fem zabaras cornell university strongandweakformsfor2dbvp

8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP

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MAE 4700 FE Analysis for Mechanical & Aerospace Design

N. Zabaras (11/06/2009)

MAE4700/5700Finite Element Analysis for

Mechanical and Aerospace DesignCornell University, Fall 2009

Nicholas ZabarasMaterials Process Design and Control Laboratory

Sibley School of Mechanical and Aerospace Engineering101 Rhodes Hall

Cornell UniversityIthaca, NY 14853-3801
http://mpdc.mae.cornell.edu/Courses/MAE4700/MAE4700.htmlhttp://mpdc.mae.cornell.edu/http://mpdc.mae.cornell.edu/http://mpdc.mae.cornell.edu/Courses/MAE4700/MAE4700.html


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Mathematical preliminaries

Let us consider a domain with interior andboundary

Usually the boundary is defined

parametrically as follows: , where is thearc length along

For a function f(x,y) on the boundary we canwrite:

We denote the primary (scalar) variable in 2D as

u(x,y). Until further notice, we assume that allfunctions used here are smooth enough for theoperations shown to be valid.

.

( ), ( )x s y s

( ) ( ( ), ( )), .f s f x s y s s=

,

.s


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Mathematical preliminaries

Recall that the gradient is the rate of change of uas we

move from (x,y) to nearby locations:

Many times we will interpret this as a mathematical operation of thedifferential operator on

The gradient determines the rate of change of uat (x,y) inany direction

The rate of change of u in a given direction

is given as:

( , ).u x y

( , ) ( , )( , )

u x y u x yu x y i j

x y

= +

i jx y

= +

( , ) ( , ) ( , )( , ) cos sin

du x y u x y u x yu x y t

dt x y

= = +

i

( , )u x y

cos sint i j = +

( , )u x y


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Flux

Similarly to the gradient the flux is a vector-valued

function or vector field.

The flux at the boundary can be decomposed in normal andtangential components:

( ) ( ) ( )n

s s n s =

i

( ) ( ) ( )s s s =

( , )x y

( , )x y

( )s

( , )u x y


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If you divide by the area of the subregion, we can view theresult as the amount of flowing into per unit area.

Flux

Consider an arbitrary part containing point of the

domain . Consider the normal flux

across the boundary The total flux across this boundary is

0 0( , )x y

( )n s ds

=

( , )x y

.

( ) ( ) ( )n s s n s = i

A


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Taking as the square region shown,then:

Dividing by the area and taking thelimit :

Divergence of the flux

The limit of as decreases in size always

containing point is called the divergence of the flux at :0 0( , )x y

0 0( , )div x y

= + x y

y x

( )( , )( , )

( , ) yx x yx y

x ydiv x y i j i j x y x y

= + = + +

( , )x y

0P

So is the density of the net flux per unit area at a point

/ ( ) / n A s ds A

=

, 0x y

x y


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Gauss divergence theorem

The total flux out of the region is then:

We can now write using an earlier expression:

This is the Gauss divergence theorem.

dxdy

=

dxdy nds

= =


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Constitutive equation and balance law

Constitutive equation: We consider problems with the

following constitutive equation between the fluxand the state variable u(x,y).

Conservation principle: Within any portion of thedomain, the net flux across the boundary of that part

must be equal to the total quantity

produced by internal sources.

nds fdxdy

=

mod

( , ) 0

( , ) ( , ) ( , )material ulus

k x y in

x y k x y u x y

>

=

f is sourceper unit area


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Balance law

Using the divergence theorem:

Since is arbitrary, we conclude thatthe local form of the balance equation is:

To make things more interesting, we assume an

additional source term proportional to u(x,y):

( ) 0 . f dxdy for all

=

( , ) ( , )x y f x y =

( , ) ( , ) ( , ) ( , )x y b x y u x y f x y + =

nds fdxdy

=


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Conservation principle at interfaces

Consider an interface separating two materials withdifferent modulus k1 and k2. Take a thin strip of thebody around a point on this interface. The balance law

takes the form:

The local balance at the interfacereduces to:

2

1

( ) ( )

( ) 0s

sn n ds

+ = + =

( ) ( )( ) ( ) ( ) 0,n s s n s n s

+ = =


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Boundary conditions

Let us assume that in the boundary

we apply natural boundary

conditions:

On the boundary , we assume essential

boundary conditions:1u

2

( ) ( ) ( ) ( )

( )( ( ) ( )),

n s s n s s

p s u s u s

s

= =

=

1

( ) ( ),u s u s s=

2q


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Summary of the problem of interest

What we are given:

and the interface

The source

The material moduliThe boundary coefficients

or

( , ) , 1, 2i

f f x y in i= =( , ), ( , ) , 1, 2

( , ), ( , ) , 1, 2

i i i

i i i

k k x y x y i

b b x y x y i

= =

= =

2( ), ( )p s u s in

1 2,

2( )s in

With this data we want to compute u(x,y) in :

( ( , ) ( , )) ( , ) ( , ) ( , ), ( , ) , 1, 2ik x y u x y b x y u x y f x y x y i + = =

2

( )( ) ( )[ ( ) ( )],u sk s p s u s u s s or n

=

2

( )( ) ( ),u sk s s sn

=

1( ) ( ),u s u s s= 0,k u n s =


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A note regarding the boundary conditions

The special case in which b=0and when only natural

BC of the form is applied and when, needs special attention:

The solution u(x,y) can only be determined up to aconstant

For u to exist, the following compatibility conditionneeds to be satisfied:

fdxdy ds

=

2

( )( ) ( ),

u sk s s s

n =

2


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Variational problem: Weak form

We start by multiplying the residual r(x,y) by asufficient smooth function w(x,y) and integrate overeach domain in which rw is smooth and set theresulting weighted average equal to zero:

We need to integrate by parts the first terms in eachof these integrals

( , ) ( ( , ) ( , )) ( , ) ( , ) ( , )r x y k x y u x y b x y u x y f x y= +

1

2

( ( , ) ( , )) ( , ) ( , ) ( , ) ( , )

( ( , ) ( , )) ( , ) ( , ) ( , ) ( , ) 0

k x y u x y b x y u x y f x y w x y dxdy


+ +

+ =


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Variational problem

Integration by parts gives:

The last 2 integrals can be simplified with use of thedivergence theorem.

1

2

( ( , ) ( , )) ( , ) ( , ) ( , ) ( , )

( ( , ) ( , )) ( , ) ( , ) ( , ) ( , ) 0



+ +

+ =

1

2

1 2

( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( , )

( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( , )

( ) ( ) 0

k x y u x y w x y b x y u x y w x y f x y w x y dxdy


wk u dxdy wk u dxdy

+ +

+

=


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Variational problem

We can separate integration over in the last 2 terms:

1

2

1 2

1

( ) ( )

( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( , )

( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( , )

Boundary of the Boundary of thedomain



u uk wds k n n

+ +

+

2

0

domain

wds

=

1 2 1 2

( ) ( ) ( ) ( )

1 2

1 2

( ) ( )

evaluated in region evaluated in region

u u u uk wds k wds k wds k wds

n n n n

u uk wds k wds

n n

=

+ +


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Variational problem

Note that the outward normal n1 to

region 1 is the negative of n2 ateach point on .

( ) ( )( ) ( )

1 2

1 2

( ) ( ) ( )


u u u uk wds k wds k k wds

n n n n

+ +

+ = +

1 2 1 2

1 2( ) ( ) ( ) ( )

1 2

( ) ( )


u u u u u uk wds k wds k wds k wds k wds k wds

n n n n n n

= + +


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Variational problem

The last integral is nothing else but: , which is 0!

Finally, we can summarize our weak form as:

( )nw s

1

2

1 2( ) ( )

( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( , )

( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( , )

0



u uk wds k wdsn n

+ +

+

=

( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( , ) 0u

k x y u x y w x y b x y u x y w x y f x y w x y dxdy k wdsn

+ =

( ) ( )

( ) ( )

1 2

1 2

( ) ( ) ( )


u u u u

k wds k wds k k wdsn n n n

+ +

+ = +


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Variational problem

Substitution of the natural boundary condition

gives:

We write:

2

1 1

( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( ) 0

( , ) ( 0 ,sin ( ) ( ), )

k x y u x y w x y b x y u x y w x y f x y w x y dxdy p u u wds

for all admissible functions w x y w on ce we require u s u s s

+ + =

= =

2 2 2 2 2

( ) p u u wds puwds puwds puwds wds

= =

( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( , ) 0

Boundary of thewhole domain

u

k x y u x y w x y b x y u x y w x y f x y w x y dxdy k wdsn

+ =

2

( )( ) ( )[ ( ) ( )], q

u sk s p s u s u s s

n

=


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Space of admissible functions: H1() Finally the weak form looks like:

What is the class of admissible functions?

We indicate this space of functions as

1 reflecting that first derivatives are square integrable

indicates the domain over which these functions are defined.

2 2

1

( , ) ( , ) ( , ) ( , ) ( , ) ( , )

( , ), 0

k x y u x y w x y b x y u x y w x y dxdy puwds f wdxdy wds

for all admissible functions w x y w on

+ + = +

=

22

2w w w dxdyx y

+ + <

1( ),H


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Summary of weak problem

Find a function such that and

the following holds:

We can repeat the above weak statement using matrix

notation as follows (more appropriate for FEMimplementation)

2 2

1

1

( , ) ( , ) ( , ) ( , ) ( , ) ( , )

( , ) ( ) 0 .

k x y u x y w x y b x y u x y w x y dxdy puwds f wdxdy wds

for all functions w x y H with w on

+ + = +

=

1( , ) ( ),u x y H 1( ) ( ),u s u s s=

1( , ) ( ),u x y H 1( ) ( ), ,u s u s s= Find such that

1

1( , ) ( ) 0w x y H with w on = the following holds:

( )

2 22 1

1 11 2

( , ) ( , ) ( , ) ( , )T

xxx

w k x y u b x y u x y w x y dxdy puwds f wdxdy wds

+ + = +


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Matrix form of the weak statement

The gradient operator here is defined as a column

vector:

We will be interchanging the 2 notations here hoping

that from the context you know which notation is

implied.

1( , ) ( ),u x y H 1( ) ( ), ,u s u s s= Find such that

1

1( , ) ( ) 0w x y H with w on = the following holds:

( )

2 22 11 11 2

( , ) ( , ) ( , ) ( , )T

xxx

w k x y u b x y u x y w x y dxdy puwds f wdxdy wds

+ + = +

( ), ,T

u

w wx xu w

u x y

y y

= = =


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Appendices

Some extra slides are included here with:

mathematics background,

proof of equivalence of weak and strong formulations, Examples of deriving weak forms for 2D BVP,

Including anisotropy (e.g. generalized Fouriers law),

etc.


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Equivalence between the weak and strong problems

Consider the following simplified BVP and the

corresponding weak form:

The weak form is: Find

We want to show that this weak form is equilavent to

the strong problem (recover the PDE + natural BCs)

( ( , )) ( , ), ( , )q x y f x y x y =

: ,where q k u=

( )( ) , , ( )q u q uu sk s q n q on and u u on withn

= =

1 1( , ) ( ) : ( ) 0 :uu x y H w H with w on =

q

q wdxdy fwdxdy qwd

+ =


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Integration by parts of the first term gives the

following:

q

q wdxdy fwdxdy qwd

+ =

q

wq nd w qd fwdxdy qwd

+ =

( ) 0u q q

wq nd wq nd w q f d qwd

+ =

( ) ( ) 10 ( ) 0q u

uw q n q d wq nd w q f d w H with w on

+ = =


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Deriving the PDE first: Select was follows:

The first equation above then results in:

( ) ( )

10 ( ) 0

q u

u

w q n q d wq nd w q f d w H with w on

+ = =

( )( ) , 0 0w x q f with on and on= = >

( )2

0 0q f d q f in

= =


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We next derive the natural BC: Select was follows:

The first equation above then results in:

( )

10 ( ) 0

q uu

w q n q d wq nd w H with w on

+ = =

( )( ) , 0 0u qw x q n q with on and on= = >

( )2

0 0

q

qq n q d q n q on

= =

G li d F i l


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Generalized Fouriers law

In general, the flux and temperature gradient are

related with the anisotropic (generalized) Fourierslaw:

For isotropic materials

For a 2D BVP with prescribed flux on andtemperature on and a source term f, can you show

that the weak form in matrix notation is: Find

2 12 2

[ ] x xx xy

y yx yy x x conductivity matrix

u

q k k xD u

uq k k

y

= =

2 2

0[ ] [ ]

0 x identity matrix

k D k I

k

= =

q

u

1 1( , ) ( ) : ( ) 0 :

uu x y H w H with w on = ( )

2 12 21 2

[ ]

q

T

xxx

w D u dxdy qwd fwdxdy

= +

q

A di G th


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Appendix: Greens theorem

If then:

Recall the one-dimensional version of this (where the

boundary is only 2 points):

0( , ) int ,u x y C egrable

ud und

=

| ( ) ( 0)du

dx un u x L u xdx

= = = =

A di Di th


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Appendix: Divergence theorem

If then:

Note that:

Explicilty, we can write the divergence theorem as:

0( , ) int ,q x y C egrable

qd q nd

=

yxqq

qx y

= +

( )yx

x x y y

qq

d q n q n d x y

+ = +

A di G th


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Appendix: Greens theorem

This is the analogue of integration by parts in 1D:

The proof of this is simple if you notice that:

This together with the divergence theorem give:

w qd wq nd w qd

=

( ) ( ) ( )yx

x y x y

qqw wwq wq wq q w q w

x y x x y y

= + = + + +

( )wq w q w q = +

( )w qd wq d w qd wq nd w qd

= =

A di G th i 1D


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Appendix: Greens theorem in 1D

The 1D analogue of Greens theorem is as follows:

or more explicitly as:

w qd wq nd w qd

=

xx x

q ww d wq nd q d x x

=

( )0| | |x

x x x x L x x x

q w ww dx wq n q dx wq wq q dx

x x x = =

= =

Recall that we had used this compact form in 1D to treat boundary conditionson the left (x=0) and the right (x=L) end points in a unified way

E l bl di th


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Example problem on divergence theorem

Consider the vector function

defined in the domain shown in the figure below.

Demonstrate the validity of the divergence theorem,

i.e.

2 3 3( . ) (3 ,3 ),

x yq q q x y y x y= = + +

qd q nd

=

E l bl di th


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From we compute the

following:

Using this, we compute as follows:

2 3 3( . ) (3 , 3 ),

x yq q q x y y x y= = + +

qd

( ) 2 26 (3 ) 6 3yxqq

q xy y xy yx y

= + = + = +

2 1 0.5 2

2 2 3

0 0 0

(6 3 ) [3 (1 0.5 ) (1 0.5 ) ] 1.5

x

qd xy y dydx x x x dx

= + = + =

dx

1 0.5y x=



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Also:

Note that on segment BC: and . Thus

dx

1 0.5y x=

(0, 1) ( 1,0)5 5(1,2)5 2

dx dy AB BC CAdx

q nd q n d q n d q n d

= + +

2 2

2

2

| |

5| |

2 2

ds dx dy

dxdx dx

= + =

+ =

0 02 3 22 3 3 3

002 10 1 0.5

5 5(3 ) (3 ) 2(3 ) ( ) ( 3 )( )

5 2x

q nd x y dx x y y x y dx x y y dy

= + + + + + +

2

2 3 3

0

1

6 (3 (1 0.5 ) (1 0.5 ) ) 2(3 (1 0.5 ) ) 0.252q nd x x x x x dx = + + + +

5

2ds dx=

(2) 5(1, 2)

5n =

6 7.75 0.25q nd

= +

6 7.75 0.25 1.5q nd

= + =

qd q nd

=

Example: Deriving the weak form


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Consider isotropic heat conduction in the domain

shown with the boundary conditions indicated. Providea complete weak statement of the problem.

We start by

multiplying with theweight function w

and integrating over the domain:

( ( , ) ( , )) ( , )k x y T x y f x y =

( ( , ) ( , )) ( , ) ( , ) 0,k x y T x y f x y w x y dxdy sufficiently smooth w

=


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The final weak form is then:

10, ( ), 0

q h

T

uk T w fw dxdy k wds w H with w on

n +

= =

1, ( ), 0

h q h

Tk T w fw dxdy hTwds qwds hT wds w H with w on

+ = + =

1( , ) ( ),

TFind T x y H with T T on such that =

In matrix form, we can also

write the weak form as:

( )

1 1 2 11 2

1( ), 0

h q h

T

x xx

T

w k T fw dxdy

hTwds qwds hT wds

w H with w on

+ = +

=

fem zabaras cornell university strongandweakformsfor2dbvp

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