MANE 4240 & CIVL 4240Introduction to Finite Elements

Numerical Integration in 2D

Prof. Suvranu De

Reading assignment:

Lecture notes, Logan 10.4

Summary:

• Gauss integration on a 2D square domain• Integration on a triangular domain• Recommended order of integration• “Reduced” vs “Full” integration; concept of “spurious” zero energy modes/ “hour-glass” modes

1D quardrature rule recap

M

iii fWdfI

1

1

1)()(

Weight Integration point

Choose the integration points and weights to maximize accuracy

Newton-Cotes Gauss quadrature

1. ‘M’ integration points are necessary to exactly integrate a polynomial of degree ‘M-1’ 2. More expensive

1. ‘M’ integration points are necessary to exactly integrate a polynomial of degree ‘2M-1’ 2. Less expensive3. Exponential convergence, error proportional to M

M

2

2

1

Example

f

-1 13/1 3/1

3/1f 3/1f

)3

1()

3

1()(

1

1 ffdf

A 2-point Gauss quadrature rule

is exact for a polynomial of degree 3 or less

dtdstsfI

1

1

1

1),(

2D square domain

s

t

1

1

1 1

3

1,

3

1

3

1,

3

1

3

1,

3

1

3

1,

3

1

M

i

M

jjiij

M

i

M

jjiji

M

jjj

tsfW

tsfWW

dstsfW

dtdstsfI

1 1

1 1

1

11

1

1

1

1

),(

),(

),(

),(

Using 1D Gauss rule to integrate along ‘t’

Using 1D Gauss rule to integrate along ‘s’

Where Wij =Wi Wj

)3

1,

3

1()

3

1,

3

1()

3

1,

3

1()

3

1,

3

1(

),(2

1

2

1

ffff

tsfWIi j

jiij

For M=2

Wij =Wi Wj=1

Number the Gauss points IP=1,2,3,4

IPIP

IP fWdtdstsfI

4

1

1

1

1

1),(

M

i

M

jjiij tsfWdtdstsfI

1 1

1

1

1

1),(),(

The rule

Uses M2 integration points on a nonuniform grid inside the parent element and is exact for a polynomial of degree (2M-1) i.e.,

121 1

1

1

1

1

MfortsWdtdstsM

i

M

jjiij

exact

A M2 –point rule is exact for a complete polynomial of degree (2M-1)

CASE I: M=1 (One-point GQ rule)

t

s

1

1

1 1 s1=0, t1=0W1= 4

)0,0(4),(1

1

1

1fdtdstsfI

is exact for a product of two linear polynomials

CASE II: M=2 (2x2 GQ rule)

)3

1,

3

1()

3

1,

3

1()

3

1,

3

1()

3

1,

3

1(

),(2

1

2

1

ffff

tsfWIi j

jiij

is exact for a product of two cubic polynomials

s

t

1

1

1 1

3

1,

3

1

3

1,

3

1

3

1,

3

1

3

1,

3

1

CASE III: M=3 (3x3 GQ rule)

is exact for a product of two 1D polynomials of degree 5

s

t

1

1

1 1

3

1

3

1

1

1

1

1),(),(

i jjiij tsfWdtdstsfI

5

3

5

3

5

3

5

3

1

23

4 5

6

7

8

9

81

4081

25

,81

64

9876

5432

1

WWWW

WWWW

W

ExamplesIf f(s,t)=1

4),(1

1

1

1

dtdstsfI

A 1-point GQ scheme is sufficient

If f(s,t)=s

0),(1

1

1

1

dtdstsfI

A 1-point GQ scheme is sufficient

If f(s,t)=s2t2

9

4),(

1

1

1

1

dtdstsfI

A 3x3 GQ scheme is sufficient

2D Gauss quadrature for triangular domains

Remember that the parent element is a right angled triangle with unit sides

s

t

1

1

s=1-tt

t

1

0

1

0dsdt),(f

t

t

stsI

The type of integral encountered

M

IPIPIP

t

t

s

fW

tsI

1

1

0

1

0dsdt),(f

Constraints on the weightsif f(s,t)=1

2

1

2

1dsdt),(f

1

1

1

0

1

0

M

IPIP

M

IPIP

t

t

s

W

W

tsI

Example 1. A M=1 point rule is exact for a polynomial

ts

tsf 1~),(

3

1,

3

1

2

1fI

s

t

1

11/3

1/3

Why?

tstsf 321),(

Assume

321

1

0

1

0 !3

1

!3

1

2

1dsdt),(f

t

t

sts

Then

But

131211321

111

1

0

1

0

!3

1

!3

1

2

1

),(dsdt),(f

tsW

tsfWtst

t

s

Hence

!3

1;

!3

1;

2

111111 tWsWW

Example 2. A M=3 point rule is exact for a complete polynomial of degree 2

22

1~),(

tsts

ts

tsf

2

1,0

6

10,

2

1

6

1

2

1,

2

1

6

1fffI

s

t

1

1

1/2

1/22

1

3

Example 4. A M=4 point rule is exact for a complete polynomial of degree 3

3223

22

1~),(

tsttss

tsts

ts

tsf

2.0,6.096

252.0,2.0

96

256.0,2.0

96

25

3

1,

3

1

96

27ffffI

s

t

1

1 2

13 4

(1/3,1/3)

(0.2,0.2)

(0.2,0.6)

(0.6,0.2)

Recommended order of integration“Finite Element Procedures” by K. –J. Bathe

“Reduced” vs “Full” integration

Full integration: Quadrature scheme sufficient to provide exact integrals of all terms of the stiffness matrix if the element is geometrically undistorted.Reduced integration: An integration scheme of lower order than required by “full” integration.

Recommendation: Reduced integration is NOT recommended.

Which order of GQ to use for full integration?

To computet the stiffness matrix we need to evaluate the following integral

1

1

1

1

T dsdt)det(BDB Jk

For an “undistroted” element det (J) =constant

Example : 4-noded parallelogram

st

tsN i

1

~

~B 1 s t

~BDBT 1 s ts2 st t2

Hence, 2M-1=2M=3/2

Hence we need at least a 2x2 GQ scheme

Example 2: 8-noded Serendipity element

~iN 1 s ts2 st t2

s2t st2

~B 1 s ts2 st t2

~BDBT 1 s t s2 st t2

s3 s2t st2 t3

s4 s3t s2t2 st3 t4

Hence, 2M-1=4M=5/2

Hence we need at least a 3x3 GQ scheme

“Spurious” zero energy mode/ “hour-glass” modeThe strain energy of an element

dVDdkdUeV

TT 2

1

2

1

Corresponding to a rigid body mode, 00 U

If U=0 for a mode d that is different from a rigid body mode, thend is known as a “spurious” zero energy mode or “hour-glass” mode

Such a mode is undesirable

Reduced integration leads to rank deficiency of the stiffness matrix and “spurious” zero energy modes

Example 1. 4-noded element

y

x

1

1

1 1 iT

NGAUSS

iiV

T DWdVDUe

12

1

Full integration: NGAUSS=4Element has 3 zero energy (rigid body) modes

Reduced integration: e.g., NGAUSS=1

004

yx

T DU

Consider 2 displacement fields

0v

xyCuxyCv

u

0

y

x

y

x

0

00

U

yxAt xyyx

We have therefore 2 hour-glass modes.

Propagation of hour-glass modes through a mesh

Example 2. 8-noded serendipity element

y

x

1

1

1 1 iT

NGAUSS

iiV

T DWdVDUe

12

1

Full integration: NGAUSS=9Element has 3 zero energy (rigid body) modes

Reduced integration: e.g., NGAUSS=4

Element has one spurious zero energy mode corresponding to the following displacement field

)3/1(

)3/1(2

2

xyCv

yxCu

y

x

Show that the strains corresponding to this displacement field are all zero at the 4 Gauss points

Elements with zero energy modes introduce uncontrolled errors and should NOT be used in engineering practice.