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MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (10/06/2009)
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 FiniteElementApproximationsFor2DBVP
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MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (10/06/2009)
Finite element approximation of 2D BVP
We approximate the domain with Efinite elements and
Nnodes placing nodes and elements in such a waythat element boundaries coincide as close as possible
to the interface with jump in k.
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MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (10/06/2009)
Finite element approximation of 2D BVP
We define an N-dimensional subspace Hhof H1(h) by
constructing appropriate global basis functions
using the elements we discussed earlier.
A typical test function in Hh is of the form:
In general the essential BC data (Dirichlet data)
is approximated as:
where the sum is over all nodes on
, 1,2,..,i
N i N =
1
( , )
( , ) ( , )
h j j
N
h j j
jw x y
w x y w N x y=
=
1
u in
( ) ( ( ), ( ))h j jj
u s u N x s y s=
1.
h
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MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (10/06/2009)
FEM problem statement
Find a function such that
and at the nodes on so the following holds:
The above equation leads to the following system ofalgebraic equations:
,h
hu H
2 2
1
( , ) ( , ) ( , )
0 .
h h h h
h h h hh h h h h h
h
h h h
u w u wk b x y u x y w x y dxdy pu w ds f w dxdy w ds
x x y y
for all w H with w on
+ + + = +
=
1
( , ) ( , )N
h j j
j
u x y u N x y=
=
jju u= 1h
1
, 1, 2,..,N
ij j i
j
K u F i N =
= =
2
( , ) , , 1,...,
h h
j ji iij i j i j
N NN NK k b x y N N dxdy pN N ds i j N
x x y y
= + + + =
2h h
i i iF fN dxdy N ds
= +
Symmetric,banded
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MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (10/06/2009)
Finite element approximation
Each of the integrals in the stiffness and load vector can becomputed as the sum of contributions from each element inthe mesh. But we need to approach this more carefully!
Let denote a typical finite element. The exact solution uon of our BVP satisfies:
Let and denote the restrictions of the approximations
and to . Then the local approximation of thevariational BVP over is:
e
e
( , ),.
e e e
n
n e
k u w buw dxdy f wdxdy wds
for all admissible functions w x ywhere is the normal component of the flux at
+ =
e
hu
e
hw
hu hw e
e
( )
e e e
e
e e e e e e
h h h h h n h
exact flux on
not known
k u w bu w dxdy f w dxdy w ds
+ =
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MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (10/06/2009)
Finite element approximation
Since , there will be no contribution in the last
integral from elements with sides that coincide with We already have in place the following approximations:
where are the shape functions in and thenumber of nodes in .
The following linear system is then obtained:
1 .h
10h hw on=
eN
e
e
( )e e e
e
e e e e e e
h h h h h n h
exact flux onnot known
k u w bu w dxdy f w dxdy w ds
+ =
1 1
( ) ( , ), ( ) ( , )e eN N
e e e e e e
h i i h j j
i j
w x w N x y u x u N x y= =
= =
( , )
e
jN x y
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Assembly process
The global system of equations is obtained by summing
over all elements Ein the mesh.
We expand the element stiffness to a matrix NxNwith zeros everywhere except those rows and columnscorresponding to nodes within and and
will be expanded to Nx1 vectors and with nonzeroentries only in those rows corresponding to nodes in
e
e
eF e
ef e
ek
1 1
, , 1, 2,...,
e
E Ej j ei i
i j ij
e e
N NN Nk bN N dxdy K i j N
x x y y= =
+ + = =
1 1
, 1,2,...,
e
E Ee
i i
e e
fN dxdy F i N = =
= =
( )1
0, 1, 2,...,E
e e e
ij j i i
e
K u F i N
=
+ = =
eK
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Boundary conditions
Note that the contributions to Kijand Fifrom boundary
conditions must enter the problem through the terms .
We note that the sum of the contour integrals can bewritten as:
( )1
0, 1, 2,...,
E
e e eij j i i
e
K u F i N =
+ = =
ei
(0) (1) ( 2)
1
, 1, 2,...,E
e
i i i i
e
S S S i N =
= + + =(0)
1e h
E
i n i
eS N ds
= = 1(1)
1h
E
i n i
eS N ds
= = 2
(2)
1h
E
i n ie
S N ds=
=
e h Portion of the boundaryof not on
(interelement boundaries)
e
e h
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Boundary conditions
This vector is defined only at interior nodes. Consider
the patch of 4 elements sharing node 1.
From the conservation lawacross an interface
where no point or line sources
are applied. Thus if fis smooth
in the patch shown
(0)
1e h
E
i n ie
S N ds=
=
1 2 3 4
(0)
1 1 1 1 1
1e
E
i n n n n n
e
S N ds N ds N ds N ds N ds=
= = + +
0n =
(0)0
iS =
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Boundary conditions
When the source fcontains a line source or
concentrated point source, then is equal to theintensity of the line source.
We can include point sources by writing f(x,y) as
We assume that the mesh is constructed so that there
is a node at the source location.
(0)
1e h
E
i n ie
S N ds=
=
n
int ( , )
( , ) ( , ) ( , )
i i h
i i
smooth part po source at x y
f x y f x y f x x y y
= +
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MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (10/06/2009)
Boundary conditions
For source at node 1 in the figure, we have:
(0)
1e h
E
i n ie
S N ds= =
int ( , )
( , ) ( , ) ( , )
i i h
i ismooth part po source at x y
f x y f x y f x x y y
= +
4 4(0)
1
1 1
1
e h m
i n i n
e m
weighted average of the jumpsat node
S N ds N ds= =
= =
(0)i
S f=
The presence of sources
leads to very singularsolutions u.
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MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (10/06/2009)
Boundary conditions
The values of uh
are prescribed at . Since isnot known on , cannot be described here.
However, once all nodal u1,u2,uNare computed, we
can evaluate from
1
(1)
1h
E
i n ie
S N ds= =
( )1
0, 1, 2,...,E
e e e
ij j i i
e
K u F i N =
+ = =
1h
n
1h (1)
iS
(1)
iS
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MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (10/06/2009)
Boundary conditions
On the natural boundary condition is prescribed.
There we set
2
(2)
1h
E
i n ieS N ds=
=
2 h
( ) ( ) ( ) ( )n hs p s u s s=
2
(2)
1 1 1h
E N N
i j j i ij j i
e j j
S p u N N ds P u
= = =
= =
2 2
1 1eh h
E Ee
i i i i
e e
N ds N ds= =
= = =
2 2
int sec 2
1 1eh h
eportion of
er ting h
E Ee
ij i j i j ij
e e
P pN N ds pN N ds P
= =
= = =
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MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (10/06/2009)
Final algebraic equations
The final system of equations is given as:
Once boundary conditions are imposed on we can
proceed solving the system of equations for the
unknown nodal values.
(1)
1
1 1
, 1, 2,...,
( ), ( )
N
ij j i i
j
E Ee e e e
ij ij ij i i i
e e
K u F S i N
K K P F F
=
= =
= =
= + = +
1h
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MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (10/06/2009)
Example
Consider the FEM
solution of the followingproblem:
( , ) ( , )u x y f x y in =
410u on=
12 25, 67 74,0 , ,u onn
=
56
u
u onn
+ =
2 12 25 56 67 74, =
1 41 =
1 1h =
2 2h =
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MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (10/06/2009)
Example
Element 1:
2 12 25 56 67 74, =
1 41 =
11
12
13
1
00
0
0
f
f
fF
=
1 1 111 12 13
1 1 121 22 23
1 1 131 32 33
1
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
k k k
k k k
k k k
K
=
1 1h =
2 2h =
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MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (10/06/2009)
Example
Element 2:
2 12 25 56 67 74, =
1 41 =
2
1
22
22
3
0
0
0
0
f
f
Ff
=
2 2 211 12 13
2 2 221 22 23
22 2 231 32 33
0 0 0 0
0 0 0 0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
k k k
k k k
K k k k
=
1 1h =
2 2h =
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MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (10/06/2009)
Example
Element 6:
2 12 25 56 67 74, =
1 41 =
61
1
62
63
00
0
0
f
Ff
f
=
6 6 611 12 13
6
6 6 621 22 23
6 6 631 32 33
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0
0 0 0 0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0 0 0 0
k k k
K
k k k
k k k
=
1 1h =
2 2h =
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MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (10/06/2009)
Example problem: Assembly
1 1
2
3
14
4
5 5
66
7
0
0
0
FF
F
FF
F
F
F
=
11 12 13 14
21 22 23 25
31 32 33 34 35 36 37
41 43 44 47
55 5652 53
65 6663 67
73 74 76 77
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
K K K K
K K K K
K K K K K K K
K K K K K
K K K K
K K K K
K K K K
=
1 1h =
2 2h =
The stiffness terms with the symbolwill be modified once natural BC are applied.
K
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MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (10/06/2009)
Natural boundary conditions
5
6
0
0
0
0
0
=
55 56
65 66
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0 0 0
P
P P
P P
=
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MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (10/06/2009)
Revisiting the FEM formulation for 2D scalar field problems
In the earlier part of this lecture, we used the stiffness
matrix and load vectors in component forms.
We will next discuss a similar example (notation is
used from heat conduction) repeating the samecalculations but in a matrix form. These calculationshave been programmed in the 2dBVP MatLab
software. Let us consider the following
2D BVP: Compute T(x,y):
( ) ( , )
T
q
D T f x y
T T on
q D T n q on
=
=
= =
D H d i k f
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MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (10/06/2009)
2D Heat conduction weak form
The weak statement for this problem takes the form:
We denote here
The weak form is written in a matrix form ready for
FEM discretization, e.g.
Find a function with such that:
{ } { }
0
[ ]
( , )
q
T T Tw D T d w f d w qd
for all w x y U
=
1( , ) ( ),T x y H ( ) ( ), TT s T s s=
1
0: ( ) 0
TU w H with w on =
{ } { }
{ }
, , [ ] , [ ]T x xx xy
y yx yyConductivityheatmatrixflux
T q k kw wxT w q D T D
T q k kx y
y
= = = = =
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2D H d i FE i l i
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MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (10/06/2009)
2D Heat conduction FE interpolation
The gradient fields of Teand weare obtained as:
Similarly for the gradient of we:
{ }
11 2 1 2
1 2
1 2 1 2
1 2
... ...
... ...
e
e ee e e eee e enen nen
nen
e
e e ee e e ee e enen nen
nen
The B matrix
TN N N N N N TT T T
x x x x x x xT
T N N N N N N T T T
y y y y y y y
+ + + = = =
+ + +
{ }
{ }
{ }
2( , )
...
e
e
e
e e e e
globalscatternodale matrixtemperaturesnen
d
T B x y d B L d
T
= =
{ } { }{ }
{ }
1 1
2 2
1 2( , ) ..... ...Te
e e
e e
e eT T T T TTe e e e e e e e
nen
scatterw matrix
e e
nen nen
N N
x y
N Nw w
w w B x y w w w w L Bx yx y
N N
x y
= = = =
Here, nen: number element nodes
2D H t d ti k f
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MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (10/06/2009)
2D Heat conduction weak form
We can substitute these expressions into the weak
form
Note as we did earlier for the 1D BVP, the vector wwas
partitioned as follows:
{ } { }1 1 1
[ ]T T
e e eq
nel nel nelT
e e e e e
e e e
w D T d w f d w qd = = =
=
{ } { } { }1
[ ] 0e e e
q
nelT T T T T e e e e e e e
F
e
w L B D B d L d N f d N qd d w=
+ =
{ } { } 0 0,E E
F FF
wdd ww wd
= = = =
2D H t d ti k f
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MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (10/06/2009)
2D Heat conduction weak form
From this equation, we can easily identify:
We now can write the weak form as:
{ } { } { }1
[ ] 0e e e
q
nelT T T T T e e e e e e e
F
ew L B D B d L d N f d N qd w
=
+ =
[ ]e
Te e e e
K B D B d
= { }
{ } { }
e eq
ee
T Te e e
f f
f N f d N qd
=
{ } { }
{ }
{ }1 1
Re
0nel nel
T TT e e e e e
F
e e
sidual r
w L K L d L f w= =
=
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