Applied Finite Element Analysis, 2ed 9789389520835, 9789389447392

Table of contents :
Cover
Half Title
Title
Copyright
Dedication
Acknowledgement
Preface to the Second Edition
Preface to the First Edition
Contents
Chapter 1: Introduction
1.1 Numerical Methods
1.2 Basic Concepts of Finite Element Method
1.3 Finite Element Formulation
1.4 Basic Equations for Elasticity Problems
Chapter 2: One-dimensional Elasticity Problems
2.1 One-dimensional Static Analysis of a Stepped Bar
2.2 Formulation of Element Matrices
2.3 Assembly of Matrices
2.4 Treatment of Boundary Conditions
2.4.1 Elimination approach
2.4.2 Penalty Approach
2.5 Formulation of Load Vector Due to Uniform Temperature Change
2.6 Analysis of Rotating Link
2.7 Length Coordinates for One-dimensional Linear Elements
Problem
Chapter 3: Analysis of Truss
3.1 Formulation of Element Matrices
3.2 Formulation of Element Equations
Problems
Chapter 4: Analysis of Beams
4.1 Derivation of Shape Functions
4.2 Formulation of Element Matrices and Equations
4.3 Evaluation of Element Load Vectors
4.4 Calculation of Bending Moment, Shear Force and Flexural Stress
4.5 System Equations
4.6 Essential and Natural Boundary Conditions
4.7 Stiffness Matrix for an Inclined Beam
4.8 Plane Frame Element
Problems
Chapter 5: Steady State Conduction Heat Transfer
5.1 Governing Equations of Steady State Conduction Heat Transfer
5.2 Boundary Conditions
5.3 Two-Dimensional Heat Conduction Analysis
5.4 Formulation of Functional
5.5 Element Matrices
5.6 Essential and Natural Boundary Conditions
5.7 One-Dimensional Steady State Conduction Analysis
5.8 Analysis of a Straight Fin
5.9 Analysis of a Tapered Fin
5.10 Galerkin Formulation of Element Equations
5.11 Point Source or Line Source
5.12 Axisymmetric Heat Conduction Problem
Problems
Chapter 6: Unsteady State Heat Condition
6.1 Energy Equation for Unsteady State Three-Dimensional Problem
6.2 Finite Difference Schemes
6.3 Element Matrices for One-Dimensional Unsteady State Heat Conduction
Problems
Chapter 7: Two-Dimensional Problems of Elasticity
7.1 Plane Stress
7.1.1 Material Property Matrix [D]
7.2 Plane Strain
7.2.1 Material Property Matrix [D]
7.3 Displacement Equations
7.4 Element Matrices
7.5 Element Equations
7.6 Formulation Using Natural Coordinates
7.7 Axisymmetric Stress Analysis
7.7.1 Strain-displacement Relationship
7.7.2 Constitutive Relationship
7.7.3 Element Stiffness Matrix
7.7.4 Elements Equations
Problems
Chapter 8: High Order Elements
8.1 Two-Dimensional Four Node Quadrilateral Element
8.2 Nine-Node Quadrilateral Element
8.3 Six-Node Triangular Element
8.4 Higher Order Isoparametric Elements
8.5 Numerical Integration: Gauss Quadrature
8.6 Numerical Integration for the Evaluation of Element Matrices
Problems
Chapter 9: Steady, Inviscid, Incompressible Flows and Torsion Analysis
9.1 Stream Function (ψ) Formulation
9.2 Velocity Potential ϕ(x, y) Formulation
9.3 Torsional Analysis of a Prismatic Bar
Problems
Chapter 10: Eigenvalue Problems
10.1 Axial Vibrations
10.2 Transverse Vibrations of Beams
10.3 Shifted Stiffness Matrix
Problems
Chapter 11: Three-Dimensional Elasticity Problems
11.1 Basic Equations
11.2 Formulation of Matrices
11.3 Three-Dimensional Elements
11.4 Principal Stress in Three-Dimensional Problem
11.5 Von-Mises Stress
11.6 Volume Coordinates
11.7 Formulation of Equations for Three-Dimensional Elasticity
Problems
Chapter 12: Dynamic Analysis
12.1 Equations of Motion
12.2 Types of Analysis
12.3 Transient Problems
12.4 Time Domain Analysis
12.4.1 Direct integration methods
12.4.2 Central difference scheme
12.4.3 Newmark method
12.4.4 Average acceleration method
12.4.5 Linear acceleration method
12.4.6 Wilson-θ method
12.5 Mode (Modal) Superposition Method
12.5.1 Rayleigh damping
12.5.2 Frequency Response
12.6 Extraction of Eigenvectors for Large Systems
12.6.1 Properties of Eigenvectors
12.7 Evaluation of Eigenvalues for Large Systems
12.7.1 Characteristic Polynomial Method
12.7.2 Iteration Methods
12.7.3 Inverse Iteration Method
12.7.4 Sturm Sequence
12.7.5 Gram-Schmidt Deflation
12.7.6 Subspace Iteration Method
12.7.7 Simultaneous Iteration Method
12.8 Transformation Methods
12.8.1 Lanczos Method
12.8.2 Jacobi Transformation
12.8.3 Givens Transformation
12.8.4 Householder Transformation
12.9 Component Mode Synthesis
12.10 Guyan Reduction
Problems
Chapter 13: Non-linear Analysis
13.1 Material Non-linearity
13.1.1 Incremental Method
13.1.2 Iterative Method
13.2 Geometric Non-linearity
Problems
Chapter 14: Bending of Thin Plates
14.1 Strain-Displacement Relations
14.2 Stress, Moment and Curvature Relations
14.3 Moment-Curvature Relationship
14.4 Differential Equation for Plate Bending
14.4.1 Equilibrium Equations
14.4.2 Differential Equation
14.4.3 Boundary Conditions
14.5 Finite Element Formulation
14.5.1 Stiffness Matrix
14.6 Weak Formulation
Problems
Chapter 15: Miscellaneous Topics
15.1 Convergence Requirements of Displacement Functions
15.2 Geometric Isotropy, Spatial Isotropy or Geometric Invariance
15.3 Mesh Refinement
15.4 Element Distortion
15.5 Discretisation
15.6 Substructuring
15.7 Software
15.8 Wave Front or Frontal Method
Short Questions and Answers
Answers to Selected Problems
Appendix
A.1 Variational Calculus
A.2 Principal of Virtual Work and Theory of Minimum Potential Energy
A.3 Galerkin’s Weighted Residual Method
A.4 Rayleigh Ritz Method
References
Index
Backcover

Citation preview

Second Edition TM

Applied Finite Element Analysis G. Ramamurty

Distributed by:

APPLIED FINITE ELEMENT ANALYSIS

APPLIED FINITE ELEMENT ANALYSIS SECOND EDITION

G. Ramamurty Former Professor Department of Mechanical Engineering Osmania University Hyderabad

I.K. International Publishing House Pvt. Ltd. NEW DELHI

BANGALORE

©Copyright 2019 I.K. International Pvt. Ltd., New Delhi-110002. This book may not be duplicated in any way without the express written consent of the publisher, except in the form of brief excerpts or quotations for the purposes of review. The information contained herein is for the personal use of the reader and may not be incorporated in any commercial programs, other books, databases, or any kind of software without written consent of the publisher. Making copies of this book or any portion for any purpose other than your own is a violation of copyright laws. Limits of Liability/disclaimer of Warranty: The author and publisher have used their best efforts in preparing this book. The author make no representation or warranties with respect to the accuracy or completeness of the contents of this book, and specifically disclaim any implied warranties of merchantability or fitness of any particular purpose. There are no warranties which extend beyond the descriptions contained in this paragraph. No warranty may be created or extended by sales representatives or written sales materials. The accuracy and completeness of the information provided herein and the opinions stated herein are not guaranteed or warranted to produce any particulars results, and the advice and strategies contained herein may not be suitable for every individual. Neither Dreamtech Press nor author shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. Trademarks: All brand names and product names used in this book are trademarks, registered trademarks, or trade names of their respective holders. Dreamtech Press is not associated with any product or vendor mentioned in this book. ISBN: 978-93-89447-39-2

EISBN: 978-93-89520-83-5 Edition: 2019

Dedicated to my students

ACKNOWLEDGEMENTS

While preparing the notes for this book, I consulted several books listed in references, whose influence may be seen and I express sincere thanks to all those authors. I wish to express thanks to Prof. K. Narasimha Rao, Former Professor, Mech. Engg., Osmania University and Prof. K.N. Seetha Ramu, Former Professor, Mech. Engg., IIT, Madras, who created interest in me to learn this important subject of Finite Element Analysis. I would like to thank my colleagues Dr. B. Nagaprasada Rao, of Mech. Engg., Osmania University and colleagues in M.J. College of Engg., Hyderabed, Ms. Ishrat. M. Mirzana, Mech. Engg., and Mr. D. Ram Mohan Rao, Civil. Engg., for their constant encouragement and help extended. I would also like to thank my wife, Mrs. Ganga Sundari and son, Mr. Gowtam for their support during preparation of this manuscript. G. Ramamurty

PREFACE TO THE SECOND EDITION

In view of the feedback received from readers, it is felt necessary to include the following additional topics which will help in improving the subject contents. Dynamic analysis has been included in chapter 12, which deals with methods to solve dynamic problems essential for the design of machine components. Chapter 13 discussing the finite element formulation of non-linear problems is included to furnish details of methods necessary for analysis of non-linear problems is included to furnish details of methods necessary for analysis of non-linear problems. Bending of thin plates is discussed in chapter 14 to provide information for analysis of plates in addition to the analysis of bars, beams, given in earlier chapters. Miscellaneous topics have been presented in chapter 15 replacing chapter 12 of first edition. Short questions and answers have been given to equip the student with a better insight for solving problems. An appendix is presented with a discussion on variational calculus, principle of virtual work, theory of minimum potential energy, Galerkin’s weighted residual method and Rayleigh Ritz method. The addition of above topics is an attempt to provide an overview of essential finite element methods. G. Ramamurty

PREFACE TO THE FIRST EDITION

Problems involving structural analysis, analysis of aircraft structures, automobile components, heat transfer, thermal stress and fluid flow require large storage data, and equations governing the physical phenomenon are non-linear. Further, these components have irregular geometries, complex material properties and sometimes even involve non-linear boundary conditions. Equations with exact solutions are available only for basic configurations and selective problems with assumptions simplifying the analysis. The techniques that are used to get solutions for problems are i) Analytical solutions and ii) Experimental techniques. Many problems do not have analytical solutions which can be used at all points throughout the domain. Experimental techniques are expensive for large sized structures which require huge equipment to conduct experiments. Under these circumstances, the alternative available is numerical method. Finite Element Method is one of the numerical methods with which solutions can be obtained for problems with complex geometries, material properties and boundary conditions. Advanced computer systems have tremendously increased the possibility of carrying out several calculations in a short time which was a difficult task by manual calculations before the advent of computer systems. Latest computer configurations have provision for storing large amounts of data and fast retrieval of stored information. These developments contributed to the popular usage of Finite Element Method in Design and Analysis relating to different fields. Many authors have contributed in introducing and explaining the concepts of Finite Element Analysis in their textbooks. Finite Element Method, Finite Element Analysis and Finite Element Techniques are synonymously used by different authors. Each textbook on FEM effectively focuses on several aspects of the subjects but one may find some missing links and details in some areas. This book is an attempt to collect information and collate it systematically and make an effective presentation. It is felt necessary to explain the basic concepts clearly with a view to make the reader understand the critical points for enhancing the required capabilities essential for solving complex problems. Knowing the basic concepts on which a software can be developed, will be an added advantage. Interpretation of results is an essential requirement while using FEM analysis. Of late, several FEM softwares or codes have been made available and it is only a choice one has to make for selecting proper software. The problem one encounters with the software is that it does not explicitly convey the basic equations on which that particular code has been developed.

x

Preface

Computers loaded with a FEM software like ANSYS, NASTRAN, NISA, ABAQUS, ADINA etc., act as only facilitators to obtain solutions by inputting the data and supplying necessary information systematically as per the procedure laid down for the use of that software. It is obvious that a student who learns to use software should also know principal equations and basic concepts with which the program was written for execution and finally displaying the results on the monitor. In view of essential requirements, an attempt is made in this book to introduce the concepts in respect of several modules covering elasticity, heat transfer, eigenvalue and fluid flow problems. These areas are introduced so that one may acquire necessary inputs for solving real life problems which will be encountered in industrial environments. The basic concepts are building blocks and using them effectively will help in building the superstructure with a strong foundation. Problems are worked out in each chapter to illustrate these methods. Formulations in global and local coordinate systems are explained. G. Ramamurty

CONTENTS

Preface Acknowledgements

1. Introduction 1.1 1.2 1.3 1.4

Numerical Methods Basic Concepts of Finite Element Method Finite Element Formulation Basic Equations for Elasticity Problems

2. One-dimensional Elasticity Problems 2.1 2.2 2.3 2.4

2.5 2.6 2.7

One-dimensional Static Analysis of a Stepped Bar Formulation of Element Matrices Assembly of Matrices Treatment of Boundary Conditions 2.4.1 Elimination approach 2.4.2 Penalty approach Formulation of Load Vector Due to Uniform Temperature Change Analysis of Rotating Link Length Coordinates for One-dimensional Linear Elements Problems

3. Analysis of Truss 3.1 3.2

Formulation of Element Matrices Formulation of Element Equations Problems

4. Analysis of Beams 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8

Derivation of Shape Functions Formulation of Element Matrices and Equations Evaluation of Element Load Vectors Calculation of Bending Moment, Shear Force and Flexural Stress System Equations Essential and Natural Boundary Conditions Stiffness Matrix for an Inclined Beam Plane Frame Element Problems

vii xi 1 1 2 3 4 9 9 12 14 22 22 22 31 39 43 56 58 58 61 74 77 79 81 82 84 85 85 96 99 105

Contents

xii

5. Steady State Conduction Heat Transfer 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 5.12

Governing Equations of Steady State Conduction Heat Transfer Boundary Conditions Two-Dimensional Heat Conduction Analysis Formulation of Functional Element Matrices Essential and Natural Boundary Conditions One-Dimensional Steady State Conduction Analysis Analysis of a Straight Fin Analysis of a Tapered Fin Galerkin Formulation of Element Equations Point Source or Line Source Axisymmetric Heat Conduction Problem Problems

6. Unsteady State Heat Condition 6.1 6.2 6.3

Energy Equation for Unsteady State Three-Dimensional Problem Finite Difference Schemes Element Matrices for One-Dimensional Unsteady State Heat Conduction Problems

7. Two-Dimensional Problems of Elasticity 7.1 7.2 7.3 7.4 7.5 7.6 7.7

Plane Stress 7.1.1 Material property matrix [D] Plane Strain 7.2.1 Material property matrix [D] Displacement Equations Element Matrices Element Equations Formulation Using Natural Coordinates Axisymmetric Stress Analysis 7.7.1 Strain-displacement relationship 7.7.2 Constitutive relationship 7.7.3 Element stiffness matrix 7.7.4 Element equations Problems

8. High Order Elements 8.1 8.2 8.3 8.4

Two-Dimensional Four Node Quadrilateral Element Nine-Node Quadrilateral Element Six-Node Triangular Element Higher Order Isoparametric Elements

108 108 108 109 110 113 115 120 132 135 137 140 142 148 151 151 153 156 166 168 168 169 170 170 172 174 175 176 180 182 183 183 186 205 208 208 212 215 216

Contents

8.5 8.6

Numerical Integration: Gauss Quadrature Numerical Integration for the Evaluation of Element Matrices Problems

9. Steady, Inviscid, Incompressible Flows and Torsion Analysis 9.1 9.2 9.3

Stream Function (ψ) Formulation Velocity Potential ϕ(x, y) Formulation Torsional Analysis of a Prismatic Bar Problems

10. Eigenvalue Problems 10.1 Axial Vibrations 10.2 Transverse Vibrations of Beams 10.3 Shifted Stiffness Matrix Problems

11. Three-Dimensional Elasticity Problems 11.1 11.2 11.3 11.4 11.5 11.6 11.7

Basic Equations Formulation of Matrices Three-Dimensional Elements Principal Stress in Three-Dimensional Problem Von-Mises Stress Volume Coordinates Formulation of Equations for Three-Dimensional Elasticity Problems

12. Dynamic Analysis 12.1 12.2 12.3 12.4

Equations of Motion Types of Analysis Transient Problems Time Domain Analysis 12.4.1 Direct integration methods 12.4.2 Central difference scheme 12.4.3 Newmark method 12.4.4 Average acceleration method 12.4.5 Linear acceleration method 12.4.6 Wilson-θ method 12.5 Mode (Modal) Superposition Method 12.5.1 Rayleigh damping 12.5.2 Frequency response 12.6 Extraction of Eigenvectors for Large Systems 12.6.1 Properties of eigenvectors

xiii

220 222 228 230 230 232 235 243 244 247 252 253 259 261 261 263 265 266 267 268 269 276 277 277 278 278 279 279 279 284 286 287 288 289 291 292 299 300

Contents

xiv

12.7 Evaluation of Eigenvalues for Large Systems 12.7.1 Characteristic polynomial method 12.7.2 Iteration methods 12.7.3 Inverse iteration method 12.7.4 Sturm sequence 12.7.5 Gram-Schmidt deflation 12.7.6 Subspace iteration method 12.7.7 Simultaneous iteration method 12.8 Transformation Methods 12.8.1 Lanczos method 12.8.2 Jacobi transformation 12.8.3 Givens transformation 12.8.4 Householder transformation 12.9 Component Mode Synthesis 12.10 Guyan Reduction Problems

13. Non-linear Analysis 13.1 Material Non-linearity 13.1.1 Incremental method 13.1.2 Iterative method 13.2 Geometric Non-linearity Problems

14. Bending of Thin Plates 14.1 14.2 14.3 14.4

Strain-Displacement Relations Stress, Moment and Curvature Relations Moment-Curvature Relationship Differential Equation for Plate Bending 14.4.1 Equilibrium equations 14.4.2 Differential equation 14.4.3 Boundary conditions 14.5 Finite Element Formulation 14.5.1 Stiffness matrix 14.6 Weak Formulation Problems

15. Miscellaneous Topics 15.1 15.2 15.3 15.4

Convergence Requirements of Displacement Functions Geometric Isotropy, Spatial Isotropy or Geometric Invariance Mesh Refinement Element Distortion

302 302 303 303 309 311 312 313 314 315 318 320 322 324 331 332 334 334 334 337 341 349 350 350 352 353 353 353 354 354 354 357 357 359 360 360 361 361 362

Contents

15.5 15.6 15.7 15.8

Discretisation Substructuring Software Wave Front or Frontal Method

xv

362 363 365 367

Short Questions and Answers

375

Answers to Selected Problems

401

Appendix A.1 Variational Calculus A.2 Principal of Virtual Work and Theory of Minimum Potential Energy A.3 Galerkin’s Weighted Residual Method A.4 Rayleigh Ritz Method

409 409 416 419 423

References

425

Index

427

CHAPTER

INTRODUCTION

1

Engineering problems requiring solutions are mathematically described by differential equations. These differential equations governing the physical phenomena are sometimes non-linear, involving complex geometries and material properties. Closed form or exact solutions cannot be obtained in many situations which will compel us to resort to a numerical method. A closed form solution gives solution valid at all points throughout the computational domain but a numerical method gives an approximate value of the desired variable which is valid at only discrete points. A numerical method has been important due to the reasons that an experiment could not be conducted because of the huge size of the component and in other situations closed form solutions are not possible. Instead of being pessimistic with no solution, it is better to have an approximate solution within acceptable accuracy. The accuracy required has to be specified and one should have physical insight into the phenomena for checking whether the solution obtained is realistic.

1.1

NUMERICAL METHODS

The following numerical methods are generally used for solving engineering problems. i) ii) iii) iv)

Finite difference method Finite element method Control volume or finite volume method Boundary element method

Finite difference scheme expresses the derivatives of the governing differential equation, in terms of the variables at the selected grid points using truncated Taylor series expansion. The computational domain is filled up with number of grids and grid points selected at the centre of the corresponding grids. Finite difference scheme has the following disadvantages. a) Specification of boundary conditions for irregular geometries. b) Incorporation of complex material properties. Finite element method is a popular computer aided numerical method based on the discretisation of the domain, structure or continuum into number of elements and obtaining the solution. The details of finite element method will be discussed subsequently as this text-book mainly focusses on finite element analysis.

Applied Finite Element Analysis

2

Control volume analysis or finite volume method formulates the discretised equations at each grid point of the domain by integrating the terms of the differential equations over the control volume and assuming the variation of the unknown variable between grid points. This method is mainly used for fluid flow problems. Boundary element method is based on the boundary integral equation formulation of the problem. It needs only a boundary discretisation in contrast to domain methods. This method has not become popular compared to finite difference, finite element and control volume formulations.

1.2

BASIC CONCEPTS OF FINITE ELEMENT METHOD

Basically, any problem can be divided into subproblems and solution is obtained by summing up the solutions of each individual subproblem. In finite element method, this is achieved by dividing the given geometry into a number of simpler shapes. Normally, this concept is introduced by many authors by considering a circle whose area can be computed by dividing the circle into a number of simple regions by using triangles. a) The circle shown in Fig. 1.1 has been divided by using eight triangles. This step of dividing the given area into subregion is called discretisation. Each subregion is called an element. b) The area of each triangle can be calculated and let this be A(e) for each triangle ‘e’. c) Knowing the area of each triangular element A(e) , the area of the circle is 8  A= A(e) . e=1

d) The discretisation by eight triangles cannot cover the whole area of the circle and because of the uncovered region shown shaded in Fig. 1.1 an error is introduced resulting in an approximation. e) An accurate solution for the area of the circle can be obtained by increasing the number of elements (e), the solution converges to an exact one depending on the number of elements considered.

By increasing the number of elements, the computer time required will increase, indirectly contributing to the cost. The solution that can be obtained by finite element method

8

1

7

2

6

3 5

4

FIGURE 1.1 Division with triangles.

Introduction

3

should be cost effective and hence the minimum number of elements are desirable to get cost effective solution without affecting the accuracy. The above example is a simple problem to introduce the concept but actual application of FEM involves additional requirements.

1.3

FINITE ELEMENT FORMULATION

Steps involved for finite element formulation are briefly discussed below: a) Discretisation: It is the process of dividing the domain, structure or continuum into subregions or subdivisions called finite elements. Element having a simple shape with which a domain or body or structure is discretised. Nodes are defined for each element and nodes are the locations or discrete points at which the unknown variables are to be determined. These unknown variables are called field variables as the unknown variable may be displacement, temperature or velocity depending on the type of problem under consideration. Collection of elements is called mesh. Elements are connected at the nodes. b) Approximation of field variable in an element: Polynomial expressions are normally employed to define the variation of the field variable. It is easy to differentiate and integrate the terms in polynomial expressions. c) Formulation of element equations: Element equations will be derived by minimisation of a functional. The functional is an integral expression and minimisation with respect to nodal variables yields the element equations. The formulation of element equations can be accomplished by using one of the following methods. i) Variation formulation using calculus of variation. ii) Weighted residual methods out of which Galerkin’s method is extensively used. For structural mechanics problems, the potential energy expression written in an integral form provides the necessary functional. Minimisation of potential energy expression with respect to nodal displacements will yield element equations which are known as equations of equilibrium. d) Principle of minimum potential energy: Among all admissible configurations of a conservative system, those that satisfy the equations of equilibrium make the potential energy stationary with respect to small admissible variations of displacement. e) Assembly of matrices to form global or system equations: The stiffness matrix and load vector (also called force vector) for each element are formulated. Assembling these matrices for all elements will give global stiffness matrix and global load vector to formulate the global or system equations. [K (e) ]{q (e) } = {F (e) } n    [K (e) ]{q (e) } − {F (e) } = 0 e=1

(1.1) (1.2)

Applied Finite Element Analysis

4

[K] {Q} = {F }

(1.3)

where n is number of elements [K (e) ] - element stiffness matrix [K] - global stiffness matrix {q (e) } - element nodal displacement vector {Q} - global nodal displacement vector (e) {F } - element load vector {F } - global load vector f) Solving the global or system equations will provide the solution for field variable i.e., displacements for a solid or structural mechanics problem. {Q} = [K]−1 {F }

(1.4)

g) Secondary quantities like strain and stress can be calculated once the nodal displacements are known. The description of finite element method given above is based on displacement formulation for structural mechanics problem. In addition to displacement formulation, other methods employed are i) equilibrium method ii) mixed method. In equilibrium method, stress is field variable and displacement is to be derived from stress. When the formulation is based on stress, compatibility equations need to satisfied. In mixed method, part of the domain is solved using displacement formulation and remaining part with equilibrium method. At the outset finite element analysis relies on two important functions. i) A continuous piecewise smooth function is needed to prescribe the field variable within the element. ii) An integral expression called functional is used to generate element equations. Even though the finite element method was originally used for structural mechanics problems, presently it is used in many areas including electrostatic, magnetic field problems as well as problems encountered in Biomedical engineering. Since finite element method (FEM) has versatility to provide solutions pertaining to several fields. Application of FEM is well known in the field of automobile, aerospace, thermal, heat transfer and fluid flow areas.

1.4

BASIC EQUATIONS FOR ELASTICITY PROBLEMS

In elasticity problems, the state of stress, strain and displacements are needed to establish that the designed components are safe to withstand loads without excessive deflection. It is also important to optimise the dimensions of the components with a view to reduce cost.

Introduction

5

The state of stress in a three-dimensional problem is described by six components of stress and six components of strain. The stress-strain components are related by elastic constants of the material and strain components are specified in terms of displacement components. Displacements are three in number which are related to strains by strain-displacement relations. Thus, we require 15 equations to determine six components of stress, six components of strain and three components of displacement at a point. These equations are provided by three equilibrium equations, six strain-displacement relations and six stress-strain relations. Further, we have to satisfy the compatibility conditions to ensure the continuity of displacements. Generally, three kinds of forces which can be considered to act on a body, namely, i) body forces ii) surface tractions and iii) point or concentrated loads. Equilibrium equations in terms of stresses: Let the components of stress be expressed in a matrix form as {σ} = [σx , σy , σz , τxy , τyz , τzx ]T

(1.5)

An infinitesimal element isolated from a body with its faces parallel to the coordinate planes is shown in Fig. 1.2 to describe the stresses at a point within a body. σx , σy and σz are normal stresses. τxy , τyz and τzx are shear stresses.

Z τxz

σz

σx

G

F

τyz

τxy

τxz D

τyz

C τxy

σy

σy

τxz τyz

τxy

O

E

τxy σx τyz

A

B

τxz σz X

FIGURE 1.2 Stress components.

Y

Applied Finite Element Analysis

6

Equilibrium equations are expressed using operator matrix

 



σx

τxy σy

symmetric

     τxz ∂/∂x fx τyz   ∂/∂y  +  fy  = 0 σz fz ∂/∂z

(1.6)

 ∂ ∂ ∂ T is the operator matrix and fx , fy and fz are components of body forces ∂x ∂y ∂z acting on the body. The equilibrium equations are expressed in terms of stresses and body forces. Stress-strain relations or Hooke’s law for linear isotropic materials are x = y = z = γxy = γyz = γzx =

1 [σx − ν(σy + σz )] E 1 [σy − ν(σz + σx )] E 1 [σz − ν(σx + σy )] E 1 τxy G 1 τyz G 1 τzx G

(1.7)

(1.8)

where E is modulus of elasticity and G is modulus of rigidity and components of strain are {} = [x , y , z , γxy , γyz , γzx ]T

(1.9)

These equations are also known as constitutive relations between stress and strain. It is also well known that the two elastic constants are required to relate stress-strain relations and there is no coupling between normal stresses and shear strains, and shear stress and normal strains. Strain-displacement relations: ∂u ; ∂x ∂v ; y = ∂y ∂w ; z = ∂z x =

∂u ∂v + ∂y ∂x ∂v ∂w + = ∂z ∂y ∂w ∂u + = ∂x ∂z

γxy = γyz γzx

(1.10)

Introduction

7

Z

R σnz

n σny

O σnx

Q

Y

P X FIGURE 1.3 Stress components on an oblique plane.

where u, v and w are displacements. Stress on a oblique plane through a point: When we know the stress components at a point, the stresses on any oblique plane passing through this point can be expressed using direction cosines. Consider the equilibrium of a tetrahedron OP QR shown in Fig. 1.3. Let the stresses acting on plane OQR, OP R and OP Q be specified as σx , τxy , τxz , on plane OQR. σy , τyz , τyx , on plane OP R. σz , τzx , τzy , on plane OP Q. Let the direction cosines of the normal to the oblique plane P QR with respect to x, y, z axes be l, m, n respectively and σnx , σny , σnz be stresses on plane P QR. Considering the force balance it can be shown that



  σnx σx τxy  σny  =  σy σnz symmetric {σn } = [σ]{l}

  τxz l τyz   m  σz n

(1.11) (1.12)

The Eq.(1.12) provides the expression for Cartesian components of stress on the inclined plane P QR. Boundary Conditions: σnx , σny and σnz expressed in the Eq.(1.11) represent the surface tractions if any acting on the boundary. Using six stress-strain relations, six strain-displacements and eliminating six strain components will result in six stress-displacement equations. These equations along with three

8

Applied Finite Element Analysis

equilibrium equations provide solution for three displacements and six stresses. Boundary conditions specified at the boundary have to be used to obtain the particular solution. Compatibility equations: Compatibility equations have to be satisfied when the formulation is in terms of stresses. The compatibility conditions ensure continuity of displacements in the continuum. If the formulation is in terms of displacements, compatibility conditions do not play any significant role since the displacement solution ensures continuity.

CHAPTER

ONE-DIMENSIONAL ELASTICITY PROBLMES

2

Having learnt the basics required to implement finite element method, it becomes simpler to understand the procedure for obtaining solution by considering one-dimensional problems. One-dimensional problems can be identified with elasticity problems, heat conduction problems and other engineering fields. Let us take up the analysis of a stepped bar subjected to an axial load.

2.1

ONE-DIMENSIONAL STATIC ANALYSIS OF A STEPPED BAR

Let us introduce a stepped bar subject to axial load P as shown in Fig. 2.1. E1 , E2 A1 , A2 L1 , L 2 P

- Modulus of elasticity of each step - Areas of cross-section - Lengths - Point load. A1, E1

A2, E2 P

O

L1

X

L2

FIGURE 2.1 Configuration.

Coordinate system has been defined with the origin ‘O’ at the left face of the bar where it is fixed and x-axis along the axis of the bar. If one resorts to conventional method of solving the above mentioned problem, differential equation that governs the deformation (displacement) ‘u’ which is a function of x is given by    du d EA =0 (2.1) dx dx Solving the above equation and using boundary conditions i) at x = 0, u = 0 and du ii) x = L1 + L2 , EA = P , the solution for u is obtained. dx

10

Applied Finite Element Analysis

The assumption of one-dimension analysis of the stepped bar is justified since the displacement is in the axial direction due to axial load P . Our interest is not to obtain solution by conventional method but through finite element method. Steps involved in finite element analysis for the stepped bar problem shown in Fig. 2.1: a) Discretisation: The lengths, areas and moduli of the bar are changing from step 1 to step 2. Therefore, the bar is discretised with two discrete elements. The type of element with which discretisation of the bar has been considered is one-dimensional linear element—also called bar element. q1

1 node

q2

1 l1

2 node

2

q3

l2

3 node

FIGURE 2.2 Elements.

Type of element: One-dimensional linear element has two nodes per element and the variation of field variable is approximated by first order polynomial u(x) = a1 + a2 x

(2.2)

At each node, the nodal displacement is denoted by letter ‘q’. Axial displacement is possible at each node under the given system of loading. Hence, it is said that there is one degree of freedom at each node and for this linear element with two nodes, there are two degrees of freedom for the element. qi O

X

e A

node i xi

qj

u

x

j node xj

FIGURE 2.3 General element (e).

Figure 2.3 shows the general element with nodal displacements and nodal coordinates. ‘O’ is the origin of the global coordinate system which is defined for the complete stepped bar. The formulation of problem is presented using global coordinate system. b) Shape functions: Now, we shall derive shape functions also called interpolation functions. The Eq.(2.2) becomes u = a1 + a2 x expressing the variation of displacement at any point which is at a distance x from the origin of the global coordinate system. a1 and a2 are constants evaluated using the boundary conditions available at node i and node j.

One-Dimensional Elasticity Problems

11

The boundary conditions are: i) at x = xi , nodal displacement is u = qi ii) at x = xj , nodal displacement is u = qj Substituting the above conditions in Eq.(2.2), it can be shown as a1 =

q i xj − q j xi xj − x i

and a2 =

qj − q i x j − xi

(2.3)

qi , qj are nodal displacements at the nodes i and j respectively. xi , xj are nodal coordinates of nodes i and j in global coordinate system and l(e) = (xj − xi ) is length of the element. Substituting the expressions for a1 and a2 in Eq.(2.2) the displacement at any point x is u(x) =

qi xj − qj xi (qj − qi )x + l(e) l(e)

(2.4)

Separating the terms containing qi and qj , the Eq.(2.4) can be written as



 (xj − x) (x − xi ) qi + qj u(x) = (e) l l(e)

(2.5)

Defining the shape functions Ni and Nj associated with nodes i and j as Ni =

(xj − x) (x − xi ) , Nj = l(e) l(e)

(2.6)

The shape functions Ni and Nj are also known as interpolation functions. The shape functions depend on the shape of the element used in the formulation and this is the reason why they are known as shape functions and we can observe that these are functions of the spatial coordinate ‘x’. The shape functions are also called interpolation functions because these are used to interpolate the displacement within the element at any point x knowing the nodal displacements qi and qj , quite obvious from the Eq.(2.6). c) Displacement equation: The Eq.(2.5) can be written as u(x) = Ni qi + Nj qj

(2.7)

In finite element analysis, the equations are written in matrix form as matrices can be handled easily by the computer. Defining the following matrices as Shape function matrix

[N ]1×2 = [Ni

Nodal displacement vector

{q}2×1 = [qi

Nj ]

(2.8)

qj ] T

(2.9)

Applied Finite Element Analysis

12

Superscript T indicates the transpose of the matrix. The Eq.(2.7) in newly defined matrices is u(x) = [N ]1×2 {q}2×1

(2.10)

At this stage one can appreciate that there is correspondence between number of nodes per element and number of constants a1 and a2 in the first order polynomial expression used to specify the linear variation of the field variable. It can be seen that we have assumed linear variation of the field variable over the element but the nodal displacements are not known and yet to be calculated. Properties of shape functions Ni and Nj :

1

1

i

j

FIGURE 2.4 Variation of

Ni .

i) At any point x; ii) At node i; x = xi ; iii) At node j; x = xj ;

2.2

j

i FIGURE 2.5 Variation of

Nj .

Ni + Nj = 1 Ni = 1, Nj = 0 Ni = 0, Nj = 1

FORMULATION OF ELEMENT MATRICES

For formulation of element matrices, as explained in Chapter 1 we need to minimise the relevant functional with respect to nodal displacements. Potential energy expression written in an integral form provides the necessary functional for a structural or solid mechanics problem. Functional π = Potential energy for structural mechanics problem Potential energy = Strain energy (U ) − Work done by external loads (We ) 1 π= 2 {} {σ} {f } {T } [P ]

 v

T

{} {σ}dv −

 v

T

{u} {f }dv −

 s

{u}T {T }ds − [P ]{q}

- Strain - Stress - Body force acting on the element (N/m3 ) - Surface traction acting on the surface of element (N/m2 ) - Point loads = [Pi Pj ]

(2.11)

One-Dimensional Elasticity Problems

13

There are three types of loading encountered normally in a structural mechanics problem if there is no temperature change. i) Body force per unit volume {f } ii) Surface traction per unit area {T } iii) Point loads [P ] if any. The above Eq.(2.11) shows that we have to write expressions for strain  and stress σ. Since this is a one-dimensional problem, d du = [Ni qi + Nj qj ] dx dx    dNj dNi dNi (qi ) + (qj ) = = dx dx dx

Strain =  =

(2.12) dNj dx



qi qj



(2.13)

Shape functions are functions of x, we need to differentiate only Ni and Nj expressions shown in Eq.(2.6).

∴

dNi 1 = − (e) dx l

and

dNj 1 = (e) dx l

Defining a strain-displacement matrix,   l dNi dNj [B]1×2 = = (e) [−1 1]1×2 dx dx 1×2 l Hence, Eq.(2.12) for strain can be expressed using Eq.(2.13) and Eq.(2.15) as       qi 1  qi dNi dNj = [B]1×2 {q}2×1 = (e) −1 1 = qj qj l dx dx

(2.14)

(2.15)

(2.16)

Now stress σ can be written as

σ = E = E[B]{q} E - Modulus of elasticity will be written as material property matrix [D]1×1 Therefore, the stress σ = [D]1×1 [B]1×2 {q}2×1

(2.17)

(2.18)

From Eq.(2.12), it can be inferred that for obtaining strain-displacement matrix, the expression for strain is essential. Similarly, for obtaining material property matrix [D], the expression for stress σ is necessary. The matrices [D] and [B] are important in finite element formulations. Substituting for strain and stress from Eqs.(2.16) and (2.18) respectively in Eq.(2.11) for the evaluation of the functional yields,

Applied Finite Element Analysis

14

      1 π = {q}T  [B]T [D][B]dV  {q} − {q}T  [N ]T {f }dV  2 V V  −{q}T [N ]T {T }ds − [P ]{q}

(2.19)

S

Minimisation of functional π (Potential energy) given in Eq.(2.19) with respect to nodal displacements.   ∂π       ∂qi   ∂π = =0 (2.20) ∂{q}  ∂π        ∂qj reduces  to      [B]T [D][B]dV  {q} − [N ]T {f }dV − [N ]T {T }ds − {P } = 0 V

V

(2.21)

S

Defining the element stiffness matrix  [K (e) ] = [B]T [D][B]dV

(2.22)

V

element load vector {F

(e)

}=



T

[N ] {f }dV +

V



[N ]T {T }dS + {P }

(2.23)

The element equations in matrix form are [K (e) ]{q (e) } = {F (e) }

(2.24)

The element stiffness matrix [K (e) ] and load vector {F (e) } are essential for the analysis.

2.3

ASSEMBLY OF MATRICES [K] = {F } =

n 

K (e) is called global stiffness matrix

(2.25)

{F (e) } is called global load vector or force vector

(2.26)

e=1 n  e=1

One-Dimensional Elasticity Problems

15

Global or system equations: Equations can be formed by assembling element equations. n   e=1

 [K (e) ]{q (e) } − {F (e) } = 0

(2.27)

[K]n×n {Q}n×1 = {F }n×1

(2.28)

{Q}n×1 is called global nodal displacement vector. The size of the global stiffness matrix [K]n×n depends on the number nodes and degrees of freedom at each node. Equation (2.28) can also be identified with equilibrium equations for the structural mechanics problem. Before solving Eq.(2.28), the boundary conditions for the displacement have to be imposed. Otherwise, the global stiffness matrix will be a singular matrix and it indicates that the body is not constrained allowing rigid body motion under the applied load. After imposing boundary conditions, the reduced global equations are solved for nodal displacements. We can write the important equations for finite element formulation as  u(x) = [N ]1×2 {q}2×1    = [B]1×2 {q}2×1 for one-dimensional problem (2.29)   σ = [D]1×1 [B]1×2 {q}2×1

Let us return to the static analysis of the stepped bar shown in Fig. 2.1. The necessary matrix formulation has already been explained with reference to the stepped bar problem. The following can be explicitly stated. q1

q2 2

1 1

2

q3

3

FIGURE 2.6 Nodal degrees of freedom.

i) Discretisation of the stepped bar: No. of elements Type of element

: (2) : One-dimensional linear element (line segment with nodes at the ends) No. of nodes per element :2 No. of degrees of freedom per node : One translational degree of freedom. Total number of nodes for the bar : 3 (Node No: 2 is common to both elements) Field variable (u) : Displacement which is a function of spatial coordinate x. ii) Field variable variation: The variation over the element is assumed to be a linear variation w.r.t x hence expressed by the first order polynomial. u = a1 + a2 x.

16

Applied Finite Element Analysis

iii) Nodal displacements and shape functions: Nodal displacement vector for the element = [qi qj ]T xj − x Shape function associated with node i, Ni = (e) l x − xi Shape function associated with node j, Nj = (e) l

(2.30) (2.31) (2.32)

iv) Displacement function: u=



Ni

Nj





qi qj



= [N ]{q}

(2.33)

qj ] T

where nodal displacement vector {q} = [qi

Strain-displacement matrix and stiffness matrix for the element: [B]1×2 = [K (e) ] =

Stiffness matrix

1 l(e)



[−1 1]

(2.34)

[B]T [D][B]dV

(2.35)

V

[D]1×1 = E

Material property matrix

dV = A dx

(2.36)

Stiffness matrix is evaluated as [K

(e)

]2×2

  K (e)

2×2

E (e) A(e) =  2 l(e) =

xj 

−1 1



i

j

1 −1

−1 1

xi

E (e) A(e) l(e)



 

−1 1

i j



dx

(2.37)

(2.38)

The size of the stiffness matrix is 2 × 2 because each element has two nodes and at each node one translational degree of freedom. The rows and columns are identified with reference to i and j for the element. vi) Element load vector: It is known that the load P is applied at node 3 (three) only, and the bar has not been subjected to any other loading     Pi i (e) F = (2.39) j Pj

One-Dimensional Elasticity Problems

17

vii) Assembling the matrices to obtain global matrices: Stiffness matrix for element (1)



K (1)



2×2

=

E1 A1 L1



1 2  1 −1 1 −1 1 2

(2.40)

E2 A2 L2



2 3  1 −1 2 −1 1 3

(2.41)

Stiffness matrix for element (2)



K (2)



2×2

=

The size of global stiffness matrix is 3 × 3 since the bar has been discretised with three nodes and each node is having one translational degree of freedom. Therefore, global stiffness matrix [K]3×3 is obtained by filling the corresponding elements from Eqs.(2.40) and (2.41) into 3 × 3 matrix.



[K]3×3

1

2

E1 A1  L1    E1 A1 =  − L 1    0



−

E1 A1 L1

E1 A1 E2 A2 + L1 L2 −

← ↓ Global d.o.f  0 1   E2 A2  2 − L2    E2 A2  3 L2 3



E2 A2 L2

(2.42)

The global stiffness matrix given by Eq.(2.42) is a symmetric and banded matrix.

Global load vector

{F }3×1



(1)



  0 1  (1)  (2)     0 2 =  P2 + P2  = P 3 (2) P3 P1

(2.43)

(2)

The load is applied at node 3 only, hence P3 = P . Labelling the rows and columns of global stiffness matrix with global degrees of freedom (d.o.f) enables us to identify the elements of the matrix. viii) Global or system equations: System equations for the bar can be written as [K]3×3 {Q}3×1 = {F }3×1

(2.44)

18

Applied Finite Element Analysis

where {Q} is global nodal displacement vector {Q} =



q1

q2

q3

T

The Eq.(2.44) is as follows.  E1 A1 E1 A1 − 0  L1 L1    E1 A1  E1 A1 E2 A2  E2 A2  − + −  L1 L1 L2 L2    E2 A2 E2 A2 0 − L2 L2

(2.45)



      q1 0    q2  =  0    q3 P  

(2.46)

Elements of the global stiffness matrix on the left hand side and elements of the load vector on the right hand side of the above equations are known. Nodal displacements q2 , q3 are unknown. The global stiffness matrix shown on the left hand side of Eq.(2.46) is singular hence the equation should not be solved without imposing known boundary conditions. The singular matrix implies that the body was not constrained. The available boundary condition for stepped bar problem under consideration is q1 = 0. For imposing this boundary condition, the first row of Eq.(2.46) and first column of the global stiffness matrix are to be deleted.      E1 A1 E1 A1 q 0     1 − 0          L L     1 1              E1 A1    E1 A1 E2 A2 E2 A2  0  q2  − + − =  L1 L1 L2 L2                    q 0       3     E2 A2 E2 A2     0 − L2 L2 The above approach of incorporating boundary condition is known as elimination approach. The reduced equations after imposing the boundary condition, we are left with     E1 A1 E2 A2 E2 A2 



+ −   q2 0 L L L 1 2 2   (2.47) =  P E2 A2  q3 E2 A2 − L2 L2 Let the determinant of stiffness matrix shown above be defined as ∆. 

E2 A2 E1 A1 ∴ ∆= L2 L1

One-Dimensional Elasticity Problems

 

  E1 A1 E2 A2 E2 A2 −1   + −  q2 0 L1 L2 L2    = P q3 E2 A2  E2 A2 − L2 L2   E2 A2 E2 A2      L2 L2 1  0 q2     ∴ = q3 E1 A1 E2 A2  P ∆  E2 A2 + L2 L1 L2 



19

(2.48)

(2.49)

Nodal displacements:

  P E2 A2 q2 = ∆ L2   P E1 A1 E2 A2 q3 = + ∆ L1 L2

(2.50)

Strain in the element (1) is 1 = [B] Since,

q1 = 0,





q1 q2

1 =

=

q2 l(1)

1 

l(1)

−1 1





q1 q2

Similarly, strain in element (2) is      q2 1  q3 − q2 q2 2 = [B] = (2) −1 1 = q3 q3 l l(2)



(2.51) (2.52)

(2.53)

Stress in element (1) is



q1 q2





q2 q3



σ1 = [D] [B]

= E1 =

Eq2 L1

(2.54)

= E2 =

E(q3 − q2 ) L2

(2.55)

Similarly, stress in element (2) is σ2 = [D] [B]

The stress and strain in each element do not vary from point to point and are constant in the element. This is due to the linear variation of displacement within the element and derivative of displacement which gives us strain is also constant and hence the stress. Let us assign numerical values to the above problem. E1 = E2 = 200 GPa; A1 = 2A2 = 900 mm2 ; L1 = 2L2 = 90 mm; P = 50 kN

20

Applied Finite Element Analysis

First decision that is to be taken before solving the problem is the unit of length whether it is in metres or cm or mm. Let us choose mm since areas and lengths are expressed in mm. ∴ E1 = E2 = 2 × 105 MPa or N/mm2 , A1 = 900 mm2 , A2 = 450 mm2 L1 = 90 mm, L2 = 45 mm. Element (1)

Stiffness matrix



Load vector





K (1) =

F (1)



=

Element (2)

Stiffness matrix



Load vector



Global Matrices



K (2) =

F (2)



=

E1 A1 L1



1 1 −1

2 1 2    −1 1 1 −1 1 = 2 × 106 1 2 −1 1 2

 (1)   P1  1 

(1)

P2

E2 A2 L2



(2.57)

 2

2 1 −1

3 2 3    −1 2 1 −1 2 6 = 2 × 10 1 3 −1 1 3

 (2)   P2  2 

(2)

P3

(2.56)

(2.58)

(2.59)

 3

Global stiffness matrix size is 3 × 3 since the number of rows and columns are each equal to the product of number of nodes and number of degrees of freedom at each node. No. of rows = (3 nodes)(1 translational degree of freedom at each node) No. of rows =3 Similarly, No. of columns = 3.

Global stiffness matrix

[K]3×3



(1)

K11

  (1) =  K12  0

(1)

K12 (1) (K22

+

(2) K22 )

(2)

K23

0 (2) K23

0

     

(2.60)

We observe that there is a summation of two terms since both elements contribute to node 2 and the node is common for both the elements.

One-Dimensional Elasticity Problems

1 2 3  1 −1 0 1 2 −1  2 [K] = 2 × 106  −1 0 −1 1 3



21

(2.61)

Global load vector or force vector: Tensile load is applied to third node only.

{F } =

          

(1)

P1 (1)

P2

(2)

P3

Global nodal displacement vector Global or system equations:  1 2 × 106  −1 0

(2)

+ P2

          

{Q} =



=

    

    

0

0       50000  

q1

q2

q3

T

[K]{Q} = {F }     −1 0  q1   0  q2 2 −1  0 =     q3 −1 1 50000

(2.62)

(2.63)

(2.64)

Imposing the boundary condition q1 = 0 in the above equation by elimination approach which was already explained.      1 −1 0  q1   0  0 q2 2 × 106  −1 2 −1  = (2.65)     50000 1 q3 0 −1 The reduced equations are       2 −1 q2 0 6 2 × 10 = −1 1 2×2 q3 2×1 50000 2×2

(2.66)

It can be observed from the above that the elimination approach of imposing boundary conditions, the stiffness matrix size is reduced to 2 × 2 from 3 × 3 as only one boundary condition is available in the given stepped bar problem.   2 −1 Solving Eq.(2.66) and using determinant of matrix which is 1. −1 1      1 1 1 0 q2 ∴ = (2.67) q3 50000 2 × 106 × 1 1 2 q2 = 0.025 mm,

q3 = 0.050 mm

(2.68)

Applied Finite Element Analysis

22

Element stresses: From Eq.(2.54) σ1 = From Eq.(2.55) σ2 =

Eq2 = 55.5 MPa L1 E(q3 − q2 ) = 111.0 MPa L2

(2.69)

Reaction R1 : The first row of the left hand side of matrix in Eq.(2.64) should be equated to R1 . −2 × 106 q2 = R1

∴

2.4

R1 = −0.05 × 106 N

TREATMENT OF BOUNDARY CONDITIONS

There are two approaches that can be used for incorporating boundary conditions in global equations. i) Elimination approach ii) Penalty approach.

2.4.1

Elimination Approach

This approach was already explained in the previous section and this method reduces the size of the matrices which will be useful for solving the problem using hand calculation but by writing a computer program, the program should include the reduction of size of matrices depending on the number of boundary conditions available.

2.4.2

Penalty Approach

This method is based on the concept of considering the support as a spring having large stiffness so that the deflection of the spring is very small. A spring of large stiffness C is included and nodal displacements of the spring are a1 and q1 as shown in Fig. 2.7. The spring which is included will contribute to the strain energy 1 term in the functional. πs = C(q1 − a1 )2 (2.70) 2 when differentiated with respect to q1 for the process of minimisation ∂πs = C(q1 − a1 ) ∂q1

(2.71)

Hence, by considering a large stiffness spring to model a support, one term Cq1 should be added to the stiffness matrix and another term Ca1 be added to the load vector corresponding to degree of freedom 1. The choice of the value C is max |Kij |×104 . Normally, one of the diagonal terms in the stiffness matrix will be maximum. If we consider the stepped bar problem discussed in previous sections, the penalty approach applied to model the left fixed support, the following modifications need to be incorporated.

One-Dimensional Elasticity Problems

a1

q2

q1

C

23

1

1 node

2 node

FIGURE 2.7 Spring included to model support.

Element stiffness matrix [K (1) ]:    K11 K12 (K11 + C) is to be modified as K12 K22 K12 Element load vector {F (1) }:     F1 F1 + Ca1 is to be modified as F2 F2

K12 K22



(2.72)

(2.73)

If the support is a fixed support a1 = 0. The advantage that we gain by employing penalty approach instead of elimination approach for the treatment of boundary condition makes the final matrices size will remain same and no specific instruction be included in the computer program. But it will not be any help to us while solving the problem by hand calculation because the size of the matrix is not reduced. Another advantage of penalty approach is that it can incorporate the multipoint constraint conditions into the formulation effectively.

Multipoint Constraints This type of conditions do occur in engineering problems. To name a few i) an inclined roller support ii) a rigid bar connected to more than one elastic body. These conditions contribute to the relation between nodal displacements at the supports, causing the nodal displacements to be interdependent. q2

q′1

θ 1

FIGURE 2.8

q1

24

Applied Finite Element Analysis

3

q2

q1 l1

P

2

1

4 l2

FIGURE 2.9

q1 cos θ + q2 sin θ = q1

(2.74)

q1 (l1 + l2 ) + q2 l1 = 0

(2.75)

Equations (2.74) and (2.75) are constraint equations for above two cases. The general constraint equation for multipoint constraints can be written as Q 1 α 1 + Q2 α 2 = α 0

(2.76)

Thus, the stiffness matrix and load vector for the corresponding nodes to be modified as given below     K11 K12 (K11 + Cα12 ) (K12 + Cα1 α2 ) ⇒ , K21 K22 (K22 + Cα22 ) (K12 + Cα1 α2 )     F1 F11 + Cα0 α1 (2.77) ⇒ F2 F12 + Cα0 α2 Comparing the equations with general equation the constants α1 , α2 and α0 can be calculated and the corresponding element in stiffness and load vector modified. The modified matrices are used for obtaining solution following the procedure laid down in stepped bar problem. To illustrate the application of penalty approach the following problem is considered. E = 200 GPa; P = 50000 N. The bar is of constant cross-section and of the same material. The question arises whether we can take only one element for obtaining the solution. The point load P (concentrated load) is applied at the mid point of the bar. Whenever a point load is applied, necessarily a node has to be fixed. Therefore the bar is discretised with minimum number of two elements. The following data can be identified with the above problem. Discretisation: Number of elements : (2) Type of element : Linear one-dimensional element. Total number of nodes :3

One-Dimensional Elasticity Problems

25

Number of degrees of freedom per node : 1 Total number of degrees of freedom :3 A = 500 mm2 O

X

P 500 mm 1000 mm q1

q2 1

q3 2

FIGURE 2.10 Uniform bar with fixed ends.

Hence the size of global stiffness matrix is 3 × 3. Two boundary conditions are known q1 = 0 and q3 = 0. a) Elimination approach: If the boundary conditions q1 = q3 = 0 are imposed, we are left out with [K]1×1 which facilitates us to get solution easily.

Element (1)

Element (2)

Global stiffness matrix

2×



105

1 1 −1

2  −1 1 1 2

× 500 500 1 2   1 −1 1 5 = 2 × 10 −1 1 2   (1)   P1 F (1) = (1) P2 [K (1) ] =

2 3  1 −1 2 K (2) = 2 × 105 −1 1 3   (2)   P2 F (2) = (2) P3



1 2 3  1 −1 0 1 5  −1 2 −1 2 [K] = 2 × 10 0 −1 1 3



(2.78) (2.79)

(2.80) (2.81)

(2.82)

26

Applied Finite Element Analysis

{F } =

Global force as vector

      

(1)

P1 (1)

P2

(2)

+ P2

(2)

P3

   

 

 0  1 50000 2 =     0 3 

(2.83)

The global force vector indicates that element load vectors need not be written and assembled to get global load vector when only concentrated load exists in the problem. Simply the concentrated load can be included directly at the respective node while writing the global load vector. Element load vectors need not be written. Incorporation of boundary conditions q1 = q3 = 0 by eliminating approach result in      1 −1 0  q1   0  q2 50000 2 × 105  −1 2 −1  =     0 −0 −1 1 q3 2q2 =

50000 = 0.25, 2 × 105

q2 = 0.125 mm

(2.84)

b) Penalty approach: The maximum value of Kij occurs in K22 element of the global stiffness matrix. Therefore, the choice of C = (2 × 105 × 2) × 104

∴ C = 4 × 109 N/mm By applying penalty approach the elements K11 and K33 are to be modified to model the fixed supports by incorporating large stiffness springs. Since the supports are fixed a1 = a 3 = 0   (1 + 2 × 104 ) −1 0 −1 2 −1  (2.85) ∴ Modified [K] = 2 × 105  0 −1 (1 + 2 × 104 ) Since,

a1 = a3 = 0    0  50000 {F } =   0

The global equations are  20001 −1 2 × 105  0

    0  −1 0  q1   50000 2 −1  q2 =     0 −1 20001 q3

Solving for q1 , q2 and q3 , we get q1 = q3 = 6 × 10−6 mm and q2 = 0.125 mm

(2.86)

(2.87)

One-Dimensional Elasticity Problems

27

The values of q1 , and q3 are so small that they can be taken as zero satisfying the boundary conditions. The value q2 = 0.125 mm which tallys with the solution obtained by elimination approach. The problem looks to be simple when we have three degrees of freedom but problem tends to be formidable as the number of degrees of freedom increases to a large number when more number of elements are necessary for a particular problem. Without the aid of computer, the finite element analysis settles down to be a formidable task to obtain solution to a large sized problem. Definitely, one should practise to write a computer program in any language in which he/she is familiar with.

Analysis of a tapered bar Real life problems are not restricted to bars having uniform cross-section but variable crosssections do exist. Let us consider a bar with linear variation of area with respect to spatial coordinate x. A(x) = a1 + a2 x

(2.88)

We have already learnt that linear variation of field variable could be expressed using shape functions Ni and Nj . Same shape functions can also be used to specify the variation of area.

∴ A(x) = Ni Ai + Nj Aj

(2.89)

where Ai is cross-sectional area at node i Aj is cross-sectional area at node j and inconsistent with earlier expression Ni =

xj − x ; l(e)

Nj =

x − xi l(e)

Modification needed for the evaluation of stiffness matrix is [K

(e)

]=



E (e) [B] [D][B]dx = (e) 2 (l ) T

V

Ai

xj

xi

Hence,

xi

Ai l(e) ; Ni dx = 2

[K (e) ] =

xj 

−1 1

1 −1 Aj

xj



(Ni Ai + Nj Aj )dx

Nj dx =

xi

E (e) (Ai + Aj ) 2 l(e)



1 −1

−1 1

Aj l(e) 2



(2.90)

(2.91)

(2.92)

Applied Finite Element Analysis

28

(e)

Defining the mean area

Am =

Ai + Aj 2 (e)

[K

(e)

E (e) Am ]= l(e)



1 −1

−1 1



(2.93)

If linear variation of area exists in a problem, Eq.(2.93) shows that mean area is to be considered.

Formulation of body force term in element load vector So far, load vector due to point load or concentrated load P is considered and formulation of element load vector is straightforward and simple. If a body force like the weight of the body acting on each particle of body is to be considered, the body force contributes to the load vector. Let f be the body force, its components expressed as force per unit volume, the contribution to element load vector is   xj  Ni T e {f }1×1 Adx {F }due to body force = {N } {f }1×1 dv = Nj v



F

(e)



=f

(e)

(e)

A

xj 

xi

Ni Nj

xi

     f A(e) l(e) 1 i 1/2 (e) (e) (e) dx = f A l = 1 j 1/2 2

(2.94)

Equation (2.94) indicates that the total body force acting on the body should be distributed to the nodes i and j equally, the magnitude being half of the total body force. Let us consider a plate shown in Fig. 2.11 which is acted upon by a concentrated load P as well as body force due to its weight. L1 O 0.5 m

P

0.5 m

L2

X FIGURE 2.11 Tapered plate.

One-Dimensional Elasticity Problems

29

E = 80 GPa L1 =1m = 0.5 m L2 Thickness = 10 mm ρg = 7.85 × 104 N/m3 P = 100 N Discretisation: The tapered plate has been replaced by equivalent stepped bar shown in Fig. 2.12 with mean areas A1 and A2 . O

q1

1

0.5 m

A1

2 2

q2

0.5 m A2

P

3

X

q3

FIGURE 2.12 Equivalent system.

Area at node 1 = 0.01 m2 Area at node 2 = 0.0075 m2 Area at node 3 = 0.005 m2 0.01 + 0.0075 = 8.75 × 10−3 m2 Mean area A1 = 2 0.0075 + 0.005 = 6.25 × 10−3 m2 Mean area A2 = 2 Replacing the tapered plate with a stepped bar with mean area A1 and A2 , the procedure laid down for a stepped bar can be followed. No. of elements : (2) Type of element : One-dimensional linear element No. of degrees of freedom at each node : 1 (translational) Total degrees of freedom for the bar :3 Element (1) Stiffness matrix

80 ×

109

10−3



1 2  1 −1 1 N/m −1 2 2

× 8.75 × 0.5 1 2   1 −1 1 N/m = 1.4 × 109 −1 1 2

[K (1) ] =

(2.95)

30

Applied Finite Element Analysis

Load vector due to body force (1) FB.F



F (1)

= 171.7





due to body force



1 1

f (1) A(1) L(1) = 2



1 1



1 newtons 2

1 newtons 2 (2.96)

Element (2) Stiffness matrix

[K (2) ] =

Load vector due to body force



80 ×

109

× 6.25 × 0.5

10−3



2 3  1 −1 2 −1 1 3

2 3  1 −1 2 −1 1 3   f (2) A(2) L(2) 24.5 = = 24.5 2

   (2) 9 K = 1 × 10 F (2)



B.F

(2.97)

(2.98) (2.99)

Global stiffness matrix [K]3×3 : 1 1.4 [K] = 1 × 109  −1.4 0



2 −1.4 2.4 −1

3  0 1  −1 2 1 3

(2.100)

Global load vector {F} has two terms, i) due to body ii) due to point hold {F } = {F }BF + {F }P L       171.7   0   171.7  1  (171.7 + 24.5) 100 296.2 2 = {F } = +       24.5 0 24.5 3

Boundary condition q1 = 0 Incorporating boundary condition by elimination approach 1 2 3       1.4 −1.4 0 1  q1   171.7  1 × 10  296.2 −1.4 2.4 −1  2 q2 =     24.5 3 q3 1 0 −1      2.4 −1 q2 296.2 9 = 1 × 10 −1 1 24.5 q3

q2 = 0.23 × 10−6 m, q3 = 0.25 × 10−6 m

(2.101)

(2.102)

(2.103) (2.104)

One-Dimensional Elasticity Problems

2.5

31

FORMULATION OF LOAD VECTOR DUE TO UNIFORM TEMPERATURE CHANGE

Let the body be subjected to uniform temperature ∆T . Considering the strain due to uniform temperature change ∆T as causing initial strain 0 in the body 0 = α(∆T )

(2.105)

where α is coefficient of thermal expansion. Let  be total strain due to mechanical as well as thermal strain. Hence, strain due to exclusively mechanical loading is related to stress σ by Hooke’s law

∴ σ = D( − 0 )  1 U = ( − 0 )T D( − 0 )dv 2 v

The strain energy term

(2.106) (2.107)

While minimising the strain term, the initial strain contributes to load vector without affecting the stiffness matrix.    (e) F = [B]T [D]0 dv (2.108) Thermal load

V

The expression for thermal load vector is similar to integral expression for stiffness matrix Eq.(2.22) except 0 term replaces B term. The element load vector considering the thermal strain can be written as      F (e) = [N ]T {f } dv + [N ]T {T } ds + {P } + [B]T [D] [0 ] dv (2.109)  V    S   Point load V  



V

Body force

[B]T [D] 0 dv = EA0

Surface traction



−1 1



Thermal load

for one-dimensional linear element

(2.110)

Let us consider the stepped bar problem with the following data which include thermal load. A2 = 450 mm2 A1 = 900 mm2 ; L1 = L2 = L3 = 45 mm; E = 2 × 105 MPa α = 0.00001 per ◦ C; P = 50 kN ∆T = 50 ◦ C Discretisation: Number of elements = (3) Type of element = One-dimensional linear element Total no. of nodes =4

32

Applied Finite Element Analysis

Total no. of degrees of freedom = 4 (Translational) The size of global stiffness matrix [K] is 4 × 4. A1

A2

P

O

45 mm

45 mm

1 node 1

x

3

2 2

q1

45 mm

node 4 q4

3 q2

q3

FIGURE 2.13 Stepped bar.

Element matrices Element (1)



 K (1) = 4 × 106



1 1 −1

2  −1 1 1 2

(2.111)

Load vector due to temperature change ∆T , using Eq.(2.110)     −2 1 F (1) = 4.5 × 104 2 2 (∆T ) Element (2)

 Element (3)

(2.112)

2 3  1 −1 2 −1 1 3   −2 2 = 4.5 × 104 2 3

   (2) 6 K = 4 × 10 F (2)



(∆T )

(2.113) (2.114)

3 4  1 −1 3 −1 1 4     −1 3 F (3) = 4.5 × 104 1 4

   (3) 6 K = 2 × 10

(2.115) (2.116)

Global or system equations:        0 −2  2 −2 0 0  q1                4   q 0 −2 4 −2 0 −5 × 10 2 4 6  = 4.5 × 10 + 2 × 10  1 0 −2 3 −1   q3  0                1 0 0 −1 1 q4 0       Temperature change

Point load

(2.117)

One-Dimensional Elasticity Problems

33

Boundary conditions q1 = q4 = 0 Eliminating corresponding rows and columns by elimination approach for imposing boundary conditions      4 −2 q2 −5 = 104 2 × 106 −2 3 4.5 q3

∴ q2 = (−)0.00375 mm q3 = (+)0.005 mm.

(2.118)

Shape functions in local and global coordinates Coordinate system defined for the whole body or structure is called global coordinate system. Local coordinate system is defined locally with respect to the element. Coordinates of any point on the element can be defined in local and global coordinates. Normalised local coordinate of a point is natural coordinate which is non-dimensional. Formulations using natural coordinates simplify the computations especially with higher order elements but it does not make any significant impact in one-dimensional problems. Having gained insight into the formulations with global coordinates, natural coordinates are being introduced to have a smooth transition. xj x xi

ζ A

O

i (xi+xj)/2

l(e) O′ l(e) 2 2

j

FIGURE 2.14 Local coordinate system.

‘O’  - origin of global coordinate system. xi - nodal coordinates in global coordinate system. xj x - global coordinate of point A. - origin of local coordinate system defined at the midpoint of element. ‘O ’ ζ - local coordinate of point A. Relation between local and global coordinates of point A: x=

(xi + xj ) +ζ 2

(2.119)

34

Applied Finite Element Analysis

Dividing Eq.(2.119) by

l(e) to normalise the equation 2     (xi + xj ) x 2 2 = + ζ (e) l(e) 2 l(e) l 2

(2.120)

Defining the natural coordinate ξ=



2ζ l(e)



(2.121)

Equation (2.120) can be written as (xi + xj ) 2x = + ξ (e) l l(e)

(2.122)

Equation (2.122) gives the relation between natural coordinate ξ and global coordinate x of any point A shown in Fig. 2.14. Differentiating the Eq.(2.123) with respect to x 2 dξ = (e) dx l   (e)   2 l dξ = dx or dx = dξ (e) 2 l

(2.123) (2.124)

The expressions shown in Eq.(2.123) and Eq.(2.124) are very important and these will be used in different calculations in ensuing chapters. Natural coordinates at node i and j for one-dimensional linear element shown in Fig. 2.15 are evaluated as given below. ξ node i

l(e) 2

O′

A

l(e) 2

node j

FIGURE 2.15 Natural coordinate system.

From Eq.(2.122) ξ=

(xi + xj ) 2x − (e) l l(e)

Value of ξ at node i: Node i is located at x = xi . Substituting in Eq.(2.125) xi for x, ξ = −1

(2.125)

One-Dimensional Elasticity Problems

35

Value of ξ at node j: Node j is located at x = xj . Substituting in Eq.(2.125) xj for x, ξ = +1. Figure 2.16 shows the values of natural coordinate at node i and node j. Shape functions associated with node i and node j can be derived using the properties given in Table 2.1. TABLE 2.1 Values of shape functions

Node

Ni

Nj

i j

1 0

0 1

At node i, Ni = 1 and Nj = 0. Coordinate of node j expressed in natural coordinate ξ is 1. Hence (1 − ξ) at node j will be zero. Therefore in the expression for node Ni by ξ ξ = –1 ξ = +1 including the term (1 − ξ), the value of Ni can be i node node j A made equal to zero. The order of shape function O′ (1 + ξ) = 0 (1 – ξ) = 0 is first order in x similar to the linear displacement function u chosen. It indicates by logic that FIGURE 2.16 Natural coordinates. Ni = Ci (1 − ξ) can be written where Ci is a constant to be evaluated by the second property that Ni = 1 at node i that is at ξ = −1. By substituting of ξ = −1 in the expression for Ni 1 = Ci (2);

∴ Ci = 1/2

1−ξ . 2 Similar arguments that Nj = 0 at node i (at ξ = −1) and Nj = 1 at node j (at ξ = +1), 1+ξ . the expression for Nj can easily be written as Nj = 2 The expressions for Ni , Nj expressed in natural coordinate ξ are: Now, the expression for Ni =

(1 − ξ) 2 (1 + ξ) Nj = 2 Ni =

(2.126)

It can be seen that shape functions can be easily written by inspection. Another advantage is the limits of integration, when ξ is included, limits will be −1 and +1 making evaluation simpler. The integration of shape functions occurs in the evaluation of load vector terms for body force and surface traction.

Applied Finite Element Analysis

36

Evaluation of load vector for body force: {F

(e)

}=



T

[N ] {f }dV = f

V

From Eq.(2.124) dx =

∴



[N ]T {f }dV = F (e)



=

(e)

A

 

Ni Nj

 dx

(2.127)

l(e) · dξ 2

 +1 (1 − ξ)  (e) (e) (e) f A (l )  ξ − dξ =  ξ+ 4 (1 + ξ) −1     f (e) A(e) l(e) 2 1 i = 2 1 j 2

f (e) A(e) (l(e) ) 4

V



(e)

f (e) A(e) l(e) 4

ξ2 2 ξ2 2

1  

−1

(2.128)

Evaluation of load vector due to surface traction: i) Constant surface traction T (Force/unit area) acting on the element is shown in Fig. 2.17. T j

i FIGURE 2.17 Constant traction force



F

(e)



=T

F (e) =

 

T δl(e) 2

Ni Nj



1 1

T.

  +1 (1 − ξ) T δl(e) dξ δdx = 4 (1 + ξ) 

−1

i j

(2.129)

where δ is the perimeter of the surface. ii) Linear variation of T shown in Fig. 2.18 can be expressed as

Tj

i FIGURE 2.18 Linear variation of

j

T.

One-Dimensional Elasticity Problems

37

T = Nj Tj      δl(e) Ni (e) dξ F = (Nj Tj ) Nj 2

Load vector

Tj δl(e) = 2

+1

−1

1 Ni Nj dξ = 4

+1

Nj2 dξ

−1



F

(e)



1 = 4

+1

−1

+1

−1

Ni Nj Nj2



dξ

 1 ξ3 1 ξ− = 3 3 −1

1 (1 − ξ )dξ = 4 2

 +1 +1 1 ξ3 2 2 2 (1 + 2ξ + ξ )dξ = ξ+ξ + = 4 3 −1 3

−1

T δl(e) = 6



1 2



i j

(2.130)

When traction force varies linearly from 0 to T the distribution of traction force to i and j is not same but as indicated in Eq.(2.130). Formulation of matrices using quadratic element. Field variable variation is expressed by second order polynomial. ξ=–1 node i

ξ=0 j O′

ξ=+1 k

FIGURE 2.19 Quadratic element.

We use natural coordinates. u = a1 + a2 ξ + a3 ξ 2 Shape functions can be written by inspection.   −ξ(1 − ξ) Ni = 0 at j and k nodes Ni = Ni = 1 at ith node 2   Nj = 0 at i and k nodes 2 Nj = (1 − ξ ) Nj = 1 at jth node   ξ(1 + ξ) Nk = 0 at i and j nodes Nk = Nk = 1 at kth node 2

(2.131)

(2.132)

38

Applied Finite Element Analysis

Writing shape functions in natural coordinates are so simple. u= Strain can be written as



Ni

Nj

Nk



1×3

{q}3×1 = [N ]1×3 {q}3×1

(2.133)

  dNi dNj dNk dξ du 2 du = = (e) {q} = dξ dξ dξ dx dx dξ l dNj 2ξ − 1 dNk 2ξ + 1 dNi = ; = −2ξ; = dξ 2 dξ dξ 2  1   = (e) (2ξ − 1) −4ξ (2ξ + 1) 1×3 {q}3×1 l

(2.134) (2.135) (2.136)

Defining strain-displacement matrix as [B] [B]1×3 =

1 [ (2ξ − 1) l(e)

−4ξ

(2ξ + 1) ]

(2.137)

[B] matrix is not constant since linear terms of natural coordinate are present. Hence strain and stress within the element varies linearly. {}1×1 = [B]1×3 {q}3×1

(2.138)

σ = [D]1×1 {}1×1 = [D]1×1 [B]1×3 {q}3×1

Stress

[D] = E for one-dimensional problem σ = E [B] {q}

(2.139)

Evaluation of stiffness matrix :

[K] =



E (e) A(e) l(e) [B] [D] [B] (Adx) = 2 T

V

[B]T [B] =

=

1 (l(e) )2  1   l(e)2 

 



(2ξ − 1)  −4ξ  (2ξ − 1) (2ξ + 1) (2ξ − 1)2

symmetric

4ξ(1 − 2ξ) 16ξ 2

+1 [B]T [B]dξ

−1

−4ξ

(2ξ + 1)

(4ξ 2 − 1)



 −4ξ(2ξ + 1)   2 (2ξ + 1)



One-Dimensional Elasticity Problems

E (e) A(e) l(e) 2

+1

[B]T [D] [B] dξ =

−1

E (e) A(e) 3l(e)

39

1 2 3  7 −8 1 1  −8 16 −8  2 1 −8 7 3



Stiffness matrix for quadratic element is



K (e)



3×3

=

E (e) A(e) 3l(e)

1 7  −8 1



2 3  −8 1 1  16 −8 2 −8 7 3

(2.140)

Element load vector due to body force f (force per unit volume)



F (e)



BF

=



V

  Ni  +1     f (e) A(e) l(e)  T Nj [N ] f dV = dξ   2     −1 Nk

 3 +1 +1 2 ξ − (ξ − ξ) dξ = 3 Ni dξ = 2 2

−1

−1

−1

−1

ξ2 2

+1

=

−1

1 3

+1 +1 +1 ξ3 4 2 Nj dξ = (1 − ξ )dξ = ξ − = 3 −1 3

+1

Nk dξ =

−1



F (e)

+1

−1



BF

ξ2 + ξ 2



dξ =



ξ3 3

+ 2

ξ2 2



=

1 3

  i 1 f (e) A(e) l(e)   j 4 =   6 k 1

(2.141)

Uneven distribution of body force at the nodes is clearly seen. One-sixth of total body force is assigned at i and k nodes and two-thirds at internal node j.

2.6

ANALYSIS OF ROTATING LINK

A link of 2 m is rotating at angular velocity 5 rad/sec shown in Fig. 2.20.

40

Applied Finite Element Analysis

E = 200 GPa l =2m A = 2 × 10−3 m2 ρ = 7850 kg/m3 The problem can be reduced to static analysis by applying centrifugal force using D’Alemberts principle.

l w 30°

FIGURE 2.20 Rotating link.

i) Analysis with 2 linear elements:

1

m

3

1

m

2

node 1

30°

FIGURE 2.21 Discretisation two linear elements.

Element (1)

1 2  1 −1 1 (1) 8 K = 4 × 10 N/m −1 1 2     f (1) A(1) l(1) l(1) 1 1 (1) = 0.5 m F = r¯1 = 1 2 2 2







m¯ r1 w 2 = ρ¯ r1 w2 = 7850 × 0.5 × 25 = 98,125 N/m3 volume       196.25 1 1 1 (1) F = = 98.125 newtons 1 1 2 2 f (1) =

∴

Element (2)    K (2) = 4 × 108

2 3  1 −1 2 N/m −1 1 3     f (2) A(2) l(2) 1 (2) F = r¯2 = 1.5 metres 1 2

ρ¯ r2 w 2 m¯ r2 w2 = = 2, 94, 375 N/m3 Volume Volume     1 2 F (2) = 294.375 newtons 1 3 f (2) =

One-Dimensional Elasticity Problems

Global or system equations:

      1 −1 0 98.125    98.125   q1  q2 2 −1  98.125 + 294.375 392.5 = = 4 × 108  −1       0 −1 1 294.375 294.375 q3 

Boundary condition q1 = 0 Using elimination approach 8

4 × 10



2 −1

−1 1



q2 q3



=



392.5 294.375



q2 = 171.71 × 10−8 m

q3 = 245.31 × 10−8 m. ii) Analysis with 2 quadratic elements: 5 4 3

2

2 1 node 1 FIGURE 2.22 Discretisation with two quadratic elements.

Element (1) 2  1 4 7 −8 8 [K ] = × 10  −8 16 3 1 −8   1  f (1) A(1) l(1)  (1) 4 F =   6 1 (1)

3 1 1 −8  2 7 3

N/m

1 2 3

f (1) = ρr1 w2 = 7850 × 0.5 × 25 = 98, 125 N/m2     1   1  1  196.25 (1) 4 2 newtons 4 = 32.7 {F } =   6  1  1 3

41

42

Applied Finite Element Analysis

Element (2) 4  3 4 7 −8 [K (2) ] = × 108  −8 16 3 1 −8   1 f (2) A(2) l(2)   (2) 4 {F } =   6 1

5  1 3 −8  4 7 5

N/m

3 4 N/m3 5

f (2) = 7850 × 1.5 × 25 = 2, 94, 375     1    1  3 588.75 (2) 4 4 4 newtons {F } = = 98.125   6  1  1 5

Global or system equations:  7 −8 1 0  −8 16 −8 0  4 × 108   1 −8 14 −8 3  0 0 −8 16 0 0 1 −8

0 0 1 −8 7

           

q1 q2 q3 q4 q5

          

 32.7      130.8 130.825 =   392.5    98.125

Boundary condition q1 = 0. Imposing boundary condition by elimination approach      16 −8 0 0 q2  130.8             4 q 14 −8 1  130.825 3 8  −8  × 10  = q4  0 −8 16 −8    392.5  3         q5 0 1 −8 7 98.125

          

Solving the above equation, the following values are obtained. q1 = 0,

q2 = 91.9 × 10−8 m

q3 = 171.69 × 10−8 m q4 = 226.88 × 10−8 m q5 = 245.28 × 10−8 m

The calculations should be carried out with more number of elements till the convergence of the solution was established. Finite element method is a numerical method which gives approximate solution and it gives an acceptable solution by achieving convergence with more number of elements.

One-Dimensional Elasticity Problems

2.7

43

LENGTH COORDINATES FOR ONE-DIMENSIONAL LINEAR ELEMENTS

Length coordinates are non-dimensional coordinates denoted by letter ‘L’ which are ratios of lengths. Length coordinates associated with node i is defined as ξ node i ξ=–1

O′

A

node j ξ=1

FIGURE 2.23 Natural coordinates.

Li =

1−ξ Length Aj = = Ni Length ij 2

(2.142)

Length of the element varies from ξ = −1 to ξ = 1. ∴ Length ij =2 Similarly length coordinate associated with node j 1+ξ Length iA = = Nj Length ij 2 Li = Ni and Lj = Nj

Lj = It can be seen that

(2.143) (2.144)

In defining length coordinates, A is the reference point and corresponding lengths are measured from A to the respective nodes. In terms of length coordinates, evaluation of integrals are easier as given below  a! b! (Γ) (2.145) Lai Lbj dΓ = (a + b + 1)! Γ

While evaluating load vectors, integral evaluation because essential.   l 1! (l) = (∵ a = 1) ∴ Ni dl = Li dl = (1 + 1)! 2   l 2! 2 (l) = (∵ a = 2) Similarly, Ni dl = L2i dl = (2 + 1)! 3   1!1!(l) Ni Nj dl = Li Lj dl = (1 + 1 + 1)! l l = = (∵ a = 1, b = 1) 3! 6 The above integrals are evaluated by comparing with Eq. (2.145).

(2.146) (2.147) (2.148)

Applied Finite Element Analysis

44

Example 2.1 Find the nodal displacements, displacement at 1200 mm and 3200 mm from the left hand support and element stresses for the stepped bar shown in figure. 600 mm2

400 mm2 2

1 O 1

2

3

q1

q2 1

q3 2

2 qi

i

X

1600 mm

2400 mm

1

P = 40 kN

3 qj

u A

j

FIGURE 2.24 Discretisation.

E = 200 GPa Discretisation: No. of elements Type of element Total no. of nodes Total d.o.f u(x) = [N ]{q} 

= [B]{q}

σ = [D][B]{q}

[K (e) ] = Element (1)



K

 (1)



: (2) : One-dimensional linear element. = 3, No. of nodes per element : 2 = 3, No. of degrees of freedom per element :1 (Translational) [N ] - Shape function matrix Ni  Nj  1 1 [B] - Strain-displacement matrix - − (e) (e) l I [D] - Material property matrix - [E]|1×1

E (e) A(e) [B]T [D] [B] dV = = l(e)

200 × 103 × 600 = 2400



i 1 −1

j    −1 i  (e)  pi = ; F pj 1 j



1 2   1 1 −1 1 1 = 5 × 104 −1 1 2 −1



2 1 −1

2 −1 1 1 2

Element (2)



K

 (2)

=

103

× 400 200 × 1600

3  3   2 −1 2 1 −1 2 = 5 × 104 1 3 −1 1 3

One-Dimensional Elasticity Problems

Global stiffness matrix : 

1 4  −1 [K] = 5 × 10 0

 

 0  0 {F } = ,   40000

 −1 0 2 −1  , −1 1

Boundary condition: q1 = 0

5 × 104



2 −1

−1 1



q2 q3





=



1 5 × 104 × 1

= q2 = 0.8 mm

0 40000



q2 q3



1 1

1 2

   q1  q2 {Q} =   q3



0 40000



q3 = 1.6 mm u|x=1200 = 0.4 mm u|x=3200 = 1.2 mm 200 × 103 [0.8] = 66 MPa 2400 200 × 103 × 0.8 = 100 MPa = 1600

σ (1) = σ (2)

Example 2.2 Find the nodal displacements and element stresses for the stepped bar shown in Fig. 2.25. O 600 mm2

1 1

2400 mm

2 2

400 mm2

1600 mm

3 P = 75 N X FIGURE 2.25 Stepped bar.

45

46

Applied Finite Element Analysis

ρ = 7850 kg/m3 ; g ≈ 10 m/sec2 ; E = 200 GPa No. of elements : (2) Type of element = One-dimensional linear element. Total no. of d.o.f = 3 Element (1) 2  −1 1 1 2   78500 × 600 × 2400 1 1 = 9 1 2 2 × 10   56.52 1 = 56.52 2



[K (1) ] = 5 × 104 {F (1) }due to BF {F (1) }due to BF

1 1 −1

Element (2) 3  −1 2 1 3     78500 × 400 × 1600 1 1 = = 25.12 9 1 1 2 × 10

[K (2) ] = 5 × 104 {F (2) }due to BF



2 1 −1

Global stiffness matrix :

     5 −5 0 56.52  q1    q2 56.52 + 25.12 [K] = 104  −5 10 −5  , {Q} = , {F } =     0 −5 5 25.12 + 75 q3 

Boundary condition q1 = 0 4

5 × 10 q2 = 2.2 mm



2 −1

−1 1



q2 q3



=



81.64 100.12



q3 = 3.5 mm

σ (1) =

200 × 103 200 × 103 × 2.2 MPa σ (2) = × 1.3 MPa 2400 1600

σ (1) = 183.33 MPa

σ (2) = 162.5 MPa

One-Dimensional Elasticity Problems

47

Example 2.3 Find the nodal displacements and element stresses for the tapered plate shown in Fig. 2.26.

A(1) =

(70 + 50) 10 = 600 mm2 ; 2

A(2) =

(50 + 30) 10 = 400 mm2 2

Thickness of the plate: 10 mm ρg = 78500 N/m3 ; E = 100 GPa Element (1)

[K

(1)

1 100 × 103 × 600  1 ]= 1000 −1

2  −1 1 1 2

N/mm

1 2  1 −1 1 −1 1 2   1 = 23.55 newtons 1

[K (1) ] = 6 × 104 {F (1) }BF



N/mm

70 mm O

1 q1 1

1

A(1)

P1 2

2

A(2)

2

1000

q2 3 q3

P1 = 60 N

1000

P2 = 150 N

P2 X

30 mm FIGURE 2.26 Equivalent system for a tapered plate.

Element (2) [K

(2)

4

] = 4 × 10



2 3  1 −1 2 −1 1 3

,

{F

(1)

}BF = 15.7



1 1



2 3

Applied Finite Element Analysis

48

Global equations:

      6 −6 0 23.55    0   q1  23.55 + 15.70 − 60 q2 = + 104  −6 (6 + 4) −4        0 −4 4 15.70 + 150 q3 

Boundary condition q1 = 0 4

10



10 −4 −4 4





q2 q3 q2 q3

 

= =

q2 = q3 =



 − 20.75 165.70    1 4 4 − 20.75 4 10 165.70 104 × 24 579.8 662.8 − 83 = 4 = 0.0025 mm 4 10 × 24 10 × 24 1574 1657 − 83 = 4 = 0.0060 mm 4 10 × 24 10 × 24

Example 2.4 Find the nodal displacements and element stresses for the stepped bar shown in Fig. 2.27, when the bar is heated by 50 ◦C. P = 50000 N; E = 70 GPa; α = 20 × 10−6 per ◦ C ∆t = 50 ◦ C Discretisation: No. of elements : (2) Type of element : One-dimensional linear element No. of degrees of freedom per element : 2 (translational) Total no. of nodes :3 Total no. of degrees of freedom :3 800 mm2

600 mm2

P

O

X

0.5 m

q1 1

0.5 m

q2

1

q3

2

2 FIGURE 2.27 Elements.

3

Element connectivity

Element(e)

i

j

(1) (2)

1 2

2 3

One-Dimensional Elasticity Problems

u = [N ]{q},   qi q= qj E = [B]{q} σ = [D][B]{q}

1 l(e)



i E (e) A(e)  1 l(e) −1 V   −1 i = E (e) A(e) (α∆T ) 1 j

[K (e) ] = {F (e) }due to ∆T



1−ξ 2 1+ξ Nj = 2  1 [B] = − (e) l

Ni =

[B]T [D] [B] dV =

j  −1 i 1 j

Element (1) 1 2  2    1 70 × 109 800  1 −1 1 1 −1 1 × 6 K (1) = = 112 × 106 0.5 10 −1 1 2 −1 1 2      −1 800  F (1) = 70 × 109 × 6 × 20 × 10−6 × 50 1 10 due to ∆T   −56 1 = 103 56 2 Element (2) 2 3  3     2 70 × 109 600  1 −1 2 1 −1 2 × 6 K (2) = = 84 × 106 0.5 10 −1 1 3 −1 1 3         −1 −1 2 F (2) = 70 × 109 × 6 × 10−4 × 20 × 10−6 × 50 = 42 × 103 1 1 3 Global or system equations:

      −56  112 −112 0  q1   0   q2 50 56 − 42 + 103 = 103 106  −112 (112 + 84) −84        q3 0 42 0 −84 84 

49

50

Applied Finite Element Analysis

Boundary conditions q1 = 0,

q3 = 0

64 0.33 ≈ m ≈ 0.33 mm 196 × 103 103     q2 0.33 70 × 109 70 × 109 −3 − 10 = − α∆T = MPa 106 0.5 × 103 106 103 × 0.5

q2 = σ (1)

σ (1) = −23.8 MPa   −q2 70 × 109 −1.33 × 70 × 106 (2) −3 σ = − 10 = = −93.1 MPa 106 0.5 × 103 106

Example 2.5 Fig. 2.28.

Find the nodal displacements and element stresses for the bar shown in

A = 24 cm2

A = 15 cm2 10 kN 75 cm q1

20 kN

75 cm q2

1

1

2

2

60 cm q3 3

qi

q4

3 4 u j

x xj FIGURE 2.28 Discretisation of bar.

E = 200 GPa; α = 11 × 10−6 per ◦ C; δT = 10 ◦ C Discretisation: No. of elements : (3) Type of elements : One-dimensional linear element No. of d.o.f at each node : 1(Translational) Total. no. of nodes :4 Total d.o.f :4 u = [N ]{q}  = [B]{q} σ = [D][B]{q}

One-Dimensional Elasticity Problems

j    E (e) A(e)  i 1 −1 i K (e) = l(e) −1 1 j Element (1)

  2     1 20 × 106 × 15 1 −1 6 (1) 1 −1 1 K = = 4 × 10 −1 1 75 −1 1 2     −1 1 F (1) = 20 × 106 × 15 × (11 × 10−6 × 10) 1 2 due to δT   1 −1 newtons = 33 × 103 2 1 Element (2)



F (2)



3  −1 2 N/cm 1 3   −1 2 3 = 33 × 10 newtons 1 3

   K (2) = 4 × 106 due to δT

Element (3)    20 × 106 × 24 1 (3) K = −1 60



F

(3)



−1 1



= (20 × 10 ) × 24 × (11 × 10 = 52.8 × 10



−1 1



2 1 −1

= 8 × 106

−6

6

3

× 10)



 −1 1

3 1 −1 

3 newtons 4

Global stiffness matrix : 1 2 3 4 4 −4 0 0 8 −4 0 [K] = 106   −4  0 −4 12 −8 0 0 −8 8



N/cm



1  2   3 N/cm 4

4  −1 3 1 4

N/cm

51

52

Applied Finite Element Analysis

Global force vector :

   

−33 33 − 33 + 10 {F } = 103 33 − 52.8 − 20    52.8

   

   

−33 10 = 103 −39.8       52.8

 1    2 newtons 3    4

Boundary conditions: q1 = q4 = 0 Reduced global or system equation after applying boundary conditions. 6

10



8 −4

−4 12



 10 = 10 −39.8      1 q2 12 4 10 = 3 q3 4 8 −39.8 10 × 80 q2 q3



3



q2 = −0.49 × 10−3 cm q3 = −3.48 × 10−3 cm

σ

(1)

σ (1)

 q2 −5 = E ([B] {q} − 0 ) = 20 × 10 − 11 × 10 l1   −3 6 −0.49 × 10 −5 − 11 × 10 = 20 × 10 N/cm2 75 6



σ (1) = −23.3 MPa   (2) 6 (q3 − q2 ) −5 σ = 20 × 10 − 11 × 10 l2   −3 (2) 6 (−3.48 + 0.49) × 10 −5 − 11 × 10 σ = 20 × 10 N/cm2 75 σ (2) = −29.96 MPa   −3 (3) 6 3.48 × 10 −5 − 11 × 10 σ = 20 × 10 N/cm2 60 σ (3) = −10.4 MPa (N/mm2 )

Example 2.6 Determine the nodal displacements for the bar shown in Fig. 2.29 where surface traction is applied.

One-Dimensional Elasticity Problems

6 × 104 N/M T

0.75 m

0.75 m E = 200 GPa

q1 O

1

A = 5 × 10–4 m2 q2 2

q3 3 X

2

1

FIGURE 2.29 Surface traction variation.

Element (1) [K

(1)

200 × 109 × 5 × 10−4 ]= 0.75



1 −1

−1 1



2   1 4 8 1 −1 1 [K ] = × 10 3 −1 1 2         Tl 1 Ni Nj 1 (1) T F = [N ] T Nj dl = T dl = Nj2 2 2 6       6 × 104 × 0.75 1 3 × 104 1 1 1 F (1) = = 2 2 2 2 6 4 (1)

Element (2) 3   2 4 1 −1 2 × 108 3 −1 1 3         Tl 1 6 × 104 × 0.75 9 × 104 1 1 (2) F = = = 1 1 2 2 2 4 [K (2) ] =

Global stiffness matrix :

2 3  1 4 1 −1 0 1 [K] = × 108   −1 2 −1 2 3 0 −1 1 3

53

54

Applied Finite Element Analysis

Global force vector :

 3  1 6+9 2 {F } = 4  18  3 104

 

   q1  q2 Global equations: [K]{Q} = {F }, where {Q} =   q3 Boundary conditions: q1 = q3 = 0 Imposing boundary conditions in Global equations, we get 4 104 × 108 (2q2 ) = (15) 3 4 1 45 1 3 104 q2 = × × 15 = × × 4 2 × 108 4 32 104 q2 = 0.133 mm

Example 2.7 Determine the nodal displacements for two rails separated by a gap of 0.5 mm shown in Fig. 2.30. P = 100 kN,

E = 80 × 103 MPa 300 mm2 P

250 mm

250 mm

0.5 mm

2

1

O q1

q3

q2

X

FIGURE 2.30 Right hand support at a distance.

Element (1) [K (1) ] =

80 × 103 × 300 250



1 −1

−1 1



One-Dimensional Elasticity Problems

[K (1) ] = 96 × 103



1 1 −1

2  −1 1 N/mm 1 2

Element (2) [K Global equations:



(2)

] = 96 × 10

1 96 × 103  −1 0

−1 2 −1

3



2 1 −1

3  −1 2 N/mm 1 3

    0 0  q1    −1  q2 100 × 103 =     1 q3 0

Boundary condition: q1 = 0, solving global equations      2 −1 q2 100 3 3 96 × 10 = 10 q3 −1 1 0      1 q2 1 1 100 = q3 0 96 1 2 q2 = 1.042 mm q3 = 1.042 mm But q3 cannot be more than 0.5 mm hence penalty approach is used.      0 (1 + 2 × 104 ) −1 0  q1    q2 −1 2 −1  100 × 103 = 96 × 103      q3 0.5 × 2 × 96 × 107 0 −1 (1 + 2 × 104 )       0 20001 −1 0  q1    q2 −1 2 −1  100 × 103 96 × 103  =     q3 0 −1 20001 96 × 107      0 20001 −1 0  q1    1  100 q2 −1 2 −1  =   96  96 × 104  0 −1 20001 q3 q2 20001q1 = q2 , q1 = 20001 4 −q2 + 20001q3 = 10 q3 =

q2 + 104 20001

55

56

Applied Finite Element Analysis

q2



 1 1 +2− − = 1.542 20001 20001 q2 = 0.771 mm q3 = 0.5 mm q1 = 3.855 × 10−5 mm

PROBLEMS 2.1 Determine the nodal displacements and element stresses and reactions at the supports in the stepped bar shown in Fig. P2.1. E2 = 100 GPa; A1 = 1000 mm2 ; A2 = 800 mm2 E1 = 70 GPa; P1 = 100 kN; P2 = 50 kN A1, E1 A2, E2

P2

P1

100 cm

100 cm

100 cm

FIGURE P2.1

2.2 Find the nodal displacements and element stresses in the bar shown in Fig. P2.2. A2, E2

A1, E1

P 25 cm

25 cm

50 cm

FIGURE P2.2

E1 = 100 GPa; P = 50 kN; ∆T = 50 ◦ C

E2 = 200 GPa; A1 = 600 mm2 ; A2 = 1200 mm2 α1 = 20 × 10−6 per ◦ C; α2 = 10 × 10−6 per ◦ C

2.3 A steel plate having a thickness of 5 mm is subjected to a loading of P = 500 N as shown in Fig. P2.3. Find nodal displacements and element stresses. E = 200 GPa; ρ = 7800 kg/m3 ; P = 500 N; g ≈ 10 m/sec2

One-Dimensional Elasticity Problems

57

20 mm

40 mm

P

40 mm

10 mm FIGURE P2.3

2.4 Calculate nodal displacements and element stresses in the bar shown in Fig. P2.4. E = 200 GPa; A1 = 500 mm2 ; A2 = 250 mm2 ; P = 50 kN α = 10 × 10−6 per ◦ C; ∆T = 50 ◦ C A1

A2

0.01 mm

P

400 mm

800 mm FIGURE P2.4

2.5 Calculate the nodal displacements and element stresses shown in Fig. P2.5 by using i) 2 linear elements ii) 1 quadratic element. E = 200 GPa; A = 500 mm2 ; P1 = 60 kN; P2 = 30 kN; ρg = 0.08 N/cm3

500 mm P1

500 mm

P2 FIGURE P2.5

CHAPTER

3

ANALYSIS OF TRUSS

A truss comprises straight members connected by pin joints in such a way the members form a combination of triangles. Smooth pin joints transfer loads and cannot support moments. The external loads are applied at the joints only. In this analysis, a plane truss is considered. A truss is restrained in such a way that it does not undergo free body motion when loaded. A plane truss is shown in Fig. 3.1. P1

P2 7

3

2

1

6 4

5

FIGURE 3.1 Plane truss.

For the analysis of the truss the following assumptions have been considered: i) All members are pin jointed. ii) Gravitational forces are neglected. iii) Loads are applied at the joints only. In light of the above assumptions, the members are regarded as two force members. That is each member is under tension or compression.

3.1

FORMULATION OF ELEMENT MATRICES

Consider an element of truss with two nodes i and j. At each node, one translational degree of freedom is considered in the coordinate system O X  shown in Fig. 3.2. When viewed in global coordinate system, xoy, two degrees of freedom have to be considered at each node.

Analysis of Truss

59

Nodal displacement vector in local coordinate system {q  } = [q  i q  j ]T

(3.1)

Y

X′ q2 j

qj′ θ

q2j – 1

L

j

q2i qi′ θ 0′ i

q2i – 1 X

O FIGURE 3.2 Inclined member.

Nodal displacement vector in global coordinate system  T {q}4×1 = q2i−1 , q2i , q2j−1, q2j

(3.2)

If the nodes are numbered as i = 1 and j = 2, the nodal displacement vector in global coordinate system is  T (3.3) {q}4×1 = q1 , q2 , q3 , q4 q1 and q3 are nodal displacements in horizontal direction. q2 and q4 are nodal displacements in vertical direction. Relation between global and nodal displacements: qi = q2i−1 l + q2i m

(3.4)

qj = q2j−1 l + q2j m

(3.5)

where l = cos θ and m = sin θ which are direction cosines of the inclined member of the truss shown in Fig. 3.2. The Eqs.(3.4) and (3.5) can be expressed as product of two matrices as   q2i−1           l m 0 0 qi   q2i q 2×1 = = (3.6) q2j−1  qj 0 0 l m 2×4      2×1 q2j 4×1

60

Applied Finite Element Analysis

  q 2×1 = [L]2×4 {q}4×1

(3.7)

where [L]2×4 is a transformation matrix used to transform the nodal displacements expressed in global coordinate system to nodal displacements expressed in local coordinate system. Strain energy of inclined truss member expressed in local coordinate system 1   T  (e)     q K q (3.8) ∪= 2     E (e) A(e) 1 −1 where K (e) = (3.9) −1 1 L(e) Strain energy expressed in global coordinate system substituting for {q  } from Eq.(3.7) into Eq.(3.8)     1 ∪ = {q}T [L]T K (e) [L] {q} (3.10) 2 Element stiffness matrix of the truss member     K (e) = [L]T K (e) [L] 4×4  2 l E (e) A(e)  lm  = 2 L(e)  −l −lm

is

lm m2 −lm −m2

−l2 −lm l2 lm

For simplifying the expression for element stiffness matrix. Let us consider matrix  2  l lm [G]2×2 = lm m2

 −lm −m2   lm  m2

Thus, the element stiffness matrix [K (e) ] can be expressed as     E (e) A(e) [G]2×2 − [G]2×2 (e) K = L(e) − [G]2×2 [G]2×2

(3.11)

(3.12)

(3.13)

where E (e) - Modulus of elasticity of the element A(e) - Cross-sectional area of the element L(e) - Length of the element. The element load vector due to point loads acting at the nodes of the element is   T P (e) = P2i−1 P2i P2j−1 P2j (3.14)

Analysis of Truss

3.2

61

FORMULATION OF ELEMENT EQUATIONS

Potential energy = π = Strain energy (∪) − Work done due to point loads (We )  T 1 π = {q}T [K (e) ] {q} − P (e) {q} 2

(3.15)

Minimizing the functional π with respect to nodal displacements yields the element equations.      K (e) q (e) = P (e) (3.16) Assembling the matrices, the global or system equations are [K] {Q} = {P }

(3.17)

Incorporating the boundary conditions, the reduced system equations can be solved for unknown nodal displacements. q2j

j

u A

q2i

θ i

q2j – 1

q2i – 1

FIGURE 3.3 Nodal degrees of freedom.

The displacement at any point A on the truss elements in the axial direction be expressed as u as shown in Fig. 3.3.     qi (3.18) u = Ni Nj qj

Ni and Nj are shape functions which were defined in Chapter 2. Using Eq.(3.6) and substituting for {q  } in Eq.(3.18).   q2i−1         l m 0 0  q2i u = Ni Nj (3.19) q2j−1  0 0 l m      q2j   u = Ni l Ni m Nj l Nj m 1×4 {q}4×1 (3.20)

62

Applied Finite Element Analysis

Strain  is =

dξ   dNi  du = l dξ dx dx



dNi dξ



m



dNj dξ





l

dNj dξ



m



{q}4×1

(3.21)

ξ is the natural coordinate in the local coordinate system for the truss element. 2 dNi −1 dNj 1 dξ = (e) , = , = and substituting in Eq.(3.21). It is known that dx dξ 2 dξ 2 L =

1  −l L(e)

−m l

m

Defining strain-displacement matrix :

1  −l −m L(e) {} = [B]1×4 {q}4×1

[B]1×4 =

l



m

{q}



(3.22)

(3.23) (3.24)

Stress in the element: σ = [D] {} = E [B]1×4 {q}4×1

(3.25)

If any truss member is subjected to a change in temperature ∆T , the load vector can be calculated using the following integral expression.    F (e) = [B]T [D] {0 } dV (3.26) V

where 0 = α∆T , α = coefficient of thermal expansion.  −l      −m F (e) = E (e) A(e) (α∆T ) l    m

      

Stiffness matrix [K (e) ] can be derived from the following equation.

[K

(e)

]=

L

[B]T [D][B]AdL

0

From Eq.(3.23)

[B] =

1  −l L(e)

−m l

m



and [D] = E.

(3.27)

Analysis of Truss



−l   E  −m   −l −m l m [B] [D][B] = (L(e) )2  l  m  2  −l2 −lm l lm −lm −m2  lm m2 E    = (L(e) )2  −l2 −lm l2 lm  lm m2 −lm −m2  2  l lm −l2 −lm   −lm −m2  lm m2 E (e) A(e)    K (e) = L(e)  −l2 −lm l2 lm  lm m2 −lm −m2 T

63



(3.28)

(3.29)

which is same as Eq.(3.11). It is obvious that deriving element stiffness matrix from Eq.(3.18) is simpler than deriving from strain energy expression given by Eq.(3.10).

Example 3.1 Calculate the nodal displacements and element stresses for the truss shown in Fig. 3.4. E = 70 GPa. Cross-sectional area A = 2 cm2 for all truss members.

Y

q6 3

q5

3

80 cm

2 q4

q2 O 1

1 60 cm

2

X

1

q1

2

30° P = 50 kN

FIGURE 3.4 Elements showing nodal degrees of freedom.

i) Discretisation: No. of elements: (2) Type of element: Truss element No. of nodes per elements: 2

q3

64

Applied Finite Element Analysis

No. of degrees of freedom per node: 2 (Translational) Total number of nodes: 3 Total degrees of freedom (d.o.f): 6 Size of global stiffness matrix: 6 × 6 ii) Coordinate system XOY is shown in Fig. 3.4 to define nodal coordinates. Nodal degrees of freedom are also shown in Fig. 3.4. Nodal coordinates in cm

Node

X

Y

1 2 3

0 60 0

0 0 80

iii) Element connectivity: Element (e)

Local nodal numbers i j

1 2

1 2

2 3

1, 2, 3 are global nodal numbers. iv) Direction cosines depend on the choice of element connectivity. Hence the element connectivity is to be decided before preparing direction cosine table. Direction cosines are calculated from the following equations: l = cos θ =

Xj − Xi ; L(e)

m=

Yj − Yi L(e)

where L(e) is the length of the element. Element (e)

Length L(e)

1 2

60 cm 100 cm

Analysis of Truss

65

Direction cosines

Element (e)

l

m

1 2

1 −3/5

0 4/5

v) Element stiffness matrices:

 Element (1)

K (e)

4×4

=

E (e) A(e) L(e)

l2  lm   −l2 −lm

lm m2 −lm −m2

 −lm −m2   lm  m2

−l2 −lm l2 lm

E = 70 × 105 N/cm2 , A = 2 cm2

  K (1)

4×4

Element (2)







K (2)



4×4

1 1 70 × 105 × 2   0 =  −1 60 0



2 3 4  0 −1 0 1 0 0 0   2 0 1 0  3 0 0 0 4

N/cm

3 4 5 6  9 −12 −9 12 3  4 70 × 105 × 2  −12 16 12 −16   = 100 × 25  −9 +12 +9 −12  5 +12 −16 −12 +16 6



N/cm

vi) Global stiffness matrix is obtained by assembling element stiffness matrices: 1 2 3 4 5 6

[K] = 105

2.333 0 –2.333 0 0 0

0 0 0 0 0 0

–2.333 0 2.333 + 0.504 0.0 –0.672 –0.504 0.672

0 0 0.0 +0.672 0.0 +0.896 0.672 –0.896

0 0 –0.504 0.672 0.504 –0.672

0 0 +0.672 –0.896 –0.672 0.896

1 2 3 4 5 6

Incorporating the boundary conditions q1 = q2 = q5 = q6 = 0 by deleting corresponding rows and columns, the reduced stiffness matrix is

66

Applied Finite Element Analysis

[K]2×2 =



3 K33 K43

4 3   K34 2.837 3 = 105 4 −0.672 K44

4  −0.672 3 4 0.896

vii) Global load vector : Resolving the point load P acting at the node 2 in horizontal and vertical directions, the global load vector is

{F }6×1 =



1

2

0 0

{F }6×1 = 10

5



3

4

P sin 30◦

1

2

3

0

0

0.25

5

−P cos 30◦ 4

(−)0.433

6

0 0

5

6

0

0

T

T

Incorporating the boundary conditions q1 = q2 = q5 = q6 = 0, the reduced load vector is     F3 0.25 3 5 = 10 newtons {F }2×1 = (−)0.433 4 F4 viii) Global equations: 105



2.837 −0.672 −0.672 0.896





q3 q4

= 105



0.25 −0.433

Solving for q3 and q4 , the nodal displacements are q3 = (−)0.032 cm;

q4 = (−)0.507 cm

ix) Element stresses: Element (1)

σ1 =

E (−q1 l − q2 m + q3 l + q4 m) L1

q1 = q2 = 0;

l = 1,

m=0

σ1 = (−)0.037 × 105 N/cm2 σ1 = (−)37 MPa

Element (2)

σ2 =

E (−q3 l − q4 m + q5 l + q6 m) L2

q5 = q6 = 0;

l = −0.6, 5

σ2 = (+)0.27 × 10 N/cm

σ2 = (+)270 MPa

m = 0.8 2



Analysis of Truss

67

Example 3.2 Determine nodal displacements and element stresses of the truss shown in Fig. 3.5. The truss members are subjected to a temperature rise of 50 ◦ C.

Y

400 mm q6

3

q4 q5

q3

2

2

60° P

1

300 mm q2 O

1

X

q1 FIGURE 3.5 Truss.

E = 200 GPa, α = 10 × 10−6 per ◦ C ∆ T = 50 ◦ C

A = 500 mm2 for all members. P = 50 kN

i) Discretisation: Number of elements: (2) Type of element: Truss element Number of nodes per element: 2 Number of degrees of freedom (d.o.f) per node: 2 (Translational) Total number of nodes: 3 Total number of d.o.f: 6 The size of the global stiffness matrix [K] is 6 × 6. ii) Coordinate system XOY is shown in Fig. 3.5 for defining nodal coordinates. Nodal degrees of freedom are also shown. Nodal coordinates in mm Node

X

Y

1 2 3

0 400 0

0 300 300

68

Applied Finite Element Analysis

iii) Element connectivity: Element (e)

Local nodal numbers i j

1 2

1 3

2 2

1, 2, 3 are global nodal numbers. iv) Direction cosines: l = cos θ =

Yj − Yi Xj − Xi ; m = sin θ = L(e) L(e)

Element (e)

Length L(e) (mm) Area (mm2 )

1 2 -

500 400 -

500 500 -

Direction cosines

Element (e)

l

m

1 2

4/5 1

3/5 0

v) Element stiffness matrices:

Element (1)

Element (2)

1 2 3 4  16 12 −16 −12 1   5 × 500   2 × 10 12 9 −12 −9   2 K (1) =   3 −16 −12 16 12 500 × 25 −12 −9 12 9 4



5 1   2 × 105 × 500  (2)  0 K =  −1 400 0



6 3 4  0 −1 0 5 0 0 0   6 0 1 0  3 0 0 0 4

Analysis of Truss

69

vi) Global stiffness matrix :

[K]6×6 =

106 125

1

2

3

4

5

6

16 12

12 9

–16 –12

–12 –9

0 0

0 0

1 2

–16

–12

(16+31.25)

(12+0)

–31.25

0

3 N/mm

–12 0 0

–9 0 0

(12+0) –31.25 0

(9+0) 0 0

0 31.25 0

0 0 0

4 5 6

The boundary conditions are q1 = q2 = q5 = q6 = 0 and after incorporating boundary conditions into the stiffness matrix by deleting corresponding rows and columns, the reduced stiffness matrix is   K33 K34 [K]2×2 = K43 K44 =

106 125



3 4  47.25 12 3 4 12 9

Elements K33 , K34 , K43 , K44 are contributed by elements 1 and 2, since q3 , q4 are common to both elements. While formulating the global stiffness matrix [K]6×6 proper care should be taken to post proper elements in correct locations. vii) Elements load vector for temperature change:

Element (1)

0 = α∆T = 10 × 10−6 × 50 = 5 × 10−4   −l         −m (1) F = EA(α∆T ) l       m   −4     2 × 105 × 500 × 5 × 10−4  −3  = 4   5     3   −4  1      −3 2 = 104 4  3      3 4

70

Applied Finite Element Analysis

Element (2)

   

−l   −m F (2) = EA(α∆T ) l    m

Global load vector for temperature change:

{F }∆T = 104

              

−4 −3 (4 + 5) (3 + 0) −5 0

   

  −5       0 4 = 10 5          0

5 6 3 4

       

  −4     1     −3 2       +9 3 = 104     +3   4       −5    5     0   6

Global load vector : Thermal load and point load acting on the truss contribute for formulation of global load vector.

∴ {F } = {F }∆T + {F }point load     −4  0              −3  0              +9 4.33 4 4 = 10 + 10 +3  −2.5               −5 0          0   0     −4  1        −3 2       +13.33 3 ∴ {F } = 104 newtons +0.5 4       −5   5       6 0

Incorporating boundary conditions q1 = q2 = q5 = q6 = 0 by elimination approach, the reduced load vector {F } = 104



13.33 0.5



3 4

Analysis of Truss

Reduced global equation: 106 125



47.25 12

12 9



 13.33 = 10 0.5   −1    125 q3 47.25 12 13.33 = −2 q4 12 9 0.5 10 q3 q4



4



q3 = 0.5025 mm

Element stresses:

Element (1)

q4 = (−)0.6005 mm

σ = E[B]{q} − E0 σ (1) =

σ (1)

2 × 105  −l 500

−m l

     q1    q2  m   q3     q4

−2 × 105 × 5 × 10−4     2 3 3 4 2 × 10 q3 + q4 − 10 = 5 5 5   2 3 2 × 10 (4q3 + 3q4 ) − 10 = 25 σ (1) = (−)83.333 MPa

Element (2)

σ (2) =

2 × 105  −l 400

−m l

−2 × 105 × 5 × 10−4

     q5    q6  m q3       q4

σ (2) = {100 (4q3 + 3q4 ) − 100}

σ (2) = (−) 79.15 MPa

Example 3.3 Formulate element stiffness matrices and global load vector for the truss shown in Fig. 3.6 and also state the boundary conditions. E = 200 GPa A = 100 mm2 P = 10 kN.

71

72

Applied Finite Element Analysis

Y

q6

q8

400 mm q5 4

3

q7

4 30° P

3

300 mm

2

5 q4

q2 O

1

1

2

q1

q3

X

800 mm FIGURE 3.6 Elements with nodal degrees of freedom.

i) Discretisation: No. of elements : (5) Type of element : Truss element No. of nodes per element : 2 No. of d.o.f per node : 2 (Translational) Degrees of freedom for an element : 4 Total number of nodes : 4 Total number of d.o.f : 8 The global stiffness matrix size is [K]8×8 . ii) Nodal coordinates (mm): Node

X

Y

1 2 3 4

0 800 400 800

0 0 300 300

iii) Element connectivity: Element (e) 1 2 3 4 5

i j 1 2 1 3 2

2 3 3 4 4

Analysis of Truss

iv) Direction cosines:

l = cos θ =

Xj − Xi ; L(e)

m = sin θ =

Element length

Y j − Yi L(e)

L(e)

Element (e)

Length L(e) (mm)

1 2 3 4 5

800 500 500 400 300

Direction cosines

Element (e)

l

m

1 2 3 4 5

1 −4/5 4/5 1 0

0 3/5 3/5 0 1

v) Element stiffness matrices: Element (1) 1 1   2 × 105 × 100   0 K (1) =  −1 800 0



2 3 4  0 −1 0 1  0 0 0  2 0 1 0  3 0 0 0 4

N/mm

Element (2) 3 4 5 6 16 −12 −16 12   2 × 105 × 100  −12 9 12 −9 (2)  K = 12 16 −12 500 × 25  −16 12 −9 −12 9





3  4   5 6

N/mm

73

74

Applied Finite Element Analysis

Element (3)

1 2 5 6  16 12 −16 −12 1   2 × 105 × 100   12 9 −12 −9   2 K (3) =   5 −16 −12 16 12 500 × 25 −12 −9 12 9 6

Element (4)



5 1   2 × 105 × 100   0 K (4) =  −1 400 0



Element (5)

3 0   2 × 105 × 100   0 K (5) =  0 300 0



6 7 0 −1 0 0 0 1 0 0 4 0 1 0 −1

{F }8×1 = 10 vii) Boundary conditions:





5  6   7 8

7 8  0 0 3 0 −1   4 0 0  7 0 1 8

vi) Global load vector :

3

8 0 0 0 0

1 2

3

4

5

6

7

0 0

0

0

0

0

5 −8.66

q1 = q2 = q4 = 0.

8

T

N/mm

N/mm

N/mm

newtons

PROBLEMS 3.1 Calculate nodal displacements and element stresses for the members of the truss shown in Fig. P3.1. E = 200 GPa, Cross-sectional area of each member = 500 mm2 and P = 25 kN

3c

m

60° P

30°

60°

FIGURE P3.1

Analysis of Truss

75

3.2 Find nodal displacements and element stress in the truss shown in Fig. P3.2. E = 80 GPa 3

Element Area (mm2 ) Length (mm) 3

1 5 kN

2

1

600 mm2 600 mm2 600 mm2

1 2 3

500 mm 600 mm 500 mm

2 FIGURE P3.2

3.3 Find the nodal displacements in the truss shown in Fig. P3.3. E = 80 GPa Cross-sectional area of each member = 500 mm2 30° P = 50 kN

4 4

3

2

3

800 mm

2

1 1 600 mm

FIGURE P3.3

3.4 Find the nodal displacements of the truss shown in Fig. P3.4. E = 200 GPa; Cross-sectional area of each member = 600 mm2 1000 mm

60° 900mm

1200 mm FIGURE P3.4

P =50 kN

76

Applied Finite Element Analysis

3.5 Find the stress in each member of the truss shown in Fig. P3.5. E = 100 GPa; Cross-sectional area of each member = 500 mm2 ; α = 15 × 10−6 per ◦ C; ∆T = 50 ◦ C. 400 mm

∆T = 50 °C 45° C

300 mm ∆T

=

° 50

FIGURE P3.5

P = 30 kN

CHAPTER

4

ANALYSIS OF BEAMS

Beam is regarded as a slender member subjected to transverse loading. Beams can be described by depth and cross-sectional shape. Apart from a member having constant or varying depth, these members can be straight or curved or may possess axis of symmetry or no axis of symmetry. The loading that can act on a beam may be a point load, a moment or distributed loading or a combination of loads. Applications to mechanical engineering components are shafts loaded with pulleys or rotors or any slender member subjected to transverse loading in addition to beams commonly used in civil engineering designs. This chapter is an introductory analysis of a beam with uniform depth and loads lying in the plane of symmetry. Y P1

p (N/m) M2

A

O 1

2

3

X

4

FIGURE 4.1 Beam configuration.

A beam shown in Fig. 4.1 where point load P1 , moment M2 and a uniformly distributed load p (N/m) act on the beam. The left end of the beam is simply supported and right end is supported on roller. The definition of the coordinate system adopted is shown in Fig. 4.1. The y-axis is defined positive upward and x-axis positive towards right. With reference to the coordinate system, the application of load, moment and distributed loading are in the positive sense. The positive moment M2 results in compressive stresses for positive values of y. Using elementary beam theory and neglecting transverse shear stress and assuming that loading is symmetric with respect to the area of cross-section, the following equations relate the moment M, shear force V and lateral load p. Free body diagram of the beam element is shown in Fig. 4.2. d2 v Bending moment M = EI 2 (4.1) dx dM (4.2) Shear force V = dx

78

Applied Finite Element Analysis

Distributed load

p=

dV dx

(4.3)

p

M + dM

M

V + dV

V

dx FIGURE 4.2 Free body diagram.



Stress

σ=−

M (y) = (−)E I

Strain

ε=−

d2 v σ = (−) 2 (y) E dx

d2 v dx2



(y)

(4.4) (4.5)

where v represents the deflection in y-direction and EI is the product of modulus elasticity E and moment of inertia I of the cross-section about an axis passing through neutral plane of the beam. The product EI is known as flexural rigidity of the beam. The moment of inertia is  I = y 2 dA A

where A is the cross-sectional area of the beam. The slope of the beam θ is related to the vertical deflection of the beam as dv (4.6) θ= dx The differential equation is expressed in terms of lateral deflection v, relating the distributed loading   d2 d2 v EI 2 = p(x) (4.7) dx2 dx Obtaining the solution of the above equation is the classical method for beam analysis. The finite element analysis for the beam under consideration will be discussed below with essential features which are specific to the beam.

Analysis of Beams

79

dv . The field variables for the beam are vertical deflection v and slope θ. The slope θ is dx Hence, only the vertical displacement is the unknown variable. Two degrees of freedom are considered at each node. i) Vertical displacement (Translational degree of freedom) dv ii) Slope θ = (Rotational degree of freedom) dx vi

vj ξ

O

θi node i

O′

A

θj node j

FIGURE 4.3 General beam element.

The beam element shown in Fig. 4.3 is having two nodes (i, j) and 4 degrees of freedom for each element. Hence, the polynomial to be chosen will have four constants which are evaluated in terms of displacement and slope at the two nodes.

4.1

DERIVATION OF SHAPE FUNCTIONS

A cubic polynomial expressed in natural coordinate ξ is given as v(ξ) = a1 + a2 ξ + a3 ξ 2 + a4 ξ 3 dv = a2 + 2a3 ξ + 3a4 ξ 2 dξ

(4.8) (4.9)

The constants are evaluated using the following conditions. v = vi = a1 − a2 + a3 − a4

At node i, ξ = −1;

θ = θi =

dv dξ dv = · dx dξ dx

(4.10) (4.11)

It is to be noted that the slope θ is related to the derivative of vertical deflection v with respect to global coordinate x. l(e) dx = where l(e) is length of element. dξ 2 From the above relations,

We know that

θi =

2 2 dv = (e) (a2 − 2a3 + 3a4 ) dξ l

l(e)

(4.12)

80

Applied Finite Element Analysis

At node j, ξ = +1;

v = vj = a1 + a2 + a3 + a4 2 θ = θj = (e) (a2 + 2a3 + 3a4 ) l

(4.13) (4.14)

Solving the following equations, the constants a1 , a2 , a3 and a4 can be evaluated. a1 − a2 + a3 − a4 = vi

l(e) θi 2 a1 + a2 + a3 + a4 = vj a2 − 2a3 + 3a4 =

a2 + 2a3 + 3a4 =

l(e) θj 2

(4.15)

Writing the equations using natural coordinate ξ simplifies the calculations. Substituting the expressions of a1 , a2 , a3 and a4 in Eq.(4.8) it can be shown that v(ξ) = H1 vi +

l(e) l(e) H2 θi + H3 vj + H4 θj 2 2

(4.16)

where H1 , H2 , H3 and H4 are called Hermite shape functions which consider both nodal displacements and slopes for an element. Expressing vi = q2i−1 , θi = q2i , vj = q2j−1 and θj = q2j the nodal degrees of freedom are shown in Fig. 4.4. q2i – 1

q2i

q2j – 1

O′

node i

q2j node j

FIGURE 4.4 Nodal degrees of freedom.

To simplify expressions, let i = 1 and j = 2 q2i−1 = q1 ; q2i = q2 ; q2j−1 = q3 and q2j = q4 Hermite shape functions are expressed in terms of natural coordinate ξ as 1 (2 − 3ξ + ξ 3 ); 4 1 H2 = (1 − ξ − ξ 2 + ξ 3 ); 4 H1 =

1 H3 = (2 + 3ξ − ξ 3 ); 4 1 H4 = (−1 − ξ + ξ 2 + ξ 3 ) 4

(4.17)

It can be observed that H1 and H3 consist of similar terms but a change in sign for two terms. Similar observations can also be made for H2 and H4 .

Analysis of Beams

81

The displacement v at any point can be written as



l(e) H2 2

v(ξ) = H1 where,

4.2

{q}4×1 = [q1

q2

q3

l(e) H4 2

H3



1×4

{q}4×1 = [H]1×4 {q}4×1

q4 ] T

(4.18) (4.19)

FORMULATION OF ELEMENT MATRICES AND EQUATIONS

The functional π is expressed in terms of potential energy of the beam. Potential energy = Strain energy (∪) – Work done due to external loads (We )  1 Strain energy ∪=  σ dV 2

(4.20)

V

From Eqs.(4.4) and (4.5)

It is known that Thus,



1 ∪= 2



E



d2 v dx2

2

(y 2 dA) dx

y 2 dA = I EI ∪= 2

(4.22)

 

d2 v dx2

2

dx

(4.23)

It can be seen that Eq.(4.23) consists of square of second order derivative of respect to x complicates the process of evaluation of strain energy. The differential equation Eq.(4.7) is a fourth order equation, hence the strain term involves second order derivative. We know that     dv dv 2 2 dH = (e) = (e) {q} dx dξ dξ l l   l(e) l(e) H2 H 3 H4 where, [H] = Shape function matrix = H1 2 2 {q} = Nodal displacement vector = [q1

Hence

(4.21)

q2

 2  d2 v d H 4 = (e) 2 {q} 2 dx dξ 2 (l )  T  2   2 2 d v d2 H d H 16 T = (e) 4 {q} {q} 2 dx2 dξ dξ 2 (l )

q3

q4 ] T

v with energy

(4.24) (4.25) (4.26) (4.27) (4.28)

Applied Finite Element Analysis

82

Strain energy

v=



EI  {q}T  2

ξ=+1 

ξ=−1

d2 H

and

dξ 2



l(e)



16 4

l(e)

(3ξ − 1) 2 2

3 ξ = 2



d2 H dξ 2

3 − ξ 2

T 





d2 H l(e)  dξ  {q} · 2 dξ 2

l(e) (3ξ + 1) 2 2



(4.29)

(4.30)

The expression for strain energy can be expressed on 1 v = {q}T [K (e) ]{q} 2

(4.31)

where element stiffness matrix

[K (e) ]4×4



6l(e)  2 4 l(e)

12

  6l(e) EI  = (e) 3  −6l(e) (l )   −12   2 6l(e) 2 l(e)

 6l(e)  2  −6l(e) 2 l(e)    (e)  + 12 −6l   (e) 2  (e) −6l 4 l −12

(4.32)

4×4

The evaluation of the integral for obtaining stiffness has been simplified because the formulation utilised natural coordinate ξ and limits of integration are – 1 to + 1. The size of the stiffness matrix [K (e) ] is now (4 × 4) which is determined by the number of degrees of freedom for the beam element.

4.3

EVALUATION OF ELEMENT LOAD VECTORS

Work done due to distributed loading p(x):

Work done =



+1 l(e) dξ pv dx = p [H] {q} 2

(4.33)

−1

If the intensity of loading p is constant



pl(e) 2

+1

−1

Work done = 



pl(e) [H] dξ = 2



pl(e) 2

 (e) 2

p l 12

 +1 [H] dξ  {q}

−1

pl(e) 2

 2 −p l(e) 12

(4.34)

(4.35) 1×4

Analysis of Beams

83

Note that direction of p is upwards and positive as the y-axis is pointed upwards. If the direction of p is downwards, the sign of p in the above expression should be negative.

Work done by applied external moments If any external moment M is applied, a node is to be taken at the point of application and work done is the product of moment (M ) and slope (θ). Anticlockwise moment is positive. Work done by external moment,

M=



Mr θr

(4.36)

r

where suffix ‘r’ denotes the number of locations where moments are applied.

Work done by point load If any point load p is applied, a node is to be fixed at the point of application and work done is obtained by the product of load P and deflection v at the point of application. P is positive if it acts in the upward direction. Work done by external moment P =



Ps vs

(4.37)

s

where suffix ‘s’ denotes the number of locations where point loads are applied. Potential energy expression which gives us the required functional π

 1      (e)  l    (e) pl 1 6 π = {q}T [K (e) ] {q} − 2 2  1      l(e)    − 6

T                  

{q} −

 r

Mr θr −



Ps vs

(4.38)

s

Minimising the above functional with respect to nodal variables, the element equations for the beam element can be obtained. ∂π =0 ∂{q}

yields

     K (e) q (e) = F (e)

(4.39)

Applied Finite Element Analysis

84

The element load vector is  pl(e)      2    2    p l(e)       12 (e) F =  pl(e)     2     (e) 2   p l     − 12

              

   Pi          0 + +     Pj          0          

0 Mi 0 Mj

      

(4.40)

The first and third rows of the elements equations refer to the forces acting on the element at the corresponding nodes. The second and fourth rows of the element equations refer to moments applied to the element at the corresponding nodes. This shows that element equations contain nodal equations corresponding to forces and moments expressed in a matrix form.

4.4

CALCULATION OF BENDING MOMENT, SHEAR FORCE AND FLEXURAL STRESS  2  d v d2 v 4 = EI dx2 (l(e) )2 dξ 2  EI  M = (e) 2 6ξq1 + (3ξ − 1)l(e) q2 − 6ξq3 + (3ξ + 1)l(e) q4 (l )

Bending moment M = EI

Shear force

Bending stress

d3 v dM 8 d3 v = EI 3 = EI (e) 3 3 dx dx (l ) dξ   6EI (e) (e) V = (e) 3 2q1 + l q2 , −2q3 , +l q4 (l )  2   2  d v d v 4E (y) = (−) (e) 2 (y) σ = −E 2 dx dξ 2 (l )  Ey  = − (e) 2 6ξq1 + (3ξ − 1)l(e) q2 , −6ξq3 + (3ξ + 1)l(e) q4 (l ) V =

(4.41)

(4.42) (4.43) (4.44) (4.45)

The stress depends on y and also the natural coordinate ξ which describes the location of the point. Shear force is constant within the element because displacement was expressed by a cubic polynomial and shear force is calculated by the third order derivative of displacement.

Analysis of Beams

4.5

85

SYSTEM EQUATIONS

Assembling the element matrices given by Eq.(4.39) the global equations can be formulated.   [K (e) ]{q (e) } − F (e) = 0 (4.46) e

[K] {Q} = {F }

(4.47)

[K] - Global stiffness matrix, {Q} - Global displacement vector, and {F } - Global force vector. Boundary conditions have to be incorporated into the Eq.(4.47) and if elimination approach is used, the reduced equation is to be solved for obtaining nodal displacements.

4.6

ESSENTIAL AND NATURAL BOUNDARY CONDITIONS

Boundary conditions depend on the type of supports provided to the beam. In a beam problem we can clearly distinguish between essential and natural boundary conditions. Essential boundary conditions: Field variables specified at the supports become essential boundary conditions. The vertical deflection v and slope θ are the field variables considered in beam analysis hence boundary conditions specifying either v or θ, and both v and θ at some supports are essential boundary conditions. Why these boundary conditions are essential because they have to be imposed on global equations before solving for obtaining solution. Natural boundary conditions: These are the higher order derivatives of the field varid2 v ables considered in the analysis. For the beam analysis, the higher order derivatives dx2 d3 v implicitly moment and (shear force implicitly) are natural boundary conditions since dx3 dv are field variables. Natural boundary conditions manifest themselves in the v and slope dx terms included in load vector hence do not need to be imposed. The local vector for beam consists of terms for moment and shear force. This is the major difference between classical method of obtaining solution and application of finite element method. A cantilever beam shown in Fig. 4.5 will provide an example for essential and natural boundary conditions. Essential boundary conditions: At x = 0 (fixed end),

v=0

(Displacement), θ = 0 (Slope)

Natural boundary conditions: At x = L (free end),

M = EI

d2 v =0 dx2

(Moment is zero)

86

Applied Finite Element Analysis

Y

o

x

L

FIGURE 4.5 Cantilever.

V = EI

d3 v =0 dx3

(Shear force is zero)

(4.48)

Example 4.1 A shaft is supported in a long bearing at the left end and another bearing at the right side which has vertical stiffness Ks . Shaft is subjected to a distributed load whose intensity is p equal to 5000 N/m and a point load P of 2500 N as shown in Fig. 4.6. The flexural rigidity EI of the shaft is 50 × 103 Nm2 . The shaft is made of steel with modulus of elasticity E equal to 200 GPa. Determine i) the slope and deflection at the ends and midspan of the shaft. ii) the moment and shear force at the middle of the distributed load. p

P Ks 1m

1m

FIGURE 4.6 Shaft supported in bearings.

p = 5000 N/m; Ks = 250 kN/m; P = 2500 N The left end bearing is long which is to be considered a fixed support. Discretisation No. of elements : (3) Type of element : 1 and 2 are beam elements and 3 is one-dimensional linear element No. of nodes :4 No. of degrees of freedom : 4 Translational degrees of freedom and 3 rotational degrees of freedom. Size of global stiffness matrix is [K] is 7 × 7.

Analysis of Beams

p

Ks

P 1m

1m

Y q1

q3

1

1 q4

q2

q5

2

3

2

X

q6 3 q7 4

FIGURE 4.7 Elements.

Element stiffness matrices : Beam Element (1) [K (1) ]4×4 =

[K (1) ]4×4 =

EI L1 3

    

12 6L1 −12 6L1

[K (2) ]4×4 =

   



103

Bar Element (3)



3 4 5 6 12 6 (−)12 6  6 4 (−)6 2   (−)12 (−)6 (+)12 (−)6 6 2 (−)6 4



50 × (1)3

6L1 2(L1 )2 −6L1 4(L1 )2



103

Beam Element (2)

−12 −6L1 (+)12 −6L1

1 2 3 4 12 6 (−)12 6  6 4 (−)6 2   (−)12 (−)6 (+)12 (−)6 6 2 (−)6 4



50 × (1)3

6L1 4(L1 )2 −6L1 2(L1 )2

1  2   3 4

3  4   5 6

Ks = 250 kN/m [K

(3)

3

] = 250 × 10



5 7  1 −1 5 −1 1 7

87

Applied Finite Element Analysis

88

Global stiffness matrix is obtained by assembling element stiffness matrices. 1 2 3 4 5 6 7 12

6

(–)12

6

0

0

0

1

6

4

(–)6

2

0

0

0

2

(–)12

(–)6

(+)12 + 12

(–)6 + 6

–12

6

0

3

6

2

(–)6 + 6

4+4

–6

2

0

4

0

0

–12

–6

12 + 5

–6

–5

5

0

0

6

2

–6

4

0

6

0

0

0

0

–5

0

5

7

[K]7×7 = 50 × 103

Global load vector:  pL1   −   2    2  p(L 1)   −   12     pL1     − 2 {F } = p(L1 )2  +    12    0       0       0 

                    

    −2500  0       5000       −    0       12    −2500 − 2500       −2500    5000 0 + = +         12     0         0        0          0           0    0      

             

  −2500           −416.6         −5000     416.6 =          0             0              0 

Boundary conditions: q1 = q2 = q7 = 0

Incorporating boundary conditions using elimination approach, the reduced system equations are, 4 5 6   3     24 0 −12 6 3 q3 −5000     0 8 −6 2  50 × 103    4  q4  =  416.6   −12 −6  17 −6  5  q5   0 6 2 −6 4 6 0 q6



 q3  q4  10−3 × 103    q5  = 50 × 103 q6



  63.9 58.1 −81.4 −5    215.1 104.6 −46.5     0.4166      186 139.5 0 symmetric 604.6 0 91

Analysis of Beams



 q3  q4  1    q5  = 5 × 103 q6



 −42.878  −23.015  1    −24.71  = 103 +38.76

At midpoint of distributed load:



 −8.575 - deflection at mid-span (m)  −4.603 - slope at mid-span (rad)    −4.942  - deflection at end (m) +7.752 - slope at end (rad)

First beam element: At ξ = 0 Moment

M =

q2 = 0

M =

50 × 103 50 × 103 (L1 q4 ) = × (−4.603) 2 (1) 103 = −230.15 N.m.

Shear force q 1 = q2 = 0

EI [0, −L1 q2 , 0, L1 q4 ] (L1 )2

V =

∴

6EI [2q1 + L1 q2 − 2q3 + L1 q4 ] (L1 )3

6 × 50 × 103 (17.150 − 4.603) 103 V = 37.641 × 102 (N )

V =

Example 4.2 Determine the deflection and slope at 2 m from the left end of a shaft mounted in bearings. A load of 1 kN and a moment of 2 kN·m act on the shaft. Model the bearings as fixed supports. Given E = 200 GPa;

I1 = 4 × 10−6 m4 ;

I2 = 2 × 10−6 m4 1 kN 2 kN.m I2

I1 2m

1m

q1

q3 2

1 1

q2

q5

2

q4

3

FIGURE 4.8 Beam with fixed supports.

q6

89

90

Applied Finite Element Analysis

i) Discretisation: Number of elements : (2) Type of element : Beam element No. of d.o.f per node : 2 (1 Translational + 1 Rotational) No. of nodes per element : 2 No. of d.o.f per element : 4 Total number of nodes :3 Total degrees of freedom : 6 The size of the global stiffness matrix [K] is 6 × 6.

ii) Element stiffness matrices: Element (1) [K (1) ]4×4 =

Element (2) [K (2) ]4×4 =

200 ×

200 ×

109

×4× (2)3

109

×2× (1)3

1 2 3 12 12 (−) 12  12 16 (−) 12   (−) 12 (−) 12 (+) 12 12 8 (−) 12

10−6



10−6



3 4 5 12 6 (−) 12  6 4 (−) 6   (−) 12 (−) 6 (+) 12 6 2 (−) 6

4 12 8 (−) 12 16 6 6 2 (−) 6 4



1  2   3 4



3  4   5 6

iii) Global stiffness matrix is obtained by assembling the element stiffness matrices. 1 2 3 4 5 6

[K]6×6 = 105

12

12

–12

12

0

0

1

12

16

–12

8

0

0

2

–12

–12

(+)12 + 48

– 12 + 24

–48

24

3

12

82

–12 + 24

16 + 16

–24

8

4

0

0

–48

–24

+48

–24

5

0

0

24

8

–24

16

6

After incorporating the boundary conditions q1 = q2 = q5 = q6 = 0 and deleting the corresponding rows and columns, the reduced stiffness matrix [K] is [K]2×2

4  3 60 12 3 = 10 12 32 4 5

Analysis of Beams

iv) Global load vector:

{F } = 103

              

0 0 (−)1 (+)2 0 0

 1     2    3 4     5    6

After incorporating boundary conditions, the reduced load vector is   −1 3 3 {F }2×1 = 10 2 4 v) Global equation: 10

5



60 12

12 32





q3 q4

3

= 10



−1 2



Solving the above equations, the values of q3 , q4 are q3 = (−)0.000315 m,

q4 = +0.000743 rad.

Example 4.3 Determine the deflection and slope under the point load for the beam shown in Fig. 4.9. Y p = 24 kN/M p = 50 kN O

2m q1

X

I2

I1 1m q3

1

2 q4

1 q2

2

1m q5

q7

3

3 q6

4 q8

FIGURE 4.9 Elements.

Given, E = 200 GPa; I1 = 4 × 10−6 m4 ;

I2 = 2 × 10−6 m4

91

Applied Finite Element Analysis

92

i) Discretisation: Number of elements : (3) Type of element : Beam element No. of nodes per element : 2 No. of d.o.f per node : 2 (1 Translational + 1 Rotational) No. of d.o.f per element : 4 Total number of nodes : 4 Total degrees of freedom : 8 The size of the global stiffness matrix [K] is 8 × 8

ii) Element stiffness matrices: [K (1) ] =

Element (1)

[K (2) ] =

Element (2)

1 2 3 4 12 12 −12 12  12 16 −12 8   −12 −12 +12 −12 12 8 −12 16

200 ×

109

10−6



200 ×

109

10−6



×4× (2)3

×2× (1)3

3 4 5 6 12 6 −12 6  6 4 −6 2   −12 −6 +12 −6 6 2 −6 4





1  2   3 4

3  4   5 6

5 6 7 8  12 6 −12 6 5 ×2× 200 ×   6 6 4 −6 2 Element (3) [K (3) ] =   (1)3  −12 −6 +12 −6  7 6 2 −6 4 8 iii) Global stiffness matrix is obtained by assembling the element stiffness matrices. [K]8×8 1 2 3 4 5 6 7 8 109

= 105



10−6

12

12

– 12

12

0

0

0

0

1

12

16

–12

8

0

0

0

0

2

– 12

– 12

+ 12 + 48

– 12 + 24

– 48

24

0

0

3

12

8

– 12 + 24

16 + 16

– 24

8

0

0

4

0

0

– 48

– 24

+ 48 + 48

– 24 + 24

– 48

24

5

0

0

24

8

– 24 +24

+ 16 + 16

– 24

8

6

0

0

0

0

– 48

– 24

+48

– 24

7

0

0

0

0

24

8

– 24

16

8

Analysis of Beams

93

After incorporating the boundary conditions q1 = q2 = q3 = q7 = q8 = 0, the reduced stiffness matrix is 4 5 32 −24 = 105  −24 96 8 0



[K]3×3

iv) Global load vector:

6  8 4 0  5 32 6

  −24  1     2 −8          3 −24       4 8 {F } = 103 5 −50         6 0         0    7  8 0

Incorporating boundary conditions, the reduced load vector is   8  4  −50 5 {F } = 103   0 6

v) Global equations:



32 −24 96 105  −24 8 0

Solving the above equation,

    8  8  q4   q5 −50 0  = 103     q6 0 32

q4 = (−) 0.0018 rad q5 = (−) 0.0056 m

q6 = (+) 0.00047 rad

Example 4.4 Determine the deflection and slope of beam shown in Fig. 4.10 at 1 m from right support. Given, E = 200 GPa; I = 4 × 10−6 m4 ; Discretisation: Number of beam elements : (3) Total number of d.o.f :8

p0 = 50 kN/m

94

Applied Finite Element Analysis

p0 (kN/m)

1m

1m q1 1

q2

q5

q3

1 q4

1m

2

q6

q7

3

2 3

q8

4

FIGURE 4.10 Beam with varying load.

Element matrices: Element (1)

[K (1) ] =

200 × 109 × 4 × 10−6 (1)3

     

1 12

2 6

3 (−)12

4  6 1 6 4 (−)6 2   2  (−)12 (−)6 (+)12 (−)6   3 6 2 (−)6 4 4 3 12

4 6

5 (−)12

5 12

6 7 6 (−)12

Element (2)

[K (2) ] =

200 × 109 × 4 × 10−6 (1)3

     

6  6 3 6 4 (−)6 2   4  (−)12 (−)6 (+)12 (−)6   5 6 2 (−)6 4 6

Element (3)

[K (3) ] =

200 × 109 × 4 × 10−6 (1)3

     

8  6 5  6 4 (−)6 2  6  (−)12 (−)6 (+)12 (−)6   7 6 2 (−)6 4 8

Analysis of Beams

95

Global stiffness matrix:

[K]8×8 = 8 ×

105

1

2

3

4

5

6

7

8

12

6

–12

6

0

0

0

0

1

6

4

–6

2

0

0

0

0

2

–12

–6

+ 12 + 12

–6+6

–12

6

0

0

3

6

2

–6 + 6

4+4

–6

2

0

0

4

0

0

–12

–6

+ 12 + 12

–6 + 6

–12

6

5

0

0

6

2

–6 +6

4+4

–6

2

6

0

0

0

0

–12

–6

+12

–6

7

0

0

0

0

6

2

–6

4

8

Boundary conditions: q1 = q2 = q3 = q7 = q8 = 0 Incorporating boundary conditions, the reduced stiffness matrix is 5 6  4 8 −6 2 4 [K]3×3 = 8 × 105  −6 24 0  5 2 0 8 6

Load vector for linearly varying distributed load: Since linearly varying load can be described by using shape function Nj , p = Nj p0 (1 + ξ) where ξ is natural coordinate. 2   H1    (e)  +1  l 2 H2  (e)    (Nj p0 ) l dξ {F (e) } =  H3  2   ξ=−1    l(e)  2 H4

Shape function Nj =

Load vector

1 (2 − 3ξ + ξ 3 ); 4 1 H2 = (1 − ξ − ξ 2 + ξ 3 ); 4 H1 =

1 H3 = (2 + 3ξ − ξ 3 ); 4 1 H4 = (−1 − ξ + ξ 2 + ξ 3 ) 4

Applied Finite Element Analysis

96

(−) (−) (e) Load vector {F } = (−) (+)

   

(0.15) p0 l(e) (0.033) p0 (l(e) )2  (0.35) p0 l(e)   (0.05) p0 (l(e) )2

   

  (−)7.5       (−)1.65 3 = 10 (−)17.5          +2.5

3 4 5 6

Incorporating boundary conditions, the reduced load vector    −1.65  4 −17.5 5 {F } = 103   +2.5 6 Reduced global equations: 

8 8 × 105  −6 2

    −6 2 q4 −1.65 24 0   q5  = 103  −17.5  q6 0 8 +2.5



  −1   q4 8 −6 2 −1.65 1  q5  =  −6 24 0   −17.5  2 8 × 10 2 0 8 +2.5 q6     q4 −0.00138  q5  =  −0.00125  q6 +0.00073

Deflection and slope at 1 m from the right support: q5 = −0.00125 m

q6 = +0.00073 rad.

4.7

STIFFNESS MATRIX FOR AN INCLINED BEAM

Consider a beam element which is inclined at an angle θ with the horizontal. In the local coordinate system X  O Y  shown in Fig. 4.11, the degrees of freedom considered at each node are i) deflection normal to the element ii) slope.

Local coordinate system X  O Y  Number of nodes in the beam element : (2) Number of degrees of freedom per node : 2 (1 Translational + 1 Rotational) Let the node i = 1 and j = 2 for simplifying our analysis. Nodal displacement vector

{q  }4×1 = [q1

q2

q3

q4 ]T

(4.49)

Analysis of Beams

q′3

Y

97

X′

q′4 j=2

Y′

l(e) q′1

θ q′ Ο′ 2 i=1 X

O FIGURE 4.11 Local coordinate system.

q1 − Displacement normal to the element at node i = 1

q3 − Displacement normal to the element at node j = 2 q2 − Slope at node i = 1

q4 − Slope at node j = 2 If the element is viewed in global coordinate system XOY , the number of degrees of freedom at each node is three i) deflection in the horizontal direction, ii) deflection in the vertical direction and iii) slope.

Global coordinate system XOY Number of nodes in the beam element : 2 Number of degrees of freedom per node : 3 (2 Translational and 1 Rotational) Nodal displacement vector {q}6×1 = [q1 q2 q3 q4 q5  q1 deflections in the horizontal direction q4  q2 deflections in the normal direction q5  q3 slopes. q6

q6 ]T

(4.50)

Relation between nodal displacement vectors can be established by identifying the transformation matrix. q1 = −q1 sinθ + q2 cosθ = −q1 m + q2 l q2 = q3

where l = cosθ, m = sinθ

(4.51) (4.52)

98

Applied Finite Element Analysis

Y

q5

q2

i=1

q6 j=2 q3

q4

q1 X

O

FIGURE 4.12 Global coordinate system.

q3 = −q4 sinθ + q5 cosθ = −q4 m + q5 l

     

q4 = q6  q1     q2 

q3         q   4

4×1



 = 

−m l 0 0 0 0 0 0

0 0 0 0 1 0 0 0 0 −m l 0 0 0 0 1

{q  }4×1 = [L]4×6 {q}6×1  −m l 0 0 0 0  0 0 1 0 0 0 Transformation matrix [L]4×6 =   0 0 0 −m l 0 0 0 0 0 0 1

(4.53)

  q1         q2         q 3   q4        q5    4×6   q   6 6×1

   

Element stiffness matrix defined in local coordinate system  12 6L (−)12 6L 2 (−)6L 2 EI  2L 6L 4L (e)  [K ] 4×4 = 3  L  (−)12 (−)6L (+)12 (−)6L 4L2 6L 2L2 (−)6L Element stiffness matrix in global coordinate system is expressed as [K (e) ]6×6 = [L]T6×4 [K (e) ] 4×4 [L]4×6

(4.54)

(4.55)

(4.56) (4.57)

   

(4.58)

(4.59)

Similarly, if the element load vector in local coordinate system is {F  }4×1 , the element global load vector is {F }6×1 = [L]T6×4 {F  }4×1

(4.60)

Analysis of Beams

4.8

99

PLANE FRAME ELEMENT

When the members are connected rigidly in a plane structure, the element used to describe the nodal degrees of freedom includes 2 translational degrees of freedom and 1 rotational degree of freedom. Thus, at each node three degrees of freedom will be used. vi

vi

θi

ui

ith node

(e)

θj jth node

uj

FIGURE 4.13 Plane frame element.

The frame element shown in Fig. 4.13 consists of 6 degrees of freedom and element stiffness matrix is [K (e) ]6×6 . The frame element shown in Fig. 4.13 suggests that it can be considered a combination of a bar element (with 2 nodes and one translation d.o.f at each node) and a beam element (with 2 nodes and one translational along with one rotational d.o.f). The element stiffness matrix

[K (e) ]6×6 =

1

2

3

4

5

6

EA L

0

0

−EA L

0

0

12EI L3

6EI L2

0

(−) 12EI L3

6EI L2

2

0

6EI L2

4EI L

0

(−) 6EI L2

2EI L

3

−EA L

0

0

EA L

0

0

4

0

(−) 12EI L3

(−) 6EI L2

0

(+) 12EI L3

(−) 6EI L2

5

0

6EI L2

2EI L

0

(−) 6EI L2

4EI L

6

1

(4.61)

Inclined plane frame element If the frame element is inclined at an angle θ with the horizontal as shown in Fig. 4.14 and the direction cosines are l = cosθ, m = sinθ

100

Applied Finite Element Analysis

vj

uj

L

θj

ui

vi

θ i θi FIGURE 4.14 Inclined frame element.

The transformation matrix [L]6×6 is 1 2

[L]6×6 =

3

4

5

6

l

m

0

0

0

0

1

−m

l

0

0

0

0

2

0

0

1

0

0

0

3

0

0

0

l

m

0

4

0

0

0

l

0

5

0

0

0

−m

0

1

6

0

(4.62)

If [K (e) ] is the element stiffness matrix given in local coordinate system which is given by above Eq.(4.61), the element stiffness matrix expressed in global stiffness matrix is [K (e) ]6×6 = [L]T [K (e) ] [L] and global load vector

{F (e) } = [L]T {F (e) }

(4.63)

Example 4.5 A rigid frame is shown in Fig. 4.15 and subjected to a vertical point load at node 2. Find the displacement vector and calculate the nodal forces for the inclined member in its local coordinate system. Given, A = 50 cm2 ; I = 500 cm4 ; E = 200 Gpa The frame is discretised with two elements.

Analysis of Beams

P = 100 kN 500 cm 2 M = 250 N.m

0c m

2

50

400 cm

3

1

1 300 cm FIGURE 4.15 Frame.

Y q

q8

X′

5

θ q6

q2

Y′

q4

2

q9

q7

1 q1 O

X

q3

FIGURE 4.16 Element with global degrees of freedom.

q′5

q′4 q′6

q′2 θ q′3 q′1 FIGURE 4.17 Local degrees of freedom.

No. of nodes per element : (2) No. of d.o.f per node : 3 (2 Translational + 1 Rotational)

101

102

Applied Finite Element Analysis

Analysis of inclined member (element 1): EI 200 × 105 × 500 = = 80 N/cm L3 (500)3 AL2 50 × (500)2 = = 25000 I 500 6L = 3000 cm, 2L2 = 2(500)2 = 5 × 105 cm2 ,

4L2 = 106 cm2

Element stiffness matrix expressed in local coordinate system X  OY 



K

 (1) 

AL2 I 0

0

0

12

0

EI = 3 L −

AL2 I 0 0



K (1)



= 80

0

0

6L

AL2 I 0

–12

6L

6L

4L2

0

–6L

2L2

0

0

–12 6L

−

0

0

–6L

AL2 I 0

12

–6L

2L2

0

–6L

4L2

1

2

3

4

5

6

25000

0

0

–25000

0

0

1

0

12

3000

0

–12

3000

2

0

3000

106

0

–3000

3

–25000

0

0

25000

0

5 × 105 0

4

0

–12

–3000

0

12

–3000

5

0

3000

5 × 105

0

–3000

106

6

The displacement vector for element (1) written in global coordinate system XOY {q (1) }6×1 = [q1

q2

q3

q4

q5

q6 ] T

The displacement vector for element (1) in local coordinate system {q (1) }6×1 = [q1

q2

q3

q4

q5

q6 ]T

Relation between {q (1) } and {q (1) } is {Q(1) }6×1 = [L(1) ]6×6

{q (1) }6×1

Analysis of Beams

103

where the transformation matrix [L(1) ] is



L(1)



6×6

=

l

m

0

0

0

0

−m

l

0

0

0

0

0

0

1

0

0

0

0

0

0

l

m

0

0

0

0

l

0

0

0

0

−m

0

1

0

4 3 For element (1), direction cosines l = , and m = . 5 5 Element stiffness matrix for element (1) in global coordinate system [K (1) ] = [L(1) ]T [K (1) ] [L(1) ]



K (1)



6×6

=

103

1

2

3

4

5

6

720.614

959.539

–192

–720.614

–959.539

–192

1

959.539

1280.346

144

–959.539

–1280.346

144

2

–192

144

80000

192

–144

40000

3

–720.614

–959.539

192

720.614

959.539

192

4

–959.539

–1280.346

–144

959.539

1280.346

–144

5

–192

144

40000

192

–144

80000

6

Element stiffness matrix for element (2) which is a horizontal member written in local coordinate system is 4 5 6 7 8 9



K

 (2) 

= 80

25000

0

0

–25000

0

0

4

0

12

3000

0

–12

3000

5

0

3000

106

0

–3000

6

–25000

0

0

25000

0

5 × 105

0

–12

–3000

0

0

3000

5 × 105

0

0

7

12

–3000

8

–3000

106

9

104

Applied Finite Element Analysis

Transformation matrix for element (2) : [L(2) ] Direction cosines l = 1, m = 0



1 0 0 0 0 0

 L(2) =

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 1 0 0

0 0 0 0 1 0

0 0 0 0 0 1

Element stiffness matrix [K (e) ] for element (2) expressed in global coordinate system is [K (2) ] = [L(2) ]T [K (2) ] [L(2) ]



K (2)



= 80

4

5

6

7

8

9

25000

0

0

–25000

0

0

4

0

12

3000

0

–12

3000

5

0

3000

106

0

–3000

–25000

0

0

25000

0

0

–12

–3000

0

0

3000

5 × 105

0

5×

105

6

0

7

12

–3000

8

–3000

106

9

It can be noted that [K (2) ] = [K (2) ] because the member (2) of the frame is horizontal and aligned parallel to global coordinate system. Assembling element stiffness matrices and imposing essential boundary conditions, q1 = q2 = q3 = q7 = q8 = q9 = 0      We get, 2720.614 959.539 192 q4 0 96   q5  = 103  −100  103  959.539 1281.306 q6 192 96 160000 25

where the matrix on the right hand side of the equation is the load vector. Solving the above equations, the following nodal displacements are obtained. q4 = 0.03736 cm, q5 = −0.10606 cm, q6 = 0.000175 rad

Nodal forces and moment acting on the inclined member (1) calculated in local coordinate system is

Analysis of Beams

105

{F (1) } = [K (1) ] [L(1) ] {q (1) } where {q (1) } is the nodal displacement vector expressed in global coordinate system. {q (1) } = 

1 2 3 4 5 0 0 0 0.03736 −0.10606

6 T 0.000175

Hence the nodal forces and moment {F (1) }

=



1 2 3 4 124830.4 131.8141 29453.54 −124830.4

5 −131.8141

6 T 36453.54

The nodal forces and moment acting on member 1 of the frame are shown in Fig. 4.18. 124830.4 N

36,453.54 N.cm 131.8141 N 1

131.8141 N

29,453.54 N.cm 124830.4 N FIGURE 4.18 Forces and moments on member 1.

PROBLEMS 4.1 Find the nodal displacements and slope at the mid-point of element (2) of the shaft shown in Fig. P4.1. The shaft is assumed to be supported in bearings regarded as fixed support at A and B. p =15 kN/m

A

B

1

2

0.5 m

0.5 m FIGURE P4.1

E = 200 GPa; I = 5 × 104 mm4

106

Applied Finite Element Analysis

4.2 Determine the nodal displacements and slopes for the beam shown in Fig. P4.2. Find the moment at the mid-point of element (1). P = 3 kN M 3

1

1

2

2m

2 2m

FIGURE P4.2

E = 200 GPa; I = 5 × 104 mm4 ; M = 6 kNm 4.3 Find the nodal displacements and slopes at the position of one-fourth distance from the left support of shaft. The shaft is simply supported at A and B. 5 kN

A

B 1

2

1m

1m FIGURE P4.3

E = 200 GPa; I = 6 × 104 mm4 4.4 Formulate the system equations for the shaft shown in Fig. P4.4. Consider the shaft to be simply supported at bearings A and B. 10 kN/m

I2

I1 A

B I1 0.5 m

1m

1m

FIGURE P4.4

E = 200 GPa; I1 = 6 × 104 mm4 ; I2 = 3 × 104 mm4

Analysis of Beams

107

4.5 Analyse the rigid frame shown in Fig. P4.5 and find shear force and beading moment for the individual member. Every joint is a rigid joint where members are connected. 500 cm 2

3

2

200 cm

3

200 cm

1 4 50 kN 200 cm

FIGURE P4.5

E = 200 GPa; A = 200 cm2 ; I = 50 × 103 cm4

for all members.

CHAPTER

STEADY STATE CONDUCTION HEAT TRANSFER

5

Steady state temperature distribution within a heated body is required in several situations. The non-uniform temperatures at different points may cause thermal stresses when the body is restrained. The evaluation of thermal stresses in machine components is needed for design and checking the components for their effective performance. Further the temperature variation in a body may create changes in metallurgical structure which may result in considerable changes in hardness. The hardness is very important for the components of antifriction bearings and high hardness has to be obtained to prevent failure. In view of the possible undesirable effects which can lead to failure of the components because of temperature variation, it is essential to obtain the temperature distribution within a body. In this chapter, we focus upon the finite element method for obtaining temperature distribution in a heated body and ultimately the rate of heat transfer.

5.1

GOVERNING EQUATIONS OF STEADY STATE CONDUCTION HEAT TRANSFER

The energy equation or differential equation governing temperature distribution in a three-dimensional orthotropic body by writing the energy balance over a differential element can be expressed as:       ∂ ∂ ∂ ∂Φ ∂Φ ∂Φ kx + ky + kz +Q=0 (5.1) ∂x ∂x ∂y ∂y ∂z ∂z where Φ is the field variable which is temperature in a heat transfer problem. The temperature is a scalar quantity hence at each node there is only one unknown, that is, the magnitude of temperature. In other words, comparing with the terminology used in structural mechanics it can be stated as one degree of freedom at each node. Q in the above Eq.(5.1) represents the rate of heat generation (Watts/m3 ) due to a uniformly distributed heat source present in the heat conduction region. kx , ky and kz are thermal conductivities (Watt/m ◦ C).

5.2

BOUNDARY CONDITIONS

Boundary conditions associated with a heat conduction problem can be presented in a general form as:

Steady State Conduction Heat Transfer

109

Y

φ1 S1 Q h, φf

S3

S2 q′′

X

O FIGURE 5.1 Boundary conditions.

i) Specified temperature:Φ =  Φ1 acting 1.  on surface S  ∂Φ ∂Φ ∂Φ l + ky m + kz n + q  = 0 on surface S2 , and ii) Specified heat flux: kx ∂x ∂y ∂z q  is heat flux moving out of the surface (W/m2 ). l, m, n are direction cosines of the outward normal to the surface. iii) Specified convection due to interaction with an adjacent fluid       ∂Φ ∂Φ ∂Φ l + ky m + kz n + h(Φ − Φ∞ ) = 0 (5.2) kx ∂x ∂y ∂z On surface S3 , h is convection heat transfer coefficient between the surface and adjacent fluid maintained at a temperature of Φf . In the present discussion, radiation boundary condition has not been considered which requires special treatment due to its non-linear nature.

5.3

TWO-DIMENSIONAL HEAT CONDUCTION ANALYSIS

Two-dimensional heat conduction equations can be written from the above equations eliminating the terms referred to the Z-direction.     ∂Φ ∂Φ ∂ ∂ kx + ky +Q=0 (5.3) Energy equation is ∂x ∂x ∂y ∂y Boundary conditions: Φ = Φ1 on surface S1     ∂Φ ∂Φ kx l + ky m + q  = 0 on the surface S2 ∂x ∂y     ∂Φ ∂Φ l + ky m + h(Φ − Φf ) = 0 on the surface S3 (5.4) kx ∂x ∂y

110

Applied Finite Element Analysis

For applying finite element method, we require a functional and minimise it with respect to nodal temperatures. For structural mechanics or elasticity problems, the functional was written in terms of potential energy which was in an integral form. But the energy equation for heat transfer problem is in differential form which cannot be used as a functional. Hence, it is necessary to use a method by which a functional pertaining to heat transfer problem can be formulated. Several methods are employed but in this chapter, we introduce variational formulation and one of the popular weighted residual methods, Galerkin’s method.

5.4

FORMULATION OF FUNCTION

Variational formulation: Without going through the mathematical details of the calculus of variation, the functional for two-dimensional heat conduction problem can be stated as     2      1 ∂Φ 2 ∂Φ 1 = kx + ky dV − QΦdV + q  ΦdS + h(Φ − Φf )2 dS (5.5) 2 ∂x ∂y 2 V

V

S2

S3

The above functional expression suggests that if you have second order derivatives in the governing differential equation, the functional consists of squares of first order derivatives in the volume integral which is similar to the strain energy expression of a structural mechanics problem. For heat generation term Q, it should be multiplied by the field variable Φ and integrate over the volume but it is associated with a minus sign. Another significant difference, we observe is that the heat flux and convective boundary conditions appear in the functional and they are natural boundary conditions for a heat transfer problem which need not be imposed unlike essential boundary conditions i.e., specified temperature Φ on surface S1 . The heat flux q  term is to be multiplied by Φ and integrated over the surface S2 on which it acts and convection term 1/2h(Φ − Φf )2 is to be integrated over the surface S3 since the convection boundary condition consists of (Φ − Φf ) term in it. The functional can be written using the inferences mentioned above because we have not gone through the rigorous mathematical analysis of calculus of variation. An important point to note is that the variational formulation cannot be applied if we have first order derivative in the governing differential equation. Alternate method in this case is Galerkin’s method which we will discuss later. The functional given by Eq.(5.5) is to be minimised with respect to nodal temperatures to obtain element equations. Before minimisation process is employed, we have to discretise the heat conduction domain into finite elements and express the field variable Φ in terms of shape functions and nodal temperatures.

Discretisation of two-dimensional domain A rectangular plate whose thickness is small compared to other dimensions is chosen for analysis. The domain is discretised using two-dimensional linear element that is triangular element which consists of three nodes placed at the vertices of a triangle Fig. 5.2.

Steady State Conduction Heat Transfer

111

Y

k (e) i

j

X

O FIGURE 5.2 Discretisation.

One triangular element with nodes i, j and k numbered in an anticlockwise direction is shown in Fig. 5.3. Numbering of nodes must be in an anticlockwise direction to see that the calculation of the area of the triangle yields positive value. Y

Φk k(xk , yk)

P(x, y)

Φj j(xj, yj)

Φi i(xi , yi) X

O FIGURE 5.3 Linear triangle element.

Let us consider the point P (x, y) within the element, and the nodal coordinates (xi , yi ), (xj , yj ), and (xk , yk ) are shown in Fig. 5.3. Φi , Φj and Φk are nodal temperatures and Φ is the temperature at the point P (x, y) within the element. The field variable approximation for the linear triangle element is Φ(x, y) = a1 + a2 x + a3 y

(5.6)

For evaluating the constants a1 , a2 and a3 in the polynomial expression, we use the following conditions.

112

Applied Finite Element Analysis

i) At node i (x = xi , y = yi ); Φ = Φi ii) At node j (x = xj , y = yj ); Φ = Φj iii) At node k (x = xk , y = yk ); Φ = Φk and substituting in Eq.(5.6)



  Φi  Φj  =  Φk    a1 1  a2  =  1 a3 1

1 1 1

  yi a1 yj   a 2  yk a3 −1   yi Φi yj   Φj  yk Φk

xi xj xk

xi xj xk

(5.7)

(5.8)

After evaluating values for a1 , a2 and a3 and substituting in Eq.(5.6) Φ(x, y) = Ni Φi + Nj Φj + Nk Φk = Φ(x, y) = [N ] {Φ}



Ni

Nj

Nk





 Φi  Φj  Φk

(5.9) (5.10)

where shape functions associated with i, j and k nodes are 1 (ai + bi x + ci y) 2A 1 (aj + bj x + cj y) Nj = 2A 1 (ak + bk x + ck y) Nk = 2A    1 xi yi    1 and A = area of the triangular element =  1 xj yj  2 1 x y  k k Ni =

ai = xj yk − xk yj ;

bi = (yj − yk ); ci = (xk − xj )

ak = xi yj − xj yi ;

bk = (yi − yj ); ck = (xj − xi )

aj = xk yi − xi yk ;

bj = (yk − yi ); cj = (xi − xk )

(5.11)

(5.12)

it is to be noted that ai + aj + ak = 2A;

bi + bj + bk = 0

ci + cj + ck = 0;

Ni + Nj + Nk = 1

(5.13)

Steady State Conduction Heat Transfer

5.5

113

ELEMENT MATRICES

Defining a vector of gradients of field variable as  ∂Ni  T  ∂Φ ∂Φ ∂x = {g} =  ∂Ni ∂x ∂y ∂y {g} = [B] {Φ}   1 bi bj bk where [B] = 2A ci cj ck 2×3

∂Nj ∂x ∂Nj ∂y

   ∂Nk Φi  ∂x   Φj  ∂Nk  Φk ∂y

(5.14) (5.15) (5.16)

{g} matrix appears to be similar to {ε} strain matrix in elasticity problems.       ∂Φ 2 ∂Φ 2 The first term in the functional, kx + ky which is in quadratic form ∂x ∂y be expressed in terms of {g} matrix as   ∂Φ  2  2     ∂Φ ∂Φ ∂φ ∂φ kx 0   ∂x  = {g}T [D] {g} (5.17) + ky = kx  0 ky ∂Φ  ∂x ∂y ∂x ∂y ∂y Substituting for {g} from the Eq.(5.15) into Eq.(5.17) kx



∂Φ ∂x

2

+ ky



∂Φ ∂y

2

= {Φ}T [B]T [D] [B] {Φ}

where the material property matrix [D] is



[D] = The functional



=



1 {Φ}T  2



V





kx 0

0 ky

(5.18)



(5.19)

can now be written as







[B]T [D] [B] dV  {Φ} − 

V





Q [N ] dV  {Φ} + 

S2

S3

S3

Minimizing Eq.(5.20) with respect to nodal temperature Φi , Φj , Φk



q  [N ] dS  {Φ}

     1 1 T  T    + {Φ} hΦf [N ]dS {Φ} + h [N ] [N ]dS {Φ} − hΦ2f dS 2 2 S3





(5.20)

114

Applied Finite Element Analysis



    ∂Π = ∂ {Φ}    

∂Π ∂Φi ∂Π ∂Φj ∂Π ∂Φk



    =0    

yields the element equations      K (e) Φ(e) = F (e)

(5.21)

where element conductance matrix [K (e) ] is given by



    (e) (e) K = K1 + K2    (e) K1 = [B]T [D] [B] dV due to conduction within the element 

(e)

(e) K2





(5.22)

V

=



h [N ]T [N ] dS due to convection

S3

The Eq.(5.22) shows that the element conductance matrix consists of two matrices. i) Due to conduction within the element and ii) Contribution due to convection if convective mode of heat transfer is present in the given problem.      T Element load vector F (e) = Q [N ]T dV − q  [N ] dS + hΦf [N ]T dS V

Element nodal temperature vector    Φ(e) = Φi

S2

Φj

Φk

S3

T

(5.23)

One important observation we have to make at this stage is that heat generation Q, heat flux q  and convection term hΦf , contribute to load vector. We find that if convection is present in the problem, it contributes to conductance matrix as well as load vector. The minus in heat flux term of load vector appears because q  is regarded as heat flux that leaves the surface in the direction of outward normal from the boundary. Care should be taken while writing the terms due to heat flux as the minus and plus sign affect the calculations significantly.

Steady State Conduction Heat Transfer

115

The global or system equations are obtained by assembling the element equations. n       K (e) Φ(e) − F (e) = 0 where n is the number of elements

e=1

[K] {Q} = {F }

(5.24)

Essential boundary conditions are imposed before solving the system equations.

5.6

ESSENTIAL AND NATURAL BOUNDARY CONDITIONS

Essential boundary conditions are field variables specific to the problem and natural boundary conditions are higher order derivatives of the field variables under consideration. Essential boundary conditions need to be imposed on system equations before solving them. Natural boundary conditions need not be imposed on system equations because these are included in the load vector. Hence, in a heat conduction problem, i) Essential boundary condition is specified temperature Φ = Φ1 on the surface S1 . ii) Natural boundary conditions are derivatives of field variable (temperature) which are heat flux q  and connective boundary condition acting on surface S2 and S3 respectively. At this stage we can make a mention of the solution techniques employed to obtain a solution. The reduced system equations which include global stiffness matrix are solved by Gaussian elimination or Cholesky decomposition in banded solution techniques. While formulating the global stiffness, it is essential to minimise the bandwidth of the matrix. For minimizing the bandwidth, nodal numbering in an element plays an important role on bandwidth. Two types of numbering are shown in Figs. 5.4 and 5.5. Y

8

12

16

20

24

4

23

3 7

11

15

19

6

10

14

18

2

22

1 1 O

5

9

13

17

21

FIGURE 5.4 Numbering of nodes in vertical direction.

X

116

Applied Finite Element Analysis

Y

20

22

21

23

24

19

13 15

14

10

9

8

7

16

18

17 11

12

1 O 1

2

3

4

6

5

X

FIGURE 5.5 Numbering of nodes in horizontal direction.

In Fig. 5.4 nodes are numbered first along the short dimension and then along the longer dimension. In Fig. 5.5, nodes are numbered first along the longer dimension. The semi-bandwidth is calculated by the formula. B = (D + 1)N

(5.25)

D = maximum difference of nodal numbers in an element N = Number of degrees of freedom at each node. Let us consider the element (1) in Figs. 5.4 and 5.5

6

In Fig. 5.4 for element (1) D = 5,

∴

1

N =1

B = (5 + 1)1 = 6

5

1

In Fig. 5.5 for element (1)

8

D = 7,

∴

N =1

B = (7 + 1)1 = 8 1 1

2

The bandwidth is given by (2B − 1), hence the bandwidth is smaller while numbering along the shorter dimension first. Minimisation of bandwidth is useful in two ways: i) it reduces the storage requirements. ii) it reduces computing time.

Steady State Conduction Heat Transfer

117

Many commercial finite element software use a technique known as wave front or frontal solution method. In this technique, the stiffness matrix is not completely assembled to obtain solution. Element numbering is critical so as to reduce computing time in wave front method. Wave front technique will be discussed later. Formulation of matrices for two-dimensional problems:    (e) K1 = [B]T [D] [B] dV V

1 [B] = 2A

  kx δ (e) K1 = 4A



    

bi ci

bj cj b2i

bk ck



,

[D] = bi bj b2j

symmetric



bi bk

kx 0 

0 ky

(5.26)



 k δ  y bj bk  +  4A b2k

(5.27)

    

c2i

ci cj c2j

symmetric

ci ck



  cj ck  (5.28)  2 ck

where δ is the thickness of the plate. Let us consider a triangular element shown in Fig. 5.6 with Q be the rate of heat generation within the element, q  heat flux leaving the side jk and convection on the side ij. Y k q′′

P (x,y)

i O

j

h, Φf

X

FIGURE 5.6 Element with heat flux and convection.

118

Applied Finite Element Analysis



(e)

K2



=



h[N ]T [N ]dS =

S



Lij



  h 

Ni2

Ni Nj Nj2

symmetric

Ni Nk



  Nj Nk  δdLij  2 Nk

Nk = 0 on the side joining nodes i and j



Ni2

     (e) ∴ K2 = hδ  Ni Nj  0

Ni Nj Nj2 0

0



  0 dLij  0

(5.29)

Using the factorial integration formula Eq.(2.145)





Ni2 dLij

=

Lij

Lij

∴



Lij 2! L2i dLij = Lij = 3! 3

Lij

Ni Nj dLij =



Lij 1! Li Lj dLij = Lij = 3! 6

(5.30)

Lij

  2 1 0   hδL ij  (e) 1 2 0  K2 = 6 0 0 0

(5.31)

Before evaluating the integrals involving the area A, let us define area coordinates. Area coordinates: k

P(x,y)

i FIGURE 5.7

j

Steady State Conduction Heat Transfer

119

Area of ∆P jk Area of ∆ijk   1 x y 1 1 Area of ∆P jk = det  1 xj yj  = (ai + bi x + ci y) 2 2 1 xk yk Li =

Area of ∆ijk = A 1 Li = (ai + bi x + ci y) = Ni 2A

(5.32)

Similarly, it can be shown Area Area Area Lk = Area Lj =

of of of of

1 ∆ipk = (aj + bj x + cj y) = Nj ∆ijk 2A 1 ∆ijp = (ak + bk x + ck y) = Nk ∆ijk 2A

we can utilise now the integral equations involving area coordinates Li , Lj , Lk Factorial integration formula:  a!b!c! (2A) Lai Lbj Lck dA = (a + b + c + 2)!

(5.33) (5.34)

(5.35)

A

Using the above identity, we can evaluate the following integrals.   2 Ni dA = L2i dA

(5.36)

A

Comparing the Eq.(5.35) with (5.36) a = 2, b = c = 0 A 2! ∴ Ni2 dA = (2A) = 4! 6   A 1!1!(2A) = Ni Nj dA = Li Lj dA = 4! 12   A 1!(2A) = Ni dA = Li dA = 3! 3



Similarly, and

(5.37) (5.38) (5.39)

Considering the element shown in Fig.5.6, let us evaluate the element load vector {F (e) } :      T T  (e) (5.40) F = Q [N ] dV − q [N ] dS + hΦf [N ]T dS V

S2

S3

Applied Finite Element Analysis

120



Q [N ]T dV =

V



Q [N ]T δdA = Qδ

A



From Eq.(5.39)







Ni



   Nj dA

(5.41)

Nk



 1 i QδA  T  1 j Q [N ] dA = 3 1 k

(5.42)

 Ni T q  [N ] dS = q  δ  Nj dLjk where heat flux q  leaving the side jk on the side Nk   Ljk Ljk and jk, Ni = 0 and from factorial integration formula Nj dLjk = Nk dLjk = 2 2    0 i  δL q jk  T 1  j (5.43) δ q  [N ] dLjk = 2 1 k 



and when convection occurs on side ij



hΦf [N ]T δdLij = hΦf δ

but Nk = 0 on the side ij

∴ hΦf δ









Ni



   Nj dLij Nk



  1 i hΦ δL   ij  f 1  j  Nj dLij = 2 0 k Nk Ni

(5.44)

Similarly equations have to be suitably modified considering the edges on which heat flux q  and convection are specified.

5.7

ONE-DIMENSIONAL STEADY STATE CONDUCTION ANALYSIS

The computational domain is discretised, for example, with one-dimensional linear element with two nodes.     Φi Φ(x) = [Ni Φi + Nj Φj ] = Ni Nj (5.45) Φj  1  {g} = [B] {Φ} , [B] = (e) −1 1 (5.46) l

Steady State Conduction Heat Transfer

121

Element conductance matrix :

  

(e) K1 (e) K2

 

= =

 

     (e) (e) K (e) = K1 + K2

k (e) A(e) [B] [D] [B] dV = l(e) T

i



1 −1

j

l(e)



−1 1

(5.47)

h [N ]T [N ] dS

If the convection occurs at ith node



(e) K2



= hA



1 0 0 0



since Ni = 1 and Nj = 0.

    0 0 (e) since Ni = 0 and Nj = 1. If the convection occurs at jth node K2 = hA 0 1 Element load vector F (e) :      T F (e) = Q [N ]T dV − q  [N ] dS + hΦf [N ]T dS (5.48)  

F

F

(e)

(e)



QAl(e) = 2



QAl(e)

=

2

 

1 1



1 1



i − q  A j



1 0



i j

+ hAΦf

heat flux at ith node

i − q  A j



0 1



i j

heat flux at jth node



1 0



i j

(5.49)

convection at ith node

+ hAΦf



0 1



i j

(5.50)

convection at jth node

While formulating load vector, the relevant equations should be used with respect to a specific problem. The above equations indicate the method to write when boundary conditions are specified at a node.

Example 5.1 A plate of thickness 10 mm is subjected to boundary conditions shown in Fig. 5.8. Find the nodal temperatures. Thickness δ = 10 mm, δ