Higher Order FEM

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!