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Introduction to FEM

1 8Shape FunctionMagic

IFEM Ch 18 Slide 1

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'Magic' MeansDirect("by inspection")

Do in 15 minutes what took smart people several months

(and less gifted, several years)

But ... it looks like magic to the uninitiated

Introduction to FEM

IFEM Ch 18 Slide 2

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Shape Function Requirements

(A) Interpolation

(B) Local Support

(C) Continuity (Intra- & Inter-Element)

(D) Completeness

See Sec 18.1 for more detailed statement of (A) through (D).

Implications of the last two requirements as

regards convergence are discussed in Chapter 19.

Introduction to FEM

IFEM Ch 18 Slide 3

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Direct Construction of Shape Functions:

Are Conditions Automatically Satisfied?

(A)Interpolation Yes: by construction except scale factor

(B)Local Support Often yes, but not always possible

(C) Continuity No: a posteriori check necessary

(D) Completeness Satisfied if (B,C) are met and the sum

of shape functions is identically one.

Section 16.6 of Notes (advanced

material) provides details

Introduction to FEM

IFEM Ch 18 Slide 4

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Direct Construction of Shape Functions

as "Line Products"

N

e

i = ci L1 L2 . . . Lmguess

where L = 0 are equations of "lines" expressed in

natural coordinates, thatcross all nodes except ik

Introduction to FEM

IFEM Ch 18 Slide 5

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The Three Node Linear Triangle

1

2

3

1

2

3

N

1 1

guess= =c

11

L cL2-3

At node 1, N = 1 whence c = 1

and N = Likewise for N and N

Introduction to FEM

2 3

1 = 0

e

e e

e

e

IFEM Ch 18 Slide 6

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Three Node Triangle Shape Function Plot

1

2

3

N1 = 1

Introduction to FEM

IFEM Ch 18 Slide 7

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The Six Node Triangle - Corner Node

1

4

56

2

3

1

4

56

2

31 = 0

1 = 1/2

Ne

1

guess= c1 L2-3 L4-6

Introduction to FEM

For rest of derivation, see Notes

IFEM Ch 18 Slide 8

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The Six Node Triangle - Midside Node

14

56

2

3

14

56

2

3

Ne

1

guess= c1 L2-3 L4-6

Introduction to FEM

1 = 0

2 = 0

For rest of derivation, see Notes

IFEM Ch 18 Slide 9

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The Six Node Triangle: Shape Function Plots

1

4

5

6

2

3

1

4

5

6

2

3

Ne

1 = 1(21 1) Ne

4 = 412

Introduction to FEM

IFEM Ch 18 Slide 10

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Ne

1

guess= c1 L2-3 L3-4

1

2

34

1

2

34 = 1

= 1

The Four Node Bilinear Quad

Introduction to FEM

For rest of derivation, see Notes

IFEM Ch 18 Slide 11

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The Four Node Bilinear Quad:Shape Function Plot

1

2

3

4

Ne

1 =14

(1 )(1 )

Introduction to FEM

IFEM Ch 18 Slide 12

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The Nine Node Biquadratic Quad

Corner Node Shape Function

1

2

3

4

89

5

7

6

1

2

3

4

89

5

7

6 = 1

= 0

= 1

= 0

Ne

1 = c1L2-3L3-4L5-7L6-8 = c1( 1)( 1)guess

Introduction to FEM

For rest of derivation, see Notes

IFEM Ch 18 Slide 13

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The Nine Node Biquadratic Quad

Internal Node Shape Function

1

2

3

4

89

5

7

6

1

2

3

4

89

5

7

6 = 1

= 1

= 1

= 1

Ne

9 = c9 L1-2L2-3L3-4L4-1 = c9 ( 1)( 1)( + 1)( + 1)

Introduction to FEM

For rest of derivation, see Notes

IFEM Ch 18 Slide 14

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The Nine-Node Biquadratic Quad:

Shape Function Plots

Introduction to FEM

(c) (d)

(a) (b)

Ne

1 =14( 1)( 1)

Ne

5 =12(1 2)( 1)

Ne

5

=1

2

(1 2)( 1)

Ne

9 = (1 2)(1 2)(back view)

1

2

3

4

8

9

5

7

6 1

2

3

4

8

9

5

7

6

1

1

2

2

3

3

4

4

88

9

5

5

7

7 66

9

IFEM Ch 18 Slide 15

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The Eight-Node "Serendipity" Quad

Corner Node Shape Function

1

2

3

4

8

5

7

6

1

2

3

4

8

5

7

6

= 1

+ = 1

= 1

= 1

Ne

1 = c1L2-3L3-4L5-8 = c1( 1)( 1)(1 + + )

Introduction to FEM

For rest of derivation, see Notes

IFEM Ch 18 Slide 16

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Can the Magic Wand Fail? Yes

1

2

34

5

N1 N5

(Exercise 18.6)

Introduction to FEM

Method also needs modifications intransition elements.

One example is covered in the next two slides.

e e

IFEM Ch 18 Slide 17

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Transition Element Example

Introduction to FEM

4

1

2

3

Ne e

e

1

guess= c11(1 2) N1(1, 0, 0) = 1 = c1

No good: fails

compatibility over side 1-2

For N try the magic wand: product of side 2-3 ( = 0)and median 3-4 ( = ):

1

1

2

1

IFEM Ch 18 Slide 18

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Transition Element Example (cont'd)Introduction to FEM

1

24

3

works

1

1

Ne

1

guess= 1 + c112

Ne

1 = 1 212

Next, try the shape function of the linear 3-node triangle

plus a correction:

Coefficient c is determined by requiring this shape function

vanish at midside node 4:

whence c = 2 andN

e

1 (12, 1

2, 0) = 1

2+ c1

14= 0,

IFEM Ch 18 Slide 19