Washkewicz College of Engineering Higher Order Triangular ...We next consider shape functions for higher order elements. To do this in an orderly fashion we introduce the concept of

Section 11: HIGHER ORDER TWO DIMENSIONAL SHAPE FUNCTIONS

Washkewicz College of Engineering

We next consider shape functions for higher order elements. To do this in an orderly fashion we introduce the concept of area coordinates. Consider a series of triangular elements depicted in the figure below

As the number of nodes in each triangular element increases there is a need to insure displacement compatibility along the edges of adjacent elements between the edge nodes –keeping in mind that nodal displacements are the same for each element that shares a node. Using complete polynomials for displacement interpolation through the element will guarantee compatibility.

Higher Order Triangular Elements

1



Complete polynomials can be identified through the use of Pascal’s triangle.

Note for the three node triangular element, the polynomial needed to define displacement as a function of position within the element is linear (order one). For the six node triangular element the polynomial needed to define displacement as a function of position in the element is of order two, and for the ten node element the polynomial is of order three.

2



Hence, for the three node element the complete polynomial is

For the six node element the complete polynomial is

For the ten node element the complete polynomial is

yaxaayxv

yaxaayxu

654

321

,,

2

12112

10987

265

24321

,

,

yaxyaxayaxaayxv

yaxyaxayaxaayxu

3

202

192

183

17

21615

214131211

310

29

28

37

265

24321

,

,

yaxyayxaxa

yaxyaxayaxaayxv

yaxyayxaxa

yaxyaxayaxaayxu

3



What is needed is a simple method to generate shape functions for higher order elements. To do this for triangular elements we introduce the concepts of area coordinates. We want to define the location of a point within the following three node triangular element

using information from the nodal coordinate geometry. We could use the following two expression in terms of unknown coefficients L1, L2, and L3

These two expressions are interpolating functions and the physical interpretations of L1, L2, and L3 are defined momentarily. We won’t be surprised to find

These three expressions are used to define L1, L2, and L3.

332211

332211

yLyLyLyxLxLxLx

3211 LLL

4



Using Cramer’s rule on the previous three expressions, the parameters are the ratios of the following determinants

Focusing on the denominator, this determinant is

If we take 1 = i, 2 = j and 3 = m then from the Constant Strain Triangle notes

111

111

321

321

32

32

1

yyyxxx

yyyxxx

L

111

111

321

321

31

31

2

yyyxxx

yyyxxx

L

111

111

321

321

21

21

3

yyyxxx

yyyxxx

L

213132321

211332133221

yyxyyxyyxxyxyxyyxyxyx

Ayyxyyxyyx

yyxyyxyyx

jimimjmji 2213132321

5



Focusing initially on L1

Aycxba

Ayxxxyyxyyx

Ayxxyxyyxyxxy

yyyxxx

yyyxxx

L

2

2

2

111

111

111

23323232

23323322

321

321

32

32

1

6



Solving the system of equations for L2

Aycxba

Ayxxxyyxyyx

Axyxyyxyxxyyx

yyyxxx

yyyxxx

L

2

2

2

111

111

222

31131312

11331331

321

321

31

31

2

7



Finally for L3

Aycxba

Ayxxxyyxyyx

Axyyxxyxyyxyx

yyyxxx

yyyxxx

L

2

2

2

111

111

333

12212121

21121221

321

321

21

21

3

8



Here

and if we take 1 = i, 2 = j and 3 = m then

If we take a = , b = and c = then

12321321213

31213213132

23132132321

xxcyybxyyxaxxcyybxyyxaxxcyybxyyxa

ijmjimjijim

mijimjimimj

jmimjimjmji

xxcyybxyyxa

xxcyybxyyxa

xxcyybxyyxa

ijmjimjijim

mijimjimimj

jmimjimjmji

xxyyxyyx

xxyyxyyx

xxyyxyyx

9



yxA

NLL

yxA

NLL

yxA

NLL

mmmmm

jjjjj

iiiii

21

21

21

3

2

1

and

Thus L1, L2, and L3 are shape functions.

10



We now seek a physical interpretations of L1, L2, and L3. Consider the following triangular element.

x

y

2

3

P(x,y)

A1

A2

A3

1

then

The numerator is twice the area of A1 in the same manner that the denominator is twice the area of the entire triangle

AA

APArea

APArea

yyyxxx

yyy

xxx

L

p

p

1

321

321

32

32

1

322

322

111

111

11

Note that from the figure

If we focus in on point P(x,y) and identify

p

p

yy

xx

321 AAAA



12

Clearly from the last expession L1 is the ratio of the orange area to the overall area of the triangle. Similarly L2 is the ratio of the pink area to the are of the triangle and L3 is the ratio of the blue area to the area of the triangle, i.e.,

and

This is the third equation presented earlier to establish a system of three equations in three unknowns (L1, L2, and L3).

321

321

321

1

1

LLL

AA

AA

AA

AAAA

AAL

AAL

AAL 3

32

21

1



With the physical interpretation of L1, L2, and L3 we now turn to the isoparametric formulation of shape functions for the three node constant strain triangular element. In the s-t coordinate system the element has vertical and horizontal sides equal to unity.

Note from the geometry of the previous page that the number of the area segment corresponds to the node number opposite to the area segment, i.e.,

s

t

2

31

1

1

Pst

21

2221

21

21

21

21

2111

21

12

23

13

123

ststA

ssA

ttA

A

P

P

P

123132231 PPP AAAAAA

13



Since the area coordinates are the ratios of the area segments to the overall areas, then

and in turn, the shape functions in the s-t coordinate system are:

stAA

AAL

tAA

AAL

sAA

AAL

P

P

P

1123

1233

123

1322

123

2311

stNtNsN

13

2

1

14



For the following isoparametric mapping

the transformation equations from the s-t coordinate system to the x-y coordinate system are as follows:

y

xs

t

2

3 1

1

1 (x1,y1)s

t2 (x2,y2)

3 (x3,y3)1

332211 ,,, xtsNxtsNxtsNx

332211 ,,, ytsNytsNytsNy 15

),(),(

tsyytsxx

),(),(

yxttyxss



Now let’s recap the formulations for the shape functions in the s-t coordinate system for line elements:

Two node line element

Three node line element (reference Example 10.6 in Logan’s text book)

23

2

2

2

1

1)(22

1)(

221)(

ssN

sssssN

sssssN

21)(

21)(

2

1

ssN

ssN

16



Now let’s recap the formulations for the shape functions in the s-t coordinate system two dimensional planar elements:

Three node triangular element

Four node quadrilateral element (Section 10 of class notes)

sttsN

ttsNstsN

1,,,

3

2

1

41

4)1)(1(),(

41

4)1)(1(),(

41

4)1)(1(),(

41

4)1)(1(),(

4

3

2

1

ststtstsN

ststtstsN

ststtstsN

ststtstsN

17



The preceding discussion should indicate that we are searching for a pattern(s) to utilize in the formulation of the shape function for each type of element. At this point there does not seem to be anything jumping out of what we have already done. However, let’s focus on quadrilateral elements and categorize the higher order elements into several families. They are referred to as the “Serendipity” family of elements represented by the element in the left hand side of the attending picture and the “Lagrangian” family of elements represented by the element in the right hand side (with a central node).

Higher Order Quadrilateral Elements

18

Serendipity Lagrangian



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:

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.

nn xaxaxaaxf 2

210

19



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:

i

n

iin xfxHxf

0 20



where

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

and the second order version is

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 (keep in mind the index i starts at 0). 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.

n

ijj ji

ji xx

xxxH

0

1

01

00

10

1

0 01 xf

xxxxxf

xxxxxf

xxxx

xf i

n

i

n

ijj ji

j

2

12

1

02

01

21

2

01

00

20

2

10

12 xf

xxxx

xxxxxf

xxxx

xxxxxf

xxxx

xxxxxf

21



In two dimensions the interpolation function has the form

where

Three dimensional Lagrange interpolation has the form

where

i

m

iii

n

iip ygyVxfxHyxf

00,

mnp

i

l

iii

m

iii

n

iip zqyQygyVxfxHzyxf

000,,

lmnp

22



Lagrange elements have an order of n with (n+1)2 nodes arranged in square-symmetric pattern. These elements require internal nodes.

Shape functions are products of nth order polynomials in each direction. The bilinear quadrilateral element (four nodes) is a Lagrangian element of order n = 1.

Lagrangian Quadrilateral Elements

t

s

t

s

t

s

23



We can easily make use of Lagrange interpolating polynomials by presenting two formulations for s-t coordinate system. The first polymial for s (identified as H for horizontal) is

and a second polynomial in t (identified as V for vertical)

In general, the shape function in the s-t coordinate system for a given node is the product of these two expressions

where m is the order of interpolation and k is the node number.

m

kii ik

i

mkkkkkkk

mkkmk ss

ssssssssssss

sssssssssssH01110

1110

m

kii ik

i

mkkkkkkk

mkkmk tt

tttttttttttt

tttttttttttV01110

1110

tVsHtsN mk

mkk ,

24



Using this procedure, consider the 16 node quadrilateral element. We wish to develop an expression for the shape function at node #6.

14610626

1410236 tttttt

tttttttV

867656

87536 ssssss

sssssssH

tVsHtsN 3

63

66 ,

t

s

25



6542332456

54322345

432234

3223

22

1

tsttstststsststtststss

tsttstsststtss

tststs

Lagrange polynomials for shape functions are complete polynomial expansions. From Pascal’s triangle we can see how many nodes are required for the representation of displacement fields of any order and completeness:

Zero Order

First Order

Second Order

Third Order

Fourth Order

Fifth Order

Sixth Order

Bilinear Quad (4 nodes)

Quad (9 nodes)

Quad (16 nodes)

26



Serendipity Quadrilateral ElementsIn general serendipity elements only use boundary nodes. Internal nodes are avoided. Serendipity elements are not as accurate as Lagrangian elements. However, they are more efficient than Lagrangian elements and they avoid certain types of instabilities.

The four node serendipity element is the same as the four node Lagrangian element.

4

1

s

t

1 1

3

2

1

1

ststtstsN

ststtstsN

ststtstsN

ststtstsN

141

4)1)(1(),(

141

4)1)(1(),(

141

4)1)(1(),(

141

4)1)(1(),(

4

3

2

1

27



For the eight node serendipity element, start by formulating the shape functions at the mid side nodes.

4

1

s

t

1 1

3

2

1

1

7

8

5

6

Shape functions for mid-side nodes are products of a second order polynomial parallel to side and a linear function perpendicular to the side.

222

8

222

7

222

6

222

5

121

2)1)(1(),(

121

2)1)(1(),(

121

2)1)(1(),(

121

2)1)(1(),(

ststtstsN

tssttstsN

ststtstsN

tssttstsN

N5 and N7 are linear in t, where N6 and N8 are linear in s.

Fold the coordinate picture down and over the shape function figure.

28



Shape functions for corner nodes are modifications of the shape functions of the bilinear quadrilateral element.

Step #1: start with appropriate bilinear quad shape function

Step #2: subtract out mid-side shape function N5 with appropriate weight, i.e., ½, see figure next overhead

Step #3: repeat Step #2 using mid-side shape function N8 and weight

4)1)(1(ˆ

1tsN

22ˆ 85

11NNNN

2ˆ 5

11NNN

29



Hence

2222

2222

22

22

22

1

241

11222241

141

1411

41

4)1)(1(

4)1)(1(

4)1)(1(

stttssstst

ststtsststst

stst

tsststst

tststsN

30



Graphically the process appears as follows

31



6542332456

54322345

432234

3223

22

1

tsttstststsststtststss

tsttstsststtss

tststs

Once shape functions have been identified, there are no procedural differences in the formulation of higher order quadrilateral elements and the bilinear quad.

Pascal’s triangle for the serendipity quadrilateral elements:

Bilinear/Serendipity Quad (4 nodes)

Serendipity Quad (8 nodes)

Serendipity Quad (12 nodes)

32



Shape Functions for Triangular Elements

We now turn our attention back to triangular elements. First define a coordinate system to identify nodes and specify location of points by identifying distances measured perpendicular from each side of the triangle. The coordinates of each node depends on the order of the element.

3-Node () (first order)#1 (1,0,0)#2 (0,1,0)#3 (0,0,1)

6-Node () (second order)#1 (2,0,0) #4 (1,1,0)#2 (0,2,0) #5 (0,1,1)#3 (0,0,2) #6 (1,0,1)

10-Node Element () (third order)#1 (3,0,0) #4 (2,1,0)

#5 (1,2,0)#2 (0,3,0) #6 (0,2,1)

#7 (0,1,2)#3 (0,0,3) #8 (1,0,2)

#9 (2,0,1)#10 (1,1,1) 33



Sylvester (1969) established the following approach to define shape functions for triangular elements using area coordinates L1, L2, and L3

Here n is the order of the element, p is the node number and q is the number of nodes in a triangular element, and

If we want the interpolation function (shape function) for node #1 in the 6 node triangular element, then the interpolation function will be quadratic, and

qpLLLLLLN p ,,2,1,, 321321

01

111

11

i ii

iLnL

122

1221

112

11

110,0,2

1111

2

1

1

1

1

11

LLLLi

iLni

iLn

LN

i ii i

34



For node #6

The two shape functions are depicted graphically as follows:

3131

1

3

1

1

316

41

1121

112

11

1,0,1

LLLLi

iLni

iLn

LNLNN

i ii i

35



The interpolation expressions given for the shape functions do not appear to be Lagrangian polynomial interpolating expressions. Note that the area coordinates (L1, L2, L3) are coordinates in the same sense that (x, y) are coordinates in a two dimensional Cartesian space. Both sets of coordinates identify the position of a point.

x

y

2

3

PA1

1L1= constant

P*

Here

Also notice that lines parallel to the sides of the triangular element denote lines of constant L1, L2, or L3. This is clearly shown in the next figure.

yxPLLLP

yxPLLLP,*,,*

,,,

321

321

36



In this figure lines of constant values of the area coordinates are depicted for two values, zero and one:

x

y2

3

PA1

1

L1= 0

L1= 1

L2= 0

L2= 1L3= 0

L3= 1

37



Halfway between the lines of the constant values of 0 and 1 should be a line of constant value equal to 1/2 (mid-side nodes added for clarity)

x

y 2

3

1

L1= 0

L1= 1L2= 0

L2= 1

L3= 0 L3= 1

L1= 1/2L2= 1/2

L3= 1/2

4

5

6

38



We now have functional values of each area coordinate along the sides of a triangular element equal to one (1), one half (1/2) and zero (0). We should be able to construct Lagrangian interpolating functions with this information, and we can. Consider again the shape function for node #1 in the six node triangular element. With three data values we should be able to establish a quadratic Lagrangian coefficient such that

where

1210

121

1110

111

2

0 11

111

LLLL

LLLL

LLLL

Llij

j ji

ji

021

1

12

11

10

LLL

39



This yields

as before. The problem is that the published versions of the Lagrange polynomials do not work well in identifying the various values of the area coordinates.

So the previous interpolation function used to establish shape functions for various nodes in a triangular element works well, i.e., it is robust. And it is a form of Lagrangian interpolation as demonstrated above.

1

11

11

1210

121

1110

1111

12010

5.015.0

NLL

LLLLLL

LLLLLli

40



The discussion above and the general interpolation function can be represented as follows:

Here it is clear that a position can represented by area coordinates, and that area coordinates will take on different values at various nodes.

qpLLLLLLN p ,,2,1,, 321321

At each node:

41



Integration Within a Triangular Domain

42

As we have seen throughout this discussion shape functions for higher order triangular elements have been developed using area coordinates rather than an r – s coordinates used in traditional isoparametric element formulations. Area coordinates can be referred to as areal, triangular, or trilinear coordinates in the literature. Although known in mathematics from many years, area coordinates were seemingly reinvented when finite element theory found a need for them.

Earlier we saw that that for three node triangular elements

We started to specify the shape functions for triangular elements with mid-side nodes. With the relationships above we can specify the shape functions for triangular nodes in terms of r – s coordinates.

stLtLsL

tsLLL

1

,,,

3

2

1

321



43

1121

12

1212

1212

333

222

111

ststLLN

ttLLN

ssLLN

t

s

tsLLLLLL ,,,,,, 654321

stsLLN

sttLLN

tsLLN

144

144

44

316

325

214

Earlier we arrived at expressions for N1 and N6. The complete functional relationships for all six nodes and their attending shape functions are as follows:



44

First consider integration along a side of the element depicted below. This integration is useful when a traction is applied to the side of an element and we wish to convert the traction to equivalent nodal forces. Specifically consider side #3 which is opposite node #3. Along side #3 the shape function L3 is zero between node #1 and node #2, i.e.,

03 L

The variation of any quantity along this side will only be a function of L1 and L2. Consider the following general integration expression along a line

Here k and l are nonnegative integers and L is the length of the side of the element. The formula yields L if both k and l are zero.

!1

!!3

1121 lk

lkLlklkLdLLL

L

lk

A similar integration over the area of the triangle yields a similar expression, i.e.,

!2!!!2321 mlk

mlkAdALLLA

mlk



The previous formula yields A for k = l = m = 0, i.e.,

Substituting for L1, L2 and L3 with their corresponding relationships in terms of s and twe obtain

Finally, with m = 0

45

AAdAdALLLAA

mlk

!0002!0!0!021110

30

20

1

!2!!!

121

21

321

mlkmlk

dtdssttsA

dALLLA

mlk

A

mlk

!2!!

21

lklkdtdsts

Alk



This is a convenient “back door” (“small rabbit out of the hat”) kind of integration expression. For example, for various values of k and l we obtain the following

The integrals on the left are strictly equal to the values on the right. Using a matrix expression

46

121

!202!2!0

21

241

!112!1!1

21

121

!022!0!2

21

61

!102!1!0

21

61

!012!0!1

21

20

11

02

10

01

dtdstsA

dtdstsA

dtdstsA

dtdstsA

dtdstsA

lk

lk

lk

lk

lk

20

1,601,

601,

201,

121,

241,

121,

61,

61,

21,,,,,,,,,1

21 322322 dtdststtsstststsA



The objective is numerically integrating the following expression

to obtain the element stiffness matrix. If we let

then

This is the Gauss quadrature expression that we developed earlier. Once again the civalues are weights and (si, ti) are the Gauss points. 47

n

iiii

nnn

k

tsgc

tsgctsgctsgcIk

1

222111

,

,,,

dtdstsg

dtdsttsJtsBDtsBk TT

,

,,,1

1

1

1

ttsJtsBDtsBtsg TT ,,,,



48

The Gauss quadrature is exact for cubic integrations. This includes integrations of 1, s, t, s2, st and t2. So for integrations of 1, s and t

We can easily solve this system for N =1 to find that

A one point integration would use the centroid of the element as the Gauss point.

N

iii

N

iiii

N

iii

N

iiii

N

ii

N

iiii

tctsgcttsg

sctsgcstsg

ctsgctsg

113

112

111

,61,

,61,

,211,

31311

1

1

1

t

s

c



49

The Gauss quadrature is exact for cubic integrations. This includes integrations of 1, s, t, s2, st and t2. For an integration of 1, s t , s2, st and t2 .

N

ii

N

iiii

N

iiii

N

iiii

N

ii

N

iiii

N

iii

N

iiii

N

iii

N

iiii

N

ii

N

iiii

tctsgcttsg

tsctsgcsttsg

sctsgcstsg

tctsgcttsg

sctsgcstsg

ctsgctsg

1

2

1

26

115

1

2

1

24

113

112

111

,121,

,241,

,121,

,61,

,61,

,211,



50

It is obvious that one Gauss point will not provide enough information to solve this system of equations. We will need at least three Gauss points (s1, t1), (s2, t2), (s3, t3 )and three weights c1, c2 and c3.

Here N = 3 and the relative location of the Gauss points within the element are depicted above.

32

31

6161

31

32

61

31

61

3

33

2

22

1

11

t

cs

t

cs

t

cs



Using the three point Gauss quadrature the element stiffness matrix is given by

The notation for the Gauss points is slightly different from the expression developed in the last set of class notes. Higher order quadratures are simple extensions of this expression.

51

3333333

2222222

1111111

,,,

,,,

,,,

cttsJtsBDtsB

cttsJtsBDtsB

cttsJtsBDtsBk

TTT

TTT

TTT



52



53

Location of Gauss points for the isoparametric version of a six node triangle. Weights are written to 5 significant figures near each Gauss point.



54

Higher order polynomials can be utilized in a Gauss quadrature rule. However the precision of the computation can be limited since most references provide points and weights to 20 significant figures at most. The primary disadvantage is inefficiency since for high N (defined as number of Gauss point pairs) other quadrature rules are available with fewer points. In addition, the location of the Gauss points become somewhat unsymmetrical and tend to cluster in regions around the vertices of the triangle. This is depicted in the following figure.

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Washkewicz College of Engineering Higher Order Triangular ...We next consider shape functions for higher order elements. To do this in an orderly fashion we introduce the concept of