ME 525 (Sullivan) Bilinear Elements and Isoparametric Mapping 1

Bilinear Element: Quadrilateral. 4 straight element sides with 4 nodes - 1 at each corner.

y

x

k

ji

m

Note: m is used instead of lowercase L to eliminate confusion with number 1. Νi = (aix + bi)(ciy + di) = αi + βix + γiy + δixy Note the Bilinear term. αi = bidi; βi = aidi; γi = bici; δi = aici As with triangles: Νi = 1 at node i = 0 at nodes j, k, m. Solve for coefficients:

1

1

1

1

xi

xj

xk

xm

yi

yj

yk

ym

xyi

xyj

xyk

xym

α i

β i

γ i

δ i

=

1

0

0

0

Permutate for i, j, k, m in CCW order.

ME 525 (Sullivan) Bilinear Elements and Isoparametric Mapping 2

Consider an irregularly shaped linear quadrilateral:

x

yi

kj

m

In general one wants to evaluate:

∇ . ( Κ ∇U ) + V . ∇U + f U = g or, numerically

{ }

{ }

( ) ( )j j j ji ix y i j i j

i iS

N N N NW WK V V W fN W Ux x y y x y

gW K U WdS

∂ ∂ ∂ ∂ ∂ ∂< − + > + < + >+ < > ∂ ∂ ∂ ∂ ∂ ∂

< > − ∇∫ ni

However this ∫ ( )dxdy over each element E will present problems due to the irregular geometry. SOLUTION: Transform x, y space to local coordinates (ξ, η) and integrate numerically.

A Bilinear Basis function implies that for any constant x or y the function is

linear in the other direction. let x = xc = constant then Νi = Ai + Biy where Ai = (αi + βixc) Bi = (γi + δixc)

ME 525 (Sullivan) Bilinear Elements and Isoparametric Mapping 3

Examine a Bilinear local coordinate system:

η

ξ

η = 1

ξ = −1

η = −1

ξ = 1

∆ξ = ∆η = 2; ξ = η = 0 at center of element. Νi = 1/4 (1 + ξiξ)(1 + ηiη) which is the bilinear solution of local system.

α

β

γ

δ

=

1

0

0

0

Coefficient Matrix

Note: for i = local node 1 Ν1 = 1/4 (1+(-1)(-1))(1+(-1)(-1)) = 1 at ξ = -1, η = -1 = 1/4 (1+(-1)(-1))(1+(-1)(1)) = 0 at ξ = -1, η = +1 = 1/4 (1+(-1)(1))(1+(-1)(1)) = 0 at ξ = 1, η = 1 = 1/4 (1+(-1)(1))(1+(-1)(-1)) =0 at ξ = 1, η = −1 We need coordinate transformation => use trial function approach.

x = ∑

4

k = 1x k ψ k ( ξ , η )

y = ∑

4

k = 1y k ψ k ( ξ , η )

with (x, y) being any point within the element and xk, yk being the nodal coordinates in (x, y) space. However, ψk is formulated in (ξ, η) space.

ME 525 (Sullivan) Bilinear Elements and Isoparametric Mapping 4

CONSTRAINTS: 1.) 1 : 1 mapping of nodes between 2 coordinate systems. 2.) linear variation in x and y along sides of (ξ, η) element. => x = a1 + b1ξ + c1η + d1ξη y = a2 + b2ξ + c2η + d2ξη Note that solution of ψi yields the basis functions Νi when the transformation (mapping) function = basis function the element is an Isoparametric element when transformation function is of higher degree than basis function => Superparametric and when transformation function is of lower degree than basis function => Subparametric.

We need terms such as:

∂N∂x

∂W∂x

+ .....

i.e express ( )

x or y∂

∂ ∂ in terms of ξ, η.

Use chain rule:

( , ) ( , ) ( , )N N x N yx y

ξ η ξ η ξ ηξ ξ ξ

∂ ∂ ∂ ∂ ∂= +

∂ ∂ ∂ ∂ ∂

( , ) ( , ) ( , )N N x N y

x yξ η ξ η ξ ηη η η

∂ ∂ ∂ ∂ ∂= +

∂ ∂ ∂ ∂ ∂

or:

ME 525 (Sullivan) Bilinear Elements and Isoparametric Mapping 5

∂ N∂ ξ

∂ N∂ η

=

∂ x∂ ξ

∂ x∂ η

∂ y∂ ξ

∂ y∂ η

∂ N∂ x

∂ N∂ y

∂ N∂ ξ

∂ N∂ η

= [ J ]

∂ N∂ x

∂ N∂ y

[ J ] ≡ J a c o b i a n t r a n s f o r m a t i o n m a t r i x .

What we need for evaluation of:

∂N∂x

∂W∂x

+ .....

, is:

∂N∂x

∂N∂y

= [J]− 1

∂N∂ξ

∂N∂η

Recall: Νi = 1/4 (1+ ξiξ)(1+ ηiη)

∴

∂ N i

∂ ξ ,

∂ N i

∂ η ⇒ n o p r o b l e m .

In [ ] , ?xJ etcξ

∂⇒ =

∂

ME 525 (Sullivan) Bilinear Elements and Isoparametric Mapping 6

x = ∑4

k = 1 x k ψ k a n d ψ k = N k ( I s o p a r a m e t r i c )

∂ x∂ ξ

= ∂

∑ x k ψ k

∂ ξ = ∑

k

∂ ( x k ψ k )∂ ξ

= ∑k

x k∂ ψ k

∂ ξ = ∑

k x k

∂ N k

∂ ξ

∂ y∂ ξ

= ∑k

y k∂ N k

∂ ξ

η = 1

ξ = 1

η = −1

ξ = −1

1

34

41

32

2

Choose any side; Ex.: η = 1 = top side in (ξ,η) = left side (x,y) i.e. node 1 in (x,y) does not need to be lower left. Νi = 1/4 (1+ξiξ)(1+ηi(1)) = Ai + Biξ

∂x∂ξ

= ∑xi∂Ni

∂ξ = Constant

⇒ dxdy

= Constant

∂y∂ξ

= ∑yi∂Ni

∂ξ = Constant

ME 525 (Sullivan) Bilinear Elements and Isoparametric Mapping 7

n Lines of constant

ξ

Lines of constant η

Along any line: η = constant => dx/dy = constant and ξ = constant => same.

∂ ( )∂ x

∂ ( )∂ y

= [ J ] − 1

∂ ( )∂ ξ

∂ ( )∂ η

= 1

[ J ]

∂ y∂ η

− ∂ x∂ η

− ∂ y∂ ξ

∂ x∂ ξ

∂ ( )∂ ξ

∂ ( )∂ η

∂ ( )∂ x

= 1 J

∂ y∂ η

∂ ( )∂ ξ

− ∂ y∂ ξ

∂ ( )∂ η

∂ ( )

∂ y = 1

J

− ∂ x

∂ η∂ ( )

∂ ξ + ∂ x

∂ ξ∂ ( )∂ η

For integration: ∫b

a ∫

d

c ( ) dx dy = ∫1

−1 ∫1

−1 ( )Jdξ dη

and

[J] =

∑ x i∂N i

∂ξ

∑ x i∂N i

∂η

∑ y i

∂N i

∂ξ

∑ y i∂N i

∂η