introduction to fem ppt

7/29/2019 introduction to fem ppt

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Introduction to the Finite Element MethodLecture 18

P.S. Koutsourelakis

[email protected] Hollister Hall

November 22 2010

Last Updated: November 22, 2010

[email protected] Lecture 18


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Higher-order isoparametric elements

In many problems there might be a need to increase the numberof shape functions (without increasing the number of elements)and/or represent more accurately the boundary of the domain.

We discuss some higher order isoparametric elements thatachieve this goal.

There is no free lunch! Every time you increase the number ofshape functions you will also increase the number of unknownsand the size of the system you need to solve.



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Serendipity Elements

e

(xe1 , ye1 ) (xe2 , y

e2 )

(xe3 , ye3 )

(xe4 , ye4 )

(xe5 , ye5 )

(xe6, ye

6)

(xe7 , ye7 )

(xe8 , ye8 )

x

y

(1,1) (1,1)

(1, 1)(1, 1)

(0,1)

(1, 0)

(0, 1)

(1, 0)

x = x()

= (x)

11 22

334

4

55

66

77

88



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Serendipity elements consist of additional nodes along the

boundaryObserve the numbering!

Let the mapping be defined as follows:

x(, ) =

8a=1 Na(, )x

ea

y(, ) = 8a=1 Na(, )yea x() =

8

a=1 Na()xea

To obtain the functions Na we first assume a form:x(, ) = b0 + b1 + b2 + b3 + b4

2 + b52 + b6

2 + b72

y(, ) = c0 + c1 + c2 + c3 + c42 + c5

2 + c62 + c7

2

where the parameters b and c should be determined by satisfying:

x(a, a) = xea

y(a, a) = yea

for a= 1, . . . , 8



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We obtain the following shape functions:

N1 =14

(1 )(1 ) 12

(N8 + N5)N2 =

14

(1 + )(1 ) 12

(N5 + N6)

N3 =14

(1 + )(1 + ) 12

(N6 + N7)

N4 =14

(1 )(1 + ) 12

(N7 + N8)

N5 =

1

2 (1

2

)(1

)N6 =12

(1 + )(1 2)N7 =

12

(1 2)(1 + )

N8 =12

(1 )(1 2)

Note that N1 arises from the linearshape function of the 4-nodequad (which is equal to 1/2 at nodes 5 and 8) by subtracting1/2 N5 and 1/2 N8 in order to become zero at nodes 5 and 8.

Similarly for N2, N3, N4.



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To find the local stiffness matrix ke:

ke =

e

(Be)T D Be d

=

11

11

(Be(, ))T D Be(, ) j(, ) dd

The strain displacement matrix Be:

31

=B

e316

de

161=

Be

dex,1dey,1.

.dex,16dey,16


S di i El


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Lets break Be in two parts:

=

uxxuyyuxy

+uyx

= 1

j(, )

y, y, 0 0

0 0 x, x,x, x, y, y,

uxuxuyuy

or:

= Be1

uxuxuy

uy


S di it El t


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and the second part:

uxuxuyuy

=

N1, 0 . . . . . . N9, 0

N1, 0 . . . . . . N9, 0

0 N1, . . . . . . 0 N9,0 N1, . . . . . . 0 N9,

de

or:

uxuxuyuy

= B

e2 d

e

Hence:B

e316

= Be134

Be2

416


S di it El t


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When is the Jacobian determinant not zero?

acceptable position for node 5L/4L/4

12

3

4

5

6

7

8


S di it El t


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Example: Find local force vector

q3q4

1 2

34

5

6

7

8


Lagrange Elements


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Lagrange Elements

(xe2 , ye2 )

x

y

x = x()

= (x)

11 22

334

4

55

66

77

88 99


Lagrange Elements


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Lagrange Elements

Lagrange elementsconsist of additional nodes in the interior ofthe element

Observe the numbering!

Lagrange polynomials. Suppose one is given npairs of values

(i, i = (i)). The the function can be approximated as follows:

L11 + L22 + . . . + Lnn

where:L1 =

(2)(3)...(n)(21)(31)...(n1)

L2 =(1)(2)...(n)

(12)(33)...(n3)

. . .

Ln =(1

)(2

)...(n1

)(1n)(2n)...(n1n)

Note: Li(j) = i,j


Lagrange Elements


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Lagrange Elements

For example, for n = 2, 1 = 1, 2 = +1:

L(2)1 =

(2)(21)

= 12

(1 )

L(2)2 =

(1)(12)

= 12

(1 + )

For n = 3 and 3 = 0:

L(3)1 =

(2)(3)(21)(31)

= 12(1 )

L(3)2 =

(1)(3)(12)(32)

= 12(1 + )

L(3)3 = (1

)(2

)(13)(23) = 12 (1 2)


Lagrange Elements


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Lagrange Elements

1 2

34

5

6

7

89

To generate the shape functions for plane isoparametric elements, it

suffices to multiply the Lagrange polynomials w.r.t. to and of

appropriate order. For example, for the 9 node element:

N1(, ) = L(3)1 () L

(3)1 () =

14

(1 )(1 )

N6(, ) = L(3)

2 () L(3)

3 () =1

4 (1 + )(1

2

)

or :

N9(, ) = L(3)2 () L

(3)2 () = (1

2)(1 2) bubble function


Lagrange Elements


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Lagrange Elements


Lagrange Elements


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Lagrange Elements

Naturally, we could derive the same shape functions if we follow the

procedure discussed for serendipity elements and use interpolating

functions of the form:

b0 + b1 + b2 + b3 + b42 + b5

2 + b62

+ b72

+ b82

2

Conversely, we can use the Lagrange polynomials to re-derive the

shape functions of the isoparametric elements we have seen before.

For example, the 4-node quadrilateral:

N1(, ) =1

4(1 )(1 ) = L

(2)1 ()L

(2)1 ()

Or the shape functions of lateral nodes of the 8-node serendipity

element, e.g.:

N5(, ) = L(3)3 ()L

(2)1 () =

1

2(1 2)(1 )

One has to be more careful for the shape functions of the corner nodes!


Isoparametric Quadrilateral Elements with Variable


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Number of nodes




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Number of nodes

5-node quadrilateral:

x

y

x = x()

= (x)

11 22

33 44

55




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Number of nodes

The shape function of lateral nodecan be found from Lagrange

polynomials:

N5 = L(3)3 () L

(2)1 () =

1

2(1 2)(1 )

To find the shape functions of the corner nodeswe use the onesfrom the 4-node quadrilateral and subtract 1/2N5 IF NEEDED sothat they become 0 at all other nodes. Hence:

N1 =14

(1 )(1 ) 12

N5N2 =

1

4

(1 + )(1 ) 1

2

N5

and:N3 =

14

(1 + )(1 + )

N4 =14

(1 )(1 + )


Summary of Shape functions for Isoparametric


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Summary of Shape functions for Isoparametric

Quadrilateral Elements

i = 5 i = 6 i = 7 i = 8 i = 9N1 =

14

(1 )(1 ) 12

N5 12

N814

N9N1 =

14

(1 )(1 ) 12

N5 12

N614

N9N1 =

14

(1 )(1 ) 12

N6 12

N714

N9N1 =

14

(1 )(1 ) 12

N7 12

N814

N9

N5 = 12 (1 2)(1 ) 12 N9N5 =

12

(1 + )(1 2) 12

N9N5 =

12

(1 2)(1 + ) 12

N9N5 =

12

(1 )(1 2) 12

N9

N9 = (1 2)(1 2)

The shape functions in columns i = 5 9 are activated only if therespective node appears in the element and they are added tothe appropriate shape function in the first column.


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