families of shape functions, numerical integration, physical … of shape... · families of shape...

Families of Shape Functions, Numerical Integration, Physical and Master Element, Concept and Mapping,

Element Stiffness and Load Vector Calculations

PM Mohite

Department of Aerospace Engineering Indian Institute of Technology Kanpur

TEQIP School on Computational Methods in Engineering Application

Knowledge Incubation for TEQIP

Two Dimensional Problems

Single variable or multivariable problems

The phenomenon is represented through partial differential equations

Finite element formulation involves same steps as one dimensional case

Boundary of a two dimensional domain Ω, in general, is a curve

Two dimensional shapes like triangle and rectangle/quadrilaterals are used to approximate the geometry.

Two Dimensional Problems:

• Poisson equation

gradient operator

These equation represent

• Heat conduction

• Electrostatics

• Stream function

• Magnetic statics

• Torsion of non-circular sections, Transverse deflection of elastic membranes

fuk in

yj

xi

ˆˆ

Sample Boundary Value Problems:

it becomes Poisson’s equation.

Finite element discretization:

• The number, shape and type (linear, quadratic,……….) of elements should be such that the geometry of the domain is accurately represented. • Density of elements should be such that the regions of large gradients of the solution are adequately modeled.

11 12 21 22 , in u u u u

k k k k f x yx x y y x x

12 21 11 22with 0andk k k k k

Sample Boundary Value Problems: Heat Conduction

Weighted residual form:

We know that,

Therefore, Similarly,

x

wqw

x

qwq

xx

xx

dAwdAw

y

uk

x

uk

yy

uk

x

uk

x22221211

x

wqwq

xwq

xxxx

Heat Conduction Equation: Weak Formulation

y

wqwq

ywq

yyyy

Now using the Divergence Theorem, and where nx and ny are the direction cosines or the components of the unit normal vector n

dsnwqdAwq

xxxx

dsnwqdAwq yyy

y


ˆ ˆ ˆ ˆcos sinx yn i n j i j n

writing the weak form over an element

Looking at boundary term:

• Primary variable – essential BC and

• coefficient of weight function in boundary term form the natural BC.

dAwfy

uk

x

uk

y

w

y

uk

x

uk

x

w

e

222112110

dsy

uk

x

ukn

y

uk

x

uknw

e yx

22212211

uw


qn - is the secondary variable.

qn - projection of the vector along the unit normal n

- positive outward from the surface as the as we more counter-clock wise along the boundary

Thus,

k u

on e

11 12 21 220e e

n

w u u w u uk k k k wf dA wq ds

x x y y x y

y

uk

x

ukn

y

uk

x

uknq yxn 22211211


Or where, - bilinear from, is not symmetric. It is symmetric only when - linear from Quadratic Potential:

wFuwB ,

eedswqdAfwwF n

uwB , 2112 kk

e

dAy

uk

x

uk

y

w

y

uk

x

uk

x

wuwB 22211211,


F w

,I u B u u F u

Approximation of over as

- are the nodal values of at node.

- are the Lagrange interpolation functions with the property

Substitute in the weak form and to get equation

replace w by to give

u e

1

, , ,

NODEee

h j jj

u x y u x y u x y

eju

ej ,e

i j j ijx y

1

, ,

NODEe e

h j j

j

u x y u x y

ei

11 12 21 22

1

0e

NODEj j j ji i

j

k k k k dAx x y y x y

ee

dsqdAf nii

Heat Conduction Equation: Finite Element Formulation

, 1,2, ,i j NODE

This leads to: where, and In matrix notation, is symmetric only when

1

NODEe e e eij j i i

j

k u f Q

dAy

kx

kyy

kx

kx

ke

jiijjie

ij

22211211

dsqQdAff e

in

e

i

e

i

e

iee

,

eeee Qfuk

2112 kk ek

Heat Conduction Equation: Finite Element Formulation

Interpolation Functions in Two Dimensions

The approximate solution over an element, for the assurance of convergence to the actual solution as the number of elements is increased and their size is decreased, must satisfy: 1) It must be continuous over the element and differentiable as required by the

weak form. - It ensures a nonzero coefficient matrix 2) It must not allow a strain to appear if the nodal displacements are compatible with a rigid-body displacements.

yxuh ,

Interpolation Functions: Requirements

3) If the nodal displacements are compatible with a uniform strain in the elements, then the interpolation functions must yield this strain for the nodal displacements. In other fields: If the nodal variables are compatible with uniform states of the variable and any of its derivatives up to the highest in the appropriate quadratic functional , then these uniform states must be preserved in the element as it shrinks to zero size. This condition is called completeness. - This requirement is necessary in order to capture all possible states of the actual solution


I

Example: If a linear polynomial without the constant term is used to represent the temperature distribution in a one-dimensional system, the approximate solution can never be able to represent a uniform state of temperature in the element.


4) The interpolation functions should be chosen so that the strain remains finite at the boundary. Note, however, that the strain can be indeterminate at a boundary. For example, if the strains involve only the first derivatives, then the displacement field must be continuous across a boundary. (strain energy must be finite) In other fields: The dependent variable and all its derivatives up to a one less than the highest order derivative in the appropriate quadratic functional should be continuous at the element interfaces. (The quadratic potential must be finite) This requirement is called compatibility condition. Elements satisfying these requirements are called conforming elements.


Degree of Compatibility achieved by interpolation functions at the element: • continuity is achieved when the field variable only and none of its derivatives maintain continuity at an interelement interface. • continuity is achieved when the field variable and its first derivatives maintain continuity at an interelement interface. • continuity is achieved when the field variable, its first derivatives and second derivatives all maintain continuity at an interelement interface.


0C

1C

2C

Continuity Requirement for completeness and compatibility: For assurance of finite element convergence for a functional having as the highest order of a derivative – • continuity – requirement for completeness • continuity – requirement for compatibility


p

pC

1pC

Lagrange Family: • Interpolation functions derived using the dependent unknown only and not its derivatives, at nodes. • - continuity

Hermite Family: • Interpolation functions derived using the dependent unknown and its derivatives as well, at nodes. • continuity – continuity of the first derivative of the dependent unknown. Cubic interpolation functions • continuity – continuity of the first and second derivatives of the dependent unknown. Quintic interpolation functions

Interpolation Functions: Types of Families

0C

1C

2C

must be at least linear in x and y.

Let

Three linearly independent terms, linear in both x and y.

to be expressed terms of nodal values of . Hence, we must have three nodes.

This is possible for a triangle with its vertices as nodes.

ycxccyxuh 321,

ic

e

i

Linear Interpolation Functions:

hu

• Consider a polynomial

• Four linear independent terms, linear in x, y with a bilinear term in x and y.

• This requires an element with four nodes.

• Two possible geometries:

- A triangle with the three nodes at its vertices and the fourth node at its centre or centroid.

- A rectangle with the nodes at the vertices

xycycxccyxuh 4321,


• A triangle with the fourth node at its centre does not provide a single valued variation of primary variable at interelement boundaries.

• This results in incompatible variations of primary variable there. Therefore, this is not admissible.

• Thus, the only possible element with assumed polynomial approximation is rectangle with the nodes at the four vertices


• Consider a polynomial with five constants

• This is an incomplete quadratic polynomial

• Note that the terms x2 and y2 can not be varied independently.

• The polynomial requires an element with five nodes.

• The possible geometry is a rectangle with a node at each vertex and the fifth node at its centre.

• This again does not give single valued variation of primary variable.

22

54321, yxcxycycxccyxuh

Quadratic Interpolation Functions:

• Consider a polynomial with six constants

• This is a complete quadratic polynomial

• The polynomial requires an element with six nodes.

• The possible geometry is a triangle with a node at each vertex and a node at the midpoint of each side.

2

6

2

54321, ycxcxycycxccyxuh

Quadratic Interpolation Functions:

Lagrange Interpolation Functions

in Two Dimensions

Linear Interpolation Functions

in Two Dimensions

A triangular element with three nodes at vertices

Approximate solution over an element is of the form

We need three conditions to find the unknowns Ci ‘s.

Using nodal values of solution we get three conditions

ycxccyxuh 321,

Interpolation Functions: Linear Triangular Element

1 2

3

Thus, we have

Or in a matrix form

unknowns Ci ‘s are found as

where,


33321333

23221222

13121111

,

,

,

ycxccyxuu

ycxccyxuu

ycxccyxuu

h

h

h

3

2

1

3

2

1

33

22

11

3

2

1

1

1

1

c

c

c

A

c

c

c

yx

yx

yx

u

u

u

3

2

1

3

2

1

1

c

c

c

u

u

u

A

321

321

321

1

2

1

eAA

3212 eA

Solving for Ci ‘s:

- determinant of

- area of the triangle

where,


3322113

3322112

3322111

2

1

2

1

2

1

uuuA

c

uuuA

c

uuuA

c

e

e

e

order natural ain permute ,, and kjikji

xx

yy

yxyx

kji

kji

jkkji

eA2

eA

A

Substituting values of Ci ‘s:

where,

With the properties:

Interpolation Functions: Linear Rectangular Element

3

1

332211332211332211

,

2

1,

i

e

i

e

i

e

h

yxu

yuuuxuuuuuuA

yxu

3,2,1 2

1, i

Ayx e

i

e

i

e

i

e

e

i

,, ij

e

j

e

j

e

i yx

3

1

,1i

e

i

,03

1

i

e

i

x

0

3

0

i

e

i

y

Lagrange Interpolation Functions: Linear Triangular Element

Linear interpolation functions for the three-noded triangular element

1 2

3

1

ψ1

1 2

3

1

ψ2

1 2

3

1

ψ3

Lagrange Interpolation Functions: Triangular Geometries to Avoid

• The nodes are almost in-line.

• The resulting coefficient matrix is nearly singular or non-invertible.

L

a L>>a

L

h L>>h

Lagrange Interpolation Functions: Linear Triangular Element

Shapes that should be avoided in the meshing.

A rectangular element with four nodes at the corner and of sides a and b.

Approximate solution over an element is of the form

We need four conditions to find the unknowns Ci ‘s.

yxcycxccyxuh 4321,


1 2

4 3

y

x

x

y

b

a

Using the nodal solution values, we get four conditions: Gives the values of constants as Putting these back and re-arranging,

110,0 cuuh 2120, cacuauh

abcbcaccubauh 43213, bccubuh 314,0

ab

uuuuc

b

uuc

a

uucuc 2143

414

312

211 ,,,

b

y

a

x

b

yu

b

y

a

xu

b

y

a

x

a

xu

ab

yx

b

y

a

xuyxuh 4321 1,


44332221, uuuuyxuh

b

x

a

x

b

y

a

x ee 1,11 21

b

y

a

x

b

y

a

x ee

1, 43

b

yy

a

xxyx iii

111,1e

i

,,e

i ijjj yx

4

1

1i

e

i

where, And can be given as With the properties:


Lagrange Interpolation Functions: Linear Rectangular Element

Linear interpolation functions for the four-noded rectangular element

ψ4

1

2

3

4 1

1

1

2

3

4

ψ2

1

2

3

1 4

ψ1

1 1

2

3

4

ψ3

b

y

a

x

b

y

a

x

b

x

a

x

b

y

a

x eeee

1,,1,11 4321

• Can also be obtained from tensor product of 1D linear interpolation functions associated with sides 1-2 and 2-3

32

411

1

b

y

b

y

a

xa

x


a

a

x1

x

a

x

b

b

y1

y

b

y

Another approach: Derivation of We know that and And zero on the lines Therefore, is of the form: Putting first condition that Hence, is In a similar way, the other interpolation functions can be derived.

4,3,2for 0,1 iyx ii

e 1, 111 yxe

e

1

byax and

yxe ,1

ybxacyxe 11 ,

0,0at 1,1 yxyxeab

c1

1

b

y

a

xybxa

abyxe 11

1,1


e

1

Higher Order Interpolation Functions

For Triangular Element

• The interpolation functions of higher degree can be systematically developed with the help of Pascal’s Triangle.

• It contain the terms of polynomials of various degrees in the two coordinates x and y.

• x and y denote the local coordinate (and not the global coordinates of the problem).

• Shape of the triangle need not be equilateral triangle.

• order triangular element has n nodes given by the relation

• A complete polynomial of degree is given by

thp

212

1 ppn

n

i

n

j

jj

sr

i uyxayxu1 1

),( psr with

Interpolation Functions: Higher Order Element

thp

Interpolation Functions: Pascal’s Triangle

1

x y

2x 2y

3y

4y

5y

6y

7y

3x

4x

5x

6x

7x

xy

yx2 2xy

yx3 22 yx3xy

yx4 23 yx32 yx

4xy

yx5 24 yx 33 yx42 yx

5xy

yx6 25 yx 34 yx 43 yx52 yx 6xy

Polynomial Degree Element

We choose the terms

in the polynomial


1

x y

2x 2y

3y

4y

5y

6y

7y

3x

4x

5x

6x

7x

xy

yx2 2xy

yx3 22 yx3xy

yx4 23 yx32 yx

4xy


5xy


0 1

1, cyxuh



1

x y

2x 2y

3y

4y

5y

6y

7y

3x

4x

5x

6x

7x

xy

yx2 2xy

yx3 22 yx3xy

yx4 23 yx32 yx

4xy


5xy


1 3

ycxccyxuh 321,



1

x y

2x 2y

3y

4y

5y

6y

7y

3x

4x

5x

6x

7x

xy

yx2 2xy

yx3 22 yx3xy

yx4 23 yx32 yx

4xy


5xy


2

6

2

6

2

54

321,

ycxcxyc

ycxccyxuh


We choose the terms

in the polynomial


1

x y

2x 2y

3y

4y

5y

6y

7y

3x

4x

5x

6x

7x

xy

yx2 2xy

yx3 22 yx3xy

yx4 23 yx32 yx

4xy


5xy


0

1

2

3

4

5

6

7

1

3

6

10

15

21

28

212

1 ppn

Lagrange Interpolation Functions: Quadratic Triangular Element

Geometric variation of Lagrangian interpolation functions

Interpolation Functions For Triangular Element

using Area Coordinates

P is an arbitrary point inside a linear triangular element Let Ai, Aj and Ak be the area of triangles formed by point P with an edge of the triangle. For example, Ak is the area formed by edge i-j and point P. Define: Natural /triangular/ barycentric coordinate and A

AL

A

AL

A

AL k

k

j

ji

i ,,

Interpolation Functions: Using Area Co-ordinates

P i

j

k

(i)

(j)

(k)

iL

kji AAAA

• For P at node i: • For P at node j: • For P at node k: Connection with triangular element:


0,0,1 kji LLL

0,1,0 kji LLL

1,0,0 kji LLL

P 1

2

3

(1)

(2)

(3)

0,0,1 321 LLL

0,0,1 312 LLL

0,0,1 213 LLL3L

2L

1L

• We needed to invert coefficient matrix when we derived interpolation functions. This can be avoided in this approach. Consider a quadratic triangular element We express natural co-ordinate Li at any point by using superscript to identify the nodes. Example: Point P moves to node 6.


P 1

2

3

(1)

(2)

(3) 4

5

6

,2

116

1 A

AL

P 1

2

3

(1) (3) 4

5

6

,0026

2 AA

AL

2

136

3 A

AL

• Point P moves to node at 4. • Point P moves to node at 5.


,2

114

1 A

AL

,2

124

2 A

AL 0

034

3 AA

AL

P

1

2

3

(1) (2)

4 5

6

,0015

1 AA

AL

,2

125

2 A

AL

2

135

3 A

AL

P 1

2

3 (2)

(3) 4

5

6

• For a cubic triangular element Example: Point P moves to node 10. Example: Point P moves to node 4.


,3

1110

1 A

AL ,

3

1210

2 A

AL

3

1310

3 A

AL

P 1

2

3

(1)

(2)

(3) 4

5 6

7

8 9

10

P 1

2

3

(1)

(2)

(3) 4

5 6

7

8 9

10

P

1

2

3

(1)

(2)

4

5 6

7

8 9

10 ,

3

214

1 A

AL ,

3

124

2 A

AL 0

034

3 AA

AL

To get the interpolation functions, we first transform the Lagrangian interpolation formula from Cartesian to natural coordinates. where j = one of the three natural coordinates n (=p) = order of interpolation polynomial r = free index from 1 up to the total number of nodes Thus, for each node:

1

1 for 1

r

jnL

rj

j jr

i

nL iL L nL

i


1 for 0

r

j jr

L L nL

1 2 3r r r rN L L L L L L

Example: For quadratic triangular element (n=p=2) For N1, put r=1 and considering j=1 first Thus, and . Therefore,


1

12 2 1 2r

jnL L

1

12 21 1

1 11 1

1 11 1

2 1 2 1

2 1 1 2 2 1 2 1

1 2

L

i i

L i L iL L

i i

L LL L

2 31 10, 0L L L L 1 1 12 1N L L

Example: For quadratic triangular element (n=p=2) For N4, put r=4 and considering j=1 first Thus, Now, considering j=2


4

1

12 2 1

2

r

jnL L

4

12 11 1 1

1 141 1

2 1 2 1 2 1 12

1

L

i i

L i L i LL L L

i i

4

2

12 2 1

2

r

jnL L

2 242L L L

Example: For quadratic triangular element (n=p=2) For N4, considering j=3 Thus, For remaining nodes,


4

32 2 0 0r

jnL L 3 41L L

4 1 24N L L

2 2 2 3 3 3

5 2 3 6 1 3

2 1 , 2 1 ,

4 , 4

N L L N L L

N L L N L L

Higher Order Interpolation Functions

For Rectangular Element

Polynomial Degree = 1 1


x y

2x 2y

3y

4y

5y

6y

7y

3x

4x

5x

6x

7x

xy

yx2 2xy

yx3 22 yx 3xy

yx4 23 yx32 yx

4xy

yx5 24 yx 33 yx42 yx 5xy

yx6 25 yx34 yx 43 yx 52 yx 6xy

xycycxccyxuh 4321,

Polynomial Degree = 2

1


x y

2x 2y

3y

4y

5y

6y

7y

3x

4x

5x

6x

7x

xy

yx2 2xy

yx3 22 yx 3xy

yx4 23 yx32 yx

4xy



22

9

2

8

2

7

2

6

2

54

321,

yxcxycyxc

ycxcxyc

ycxccyxuh

Polynomial terms Polynomial Degree = 3

can be chosen 1


x y

2x 2y

3y

4y

5y

6y

7y

3x

4x

5x

6x

7x

xy

yx2 2xy

yx3 22 yx 3xy

yx4 23 yx32 yx

4xy



32

16

3

15

3

14

33

13

23

12

3

11

3

10

22

9

2

8

2

7

2

6

2

54

321,

yxcxycycyxc

yxcyxcxc

yxcxycyxc

ycxcxyc

ycxccyxuh

Polynomial terms Polynomial Degree Element

can be chosen 1


x y

2x 2y

3y

4y

5y

6y

7y

3x

4x

5x

6x

7x

xy

yx2 2xy

yx3 22 yx 3xy

yx4 23 yx32 yx

4xy



1

2

3

4

2

No. of Nodes 1n p


Rectangular array

Interpolation Functions: Pascal’s Triangle in Rectangular Array Form

1

2

3

2

No. of Nodes 1n p

333233

232222

32

321

yxyxxyy

yxyxxyy

yxyxxyy

xxx

0

The Lagrange interpolation functions associated with rectangular elements can be obtained from corresponding one-dimensional Lagrange interpolation functions by taking tensor product.

T

b

b

b

b

b

a yy

byy

ayy

aa

xx

aaa

axx

aa

axa

x

2

2

4

2

2

21

2

21

21

963

852

741

2

2

2

)(

)(

)(

)(

)(

)(

)(

)(2

)(2

Interpolation Functions: Higher Order Rectangular Elements

114

1,11

4

1

,114

1,11

4

1

41

21

2 2 2 2 2 21 2 3

2 2 2 2 2 24 5 6

2 2 2 2 2 27 8 9

1 1 1, 1 , ,

4 2 4

1 1 11 , 1 1 , 1 ,

2 4 2

1 1 1, 1 ,

4 2 4

Interpolation Functions: Master Rectangular Elements

1,1 1,1

1,1 1,1

1 2

34

1,1 1,1

1,1 1,1

1 2 3

4 56

7 8 9

Tensor product of 1D functions:

Lagrange Interpolation Functions: Quadratic Rectangular Element

Geometric variation of Lagrangian interpolation functions

Lagrange Interpolation Functions: Quadratic Rectangular Element

Geometric variation of

Lagrange interpolation

functions:

• The shape function presented earlier are for the interpolation of primary variables.

• The Hermite cubic interpolation functions based on the interpolation of can be generated.

Interpolation for

Variable u

derivative

2

, , ,u u u

ux y x y

2 21

1 216

i i i i i

Hermite Interpolation Functions: Master Rectangular Element

u

2 21

2 216

i i i i

Derivative

Derivative

where, i=1,….4 are the nodes of the element.

2 2

1

11 1

16i i i i i

2 21

2 116

i i i i i

Hermite Interpolation Functions: Master Rectangular Element

u

2u

Serendipity Family of Interpolation Functions



Terms inside this

cone are not

included

1


2

3

4

x y

2x 2y

3y

4y

5y

6y

7y

3x

4x

5x

6x

7x

xy

yx2 2xy

yx3 22 yx 3xy

yx4 23 yx32 yx

4xy




1


2

8

2

7

2

6

2

54

321,

xycyxc

ycxcxyc

ycxccyxuh

x y

2x 2y

3y

4y

5y

6y

7y

3x

4x

5x

6x

7x

xy

yx2 2xy

yx3 22 yx 3xy

yx4 23 yx32 yx

4xy




1


3

12

3

11

3

10

3

9

2

8

2

7

2

6

2

54

321,

xycycyxc

xcxycyxc

ycxcxyc

ycxccyxuh

x y

2x 2y

3y

4y

5y

6y

7y

3x

4x

5x

6x

7x

xy

yx2 2xy

yx3 22 yx 3xy

yx4 23 yx32 yx

4xy



Polynomial terms Polynomial Degree Element

can be chosen 1


2

3

4

x y

2x 2y

3y

4y

5y

6y

7y

3x

4x

5x

6x

7x

xy

yx2 2xy

yx3 22 yx 3xy

yx4 23 yx32 yx

4xy



Derivation of : vanishes on the lines

Interpolation Functions: Derivation of Serendipity Functions

1,1 1,1

1,1 1,1

1 2 3

4 5

6 7 8

1 0 1 0

1 0

1 0

1 0

1 0

1 0

1 0

1

1

1 0, 1 0,

1 0

Therefore, is of the from

It has a value of 1 at

vanishes along the lines

Let

1

1 , 1 1 1C

1

11 1 1

4

2 1 0,1 0,1 0

2 1 1 1c

2 1 at 0,-1 12c


1, 1 14

c

22

11 1

2

13 4

214 4

215 2

1 1 1

1 1

1 1

16 4

217 2

18 4

1 1 1

1 1

1 1 1

Similarly, Note: The serendipity interpolation functions are different than Lagrange interpolation functions as internal nodes are not present.


Hierarchic Interpolation Functions


• The set of interpolation functions of polynomial degree p should be in the set of interpolation functions of polynomial degree (p+1). • The number of interpolation functions which do not vanish at vertices and sides should be smallest possible. • Example: 1D Hierarchic interpolation functions

Interpolation Functions: Hierarchic Functions

• Nodal Interpolation functions:

- Four nodal interpolation functions

- Exactly same as the functions for four noded quadrilateral element

Hierarchic Shape Functions: Rectangular Elements

114

1,11

4

1

,114

1,11

4

1

41

21

• Side Nodes Interpolation functions:

- There are interpolation functions associated with the sides of finite elements with the order of approximation

- For Side 1

- For Side 2

- For Side 3

- For Side 4


piii ,,2 12

11

14 p

2p

piii ,,2 12

12

piii ,,2 12

13

piii ,,2 12

14

• Internal Nodes Interpolation functions:

- There are interpolation functions associated with the internal nodes of finite elements with the order of approximation

and so on.


0 0

2 2 3 21 2

0 0

2 3 4 23 4

0 0

3 3 2 45 6

, ,

, ,

, ,

2 3 2p p 4p

where, polynomial of degree j is:

- is Legendre polynomial.

Using the properties of Legendre polynomials


1

1

2 1, 2,3,

2j j

jP t dt j

2

1

2 2 1j j jP P

j

jP

Legendre polynomial is given by Bonnet Recursion formula:

with

and


1 11 2 1 , 1,2,n n nn P n P nP n

1

1

2 for

2 1

0 for i j

i jP P d i

i j

0 11,P P

Hierarchic Interpolation Functions

For Triangular Element

• Nodal Interpolation functions:

- Three nodal interpolation functions -

- Exactly same as the functions for three noded triangular element

• Side Modes:

- There are side modes and vanish on the other two sides

Sides 1:

where,


1 2 3, ,L L L

3 1p

1

2 1 2 1 , 2, ,i iL L L L i p

211 , 2, ,

4j j j p

For example:

• Internal Modes:

- There are internal modes

Mode 1:

Other modes:

and so on.


1 2 2p p

0

1 2 31 L L L

22 3 4

76, 10 , 5 1

8

0 0

1 2 3 1 2 1 1 2 3 1 32 3, 2 1L L L P L L L L L P L

Physical to Master Element Mapping

• Non rectangular domains cannot be represented using rectangular elements. However, it can be represented more accurately by quadrilateral elements.

• Interpolation functions are easily available for a rectangular element

• And it is easy to evaluate integral over rectangular geometric

• Therefore, we transform the finite element integral statement over quadrilaterals to a rectangular.

• Similarly, the finite element integral statement over a triangular element is transformed to an isosceles triangle.

• Caution: Transformation is for numerical integration purpose only ! The actual domain is not mapped to another domain.

Mapping:

• Then numerical integration schemes like Gauss-Legendre are used to evaluate the integrals.

• These schemes require that the integral be evaluated on a specific domain or with respect to a specific coordinate system.

• In general it is a coordinate system such that

• Transformation between physical element and master element

where, interpolation functions over master element

, 1,1

e

1 1

ˆ ˆ, , ,

m me e e ej j j j

j j

x x y y

e

j

Mapping:

In general, the dependent variables(s) are approximated by expressions of the form

The degree of approximations used for the geometry and the dependent variables(s) are different.

Depending upon it, the finite element formulations are classified as:

- Sub-parametric

- Iso-parametric

- Super parametric

1

, ,

ne ej i

j

u x y u x y

Mapping:

a) super- parametric : The approximation used for the geometry is of

higher order than used for the dependent variable (s).

b) iso-parametric : Equal degree of approximation for geometry and

dependent variable(s).

- This is extensively used in h- version of finite element programs.

nm

nm

Mapping:

c) Sub-parametric : Approximation order used for geometry is lower than that used for primary variable (s).

• Coordinate transformations is only for the purpose of numerical integration.

• When an element is transformed to its master element for the purpose of numerical integration, the integrand must also be expressed in terms of the coordinates of the master element.

nm

,

Mapping:

For example, consider the stiffness coefficient Here, we need to relate with and using the transformation. • can be expressed in terms of local coordinate and . Hence, the chain rule for partial differentiation gives In matrix form,

, ( , ) ( , )e

e ee ej je e ei i

ij i jk a x y b x y c x y dxdyx x y y

,e ei i

x y

e

i

e

i

),( yxe

i

e e ei i i

e e ei i i

x y

x y

x y

x y

e ei i

eeii

x y

x

x y

Jacobian Matrix

Mapping :

In matrix form, where, the Jacobian matrix is given as

e ei i

eeii

x y

x

x y

Mapping :

ex y

Jx y

We need to relate with

Hence, we need the inverse of the Jacobian matrix.

must be non singular.

Now, using the transformation, we can write

,e ei i

x y

ˆ ˆand

e ei i

1

ˆ

ˆ

eeii

e ei i

xJ

y

J

1 1

1 1

ˆ ˆ, ,

ˆ ˆ,

em m ej i

j j

j j

em m ej i

j j

j j

x yx y

x yx y

Mapping :

1 1

1 1

ˆ ˆ

ˆ ˆ

m me ei i

j j

j j

m me ei i

j j

j j

x y

J

x y

1 11 2

2 2

1 2

ˆˆ ˆ

ˆˆ ˆ

m

m

m m

x y

x y

x y

A necessary and sufficient condition for to exist is that the determinant , called the Jacobian, be non negative at every point in . Thus, the function must be continuous, differentiable and invertible.

1J J

,

det 0x y x y

J J

),(),,( yxyx

Mapping :

• In case of higher order triangular and rectangular elements the placement of edge and interior nodes is restricted.

• It can be shown that the side nodes should be placed at a distance greater than a quarter of the length of the side form either corner node.

)21(4

)21(4

by

ax

1 2 2 1 2 2 1 0J b a

Mapping :

0,0 1,0

0,1

1 2

3

0

0

1

4

56

x

y

1 2

3

4

56 ,a b

0.25

0.25

Mapping the elemental area dA from x,y to We have Now, the derivatives of the shape functions with respect to x,y can be given as

ˆ

ˆˆ ˆ ˆ ˆ

dA dx dy dx dy k

x x y yd i d j d i d j k

x y y yd d

J d d

1 *

ˆ ˆ

ˆ ˆ

e eei ii

e e ei i i

xJ J

y

Mapping : d d

Now, using this we can write the element calculations over master element as or

11 12 11 12

ˆ

21 22 21 22

ˆ ˆˆ ˆˆ

ˆ ˆ

e

j je i iij i j

j ji i

i i i ii j

k a b c dxdyx x y y

a J J J J

b J J J J c J d d

Mapping :

ˆ

,eijk F d d

Mapping of Physical Elements to Master Elements

Affine Mapping

Linear Master Element Linear Physical Element

• The affine mapping essentially stretches, translates, and rotates the triangle.

• Straight or planar faces of the reference cell are therefore mapped onto straight or planar faces in the physical coordinate system.

• Preferred when all the edges of an element are straight.

Physical to Master Element Mapping: Linear Triangular Element

0,0 1,0

0,1

1 2

3

0

0

1

x

y

1

2

3

3

1

3

1

ˆ ,ˆj

e

jj

j

e

jj yyxx

For three noded triangular element the transformation of co-ordinates is given by

• where, are the interpolation functions over the master three noded triangular element.

• Master triangle is an isosceles triangle with unit length and right angle

3 3

1 1

ˆ ˆ, , ,i i i i

i i

x x y y

i

1

2

3

ˆ 1

ˆ

ˆ

Numerical Integration: Triangular Element

The Jacobian matrix, [J] for the linear triangular element is

Inverse transformation from the element to is:

where, A the area of the element

e

1 3 1 1 3 1

1 1 2 1 2 1

1

2

1

2

x x y y y y x xA

x x y y y y x xA

e

Mapping: Triangular Element

1313

1212

yyxx

yyxxJ

Linear Master Element Linear Physical Element

Physical to Master Element Mapping: Linear Rectangular Element

4

1

4

1

ˆ ,ˆj

e

jj

j

e

jj yyxx

1,1 1,1

1,1 1,1

1 2

34

x

y

1

2

3

4


Iso-parametric Mapping

Physical to Master Element Mapping: Quadratic Triangular Element

Quadratic Master Element Quadratic Physical Element

• The straight faces of the reference triangle are mapped onto curved faces of parabolic shape in the physical coordinate system.

0,0 1,0

0,1

1 2

3

0

0

1

4

56

x

y

1

2

3

4

5

6

6

1

6

1

ˆ ,ˆj

e

jj

j

e

jj yyxx


Physical to Master Element Mapping: Quadratic Rectangular Element

8

1

8

1

ˆ ,ˆj

e

jj

j

e

jj yyxx

x

y

1

2

3

4

56

78

1,1 1,1

1,1 1,1

1 2

34

67

8

5


Blending Function Mapping

Curved Quadratic Physical Element with

Side 2 is the only curved side.

Curve given in parametric

form so that

are the coordinates of node i


,i iX Y

x

y

1

2

3

4

2

2

x x

y y

2 2,x x y y

2 2 2 2 2 3 2 31 , 1 , 1 , 1x X y Y x X y Y

We have

The first four terms are the linear mapping terms.

The fifth term is the product of two functions:

The bracketed term represents the difference between

and the x-coordinates of the chord that connects and

The other is the linear blending function , which is unity along side 2 and zero along side 4.


1 2 3 4

2 2 3

1 1 1 11 1 1 1 1 1 1 1

4 4 4 4

1 1 1

2 2 2

x X X X X

x X X

2x

2 2,X Y 3 3,X Y

1 2

Therefore, we simplify as

Similarly,

In the general case all sides may be curved. We write the curved sides in the parametric form:

Mapping functions are:


1 4 2

1 1 11 1 1 1

4 4 2x X X x

, , 1 1 1,2,3,4i ix x y y i

1 4 2

1 1 11 1 1 1

4 4 2y Y Y y

1 2 3 4

1 2 3 4

1 1 1 11 1 1 1 1 1 1 1

4 4 4 4

1 1 1 1

2 2 2 2

x X X X X

x x x x

Mapping functions for y coordinate are:

Inverse mapping cannot be given explicitly.

Newton-Raphson method or similar procedure is used.

One can see the quadratic iso-parametric mapping as a special application of the blending function method.


1 2 3 4

1 2 3 4

1 1 1 11 1 1 1 1 1 1 1

4 4 4 4

1 1 1 1

2 2 2 2

y Y Y Y Y

y y y y

Numerical Integration over a

Master Rectangular Element

1 1 1

1ˆ 1 1 1

, , ,

M

j j

j

F d d F d d w d

ji

M

i

N

j

ji wwF

1 1

,

M,N - Number of quaddrature points in and directions. - Gauss points, - corresponding weights. The Gauss point locations are given by the tensor product of one dimensional Gauss points.

,i i

ji ww ,

M N

Numerical Integration:

ˆ

,eijk F d d

Numerical Integration: Rectangular Element

Selection of integration order

and location of the Gauss

points:

Numerical Integration over a

Master Triangular Element

1 1

1ˆ 0 0

, , ,

NINT

j j j

j

F d d F d d w

NINT – Number of quadrature points - Gauss points, - corresponding weight.

,i i

iw

Numerical Integration:

ˆ

,eijk F d d

Numerical Integration: Triangular Element

Line integration: Area integration: Transformation of the co-ordinates to natural co-ordinates: where, are the global coordinates of ith node.

Numerical Integration: Using Area Co-ordinates

1 2

! !

1 !

bm k

a

m kL L ds b a

m k

1 2 2 2! ! !

2 !

m k l

A

Am k l

L L L dAm k l

1 1

,

n n

i i i i

i i

x x L y y L

,i ix y

Thank you.


• This is also an Iso-parametric mapping of quadratic element.


8

1

8

1

ˆ ,ˆj

e

jj

j

e

jj yyxx

1,1 1,1

1,1 1,1

1 2 3

4 56

7 8 9

x

y

1

2

3

4

56

78

Physical to Master Element Mapping: Quadratic Triangular Element


• This is also an Iso-parametric mapping of quadratic element.

0,0 1,0

0,1

1 2

3

0

0

1

4

56

x

y

1

2

3

45

6

6

1

6

1

ˆ ,ˆj

e

jj

j

e

jj yyxx

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