the finite element method
DESCRIPTION
NBCR Summer Institute 2006: Multi-Scale Cardiac Modeling with Continuity 6.3 Wednesday: Finite Element Discretization and Anatomic Mesh Fitting Andrew McCulloch and Fred Lionetti. The Finite Element Method. Evolved first from the matrix methods of structural analysis in the early 1960’s - PowerPoint PPT PresentationTRANSCRIPT
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NBCR Summer Institute 2006:Multi-Scale Cardiac Modeling with
Continuity 6.3
Wednesday:Finite Element Discretization and
Anatomic Mesh Fitting
Andrew McCulloch and Fred Lionetti
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• Evolved first from the matrix methods of structural
analysis in the early 1960’s• Uses the algorithms of linear algebra• Later found to have a more fundamental foundation• The essential features are in the formulation• There are two alternative formulations that are broadly
equivalent in most circumstances– Variational formulations, e.g. the Rayleigh-Ritz method– Weak or weighted residual formulations, e.g.the
Galerkin method• Both approaches lead to integral equations instead of
differential equations (the strong form)
The Finite Element Method
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The Finite Element Method• Solution is discretized using a finite number of functions
– Piecewise polynomials (elements)– Continuity across element boundaries ensured by
defining element parameters at nodes with associated basis functions,
12 13
14 15
21 22
23 24
• FE equations are derived from the weak form of the governing equations
R = 0Finite differences:Finite elements:
R = 0
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The Finite Element Method• Integrate governing equations in each element
• Assemble global system of equations by adding contributions from each element
1 2
5 6
7 8
3 4
Element equationsk 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 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
u
u
u
u
u
u
u
u
11 12 13 14 15 16 17 18
21 22 23 24 25 26 27 28
31 32 33 34 35 36 37 38
41 42 43 44 45 46 47 48
51 52 53 54 55 56 57 58
61 62 63 64 65 66 67 68
71 72 73 74 75 76 77 78
81 82 83 84 85 86 87 88
1
2
3
4
5
6
7
8
L
N
MMMMMMMMMMM
O
Q
PPPPPPPPPPP
L
N
MMMMMMMMMMM
O
Q
PPPPPPPPPPP
L
N
MMMMMMMMMMM
O
Q
PPPPPPPPPPP
f
f
f
f
f
f
f
f
1
2
3
4
5
6
7
8
12 13
14 15
21 22
23 24
Global equations
L
N
MMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMM
O
Q
PPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP
L
N
MMMMMMMMMMMM
MMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMM
O
Q
PPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP
L
N
MMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMM
O
Q
PPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP
PPPP
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Consider the strong form of a linear partial differential equation, e.g. 3-D Poisson’s equation with zero boundary conditions:
0
),,(2
2
2
2
2
2
u
zyxfz
u
y
u
x
uOn region R
on boundary S
Strong Form Lu = f
Variational Principle, e.g. minimum potential energy
Rv
vfLvu Vd)2(min
Weighted Residual (weak) form, e.g. virtual work
0Vd)( R
wfLu
Integral Formulations
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0
),(2
2
2
2
u
yxfy
u
x
uOn region S
on boundary C
Weak form
SS
yxwfyxwy
u
x
udddd
2
2
2
2
Integrate by parts
d d d d d dS C C S
u w u w u ux y w x w y f w x y
x x y y y x
Where, u and w vanish at the boundary
0 0
Weak Form for 2-D Poisson’s Equation
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• Choose a finite set of approximating (trial) functions, i(x,y), i = 1, 2, …, N
• Allow approximations to u in the formU(x,y) = U11 + U22 + U33 + … + UNN
(that can also satisfy the essential boundary conditions)
• Solve N discrete equations for U1, U2, U3, …, UN
ij
jij
si
S
iNN
iNN
FUK
yxf
yxyy
Uy
Uxx
Ux
U
dd
dd...... 11
11
Galerkin’s Method for 2-D Poisson’s Equation
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yxfF
Kyxyyxx
K
Sii
jiS
jijiij
dd
dd
[K]U = F
[K] is the stiffness matrix and F is the load (RHS) vector
[K] is symmetric and positive definite
Galerkin’s Method for 2-D Poisson’s Equation
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• Galerkin is more general than Rayleigh-Ritz. If we add u/x, symmetry & the variational principle are lost, but Galerkin still works
• If w is chosen as Dirac delta functions at N points, weighted residuals reduces to the collocation method
• If w is chosen as the residual functions Lu-f, weighted residuals reduces to the least squares method
• By choosing w to be the approximating functions, Galerkin’s method requires the error (residual) in the solution to be orthogonal to the approximating space.
• The integration by parts (Green-Gauss theorem) automatically introduces the Neumann (natural) boundary conditions
• The Dirichlet (essential) boundary conditions must be satisifed explicitly when solving [K]U=F
• Since discretized integrals are sums, contributions from many elements are assembled into the global stiffness matrix by addition.
• The Ritz-Galerkin FEM finds the approximate solution that minimizes the error in the energy
Comments on Galerkin’s Method
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1. Formulate the weighted residual (weak form)
2. Integrate by parts (or Green-Gauss Theorem) reduces derivative order of differential operator naturally introduces derivative (Neumann) boundary
conditions, e.g. flux or traction. Hence called that
natural boundary condition
3. Discretize the problem discretize domain into subdomains (elements) discretize dependent variables using finite
expansions of piecewise polynomial interpolating
functions (basis functions) weighted by parameters
defined at nodes
Steps in the Finite Element Method
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4. Derive Galerkin finite element equations substitute dependent variable approximation in
weighted residual integral Choose weight functions to be interpolating
functions — the Galerkin assumption (Galerkin,
1906)
5. Compute element stiffness matrices and RHS integrate Galerkin equations over each element
subdomain integrate right-hand side to obtain element load
vectors which also include any prescribed Neumann
boundary conditions
Steps in the Finite Element Method (…cont’d)
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6. Assemble global stiffness matrix and load vector Add element matrices and RHS vectors into global
system of equations Structure of global matrix depends on node ordering
7. Apply essential (i.e. Dirichlet) boundary conditions at least one is required (essential) for a solution prescribed values of dependent variables at specified
boundary nodes, e.g. prescribed displacements eliminate corresponding rows and columns from
global stiffness matrix and transfer column effects of
prescribed values to Right Hand Side the constraint reduced system
Steps in the Finite Element Method (…cont’d)
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8. Solve global equations for unknown nodal dependent variables using algorithms for Ax = b or Ax = x
9. Evaluate element solutions interpolate dependent variables evaluate derivatives, e.g. fluxes derived quantities, e.g. stresses or strain energy graphical visualization; post-processing
10.Test for convergence refine finite element mesh and repeat solution
Steps in the Finite Element Method (…cont’d)
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2
2
d2
d(1) 0
(4) 9
u
xu
u
solution
u x x( ) ( ) 1 2
1 2 3 4U1=0
2
4
6
8
x
u
U4=9
U3 =?
U2=?
Galerkin FEM: Simple 1-D Example
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0xd)( R
wfLu
2. Integrate by parts (or Green-Gauss Theorem)
0xd24
12
2
wwdx
ud
0xd24
1
4
1
w
dxdu
wdxdw
dxdu
1. Formulate the weighted residual (weak) form
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• 4 global nodal parameters U1, U2, U3, U4
• 3 linear elements each with 2 element nodal
parameters u1, u2.
• Adjacent elements share global nodal parameters,
e.g., global parameter U2 is element parameter u2 of
element 1 and u1 of element 2.
• Two (linear) element interpolation functions for each
element, i(x), i = 1, 2
• Allow element approximations to u in the form
u(x) = u1 1 + u2 2 = ui i i=1,2
3. Discretize the problem
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0 0.5 10
0.5
1
x
21
element basis functions
Element Basis Functions
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0xd2
xd2xd2
4
1
4
3
3
2
2
1
wdx
duw
dx
dw
dx
du
wdx
dw
dx
duw
dx
dw
dx
du
In each element, let
u(x) u1 1 + u2 2 = ui i (x)
and
w(x) i (x)
4. Derive Galerkin equations for each element
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i
jjij
2
1
2
1
2
2
1
1
fuk
d2ddd
dd
dd
xx
xxu
xu
i
i
e.g. for Element 1 (no derivative boundary conditions):
[k] = [(kij)] is the element stiffness matrix
f = (fi) is the element load vector
4. Derive Galerkin equations for each element (… cont’d)
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x
xxxd2f
kd k
ii
jiij
ji
[k]u = f
Element stiffness matrix, [k] and load (RHS) vector, f
1
2
:1 Element
2
1
x
x
11
11
d11
11
)1 ele(
2
1
2
1)1 ele(
[k]
[k]xx
xxx
5. Compute element stiffness matrices
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xd2f ii
1
2
:1 Element
2
1
x
x
1
1
)21()44(
)14()48(
2
4d
22
24
)1 ele(
2
2 2
1
2
1)1 ele(
f
fxx
xxx
x
x
In this problem, each element is the same size and thus:
[k](ele 1) = [k](ele 2) = [k](ele 3)
and:
f(ele 1) = f(ele 2) = f(ele 3)
5. Compute element RHS matrices
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1100
1210
0121
0011
11
1111
1111
11
[K]
1
2
2
1
1
11
11
1
F
6. Assemble global stiffness matrix and load vector
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1
2
2
1
1100
1210
0121
0011
4
3
2
1
U
U
U
U
F[K]U
u U
u U
( )
( )
1 0
4 91
4
That leaves global equations 2 and 3
2 3
2 3
0 2 2
2 9 2
U U
U U
7. Apply essential (i.e. Dirichlet) boundary conditions
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2 3 2
2 3 3
0 2 2 1
2 9 2 4
U U U
U U UExact!
8. Solve global equations (constraint-reduced)
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Representing a One-Dimensional Field
• Polynomials are convenient, differentiated and integrated readily• For low degree polynomials this is satisfactory• If the polynomial order is increased further to improve the accuracy,
it oscillates unacceptably• Divide domain into subdomains and use low order piecewise
polynomials over each subdomain – called elements
2 3Use a polynomial expression ( ) ...
and estimate the monomial coefficients a, b, c and d
u x a bx cx dx
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Making Piecewise Polynomials Continuous
• constrain the parameters to ensure continuity of u across the element boundaries
• or better, replace the parameters a and b in the first
element with parameters u1 and u2, which are the
values of u at the two ends of that element:
where is a normalized measure of distance along the curve
u u u( ) ( ) 1 1 2
( )0 1
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u = u(x)
++
++
+ ++ + + +
++
+x
u
u = a + bx u = c + dx u = e + fx
++
+++ +
++ + +
++
+
x
u
u1
u2
u3
u4
0
1
u = (1- ) u1 + u2
element 1 element 2 element 3
nodes
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u
u1
u2
u = (1- ) u1 + u2
0 1
0 1
1
0
1
1
= 1-
=
Linear Lagrange Interpolation
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Global-Element Mapping
• Associate the nodal quantity un with element node n• Map the value U defined at global node onto local node n of
element e by using a connectivity matrix (n, e),
u Un n e ( , )
Thus, in the first element
u u u( ) ( ) ( ) 1 1 2 2
with u1=U1 and u2=U2..In the second element u is interpolated by
u u u( ) ( ) ( ) 1 1 2 2
With u1=U2 and u2=U3.
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We have u () but to define u (x) we need x ().Define x as an interpolation of nodal values, e.g.
u u
x x
n nn
n nn
( ) ( )
( ) ( )
Isoparametric Interpolation
u
x
u1
u2
x2
x1
1
1
u1
u2
u
x2x1x
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Quadratic Lagrange Basis Functions
Use three nodal parameters u1, u2 and u3
are the quadratic Lagrange basis functions.
1 1 2 2 3 3
1
2
3
where
2 1 0.5
4 1
2 0.5
u u u u
0 0.5 1.0
1.0
0 0.5 1.0
1.0
0 0.5 1.0
1.0
12
3
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Cubic Hermite Basis Functions
1
1
1
1
1
1
0
0
0
0
3211 231
23212
221 1
1222
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Scaling Factors
=0 =1=0 =0
s1 s2 s3
n e n,e
n en e n,e i
i i i (no sum on i)
U U
U U s
s
Global to local mapping:
Scaling Factors arc lengths
arc length
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Two-DimensionalTensor-Product Elements
1 1 2 1 1 1 2 1 2
2 1 2 2 1 1 2 1 2
3 1 2 1 1 2 2 2 2
4 1 2 2 1 2 2 1 2
( , ) ( ) ( ) (1 )(1 )
( , ) ( ) ( ) (1 )
( , ) ( ) ( ) (1 )
( , ) ( ) ( )
u( 1 , ) ( , ) ( , ) ( , ) ( , )2 1 1 2 1 2 1 2 2 3 1 2 3 4 1 2 4 u u u u
Bilinear interpolation can be constructed
where
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1
0 11
2
1
1 1
2
u
y
x
1
x = n xn
u = n un
y = n yn0
Bilinear Tensor-Product Basis Functions
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A Six-NodedQuadratic-Linear Element
1
3
5
( , ) ( ) ( )
( , ) ( )
( , ) ( )
1 2 1 1 2
1 2 1 1 2
1 2 1 1 2
2 11
21
21
2
1
21
4 1
FH IK
FH IK FH IK
2
4
6
( , ) ( )( )
( , )
( , )
1 2 1 1 2
1 2 1 1 2
1 2 1 1 2
4 1 1
2 11
2
21
2
FH IK FH IK
b g
1
2
1.0
1.000
0.5
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Three-dimensional Linear Basis Functions
e.g. trilinear element has eight nodes with basis functions:
21 ;1
1
2
3
4
5
6
7
8
1
2
3
.,,
;,,
;,,
;,,
;,,
;,,
;,,
;,,
3222123218
3222113217
3221123216
3321113215
3122123214
3122113213
3121123212
3121113211
8
1321 ,,
iii uu
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1
2
3
5
6
7
1
2
3
In each node we define:
221
3
32
2
31
2
321
2
21
,,
,,,,,
uuu
uuuuu
Tri-Cubic Basis Functions
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321
3
3218
32
2
3217
31
2
3216
3321
5
21
2
3214
8
1 2321
3
1321
2321
1321
,,,,
,,,,,,
,,,,,,,,
ii
ii
ii
ii
ii
i
ii
iiii
uu
uuu
uuuu
8,...2,1,;2,1,,,,,
;,, 321321
jirqnmlk
rq
nm
lk
ji
Tri-Cubic Basis Functions (Cont’d)
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Scaling Factors
=0 =1=0 =0
s1 s2 s3
n e n,e
n en e n,e i
i i i
n e n e n en,e2 i j
i j i j i j
n e n,e3
i j k i j k
(no sum on i)
(no sum on i,j)
U U
U U s
s
U U s s
s s
U U
s s s
n e n e n ei j k
i j k(no sum on i,j,k)
s s s
Global to local mapping:
Scaling Factors arc lengths
arc length
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Coordinate Systems• Rectangular Cartesian global reference
coordinate system • Orthogonal curvilinear coordinate
system to describe geometry and deformation
• Curvilinear local finite element coordinates
• Locally orthonormal body coordinates define material symmetry and structure, related to the finite element coordinates by a rotation about the -normal axis through the "fiber angle" ,
1 2 3{ }Y ,Y ,Y
1 2 3{ }, ,
1 2 3{ }, ,
1 2 3{ }X ,X ,X
1 2( , ) 1
From Costa et al, J Biomech Eng 1996;118:452-463
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Curvilinear World
Coordinates
R
Y1
Y2
Y3
A) Rectangular Cartesian Coordinates: {A}=(X,Y,Z)
C) Spherical Polar Coordinates: {A}=(R,,)
Y1 = R cos cosY2 = R cos sinY3 = R sin
B) Cylindrical Polar Coordinates: {A}=(R,,Z)
Y1 = R cosY2 = R sinY3 = Z
Y3=ZY2=Y
Y1=X
Y1
R
Y2
Y3=Z
+
Y2
Y1
Y3
d=focus
= d coshcos= d sinhsincos= d sinhsinsin
Y2
Y1
Y3
D) Prolate Spheroidal Coordinates ()
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Fiber/Sheet Coordinates
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Coordinate System
Notations
* R e p r e s e n t s a L a g r a n g i a n d e s c r i p t i o n o f t h e d e f o r m a t i o n f r o m B t o B .
S t a t e o f B o d y
I n d i c e s
C o o r d i n a t e s
C o v a r i a n t B a s e V e c t o r s
C o v a r i a n t M e t r i c T e n s o r s
A ) r e c t a n g u l a r C a r t e s i a n r e f e r e n c e c o o r d i n a t e s
B R , S RY Re R , S
B r , s Ry re r , s
B ) c u r v i l i n e a r w o r l d c o o r d i n a t e s
B A , B A R
A RA
Y
G e
R R
A B A B
Y YG
B , r
ry
g e
r ry yg
C ) n o r m a l i z e d f i n i t e e l e m e n t c o o r d i n a t e s ( L a g r a n g i a n )
B K , L K K AK
G G A B
K L A BK LG G
B * K K
g g * K L K L
g g
D ) l o c a l l y o r t h o n o r m a l b o d y / f i b e r c o o r d i n a t e s
B I , J IX K
xI KIX
G G
K L
xI J K L I JI J
G GX X
B i , j ix K
xi Kix
g g
K Lx
i j K L i ji jg g
x x
B * K
xI KI
x
X
g g * K L
xI J K LI J
x xg g
X X
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Fitting with Linear Lagrange 1-D Elements
Two linear Lagrange elements fit the data with a root-mean-squared-error (RMSE) of 0.614892.
Result of twice refining the mesh (yielding 8
elements) and re-fitting: RMSE = 0.0930764
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2
1,
2 2
1 21 2
2 2 22 2 2
3 4 52 21 2 1 2
d
d d dd D
d d d d
d d d d d d
F
d
X X X
X X X X
X X X X X X
The least squares fit minimizes the objective function:
dX dX
i
ξd
where is measured coordinate or field variable;
are smoothing weights
is the interpolated value at
Least Squares Fitting
d are weights applied to the data points
i id d
Nd j j N
NjN
j
0 a linear system of equations for nodal parameters
X X
FX
X
X
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Fitting a Coronary Vascular Tree with Quadratic Lagrange 1-D Elements
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• anesthetized & ventilated New Zealand White rabbit
• heart arrested in diastole, excised
• pulmonary vessels removed, aorta cannulated
• heart suspended in Ringers lactate, perfused in unloaded state with buffered formalin at 80 mm Hg for 4 minutes
• heart cast in polyvinylsiloxane
plunger
tube
knife
heart cast in rubber
Rabbit Ventricular Anatomy
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plunger
knife
Rabbit Ventricular Anatomy
BASE
APEX
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data point projects onto
surface at
ddd
Bicubic Hermiteisoparametric interpolation
(1, 2) {ii1
i
4
(1, 2)
i
1
i2(1, 2)
i
2
i3(1, 2)
2i
12
i4(1, 2)}
1
x = d cosh cos y = d sinh sin cos z = d sinh sin sin
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endo
epi
RIGHT VENTRICLE
LEFT VENTRICLE
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• 8,351 geometric points
• 14,368 fiber angles• 36 elements• 552 geometric DOF
RMSE = ±0.55 mm• 184 Fiber angle DOF
RMSE = ±19°
Anatomic Model
Vetter & McCulloch Prog Biophys & Mol Biol 69(2/3):157 (1998)
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Strain Analysis
X2, longitudinal
X1, circumferential
X 3, radial
Xc , crossfiber
X f , fiber
X r , radial
T12
2 2 12
=
d d
i i kij i j i j i j
j k j
ij i j
x xF
X X
s S E dX dX
F e e e e e e
E F F I
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A/P View Lateral View
Reconstructed3D Coordinates
Transform
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Baseline 2 minutes ischemia
End-Systolic Circumferential Strain
0.04
0.00
-0.04
-0.07
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x
x x
x x xx
2
1,
2 2
1 2
2 2 22 2 22
2 21 2 1 2
d
d d dd D
d d d d
d d d d d d
F
dx
X X X
X X X X
X X X X X X
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7R
MS
Fit
tin
g E
rro
r (m
m)
10-7 10-6 10-5 10-4 10-3 10-2 10-1 100
Smoothing Weight
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0.018
0.02
0.022
0.0240.026
0.030.028
10-6
10-5 10-4 10-3 10-2
10-6
10-5
10-4
10-3
10-20.032
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Fiber Strain Cross-fiber StrainMyocardial Blood Flow
Co
ntr
ol
LA
D O
cclu
sio
n
-0.05 0.00 0.050.0 1.5 3.0mL/min/g
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SE
PT
AL
LA
TE
RA
L
3months post-surgeryPre-surgery