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MANE 4240 & CIVL 4240 Introduction to Finite Elements. Prof. Suvranu De. Numerical Integration in 2D. Reading assignment: Lecture notes, Logan 10.4. Summary: Gauss integration on a 2D square domain Integration on a triangular domain Recommended order of integration

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MANE 4240 & CIVL 4240 Introduction to Finite Elements


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mane 4240 civl 4240 introduction to finite elements

MANE 4240 & CIVL 4240Introduction to Finite Elements

Prof. Suvranu De

Numerical Integration in 2D

slide2

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
slide3

Integration point

Weight

1D quardrature rule recap

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 ‘2M-1’

2. Less expensive

3. Exponential convergence, error proportional to

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

2. More expensive

slide4

Example

f(x)

x

-1

1

A 2-point Gauss quadrature rule

is exact for a polynomial of degree 3 or less

slide5

2D square domain

t

1

1

1

s

1

Using 1D Gauss rule to integrate along ‘t’

Using 1D Gauss rule to integrate along ‘s’

Where Wij=Wi Wj

slide6

For M=2

Wij=Wi Wj=1

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

slide7

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.,

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

slide8

1

1

s1=0, t1=0

W1= 4

1

s

1

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

is exact for a product of two linear polynomials

t

slide9

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

t

1

1

1

s

1

is exact for a product of two cubic polynomials

slide10

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

t

1

1

3

6

2

1

1

9

7

s

4

8

5

1

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

examples
Examples

If f(s,t)=1

A 1-point GQ scheme is sufficient

If f(s,t)=s

A 1-point GQ scheme is sufficient

If f(s,t)=s2t2

A 3x3 GQ scheme is sufficient

slide12

2D Gauss quadrature for triangular domains

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

The type of integral encountered

t

1

t

s=1-t

t

s

1

slide15

Why?

Assume

Then

But

Hence

slide16

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

t

1/2

1

1

2

1/2

s

3

1

slide17

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

t

(0.2,0.6)

(1/3,1/3)

1

2

1

3

4

s

(0.2,0.2)

(0.6,0.2)

1

slide18

Recommended order of integration

“Finite Element Procedures”

by K. –J. Bathe

slide19

“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

1

s t

1

s t

s2 st t2

Which order of GQ to use for full integration?

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

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

Example : 4-noded parallelogram

slide21

1

s t

s2 st t2

s2t st2

1

s t

s2 st t2

Hence, 2M-1=2

M=3/2

Hence we need at least a 2x2 GQ scheme

Example 2: 8-noded Serendipity element

slide22

1

s t

s2 st t2

s3 s2t st2 t3

s4 s3t s2t2 st3 t4

Hence, 2M-1=4

M=5/2

Hence we need at least a 3x3 GQ scheme

slide23

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

“Spurious” zero energy mode/ “hour-glass” mode

The strain energy of an element

Corresponding to a rigid body mode,

If U=0 for a mode d that is different from a rigid body mode, then

d is known as a “spurious” zero energy mode or “hour-glass” mode

Such a mode is undesirable

slide24

Example 1. 4-noded element

y

1

1

Full integration: NGAUSS=4

Element has 3 zero energy (rigid body) modes

Reduced integration: e.g., NGAUSS=1

1

x

1

slide25

Consider 2 displacement fields

y

y

x

x

We have therefore 2 hour-glass modes.

slide27

Example 2. 8-noded serendipity element

y

1

1

Full integration: NGAUSS=9

Element has 3 zero energy (rigid body) modes

Reduced integration: e.g., NGAUSS=4

1

x

1

slide28

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

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

y

x

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