# Higher Order Elements - Introduction to Finite Elements - Lecture Slides, Slides for Mathematical Methods for Numerical Analysis and Optimization. Shree Ram Swarup College of Engineering & Management

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1

MANE 4240 & CIVL 4240 Introduction to Finite Elements

Higher order elements

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

⎩ ⎨ ⎧ =

nodesotherallat inodeat

Ni 0 1

...2211 ++= uNuNu

Inside an element

At node 1, N1=1, N2=N3=…=0, hence

11 uu

node =

Facilitates the imposition of boundary conditions

2. Polynomial completeness

yyN

xxN

N

i i

i

i i

i

i i

=

=

=

∑ 1

yxuIf 321 ααα ++≈

Then

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2

Higher order elements in 1D

2-noded (linear) element:

1 2

x2x1

12

1 2

21

2 1

xx xx

N

xx xx

N

− −

=

− −

=

In “local” coordinate system (shifted to center of element)

1 2

x

a a a xaN

a xaN

2

2

2

1

+ =

− =

x

1 2

x2x1

( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( )2313

21 3

3212

31 2

3121

32 1

xxxx xxxxN

xxxx xxxx

N

xxxx xxxx

N

−− −−

=

−− −−

=

−− −−

=

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

321 4

432313

421 3

423212

431 2

413121

432 1

xxxxxx xxxxxx

N

xxxxxx xxxxxx

N

xxxxxx xxxxxx

N

xxxxxx xxxxxx

N

−−− −−−

=

−−− −−−

=

−−− −−−

=

−−− −−−

=

In “local” coordinate system (shifted to center of element)

1 2

2a/3 x

a a

)3/)(3/)(( 16

27

))(3/)(( 16

27

)3/)(3/)(( 16

9

)3/)(3/)(( 16

9

34

33

32

31

axaxax a

N

xaxaxa a

N

axaxax a

N

axaxax a

N

+−−−=

+−−=

+−+=

+−−−=

3

x3 x

x4

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

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3

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 A

L

A A

L

A A

L

3 3

2 2

1 1

=

=

=

Note: Only 2 of the three area coordinates are independent, since

x

y

2

3

P A1

A2A3

L1+L2+L3=1

A ycxbaL iiii 2

++ =

12321312213

31213231132

23132123321

33

22

11

x1 x1 x1

det 2 1

xxcyybyxyxa xxcyybyxyxa xxcyybyxyxa

y y y

triangleofareaA

−=−=−= −=−=−= −=−=−=

⎥ ⎥ ⎥

⎢ ⎢ ⎢

⎡ ==

Check that

yyLyLyL xxLxLxL

LLL

=++ =++

=++

332211

332211

321 1

x

y

2

3

P A1

1 L1= constant

P’

Lines parallel to the base of the triangle are lines of constant ‘L’

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4

We will develop the shape functions of triangular elements in terms of the area coordinates

x

y

2

3

P A1

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

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5

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= 1/3L3= 2/3

L2= 1/3 L1= 1/3

5

4

6 7

8 9

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

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6

2. Integration on triangular domain

)!1( !!.2

)!2( !!!2.1

2121 21

321

mk mkldSLL

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.,

But A

b x L

A b

x L

A b

x L

2 ;

2 ;

2 332211 =

∂ ∂

= ∂ ∂

= ∂ ∂

e.g., for the 6-noded triangle

A bL

A bL

x N

LLN

2 4

2 4

4

1 2

2 1

4

214

+= ∂ ∂

=

Rectangular elements Lagrange family Serendipity family

Lagrange family

4-noded rectangle

x

y a 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

−+ =

−− =

+− =

++ =

x

y a 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

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7

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:

22 ˆ

4 ))((ˆ

85 11

1

NNNN

ab ybxaN

−−=

++ =

x

y a 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 ))((

78 4

76 3

65 2

85 1

NN ab

ybxaN NN

ab ybxaN

NN ab

ybxaN NN

ab ybxaN

−− −+

=−− −−

=

−− +−

=−− ++

=

x

y a 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!

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