Approximate Methods - Weighted Residual Methods

8/12/2019 Approximate Methods - Weighted Residual Methods

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Approximate Methods in StructureMechanics

Mohammad Tawfik

19 February 2014


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Introduction 2

Contents

1 Introduction ....................................................................................................................... 3

1.1 Objectives .................................................................................................................... 3

1.2 Why Approximate?...................................................................................................... 3 1.3 Classification of Approximate Solutions of D.E.s ....................................................... 3

2 Weighted Residual Methods.............................................................................................. 5

2.1 Basic Concepts ............................................................................................................. 5

2.2 General Weighted Residual Method ........................................................................... 5

2.3 Collocation Method ..................................................................................................... 8

2.4 The Subdomain Method ............................................................................................ 11

2.5 The Galerkin Method ................................................................................................ 13

3 Stationary Functional Approach ...................................................................................... 16

3.1 Some Definitions ....................................................................................................... 16

3.2 Applications ............................................................................................................... 17

3.2.1 The bar tensile problem ..................................................................................... 17

3.2.2 Beam Bending Problem ..................................................................................... 19

3.3 Plane Elasticity ........................................................................................................... 21

3.3.1 Strain-Displacement Relations ........................................................................... 22

3.3.2 Strain Energy ...................................................................................................... 22 3.4 Finite Element Model of Plates in Bending ............................................................... 25

3.4.1 Displacement Function ...................................................................................... 25

3.4.2 Strain-Displacement Relation ............................................................................ 26

3.4.3 Constitutive Relations of Piezoelectric Lamina .................................................. 27

3.4.4 Stiffness and Mass Matrices of the Element ..................................................... 28


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Approximate Methods in Structure Mechanics 3

1 Introduction

1.1 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 residualmethod. Derivation of a system of linear equations to approximate the solution of an ODEwill be presented using different techniques as an introduction to the finite elementmethod.

1.2 Why Approximate?

The question that usually rises in the minds of engineers and students alike is, why do westudy approximate methods? The main answer that should reply to that question is theignorance of the humans! Up to this moment, scientists and engineers have been able topresent a vast amount of mathematical models for physical phenomena, unfortunately, avery small percentage of those models, which are usually in the form of differentialequations, have close form solutions! Thus, the necessity of solving those problems impliesthe use of approximate methods to get solutions for some specific problems of certaininterest.

In the modern engineering life, packages that present solutions for problems using digitalcomputers are everywhere. The understanding of how those packages perform approximatesolutions 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 goingto be in order to be able to distinguish the right results from errors that may occur due tobugs 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 ofan advanced, or a new, problem, has to have a good background on how the old problemswere solved.

1.3 Classification of Approximate Solutions of D.E.s

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

Discrete coordinate methods depend on solving the differential relations at pre-specifiedpoints in the domain. When those points are determined, the differential equation may beapproximately presented in the form of a difference equation. The difference equationpresents a relation, based of the differential equation, between the values of the dependentvariables at different values of the independents variables. When those equations aresolved, 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


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Introduction 4

methods are finite difference methods and the Runge-Kutta methods. Discrete coordinatemethods 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. Thesemethods, 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 theproposed functions is multiplied by a coefficient, generalized coordinate, that is thenevaluated by a certain technique that identifies different methods from one another. Afterthe solution of the problem, you will obtain a function that represents, approximately, thesolution of the problem at any point in the domain.

Stationary functional methods are part of the distributed coordinate methods family. Thesemethods depend on minimizing/maximizing the value of a functional that describes acertain property of the solution, for example, the total energy of the system. Using thestationary 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 interactionproblems or structure dynamics involving control mechanisms.

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


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Weighted Residual Methods 6

functions are any set of functions that are continuous over the domain of the differentialequation. A set functions may be polynomial, sinusoidal, hyperbolic, or any combination offunctions. The number of functions needed should be equal to the unknown generalizedcoordinates to produce a set of equations that are solvable in the unknowns. Also, theweighting functions need to be linearly independent for the equations to be solvable.

Expanding the series of proposed solution functions, we get:

x R x g x La x La x La nn ...2211

Multiplying by the weighting function and integrating, we get:

01

Doma in

n

iii j

Domain j dx x g x La xwdx x R xw

Domain

nn j Domain

j dx x g x La x La x La xwdx x R xw ...2211

In matrix form

Domain

ji

nninn

njij j

ni

dx x g xwa

k k k

k k k

k k k

1

1

1111

Where

Domain

i jij dx x L xwk

Example Problem

The bar tensile problem is a classical problem that

describes the relation between the axiallydistributed loads and the displacement of a bar.Lets consider the bar in Figure 2.1 with constantmodulus of elasticity and cross section area. Theforce displacement relation is given by: Figure 2.1. Sketch of a bar with distributed axial forces

022

F x

u EA

Subject to the boundary conditions

0/&00 dxdul xu x


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Now, lets use the approximate solution

n

iii xa xu

1

Substituting it into the differential equation, we get

x R F dx

xd a EA

n

i

ii

1

2

2

Selecting weighting functions, W i , and applying the method, we get:

l

ji

l i

j dx xw F adxdx xd

xw EA00

2

2

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

l x

Sin x2

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

l l fdxadx

l xSin

l EA

01

0

2

22

Performing the integration, we get:

fl al

xCos

l EA

l

1022

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

EA f l

EA f l

l EA fl

a22

1 637.02

2

Then, the approximate solution for this problem becomes

l x

Sin EA

f l xu

2637.0

2


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

5.0637.02

exact EA f l

l u

And maximum strain is:

0.10.10 exact EAlf

u x

2.3 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.

0 j x R

The collocation method may be seen as one of the weighted residual family when theweighting function becomes the delta function. The delta function is one that may be

described as:

j j

j

j j

x F dx x F x x

x x

x x x x

0

1

Now, if we select a set of points x j inside the domain of the problem, we may write downthe integral of the residue, multiplied by the delta functions, as follows:

01

j

n

i jii j x F x La x R

Which gives


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n

ji

nninn

njij j

ni

x g

x g

x g

a

k k k

k k k

k k k 1

1

1

1111

Where

jiij x Lk

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:

x R x F dx

xd a EA

n

i

ii

1

2

2

Evaluating the residue at the collocation points, we get

01

2

2

j

n

i

jii x F dx

xd a EA

In matrix form


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nnnnnn

n

n

x F

x F

x F

a

a

a

k k k

k k k

k k k

2

1

2

1

21

22212

12111

...

...

...

Where

j x x

iij dx

xd EAk 2

2

Solve the above system for the generalized coordinates a i 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 x

Sin x2

Using the function into the DE:

l x

Sinl

EAdx

xd EA

22

2

2

2

A natural selection for the collocation point may be the central point of the bar. Substitutingby the value of x=l/2, we get

EA f l

EA f l

Sinl EA f

a2

2

2

21 57.024

42

Then, the approximate solution for this problem is:

l x

Sin EA

f l xu

257.0

2

Which gives the maximum displacement to be

5.057.02

exact EA

f l l u

And maximum strain to be:


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0.19.00 exact EAlf

u x

2.4 The Subdomain Method

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

1

11

0

1

0

1

j j

j j j j

j

j j

x xor x x

x x x x xU x xU

x x

x x x xU

Hence the integral of the weighted residual method becomes

01 j

j

x

x

dx x R

Substituting using the series solution

011

1

j

j

j

j

x

x

n

i

x

x ii dx x g dx x La

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

For the bar application


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x R x F dx

xd a EA

n

i

ii

1

2

2

Performing the integration and equating by zero

11

12

2 j

j

j

j

x

x

n

i

x

x

ii dx x F dxdx

xd a EA

Which gives the equation in matrix form as

11

2

2 j

j

j

j

x

xi

x

x

i dx x F adxdx

xd EA

Using Admissible Function

l x

Sin x2

The differentiation will give

l x

Sinl

EAdx

xd EA

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

l l

fdxadxl

xSin

l EA

01

0

2

22

Performing the integral

fl al x

Cosl EA

l

1022

Evaluating the generalized coordinate

EA f l

EA f l

l EA fl

a22

1 637.02

2

Then, the approximate solution for this problem is:


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Substituting with the approximate solution:

Domain

j

n

i Domain

i ji dx x F xdxdx

xd xa EA

12

2

We have

l l

fdxl

xSina

l EAdx

l x

Sinl

xSina

l EA

0

21

2

01

2

22222

Which gives

l l

al

EA 2

22 12

Substituting and solving for the generalized coordinate, we get

EA fl l

EA f

a2

3

2

1 52.016

In most structure mechanics problems, the differential equation involves second derivativeor higher for the displacement function. When Galerkin method is applied for suchproblems, you get the proposed function multiplied by itself or by one of its function family.This suggests the use of integration by parts. Lets examine this for the previous example.Substituting with the approximate solution: (Int. by Parts)

Domain

i jl

i j

Domain

i j dxdx

xd dx

xd

dx xd

xdxdx

xd x

02

2

But the boundary integrals are equal to zero since the functions were already chosen tosatisfy the boundary conditions. Evaluating the integrals will give you the same results.

l l a

l

EA 2

22 1

2

EA fl l

EA f

a2

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!


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

Homework #1

Figure 2.4. A simply supported beam

)(44

x F dx

wd

subject to0

)()0(0)()0( 2

2

2

2

dxl wd

dxwd

and l ww

Exact Solution for this problem is

12/1103

157

412

2/1060

1312

)(

23

3

x x x x

x x x

xw

Solve the beam bending problem, for beam displacement, for a simply supportedbeam 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 BCs

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


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Stationary Functional Approach 16

3 Stationary Functional Approach

In this section, the stationary functional approach will be presented as a method by whichthe finite element model may be derived. The approach will depend on some definitionsthat are presented in section 3.1 then some applications will be presented in the followingsections.

3.1 Some Definitions

A Fu nc tiona l: Si mple Defin ition

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

Domain

nnmnnmn dxdx x x f x x f G x x f x x f I ...,...,,...,,...,,...,,...,,..., 1111111

Variation: Another simple definition

Variation of a functional is the differentiation of the functional with respect to one ormore of its entries (functions). Note that the Variation of the functional with respect to theindependent variables is always equal to zero.

Domainnm

mm dxdx f df

dG f df dG f

df dG f f f I ......,...,, 12

21

121

Stress-Strain Relation

Stresses in structures are related to the strains through constitutive relations . The maincomponents of the constitutive relations is the modulus of elasticity, Hooks constants . For1-D structures, we may write

E

Strain Displacement Relations

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

dxdu

x

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


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2

2

dxwd

z x

Where w is the transverse deflection of the beam and z is the location above the neutralaxis. Other relations exist for different theories, but they will be mentioned in theirrespective 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

dV U 21

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

3.2 Applications

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

3.2.1 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

dx x F u xu

EA .21

2

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 theextreme points are found when the derivative is equal to zero. Obtaining the variation ofthe total energy, we get

0.

BarLength

dx x F u xu

xu

EA

Now, let us perform integration by parts, we get


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0.22

0

BarLength

l

dx x F u x

uu EA

xu

u EA

Which indicates that

l x x

l

xu

u EA xu

u EA xu

u EA

&0000

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

0.22

BarLength

dx x F u x

uu EA

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

022

x F

xu

EAu

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

zero everywhere which yields

022

x F

xu

EA

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

eu x N xu

eu x N xu

Substituting into the energy variation relation:

0 gth ElementLen

T ee x x

T e dx x F N uu N N u EA

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


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00

l e

x x

T e dx x F N u N N EAu

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

00

l e

x x dx x F N u N N EA

Now we may write

Or

ee f uk Where

l

el

x x dx x F N f dx N N EAk 00

&

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

3.2.2 Beam Bending ProblemObtaining the strain energy expression for the beam under transverse loading, we get

l dx x F w

dxwd

EI 0

2

2

2

.21

The expression for the variation of the total energy becomes

0.0

2

2

2

2

l dx x F w

dxwd

dxwd

EI

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

ew x N xw And

l

el

x x dx x F N udx N N EA00


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3.3 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 aresubjected 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 componentassociated with them. All variables are assumed to constant across the thickness.

Figure 3.1. A sketch presenting a plain element with the stresses applied on it.

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

xy xy

y x y

y x x

G

D D D D

2

Where

12

1 2

E G

E D

In matrix form

xy

y

x

xy

y

x

G

D D

D D

200

0

0

Or

Q


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3.3.1 Strain-Displacement RelationsThe strain displacement relation in the 2-D problem is slightly different taking into accountthe displacement in the y-direction as well

dxdv

dydu

dydvdxdu

xy

y

x

21

Or, in matrix form

dxdv

dydu

dydvdxdu

xy

y

x

21

3.3.2 Strain EnergyThe strain energy should take all stresses and strains into account. Thus, we get theexpression as

Volume

T

Volume

dV QdV U 21

21

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

Area

T dAQhU 21

A Rectangular Element

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


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Figure 3.2. A sketch of the plate element

xya ya xaa y xu 4321,

Recall General 2-D Elements

eu y x N a y x H y xu ,,,

ab

xy

b

yab xy

ab xy

a x

ab xy

b y

a x

y x N y x N T

1

,,

In the 2-D elasticity problem, we displacements in both the x and y-directions at every pointof 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

N N N N

N N N N

y xv

y xu

Strain-Displacement Relations


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4

1

4

1

43214321

4321

4321

,,,,,,

,,,0,0,0,0

0,0,0,0,,,

v

v

u

u

N N N N N N N N

N N N N

N N N N

dxdv

dydu

dydvdxdu

x x x x y y y y

y y y y

x x x x

xy

y

x

mmm w B

Strain Energy

Area

T dAQhU 21

Area

mmT

mT

m dAw BQ BwhU 21

mmT m Area

mmT

mT

m wk wdAw BQ BwhU


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3.4 Finite Element Model of Plates in Bending

3.4.1 Displacement Function

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

( 3-1)

where w H is a 64 element row vector and { a } is the vector of unknown coefficients. Forthe plate element under consideration, the bending degrees of freedom associated witheach node are

16

2

1

2,,

,

,

a

a

a

H

H

H

H

y xw

yw xw

w

y x

y

x

w

w

w

w

( 3-2)

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

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

( 3-3)

where

b H

H

H

H

H

T

y xw

y xw yw xww

w

y x

y x

y

x

w

w

w

w

w

bb

,0

0,0

0,0

0,0

0,0

][&

,,

,,

,

,

42

12

1

1

1

( 3-4)

From equation (14), we can obtain

( 3-5)

Substituting equation (16) into equation (12) gives

a H y xw w),(

aT w bb

bb wT a 1


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( 3-6)

where [ N w] is the shape function for bending given by

( 3-7)

3.4.2 Strain-Displacement Relation

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

z

xy

y

x

( 3-8)

where z is the vertical distance from the neutral plane and { } is the curvature vector whichcan be written as,

( 3-9)

where

( 3-10)

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

( 3-11)

where

( 3-12)

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

bwbbw w N wT H y xw 1),(

1

bww T H N

}{

222

2

2

2

aC

y xw

y

w x

w

b

xy

yy

xx

w

w

w

b

H

H

H

C

,

,

,

2

}{}{1 bbbbb w BwT C

1 bbb T C B


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( 3-13)

3.4.3 Constitutive Relations of Piezoelectric Lamina

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

( 3-14)

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/m 2), 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 Youngs modulus of elasticity at constant electric field, and is the Poissonsratio.

Equation (28) can be rearranged as follows

De

eeeQ

E xy

y

x

T

T E

xy

y

x

( 3-15)

bb w B z z }{

E e

eQ

D xy

y

x

T

E

xy

y

x

xy y x ,,

xy y x ,,

E e

E Q E Q

1200

011

011

22

22

E

E E E E

Q E


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or

( 3-16)

and

( 3-17)

where .

3.4.4 Stiffness and Mass Matrices of the Element

The principal of virtual work states that

( 3-18)

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.

3.4.4.1 The Potential Energy

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

( 3-19)

where V is the volume of the structure. Substituting equation (30) and (31) into equation(33) gives,

( 3-20)

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

( 3-21)

DeQ xy

y

x D

xy

y

x

De E xy

y

xT

1

0 W T U

V V

T dV E DdV U

V

T

V

DT dV D z e DdV De z Q z U

V

D DbbT T

D D

V

D Dbb DT

bb

dV w N w B z ew N

dV w N ew B z Qw B z U


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

,

,

,

and ;

where [ k b] is bending stiffness matrix, [ k bD] is bending displacement-electric displacementcoupling matrix, and [ k D] is the electric stiffness matrix.

3.4.4.2 The Kinetic Energy

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

(

3-22)

where is the density/equivalent density and h is the thickness of the element. The aboveequation can be rewritten in terms of nodal displacements as follows

( 3-23)

where [ m b] is the element bending mass matrix.

3.4.4.3 The external work

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

A

dAq DLW ( 3-24)

bbT bV

bb DT

bb wk wdV w BQw B z 2

DbDT bV

D DT

bb wk wdV w N ew B z

bT bDT Db DbT DV

bbT T

D D wk wwk wdV w B z ew N

D DT DV

D DT

D D wk wdV w N w N

AdAt

whwT 2

2

bbT b A

bwT

wT

b

A

wmwdAw N N whdAt w

hw

2

2


30/30

where A is the element area, L is the shunted inductance, and q is the charge flowing in thecircuit. But, as the charge is the integral of the electric displacement over the element area;then equation (38) reduces to,

A AdA D L DdAW

(

3-25)

Substituting from equation (20), gives

A

D D A

T D

T D dAw L N dA N wW ( 3-26)

which can be recast in the following form,

D DT D wmwW ( 3-27)

where [ m D] is the element electric mass matrix.

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

0

00

0

D

b

D Db

bDb

D

b

D

b

w

w

k k

k k

w

w

m

m ( 3-28)

Documents

Approximate Methods - Weighted Residual Methods