FINITE ELEMENT ANALYSIS

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FINITE ELEMENT ANALYSIS

A Primer

By

S. M. Musa
A. V. Kulkarni
V. K. Havanur

Copyright ©2014 by MERCURY LEARNING AND INFORMATION LLC. All rights reserved.
Reprinted and revised with permission.

Original title and copyright: A Primer on Finite Element Analysis. Copyright © 2011 by Laxmi Publications Pvt. Ltd. All rights reserved. ISBN: 978-93-81159-10-1

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S. M. Musa, A. V. Kulkarni, V. K. Havanur. FINITE ELEMENT ANALYSIS: A Primer.
ISBN: 978-1-938549-34-2

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Dedicated to my late father, Mahmoud, my mother, Fatmeh, and my wife, Lama.

CONTENTS

Preface

Acknowledgments

Chapter 1: Mathematical Preliminaries

1.1 Introduction

1.2 Matrix Definition

1.3 Types of Matrices

1.4 Addition or Subtraction of Matrices

1.5 Multiplication of a Matrix by Scalar

1.6 Multiplication of a Matrix by Another Matrix

1.7 Rules of Matrix Multiplications

1.8 Transpose of a Matrix Multiplication

1.9 Trace of a Matrix

1.10 Differentiation of a Matrix

1.11 Integration of a Matrix

1.12 Equality of Matrices

1.13 Determinant of a Matrix

1.14 Direct Methods for Linear Systems

1.15 Gaussian Elimination Method

1.16 Cramer’s Rule

1.17 Inverse of a Matrix

1.18 Vector Analysis

1.19 Eigenvalues and Eigenvectors

1.20 Using MATLAB

Problems

References

Chapter 2: Introduction to Finite Element Method

2.1 Introduction

2.2 Methods of Solving Engineering Problems

2.2.1 Experimental Method

2.2.2 Analytical Method

2.2.3 Numerical Method

2.3 Procedure of Finite Element Analysis (Related to Structural Problems)

2.4 Methods of Prescribing Boundary Conditions

2.4.1 Elimination Method

2.4.2 Penalty Method

2.4.3 Multipoint Constrains Method

2.5 Practical Applications of Finite Element Analysis

2.6 Finite Element Analysis Software Package

2.7 Finite Element Analysis for Structure

2.8 Types of Elements

2.9 Direct Method for Linear Spring

Problems

References

Chapter 3: Finite Element Analysis of Axially Loaded Members

3.1 Introduction

3.1.1 Two-Node Bar Element

3.1.2 Three-Node Bar Element

3.2 Bars of Constant Cross-Section Area

3.3 Bars of Varying Cross-Section Area

3.4 Stepped Bar

Problems

References

Chapter 4: Finite Element Analysis Trusses

4.1 Introduction

4.2 Truss

Problems

References

Chapter 5: Finite Element Analysis of Beams

5.1 Introduction

5.2 Simply Supported Beams

5.3 Cantilever Beams

Problems

References

Chapter 6: Stress Analysis of a Rectangular Plate with a Circular Hole

6.1 Introduction

6.2 A Rectangular Plate with a Circular Hole

Problems

References

Chapter 7: Thermal Analysis

7.1 Introduction

7.2 Procedure of Finite Element Analysis (Related to Thermal Problems)

7.3 One-Dimensional Heat Conduction

7.4 Two-Dimensional Problem with Conduction and with Convection Boundary Conditions

Problems

References

Chapter 8: Fluid Flow Analysis

8.1 Introduction

8.2 Procedure of Finite Element Analysis (Related to Fluid Flow Problems)

8.3 Potential Flow Over a Cylinder

8.4 Potential Flow Around an Airfoil

Problems

References

Chapter 9: Dynamic Analysis

9.1 Introduction

9.2 Procedure of Finite Element Analysis (Related to Dynamic Problems)

9.3 Fixed-Fixed Beam for Natural Frequency Determination

9.4 Transverse Vibrations of a Cantilever Beam

9.5 Fixed-Fixed Beam Subjected to Forcing Function

9.6 Axial Vibrations of a Bar

9.7 Bar Subjected to Forcing Function

Problems

References

Chapter 10: Engineering Electromagnetics Analysis

10.1 Introduction to Electromagnetics

10.2 Maxwellfs Equations and Continuity Equation

10.2.1 Maxwellfs Equations and Continuity Equation in Differential Form

10.2.2 Maxwellfs Equations and Continuity Equation in Integral Form

10.2.3 Divergence and Stokes Theorems

10.2.4 Maxwellfs Equations and Continuity Equation in Quasi-Statics Case

10.2.5 Maxwellfs Equations and Continuity Equation in Statics Case

10.2.6 Maxwellfs Equations and Continuity Equation in Source-Free Regions of Space Case

10.2.7 Maxwellfs Equations and Continuity Equation in Time-Harmonic Fields Case

10.3 Lorentz Force Law and Continuity Equation

10.4 Constitutive Relations

10.5 Potential Equations

10.6 Boundary Conditions

10.7 Laws for Static Fields in Unbounded Regions

10.7.1 Coulomb’s Law and Field Intensity

10.7.2 Bio-Savart’s Law and Field Intensity

10.8 Electromagnetic Energy and Power Flow

10.9 Loss in Medium

10.10 Skin Depth

10.11 Poisson’s and Laplace’s Equations

10.12 Wave Equations

10.13 Electromagnetic Analysis

10.13.1 One-Dimensional Elements

10.13.1.1 The Approach to FEM Standard Steps Procedure

10.13.1.2 Application to Poisson’s Equation in One-Dimension

10.13.1.3 Natural Coordinates in One-Dimension

10.13.2 Two-Dimensional Elements

10.13.2.1 Applications of FEM to Electrostatic Problems

10.13.2.1.1 Solution of Laplace’s Equation ▿2V = 0 FEM

10.13.2.1.2 Solution of Passion’s Equation ▿2V = with FEM

10.13.2.1.3 Solution of Wave’s Equation ▿2Φ+k^2Φ = g with FEM

10.14 Automatic Mesh Generation

10.14.1 Rectangular Domains

10.14.2 Arbitrary Domains

10.15 Higher Order Elements

10.15.1 Pascal Triangle

10.15.2 Local Coordinates

10.15.3 Shape Functions

10.15.4 Fundamental Matrices

10.16 Three-Dimensional Element

10.17 Finite Element Methods for External Problems

10.17.1 Infinite Element Method

10.17.2 Boundary Element Method

10.17.3 Absorbing Boundary Conditions

10.18 Modeling and Simulation of Shielded Microstrip Lines with COMSOL Multiphysics

10.18.1 Rectangular Cross-Section Transmission Line

10.18.2 Square Cross-Section Transmission Line

10.18.3 Rectangular Line with Diamondwise Structure

10.18.4 A Single-Strip Shielded Transmission Line

10.19 Multistrip Transmission Lines

10.19.1 Double-strip Shielded Transmission Line

10.19.2 Three-strip Line

10.19.3 Six-strip Line

10.19.4 Eight-strip Line

10.20 Solenoid Actuator Analysis with ANSYS

Problems

References

Appendix A: ANSYS

Appendix B: MATLAB

Appendix C: Color Figures

Index

PREFACE

Today, the finite element method (FEM) has become a common tool for solving engineering problems in industries for the obvious reasons of its versatility and affordability. To expose an undergraduate student in engineering to this powerful method, most of the universities have included this subject in the undergraduate curriculum. This book contains materials applied to mechanical engineering, civil engineering, electrical engineering, and physics. It is written primarily for students and educators as a simple introduction to the practice of FEM analysis in engineering and physics. This book contains many 1D and 2D problems solved by the analytical method, by FEM using hand calculations, and by using ANSYS 11 academic teaching software and COMSOL. Results of all the methods have been compared. This book compromises 10 chapters and 3 appendices.

Chapter 1 contains mathematical preliminaries needed for understanding the chapters of the book. Chapter 2 provides a brief introduction to FEA, a theoretical background, and its applications. Chapter 3 contains the linear static analysis of bars of a constant cross-section, tapered cross-section, and stepped bar. In each section, a different variety of exercise problems is given. Chapter 4 contains the linear static analysis of trusses. Trusses problems are also selected in such a way that each problem has different boundary conditions to apply. Chapter 5 provides the linear static analysis of simply supported and cantilever beams. In Chapters 3 to 5, all the problems are considered as one dimensional in nature. Indeed, stress analysis of a rectangular plate with a circular hole is covered in Chapter 6. In this chapter, emphasis is given on the concept of exploiting symmetric geometry and symmetric loading conditions. Also, stress and deformation plots are given. Chapter 7 introduces the thermal analysis of cylinders and plates. Here both one dimensional and two dimensional problems are considered. Chapter 8 contains the problems of potential flow distribution over a cylinder and over an airfoil. Chapter 9 provides the dynamic analysis (modal and transient analysis) of bars and beams. Chapter 10 provides the engineering electromagnetics analysis. The chapter gives an overview of electromagnetics theory and provides the finite element method analysis toward the electromagnetics, some models are demonstrated using the COMSOL multiphysics application and also ANSYS.

The appendices are located on the companion disc in the back of the book. Appendix A contains the introduction to Classic ANSYS and ANSYS Workbench. Appendix B contains an overview of computational MATLAB. Appendix C contains the color figures in the book.

Sarhan M. Musa

Houston, Texas

ACKNOWLEDGMENTS

Thank you to Brain Gaskin and James Gaskin for their wonderful hearts and for being great American neighbors. It is my pleasure to acknowledge the outstanding help and support of the team at Mercury Learning and Information in preparing this book, especially from David Pallai. My gratitude to Dr. John Burghduff and Professor Mary Jane Ferguson for their support, understanding, and for being great friends. I would also like to thank Dr. Kendall Harris, my college dean, for his constant support. Finally, the book would never have seen the light of day if not for the constant support, love, and patience of my family.

Chapter 1

MATHEMATICAL PRELIMINARIES

1.1 INTRODUCTION

This chapter introduces matrix and vector algebra that is essential in the formulation and solution of finite element problems. Finite element analysis procedures are most commonly described using matrix and vector notations. These procedures eventually lead to the solutions of a large set of simultaneous equations. This chapter will be a good help in understanding the remaining chapters of the book.

1.2 MATRIX DEFINITION

A matrix is an array of numbers or mathematical terms arranged in rows (horizontal lines) and columns (vertical lines). The numbers, or mathematical terms, in the matrix are called the elements of the matrix. We denote the matrix through this book, by a boldface-letter, a letter in brackets [], or a letter in braces {}. We sometimes use {} for a column matrix. Otherwise, we define the symbols of the matrices.

Example 1.1

The following are matrices.

The size (dimension or order) of the matrices varies and is described by the number of rows (m) and the number of columns (n). Therefore, we write the size of a matrix as m × n (m by n). The sizes of the matrices in Example 1.1 are 2 × 2, 3 × 3, 2 × 1, 1 × 2, and 1 × 1, respectively.

We use aij to denote the element that occurs in row i and column j of matrix A. In general, matrix A can be written

Example 1.2

Location of an element in a matrix.

Find (a) size of the matrix A

(b) location of elements a_11, a_12, a_32, and a_33

Solution

(a)  Size of the matrix A is 3 × 3

(b)  a_11 is element a at row 1 and column 1

a_12 is element a at row 1 and column 2

a_32 is element a at row 3 and column 2

a_33 is element a at row 3 and column 3

Note that, two matrices are equal if they have the same size and their corresponding elements in the two matrices are equal. For example,

are not the same size. Also, [B] ≠ [C] since the corresponding elements are not all equal.

1.3 TYPES OF MATRICES

The types of matrices are based on the number of rows (m) and the number of columns (n) in addition to the nature of elements and the way the elements are arranged in the m atrix.

(a)

Rectangular matrix is a matrix of different number of rows and columns, that is, m ≠ n. For example, the matrix

(b)

Square matrix is a matrix of equal number of rows and columns, that is, m = n. For example, the matrix

(c)

Row matrix is a matrix that has one row and has more than one column, that is, m = 1 and n > 1. For example, the matrix

[F] = [x y z], is row matrix.

(d)

Column matrix is a matrix that has one column and has more than one row, that is, n = 1 and m > 1. For example, the matrix

(e)

Scalar matrix is a matrix that has the number of columns and the number of rows equal to 1, that is, m = 1 and n = 1. For example, the matrix

[M] = [7], is a scalar matrix; we can write it as 7 without bracket.

(f)

Null matrix is a matrix whose elements are all zero. For example, the matrix

(g)

Diagonal matrix is a square matrix that has zero elements everywhere except on its main diagonal. That is, for diagonal matrix aij = 0 when i ≠ j and not all are zero for aii when i = j. For example, the matrix

Main diagonal elements have equal row and column subscripts. The main diagonal runs from the upper-left corner to the lower-right corner. The main diagonal of the matrix here is a_11, a_22, and a_33.

(h)

Identity (unit) matrix [I] or I, is a diagonal matrix whose main diagonal elements are equal to unity (1’s) for any square matrix. That is, if the elements of an identity matrix are denoted as eij, then

For example, the matrix

(i)

Banded matrix is a square matrix that has a band of nonzero elements parallel to its main diagonal. For example, the matrix

(j)

Symmetric matrix is a square matrix whose elements satisfy the condition aij = aji for i ≠ j. For example, the matrix

(k)

Anti-symmetric (Skew-symmetric) matrix is a square matrix whose elements aij = −aji for i ≠ j, and aij = 0. For example, the matrix

(l)

Triangular matrix is a square matrix whose elements on one side of the main diagonal are all zero. There are two types of triangular matrices; first, an upper triangular matrix whose elements below the main diagonal are zero, that is, aij = 0 for i >j; second, a lower triangular matrix whose elements above the main diagonal are all zero, that is aij = 0 for i< j. For example, the matrix

While, the matrix

(m)

Partitioned matrix (Super-matrix) is a matrix that can be divided into smaller arrays (sub-matrices) by horizontal and vertical lines, that is, the elements of the partitioned matrix are matrices. For example, the matrix

is partitioned matrix with four smaller matrices,

where

1.4 ADDITION OR SUBTRACTION OF MATRICES

Addition and subtraction of matrices can only be performed for matrices of the same size, that is, the matrices must have same number of rows and columns. The addition is accomplished by adding corresponding elements of each matrix. For example, for addition of two matrices A and B, can give C matrix, that is, C=A+ B implies that cij = aij + bij. Where cij, aij, and bij are typical elements of the C, A, and B matrices, respectively.

Now, the subtraction of matrices is accomplished by subtracting corresponding elements of each matrix. For example, for subtraction of two matrices A and B, can give you C matrix, that is, C = A − B implies that cij = aij − bij. Note that, both A and B matrices are the same size, m × n, then the resulting matrix C is also of size m × n.

For example, let

Matrices addition and subtraction are associative; that is

For example,

Therefore, (A+B) + C = A + (B + C).

Matrices addition and subtraction are commutative; that is

For example,

1.5 MULTIPLICATION OF A MATRIX BY SCALAR

A matrix is multiplied by a scalar, c, by multiplying each element of the matrix by this scalar. That is, the multiplication of a matrix [A] by a scalar c is defined as

The scalar multiplication is commutative.

For example,

1.6 MULTIPLICATION OF A MATRIX BY ANOTHER MATRIX

The product of two matrices is C = AB , if and only if, the number of columns in A is equal to the number of rows in B. The product of matrix A of size m×n and matrix B of size n×r, the result in matrix C has size m×r.

where, the (ij)th component of matrix C is obtained by taking the dot product Cij = (i th row of A). (j th column of B).

That is, to find the element in row i and column j of [A][B], you need to single out row i from [A] and column j from [B], then multiply the corresponding elements from the row and column together and add up the resulting products.

For example,

Size of [A][B] = 2 × 3.

1.7 RULES OF MATRIX MULTIPLICATIONS

Matrix multiplication is associative; that is

Matrix multiplication is distributive; that is

or

Matrix multiplication is not commutative; that is

A square matrix multiplied by its identity matrix is equal to same matrix; that is

A square matrix can be raised to an integer power n; that is

A same square matrix multiplication with different integer power n and m can be given as

Transpose of product of matrices rule is given as

Example 1.3

Given matrices

Find the following:

a.  [B]+[C]

b.  [B]−[C]

c.  5{A}

d.  [B]{A}

e.  [D]2

f.  show that [D][I]=[I][D]=[D]

Solution

1.8 TRANSPOSE OF A MATRIX MULTIPLICATION

The transpose of a matrix A = [aij] is denoted as AT = [aji]. It is obtained by interchanging the rows and columns in matrix A. Thus, if a matrix A is of order m × n, then AT will be of order n × m.

For example,

Note that it is valid that, (AB)T = BT AT, (A + B)T = AT + BT, (cB)T = cBT, and (AT)T = A. Also note, if AT = A, then A is a symmetric matrix.

Example 1.4

Consider that matrix

Show that ([A][B])T = [B]T[A]T.

Solution

Therefore, ([A][B])T = [B]T[A]T.

1.9 TRACE OF A MATRIX

A trace of a matrix A, tr(A), is a square matrix and is defined to be the sum of the elements on the main diagonal of matrix A.

For example, let,

Example 1.5

Consider that matrix

Find the tr(A).

Solution

tr(A) = 5 + 6 = 11.

1.10 DIFFERENTIATION OF A MATRIX

Differentiation of a matrix is differentiation of every element of the matrix separately. For example, if the elements of the matrix A are a function of t, then

Example 1.6

Consider the matrix

Solution

1.11 INTEGRATION OF A MATRIX

Integration of a matrix is integration of every element of the matrix separately. For example, if the elements of the matrix A are a function of t, then

Example 1.7

Consider the matrix

Solution

1.12 EQUALITY OF MATRICES

Two matrices are equal if they have the same sizes and their corresponding elements are equal.

Example 1.8

If the matrices A and B are equal, find the value of x, w, z, and k.

Solution

1.13 DETERMINANT OF A MATRIX

The determinant of a square matrix A is a scalar number denoted by |A| or det [A]. It is the sum of the products (−1)i+j aijMij, where aij are the elements along any one row or column and Mij are the deleted elements of i th row and j th column from the matrix [A].

For example, the value of the determinant of matrix [A] is a and can be obtained by expanding along the first row as:

where the minor Mij is a (n − 1) × (n − 1) determinant of the matrix formed by removing the i th row and j th column.

Also, the value a can be obtained by expanding along the first column as:

Now, the value of a second-order determinant of (2 × 2) matrix is calculated by

The value of a third-order determinate of (3 × 3) matrix is calculated by

Example 1.9

Find the value, a, of the following determinants:

Solution

An alternative method of obtaining the determinant of a (3 × 3) matrix is by using the sign rule of each term that is determined by the first row in the diagram as follows:

or by repeating the first two rows and multiplying the terms diagonally as follows:

1.14 DIRECT METHODS FOR LINEAR SYSTEMS

Many engineering problems in finite element analysis will result in a set of simultaneous equations represented by [A]{X} = {B}.

For a set of simultaneous equations having the form

where there are n unknown x_1, x_2, x_3,…, xn to be determined. These equations can be written in matrix form as

This matrix equation can be written in a compact form as

where A is a square matrix with order n × n, while X and B are column matrices defined as

There are several methods for solving a set of simultaneous equations such as by substitution, Gaussian elimination, Cramer’s rule, matrix inversion, and numerical analysis.

1.15 GAUSSIAN ELIMINATION METHOD

In the argument matrix of a system, the variables of each equation must be on the left side of the equal sign (vertical line) and the constants on the right side. For example, the argument matrix of the system

The argument matrix is used in Gaussian elimination method. The Gaussian elimination method is summarized by the following steps:

1.  Write the system of equations in the argument matrix form.

2.  Perform elementary row operations to get zeros below the main diagonal.

a.  interchange any two rows

b.  replace a row by a nonzero multiply of that row

c.  replace a row by the sum of that row and a constant nonzero multiple of some other row

3.  Use back substitution to find the solution of the system.

We demonstrate the Gaussian elimination method in Example 1.10.

Example 1.10

Solve the linear system using the Gaussian elimination method.

Solution

We use Ri to represent the i th row. Write the argument matrix of the system as:

R_3 gives x_3 = 1, substitute the value of x_3 in R_2 and R_3, this gives x_2 = 1, and x_1 = 1, respectively.

1.16 CRAMER’S RULE

Cramer’s rule can be used to solve the simultaneous equations for x_1, x_2, x_3,…, Xn as

where the α’s are the determinations expressed as

It is worth noting that a is the determinant of matrix A and an is the determinant of the matrix formed by replacing the nth column of A by B. Also, Cramer’s rule applies only when a ≠ 0, but when a = 0, the set of questions has no unique solution, because the equations are linearly dependent.

Summary of Cramer’s Rule

1.  Form the coefficient matrix of A and column matrix B.

2.  Compute the determinant of matrix of A. If det[A] = 0, then the system has no solution; otherwise, go to the next step.

3.  Compute the determinant of the new matrix [Ai], by replacing the i th matrix with the column vector B.

4.  Repeat Step 3 for i= 1, 2,…, n.

5.  Solve for the unknown variable Xi using

Example 1.11

Solve the simultaneous equations

Solution

The matrix form of the given equations is

The determinants are calculated as

Thus,

Example 1.12

Solve the simultaneous equations

Solution

In matrix form, the given set of equations becomes

The determinants are calculated as

Thus,

1.17 INVERSE OF A MATRIX

Matrix inversion is used in many applications including the linear system of equations.

For the matrix equation AX = B, we can invert A to obtain X, that is,

where A−1 is the inverse matrix of A. The inverse matrix satisfies

where

where Adj [A] is the adjoint of matrix A. The Adj [A] is the transpose of the cofactors of matrix A. For example, let the n × n matrix A be presented as

The cofactors of the matrix A are written in matrix F as

where fij is the product of (−1)i+j and the determinant of the (n − 1)×(n − 1) submatrix is obtained by removing the i th row and j th column from matrix A.

For instance, by removing the first row and the first column of matrix A, we find the cofactor f_11 as

Now the adjoint of matrix A can be obtained as

So, the inverse of A matrix can be written as

A matrix that possesses an inverse is called invertible matrix (nonsingular matrix).

A matrix without an inverse is called a non-invertible matrix (singular matrix).

Consider a 2 × 2 matrix, if

The inverse of product of matrices rule can be presented as

Example 1.13

Let matrix

Solution

Example 1.14

Let matrix, , find its inverse matrix, A−1.

Solution

Using the concept of equation (1.27), we get

Therefore,

1.18 VECTOR ANALYSIS

A vector is a special case of a matrix with just one row or one column. A vector is a quantity (mathematical or physical) that has both magnitude and direction. Examples of vectors are force, momentum, acceleration, velocity, electric field intensity, and displacement. A scalar is a quantity that has only magnitude. Examples of scalars are mass, time, length, volume, distance, temperature, and electric potential.

A vector A has both magnitude and direction. A vector A in Cartesian (rectangular) coordinates can be written as (Ax, Ay, Az) where Ax, Ay, and Az are components of vector A in the x, y, and z directions, respectively. The magnitude of vector A is a scalar written as |A| or A and given as

A unit vector aA along vector A is defined as a vector whose magnitude is unity (i.e., 1) and its direction is along vector A, that is,

Figure 1.1 (a) illustrates the components of vector A and Figure 1.1 (b) shows the unit vectors.

Figure 1.1. (a) Components of vector A (b) Unit vectors.

(a) Vectors equality

Two vectors are equal if they are the same type (row or column) and their corresponding elements are equal to each other.

(b) Vector addition and subtraction

Two vectors can be added or subtracted only if they are of the same type, (i.e., both row vectors or both column vectors) and they are of the same number of components (elements).

Two vectors A = (Ax, Ay, Az) and B = (Bx, By, Bz) can be added together to give another vector C, that is,

Vector subtraction is similarly presented as

(c) Multiplication of a scalar by a vector

When a vector is multiplied by a scalar, each element is manipulated by the scalar. Let, vector A = (Ax, Ay, Az) and scalar k, then

There are three basic laws of algebra for given vectors A, B, and C when k and l are scalars, summarized in Table 1.1.

Table 1.1 Three Basic Laws of Vector Algebra

Law

Addition

Multiplication

Commutative

A + B = B + A

k A = A k

Associative

A + (B + C) = (A + B) + C

k(l A) = (k l) A

Distributive

k(A + B) = kB + kA

(d) Vector multiplication

There are two types of vector multiplication:

1.  Scalar (dot) product, A·B

2.  Vector (cross) product, A×B

1.  The dot product of two vectors A = (Ax, Ay, Az) and B = (Bx, By, Bz) written as A·B is defined as

where θAB is the smallest angle between vectors A and B. Also, the dot product is defined as,

It is worth it to know that, two vectors A and B are perpendicular (orthogonal) if and only if A·B = 0. Also, two vectors A and B are parallel if and only if B = kA. For vectors A, B, C and k scalar, the following prosperities dot product hold:

2.  The cross product of two vectors A = (Ax, Ay, Az) and B = (Bx, By, Bz) written as A × B, is defined as

where an is a unit vector normal to the plane containing vectors A and B. The direction of an is taken as the direction of the right thumb when the fingers of the right hand rotate from vector A to vector B as shown in Figure 1.2.

Figure 1.2. Right-hand rule for the direction of A × B and an.

Also, the cross product is defined as,

Because of the direction requirement of the cross product, the commutative law does not apply to the cross product. Instead,

Example 1.15

Given A = (3, −2, 5) and A = (2, 4, −6), find

a. A + B

b. A − B

c. |A − B|

Solution

Example 1.16

Given A = 3ax + 2ay − az and B = ax + ay, find A·B and A × B.

Solution

(e) The Del (▿) operator

The Del (▿) operator is a vector differential operator and known as gradient operator.

We obtain ▿ in Cartesian coordinates (x,y,z) as,

We obtain ▿ in cylindrical coordinates (ρ,ϕ,z) as,

We obtain ▿ in spherical coordinates (r,θ,ϕ) as,

The Del (▿) operator is useful in defining the following operations on a scalar or a vector:

1.  ▿A is the gradient of a scalar A (the result of this operation is a vector)

Considering A and B are scalars and n is an integer, the following formulas are true on gradient:

2.  ▿· A is the divergence of a vector A (the result of this operation is a scalar)

Considering A and B are vectors and k is a scalar, the following formulas are true on divergence of a vector:

3.  ▿ × A is the curl of a vector A (the result of this operation is a vector)

or

or

or

Considering A and B are vectors and k is a scalar, the following formulas are true on curl of a vector:

4.  ▿2A-Laplacian of a scalar A (the result of this operation is a scalar)

(c)  For spherical coordinates,

The Laplacian of a vector A, can be defined as

Example 1.17

Find the gradient of the scalar field A = e−z sin 3x cosh y.

Solution

▿A = 3e−z cos 3x cosh yax + e−z sin 3x sinh yay −e−z sin 3x cosh yaz

1.19 EIGENVALUES AND EIGENVECTORS

Eigenvalues problems arises from many branches of engineering especially in analysis of vibration of elastic structures and electrical systems.

The eigenvalue problem is presented in linear equations in the form

Where [A] is a square matrix; λ is a scalar and called eigenvalue of matrix [A] ; {X} is eigenvector of matrix [A] corresponding to λ.

To find the eigenvalues of a square matrix [A] we rewrite the equation (1.55) as

or

There must be a nonzero solution of equation (1.87) in order for λ to be an eigenvalue. However, equation (1.87) can have a nonzero solution if and only if

Equation (1.88) is called the characteristic equation of matrix [A] and the scalars satisfy the equation (1.88) are the eigenvalues of matrix [A]. If matrix [A] has the form

then equation (1.88) can be written as

The equation (1.89) can be expanded to a polynomial equation in λ as

Thus, the nth degree polynomial is

Equation (1.91) is called a characteristic polynomial of n × n matrix [A]. Indeed, the nth roots of the polynomial equation are the nth eigenvalues of matrix [A]. The solutions of equation (1.87) with the eigenvalues substituted on the equation are called eigenvectors.

Example 1.18

Find the eigenvalues and eigenvectors of the 2 × 2 matrix

Solution

Since

the characteristic polynomial of matrix [A] is

and the characteristic equation of matrix [A] is

The solutions of this equation are λ_1 = 2 and λ_2 = 9; these values are the eigenvalues of matrix [A].

The eigenvectors for each of the above eigenvalues are calculated using equation (1.87).

For λ_1 = 2, we obtain

The above equation yields to two simultaneous equations for x_1 and x_2, as follows:

Thus, choosing x_2 = 4, we obtain the eigenvector where k is an arbitrary constant.

For λ_2 = 9 We obtain

The above equation yields to two simultaneous equations for x_1 and x_2, as follows:

Thus, choosing x_1 = −1, we obtain the eigenvector where k is an arbitrary constant.

1.20 USING MATLAB

MATLAB is a numerical computation and simulation tool that uses matrices and vectors. Also, MATLAB enables users to solve wide analytical problems. The majority of engineering systems are presented by matrix and vector equations. Therefore, MATLAB becomes essential to reduce the computational workload.

All MATLAB commands or expressions are entered in the command window at the MATLAB prompt “”. To execute a command or statement, we must press return or enter at the end. If the command does not fit on one line, we can continue the command on the next line by typing three consecutive periods (…) at the end of the first line. A semicolon (;) at the end of a command suppresses the screen output and the command is carried out. Typing anything following a % is considered as comment, except when the % appears in a quote enclosed character string or certain I/O format statements. Comment statements are not executable. To get help on a topic (such as matrix), you can type the command helpmatrix. Here, we introduce basic ideas of matrices and vectors operations. For more details, see Appendix B.

Elements of a matrix are enclosed in brackets and they are row-wise. The consecutive elements of a row are separated by a comma or a space and are entered in rows separated by a space or a comma, and the rows are separated by semicolons (;) or carriage returns (enter).

A vector is entered in the MATLAB environment the same way as a matrix.

For example, matrix A,

The basic scalar operations are shown in Table 1.2. In addition to operating on mathematical scalar, MATLAB allows us to work easily with vectors and matrices. Arithmetic operations can apply to matrices and Table 1.3 shows extra common operations that can be implemented to matrices and vectors.

Table 1.2 MATLAB Common Arithmetic Operators

Operators Symbols

Descriptions

+

addition

−

subtraction

*

multiplication

/

right division (means )

\

left division (means )

^

exponentiation (raising to a power)

‘

converting to complex conjugate transpose

( )

specify evaluation order

Table 1.3 Matrix Operations

Operations

Descriptions

A’

Transpose of matrix A

det (A)

Determinant of matrix A

inv (A)

Inverse of matrix A

eig (A)

Eigenvalues of matrix A

diag (A)

Diagonal elements of matrix A

rank (A)

Rank of matrix A

cond (A)

Condition number of matrix A

eye (n)

The n × n identity matrix (1’s on the main diagonal)

eye (m, n)

The m × n identity matrix (1’s on the main diagonal)

trace (A)

Summation of diagonal elements of matrix A

zeros (m, n)

The m × n matrix consisting of all zeros

ones (m, n)

The m × n matrix consisting of all ones

rand (m, n)

The m × n matrix consisting of random numbers

randn (m, n)

The m × n matrix consisting of normally distributed numbers

diag (A)

Extraction of the diagonal matrix A as vector

diag (A,1)

Extracting of first upper off-diagonal vector of matrix A

diag (u)

Generating of a diagonal matrix with a vector u on the diagonal

expm (A)

Exponential of matrix A

ln (A)

LU decomposition of matrix A

svd (A)

Singular value decomposition of matrix A

qr (A)

QR decomposition of matrix A

min (A)

Minimum of vector A

max (A)

Maximum of vector A

sum (A)

Sum of elements of vector A

std (A)

Standard deviation of the data collection of vector A

sort (A)

Sort the elements of vector A

mean (A)

Means value of vector A

triu (A)

Upper triangular of matrix A

triu (A, I)

Upper triangular with zero diagonals of matrix A

tril (A)

lower triangular of matrix A

tril (A, I)

lower triangular with zero diagonals of matrix A

Example 1.19

Given the following matrices:

Use MATLAB to perform the following operations:

a. [A] + [B]

b. [A] − [B]

c. 5[B]

d. [A][B]

e. [A][C]

f. [A]2

g. [A]T

h. [B]−1

i. tr (A)

j. |B|

Solution

a. [A] + [B]

b. [A] − [B]

c. 5[B]

d. [A][B]

e. [A][C]

f. [A]2

g. [A]T

h. [B]−1

i. tr (A)

j. |B|

Example 1.20

Solve the following system of three equations:

using the following methods:

a. the matrix inverse

b. Gaussian elimination

c. Reverse Row Echelon Function

Solution

a.  Since we know A−1A = 1, we can find the solution of the system of linear equations AX = B by using X = A−1B.

Now, we write the system of equations by using the following matrices:

Generally, using the matrix inverse to solve linear systems of equations should be avoided due to the excessive round-off errors.

b.  We use the left division operator in MATLAB X = A\B to solve linear systems of equations using Gaussian elimination.

c.  The reduced row echelon function use, rref, to solve the system of linear equations. The rref function requires an expanded matrix as input, representing the coefficients and results. The last column in the output array represents the solution of equations.

Example 1.21

Solve the following set of equations using the Cramer’s rule:

Solution

PROBLEMS

1.  Identify the size and the type of the given matrices. Identify if the matrix is a square, column, diagonal, row, identity, banded, symmetric, or triangular.

2.  Given the matrices

a. A+B

b. A−B

c. 4A

d. AB

e. A{C}

f. A^2

g. IA

h. AI

3. Given the matrices

a. AT

b. BT

c. |A|

d. |A|−1

e. |B|−1

4. What is the 3 × 3 null matrix and the 5 × 5 identity matrix?

5.  Express the following systems of equations in matrix form AX = B.

6.  Solve the system using the Gaussian elimination method.

7.  Solve the simultaneous equations using Cramer’s rule.

8.  Show that know that A · B = AxBx + AyBy AzBz, know that ax · ax = ay · ay = az · az = 1.

9.  Given A = 2ax +4ay − 5az and B = 3ax −ay +az, find A · B and A × B.

10.  Show that if A = 5ax − 4ay −az and B = ax + 2ay + 2az, then they are perpendicular or not.

11.  Determine the gradient of the scalar fields A = x^3 y + xyz.

12.  Find the eigenvalues and eigenvectors of the 2 × 2 matrix

13.  Solve problem 2 using MATLAB.

14.  Solve problem 3 using MATLAB.

REFERENCES

1.  R. Butt, “Applied Linear Algebra and Optimization using MATLAB,” Mercury Learning and Information, 2011.

2.  H. Anton, “Elementary Linear Algebra, 6th Edition,” John Wiley and Sons, INC., 1991.

3.  S. Nakamura, “Applied Numerical Methods with Software,” Prentice-Hall, 1991.

4.  B. Kolman, “Introductory Linear Algebra with Applications, 6th Edition,” John, Prentice Hall, 1997.

Chapter 2

INTRODUCTION TO FINITE ELEMENT METHOD

2.1 INTRODUCTION

Finite element analysis method is a numerical procedure that applies to many areas in real-world engineering problems, including structural/stress analysis, fluid flow analysis, heat transfer analysis, and electromagnetics analysis. Indeed, finite element has several advantages and features such as the capability of solving complicated and complex geometries, flexibility, strong mathematical foundation, and high-order approximation. Therefore, finite element analysis (FEA) has become an important method in the design and modeling of a physical event in many engineering disciplines. The actual component in the FEA method is placed by a simplified model that is identified by a finite number of elements connected at common points called nodes, with an assumed response of each element to applied loads, and then evaluating the unknown field variable (displacement) at these nodes.

2.2 METHODS OF SOLVING ENGINEERING PROBLEMS

There are 3 common methods to solve any engineering problem:

1.  Experimental method

2.  Analytical method

3.  Numerical method

2.2.1 Experimental Method

This method involves actual measurement of the system response. This method is time consuming and needs expensive set up. This method is applicable only if physical prototype is available. The results obtained by this method cannot be believed blindly and a minimum of 3 to 5 prototypes must be tested. Examples of this method are strain photo elasticity, heat transfer for a gas turbine engine, static and dynamic response for aircraft and spacecraft, amount of water which is lost for groundwater seepage, etc.

2.2.2 Analytical Method

This is a classic approach. This method gives closed form solutions. The results obtained with this method are accurate within the assumptions made. This method is applicable only for solving problems of simple geometry and loading, like cantilever and simply supported beams, etc. Analytical methods produce exact solutions of the problem. Examples of this method are integral solutions (such as Laplace and Fourier transforms), conformal mapping, perturbation methods, separation of variables, and series expansion.

2.2.3 Numerical Method

This approximate method is resorted to when analytical method fails. This method is applicable to real-life problems of a complex nature. Results obtained by this method cannot be believed blindly and must be carefully assessed against experience and the judgment of the analyst. Examples of this method are finite element method, finite difference method, moment method, etc.

2.3 PROCEDURE OF FINITE ELEMENT ANALYSIS (RELATED TO STRUCTURAL PROBLEMS)

Step (i). Discretization of the structure

This first step involves dividing the structure or domain of the problem into small divisions or elements. The analyst has to decide about the type, size, and arrangement of the elements.

Step (ii). Selection of a proper interpolation (or displacement) model

A simple polynomial equation (linear/quadratic/cubic) describing the variation of state variable (e.g., displacement) within an element is assumed. This model generally is the interpolation/shape function type. Certain conditions are to be satisfied by this model so that the results are meaningful and converging.

Step (iii). Derivation of element stiffness matrices and load vectors

Response of an element to the loads can be represented by element equation of the form

where, [k] = Element stiffness matrix,

{q} = Element response matrix or element nodal displacement vector, or nodal degree of freedom,

{Q} = Element load matrix or element nodal load vector.

From the assumed displacement model, the element properties, namely stiffness matrix and the load vector are derived. Element stiffness matrix [k] is a characteristic property of the element and depends on geometry as well as material. There are 3 approaches for deriving element equations. They are

(a)  Direct approach,

(b)  Variational approach,

(c)  Weighted residual approach.

(a)  Direct approach: In this method, direct physical reasoning is used to establish the element properties (stiffness matrices and load vectors) in terms of pertinent variables. Although this approach is limited to simple types of elements, it helps to understand the physical interpretation of the finite element method.

(b)  Variational approach: This approach can be adopted when the variational theorem (extremum principle) that governs the physics of the problem is available. This method involves minimizing a scalar quantity known as functional that is typical of the problem at hand (e.g., potential energy in stress analysis problems).

(c)  Weighted residual approach: This approach is more general in the sense that it is applicable to all situations where the governing differential equation of the problem is available. This method involves minimizing error resulting from substituting trial solution in to the differential equation.

Step (iv). Assembling of element equations to obtain the global equations

Element equations obtained in Step (iii) are assembled to form global equations in the form of

where, [K] is the global stiffness matrix,

{r} is the vector of global nodal displacements, and

{R} is the global load vector of nodal forces for the complete structure.

Equation (2.2) describes that the behavior of entire structure.

Step (v). Solution for the unknown nodal displacements

The global equations are to be modified to account for the boundary conditions of the problem. After specifying the boundary conditions, the equilibrium equations can be expressed as

For linear problems, the vector {r_1} can be solved very easily.

Step (vi). Computation of element strains and stresses

From the known nodal displacements {r_1}, the elements strains and stresses can be computed by using predefined equations for structure.

The terminology used in the previous 6 steps has to be modified if we want to extend the concept to other fields. For example, put the field variable in place of displacement, the characteristic matrix in place of stiffness matrix, and the element resultants in place of element strains.

2.4 METHODS OF PRESCRIBING BOUNDARY CONDITIONS

There are 3 methods of prescribing boundary conditions.

2.4.1 Elimination Method

This method is useful when performing hand calculations. It poses difficulties in implementing in software. This method has been used in this book for solving the problems by finite element method using hand calculations and results in reduced sizes of matrices thus making it suitable for hand calculations. The method is explained below in brief. Consider the following set of global equations,

Let u_3 be prescribed, i.e., u_3 = s.

This condition is imposed as follows:

1.  Eliminate the row corresponding to u_3 (3rd row).

2.  Transfer the column corresponding to u_3 (3rd column) to right-hand side after multiplying it by “s”. These steps result in the following set of modified equations,

This set of equations now may be solved for non-trivial solution.

2.4.2 Penalty Method

This is the method used in most of the commercial software because this method facilitates prescribing boundary conditions without changing the sizes of the matrices involved. This makes implementation easier.

2.4.3 Multipoint Constrains Method

This method is commonly used in functional analysis between nodes. For example, there are many applications in trusses where the end supports are on an inclined plane and do not coincide with the coordinate system used to describe the truss. Another application of the method is the functional relationship between the temperature at one node and temperature at one or more other nodes.

2.5 PRACTICAL APPLICATIONS OF FINITE ELEMENT ANALYSIS

There are 3 practical applications of finite element analysis:

•  Analysis of new design

•  Optimization projects

•  Failure analysis

2.6 FINITE ELEMENT ANALYSIS SOFTWARE PACKAGE

There are 3 main steps involved in solving an engineering problem using any commercial software:

Step (i). Preprocessing

In this step, a CAD model of the system (component) is prepared and is meshed (discretized). Boundary conditions (support conditions and loads) are applied to the meshed model.

Step (ii). Processing

In this step, the software internally calculates the elements stiffness matrices, element load vectors, global stiffness matrix, global load vector, and solves after applying boundary conditions for primary unknowns (e.g., displacements/ temperatures etc.) and secondary unknowns (e.g., stress/strain/heat flux etc.).

Step (iii). Post-processing

Post-processing involves sorting and plotting the output to make the interpretation of results easier.

2.7 FINITE ELEMENT ANALYSIS FOR STRUCTURE

There are several common methods in finite element analysis used for evaluating displacements, stresses, and strains in any structure under different boundary conditions and loads. They are summarized below:

1.  Displacement Method: This method is the most commonly used method. The structure is subjected to applied loads or/and specific displacements. The primary unknowns are displacements found by using an inversion of the stiffness matrix, and the derived unknowns are stresses and strains. Indeed, the stiffness matrix for any element can be calculated by the variational principle.

2.  Force Method: The structure is subjected to applied loads or/and specific displacements. The primary unknowns are member forces, found by using an inversion of the flexibility matrix, and the derived unknowns are stresses and strains. Indeed, the calculation of the flexibility matrix is possible only for discrete structural elements (e.g., piping, beams, and trusses).

3.  Mixed Method: The structure is subjected to applied loads or/and specific displacements. This method uses very large stiffness coefficients and very small flexibility coefficients in the same matrix.

4.  Hybrid Method: The structure is subjected to applied loads and stress boundary conditions. This hybrid method has the merit of the FEA method, i.e., the flexibility and sparse matrix of FEM for complicated inhomogeneous scatterers.

2.8 TYPES OF ELEMENTS

In general, the region in space is considered nonregular geometric. However, the FEA method divides the nonregular geometric region to small regular geometric regions. There are 3 types of elements in finite element.

1.  One-Dimensional Elements: The objects are subdivided into short-line segments. A one-dimensional finite element expresses the object as a function of one independent variable such as one coordinate x. Finite elements use one-dimensional elements to solve systems that are governed by ordinary differential equations in terms of an independent variable. The number of node points in an element can vary from 2 up to any value needed. Indeed, increasing the number of nodes for an element increases the accuracy of the solution, but it also increases the complexity of calculations. When the elements have a polynomial approximation higher than first order, we call that higher order elements. Figure 2.1 shows one-dimensional elements. For example, the one-dimensional element is sufficient in dealing with heat dissipation in cooling fins.

Figure 2.1. One-dimensional elements.

2.  Two-Dimensional Elements: The objects can be divided into triangles, rectangles, quadrilaterals, or other suitable subregions. A two-dimensional finite element expresses the object as a function of 2 variables such as the 2 coordinates x and y. A finite element uses two-dimensional elements to solve systems that are governed by partial differential equations. The simplest two-dimensional element is the triangular element. Figure 2.2 shows two-dimensional elements. For example, a two-dimensional element is sufficient in plane stress or plane strain.

Figure 2.1. Tow-dimensional elements.

3.  Three-Dimensional Elements: The objects can be divided into tetrahedral elements, rectangular prismatic elements, pie-shaped elements, or other suitable shapes of elements. A three-dimensional finite element expresses the object as a function of 3 variables such as the 3 coordinates x, y, and z. A finite element uses three-dimensional elements to solve systems that are governed by differential equations. The simplest three-dimensional element is the tetrahedral element. Figure 2.3 shows three-dimensional elements.

Figure 2.3. Three-dimensional elements.

These three types of elements are applied and discussed in the electromagnetics analysis chapter of this book.

2.9 DIRECT METHOD FOR LINEAR SPRING

Here, we will use the direct method in a one-dimension domain to derive the stiffness matrix for the linear spring element shown in Figure 2.4. Reference points 1 and 2 located at the ends of the linear spring element are the nodes. The symbols f_1 and f_2 are local nodal forces (or axial loads) associated with the local axis x. The symbols u_1 and u_2 are local nodal displacements (or degree of freedom at each node) for the spring element. ui is the displacement of the spring due to a load fi. The symbol k is the stiffness of the spring (or spring constant). k is load required to give the spring a unit displacement. The symbol L is the bar length. The local axis x acts in the same direction of the spring that can lead to direct measurement of forces and displacements along the spring.

Figure 2.4. Linear spring element.

The displacements can be defined related to forces as

The equilibrium of forces gives

Based on equation (2.7), the above equation becomes

By combining equations (2.7) and (2.9) and writing the resulting equations in matrix form we get

or

where

{fi} = a vector of internal nodal forces

[ki] = the elemental stiffness matrix

{ui} = a vector of nodal displacements

For many interconnected spring elements, we can use the following:

where

{Qi} = a vector of external nodal forces = Σ{fi}

[K] = the structural stiffness matrix = Σ[k]

{u} = a vector of nodal displacements of the structure.

PROBLEMS

1.  Define finite element analysis?

2.  What are the advantages and features of finite element analysis?

3.  What are the 3 common methods to solve any engineering problem?

4.  What is procedure of finite element analysis (related to structural problems)?

5.  What are the 2 methods for prescribing the boundary conditions?

6.  Give the 3 practical applications of finite element analysis?

7.  What are the 3 main steps involved in solving an engineering problem using any commercial software?

8.  What are the 4 common methods in finite element analysis used for evaluating displacements, stresses, and strains in any structure under different boundary conditions and loads?

9.  What is the primary variable in finite element method structural analysis?

10.  Calculate the structural stiffness matrix of the system as shown in Figure 2.5.

Figure 2.5. Two springs in series structure.

REFERENCES

1.  P. P. Silvester and R. L. Ferrari, “Finite Elements for Electrical Engineering,y” Cambridge Press, 1983.

2.  B. Szabo and I. Babuska, “Finite Element Analysis,” John Wiley & Sons, 1991.

3.  R. D. Cook and D. S. Malkus, and M. E. Plesha, “Concepts and Applications of Finite Element Analysis,” John Wiley and Sons, 1989.

4.  O. C. Zienkiewicz, “The Finite Element Method, 3rd. ed.,” McGraw-Hill, 1979.

5.  W. B. Bickford, “A First Course in the Finite Element Method,” Richard D. Irwin Publisher, 1989.

6.  G. L. Narasaiah, “Finite Element Analysis,” CRC Press, 2009.

7.  P. E. Allaire, “Basics of the Finite Element Method: Solid Mechanics, Heat Transfer, and Fluid Mechanics,” Wm. C. Brown Publishers, 1985.

8.  CTF Ross, “Finite Element Methods in Structural Mechanics,” Ellis Horwood Limited Publishers, 1985.

Chapter 3

FINITE ELEMENT ANALYSIS OF AXIALLY LOADED MEMBERS

3.1 INTRODUCTION

In this chapter, we will use the bar element in the analysis of rod-like axially loaded members. We start with the two popular bar elements using a two-node element and a three-node element as well as bars of constant cross-section area, bars of varying cross-section area, and the stepped bar.

Stress is an internal force that has been distributed over the area of the rod’s cross section and it is defined as

where σ is the stress, F is the force, and A is the cross-sectional area.

Thus, stress is a measure of force per unit area. When the stress tends to lengthen the rod, the stress is called tension, and σ > 0. When the stress tends to shortened the rod, the stress is called compression, and σ < 0. The orientations of forces in tension and compression are shown in Figure 3.1.

Figure 3.1. Directions of tensile and compressive forces.

The derived unit of stress is the pascal (Pa), where pascal is equal to newtons per square meter (N/m^2), 1 Pa = 1 N/m^2, pascal is used in the SI units. The derived unit of stress is the dimension pound-per-square-inch (psi), where 1 psi = 1 lb/in^2. Psi is used in the USCS (U.S. Customary) units. In stresses, calculations are generally very large, therefore, they often use the prefixes kilo- (k), mega- (M), and giga- (G) for factors of 10^3, 10^6, and 10^9, respectively. Thus,

1 kPa = 10^3 Pa, 1 MPa = 10^6 Pa, 1 GPa = 10^9 Pa.

The numerical values for stresses unit conversion between the USCS and SI can be presented as

1 psi = 6.895 × 10−3 MPa.

Strain (ε) is the amount of elongation that occurs per unit of the rod’s original length and is calculated as

where, ΔL is the change in length of the rod (elongation).

Strain is a dimensionless quantity and is generally very small.

For each individual rod, the applied force and elongation are proportional to each other based on the following expression

where k is the stiffness.

Figure 3.2 shows force and elongation behaviors of rods at various cross-sectional areas and lengths.

Figure 3.2. Force and elongation behaviors of rods at various cross-sectional areas and lengths.

Stress and strain are useful in mechanical engineering because they are scaled with respect to the rod’s size.

Stress and strain are proportional to each other and presented as

where E is the elastic modulus (or Young’s modulus).

The elastic modulus has the dimensions of force per unit area. The elastic modulus is a physical material property, and is the slope of the stress-strain curve for low strain.

By combining equations (3.1) and (3.2), we get

With the stiffness in equation (3.3), it can be written as

Each rod formed of same material has similar stress-strain behavior as presented in Figure 3.3.

Figure 3.3. All rods formed of same material have similar stress-strain behavior.

When a system is motionless or has constant velocity, then the system has zero acceleration, and the system is to be in equilibrium. The static equilibrium is used for a system at rest. For equilibrium, the resultant of all forces and all moments acting on the system is balanced to zero resultant. That is, the sum of all force vectors (F) acting upon a system is zero and the sum of all moment vectors (M) acting upon a system is zero, and they can be written as

The total extension (or contraction) of a uniform bar in pure tension or compression is defined as

The equation (3.9) does not apply to a long bar loaded in compression if there is a possibility of bucking.

3.1.1 Two-Node Bar Element

Figure 3.4. Two-node bar for rod-like axially loaded members.

This element has 2 end nodes and each node has 1 degree of freedom, namely translation along its length. Its formulation is based on linear interpolation. It gives accurate results only if loads are applied at nodes and the area is constant over the element. However, required accuracy for practical purposes in other cases can be obtained by taking a larger number of smaller elements. The interpolation equation, element stiffness matrix, strain-displacement matrix, element strain, and element stress for 2-node (linear) bar element are given by

where

u_1 and u_2 = nodal (unknown) displacements (degree of freedom)at node 1 and 2, respectively

{u} = displacement matrix at the nodes

A = cross section of the area of the bar

L = x_2 − x_1 = length of the bar

E = Young’s modulus (modulus of elasticity)

[k] = stiffness matrix of the element

[B] = strain-displacement matrix

{ε} = strain matrix

{σ} = stress matrix.

Uniformly distributed load per unit length w acting on the element can be converted into equivalent loads using,

{W} = the potential energy of load system.

Thermal loads due to a change in temperature ΔT can be converted into equivalent nodal loads using

where

α is the coefficient of thermal expansion.

3.1.2 Three-Node Bar Element

Figure 3.5. Three-node bar for rod-like axially loaded members.

This element has a midside node, in addition to 2 end nodes. Each node has 1 degree of freedom, namely translation along its length. Its formulation is based on quadratic interpolation and this element gives accurate results even with distributed loads and a linearly varying cross-sectional area. Coarse mesh with fewer of these elements can give the desired accuracy as compared to a fine mesh of 2-node bar element. The interpolation equation, element stiffness matrix, strain-displacement matrix, element strain, and element stress for the quadratic bar element are given by,

Uniformly distributed load per unit length w, acting on the element, can be converted into equivalent loads using,

3.2 BARS OF CONSTANT CROSS-SECTION AREA

This section will demonstrate examples on bars of constant cross-sectional area using FEA.

Example 3.1

Consider a 2 m long steel bar of 50 mm^2 cross-sectional areas as shown in Figure 3.6. Use a two element mesh to model this problem. Find nodal displacements, element stresses, and reaction.

Take Young’s modulus,

Figure 3.6. Bar with tip load for Example 3.1.

Solution

(I) Analytical method [Refer to Figure 3.6(a)]

Figure 3.6(a). Analytical method for bar with tip load for Example 3.1.

Displacement calculation

Displacement at section a–a,

Displacement at node 2,

δx = 1000 =1×10−5 ×1000=0.01 mm.

Displacement at node 3,

δx = 2000 =1×10−5 ×2000=0.02 mm.

Stress calculation

Maximum stress in the bar

Reaction calculation

(II)  FEM by hand calculations

Figure 3.6(b). Finite element model for Example 3.1.

Stiffness matrix for element 1 is,

Global equation is,

Boundary conditions are, at node 1, u_1 = 0.

By using elimination method, the above matrix reduces to,

By matrix multiplication, we get

By solving equations (3.26) and (3.27), we get

u_2 = 0.01 mm

u_3 = 0.02 mm.

Stress (σ) calculation

Stress for element 1 is,

Stress for element 2 is,

Reaction calculation

From equation (3.25)

(III)  Software results

Figure 3.6(c). Deflection pattern for a bar (refer to Appendix C for color figures).

Deflection values as node (Computer generated output)

The following degree of freedom results are in global coordinates:

Maximum absolute values

Figure 3.6(d). Stress pattern for a bar (refer to Appendix C for color figures).

Stress values at elements (Computer generated output)

STAT

CURRENT

ELEM

LS1

1

2.0000

2

2.0000

Reaction value (Computer generated output)

The following X, Y, Z solutions are in global coordinates

NODE

FX

FY

1

−100.00

0.0000

Answers for Example 3.1

Example 3.2

Bar under distributed and concentrated forces. Consider the bar shown in Figure 3.7 subjected to loading as shown below. Use 4 element mesh models and find nodal displacements, element stresses, and reaction at the fixed end. Take E=2×10^5 N/mm^2, A = 50 mm^2, P = 100 N.

Figure 3.7. Bar under distributed and concentrated forces for Example 3.2.

Solution

(I)  Analytical method [Refer Figure 3.7(a)]

Figure 3.7(a). Analytical method for the Bar under distributed and concentrated forces for Example 3.2.

Reaction calculation

Stress calculation

To find σBD, consider section XX

Displacement calculation

Displacement at E,

Displacement at B,

Displacement at D,

Displacement at C,

(II) FEM by hand calculations

Figure 3.7(b). Finite element model for Example 3.2.

Stiffness matrix for elements is,

Nodal load calculation for elements 2 and 3,

Global equation is,

Boundary conditions are at node 1, u_1 = 0

By using the elimination method the above matrix reduces to,

By solving the above matrix and equations, we get

Stress (σ) calculation

Reaction calculation: from equation (3.29)

Figure 3.7(c). Deflection pattern for a bar (refer to Appendix C for color figures).

Deflection values at nodes (Computer generated output)

The following degree of freedom results are in global coordinates

Maximum absolute value

Figure 3.7(d). Stress pattern for a bar (refer to Appendix C for color figures).

Stress values at elements (Computer generated output)

STAT

CURRENT

ELEM

LS1

1

3.0000

2

2.7500

3

2.2500

4

2.0000

Reaction value (Computer generated output)

The following X, Y, Z SOLUTIONS are in global coordinates

NODE

FX

FY

1

−150.00

0.0000

Answers to Example 3.2

Example 3.3

A and P = 80 kN is applied as shown in Figure 3.8. Determine the nodal displacements, element stresses, and support reactions in the bar. Take E = 20 × 10^3 N/mm^2.

Figure 3.8. Example 3.3.

Solution

(I) Analytical method [Refer to Figure 3.8(a)]

Figure 3.8(a). Analytical method for Example 3.3.

Let R_3 be the reaction developed at the wall after contact.

Solving equations (3.30) and (3.31)

R_1 = 58018 N

R_3 = 21982 N

Stresses are,

Deflections are,

(II)  FEM by hand calculations

Figure 3.8(b). Finite element model for Example 3.3.

L_1 = 200 mm, L_2 = 200 mm

First, we should check whether contact occurs between the bar and the wall. For this, assume that the wall does not exist. The solution to the problem is as below. (Consider the two element model.)

Stiffness matrices are,

Global equation is,

Boundary conditions are at node 1, u_1 = 0.

By using the elimination method, the above matrix reduces to,

By matrix multiplication, we get

By solving equations (3.33) and (3.34), we get, u_2 = 2.67 mm and u_2 = 2.67 mm.

Since displacement at node 3 is 2.67 mm (greater than 1.2 mm), we can say that contact does occur. The problem has to be resolved since the boundary conditions are now different. The displacement at B’ is specified to be 1.2 mm as shown in Figure 3.8.

Global element equation is,

Boundary conditions are at node 1, u_1 = 0 and at node 3, u_2 = 1.2.

By using the elimination method, the above matrix reduces to,

Stress (σ) calculation: stress for element 1 is,

Stress for element 2 is,

Reaction calculation: from equation (3.36)

We know that,

(III)  Software results

Figure 3.8(c). Deflection pattern for a bar (refer to Appendix C for color figures).

Deflection values at nodes

The following degree of freedom results are in global coordinates

Maximum absolute values

Figure 3.8(d). Stress pattern for a bar (refer to Appendix C for color figures).

Stress values at elements

STAT

CURRENT

ELEM

LS1

1

193.33

2

−73.333

Reaction value

The following X, Y, Z solutions are in global coordinates

NODE

FX

FY

1

−58000

0.0000

3

−22000

Answers to Example 3.3

Example 3.4

A bar is subjected to self weight. Determine the nodal displacement for the bar hanging under its own weight as shown in Figure 3.9. Use two equal length elements. Let E = 2 × 10^11 N/mm^2, mass density ρ = 7800 kg/m^3, Area A = 1000 mm^2.

Consider length of rod L= 2 m.

Figure 3.9. Bar under self weight for Example 3.4.

Solution

(I) Analytical method [Refer Figure 3.9(a)]

Figure 3.9(a). Analytical method for Example 3.4.

(II) FEM by hand calculations

Figure 3.9(b). Finite element model for Example 3.4.

The element stiffness matrices are,

For element 1,

For element 2,

Nodal load vector due to weight is,

Global equation is,

Boundary conditions are at node 1, u_1 = 0.

By using the elimination method, the above matrix reduces to,

By matrix multiplication, we get

By solving equations (3.39) and (3.40),

we get

(III) Software results

Figure 3.9(c). Deflection pattern for a bar (refer to Appendix C for color figures).

Deflection values at nodes

The following degree of freedom results are in global coordinates

Maximum absolute values

Answers of Example 3.4

Example 3.5

A rod rotating at a constant angular velocity ω = 45 rad/sec is shown in Figure 3.10. Determine the nodal displacements and stresses in the rod. Consider only the centrifugal force. Ignore the bending of the rod. Use two quadratic elements. Take A = 350 mm^2, E = 70 GPa, Mass density ρ = 7850 kg/m^3, Length of the rod L = 1 m.

Figure 3.10. Rod rotation at a constant angular velocity for Example 3.5.

Solution

(I) Analytical method [Refer to Figure 3.10(a)]

Figure 3.10(a). Analytical method for rod rotation at a constant angular velocity for Example 3.5.

L = 1 m

Stress calculation: Stress at section a-a,

Displacement at section a–a = change in length of x,

Reaction calculation

(II) FEM by hand calculations

Figure 3.10(b). Finite element model for Example 3.5 (with two quadratic elements).

A finite element model of the rod, with two quadratic elements, is shown in Figure 3.10(b). The element stiffness matrices are,

Thus, the global stiffness matrix is,

The centrifugal force or body force Fc (kg/m^3) is given by,

Note that F is a function of the distance r from the pin. Taking the average values of F over each element, we have,

Thus, the element body force vectors are,

Assembling f_1 and f_2, we obtain,

The global equation is,

Boundary conditions are at node 1, u_1 = 0.

By using the elimination method, the above matrix reduces to,

By solving the above matrix and equations, we get

The stress at node 1 in element 1 is given by,

The stress at node 2 in element 1 is given by,

The stress at node 3 in element 1 is given by,

The stress at node 1 in element 2 is given by,

The stress at node 2 in element 2 is given by,

The stress at node 3 in element 2 is given by,

(III) Software results

While solving the problem using software, 4 linear bar elements are taken instead of 2 quadratic elements.

Figure 3.10(c). Deflection pattern for a rod (refer to Appendix C for color figures).

Deflection values at nodes (in m)

The following degree of freedom results are in global coordinates

Maximum absolute values

Reaction value

The following X, Y, Z solutions are in global coordinates

NODE

FX

FY

1

−2781.8

−3.3691

Answers of Example 3.5

Each problem given in this book uses a different procedure for solving using software. For familiarizing, procedure for one problem is given from each chapter using software. Other problems are left to the user to explore the software for solving the problems.

Procedure for solving the problem using ANSYS® 11.0 academic teaching software

For Example 3.3

PREPROCESSING

1.  Main Menu > Preprocessor > Element Type > Add/Edit/Delete > Add > Structural Link > 2D spar 1 > OK > Close

Figure 3.11. Element selection.

2.  Main Menu > Preprocessor > Real Constants > Add/Edit/Delete > Add > OK

Figure 3.12. Enter the cross-sectional area.

Cross-sectional area AREA > Enter 300 > OK > Close

Enter the material properties

3.  Main Menu > Preprocessor > Material Props > Material Models

Material Model Number 1, click Structural> Linear > Elastic > Isotropic

Enter EX= 2E4 and PRXY=0.3 > OK

(Close the Define Material Model Behavior window.)

Create the nodes and elements. Create 3 nodes 2 elements.

4.  Main Menu > Preprocessor > Modeling > Create > Nodes > In Active CS

Enter the coordinates of node 1 > Apply

Enter the coordinate of node 2 > Apply

Enter the coordinates of node 3 > OK

Node locations

Node number

X coordinate

Y coordinate

1

0

0

2

200

0

3

400

0

Figure 3.13. Enter the node coordinates.

5.  Main Menu > Preprocessor > Modeling > Create > Elements > Auto Numbered > Thru node Pick the 1st and 2nd node > Apply Pick 2nd and 3rd node > OK

Figure 3.14. pick the nodes to create elements.

Apply the displacement boundary conditions and loads.

6.  Main Menu > Preprocessor > Loads > Apply > Structural > Displacement > On Nodes Pick the 1st node > Apply > All DOF=0 > OK

7.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Structural > Displacement > On Nodes Pick the 3rd node > Apply > Select UX and enter displacement value = 1.2 > OK

8.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Structural > Force/Moment > On Nodes Pick the 2nd > OK > Force. Moment value = 80e3 > OK

Figure 3.15. Model with loading and displacement boundary conditions.

The model-building step is now complete, and we can proceed to the solution. First to be safe, save the model.

Solution

The interactive solution proceeds.

9.  Main Menu > Solution > Solve >  Current LS > OK

The STATUS Command window displays the problem parameters and the Solve Current Load Step window is shown. Check the solution options in the /STATUS window and if all is OK, select File > Close

In the Solve Current Load Step WINDOW, Select OK, and the solution is complete, close the ‘Solution is Done!’ window.

POST-PROCESSING

We can now plot the results of this analysis and also list the computed values.

10.  Main Menu > General Postproc > Plot Results > Contour Plot > Nodal Solu > DOF Solution > Displacement vector sum > OK

The result is shown in Figure 3.8(c).

To find the axial stress, the following procedure is followed.

11.  Main Menu > General Postproc > Element Table > Define Table > Add

Figure 3.16. Define the element table.

Select By sequence num and LS and type 1 after LS as shown in Figure 3.17.

Figure 3.17. Selecting options in element table.

OK > Close

12.  Main Menu > General Postproc > Plot Results > Contour Plot > Elem Table > Select > LS1 > OK

Figure 3.18. Selecting options for finding out axial stress.

The result is shown in Figure 3.8(d).

3.3 BARS OF VARYING CROSS-SECTION AREA

This section will demonstrate thorough examples explaining FEA on bars of varying cross-section area.

Example 3.6

Solve for displacement and stress given in Figure 3.19 using 2 finite elements model. Take Young’s modulus E = 200 GPa.

Figure 3.19. Example 3.6.

Solution

(I) Analytical method [Refer Figure 3.19(a)]

Figure 3.19(a). Analytical method for Example 3.6.

L = 2 m.

Stress calculation

Displacement Calculation

Displacement at section a-a = change in length of (L-x)

(II) FEM by hand calculations

Using 2 elements each of 1 m length, we obtain the finite element model as shown in Figure 3.19(c). We can write the equivalent model as shown in Figure 3.19(b). At the middle of the bar width is,

Figure 3.19(b). Equivalent model of Finite element model for Example 3.6.

Figure 3.19(c). Finite element model for Example 3.6.

Stiffness matrix for element 1 is,

Stiffness matrix for element 2 is,

Distributes load calculation for elements 1 and 2,

Global equation is,

Boundary conditions are at node 1, u_1 = 0.

By using the elimination method, the above matrix reduces to,

By matrix multiplication, we get

By solving equations (3.47) and (3.48), we get

u_2 = 9.5×10−5 m= 0.095 mm

u_3 = 2.37×10−4 m= 0.237 mm

Stress (σ) calculation

Stress in element 1,

Stress in element 2,

(III) Software results

Figure 3.19(d). Deflection pattern for a tapered bar (refer to Appendix C for color figures).

Deflection values at node

The following degree of freedom results are in global coordinates

Maximum absolute values

Figure 3.19(e). Stress pattern for a tapered bar (refer to Appendix C for color figures).

Stress values at elements

STAT

CURRENT

ELEM

LS1

3

19.000

4

28.333

Reaction value

The following X, Y, Z solutions are in global coordinates

NODE

FX

FY

1

0.0000

0.25000E+06

Answers of Example 3.6

In the above example, 2 elements are used for solving the problem by hand calculation and by software. To get the convergence of the solution with the analytical method a higher number of elements are to be used.

Example 3.7

Find the displacement and stress distribution in the tapered bar shown in Figure 3.20 using 2 finite elements under an axial load of P = 100 N.

Cross-sectional area at fixed end = 22 mm^2

Cross-sectional are at free end =100 mm^2

Young’s modulus E = 200 GPa

Figure 3.20. Example 3.7

Solution

(I) Analytical method [refer to Figure 3.20(a)]

Figure 3.20(a). Analytical method for Example 3.7.

Assume, b = thickness = 10 mm

Area at section a-a = b × hx

Stress Calculation

Displacement calculation

(II) FEM by hand calculations

Using 2 elements each of 50 mm length, we obtain the finite element model as shown in Figure 3.20(c). We can write the equivalent model as shown in Figure 3.20(b) at the middle, area of cross-section of bar is

Figure 3.20(b). Equivalent model of finite element model for Example 3.7.

Figure 3.20(c). Finite element model for Example 3.7.

Stiffness matrix for element 1 is,

Stiffness matrix for element 2 is,

Global equation is,

Boundary conditions are at node 1, u_1 = 0.

By using the elimination method, the above matrix reduces to,

By matrix multiplication, we get

By solving equations (3.51) and (3.52), we get

u_2=1.429×10−4 mm

u_2=3.429×10−4 mm.

Stress calculation

Stress in element 1,

Stress in element 2,

(III) Software results

Figure 3.20(d). Deflection pattern for a tapered bar (refer to Appendix C for color figures).

Deflection values at nodes

The following degree of freedom results are in global coordinates

Figure 3.20(e). Stress pattern for a tapered bar (refer to Appendix C for color figures).

Stress values at elements

STAT

CURRENT

ELEM

LS1

1

0.57143

2

0.80000

Answer for Example 3.7

Example 3.8

Find the nodal displacements, element stresses, and reaction in the tapered bar subjected to a load of 6000 N as shown in Figure 3.21. Further the member experiences a temperature increase of 30°C. Use 3 equal length elements for finite element model. Take E = 200 GPa, v = 0.3, and α = 7 × 10−6/°C.

Figure 3.21. Example 3.8

Solution

(I) FEM by hand calculations

We obtain the finite element model as shown in Figure 3.21(b). We can write the equivalent model as shown in Figure 3.21(a).

Figure 3.21(a). Equivalent model of the finite element model for Example 3.8.

Figure 3.21(b). Finite element model for Example 3.8.

Element stiffness matrices are,

Nodal loads due to thermal effect are,

Global forced vector

Global equation is,

Using the elimination method of applying boundary conditions, i.e., u_1 = 0.

The equation (3.53) reduces to,

Solving the above matrix and equations, we get

u_2 =0.1125 mm

u_3 =0.2275 mm

u_4 =0.3475 mm

Stress calculation

Reaction Calculation: from equation (3.53),

(III) Software results

Figure 3.21(c).Deflection pattern for a tapered bar (refer to Appendix C for color figures).

Deflection values at nodes

The following degree of freedom results are in global coordinates

Figure 3.21(d). Stress pattern for a tapered bar (refer to Appendix C for color figures).

Stress values at elements

STAT

CURRENT

ELEM

LS1

1

3.0000

2

4.0000

3

6.0000

Reaction value

The following X, Y, Z solutions are in global coordinates

NODE

FX

FY

1

−6000.0

0.0000

Answers for Example 3.8

Parameter

FEM-hand calculations

Software results

Displacement at node 2

0.1125 mm

0.1125 mm

Displacement at node 3

0.2275 mm

0.2275 mm

Displacement at node 4

0.3475 mm

0.3475 mm

Stress in node 1 of element 1

3 MPa

3 MPa

Stress in node 2 of element 1

4 MPa

4 MPa

Stress in node 3 of element 1

6 MPa

6 MPa

Reaction at fixed end

−6000 N

−6000 N

Procedure for solving the problem using ANSYS® 11.0 academic teaching software

For Example 3.6

PROCESSING

1.  Main Menu > Preprocessor > Element Type > Add/Edit > Delete > Add > Structural Link > 2D spar 1 > OK > Close

Figure 3.22. Element selection.

2.  Main Menu > Preprocessor > Real Constants > Add/Edit/Delete > Add > OK

Figure 3.23. Enter the cross-sectional area of 1st element.

Cross-sectional area > Enter 12500 > OK > Add > OK

Figure 3.24. Enter the cross-sectional area of 2nd element.

Cross-sectional area AREA > Enter 7500 > OK > Close

Enter the material properties.

3.  Main Menu > Preprocessor > Material Props > Material Models

Material Model Number 1, click Structural > Linear > Elastic > Isotropic

Enter EX=200E3 and PRXY=0.3 > OK

(Close the Define Material Model Behavior window.)

Create the nodes and element. As stated in the example, use 2 element model.

Hence create 3 nodes and 2 elements.

4.  Main Menu > Preprocessor > Modeling > Create > Nodes > In Active CS

Enter the coordinates of node 1 > Apply Enter the coordinates of node 2 > Apply Enter the coordinates of node 3 > OK.

Node locations

Node number

X coordinate

Y coordinate

1

0

0

2

0

−1000

3

0

−2000

Figure 3.25. Enter the node coordinates.

5.  Main Menu > Preprocessor > Modeling > Create > Elements > Elem Attributes > OK > Auto Numbered > Thru nodes Pick the 1st and 2nd node > OK

Figure 3.26. Assigning element attributes to element 1 and creating element 1.

Elem Attributes > change the Real constant set number to 2 > OK > Auto Numbered > Thru nodes Pick the 2nd and 3rd node > OK

Figure 3.27. Assigning element attributes to element 2 and creating element 2.

Apply the displacement boundary conditions and loads.

6.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Structural > Displacement > On Nodes Pick the 1st node > Apply > All DOF=0 > OK

7.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Structural > Force/Moment > On Nodes Pick the 2nd node > OK > Force/Moment value=−25e3 in FY direction > OK > Force/Moment > On Nodes Pick the 1st node > OK > Force/Moment value=−12.5e3 in FY direction > OK

Figure 3.28. Model with loading and displacement boundary conditions.

The mode-building step is now complete, and we can proceed to the solution.

First to be safe, save the model.

Solution. The interactive solution proceeds.

8.  Main Menu > Solution > Solve > Current LS > OK

The /STATUS Command window displays the problem parameters and the Solve Current Load Step window. Select OK, and when the solution is complete, close the ‘Solution is DONE!’ window.

POST-PROCESSING

We can now plot the results of this analysis and also list the computed values.

9.  Main Menu > General Postproc > Plot Results > Contour Plot > Nodal Solu > DOF Solution > Displacement vector sum > OK

This result is shown in Figure 3.16(d).

To find the axial stress, the following procedure is followed.

10.  Main Menu > General Postproc > Element Table > Define Table > Add

Figure 3.29. Defining the element table.

Select By sequence num LS and type 1 after LS as shown in Figure 3.27.

Figure 3.30. Selecting options in the element table.

OK

11.  Main Menu > General PostProc > Plot Results > Contour Plot > Elem Table > Select LS1 > OK

Figure 3.31. Selecting options for finding out axial stress.

This result is shown in Figure 3.19(e).

3.4 STEPPED BAR

This section will demonstrate examples on stepped bar using FEA.

Example 3.9

Find the nodal displacements, stresses in each element, and reaction at the fixed end for the Figure 3.32 shown below. Take A_1 = 200 mm^2, A_2 = 200 mm^2, and E_1 = E_2 = 200 GPa.

Figure 3.32. Example 3.9

Solution

(I) Analytical method [refer to Figure 3.32]

Displacement Calculation

Stress calculation

Reaction calculation

(II) FEM by hand calculations

Figure 3.32(a). Finite element model for Example 3.9.

L_1 = 200 mm

L_2 = 100 mm

Displacement calculation

Stiffness matrices for elements 1 and 2 are,

Global equation is,

Using the elimination method and applying boundary conditions at node 1, u_1 = 0.

The equation (3.54) reduces to

By solving the above matrix and equations,

we get,

Stress calculations

Reaction calculation

From equation (i)

(III) Software results

Figure 3.32(b). Deflection pattern for a stepped bar (refer to Appendix C for color figures).

Deflection values at nodes

The following degree of freedom results are in global coordinates

Figure 3.32(c). Stress for a stepped bar (refer to Appendix C for color figures).

Stress value at elements

STAT

CURRENT

ELEM

LS1

1

−2.5000

3

5.0000

Reaction value

The following X, Y, Z solutions are in global coordinates

NODE

FX

FY

1

500.00

0.0000

Answers for Example 3.9

Example 3.10

Find the nodal displacements, stress in each element, and reaction of the fixed end for Figure 3.33 shown below. Take E_1 = 2 × 10^5 N/mm^2 and E_2 = 1 × 10^5 N/mm^2.

Figure 3.33. Example 3.10

Solution

(I) Analytical method [refer to Figure 3.33]

In the absence of the right wall,

Hence, the contact does occur with the right will since u_3 = 1.415 mm.

Let RA and RC be the reactions developed due to constraint.

By solving equations (3.55) and (3.56),

we get

RA = 776547.49 N

RC = 23452.51 N.

Displacement calculation

Stress calculation

(II) FEM by hand calculations

Figure 3.33(a). Finite element model for Example 3.10.

L_1 = 1000 mm

L_2 = 2000 mm.

In this example, first determine whether contact occurs between the bar and the wall. To do this, assume that the wall does not exist. Then the solution to the problem is (consider the 2 element model),

Stiffness matrix for element 1 is,

Stiffness matrix for element 2 is,

Global equation is,

Boundary conditions are at node 1, u_1 = 0.

By using the elimination method, the above matrix reduces to,

By matrix multiplication, we get

By solving equations (3.59) and (3.60)

we get

u_2 = 1.415 mm and u_3 = 1.415 mm.

Since the displacement of node 3 is 1.415 mm, we can say that contact does occur.

The problem has to be resolved since the boundary conditions are now different.

The displacement at node 3 is given as 1 mm.

Global equation is,

Boundary conditions at node 1, u_1 = 0 and at node 3, u_2 = 1 mm.

By using the elimination method, the above matrix reduces to,

Stress calculation

Reaction calculation

From equation (3.60)

We know that,

(III) Software results

Figure 3.33(b). Deflection pattern for a stepped bar (refer to Appendix C for color figures).

Deflection values at nodes

The following degree of freedom results are in global coordinates

Figure 3.33(c). Stress pattern for a stepped bar (refer to Appendix C for color figures).

Stress values at elements

STAT

CURRENT

ELEM

LS1

1

247.65

2

−18.662

Reaction value

The following X, Y, Z solutions are in global coordinates

NODE

FX

FY

1

−0.77655E+06

0.0000

3

−23451.

Answers for Example 3.10

Example 3.11

Find the nodal displacement, stress in each element, and reaction at fixed ends for Figure 3.34 as shown below. If the structure is subjected to an increase in temperature, ΔT = 75°C, P_1 = 50 kN, P_2 = 75 kN.

Figure 3.34 Example 3.11.

Bronze

Aluminum

Steel

A = 2400 mm^2

1200 mm^2

600 mm^2

E = 83 GPa

70 GPa

200 GPa

a = 18.9 × 10−6/°C

23 × 10−6/°C

11.7 × 10−6/°C

Solution

(I) Analytical method [refer to Figure 3.34]

Problem can be solved by method of superposition by considering load and temperature separately.

Step 1: Consider only the loads P_1, P_2, and neglect rise in temperature.

Solving equations (3.61) and (3.62)

Step 2: Consider only the rise in temperature and neglect P_1 and P_2. Free expansions due to ΔT = 75°C are

For equilibrium

Solving,

Step 3: Use method of superposition and combine steps (1) and (2).

Stresses are,

Similarity,

Change in lengths are,

Reactions are,

(II) FEMby hand calculations

Figure 3.34(a). Finite element model for Example 3.11.

Element stiffness matrices are,

Effect of temperature and thermal loads are,

Global Force vector

Using the elimination method and applying boundary conditions,

i.e.,   u_1 = u_4 = 0.

The equation (3.63) reduces to,

By solving the above matrix and equation,

we get   u_2 = 0.2212 mm and u_3 = −0.0101 mm.

Stress calculation

Reaction calculation

(III) Software results

Figure 3.34(b). Deflection pattern for a stepped bar (refer to Appendix C for color figures).

Deflection values at nodes

The following degree of freedom results are in global coordinates

Figure 3.34(c). Stress pattern for a stepped bar (refer to Appendix C for color figures).

Stress values at elements

STAT

CURRENT

ELEM

LS1

1

−94.700

2

−147.73

3

−170.47

Reaction values

The following X, Y, Z solutions are in global coordinates

NODE

FX

FY

1

0.22728E +06

0.0000

4

−0.10228E+06

0.0000

Answers for Example 3.11

Procedure for solving the example using ANSYS® 11.0 academic teaching software

For Example 3.11

PREPROCESSING

1.  Main Menu > Preprocessor > Element Type > Add/Edit/Delete > Add > Structural Link > 2D spar 1 > Ok > Close

Figure 3.35. Element selection.

2.  Main Menu > Preprocessor > Real Constants > Add/Edit/Delete > Add > OK

Figure 3.36. Enter the cross-sectional area of 1st element.

Cross-sectional area AREA > Enter 2400 > OK > Add > OK

Figure 3.37. Enter the cross-sectional area of 2nd element.

Cross-sectional area AREA > Enter 1200 > OK > Add > OK

Figure 3.38. Enter the cross-sectional area of 3rd element.

Cross-sectional area AREA > Enter 600 > OK > Add > OK > Close

Enter the material properties.

3.  Main Menu > Preprocessor > Material Props > Material Models

Material Model Number 1,

Click Structural > Linear > Elastic > Isotropic

Enter EX = 0.83E5 and PRXY = 0.34 > OK

Enter the coefficient of thermal expansion α

Click Structural > Thermal Expansion > Secant coefficient > Isotropic

Enter ALPX − 18.9E-6 > OK

Then in the material model window click on Material menu > New Model > OK

Material Model Number 2,

Click Structural > Linear > Elastic > Isotropic

Enter EX = 0.7E5 AND PRXY -0.35 > OK

Enter the coefficient of thermal expansion α

Click Structural > Thermal Expansion > Secant coefficient > Isotropic

Enter ALPX = 23E-6 > OK

Then in the material model window click on Material menu > New Model > OK

Material Model Number 3,

Click Structural > Linear > Elastic > Isotropic

Enter EX =2E5 and PRXY = 0.3 > OK

Enter the coefficient of thermal expansion α

Click Structural > Thermal Expansion > Secant coefficient > Isotropic

Enter ALPX = 11.7E-6 > OK

(Close the Define Material Model Behavior window.)

Create the nodes and elements. Use 3 element models. Hence create 4 nodes and 3 elements.

4.  Main Menu > Preprocessor > Modeling > Create > Nodes > In Active CS Enter the coordinates of node 1 > Apply Enter the coordinates of node > Apply Enter the coordinates of node 3 > Apply > Enter the coordinates of node 4 > OK.

Node locations

Node number

X COORDINATE

Y COORDINATE

1

0

0

2

800

0

3

1400

0

4

1800

0

Figure 3.39. Enter the node coordinates.

5.  Main Menu > Preprocessor > Modeling > Create > Elements > Elem Attributes > OK > Auto Numbered > Thru nodes Pick the 1st and 2nd node > OK

Figure 3.40. Assigning element attributes to element 1 and creating element 1.

Elem Attributes > change the material number to 2 > change the Real constant set number to 2 > OK > Auto Numbered > Thru nodes Pick the 2nd and 4th node > OK

Figure 3.41. Assigning element attributes to element 2 and creating element 2.

Elem Attributes > change the material number to 3 > change the Real constant set number to 3 > OK > Auto Numbered > Thru nodes Pick the 3rd and 4th node > OK

Figure 3.42. Assigning element attributes to element 3 and creating element 3.

Apply the displacement boundary conditions, load, and temperature.

6.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Structural > Displacement > On Nodes Pick the 1st and 4th node > Apply > All DOF = 0. > OK

7.  Main Menu > preprocessor > Loads > Define Loads > Apply > Structural > Force/Moment > On Nodes Pick the 2nd node > OK > Force/Moment value = −50e3 in FX direction > OK > Force/Moment > On Nodes Pick the 3rd node > OK > Force/Moment value =−75e3 in FX direction > OK

8.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Structural > Temperature > On Elements Pick the element, 2nd element and 3rd element > OK

Enter Temperature at location N = 75 as shown in Figure 3.38.

Figure 3.43. Enter the rise in temperature on elements.

Figure 3.44. Model with loading and displacement boundary conditions.

The model-building step is now complete, and we can proceed to the solution. First to be safe, save the model.

Solution

The interactive solution proceeds.

9.  Main Meni > Solution > Solve > Current LS > OK

The /STATUS Command window displays the problem parameters and the Solve Current Load Step window and if all is OK, select FILE > CLOSE

In the Solve Current Load Step window, Select OK, and when the solution is complete, close the ‘Solution is Done!’ window.

POST-PROCESSING

We can now plot the results of this analysis and also list the computed values.

10.  Main Menu > General Postproc > Plot Results > Contour Plot > Nodal Solu >DOF Solution > Displacement vector sum > OK

This result is shown in Figure 3.34(b).

To find the axial stress, the following procedure is followed.

11.  Main Menu > General Postproc > Element Table > Define Table > Add

Figure 3.45. Defining the element table.

Select By sequence num and LS and type 1 after LS as shown in Figure 3.43.

Figure 3.46. Selecting options in element table.

>OK

12.  Main Menu > General Postproc > Plot Results > Contour Plot > Elem Table > Select LS1 > OK

Figure 3.47. Selecting options for finding out axial stress.

This result is shown in Figure 3.34(c).

PROBLEMS

1.  Determine the nodal displacement and element stress for the bar shown in Figure 3.48. Take 3 elements finite element model. Take E = 70 GPa.

Figure 3.48. Problem 1

2.  Determine the nodal displacements and stresses in the element for the axial distributed loading shown in Figure 3.49. Take one element model. Take E = 200 GPa, A = 5 × 10−4 m^2.

Figure 3.49. Problem 2

3.  For the bar assembly shown in Figure 3.50, determine the nodal displacements, stresses in each element, and reactions. Take E = 210 GPa, A = 5 × 10−4 m^2.

Figure 3.50. Problem 3

4.  Find the deflection at the free end under its own weight for a tapered bar shown in Figure 3.51. Use 2 element models. Take E = 200 GPa, weight density ρ = 7800 kg/m^3.

Figure 3.51. Problem 4

5.  Determine the displacement, element stresses, and reactions for the tapered bar shown in Figure 3.52. Use 2 elements finite element models. Take E = 200 GPa, A_1 = 2000 mm^2, A_2 = 4000 mm^2.

Figure 3.52. Problem 5

6.  Consider the bar shown in Figure 3.53. An axial load P = 500 kN is applied as shown. Determine the

(a) Nodal displacement (b) Stresses in each material (c) Reaction forces.

Figure 3.53. Problem 6

Aluminum

Steel

A_1 = 300 mm^2

A_2 = 1000 mm^2

E_1 = 70 GPa

E_2 = 200 GPa

7.  In Figure 3.54, determine displacements at 2 and 3 stresses in the members and reactions if the temperature is increased by 60°.

Figure 3.54. Problem 7

8.  For the vertical bar shown in Figure 3.55, for the deflection at 2 and 3 and stress distribution. Take E = 25 GPa and density, ρ = 2100 kg/m^3. Take self-weight of the bar into consideration and solve the problem using 2 elements.

Figure 3.55. Problem 8

9.  Find displacement and stresses shown in Figure 3.56. Take E = 200 GPa.

Figure 3.56. Problem 9 (all dimensions are in mm).

REFERENCES

1.  J. Wickert, “An Introduction to Mechanical Engineering, Second Edition,” Thomson Publisher, 2006.

2.  J. M. Gere and S. P. Timoshenko, “Mechanics of Materials, Fourth Edition,” PWS Publishing, 1997.

3.  R. G. Budynas and J. K. Nisbett, “Shigley’s Mechanical Engineering Design, Eighth Edition,” McGraw-Hill Higher Education, 2008.

4.  H. C. Martin and G. F. Carey, “Introduction to Finite Element Analysis: Theory and Applications,” McGraw-Hill Book Company, 1973.

5.  W. B. Bickford, “A First Course in the Finite Element Method, Second Edition,” Richard D. Irwin, INC., 1990.

Chapter 4

FINITE ELEMENT ANALYSIS TRUSSES

4.1 INTRODUCTION

This chapter introduces the basic concepts in finite element formulation of trusses and provides the illustration of its ANSYS program.

4.2 TRUSS

Truss, by definition, is a load bearing structure formed by connecting members using pin joints. Truss element is used in the analysis of 2-D trusses.

Figure 4.1. A 2-D Truss.

The element has two nodes, each having two degrees of freedom namely translations along the x- and y-axes.

The element stiffness matrix and element stress for a truss element are given by,

θ = angle of truss element at node 1 with positive x-axis (in degrees).

Example 4.1

Determine the nodal displacements, element stresses, and support reactions for the 3 member truss shown in Figure 4.2. Take A = 800 mm^2 and E = 200 GPa for all members.

Figure 4.2. Example 4.1

Solution

(I) Analytical method [refer to Figure 4.2]

Consider equilibrium of joint B,

Figure 4.2(a). Analytical method for joint B in Example 4.1.

Solving equations (4.3) and (4.4)

PAB= −5 kN and PBC = −15 kN (PAB and PBC are compressive).

Consider equilibrium of joint A,

Figure 4.2(b). Analytical method for joint A in Example 4.1.

R1y−PABsinθ = 0

Consider equilibrium of joint C,

Figure 4.2(c). Analytical method for joint C in Example 4.1.

For equation (4.5)

Calculation of nodal displacements u_2, v_2, and u_3

Figure 4.2(d). Analytical method for Calculation of nodal displacements u_2, v_2, and u_3 in Example 4.1.

From geometry [refer to Figure 4.2(d)].

Substituting in equations (4.6) and (4.7)

Solving equations (4.8) and (4.9), we get,

v_2 = 0.4604 mm (since point B moves downwards). Hence v_2 = −0.6404 mm.

(III) FEM by hand calculation [refer to Figure 4.2]

Angle calculation

For element 2

Figure 4.2(e). Angle calculation for element 2 in Example 4.1.

For element 3

Figure 4.2(f). Angle calculation for element 3 in Example 4.1.

Element stiffness matrix for element 1 is,

Element stiffness matrix for element 2 is,

Element stiffness matrix for element 3 is,

Global stiffness matrix is,

Global equation is,

Using the elimination method for applying boundary conditions,

u_1 = v_1 = v_3 = 0.

Then the above matrix reduces to,

Solving the above matrix and equations,

we get           u_2 = 0.2477 mm, v_2 = 0.4604 mm, and v_3 = 0.3 mm.

Stress calculation

Stress in element 1 is,

Stress in element 2 is,

Stress in element 3 is,

Reaction calculation

From global equation,

(III) Software results

Figure 4.2(g). Deflection pattern for a truss for Example 4.1 (refer to Appendix C for color figures).

Deflection value at nodes

The following degree of freedom results are in global coordinates

Figure 4.2(h). Stress pattern for a truss for Example 4.1 (refer to Appendix C for color figures).

Stress values of elements

STAT

CURRENT

ELEM

LS1

1

15.000

2

−6.2500

3

−18.750

Reaction values

The following X, Y, Z solutions are in global coordinates

NODE

FX

FY

1

−8000.0

3000.0

3

9000.0

Answers for Example 4.1

Example 4.2

For the truss shown in Figure 4.3, determine nodal displacements and stresses in each member. All elements have E = 200 GPa and A = 500 mm^2.

Figure 4.3. Example 4.2

Solution

(I) FEM by hand calculation

Stiffness matrices for elements 1, 2, and 3 are,

Global stiffness matrix is,

Global equation is,

Using the elimination method for applying boundary conditions,

u_2 = v_2 = u_3 = v_3 = u_4 = v_4 = 0.

Then the above matrix reduces to,

Solving the above matrix and equations,

we get,

Stress calculation

Stress in element 1 is,

Stress in element 2 is,

Stress in element 3 is,

(II) Software results

Figure 4.3(a). Deflection pattern for a truss for Example 4.2 (refer to Appendix C for color figures).

Deflection value at nodes

The following degree of freedom results are in global coordinates system

Figure 4.3(b). Stress pattern for a truss for Example 4.2 (refer to Appendix C for color figures).

Stress values of elements

STAT

CURRENT

ELEM

LS1

1

14.995

2

35.363

3

34.995

Answers for Example 4.2

Parameter

FEM-hand calculations

Software results

Displacement of node 1 in

x-direction

−0.375 mm

−0.37486 mm

y-direction

−0.875 mm

−0.87486 mm

Stress in Element 1

15 MPa

14.995 MPa

Stress in Element 2

35.352 MPa

35.363 MPa

Stress in Element 3

35 MPa

34.995 MPa

Example 4.3

The bar truss shown in Figure 4.4, determine the displacement of node 1 and the axial stress in each member. Take E = 210 GPa and A = 600 mm^2. Solve the problem if node 1 settles an amount of δ = 25 mm in the negative y-direction.

Figure 4.4. Example 4.3

Solution

(I) FEM by hand calculation

Angle calculation

For 2nd element,

Figure 4.4.(a). Angle calculation for 2nd element for Example 4.3.

Stiffness matrices for element 1 and 2 are,

Global stiffness matrix is,

Global equation is,

Using the elimination method for applying boundary conditions,

i.e.,                  u_2 = v_2 = u_3 = v_3 = 0.

Then the above matrix reduces to,

We know that v_1 = −25 mm, substitute this in the above matrix, then,

Solving the above matrix and equations we get,

u_1 = −48.34 mm.

Stress calculation

Stress in element 1 is,

Stress in element 2 is,

(II) Software results

Figure 4.4(b). Deflection pattern for a truss for Example 4.3 (refer to Appendix C for color figures).

Deflection value at nodes

The following degree of freedom results are in global coordinates system

Maximum absolute values

Figure 4.4(c). Stress pattern for a truss for Example 4.3 (refer to Appendix C for color figures).

Stress values of elements

STAT

CURRENT

ELEM

LS1

1

−2537.9

2

−994.27

Answers for Example 4.3

Parameter

FEM-hand calculations

Software results

Displacement of node 1 in

x-direction

−48.34 mm

−48.341 mm

y-direction

−25 mm

−25 mm

Stress in Element 1

2537.85 MPa

2537.9 MPa

Stress in Element 2

994.22 MPa

994.27 MPa

Procedure for solving the problems using ANSYS® 12.0 academic teaching software

For Example 4.3

PREPROCESSING

1.  Main Menu > Preprocessor > Element Type > Add/Edit/Delete > Add > Structural Link > 2D spar 1 > OK > Close

Figure 4.5. Element selection.

2.  Main Menu > Preprocessor > Real Constants > Add/Edit/Delete > Add > OK

Figure 4.6. Enter the cross-sectional area.

Cross-sectional area AREA > Enter 600 > OK > Close

Enter the material properties.

3.  Main Menu > Preprocessor > Material Props > Material Models

Material Model Number 1, Click Structural > Linear > Elastic > Isotropic

Enter EX = 2.1E5 and PRXY = 0.3 > OK

(Close the Define Material Model Behavior window.)

Create the nodes and elements as shown in the figure.

4.  Main Menu > Preprocessor > Modeling > Create > Nodes > In Active CS

Enter the coordinates of node 1 > Apply Enter the coordinates of node 2 >

Apply Enter the coordinates of node 3 > OK

Node locations

Node number

X-coordinate

Y-coordinate

1

0

0

2

−4000

3000

3

−4000

0

Figure 4.7. Enter the node coordinates.

5.  Main Menu > Preprocessor > Modeling > Create > Elements > Auto Numbered > Thru nodes Pick the 1st and 2nd node > Apply Pick the 1st and 3rd node > OK

Figure 4.8. Pick the nodes to create elements.

Apply the displacement boundary conditions and loads.

6.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Structural > Displacement > On Nodes Pick the 2nd and 3rd node > Apply > All DOF=0 > OK

7.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Structural > Displacement > On Nodes Pick the 1st node > Apply > UY=-25 > OK

8.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Structural > Force/Moment > On Nodes Pick the 1st node > OK > Force/Moment value= -2000e3 > OK

Figure 4.9. Model with loading and displacement boundary conditions.

The model-building step is now complete, and we can proceed to the solution. First to be safe, save the model.

Solution

The interactive solution proceeds.

9.  Main Menu > Solution > Solve > Current LS > OK

The /STATUS Command window displays the problem parameters and the Solve Current Load Step window is shown. Check the solution options in the /STATUS window and if all is OK, select File > Close.

In the Solve Current Load Step window, select OK, and the solution is complete, close the ‘Solution is Done!’ window.

POST-PROCESSING

We can now plot the results of this analysis and also list the computed values.

10.  Main Menu > General Postproc > Plot Results > Contour Plot > Nodal Solu > DOF Solution > Displacement vector sum > OK

This result is shown in Figure 4.4(b).

To find the axial stress, the following procedure is followed.

11.  MAIN Menu > General Postproc > Element Table > Define Table > Add

Figure 4.10. Defining the element table.

Select By sequence num and LS and type 1 after LS (as shown in Figure 4.11) >OK

Figure 4.11. Selecting options in element table.

12.  Main Menu > General Postproc > Plot Results > Contour Plot > Elem Table > Select LS1 > OK

Figure 4.12. Selecting options for finding out axial stress.

This result is shown in Figure 4.4(c).

PROBLEMS

1.  For a 5 bar truss shown in Figure 4.13, determine the following:

(a) nodal displacements

(b) stresses in each element

(c) reaction forces.

Take E = 200 GPa and Area A = 750 mm^2 for all elements.

Figure 4.13. Problem 1

2.  For the 3 bar truss shown in Figure 4.14, determine the displacement of node 1 and the stresses in elements. Take A = 300 mm^2 and E = 210 GPa.

Figure 4.14. Problem 2

3.  Consider the truss shown in Figure 4.15, determine the nodal displacements, element stresses, and reactions. Take E = 200 GPa. A_1 = A_2 = A_3 = 500 mm^2, P_1 = 300 kN, P_2 = 200 kN.

Figure 4.15. problem 3

4.  Consider the truss structure shown in Figure 4.16, determine the stresses of the truss structure. Take all members have elastic modulus (E) of 210 GPa and cross-sectional area (A) of 250 mm^2.

Figure 4.16. Problem 4

5.  Consider the truss structure shown in Figure 4.17, derive the finite element matrix equations using 2 elements. Determine the displacements and the stresses in the member. Assume all members have elastic modulus (E) of 200 GPa and cross-sectional area (A) of 300 mm^2.

Figure 4.17. Problem 5

6.  Consider the truss structure shown in Figure 4.18, determine the nodal displacement and the element forces assuming that all elements have the same AE.

Figure 4.18. Problem 6

7.  Determine the nodal displacements, element stresses, and support reactions for the 3 member truss shown in Figure 4.19. Take A_1 = 10 in^2, A_2 = 15 in^2, A_3 = 10 in^2 and E = 20 msi for all members.

Figure 4.19. Problem 7

8.  Determine the nodal displacements, element stresses and support reactions for the three member truss shown in Figure 4.20. Take A_1 = 1 in^2, A_2 = 2 in^2, A_3 = 3 in^2, and E = 30 Mlb/in^2 for all members.

Figure 4.20. Problem 8

9.  Determine the nodal displacements, element stresses, and support reactions for the 3 member truss shown in Figure 4.21. Take A_1 = 6 cm^2, A_2 = 8 cm^2, A_2 = 8 cm^2, and E = 20 MN/cm^2 for all members.

Figure 4.21. Problem 9

REFERENCES

1.  Y. W. Hwon and H. Bang, “The Finite Element Method Using MATLAB, Second Edition,” CRC Press, 2000.

2.  D. L. Logan, “A First Course in the Finite Element Method, Fifth Edition,” Cengage Learning, 2012.

3.  S. Moaveni, “Finite Element Analysis: Theory and Application with ANSYS, Third Edition,” Prentice Hall, 2008.

4.  J. N. Reddy, “An Introduction to the Finite Element Method, Third Edition,” McGraw Hill Higher Education, 2004.

5.  C. T. F. Ross, “Finite Element Method in Structural Mechanics,” Ellis Horwood Limited Publishers, 1985.

6.  F. L. Stasa, “Applied Finite Element Analysis for Engineering,” Holt, Rinehart and Winston, 1985.

7.  L. J. Segerlind, “Applied Finite Element Analysis, Second Edition,” John Wiley and Sons, 1984.

Chapter 5

FINITE ELEMENT ANALYSIS OF BEAMS

5.1 INTRODUCTION

Beam is very common structure in many engineering applications because of its efficient load carrying capability. Beam by definition is a transversely loaded structural member. Beam element is used in the analysis of beams.

Figure 5.1. Beam element.

This element has 2 end nodes each having 2 degrees of freedom, namely transverse displacement and slope. Beam element gives accurate results if acted upon by nodal forces and moments. A greater number of small elements will be necessary in the case of a beam acted upon by distributed loads in order to get good results. The interpolation equation and element stiffness matrix for beam element are given by

5.2 SIMPLY SUPPORTED BEAMS

Example 5.1

For the beam shown in Figure 5.2, determine the nodal displacements, slope, and reactions. Take E = 210 GPa and I = 4 × 10−4 m^4.

Figure 5.2. The beam for Example 5.1.

Solution

(I)  Analytical method [refer to Figure 5.2]

L = 10 m

P = 8 kN.

Deflection,

θC=0, by symmetry.

Reaction,

(II)  FEM by hand calculations [refer to Figure 5.2(a)]

Figure 5.2(a). Finite element model for Example 5.1.

Element stiffness matrices are,

Due to symmetry,

Global equation is,

Using the elimination method for applying boundary conditions,

w_1 = w_3 = 0.

The above matrix reduces to

By solving the above equations, we get,

w_2 = −0.002 m = −2 mm,

θ_1 = −0.0006 rad, θ_2 = 0 rad, and θ_3 = 0.0006 rad.

Reaction calculation

(III) Software results

Figure 5.2(b). Deflection pattern for a simply supported beam (refer to Appendix C for color figures).

Deflection values at nodes (in meters)

The following degree of freedom results are in global coordinates

The following degree of freedom results are in global coordinates

NODE

ROTZ

1

−0.59524E-03

2

0.0000

3

0.59524E-03

Reaction values

The following X, Y, Z solutions are in global coordinates

Answers for Example 5.1

Example 5.2

For the beam shown in Figure 5.3, determine displacements, slopes, reactions, maximum bending moment, shear force, and maximum bending stress. Take E = 210 GPa and I = 2 × 10−4 m^4. The beam has rectangular cross-section of depth h = 1 m.

Figure 5.3. The beam for Example 5.2.

Solution

(I)Analytical method [refer to Figure 5.3]

Reaction,

θC = 0, by symmetry.

Maximum bending moment,

Shear force,

Maximum bending stress,

Figure 5.3(a). Bending moment diagram.

Figure 5.3(b). Shear force diagram.

(II)  FEM by hand calculations [refer to Figure 5.3(c)]

Figure 5.3(c). Finite element model for Example 5.2.

Stiffness matrices are,

Due to symmetry,

Nodal force calculation

For element 1,

Figure 5.3(d). Nodal force calculation for element 1 in Example 5.2.

Nodal forces and moments for element 1 is,

For element 2,

Figure 5.3(e). Nodal force calculation for element 2 in Example 5.2.

Due to symmetry,

Global equation is,

Using the elimination method for applying boundary conditions,

w_1 = w_3 = 0.

The above matrix reduces to

By solving the above equations, we get,

w_2 = −0.0102 m,

θ_1 = −0.0036 rad, θ 2 = 0 rad, and θ 3 =0.0036 rad.

Reactions are calculated from 1st and 5th rows of global matrix.

Similarly from 5th row

R3y = 22.865 kN.

(III)   Software results

Figure 5.3(f). Deflection pattern for a simply supported beam (refer to Appendix C for color figures).

Deflection values at nodes (in meters)

The following degree of freedom results are in global coordinates

Slope values at nodes

The following degree of freedom results are in global coordinates

NODE

ROTZ

1

−0.36161E-02

2

0.0000

3

0.36161E-02

Reaction values

The following X, Y, Z solutions are in global coordinates

Total values

Figure 5.3(g). Bending moment diagram for a simply supported beam (refer to Appendix C for color figures).

Figure 5.3(h). Shear force diagram for a simply supported beam (refer to Appendix C for color figures).

Figure 5.3(i). Bending stress for a simply supported beam (refer to Appendix C for color figures).

Answers for Example 5.2

Example 5.3

For the beam shown in Figure 5.4, determine displacements, slopes, and reactions. Take E = 200 GPa and I = 6.25 × 10−4 m^4.

Figure 5.4. The beam for Example 5.3.

Solution

(I) Analytical method[refer to Figure 5.4]

Reaction,

(II) FEM by hand calculations [refer to Figure 5.4(a)]

Figure 5.4(a). Finite element model for Example 5.3.

Stiffness matrices are,

Due to symmetry,

Global stiffness matrix is,

Load vector,

For element 1,

For element 2,

Global load vector is,

Global equation is,

Using the elimination method for applying boundary conditions,

w_1 = w_3 = 0.

The above matrix reduces to

By solving the above equations, we get,

Reaction calculation

(III) Software results

Figure 5.4(b). Deflection pattern for a simply supported beam (refer to Appendix C for color figures).

Deflection values at nodes (in meters)

The following degree of freedom results are in global coordinates

Maximum absolute values

Slope values at nodes

The following degree of freedom results are in global coordinates

NODE

ROTZ

1

−0.39822E-02

2

−0.24889E-03

3

0.45511E-02

Reaction values

The following X, Y, Z solutions are in global coordinates

Total values

Answers for Example 5.3

Example 5.4

Calculate the maximum deflection in the beam shown in Figure 5.5. Take E = 200 GPa.

Figure 5.5. The beam for Example 5.4.

Solution

(I) Analytical method [refer Figure 5.5(a)]

Figure 5.5(a). Analytical method for Example 5.4.

(II) FEM by hand calculations [refer to Figure 5.5(b)]

Figure 5.5(b). Finite element model for Example 5.4.

For beam,

For element 1 and 2, L = 40 mm

Due to symmetry,

Global stiffness matrix is,

Global load vector is,

Global equation is,

Using the elimination method for applying boundary conditions,

w_1 = w_3 = 0.

The above matrix reduces to

By solving the above equations, we get,

At θ_2 = 0, max deflection between supports is 0.0128 mm.

Deflection at ends (overhang) = 4.8×10−4×10 = 4.8 ×10−3 mm.

(III) Software results

Figure 5.5(c). Deflection pattern for a simply supported beam (refer to Appendix C for color figures).

Deflection values at nodes (in mm)

The following degree of freedom results are in global coordinates

Maximum absolute values

Slope values at nodes

The following degree of freedom results are in global coordinates

NODE

ROTZ

1

−0.48000E-03

2

−0.48000E-03

3

0.0000

4

0.48000E-03

5

0.48000E-03

Answers for Example 5.4

Procedure for solving the problems using ANSYS® 11.0 academic teaching software.

For Example 5.2

PREPROCESSING

1.  Main Menu > Preprocessor > Element Type > Add/Edit/Delete > Add > Beam > 2D elastic 3 > OK > Close

Figure 5.6. Element selection.

2.  Main Menu > Preprocessor > Real Constants > Add/Edit/Delete > Add > OK

Figure 5.7. Enter the area, moment of inertia, and height of beam.

Cross-sectional area AREA > Enter 1

Area moment of inertia IZZ > Enter 2e-4

Total beam height HEIGHT > Enter 1 > OK > Close

Enter the material properties.

3.  Main Menu > Preprocessor > MATERIAL Props > Material Models Material Model Number 1, Click Structural > Linear > Elastic > Isotropic Enter EX = 210E9 and PRXY = 0.3 > OK (Close the define material model behavior window.) Create the nodes and elements as shown in the table below and Figure 5.8.

4.  Main Menu > Preprocessor > Modeling > Create > Nodes > In Active CS Enter the coordinates of node 1 > Apply Enter the coordinates of node 2 > Apply Enter the coordinates of node 3 > OK.

Node locations

Node number

X-coordinate

Y-coordinate

1

0

0

2

4.5

0

3

9

0

Figure 5.8. Enter the node coordinates.

5.  Main Menu > Preprocessor > Modeling > Create > Elements > Auto Numbered > Thru nodes Pick the 1st and 2nd node > Apply Pick the 2nd and 3rd node > OK

Figure 5.9. Pick the nodes to create elements.

Apply the displacement boundary conditions and loads.

6.   Main Menu > Preprocessor > Loads > Define Loads > Apply > Structural > Displacement > On Nodes Pick the 1st node and 3rd node > Apply > Select UX and UY and Enter displacement value = 0 > OK

Figure 5.10. Apply the displacement constraint.

7.   Main Menu > Preprocessor > Loads > Define Loads > Apply > Structural > Pressure > On Beams Pick the 1st element > OK > Enter Pressure values at node I = 5000 > OK

Figure 5.11. Applying loads on element 1.

8.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Structural > Pressure > On Beams Pick the 2nd element > OK > Enter Pressure value at node I = 5000 > OK

Figure 5.12. Model with loading and displacement boundary conditions.

The model-building step is now complete, and we can proceed to the solution. First, to be safe, save the model.

Solution

The interactive solution proceeds.

9.  Main Menu > Solution > Solve > Current LS > OK

The /STATUS Command window displays the problem parameters and the Solve Current Load Step window is shown. Check the solution options in the /STATUS window and if all is OK, select File > Close.

In the Solve Current Load Step window, select OK, and when the solution is complete, close the ‘Solution is Done!’ window.

POST-PROCESSING

We can now plot the results of this analysis and also list the computed values.

10.  Main Menu > General Postproc > Plot Results > Contour Plot > Nodal Solu > DOF Solution > Displacement vector sum > OK

This result is shown in figure 5.3 (f).

11.  Main Menu > General Postproc > List Results > Nodal Solution > Select Rolation vector sum > OK

12.  Main Menu > General Postproc > List Results > Reaction Solu > OK To find the bending moment diagram, the following procedure is followed.

13.  Main Menu > General Postproc > Element Table > Define Table > Add as shown in Figure 5.13.

Figure 5.13. Define the element table.

Select By sequence num and SMISC and type 6 after SMISC (as shown in Figure 5.14) > APPLY

Figure 5.14. Selecting options in element table.

Then again select By sequence num and SMISC and type 12 after SMISC > OK

14.  Main Menu > General Postproc > Plot Results > Contour Plot > Line Elem Res > Select SMIS 6 and SMIS 12 in the rows of LabI and LabJ respectively as shown in Figure 5.15 > OK

Figure 5.15. Selecting options for finding out bending moment.

This result is shown in Figure 5.3(g).

To find the shear force diagram the following procedure is followed.

15.  Main Menu > General Postproc > Element Table > Define Table > Add

Select By sequence num and SMISC and type 2 after SMISC > APPLY

Then again select By sequence num and SMISC and type 8 after SMISC > OK

16.  Main Menu > General Postproc > Plot Results > Contour Plot > Lone Elem Res > Select SMIS 2 and SMIS 8 > OK

This result is shown in Figure 5.3(h).

To find the bending stress, the following procedure is followed.

17.  Main Menu > General Postproc > Element Table > Define Table > Add Select By sequence num and LS and type 3 after LS > APPLY Then again select By sequence num and LS and type 6 after LS > OK

18.  Main Menu > General Postproc > Plot Results > Contour Plot > Line Elem Res > Select LS 3 and LS 6 > OK This result is shown is Figure 5.3(i).

5.3 CANTILEVER BEAMS

Example 5.5

Beam subjected to concentrated load. For the beam shown in Figure 5.16, determine the deflections and reactions. Let E = 210 GPa and I = 2 × 10−4 m^4. Take 2 elements.

Figure 5.16. Beam subjected to concentrated load for Example 5.5.

Solution

(I) Analytical method [refer to Figure 5.16(a)]

Figure 5.16(a). Analytical method for Example 5.5.

The solution is obtained by Macaulay’s method. The number within the brackets <> is to be neglected whenever it is less than zero.

At section a-a

Similarly,

(II) FEM by hand calculations

Figure 5.16(b). Finite element model.

Element stiffness matrix for element 1 is,

Element stiffness matrix for element 2 is,

Global stiffness matrix is,

The global equations are,

By using the elimination method for applying boundary conditions,

w_1 = θ_1 = 0.

The above matrix reduces to,

By solving the above matrix and equations, we get,

Deflections and slopes as

Reaction calculation

(III) Software results

Figure 5.16(c). Deflection pattern for a cantilever beam (refer to Appendix C for color figures).

Deflection values at nodes (in meters)

The following degree of freedom results are in global coordinates

Maximum absolute values

Rotational deflection values at nodes

The following degree of freedom results are in global coordinates

NODE

ROTZ

1

0.0000

2

−0.23810E-03

3

−0.23810E-03

Reaction values

The following X, Y, Z solutions are in global coordinates

Answers for Example 5.5

Example 5.6

Propped cantilever beam with distributed load. Find nodal displacements and support reactions for the beam shown in Figure 5.17. Let E = 70 GPa and I = 6 × 10−4 m^4.

Figure 5.17. propped cantilever beam with distributed load for Example 5.6.

Solution

(I) Analytical method [refer to Figure 5.17(a)]

Figure 5.17(a). Analytical method for Example 5.6.

The solution is obtained by Macaulay’s method. The number within the brackets <> is to be neglected whenever it is less than zero.

At section a-a

Boundary conditions are,

Substituting in equations (5.9) and (5.10)

and

yA = y_1 = 0 and yB = y_2 = 0 (Given boundary conditions).

(II) FEM by hand calculations

Figure 5.17(b). Finite element model for Example 5.6.

E = 70 × 10^3 N/mm^2 and I = 6 ×10^8 mm^4

Stiffness matrix for element 1 is,

Due to symmetry,

Nodal force calculation

For element 2,

Figure 5.17(c). Nodal force calculation for element 2 in Example 5.6.

The nodal forces and moments for element 2 is,

The global equations are,

By using the elimination method for applying boundary conditions,

w_1 = θ_1 = w_2 = 0.

The above matrix reduces to

Solving the above matrix and equations, we get,

Reaction calculation

(III) Software results

Figure 5.17(d). Deflection pattern for a cantilever beam (refer to Appendix C for color figures).

Deflection values at nodes (in meters)

The following degree of freedom results are in global coordinates

Rotational deflection values at nodes

The following degree of freedom results are in global coordinates

NODE

ROTZ

1

0.0000

2

−0.15238E-02

3

−0.35556E-02

Reaction values

The following X, Y, Z solutions are in global coordinates

Answer for Example 5.6

Example 5.7

Propped cantilever beam with varying load. For the beam shown in Figure 5.18, determine the nodal displacements, slopes, reactions, maximum bending moment, shear force, and maximum bending stress. Take E = 200 GPa.

Figure 5.18. Propped cantilever beam with varying load for Example 5.7.

Solution

(I) FEM hand calculations

Figure 5.18(a). Finite element model for Example 5.7.

Stiffness matrices for element 1 and 2 are,

Due to symmetry,

Nodal force calculation

For element 1,

Figure 5.18(b). Nodal force calculation for element 1 for Example 5.7.

The nodal forces and moments for element 1 is,

For element 2,

Figure 5.18(c). Nodal force calculation for element 2 for Example 5.7.

The nodal forces and moments for element 2 is,

The combined nodal forces and moments matrix is,

The global equations are,

By using the elimination method for applying boundary conditions, w_1 = θ_1 = w_2 = 0. The above matrix reduces to,

By solving the above matrix and equations, we get

θ_2 = −0.0128 rad, w_3 = −0.0811 m, and θ = −0.0319 rad.

Reaction calculation

(II) Software results

Figure 5.18(d). Deflection pattern for a cantilever beam (refer to Appendix C for color figures).

Deflection values at nodes (in meters)

The following degree of freedom results are in global coordinates

Maximum absolute values

Rotational deflection values at nodes

The following degree of freedom results are in global coordinates

NODE

ROTZ

1

0.0000

2

−0.12834E-01

3

−0.31948E-01

Reaction values

The following X, Y, Z solutions are in global coordinates

Total values

Figure 5.18(e). Bending moment diagram for a propped cantilever beam (refer to Appendix C for color figures).

Figure 5.18(f). Shear force diagram for a propped cantilever beam (refer to Appendix C for color figures).

Figure 5.18(g). Bending stress diagram for a propped cantilever beam (refer to Appendix C for color figures).

Figure 5.18(h). Maximum stress diagram for a propped cantilever beam (refer to Appendix C for color figures).

Answers for Example 5.7

Parameter

FEM-hand calculations

Software results

Deflection at node 3

−0.0811 m

−0.081098 m

Rotational deflection at node

2

−0.0128 rad

−0.012834 rad

3

−0.0319 rad

−0.031948 rad

Reaction force at

1

−91.97 kN

−92.25 kN

2

272.65 kN

272.25 kN

Reaction moment at node 1

−96.47 kN-m

−96.75 kN-m

Maximum bending moment

.. …

96750 N-m

Shear force

.. …

137250 N

Maximum bending stress

.. …

129 MPa

Maximum stress (bending stress + direct stress)

.. …

300 MPa

Example 5.8

Propped cantilever beam with stepped loading. Analyze the beam in Figure 5.19 by finite element method and determine the reactions. Also, determine the deflections.

Given         E = 200 GPa and I = 5 × 10−4 m^4.

Figure 5.19. Propped cantilever beam with stepped loading for Example 5.8.

Solution

(I) FEM by hand calculations

Figure 5.19(a). Finite element model for Example 5.8.

Stiffness matrix for element 1 and 2 are,

Nodal force calculation

For element 1,

Figure 5.19(b). Nodal force calculation for element 1 in Example 5.8.

The nodal forces and moments for element 1 is,

For element 2,

Figure 5.19(c). Nodal force calculation for element 2 in Example 5.8.

The nodal forces and moments for element 2 is,

The combined nodal forces and moments is,

The global equations are,

By using the elimination method for applying boundary conditions,

w_1 = θ_1 = w_2 = 0,

the above matrix reduces to

By solving the above matrix and equations, we get

Reaction calculation

(II) Software results

Figure 5.19(d). Deflection pattern for a cantilever beam (refer to Appendix C for color figures).

Deflection values at nodes (in meters)

The following degree of freedom results are in global coordinates

Maximum absolute values

Rotational deflection values at nodes

The following degree of freedom results are in global coordinates

NODE

ROTZ

1

0.0000

2

−0.34375E-02

3

−0.84375E-02

Reaction values

The following X, Y, Z solutions are in global coordinates

Answers for Example 5.8

parameter

FEM-hand calculations

Software results

Deflection at node 3

−0.035938 m

−0.035938 m

Rotational deflection at node

2

−0.003438 rad

−0.0034375 rad

3

−0.003438 rad

−0.0034375 rad

Reaction force at

1

−52.512 kN

−52.5 kN

2

232.493 kN

232.25 kN

Reaction moment at node 1

−112.52 kN-m

−112.5 kN-m

Procedure for solving the problems using ANSYS® 11.0 academic teaching software.

For Example 5.7

PREPROCESSING

1.  Main Menu > Preprocessor > Element Type > Add/Edit/Delete > Add > Beam > 2D elastic 3 > OK > Close

Figure 5.20. Element selection.

2.  Main Menu > Preprocessor > Sections > Beam > Common sections, following dialog box appears

Figure 5.21. Choose cross-section of the beam.

In that dialog box, select Sub-Type, choose Square Cross-Section, then Enter value of B = 0.165 and H = 0.165 as shown in Figure 5.21.

Click on Preview > OK

The following figure appears on the screen.

Figure 5.22. Details of geometrical properties of the beam.

From Figure 5.22, note down the values of Area A = 0.027225 m^2 and moment of inertia Izz = 0.681 × 10−4 m^4.

3.  Main Menu > Preprocessor > Real Constants > Add/Edit/Delete > Add > OK

Cross-sectional area AREA > Enter 0.027225

Area moment of inertia IZZ > Enter 0.618e-4

Total beam height HEIGHT > Enter 0.165 > OK > Close

Enter the material properties.

4.  Main Menu > Preprocessor > Material Props > Material Models

Material Model Number 1, click Structural > Linear > Elastic > Isotropic

Enter EX = 200E9 and RRXY = 0.3 > OK

(Close the Define Material Model Behavior window.)

Create the nodes and elements as shown in the figure.

5.  Main Menu > Preprocessor > Modeling > Create > Nodes > In Active CS Enter the coordinated of node 1 > Apply Enter the coordinates of node 2 > Apply Enter the coordinate of node 3 > OK.

Node locations

Node number

X-coordinate

Y-coordinate

1

0

0

2

3

0

3

6

0

Figure 5.24. Enter the node coordinate.

6.  Main Menu > Preprocessor > Modeling > Create > Elements > Auto Numbered > Thru nodes Pick the 1st and 2nd node > Apply Pick the 2nd and 3rd node > OK

Figure 5.25. Pick the nodes to create elements.

Apply the displacement boundary conditions and loads.

7.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Structural > Displacement > On Nodes Pick the 1st node > Apply > All DOF = 0 > OK

8.  Main Menu > Preprocessor > loads > Define Loads > Apply > Structural > Displacement > On Nodes Pick the 2nd node > Apply > Select UX and UY = 0 > OK

Figure 5.26. Applying boundary conditions on node 2.

9.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Structural > Pressure > On Beams Pick the 1st element > OK > Enter Pressure value at node I = 0 and Pressure value at node J = 30e3 > OK

Figure 5.27. Applying loads on element 1.

10.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Structural > Pressure > On Beams Pick the 2nd element > OK > Enter Pressure value at node I = 30e3 and Pressure value at node J = 60e3 > OK

Figure 5.28. Model with loading and displacement boundary conditions.

The model-building step is now complete, and we can proceed to the solution. First to be safe, save the model.

Solution

The interactive solution proceeds.

11.  Main Menu > Solution > Solve > Current LS > OK

The /STATUS Command window displays the problem parameters and the Solve Control Load Step window is shown. Click the solution options in the /STATUS window and if all is OK, select File > Close.

In the Solve Current Load Step window, select OK, and when the solution is complete, close the “Solution is Done!’ window.

POST-PROCESSING

We can now plot the results of this analysis and also list the computed values.

12.  Main Menu > General Postproc > Plot Results > Contour Plot > Nodal Solu > DOF Solution > Displacement vector sum > OK

This result is shown in Figure 5.18(d).

13.  Main Menu > General Postproc > List Results > Nodal Solu > Select Roatation vector sum > OK

14.  Main Menu > General Postproc > List Results > Reaction Solu > PL

To find the bending moment diagram following procedure is followed.

15.  Main Menu > General Postproc > Element Table > Define Table > Add

Figure 5.29. Define the element table.

Select By sequence num and SMISC and type 6 after SMISC (as shown in Figure 5.30) > APPLY

Figure 5.30. Selecting options in element table.

Then again select By sequence num and SMISC and type 12 after SMISC > OK

16.  Main Menu > General Postproc > Plot Results > Contour Plot > Line Elem Res > Select SMIS 6 and SMIS 12 in the rows of LabI and LabJ, respectively as shown in Figure 5.31 > OK

Figure 5.31. Selecting options for finding out bending moment.

This result is shown in Figure 5.18(e).

To find the shear force diagram following procedure is followed.

17.  Main Menu > General Postproc > Element Table > Define Table > Add Select By sequence num and SMISC and type 2 after SMISC > APPLY Then again select By sequence num and SMISC and type 8 after SMISC > OK > Close

18.  Main Menu > General Postproc > Plot Results > Contour Plot > Line Elem

Res > Select SMIS 2 and SMIS 8 > OK

This result is shown in Figure 5.18(f).

To find the bending stress the following procedure is followed.

19.  Main Menu > General Postproc > Element Table > Define Table > Add Select By sequence num and LS and type 3 after LS > APPLY

Then again select By sequence num and LS and type 6 after LS > OK

20.  Main Menu > General Postproc > Plot Results > Contour Plot > Line Elem Res > Select LS 3 and LS 6 > OK

This result is shown in Figure 5.18(g).

To find the maximum stress (direct stress + bending stress) following procedure is followed.

21.  Main Menu > General Postproc > Element Table > Define Table > Add Select By sequence num and NMISC and type 1 after NMISC > APPLY Then again select By sequence num and NMISC and type 3 after NMISC > OK

22.  Main Menu > General Postproc > Plot Results > Contour Plot > Line Elem Res > Select NMISC 1 and NIMS 3 > OK

PROBLEMS

1.  For the bean shown in Figure 5.32, determine the deflection, slopes, reactions, maximum bending moment, shear force, and maximum bending stress. Take E = 210 GPa and I = 7 × 10−4 m^4.

Figure 5.32. problem 1

2.  Find the deflection, slopes, reactions, maximum bending moment, shear force, and maximum bending stress for the aluminum beam shown in Figure 5.33. Take E = 200 GPa and I = 3 × 10−4 m^4.

Figure 5.33. Problem 2

3.  Find the deflection at the load and the slopes at the end for the shaft shown in Figure 5.34. Also find the maximum bending moment, maximum bending stress, and reactions developed in the bearings. Consider the shaft to be simply supported at bearings A and B. Take E = 200 GPa.

Figure 5.34. Problem 3

4.  Find the deflection of the bean shown in Figure 5.35 under self-weight. Take E = 200 GPa and mass density ρ = 7800 kg/m^3.

Figure 5.35. Problem 4

5.  Find the deflection and bending stress distribution for the cantilever beam shown in Figure 5.36 under combined loading. Take E = 200 GPa.

Figure 5.36. Problem 5

6.   For the beam shown in Figure 5.37, determine the deflection at nodes and reaction. Also, plot the bending moment diagram, shear force diagram and find the bending stress. Take E = 200 GPa and I = 8 × 10−4 m^4.

Figure 5.37. Problem 6

7.  A cantilever beam is shown in Figure 5.38. Using 2 beam elements determine the nodal deflection and reaction. Take E = 0.25 × 10^5 N/mm^2 and I = 8 × 10−4 m^4.

Figure 5.38. problem 7

[Pmax = ρ×g×h]

8.  Determine the deflection, reaction, and bending stress for the beam shown in Figure 5.39. Also, plot the bending moment and shear force diagram. Take E = 207 GPa, W = 150 N/mm, h = 800 mm, b = 400 mm, t_1 = 40 mm, t_2 = 40 mm, and t_3 = 50 mm.

Figure 5.39. Problem 8

9.  Figure 5.40 presents a beam fixed at one end, supported by a cable at the other end, subjected to a uniformly distributed load of 70 lb/in. TakeE = 30 × 10^6 psi, Beam cross-section = 4 in × 4 in, and cable cross-section = 1 in^2. Determine the finite element equilibrium equations of the system by using one finite element for the beam and one finite element for the cable, the displacement of nodes 1 and 2, and the stress distribution in the beam and in the cable.

Figure 5.40. Problem 9

REFERENCES

1.  Y. W. Hwon and H. Bang, “The Finite Element Method Using MATLAB, Second Edition,” CRC Press, 2000.

2.  D. L. Logan, “A First Course in the Finite Element Method, Fifth Edition,” Cengage Learning, 2012.

3.  S. Moaveni, “Finite Element Analysis: Theory and Application with ANSYS, Third Edition,” Prentice Hall, 2008.

4.  J. N. Reddy, “An Introduction to the Finite Element Method, Third Edition,” McGraw Hill Higher Education, 2004.

5.  C. T. F. Ross, “Finite Element Method in Structural Mechanics,” Ellis Horwood Limited Publishers, 1985.

6.  F. L. Stasa, “Applied Finite Element Analysis for Engineering” Holt, Rinehart and Winston, 1985.

7.  L. J. Segerlind, “Applied Finite Element Analysis, Second Edition,” John Wiley and Sons, 1984.

8.  S. S. Rao, “The Finite Element Method in Engineering, Fifth Edition,” Butterworth-Heinemann, 2011.

9.  N. Willems and W. M. Lucas, Jr., “Structural Analysis for Engineering,” McGraw-Hill, 1978.

10.  R. C. Hibbleer, “Mechanics of Materials, Second Edition,” Macmillan, 1994.

Chapter 6

STRESS ANALYSIS OF A RECTANGULAR PLATE WITH A CIRCULAR HOLE

6.1 INTRODUCTION

Two dimensional problems in structural analysis are dealt with in this chapter. Hand calculations, even with 2 elements, become too long and hence are not given for these problems: only analytical method solutions and software solutions using ANSYS have been provided.

Two dimensional problems can either be plane stress or plane strain problems. Method of analysis is the same for both, except that stress strain matrix is different in 2 cases.

Plane bodies that are flat and of constant thickness that are subjected to in-plane loading fall under the category of plane stress problems. Stress components σz, τxz, and τyz assume zero values in these problems.

Some of the elements used in the analysis of 2 dimensional problems are constant strain triangles (CST), linear strain triangle (LST), linear quadrilateral, isoparametric quadrilateral, etc. Each of these elements has 2 degrees of freedom per node namely the translation in x and y directions.

Stress within the element may be calculated using the equation,

6.2 A RECTANGULAR PLATE WITH A CIRCULAR HOLE

The stress analysis of a rectangular plate with a circular hole problem is assumed as a 2 dimensional plane stress problem. Plane stress is defined as a state of stress in which the normal stress and the shear stress directed perpendicular to the plane are assumed to be negligible.

The above problem can be categorized into 3 sub cases.

Sub Case 1

A rectangular plate with a very small circular hole at the center with one vertical edge fixed and the other vertical edge is acted upon by a horizontal tensile load in the form of pressure.

Figure 6.1. Rectangular plate with a very small circular hole subjected to tensile load at one edge.

Sub Case 2

A rectangular plate with a small circular hole at the center and a horizontal tensile force in the form of pressure is acting on both the vertical edges of the plate.

Figure 6.2. Rectangular plate with a hole subjected to tensile load at both the edges.

The above problem is solved by exploiting the symmetric geometry and symmetric loading boundary conditions. Now we can draw the above Figure 6.2 as below for the analysis purpose (refer Figure 6.3).

Place the origin of x-y coordinates at the center of the hole and pull on both ends of the plate. Then points on the centerlines will not move perpendicular to them but move along the centerlines. This indicated the appropriate displacement conditions to use as shown in Figure 6.3.

Figure 6.3. Finite element model of one-quarter of the plate.

Sub Case 3

A rectangular plate with a large circular hole at the center and a uniform pressure acts on the boundary of the hole.

Figure 6.4. Rectangular plate with a hole subjected to uniform pressure at the boundary of the hole.

The above problem can be solved considering one quarter of the plate and by exploiting the symmetric geometry and loading conditions. The finite element model is shown below.

Figure 6.5. Finite element model of one-quarter of the plate.

Example 6.1

A rectangular plate of size 1000 mm × 500 mm is subjected to uniform pressure as shown in Figure 6.6. The plate has a thickness of 10 mm and has a central hole 50 mm in diameter. The material of the plate is steel with Young’s modulus E = 210 GPa and Poisson’s ratio, v = 0.3. Assume a case of plane stress. Plot the Von Mises stress distribution and compare result with analytical method.

Figure 6.6. Rectangular plate with very small circular hole at the center of the plate.

Solution

(I) Analytical method

Comparing the above case with the infinite plate with a very small circular hole, for this, the stress concentration factor is (SCF) = 3.

Hence,

Maximum stress = SCF × Nominal stress = 3 × 1 = 3 Pa.

(II) Software results

Figure 6.6(a). Von Mises stress distribution pattern (refer to Appendix C for color figures).

From the software, we got, Maximum stress (Von Mises stress) = 3.172 Pa.

Answers for Example 6.1

Example 6.2

A rectangular plate with hole at the center is subjected to uniform pressure as shown in Figure 6.7. The plate is under plane stress. Find the maximum deflection and maximum stress distribution. Also find the deformed shape of the hole. Assume plate thickness, t = 25 mm, E = 207 GPa, and v = 0.3.

Figure 6.7. Rectangular plate with a hole with symmetrical loading.

Solution

(I) Analytical method

From the design data handbook,

for of 0.4 the stress concentration factor (SCF) = 2.25

Hence,

Tensile force = Pressure × Cross-sectional area = 14 × 100 × 25 = 35000 N

Maximum stress = SCF × Nominal stress = 2.25 × 23.333 = 52.5 MPa.

(II) Software results

For the analysis using software, one quarter of the plate is modeled and analyzed.

Figure 6.7(a). Deformed shape of the hole (refer to Appendix C for color figures).

Figure 6.7(b). Von Mises stress distribution pattern (refer to Appendix C for color figures).

From the software, we got, maximum stress = 53.3 MPa.

Figure 6.7(c). Deflection pattern (refer to Appendix C for color figures).

Answers for Example 6.2

Example 6.3

Determine the stress distribution and displacement for a rectangular plate with a hole at the center of the plate with uniform thickness of 10 mm. A uniform pressure of p = 10 MPa acts on the boundary of the hole as shown in Figure 6.8. Assume Young’s modulus E = 120 GPa and the Poisson’s ratio is 0.28. Assume plane stress condition.

Figure 6.8. Rectangular plate with a hole subjected to uniform pressure at the boundary of the hole.

Solution

(I) Software results

For the analysis using software, one quarter of the plate is modeled and analyzed.

Figure 6.8(a). First principal stress distribution pattern (refer to Appendix C for color figures).

From the software, we got maximum stress = 50.727 MPa.

Figure 6.8(b). Deflection pattern (refer to Appendix C for color figures).

Validation of the Results

The reactions at the supports must balance the applied forces. Therefore, from the software, the total reaction force in the x-direction is − 7500 N.

Applied force = 10 × 75 × 10 = 7500 N in positive x-direction.

So the reaction cancels out the applied force in the x-direction.

Answers for Example 6.3

Parameter

Software results

Maximum stress

50.727 MPa

Maximum deflection

0.036478 mm

Procedure for solving the problem using ANSYS® 11.0 academic teaching software.

For Example 6.2

PREPROCESSING

1.  Main Menu > Preprocessor > Element Type > Add/Edit/Delete > Add > Structural Solid > Quad 4 node 42 > OK

Figure 6.9. Element selection.

Select the option where you define the plate thickness.

2.  Options (Element behavior K3) > Plane strs w/thk > OK > Close

Figure 6.10. Element options.

3.  Main Menu > Preprocessor > Real Constants > Add/Edit/Delete > Add > OK

Figure 6.11. Real constants.

(Enter the plate thickness of 0.025 m) > Enter 0.025 > OK > Close

Figure 6.12. Enter the plate thickness.

Enter the material properties.

4.  Main Menu > Preprocessor > Material Props > Material Models Material Model Number 1, click Structural > Linear > Elastic > Isotropic Enter EX = 2.07E11 and PRXY = 0.3 > OK (Close the Define Material Model Behavior window.)

Create the geometry for the upper-right quadrant of the plate by subtracting a 0.04 m diameter circle from a 0.075 × 0.05 m rectangle. Generate the rectangle first.

5.  Main Menu > Preprocessor > Modeling > Create > Areas > Rectangle > By 2 Corners

Enter (lower left corner) WP X = 0.0, WP Y = 0.0 and Width = 0.075, Height = 0.05 > OK

6.  Main Menu > Preprocessor > Modeling > Create > Areas > Circle > Solid Circle Enter WP X = 0.0, WP Y = 0.0 and Radius = 0.02 > OK

Figure 6.13.Create areas.

Figure 6.14. Rectangle and circle.

Now subtract the circle form the rectangle. (Read the messages in the window at the bottom of the screen as necessary.)

7.  Main Menu > Preprocessor > Modeling > Operate > Booleans > Subtract > Areas

Pick the rectangle > OK, then pick the circle > OK

Figure 6.15. Geometry for quadrant of plate.

Create a mesh triangular element over the quadrant area.

8.  Main Menu > Preprocessor > Meshing > Mesh Tool

The Mesh Tool dialog box appears. In that dialog box, click on the Smart Size and move the slider available below the Smart Size to 2 (i.e., towards Fine side).

Then close the Mesh Tool box.

Figure 6.16. Mesh tool box.

9.  Main Menu > Preprocessor > Meshing > Mesh > Areas > Free Pick the quadrant > OK

Figure 6.17. Quad element mesh.

Apply the displacement boundary conditions and loads.

10.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Structural > Displacement > On Lines Pick the left edge of the quadrant > OK > UX = 0 > OK

11.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Structural > Displacement > On Lines pick the bottom edge of the quadrant > OK > UY = 0 > OK

12.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Structural > Displacement > On Lines. Pick the right edge of the quadrant > OK > Pressure = –14E6 > OK

(A positive pressure would be a compressive load, so we use a negative pressure. The pressure is shown as a single arrow.)

Figure 6.18. Model with loading and displacement boundary conditions.

The model-building step is now complete, and we can proceed to the solution. First to be safe, save the model.

Solution

The interactive solution proceeds

13.  Main Menu > Solution > Solve > Current LS > OK

The /STATUS Command window displays the problem parameters and the Solve Current Load Step window is shown. Check the solution options in the /STATUS window and if all is OK, select File > Close.

In the Solve Current Load Step window, select OK, and when the solution is complete close the ‘Solution is Done!’ window.

POST-PROCESSING

We can now plot the results of this analysis and also list the computed values. First examine the deformed shape.

14.  Main Menu > General Posrproc > Plot Results > Deformed Shape > Def. + Undeformed > OK

This result is shown in Figure 6.7(a).

15.  Main Menu > General Posrproc > Plot Results > Contour Plot > Nodal Solu > Stress > Von Mises stress > OK

This result is shown in Figure 6.7(b).

16.  Main Menu > General Posrproc > Plot Results > Contour Plot > Nodal Solu > DOF Solution > Displacement vector sum > OK

This result is shown in Figure 6.7(c).

PROBLEMS

1.  Find the maximum stress in the aluminum plate shown in Figure 6.19. Consider an aluminum plate 10 mm thick with a hole at the center of the plate. Assume plane stress condition. Take E = 70 GPa and v = 0.35. Also, calculate the maximum stress by analytical method and compare the results.

Figure 6.19. Problem 1

2.  Find the maximum stress for the plate shown in Figure 6.20 if the hole is located halfway between the center line and the top edge as shown. Take E = 70 GPa and v = 0.35. Assume plane stress condition.

Figure 6.20. problem 2

[Model half of the plate by taking symmetry about y-axis.]

3.  For the plate shown in Figure 6.21, find the maximum stress. Take Young’s modulus E = 210 GPa, Poisson’s ratio = 0.3. Assume plane stress condition. Thickness of the plate = 10 mm with hole at the center of the plate.

Figure 6.21. problem 3

4.  For the plate shown in Figure 6.22, find the maximum stress. Plate is made up of two materials.

For Material 1, E = 210 GPa and v = 0.3.

For Material 2, E = 70 GPa and v = 0.35.

Assume plane stress condition.

Thickness of the plate = 10 mm with a hole at the center of the plate.

Figure 6.22. problem 4

5.  For the plate with a hole at the center shown in Figure 6.23, find the maximum stress. Take E = 210 GPa and v = 0.3, thickness of plate t = 10 mm. Assume plane stress condition.

Figure 6.23. problem 5

[To find the pressure, divide distributed load by thickness of plate.]

6.  Determine the stresses in the plate with the round hole subjected to the tensile stresses in Figure 6.24. Find the maximum stress. Take E = 210 GPa and v = 0.25, thickness of plate t = 10 mm. Assume plane stress condition.

Figure 6.24. Problem 6

7.  For the plate with a hole at the center shown in Figure 6.25, find the maximum stress. Take E = 210 GPa and v = 0.3, thickness of plate t = 0.375 in. Assume plane stress condition.

Figure 6.25. problem 7

8.  For the plate with a hole at the center shown in Figure 6.26, find the maximum stress. Take E = 30 × 10^6 psi and v = 0.25, thickness of plate t = 0.1 in. Assume plane stress condition.

Figure 6.26. problem 8

9.  Find the maximum stress for the plate shown in Figure 6.27 if the hole is located halfway between the center line and the top edge as shown. Take E = 20 × 10^6 N/cm^2 and v = 0.25. Assume plane stress condition.

Figure 6.27. problem 9

[Model half of the plate by taking symmetry about y-axis.]

REFERENCES

1.  C. S. Desai and J. F. Abel, “Introduction to the Finite Element Method,” Van Nostrad Reinhold, 1972.

3.  M. Asghar Bhatti, “Fundamental Finite Element Analysis and Applications with Mathematica and MATLAB Computations,” John Wiley and Sons, 2005.

4.  H. Grandin, Jr., “Fundamentals of the Finite Element Method,” Macmillan Publishing Company, 1986.

5.  D. L. Logan, “A First Course in the Finite Element Method, Fifth Edition,” Cengage Learning, 2012.

6.  F. L. Stasa, “Applied Finite Element Analysis for Engineering,” Holt, Rinehart, and Winston, 1985.

7.  N. Troyani, C. Gomes, and G. Sterlacci, “Theoretical Stress Concentration Factors for Short Rectangular Plates with Central Circular Holes,” Journal of Mechanical Design, ASME, Vol. 124, pp. 126–128, 2002.

8.  T. Hayashi, “Stress Analysis of a Rectangular Plate with a Circular Hole under Uniaxial Loading,” Journal of Thermoplastic Composite Materials, Vol. 2, 1989.

Chapter 7

THERMAL ANALYSIS

7.1 INTRODUCTION

The computation of temperature distribution within a body will be used in this chapter due to its importance in many engineering applications. Conduction (q) is the transfer of heat through materials without any net motion of the mass of the material. The rate of heat flow in x-direction by conduction (q) is given by

where

k is the thermal conductivity of the material, A is the area normal to x-direction through which heat flows, T is the temperature, and x is the length parameter.

Convection is the process by which thermal energy is transferred between a solid and a fluid surrounding it. The rate of heat flow by convection (q) is given by

where

h is the heat transfer coefficient, A is the surface area of the body through which heat flows, T is the temperature of the surface of the body, and T∞ is the temperature of the surrounding medium.

Thermal analysis is one of the scalar field problems. These problems have only 1 degree of freedom per node namely temperature. In this chapter, one-dimensional and two-dimensional heat conduction problems are dealt with. In these problems, a bar element with 2 end nodes each having temperature (T) as sole degree of freedom is useful. Nodal heat flow rates (Q) or heat fluxes are analogous quantities to nodal forces, in structural bar element.

The governing equation for this element is given by,

where,

q = heat generation rate per unit length

7.2 PROCEDURE OF FINITE ELEMENT ANALYSIS (RELATED TO THERMAL PROBLEMS)

Step 1. Select element type.

Step 2. Select temperature distribution function.

Step 3. Define the temperature gradient/temperature and heat flux/temperature gradient relationships.

Step 4. Derive the element conduction matrix and heat flux matrix.

Step 5. Assemble the element equations to obtain the global equations and introduce boundary conditions.

Step 6. Solve for the nodal temperatures.

Step 7. Solve for the element temperature gradients and heat fluxes.

7.3 ONE-DIMENSIONAL HEAT CONDUCTION

Example 7.1

A composite wall consists of 3 materials. The outer temperature is T_0 = 20°C. Convection heat transfer takes place on the inner surface of the wall with T∞ = 800°C and h = 25 W/m^2°C. Determine the temperature distribution in the wall. Take k_1 = 30 W/m°C, k_2 = 50 W/m°C, k_3 = 20 W/m°C.

Figure 7.1. A composite wall consists of 3 materials for Example 7.1.

Solution

(I) Analytical method

Figure 7.1(a). Analytical method for Example 7.1.

Heat flow rate per unit area,

Now,

By solving the above, we get

T_1 = 284.3°C

T_2 = 155.37°C

T_3 = 116.7°C.

(II) FEM by calculations [refer to Figure 7.1(b)]

Figure 7.1(b). Finite element model for Example 7.1.

Governing equation is,

Since there is no heat generation specified, q = 0. For element 1,

For element 2,

For element 3,

Global equation after assembly,

Boundary conditions are T_4 = 20°C and

Q_1 = −h(T∞ − T_1)

Q_1 = 25(800 −T_1) = 20000−25T_1.

So modified equation,

After solving the matrix and simultaneous equations, we get,

T_1 = 284.3°C

T_2 = 155.37°C

T_3 = 116.7°C.

(III) Software results

Temperature values

NODE

TEMP

1

284.30

2

155.37

3

116.69

4

20.000

5

800.00

Figure 7.1(c). Temperature distribution in a composite wall (refer to Appendix C for color figures).

Answers for Example 7.1

Procedure for solving the problem using ANSYS® 11.0 academic teaching software.

For Example 7.1

PREPROCESSING

1.  Main Menu > Preferences, then select Thermal > OK

Figure 7.2. Selecting the preferences.

2.  Main Menu > Preprocessor > Element Type > Add/Edit/Delete > Add > Click on Link > then on 2d conduction 32 > OK > Add > Click on Link > then on 3D convection 34 > OK > Close

Figure 7.3. Selecting the element for conduction.

Figure 7.4. Selecting the element for convection.

3.  Main Menu > Preprocessor > Real Constants > Add/Edit/Delete > Add > Click on Link 32 > OK

Figure 7.5. Enter the cross-sectional area for Link 32.

Enter cross-sectional area AREA > Enter 1 > OK Add > Click on Link 34 > OK

Enter cross-sectional area AREA > Enter 1 > OK > Close

Figure 7.6. Enter the cross-sectional area for Link 34.

Enter the material properties.

4.  Main Menu > Preprocessor > Material Props > Material Models

Material Model Number 1,

click Thermal > Conductivity > Isotropic

Enter KXX = 30 > OK

Then in the material model window, click on Material menu > New Model > OK

Material Model Number 2,

click Thermal > Conductivity > Isotropic

Enter KXX = 50 > OK

Then in the material model window, click on Material menu > New Model > OK

Material Model Number 3,

click Thermal > Conductivity > Isotropic

Enter KXX = 20 > OK

Then in the material model window, click on Material menu > New Model > OK

Material Model Number 4,

click Thermal > Convection or Film Coef.

Enter HF = 25 > OK

(Close the Define Material Model Behavior window.)

Create the nodes and elements.

5. Main Menu > Preprocessor > Modeling > Create > Nodes > In Active CS Enter the coordinates of node 1 > Apply Enter the coordinates of node 2 > Apply

Enter the coordinates of node 3 > Apply Enter the coordinates of node 4 > Apply Enter the coordinates of node 5 > OK

Node Locations

Node number

X coordinates

Y coordinates

1

0

0

2

0.3

0

3

0.45

0

4

0.6

0

5

−0.1

0

Figure 7.7. Enter the node coordinates.

6.  Main Menu > Preprocessor > Modeling > Create > Elements > Elem Attributes > OK > Auto Numbered > Thru nodes Pick the 1st and 2nd node > OK

Figure 7.8. Assigning element attributes to element 1 and creating element 1.

Elem Attributes > change the material number to 2 > OK > Auto Numbered > Thru nodes Pick the 2nd and 3rd node > OK

Figure 7.9. Assigning element attributes to element 2 and creating element 2.

Elem Attributes > change the material number to 3 > OK > Auto Numbered > Thru nodes Pick the 3rd and 4th node > OK

Figure 7.10. Assigning element attributes to element 3 and creating element 3.

Elem Attributed > change the element type to Link 34 > change the material number to 4 > change the Real constant set number to 2 > OK > Auto Numbered > Thru nodes Pick the 1st and 5th node > OK

Figure 7.11. Assigning element attributes to element 4 and creating element 4.

Apply the boundary conditions and temperature.

7.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Thermal > Temperature > On Nodes Pick the 4th node > Apply > Click on TEMP and Enter Value = 20 > OK

Figure 7.12. Applying temperature on node 4.

8.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Thermal > Temperature > On Nodes Pick the 5th node > Apply > Click on TEMP and Enter Value = 800 > OK

Figure 7.13. Applying temperature on node 5.

Solution

The interactive solution proceeds.

9.  Main Menu > Solution > Solve > Current LS > OK

The /STATUS Command window displays the problem parameters and the Solve Current Load Step window is shown. Check the solution options in the /STATUS window and if all is OK, select File > Close.

In the Solve Current Load Step window, select OK, and when the solution is complete, close the ‘Solution is Done!’ window.

POST-PROCESSING

We can now plot the results of this analysis and also list the computed values.

10.   Main Menu > General Posrproc > Plot Results > Contour Plot > Nodal Solu > DOF Solution > Temperature > OK

This result is shown in Figure 7.1(c).

11.  Main Menu > General Postproc > List Results > Nodal Solu > Select Temperature > OK

Example 7.2

Heat is generated in a large plate with k = 0.75 W/m°C at the rate 6000 W/m^3. The plate is 40 cm thick. The outside surfaces of the plate are exposed to fluid at 35°C with a convective heat transfer coefficient of 15 W/m^2°C. Determine temperature distribution in wall. The 2 element model to be used for solution.

Figure 7.14. Example 7.2

Solution

(I) Analytical method

Governing equation is,

Boundary conditions are,

We know at x = 0.2, T_3 = 115°C.

Substituting this in equation (7.5),

115 = −4000(0.2)2 + c_2 ⇒ c_2 = 275.

Substituting c_2 in equation (7.5),

(II) FEM by hand calculations [refer to Figure 7.14(a)]

Figure 7.14(a). Symmetric finite element model for Example 7.2.

Given: t = 40 cm = 0.4 m, T∞ = 35°C, h = 15 W/m^2°C, k = 0.75 W/m°C

Governing equation is,

For element 1,

For element 2,

Boundary conditions are, Q_1 = 0 and Q_3 = −h(T_3 − T∞) => Q_3 = −15(T_3 − 35) = −15T_3 + 525

Now,

By solving the above matrix and simultaneous equations, we have temperature distribution as,

Therefore,

T_1 = 275°C

T_2 = 235°C

T_3 = 115°C.

(III) Software results

Due to symmetry of the geometry, only half of the finite element model is created for software analysis.

Figure 7.14(b). Temperature distribution in a large plate (refer to Appendix C for color Figures).

Temperature values

NODE

TEMP

1

275.00

2

235.00

3

115.00

4

35.000

Answers for Example 7.2

Procedure for solving the problem using ANSYS® 11.0 academic teaching software.

For Example 7.2

PREPROCESSING

1.  Main Menu > Preferences, then select Thermal > OK

Figure 7.15. Selecting the preferences.

2.  Main Menu > Preprocessor > Element Type > Add/Edit/Delete > Add > Click on Link > then on 2D conduction 32 > OK > Add > Click on Link > then on 3D convection 34 > OK > Close

Figure 7.16. Selecting the element for conduction.

Figure 7.17. Selecting the element for convection.

3.  Main Menu > Preprocessor > Real Constants > Add/Edit/Delete > Add > Click on Link 32 > OK

Figure 7.18. Enter the cross-sectional area for Link 32.

Enter cross-sectional area AREA > Enter 1 > OK

Add > Click on Link 34 > OK

Enter cross-sectional area AREA > Enter 1 > OK

Figure 7.19. Enter the cross-sectional area for Link 34.

Enter the material properties.

4.  Main Menu > Preprocessor > Material Props > Material Models

Material Model Number 1,

click Thermal > Conductivity > Isotropic

Enter KXX = 0.75 > OK

Then in the material model window, click on Material menu > New Model > OK

Material Model Number 2,

Click Thermal > Convection or Film Coef.

Enter HF = 15 > OK

(Close the Define Material Model Behavior window.)

Create the nodes and elements. Due to geometric symmetry, only half of the model is created.

5.  Main Menu > Preprocessor > Modeling > Create > Nodes > In Active CS Enter the coordinate of node 1 > Apply Enter the coordinates of node 2 > Apply Enter the coordinates of node 3 > Apply Enter the coordinates of node 4 > OK

NODE LOCATIONS

NODE NUMBER

X COORDINATES

Y COORDINATES

1

0

0

2

0.1

0

3

0.0

0

4

0.3

0

Figure 7.20. Enter the node coordinates.

6.  Main Menu > Preprocessor > Modeling > Create > Elements > Elem Attributes > OK > Auto Numbered > Thru nodes Pick the 1st and 2nd node > Apply > then Pick the 2nd and 3rd node OK

Figure 7.21. Assigning element attributes to elements 1 and 2 and creating elements 1 and 2.

Elem Attributes > change the element type to Link 34 > change the material number to 2 > change the Real constant set number to 2 > OK > Auto Numbered > Thru nodes Pick the 3rd and 4th node > OK

Figure 7.22. Assigning elements attributes to element 3 and creating element 3.

Apply the boundary conditions and temperature.

7.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Thermal > Temperature > On Nodes Pick the 4th node > Apply > Click on TEMP and Enter Value = 35 > OK

Figure 7.23. Applying temperature on node 4.

8.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Thermal > Heat Generat > On Nodes Pick the 1st, 2nd, and 3rd nodes > Apply > Enter HGEN Value = 6000 > OK

Figure 7.24. Assigning heat generation on nodes 1, 2, and 3.

Solution

The interactive solution proceeds.

9.  Main Menu > Solution > Solve > Current LS > OK

The /STATUS Command window displays the problem parameters and the Solve Current Load Step window is shown. Check the solution options in the /STATUS window and if all is OK, select File > Close.

In the Solve Current Load Step window, select OK, and when the solution is complete, close the ‘Solution is Done!;’ window.

POST-PROCESSING

We can now plot the results of this analysis and also list the computed values.

10.  Main Menu > General Posrproc > Plot Results > Contour Plot > Nodal Solu > DOF Solution > Temperature > OK

This result is shown in Figure 7.14(b).

11.  Main Menu > General Postproc > List Results > Nodal Solu > Select Temperature > OK

Example 7.3

Compute the temperature distribution in a long steel cylinder with an inner radius of 125 mm and an outer radius of 250 mm. The interior of the cylinder is kept at 300°K and heat is lost on the exterior by convection to a fluid whose temperature is 280°K. The convection heat transfer coefficient h is 0.994 W/m^2°K and the thermal conductivity for steel k is 0.031 W/m°K.

Figure 7.25. Example 7.3

Solution

(I) Analytical method

Here the problem is solved considering heat flow in radial direction.

Figure 7.25(a). Analytical method for Example 7.3.

r_0 = 250 mm = 0.25 m and r_1 = 125 mm = 0.125 m

Assume unit length of the cylinder

Now,

Let

(II) FEM by hand calculations

Figure 7.25(b). Finite element model for Example 7.3.

r_1 = 125 mm, r_2 = 187.5 mm, and r_3 = 250 mm

Element matrices are,

Global conduction matrix is,

Global equation is,

Applying boundary conditions, T_1 = 300°K and Q_3 = −hA_0 (T_3 − T∞)

A_0 = 2πr_3 2π×0.25 = 1.57 m^2.

Solving the above equation, we get T_3 = 283.16°K and T_2 = 290.13°K.

(III) Software results

Due to the symmetry of the cylinder geometry, only a quarter of the geometry is drawn for finite element analysis.

Figure 7.25(c). Temperature distribution in a long cylinder (refer to Appendix C for color figures).

The temperature in the interior is 300°K and on the outside wall, it is found to be 280.004°K.

Answers of Example 7.3

Procedure for solving the problem using ANSYS® 11.0 academic teaching software.

For Example 7.3

PREPROCESSING

1.  Main Menu > Preferences, then select Thermal > OK

Figure 7.26. Selecting the preferences.

2.  Main Menu > Preprocessor > Element Type > Add/Edit/Delete > Add > Click on Solid > then on Quad 8 node 77 > OK > Close

Figure 7.27. Selecting the element.

3.  PLANE 77 does not require any real constant

Enter the material properties.

4.  Main Menu > Preprocessor > Material Props > Material Models

Material Model Number 1,

Click Thermal > Conductivity > Isotropic

Enter KXX = 0.031 > OK

(Close the Define Material Model Behavior window.)

Recognize symmetry of the problem, and a quadrant of a section through the cylinder is created.

5.  Main Menu > Preprocessor > Modeling > Create > Areas > Circles > Partial Annulus

Enter the data as shown below.

Figure 7.28. Create partial annular area.

Figure 7.29. Quadrant of a cylinder.

6.  Main Menu > Preprocessor > Meshing > Mesh Tool

The Mesh Tool dialog box appears. In that dialog box, click on the Smart Size and move the slider available below the Smart Size to 2 (i.e., towards Fine side). Then close the Mesh Tool box.

Figure 7.30. Mesh tool box.

7.  Main Menu > Preprocessor > Meshing > Mesh > Areas > Free. Pick the quadrant > OK

Figure 7.31. Quad element mesh.

8.  Main Menu > Preprocessor > Loads > Define loads > Apply > Thermal > Temperatures > On Lines

Select the line on the interior and set the temperature to 300.

Figure 7.32. Setting the temperature on the interior of the cylinder.

9.  Main Menu > Preprocessor > Loads > Apply > Convection > On Lines

Select the lines defining the outer surface and set the convection coefficient to 0.994 and the fluid temp to 280.

Figure 7.33. Setting the convection coefficient on outer surface.

10.  Main Menu > Preprocessor > Loads > Apply > Heat Flux > On Lines

To account for symmetry, select the vertical and horizontal lines of symmetry and set the heat flux to zero.

Figure 7.34. Setting the heat flux.

Figure 7.35. Model with boundary conditions.

Solution

The interactive solution proceeds.

11.  Main Menu > Solution > Solve > Current LS > OK

The /STATUS Command window displays the problem parameters and the Solve Current Load Step window is shown. Check the solution options in the /STATUS window and if all is OK, select File > Close.

In the Solve Current Load Step window, select OK, and when the solution is complete, close the ‘Solution is Done!’ window.

POST-PROCESSING

We can now plot the results of this analysis and also list the computed values.

12.  Main Menu > General Posrproc > Plot Results > Contour Plot > Nodal Solu > DOF Solution > Temperature > OK

This result is shown in Figure 7.25(b).

13.  Main Menu > General Postproc > List Results > Nodal Solu > Select Temperature > OK

7.4 TWO-DIMENSIONAL PROBLEM WITH CONDUCTION AND WITH CONVECTION BOUNDARY CONDITIONS

Example 7.4

A body having rectangular cross-section is subjected to boundary conditions as shown in Figure 7.36. The thermal conductivity of the body is 1.5 W/m°. On one side of the body, it is insulated and on the other side, convection takes place with h = 50 W/m^2°C and T∞ = 35°C. The top and bottom sides are maintained at a uniform temperature of 180°C. Determine the temperature distribution in the body.

Figure 7.36. Example 7.4

Solution

(I) Software results

The temperature at the top and bottom edges is found to be 180°C and at the right edge the temperature is found to be 46.802°C.

Figure 7.36(a). Temperature distribution in a body of rectangular cross-section (refer to Appendix C for color figures).

PROBLEMS

1.  Define conduction and convection.

2.  Write the formulas for the rate of heat flow in x-direction by conduction and the rate of heat flow by convection.

3.  Determine the temperature distribution for the two-dimensional body shown in Figure 7.37, subjected to boundary conditions as shown in the figure. The top and bottom edges are insulated. The left side of the body is maintained at a temperature of 45°C. On the right side, the convection process takes place with heat transfer coefficient h = 100 W/m^2°C and T∞ = 20°C. The thermal conductivity of the body is k = 45 W/m°C.

Figure 7.37. Problem 3

4.  Determine the temperature distribution for the two dimensional body shown in Figure 7.38. The temperature of 200°C is maintained at the top and bottom edges. The left and right edges are insulated. Heat is generated at the rate of q = 2000 W/m^3 in a body as shown in the figure. Let k = 35 W/m°C.

Figure 7.38. Problem 4

5.  Determine the temperature distribution for the two-dimensional body shown in Figure 7.39, subjected to boundary conditions as shown in the figure. The top and bottom edges are insulated. The left side of the body is maintained at a temperature of 50°C. On the right side, the convection process takes place with heat transfer coefficient h = 150 W/m^2°C and T∞ = 25°C. The thermal conductivity of the body is k = 50 W/m°C.

Figure 7.39. Problem 5

6.  Determine the temperature distribution for the two dimensional body shown in Figure 7.40. The temperature of 200°C is maintained at the top and bottom edges. The left and right edges are insulated. Heat is generated at the rate of q = 2100 W/m^3 in a body as shown in figure. Let k = 45 W/m°C.

Figure 7.40. Problem 6

7.  Determine the load matrix and the global load matrix for Figure 7.41. The top and bottom edges are insulated.

Figure 7.41. Problem 7

8.  Consider the rectangular plate shown in Figure 7.42. The outer temperature is T_0 = 30°C. Convection heat transfer takes place on the inner surface of the wall with T∞= 80°C and h = 50 W/m^2°K. Determine the temperature distribution in the wall. Take the thermal conductivity value k = 160 W/m°K.

Figure 7.42. Problem 8

9.  Consider a composite wall consisting of 2 materials shown in Figure 7.43. The outer temperature is T_0 = 30°C. Convection heat transfer takes place on the inner surface of the wall with T∞ = 80°C and h = 50 W/m^2°K. Determine the temperature distribution in the wall. Take the thermal conductivity value k_1= 40 W/m°C and k_2 = 60 W/m°C.

Figure 7.43. Problem 9

REFERENCES

1.  L. Segrlind, “Applied Finite Element Analysis, Second Edition,” Jon Wiley and Sons, 1984.

2.  S. Moaveni, “Finite Element Analysis: Theory and Application with ANSYS, Third Edition,” Prentice Hall, 2008.

3.  D. L. Logan, “A First Course in the Finite Element Method, Fifth Edition,” Cengage Learning, 2012.

4.  F. P. Incropera and D. P. DeWitt, “Fundamentals of Heat and Mass Transfer, Fourth Edition,” Wiley, 1996.

5.  F. P. Incropera, D. P. DeWitt, T. L. Bergman, and A. S. Lavine, “Introduction to Heat Transfer, Fifth Edition,” Wiley, 2007.

6.  S. S. Rao, “The Finite Element Method in Engineering, Fifth Edition,” Butterworth Heinemann, 2011.

Chapter 8

FLUID FLOW ANALYSIS

8.1 INTRODUCTION

A substance (liquid or gas) that will deform continuously by applied surface (shearing) stresses is called a fluid. The magnitude of shear stress depends on the magnitude of angular deformation. Indeed, different fluids have different relations between stress and the rate of deformation. Also, fluids are classified as compressible (usually gas) and incompressible (usually liquid).

The terms of velocities and accelerations of fluid particles at different times and different points throughout the fluid filled space are used to describe the flow field. The fluid is called ideal when the fluid has zero viscosity and is incompressible. A fluid is said to be incompressible if the volume change is zero (i.e., ρ = constant).

▿ · v = 0,

where,

v is velocity vector.

Depending on the importance of the viscosity of the fluid in the analysis, a flow can be termed as inviscid or viscous. An inviscid flow is a frictionless flow characterized by zero viscosity, that is, there is no real fluid. In other words, a fluid is called inviscid if the viscosity is zero (i.e., µ = 0).

A viscous flow is a flow in which the fluid is assumed to have nonzero viscosity. An irrotational flow is a flow in which the particles of the fluid are not rotating, the rotation is zero. In other words, an irrotational flow is a flow with negligible angular velocity, if

▿ × v = 0.

On the other hand, a potential flow is an irrotational flow of an ideal fluid (i.e., ρ = constant and µ = 0).

A line that connects a series of points in space at a given instant where all particles falling on the line at that instant have velocities whose vectors are tangent to the line is called a streamline.

The flow is steady which means that the flow pattern or streamlines do not change over time and the streamlines represent the trajectory of the fluid’s particles. But, when the flow is ideal that means that the fluid has zero velocity.

This chapter covers the finite element solution of ideal or potential flow (inviscid, incompressible flow) problems. Typical examples of potential flow are flow over a cylinder, flow around an airfoil, and flow out of an orifice.

The two-dimensional potential flow (irrotational flow) problems can be formulated in terms of a velocity potential function (φ) or a stream function (Ψ) The selection between velocity and stream function formulations in the finite element analysis depends on the ease of applying boundary conditions. If the geometry is simple, any one function can be used.

Fluid elements (e.g., FLUID141) are used in the steady state or transient analysis of fluid systems. Pressure, velocity, and temperature distributions can be obtained using these elements.

Two-dimensional fluid elements are defined using 3 (triangular element) or 4 (quadrilateral element) nodes added by isotropic properties. Inputs to these elements are nodal coordinates, real constants, material properties, surface and body loads, etc. Outputs of interest are nodal values of pressure and velocity.

8.2 PROCEDURE OF FINITE ELEMENT ANALYSIS (RELATED TO FLUID FLOW PROBLEMS)

Step 1. Select element type—the basic 3 node triangular element can be used.

Step 2. Choose a potential function.

Step 3. Define the gradient/potential and velocity/gradient relationships.

Step 4. Derive the element stiffness matrix and equations.

Step 5. Assemble the element equations to obtain the global equations and introduce boundary conditions.

Step 6. Solve for the nodal potentials.

Step 7. Solve for the element velocities and volumetric rates.

The finite element solution using software of potential flow problems is illustrated below. Only potential function formulation is considered. Two cases are considered in this chapter.

8.3 POTENTIAL FLOW OVER A CYLINDER

Figure 8.1. potential flow over a cylinder.

The previous figure depicts the steady-state irrotational flow of an ideal fluid over a cylinder, confined between 2 parallel plates. We assume that, at the inlet, velocity is uniform, say u_0. Here, we have to determine the flow velocities near the cylinder.

Flow past a fixed circular cylinder can be obtained by combining uniform flow with a doublet.

Figure 8.2. Superposition of a uniform flow and a doublet.

The superimposed stream function and velocity potential are given by,

and

where, U is velocity.

Because the streamline that passes through the stagnation point has a value of zero, the stream function on the surface of the cylinder of radius ‘a ’ is then given by,

which gives the strength of the doublet as,

The stream function and velocity potential for flow past a fixed circular cylinder becomes

and

The plot of the streamlines is shown in Figure 8.3.

Figure 8.3. Streamlines for flow past a fixed cylinder.

The velocity components can be determined by,

Along the cylinder (r = a), the velocity components reduce to vr = 0 and vθ = −2 U sin θ.

The radial velocity component is always zero along the cylinder while the tangential velocity component varies from 0 at the stagnation point (θ = π) to a maximum velocity of 2 U at the top and bottom of the cylinder .

8.4 POTENTIAL FLOW AROUND AN AIRFOIL

Figure 8.4. Potential flow around an airfoil.

The x- and y-components of fluid’s velocity respectively can be expressed in a stream function Ψ (x , y) as

The x- and y-components of fluid’s velocity with irrotational flows respectively can be expressed in a potential function Φ (x, y) as

Example 8.1

Flow over a circular cylinder between 2 parallel plates is shown in Figure 8.5. Assume unit thickness. Find the velocity distribution for the flow over a circular cylinder. Consider the flow of a liquid over a circular cylinder. Take liquid as water. Water density = 1000 kg/m^3 and viscosity = 0.001 N-s/m^2.

Figure 8.5. Flow over a cylinder.

Solution

(I) Software results

Procedure for solving the problem using ANSYS® 11.0 academic teaching software.

PREPROCESSING

1.  Main Menu > Preferences, then select FLOTRAN CFD > OK

Figure 8.6. Selecting the preferences.

2.  Main Menu > Preprocessor > Element Type > Add/Edit/Delete > Add > FLOTRAN CFD > 2D FLOTRAN 141 > OK

Figure 8.7. Element selection.

3.  Main Menu > Preprocessor > Modeling > Create > Areas > Rectangle > By 2 Corners

Enter (lower left corner) WP X=0.0, WPY=0.0 and Width = 2, Height = 1 > OK

4.  Main Menu > Preprocessor > Modeling > Create > Areas > Circle > Solid Circle. Enter WP X = 1, WP Y = 0.5 and Radius = 200e-3 > OK

Figure 8.8. Create areas.

Figure 8.9. Rectangle and circle.

Now subtract the circle from the rectangle. (Read the messages in the window at the bottom of the screen as necessary.)

5.  Main Menu > Preprocessor > Modeling > Operate > Booleans > Subtract > Areas >

Pick the rectangle > OK, then pick the circle > OK

Figure 8.10. Geometry for the flow over a cylinder.

Create a mesh of quadrilateral elements over the area.

6.  Main Menu > Preprocessor > Meshing > Mesh Tool

The Mesh Tool dialog box appears. Close the Mesh Tool box.

7.  Main Menu > Preprocessor > Meshing > Mesh > Areas > Free Pick the area > OK

Figure 8.11. Quadrilateral element mesh.

Apply the velocity boundary conditions and pressure.

8.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Fluid/CFD > Velocity

> On Lines Pick the left edge of the plate > OK > Enter VX = 1 > OK (VX = 1 means an initial velocity of 1 m/s^2)

9.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Fluid/CFD > Velocity

> On Lines Pick the edges around the cylinder > OK > Enter VX = 0 and VY = 0 > OK

10.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Fluid/CFD > Pressure

DOF > On Lines Pick the top, bottom and right edges of the plate > OK > OK Once all the boundary conditions are applied, the cylinder with plate will look like Figure 8.12.

Figure 8.12. Model with boundary conditions.

The model-building step is now complete, and we can proceed to the solution. First, save the model.

Solution

The interactive solution proceeds.

11.  Main Menu > Solution > FLOTRAN Set Up > Fluid Properties > A dialog in that select against density as liquid and against viscosity as liquid > OK

Then another dialog box appears and, in that, enter the value of density = 1000 value = 0.001 > OK

12.  Main Menu > Solution > FLOTRAN Set Up > Execution Ctrl > a dialog in that Enter in the first row “Global iterations EXEC” = 200

13.  Main Menu > Solution > Run FLOTRAN

When the solution is complete, close the ‘Solution is Done!’ window.

POST-PROCESSING

We can now plot the results of this analysis and also list the computed values.

14.  Main Menu > General Postproc > Read Results > Last Set

15.  General Postproc > Plot Results > Contour Plot > Nodal Solu

Select DOF Solution and Fluid Velocity and click OK

This is what the solution should look like:

Figure 8.13. Velocity distribution over a cylinder (refer to Appendix C for color figures).

16.  Next, go to Main Menu > General Postproc > Plot Results > Vector Plot > Predefined. One window will appear then click OK

Figure 8.14. Vector plot of the fluid velocity (refer to Appendix C for color figures).

17.  General Postproc > Plot Results > Contour Plot > Nodal Solu Select DOF Solution and Pressure and Click OK

Figure 8.15. pressure distribution over a cylinder (refer to Appendix C for color figures).

PROBLEMS

1.  Define a fluid, inviscid flow, viscous flow, and irrotational flow.

2.  What are the 2 fluid classifications?

3.  Define streamline in a graphic of fluid motion?

4.  What we mean when we say the flow is steady and ideal?

5.  Define irrotational flow and potential flow?

6.  Compute and plot velocity distribution over the airfoil as shown in Figure 8.16. Assume unit thickness. Take density of air = 1.23 kg/m^3 and viscosity = 1.79 × 10-5N-s/m^2.

Figure 8.16. Flow over an airfoil.

7.  Flow over a circular cylinder between 2 parallel plates is as shown in Figure 8.17. Assume unit thickness. Find the velocity distribution for the flow over a circular cylinder. Consider the flow of a liquid over a circular cylinder. Take liquid as water. Water density = 1000 kg/m^3 and viscosity = 0.001 N-s/m^2, u = u_0 = 2 m/s^2, h = 2 m, and L = 4m.

Figure 8.17. Flow over a circular cylinder between 2 parallel plates.

8.  Compute and plot velocity distribution over the airfoil as shown in Figure 8.18. Assume unit thickness. Take density of air = 1.23 kg/m^3 and viscosity = 1.79 X 10−5N-s/m^2, L = 4 m, L_1 = 20 m, h_1 = 18 m, and u = u_0 = 3 m/s^2.

Figure 8.18. Flow over an airfoil.

9.  Flow over an elliptical cylinder between 2 parallel plates is shown in Figure 8.19. Assume unit thickness. Find the velocity distribution for the flow over a circular cylinder. Consider the flow of a liquid over a circular cylinder. Take liquid as water. Water density = 1000 kg/m^3 and viscosity = 0.001 N-s/m^2, u = u_0 = 1 m/s^2, D = 2 m, b = 1 m, h = 4 m, and L = 8 m.

Figure 8.19. Flow over an elliptical cylinder between 2 parallel plates.

REFERENCES

1.  J. G. Knudsen and D. L. Katz, “Fluid Dynamics and Heat Transfer,” McGraw-Hill, 1958.

2.  J. W. Daily and D. R. F. Harleman, “Fluid Dynamics,” Addison-Wesley, Reading, 1966.

3.  S. S. Rao, “The Finite Element Method in Engineering, Fifth Edition,” Butterworth-Heinemann, 2011.

4.  T. J. Chung, “Finite Element Analysis in Fluid Dynamics,” McGraw-Hill, 1978.

5.  D. L. Logan, “A First Course in the Finite Element Method, Fifth Edition,” Cengage Learning, 2012.

6.  J. N. Reddy, “An Introduction to the Finite Element Method, Third Edition,” McGraw-Hill, 2004.

Chapter 9

DYNAMIC ANALYSIS

9.1 INTRODUCTION

A dynamic system is a system that has mass and components, or parts, that are capable of relative motion. Structural dynamics encompass modal analysis, harmonic response analysis, and transient response analysis. Modal analysis consists of finding natural frequencies and corresponding modal shapes of structures. Finding amplitude of vibration when the loads vary sinusoidal with time is known as harmonic response analysis. Finding the structural response to arbitrary time dependent loading is referred to as transient response analysis.

In this chapter, one-dimensional problems relating to these topics are covered. In vibration analysis, mass matrix and damping matrix will also be discussed in addition to stiffness matrix.

Governing equation of undamped free vibration assumes the form,

The nontrivial solution of equation (9.1) is the determinate,

where

ω = radian (or natural) frequency.

The solution of equation (9.2) gives natural frequencies (ω). Substituting the value of ω back into the governing equation gives modal shapes (or amplitudes of the displacements) defined by {q}.

The governing equation for the complete structure in global coordinate is

([K]-ω^2 [M]){q} =0.

Mass matrices for bar element and beam elements are given by,

where

p = density of the element material

A = cross-sectional area

L = length.

9.2 PROCEDURE OF FINITE ELEMENT ANALYSIS (RELATED TO DYNAMIC PROBLEMS)

Step 1. Select element type.

Step 2. Select a displacement function.

Step 3. Define the strain/displacement and stress/strain relationships.

Step 4. Derive the element stiffness and mass matrices and equations.

Step 5. Assemble the element equations to obtain the global equations and introduce boundary conditions.

Step 6. Solve for the natural requencies and mode shapes.

9.3 FIXED-FIXED BEAM FOR NATURAL FREQUENCY DETERMINATION

Example 9.1

Determine the first 2 natural frequencies for the fixed-fixed beam shown in Figure 9.1. The beam is made of steel with modulus of elasticity E = 209 GPa, Poisson’s ratio = 0.3, length L = 0.75 m, cross-section area A = 625 mm^2, mass density p = 7800 kg/m^3, moment of inertia I = 34700 mm^4.

Figure 9.1. Fixed-fixed beam for Example 9.1.

Solution

(I) Analytical method

Frequency,

Frequency,

(II) FEM by hand calculations

Figure 9.1(a). Finite element model.

Mass matrices are,

Stiffness matrices are,

Global mass matrix,

Global stiffness matrix,

Governing equation is,

For nontrivial solution,

Solving, we get

Given:

Substituting, we get

Frequency,

Frequency,

(III) Software results

Figure 9.1(b). Deflection pattern for a fixed-fixed beam for mode 1 (refer to Appendix C for color figures).

Frequency values (in Hz)

The following are the mode shapes:

Figure 9.1(c).Mode 1 for fixed-fixed beam (refer to Appendix C for color figures).

Figure 9.1(d). Mode 2 for fixed-fixed beam (refer to Appendix C for color figures).

Answers for Example 9.1

Procedure for solving the problem using ANSYS® 11.0 academic teaching software For Example 9.1

PREPROCESSING

1.  Main Menu > Preprocessor > Element Type > Add/Edit/Delete > Add > Beam > 2D elastic 3 > OK > Close

Figure 9.2. Element selection.

2.  Main Menu > Preprocessor > Real Constants > Add/Edit/Delete > Add > OK

Figure 9.3. Enter the area and moment of inertia.

Cross-sectional area AREA > Enter 625e-6

Area moment of inertia IZZ > Enter 34700e-12

Total beam height HEIGHT > Enter 1 > OK > Close

Enter the material properties.

3.  Main Menu > Preprocessor > Material Props > Material Models

Material Model Number 1, click Structural > Linear > Elastic > Isotropic

Enter EX = 209E9 and PRXY = 0.3 > OK

click Structural > Linear > Density

Enter DENS = 7800 > OK

(Close the Define Material Model Behavior window.)

Create the keypoints and lines as shown in the figure.

4.  Main Menu > Preprocessor > Modeling > Create > Keypoints > In Active CS, Enter the coordinates of keypoint 1 > Apply Enter the coordinates of keypoint 2 >OK

Keypoint locations

Keypoint number

X COORDINATE

Y coordinate

1

0

0

2

0.75

0

Figure 9.4. Enter the keypoint coordinates.

5.  Main Menu > Preprocessor > Modeling > Create > Lines > Lines > Straight Line, Pick the 1st and 2nd keypoint > OK

Figure 9.5. Pick the keypoints to create lines.

6.  Main Menu > Preprocessor > Meshing > Size Cntrls > Manual Size > Lines > All Lines > Enter NDIV No. of element divisions = 10

Figure 9.6. Specify element length.

7.  Main Menu > Preprocessor > Meshing > Mesh > Lines > Click Pick All

Figure 9.7. Create elements by meshing.

8.  Main Menu > Solution > Analysis Type > New Analysis > Select Modal > OK

Figure 9.8. Define the type of analysis.

9.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Structural > Displacement > On Nodes Pick the left most node and right most node > Apply > Select All DOF > OK

Figure 9.9. Apply the displacement constraint.

10.  Main Menu > Solution > Analysis Type > Analysis Options > Select PCG Lanczos option

Enter No. of modes to extract = 2

NMODE No. of modes to expand = 2 > OK

After OK one more window will appear, for that also click OK

Figure 9.10. Select the number of modes to extract.

Solution

The interactive solution proceeds.

11.  Main Menu > Solution > Solve > Current LS > OK

The / STATUS Command window displays the problem parameters and the Solve Current Load Step window is shown. Check the solution options in the /STATUS window and if all is OK, select File > Close.

In the Solve Current Load Step window, select OK, and when the solution is complete, close the ‘Solution is Done!’ window.

POST-PROCESSING

12.  Main Menu > General Postproc > Results Summary

This result is shown as frequency values in Hz.

13.  Main Menu > General Postproc > Read Results > First Set

14.  Main Menu > General Postproc > Plot Results > Deformed Shape > Click Def + undeformed > OK

This result is the first mode shown in Figure 9.1(c).

15.  Main Menu > General Postproc > Read Results > Next Set

16.  Main Menu > General Postproc > Plot Results > Deformed Shape > Click Def + undeformed > OK

This result is the second mode shown in Figure 9.1(d).

9.4 TRANSVERSE VIBRATIONS OF A CANTILEVER BEAM

Example 9.2

Determine the first 4 natural frequencies for the cantilever beam shown in Figure 9.11. The beam is made of steel with modulus of elasticity, E = 207 GPa, Poisson’s ratio = 0.3, length L = 0.75 m, cross-section area A = 625 mm^2, mass density p = 7800 kg/m^3, moment of inertia I = 34700 mm^4.

Figure 9.11. Cantilever beam for Example 9.2.

Solution

(I) Analytical solution

Frequency,

Frequency,

Frequency,

Frequency,

(II) FEM by hand calculations

Figure 9.11(a). Finite element model.

Stiffness matrices are,

Global stiffness matrix,

Mass matrices are,

Global mass matrix is,

Governing equation is,

For a nontrivial solution

Det ([K]−ω^2[M])= 0 ⇒∣([K]−ω^2[M])∣ = 0.

Substituting and solving, we get

(III) Software results

Figure 9.11(b). Deflection pattern for a fixed-fixed beam for mode 1 (Refer to Appendix C for color figures).

Frequency values (in Hz)

The following are the mode shapes:

Figure 9.11(c). Mode 1 for cantilever beam (refer to Appendix C for color figures).

Figure 9.11(d). Mode 2 for cantilever beam (refer to Appendix C for color figures).

Figure 9.11(e). Mode 3 for cantilever beam (refer to Appendix C for color figures).

Figure 9.11(f). Mode 4 for cantilever beam (refer to Appendix C for color figures).

Answers of Example 9.2

9.5 FIXED-FIXED BEAM SUBJECTED TO FORCING FUNCTION

Example 9.3

For the fixed-fixed beam subjected to the time dependent forcing function shown in Figure 9.12, determine the displacement response for 0.2 seconds. Use time step integration of 0.01 sec. Let E = 46 GPa, Poisson’s ratio = 0.35, length of beam L = 5 m, cross-section area A = 1 m^2, mass density, p = 1750 kg/m^3, moment of inertia I = 4.2 × 10−5 m^4.

Figure 9.12. Fixed-fixed beam subjected to the time dependent forcing function for Example 9.3

Solution

(I) Software results

Figure 9.12(a). Displacement response for 0.2 sec for node 2 (refer to Appendix C for color figures).

Figure 9.12(b). Displacement response for 0.2 sec for node 3 (refer to Appendix C for color figures).

Displacement values (in meters) for node 2

TIME

2 UY

UY_2

0.0000

0.00000

0.10000E−01

0.421220E−05

0.20000E−01

0.284618E−03

0.50000E−01

0.602161E−02

0.80000E−01

0.121677E−01

0.10000

0.153042E−01

0.12000

0.148820E−01

0.15000

0.979873E−02

0.18000

0.868368E−04

0.20000

−0.649350E−02

Displacement values (in meters) for node 3

TIME

3 UY

UY_3

0.0000

0.00000

0.10000E−01

0.505126E−03

0.20000E−01

0.218959E−02

0.50000E−01

0.113766E−01

0.80000E−01

0.241211E−01

0.10000

0.286233E−01

0.12000

0.292504E−01

0.15000

0.183799E−01

0.18000

−0.205644E−03

0.20000

−0.117477E−01

Procedure for solving the problem using ANSYS® 11.0 academic teaching software.

For Example 9.3

PREPROCESSING

1.  Main Menu > Preferences > Select Structural > OK

Figure 9.13. Selecting the preferences.

2.  Main Menu > Preprocessor > Element Type > Add/Edit/Delete > Add > Beam > 2D elastic 3 > OK > Close

Figure 9.14. Element selection.

3.  Main Menu > Preprocessor > Real Constants > Add/Edit/Delete > Add > OK

Figure 9.15. Enter the area and moment of inertia.

Cross-sectional area AREA > Enter 1

Area moment of inertia IZZ > Enter 4.2e–5

Total beam height HEIGHT > Enter 1 > OK > Close

Enter the material properties.

4.  Main Menu > Preprocessor > Material Props > Material Models

Material Model Number 1, click Structural > Linear > Elastic > Isotropic

Enter EX = 46E9 and PRXY = 0.35 > OK

Click Structural > Linear > Density

Enter DENS = 1750 > OK

(Close the Define Material Model Behavior window.)

Create the nodes and elements as shown in the figure.

5.  Main Menu > Preprocessor > Modeling > Create > Nodes > In Active CS Enter the coordinates of node 1 > Apply > Enter the coordinates of node 2 > Apply > Enter the coordinates of node 3 > Apply > Enter the coordinates of node 4 > Apply Enter the coordinates of node 5 > OK

Node locations

Node number

x coordinate

Y coordinate

1

0

0

2

1.25

0

3

2.5

0

4

3.75

0

5

5

0

Figure 9.16. Enter the node coordinates.

6.  Main Menu > Preprocessor > Modeling > Create > Elements > Auto Numbered > Thru

Nodes Pick the 1st and 2nd node > Apply > Pick the 2nd and 3rd node > Apply > Pick the 3rd and 4th node > Apply > Pick the 4th and 5th node > OK

Figure 9.17. Pick the nodes to create elements.

7.  Main Menu > Solution > Analysis Type > New Analysis > Select Transient > OK

Figure 9.18. Define the type of analysis.

then select > Reduced > OK

Figure 9.19. Define the type of transient analysis.

8.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Structural > Displacement > On Nodes > Pick the left most node and right most node > Apply > Select All DOF > OK

Figure 9.20. Apply the displacement constraint.

9.  Main Menu > Solution > Master DOFs > User Selected > Define > Pick 2nd, 3rd, and 4th node > Apply > Select UY from Lab 1 1st degree of freedom > OK

Figure 9.21. Defining master DOF.

10.  Main Menu > Solution > Load Step Opts > Time/Frequenc > Time-Time Step

Enter [TIME] Time at end of load step – 0

Enter [DELTIM] Time step size – 0.01 < OK

Figure 9.22. Defining time step size.

11.  Main Menu > Solution > Load Step Opts > Write LS File Enter LSNUM Load step fine number n = 1 > OK

Figure 9.23. Creating LS file.

12.  Main Menu > Solution > Define Loads > Apply > Structural > Force/Moment > On Nodes > Pick the middle or 3rd node Apply > Enter FY = 45e3 > OK

Figure 9.24. Applying force on node.

13.  Main Menu > Solution > Load Step Opts > Time/Frequenc > Time-Time Step

Enter [TIME] Time at end of load step – 0.01 > OK

Figure 9.25. Defining time at the end of 1st load step.

14.  Main Menu > Solution > Load Step Opts > Write LS File Enter LSNUM Load step fine number n = 2 > OK

Figure 9.26. Creating LS file for 1st load step.

Similarly repeat Steps 13 and 14 for Time at end of load step of 0.02, 0.05, 0.08, and 0.1 and each time create a LS file with next numbers (n), i.e., 3, 4, 5, and 6.

15.  Main Menu > Solution > Define Loads > Delete > Structural > Force/ Moment > On Nodes > Pick the middle or 3rd node Apply > OK

16.  Main Menu > Solution > Define Loads > Apply > Structural > Force/Moment > On Nodes > Pick the middle or 3rd node Apply > Enter FY = 36e3 > OK

Figure 9.27. Applying force on node.

17.  Main Menu > Solution > Load Step Opts > Time/Frequenc > Time-Time Step

Enter [TIME] Time at end of load step– 0.12 > OK

Figure 9.28. Defining time at the end of 6th load step.

18.  Main Menu > Solution > Load Step Opts > Write LS File Enter LSNUM Load step fine number n = 7 > OK

Figure 9.29. Creating LS file for 6th load step.

Repeat Step 15 and delete the force.

Then apply the force of 22.5 kN (i.e., 22.5e3) and define the Time at end of load step of 0.15 and create a LS file with number (n) = 8.

Again, repeat Step 15 and delete the force.

Then apply the force of 9 kN (i.e., 9e3) and define the Time at end of load step of 0.18 and create a LS file with number (n) = 9.

Again, repeat Step 15 and delete the force.

Define the Time at end of load step of 0.2 and create a LS file with number (n) = 10.

19.  Main Menu > Solution > Solve > From LS Files

Enter LSMIN Starting LS file number = 1

Enter LSMAX Ending LS file number = 10

LSINC File number increment = 1

Figure 9.30. Solving from LS files.

20.  Main Menu > Time Hist Post pro

The following dialog box will appear

Figure 9.31. Time hist dialog box.

In that dialog box click on the first icon, i.e., on Add data, one more dialog box will appear as shown below. Then click on DOF Solution > y-Component of displacement > OK.

Figure 9.32. Selecting the displacement in y-direction.

Figure 9.33. Selecting the node.

It asks for the node to pick, so pick the node 3 or middle node < OK.

Then, in the Time hist dialog box, click on 4th icon, i.e., List Data (refer to Figure 9.31).

This result is shown as displacement values for node 3 in the software results of the problem. Then, in the Time hist dialog box, click on 3rd icon,i.e., Graph Data (refer to Figure 9.31).

The result is shown in Figure 9.12(b) for node 3 in the software results of the problem.

Maximum displacement values (in meters)

Figure 9.34. Values of displacement.

9.6 AXIAL VIBRATIONS OF A BAR

Example 9.4

For the bar shown in Figure 9.35, determine the first 2 natural frequencies. Let E = 207 GPa, Poisson’s ratio = 0.3, length L = 2.5 m, cross-section area A = 1 m^2, mass density p = 7800 kg/m^3.

Figure 9.35. The Bar for Example 9.4.

Solution

(I) Analytical method

Frequency,

Frequency,

(II) FEM by hand calculations

Figure 9.35(a). Finite element model.

Mass matrices are,

Stiffness matrices are,

Global mass matrix is,

Global stiffness matrix is,

Governing equation,

Boundary conditions are, u1 = 0.

Applying boundary conditions and for a nontrivial solution,

i.e.,

where

Frequency,

Frequency,

(III) Software results

Figure 9.35(b). Deflection pattern for a bar (refer to Appendix C for color figures).

Frequency values (in Hz)

Answers for Example 9.4

Procedure for solving the problem using ANSYS® 11.0 academic teaching software.

For Problem 9.4

PREPROCESSING

1.  Main Menu > Preprocessor > Element Type > Add/Edit/Delete > Add > Link > 2D spar 1 > OK > Close

Figure 9.36. Element selection.

2.  Main Menu > Preprocessor > Real Constants > Add/Edit/Delete > Add > OK

Figure 9.37. Enter the cross-sectional area.

Cross-sectional area AREA > Enter 1 > OK > Close

Enter the material properties.

3.  Main Menu > Preprocessor > Material Props > Material Models

Material Model Number 1, click Structural > Linear > Elastic > Isotropic

Enter EX = 207E9 and PRXY = 0.3 > OK

Click Structural > Linear > Density

Enter DENS = 7800

(Close the Define Material Model Behavior window.)

Create the keypoints and lines as shown in the figure.

4.  Main Menu > Preprocessor > Modeling > Create > Keypoints > In Active CS Enter the coordinates of keypoint 1 > Apply Enter the coordinates of keypoint 2 > OK

Keypoint locations

Keypoint number

X coordinate

Y coordinate

1

0

0

2

2.5

0

Figure 9.38. Enter the keypoint coordinates.

5.  Main Menu > Preprocessor > Modeling > Create > Lines > Lines > Straight Line Pick the 1st and 2nd keypoint > OK

Figure 9.39. Pick the keypoints to create lines.

6.  Main Menu > Preprocessor > Meshing > Size Cntrls > Manual Size > Lines > All Lines > Enter NDIV No. of element divisions = 10

Figure 9.40. Specify element length.

7.  Main Menu > Preprocessor > Meshing > Mesh > Lines > Click Pick All

Figure 9.41.Create elements by meshing.

8.  Main Menu > Solution > Analysis Type > New Analysis > Select Modal > OK

Figure 9.42. Define the type of analysis.

9.  Main Menu   > Preprocessor > Loads > Define Loads > Apply > Structural > Displacement > On Nodes Pick the left most node > Apply > Select All DOF > OK

Figure 9.43. Apply the displacement constraint.

10.  Main Menu > Solution > Analysis Type > Analysis Options > Select Reduced option

Enter No. of modes to extract = 2

NMODE No. of modes to expand = 2 > OK

Figure 9.44. Select the number of modes to extract.

Enter FREQE Frequency range 0 2500 > OK

Figure 9.45. Enter the frequency range.

11.  Main Menu > Solution > Master DOFs > User Selected > Define > Pick all nodes except left most node > OK > Select UX from Lab 1 1st degree of freedom > OK

Figure 9.46. Defining the master degree of freedom.

Figure 9.47. Model with master DOF applied.

Solution

The interactive solution proceeds.

12.  Main Menu > Solution > Solve > Current LS > OK

The /STATUS Command window displays the problem parameters and the Solve Current Load Step window is shown. Check the solution options in the /STATUS window and if all is OK, select File > Close.

In the Solve Current Load Step window, select OK, and when the solution is complete, close the ‘Solution is Done!’ window.

POST-PROCESSING

13.  Main Menu > General Postproc > Results Summary

This result is shown as frequency values in Hz.

14.  Main Menu > General Postproc > Read Results > First Set

15.  Main Menu > General Postproc > Plot Results > Contour Plot > Nodal Solu > DOF Solution, click on Displacement vector sum > OK

This result is shown in Figure 9.35(b).

9.7 BAR SUBJECTED TO FORCING FUNCTION

Example 9.5

The bar shown in Figure 9.48 is subjected to time dependent forcing function as shown, determine the nodal displacements for 5 time steps using 2 finite elements. Let E = 207 GPa, Poisson’s ratio = 0.3, length of beam L = 5 m, cross-section area A= 625 × 10−6 m^2, mass density p = 7800 kg/m^3. Use time step of integration 0.00025 seconds.

Figure 9.48. The bar for Example 9.5.

Solution

(I) Software results

Figure 9.48(a). Displacement response for 0.00025 sec for node 2 (refer to Appendix C for color figures).

Figure 9.48(b). Displacement response for 0.00025 sec for node 3 (refer to Appendix C for color figures).

Displacement values (in meters) for node 2

TIME

2 UX

ux_2

0.0000

0.00000

0.25000E−03

−0.467370E−06

0.50000E−03

−0.821457E−06

0.75000E−03

0.396081E−05

0.10000E−02

0.210563E−04

0.12500E−02

0.535055E−04

0.15000E−02

0.950064E−04

0.17500E−02

0.128841E−03

0.20000E−02

0.138387E−03

Displacement values (in meters) for node 3

TIME

3 UX

ux_3

0.0000

0.00000

0.25000E−03

0.375512E−05

0.50000E−03

0.191517E−04

0.75000E−03

0.488709E−04

0.10000E−02

0.889759E−04

0.12500E−02

0.130597E−03

0.15000E−02

0.161991E−03

0.17500E−02

0.179673E−03

0.20000E−02

0.184097E−03

Maximum displacement values (in meters)

Figure 9.48(c). Values of displacement.

PROBLEMS

1.  What is the governing equation of undamped free vibration and its nontrivial solution?

2.  What are the mass matrices for bar element and beam elements?

3.  Determine the first 5 natural frequencies for the fixed-fixed beam shown in Figure 9.49. The beam is made of steel with E = 200 GPa, Poisson’s ration = 0.3, length = 2 m, cross-section area = 60 cm^2, mass density p = 7800 kg/m^3, moment if inertia I = 200 mm^4.

Figure 9.49. Fixed-fixed beam for problem 3.

4.  For the bar shown in Figure 9.50, determine nodal displacements for the 5 time finite elements. Let E = 70 GPa, p = 2700 kg/m^3, A = 645 mm^2, and L = 2.5 m.

Figure 9.50. The bar for problem 4.

5.  The beam shown in Figure 9.51 is subjected to the forcing functions shown, determine the maximum deflections. Let E = 207 GPa, p = 7800 kg/m^3, A = 0.0194 m^2, I = 8.2 × 10−5 m^4, L = 6 m. Take time step of 0.05 seconds.

Figure 9.51. The beam for problem 5.

6.  Determine the natural frequencies of vibrations for the cantilever beam shown in Figure 9.52.

Figure 9.52. Cantilever beam for problem 6.

7.  For the bar shown in Figure 9.53, determine nodal displacements for the 5 time finite elements. Let E = 210 GPa, p = 2800 kg/m^3, A = 825 mm^2 and L = 3 m.

Figure 9.53. The bar for problem 7.

8.  For the beam shown in Figure 9.54, determine the mode shapes. Let E = 310 × 10^6 psi, p = 0.283 lbf/in^3, A = 1 in^2, v = 0.3, and L = 30 in.

Figure 9.54. The beam for problem 8.

9.  For the bar shown in Figure 9.55, subjected to the forcing functions shown, determine the nodal displacement, velocities, acceleration, and the maximum deflections for 5 time steps using 2 finite elements. Let E = 2 × 10^6 psi, p = 2 lb-s^2 in^4, A = 2 in^2, I = 322.83 in^4, L = 10 in.

Figure 9.55. The bar for problem 9.

REFERENCES

1.  D. L. Logan, “A First Course in the Finite Element Method, Fifth Edition,” Cengage Learning, 2012.

2.  S. Moaveni, “Finite Element Analysis: Theory and Application with ANSYS, Third Edition,” Prentice Hall, 2008.

3.  C. T. F. Ross, “Finite Element Method in Structural Mechanics,” Ellis Horwood Limited Publishers, 1985.

4.  S. S. Rao, “The Finite Element Method in Engineering, Fifth Edition,” Butterworth-Heinemann, 2011.

5.  G. L. Narasaiah, “Finite Element Analysis,” CRC Press, 2009.

6.  W. T. Thompson and M. D. Dahleh, “Theory of Vibrations with Applications, Fifth Edition,” Prentice–Hall, 1998.

Chapter 10

ENGINEERING ELECTROMAGNETICS ANALYSIS

10.1 INTRODUCTION TO ELECTROMAGNETICS

Electromagnetics (EM) govern many applications in engineering such as the transmission lines system. Therefore, it is essential to understand the fundamental concepts of EM in order to properly design and model electrical systems and devices using the finite element method (FEM). Furthermore, EM becomes more useful in designing engineering systems with recent technologies, especially due to the increasing speeds of digital devices and the increased use of modern electronics circuits such as printed-circuit-board and communications systems such as cellular phones. The most important equations in EM theory are Maxwell’s equations, which are known as the foundation of EM theory.

10.2 MAXWELL’S EQUATIONS AND CONTINUITY EQUATION

In electromagnetic analysis on a macroscopic level, it is based on solving the Maxwell’s equations issue on certain boundary conditions. Also, there is another fundamental equation that can specify the conservation (indestructibility) of electric charge know as the equation of continuity. Maxwell’s equations and continuity equation can be written in both differential and integral forms. We choose to start here with the differential form because it leads to differential equations that the finite element method (FEM) can handle.

10.2.1 Maxwell’s Equations and Continuity Equation in Differential Form

Now, we can present the 4 Maxwell’s equations in differential form in time-varying EM fields as:

where

E = Electric field intensity, (in volt/meter) −V/m^2

D = Electric flux density (or electric displacement), (in coulomb/meter^2) −C/m^2

H = Magnetic field intensity, (in ampere/meter) −A/m^2

B = Magnetic flux density, (in tesla or weber/meter^2)−T or Wb/m^2

Je = Electric Current density or charge flux (surface), (in ampere/meter^2) −A/m^2

Jm = The magnetic conductive current density, (in volt/meter^2) −V/m^2, where Jm = σm H

σm = The magnetic conductive resistivity (in ohm/meter) −Ω/m

ρV = Electric charge density (volume), (in coulomb/meter^2) −C/m^2.

Now, the equation of continuity can be written in differential form as

There are 3 independent equations from the above 5 equations. They are either equations 1, 2, and 3, or equations 1, 2, and 5. The other two equations 4 and 5, or equations 3 and 4 can be derived from the independent equations, and therefore are called dependent equations. Additionally, equation 5 can be derived the divergence of equation 1 and using equation 3.

10.2.2 Maxwell’s Equations and Continuity Equation in Integral Form

Furthermore, let us now look to the 4 Maxwell’s equations and the continuity equation in integral form in time-varying EM fields. The integrals are taken over in an open surface S or its boundary contour L as shown in Figure 10.1, where I is the electric current that flows through the path L.

Figure 10.1. The surface S and contour L for the integral form of Maxwell’s equations.

where the surface S encloses the volume v, while the contour L encloses the surface S. l is the line vector over the contour L and S is the surface vector. Note that, the direction of dl must be consistent with the direction of the dS in agreement with the right-hand rule.

10.2.3 Divergence and Stokes Theorems

Indeed, equations 6 through 10, the integral forms can be derived from the differential forms or vice versa. This can be done by using either divergence (Gauss’s) theorem or Stokes’s theorem,

where F is any arbitrary vector field.

10.2.4 Maxwell’s Equations and Continuity Equation in Quasi-Statics Case

So far, we did the Maxwell’s equations in fully dynamic case. Now, we can express Maxwell’s equations in quasi-statics case which the displacement current (D) is neglected. That is,

Whereas equations (10.2), (10.3), and (10.4) remain the same. Also, we can write the continuity equation (10.5) in quasi-statics case as

Indeed, the quasi-static approximation is mainly used for time-varying fields in various conducting media. This is due to that, for good conductors, the conduction current greatly exceeds the displacement current, D, for the frequencies.

10.2.5 Maxwell’s Equations and Continuity Equation in Statics Case

In the statics field case, the displacement current term and the time-varying magnetic flux density term are neglected (the field quantities do not vary with time). Therefore, the Maxwell’s equations in static form are expressed as

whereas equations (10.3), (10.4), and (10.13) still hold. Also, the continuity equation (10.14) remains the same.

To emphasize, there is no interaction between the electric and the magnetic fields. Thus, the static case can be divided into 2 separate cases, electrostatic case and magnetostatic case.

In electrostatic case, it can be described by equations (10.3) and (10.15), while, for magnetostatic case, it can be described by equations (10.4) and (10.13).

10.2.6 Maxwell’s Equations and Continuity Equation in Source-Free Regions of Space Case

The sources of the electromagnetic fields can be the volume charge density (ρv) and the electric current density (Je). In fact, these densities are localized in space. Also, these sources can make the generated electric and magnetic fields to radiate away from them and they can make the generated electric and magnetic fields to propagate to larger distances to the receiving destination. Therefore, Maxwell’s equations can be written in source-free regions of space (away from the source) as:

whereas equation (10.4) remains the same. With this in mind, the continuity equation (10.14) also remains the same.

10.2.7 Maxwell’s Equations and Continuity Equation in Time-Harmonic Fields Case

So far, we considered the arbitrary time variation of electromagnetic fields. Here, we consider only the steady-state (equilibrium) solution of electromagnetic fields when produced by sinusoidal currents. The time-harmonic (sinusoidal steady-state) field for Maxwell’s equations exists when the field quantities in the equations are harmonically oscillating functions with a single sinusoidal frequency ω. The time-harmonic fields case is the most regularly used in electrical engineering. Now, an arbitrary time-dependent field F (x, y, z, t) or F (r, t) can be written as

where ejwt is the time convention, ω is the angular frequency (rad/s) of the sinusoidal excitation, Fs (r) = Fs (x, y, z) is the phasor form of F (r, t) and it is in general complex, and Re ( ) indicates taking the real part of quantity in the parenthesis. Furthermore, the electromagnetic field quantities can be expressed in phasor notation as

For example, the fields can be expresses in time dependent ejwt, as in equation (10.20), H(r, t) = H(r)ejwt and E(r, t) = E(r)ejwt, etc.

As a result, using the phasor representation can allow us to replace the time derivations by jω , because

Therefore, the Maxwell’s equations can be expressed in time-harmonic as

Now, the continuity equation can be presented as

On the other hand, a nonsinusoidal field can be presented as

Therefore, the solutions to Maxwell’s equations for a nonsinusoidal field can be found by assuming that all the Fourier components Fs(r, ω) over ω.

10.3 LORENTZ FORCE LAW AND CONTINUITY EQUATION

The Lorentz Force F is the force on a charge q with a vector velocity u in the present electric filed E and magnetic field B and can be obtained as

In addition, the volume charge ρv and the current distribution J can be subjected to the forces in the presence of fields. Thus, Lorentz Force F per unit volume acting on the volume charge and the current distribution can be expressed as

However, if the current distribution J occurs from the motion of the charges q within the volume charge ρv, then current distribution J can be formed as J = ρ vv.

This can make the Lorentz Force F as

Moreover, the Lorentz Force law is essential to understand the interaction between EM fields and matter. Indeed, the law is used in the design of many electrical devices.

Furthermore, the continuity equation which expresses the conservation of electric charge can be written as

Equation (10.31) is implicit in Maxell’s equations.

10.4 CONSTITUTIVE RELATIONS

In addition to the Maxwell’s equations and the continuity equation, there are constitutive relations which describe the macroscopic properties of the medium in which the fields exist. In other words, constitutive relations describe the relationship between the EM fields through the properties of the medium. Indeed, Maxwell’s equations and constitutive relations are used to obtain the solutions of EM fields that exist in any microwave structures. The constitutive relations can be presentenced in vacuum (free space) as

where

ɛ_0 = the permittivity of vacuum

µ_0 = the permeability of vacuum

σe = the electrical conductivity

M = magnetization field.

The numerical values of ε_0 and µ_0 are written as

We can use these 2 quantities to define the speed of light (c_0) and the characteristic impedances in vacuum (η_0) as:

To emphasis, the constitutive relations are needed to solve for EM fields quantities using Maxwell’s equations.

For simple homogenous isotropic dielectric and for magnetic material (linear and isotropic media), the constitutive relations are given as

where as equations (10.34) and (10.35) remain the same.

Where, ɛ is the permittivity of the material, and µ is the permeability of the material.

For inhomogeneous media, the constitutive relations are functions of the position.

The permittivity of the material ɛ and the permeability of the material µ can be presented as

where χe is the electric susceptibility of the material which is the measure of the electric polarization property of material (dimensionless scalar), and χm is the magnetic susceptibility of the material which is the measure of the magnetic polarization property of material (dimensionless scalar).

Moreover, the speed of light in the material c and the characteristic impedance of the material η is expressed as

The relative permittivity εr of a material, the relative permeability µr of a material, and the refractive index n of a material are formed as

By using equation (10.41) and (10.42), we get

It is good to know that for nonmagnetic material μr = 1 or μr = μ_0, and

Now, the constitutive relations for time-harmonic fields in a simple media are:

Furthermore, both the electric polarization P (Coulomb/m^2) which describes how the material is polarized when an electric field E is present and the magnetization M (Ampere/m) which describes how the material is magnetized when a magnetic field H can be included in the constitutive relations in any material as

where P = ε_0χeE and M = χm H.

Next, for nonlinear material, the constitutive relationships can be presented as

where Dre is the remanent displacement that is the displacement when the electric field is not present, Bre is the remanent magnetic flux density that is the magnetic flux density when the magnetic field is not present, and Jex is an externally generated current.

It is beneficial to know that the Maxwell’s equations can be expressed in an approach that ensures the contribution of the medium in terms of the fields E and B as

Example 10.1

Given H = Hej(ωt +βz)ax in free space, calculate E.

Solution

10.5 POTENTIAL EQUATIONS

Often under certain circumstances, it can be essential to formulate EM problems in terms of potential functions, that is, the scalar electric potential Ve and vector magnetic potential A. These potential functions are arbitrary and they are required to satisfy Maxwell’s equations. They are described by

In fact, equation (10.55) is a direct consequence of the magnetic Gauss’ law and equation (10.55) is a result from Faraday’s law. In the magnetostatic case (there are no currents present), Ampere’s law reduces to

Indeed, when equation (10.57) holds, we can present the scalar magnetic potential Vm by

It is clear that, equations (10.58) and (10.59) satisfy the Maxwell’s equations (1.2) and (1.4). Now, to relate the potential functions to other two Maxwell’s equations (1.1) and (1.3), by assuming the Lorentz condition hold, that is,

These equations can be written in the case of linear and homogenous medium as

Equations (10.63) and (10.64) as wave equations and the integral solutions to these equations are known as the retarded potential solutions, i.e.,

where R is the distance from the source point to the field point at which the potential is required, and the square brackets [ ] denote that ρv and J are specified at a time earlier than for which Ve or A is being formed.

10.6 BOUNDARY CONDITIONS

The material medium in which an electromagnetic field exists is usually characterized by its constitutive parameters σ, ɛ, and µ. If σ, ɛ, and µ are independent of E and H, the medium is linear. Also, if σ, ε, and µ are dependent of E and H, the medium is nonlinear. Now, if σ , ε, and µ are functions of space variables, the medium is inhomogeneous. But, if σ, ε, and µ are not functions of space variables, the medium is homogeneous. Additionally, if σ, ε, and µ are independent of direction (scalars), the medium is isotropic. If σ, ε, and µ are dependent of direction (vectors), the medium is anisotropic. Indeed, most of substrates used in electronic circuits are homogenous, isotropic, and linear.

The boundary conditions at the interface separating 2 different media 1 and 2, with parameters (ɛ_1, µ_1, σ_1) and (ɛ_2, µ_2, σ_2), respectively, as shown in Figure 10.2.

Figure 10.2. Interface between two media.

The boundary conditions for the EM fields across material boundaries are derived from the integral form of Maxwell’s equations. They are given by

where n is a unit normal vector directed from medium 1 to medium 2, subscript t and n denote tangent and normal components of the fields, respectively, ρs and Js are surface electric charge density (coulomb/m^2) and surface current density (ampere/m), respectively. Furthermore, equations (10.67) and (10.70) state that the tangential components of E and the normal components of B are continuous across the boundary. But, equation (10.68) states that the discontinuity in the normal component D is the same as the surface electric charge density ρs on the boundary. However, equation (1.69) states that the tangential component of H is discontinuous by the surface current density Js on the boundary. In many interface problems, only 2 of Maxwell’s equations are used, equations (10.68) and (10.70), when a medium is source free (J = 0, ρv = 0), since the other 2 boundary conditions are implied. In such a case, the boundaries conditions may be written as

Moreover, Maxwell’s equations under the source free condition are applicable to passive microwave structures such as transmissions lines.

However, when one of the media is a perfect conductor, boundary conditions are different. A perfect conductor has infinite electrical conductivity and thus no internal electric field (full of free charges). Or else, it would produce an infinite current density according to the third constitutive relations. When an EM field is applied to a perfect conductor medium, the free charges which are pushed to the applied EM field, move themselves in such a way that they produce an opposite EM field that completely cancels the applied EM field. Indeed, this causes the creation of the surface charges and currents on the boundary of the perfect conductor. At an interface between a dielectric and a perfect conductor, the boundary conditions for E and D fields are simplified. Now, assume that medium 1 is a perfect conductor, then E_1 = 0 and D_1 = 0. Also, if it is a time-varying case, then H_1 = 0 and B_1 = 0, and, in addition, as a correspondence of Maxwell’s equations. Therefore, the boundary conditions for the fields in the dielectric medium for the time-varying at the surface are

Furthermore, we can apply the integral form of the continuity equation (10.10) to the surface at the interface between lossy media (i.e., σ_1 ≠ 0, σ_2 ≠ 0) or lossy dielectric (i.e., σ_1 ≠ σ_2 and ɛ_1 ≠ ɛ_2), or perfect conductor (i.e., no fields inside the media). Therefore, the interface condition for current density J can be obtained as

Equation (10.79) states that the normal component J is continuous, except where the time-varying surface electric charge density ρs on the boundary may exist.

10.7 LAWS FOR STATIC FIELDS IN UNBOUNDED REGIONS

Coulomb’s law and Biot-Savart’s law are the 2 fundamental laws governing the static fields in unbounded regions.

10.7.1 Coulomb’s Law and Field Intensity

Coulomb’s law is an experimental law that deals with the force a point charge exerts on another point charge. In other words, Coulomb’s law states that the force F (in newtons) between two points charges Q_1 (in coulombs) and Q_2 is

where R (in meter) is the distance between the 2 charges. We can define the electrostatic field intensity E as the force F applied by 1 charge Q on a unit positive point charge as

Knowing that, the point at which the charge Q is located is called the source point, and the point at which the electrostatic field intensity E is taken is called the field point. Thus, here aR is the unite vector in the direction from the source point toward the field point, and R is the distance between the source point and the field point.

Now, it is possible to obtain a continuous charge along a line, on a surface, or in a volume, respectively as

where L is the line along which the charge is distributed, S is the surface which the charge is distributed, v is the volume enclosed by a surface S. ρl, ρs, and ρv, are the line, surface, and volume charge density, respectively.

10.7.2 Bio-Savart’s Law and Field Intensity

The Bio-Savart’s law is a magnetostatic law used to express the static magnetic field as a summation over elementary current sources. Now, we can obtain the Bio-Savart law for the line current, surface current, and volume current, respectively in terms of the distributed current sources as

where I is the line current density, Js is the surface charge density, Jv is the volume charge density, and aR is a unit vector pointing from the differential elements of current to the point of interest. Indeed, the source elements are related as

10.8 ELECTROMAGNETIC ENERGY AND POWER FLOW

The electric energy We is defined as

where D is the magnitude of electric displacement, and T is the period.

The electrostatic energy present in an assembly of charges can be written as

where V is the potential, and Q is the point charge. Now, instead of point charges, the region has a continuous charge distribution, the summation equation (10.90) becomes integrations for line charge, surface charge, and volume charge, respectively as

In the meantime, ρv = ▿.D, E = −▿V, and D = ε_0E, and by using the identity for vector and scalar rules and applying divergence theorem, and knowing that in a simple medium, whose constitutive parameters (µ, ε, and σ) do not change with time, we have

We can obtain electrostatic energy as

Also, the electrostatic energy density we (in J/m^2) can be obtained as

When a wave propagates in a medium, it carries the electric field and power. However, the time derivatives of equation (10.89) is the electric power which is written as

Furthermore, the magnetic energy can be defined as

where B is the magnitude of magnetic flux density, and T is the period.

The magnetostatic energy present in an assembly of currents k can be written as

where Ik is the kth current, and Φk is kth magnetic flux.

Note, knowing that in a simple medium, whose constitutive parameters (µ, ɛ, and σ) do not change with time, we have

We can obtain magnetostatic energy as

Also, the magnetostatic energy density wm (in J/m^2) can be obtained as

When a wave propagates in a medium, it carries the magnetic field and power. However, the time derivatives of equation (10.98) is the magnetic power that is written as

The instantaneous power density vector associated with the electromagnetic field at a given point is known as the Poynting vector Pov (in W/m^2), which is written as

For more practical value than Pov, we determine the time-average instantaneous Poynting vector (or power average density) (in W/m^2) over the period as

In addition, for time-harmonic fields, we can defined a phasor Poynting vector as

Where Hs* is the complex conjugate of Hs. Now, for a phasor Poynting vector, we can define the time-average power which is equivalent of equation (10.106) as

where Re( ) stands for the real part of a complex quantity. Furthermore, the total time-average power crossing a given surface S is given by

The electric and magnetic powers quantities are related through Poynting’s theorem as

where

is called resistive losses which result in heat dissipation in the material.

is called the radiative losses.

However, the Poynting’s theorem as presented in equation (10.109) can be written as

where

is the total power leaving the volume.

is the rate of decrease in energy stored in electric and magnetic fields.

is the decrease in ohmic power density (dissipated).

Indeed, under the material is linear and isotropic, it holds that

Therefore, based on equations (10.111) and (10.112), the equation (10.109) can be written as

Now, by integrating the left-hand side of equation (10.113) is the total electromagnetic energy density wt

10.9 LOSS IN MEDIUM

The electronic circuits have dielectrics that are always not perfect. Thus, there is always loss in any practical nonmagnetic dielectrics that is known as dielectric loss. This dielectric loss is due to a nonzero conductibility of the medium. Now, we can write the time harmonic Maxwell’s equation (10.22), making use the time-harmonic constitutive relations (10.44) and (10.46), as

or

where

Equation (10.117) is called the loss tangent of the medium, which is usually used to characterize the medium’s loss. In addition, now we can define a complex dielectric constant of a lossy medium as

where the real part ε' of the complex dielectric constant is the dielectric property that contributes to the stored electric energy in the medium and it is defined as

and the imaginary part ɛ" contains the finite conductivity and results in loss in the medium which is defined as

For example, the loss tangent for GaAs material is 0.006 at frequency 10Ghz, relative dielectric constant equal to 12.9, and temperature 25°C. Also, the loss tangent for silicon material is 0.004 at frequency 10Ghz, relative dielectric constant equal to 11.9, and temperature 25°C.

10.10 SKIN DEPTH

The measure of the depth to which the electromagnetic wave can penetrate the medium is known as skin depth (or depth of penetration). Skin depth is one of the most important parameters of a medium, because it presents the distance from the medium surface over which the magnitude of the fields of a wave traveling in the medium are reduced to e−1 (or 0.368) of those at the medium’s surface. The skin depth d of a good conductor is approximately written as

where ω = 2πf.

It is essential to know that the skin depth of good conductors is very small, especially at high frequencies. Thus, it results a low conduction loss.

Example 10.2

Calculate the skin depth, δ, for aluminum in 1.6 × 10^6 Hz field (σ = 38.2 × 10^6 S/m and µ = 1).

Solution

10.11 POISSON’S AND LAPLACE’S EQUATIONS

Poisson’s and Laplace’s equations are derived from Gauss’s law (for a linear, isotropic material medium)

and

By substituting equation (10.123) into equation (10.122), we get

for an inhomogeneous medium. Equation (10.124) can be obtained for a homogeneous medium as

Equation (10.125) is known as Poisson’s equation.

Now, Laplace’s equation is a special case of Poisson’s equation when ρv = 0 (i.e., for a charge free region), and it can be described as

Laplace’s equation is used to determine the static or quasi-static characteristic impedance and effective relative dielectric constant of a transmission line.

10.12 WAVE EQUATIONS

We used so far Maxwell’s equations and constitutive relations directly to determine the EM fields. However, it can very convenient to obtain the EM fields by solving wave equations.

When the electromagnetic wave is in a simple (linear, isotropic, and homogenous) nonconducting medium (ɛ, µ, and σ = 0), the homogenous vector wave equations can be presented as

and

On the other hand, the relation between scalar potential V and vector potential A is called the Lorentz condition (or Lorentz gauge) for potentials that is expressed as

The nonhomogenous wave equation for vector potential A is given by

But, the nonhomogenous wave equation for scalar potential V is given by

The time-harmonic wave equations for vector potential A and scalar potential V equations can be obtained, respectively as

and

where

Equation (10.134) is called wave number, and equations (10.132) and (10.133) are known as nonhomogenous Helmholtz’s equations.

However, when the EM wave in a simple, nonconducting source free medium (characterize by ρ = 0, J = 0, σ = 0) and the time-harmonic wave equations can be obtained as

and

Equations (10.135) and (10.136) are known as the homogenous vector Helmholtz’s equations.

10.13 ELECTROMAGNETIC ANALYSIS

Due to the cost effectiveness of experiments and testing, the development of transmission lines in integrated circuit systems is time consuming. Today, researchers, designers, and engineers used several numerical and analytical methods to study and investigate the parameters variations and properties of designing high-speed integrated circuits (microwave circuits) and electromagnetic (EM) problems. The most common analytical methods used for exact solutions in electromagnetic are conformal mapping, integral solutions, separation of variables, and series expansion. Also, the most popular numerical methods used for approximate solutions are methods called moment methods, methods of line, finite difference methods, and finite element methods (FEM).

Finite element method has a great success in electromagnetic analysis compared to other methods. In contrast to other numerical methods, it is very useful for solving problems in complex geometries and inhomogeneous media. In this chapter, we show an overview of the finite element method. FEM requires that any problem involved in the geometrical region to be subdivided into finite number of smaller regions or elements. An approximate solution for the partial differential equation can be developed for each of these elements. In addition, the total solution is generated by assembling together the individual solutions taking care in order to ensure continuity at the interelement boundaries. Basically, there are four steps used in FEM: first, creating and discretizing the solution region (domain) into a finite number of subregions or elements; that is, divide the problem into nodes and elements and assume a shape function to represent the physical behavior of an element; second, developing equations for an element; third, assembling all the elements to represent in solution region, constructing the global coefficient matrix and applying boundary conditions and initial conditions; fourth, solving the system of equations to obtain the important information of the problem.

10.13.1 One-Dimensional Elements

10.13.1.1 The Approach to FEM Standard Steps procedure

The first step is the discretization step, that is, the solution domain is divided into finite elements. Figure 10.3 provides an example of elements employed in one dimension. It shows the points of intersection of the lines that make up the sides of the elements called nodes and the sides themselves are known as nodal lines.

Figure 10.3. Example of elements in one-dimensional (1D).

The second step is the development of equations to approximate the solution for each element. It can be done by choosing an approximate function with unknown coefficients that will be used to approximate the solution. We use a first-order polynomial (straight line) as a linear variation of potential between the nodes over element m, i.e.,

where V(x) is the dependent variable (potential function); a and b are constants; x is the independent variable.

We can find the two constants a and b by using the two nodes to satisfy the equation at the location of the two nodes as:

and

where V_1(m) = V(m) (x_1) and V_2(m) = V(m) (x_2). By using Cramer’s rule, we can solve equations (10.138) and (10.139), i.e.,

Equations (10.140) and (10.141) can be substituted into equation (10.137) to give the approximate (or shape) function V(x) in terms of the interpolation functions, H_1 and H_2 over element m, that is,

where

Indeed, equation (10.142) is a first-order interpolating polynomial. In addition, it provides a means to calculate intermediate values between the given values V_1 and V_2 at the nodes.

The shape function, along with the corresponding interpolation functions, is presented in Figure 10.4. Moreover, the sum of the interpolation functions, a_1 and a_2, that is,

Figure 10.4. (a) a line element, (b) a linear approximation (or shape) function, (c) the corresponding interpolation function α_1(m) (X) for V(m) (X), and (d) the corresponding interpolation function α_2(m) (X) for V(m) (X).

Furthermore, it follows that,

and

Thus,

Now, the integral of V(m) is:

Now, we evaluate the coefficients so that the function approximates the solution in a best approach. The most common methods used for this propose are the variational approaches, the weighted residuals, and the direct approaches. These methods can specify the relationships between the unknowns in equation (10.142) that satisfy the partial differential equation in an optimal approach. The resulting element equations can be expressed in a set of linear equations in matrix form, i.e.,

where

[C(m)] is element property (stiffness) matrix; {Vc(m)} is a column vector of unknowns at the nodes over element m; and {ψc(m)} is a column vector reflecting the effect of any external influences applied at the node over element m.

Third, we assemble all the elements to represent in the solution region. The solutions for closest elements are matched so that the unknown values at their common nodes are equivalent. Therefore, the total sum will be continuous. Then, the assembled system need to be modified for its boundary condition. The system can be expressed as:

where

[Ca(m)] is the assemblage element property (stiffness) matrix; {Vac(m)} is the assemblage column vector of unknowns at the nodes over element m; and {ψac(m)} is the assemblage column vector reflecting the effect of any external influences applied at the node over element m.

Fourth, solving the system of equations (10.151) to obtain the important information of the problem, it can be obtained by LU decomposition technique.

10.13.1.2 Application to Poisson’s Equation in One-Dimension

In this section, we solve the one-dimensional (1D) Poisson’s equation for the potential distribution V (x)

with boundary conditions (BCs) V(a) = v_1, V(b) = v_2.

Using the same essential four steps as in the previous section with FEM, we focus here on the source term and only the major differences.

We will use the variational principle and the weighted residuals method to obtain the solution of one-dimensional (1D) Poisson’s equation.

(1) Variational Approach

The deriving element governing equations step. We look for the potential distribution V(x) that can minimize an energy function F(V) as

Two nodal values of V(x) are required to define uniquely a line variation of V(m) (x) over an element (m). Hence, the linear variation of V(m) (x) can be presented as

where the interpolation functions a_1(x) and a_2(x) are presented as

The resulting element equation (10.154) can be expressed in a set of linear equations in matrix form:

The energy function can be written as

where N is the number of elements with the domain a ≤ x ≤ b. Now, substituting equation (10.156) into (10.153) can give

By minimizing the Fm) (V(m)) with respect to the nodal values of V, we obtain the following equations for an element (m)

and

These equations can be expressed in matrix form as

where

where the elements of {ψc(m)} are the nodal forcing functions. The equations in (10.161) can give the characteristics of the Poisson’s equation in 1D. Indeed, in spite of the type of element we choose to formulate the Poisson’s equation in 1D, the element equations will have the form as equation (10.161). For the solution of the Poisson’s equation in 1D, it is essential to derive the equations for all the elements in the assemblage and then to assemble these algebraic equations.

(2) Weighted Residuals Method

In the variational approach for 1D Poisson equation with boundary condition, we derive the element matrices [C(m)] and {ψc(m)} for a linear variation of potential V (x) over element (m) with two nodes. Now, we will use Galerkin’s method with weighting functions Wi = ai to derive the element matrices. We approximate the unknown exact solution Vm(x) by

where

N is number of nodes (here N = 2), Vmi is the unknown nodal values, i = 1, 2.

Note that, we do not consider the fixed boundary conditions at the element level, but these are included after the assembly process as in the previous method. Now, by applying Galerkin’s method we get:

where

x_1 and x_2 are the coordinates of the end nodes of the line element.

By using integration by parts to the term with the derivatives of Vm (x), that is,

Taking the derivative of equation (2.29) as

where

{Vc(m)} is the column vector of nodal unknowns for the element m.

Thus, equation (2.30) becomes

Furthermore, the first term on the right-hand side of equation (10.169) represents natural boundary conditions for the element m. We obtain these as

because a_1(x_2) = 0, a_1(x_1) = 1,

and

because a 2(x_2) = 1, a 2(x 2) = 0.

We use the end-point values of ai shown in Figure 10.4. Thus, the element equations are presented as

where

The extension to an element with N nodes follows the same steps, but with i = 1, 2,…, N. In addition, the matrices for an element with N nodes contain terms similar to the equation (10.172), but with additional rows and columns to account for N element equations.

10.13.1.3 Natural Coordinates in One-Dimension

We use natural (length) coordinates in deriving interpolation functions that can be used to evaluate the integrals in the element equations. In addition, we use the natural coordinate system in describing the location of a point inside an element in terms of the coordinates associated with the nodes of the element. Let ηi be the natural coordinates, where i = 1, 2,…, N; N is the number of external nodes of the element. Knowing that, natural coordinates are functions of the global Cartesian coordinate system in which the element is defined, the one coordinate is associated with node i and has unit value there.

Figure 10.5 shows a line element with natural coordinates η_1, η_2 and location point xl.

Figure 10.5. Example of two-node line element in one-dimensional (1D) with global coordinate Xl.

The global coordinate x_1 can be expressed as

We can interpret natural (length) coordinates η_1 and η_2 as weighting functions relating the coordinates of the end modes to the coordinate of any interior point. As we know that,

although, the weighting functions are not independent. Let us consider x_1 = x and solving for η_1 and η_2 from equations (10.174) and (10.175), we get

The linear interpolation used for the potential distribution variable V(x) in the previous section, which can be written as

By differential of V(x) using the chain rule, we get

where

Now, taking the integration of length coordinates over the length of an element, that is,

where i and j are integer exponents.

10.13.2 Two-Dimensional Elements

10.13.2.1 Applications of FEM to Electrostatic problems

It’s often known that FEM is a numerical method used to find the approximate solutions either for partial differential equations or integral equations. These equations are most involved in electromagnetic problems. We illustrate the four steps above used to find the solution in FEM through three different types of differential equations, Laplace’s equation, Poisson’s equation, and wave equation.

10.13.2.1.1 Solution of Laplace’s Equation ▿2V = 0 with FEM

To find the potential distribution, V(x, y), for the two-dimensional (2-D) solution region, as shown in Figure 10.6. We illustrate the following steps to get the solution of Laplace equation, ▿2V = 0.

Figure 10.6. The solution region of the problem showing domain for the 2-D boundary value.

First step, using finite element discretization to find the potential distribution for the two-dimensional solution, V(x, y) as shown in Figure 10.7, where the solution region is subdivided into seven nonoverlapping finite elements of triangles. It is always preferable in computation to have the same type of elements through the solution region which in our case is the triangle.

Figure 10.7. The finite element discretization of the solution.

We look for an approximation solution for the potential Vm(x, y) within an element m and then interrelate the potential distribution in various elements such that the potential is continuous across interelement boundaries. We can express the approximation solution for the whole region as

where N is the number of triangle elements into which the solution region is divided.

The most common form of approximation for Vm(x, y) within an element is polynomial approximation for a triangle element, that is,

where the constants a, b and c are to be determined. The potential Vm(x, y) in general is nonzero within element m, but zero outside m. Furthermore, our assumption of linear variation of potential within the triangle element as in equation (2.46) is the same as assuming that the electric field is uniform within the element, that is to say,

Second step, developing equations for the element. Let us choose a typical triangle element shown in Figure 10.8.

Figure 10.8. Typical triangle element; local node numbering 1−2−3 must proceed counter-clockwise as indicated by the arrow.

The potential Vm1(x_1, y_1), Vm2((x_2, y_2), and Vm3(x_3, y_3) at nodes 1, 2, and 3, respectively, are obtained using equation (10.183), namely

The coefficients a, b, and c are determined from equation (10.185) as

Therefore, equation (10.183) can be rewritten by substituting for a, b, and c, i.e.,

Equation (10.187) can be written as

where ai (x, y) is given by

and ai, bi, and ci are given by

where i, j, and k are cyclical, that is, (i = 1, j = 2, k = 3), (i = 2, j = 3, k = 1), and (i = 3, j = 1, k = 2).

Note that, by substituting equations (10.190), (10.191), and (10.192) into equation (10.187) gives

Also, using equations (10.190), (10.191), and (10.192) into equation (10.188), gives

and A is given by

or

where A is the area of the element m.

The value of A is positive if the nodes are numbered counterclockwise, starting from any nodes, as shown by the arrow in Figure 10.5.

Furthermore, equation (10.188) gives the potential at any point (x, y) within the element provided that the potentials at the vertices are known. In addition, ai(x, y) are linear interpolation functions. They are called the element shape functions and they have the following properties:

The shape of functions a_1(x, y), a_2(x, y), and a_3(x, y), for example, are illustrated in Figure 10.9.

Figure 10.9. Shape functions α_1(x, y), α_2(x, y), and α_3(x, y) for a triangle element.

The functional, Wm, corresponding to Laplace’s equation, which physically is the energy per unit length associated with element m, is given by

But from equation (10.188),

By substituting equation (10.198) into equation (10.197), gives

If we define the term in brackets as

now, we can write equation (10.199) in matrix form as

where the superscript t denotes the transpose of the matrix,

and

The matrix [C(m)] is usually called the element coefficient matrix (or stiffness matrix in structural analysis). The element of the coefficient matrix may be regarded as the coupling between nodes i and j; its value is obtained from equations (10.194) and (10.200). For instance,

Similarly,

Additionally,

Now, for the third step, after having considered a typical element, the next step is to assemble all such elements in the solution region. The energy associated with the assemblage of elements assuming that the whole solution region is homogeneous so that ε is constant, i.e.,

where

where

n is the number of nodes, N is the number of elements, and [C] is called the overall or global coefficient matrix, which is the assemblage of individual element coefficient matrices.

For an inhomogeneous solution region such as that shown in Figure 10.10, the region is discretized with triangle elements such that each finite element is homogeneous. In this case, equation (10.197) still holds, however, equation (10.205) does not apply since ε (ε = εr ε_0) or simply εr varies from element to element. To apply equation (10.205), we need to replace ε by ε_0 and multiply the integrand in equation (10.200) by εr.

Figure 10.10. Discretization of an inhomogeneous solution region with triangle elements.

We use an example to illustrate the process by which individual element coefficient matrices are assembled to obtain the global coefficient matrix. In this example, we consider the finite element mesh consisting of three finite elements as shown in Figure 10.11. Observe the numberings of the mesh.

Figure 10.11. Assembly of three elements; i − j − k corresponding to local numbering (1-2-3) of the element in Figure 10.5.

The numbering of nodes 1, 2, 3, 4, and 5 is called global numbering. The numbering i − j − k is called local numbering, and it corresponds with 1-2-3 of the element in Figure 10.5, the local numbering must be in counterclockwise sequence starting from any node of the element. For element 1, we could choose 2-1-4 instead of 1-4-2 to correspond with 1-2-3 of the element to Figure 10.8. Thus, the numbering in Figure 10.8 is not unique. But whichever numbering is used, the global coefficient matrix remains the same. Assuming the particular numbering in Figure 10.11, the global coefficient matrix is expected to have the form

which is a 5 × 5 matrix since five nodes (n = 5) are involved. As we know, Cij is the coupling between nodes i and j. The Cij can be obtained by using the fact that the potential distribution must be continuous across interelement boundaries. The contribution to the i, j position in [C] comes from all elements containing nodes i and j. For instance, in Figure 10.11, elements 1 and 2 have node 1 in common; therefore

Node 2 belongs to element 1 only; therefore

Node 4 belongs to element 1, 2, and 3; accordingly

Nodes 1 and 4 belong simultaneously to element 1 and 2; as a result

Since there is no coupling (or direct link) between nodes 2 and 3; hence

By continuing in this approach, we can obtain all the terms in the global coefficient matrix by inspection of Figure 10.11 as

Note that element coefficient matrices overlap at nodes shared by elements and that are 27 terms (9 for each of the 3 elements) in the global coefficient matrix [C]. Also note the following properties of the matrix [C]:

1.  It is symmetric (Cij = Cji) just as the element coefficient matrix.

2.  Since Cij = 0 if no coupling exists between nodes i and j, it is expected that for a large number of elements [C] becomes sparse. Matrix [C] is also banded if the nodes are carefully numbered. It can be shown using equation (10.203), i. e.,

3.  It is singular. Although this is not obvious, it can be shown using the finite element coefficient matrix of equation (10.202c).

Finally, fourth step, by solving the resulting equations. It can be shown that, from variational calculus, it is known that Laplace’s (or Poisson’s) equation is satisfied when the total energy in the solution region is minimum. Therefore, we require that the partial derivatives of W with respect to each nodal value of the potential be zero; that is,

or

For instance, to get = 0 for the finite element mesh of Figure 2. 9, we substitute equation (10.207) into equation (10.205) and take the partial derivative of W with respect to V_1. We obtain

or

In general case,

where n is the number of nodes in the mesh. By writing equation (10.213) for all nodes k = 1, 2, 3,.., n, we obtain a set of simultaneous equations from which the solution of the transpose matrix for the potential distribution, [V]t = [V_1 V_2…Vn], can be found. This can be done in two ways:

(1) Iteration Method

Suppose node 1 in Figure 10.11, for example, is a free node. The potential at node 1 can be obtained from equation (10.212) as

Thus, in general case, the potential at a free node k in a mesh with n nodes is obtained from equation (10.213) as

Since Cki = 0 is not directly connected to node i, only nodes that are directly linked to node k contribute to Vk in equation (10.215). Note equation (10.215) can be applied iteratively to all the free nodes. The iteration process begins by setting the potentials of fixed nodes (where the potentials are prescribed or known) to their prescribed values and the potentials at the free nodes (where the potentials are known) equal to zero or to the average potential

where Vmin and Vmax are the minimum and maximum values of the prescribed potentials at the fixed nodes, V, respectively. With these initial values, the potentials at the free nodes are calculated using equation (10.215). At the end of the first iteration, when the new values have been calculated for all the free nodes, they become the old values for the second iteration. Indeed, the procedure is repeated until the change between subsequent iteration is negligible enough.

(2) Band Matrix Method

If all free nodes are numbered first and the fixed nodes last, equation (10.205) can be written such that

where subscripts f and p, refer to nodes with free and fixed (or prescribed) potentials, respectively. Since Vp is constant (it consists of known, fixed values), we only differentiate with respect to Vf so that applying equations (10.211) to (10.217) which yields to

or

This equation can be written as

or

where [V] = [Vf],[A] = [Cff], [B] = −[Cfp][Vp]. Since, in general, nonsingular, the potential at the free nodes can be found using equation (10.219). Note we can solve for [V] in equation (10.219a) using Gaussian elimination technique. Also, we can solve for [V] in equation (10.219b) using matrix inversion if the size of the matrix to be inverted is not large.

In fact, it is sometimes necessary to impose the Neumann condition as a boundary condition or at the line of symmetry when we take advantage of the symmetry of the problem. Indeed, suppose that for concreteness, a solution region is symmetric along the y-axis as in Figure 10.12. We impose condition along the y-axis by making

Figure 10.12. A solution region that is symmetric along the y—axis.

With this in mind that as from equation (10.197) onward; the solution has been restricted to a two-dimensional problem involving Laplace’s equation, Δ^2 V = 0.

10.13.2.1.2 Solution of Passion’s Equation with FEM

In this section, we solve the two-dimensional (2D) Poisson’s equations

using the same essential four steps as in previous section with FEM, we focus here on the source term and only the major differences.

The deriving element governing equations step. We divide the solution region into triangles, and then we approximate the potential distribution Vm (x, y) and the source term ρvm over each triangle element by linear combinations of the local interpolation polynomial ai, namely,

where

Vmi is the values of V at vertex i of element m; ρmi is the values of ρv at vertex i of element m. The values of ρmi are known since ρv (x, y) is prescribed, while the values of Vmi are to be determined.

An energy functional which associated Euler equation with equation (10.221) is

where

F(Vm) is the total energy per length within element is the energy density in the electrostatic system and it is equal to is the work done in moving the charge ρvmdS to its location at potential Vm.

Now, by substituting equations (10.222) and (10.223) into equation (10.224) we get

Equation (10.225) can be applied to every element in the solution region.

Also, it can be written in matrix form as

where

we know that equation (10.226) is already defined in equation (10.203) and

Also, can be written as

where A is the area of the triangle element.

We can obtain the discredited functional for the whole solution region, with N elements and n nodes, as the sum of the functional for the individual elements, that is, from equation (10.229),

where

t is the transposition symbol. In equation (10.230), the column matrix [V] consists of the values of Vmi, while the column matrix [ρ] contains n values of the source function, ρv, at the nodes. The functional in equation (10.230) is now minimized by differentiating with respect toVmi and setting the result equal to zero.

Now, we work on the solving the resulting equations step. We can solve the resulting equations by using either the iteration method or the band matrix method.

(1) Iteration Methods

By considering the solution region in Figure 10.8 which has five nodes, n = 5 and from the equation (10.230), we can get the energy functional as

The energy can be minimized by applying

For example, from equation (10.231), we get

or

Therefore, in general, for a mesh with n nodes

where

node k is assumed to be a free node.

By fixing the potential at the prescribed nodes and setting the potential at the free nodes initially equal to zero, we apply equation (10.234) iteratively to all free nodes until convergence is reached.

(2) Band Matrix Method

In this method, we let the free nodes be numbered first and the prescribed nodes last. In doing this, equation (10.230) can be written as

where

subscript f is the free node; subscript ρ is the prescribed node;

ρf is the submatrix containing the values of p at free node; ρp is the submatrix containing the values of ρ at the prescribed node.

Minimizing F(V) with respect to Vf, namely,

gives

or

Indeed, equation (10.236) can be written

where

[A] = [Cff], [V] = [Vf], and [B] is the right-hand side of equation (10.236). Equation (10.237) can be solved to determine [V] either by matrix inversion or Gaussian elimination technique.

10.13.2.1.3 Solution of Wave’s Equation ▿2Φ + k^2Φ = g with FEM

A typical wave equation is the inhomogeneous scalar Helmholtz’s equation

where

Φ is the potential (for waveguide problem, Φ = Hz for TE mode or Ez for TM mode) to be determined, g is the source function, and is the wave number of the medium. The following three distinct special cases of equation (10.238) should be noted:

1.  k = 0 and g = 0; Laplace’s equation;

2.  k = 0; Poisson’s equation; and

3.  k is an unknown, g = 0; homogeneous, scalar Helmholtz’s equation.

It is known that the variational solution to the operator equation

is obtained by examining the functional

where L is an operator (differential, integral, or integro-differential), g is the unknown excitation or source, and Φ is the unknown function to be determined (here is the potential).

Therefore, the solution of equation (10.238) is equivalent to satisfying the boundary conditions and minimizing the functional

Note that, if other than the natural boundary conditions (i. e., Dirichlet of homogenous Neumann conditions) must be satisfied, appropriate terms must be added to the functional. The potential Φ and the source function g can be expressed now in terms of the shape functions ai over a triangle element as

where

Φmi is the value of Φ at the nodal point i of element m.

And

gmi is the value of g at the nodal point i of element m.

Substituting equations (10.242) and (10.243) into equation (10.241) gives

where

are defined in equations (10.158) and (10.185), respectively.

The equation (10.244) is for a single element, but it can be applied for all N elements in the solution region. Therefore,

From equations (10.244) and (10.245), I(Φ) can be expressed in matrix form as

where

[C], and [T] are global matrices consisting of local matrices [C(m)] and [T(m)], respectively.

Now, if free nodes are numbered first and the prescribed nodes last, and considering the source function g = 0, we can write equation (2.109) as

By setting , gives

For TM modes, Φp = 0 and hence

Premultiplying equation (10.250) by , gives

By letting

and U is a unit matrix,

we can obtain the standard eigenvalue problem

any standard procedure may be used to obtain some or all of the eigenvalues ß_1, ß_2,…, ßnf and eigenvectors X_1, X_2,... Xnf, where nf is the number of free nodes. The eigen values are always real since C and T are symmetric.

The solution of the algebraic eigenvalue problems in equation (10.252) furnishes eigenvalues and eigenvectors, which form good approximations to the eigenvalues and eigenfunctions of the Helmholtz problem, i. e., the cutoff wavelengths and field distribution patterns of the various modes possible in a given waveguide.

The solution of the problem of equation (10.238) is summarized in equation (10.251), and can be viewed as the finite element solution of homogeneous waveguides. The idea can be extended to handle inhomogeneous waveguide problems. However, in applying FEM to inhomogeneous problems, a serious difficulty is the appearance of spurious, nonphysical solutions. There are several techniques that have been proposed to overcome the difficulty.

10.14 AUTOMATIC MESH GENERATION

It is a fact that, one of the major difficulties encountered in the finite element analysis of continuum problems is the tedious and time-consuming effort required in data preparation. Indeed, efficient finite element programs must have node and element generating schemes, referred to collectively as mesh generators. Automatic mesh generation minimizes the input data required to specify a problem. In fact, it not only reduces the time involved in data preparation, it eliminates human errors that are introduced when data preparation is preformed manually. Furthermore, combining the automatic mesh generation program with computer graphics is particularly valuable since the output can be monitored visually.

A number of mesh generation algorithms of varying degrees of automation have been proposed. In this section, we focus on two types of domains, rectangular domains and arbitrary domains.

10.14.1 Rectangular Domains

Since some applications of FEM to EM problems involve simple rectangular domains, we consider the generation of simple meshes. Now, let us consider a rectangular solution region of a size a × b as shown in Figure 10.13. The goal here is to divide the region into rectangular elements, each of which is later divided into two triangular elements.

Figure 10.13. Discretization of a rectangular region into a nonuniform mesh.

Supposen nx and ny are the number of divisions in x and y directions, the total number of elements and nodes are, respectively, given by

As a result, it is easy to figure out from Figure 10.13 a systematic way of numbering the elements and nodes. Indeed, to obtain the global coordinates (x, y) for each node, we need an array containing Δxi, i = 1, 2,…,nx and Δyj, j = 1, 2,…, ny, which are, respectively, the distances between nodes in the x and y directions. If the order of node numbering is from left to right along horizontal rows and from bottom to top along the vertical rows, then the first node is the origin (0, 0). The next node is obtained as x → x + Δx_1 while y = 0 remains unchanged. The following node x → x + Δx_2, y = 0, and so on until Δxi are exhausted. We start the second next horizontal row by starting with x = 0, y → y + Δy_1 and increasing x until Δxi are exhausted. We repeat the process until the last node (nx + 1) (ny + 1) is reached, i. e., when Δxi, and Δyi are exhausted simultaneously.

The procedure presented here allows for generating uniform and nonuniform meshes. A mesh is uniform if all Δxiare equal and all Δyi are equal; it is nonuniform otherwise. A nonuniform mesh is preferred if it is known in advance that the parameter of interest varies rapidly in some parts of the solution domain. This allows a concentration of relatively small elements in the regions where the parameter changes rapidly, particularly since these regions are often of greatest interest in the solution. Additionally, without the preknowledge of the rapid change in the unknown parameter, a uniform mesh can be used. In that case, we set

where

hx = a/nx and hy = a/ny.

In some cases, we also need a list of prescribed nodes. If we assume that all boundary points have prescribed potentials, the number np of prescribed nodes is given by

A simple way to obtain the list of boundary points is to enumerate points on the bottom, right, top, and left of the rectangular region in that order.

10.14.2 Arbitrary Domains

The basic steps involved in a mesh generation are as follows:

A.  subdivide solution region into few quadrilateral blocks,

B.  separately subdivide each block into elements,

C.  connect individual blocks.

A. Definition of Blocks

The solution region is subdivided into quadrilateral blocks. Subdomains with different constitutive parameters (σ, μ, ε) must be represented by separate blocks. As input data, we specify block topologies and the coordinates at eight points describing each block. Each block is represented by an eight-node quadratic isoparametric element. With natural coordinate system (ζ, η), the x and y coordinates are represented as

where ai(ζ, η) is a shape function associated with node i, and (xi, yi) are the coordinates of node i defining the boundary of the quadrilateral block as shown in Figure 10.14.

Figure 10.14. Typical quadrilateral block.

The shape functions are expressed in terms of the quadratic or parabolic isoparametric elements shown in Figure 10.15.

Figure 10.15. Eight-node serendipity element.

They are given by:

For corner nodes,

For midside nodes, note the following properties of the shape functions:

1.  They satisfy the conditions.

2. They become quadratic along element edges(ζ = ±1,η = ±1).

B. Subdivision of Each Block

Furthermore, for each block, we specify N DIV X and N DIV Y, the number of element subdivisions to be made in the ζ and η directions, respectively. In addition, we specify the weighting factors (Wζ)i and (Wη)i allowing for graded mesh within a block. It is essential to know that, in specifying N DIV X, N DIV Y, (Wζ)i and (Wη)i care must be taken to ensure that the subdivision along block interfaces (for adjacent blocks) are compatible. We initialize ζ and η to a value of −1 so that the natural coordinates are incremented according to

where

and

Now, there are three elements types permitted: (1) linear four-node quadrilateral elements, (2) linear three-node triangle elements, and (3) quadratic eight-node isoparametric elements.

C. Connection of Individual Blocks

After subdividing each block and numbering its nodal points separately, it is necessary to connect the blocks and have each node numbered uniquely. This is accomplished by comparing the coordinates of all nodal points and assigning the same number to all nodes having identical coordinates. In other words, we compare the coordinates of node 1 with all other nodes, and then node 2 with other nodes, etc., until all repeated nodes are eliminated.

10.15 HIGHER ORDER ELEMENTS

Finite elements use higher order elements. The shape function or interpolation polynomial of the order two or more is called higher order element. To emphasize, the accuracy of a finite element solution can be improved by using finer mesh or using higher order elements or both. Desai and Abel studied mesh refinement versus higher order elements in [44]. Generally, fewer higher order elements are needed to achieve the same degree of accuracy in the final results. Moreover, the higher order elements are particularly useful when the gradient of the field variable is expected to vary rapidly.

10.15.1 pascal Triangle

High order triangular elements can be systematically developed with the aid of the so-called Pascal triangle given in Figure 10.16. The family of finite elements generated in this matter with distribution of nodes illustrated in Figure 10.17. Note that in higher order elements, some secondary (side and/or interior) nodes are introduced in addition to the primary (corner) nodes so as to produce exactly the right number of nodes required to define the shape function of that order.

Figure 10.16. The pascal triangle (2D).

Figure 10.17. pascal triangle (2D) and the associated polynomial basis functions degree n = 1 to 3.

Indeed, the Pascal triangle contains terms of the basic functions of various degrees in variable x and y. An arbitrary function Φi (x, y) can be approximated in an element in terms of a complete n th order polynomial as

where

r is the number of terms in complete polynomials (also the number of nodes in the triangle). For example, for the third order (n = 3) or cubic (ten-node) triangular elements,

Equation (10.266) has ten coefficients, and hence the element must have ten nodes. It is also complete through the third order terms. A systematic derivation of the interpolation function a for the higher order elements involves the use of the local coordinates.

10.15.2 Local Coordinates

Now, the triangular local coordinates (η_1, η_2 η_3) are related to Cartesian coordinates (x, y) as

The local coordinates are dimensionless with values ranging from 0 to 1. Furthermore, by definition, ηi at any point within the triangle is the ratio of the perpendicular distance from the point to the side opposite to vertex i to the length of the altitude drawn from vertex i. Therefore, from Figure 10.18 the value of ηi at P, for example, is given by the ratio of the perpendicular distance d from the side opposite vertex l to the altitude h of that side, namely,

Alternatively, from Figure 10.15, ηi at P can be defined as

so that

Since A_1 +A_2 + A_3 = A. The local coordinates ηi in equation (10.269) are also called area coordinates. The variation of(η_1, η_2, η_3) inside an element is shown in Figure 10.19.

Figure 10.18. Definition of local coordinates.

Figure 10.19. Variation of local coordinates.

Although the coordinates η_1, η_2, and η_3 are used to define a point P, only two are independent since they must satisfy equation (10.270). The inverted form of equations (10.267) and (10.268) is

where

and (i, j, k) is an even permutation of (1, 2, 3). The differentiation and integration in local coordinates are carried out using [47]:

10.15.3 Shape Functions

Now, we may express the shape function for higher order elements in terms of local coordinates. Indeed, sometimes, it is convenient to label each point in the finite elements in Figure 10.17 with three integers, i, j, and k from which its local coordinates (η_1, η_2, η_3) can be found or vice versa. For instance, at each point Pijk

Thus, if a value of Φ, say Φijk, is prescribed at each point Pijk, equation (10.264) can be expressed as

where

and e ∈ (i, j, k). Further, pe(η) may also be written as

where p_0(η) = 1.

The relationships between the subscript q ∈ { 1, 2, 3 } on ηq, l∈ {1, 2,.., r} on a_1, and e ∈ {i, j, k} on pe and Pijk in equations (10.276) to (10.278) are illustrated in Figure 10.20 for n ranging from 1 to 3. Furthermore, point Pijk will be written as Pn for conciseness.

Figure 10.20. Distribution of nodes over triangles for n = 1 to 3. The triangles are in standard position.

Notice from equations (10.277) or (10.278) that

Indeed, by substituting equation (10.279) into equation (10.276) gives the shape functions al for nodes l = 1, 2,…, r, as shown in Table 10.1 for n = 1 to 3. In addition, observe that each al takes the value of 1 at node l and value 0 at all other nodes in the triangle. It can be verified by using equation (10.274) in conjunction with Figure 10.20.

Table 10.1. polynomial Basic Functions αl (η_1, η_2, η_3) for First, Second, and Third

10.15.4 Fundamental Matrices

The fundamental matrices [T] and [Q] for triangle elements can be derived using the shape functions in Table 10.1. The matrix T is defined as

From Table 10.1, we substitute al in equation (10.280) and apply equations (10.273f) and (10.273g) to obtain elements of T. For example, for n = 1,

Furthermore, when i ≠ j, Tij can be written as

but, when i = j,

Thus,

Now, by following the same procedure, higher order T matrices can be obtained. The T matrices for orders up to n = 3 are tabulated in Table 10.2 where the factor A, the area of the element, has been expressed. The actual matrix elements are obtained from Table 2.2 by multiplying the tabulated numbers by A and dividing by the indicated common denominator. Indeed, the following properties of the T matrix are worth knowing:

1.  T is symmetric with positive elements;

2.  elements of T all add up to the area of the triangle, that is, since by definition at any point within the element;

3.  elements for which the two triple subscripts from similar permutations are equal, that is, Tijk,peq = Tikj,peq = Tkij,epq = Tkij,epq = Tjki,qep = Tijk,qpe; this should be obvious from equations (2.280) and (10.276).

As a result, the above properties are not only useful in checking the matrix; they have proved useful in saving computer time and storage.

Table 10.2. Table of T Matrices for n = 1 to 3

In equation (10.227), elements of [C] matrix are defined by

By applying equations (10.273a) to (10.273d) to equation (2.147), it can be shown that:

or

where θq is the include angle of vertex q ∈ {1, 2, 3} of the triangle and

It is clear that matrix C depends on the triangle shape, whereas the matrices Q(q) do not. The Q(1) matrices for n = 1 to 3 are tabulated in Table 10.3.

The following properties of Q matrices should be noted as:

1.  they are symmetric;

2.  the row and column sums of any Q matrix are zero, that is, so that the C matrix is singular.

Q(2) and Q(3) are easily obtained from Q(1) by row and column permutations so that the matrix C for any triangular element is constructed easily if Q(1) is known.

Table 10.3. Table of Q Matrices for n = 1 to 3

For example, for n = 1, the rotation matrix is basically derived from Figure 10.21 as

where Rij = 1 node i is replaced by node j after one counterclockwise rotation, or Rij = 0 otherwise.

Figure 10.21. One counterclockwise rotation of the triangle in (a) gives the triangle in (b).

Moreover, Table 10.4 presents the R matrices for n = 1 to 3. Note that each row or column of R has only one nonzero element since R is essentially a unit matrix with rearranged elements.

Table 10.4. Table of R Matrices for n = 1 to 3

Now, once the R is known, we can obtain Q(2) and Q(3) as

where Rt is the transpose of R.

10.16 THREE-DIMENSIONAL ELEMENT

In this section, we will discuss the finite element analysis of Helmholtz’s equation in three dimensions, i. e.,

First, we divide the solution region into tetrahedral or hexahedral (rectangular prism) elements as in Figure 10.22.

Figure 10.22. Three-dimensional elements: (a) Four-node or linear-order tetrahedral, (b) eight-node or linear-order hexahedral.

Now, assuming a four-node tetrahedral element, the function Φ is represented within element by

The same applies to the function g. Since equation (10.290) must be satisfied at the four nodes of the tetrahedral elements,

Therefore, we have four simultaneous equations with the potentials Vm1, Vm2, Vm3, and Vm4 at nodes 1, 2, 3, and 4, respectively, i. e.,

The coefficients a, b, c, and d are determined from equation (10.291)

The determinant of the system of equations is

where v is the volume of the tetrahedron. By finding a, b, c, and d, we can express Φm as

where

Indeed, for higher order approximation, the matrices for as become large in size and we resort to local coordinates. The existence of integration equations for local coordinates can simplify the evaluation of the fundamental matrices T and Q.

Now, for the tetrahedral element, the local coordinates are η_1, η_2, η_3, and η_4, each perpendicular to a side. They are defined at a given point as the ratio of the distance from that point to the appropriate apex to the perpendicular distance from the side to the opposite apex. In addition, they can be interpreted as volume ratios, that is, at a point P

where vi is the volume bound by P and face i. It is evident that

Note that, the following properties are useful in evaluating integration involving local coordinates:

In terms of the local coordinates, an arbitrary function Φ (x, y) can be approximated within an element in terms of a complete nth order polynomial as

where is the number of nodes in the tetrahedron or number of terms in the polynomial. The terms in a complete three-dimensional polynomial may be arrayed for polynomial basic functions degree n = 1 to 3 as shown in Figure 10.23.

Figure 10.23. pascal tetrahedral (3D) and the associated polynomial basic functions degree n = 1 to 3.

Each point in the tetrahedral element is represented by four integers, i, j, k, and l which can be used to determine the local coordinates (η_1, η_2, η_3, η_4). That is at point Pijkl,

Thus, at each node,

where q = 1, 2,…, r and pe is defined in equation (10.277) or (10.278). Figure 10.22 illustrates the relationship between the node numbers q and ijkl for the second order tetrahedron (n = 2). The shape functions obtained by substituting equation (10.277) into (10.293) are presented in Table 2.5 for n = 3.

Figure 10.24. Numbering scheme for second-order tetrahedral.

Table 10.5. Polynomial Basic Functions aq (η_1, η_2, η_3) for n = 1 to 3

The fundamental matrices [T] and [Q] are involved triple integration. For Helmholtz equation, for example, equation (10.250) applies, namely,

except that

10.17 FINITE ELEMENT METHODS FOR EXTERNAL PROBLEMS

We can apply the finite element to exterior or unbounded problems such as open-type transmission lines (e. g., microstrip). They pose certain difficulties. In this section, we will consider three common approaches: first, the infinite element method; second, the boundary element method; and third, the absorbing boundary condition.

10.17.1 Infinite Element Method

Let us consider the solution region shown in Figure 10.25. We can divide the entire domain into a near field region, which is bounded, and a far field region, which is unbounded. The near field region is divided into finite triangular elements as usual, while the far field region is divided into infinite elements. Knowing that, each infinite element shares two nodes with a finite element. We mainly will be focusing on the infinite elements.

Figure 10.25. Division of solution region into finite and infinite elements.

Now, consider the infinite element in Figure 10.26 with nodes 1 and 2 and radial sides intersecting at point (x_0, y_0).

Figure 10.26. Typical infinite element.

We can relate triangular polar coordinates(ρ, η) to the global Cartesian coordinates (x, y) as:

where 1 ≤ ρ < ∞, 0≤ η ≤ 1. The potential distribution within the element is approximated by a linear variation as

or

where V_1 and V_2 are potentials at nodes 1 and 2 of the infinite elements, a_1 and a_2 are the interpolation or shape functions, that is,

Moreover, the infinite element is compatible with the ordinary first order finite element and satisfies the boundary condition at infinity. Indeed, with the shape functions in equation (10.308), we can obtain the [C(m)] and [T(m)] matrices. We obtain solution for the exterior problem by using a standard finite element program with the [C(m)] and [T(m)] matrices of the infinite elements added to the [C] and [T] matrices of the near field region.

10.17.2 Boundary Element Method

The boundary element method is a finite element approach for handling exterior problems. It involves obtaining the integral equation formulation of the boundary value problem, and solving this by a discretization procedure similar to that used in regular finite element analysis. But, since the boundary element method is based on the boundary integral equivalent to the governing differential equation, only the surface of the problem domain needs to be modeled. Moreover, for the dimension of 2D problems, the boundary elements are taken to be straight line segments, whereas for 3D problems, they are taken as triangular elements.

10.17.3 Absorbing Boundary Conditions

To apply the finite element approach to open region problems, an artificial boundary is introduced in order to bound the region and limit the number of unknowns to a manageable size. It can be expected that, as the boundary approaches infinity, the approximate solution tends to the exact one. However, the closer the boundary to the modeled object, the less computer memory is required. To avoid the error caused by this truncation, an absorbing boundary condition (ABC) can be imposed on the artificial boundary S, as typically portrayed in Figure 10.27.

Figure 10.27. An object surrounded by an absorbing boundary.

Indeed, the ABC minimizes the nonphysical reflections from the boundary. The major challenge of these ABCs is to bring the truncation boundary as close as possible to the object without sacrificing accuracy and to absorb the outgoing waves with little or no reflection.

Furthermore, the sequence of BGT operators can be obtained by the recursion relation, i. e.,

Now, since Φ satisfies the higher-order radiation condition

By imposing the mth-order boundary condition

will compel the solution Φ to match the first 2m terms of the expansion in equation (10.309). Equation (10.312) along with other appropriate equations is solved for Φ using the FEM.

10.18 MODELING AND SIMuLATION OF SHIELDED MICROSTRIP LINES WITH COMSOL MULTIPHYSICS

In today’s fast-paced research and development culture, simulation power gives you the competitive edge. COMSOL Multiphysics delivers the ideal tool to build simulations that accurately replicate the important characteristics of your designs. Its unparalleled ability to include all relevant physical effects that exist in the real world is known as multiphysics. This approach delivers results—tangible results that save precious development time and spark innovation. COMSOL Multiphysics brings you this remarkable technology in an easy-to-use, intuitive interface, making it accessible to all engineers including designers, analysts, and researchers.

Today, electromagnetic propagation on multiple parallel transmission lines has been a very attractive area in computational electromagnetics. Multiple parallel transmission lines have been successfully applied and used by designers in compact packaging, semiconductor device, high speed interconnecting buses, monolithic integrated circuits, and other applications. Microstrip lines are the most commonly used in all planar circuits despite the frequencies ranges of the applied signals. Microstrip lines are the most commonly used transmission lines at high frequencies. Quasi-static analysis of microstrip lines involves evaluating them as parallel plates transmission lines, supporting a pure “TEM” mode. Development in microwave circuits using rectangular coaxial lines as a transmission medium has been improving over the past decades. Reid and Webster used rectangular coaxial transmission lines to fabricate a 60 GHz branch line coupler. The finite difference time domain method has been used for analyzing a satellite beamforming network consisting of rectangular coaxial line.

Advances in microwave solid-state devices have stimulated interest in the integration of microwave circuits. Today, microstrip transmission lines have attracted great attention and interest in microwave integrated circuit applications. This creates the need for accurate modeling and simulation of microstrip transmission lines. Due to the difficulties associated with analytical methods for calculating the capacitance of shielded microstrip transmission lines, other methods have been applied. Such methods include finite difference technique, extrapolation, point-matching method, boundary element method, spectral-space domain method, finite element method, conformal mapping method, transverse modal analysis, and mode-matching method.

In this book, we consider systems of rectangular coaxial lines as well as singlestrip, double-strip, three-strip, six-strip, and eight-strip (multiconductor) shielded microstrip lines. Using COMSOL, a finite element package, we performed the simulation of these systems of microstrip lines. We compared the results with other methods and found them to be in good agreement.

The rectangular coaxial line consists of a two-conductor transmission system along which TEM wave propagates. The characteristic impedance of such a lossless line is given by

where

Z = characteristic impedance of the line

L = inductance per unit length of the line

C = capacitance per unit length of the line

c = 3 × 10^8 m/s (the speed of light in vacuum).

As shown in Figure 10.28, a rectangular coaxial line consists of inner and outer rectangular conductors with a dielectric material separating them.

Figure 10.28. Cross-section of the rectangular coaxial line.

Using COMSOL for each type of the rectangular lines involves taking the following steps:

1.  Develop the geometry of the inner and outer conductors, such as shown in Figure 10.28(a).

Figure 10.28(a). geometry of the rectangular coaxial line model.

2.  Select both conductors/rectangle and take the difference.

3.  We select the relative permittivity as 1 for the difference in Step 2. For the boundary, we select the outer conductor as ground and inner conductor as port.

4.  We generate the finite element mesh as in Figure 10.29.

Figure 10.29. Mesh of the rectangular coaxial line.

5.  We solve the model and obtain the potential shown in Figure 10.30.

Figure 10.30. 2D for the potential distribution of the rectangular coaxial line.

6.  As post-processing, we select Point Evaluation and choose capacitance element 11 to find the capacitance per unit length of the line.

We now consider the following three models.

10.18. 1 Rectangular Cross-Section Transmission Line

For COMSOL, we use the following values.

Dielectric material:

ɛr = 1, μr = 1, σ = 0 S/m (air)

Conducting material:

ɛr = 1, μr = 1, σ = 5.8 × 10^7 S/m (copper)

where

ε_0 = permittivity of free space

εr = dielectric constant

μr = relative permeability

μ_0 = permeability of free space = 4 π × 10−7 = 1. 257 × 10−6H/m

σ = conductivity of the conductor

a = width of the inner conductor = 1 mm

b = height of the inner conductor = 0. 8 mm

A = width of the outer conductor = 2. 2 mm

B = height of the outer conductor = 2 mm

From the COMSOL model, we obtained the capacitance per unit length (based on the dimensions given above) as 72.94 pF/m. Using the finite difference (FD) method, we obtained the capacitance per unit length of the line as 71.51 pF/m. Table 10.6 shows the comparison of the characteristic impedance using equation (10.313) of several models. It is evident from the table that the results are very close.

Table 10.6. Comparison of Characteristic Impedance Values of Rectangular Coaxial Line

Name

Z_0

Zheng

45.789

Chen

45.759

Costamagna and Fanni

45. 767

Lau

45.778

Finite difference (FD)

46.612

COMSOL

45.70

10.18.2 Square Cross-Section Transmission Line

This is only a special case of the rectangular line. We used the same values for the dielectric and conducting materials. We used the following dimensions for the line.

a = width of the inner conductor = 2 mm

b = height of the inner conductor = 2 mm

A = width of the outer conductor = 4 mm

B = height of the outer conductor = 4 mm

From the COMSOL model, we obtained the capacitance per unit length as 90. 696 pF/m. Using the finite difference (FD) method, we obtained the capacitance per unit length of the line as 90. 714 pF/m. Table 10.7 presents the comparison of the characteristic impedance of several models. It is evident from the table that the results are in good agreement.

Table 10.7. Comparison of Characteristic Impedance Values of Square Coaxial Line

Name

Z0

Zheng

36.79

Lau

36.81

Cockcroft

36.80

Bowan

36.81

Green

36.58

Ivanov and Djankov

36.97

Costamagna and Fanni

36.81

Riblet

36.80

Finite difference (FD)

36.75

COMSOL

36.75

10.18.3 Rectangular Line with Diamondwise Structure

The geometry of the cross-section of this line is shown in Figure 10.31. The same dielectric and conducting materials used for the rectangular line are used for this line.

Figure 10.31. Cross-section of the Diamondwise (or Rhombus) structure with 45o offset angle.

The following values are used for the COMSOL model of the line.

d = 1 mm

A = width of the outer conductor = 4 mm

B = height of the outer conductor = 4 mm

For the COMSOL model, we obtained the capacitance per unit line as 57. 393 pF/m.

Table 10.8 displays the comparison of the characteristic impedance of several models. It is evident from the table that the results are in good agreement.

Table 10.8. Comparison of Characteristic Impedance Values of Diamondwise Structure

Name

Z_0

Zheng et al.

56.742

Bowan

56.745

Riblet

56.745

COMSOL

58. 079

10.18.4 A Single-Strip Shielded Transmission Line

Figure 10.32 presents the cross-section of a single-strip shielded transmission line.

Figure 10.32. Cross-section of the Single-strip Shielded Transmission Line.

The following parameters are used in modeling the line. The characteristic impedance of such a lossless line is given by

where

Z = characteristic impedance of the line

Co= capacitance per unit length of the line when the substrate is replaced with air

C = capacitance per unit length of the line when the substrate is in place

c = 3 × 10^8 m/s (the speed of light in vacuum).

For COMSOL, the simulation was done twice on Figure 10.32 (to find Co and C) using the following values. Air:

ɛr = 1, μr = 1, σ = 0 S/m

Dielectric material:

ɛr = 1, μr = 1, σ = 0 S/m

Conducting material:

ɛr = 1, μr = 1, σ = 5.8 × 10^7 S/m (copper)

w = width of the inner conductor = 1 mm

t = height of the inner conductor = 0. 1 × 10−4 m

h = height of dielectric material = 1 mm

a = width of the outer conductor = 19 mm

b = height of the air-filled region = 9 mm

Using COMSOL for modeling and simulation of the lines involves taking the following steps:

1.  Develop the geometry of the line, such as shown in Figure 10.33.

Figure 10.33. Geometry of a Single-strip Shielded Transmission Line at Air.

2.  We take the difference between the conductor and dielectric material.

3.  We select the relative permittivity as 1 for the difference in Step 2.

4.  For the boundary, we select the outer conductor as ground and inner conductor as port.

5.  We generate the finite element mesh, and then we solve the model and obtain the potential.

6.  As post-processing, we select Point Evaluation and choose capacitance element 11 to find the capacitance per unit length of the line.

7.  We add a dielectric region under the inner conductor with relative permittivity as 8. 8, as in Figure 10.33. Then we take the same steps from 3 to 6 to generate the mesh as in Figure 10.34 and the potential distribution as in Figure 10.35.

Figure 10.34. Mesh of a Single-strip Shielded Transmission Line.

Figure 10.35. The potential distribution along y = 0. 002.

Table 10.9 shows the comparison between our method using COMSOL and other methods. It is evident that the results are very close.

Table 10.9. Comparison of Capacitance Values for a Single-strip Shielded Transmission Line

Methods

C_0(pF/m)

C(pF/m)

Finite difference method

26.79

1405.2

Extrapolation

26.88

1393.6

Analytical derivation

27.00

1400.9

COMSOL

26.87

1574.0

10.19 MULTISTRIP TRANSMISSION LINES

Recently, with the advent of integrated circuit technology, the coupled microstrip transmission lines consisting of multiple conductors embedded in a multilayer dielectric medium have led to a new class of microwave networks. Multiconductor transmission lines have been utilized as filters in the microwave region which make it interesting in various circuit components. For coupled multiconductor microstrip lines, it is convenient to write:

where Qi is the charge per unit length, Vj is the voltage of jth conductor with reference to the ground plane, Csij is the short circuit capacitance between ith conductor and jth conductor. The short circuit capacitances can be obtained either from measurement or from numerical computation. From the short circuit capacitances, we obtain

where Cii is the capacitance per unit length between the i th conductor and the ground plane. Also,

where Cij is the coupling capacitance per unit length between the ith conductor and jth conductor. The coupling capacitances are illustrated in Figure 10.36.

Figure 10.36. The Per-unit Length Capacitances of a General m-conductor Transmission Line.

For m-strip line, the per-unit-length capacitance matrix is given by

Also, we can determine the characteristic impedance matrix for m-strip line by using

where Zo is the characteristic impedance per unit length.

Using COMSOL for modeling and simulation of the lines involves taking the following steps:

1.  Develop the geometry of the line.

2.  We take the difference between the conductor and dielectric material.

3.  We select the relative permittivity as 1 for the difference in Step 2.

4.  We add a dielectric region under the inner conductors with specified relative permittivity.

5.  For the boundary, we select the outer conductor as ground and the inner conductors as ports.

6.  We generate the finite element mesh, and then we solve the model.

7.  As post-processing, we select Point Evaluation and choose capacitance elements to find the coupling capacitance per unit length of the line.

These steps were taken for the following four cases.

10.19. 1 Double-strip Shielded Transmission Line

Figure 10.37 presents the cross-section of double-strip shielded transmission line, which consists of two inner conductors.

Figure 10.37. Cross-section of the Double-strip Shielded Transmission Line.

For COMSOL, the simulation was done twice on Figure 10.36 (one for Co and other for C) using the following values.

Air:

εr = 1, μr = 1, σ = 0 S/m

Dielectric material:

εr = 2, μr = 1, σ = 0 S/m

Conducting material:

εr = 1, μr = 1, σ = 5.8 − 10_7 S/m (copper)

For the geometry (see Figure 10.37), we followed the following values:

w = width of each of the inner conductors = 3 mm

t = height (or thickness) of the inner conductors = 1 mm

s = distance between the inner conductors = 2 mm

h = height of dielectric material = 1 mm

a = width of the outer conductor = 11 mm

b = height of the air-filled region = 2. 7 mm

From the COMSOL model, the simulation was done twice, one for the case in which the line is air-filled (the dielectric was replaced by air) and the other case in which the dielectric is in place as shown in Figure 10.37. Figure 10.38 shows the finite element mesh while Figure 10.39 depicts the potential distribution for the dielectric case. The potential distribution for y = 1 mm is portrayed in Figure 10.10.

Figure 10.38. Mesh the of Double-strip Shielded Transmission Line.

Figure 10.39. Potential distribution.

Figure 10.40. potential distribution at y =1 mm.

We obtained the capacitances per unit length (Co and C) by taking steps enumerated above for the single-strip transmission line. The results are shown in Table 10.9. Table 10.10 is for the case which the line is air-filled, i. e., the dielectric in Figure 10.37 is replaced by air. Table 10.11 is for the case in which the dielectric is in place. The results in Table 10.11 are compared with other methods and found to be close.

Table 10.10. Capacitance Values for Double-strip Air-filled Shielded Transmission Line

Methods

C_11 = C_22 (pF/m)

C_12 = C_21 (pF/m)

COMSOL

72. 9

−4. 591

Table 10.11. Comparison of Capacitance Values for Double-strip Shielded Transmission Line Shown in Figure 10.36

Methods

C_11 = C_22 (pF/m)

C_12 = C_21 (pF/m)

Spectral-space domain method

108. 1

−4. 571

Finite element method

109. 1

−4. 712

Point-matching method

108. 8

−4. 683

COMSOL

108. 5

−4. 618

10.19.2 Three-strip Line

Figure 10.40(a) shows the cross-section for three-strip transmission line. For COMSOL, the simulation was done twice on Figure 10.40 (one for Co and other for C) using the following values:

Air:

εr = 1, μr = 1, σ = 0 S/m

Dielectric material:

εr = 8.6, μr = 1, σ = 0 S/m

Conducting material:

εr = 1, μr = 1, σ = 5.8 − 10_7 S/m (copper)

For the geometry (see Figure 10.40(a)), we used the following values:

a = width of the outer conductor = 13 mm

b = height of the free space region (air) = 4 mm

h = height of the dielectric region = 2 mm

w = width of each inner strip = 2 mm

t = thickness of each inner strip = 0. 01 mm

D = distance between the outer conductor and the first strip = 2. 5 mm s = distance between two consecutive strips = 1 mm

Figure 10.40(a). Cross-section of the Three-strip Transmission Line.

Figure 10.41 shows the finite element mesh, while Figure 10.42 illustrates the potential distribution along line y = h.

Figure 10.41. Mesh for the Three-strip Transmission Line.

Figure 10.42. Potential distribution along the Air-dielectric interface (y = h) for the Three-strip Transmission Line.

Table 10.12 shows the finite element results for the three-strip line. Unfortunately, we could not find any work in the literature to compare our results.

Table 10.12. Capacitance Values (in pF/m) for Three-strip Shielded Microstrip Line

10.19.3 Six-strip Line

Figure 10.43 shows the cross-section for six-strip transmission line. For COMSOL, the simulation was done twice on Figure 10.42 (one for Co and other for C) using the following values:

Air:

εr = 1, μr = 1, σ = 0 S/m

Dielectric material:

εr =6, μr = 1, σ = 0 S/m

Conducting material:

εr = 1, μr = 1, σ = 5.8 − 10_7 S/m (copper)

For the geometry (see Figure 10.43), we used the following values:

a = width of the outer conductor = 15 mm

b = height of the free space region (air) = 2 mm

h = height of the dielectric region = 8 mm

w = width of each inner strip = 1 mm

t = thickness of each inner strip = 0. 01 mm

D = distance between the outer conductor and the first strip = 2 mm s = distance between two consecutive strips = 1 mm

Figure 10.43. Cross-section of the Six-strip Transmission Line.

Figure 10.44 shows the finite element mesh, while Figure 10.45 depicts the potential distribution along line y = h.

Figure 10.44. Mesh for the Six-strip Transmission Line.

Figure 10.45. potential distribution along the Air-dielectric Interface (y = h) for the Six-strip Transmission Line.

The capacitance values for six-strip shielded microstrip line are compared with other methods as shown in Table 10.13, where “iterative” refers to an iterative method and ABC refers to the asymptotic boundary condition. It is evident from the table that the finite element methods based closely agree. The finite element methods seem to be more accurate than the iterative and ABC techniques. (The negative capacitances are expected from equation (10.318). )

Table 10.13. Capacitance Values (in pF/m) for Six-strip Shielded Microstrip Line

10.19.4 Eight-strip Line

Figure 10.46 shows the cross-section for eight-strip transmission line. For COMSOL, the simulation was done twice on Figure 10.45 (one for Co and other for C) using the following values:

Air:

εr = 1, μr = 1, σ = 0 S/m

Dielectric material:

εr = 12.9, μr = 1, σ = 0 S/m

Conducting material:

εr = 1, μr = 1, σ = 5.8 − 10_7 S/m (copper)

For the geometry (see Figure 10.46), we used the following values:

a = width of the outer conductor = 175 mm

b = height of the free space region (air) = 100 mm

h = height of the dielectric region = 16 mm

w = width of each inner strip = 1 mm

t = thickness of each inner strip = 0. 01 mm

D = distance between the outer conductor and the first strip = 80 mm

s = distance between two consecutive strips = 1 mm

Figure 10.46. Cross-section of the Eight-strip Transmission Line.

Figure 10.47 shows the finite element mesh, while Figure 10.48 depicts the potential distribution along line y = 20 mm.

Figure 10.47. Mesh for the Eight-strip Transmission Line.

Figure 10.48. Potential distribution along the Air-dielectric interface (y = 20 mm) for the Eight-strip Transmission Line.

The capacitance values for eight-strip shielded microstrip line are compared with other methods as shown in Table 10.14, where other authors used the analytic approach and Fourier series expansion. It is evident from the table that the results from the finite element method (COMSOL) closely agree with the analytic approach.

Table 10.14. Capacitance Values (in pF/m) for Eight-strip Shielded Microstrip Line

10.20 SOLENOID ACTUATOR ANALYSIS WITH ANSYS

We use Ansys to do magnetic analysis (linear static) of a solenoid actuator. A solenoid actuator is to be analyzed as a 2D axisymmetric model as shown in Figure 10.49. For the given current, we determine the force on the armature.

Figure 10.49. Cross-section of the Solenoid Actuator.

The dimensions of the solenoid actuator are in centimeters. The armature is the moving component of the actuator. The back-iron is the stationary iron component of the actuator that completes the magnetic circuit around the coil. The stranded, wound coil of 650 windings with 1 amp/turn supplies the predefined current. The current per winding is 1 amp. The air-gap is the thin rectangular region of air between the armature and the pole faces of the back-iron.

The magnetic flux produced by the coil current is assumed to be so small that no saturation of the iron occurs. This allows a single iteration linear analysis. The flux leakage out of the iron at the perimeter of the model is assumed to be negligible. This assumption is made simple to keep the model small. The model would normally be created with a layer of air surrounding the iron equal to or greater than the maximum radius of the iron.

The air gap is modeled so that a quadrilateral mesh is possible. A quadrilateral mesh allows for an uniform thickness of the air elements adjacent to the armature where the virtual work force calculation is performed. This is desirable for an accurate force calculation. The program requires the current to be input in the form of current density (current over the area of the coil). The assumption of no leakage at the perimeter of the model means that the flux will be acting parallel to this surface. This assumption is enforced by the “flux parallel” boundary condition placed around the model. This boundary condition is used for models in which the flux is contained in an iron circuit. Forces for the virtual work calculation are stored in an element table and then summed. The force is also calculated by the Maxwell Stress Tensor method and the two values are found to be relatively close. Table 10.15 summarizes the parameters of the model for the actuator geometry.

Table 10.15. Parameters of the Model for the Actuator Geometry

Parameter

Value

Number of turns in the coil; used in post-processing

n = 650

Current per turn

I = 1. 0

Thickness of inner leg of magnetic circuit

ta = 0. 75

Thickness of lower leg of magnetic circuit

tb = 0. 75

Thickness of outer leg of magnetic circuit

tc = 0. 50

Armature thickness

td = 0. 75

Width of coil

Wc = 1

Height of coil

hc = 2

Air Gap

gap = 0. 25

Space around coil

space = 0. 25

Ws

Ws = Wc + 2* space

hs

hs = hc + 0. 75

Total width of model

W = ta + Ws + tc

hb

hb = tb + hs

Total height of model

h = hb + gap + td

Coil area

acoil = Wc * hc

Current density of coil

idens = n*i/acoil

The below steps are a guideline in solving the above model.

1.  Input the geometry of the model

We use the information in the problem description to make Figure 10.50.

Figure 10.50. The 2D geometry of the Solenoid Actuator model.

2.  Define the materials

(a) Set preferences

You will now set preferences in order to filter quantities that pertain to this discipline only.

1.  Main Menu > Preferences

2.  Check “Magnetic-Nodal” as in Figure 10.50(a)

3.  OK

Figure 10.50(a). Preferences for GUI filtering.

(b) Specify material properties

Now specify the material properties for the magnetic permeability of air, back-iron, coil, and armature. For simplicity, all material properties are assumed to be linear. (Typically, iron is input as a nonlinear B-H curve. ) Material 1 will be used for the air elements. Material 2 will be used for the back-iron elements. Material 3 will be used for the coil elements. Material 4 will be used for the armature elements.

1.  Main Menu > Preprocessor > Material Props > Material Models

2.  Double-click “Electromagnetics”, then “Relative Permeability”, then “Constant”

3.  “MURX” = 1

4.  OK

5.  Edit > Copy

6.  OK to copy Material Model Number 1 to become Material Model Number 2.

7.  Double-click “Material Model Number 2”, then “Permeability (Constant)”

8.  “MURX” = 1000 as shown in Figure 10.51

9.  OK

10.  Edit > Copy

11.  “from Material Number” = 1

12.  “to Material Number” = 3

13.  OK

14.  Edit > Copy

15.  “from Material Number” = 2

16.  “to Material Number” = 4

17.  OK

18.  Double-click “Material Model Number 4”, then “Permeability (Constant)”

19.  “MURX” = 2000 as shown in Figure 10.52

20.  OK

21.  Material > Exit

22.  Utility Menu > List > Properties > All Materials

23.  Review the list of materials, then: as shown in Figure 10.53 File > Close (Windows)

Figure 10.51. Definition of material model behavior for Number 1 and 2.

Figure 10.52. Definition of material model behavior for Number 1, 2, 3, and 4.

Figure 10.53. Review the list of materials of the model.

3.  Generating the mesh

(a) Define element types and options

In this step, you will define element types and specify options associated with these element types.

The higher-order element PLANE53 is normally preferred, but to keep the model size small, use the lower-order element PLANE13.

1.  Main Menu > Preprocessor > Element Type > Add/Edit/Delete

2.  Add...

3.  “Magnetic Vector” (left column)

4.  “Vect Quad 4nod13 ( PLANE13” (right column)

5.  OK

6.  Options...

7.  (drop down) “Element behavior” = Axisymmetric, as shown in Figure 10.54

8.  OK

9.  Close

Figure 10.54.Element type PLANE 13.

(b) Assign material properties.

Now assign material properties to air gaps, iron, coil, and armature areas.

1.  Main Menu > Preprocessor > Meshing > MeshTool

2.  (drop down) “Element Attributes” = Areas; then [Set] as in Figure 10.55

3.  Pick four areas of air gaps, A13, A14, A17, and A18 (the picking “hot spot” is at the area number label).

Figure 10.55. Element attribute for MeshTool.

4.  OK

5.  (drop down) “Material number” = 1

6.  Apply

7.  Pick the five back-iron areas, A7, A8, A9, A11, A12 as in Figure 10.56

Figure 10.56. Five Back-iron areas, A7, A8, A9, A11, A12.

8.  OK

9.  (drop down) “Material number” = 2

10.  Apply

11.  Pick coil area, A4

12.  OK

13.  (drop down) “Material number” = 3

14.  Apply

15.  Pick armature area, A10, A15, A16

16.  OK

17.  (drop down) “Material number” = 4

18.  OK

19.  Toolbar: SAVE_DB

(c) Specify meshing-size controls on air gap

Adjust meshing size controls to get two element divisions through the air gap.

1.  Main Menu > Preprocessor > Meshing > Size Cntrls > ManualSize > Lines > Picked Lines

2.  Pick four vertical lines through air gap

3.  OK

4.  “No. of element divisions“ = 2

5.  OK

(d) Mesh the model using the MeshTool

1.  “Size control global” = [Set]

2.  “Element edge length” = 0. 25

3.  OK as in Figure 10.57

4.  (drop down) “Mesh” = Areas

5.  Mesh

6.  Pick All

7.  Close

Figure 10.57. Global element sizes.

8.  Utility Menu > PlotCtrls > Numbering as in Figure 10.58

9.  (drop down) “Elem / attrib numbering” = Material numbers as in Figure 10.59

10.  OK as in Figure 10.60

Figure 10.58. Numbering after PlotCtrls.

Figure 10.59. Plot numbering control.

Figure 10.60. Numbering of the model.

(e) Scale model to meters for solution

For a magnetic analysis, a consistent set of units must be used. In this tutorial, MKS units are used, so you must scale the model from centimeters to meters.

1.  Main Menu > Preprocessor > Modeling > Operate > Scale > Areas

2.  Pick All

3.  “RX, RY, RZ Scale Factors” = 0. 01, 0. 01, 1

4.  (drop down) “Existing areas will be” = Moved

5.  OK as in Figure 10.61

6.  Toolbar: SAVE_DB

Figure 10.61. Scale area of the model.

4.  Apply Loads

(a) Define the armature as a component

The armature can conveniently be defined as a component by selecting its elements.

1.  Utility Menu > Select > Entities

2.  (first drop down) “Elements”

3.  (second drop down) “By Attributes”

4.  “Min, Max, Inc” = 4

5.  OK as in Figure 10.62

Figure 10.62. The entities of the model.

6.  Utility Menu > Plot > Elements as in Figure 10.63

Figure 10.63. The Armature as a component of the model.

7.  Utility Menu > Select > Comp/Assembly > Create Component

8.  “Component name” = ARM

9.  (drop down) “Component is made of” = Elements

10.  OK

(b) Apply force boundary conditions to armature

1.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Magnetic > Flag > Comp. Force/Torq

2.  (highlight) “Component name” = ARM

3.  OK

4.  Utility Menu > Select > Everything

5.  Utility Menu > Plot > Elements as in Figure 10.64

Figure 10.64. Plot of apply Force Boundary Conditions to Armature.

(c) Apply the current density

The current density is defined as the number of coil windings times the current, divided by the coil area. This equals (650)(1)/2, or 325. To account for scaling from centimeters to meters, the calculated value needs to be divided by. 01**2.

1.  Utility Menu > Plot > Areas

2.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Magnetic > Excitation > Curr Density > On Areas

3.  Pick the coil area, which is the area in the center

4.  OK

5.  “Curr density value” = 325/. 01**2

6.  OK

Close any warning messages that appear.

(d) Obtain a flux parallel field solution

Apply a perimeter boundary condition to obtain a “flux parallel” field solution. This boundary condition assumes that the flux does not leak out of the iron at the perimeter of the model. Of course, at the centerline, this is true due to axisymmetry.

1.  Utility Menu > Plot > Lines

2.  Main Menu > Preprocessor > Loads > Define Loads > Apply > Magnetic > Boundary > Vector Poten > Flux Par’l > On Lines

3.  Pick all lines around perimeter of model (14 lines) as in Figure 10.65

Figure 10.65. Plot of lines for Flux Parallel field of the model.

4.  OK

5.  Toolbar: SAVE_DB

5.  Obtain solution

(a) Solve

1.  Main Menu > Solution > Solve > Electromagnet > Static Analysis > Opt & Solve

2.  OK to initiate the solution

3.  Close the information window when solution is done

6. Review results

(a) Plot the flux lines in the model

Note that a certain amount of undesirable flux leakage occurs out of the back-iron.

1.  Main Menu > General Postproc > Plot Results > Contour Plot > 2D Flux Lines

2.  OK as in Figure 10.66

Figure 10.66. Contour Plot for 2D Flux Lines of the Model.

Your results may vary slightly from what is shown here due to variations in the mesh.

(b) Summarize magnetic forces

1.  Main Menu > General Postproc > Elec & Mag Calc > Component Based > Force

2.  (highlight) “Component name(s)” = ARM

3.  OK

4.  Review the information, then choose:

File > Close (Windows),

or

Close (X11/Motif) to close the window.

(c) Plot the flux density as vectors

1.  Main Menu > General Postproc > Plot Results > Vector Plot > Predefined

2.  “Flux & gradient” (left column)

3.  “Mag flux dens B” (right column)

4.  OK as in Figure 10.67

Figure 10.67. Plot the Flux Density as Vectors of the model.

(d) Plot the magnitude of the flux density

Plot the magnitude of the flux density without averaging the results across material discontinuities.

1.  Main Menu > General Postproc > Plot Results > Contour Plot > Nodal Solu

2.  Choose “Magnetic Flux Density,” then “Magnetic flux density vector sum”

3.  OK as in Figure 10.68

Figure 10.68. Contour Plot of the model.

Next, you will see how the flux density is distributed throughout the entire actuator. Up to this point, the analysis and all associated plots have used the 2D axisymmetric model, with the axis of symmetry aligned with the left vertical portion of the device. ANSYS will continue the analysis on the 2D finite element model, but will allow you to produce a three-quarter expanded plot representation of the flux density throughout the device, based on the defined axisymmetry. This function is purely graphical. No changes to the database will be made when you produce this expanded plot.

4.  Utility Menu > PlotCtrls > Style > Symmetry Expansion > 2D Axi-Symmetric

5.  (check) “3/4 expansion” as in Figure 10.69

6.  OK as in Figure 10.70

Figure 10.69. 2D Axi-symmetric expansion with Amount 3/4.

Figure 10.70. 2D Axi-symmetric plot of the model.

7.  Utility Menu > PlotCtrls > Pan, Zoom, Rotate, as in Figure 10.71

8.  Iso

9.  Close

Figure 10.71. Rotation of the model.

(e) Exit the ANSYS program

1.  Toolbar: QUIT

2.  (check) “Quit − No Save!”

3.  OK

PROBLEMS

1.  Given H = Hej(ωt+2ßz)ax in free space, known that,

2.  Calculate the skin depth, δ, for a copper conductor in 50 Hz field (σ = 56 × 10^6S/m).

3.  EM problems and examples.

4.  For the axisymmetric coaxial cable illustrated in Figure 10.71. Determine one dimension finite element general solution based on the following:

a.  Obtain and solve the governing differential solution for the coaxial cable,

b.  Obtain the boundary conditions and continuity conditions, hint: ϕ_1 (r = a) = ϕω ϕ_2 (r = c) = 0, and the electric potential and the electric displacement are continuous at r = b.

c.  Formulate the equations of part (b) as a matrix equation that can be solved for the constants of integrations.

d.  Determine the shape functions for a general three-node quadratic element in terms of x_1, x_2, and x_3.

e.  Determine the shape functions for a general three-node quadratic element when x_1 = −L, x_2 = 0, and x_3 = L.

f.  Find the local stiffness matrix for an element of length 2L with coordinates (−L, 0, L).

Figure 10.71(a). Axisymmetric radial element.

Figure 10.71(b). Coaxial cable.

Figure 10.71(c). Three-node element.

5.  Determine the variational function for two-dimensional axisymmetric heat conduction in r, z coordinate and formulate the corresponding local finite stiffness matrix using three-node triangular elements.

6.  Use COMSOL in modeling of the four-conductor transmission lines with the following parameters as in Figure 10.72:

εr1 = dielectric constant of the dielectric material = 4. 2

εr2 = dielectric constant of the free space = 1. 0

W = width of the dielectric material = 10 mm

w = width of a single conductor line = 1 mm

H_1 = distance of conductors 1 and 2 from the ground plane = 3 mm

H_2 = distance of conductor 4 from the ground plane = 1 mm

H_3 = distance of conductor 3 from the ground plane = 2 mm

s = distance between the two coupled conductors = 1 mm

t = thickness of the strips = 0. 01 mm

Figure 10.72. Cross-section of the four-conductor transmission lines.

The geometry is enclosed by a 10 × 10 mm shield. Find the capacitances per unit length, C_11, C_12, C_13 C_14 C_22 C_23, C_24, C_33, C_34, and C_44.

7.  Use COMSOL in modeling of the shielded two vertically coupled striplines geometry is enclosed by a 3.4 × 1 mm shield with the following parameters as in Figure 10.73:

Figure 10.73. Cross-section of the two shielded vertically coupled striplines embedded in dielectric material.

εr = dielectric constant = 1 and 7.5

W_1 = width of the stripline 1 = 1.4 mm

W_2 = width of the stripline 2 = 1 mm

H_1 = height from stripline 1 and stripline 2 to the upper side and lower side of the shield respectively = 0.4 mm

H_2 = distance between the two striplines = 0.2 mm

S = distance between the stripline 1 and right/left side of the shield = 1 mm

a = (W_1−W_2)/2 = 0.2 mm

t = thickness of the striplines = 0. 01 mm

Find the capacitances per unit length, C_11, C_21 and C_22.

8.  Use ANSYS Modeling of harmonic high-frequency electromagnetic of a coaxial waveguide as shown in Figure 10.74. The properties of the model is summarized as

Material property:

μr = 1.0, εr = 1.0,

Geometric property:

ri = 0.025 m, r_0 = 0.075 m, I = 0.375 m,

Load used

Port voltage = 1.0

Ω = 0.8 GHz

Figure 10.74. Cross-section of a coaxial waveguide.

Find S11, S12, ZRe, Zim, RL

9.  Use ANSYS Modeling of electrostatic of a shielded microstrip transmission line consisting of a substrate, microstrip, and a shield. The strip is at potential V_1 and the shield is at a potential V_0. Find the capacitance of the transmission line as shown in Figure 10.75.

The properties of the model is summarized as

Material property:

Air: εr = 1

Substrate: εr = 12

Geometric property:

a = 10 cm

b = 1 cm

w = 2 cm

Loading property:

V_0 = 1 V

V_1 = 10 V

Knowing that the electrostatic energy, We is defined as

Also, you need to type the following values in scalar parameters as:

C = (w*2)/((V_1−V_0)**2) and C = ((C*2)*1e12).

Figure 10.75. Cross-section of shielded microstrip line.

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Appendix A

ANSYS

A.1 ANSYS (CLASSIC OR TRADITIONAL)

ANSYS is a finite element modeling package for numerically solving variety engineering problems such as structural, thermal, fluid, static and dynamic, linear and nonlinear, acoustic, electromagnetics, and multiphysics, etc.

ANSYS has two methods. The graphical user interface (GUI) and the command files. The focus in this section as in the book only on the ANSYS GUI environment (traditional or classic ANSYS) and in the next section we focus on the ANSYS workbench environment.

To start ANSYS, double click on the ANSYS icon or you can use start > Programs > ANSYS 11. 0 > ANSYS.

Then the ANSYS main window will be shown with the ANSYS output window.

The ANSYS GUI environment has two windows, the main window and the output window.

The ANSYS main window has the following components:

Figure A.1. ANSYS GUI main window.

1.  Utility menu: contains functions that are available through out the ANSYS sessions.

(a)  File menu:

File menu contains file and database related functions such as, options to clear the database, change, resume, and save the current model.

(b)  Select menu

Select menu provides functions that allow the user to select subsets of entities and to create components.

(c)  List menu

List menu allows the user to view any data item stored in the ANSYS database, view log and error files, obtain listing of geometries entities and their components, elements and their properties, nodes, and boundary conditions and loads. You can obtain information about different areas of the program and list the contents of files residing of the system.

(d)  Plot menu

Plot menu allows you to plot various components of the model such as, keypoints, lines, areas, volumes, nodes, elements, and other data can be plotted.

(e)  PlotCtrls menu

PlotCtrls menu includes functions to control the view and pan/zoom/rotate the model, select numbering options, change styles, and make hardcopies of the plots.

(f)  WorkPlane menu

WorkPlane menu allows the user to toggle the working plane on or off and to move and rotate, and you can create, delete, and switch coordinate systems.

(g)  Parameters menu

Parameters menu allows the user to define, edit, and delete scalar and array parameters.

(h)  Macro menu

Macro menu allows the user to create and execute macros and execute data blocks. Also, the user can create, edit, and delete abbreviations.

(i)  MenuCtrls menu

MenuCtrls menu allows the user to create, edit, and delete abbreviations on the ANSYS toolbar and modify the colors and fonts in the GUI display.

(j)  Help menu

Help menu allows the user to find all the manuals and tutorials available on ANSYS.

2.  Comments input: shows program prompt messages and allows the user to type in commands directly.

3.  Toolbar menu: contains graphics buttons that execute frequently used ANSYS commands, that is, collects commands that are frequently used.

4.  Main menu: contains primary ANSYS functions.

5.  Graphics area: displays geometrics, finite elements, and simulations.

The ANSYS output window dynamically provides important information during the preprocessor, solution, and post-processor. The warnings in the output window should be considered to prevent future errors. The output window is usually positioned behind the main window and can be put to the front, if necessary.

Figure A.2. ANSYS GUI output window.

To save the model, select from ANSYS Utility Menu, File > Save as Jobname.db.

Or you can save your model by selecting File > Save as, as shown below.

Figure A.3. Save database of the model.

To start up ANSYS, recall and continue previous job by starting ANSYS and then select Utility Menu: File > Resume from and click on the model from the list that appears.

To quickly save an image of the entire screen or the current Graphics window, go to ANSYS Utility Menu > PlotCtrls > Hard copy. There are two options to choose: To Printer and To File.

Figure A.4. Print to printer.

Figure A.5. Print to file.

To finish ANSYS, choose File > Exit command. Then Exit from ANSYS window will show up as shown in Figure A.6. Choose a proper option and click OK to finish the ANSYS program.

Figure A.6. Exit from ANSYS.

A. 2 ANSYS WORKBENCH

The ANSYS workbench provides unified data sharing and project file management across the range of ANSYS products. Workbench is not an application on its own. It is a group of technologies for developing simulating tools. This has been created by ANSYS incorporation to help developers develop robust tools to meet the needs of designers and analysts. It provides a common interface for the user to access the wealth of technology that ANSYS has with it.

Simulation data is displayed in a hierarchical tree that separates out the various groups of inputs and outputs. Workbench makes use of a project file storage method to assist with the organization of data, and each can contain all types of information. Workbench stores the model information at the highest level and this contains:

1.  geometry that your mesh is built on,

2.  contacts (which define how parts interact),

3.  mesh and

4.  environment.

The mesh is the core element-based representation of the parts, on which all FEA/CFD tasks are based. Workbench allows you to intelligently refine the mesh.

The Workbench user interface: Tabs based user interface, that is familiar to most users, is used in Workbench. We start with the Project Page tab and this provides us access to the tools we need to read geometry in, set-up the boundary conditions, create results visualization assets, and report on results.

Geometry definition: Data pertaining to geometry can be read from the majority of leading software, including Pro/Engineer, Solid Works, NX/Unigraphics, Catia, etc. The system reads the data into the Workbench user interface and preserves part names and material definitions, if any.

Design Modeler: Design Modeler provides us with many tools that allow us to take data from a variety of sources, integrate it into a single pack, and use it.

Finite Element Modeler: It allows you to read meshed, element-based data from a wide variety of sources. It performs checks for integrity, extracts the information, and places it into the appropriate folders of the simulation tree.

Materials definition: ANSYS Workbench is supplied with a range of standard materials. Materials (if present) can be extracted from the CAD file or may be adapted or enter new materials.

Reporting: Reports can be created based on a wizard style workflow. Various inputs (title, description, etc. ) may be defined and the system handles rest of the work. Workbench allows you to create diagrams at any point, so the graphics window and a caption is stored and then added to the report.

Design optimization-Design Xplorer: Workbench has inherent links with CAD system, and bi-directionality can be used to our advantage to carry out optimization to the maximum extent.

Mechanisms: Workbench also makes it possible to carry out in-depth studies of mechanisms intelligently. It allows us to model both rigid and flexible bodies in one environment.

Lastly, it can be said that ANSYS provides a single unified environment where user has access to a good functionality within the simulation arena. It helps us to build high quality products that cost less and serve better their intended purpose.

Method of solving problem using workbench is illustrated using a simple example.

Example A. 1

For the cantilever beam shown in Figure A.7, determine the total deflection, maximum stress, reaction force, and reaction moment. Take Young’s modulus E = 210 GPa and Poisson ratio v = 0. 3. Force P = 5000 N. Assume plane stress condition.

Figure A.7. The cantilever beam for Example A.1.

solution

(I) Software method

Procedure for solving the problems using ANSYS® 11. 0 academic teaching software using ANSYS Workbench.

Step 1. Selecting Analysis System and Specifying Material Properties Start ANSYS Workbench Start > Programs > ANSYS 11. 0 > Workbench In Toolbox Customization, click Static Structural (ANSYS)

Figure A.8. Selecting type of analysis.

In Toolbox, left click (and hold) on Static Structural (ANSYS) and drag the icon to the empty space in the Project Schematic. In Project Schematic space, right click on Static Structural (ANSYS) and Rename the project to Cantilever_beam.

Figure A.9. Creating cantilever beam cell.

Specify Material Properties

In the Cantilever beam cell, double click on Engineering Data. A new page will be opened. The default material given is Structural Steel. This will be seen in the Outline of Schematic.

Example A2: Engineering Data.

In the Properties of Outline Row 3: Structural Steel window, double click on Isotropic Elasticity and change the Young’s Modulus value. To change the value, double click on the numerical value and enter E = 2. 1e11 Pa and Poisson ratio = 0. 3.

Figure A.10. Specifying material properties.

Next, click on Return to Project to return to Workbench Project Schematic window.

Figure A.11. Return to Workbench.

Step 2. Geometry Creation

In the Project Schematic contilever_beam cell, right click on Geometry and select Properties.

A window named Properties of Schematic A3: Geometry will be seen on the right hand side of the Workbench window.

In that window, under Basic Geometry Options, select Line bodies.

Figure A.12. Selecting line bodies.

Then in the Project Schematic window, under the cell, double left click on Geometry to start preparing geometry. At this point, a new window ANSYS Design Modeler will be opened. Select meter unit as desired length unit and click OK.

Creating a Sketch

Use XY Plane for creation of sketch. On the left hand side in Tree Outline, click on XY Plane and then click on Sketching, next to Modeling tab.

Figure A.13. Selecting XY plane for sketching.

Figure A.14. Sketching toolbox.

Note: In sketching, undo features can be used.

On the right hand side, there is a Graphics window. At the lower right hand corner of the Graphics window, click on +Z axis. Then Graphics window will appear as shown below.

Figure A.15. Graphics window.

In the Sketching toolbox, select Line and, in the Graphics window, create one rough line starting from the origin in the positive XY-direction. (The letter P should be seen at the origin before creating the Line. The letter P indicates geometry is constrained at the origin. )

Figure A.16. Creating a line.

Dimensions

Under the Sketching Toolbox, select the Dimensions tab. (Use the default dimensioning tools.)

Then left click on the line in the Graphics window and drag the dimension line. The dimension line shows H1.

On lower left hand corner under Details View, change H1 to 5. This 5 is the length of the line drawn. Now sketching is done.

Figure A.17. Dimensioning the line and entering the length value.

Surface Creation

On top menu of the Design Modeler window, click on

Concept > Lines from Sketches.

This will create a new line, Line1. In Tree Outline, double click on XY Plane then select Sketch 1, and under Details View, in Base Objects, click Apply. Finally on the top, click Generate to generate the surface.

Create Cross-Section

Concept > Cross-Section > Rectangular

Under Details View, input B = 1 m and H = 1 m.

Figure A.18. Entering the breadth and height values of the beam.

Now under Tree Outline > 1 Part, 1 Body > Line Body, attach Rect1 to Cross-Section under Details View as shown below.

Figure A.19. Attaching a rectangular cross-section.

Now geometry is done. Close the Design Modeler and go back to Workbench. Step 3.Meshing

Save the work in the Workbench window.

In the Cantilever_beam cell, right click on Model and click Edit. A new ANSYS Model window will open.

Figure A.20. ANSYS model window for meshing, for applying boundary conditions and for the solution.

In Outline heading, right click on Mesh and click Generate Mesh. Then Meshed model will appear in Graphics window.

Figure A.21. Meshed model.

Step 4. Boundary Conditions

Start the setting up of boundary condition at A.

In Outline, right click Static Structural (A5) > Insert then select Remote Displacement.

In Graphics window right click and change Cursor Mode to Vertex.

Select point A in the Graphics window and click Apply next to Geometry under Details of "Remote Displacement"

Figure A.22. Selecting point A.

Now under the heading of Details of "Remote Displacement" and under Definition heading change UX, UY, UZ, ROTX, ROTY. and ROTZ to zero.

Figure A.23. Applying displacement boundary conditions.

At point C, force component needs to be applied.

In Graphics window, select point C.

Then Outline > Static Structural (A5) > Insert > Force.

Under Details of "Force", in Definition heading, next to Define By, change Vector to Components.

Enter -5000 for Y component.

Figure A.24. Applying force.

Click on Static Structural (A5) to view the model with boundary conditions in Graphics window shown below.

Figure A.25. Model with boundary conditions.

Step 5.Solution

For finding deflection, under Outline heading, right click Solution (A6), Insert > Deformation > Total.

For finding Stresses in beam, Outline > Solution (A6) > Insert > Beam Tool > Beam Tool.

For finding reaction force, Outline > Solution (A6) > Insert > Probe > Force Reaction.

Select point A and Under Details of "Force Reaction", in Definition heading next to Boundary Condition, select Remote Displacement.

For finding reaction moment, Outline > Solution (A6) > Insert > Probe > Moment Reaction.

Select point A and Under Details of "Moment Reaction", in Definition heading next to Boundary Condition, select Remote Displacement.

Then on top, click Solve button.

Step 6. Results

Under Outline and under Solution (A6), click on Total Deformation. The result is shown below.

Figure A.26. Deflection pattern for a cantilever beam.

Figure A.27. Deflection values for a cantilever beam.

Then double click on Beam Tool and click on Maximum combined stress. This result is shown below.

Figure A.28. Maximum combined stress on a cantilever beam.

Figure A.29. Maximum combined stress values.

Then click on Force Reaction, and this result is as shown below.

Figure A.30. Values of reaction force.

Then click on Moment Reaction, this result is shown below.

Figure A.31. Values of reaction moment.

(II) Analytical method (Refer Figure A.1).

Maximum bending stress occurs at the support.

Bending moment at support M = P X L = 5 X 10^3 X 5 = 25000 N-m

Parameter

Software results

Analytical method

Maximum deflection

0. 012277 mm

0. 012 mm

Maximum stress

0. 15 MPa

0. 15 MPa

Reaction force

5000 N

5000 N

Reaction moment

25000 N-m

25000 N-m

REFERENCES

1.  ANSYS Release 11. 0 Tutorials.

2.  ANSYS Tutorials of University of Alberta, http://www.mece.ualberta.ca

3.  ANSYS Tutorials, http://mae.uta.edu/~lawrence

4.  ANSYS Tutorials, http://www.andrew.cmu.edu

5.  ANSYS Tutorials, https://confluence.cornell.edu/display/SIMULATION

6.  E. M. Alawadhi, “Finite Element Simulations Using ANSYS,” CRC Press, 2010.

7.  Y. Nakasone, S. Yoshimoto, and T. A. Stolarski, “Engineering Analysis with ANSYS Software,” Butterworth-Heinemann, 2006.

Appendix B

MATLAB

MATLAB (short for Matrix Laboratory) has become the useful and dominant tool of technical professionals around the world. MATLAB is a sophisticated numerical computation and simulation tool that uses matrices and vectors. Also, MATLAB enables the users to solve a wide variety of analytical problems.

A copy of MATLAB software can be obtained from

The Mathworks, Inc.

3 Apple Hill Drive

Natick, MA 01760–2098

Phone: 508–647–7000

Website: http://www.mathworks.com

This brief overview of MATLAB is presented here to give a general idea about computation of the software. MATLAB computational applications to engineering systems are used to solve practical problems.

B.1 GETTING STARTED AND WINDOWS OF MATLAB

When you double-click on the MATLAB icon on the desktop or use the start menu to find the program, it opens as shown in Figure B.1. The command window, where the special >> prompt appears, is the main area in which the user interacts with MATLAB. To make the Command Window active, you need to click anywhere inside its border. The MATLAB prompt >> tells the user that MATLAB is ready to enter a command. To quit MATLAB, you can select EXIT MATLAB from the File menu, or by enter quit or exitat the Command Window prompt. Note, do not click on the X (close box) in the top right corner of the MATLAB window, because it may cause problems with the operating software. Figure B.1 contains four default windows: Command Window, Workplace Window, Command History Window, and Current Folder Window. Table B.1 shows a list of the various windows and their purpose of MATLAB.

Figure B.1. MATLAB default environment.

Table B.1. MATLAB Windows

MATLAB Window

Description

Command window

Main window, enter variables, runs programs

Workplace window

Gives information about the variables used

Command History window

Records commands entered in the Command window

Current Folder window

Shows the files in the current directory with details

Editor window

Makes and debugs script and function files

Help window

Gives help information

Figure window

Contains output from the graphic commands

Launch Pad window

Provides access to tools, demos, and documentation

B.2 USING MATLAB IN CALCULATIONS

Table B.2 shows the MATLAB common arithmetic operators. The order of operations is as follows: first, parentheses ( ), the innermost are executed first for nested parentheses; second, exponentiation ^; third, Multiplication * and division / (they are equal precedence); fourth, addition + and subtraction -.

Table B.2. MATLAB Common Arithmetic Operators

MATLAB Operators Symbols

Descriptions

+

Addition

-

Subtraction

*

Multiplication

/

Right division (means )

\

b Left division (means )

^

Exponentiation (raising to a power)

‘

Converting to complex conjugate transpose

( )

Specify evaluation order

For example,

MATLAB has different displays of a number formats. Table B.3 provides common numeric display formats. You can obtain more by typing help in the Command window (>> format).

Table B.3. MATLAB Common Numeric Display Formats

MATLAB Command

Descriptions

format

Default display, same as short

format short

Display 4 decimal digits

format long

Display 14 decimal digits

format short e

Display 4 decimal digits in scientific notation

format long e

Display 14 decimal digits in scientific notation

format short g

Display the best short format selected by MATLAB

format long g

Display the best long format selected by MATLAB

format rat

Display fractional form

For example,

Table B.4 provides common samples of MATLAB functions. You can obtain more by typing help in the Command window (>> help).

Table B.4. MATLAB Typical Elementary Math Functions

MATLAB Function

Description

abs (x)

Absolute value or complex magnitude of x

acos (x), acosh (x)

Inverse cosine and inverse hyperbolic cosine of x (in radians)

angle (x)

Phase angle (in radians) of a complex number x

asin (x), asinh (x)

Inverse sine and inverse hyperbolic sine of x (in radians)

atan (x), atanh (x)

Inverse tangent and inverse hyperbolic tangent of x (in radians)

conj (x)

Complex conjugate of x (in radians)

cos (x), cosh (x)

Cosine and inverse hyperbolic cosine of x (in radians)

cot (x), coth (x)

Inverse cotangent and inverse hyperbolic cotangent of x (in radians)

exp (x)

Exponential of x

fix

Round toward zero

imag (x)

Imaginary part of a complex number x

log (x)

Natural logarithm of x

log2 (x)

Natural logarithm of x to base 2

log10 (x)

Common logarithms (base 10) of x

real (x)

Real part of a complex number of x

sin (x), sinh (x)

Sine and inverse hyperbolic sine of x (in radians)

sqrt (x)

Square root of x

tan (x), tanh (x)

Tangent and inverse hyperbolic tangent of x (in radians)

For example,

In addition to operating on mathematical functions, MATLAB allows us to work easily with vectors and matrices. A vector (or one-dimensional array) is a special matrix (or two-dimensional array) with one row or one column. Arithmetic operations can apply to matrices and Table B.5 has extra common operations that can be implemented to matrices.

Table B.5. MATLAB Matrix Operations

MATLAB Operations

Descriptions

A′

Transpose of matrix A

det (A)

Determinant of matrix A

inv (A)

Inverse of matrix A

eig (A)

Eigenvalues of matrix A

diag (A)

Diagonal elements of matrix A

rank (A)

Rank of matrix A

cond (A)

Condition number of matrix A

eye (n)

The n x n identity matrix (1’s on the main diagonal)

eye (m, n)

The m x n identity matrix (1’s on the main diagonal)

trace (A)

Summation of diagonal elements of matrix A

zeros (m, n)

The m x n matrix consisting of all zeros

ones (m, n)

The m x n matrix consisting of all ones

rand (m, n)

The m x n matrix consisting of random numbers

randn (m, n)

The m x n matrix consisting of normally distributed numbers

diag (A)

Extraction of the diagonal matrix A as vector

diag (A, 1)

Extracting of first upper off-diagonal vector of matrix A

diag (u)

Generating of a diagonal matrix with a vector u on the diagonal

expm (A)

Exponential of matrix A

ln (A)

LU decomposition of matrix A

svd (A)

Singular value decomposition of matrix A

qr (A)

QR decomposition of matrix A

min (A)

Minimum of vector A

max (A)

Maximum of vector A

sum (A)

Sum of elements of vector A

std (A)

Standard deviation of the data collection of vector A

sort (A)

Sort the elements of vector A

mean (A)

Means value of vector A

triu (A)

Upper-triangular of matrix A

triu (A, I)

Upper-triangular with zero diagonals of matrix A

tril (A)

Lower-triangular of matrix A

tril (A, I)

Lower-triangular with zero diagonals of matrix A

A vector can be created by typing the elements inside brackets [ ] from a known list of numbers.

Examples shown in Table B.5.

Also, a vector can be created with constant spacing by using the command variable-name = [a: n: b], where a is the first term of the vector; n is spacing; b is the last term.

For example,

In addition, a vector can be created with constant spacing by using the command variable-name = linespace (a, b, m), where a is the first element of the vector; b is the last element; mis number of elements.

For example,

Special constants can be used in MATLAB. Table B.6 provides special constants used in MATLAB.

Table B.6. MATLAB Named Constants

Name

Content

pi

π = 3.14159...

i or j

Imaginary unit,

eps

Floating-point relative precision, 2-52

realmin

Smallest floating-point number, 2-1022

realmax

Largest floating-point number, (2-eps). 2^1023

bimax

Largest positive integer, 2^53 - 1

Inf or Inf

Infinity

nan or NaN

Not a number

rand

Random element

eye

Identity matrix

ones

An array of 1’s

zeros

An array of 0’s

For example,

Arithmetic operations on arrays are done element by element. Table B.7 provides MATLAB common arithmetic operations on arrays.

Table B.7. MATLAB Common Arithmetic Operations on Arrays

MATLAB Operators Symbols on Arrays

Descriptions

+

Addition same as matrices

-

Subtraction same as matrices

.*

Element-by-element multiplication

./

Element-by-element right division

.\

Element-by-element left division

.^

Element-by-element power

.‘

Unconjugated array transpose

For example,

B.3 SYMBOLIC COMPUTATION

In this section, MATLAB can manipulate and solve symbolic expressions that make the user compute with math symbols rather than numbers. This process is called symbolic math. Table B.8 shows some common symbolic commands.

You can practice some of symbolic expressions as below:

Table B.8. Common Symbolic Commands

MATLAB Command

Description

diff

Differentiate symbolic expression

int

Integrate symbolic expression

jacobian

Compute Jacobian matrix

limit

Compute limit of symbolic expression

symsum

Evaluate symbolic sum of series

taylor

Taylor series expansion

colspace

Return basis for column space of matrix

det

Compute determinant of symbolic matrix

diag

Create or extract diagonals of symbolic matrices

eig

Compute symbolic eigenvalues and eigenvectors

expm

Compute symbolic matrix exponential

inv

Compute symbolic matrix inverse

jordan

Compute Jordan canonical form of matrix

null

Form basis for null space of matrix

poly

Compute characteristic polynomial of matrix

rank

Compute rank of symbolic matrix

rref

Compute reduced row echelon form of matrix

svd

Compute singular value decomposition of symbolic matrix

tril

Return lower triangular part of symbolic matrix

triu

Return upper triangular part of symbolic matrix

coeffs

List coefficients of multivariate polynomial

collect

Collect coefficients

expand

Symbolic expansion of polynomials and elementary functions

factor

Factorization

horner

Horner nested polynomial representation

numden

Numerator and denominator

simple

Search for simplest form of symbolic expression

simplify

Symbolic simplification

subexpr

Rewrite symbolic expression in terms of common subexpressions

subs

Symbolic substitution in symbolic expression or matrix

compose

Functional composition

dsolve

Symbolic solution of ordinary differential equations

finverse

Functional inverse

solve

Symbolic solution of algebraic equations

cosint

Cosine integral

sinint

Sine integral

zeta

Compute Riemann zeta function

ceil

Round symbolic matrix toward positive infinity

conj

Symbolic complex conjugate

eq

Perform symbolic equality test

fix

Round toward zero

floor

Round symbolic matrix toward negative infinity

frac

Symbolic matrix elementwise fractional parts

imag

Imaginary part of complex number

log10

Logarithm base 10 of entries of symbolic matrix

log2

Logarithm base 2 of entries of symbolic matrix

mod

Symbolic matrix elementwise modulus

pretty

Pretty-print symbolic expressions

quorem

Symbolic matrix elementwise quotient and remainder

real

Real part of complex symbolic number

round

Symbolic matrix elementwise round

size

Symbolic matrix dimensions

sort

Sort symbolic vectors, matrices, or polynomials

sym

Define symbolic objects

syms

Shortcut for constructing symbolic objects

symvar

Find symbolic variables in symbolic expression or matrix

fourier

Fourier integral transform

ifourier

Inverse Fourier integral transform

ilaplace

Inverse Laplace transform

iztrans

Inverse z-transform

laplace

Laplace transform

ztrans

z-transform

B.3.1 Simplifying Symbolic Expressions

Symbolic simplification is not always straightforward; there is no universal simplification function, because the meaning of a simplest representation of a symbolic expression cannot be defined clearly. MATLAB use the sym or syms command to declare variables as symbolic variable. Then, the symbolic can be used in expressions and as arguments to many functions.

For example, to rewrite a polynomial in a standard form, use the expand function:

You can use the subs command to substitute a numeric value for a symbolic variable or replace one symbolic variable with another.

For example,

B.3.2 Differentiating Symbolic Expressions

Use diff ( ) command for differentiation.

For example,

B.3.3 Integrating Symbolic Expressions

The int(f) function is used to integrate a symbolic expression f.

For example,

B.3.4 Limits Symbolic Expressions

The limit(f) command is used to calculate the limits of function f.

For example,

B.3.5 Taylor Series Symbolic Expressions

Use taylor( ) function to find the Taylor series of a function with respect to the variable given.

For example,

B.3.6 Sums Symbolic Expressions

Use symsum ( ) function to obtain the sum of a series.

For example,

B.3.7 Solving Equations as Symbolic Expressions

Many of MATLAB’s commands and functions are used to manipulate the vectors or matrices consisting of symbolic expressions.

For example,

Use dsolve ( ) function to solve symbolic differential equations.

For example,

Appendix C

COLOR FIGURE

CHAPTER 3

Figure 3.6(c). Deflection pattern for a bar.

Figure 3.6(d). Stress pattern for a bar.

Figure 3.7(c). Deflection pattern for a bar.

Figure 3.7(d). Stress pattern for a bar.

Figure 3.8(c). Deflection pattern for a bar.

Figure 3.8(d). Stress pattern for a bar.

Figure 3.9(c). Deflection pattern for a bar.

Figure 3.10(c). Deflection pattern for a rod.

Figure 3.19(d). Deflection pattern for a tapered bar.

Figure 3.19(e). Stress pattern for a tapered bar.

Figure 3.20(d). Deflection pattern for a tapered bar.

Figure 3.20(e). Stress pattern for a tapered bar.

Figure 3.21(c). Deflection pattern for a tapered bar.

Figure 3.21(d). Stress pattern for a tapered bar.

Figure 3.32(b). Deflection pattern for a stepped bar.

Figure 3.32(c). Stress pattern for a stepped bar.

Figure 3.33(b). Deflection pattern for a stepped bar.

Figure 3.33(c). Stress pattern for a stepped bar.

Figure 3.34(b). Deflection pattern for a stepped bar.

Figure 3.34(c). Stress pattern for a stepped bar.

CHAPTER 4

Figure 4.2(g). Deflection pattern for a truss.

Figure 4.2(h). Stress pattern for a truss.

Figure 4.3(a). Deflection pattern for a truss.

Figure 4.3(b). Stress pattern for a truss.

Figure 4.4(b). Deflection pattern for a truss.

Figure 4.4(c). Stress pattern for a truss.

CHAPTER 5

Figure 5.2(b). Deflection pattern for a simply supported beam.

Figure 5.3(f). Deflection pattern for a simply supported beam.

Figure 5.3(g). Bending moment diagram for a simply supported beam.

Figure 5.3(h). Shear force diagram for a simply supported beam.

Figure 5.3(i). Bending stress for a simply supported beam.

Figure 5.4(b). Deflection pattern for a simply supported beam.

Figure 5.5(c). Deflection pattern for a simply supported beam.

Figure 5.16(c). Deflection pattern for a cantilever beam.

Figure 5.17(d). Deflection pattern for a cantilever beam.

Figure 5.18(d). Deflection pattern for a cantilever beam.

Figure 5.18(e). Bending moment diagram for a propped cantilever beam.

Figure 5.18(f). Shear force diagram for a propped cantilever beam.

Figure 5.18(g). Bending stress for a propped cantilever beam.

Figure 5.18(h). Maximum stress for a propped cantilever beam.

Figure 5.19(d). Deflection pattern for a cantilever beam.

CHAPTER 6

Figure 6.6(a). Von Mises stress distribution pattern.

Figure 6.7(a). Deformed shape of the hole

Figure 6.7(b). Von Mises stress distribution pattern.

Figure 6.7(c). Deflection pattern.

Figure 6.8(a). 1st principle stress distribution pattern.

Figure 6.8(b). Deformed pattern.

CHAPTER 7

Figure 7.1(c). Temperature distribution in a composite wall.

Figure 7.14(b). Temperature distribution in a large plate.

Figure 7.25(c). Temperature distribution in a long cylinder.

Figure 7.36(a). Temperature distribution in a body of rectangular cross-section.

CHAPTER 8

Figure 8.13. Velocity distribution over a cylinder.

Figure 8.14. Vector plot of the fluid velocity.

Figure 8.15. Pressure distribution over a cylinder.

CHAPTER 9

Figure 9.1(b). Deflection pattern for a fixed-fixed beam for node 1.

Figure 9.1(c). Mode 1 for a fixed-fixed beam.

Figure 9.1(d). Mode 2 for a fixed-fixed beam.

Figure 9.11(b). Deflection pattern for a cantilever beam for node 1.

Figure 9.11(c). Mode 1 for a cantilever beam.

Figure 9.11(d). Mode 2 for a cantilever beam.

Figure 9.11(e). Mode 3 for a cantilever beam.

Figure 9.11(f). Mode 4 for a cantilever beam.

Figure 9.12(a). Displacement response for 0.2 sec for node 2.

Figure 9.12(b). Displacement response for 0. 2 sec for node 3.

Figure 9.35(b). Deflection pattern for a bar.

Figure 9.48(a). Displacement response for 0. 00025 sec for node 2.

Figure 9.48(b). Displacement response for 0. 00025 sec for node 3.

INDEX

A

ABC. See Absorbing boundary conditions

Absorbing boundary conditions (ABC), 422–423

Airfoil, potential flow

post-processing, 301–302

preprocessing, 297–300

software results, 297

Analytical method, 48

axial vibrations, 335

cantilever beams, 202–203, 208–209

constant cross-section area, 65–66, 70–72, 76–77, 82, 85–87

engineering problem, 48

natural frequency determination, 307

one-dimensional heat conduction problems, 259

simply supported beams, 176, 180–181, 187–188, 192–193

stepped bar, 118, 122–123, 127–129

stress analysis, rectangular plate with circular hole, 240–241, 242–243

truss, 147–150

varying cross-section area, 97–98, 103–104

ANSYS

classic/traditional, 473–482

design modeler, 483

design optimization-design, 484

engineering data, 486–501

finite element modeler, 483

geometry definition, 483

graphical user interface (GUI), 473

materials definition, 483

mechanisms, 484

reporting, 483

software method, 485–486

Solenoid actuator analysis, 441–459

workbench, 483–501

workbench user interface, 483

ANSYS program, 145

Anti-symmetric (Skew-symmetric) matrix, 5

Argument matrix, 19

Automatic mesh generation

arbitrary domains, 402–405

rectangular domains, 401–402

Axially loaded members

constant cross-section area, 65–96

stepped bar, 117–144

three-node bar element, 64

two-node bar element, 62–63

varying cross-section area, 97–117

Axial vibrations

analytical method, 335

FEM by hand calculations, 336–337

frequency values, 338

post-processing, 344

preprocessing, 339–344

software results, 338

Axisymmetric radial element, 459

B

Banded matrix, 5

Beam

beam element, 175

cantilever beams, 202–223

definition, 175

simply supported beams, 176–202

Bio-Savart’s law and field intensity, 365

Boundary conditions, 362–363

Boundary element method, 422

C

CAD model, 52

Cantilever beam, 347

Cantilever beams

analytical method, 202–203, 208–209

bending moment diagram, 218

bending stress diagram, 219

deflection values at nodes, 206, 213, 218, 224

FEM by hand calculations, 203–206, 209–212, 214–217, 220–225

maximum stress diagram, 219

nodal force calculation, 210–212, 215–217, 221–223

post-processing, 230–232

preprocessing, 226–230

reaction calculation, 212, 217, 223–224

reaction values, 207, 213, 218, 225

rotational deflection values at nodes, 206, 213, 218, 225

shear force diagram, 219

software results, 206–207, 212–214, 217–220

total values, 218

Cantilever beam, transverse vibrations

analytical solution, 317–318

FEM by hand calculations, 318–320

of software results, 320–322

Coaxial cable, 460

Coaxial waveguide, 462

Column matrix, 4

Compressible fluid, 291

Compressive forces, 59

Conduction boundary conditions, two-dimensional problem, 285–286

Constant cross-section area

analytical method, 65–66, 70–72, 76–77, 82, 85–87

deflection values as node, 68–69, 75, 80, 84, 92–93

displacement calculation, 65–66, 71–71

FEM by hand calculations, 66–68, 72–76, 77–79, 82–84, 87–91

post-processing, 95–96

preprocessing, 93–95

reaction calculation, 66, 70, 79, 86–87

reaction value, computer generated output, 69, 76

software results, 68–70, 80–81, 84–85, 91–93

stress calculation, 66, 67, 70–71, 74, 79, 86

stress values, computer generated output, 69

Young’s modulus, 65

Constitutive relations, 357–360

Convection boundary conditions, two-dimensional problem, 285–286

Coulomb’s law and field intensity, 364

Cramer’s rule, 21–23

D

Diagonal matrix, 4

Dielectric loss, 369

Direct approach, 49

Displacement method, 52

Double-strip shielded transmission line, 433–435

Dynamic analysis

bar, axial vibrations of, 335–344

cantilever beam, transverse vibrations of, 316–322

forcing function, bar subjected, 345–346

forcing function, fixed-fixed beam subjected to, 323–334

natural frequency determination, fixed-fixed beam for, 307–316

procedure of finite element analysis, 306

E

Eigenvalues, 34–36

Eigenvectors, 34–36

Eight-strip Line, 439–441

Electromagnetic analysis, 373–400

Electromagnetic energy and power flow, 365–369

Elementary row operations, 19

Element stiffness matrix, 146, 151

Element strains and stresses, 50

Elimination method, 50–51

Engineering electromagnetics analysis

automatic mesh generation, 401–405

Bio-Savart’s law and field intensity, 365

boundary conditions, 362–363

constitutive relations, 357–360

Coulomb’s law and field intensity, 364

electromagnetic analysis, 373–400

electromagnetic energy and power flow, 365–369

external problems, 420–423 FEM by hand calculations (See Hand calculations, FEM by)

higher order elements, 405–415

Lorentz force law and continuity equation, 356–357

loss in medium, 370

Maxwell’s equations and continuity equation, 351–356

multistrip transmision lines, 432–441

Poison’s and Laplace’s equations, 371

potential equations, 360–361

shielded microstrip lines, modeling and simulation of, 423–431

skin depth, 370–371

solenoid actuator analysis, with ANSYS, 441–459

three-dimensional element, 415–420

wave equations, 372–373

Engineering problem

analytical method, 48

experimental method, 48

numerical method, 48

post-processing, 52

preprocessing, 52

processing, 52

Equation of continuity, 352

Equations

of continuity, 352

Laplace’s equation, 371, 382–392

Lorentz force law and continuity equation, 356–357

Maxwell’s equations and continuity equation, 351–356, 352–356

Poison’s and Laplace’s equations, 371

Poisson’s equation, 377–381

potential equations, 360–361

symbolic expressions, solving equations, 519–520

wave equations, 372–373

Equilibrium, 61

Experimental method, 48

F

FEA. See Finite element method

Finite element method (FEA)

dynamic problems, procedure of, 306

fluid flow problems, procedure of, 292–293

linear spring, direct method for, 55–56

practical applications of, 51

prescribing boundary conditions, 50–51

procedure of, 48–50

software package, 52

solving engineering problems, 47–48

structural problems, procedure of, 48–50

for structure, 52–53

thermal problems, procedure of, 258

types of, 53–54

Fixed-fixed beam, 347

forcing function, 323–334

natural frequency determination, 307–316

Fluid flow analysis

airfoil, potential flow, 296–304

cylinder, potential flow, 293–295

procedure of finite element analysis, 292–293

Force and elongation behaviors, 60

Force method, 52

Forcing function, bar

displacement values, 346

maximum displacement values, 346

software results, 345–346

Forcing function, fixed-fixed beam

maximum displacement values, 334

preprocessing, 325–334

software results, 322–324

Four-conductor transmission lines, 461

G

Gaussian elimination method, 19–20

Gauss’s law, 371

Global stiffness matrix, 152

Gradient operator, 31

H

Hand calculations, FEM by

axial vibrations, 336–337

cantilever beams, 203–206, 209–212, 214–217, 220–225

cantilever beam, transverse vibrations of, 318–320

constant cross-section area, 66–68, 72–76, 77–79, 82–84, 87–91

natural frequency determination, 308–310

simply supported beams, 177–178, 181–184, 188–191, 193–195

stepped bar, 118–121, 123–125, 130–132

truss, 150–160, 162–165

varying cross-section area, 98–103, 104–106, 108–111

Higher order elements

fundamental matrices, 411–415

local coordinates, 406–408

pascal triangle, 405–406

shape functions, 409–410

Homogeneous, 362

Hybrid method, 53

I

Identity (unit) matrix, 4–5

Incompressible fluid, 291

Infinite element method, 420–421

Inhomogeneous, 362

Integration of matrix, 15

Inverse of matrix, 24–26

Invertible matrix (nonsingular matrix), 25

Irrotational flow, 291–292

L

Laplace’s equations, 371

Linear spring element, 55–56

Linear systems, direct methods for, 19

Lorentz force law and continuity equation, 356–357

Loss tangent, 370

M

Magnetostatic energy, 367

MATLAB, 36–43

MATLAB (Matrix Laboratory)

calculations, 505–512

differentiating symbolic expressions, 516

integrating symbolic expressions, 517

limits symbolic expressions, 517

simplifying symbolic expressions, 515

sums symbolic expressions, 518

symbolic computation, 512–514

symbolic expressions, solving equations, 519–520

Taylor series symbolic expressions, 518

windows, 503–505

Matrix

addition of, 6–7

Cramer’s rule, 21–23

definition, 1–3

determinant of, 16–17

differentiation of, 14

eigenvalues and eigenvectors, 34–36

equality of, 15

Gaussian elimination method, 19–20

integration of, 15

inverse of, 24–26

linear systems, direct methods for, 19

MATLAB, 36–43

multiplication of, 8–9

multiplied by scalar, 8

operations, 38

rules of multiplications, 9–12

subtraction of, 6–7

trace of, 14

transpose of, 12–13

types of, 3–6

vector analysis, 27–34

Matrix Laboratory. See MATLAB

Matrix multiplications

by another matrix, 8–9

rules of, 9–12

by scalar, 8

Maximum bending moment, 232–233

Maxwell’s equations and continuity equation

differential form, 352

divergence and Stokes theorems, 353

integral form, 353

quasi-statics case, 354

space case, source-free regions of, 354–355

statics case, 354

time-harmonic fields case, 355–356

Mixed method, 52

Multipoint constrains method, 51

Multistrip transmission lines, 432–441

N

Natural frequency determination

analytical method, 307

FEM by hand calculations, 308–310

post-processing, 316

preprocessing, 312–316

software results, 310–311

Nodal displacements, 50, 141, 142, 149, 156

Nodal lines, 374

Nodes, 47

Non-invertible matrix (singular matrix), 25

Nonsingular matrix, 25

Null matrix, 4

Numerical method, 48

O

One-dimensional elements, 53

FEM standard steps procedure, 374–377

natural coordinates, 381–382

Poisson’s equation, 377–381

variational approach, 377–379

weighted residuals method, 379–381

One-dimensional heat conduction problems, 257, 258–262

analytical method, 259

FEM by calculations, 260–261

post-processing, 268–272, 277–280, 285

preprocessing, 262–267, 272–277, 281–285

software results, 261–262, 280

P

Partitioned matrix (Super-matrix), 6

Penalty method, 51

Plot velocity distribution, 303

Poison’s and Laplace’s equations, 371

Poisson’s equation, 377–381

Post-processing

axial vibrations, 344

cantilever beams, 230–232

constant cross-section area, 95–96

engineering problem, solving, 52

natural frequency determination, 316

simply supported beams, 200–202

stepped bar, 140–141

stress analysis, rectangular plate with circular hole, 251

truss, 169–170

varying cross-section area, 116–117

Potential equations, 360–361

Poynting’s theorem, 368

Poynting vector, 367

Practical applications, 51

Preprocessing

axial vibrations, 339–344

cantilever beams, 226–230

constant cross-section area, 93–95

engineering problem, solving, 52

natural frequency determination, 312–316

simply supported beams, 197–200

stepped bar, 134–139

stress analysis, rectangular plate with circular hole, 246–251

truss, 167–169

Prescribing boundary conditions

elimination method, 50–51

multipoint constrains method, 51

penalty method, 51

Processing

engineering problem, solving, 52

varying cross-section area, 113–116

R

Rectangular cross-section transmission line, 426–427

Rectangular line, with diamondwise structure, 428–429

Rectangular matrix, 3

Rectangular plate with circular hole. See Stress analysis

Row matrix, 3

S

Scalar matrix, 4

Simply supported beams

analytical method, 176, 180–181, 187–188, 192–193

deflection values at nodes, 179, 185, 191, 196

FEM by hand calculations, 177–178, 181–184, 188–191, 193–195

nodal force calculation, 182–184

post-processing, 200–202

preprocessing, 197–200

reaction calculation, 178, 191

reaction values, 179, 185, 192

slope values at nodes, 185, 191, 196

software results, 179–180, 184–187, 191–192, 195–196

total values, 185–186, 192

Single-strip shielded transmission line, 429–431

Singular matrix, 25

Six-strip line, 437–439

Skew-symmetric matrix, 5

Skin depth, 370–371

Software package, finite element analysis, 52

Software results

airfoil, potential flow, 296

axial vibrations, 338

cantilever beams, 206–207, 212–214, 217–220

cantilever beam, transverse vibrations of, 320–322

constant cross-section area, 68–70, 80–81, 84–85, 91–93

forcing function, bar, 345–346

forcing function, fixed-fixed beam, 323–324

natural frequency determination, 310–311

one-dimensional heat conduction problems, 261–262, 280

simply supported beams, 179–180, 184–187, 191–192, 195–196

stepped bar, 125–127, 133–134

stress analysis, rectangular plate with circular hole, 241–242, 243–245

truss, 160–162, 165–167

varying cross-section area, 106–108, 111–113

Solenoid actuator analysis, with ANSYS, 441–459

Square cross-section transmission line, 427–428

Square matrix, 3

Static equilibrium, 61

Stepped bar

analytical method, 118, 122–123, 127–129

deflection values at nodes, 120, 125–126, 133

displacement calculation, 118, 119, 122

FEM by hand calculations, 118–121, 123–125, 130–132

nodal displacements, 117

post-processing, 140–141

preprocessing, 134–139

reaction calculation, 118, 120, 125, 132

reaction value, 121, 126

reaction values, 134

software results, 125–127, 133–134

stress calculation, 118, 119, 123, 124, 132

stress value at elements, 121, 126, 134

Stiffness matrix, 72, 99

Stokes theorems, 353

Strain, 60

Streamline, 292

Stress, 59

Stress analysis, rectangular plate with circular hole

analytical method, 240–241, 242–243

boundary of hole, 239

center of plate, 240

deflection pattern, 245

element options, 247

first principal stress distribution, 245

one-quarter of plate, 239, 240

plate thickness, 247

post-processing, 251

preprocessing, 246–251

real constants, 247

software results, 241–242, 243–245

sub cases, 238–240

tensile load at both edges, 238

tensile load at one edge, 238

validation of results, 246

Von Mises stress distribution, 243

Stress calculation

constant cross-section area, 66, 67, 70–71, 74, 79, 86

stepped bar, 118, 119, 123, 124, 132

truss, 153–154, 159–160, 165

varying cross-section area, 98, 101, 104, 106, 110

Stress-strain behavior, 61

Structural problems

direct approach, 49

discretization of, 48

element stiffness matrices and load vectors, 49

element strains and stresses, 50

nodal displacements, 50

proper interpolation, 48

variational approach, 49

weighted residual approach, 49–50

Super-matrix, 6

Symbolic expressions, MATLAB

differentiating symbolic expressions, 516

integrating symbolic expressions, 517

limits symbolic expressions, 517

simplifying symbolic expressions, 515

sums symbolic expressions, 518

symbolic expressions, solving equations, 519

taylor series symbolic expressions, 518

Symmetric matrix, 5

T

Temperature distribution, 286–289

Tensile forces, 59

Tension, 59

Thermal analysis

one-dimensional heat conduction problems, 257, 258–262

procedure of FEM, 258

two-dimensional heat conduction problems, 258, 285–286

Three-dimensional element, 54, 415–420

Three-node bar element, 64

Three-node element, 460

Three-strip line, 435–437

Triangular matrix, 5–6

Truss

analytical method, 147–150

angle calculation, 150–153, 162–164

definition, 145

deflection value at nodes, 155, 160–161, 166

2-D trusses, 145

element stiffness matrix, 146

element stress, 146

FEM by hand calculation, 150–160, 162–165

nodal displacements, 146

post-processing, 169–170

preprocessing, 167–169

reaction calculation, 154

software results, 160–162, 165–167

stress calculation, 153–154, 159–160, 165

stress values of elements, 155–156, 161–162, 166–167

Two-dimensional elements, 53–54

band matrix method, 392–395, 397–400

FEM to electrostatic problems, 382–400

iteration method, 392, 396–397

Laplace’s equation, 382–392

Two-dimensional problem, 285–286

Two-node bar element, 62–63

V

Variational approach, 49

Varying cross-section area

analytical method, 97–98, 103–104

deflection values at node, 101, 107, 111–112

displacement and stress, 97

displacement calculation, 98, 104

equivalent model, 99

FEM by hand calculations, 98–103, 104–106, 108–111

post-processing, 116–117

processing, 113–116

reaction calculation, 111

reaction value, 102, 112

software results, 106–108, 111–113

stress values, 102, 107, 112

Young’s modulus, 97

Vector

addition and subtraction, 28

algebra, 28

components of, 27

Del (▿) operator, 31–34

equality, 28

multiplication, 29–30

multiplication of scalar, 28

right-hand rule, 29

unit, 27

Vector analysis, 27–34

Von Mises stress distribution, 243

W

Wave equations, 372–373

Weighted residual approach, 49

Y

Young’s modulus

constant cross-section area, 65

varying cross-section area, 97