Lagrange Interpolation - Review

[Pages:60]Section 8: ISOPARAMETRIC FORMULATION

Lagrange Interpolation - Review

In data analysis for engineering designs we are frequently presented with a series of data values where the need arises to interpolate values between the given data points. Recall linear interpolation used extensively to find intermediate tabular values. Another common approach is using higher order polynomials to "curve fit" a function between data values. The polynomial usually takes the form:

f (x) = a0 + a1x + a2 x2 L an xn

For (n+1) data points, there is only one polynomial of order n that passes through all the values. For example, there is only one straight line (a first order polynomial) that passes through two data points. Similarly, only one parabola connects a set of three data points. Polynomial interpolation consists of determing the unique nth-order polynomial that fits (n+1) data points. This polynomial then provides a formula to compute intermediate values.

Section 8: ISOPARAMETRIC FORMULATION

We have been doing this all semester without the formal definition given on the previous slide. Although there is only one nth-order polynomial that fits (n+1) data points, there are a variety of methods that can be utilized to obtain the final form of the interpolating polynomial. These methods include (but are not limited to)

? Newton's Divided Difference Approach

? The Method of Lagrange Polynomials

? Regression Analysis (linear and non-linear)

? Splines

Here we focus on Lagrange interpolating polynomials because the method leads directly to the formulation of shape functions for higher order elements with an appropriate number of internal nodes.

The Lagrange interpolating polynomial is a reformulation of the Newton polynomial,

but avoids the computation of divided differences. The Lagrange polynomial has the

form:

n

fn (x) = Hi (x) f (xi ) i=0

Section 8: ISOPARAMETRIC FORMULATION

where

(( )) ( ) Hi x

=

n x- xj j=0 xi - x j

ji

For example, the linear version (n=1) is

f1(x)

=

( ) ( ) n

n

( ) i=0 j=0 ji

x- xj xi - x j

f

xi

=

(x - x1) (x0 - x1 )

f

(x0

)

+

(x - x0 ) (x1 - x0 )

f

(x1 )

and the second order version is

f2 (x)

=

(x - x1) (x0 - x1 )

(x - x2 ) (x0 - x2 )

f

(x0

)

+

(x - x0 ) (x1 - x0 )

(x - x2 ) (x1 - x2 )

f

(x1 )

+

(x - x0 ) (x2 - x0 )

(x - x1) (x2 - x1)

f

(x2

)

One can begin to see the usefulness of Lagrange polynomials by realizing that each term Hi(x) will be 1 at x = xi and 0 at all other data points. This is the quality we are looking for in a shape function, i.e., the shape function for a particular node is 1 at the node, and zero at all other nodes.

Section 8: ISOPARAMETRIC FORMULATION

In two dimensions the interpolation function has the form

where

f p (x, y)

=

n i=0

H

i

(x)

f

(xi

)

m i=0

Vi

(

y

)

g(

yi

)

p = nm

Three dimensional Lagrange interpolation has the form

f p (x, y, z)

=

n i=0

Hi

(x)

f

(xi

)

m i=0

Vi

(y)

g ( yi

)

l i=0

Qi

(y)q(zi

)

where

p = nml

Section 8: ISOPARAMETRIC FORMULATION

Recap ? Shape Functions

This is a good place to stop and remind ourselves where we are in the process of formulation solutions using finite element methods. For a component we are solving the global force displacement equation

{F} = [K]{d}

for displacements, i.e.,

{d} = {F}[K ]-1

The key to doing this is formulating the global stiffness matrix [K] properly and finding its

inverse. Once we have solved for the displacements AT THE NODES, we can interpolate

displacements (u, v) across the element through the use of shape functions. For one

dimensional elements

u^

=

d^1x

+

d^2x

- L

d^1x

x^

=

N1d^1x

+

N 2 d^2 x

=

1- x^ L

,

x^ L

dd^^12xx

where the coordinate axis was attached to the left end of the element.

Section 8: ISOPARAMETRIC FORMULATION

The linear shape functions for the one dimensional rod element are

N1

=1-

x^ L

N2

=

x^ L

relative to a local coordinate system attached to the left end. For a two dimensional constant strain triangle, once the nodal displacements were determined, the displacements across the element can be interpolated again through the use of shape functions

u (x, y) = Ni u i + N j u j + Nm u m

v (x, y) = Ni v i + N j v j + Nm vm

but these shape functions are formulated using global coordinate axes (x, y). For the constant strain triangle the shape functions are linear in x and y, i.e.,

Ni

=

1 2A

(

i

+

i x

+ i y)

( ) N j

=

1 2A

j

+

jx

+ jy

Nm

=

1 2A

(

m

+

mx

+ m y)

Section 8: ISOPARAMETRIC FORMULATION

For the linear strain triangle element displacements were quadratic functions of position. Recall that

u (x, y) = a1 + a2 x + a3 y + a4 x2 + a5xy + a6 y2 v (x, y) = a7 + a8x + a9 y + a10 x2 + a11xy + a12 y2

Here after the nodal displacements are determined recall that the coefficients above are determined through the following expression

{a} = [ ]-1 {d}

and the displacements are interpolated across the element using

u

=

{M *}{a}

v

= {M *}[ ]-1 {d} = {N}{d}

where now the shape functions are defined as

{ N} = {M *}[ ]-1

Section 8: ISOPARAMETRIC FORMULATION

For the linear strain triangle element the displacements vary quadratically across the element and the shape functions must be able to interpolate the nodal displacements quadratically across the element. Thus for the the example problem

N1

=1-

3x b

-

3y h

+

2x2 b2

+

4xy bh

+

2y2 h2

N2

=

-x b

+

2x2 b2

N3

=

-y h

+

2y2 h2

Note carefully that the shape functions are dependent upon a coordinate system whose origin was attached to the first corner node. Change the coordinate system and the shape functions change.

N4

=

4xy bh

N5

=

4y h

-

4xy bh

-

4y2 h2

N6

=

4x b

-

4x2 b2

-

4xy bh

Tracking the shape function for each individual element relative to a single coordinate system now becomes problematic. This is not something one can do by hand or track easily in computer software.

We will begin the use a local coordinate system as we formulate higher order elements.

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