Transcript MANE 4240 & CIVL 4240 Introduction to Finite Elements

MANE 4240 & CIVL 4240
Introduction to Finite Elements
Prof. Suvranu De
Higher order elements
Reading assignment:
Lecture notes
Summary:
• Properties of shape functions
• Higher order elements in 1D
• Higher order triangular elements (using area coordinates)
• Higher order rectangular elements
Lagrange family
Serendipity family
Recall that the finite element shape functions need to satisfy the
following properties
1. Kronecker delta property
 1 at nodei
Ni  
0 at all other nodes
Inside an element
u  N1u1  N 2 u 2  ...
At node 1, N1=1, N2=N3=…=0, hence
u node 1  u1
Facilitates the imposition of boundary conditions
2. Polynomial completeness
If u  1   2 x   3 y
Then
N
i
1
i
N x
i
i
x
i
N
i
i
yi  y
Higher order elements in 1D
2-noded (linear) element:
x1
x2
x  x2
N1 
x1  x 2
x
x  x1
N2 
x 2  x1
2
1
In “local” coordinate system (shifted to center of element)
x
1
a
a
2
ax
2a
ax
N2 
2a
N1 
3-noded (quadratic) element:
x1 x3
x2
x
1
3
2
x  x2 x  x3 
N1 
x1  x2 x1  x3 
x  x1 x  x3 
N2 
x2  x1 x2  x3 
x  x1 x  x2 
N3 
x3  x1 x3  x2 
In “local” coordinate system (shifted to center of element)
x
1
3
a a
2
x1  a ; x2  0 ; x3  a
x a  x 
2a 2
xa  x 
N2 
2a 2
a2  x2
N3 
a2
N1  
x  x 2 x  x3 x  x 4 
N1 
x1  x 2 x1  x3 x1  x 4 
4-noded (cubic) element:
x  x1 x  x3 x  x 4 
N2 
x1 x3 x4 x2
x 2  x1 x 2  x3 x 2  x 4 
x
x  x1 x  x 2 x  x 4 
N

3
1 3 4 2
x3  x1 x3  x 2 x3  x 4 
x  x1 x  x 2 x  x3 
N4 
x 4  x1 x 4  x 2 x 4  x3 
In “local” coordinate system (shifted to center of element)
2a/3 2a/3 2a/3
1
a
3
4
a
2
9
N1  
( x  a)(x  a / 3)(x  a / 3)
3
16a
x
9
N2 
( x  a)(x  a / 3)(x  a / 3)
3
16a
27
N3 
(a  x)(a / 3  x)(a  x)
3
16a
27
N4  
( x  a)(x  a / 3)(x  a / 3)
3
16a
Polynomial completeness
Convergence
rate (displacement)
2 node; k=1; p=2
1
x
x2
x3
3 node; k=2; p=3
4 node; k=3; p=4
x4

Recall that the convergence in displacements
u  uh 0  Ch p ; p  k 1
k=order of complete polynomial
Triangular elements
Area coordinates (L1, L2, L3)
1
Total area of the triangle A=A1+A2+A3
P A2
A3
A1
y
At any point P(x,y) inside
the triangle, we define
3
2
x
A1
L1 
A
A2
L2 
A
A3
L3 
A
Note: Only 2 of the three area coordinates are independent, since
L1+L2+L3=1
ai  bi x  ci y
Li 
2A
1 x 1
1
A  area of triangle  det 1 x 2
2
1 x 3
y1 
y 2 
y 3 
a1  x 2 y 3  x3 y 2
b1  y 2  y 3
c1  x3  x 2
a 2  x3 y1  x1 y 3
a3  x1 y 2  x 2 y1
b2  y 3  y1
b3  y1  y 2
c 2  x1  x3
c3  x 2  x1
Check that
L1  L2  L3  1
L1 x1  L2 x2  L3 x3  x
L1 y1  L2 y2  L3 y3  y
1
L1= constant
P’
P
y
A1
3
2
x
Lines parallel to the base of the triangle are lines of constant ‘L’
L1= 1
1
P
y
L1= 0
A1
3
2
L3= 0
x L =1
2
L3= 1
L2= 0
We will develop the shape functions of triangular elements in
terms of the area coordinates
For a 3-noded triangle
N1  L1
N 2  L2
N 3  L3
For a 6-noded triangle
L2= 0
L1= 1
1
L2= 1/2
L2= 1
L1= 1/2
6
y
4
L1= 0
3
2
L3= 0
5
x
L3= 1/2
L3= 1
How to write down the expression for N1?
Realize the N1 must be zero along edge 2-3 (i.e., L1=0) and at
nodes 4&6 (which lie on L1=1/2)
N1  cL1  0L1  1 / 2
Determine the constant ‘c’ from the condition that N1=1 at
node 1 (i.e., L1=1)
N1 (at L1  1)  c11  1 / 2  1
c2
 N1  2 L1 L1  1 / 2
N1  2 L1 ( L1  1 / 2)
N 2  2 L2 ( L2  1 / 2)
N 3  2 L3 ( L3  1 / 2)
N 4  4 L1 L2
N 5  4 L3 L2
N 6  4 L3 L1
For a 10-noded triangle
L1= 1
L2= 0
1
L2= 1/3
9
4
L2= 2/3
L2= 1
L1= 2/3
y
10
5
2
6
L1= 1/3
8
L1= 0
7
x
L3= 0 L3= 1/3 L3= 2/3 L3= 1
3
N1 
N2 
N3 
N4 
N5 
N6 
N7 
9
L1 ( L1  1 / 3)(L1  2 / 3)
2
9
L2 ( L2  1 / 3)(L2  2 / 3)
2
9
L3 ( L3  1 / 3)(L3  2 / 3)
2
27
L1 L2 ( L1  1 / 3)
2
27
L1 L2 ( L2  1 / 3)
2
27
L2 L3 ( L2  1 / 3)
2
27
L2 L3 ( L3  1 / 3)
2
:
N 10  2 7L1 L2 L3
NOTES:
1. Polynomial completeness
Convergence
rate (displacement)
3 node; k=1; p=2
1
x
x2
x3 x2 y

y
xy
y2
xy 2 y 3
6 node; k=2; p=3
10 node; k=3; p=4
2. Integration on triangular domain

1.
A
1
2. 
l1-2
L1 L2 L3 dA  2 A
k
1 2 edge
y
3
2
x
m
n
L1 L2 dS  l12
k
m
k! m! n!
(2  k  m  n)!
k! m!
(1  k  m)!
3. Computation of derivatives of shape functions: use chain rule
e.g.,
N i N i L1 N i L2 N i L3



x
L1 x L2 x L3 x
But
L1
b1

;
x 2 A
b3
L2
b2 L3

;

x
2 A x
2A
e.g., for the 6-noded triangle
N 4  4 L1 L2

N 4
b
b
 4 L1 2  4 L2 1
x
2A
2A
Rectangular elements
Lagrange family
Serendipity family
Lagrange family
4-noded rectangle
y
2
1
a
a
b
In local coordinate system
N1 
x
b
3
4
N2 
N3 
N4 
( a  x )(b 
4a b
( a  x )(b 
4a b
( a  x )(b 
4a b
( a  x )(b 
4a b
y)
y)
y)
y)
9-noded quadratic
y 5
a
a
2
b
6
b
3
9
7
Corner nodes
 x(a  x)   y(b  y ) 
 x(a  x)   y(b  y) 
N1  
N


2
2
2




2a 2   2b 2 
 2a   2b 

 x(a  x)   y(b  y) 
 x(a  x)   y(b  y) 
N 3  

N

4
2
2


 2a 2   2b 2 
2
a
2
b






1
8
4
x
Midside nodes
2
2
 a 2  x 2   y (b  y ) 
 x(a  x)   b  y 
N5  
N 6  



2
2

a
2
b
2a 2   b 2 




2
2
 a 2  x 2   y (b  y ) 
 x(a  x)   b  y 
N7  
N8  
 


2
2
2
2

a
2
b

 2a   b



Center node
 a2  x2 b2  y 2 
N9  


2
2
 a  b 
NOTES:
1. Polynomial completeness
1
x
y
x2
x3
x4
xy
x2 y
x3 y
y2
Convergence
rate (displacement)
4 node; p=2
9 node; p=3
xy 2 y 3
x 2 y 2 xy 3 y 4
x 5 x 4 y x 3 y 2 x 2 y 3 xy 4 y 5
Lagrange shape functions contain higher order terms but miss
out lower order terms
Serendipity family
4-noded same as Lagrange
8-noded rectangle: how to generate the shape functions?
First generate the shape functions of the
y 5
midside nodes as appropriate products of
2
1
a
a
1D shape functions, e.g.,
b
8
2
2
2
2
6
x N   a  x   (b  y)  ; N   (a  x)   b  y 
5
8
 2 

 2a   b 2 
b
a
2
b







7
4 Then go to the corner nodes. At each corner
3
node, first assume a bilinear shape function as in
a 4-noded element and then modify:
ˆ  ( a  x )(b  y )
N
1
“bilinear” shape fn at node 1:
4ab
N N
actual shape fn at node 1:
N1  Nˆ 1  5  8
2
2
8-noded rectangle
y 5
a
a
2
b
1
8
6
b
3
Midside nodes
7
x
2
2
 a 2  x 2   (b  y ) 
 (a  x)   b  y 
N5  
N6  

2

  b 2 
2
b
2
a
a







2
2
 a 2  x 2   (b  y ) 
 (a  x)   b  y 
N7  
N8  



2
2

 2a   b
 a
  2b 

4
Corner nodes
(a  x)(b  y ) N 5 N 8
(a  x)(b  y ) N 5 N 6
N1 


N2 


4ab
2
2
4ab
2
2
(a  x)(b  y ) N6 N 7
(a  x)(b  y ) N8 N 7
N3 


N4 


4ab
2
2
4ab
2
2
NOTES:
1. Polynomial completeness
1
x
y
x2
x3
x4
x
5
x2 y
x3 y
4
3
x y x y
y2
xy
8 node; p=3
xy 2 y 3
12 node; p=4
x 2 y 2 xy 3 y 4
2
2
x y
3
Convergence
rate (displacement)
4 node; p=2
xy
4
y
5
16 node; p=4
More even distribution of polynomial terms than Lagrange
shape functions but ‘p’ cannot exceed 4!