ECIV 720 A Advanced Structural Mechanics and Analysis Lecture 15: Quadrilateral Isoparametric Elements (cont’d) Force Vectors Modeling Issues Higher Order.

ECIV 720 A Advanced Structural

Mechanics and Analysis

Lecture 15: Quadrilateral Isoparametric Elements

(cont’d)Force VectorsModeling Issues

Higher Order Elements

Integration of Stiffness Matrix

1

1

1

1

det dJdt Te DBBk

B (3x8)

D (3x3)

BT(8x3)

ke (8x8)


Each term kij in ke is expressed as

1

1

1

1

1

1

1

1

3

1

3

1

,

,det

ddgt

ddJBDBtkm l

ljmlTimij

Linear Shape Functions is each Direction

Gaussian Quadrature is accurate if

We use 2 Points in each direction


311 312

311

311

2222212112121111 ,,,, gwwgwwgwwgww

1

1

1

1

, ddgt

11 w

12 w

22211211 ,,,, gggg

Choices in Numerical Integration

• Numerical Integration cannot produce exact results

• Accuracy of Integration is increased by using more integration points.

• Accuracy of computed FE solution DOES NOT necessarily increase by using more integration points.

FULL Integration

• A quadrature rule of sufficient accuracy to exactly integrate all stiffness coefficients kij

• e.g. 2-point Gauss ruleexact for

polynomials

up to 2nd order

311 312

311

311

311 312

311

311

Reduced Integration, Underintegration

Use of an integration rule of less than full order

Advantages

• Reduced Computation Times

• May improve accuracy of FE results

• Stabilization

Disadvantages

Spurious Modes (No resistance to nodal loads that tend to activate the mode)

Spurious Modes

t=1

E=1

v=0.3

8 degrees of freedom 8 modes

1

1

Consider the 4-node plane stress element

Solve Eigenproblem

Spurious Modes

01

Rigid Body Mode 02

Rigid Body Mode

Spurious Modes

03

Rigid Body Mode

Spurious Modes

495.05

Flexural Mode

495.04 Flexural Mode

Spurious Modes

769.06 Shear Mode

Spurious Modes

769.07 Stretching Mode

43.18 Uniform Extension Mode

(breathing)

Element Body Forces

ii

Ti

el

T

eA

T

eA

T

e

e

e

tdl

tdA

tdA

Pu

Tu

fu

Dεε2

1

ii

Ti

eA

T

eV

T

eV

T

e

e

e

dA

dV

dV

Pφ

Tφ

fφ

φεσ 0

Total Potential Galerkin

Body Forces

eA

T dAd detfu

Integral of the form

8

1

4321

4321

0000

0000

q

q

NNNN

NNNN

v

u

8

1

4321

4321

0000

0000

NNNN

NNNN

y

x

eA

T dAd detfφ

Body Forces

In both approaches

e

e

e

e

Ay

Ax

Ay

Ax

dAdNf

dAdNf

dAdNf

dAdNf

qqWP

det

det

det

det

4

2

1

1

81

Linear Shape Functions

Use same quadrature as stiffness maitrx

Element Traction

ii

Ti

el

T

eA

T

eA

T

e

e

e

tdl

tdA

tdA

Pu

Tu

fu

Dεε2

1

ii

Ti

eA

T

eV

T

eV

T

e

e

e

dA

dV

dV

Pφ

Tφ

fφ

φεσ 0

Total Potential Galerkin

Element Traction

Similarly to triangles, traction is applied along sides of element

4

2

3

Tx

Ty

u

v

1

4

0

12

1

12

1

0

4

3

2

1

N

N

N

N

el

TT tdlWP Tu

32

Traction

8

1

32

32

000000

000000

q

q

NN

NN

v

u

0

0

0

0

232

8132

y

x

eT T

Tlt

qqWP

For constant traction along side 2-3

Traction

components

along 2-3

Stresses

311 312

311

311

DBqσ

12221121

1121

1222

00

00

det

1

JJJJ

JJ

JJ

JA

10101010

10101010

01010101

01010101

4

1G

More Accurate at

Integration points

Stresses are calculated at any

Modeling Issues: Nodal Forces

ii

Ti

el

T

eA

T

eA

T

e

e

e

tdl

tdA

tdA

Pu

Tu

fu

Dεε2

1

A node should be

placed at the location

of nodal forces

In view of…

Or virtual potential energy

Modeling Issues: Element Shape

Square : Optimum Shape

Not always possible to use

Rectangles:

Rule of Thumb

Ratio of sides <2

Angular Distortion

Internal Angle < 180o

Larger ratios

may be used

with caution

Modeling Issues: Degenerate Quadrilaterals

Coincident Corner Nodes

1

2

3

4

32

1

4

xx

xx

x

xx

x

Integration Bias

Less accurate


Three nodes collinear

1

2

3

4

xx

xx

1

2

3

4 x

xx

x

Less accurate

Integration Bias


Use only as necessary to improve representation of geometry

2 nodes

Do not use in place of triangular elements

A NoNo Situation

3

4

1 2

x

y

(3,2) (9,2)

(7,9)

(6,4)

Parent

All interior angles < 180

J singular

Another NoNo Situation

x, y

not uniquely

defined

FEM at a glance

It should be clear by now that the cornerstone in FEM procedures is the interpolation of the displacement field from discrete values

i

m

ni zyxzyx uNu

1

,,,,

Where m is the number of nodes that define the interpolation and the finite element and N is a set of Shape Functions

FEM at a glance

1=-1 2=1

m=2

1=-1

1

3

2=1

2m=3

FEM at a glance

12

3q6

q5

q4

q3

q2

q1

vu

m=3

4

1

2

3

m=4

FEM at a glance

In order to derive the shape functions it was assumed that the displacement field is a polynomial of any degree, for all cases considered

nn xaxaxaax 2

210u

..., 3210 xyayaxaayxu

Coefficients ai represent generalized coordinates

1-D

2-D

FEM at a glance

For the assumed displacement field to be admissible we must enforce as many boundary conditions as the number of polynomial coefficients

1=-1

1

3

2=1

2e.g.

12121101 uxaxaaxx u

22222102 uxaxaaxx u

32323103 uxaxaaxx u

FEM at a glance

This yields a system of as many equations as the number of generalized displacements

nn u

u

a

a

zyx

Matrix

tCeofficien

10

),,(

nn u

u

C

a

a

1

10

that can be solved for ai

FEM at a glance

nn xaxaxaax 2

210u

Substituting ai in the assumed displacement field

and rearranging terms…

i

m

ni uxNxu

1

FEM at a glance

113N

12

11N 1

2

12N

u(-1)=a0 -a1 +a2 =u1

u(1)=a0 +a1 +a2 =u2

u(0)=a0 =u3

33

22

11

uN

uN

uNu

…

u()=a0+a1 +a2 2

1=-1

1

2=1

3 2

Let’s go through the exercise

x1

1

x2

2

220 xaaxu

Assume an incomplete form of quadratic variation

Incomplete form of quadratic variation

We must satisfy

12

1201 uxaaxu

22

2202 uxaaxu

2

1

1

0

22

21

1

1

u

u

a

a

x

x

x1

1

x2

2


2

1

1

0

22

21

1

1

u

u

a

a

x

x

And thus,

2

121

22

21

221

0

11

1

u

uxx

xxa

a


21

22

2211

22

0 xx

uxuxa

21

22

211 xx

uua

220 xaaxu

And substituting in

221

22

2121

22

2211

22 x

xx

uu

xx

uxuxxu


221

22

2121

22

2211

22 x

xx

uu

xx

uxuxxu

2

1

21

22

221

21

22

222

u

u

xx

xx

xx

xxxu

Which can be cast in matrix form as

Isoparametric Formulation

The shape functions derived for the interpolation of the displacement field are used to interpolate geometry

2

1

21

22

221

21

22

222

x

x

xx

xx

xx

xxx

x1

1

x2

2

Intrinsic Coordinate Systems

Intrinsic coordinate systems are introduced to eliminate dependency of Shape functions from geometry

1 (-1,-1) 2 (1,-1)

4 (-1,1) 3 (1,1)

The price?

Jacobian of transformation

iiiN 114

1

Great Advantage for the money!

Field Variables in Discrete Form

nNxx

Geometry

Displacement

nNuu

= DB un

Stress Tensor

= B un

Strain Tensor

)(intrinsic

)(cartesianJ

FEM at a glance

eeTeV

Te

e

dVU ukuσDε2

1

2

1

Element Strain Energy

ee V

TTeV

Tf dVdVW fNufu

Work Potential of Body Force

ee S

TTeS

Tf dSdSW TNuTu

Work Potential of Surface Traction

etc


Quadrilateral Elements

Recall the 4-node

4321, aaaau

Complete Polynomial

4 generalized displacements ai

4 Boundary Conditions for admissible displacements


Quadrilateral Elements

29

28

227

26

25

432

1,

aaaaa

aaa

au

Assume Complete Quadratic Polynomial

9 generalized displacements ai

9 BC for admissible displacements

9-node quadrilateral

9-nodes x 2dof/node = 18 dof

BT18x3 D3x3 B3x18

ke 18x18

9-node element Shape Functions

Following the standard procedure the shape functions are derived as

1 2

34

4,3,2,14

1 iN iii

Corner Nodes

5

6

7

8

8,7,6,5

11

12

1 22

i

N

iiii

iii

Mid-Side Nodes9

Middle Node

911 22 iN i

9-node element – Shape Functions

113N

12

11N 1

2

12N

Can also be derived from the 3-node axial element

1=-1

1

2=1

3 2

Construction of Lagrange Shape Functions

(1,)(1,1)

1 (-1,-1)

12

11N

12

11N

1

2

11

2

1, 111 NNN

4,3,2,14

1 iN iii

N1,2,3,4 Graphical Representation


N9 Graphical Representation

Polynomials & the Pascal Triangle n

n xaxyayaxaayx 3210, u

1

x y

x2 xy y2

x3 x2y xy2 y3

x4 x3y x2y2 xy3 y4

…….

x5 x4y x3y2 x2y3 xy4 y5

Degree

1

2

3

4

5

0

Pascal Triangle

Polynomials & the Pascal Triangle

To construct a complete polynomial

1

x y

x2 xy y2

x3 x2y xy2 y3

x4 x3y x2y2 xy3 y4

…….


etc

Q1

xyayaxaayx 3210, u

4-node QuadQ2

39

28

27

36

254

23

21

01

,

yaxyayxaxa

yaxyaxa

yaxa

a

yx

u

9-node Quad

Incomplete Polynomials

1

x y

x2 xy y2

x3 x2y xy2 y3

x4 x3y x2y2 xy3 y4

…….


yaxaayx 210, u

3-node triangular


1

x y

x2 xy y2

x3 x2y xy2 y3

x4 x3y x2y2 xy3 y4

…….


27

26

254

23

21

01

,

xyayxa

yaxyaxa

yaxa

a

yx

u


Assume interpolation

1 2

34

5

6

7

8

27

26

254

23

21

01

,

xyayxa

yaxyaxa

yaxa

a

yx

u

8 coefficients to determine for admissible displ.



BT16x3 D3x3 B3x16

ke 16x16



1 2

34

4,3,2,1

1114

1

i

N iiiii

Corner Nodes

5

6

7

8

8,7,6,5

112

1 22

i

N iiiii

Mid-Side Nodes




1

x y

x2 xy y2

x3 x2y xy2 y3

x4 x3y x2y2 xy3 y4

…….


254

23

21

01

,

yaxyaxa

yaxa

a

yx

u

6-node Triangular

Assume interpolation

1 2

3

4

56

254

23

21

01

,

yaxyaxa

yaxa

a

yx

u

6 coefficients to determine for admissible displ.

6-node triangular


BT12x3 D3x3 B3x12

ke 12x12

1 2

3

4

56



3,2,112 iLLN iii

Corner Nodes

1 2

3

214 4 LLN

Mid-Side Nodes

4

56

325 4 LLN

136 4 LLN Li:Area coordinates

Other Higher Order Elements

1

x y

x2 xy y2

x3 x2y xy2 y3

x4 x3y x2y2 xy3 y4

…….


12-node quad

1 2

34

Other Higher Order Elements


16-node quad1

x y

x2 xy y2

x3 x2y xy2 y3

x4 x3y x2y2 xy3 y4

……. x3y21 2

34

ECIV 720 A Advanced Structural Mechanics and Analysis Lecture 15: Quadrilateral Isoparametric Elements (cont’d) Force Vectors Modeling Issues Higher Order.

Documents

direction slide

eigenproblem slide

form slide

stiffness maitrx slide

e 8x8 slide

nd order slide

integration rule

reduced integration