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MANE 4240 & CIVL 4240Introduction to Finite Elements
Higher order elements
Prof. Suvranu De
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 familySerendipity family
Recall that the finite element shape functions need to satisfy the following properties
1. Kronecker delta property
nodesotherallat
inodeatN i 0
1
...2211 uNuNu
Inside an element
At node 1, N1=1, N2=N3=…=0, hence
11uu
node
Facilitates the imposition of boundary conditions
2. Polynomial completeness
yyN
xxN
N
ii
i
ii
i
ii
1
yxuIf 321
Then
Higher order elements in 1D
2-noded (linear) element:
1 2
x2x1
12
12
21
21
xx
xxN
xx
xxN
In “local” coordinate system (shifted to center of element)
1 2
x
a a a
xaN
a
xaN
2
2
2
1
x
3-noded (quadratic) element:
1 2
x2x1
2313
213
3212
312
3121
321
xxxx
xxxxN
xxxx
xxxxN
xxxx
xxxxN
In “local” coordinate system (shifted to center of element)
1 2
x
a a
2
22
3
22
21
2
2
a
xaN
a
xaxN
a
xaxN
3
x3
x
axxax 321 ;0;
3
4-noded (cubic) element:
1 2
x2x1
342414
3214
432313
4213
423212
4312
413121
4321
xxxxxx
xxxxxxN
xxxxxx
xxxxxxN
xxxxxx
xxxxxxN
xxxxxx
xxxxxxN
In “local” coordinate system (shifted to center of element)
1 2
2a/3x
a a
)3/)(3/)((16
27
))(3/)((16
27
)3/)(3/)((16
9
)3/)(3/)((16
9
34
33
32
31
axaxaxa
N
xaxaxaa
N
axaxaxa
N
axaxaxa
N
3
x3
xx4
4
3 4
2a/32a/3
Polynomial completeness
4
3
2
1
x
x
x
x 2 node; k=1; p=2
3 node; k=2; p=3
4 node; k=3; p=4
Convergence rate (displacement)
1;0
kpChuu ph
Recall that the convergence in displacements
k=order of complete polynomial
Triangular elements
Area coordinates (L1, L2, L3)1
A=A1+A2+A3Total area of the triangle
At any point P(x,y) inside the triangle, we define
A
AL
A
AL
A
AL
33
22
11
Note: Only 2 of the three area coordinates are independent, since
x
y
2
3
PA1
A2A3
L1+L2+L3=1
A
ycxbaL iiii 2
12321312213
31213231132
23132123321
33
22
11
x1
x1
x1
det2
1
xxcyybyxyxa
xxcyybyxyxa
xxcyybyxyxa
y
y
y
triangleofareaA
Check that
yyLyLyL
xxLxLxL
LLL
332211
332211
321 1
x
y
2
3
PA1
1
L1= constantP’
Lines parallel to the base of the triangle are lines of constant ‘L’
We will develop the shape functions of triangular elements in terms of the area coordinates
x
y
2
3
PA1
1
L1= 0
L1= 1
L2= 0
L2= 1L3= 0
L3= 1
For a 3-noded triangle
33
22
11
LN
LN
LN
x
y
2
3
1
L1= 0
L1= 1L2= 0
L2= 1
L3= 0 L3= 1
For a 6-noded triangle
L1= 1/2L2= 1/2
L3= 1/2
4
5
6
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)
2/10 111 LLcN
Determine the constant ‘c’ from the condition that N1=1 at node 1 (i.e., L1=1)
2/12
2
12/111)1(
111
11
LLN
c
cLatN
136
235
214
333
222
111
4
4
4
)2/1(2
)2/1(2
)2/1(2
LLN
LLN
LLN
LLN
LLN
LLN
x
y
2
3
1
L1= 0
L1= 1L2= 0
L2= 1
L3= 0 L3= 1
For a 10-noded triangle
L1= 2/3
L2= 2/3
L3= 1/3 L3= 2/3
L2= 1/3
L1= 1/3
5
4
67
89
10
32110
3327
2326
2215
1214
3333
2222
1111
27
:
)3/1(2
27
)3/1(2
27
)3/1(2
27
)3/1(2
27
)3/2)(3/1(2
9
)3/2)(3/1(2
9
)3/2)(3/1(2
9
LLLN
LLLN
LLLN
LLLN
LLLN
LLLN
LLLN
LLLN
NOTES:1. Polynomial completeness
3 node; k=1; p=2
6 node; k=2; p=3
10 node; k=3; p=4
Convergence rate (displacement)
3223
22
1
yxyyxx
yxyx
yx
2. Integration on triangular domain
)!1(
!!.2
)!2(
!!!2.1
2121 21
321
mk
mkldSLL
nmk
nmkAdALLL
edge
mk
A
nmk
1
x
y
2
3
l1-2
3. Computation of derivatives of shape functions: use chain rule
x
L
L
N
x
L
L
N
x
L
L
N
x
N iiii
3
3
2
2
1
1
e.g.,
ButA
b
x
L
A
b
x
L
A
b
x
L
2;
2;
2332211
e.g., for the 6-noded triangle
A
bL
A
bL
x
N
LLN
24
24
4
12
21
4
214
Rectangular elements
Lagrange familySerendipity family
Lagrange family
4-noded rectangle
x
ya a 12
3 4
b
b
In local coordinate system
ab
ybxaN
ab
ybxaN
ab
ybxaN
ab
ybxaN
4
))((4
))((4
))((4
))((
4
3
2
1
9-noded quadratic
x
ya a 12
3 4
b
b
Corner nodes
224223
222221
2
)(
2
)(
2
)(
2
)(
2
)(
2
)(
2
)(
2
)(
b
yby
a
xaxN
b
yby
a
xaxN
b
yby
a
xaxN
b
yby
a
xaxN5
6
7
89
Midside nodes
2
22
2822
22
7
2
22
2622
22
5
2
)(
2
)(
2
)(
2
)(
b
yb
a
xaxN
b
yby
a
xaN
b
yb
a
xaxN
b
yby
a
xaN
Center node
2
22
2
22
9 b
yb
a
xaN
54322345
432234
3223
22
1
yxyyxyxyxx
yxyyxyxx
yxyyxx
yxyx
yx
NOTES:1. Polynomial completeness
4 node; p=2
9 node; p=3
Convergence rate (displacement)
Lagrange shape functions contain higher order terms but miss out lower order terms
Serendipity family
Then go to the corner nodes. At each corner node, first assume a bilinear shape function as in a 4-noded element and then modify:
22ˆ
4
))((ˆ
8511
1
NNNN
ab
ybxaN
x
ya a 12
3 4
b
b
5
6
7
8
First generate the shape functions of the midside nodes as appropriate products of 1D shape functions, e.g.,
2
22
82
22
5 2
)(;
2
)(
b
yb
a
xaN
b
yb
a
xaN
4-noded same as Lagrange
8-noded rectangle: how to generate the shape functions?
“bilinear” shape fn at node 1:
actual shape fn at node 1:
Corner nodes
224
))((
224
))((224
))((
224
))((
784
763
652
851
NN
ab
ybxaN
NN
ab
ybxaN
NN
ab
ybxaN
NN
ab
ybxaN
x
ya a 12
3 4
b
b
5
6
7
8
Midside nodes
2
22
82
22
7
2
22
62
22
5
2
)(
2
)(
2
)(
2
)(
b
yb
a
xaN
b
yb
a
xaN
b
yb
a
xaN
b
yb
a
xaN
8-noded rectangle
54322345
432234
3223
22
1
yxyyxyxyxx
yxyyxyxx
yxyyxx
yxyx
yx
NOTES:1. Polynomial completeness
4 node; p=2
8 node; p=3
Convergence rate (displacement)
12 node; p=4
16 node; p=4
More even distribution of polynomial terms than Lagrange shape functions but ‘p’ cannot exceed 4!
