Upload
dr-ir-r-didin-kusdian-mt
View
215
Download
0
Embed Size (px)
Citation preview
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
1/38
1
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.html8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
2/38
2
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
3/38
3
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
4/38
4
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
5/38
5
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
6/386
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
7/387
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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
= =
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
8/388
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
9/389
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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
=
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
10/3810
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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
+ = =
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
11/3811
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
12/3812
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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 =
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
13/3813
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
14/3814
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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
k x y u x y b x y u x y f x y w x y dxdy
+ +
+ =
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
15/38
15
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
Variational problem
Integration by parts gives:
The last 2 integrals can be simplified with use of thedivergence theorem.
1
2
( ( , ) ( , )) ( , ) ( , ) ( , ) ( , )
( ( , ) ( , )) ( , ) ( , ) ( , ) ( , ) 0
k x y u x y b x y u x y f x y w x y dxdy
k x y u x y b x y u x y f x y w x y dxdy
+ +
+ =
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
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
+ +
+
=
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
16/38
16
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
Variational problem
We can separate integration over in the last 2 terms:
1
2
1 2
1
( ) ( )
( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( , )
( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( , )
Boundary of the Boundary of thedomain
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 x y u x y w x y b x y u x y w x y f x y w x y dxdy
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
=
+ +
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
17/38
17
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
Variational problem
Note that the outward normal n1 to
region 1 is the negative of n2 ateach point on .
( ) ( )( ) ( )
1 2
1 2
( ) ( ) ( )
evaluated in region evaluated in region
u u u uk wds k wds k k wds
n n n n
+ +
+ = +
1 2 1 2
1 2( ) ( ) ( ) ( )
1 2
( ) ( )
evaluated in region evaluated in region
u u u u u uk wds k wds k wds k wds k wds k wds
n n n n n n
= + +
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
18/38
18
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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
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 x y u x y w x y b x y u x y w x y f x y w x y dxdy
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
( ) ( ) ( )
evaluated in region evaluated in region
u u u u
k wds k wds k k wdsn n n n
+ +
+ = +
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
19/38
19
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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
=
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
20/38
20
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
21/38
21
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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
+ + = +
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
22/38
22
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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
= = =
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
23/38
23
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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.
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
24/38
24
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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
+ =
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
25/38
25
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
Equivalence between the weak and strong problems
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
+ = =
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
26/38
26
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
Equivalence between the weak and strong problems
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
= =
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
27/38
27
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
Equivalence between the weak and strong problems
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
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
28/38
28
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
29/38
29
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
30/38
30
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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
d q n q n d x y
+ = +
A di G th
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
31/38
31
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
32/38
32
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
33/38
33
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
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
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
34/38
34
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
Example problem on divergence theorem
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=
Example problem on divergence theorem
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
35/38
35
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
Example problem on divergence theorem
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
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
36/38
36
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N. Zabaras (11/06/2009)
Example: Deriving the weak form
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
=
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
37/38
Example: Deriving the weak form
8/9/2019 FEM Zabaras Cornell University StrongAndWeakFormsFor2DBVP
38/38
MAE 4700 FE Analysis for Mechanical & Aerospace Design
N Z b (11/06/2009)
Example: Deriving the weak form
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
+ = +
=