Shape Function Magic

Introduction to FEM

18

Shape Function Magic

IFEM Ch 18 – Slide 1

Introduction to FEM

'Magic' Means Direct ("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

IFEM Ch 18 – Slide 2

Introduction to FEM

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.

IFEM Ch 18 – Slide 3

Introduction to FEM

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

IFEM Ch 18 – Slide 4

Introduction to FEM

Direct Construction of Shape Functions as "Line Products"

Nie

guess

= ci L1 L2 . . . Lm

where L k = 0 are equations of "lines" expressed in natural coordinates, that cross all nodes except i

IFEM Ch 18 – Slide 5

Introduction to FEM

The Three Node Linear Triangle 3

3

2 1

ζ1 = 0 2

1

e guess

N

=

cL = cL2-3

At node 1, N 1e = 1 whence c1 = 1 and N1e = ζ 1 Likewise for N 2e and N 3e

IFEM Ch 18 – Slide 6

Introduction to FEM

Three Node Triangle Shape Function Plot

N1 = ζ1 3 1

2

IFEM Ch 18 – Slide 7

Introduction to FEM

The Six Node Triangle - Corner Node ζ1 = 0

3

3

5

6

5

6

2

2 4 1

N

4

ζ 1 =1/2 1

e guess

= c1 L 2-5-3 L 4-6 = c1 ζ 1 (ζ 1 −1/2) For rest of derivation, see Notes

IFEM Ch 18 – Slide 8

Introduction to FEM

The Six Node Triangle - Midside Node 3

ζ1 = 0

3 5

6

5

6 2

4 1

2 ζ2 = 0

4 1

e guess

N4 = c 4 L2-5-3 L 3-6-1 = c4 ζ 1 ζ 2 For rest of derivation, see Notes

IFEM Ch 18 – Slide 9

Introduction to FEM

The Six Node Triangle: Shape Function Plots

3 3

6

6

1

1

5 5

4

4 2 2

e

N1 = ζ1 (2ζ1 − 1)

N4e = 4ζ1 ζ2

IFEM Ch 18 – Slide 10

Introduction to FEM

The Four Node Bilinear Quad η=1

η 4

3

4

3

ξ=1

ξ 1

1

2

2

e N1

guess

=

c1 L2-3 L3-4

For rest of derivation, see Notes

IFEM Ch 18 – Slide 11

Introduction to FEM

The Four Node Bilinear Quad: Shape Function Plot 4

3

1

2

e

N1

= 14 (1 − ξ)(1 − η)

IFEM Ch 18 – Slide 12

Introduction to FEM

The Nine Node Biquadratic Quad Corner Node Shape Function η

η=1

3 4 8 1

7 9

6

5 2

e N1

η=0 4

7

8

9

ξ 1

3

ξ=0 6

ξ=1

5 2

guess

= c1 L2-3 L3-4 L5-7 L6-8 = c1 (ξ − 1)(η − 1)ξη For rest of derivation, see Notes

IFEM Ch 18 – Slide 13

Introduction to FEM

The Nine Node Biquadratic Quad Internal Node Shape Function η=1

η 3 4 8 1

7 9

4 6

e

ξ = −1

8 1

5 2

N9

ξ

7

η = −1

9

3 6

ξ=1

5 2

= c9 L1-2 L2-3 L3-4 L4-1 = c9 (ξ − 1)(η − 1)(ξ + 1)(η + 1) For rest of derivation, see Notes

IFEM Ch 18 – Slide 14

The Nine-Node Biquadratic Quad: Shape Function Plots 4

(a)

(b)

7

4 7

8

8

3

3

9

9

1

6

1

6

5

5 2

e

2

N1 = 14 (ξ − 1)(η − 1)ξ η (c)

Introduction to FEM

N5e

5

(d)

1

4 8

9

4

9 6

− ξ )η(η − 1) 2

7 8

2

=

1 (1 2

3

1

7

6 5

3 e

N5 = 12 (1 − ξ 2 )η(η − 1)

2 (back view)

IFEM Ch 18 – Slide 15

e

N9 = (1 − ξ 2 )(1 − η2 )

Introduction to FEM

The Eight-Node "Serendipity" Quad Corner Node Shape Function η

η=1

3 4

7 4 6

8 1

e

3

ξ + η = −1 ξ = −1

6

8

ξ=1

5 2

N1

ξ

7

1

5 2

= c1 L2-3 L3-4 L5-8 = c1 (ξ − 1)(η − 1)(1 + ξ + η) For rest of derivation, see Notes

IFEM Ch 18 – Slide 16

Introduction to FEM

Can the Magic Wand Fail? Yes η 3

4

ξ

5

(Exercise 18.6)

1 2 e

N1

N5e

Method also needs modifications in transition elements. One example is covered in the next two slides.

IFEM Ch 18 – Slide 17

Introduction to FEM

Transition Element Example 3

1 2

4 e

For N1 try the magic wand: product of side 2-3 (ζ 1 = 0) and median 3-4 (ζ 1 = ζ 2 ): e

N1

guess

=

c1 ζ1 (ζ1 − ζ2 )

e

N1 (1, 0, 0) = 1 = c1 No good: fails compatibility over side 1-2

IFEM Ch 18 – Slide 18

Introduction to FEM

Transition Element Example (cont'd) 3

1 2 4 Next, try the shape function of the linear 3-node triangle plus a correction: e

N1

guess

=

ζ1 + c1 ζ1 ζ2

Coefficient c 1 is determined by requiring this shape function vanish at midside node 4: N1e ( 1 , 1 , 0) = 1 + c1 1 = 0, 4 2 2 2 whence c 1 = −2 and e

N1 = ζ1 − 2ζ1 ζ2

IFEM Ch 18 – Slide 19

works