1

MANE 4240 & CIVL 4240

Introduction to Finite Elements

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

⎩

⎨

⎧

=nodesotherallat

inodeat

Ni0

1

...

2211 ++= uNuNu

Inside an element

At node 1, N1=1, N2=N3=…=0, hence

1

1uu node =

Facilitates the imposition of boundary conditions

2. Polynomial completeness

yyN

xxN

N

i

ii

i

ii

ii

=

=

=

∑

∑

∑

1

yxuIf 321

α

α

α

+

+

≈

Then

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Higher order elements in 1D

2-noded (linear) element:

12

x2

x1

12

1

2

21

2

1

xx

xx

N

xx

xx

N

−

−

=

−

−

=

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

12

x

aa a

xa

N

a

xa

N

2

2

2

1

+

=

−

=

x

3-noded (quadratic) element:

12

x2

x1

()

(

)

()()

()()

()()

()()

()()

2313

21

3

3212

31

2

3121

32

1

xxxx

xxxx

N

xxxx

xxxx

N

xxxx

xxxx

N

−−

−−

=

−−

−−

=

−−

−

−

=

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

12

x

aa

()

()

2

22

3

2

2

2

1

2

2

a

xa

N

a

xax

N

a

xax

N

−

=

+

=

−

−=

3

x3x

axxax

=

=−= 321 ;0;

3

4-noded (cubic) element:

12

x2

x1

()()()

()()()

()()()

()()()

()()()

()()()

()()()

()()()

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)

12

2a/3 x

aa

)3/)(3/)((

16

27

))(3/)((

16

27

)3/)(3/)((

16

9

)3/)(3/)((

16

9

3

4

3

3

3

2

3

1

axaxax

a

N

xaxaxa

a

N

axaxax

a

N

axaxax

a

N

+−−−=

+−−=

+−+=

+−−−=

3

x3x

x4

4

34

2a/3

2a/3

Polynomial completeness

#

4

3

2

1

x

x

x

x2 node; k=1; p=2

3 node; k=2; p=3

4 node; k=3; p=4

Convergence

rate (displacement)

1;

0+=≤− kpChuu p

h

Recall that the convergence in displacements

k=order of complete polynomial

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Triangular elements

Area coordinates (L1, L2, L3)

1A=A1+A2+A3

Total 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

PA1

A2

A3

L1+L2+L3=1

A

ycxba

Liii

i2

+

+

=

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

PA1

1

L1= constant

P’

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

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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= 1

L3= 0

L3= 1

For a 3-noded triangle

33

22

11

LN

LN

LN

=

=

=

x

y

2

3

1

L1= 0

L1= 1

L2= 0

L2= 1

L3= 0 L3= 1

For a 6-noded triangle

L1= 1/2

L2= 1/2

L3= 1/2

4

5

6

How to write down the expression for N1?

Realize the N1must 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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University:
Shree Ram Swarup College of Engineering & Management

Subject:
Finite Element Method

Upload date:
07/05/2013