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Finite elements with Isoparametric formulation
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MANE 4240 & CIVL 4240Introduction to Finite Elements
Mapped element geometries and shape functions: the isoperimetric formulation
Prof. Suvranu De
Reading assignment:
Chapter 10.1-10.3, 10.6 + Lecture notes
Summary:
• Concept of isoparametric mapping• 1D isoparametric mapping• Element matrices and vectors in 1D• 2D isoparametric mapping : rectangular parent elements • 2D isoparametric mapping : triangular parent elements • Element matrices and vectors in 2D
For complex geometries© 2002 Brooks/Cole Publishing / Thomson Learning™
General quadrilateral elements
Elements with curved sides
s
t1 1 12
3 4
1
1
Consider a special 4-noded rectangle in its local coordinate system (s,t)
44332211
44332211
vNvNvNvNv
uNuNuNuNu
Displacement interpolation
)1)(1(4
1),(
)1)(1(4
1),(
)1)(1(4
1),(
)1)(1(4
1),(
4
3
2
1
tstsN
tstsN
tstsN
tstsN
Shape functions in local coord system
Recall that
ttNtNtNtN
ssNsNsNsN
NNNN
44332211
44332211
4321 1 Rigid body modes
Constant strain states
Goal is to map this element from local coords to a general quadrilateral element in global coord system
s
t1 1 12
3 4
1
1
Local coordinate system
s
y 1
2
3
4x
s
t
Global coordinate system
),(
),(
tsyy
tsxx
),(
),(
yxtt
yxss
),(1 tsN ),(1 yxN
),()),(),,((),( 111 yxNyxtyxsNtsN
In the mapped coordinates, the shape functions need to satisfy
1. Kronecker delta property
nodesotherallat
inodeatN i 0
1
yyN
xxN
N
ii
i
ii
i
ii
1
Then
2. Polynomial completeness
The relationship
ii
i
ii
i
ytsNy
xtsNx
),(
),(
Provides the required mapping from the local coordinate systemTo the global coordinate system and is known as isoparametric mapping
(s,t): isoparametric coordinates(x,y): global coordinates
s
t1 1
1
1
s
t
y
x
Examples
s
t
1
1
y
x
s
t
1D isoparametric mapping
1 2
x2x1 x1 2
s1 1
Local (isoparametric) coordinates
3 noded (quadratic) element
3
3 x3
23
2
1
1)(
2
1)(
2
1)(
ssN
sssN
sssN
Isoparametric mapping
3
1
)(i
ii xsNx
32
21 12
1
2
1xsx
ssx
ssx
32
21 12
1
2
1xsx
ssx
ssx
NOTES1. Given a point in the isoparametric coordinates, I can obtain the corresponding mapped point in the global coordinates using the isoparametric mapping equation
2
3
1
;1
;0
;1
xxsAt
xxsAt
xxsAt
Question x=? at s=0.5?
2. The shape functions themselves get mappedIn the isoparametric coordinates (s) they are polynomials.In the global coordinates (x) they are in general nonpolynomialsLets consider the following numerical example
4;6;0 321 xxx
1 2
x
3
4 2
Isoparametric mapping x(s)
21 2 3
2
2
1 11
2 21 1
0 6 1 42 2
4 3
s s s sx x x s x
s s s ss
s s
Inverse mapping s(x)
2
4253 xs
Simple polynomial
Complicated function
Now lets compute the shape functions in the global coordinates
)(
4252102
1
2
42531
2
4253
2
1
2
1)(
2
2
xN
xx
xx
sssN
Now lets compute the shape functions in the global coordinates
N2(x) is a complicated function
However, thanks to isoparametric mapping, we always ensure 1. Knonecker delta property2. Rigid body and constant strain states
1 2
s1 1
3
N2(s)
1 2
x
3
4 2 11
N2(x)
N2(s) is a simple polynomial
2
1)(2
sssN
xxxN 425210
2
1)(2
Element matrices and vectors for a mapped 1D bar element
1 2
x1 2
s1 1 3
3
Displacement interpolation dNuNuNuNu 332211
Strain-displacement relation dBudx
dNu
dx
dNu
dx
dN
dx
du 3
32
21
1
Stress dBEE
The strain-displacement matrix
dx
dN
dx
dN
dx
dNB 321
The only difference from before is that the shape functions are in the isoparametric coordinates
23
2
1
1)(
2
1)(
2
1)(
ssN
sssN
sssN
And we will not try to obtain explicitly the inverse map.How to compute the B matrix?
We know the isoparametric mapping
3
1
)(i
ii xsNx
Do I know ?)(
ds
sdN i
Do I know ?dx
ds
I know
3
1
)(i
ii xsNx
Hence )()(3
1
mappingofJacobianJxds
sdN
ds
dx
ii
i
ds
sdN
Jdx
sdN ii )(1)(From (*)
dx
ds
ds
sdN
dx
sdN ii )()(
Using chain rule
(*)
What does the Jacobian do?
dsJdx
Maps a differential element from the isoparametric coordinates to the global coordinates
The strain-displacement matrix
ds
dN
ds
dN
ds
dN
J
dx
dN
dx
dN
dx
dNB
321
321
1
For the 3-noded element
321
3
1
22
12
2
12)(sxx
sx
sx
ds
sdNJ
ii
i
s
ss
JB 2
2
12
2
121
The element stiffness matrix
1
1
2
1
JdsdxJdsBBEA
dxBBEAk
T
x
x
T
NOTES1. The integral on ANY element in the global coordinates in now an integral from -1 to 1 in the local coodinates2. The jacobian is a function of ‘s’ in general and enters the integral. The specific form of ‘J’ is determined by the values of x1, x2 and x3. Gaussian quadrature is used to evaluate the stiffness matrix3. In general B is a vector of rational functions in ‘s’
Isoparametric mapping in 2D: Rectangular parent elements© 2002 Brooks/Cole Publishing / Thomson Learning™
Parent element
Isoparametric mapping
Mapped element in global coordinates
ii
i
ii
i
ytsNy
xtsNx
),(
),(
1
2
3
4
1( , ) (1 )(1 )
41
( , ) (1 )(1 )41
( , ) (1 )(1 )41
( , ) (1 )(1 )4
N s t s t
N s t s t
N s t s t
N s t s t
Isoparametric mapping
ii
i
ii
i
ytsNy
xtsNx
),(
),(
Shape functions of parent element in isoparametric coordinates
NOTES:1. The isoparametric mapping provides the map (s,t) to (x,y) , i.e.,
if you are given a point (s,t) in isoparametric coordinates, then you can compute the coordinates of the point in the (x,y) coordinate system using the equations
2. The inverse map will never be explicitly computed.
ii
i
ii
i
ytsNy
xtsNx
),(
),(
s
t1 1 34
1 2
1
1
7
8
5
6
8-noded Serendipity element© 2002 Brooks/Cole Publishing / Thomson Learning™
8-noded Serendipity element: element shape functions in isoparametric coordinates
1 5
2 6
3 7
4 8
1 1( , ) (1 )(1 )( 1) ( , ) (1 )(1 )(1 )
4 21 1
( , ) (1 )(1 )( 1) ( , ) (1 )(1 )(1 )4 21 1
( , ) (1 )(1 )( 1) ( , ) (1 )(1 )(1 )4 2
1 1( , ) (1 )(1 )( 1) ( , ) (1 )
4 2
N s t s t s t N s t t s s
N s t s t s t N s t t t s
N s t s t s t N s t t s s
N s t s t s t N s t s
(1 )(1 )t t
NOTES1. Ni(s,t) is a simple polynomial in s and t. But Ni(x,y) is a complex function of x and y.2. The element edges can be curved in the mapped coordinates3. A “midside” node in the parent element may not remain as a midside node in the mapped element. An extreme example
s
t1 1 12
3 4
1
1
5
6
7
8x
y 1
85
7
2,6,3
4. Care must be taken to ensure UNIQUENESS of mapping
s
t1 1 12
3 4
1
1
x
y
1
2
3
4
Isoparametric mapping in 2D: Triangular parent elements
s
t2
3 1
1
1
P(s,t)st
)1(2
12
2
2
1
12
23
31
123
ts
s
t
P
P
P
tsL
tL
sL
P
P
P
1123
123
123
312
123
231
Parent element: a right angled triangles with arms of unit length Key is to link the isoparametric coordinates with the area coordinates
Now replace L1, L2, L3 in the formulas for the shape functions of triangular elements to obtain the shape functions in terms of (s,t)
Example: 3-noded triangle
y
xs
t2
3 1
1
1 (x1,y1)s
t
Parent shape functions
tsN
tN
sN
13
2
1
Isoparametric mapping
332211
332211
),(),(),(
),(),(),(
ytsNytsNytsNy
xtsNxtsNxtsNx
2 (x2,y2)
3 (x3,y3)1
Element matrices and vectors for a mapped 2D element
Stress approximation
Recall: For each element
Element stiffness matrix
Element nodal load vector
dNu
dBD
dBε
eV
k dVBDBT
S
eT
b
e
f
S ST
f
V
T dSTdVXf NN
Displacement approximation
Strain approximation
STe
e
In isoparametric formulation1. Shape functions first expressed in (s,t) coordinate systemi.e., Ni(s,t)2. The isoparamtric mapping relates the (s,t) coordinates with the
global coordinates (x,y)
3. It is laborious to find the inverse map s(x,y) and t(x,y) and we do not do that. Instead we compute the integrals in the domain of the parent element.
ii
i
ii
i
ytsNy
xtsNx
),(
),(
NOTE1. Ni(s,t) s are already available as simple polynomial functions2. The first task is to find and
x
N i
y
N i
Use chain rule
t
y
y
N
t
x
x
N
t
yxN
s
y
y
N
s
x
x
N
s
yxN
iii
iii
),(
),(
In matrix form
y
Nx
N
t
y
t
xs
y
s
x
t
Ns
N
i
i
i
i
Can be computed
We want to compute these for the B matrix
This is known as the Jacobian matrix (J) for the mapping(s,t) → (x,y)
y
Nx
N
J
t
Ns
N
i
i
i
i
t
Ns
N
J
y
Nx
N
i
i
i
i
1
How to compute the Jacobian matrix?Start from
ii
i
ii
i
ytsNy
xtsNx
),(
),(
ii
i
ii
i
ii
i
ii
i
yt
tsN
t
yy
s
tsN
s
y
xt
tsN
t
xx
s
tsN
s
x
),(;
),(
),(;
),(
ii
i
ii
i
ii
i
ii
i
yt
tsNx
t
tsN
ys
tsNx
s
tsN
t
y
t
xs
y
s
x
J
),(),(
),(),(
Need to ensure that det(J) > 0 for one-to-one mapping
3. Now we need to transform the integrals from (x,y) to (s,t)
Case 1. Volume integrals
1
1
1
1
)det(),(),(),( dsdtJhtsfdAhtsfdVtsfee AV
h=thickness of element
This depends on the key result
dsdtJdxdydA )det(
s
t1 1 12
3 4
1
1
Problem: Consider the following isoparamteric map
y
12
3 4(3,1) (6,1)
(6,6)(3,6)
xISOPARAMETRIC COORDINATES
y
GLOBAL COORDINATES
44332211
44332211
vNvNvNvNv
uNuNuNuNu
Displacement interpolation
)1)(1(4
1),(
)1)(1(4
1),(
)1)(1(4
1),(
)1)(1(4
1),(
4
3
2
1
tstsN
tstsN
tstsN
tstsN
Shape functions in isoparametric coord system
1 1 2 2 3 3 4 4
1 1 2 2 3 3 4 4
( , ) ( , ) ( , ) ( , )
( , ) ( , ) ( , ) ( , )
x N s t x N s t x N s t x N s t x
y N s t y N s t y N s t y N s t y
The isoparamtric map
(1 )(1 ) (1 )(1 ) (1 )(1 ) (1 )(1 )6 3 3 6
4 4 4 43(1 )
2(1 )(1 ) (1 )(1 ) (1 )(1 ) (1 )(1 )
6 6 1 14 4 4 4
7 5
2
s t s t s t s tx
s
s t s t s t s ty
t
In this case, we may compute the inverse map, but we will NOT do that!
The Jacobian matrix
NOTE: The diagonal terms are due to stretching of the sides along the x-and y-directions. The off-diagonal terms are zero because the element does not shear.
30
25
02
x y
s sJx y
t t
3(1 )
27 5
2
sx
ty
since
1 2 / 3 0 15and det( )
0 2 / 5 4J J
Hence, if I were to compute the first column of the B matrix along the positive x-direction
I would use
1
1
1
0
N
x
N
y
B
1 1
1
1 1
112 / 3 0 64
0 2 / 5 1 1
4 10
N N ttx sJN N s sy t
1
1
1
1
60 0
1
10
N t
x
N s
y
B
Hence
The element stiffness matrix
1 1T T
1 1B D B dV B D B det( ) dsdt
eVk J h
