Introduction to the Finite Element Method

Introduction to Finite Element Analysis

Mohammad Tawfik

Introduction 2

Contents

1 Introduction ....................................................................................................................... 4

2 Numerical Solution of Boundary Value Problems ............................................................. 4

2.1 Objectives .................................................................................................................... 4

2.2 Why Approximate? ..................................................................................................... 4

2.3 Classification of Approximate Solutions of D.E.’s ....................................................... 5

2.4 Basic Concepts ............................................................................................................. 5

2.5 General Weighted Residual Method ........................................................................... 6

2.6 Collocation Method ..................................................................................................... 9

2.7 The Subdomain Method ............................................................................................ 11

2.8 The Galerkin Method ................................................................................................ 13

3 The Finite Element Method ............................................................................................. 17

3.1 Discretization ............................................................................................................. 18

3.2 Element Equations .................................................................................................... 18

3.3 Assembling elements’ equations .............................................................................. 21

3.4 Applying Boundary Conditions .................................................................................. 23

3.5 Solving ....................................................................................................................... 23

3.6 Secondary Variables .................................................................................................. 24

3.7 Generalized Procedure for Getting Interpolation Functions .................................... 24

4 Trusses ............................................................................................................................. 29

5 Beams and Frames ........................................................................................................... 33

5.1 Euler-Bernoulli Beam Theory .................................................................................... 33

5.2 Finite Element Trial Functions ................................................................................... 33

5.3 Beam Stiffness Matrix ............................................................................................... 36

6 Two Dimensional Elements .............................................................................................. 38

6.1 A Rectangular Element .............................................................................................. 38

6.2 Assembling 2-D Elements .......................................................................................... 42

7 Stationary Functional Approach ...................................................................................... 46

7.1 Some Definitions ....................................................................................................... 46

7.2 Applications ............................................................................................................... 47

7.2.1 The bar tensile problem ..................................................................................... 47

7.2.2 Beam Bending Problem ..................................................................................... 49

7.3 Plane Elasticity ........................................................................................................... 51

7.3.1 Strain-Displacement Relations ........................................................................... 52

7.3.2 Strain Energy ...................................................................................................... 52

Introduction to the finite element method 3

7.4 Finite Element Model of Plates in Bending ............................................................... 55

7.4.1 Displacement Function ...................................................................................... 55

7.4.2 Strain-Displacement Relation ............................................................................ 56

7.4.3 Constitutive Relations of Piezoelectric Lamina .................................................. 57

7.4.4 Stiffness and Mass Matrices of the Element ..................................................... 58

Introduction 4

1 Introduction

The finite element method has gained popularity for its simplicity and ability to model complex domains. The programming of the finite element method has been a task that a lot of software development firms have taken as its mission to provide the engineering market with packages that are capable of solving a multitude of problems starting from simple structure of fluid dynamics problems up to complex problems with coupled fields of electricity, mechanics of structures, fluid dynamics, and more.

In this course, we will aim at developing the necessary tools for the student to model physical problems using finite element method as well as programming simple problems of structure mechanics and some coupled problems. To achieve that aim, simple examples will be presented and the MATLAB code used to solve the problems will be listed. Further, the application of symbolic manipulators will be presented using the MATHEMATICA package.

2 Numerical Solution of Boundary Value Problems

Weighted Residual Methods

2.1 Objectives

In this section we will be introduced to the general classification of approximate methods. One of the approximate methods will receive attention, namely, the weighted residual method. Derivation of a system of linear equations to approximate the solution of an ODE will be presented using different techniques as an introduction to the finite element method.

2.2 Why Approximate?

The question that usually rises in the minds of engineers and students alike is, why do we study approximate methods? The main answer that should reply to that question is the ignorance of the humans! Up to this moment, scientists and engineers have been able to present a vast amount of mathematical models for physical phenomena, unfortunately, a very small percentage of those models, which are usually in the form of differential equations, have close form solutions! Thus, the necessity of solving those problems implies the use of approximate methods to get solutions for some specific problems of certain interest.

In the modern engineering life, packages that present solutions for problems using digital computers are everywhere. The understanding of how those packages perform approximate solutions for a certain physical problem is a necessity for the engineer to be able to use them. It is always a good idea to be able to predict how the output of the package is going to be in order to be able to distinguish the right results from errors that may occur due to bugs in the program or errors in the data given to the program.

On the other hand, an engineer who needs to develop a new technique for the solution of an advanced, or a new, problem, has to have a good background on how the old problems were solved.


2.3 Classification of Approximate Solutions of D.E.’s

Two main families of approximate methods could be identified in the literature. The discrete coordinate methods and the distributed coordinate methods.

Discrete coordinate methods depend on solving the differential relations at pre-specified points in the domain. When those points are determined, the differential equation may be approximately presented in the form of a difference equation. The difference equation presents a relation, based of the differential equation, between the values of the dependent variables at different values of the independents variables. When those equations are solved, the values of the dependent variables are determined at those points giving an approximation of the distribution of the solution. Examples of the discrete coordinate methods are finite difference methods and the Runge-Kutta methods. Discrete coordinate methods are widely used in fluid dynamics and in the solution of initial value problems.

The other family of approximate methods is the distributed coordinate methods. These methods, generally, are based on approximating the solution of the differential equation using a summation of functions that satisfy some or all the boundary conditions. Each of the proposed functions is multiplied by a coefficient, generalized coordinate, that is then evaluated by a certain technique that identifies different methods from one another. After the solution of the problem, you will obtain a function that represents, approximately, the solution of the problem at any point in the domain.

Stationary functional methods are part of the distributed coordinate methods family. These methods depend on minimizing/maximizing the value of a functional that describes a certain property of the solution, for example, the total energy of the system. Using the stationary functional approach, the finite element model of a problem may be obtained. It is usually much easier to present the relations of different variables using a functional, especially when the relations are complex as in the case of fluid structure interaction problems or structure dynamics involving control mechanisms.

The weighted residual methods, on the other hand, work directly on the differential equations. As the approximate solution is introduced, the differential equation is no more balanced. Thus, a residue, a form of error, is introduced to the differential equation. The different weighted residual methods handle the residue in different ways to obtain the values of the generalized coordinates that satisfy a certain criterion.

2.4 Basic Concepts

A differential equation in the dependent variable f with the independent variable x may be written in the form:

xgxfL or 0 xgxfL

Numerical Solution of Boundary Value Problems 6

Where L(.) is a linear differential operator and g(x) is the excitation function. The above equation has to have certain boundary conditions that render the solution unique. If we select different functions, Ψi, that satisfy all boundary conditions bud do not necessarily satisfy the differential equation, we may write the approximate solution of f(x) in the form of:

n

i

ii xaxf1

Where ai are the generalized coordinates, or the unknown coefficients. Applying the differential operator on the approximate solution, you get:

011

xgxLaxgxaLxgxfLn

i

ii

n

i

ii

Note that the right hand side of the above equation is not equal to zero. The non-zero value

of the right hand side is called the reside.

xRxgxLan

i

ii 1

The admissibility Conditions

The trial functions need to:

• Satisfy all boundary conditions!

• Be differentiable as much as the higher differentiation in the equation

2.5 General Weighted Residual Method

The general weighted residual method is based on integrating the residue multiplied by weighting functions and then equating the resulting equations by zero. The weighting functions are any set of functions that are continuous over the domain of the differential equation. A set functions may be polynomial, sinusoidal, hyperbolic, or any combination of functions. The number of functions needed should be equal to the unknown generalized coordinates to produce a set of equations that are solvable in the unknowns. Also, the weighting functions need to be linearly independent for the equations to be solvable.

Expanding the series of proposed solution functions, we get:

xRxgxLaxLaxLa nn ...2211

Multiplying by the weighting function and integrating, we get:


01

Domain

n

i

iij

Domain

j dxxgxLaxwdxxRxw

Domain

nnj

Domain

j dxxgxLaxLaxLaxwdxxRxw ...2211

In matrix form

Domain

ji

nninn

njijj

ni

dxxgxwa

kkk

kkk

kkk

1

1

1111

Where

Domain

ijij dxxLxwk

Example Problem

The bar tensile problem is a classical problem that describes the relation between the axially distributed loads and the displacement of a bar. Let’s consider the bar in Figure ‎2.1 with constant modulus of elasticity and cross section area. The force displacement relation is given by:

Figure ‎2.1. Sketch of a bar with distributed axial forces

02

2

F

x

uEA

Subject to the boundary conditions

0/&00 dxdulxux

Now, lets use the approximate solution

n

i

ii xaxu1

Substituting it into the differential equation, we get

xRF

dx

xdaEA

n

i

ii

12

2

Selecting weighting functions, Wi, and applying the method, we get:


l

ji

l

ij dxxwFadx

dx

xdxwEA

00

2

2

For the boundary conditions to be satisfied, we need a function that has zero value at x=0 and has a slope equal to zero at the free end. Sinusoidal functions are appropriate for this hence, using one-term series, we may use:

l

xSinx

2

For the weighting function, we may use a polynomial term. The simplest term would be 1.

ll

fdxadxl

xSin

lEA

0

1

0

2

22

Performing the integration, we get:

flal

xCos

lEA

l

1

022

When the equation is solved in the unknown coefficient (generalized coordinate), we get:

EA

fl

EA

fl

lEA

fla

22

1 637.02

2

Then, the approximate solution for this problem becomes

l

xSin

EA

flxu

2637.0

2

Now we may compare the obtained solution with the exact one that may be obtained from solving the differential equation. The maximum displacement and the maximum strain may be compared with the exact solution. The maximum displacement is

5.0637.02

exactEA

fllu

And maximum strain is:

0.10.10 exactEA

lfux


2.6 Collocation Method

The idea behind the collocation method is similar to that behind the buttons of your shirt! Assume a solution, and then force the residue to be zero at the collocation points.

0jxR

The collocation method may be seen as one of the weighted residual family when the

weighting function becomes the delta function. The delta function is one that may be

described as:

jj

j

jj

xFdxxFxx

xx

xxxx

0

1

Now, if we select a set of points xj inside the domain of the problem, we may write down

the integral of the residue, multiplied by the delta functions, as follows:

01

j

n

i

jiij xFxLaxR

Which gives

n

ji

nninn

njijj

ni

xg

xg

xg

a

kkk

kkk

kkk

1

1

1

1111

Where

jiij xLk


Figure ‎2.2. A sketch of the differences between the exact and approximate solutions

Example Problem

Applying this method to the bar tensile problem described before, we get:

xRxF

dx

xdaEA

n

i

ii

12

2

Evaluating the residue at the collocation points, we get

0

12

2

j

n

i

ji

i xFdx

xdaEA

In matrix form

nnnnnn

n

n

xF

xF

xF

a

a

a

kkk

kkk

kkk

2

1

2

1

21

22212

12111

...

...

...

Where

jxx

iij

dx

xdEAk

2

2

Solve the above system for the “generalized coordinates” ai to get the solution for u(x)

Using Admissible Functions

• For a constant forcing function, F(x)=f


• The strain at the free end of the bar should be zero (slope of displacement is zero). We may use:

l

xSinx

2

Using the function into the DE:

l

xSin

lEA

dx

xdEA

22

2

2

2

A natural selection for the collocation point may be the central point of the bar. Substituting

by the value of x=l/2, we get

EA

fl

EA

fl

SinlEA

fa

2

2

2

21 57.024

42

• Then, the approximate solution for this problem is:

l

xSin

EA

flxu

257.0

2

Which gives the maximum displacement to be

5.057.02

exactEA

fllu

And maximum strain to be:

0.19.00 exactEA

lfux

2.7 The Subdomain Method

The idea behind the subdomain method is to force the integral of the residue to be equal to zero on a subinterval of the domain. The method may be also seen as using the unit step functions as weighting functions. The unit step function may be described by the following relation:

1

11

0

1

0

1

jj

jjjj

j

jj

xxorxx

xxxxxUxxU

xx

xxxxU

Hence the integral of the weighted residual method becomes


0

1

j

j

x

x

dxxR

Substituting using the series solution

0

11

1

j

j

j

j

x

x

n

i

x

x

ii dxxgdxxLa

Figure ‎2.3. Sketch of the differences between the exact and approximate solutions

For the bar application

xRxF

dx

xdaEA

n

i

ii

12

2

Performing the integration and equating by zero

11

12

2 j

j

j

j

x

x

n

i

x

x

ii dxxFdx

dx

xdaEA

Which gives the equation in matrix form as

11

2

2 j

j

j

j

x

x

i

x

x

i dxxFadxdx

xdEA

Using Admissible Function

l

xSinx

2


The differentiation will give

l

xSin

lEA

dx

xdEA

22

2

2

2

Since we only have one term in the series, we will perform the integral on one subdomain; i.e. the whole domain

ll

fdxadxl

xSin

lEA

0

1

0

2

22

Performing the integral

flal

xCos

lEA

l

1

022

Evaluating the generalized coordinate

EA

fl

EA

fl

lEA

fla

22

1 637.02

2

Then, the approximate solution for this problem is:

l

xSin

EA

flxu

2637.0

2

Which gives the maximum displacement to be:

5.0637.02

exactEA

fllu

And maximum strain to be:

0.10.10 exactEA

lfux

2.8 The Galerkin Method

The Galerkin method uses the proposed solution functions as the weighting functions. Thus the solution procedure will require the selection of one set of functions. That method has proven very efficient and accurate as a weighted residual method. Many numerical solutions methods are derived from the Galerkin method. The Galerkin method may be presented by the following integral


0Domain

j dxxxR

When substituting with the series solution, the weighted residual integral will become

01

Domain

j

n

i Domain

iji dxxgxdxxLxa

Applying the Galerkin method to the bar problem, we get

Domain

j

n

i Domain

iji dxxFxdx

dx

xdxaEA

12

2

Which in matrix form becomes

Domain

ji

Domain

ij dxxFxadx

dx

xdxEA

2

2

Solve the above system for the “generalized coordinates” ai to get the solution for u(x). Let’s use the same function as in the previous methods

l

xSinx

2

Substituting with the approximate solution:

Domain

j

n

i Domain

iji dxxFxdx

dx

xdxaEA

12

2

We have

ll

fdxl

xSina

lEAdx

l

xSin

l

xSina

lEA

0

2

1

2

0

1

2

22222

Which gives

lla

lEA

2

221

2

Substituting and solving for the generalized coordinate, we get

EA

fll

EA

fa

2

3

2

1 52.016


In most structure mechanics problems, the differential equation involves second derivative

or higher for the displacement function. When Galerkin method is applied for such

problems, you get the proposed function multiplied by itself or by one of its function family.

This suggests the use of integration by parts. Let’s examine this for the previous example. Substituting with the approximate solution: (Int. by Parts)

Domain

ij

l

ij

Domain

ij dx

dx

xd

dx

xd

dx

xdxdx

dx

xdx

0

2

2

But the boundary integrals are equal to zero since the functions were already chosen to

satisfy the boundary conditions. Evaluating the integrals will give you the same results.

lla

lEA

2

221

2

EA

fll

EA

fa

2

3

2

1 52.016

So, what did we gain by performing the integration by parts?

• The functions are required to be less differentiable

• Not all boundary conditions need to be satisfied

• The matrix became symmetric!

The above gains suggested that the Galerkin method is the best candidate for the derivation of the finite element model as a weighted residual method.

Homework #1

Figure ‎2.4. A simply supported beam

)(4

4

xFdx

wd

subject to 0

)()0(0)()0(

2

2

2

2

dx

lwd

dx

wdandlww

Exact Solution for this problem is


12/110

3

15

7

412

2/1060

13

12)(

23

3

xxxx

xxx

xw

• Solve the beam bending problem, for beam displacement, for a simply supported beam with a load placed at the center of the beam using

– Any weighting function

– Collocation Method

– Subdomain Method

– Galerkin Method

• Use three term Sine series that satisfies all BC’s

• Write a program that produces the results for n-term solution.


3 The Finite Element Method

2nd order DE’s in 1-D

The finite element method is based on the weighted residual methods presented in the previous chapter. The main difference of the finite element method is that the proposed solution functions have generalized coordinates, coefficients, that are equal to the physical quantity of the unknown variable at some points inside the domain. Hence, the proposed solution becomes an interpolation function for the values intermediate to those points.

Many physical problems may be approximately presented by 1-D differential equation. Such a problem will be characterized by having all dependent variables subject in one independent variable, usually x coordinate, and in initial boundary value problems, the time will be introduced as another independent variable but the problem will be still considered a 1-D problem.

Figure ‎3.1 presents a sketch for a general 1-D 2nd order problem. The problem may be considered as one describing the steady state of a heat transfer problem with general boundary conditions. Different terms are described in the sketch.

Figure ‎3.1. A sketch of the 1-D 2nd

order problem

The Finite Element Method 18

The problem then may be stated as; solve:

Lx

fcudx

dua

dx

d

0

0

Subject to:

00 ,0 Qdx

duauu

Lx

Now, we will apply the steps of the finite element method to model this problem

approximately.

3.1 Discretization

At the first step of the finite element modeling, we discretize the domain into elements that all lie inside the domain. Those elements are connected at nodes. At each node, the value of the dependent variable will be called a degree of freedom.

Figure ‎3.2. Discretization of the domain

3.2 Element Equations

Let’s concentrate our attention to a single element. The same DE applies on the element level; hence, we may follow the procedure for weighted residual methods on the element level using the boundary conditions as the values of the dependent function at the nodes.


21

0

xxx

fcudx

dua

dx

d

Subject to

21

2211

21

,

,,

Qdx

duaQ

dx

dua

uxuuxu

xxxx

At this step, let’s think about the interpolation of the values of the function over the element, between the nodes. The most common interpolation function is the polynomial.

Figure ‎3.3. Single 1-D element with the values of the dependent function and the external excitation function determined at the nodes.

Using linear interpolation, we may write

01 bxbxu

Forcing the value of the function to be equal to the given values at the nodes, we get

01111 bxbuxu

02122 bxbuxu

When we solve for the unknown values b0 and b1, and rearranging, we get

2

12

11

12

2 uxx

xxu

xx

xxxu

Which may be written as

euxu

uuuxu

2

1

212211

It may be noticed directly that after rearranging the interpolation function and collecting

terms containing the nodal-values of the dependent variable we get two linear functions


that are used as the proposed solution for the weighted residual method. Note also that the

proposed solution satisfies the essential boundary conditions of the problem; i.e. the values

of the function at the nodes. The sketch of the functions may be seen in Figure ‎3.4

Figure ‎3.4. Sketch of the proposed solution functions

Note also that the proposed functions do not satisfy the original admissibility conditions since we are planning to use the Galerkin method as the weighted residual method of choice.

Assuming constant domain properties a and c may be drawn out of the differentiation in the differential equation to get

21

2

2

0

xxx

fcudx

uda

Using the proposed solution obtained in this section and applying the Galerkin method we get

02

2

Domain

jiijii

j dxfxuxxcudx

xdxa

Note that

Domain

ij

x

x

ij

Domain

ij dx

dx

xd

dx

xda

dx

xdxadx

dx

xdxa

2

1

2

2

And

ee hdx

xd

hdx

xd 1,

1 21

Thus, we may write the integral as


02

1

2

1

2

1

x

x

jiij

x

x

iij

x

x

ij dxfxuxxcdxu

dx

xd

dx

xda

dx

xdxa

For j=1 we may get the first equation

01111 2

1

22

121

2

22212

x

x eeeeeeee

dxfh

xxu

h

xx

h

xxcu

h

xxcu

hau

ha

ha

ha

Which gives

0263

21

ee

e

e

e

fhu

ch

h

au

ch

h

a

Repeating for using j=2, we get

0236

21

ee

e

e

e

fhu

ch

h

au

ch

h

a

In matrix form

1

1

221

12

611

11

2

1 ee

e

fh

u

uch

h

a

The above equation is called the element equation.

3.3 Assembling elements’ equations

In the previous section, we obtained the element equation that relates the element degrees of freedom to the externally applied fields. The element equation may be rewritten in compact form as

2

1

2

1

43

21

f

f

u

u

kk

kk

Now, let’s consider two adjacent elements as in Figure ‎3.5. The element equation for each element may be written as

2

2

2

1

2

2

2

1

2

4

2

3

2

2

2

1

f

f

u

u

kk

kk

1

2

1

1

1

2

1

1

1

4

1

3

1

2

1

1

f

f

u

u

kk

kk


Where the superscript indicates the element number.

Figure ‎3.5. Sketch of two adjacent elements

Note that for the continuity of the field values, the value of the function at the right-hand side of the first element should be equal to that of the left-hand value of the second element. Hence, we may add the second equation of the first element to the first equation of the second element to obtain three equations in the three unknown degrees of freedom.

3

2

1

3

2

1

3

2

1

2

4

2

3

2

2

2

1

1

4

1

3

1

2

1

1

0

0

Q

Q

Q

f

f

f

u

u

u

kk

kkkk

kk

The vector {Q} presents any external fields that are concentrated at the nodes; e.g. Support reactions in bar problems or fixed temperature in heat transfer problems.

Example

Let us reconsider the bar problem presented in Figure ‎2.1.

Divide the bar into N number of elements. The length of each element will be (L/N). Derive the element equation from the differential equation for constant properties and externally applied force:

02

1

2

x

x

jiij

e

dxfudx

d

dx

d

h

EA

The element equation will be obtained as follows

1

1

211

11

2

1 e

e

e

e

fh

u

u

h

EA

Note that if the integration is evaluated from 0 to he, where he is the element length, the same results will be obtained. That is true due to the fact that the bound integration evaluates the area under the curve, and the area does not change if the axes are transformed from the origin to x1.

Now, let us consider two–Element bar example,

1

2

1

1

1

2

1

1

11

11

f

f

u

u

h

EA

e


2

2

2

1

2

2

2

1

11

11

f

f

u

u

h

EA

e

Assembling the equations as presented above, we obtain

0

0

1

2

1

2110

121

011

3

2

1 Rfh

u

u

u

h

EA e

e

Where R is the reaction force at the support of the bar. Note that the external forces

corresponding to the second and third nodes are equal to zero as the problem does not

have any external concentrated forces.

3.4 Applying Boundary Conditions

For the bar with fixed left side and free right side, we may force the value of the left-displacement to be equal to zero; which is the given boundary condition. The obtained equation becomes

0

0

1

2

1

2

0

110

121

011

3

2

Rfh

u

uh

EA e

e

Which may be separated into two equations as follows

1

2

211

12

3

2 e

e

fh

u

u

h

EA

Rfh

uh

EA e

e

2

2

Now, we may solve the first equation in the unknown values of the nodal displacements and then substitute into the second equation to obtain the reaction force.

3.5 Solving

Solving first equation for the displacements, we get

4

3

2

2

3

2

EA

fh

u

ue


Which is the exact value of the displacement that may be obtained if the exact solution was

utilized.

3.6 Secondary Variables

Using the values of the displacements obtained, we may get the value of the reaction force by substituting into the second equation.

Rfhfh ee 22

3

efhR 2

Which is the exact value of the reaction force.

3.7 Generalized Procedure for Getting Interpolation Functions

The need for different types of elements for different structures implied the need for a generalized methodology to increase the flexibility of the finite element programming. A large number of proposed trial/interpolation functions is available in the literature, but the continuous need for development required the development of generalized procedure. In section ‎3.2, we obtained the element equations for two-node element with linear interpolation functions. Let’s re-derive the equations introducing a more general procedure then use the same procedure for the derivation of a three-node element.

The interpolation Function

xaaxu 21)(

May be written in matrix form as

axHxu

Then, forcing the function to equal the values at the nodes gives

aHuu 00 1

alHulu 2

This may be written in matrix form as

aT

a

a

la

a

lH

H

u

u

2

1

2

1

2

1

1

010

Solving for the unknown coefficients, we get


2

1

2

1 1101

u

u

lla

a

Or euTa1

Substituting into the interpolation function, we get

ee uxNuTxHxu 1

Expanding the terms, we find that

2

1

2

1111

011

u

u

l

x

l

x

u

u

ll

xxu

Or 211)( u

l

xu

l

xxu

Now, we may write the element equation using the new notation to be

000

l

e

l

xx fdxxNudxxNxNEA

Note that

ll

xx dxll

l

ldxxNxN00

11

1

1

11

111

11

11

0

22

22

i

l

ldx

ll

ll

Which, again, gives the element equation as

1

1

211

11

2

1 e

e

e

e

fh

u

u

h

EA

Hw 1, x ;

Hwx D Hw, x ;

Hwxx D Hwx, x ;


Let us now try to follow the same procedure to obtain the element matrix for an element with three nodes. Since we are considering an element with three nodes, we will need a parabolic interpolation function

Figure ‎3.6. Three-node bar element 2

321)( xaxaaxu

This function may be written in matrix form as

axHxu

Using the values of the function at the nodes, we get

alHulu 2/2/ 2

x 0; Tb1 Hw;

x A; Tb3 Hw;

Clear x ;

TB Tb1, Tb3

TBINV Inverse TB

1, 0 , 1, A

1, 0 ,1

A,

1

A

NN Hw.TBINV

1x

A,

x

A

A L

Mm Inner Times, Transpose Hw , Hw ;

Mb Integrate Mm, x, 0, A

Mb Transpose TBINV .Mb.TBINV MatrixForm

L

L,L2

2,

L2

2,

L3

3L

3

L

6

L

6

L

3

CQC Inner Times, Transpose Hwx , Hwx ;

CQC Integrate CQC, x, 0, A

Kb Transpose TBINV .CQC.TBINV MatrixForm

0, 0 , 0, L1

L

1

L

1

L

1

L

aHuu 00 1


alHulu 3

In matrix form

aT

a

a

a

lH

lH

H

u

u

u

3

2

1

3

2

1

2/

0

Giving

euTa1

Finally

ee uxNuTxHxu 1

Expanding the function, we get

3

23

2

13

12

32

3

12

11

31

3

21

2 uxx

xx

xx

xxu

xx

xx

xx

xxu

xx

xx

xx

xxxu

Which may be also written as

3212/

2/

2/2/2/

2/u

l

lx

l

xu

l

lx

l

xu

l

lx

l

lxxu

Where “l” is the element length. The proposed solution functions may be plot as in Figure ‎3.7. Note that each of the three functions is equal to one at its corresponding point and zero at the other two points.

Figure ‎3.7. Plot of the proposed solution functions for A 3-node element


Hw 1, x, x x ;

Hwx D Hw, x ;

Hwxx D Hwx, x ;

x 0; Tb1 Hw;

x A; Tb3 Hw;

x A 2; Tb2 Hw;

Clear x ;

TB Tb1, Tb2, Tb3

TBINV Inverse TB

1, 0, 0 , 1,A

2,A2

4, 1, A, A

2

1, 0, 0 ,3

A,

4

A,

1

A,

2

A2,

4

A2,

2

A2

NN Hw.TBINV

13 x

A

2 x2

A2,

4 x

A

4 x2

A2,

x

A

2 x2

A2

A L

Mm Inner Times, Transpose Hw , Hw ;

Mb Integrate Mm, x, 0, A

Mb Transpose TBINV .Mb.TBINV MatrixForm

L

L,L2

2,

L3

3,

L2

2,

L3

3,

L4

4,

L3

3,

L4

4,

L5

52L

15

L

15

L

30

L

15

8L

15

L

15

L

30

L

15

2L

15

CQC Inner Times, Transpose Hwx , Hwx ;

CQC Integrate CQC, x, 0, A

Kb Transpose TBINV .CQC.TBINV MatrixForm

0, 0, 0 , 0, L, L2, 0, L

2,

4 L3

37

3L

8

3L

1

3L

8

3L

16

3L

8

3L

1

3L

8

3L

7

3L


4 Trusses

A truss is a set of bars that are connected at frictionless joints. The bar does not undergo deformations normal to its axis but when another one that is connected to and is not coaxial with it deforms it rotates around its axis. Since the truss is mainly a set ob bars that generally oriented in the plane and connected to one another, then we have to consider its general motion. Now, the problem lies in the transformation of the local displacements of the bar, which are always in the direction of the bar, to the global degrees of freedom that are generally oriented in the plain.

Figure ‎4.1. A truss element generally oriented in the plane (a) Displacements and forces in the bar local coordinate (b) Resolution of the displacements and forces in the global coordinates

Consider the truss element presented in Figure ‎4.1. The force-displacement equations in the local coordinates of the element may be presented by

0

0

0000

0101

0000

0101

2

1

2

2

1

1

F

F

v

u

v

u

h

EA

From the geometry, the local and global coordinates may be related using the relation

DOFdTransforme

DOFLocalv

u

v

u

CosSin

SinCos

CosSin

SinCos

v

u

v

u

2

2

1

1

2

2

1

1

00

00

00

00

Or

DOF

dTransformeDOFLocal T

Trusses 30

Where and are the displacements in local and global coordinates respectively and

the transformation matrix is

CosSin

SinCos

CosSin

SinCos

T

00

00

00

00

The Equation of Motion Becomes

FTK

Transforming the forces:

FTTKT11

Note that

TTT 1

Finally

FK

where

TKTKT

Where

0000

0101

0000

0101

h

EAK

Element Stiffness Matrix in Global Coordinates becomes

CosSin

SinCos

CosSin

SinCos

CosSin

SinCos

CosSin

SinCos

h

EAK

00

00

00

00

0000

0101

0000

0101

00

00

00

00

Or


22

22

22

22

22

12

2

1

22

12

2

1

22

12

2

1

22

12

2

1

SinSinSinSin

SinCosSinCos

SinSinSinSin

SinCosSinCos

h

EAK

Example

Use the finite element analysis to find the displacements of node C and the stresses in each element.

Figure ‎4.2. The truss of the example

The stiffness matrix in global coordinates of each of the truss elements may be written as

0000

0101

0000

0101

1

L

EAK

1010

0000

1010

0000

2

L

EAK

3536.03536.03536.03536.0

3536.03536.03536.03536.0

3536.03536.03536.03536.0

3536.03536.03536.03536.0

3

L

EAK

Assembling the matrices into the global stiffness matrix, we get

3536.13536.0103536.03536.0

3536.03536.0003536.03536.0

101000

000101

3536.03536.0003536.03536.0

3536.03536.0013536.03536.1

L

EAK

Trusses 32

Similarly, the force vector is obtained as

P

P

F

F

F

F

F

F

F

F

F

F

Fy

x

y

x

y

x

y

x

y

x

2

2

2

1

1

3

3

2

2

1

1

The boundary conditions of the truss are

02211 VUVU

Removing the corresponding rows and columns, we get

P

P

V

U

L

EA

23536.13536.0

3536.03536.0

3

3

Solving for the displacements, we get

EA

PLV

EA

PLU

3 ,828.5 33

Using the secondary equations to get the reaction forces, we get

PFFPFPF yxyx 3 ,0, , 2211

Performing post-computation, we get

e

e

e

e

e

A

P

A

P 21

e

e

eee

e

u

u

L

EA

P

P

2

1

2

1

11

11

2

2

1

1

2

2

1

1

00

00

00

00

v

u

v

u

CosSin

SinCos

CosSin

SinCos

v

u

v

u

Finally

A

P

A

P2 ,

3 ,0 )3()2()1(


5 Beams and Frames

Beams are the most-used structural elements. Many real structures may be approximated as beam elements. Two main beam theories; Euler-Bernoulli beam theory and Timoshenko beam theory. In this chapter, we will be focusing out attention to the Euler-Bernoulli beam theory.

5.1 Euler-Bernoulli Beam Theory

The main assumption in the Euler-Bernoulli beam theory is that the beam’s thickness is too small compared to the beam length. That assumption resulted in that the sheer deformation of the beam may be neglected without much error in the analysis. The equation governing the deformation of and E-B beam under transverse loading may be written as follows

)(2

2

2

2

xFdx

wdxEI

dx

d

Subject to any combination of the following boundary conditions. At x=0

03

3

0

000 P

dx

wdEIorww

Indicating that either the deflection or the sheer force is specified

02

2

0' 0

00

Mdx

wdEIorw

dx

dw

Indicating that either the slope or the bending moment is specified. At x=l

ll Pdx

lwdlEIorwlw

3

3

Indicating that either the deflection or the sheer force is specified

ll M

dx

lwdlEIorw

dx

ldw

2

2'

Indicating that either the slope or the bending moment is specified.

Where w is the transverse deflection of the beam, E is the modulus of elasticity, I is the second moment of area, and F is the transverse load in the positive direction of w.

5.2 Finite Element Trial Functions The finite element model for the EB beam will depend on the knowledge of four boundary conditions, namely, the deflections and slopes at each node of the element (see Figure ‎5.1).

Beams and Frames 34

That will require the use of a third order polynomial to interpolate the deflection function in the element.

3

4

2

321)( xaxaxaaxw

Figure ‎5.1. Sketch of a beam element

Using the same procedure adopted for the bar element, we get

axHxw

Enforcing the boundary conditions

aHww 00 1

alHwlw 2

aHww x 0'0' 1

alHwlw x 2''

In matrix form

aT

a

a

a

a

lH

lH

H

H

w

w

w

w

x

x

4

3

2

1

2

2

1

1

0

0

'

'

Or

4

3

2

1

2

32

2

2

1

1

3210

1

0010

0001

'

'

a

a

a

a

ll

lll

w

w

w

w

Solving


2

2

1

1

2323

22

4

3

2

1

'

'

1212

13230010

0001

w

w

w

w

llll

llll

a

a

a

a

Which gives euxNaxHxw Where

2

32

3

3

2

2

3

32

3

3

2

2

23

2

231

l

x

l

xl

x

l

xl

x

l

xx

l

x

l

x

xNxNT

Plotting the four functions, we get the graph in Figure ‎5.2. Notice that N1 and N3 functions are equal to one at x=0 and x=1 respectively while equal to zero at x=1 and x=0 respectively which can also be related to the fact that N1 and N3 correspond to w1 and w2 respectively. Similarly, N2 and N4 have zero slopes at x=l and x=0 respectively and are corresponding to w’1 and w’2 respectively. Now we may write down the displacement function in terms of the nodal displacements as follows

ee wxNwTxHaxHxw 1

Or

4

1i

ii wxNxw

Figure ‎5.2. A plot of the beam interpolation functions.

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

N(x

)

X

N1

N2

N3

N4

Beams and Frames 36

The symbolic manipulator MATHEMATICA may be used to derive the trial functions N(x) as presented below

5.3 Beam Stiffness Matrix Recall that the governing equation is

)(2

2

2

2

xFdx

wdxEI

dx

d

Substituting with the series solution obtained in the previous section

4

1i

ii wxNxw

The governing equation becomes

)()(4

12

2

2

2

xRxFwdx

NdxEI

dx

d

i

ii

Applying Galerkin method:

In[2]:= Hw 1, x, x x, x x x ;

Hwx D Hw, x ;

Hwxx D Hwx, x ;

In[5]:= x 0; Tb1 Hw; Tb2 Hwx;

x L; Tb3 Hw; Tb4 Hwx;

Clear x ;

TB Tb1, Tb2, Tb3, Tb4

TBINV Inverse TB

Out[8]= 1, 0, 0, 0 , 0, 1, 0, 0 , 1, L, L2, L

3, 0, 1, 2 L, 3 L

2

Out[9]= 1, 0, 0, 0 , 0, 1, 0, 0 ,3

L2,

2

L,

3

L2,

1

L,

2

L3,

1

L2,

2

L3,

1

L2

In[10]:= NN Hw.TBINV

Out[10]= 13 x2

L2

2 x3

L3

, x2 x2

L

x3

L2,

3 x2

L2

2 x3

L3,

x2

L

x3

L2

In[11]:= Simplify NN

Out[11]=L x 2 L 2 x

L3

,L x 2 x

L2

,3 L 2 x x2

L3

,x2 L x

L2


ee l

j

i

ii

l

j dxNxFwdx

NdxEI

dx

ddxNxR

0

4

12

2

2

2

0

)()(

Using integration by parts, twice, we get:

0)(0

4

12

2

2

2

el

j

i

i

ji dxNxFwdx

Nd

dx

NdxEI

In matrix form

ee l

xx

e

l

xxxx dxNxFwdxNNxEI00

)(

Using MATHEMATICA, we get

Giving the stiffness matrix to be

In[2]:= Hw 1, x, x x, x x x ;

Hwx D Hw, x ;

Hwxx D Hwx, x ;

In[5]:= x 0; Tb1 Hw; Tb2 Hwx;

x L; Tb3 Hw; Tb4 Hwx;

Clear x ;

TB Tb1, Tb2, Tb3, Tb4

TBINV Inverse TB

Out[8]= 1, 0, 0, 0 , 0, 1, 0, 0 , 1, L, L2, L

3, 0, 1, 2 L, 3 L

2

Out[9]= 1, 0, 0, 0 , 0, 1, 0, 0 ,3

L2,

2

L,

3

L2,

1

L,

2

L3,

1

L2,

2

L3,

1

L2

In[10]:= NN Hw.TBINV

Out[10]= 13 x2

L2

2 x3

L3

, x2 x2

L

x3

L2,

3 x2

L2

2 x3

L3,

x2

L

x3

L2

In[11]:= Simplify NN

Out[11]=L x 2 L 2 x

L3

,L x 2 x

L2

,3 L 2 x x2

L3

,x2 L x

L2

Two Dimensional Elements 38

6 Two Dimensional Elements

More complicated and realistic problems can not be approximated by one dimensional elements. Problems such as 2-D fluid flow, heat conduction, plate bending, ocean dynamics, and others need a two dimensional element to be used. We will focus in this chapter on 2-D rectangular elements with 1 DOF per node. Later, we will introduce more general elements.

6.1 A Rectangular Element For the rectangular element presented in Figure ‎6.1, the four nodes have the coordinates shown relative to local coordinate system sat has its origin at point 1. For the problem where one dependent variable is being investigated, we may interpolate that function using

Figure ‎6.1. Sketch of a 2-D rectangular element.

xyayaxaayxf 4321,

Which is a 2-D function composed of two linear terms multiplied by each other. The function may be presented in the general form

ayxHyxf ,,

Now, relating the generalized coordinated to the degrees of freedom at each node, we get

aHff 0,00,0 1

aaHfaf ,00, 2

abaHfbaf ,, 3

abHfbf ,0,0 4

In matrix form


aT

a

a

a

a

bH

baH

aH

H

f

f

f

f

4

3

2

1

4

3

2

1

0

,

0,

0,0

Or

4

3

2

1

4

3

2

1

001

1

001

0001

a

a

a

a

b

abba

a

f

f

f

f

Inverting the transformation matrix, we get

4

3

2

1

4

3

2

1

1111

100

1

0011

0001

f

f

f

f

abababab

bb

aa

a

a

a

a

Giving the interpolation function as

efyxNayxHyxf ,,,

Where

ab

xy

b

yab

xyab

xy

a

xab

xy

b

y

a

x

yxNyxNT

1

,,

Figure ‎6.2 presents the variation of the functions N1 through N4 in the 2-D domain. Note that each of the function is equal to one at the node that corresponds to it while equals zero at the three other nodes.


(a) (b)

(c) (d)

Figure ‎6.2. Plot of the 2-D interpolation functions obtained earlier (a) presents the variation of N1, (b) presents the variation of N2, (c) presents the variation of N3, (d) presents the variation of N4.

Example: Laplace Equation

The Laplace equation is used in many engineering applications such as the 2-D potential fluid flow problem. The Laplace equation is presented by the following notation

02

Where is called the Laplace operator. In expanded form, the equation will be

02

2

2

2

yx

Introducing 2-D series solution

ei

ii yxNyxN ,,4

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

0.3

0.6

0.9

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

N1

x

y

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

0.3

0.6

0.9

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

N2

x

y

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

0.3

0.6

0.9

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

N3

x

y

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

0.3

0.6

0.9

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

N4

x

y


Applying the Galerkin method and integrating by parts, the element equation becomes

0 e

Area

yyxx

Boundary

yx dANNNNdNNNN

It can be shown that the boundary integral will be equal to zero. Finally, the equation will be

0e

Area

yyxx dANNNN

Performing the integration over the area we get

0

222

222

222

222

6

1

22222222

22222222

22222222

22222222

e

babababa

babababa

babababa

babababa

ab

This may also be obtained using symbolic manipulators as follows

Defining the interpolation function and it’s derivatives

Evaluating the transformation matrix

Evaluating the integral of the element equation

Hw 1, x, y, x y ;

Hwx D Hw, x ;

Hwy D Hw, y ;

x 0; y 0; Tb1 Hw;

x a; y 0; Tb2 Hw;

x a; y b; Tb3 Hw;

x 0; y b; Tb4 Hw;

Clear x, y ;

TB Tb1, Tb2, Tb3, Tb4 ;

TBINV Inverse TB ;

NN Hw.TBINV MatrixForm

1x

a

y

b

xy

ab

x

a

xy

ab

xy

ab

y

b

xy

ab

x Inner Times, Transpose Hwx , Hwx ;

y Inner Times, Transpose Hwy , Hwy ;

o x y;

1 Integrate Integrate o, x, 0, a , y, 0, b ;

FullSimplify Transpose TBINV . 1.TBINV ;

MatrixForm


6.2 Assembling 2-D Elements

The Logistic Problem!

In the 2-D problems, the numbering scheme, usually, is not as straight forward as the 1-D problem. For example, in 1-D problem as that shown in Figure ‎6.3, we can easily relate the nodes’ numbers to the elements’ numbers. Although the finite element method does not mandate the numbering scheme, it is usually more practical to use the simplest scheme.

Figure ‎6.3. 1-D elements’ connectivity and nummbering

Figure ‎6.4. Example of 2-D connectivity and numbering scheme

Table ‎6.1. Relation between local and global node numbering for element number 5

Local Node Number

Global Node Number

1 5

2 6

3 9

4 8

In a 2-D problem, such as that shown in Figure ‎6.4, the node and element numbering becomes more complicated than the 1-D problem, although, an easy relation may still be

a2 b2

3ab

a

6b

b

3a

a2 b2

6ab

a

3b

b

6a

a

6b

b

3a

a2 b2

3ab

a

3b

b

6a

a2 b2

6ab

a2 b2

6ab

a

3b

b

6a

a2 b2

3ab

a

6b

b

3a

a

3b

b

6a

a2 b2

6ab

a

6b

b

3a

a2 b2

3ab


obtained between the nodes’ and the elements’ numbering it is not advised to keep on applying this relation inside the program. Rather, a register should be filled in which each number is related to the nodes to which it is connected. Table ‎6.1 shows the register entry for element number 5. The register entry contains the local node number and the corresponding global node number. From the register, it is easy to determine which entry in the global matrix each of the element matrix entries should be added to. Figure ‎6.5 shows, graphically, the locations of in the global matrix to which the element matrix entries should be added.

12 11 10 9 8 7 6 5 4 3 2 1

1

2

3

4

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

2,3 2,4 2,2 2,1 6

7

4,3 4,4 4,2 4,1 8

3,3 3,4 3,2 3,1 9

10

11

12 Figure ‎6.5. Entries of the global matrix corresponding to each entry of the element matrix of element number 5.

The procedure presented above for the element number 5 is now going to be generalized for the rest of the elements. Table ‎6.2 presents the local-to-global correspondence node numbering for the 2-D problem presented in Figure ‎6.4.

Now we need to apply the procedure used for element number 5, explained above, to the rest of the elements.

Table ‎6.3 presents an algorithm that may be used for assembling the global matrix for a 2-D problem with a single degree of freedom per node. The algorithm is based on scanning the nodes of each element, reading their corresponding global number and adding the local matrix entry to the corresponding global matrix entry. Thus, two loops on the elements’ nodes are used together with the main elements’ loop.


Table ‎6.2. Element number connectivity table.

Node Number Element Number

4 3 2 1

4 5 2 1 1

7 8 5 4 2

10 11 8 7 3

5 6 3 2 4

8 9 6 5 5

11 12 9 8 6

Table ‎6.3. Algorithm for Assembling Global Matrix

1. Create a square matrix “A”; N*N (N=Number of nodes) 2. For the ith element 3. Get the element matrix “B” 4. For the jth node 5. Get its global number k 6. For the mth node 7. Get its global number n 8. Let Akn=Akn+Bjm 9. Repeat for all m 10. Repeat for all j 11. Repeat for all i

Using the symbolic manipulator MATHEMATICA, a code was created to assemble the global matrix for the above problem. The code uses the results obtained from the previous part that creates the element matrix, then, it creates the element-node registry which is then used to assemble the global matrix.

Filling the element-node correspondence table

Assembling the global matrix

Nx 2;

Ny 3;

Elements Table 0, ii, 1, Nx Ny , jj, 1, 4 ;

For ii 1, ii Nx, ii , For jj 1, jj Ny, jj , ElNm jj Ny ii 1 ;

Elements ElNm , 1 ii jj 1 Nx 1 ;

Elements ElNm , 2 ii 1 jj 1 Nx 1 ;

Elements ElNm , 3 ii 1 jj Nx 1 ;

Elements ElNm , 4 ii jj Nx 1 ; ;

Elements MatrixForm

1 2 5 4

4 5 8 7

7 8 11 10

2 3 6 5

5 6 9 8

8 9 12 11


Finally, the assembled global matrix will look as follows

KGlob Table 0, ii, 1, Nx 1 Ny 1 , jj, 1, Nx 1 Ny 1 ;

For ii 1, ii Nx Ny, ii ,

For jj 1, jj 4, jj ,

NodeJJ Elements ii, jj ;

For ll 1, ll 4, ll ,

NodeLL Elements ii, ll ;

KGlob NodeJJ , NodeLL KGlob NodeJJ , NodeLL jj , ll ;

;

Simplify 6 a b KGlob MatrixForm

2 a2

b2

a2

2 b2

0 2 a2

b2

a2

b2

0 0 0 0 0 0 0

a2

2 b2

4 a2

b2

a2

2 b2

a2

b2

4 a2

2 b2

a2

b2

0 0 0 0 0 0

0 a2

2 b2

2 a2

b2

0 a2

b2

2 a2

b2

0 0 0 0 0 0

2 a2

b2

a2

b2

0 4 a2

b2

2 a2

2 b2

0 2 a2

b2

a2

b2

0 0 0 0

a2

b2

4 a2

2 b2

a2

b2

2 a2

2 b2

8 a2

b2

2 a2

2 b2

a2

b2

4 a2

2 b2

a2

b2

0 0 0

0 a2

b2

2 a2

b2

0 2 a2

2 b2

4 a2

b2

0 a2

b2

2 a2

b2

0 0 0

0 0 0 2 a2

b2

a2

b2

0 4 a2

b2

2 a2

2 b2

0 2 a2

b2

a2

b2

0

0 0 0 a2

b2

4 a2

2 b2

a2

b2

2 a2

2 b2

8 a2

b2

2 a2

2 b2

a2

b2

4 a2

2 b2

a2

b2

0 0 0 0 a2

b2

2 a2

b2

0 2 a2

2 b2

4 a2

b2

0 a2

b2

2 a2

b2

0 0 0 0 0 0 2 a2

b2

a2

b2

0 2 a2

b2

a2

2 b2

0

0 0 0 0 0 0 a2

b2

4 a2

2 b2

a2

b2

a2

2 b2

4 a2

b2

a2

2 b2

0 0 0 0 0 0 0 a2

b2

2 a2

b2

0 a2

2 b2

2 a2

b2

Stationary Functional Approach 46

7 Stationary Functional Approach

In this section, the stationary functional approach will be presented as a method by which the finite element model may be derived. The approach will depend on some definitions that are presented in section ‎7.1 then some applications will be presented in the following sections.

7.1 Some Definitions

A Functional: Simple Definition

A functional is a “function of functions” that produces a real/complex number. The functional is presented in the form of a bound integral which, when evaluated, produces a real number. In mechanics problems, usually, the functional used is the total energy functional which contains the potential energy, the kinetic energy, and the externally work done on the system. A functional may be presented in the form

Domain

nnmnnmn dxdxxxfxxfGxxfxxfI ...,...,,...,,...,,...,,...,,..., 1111111

Variation: Another simple definition

Variation of a functional is the “differentiation” of the functional with respect to one or more of its entries (functions). Note that the Variation of the functional with respect to the independent variables is always equal to zero.

Domain

nm

m

m dxdxfdf

dGf

df

dGf

df

dGfffI ......,...,, 12

2

1

1

21

Stress-Strain Relation

Stresses in structures are related to the strains through “constitutive relations”. The main components of the constitutive relations is the modulus of elasticity, Hook’s constants. For 1-D structures, we may write

E

Strain Displacement Relations

The strain is usually related to the displacement fields in structure mechanics problems. The relation may be obtained from the theory of elasticity or an approximate theory such as the Euler-Bernoulli beam theory. For 1-D elasticity problems, the strain displacement relations are usually simple ones such as the case of a bar, where the relation is defined as

dx

dux

Where u is the axial displacement of the bar. Meanwhile, the strain displacement relation for an Euler-Bernoulli beam is given by


2

2

dx

wdzx

Where w is the transverse deflection of the beam and z is the location above the neutral axis. Other relations exist for different theories, but they will be mentioned in their respective places.

Strain Energy

Strain energy is the amount of mechanical energy stored in a structure, potential energy, due to the deflection of the structure. An expression for the strain energy may be given by

Volume

dVU 2

1

Where U is the strain energy. The concept here is defined for linear elastic structures, but may be used for nonlinear material properties as well as dissipative material properties with minor constraints.

7.2 Applications

In the following sections, we will present the application of the concepts of variation and strain energy to obtain the finite element model as well as demonstrate that the presentation is equivalent to the more commonly used differential equation presentation.

7.2.1 The bar tensile problem

The total energy of the elastic structure is given as the difference between the strain energy

and the work done by the externally applied forces. An expression for the total energy for a

bar, may be given by the following integral

BarLength

dxxFux

uEA .

2

12

For equilibrium, the total energy needs to be at a minimum value, that is to say, its variation is zero (note the analogy with the minimum of a function in one dimension where the extreme points are found when the derivative is equal to zero. Obtaining the variation of the total energy, we get

0.

BarLength

dxxFux

u

x

uEA

Now, let us perform integration by parts, we get


0.2

2

0

BarLength

l

dxxFux

uuEA

x

uuEA

Which indicates that

lxx

l

x

uuEA

x

uuEA

x

uuEA

&00

00

These are the boundary conditions; i.e. at any boundary, either the displacement is equal to

zero or the strain is equal to zero. The other term becomes

0.2

2

BarLength

dxxFux

uuEA

Since the above integral is equal to zero, then the integrand should be equal to zero

02

2

xF

x

uEAu

And, since the variation of the displacement is an arbitrary function, it can not be equal to

zero everywhere which yields

02

2

xF

x

uEA

This is the original differential equation for the displacement function of a bar subject to

distributed loading along its axis. Now, if we select the approximate solution of the problem

and substitute it into the equation representing the variation of the total energy, above, and

handling the variation of the displacement as the weighting functions, we get

euxNxu

euxNxu

Substituting into the energy variation relation:

0 gthElementLen

Tee

xx

Te dxxFNuuNNuEA

But the nodal values of the function or its variation are independent of the integration


00

l

e

xx

Te dxxFNuNNEAu

Also, the variation is arbitrary, therefore, it can not be zero; hence:

00

l

e

xx dxxFNuNNEA

Now we may write

Or

ee fuk

Where

l

e

l

xx dxxFNfdxNNEAk00

&

Which is the same model that we obtained when applying the weighted residual method to the differential equation.

7.2.2 Beam Bending Problem Obtaining the strain energy expression for the beam under transverse loading, we get

l

dxxFwdx

wdEI

0

2

2

2

.2

1

The expression for the variation of the total energy becomes

0.0

2

2

2

2

l

dxxFwdx

wd

dx

wdEI

We may continue the derivation, as for the case of the bar, to obtain the differential

equation. But using the approximate solution into the above expression, we have

ewxNxw

And

l

e

l

xx dxxFNudxNNEA00


ewxNxw

Which gives

00

l

Tee

xxxx

Te dxxFNwwNNwEI

Using the same procedure as for the bar example above, we get

ee fwk

Where

l

e

l

xxxx dxxFNfdxNNEIk00

&

Which is, again, the same model as the one obtained using the weighted residual methods.


7.3 Plane Elasticity

Now, we have enough background to extend our study to cover the plain elasticity problem. In this problem we are only concerned with the thin structures, such as thin plates, that are subjected to in-plane loading. In such a problem, the strain components we are concerned with become the axial strains in the plane of the plate and the shear strain component associated with them. All variables are assumed to constant across the thickness.

Figure ‎7.1. A sketch presenting a plain element with the stresses applied on it.

The above described stresses and strain are related through the following relations

xyxy

yxy

yxx

G

DD

DD

2

Where

12

1 2

EG

ED

In matrix form

xy

y

x

xy

y

x

G

DD

DD

200

0

0

Or

Q


7.3.1 Strain-Displacement Relations The strain displacement relation in the 2-D problem is slightly different taking into account the displacement in the y-direction as well

dx

dv

dy

du

dy

dv

dx

du

xy

y

x

2

1

Or, in matrix form

dx

dv

dy

du

dy

dvdx

du

xy

y

x

2

1

7.3.2 Strain Energy The strain energy should take all stresses and strains into account. Thus, we get the expression as

Volume

T

Volume

dVQdVU 2

1

2

1

For constant thickness, and since all the variables are constant across the thickness, we may simplify the integral over the volume to become an integral over the area

Area

TdAQhU

2

1

A Rectangular Element

For the approximation of the displacement function u(x,y) over the element, use the 2-D interpolation function


Figure ‎7.2. A sketch of the plate element

xyayaxaayxu 4321,

Recall General 2-D Elements

euyxNayxHyxu ,,,

ab

xy

b

yab

xyab

xy

a

xab

xy

b

y

a

x

yxNyxNT

1

,,

In the 2-D elasticity problem, we displacements in both the x and y-directions at every point of the plate. For a rectangular element, you get 8 DOF per element

The displacement “vector”

4

1

4

1

4321

4321

,,,0,0,0,0

0,0,0,0,,,

,

,

v

v

u

u

NNNN

NNNN

yxv

yxu

Strain-Displacement Relations


4

1

4

1

43214321

4321

4321

,,,,,,

,,,0,0,0,0

0,0,0,0,,,

v

v

u

u

NNNNNNNN

NNNN

NNNN

dx

dv

dy

du

dy

dvdx

du

xxxxyyyy

yyyy

xxxx

xy

y

x

mmm wB

Strain Energy

Area

TdAQhU

2

1

Area

mm

T

m

T

m dAwBQBwhU2

1

mm

T

m

Area

mm

T

m

T

m wkwdAwBQBwhU


7.4 Finite Element Model of Plates in Bending

7.4.1 Displacement Function

The transverse displacement w(x,y), at any location x and y inside the plate element, is

expressed by

(‎7-1)

where wH is a 64 element row vector and {a} is the vector of unknown coefficients. For

the plate element under consideration, the bending degrees of freedom associated with

each node are

16

2

1

2

,,

,

,

a

a

a

H

H

H

H

yx

w

y

wx

ww

yx

y

x

w

w

w

w

(‎7-2)

where Hw,i is the partial derivative of Hw with respect to i. Substituting the nodal coordinates

into equation (13), the nodal bending displacement vector {wb} is obtained as follows,

(‎7-3)

where

bH

H

H

H

H

T

yx

w

yx

w

y

wx

w

w

w

yx

yx

y

x

w

w

w

w

w

bb

,0

0,0

0,0

0,0

0,0

][&

,,

,,

,

,

4

2

1

2

1

1

1

(‎7-4)

From equation (14), we can obtain

(‎7-5)

Substituting equation (16) into equation (12) gives

aHyxw w),(

aTw bb

bb wTa1


(‎7-6)

where [Nw] is the shape function for bending given by

(‎7-7)

7.4.2 Strain-Displacement Relation

Consider the classical plate theory, for the strain vector {} can be written in terms of the

lateral deflections as follows

z

xy

y

x

(‎7-8)

where z is the vertical distance from the neutral plane and {} is the curvature vector which

can be written as,

(‎7-9)

where

(‎7-10)

Substituting equation (17) into equation (23), gives

(‎7-11)

where

(‎7-12)

Thus, the strain-nodal displacement relationship can be written as

bwbbw wNwTHyxw 1

),(

1 bww THN

}{

22

2

2

2

2

aC

yx

w

y

wx

w

b

xy

yy

xx

w

w

w

b

H

H

H

C

,

,

,

2

}{}{1

bbbbb wBwTC

1 bbb TCB


(‎7-13)

7.4.3 Constitutive Relations of Piezoelectric Lamina

The general form of the constitutive equation of the piezoelectric patch are written as

follows

(‎7-14)

where, are the stress in the x-direction, stress in the y-direction, and the planar

shear stress respectively; are the corresponding mechanical strains; D is the

electric displacement (Culomb/m2), is the electric field (Volt/m), piezoelectric

material constant relating the stress to the electric field, is the material dielectric

constant at constant stress (Farad/m), and is the mechanical stress-strain constitutive

matrix at constant electric field. is given by,

where E is the Young’s modulus of elasticity at constant electric field, and is the Poisson’s

ratio.

Equation (28) can be rearranged as follows

D

e

eeeQ

E

xy

y

x

T

TE

xy

y

x

(‎7-15)

bb wBzz }{

E

e

eQ

D

xy

y

x

T

E

xy

y

x

xyyx ,,

xyyx ,,

E e

EQ

EQ

1200

011

011

22

22

E

EE

EE

QE


or

(‎7-16)

and

(‎7-17)

where .

7.4.4 Stiffness and Mass Matrices of the Element

The principal of virtual work states that

(‎7-18)

where is the total energy of the system, U is the strain energy, T is the kinetic energy, W

is the external work done, and (.) denotes the first variation.

7.4.4.1 The Potential Energy

The variation of the mechanical and electrical potential energies is given by

(‎7-19)

where V is the volume of the structure. Substituting equation (30) and (31) into equation

(33) gives,

(‎7-20)

Substituting from equations (20) and (27), we get,

(‎7-21)

DeQ

xy

y

x

D

xy

y

x

DeE

xy

y

x

T

1

0 WTU

VV

TdVEDdVU

V

T

V

DTdVDzeDdVDezQzU

V

DDbb

TT

DD

V

DDbb

DT

bb

dVwNwBzewN

dVwNewBzQwBzU


The terms of the expansion of equation (35) can be recast as follows

,

,

,

and ;

where [kb] is bending stiffness matrix, [kbD] is bending displacement-electric displacement

coupling matrix, and [kD] is the electric stiffness matrix.

7.4.4.2 The Kinetic Energy

The variation of the kinetic energy T of the plate/piezo patch element is given by,

(‎7-22)

where is the density/equivalent density and h is the thickness of the element. The above

equation can be rewritten in terms of nodal displacements as follows

(‎7-23)

where [mb] is the element bending mass matrix.

7.4.4.3 The external work

The variation of the external work done exerted by the shunt circuit is given by

A

dAqDLW (‎7-24)

bb

T

b

V

bb

DT

bb wkwdVwBQwBz 2

DbD

T

b

V

DD

T

bb wkwdVwNewBz

b

T

bD

T

DbDb

T

D

V

bb

TT

DD wkwwkwdVwBzewN

DD

T

D

V

DD

T

DD wkwdVwNwN

A

dAt

whwT

2

2

bb

T

b

A

bw

T

w

T

b

A

wmwdAwNNwhdAt

whw

2

2


where A is the element area, L is the shunted inductance, and q is the charge flowing in the

circuit. But, as the charge is the integral of the electric displacement over the element area;

then equation (38) reduces to,

AA

dADLDdAW (‎7-25)

Substituting from equation (20), gives

A

DD

A

T

D

T

D dAwLNdANwW (‎7-26)

which can be recast in the following form,

DD

T

D wmwW (‎7-27)

where [mD] is the element electric mass matrix.

Finally, the element equation of motion with no external forces can be written as

0

0

0

0

D

b

DDb

bDb

D

b

D

b

w

w

kk

kk

w

w

m

m

(‎7-28)

Documents

Introduction to the Finite Element Method