∑

∑ ∑

(6.3.8)

The above can be expressed in matrix form as follows:

0 0

0 0

0

0

0

i

ix

yi i

xy iP

xi

iy

ii

Nx

Ny

N Nd

x yN Nx

N Ny

χχχεγγ

∂ ∂

∂ − ∂ ∂ ∂ −= = ∂ ∂ ∂ ∂

∂ − ∂

(6.3.9)

Here, for a four node quadrilateral element, the nodal displacement vector di will become

𝑑𝑖 = 𝑤1,𝜃𝑥1,𝜃𝑦1,𝑤2,𝜃𝑥2,𝜃𝑦2,𝑤3,𝜃𝑥3, 𝜃𝑦3,𝑤4,𝜃𝑥4,𝜃𝑦4𝑇 (6.3.10)

Thus, the strain-displacement relationship matrix will be

20

[ ]

31 2 4

31 2 4

3 31 1 2 2 4 4

1 2 431 2 43

1 2 431 2 43

0 00 0 0 0 0 0

0 00 0 0 0 0 0

0 0 00

0 0 00

0 0 00

NN N Nxx x x

NN N Nyy y y

N NN N N N N NB

x y x y x yx yN N NNN N NNx x xx

N N NNN N NNy y yy

∂ ∂ ∂ ∂ ∂∂ ∂ ∂

∂∂ ∂ ∂ −− − − ∂∂ ∂ ∂

∂ ∂∂ ∂ ∂ ∂ ∂ ∂ − − −−= ∂ ∂ ∂ ∂ ∂ ∂∂ ∂∂ ∂ ∂∂ ∂ ∂ ∂∂∂ ∂ ∂∂ − − −− ∂ ∂ ∂∂

(6.3.11) Or,

[ ] [ ]( ) [ ]( ) [ ]( ) [ ]( )1 2 3 45 3 5 3 5 3 5 3B B B B B

× × × × =

(6.3.12)

Now, eq. (6.3.5) can be expressed using above relation as

[ ] [ ] [ ] [ ] 4

1i iP P P

iC B d C B dσ

=

= = ∑ (6.3.13)

Or,

[ ] [ ] [ ]( ) [ ]( ) [ ]( ) [ ]( )1 2 3 45 3 5 3 5 3 5 3P P P PPC B C B C B C B C B

× × × × =

(6.3.14)

Considering the ith sub-matrix of the above equation,

[ ] ( )

2 2

22

22

1 10

110

12 1 220

660

66 0

i i

ii

iii

i

i

i

i

Nt Nty x

NtNtyx

Et NtNtCByx

NNx

Ny N

µµ µ

υµµ

µ

αα

αα

∂− ∂ − ∂ − ∂ ∂∂− − ∂− ∂

∂∂− = + ∂∂ ∂

∂ ∂ ∂ −

(6.3.15)

The bending and shear terms form above equation are separated and written as

21

2 2i i

22ii

i i22ii i

t N t N0 0 01 y 1 x00 0 0t Nt N0 0 01 y1 xEt 0CB 6 N12 1 t Nt N N0x

2 y2 x0 N0 0 00 0 0

i

iNy 0

(6.3.16) The above expression can be written in compact form as

i i ib sCB = CB + CB (6.3.17)

Here, the contributions of bending and shear terms to stress displacement matrix is denoted as [CBi]b and [CBi]s respectively. Generally, the contribution due to bending, [CBi]b in eq. (6.3.17) is evaluated considering 2×2 Gauss points where as the shear contribution [CBi]s is evaluated considering 1×1 Gauss point. 6.3.3 Element Stiffness Matrix The expression for element stiffness matrix is

T

pAk B C B dx dy (6.3.18)

In natural coordinate system, the stiffness matrix is expressed as

T

pAk B C B J d d (6.3.19)

Using value of [C]P form eq. 6.3.1 and [Bi] from 6.3.11, the product of T

pB C B is evaluated as

22

i i

T i iip

i ii

i 1,2,3,4

3

2

N N0 0 0x y

N Nk B C B 0 0 Ny x

N N0 N 0x y

1 0 0 0Et 1 0 0 0

l 11 0 00 0

21 0Et0 0 00 12 10 0 0

i

i

i i

ii

ii

i 1,2,3,4

N0 0x

N0 0y

N N0x y

N 0 NxN N 0y

Or in short, (6.2.20)

11 12 13 14

21 22 23 24T

p33 32 33 34

41 42 43 44

k k k k

k k k kk B C B

k k k k

k k k k

(6.2.21)

Where,

i i i i i ij j

2 2j ji i

iij i 2 2

ji ii j

N N N N N N6 6 N 6 Nx x y y y y

N Nt N t N1 y y 1 y xEt Nk 6 N

12 1 y Nt N t N6 N N2 x x 2

j

2 2j ji i

ii 2 2

j ji i

Nx y

N Nt N t N1 x y 1 x xN6 N

x N Nt N t N2 y x 2 y y

23

(6.3.22) By separating the bending and shear terms from above equation,

2 2 2 2

j j j ji i i iij

2 2j ji i

0 0 0

N N N NEt t N t N t N t Nk 01 21 1 y y 2 x x 1 y x 2 x y

N Nt N t N01 x y 2 y x

2 2j ji i

i i i i i ij j

ii i j

ij i j

N Nt N t N1 x x 2 y y

N N N N N NN Nx x y y y x

N6 N N N 0y

N N 0 N Nx

(6.3.23)

Thus, the matrix k can now be written as the sum of bending and shear contributions

b sk k k (6.2.24)

Or,

11 12 13 14 11 12 13 14b b b b s s s s

21 22 23 24 21 22b b b b s

33 32 33 34b b b b

41 42 43 44b b b b

k k k k k k k k

k k k k k kk

k k k k

k k k k

23 24s s s

33 32 33 34s s s s

41 42 43 44s s s s

k k

k k k k

k k k k

(6.2.25)

The stiffness matrix k can be evaluated from the following expression by substituting

T

pk for B C B in eq. (6.3.19) and is given as

1 1

1 1k k J d d

(6.2.26)

Here, J is the determinate of the Jacobian matrix. The Gauss Quadrature integration rule is used to

compute the stiffness matrix [k]. 6.3.4 Nodal Load Vector Considering a uniformly distributed load q on the plate, the equivalent nodal load vector can be calculated for finite element analysis from the flowing expression.

24

x 1 1

i x i iA -1 -1

x

F q qQ = M N 0 d A N 0 J dξ dη

M 0 0

(6.3.27)

Using Gauss Quadrature integration rule the above expression can be evaluated as,

in n

i ji=1 i=1

i=1,2,3,4

NQ = q w w J 0

0

(6.3.28)

The nodal load vectors from each element are assembled to find the global load vector at all the nodes.

25

Lecture 4: Finite Element Analysis of Skew Plate 6.4.1 Introduction Skew plates often find its application in civil, aerospace, naval, mechanical engineering structures. Particularly in civil engineering fields they are mostly used in construction of bridges for dealing complex alignment requirements. Analytical solutions are available for few simple problems. However, several alternatives are also available for analyzing such complex problems by finite element methods. Commonly used three discretization methods for skew plates are shown in Fig 6.4.1.

(a) Discretization using rectangular plate elements

(b) Discretization using combination of rectangular and triangular plate

elements

(c) Discretization using skew plate element

Fig 6.4.1 Discretization of a skew plate

26

If the skew plate is discretized using only rectangular plate elements, the area of continuum excluded from the finite element model may be adequate to provide incorrect results. Another method is to use combination of rectangular and triangular elements. However, such analysis will be complex and it may not provide best solution in terms of accuracy as, different order of polynomials is used to represent the field variables for different types of elements. Another alternative exist using skew element in place of rectangular element. 6.4.2 Finite Element Analysis of Skew Plate Let consider a skew plate of dimension “2a” and “2b” as shown in Fig. 6.4.2. Let the skew angle of the element be “ϕ”. It is possible for the parallelogram shown in Fig. 6.4.3 to map the coordinate from orthogonal global coordinate system to a skew local coordinate system. If the local coordinates are represented in the form of ξ, η, then the relationship can is represented as,

x cos , y sin (6.4.1) Hence,

ycosec , x ycot (6.4.2)

Fig. 6.4.2 Skew plate in global coordinate system

27

Fig. 6.4.3 Point “P” in global and local coordinate system

It is important to note that the terms (ξ, η) in the above equations represent the absolute coordinate of the point “P” in skew coordinate system, not the natural coordinates. Since the above element has four corner nodes and each node have three degrees of freedom present, a polynomial with minimum 12 independent terms are necessary for defining the displacement function w(x,y). Considering Pascal triangle, and in order to maintain geometric isotropy, the displacement function may be considered as follows:

2 2 3 20 1 2 3 4 5 6 7

2 3 3 38 9 10 11

w

+

(6.4.3)

Hence corresponding values for 𝜃𝑥,𝜃𝑦are,

2 2 3 22 4 5 7 8 9 10 11

w 2 2 3 3

(6.4.4)

2 2 2 31 3 4 6 7 8 10 11

w 2 3 2 3

(6.4.5)

Or, in matrix form,

02 2 3 2 2 3 3 3

12 2 3 2

2 2 2 3

11

w 10 0 1 0 2 0 2 3 30 1 0 2 0 3 2 0 3

(6.4.6) Or,

28

[ ] d α= Φ (6.4.7)

The value of [α] can be determined using value of 𝑤,𝜃𝜉 ,𝜃𝜂 at four nodes as

1

x1

y12 3

22

x22 3

y2

3

x3

y3

4

x4

y4

w 1 0 0 0 0 0 0 0 0 0 0 00 0 1 0 0 0 0 0 0 0 0 00 1 0 0 0 0 0 0 0 0 0 0

w 1 0 b 0 0 b 0 0 0 b 0 00 0 1 0 0 2b 0 0 0 3b 0 00 1 0 0 b 0 0 0 b 0 0 b

w 1

w

0

1

2

3

4

52 2 3 2 2 3 3 3

62 2 3 3

72 2 2 3

82 3

92 3

102

11

a b a ab b a a b ab b a b ab0 0 1 0 a 2b 0 a 2ab 3b a 3ab0 1 0 2a b 0 3a 2ab b 0 3a b b1 a 0 a 0 0 a 0 0 0 0 00 0 1 0 a 0 0 a 0 0 a 00 1 0 2a 0 0 3a 0 0 0 0 0

(6.4.8) Or,

[ ] 1idα −= Φ (6.4.9)

Further, considering eq. (6.4.2)

0, cosecx y

1, cotx y

(6.4.10)

Again,

𝜒𝑥𝜒𝑦𝜒𝑥𝑦

=

⎣⎢⎢⎢⎡−

𝜕2𝑤𝜕𝑥2

− 𝜕2𝑤𝜕𝑦2

2𝜕2𝑤𝜕𝑥𝜕𝑦 ⎦

⎥⎥⎥⎤

(6.4.11)

The values in right hand side of the eq. (6.4.11) can be calculated by using chain rule as, w w w w.x x x

(6.4.12)

Therefore, 2 2

2 2

w wx

(6.4.13)

29

Similarly,

w w w w wcos cosecy y y

Thus, further derivation provides 2 2 2 2

2 22 2 2

w w w wcosec cot 2cos cosecy

(6.4.14)

And, 2 2 2

22

w w w wcot cosecx y x y

(6.4.15)

Hence eq. (6.4.11) is converted to 22

22

2 22 2

2 22

2 2

wwx 1 0 0w wcot cosec -cot cosec

y-2cot 0 cosec

2 w 2 wx y

(6.4.16)

Or,

x,y ,H (6.4.17)

Further, by partial differentiation of eq. (6.4.3), 2

3 6 7 102

2

5 8 9 112

22 2

7 8 10 11

w 2 6 2 6

w 2 2 6 6

w2 4 4 6 6

(6.4.18)

Or, in matrix form,

30

2

2

02

22 2

112

w

0 0 0 2 0 0 6 2 0 0 6 0w 0 0 0 0 0 2 0 0 2 6 0 6

0 0 0 0 0 0 0 4 4 0 6 62 w

(6.4.19) Or,

1, iB B d

Or,

1x,y iH B d (6.4.20)

Again,

x,y x,yM D (6.4.21)

Where [D] for plane stress condition is

3

2

1 0EhD 1 0

l2 110 0

2

(6.4.22)

Using eq. (6.4.20) in eq. (6.4.21),

1x,yM D H B A d (6.4.23)

The expression for bending strain energy stored,

a b

T

x,y x,y0 0

1U M dxdy2

(6.4.24)

Hence force vector,

a bT TT1 1

0 0

a bT TT1 1

0 0

U B H D H B d dxdyd

B H D H B J d d d

(6.4.25)

Where,

31

x y1 0x, y

J sinx y cos sin,

(6.4.26)

From expression in eq. (6.4.25)

UF k dd

(6.4.27)

Hence,

a b

T T T1 1

0 0

k sin B , H D H B , d d

(6.4.28) Thus, the element stiffness matrix of a skew element for plate bending analysis can be evaluated from the above expression using Gauss Quadrature numerical integration.

32

Lecture 5: Introduction to Finite Strip Method 6.5.1 Introduction The finite strip method (FSM) was first developed by Y. K. Cheung in 1968. This is an efficient tool for analyzing structures with regular geometric platform and simple boundary conditions. If the structure is regular, the whole structure can be idealized as an assembly of 2D strips or 3D prisms. Thus the geometry of the structure needs to be constant along one or two coordinate directions so that the width of the strips or the cross-section of the prisms does not change. Therefore, the finite strip method can reduce three and two-dimensional problems to two and one-dimensional problems respectively. The major advantages of this method are (i) reduction of computation time, (ii) small amount of input (iii) easy to develop the computer code etc. However, this method will not be suitable for irregular geometry, material properties and boundary conditions. 6.5.2 Finite Strip Method To understand the finite strip method, let consider a rectangular plate with x and y axes in the plane of the plate and axis z in the thickness direction as shown in Fig. 6.5.1. The corresponding displacement components of the plate are denoted as u, v and w.

Fig. 6.5.1 Finite strip in a plate The strips are assumed to be connected to each other along a discreet number of nodal lines that coincide with the longitudinal boundaries of the strip. The general form of the displacement function in two dimensions for a typical strip is given by 𝑤 = 𝑤(𝑥,𝑦) = ∑ 𝑓𝑚(𝑦)𝑋𝑚(𝑥)𝑛

𝑚=1 (6.5.1) Here, the functions ƒm(y) are polynomials and the functions Xm(x) are trigonometric terms that satisfy the end conditions in the x direction. The functions Xm(x) can be taken as basic functions (mode shapes) of the beam vibration equation.

33

𝑑4𝑋𝑑𝑦4

− 𝜇4

𝐿4𝑌 = 0 (6.5.2)

Here L is the length of beam strip and µ is a parameter related to material, frequency and geometric properties. The general solution of the above equation will become

𝑋𝑚(𝑥) = 𝐶1 sin 𝑛𝜋𝑥𝐿 + 𝐶2 cos 𝑛𝜋𝑥

𝐿 + 𝐶3 sinh 𝑛𝜋𝑥

𝐿+ 𝐶4 cosh 𝑛𝜋𝑥

𝐿 (6.5.3)

Four conditions at the boundaries are necessary to determine the coefficients C1 to C4 in the above expression. 6.5.2.1 Boundary conditions According to different end conditions eq. (6.5.3) can be solved. Solution of the above equation is evaluated for few boundary conditions in the below. (a) Both end simply supported

For simply supported end, following conditions will arise: (i) At one end (say at x =0) displacement and moment will be zero: x 0 = x" 0 = 0

(ii) At other end (at x =L) displacement and moment will be zero: x L = x" L = 0

Thus, considering above boundary conditions, eq. (6.5.3) yields to the following mode shape function:

t hmWhere, ,2 .......upto term

mm

mn=1

nπx μ πxX x = sin = sin nl l

(6.5.4)

Since the functions Xm are mode shapes, they are orthogonal and therefore, they satisfy the following relations:

∫ 𝑋𝑚𝐿0 (𝑥)𝑋𝑛(𝑥)𝑑𝑥 = 0 𝑓𝑜𝑟 𝑚 ≠ 𝑛 (6.5.5)

And

∫ 𝑋"(𝑥)𝑚𝐿0 𝑋"(𝑥)𝑛𝑑𝑥 = 0 𝑓𝑜𝑟 𝑚 ≠ 𝑛 (6.5.6)

The orthogonal properties of Xm(x) result in structural matrices with narrow bandwidths and thus minimizing computational time and storage. Using relation in eq. (6.5.4), eq. (6.5.1) can now be written as

𝑤 = 𝑤(𝑥,𝑦) = ∑ 𝑓𝑚(𝑦).𝑚𝑛=1 sin 𝑛𝜋𝑥

𝐿 (6.5.7)

(b) Both end fixed supported In case of fixed supported end at both the side, the following boundary conditions will be adopted:

(i) At one end (say at x =0) displacement and slope will be zero: x 0 = x' 0 = 0

(ii) At other end (at x =L) displacement and slope will be zero: x L = x' L = 0

For the above boundary conditions, eq. (6.5.3) yields to the following:

m m m mm m

μ x μ y μ y μ yX x = sin - sin h -α cos - cos hL L L L

(6.5.8)

34

Where and m mm m

m m

2m+1 sinμ - sinhμμ = 4.73,7.8532,10.996........ π α =2 cosμ - coshμ

(c) One end simply supported, other end fixed (i) At simply supported end (say at x =0) displacement and moment will be zero: x 0 = x" 0 = 0

(ii) At fixed end (say at x =L) displacement and slope will be zero: x L = x' L = 0

Thus, the solution of eq. (6.5.3) will become

m mm m

μ x μ xX x = sin -α sin hl l

(6.5.9)

4m 1where, 3.9266,7.0685,10.2102....... and4

mm m

m

sinμμ α =sinhμ

(d) Both end free If both the end of the strip element is free, the following boundary conditions will be assumed:

(i) At one end (say at x =0) moment and shear will become zero: x" 0 = x"' 0 = 0

(ii) At other end (at x =L) moment and shear will become zero: x" L = x'" L = 0

Thus, for the above end conditions, eq. (6.5.3) yields to the following:

1 1 2 2

m m m mm m

2xX = 1, μ = 0 & X = 1- ,μ = 1L

μ x μ x μ x μ xX x = sin + sin h -α cos + coshl l l l

(6.5.10)

Where, and form mm m

m m

sinμ - sin hμ 2m - 3α = μ = 4.73, 7.8532, 10.996....... π, m= 3,4 ,- - -cosμ - coshμ 2

6.5.3 Finite Element Formulation In this section, finite element solution for a finite strip will be evaluated considering simply supported conditions at both the end. As a result, the functions ƒm(Y) in eq. (6.5.7) can be expressed for the bending problem as 𝑓𝑚(𝑌) = 𝑤(𝑦) = ∝0+∝1 𝑦 +∝2 𝑦2 +∝3 𝑦3 (6.5.11) Applying boundary conditions of the strip plate of width b, the following relations will be obtained.

𝑤0𝜃𝑥0𝑤0𝜃𝑥0

=

1 0 0 00 1 0 010

𝑏1

𝑏2 𝑏32𝑏 3𝑏2

𝛼0𝛼1𝛼2𝛼3

(6.5.12)

Thus, the nodal displacement can be written in short as 𝑑 = [𝐴]𝛼 (6.5.13) Thus, the unknown coefficient α are obtained from the following relations.

𝛼 = [𝐴]−1𝑑 (6.5.14)

35

The formulation of the finite strip method is similar to that of the finite element method. For example, for a strip subjected to bending, the moment curvature relation will become

𝑀𝑥𝑀𝑦𝑀𝑥𝑦

= 𝐷𝑥 𝐷1 0𝐷1 𝐷𝑦 00 0 2𝐷𝑥𝑦

⎩⎪⎨

⎪⎧−

𝜕2𝑤𝜕𝑥2

− 𝜕2𝑤𝜕𝑦2

2 𝜕2𝑤𝜕𝑥𝜕𝑦⎭

⎪⎬

⎪⎫

(6.5.15)

Where Mx , My and Mxy are moments per unit length and [D] is the elasticity matrix. From eq. (6.5.7) the following expressions are evaluated to incorporate in the above equation.

−𝜕2𝑤𝜕𝑥2

= ∑ 𝑤(𝑦) 𝑛𝜋𝐿2𝑠𝑖𝑛 𝑛𝜋𝑥

𝐿𝑚𝑛=1

−𝜕2𝑤𝜕𝑦2

= −∑ −(2𝛼2 + 6𝛼3𝑦)𝑠𝑖𝑛 𝑛𝜋𝑥𝐿

𝑚𝑛=1 (6.5.16)

2 𝜕2𝑤𝜕𝑥𝜕𝑦

= ∑ 2(𝛼1 + 2𝛼2𝑦 + 3𝛼3𝑦2) 𝑛𝜋𝐿 𝑐𝑜𝑠 𝑛𝜋𝑥

𝐿𝑚𝑛=1

The above expression is written in matrix form in the below

⎩⎪⎨

⎪⎧−

𝜕2𝑤𝜕𝑥2

− 𝜕2𝑤𝜕𝑦2

2 𝜕2𝑤𝜕𝑥𝜕𝑦⎭

⎪⎬

⎪⎫

= ∑𝑤(𝑦)𝑁2𝑠𝑖𝑛𝑁𝑥−∑−(2𝛼2 + 6𝛼3𝑦)𝑠𝑖𝑛𝑁𝑥∑2(𝛼1 + 2𝛼2𝑦 + 3𝛼3𝑦2)𝑁𝑐𝑜𝑠𝑁𝑥

(6.5.17)

Here,𝑛𝜋𝐿

is denoted as N. Rearranging the above expression, one can find the following.

⎩⎪⎨

⎪⎧−

𝜕2𝑤𝜕𝑥2

− 𝜕2𝑤𝜕𝑦2

2 𝜕2𝑤𝜕𝑥𝜕𝑦⎭

⎪⎬

⎪⎫

= sin𝑁𝑥 0 0

0 sin𝑁𝑥 00 0 cos𝑁𝑥

𝑤(𝑦)𝑁2

−(2𝛼2 + 6𝛼3𝑦)2(𝛼1 + 2𝛼2𝑦 + 3𝛼3𝑦2)𝑁

=sin𝑁𝑥 0 0

0 sin𝑁𝑥 00 0 cos𝑁𝑥

𝑁2 𝑁2𝑦 𝑁2𝑦2 𝑁2𝑦30 0 −2 −6𝑦0 2𝑁 4𝑁𝑦 6𝑁𝑦2

𝛼0𝛼1𝛼2𝛼3

(6.5.18)

Thus, in short, the curvature and moment equation will become 𝜒 = [𝐻(𝑥)][𝐵(𝑦)]𝛼 = [𝐻(𝑥)][𝐵(𝑦)][𝐴]−1𝑑 (6.5.19) 𝑀 = [𝐷]𝜒 = [𝐷][𝐻(𝑥)][𝐵(𝑦)][𝐴]−1𝑑 (6.5.20) Now, the strain energy for the bending element can be written similar to plate bending formulation.

𝑈 =12𝜒𝑇

𝑏

0

𝐿

0

𝑀𝑑𝑥 𝑑𝑦 =12𝛼𝑇

𝑏

0

𝐿

0

[𝐴−1]𝑇[𝐵(𝑦)]𝑇[𝐻(𝑥)]𝑇[𝐷][𝐻(𝑥)][𝐵(𝑦)][𝐴]−1𝑑𝑑𝑥 𝑑𝑦

(6.5.21) Thus the force vector can be derived as

36

𝐹 = 𝜕𝑈𝜕𝑑 = ∫ ∫ [𝐴−1]𝑇[𝐵(𝑦)]𝑇[𝐻(𝑥)]𝑇[𝐷][𝐻(𝑥)][𝐵(𝑦)][𝐴]−1 𝑑𝑥 𝑑𝑦 𝑑𝑏

0𝐿0 = [𝑘]𝑑

(6.5.22) Thus, the stiffness matrix of a strip element can be obtained from the following expression.

[𝑘] = ∫ ∫ [𝐴−1]𝑇[𝐵(𝑦)]𝑇[𝐻(𝑥)]𝑇[𝐷][𝐻(𝑥)][𝐵(𝑦)][𝐴]−1 𝑑𝑥 𝑑𝑦 𝑏0

𝐿0

= [𝐴−1]𝑇 ∫ ∫ [𝐵(𝑦)]𝑇[𝐻(𝑥)]𝑇[𝐷][𝐻(𝑥)][𝐵(𝑦)]𝑑𝑥 𝑑𝑦 [𝐴]−1𝑏0

𝐿0 (6.5.23)

The stiffness matrix [k] can be simplified by integrating the term∫ [𝐻(𝑥)]𝑇[𝐷][𝐻(𝑥)]𝑑𝑥𝐿0 as follows.

∫ [𝐻(𝑥)]𝑇[𝐷][𝐻(𝑥)]𝑑𝑥𝐿0

= ∫ sin𝑁𝑥 0 0

0 sin𝑁𝑥 00 0 cos𝑁𝑥

𝐿0

𝐷𝑥 𝐷1 0𝐷1 𝐷𝑦 00 0 2𝐷𝑥𝑦

sin𝑁𝑥 0 0

0 sin𝑁𝑥 00 0 cos𝑁𝑥

𝑑𝑥

= ∫ 𝐷𝑥sin𝑁𝑥 𝐷1sin𝑁𝑥 0𝐷1sin𝑁𝑥 𝐷𝑦sin𝑁𝑥 0

0 0 2𝐷𝑥𝑦cos𝑁𝑥

sin𝑁𝑥 0 00 sin𝑁𝑥 00 0 cos𝑁𝑥

𝑑𝑥𝐿0

= ∫ 𝐷𝑥sin2 𝑁𝑥 𝐷1sin2 𝑁𝑥 0𝐷1sin2 𝑁𝑥 𝐷𝑦sin2 𝑁𝑥 0

0 0 2𝐷𝑥𝑦cos2 𝑁𝑥 𝑑𝑥𝐿

0 (6.5.24)

Here, the terms Dx, Dy and Dxy are constant and not varied with x or y. Following integrations are carried out to simplify the above expression further.

Now ∫ 𝑠𝑖𝑛2𝑁𝑥 𝑑𝑥 = ∫ 1−𝑐𝑜𝑠2𝑁𝑥2

𝑑𝑥 =𝐿0

12𝑥 − 𝑠𝑖𝑛2𝑁𝑥

2𝑁0

𝐿=𝐿

012𝐿 − 𝑠𝑖𝑛2𝑁𝐿

2𝑁

Putting, 𝑁 = 𝑛𝜋𝐿

finally ∫ 𝑠𝑖𝑛2𝑁𝑥 𝑑𝑥𝐿0 will become 1

2𝐿 −

𝑠𝑖𝑛2𝑛𝜋𝐿 𝐿

2𝑛𝜋𝐿 = 1

2𝐿

Similarly; ∫ 𝑐𝑜𝑠2𝑁𝑥 𝑑𝑥 = ∫ 1+𝑐𝑜𝑠2𝑁𝑥2

𝑑𝑥 =𝐿0

12𝑥 + 𝑠𝑖𝑛2𝑁𝑥

2𝑁0

𝐿=𝐿

012𝐿

Thus,

∫ [𝐻(𝑥)]𝑇[𝐷][𝐻(𝑥)]𝑑𝑥 = 𝐿0

𝐿2𝐷𝑥 𝐷1 0𝐷1 𝐷𝑦 00 0 2𝐷𝑥𝑦

= 𝐿2

[𝐷] (6.5.25)

Using eq. (6.5.25), the expression for stiffness matrix [k] in eq. (6.5.23) is simplified as follows.

[𝑘] = [𝐴−1]𝑇 ∫ [𝐵(𝑦)]𝑇 𝐿2

[𝐷][𝐵(𝑦)]𝑑𝑦 [𝐴−1]𝑏0

= 𝐿2

[𝐴−1]𝑇 ∫

⎣⎢⎢⎡ 𝑁

2 0 0𝑁2𝑦 0 2𝑁𝑁2𝑦2

𝑁2𝑦3−2−6𝑦

4𝑁𝑦6𝑁𝑦2⎦

⎥⎥⎤𝐷𝑥 𝐷1 0𝐷1 𝐷𝑦 00 0 2𝐷𝑥𝑦

𝑁2 𝑁2𝑦 𝑁2𝑦2 𝑁2𝑦30 0 −2 −6𝑦0 2𝑁 4𝑁𝑦 6𝑁𝑦2

𝑏0 𝑑𝑦[𝐴]−1

37

=𝐿2

[𝐴−1]𝑇

⎣⎢⎢⎢⎡ 𝑁2𝐷𝑥 𝑁2𝐷1 0

𝑁2𝑦𝐷𝑥 𝑁2𝑦𝐷1 4𝑁𝐷𝑥𝑦𝑁2𝑦2𝐷𝑥 − 2𝐷1𝑁2𝑦3𝐷𝑥 − 6𝑦𝐷1

𝑁2𝑦2𝐷1 − 2𝐷𝑦𝑁2𝑦3𝐷1 − 6𝑦𝐷𝑦

8𝐷𝑥𝑦𝑁𝑦12𝑁𝑦2𝐷𝑥𝑦⎦

⎥⎥⎥⎤𝑏

𝑜

𝑁2 𝑁2𝑦 𝑁2𝑦2 𝑁2𝑦30 0 −2 −6𝑦0 2𝑁 4𝑁𝑦 6𝑁𝑦2

𝑑𝑦[𝐴]−1

=𝐿2

[𝐴−1]𝑇

⎣⎢⎢⎢⎢⎢⎢⎢⎡

(𝑁4𝐷𝑥) (𝑁4𝑦𝐷𝑥) (𝑁4𝑦2𝐷𝑥 − 2𝑁2𝐷1) (𝑁4𝑦3𝐷𝑥 − 6𝑁2𝑦𝐷1)

(𝑁4𝑦𝐷𝑥) 𝑁4𝑦2𝐷𝑥 + 8𝑁2𝐷𝑥𝑦 𝑁4𝑦3𝐷𝑥 − 2𝑁2𝑦𝐷1 + 16𝑁2𝑦𝐷𝑥𝑦 𝑁4𝑦4𝐷𝑥 − 6𝑁2𝑦2𝐷1 + 24𝑁2𝑦2𝐷𝑥𝑦

(𝑁4𝑦2𝐷𝑥 − 2𝑁2𝐷1)(𝑁4𝑦3𝐷𝑥 − 6𝑁2𝑦𝐷1)

𝑁4𝑦3𝐷𝑥 − 2𝑁2𝑦𝐷1 + 8𝑁2𝑦𝐷𝑥𝑦𝑁4𝑦4𝐷𝑥 − 6𝑁2𝑦2𝐷1 + 24𝑁2𝑦2𝐷𝑥𝑦

𝑁4𝑦4𝐷𝑥 − 4𝑁2𝑦2𝐷1+4𝐷𝑦 + 32𝑁2𝑦2𝐷𝑥𝑦

𝑁4𝑦5𝐷𝑥 − 6𝑁2𝑦3𝐷1−2𝑁2𝑦3𝐷1 + 12𝑦𝐷𝑦

+48𝑁2𝑦3𝐷𝑥𝑦

𝑁4𝑦5𝐷𝑥 − 2𝐷1𝑁2𝑦3 − 6𝑁2𝑦3𝐷1

−12𝑦𝐷𝑦 + 48𝑁2𝑦3𝐷𝑥𝑦

𝑁4𝑦6𝐷𝑥 − 6𝑁2𝑦4𝐷1−6𝑁2𝑦4𝐷1 + 36𝑦2𝐷𝑦

+72𝑁2𝑦4𝐷𝑥𝑦

⎦⎥⎥⎥⎥⎥⎥⎥⎤

𝑑𝑦[𝐴]−1𝑏

0

(6.5.26) Thus, by putting the assumed shape function, the stiffness matrix of a strip element can be evaluated numerically using Gaussian Quadrature or other numerical integration methods.

38

Lecture 6: Finite Element Analysis of Shell 6.6.1 Introduction A shell is a curved surface, which by virtue of their shape can withstand both membrane and bending forces. A shell structure can take higher loads if, membrane stresses are predominant, which is primarily caused due to in-plane forces (plane stress condition). However, localized bending stresses will appear near load concentrations or geometric discontinuities. The shells are analogous to cable or arch structure depending on whether the shell resists tensile or, compressive stresses respectively. Few advantages using shell elements are given below.

1. Higher load carrying capacity 2. Lesser thickness and hence lesser dead load 3. Lesser support requirement 4. Larger useful space 5. Higher aesthetic value.

The example of shell structures includes large-span roof, cooling towers, piping system, pressure vessel, aircraft fuselage, rockets, water tank, arch dams, and many more. Even in the field of biomechanics, shell elements are used for analysis of skull, Crustaceans shape, red blood cells, etc. 6.6.2 Classification of Shells Shell may be classified with several alternatives which are presented in Fig 6.6.1.

39

Fig 6.6.1 Classification of shells

Depending upon deflection in transverse direction due to transverse shear force per unit length, the shell can be classified into structurally thin or thick shell. Further, depending upon the thickness of the shell in comparison to the radii of curvature of the mid surface, the shell is referred to as geometrically thin or thick shell. Typically, if thickness to radii of curvature is less than 0.05, then the shell can be assumed as a thin shell. For most of the engineering application the thickness of shell remains within 0.001 to 0.05 and treated as thin shell. 6.6.3 Assumptions for Thin Shell Theory Thin shell theories are basically based on Love-Kirchoff assumptions as follows.

1. As the shell deforms, the normal to the un-deformed middle surface remain straight and normal to the deformed middle surface undergo no extension. i.e., all strain components in the direction of the normal to the middle surface is zero.

2. The transverse normal stress is neglected. Thus, above assumptions reduce the three dimensional problems into two dimensional. 6.6.4 Overview of Shell Finite Elements Many approaches exist for deriving shell finite elements, such as, flat shell element, curved shell element, solid shell element and degenerated shell element. These are discussed briefly bellow.

40

(a) Flat shell element The geometry of these types of elements is assumed as flat. The curved geometry of shell is obtained by assembling number of flat elements. These elements are based on combination of membrane element and bending element that enforced Kirchoff’s hypothesis. It is important to note that the coupling of membrane and bending effects due to curvature of the shell is absent in the interior of the individual elements. (b) Curved shell element Curved shell elements are symmetrical about an axis of rotation. As in case of axisymmetric plate elements, membrane forces for these elements are represented with respected to meridian direction as(𝑢,𝑁𝑧 ,𝑀𝜃) and in circumferential directions as(𝑤,𝑁𝜃,𝑀𝑧). However, the difficulties associated with these elements includes, difficulty in describing geometry and achieving inter-elemental compatibility. Also, the satisfaction of rigid body modes of behaviour is acute in curved shell elements. (c) Solid shell element Though, use of 3D solid element is another option for analysis of shell structure, dealing with too many degrees of freedom makes it uneconomic in terms of computation time. Further, due to small thickness of shell element, the strain normal to the mid surface is associated with very large stiffness coefficients and thus makes the equations ill conditioned. (d) Degenerated shell elements Here, elements are derived by degenerating a 3D solid element into a shell surface element, by deleting the intermediate nodes in the thickness direction and then by projecting the nodes on each surface to the mid surface as shown in Fig. 6.6.2.

(a) 3D solid element (b) Degenerated Shell element

41

6.6.2 Degeneration of 3D element This approach has the advantage of being independent of any particular shell theory. This approach can be used to formulate a general shell element for geometric and material nonlinear analysis. Such element has been employed very successfully when used with 9 or, in particular, 16 nodes. However, the 16-node element is quite expensive in computation. In a degenerated shell model, the numbers of unknowns present are five per node (three mid-surface displacements and two director rotations). Moderately thick shells can be analysed using such elements. However, selective and reduced integration techniques are necessary to use due to shear locking effects in case of thin shells. The assumptions for degenerated shell are similar to the Reissner-Mindlin assumptions. 6.6.5 Finite Element Formulation of a Degenerated Shell Let consider a degenerated shell element, obtained by degenerating 3D solid element. The degenerated shell element as shown in Fig 6.6.2(b) has eight nodes, for which the analysis is carried out. Let (𝜉, 𝜂) are the natural coordinates in the mid surface. And ς is the natural coordinate along thickness direction. The shape functions of a two dimensional eight node isoparametric element are:

1 5

2 6

3 7

4 8

1 1 1 1 1 14 2

1 1 1 1 1 14 2

1 1 1 1 1 14 2

1 1 1 1 1 14 2

N N

N N

N N

N N

(6.6.1)

The position of any point inside the shell element can be written in terms of nodal coordinates as

8

1

1 1,2 2

i i

i i ii

i itop bottom

x x xy N y yz z z

V V

(6.6.2)

Since, ς is assumed to be normal to the mid surface, the above expression can be rewritten in terms of a vector connecting the upper and lower points of shell as

1,2 2

i i i i

i i i i i

i i i itop bottom top bottom

x x x x xy N y y y yz z z z z

V

8

1i

Or,

42

8

31

,2

i

i i ii

i

x xy N y Vz z

V

(6.6.3)

Where,

12

i i i

i i i

i i itop bottom

x x xy y yz z z

and, 3

i i

i i i

i itop bottom

x xV y y

z z

(6.6.4)

Fig. 6.6.3 Local and global coordinates

For small thickness, the vector V3i can be represented as a unit vector tiv3i:

8

31

,2

i

i i i ii

i

x xy N y t vz z

V

(6.6.5)

Where, ti is the thickness of shell at ith node. In a similar way, the displacement at any point of the shell element can be expressed in terms of three displacements and two rotation components about two orthogonal directions normal to nodal load vector V3i as,

8

1 21

,2

iii

i i i ii i

i

u utv N v v v

w w

V b

(6.6.6)

Where, (𝛼𝑖,𝛽𝑖) are the rotations of two unit vectors v1i & v2i about two orthogonal directions normal to nodal load vector V3i.The values of v1i and v2i can be calculated in following way: The coordinate vector of the point to which a normal direction is to be constructed may be defined as ˆˆ ˆx xi yj zk (6.6.7)

43

In which, ˆˆ ˆ, ,i j k are three (orthogonal) base vectors. Then, 1iV is the cross product of i & V3i as shown below.

1 3ˆ

i iV i V & 2 3 1i i iV V V (6.6.8) and,

11

1

ii

i

Vv V & 22

2

ii

i

Vv V (6.6.9)

6.6.5.1 Jacobian matrix The Jacobian matrix for eight node shell element can be expressed as,

8 8 8* * *

1 1 1

8 8 8* * *

1 1 1

8 8 8* * *

1 1 1

i i ii i i i i i

i i i

i i ii i i i i i

i i i

i i ii i i

N N Nx tx y ty z tz

N N NJ x tx y ty z tz

Nx Ny Nz

(6.6.10)

6.6.5.2 Strain displacement matrix The relationship between strain and displacement is described by

B de (6.6.11)

Where, the displacement vector will become:

1 1 1 11 21 8 8 8 18 28Td u v w v v u v w v v (6.6.12)

And the strain components will be

uxvy

u vy xv wz yw ux z

e

(6.6.13)

Using eq. (6.6.6) in eq. (6.6.13) and then differentiating w.r.t. (𝜉, 𝜂, 𝜍) the strain displacement matrix will be obtained as

44

18 8

22

1 13

20

i i

i i i ii i i

i i

i

N Nu v w

N t v Nu v w u v w

u v w N

V b

V b

b

V V V

18

12

13

2

i

T T

i i i

ii i

i

N

t v N

N

V

V

(6.6.14)

6.6.5.3 Stress strain relation The stress strain relationship is given by

Ds e (6.6.15)

Using eq. (6.6.11) in eq. (6.6.15) one can find the following relation. D B ds (6.6.16)

Where, the stress strain relationship matrix is represented by

2

1 0 0 01 0 0 0

10 0 0 02

1 10 0 0 0

21

0 0 0 02

ED

(6.6.17)

The value of shear correction factor is considered generally as 5/6. The above constitutive matrix can be split into two parts ([Db] and [Ds] )for adoption of different numerical integration schemes for bending and shear contributions to the stiffness matrix.

0

0

b

s

DD

D

(6.6.18)

Thus,

2

1 01 0

110 0

2

bED

(6.6.19)

and

1 00 12 1s

ED

(6.6.20)

45

It may be important to note that the constitutive relation expressed in eq. (6.6.19) is same as for the case of plane stress formulation. Also, eq. (6.6.20) with a multiplication of thickness h is similar to the terms corresponds to shear force in case of plate bending problem. 6.6.5.4 Element stiffness matrix Finally, the stiffness matrix for the shell element can be computed from the expression

Tk B D B d (6.6.21)

However, it is convenient to divide the elemental stiffness matrix into two parts: (i) bending and membrane effect and (ii) transverse shear effects. This will facilitate the use of appropriate order of numerical integration of each part. Thus,

b sk k k (6.6.22)

Where, contribution due to bending and membrane effects to stiffness is denoted as [k]b and transverse shear contribution to stiffness is denoted as [k]s and expressed in the following form.

andT T

b b b b s s s sk B D B d k B D B d (6.6.23)

Numerical procedure will be used to evaluate the stiffness matrix. A 2 ×2 Gauss Quadrature can be used to evaluate the integral of [k]b and one point Gauss Quadrature may be used to integrate [k]s to avoid shear locking effect.

Content Module 7: Additional Applications of FEM (03)

Lecture 1: Finite Elements for Elastic Stability(Page: 1-8)

7.1.1 Introduction 7.1.2 Buckling of Truss Members 7.1.3 Buckling of Beam-Column Members 7.1.4 Buckling of Plate Bending Elements

Lecture 2: Finite Elements in Fluid Mechanics(Page: 9-15)

7.2.1 Governing Fluid Equations 7.2.2 Finite Element Formulation

7.2.2.1 Displacement based finite element formulation 7.2.2.2 Pressure based finite element formulation

7.2.3 Finite Element Formulation of Infinite Reservoir

Lecture 3: Dynamic Analysis(Page: 16-23)

7.3.1 Hamilton’s Principle 7.3.2 Finite Element Form in MDOF System 7.3.3 Solid Body with Distributed Mass

7.3.3.1 Mass matrix for truss element 7.3.3.2 Mass matrix for beam element 7.3.3.3 Mass matrix for plane frame element 7.3.3.4 Mass matrix for space frame element

7.3.4 Time History Analysis

1

Lecture 1: Finite Elements for Elastic Stability

7.1.1 Introduction There are two types of structural failure associated: namely (i) material failure and (ii) geometry or configuration failure. In case of material failure, the stresses exceed the permissible values which may lead to formation of cracks In configuration failure, the structure is unable to maintain its designed configuration under the external disturbance even though the stresses are in permissible range. The stability loss due to tensile forces falls in broad category of material instability. The loss of stability under compressive load is termed as structural or geometric instability which is commonly known as buckling. Thus, buckling is considered by a sudden failure of a structural element subjected to large compressive stress, where the actual compressive stress at the point of failure is less than the ultimate allowable compressive stresses. Buckling is a wide term that describes a range of mechanical behaviours. Generally, it refers to an incident where a structural element in compression deviates from a behaviour of elastic shortening within the original geometry and undergoes large deformations involving a change in member shape for a small increase in force. Various types of buckling may occur such as flexural buckling, torsional buckling, torsional flexural buckling etc. Here, the use of finite element technique for elastic stability analysis of various structural members will be briefly discussed. 7.1.2 Buckling of Truss Members For a truss member, the axial strain can be expressed in terms of its displacements

2 2

x1

x 2 x xu u v+

(7.1.1)

Here, u and v are the displacements in local X and Y directions respectively. Here, the strain-displacement relation is nonlinear. The total potential energy in the member with uniform cross sectional area and subjected to external forces can be written as

L

2x 1 1 2 2

0

EAΠ = ε dx - Pv +P v2 (7.1.2)

Where, P1 and P2 are the external forces in nodes 1 and 2, v1 and v2 are the vertical displacements in nodes 1 and 2, E and A are the modulous of elasticity and cross sectional area respectively (Fig. 7.1.1). The length of the member is considered to be L.

2

Fig.7.1.1 Truss element in local coordinate system

The above expression can be rewritten in terms of displacement variables as

2 21

x 2 x x

2L

1 1 2 20

EA u u vΠ = + dx - Pv +P v2

2 3 2 21

x x x x 4 x x

22L

1 1 2 20

EA u u v u u v= + + dx - Pv + P v2

(7.1.3)

Neglecting higher order terms and considering EAxu = P

the above expression can simplify to the

following form.

2 2v

x x

L L

1 1 2 20 0

EA u PΠ = dx+ dx - Pv +P v2 2

(7.1.4)

Considering nodal displacements at nodes 1 and 2 as u1, v1 and u2, v2, the displacement at any point inside the element can be represented in terms of its interpolation functions and nodal displacements. The shape function for a two node truss element is

[ ]

−=

Lx

LxN 1 (7.1.5)

Equation expressed in eq. (7.1.5) is expressed in terms of the nodal displacement vectors ui and vi with the help of interpolation function.

T T T T' ' ' 'L L

i i i i 1 1 2 20 0

1 PΠ = u N EA N u dx+ v N N v dx - Pv +P v2 2 (7.1.6)

Where,

andi i i iu v= N u = N' u = N v = N' vx x x x

(7.1.7)

Making potential energy (eq. 7.1.6) stationary one can find the equilibrium equation as

3

T T' ' ' 'L L

i i0 0

F = N EA N dx u + P N N dx v (7.1.8)

Or,

PA i G iF = k u k v (7.1.9)

Where, [kA] is the axial stiffness of the member and [kG] is the geometric stiffness of the member in its local coordinate system and can be derived as follows.

T' 'L

L

A 00

1- 1 -11 1 AELk = N EA N dx = AE - dx =1 -1 1L L LL

(7.1.10)

T' 'L

L

G 00

1- 1 -11 1 1Lk = N N dx = - dx =1 -1 1L L LL

(7.1.11)

However, in a generalised form to accommodate both the direction, these stiffness matrices can be written as

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

and1 0 1 0 0 0 0 0

0 0 0 0 0 1 0 1

A GAE 1k = k =L L

(7.1.12)

The generalised stiffness matrix of a plane truss member in global coordinate system can be derived using the transformation matrix as derived module 4, lecture 1. The transformation matrix can be recalled and written here as follows.

[ ]

cos 0 0cos 0 0

0 0 cos0 0 cos

sinsin

Tsin

sin

θ θθ θ

θ θθ θ

− = −

(7.1.13)

Thus, the stiffness matrices with respect to global coordinate system will become

andT TA A G GK = T k T K = T k T (7.1.14)

Here, [KA] and [KG] are the axial stiffness and geometric stiffness of the member in global coordinate system. The force-displacement relationship in global coordinate system can be written from eq. (7.1.9) as

4

P PA G A GF = K d K d K K d (7.1.15)

Where, d is the displacement vector in global coordinates. If the external force is absent in eq. (7.1.15), the value of P will be considered to be undetermined as.

[ ] [ ] 0A GK d P K d+ = (7.1.16)

The above equation can be solved as an eigenvalue problem to calculate buckling load P. 7.1.3 Buckling of Beam-Column Members Let consider a pin ended column under the action of compressive force P. The elastic and geometric stiffness matrices can be developed from 1st principles for a beam-column element which can be used in the linear elastic stability analysis of frameworks. Considering small deflection approximation to the curvature, the total potential energy is computed from the following.

𝛱 = 𝐸𝐼2 ∫ 𝑑

2𝑤𝑑𝑥2

2𝑑𝑥 + 𝑃

2𝐿0 ∫ 𝑑𝑤

𝑑𝑥2𝑑𝑥𝐿

0 = ∫ 𝐸𝐼2𝑑

2𝑤𝑑𝑥2

2

+ 𝑃2𝑑𝑤𝑑𝑥2 𝑑𝑥𝐿

0 (7.1.17)

Here, w is the transverse displacement and I is the moment of inertia of the member.

Fig. 7.1.2 Beam-column member

Considering nodal displacements at nodes 1 and 2 as w1, θ1 and w2, θ2, the displacement at any point inside the element can be represented in terms of its interpolation functions and nodal displacements.

𝑤 = [𝑁1 𝑁2 𝑁3 𝑁4]

𝑤1𝜃1𝑤2𝜃2

= [𝑁]𝑑 (7.1.18)

Thus, the above energy equation can be rewritten as

L TT " " ' T '

0

1Π= d N EI N d + d P N N d dx2

(7.1.19)

Where,

5

and2 2

2 2

d w d d wd= N d = N" d = N d = N' ddx dx dx dx

(7.1.20)

Applying the variational principle one can express

[ ] [ ] [ ]( ) "

0" ' '

l T TF N EI N P N N dx dd

∂Π = = + ∂ ∫ (7.1.21)

Thus, the stiffness matrix will be obtained as follows which have two terms.

[ ] [ ]( ) [ ] [ ]( ) [ ] [ ]"

0 0" ' '

l lT TF Gk N EI N dx P N N dx k P k = + = + ∫ ∫ (7.1.22)

The first term resembles ordinary stiffness matrix for the bending of a beam. So this matrix is called flexural stiffness matrix. The second matrix is known as geometric stiffness matrix as it only depends on the geometrical parameters. Thus, the flexural stiffness matrix [kF] and geometric stiffness matrix [kG] can be derived from the following expressions.

[ ] [ ]( ) [ ] [ ] [ ]( )"

0 0" and ' '

l lT TF Gk N EI N dx k N N dx = = ∫ ∫ (7.1.23)

The above matrices can be derived from the assumed interpolation function. For example, the interpolation functions for a two node beam element are expressed by the following equation.

[𝑁] = 1 − 3 𝑥2

𝐿2+ 2 𝑥3

𝐿3 , 𝑥 − 2 𝑥2

𝐿+ 𝑥3

𝐿2 , 3 𝑥2

𝐿2− 2 𝑥3

𝐿3 , − 𝑥2

𝐿+ 𝑥3

𝐿2 (7.1.24)

Now, the first and second order derivative of the above function will become

[𝑁 ′] = −6 𝑥𝐿2

+ 6 𝑥2

𝐿3 , 1 − 4 𝑥

𝐿+ 3 𝑥2

𝐿2 , 6 𝑥

𝐿2− 6 𝑥2

𝐿3 , −2 𝑥

𝐿+ 3 𝑥2

𝐿2 (7.1.25)

𝑁" = − 6𝐿2

+ 12 𝑥𝐿3 , − 4

𝐿+ 6 𝑥

𝐿2 , 6

𝐿2− 12 𝑥

𝐿3 , − 2

𝐿+ 6 𝑥

𝐿2 (7.1.26)

Using the above expressions in eq. (7.1.23), flexural and geometric stiffness matrices can be derived and obtained as follows.

[𝐾𝐹] = 𝐸𝐼𝐿3

12 6𝐿6𝐿 4𝐿2

−12 6𝐿−6𝐿 2𝐿2

−12 −6𝐿6𝐿 2𝐿2

12 −6𝐿−6𝐿 4𝐿2

(7.1.27)

[𝐾𝐺] = 130𝐿

36 3𝐿3𝐿 4𝐿2

−36 3𝐿−3𝐿 −𝐿2

36 −3𝐿3𝐿 −𝐿2

36 −3𝐿−3𝐿 4𝐿2

(7.1.28)

Due to external forces F, one can find the displacement vectors from the following equation.

[ ] [ ] [ ] F GF k d k d P k d= = + (7.1.29)

The above is the beam-column equation in finite element form. In the absence of transverse load, the member will become column and P will be considered to be undetermined.

[ ] [ ] 0F Gk d P k d+ = (7.1.30)

Denoting P as -λ we can reach to the familiar eigenvalue problem as given below.

6

[ ] [ ] F Gk d k dλ= (7.1.31)

Solving above equation, values of λ and associated nodal displacement vectors can be obtained. Mathematically this can be expressed as [ ] [ ]( ) 0F Gk k dλ− = (7.1.32)

Thus,

[ ] [ ] 0F Gk kλ− = (7.1.33)

For the matrix of size n, one can find n+1 degree polynomial in λ. The smallest root of the above equation will become the first approximate buckling load. From this value λ, one can find a set of ratios for the nodal displacement components. From this, the first buckling mode shape can be calculated. Higher mode approximations can also be found in a similar process. The procedure to determine the critical load by the above method is illustrated in the following example. Example 7.1.1 Consider a column with one end clamped and other end free as shown in Fig. 7.1.2.

Fig. 7.1.3 Column with one end fixed and other end free

Now, the finite element equation of the column considering single element will be

𝐸𝐼𝐿3

12 6𝐿6𝐿 4𝐿2

−12 6𝐿−6𝐿 2𝐿2

−12 −6𝐿6𝐿 2𝐿2

12 −6𝐿−6𝐿 4𝐿2

− λ1

30𝐿

36 3𝐿3𝐿 4𝐿2

−36 3𝐿−3𝐿 −𝐿2

36 −3𝐿3𝐿 −𝐿2

36 −3𝐿−3𝐿 4𝐿2

= 0

7

The boundary conditions for this member are given by

1At = 0, 0 and 0dwxdx

1d

Thus, according to the above boundary conditions, the first and second rows as well as columns of the equation are deleted and rewritten in the following form.

𝐸𝐼𝐿3 12 −6𝐿−6𝐿 4𝐿2 −

𝜆30𝐿

36 −3𝐿−3𝐿 4𝐿2 = 0

Or

4𝐸𝐼𝐿

−2𝜆𝐿15

12𝐸𝐼𝐿3

−6𝜆5𝐿 − −

6𝐸𝐼𝐿2

+𝜆

102

= 0

Solving the above expression, the critical value of λcr and thus the critical value of force Pcr will become as

𝜆𝑐𝑟 = 𝑃𝑐𝑟 = 2.486𝐸𝐼𝐿2

It is important to note that the exact value for such clamped-free column is

2 2

crP 2.46722 2e

EI EI EIL L2L

The finite element result has slight deviation from the exact result. This difference can be minimized by the increase of number of elements in the column as we know, more we subdivide the continuum, better we can obtain the result close to the exact one.

7.1.4 Buckling of Plate Bending Elements The elastic stability analysis of rectangular plates is discussed in this section. The total potential energy for plate are expressed as

𝜋 = 𝐷2 ∫ ∫ (∇2w)2 + 2(1 − 𝜇) 𝜕

2𝑤𝜕𝑥𝜕𝑦

2− 𝜕

2𝑤𝜕𝑥2

𝜕2𝑤𝜕𝑦2

𝑏−𝑏

𝑎−𝑎 𝑑𝑥𝑑𝑦

−12 ∫ ∫ 𝐹𝑥

𝜕𝑤𝜕𝑥2

+ 2𝐹𝑥𝑦𝜕𝑤𝜕𝑦

𝜕𝑤𝜕𝑥

+ 𝐹𝑦 𝜕𝑤𝜕𝑦2𝑏

−𝑏𝑎−𝑎 𝑑𝑥𝑑𝑦 (7.1.34)

Here, Fx, Fy, and Fxy, are the in-plane edge load and compressive load is considered as positive. For, finite element formulation the deflection in above expression needs to convert in terms of nodal displacements in the element. Following the derivations in beam-column member (eqs. 7.1.18 to 7.1.22), the flexural and geometric stiffness for the plate element can be derived. Thus, the above expression can be derived to the following form using interpolation functions.

𝜋 = 12

𝑑𝑇[𝑘𝐹]𝑑 − 𝐹𝑥2

𝑑𝑇[𝑘𝐺𝑥]𝑑 −𝐹𝑦2

𝑑𝑇𝑘𝐾𝐺𝑦𝑑 − 𝐹𝑥𝑦2

𝑑𝑇𝑘𝐺𝑥𝑦𝑑 (7.1.35)

Here [𝑘𝐹] is the flexural stiffness matrix. The other stiffness matrices are analogous to the geometric stiffness matrices of the plate and can be expressed as 𝑘𝐺𝑥 = ∬⌊𝑁′𝑥⌋ 𝑁′𝑥𝑑𝑥 𝑑𝑦

8

𝑘𝐺𝑦 = ∬𝑁′𝑦 𝑁′𝑦𝑑𝑥 𝑑𝑦 (7.1.36) 𝑘𝐺𝑥𝑦 = ∬⌊𝑁′𝑥⌋ 𝑁′𝑦𝑑𝑥 𝑑𝑦 Where [𝑁′𝑥] and 𝑁′𝑦 indicate partial derivative of [𝑁] with respect to x and y respectively. Thus, the equation of buckling becomes [𝑘𝐹]𝑑 − 𝐹𝑥[𝑘𝐺𝑥]𝑑 − 𝐹𝑦𝑘𝐺𝑦𝑑 − 𝐹𝑥𝑦𝑘𝐺𝑥𝑦𝑑 = 0 (7.1.37) If the in-plane loads have a constant ratio to each other at all time during their buildup, the above equation can be expressed as follows [𝑘𝐹]𝑑 = 𝑃∗𝛼[𝑘𝐺𝑥] + 𝛽𝑘𝐺𝑦 + 𝛾𝑘𝐺𝑥𝑦𝑑 (7.1.38) The term 𝑃∗ is called the load factor, and 𝛼,𝛽, and 𝛾 are constants relating the in-plane loads in the plate member. Solving the above expression, the buckling mode shapes are possible to determine.

9

Lecture 2: Finite Elements in Fluid Mechanics 7.2.1 Governing Fluid Equations The fluid mechanics topic covers a wide range of problems of interest in engineering applications. Basically fluid is a material that conforms to the shape of its container. Thus, both the liquids and gases are considered as fluid. However, the physical behaviour of liquids and gases is very different. The differences in behaviour lead to a variety of subfields in fluid mechanics. In case of constant density of liquid, the flow is generally referred to as incompressible flow. The density of gases not constant and therefore, their flow is compressible flow. The Navier–Stokes equations are the fundamental basis of almost all fluid dynamics related problems. Any single-phase fluid flow can be defined by this expression. The general form of motion of a two dimensional viscous Newtonian fluid may be expressed as

j i, j i, ji i1 p ,i vi v v v f

(7.2.1)

Here, = kinematic viscosity = mass density of fluid vi = Velocity components

if = Body forces p = fluid pressure

The suffix ,j and ,ji are the derivatives along j and j & i direction respectively. The dot represents the derivative with respect to time. Neglecting non linear convective terms, viscosity and body forces, eq. (7.2.1) can be simplified as:

ip, i v 0 (7.2.2) Now, the continuity equation of the fluid is expressed by

2k,kp c v 0 (7.2.3)

Here, c is the acoustic wave speed in fluid. In the above expression, two sets of variables, the velocity and the pressure are used to describe the behaviour of fluid. Now it is possible to combine equation (7.2.2) and (7.2.3) to obtain a single variable formulation. For the small amplitude of fluid motion, one can assume

i iv u (7.2.4)

Where iu is the displacement component of fluid. To obtain single variable formulation, eq. (7.2.4) may be substituted into eq. (7.2.3) and one can get

2k,kp c u 0 (7.2.5)

Integrating eq. (5) w. r. t. time we have

10

2k,kp c u (7.2.6)

Now differentiating the above expression w. r. t. xi following expression will be arrived: 2

i k,kip, c u (7.2.7)

Substituting the above in to eq. (7.2.2) one can have

2

i k,kiv c u 0 (7.2.8)

Thus, eq. (7.2.8) is expressed in terms of displacement variables only and known as displacement based equation. Similarly, it is possible to obtain the fluid equation in terms of pressure variable only. Differentiating eq. (7.2.3) w. r. t. Time, the following expression can be obtained.

2

k,kp c v 0 (7.2.9)

Again, differentiating eq. (7.2.2) w. r. t. xi, we have

ii i,i p, v 0 (7.2.10)

From eqs. (7.2.9) and (7.2.10), the pressure based single variable expression can be arrived as given below.

2iip c p, 0 (7.2.11)

The above expression is basically the Helmholtz wave equation for a compressible fluid having acoustic speed c.

22

1p p 0c

(7.2.12)

Thus, the general form of fluid equations of 2D linear steady state problems can be expressed by the Helmholtz equation. For incompressible fluid c becomes infinitely large. Hence for incompressible fluid, eq. (7.2.12) can be written as

2p 0 (7.2.13) For the ideal, irrotational fluid flow problems, the field variables are the streamline, ψ and potential φ functions which are governed by Laplace’s equations

2 2

2 2

2 2

2 2

0x y

0x y

(7.2.14)

11

Derivation of the above expression and many other related equations can be found in details in fluid mechanics related text books. 7.2.2 Finite Element Formulation The equation of motion of fluid can be expressed in various ways and some of those are shown in previous section. Finite element form of those expressions can be derived using various methods considering pressure, displacement, velocity, velocity potential, stream functions and their combinations. Here, displacement and pressure based formulations will be derived using finite element method. 7.2.2.1 Displacement based finite element formulation Consider the equation (7.2.8) which can be expressed only in terms of displacement variables.

2i k,kiu c u 0 (7.2.15)

Here, u is the displacement vector. Now, the weak form of the above equation will become

2i i k,kiw u c u d 0

(7.2.16)

Performing integration by parts on the second terms, one can arrive at the following expression:

2 2i i i k,k i,i k,k

2 2i i i,i k,ki i k,k

w u d w c u d w c u d 0

o r, w u d w c u d w c u d

(7.2.17)

Now from earlier relation (eq. 7.2.6) we have, 2

k,kp c u . Thus, the above equation may be

written as: 2

i i i,i k,ki i w u d w c u d w pd

(7.2.18)

In case of fluid filled rigid tank, the weighting function wi must satisfy the condition i iw n 0 on its rigid boundary. Therefore, the above eq. will become

p

2i i i,i k,k i iw u w c u d w n p

(7.2.19)

For finite element implementation of the above expression, let consider the interpolation function as N and u as the nodal displacement vector. Thus,

12

u Nu and w Nw (7.2.20) Now the divergence of the displacement vector can be expressed as:

i,iu Lu LNu Bu (7.2.21)

Where L x y

is the differential operator. Thus eq. (7.2.19) may be written as

p

T T T 2 T Tw N Nu B c Bu d w N np d

(7.2.22)

Or,

M u K u F (7.2.23)

Where,

p

T

T2

T

M N N d

K c B B d

F N n p d

(7.2.24)

Using eq. (7.2.23), the displacements in fluid domain can be determined under external forces applying proper boundary conditions. 7.2.2.2 Pressure based finite element formulation The Helmholtz equation (7.2.12) for a compressible fluid in two dimension can be used to determine the pressure distribution in the fluid domain using finite element technique.

2

1p,ii p 0c

(7.2.25)

The weak form of the above expression can be written as

i 2

1w p,ii p d 0c

(7.2.26)

Now, performing integration by parts on the first term, the following expression can be obtained.

i i,i i2

1w p,i d w p,i d w pd 0c

(7.2.27)

Thus,

i i,i ,i i ,i2

1 w p d w p d w p d c

(7.2.28)

13

Assuming interpolation function as N and p as the nodal pressure vector, the pressure (p) at any point can be written as: p Np and similarly, w Nw . The divergence of the pressure can be expressed

as: p, i Lp LNp Bp , where, Lx y

. Again,

i,i

T T Ti

w Lw LNw Bw

w p Nw Np w N Np

(7.2.29)

Thus, eq. (7.2.28) will become: T T T

T T T2

1 pw N Npd w B Bpd w N dc n

(7.2.30)

Or,

E p G p B (7.2.31)

Where,

T

2

T

T

1E N N dc

G B B d

pB N dn

(7.2.32)

Applying boundary conditions, eq. (7.2.31) can be solved to calculate the dynamic pressure developed in the fluid under applied accelerations on the domain. 7.2.3 Finite Element Formulation of Infinite Reservoir Let consider an infinite reservoir adjacent to a dam like structure. In such case, if the dam is vibrated, the hydrodynamic pressure will be developed in the reservoir which can be calculated using above method. For finite element analysis, it is necessary to truncate such infinite domain at a certain distance away from structure to have a manageable computational domain. The reservoir has four sides (Fig. 7.2.1) and as a result four types of boundary conditions need to be specified.

1 2 3 4B B B B B (7.2.33)

14

Fig. 7.2.1 Reservoir and its boundary conditions

(i) At the free surface (Γ1) Neglecting the effects of surface waves of the water, the boundary condition of the free surface may be expressed as

p x, H, t 0 (7.2.34)

Here, H is the depth of the reservoir. However, sometimes, the effect of surface waves of the water needs to be considered at the free surface. This can be approximated by assuming the actual surface to be at an elevation relative to the mean surface and the following linearised free surface condition may be adopted.

0 = yp + p

g1

∂∂

(7.2.35)

Thus, the above expression may be written in finite element form as

1 1p 1B R pn g

(7.2.36)

In which,

1

T1R N N d

(7.2.37)

(ii) At the dam-reservoir interface (Γ2) At the dam-reservoir interface, the pressure should satisfy

i tp (0, y, t) = aen

ωρ∂∂

(7.2.38)

15

where aeiωt is the horizontal component of the ground acceleration in which, ω is the circular frequency

of vibration and 1− = i , n is the outwardly directed normal to the elemental surface along the interface. In case of vertical dam-reservoir interface np ∂∂ may be written as xp ∂∂ as both will represent normal to the

element surface. For an inclined dam-reservoir interface xp ∂∂ cannot represent the normal to the element

surface. Therefore, to generalize the expressions np ∂∂ is used in eq. (7.2.38). If a is the vector of nodal accelerations of generalized coordinates, B2 may be expressed as

2 2B R a (7.2.39)

where,

2

T2 r d= N T N dR

(7.2.40)

Here, [T] is the transformation matrix for generalized accelerations of a point on the dam reservoir interface and [Nd] is the matrix of shape functions of the dam used to interpolate the generalized acceleration at any point on their interface in terms of generalized nodal accelerations of an element. (iii) At the reservoir bed interface (Γ3) At the interface between the reservoir and the elastic foundation below the reservoir, the accelerations should not be specified as rigid foundation because they depend on the interaction between the reservoir and the foundation. However, for the sake of simplicity, the reservoir bed can be assumed as rigid and following boundary condition may be adopted.

3pB x,0, t 0

(7.2.41)

(iv) At the truncation boundary (Γ4) The specification of the far boundary condition is one of the most important features in the finite element analysis of a semi-infinite or infinite reservoir. This is due to the fact that the developed hydrodynamic pressure, which affects the response of the structure, is dependent on the truncation boundary condition. The infinite fluid domain may truncated at a finite distance away from the structure for finite element analysis satisfying Sommerfeld radiation boundary condition. Application of Sommerfeld radiation condition at the truncation boundary leads to

4pB (L, y, t) = 0x

(7.2.42)

Here, L represents the distance between the structure and the truncation boundary. Thus, the hydrodynamic pressure developed on the dam-reservoir interface can be calculated under external excitation by the use of finite element technique.

16

Lecture 3: Dynamic Analysis The finite element method is a powerful device for analyzing the dynamic response of structures as in case of static analyses. The responses such as displacements, velocities, strains, stresses become time dependent in dynamic analysis. The dynamic equation of motion of a structure can be derived from Hamilton’s principle using Lagrange equation. 7.3.1 Hamilton’s Principle Hamilton’s principle is a simple but powerful tool to derive discritized dynamic system equations. It can be stated as “Of all the possible time histories of displacement the most accurate solution makes the Lagrangian functional a minimum provided the following conditions are satisfied.”

• the essential or the kinematic boundary conditions, • the compatibility equations, and • the conditions at initial (t1) and final time (t2).

First condition of the above ensures that the displacement constraints are fulfilled. Second condition makes sure that the displacements are continuous in the problem domain and the third condition necessitates the displacement history to satisfy the constraints at the initial and final times. Hamilton’s principle allows one to assume any set of displacements, as long as it satisfies the above three conditions. It is basically a variational formulation method. Just like in the other variational formulations, a functional is used here. Mathematically, Hamilton’s principle states:

2

1

dt 0t

t

Lδ =∫ (7.3.1)

The Langrangian functional, L, is defined as follows using a set of permissible time histories of displacement.

L T (7.3.2) Where T is the kinetic energy and is the potential energy which is basically elastic strain energy. The kinetic energy, T is defined in the integral form as

T1T d2

x x (7.3.3)

Here Ω represents the whole volume of the domain, ρ is the mass density of structure and x is the set of time histories of velocities. The strain energy in the entire domain of elastic structure can be expressed as

T T1 1d d2 2

D s (7.3.4)

Here ε are the strains obtained from the set of time histories of displacements. Here, L can be expressed in terms of the generalized variables 1 2 n 1 2 nx , x x an d x , x x

. Here, xi is the displacement and the velocity is expressed by

17

i idx xdt

(7.3.5)

Considering xi as generalised displacement, the Lagrange’s equation of motion are expressed as

i i

d L L 0 where i 1 to ndt x x

(7.3.6)

7.3.2 Finite Element Form in MDOF System Consider a two-degree of freedom system as shown in the figure below.

Fig. 7.3.1 Two Degree Freedom System

The kinetic and potential energy can be expressed by following form:

2 21 1 2 2

221 1 2 2 1

1 1T m x m x2 21 1k x k x x2 2

(7.3.7)

Now, substituting the values in Lagrangean L, we can obtain

2 2 2 2 21 1 2 2 1 1 2 2 2 1 2 2 1

1 1 1 1 1L T m x m x k x k x k x x k x2 2 2 2 2

(7.3.8)

Thus,

1 1 1 1 2 2 2 11 1

1 1 1 1 2 2 1

2 2 2 2 2 12 2

2 2 2 2 1

d L L m x k x k x k xdt x x

m x k x k x x 0

d L L m x k x k xdt x x

m x k x x 0

(7.3.9)

Above equations can be written in the matrix form as shown below:

18

1 2 2 111

222 2 2

k k k xxm 0 0xx0 m 0k k

(7.3.10)

Thus, for a multi degree freedom system, the above expression can be written as

M x K x 0 (7.3.11)

For a steady state condition, one can consider sinx= a t . Thus, sin

2 2 xx= - a t = - . As a result, the above equation can be re-written as

2K x M x 0

2K M x 0

12 M K (7.3.12)

Thus, for a single degree of freedom system, the fundamental frequency can be found as

KM

(7.3.13)

7.3.3 Solid Body with Distributed Mass The velocity vector x can be expressed in terms of the elemental nodal velocities with the help of

interpolation functions as Nxx . Here, x is the nodal velocity, and N is the interpolation function. Thus the expression of kinetic energy can be re-written as

T T T1 1T d x N N d x2 2

x x (7.3.14)

In the above equations, the mass matrix can be obtained as

T1M N N d2

(7.3.15)

Volume integral has to be taken over the volume of the element to find the mass matrix of that element. 7.3.3.1 Mass matrix for truss element In case of truss member, the degrees of freedom at each node is one because of its only axial deformation. Thus for a two node truss element, the interpolation function will become

N 1

Here, x / L and thus, dx Ld . Hence, the mass matrix can be expressed as

19

T T T

0 0

2

20 0

13 2 3

2 3 3

0

M N Nd A N Ndx AL N Nd

1 11AL 1 d AL d1

1 112 1AL3 63 2 3AL AL

1 1 1 266 32 3 3

l

l l (7.3.16)

Thus, the consistent mass matrix will be

2 1ALM1 26

(7.3.17)

Whereas the lumped mass matrix can be expressed as follows.

1 0ALM0 12

(7.3.18)

Now, to obtain a generalised mass matrix of a two dimensional truss member, consider Fig. 7.3.2.

Fig. 7.3.2 Two Degree Freedom System

The displacement vector and the shape function for such case are expressed as

1

1

2

2

uv

duv

and 1 2

1 2

N N0 0NN N0 0

(7.3.19)

Thus, the mass matrix can be derived as given below.

20

T

0

1 00 1 1 0 0

M N Nd AL d0 0 1 0

0

l

(7.3.20)

After performing integration, the mass matrix will be obtained as

2 0 1 00 2 0 1pALM 1 0 2 060 1 0 2

(7.3.21)

The lumped mass matrix of two dimensional truss member will become

1 0 0 00 1 0 0pALM =0 0 1 020 0 0 1

(7.3.22)

7.3.3.2 Mass matrix for beam element The beam element has two degrees of freedom at each node, one transverse displacement (v) and the other one is an angle of rotation (𝜃). So, a beam with two nodes has a total of four degrees of freedom. From earlier notes in module 4, lecture 3, the interpolation function of a beam is obtained as

32 3 2

1 22 3 2

2 3 2 3

3 42 3 2

3 2 2 xN 1 x x , N x x ,L L L L

3x 2x x xN and NL L L L

(7.3.23)

The consistent mass matrix can be following earlier process.

TT

0M N Nd A N N dx

l

Substitution for the interpolation functions and performing the required integrations, one can arrive the following mass matrix for the beam element.

[M] = 𝜌𝐴𝐿420

156 22𝐿 54 −13𝐿22𝐿 4𝐿2 13𝐿 −3𝐿254 13𝐿 156 −22𝐿

−13𝐿 −3𝐿2 −22𝐿 4𝐿2 (7.3.24)

Similarly, the lumped mass matrix of the beam element is

21

[M] = 𝜌𝐴𝐿2

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

(7.3.25)

7.3.3.3 Mass matrix for plane frame element The plane frame element has three degrees of freedom at each node: (i) axial displacement (u), (ii) transverse displacement (v) and (iii) angle of rotation (𝜃). So it has a total of six degrees of freedom for a two node plane frame element and its shape functions are given by

32 3 2

1 2 32 3 2

2 3 2 3

4 5 62 3 2

x 3 2 2 xN 1 , N 1 x x , N x x ,L L L L L

x 3x 2x x xN , N and NL L L L L

(7.3.26)

Similar to earlier case, one can derive the consistent mass matrix of a two node plane frame member which will become as given below.

[M] =𝜌𝐴𝐿

⎣⎢⎢⎢⎢⎢⎡

1/3 0 0 1/6 0 013/35 11𝐿/210 0 9/70 −13𝐿/420

𝐿2/105 0 13𝐿/420 −𝐿2/1401/3 0 0

𝑆𝑦𝑚. 13/35 −11𝐿/210𝐿2/105 ⎦

⎥⎥⎥⎥⎥⎤

(7.3.27)

The lumped mass matrix of such two dimensional frame element will be

[M] = 𝜌𝐴𝐿2

⎣⎢⎢⎢⎢⎡1 0 0 0 0 00 1 0 0 0 00 0 1 0 0 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 1⎦

⎥⎥⎥⎥⎤

(7.3.28)

7.3.3.4 Mass matrix for space frame element In case of space frame, each node has six degrees of freedom: (i) three displacements and (ii) three rotations. Therefore, for a two node space frame element there will be twelve degrees of freedom and the consistent mass matrix of the element in the local xyz system can be derived as

22

2

2

2 2

2 2

13

13035

130 035

J0 0 03A

11 L0 0 L 0210 105

11 L0 L 0 0 0210 105M AL

1 10 0 0 0 06 3

9 13 130 0 0 0 L 070 420 35

9 13 130 0 0 L 0 0 070 420 35

J J0 0 0 0 0 0 0 06A 3A

13 L 11 L0 0 L 0 0 0 0 L 0420 140 210 105

13 L 11 L0 L 0 0 0 0 L 0 0 0420 140 210 105

(7.3.29)

In a similar process, one can derive the mass matrix of other types of structures such as two dimensional plane stress/strain element, three dimensional solid element, plate bending element, shell element etc. 7.3.4 Time History Analysis The dynamic equation of motion of a multi-degree freedom system can be written as

M x C x K x tF (7.3.30)

Where, [C] is the damping matrix and F(t) is the time dependent force vector. The element matrices of the above equation can be expressed as follows.

23

T

T

T

1M N N d2

1C N N d2

1K B D Bd2

(7.3.31)

In a linear dynamic system, these values remain constant throughout the time history analysis. Damping is a mechanism which dissipates energy, causing the amplitude of free vibration to decay with time. The various types of damping which influences structural dynamical behaviour are viscous damping, hysteresis damping, radiation damping etc. Viscous damping exerts force proportional to velocity, as exhibited by the term. A formulation for this kind of problem was developed by Rayleigh. In this case, the energy dissipated is proportional to frequency and to the square of amplitude. Viscous damping is provided by surrounding gas or liquid or by the viscous damper attached to the structure. Radiation damping refers to energy dissipation to a practically unbounded medium, such as soil that supports structure. Among the various types of physical damping, viscous damping is easy represent computationally in dynamic equations. Fortunately, damping in structural problems is usually small enough regardless of its actual source. Its effect on structural response is modelled well by regarding it as viscous. The Rayleigh damping defines the global damping matrix [ ]C as a linear combination of the global mass and stiffness matrices. [ ] [ ] [ ]KMC βα ′+′= (7.3.32) Here, α ′ and β ′ are the stiffness and mass proportional damping constants respectively. The relationship between the fraction of critical damping ratio, ξ , α and β at frequency ω is given by

+=

ωβαωξ

21 (7.3.33)

Damping constants α′ and β ′are determined by choosing the fractions of critical damping ( 1ξ ′ and 2ξ ′ ) at two different frequencies (ω 1 and ω2) and solving simultaneously as follows

1 2 2 2 1 12 22 1

1 2 2 12 22 1

2

2

(7.3.34)

There are several numerical integration schemes are available to obtain the time history response of the structural system. Amongst, the implicit type of methods such as the Newmark’s method is one of the most popular one.