ECIV 720 A Advanced Structural

Mechanics and Analysis

Lecture 13 & 14: Quadrilateral Isoparametric ElementsStiffness MatrixNumerical IntegrationForce VectorsModeling Issues

Two Dimensional – Plane StressThin Planar Bodies subjected to in plane loading

Tvuu

Tyx fff

Tyx TTT

tdAdV

Tiyixi PPP

Stress and Strain

Txyyx σ

Txyyx ε

Stress Strain Relationship

T

y

u

y

u

y

v

x

u

ε

FEM Solution: Area Triangulation

Area is Discretized into Triangular Shapes

Constant Strain Triangle

For Every Triangular Element

•Strain-Displacement Matrix B

122131132332

211332

123123

000

000

2

1

yxyxyx

xxx

yyy

AB

1 2

3q6

q5

q4

q3

q2

q1

vu

Constant strain

11 LN

22 LN

133 LN

Nqu ,

Bqε

FEM Solution: Quadrilateral Mesh

Area is Discretized into Quadrilateral Shapes

FEM Solution: Quadrilateral Elements

X

Y

1 (x1,y1) 2 (x2,y2)

4 (x4,y4)

q8

q7

q4

q3

q1

q2

8 Degrees of Freedom

3 (x3,y3)

q6

q5v

uP (x,y)

FEM Solution: Objective

• Use Finite Elements to Compute Approximate Solution At Nodes

• Interpolate u and v at any point from Nodal values q1,q2,…q8

q8

q7

q4

q3

q1

q2

vu

q6

q5

To this end…

•Intrinsic Coordinate System•Shape Functions of 4-node quadrilateral

•From 1-D

•Direct •Jacobian of Transformation•Strain-Displacement Matrix•Stiffness Matrix

Intrinsic Coordinate System

12

112

xxxx

Map ElementDefine Transformation

x1x

x2

1=-1 2=1

Recall 1-D

Intrinsic Coordinate System

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

4 (-1,1)

Map ElementDefine Transformation

Parent

3 (1,1)

4

1

2

3

Lagrange Shape Functions

1=-1 2=1

12

11N 1

2

12N

For 1-D

N1()N1=1

baN 1

111 N

011 N

2-D Lagrange Shape Functions

12

11,1f

(1,)(1,1)

1 (-1,-1)

What is the lowest order

polynomial

f1(,) along side =-1

that satisfies

f1(-1,-1) =1 & f1(1,-1)=0?

2-D Lagrange Shape Functions

12

1,12f

What is the lowest order

polynomial

f2(,) along side =-1

that satisfies

f2(-1,-1) =1 & f2(-1,1)=0?

(1,)(1,1)

1 (-1,-1)

12

1,12f

f2=0

f2=1

Construction of Lagrange Shape Functions

(1,)(1,1)

1 (-1,-1)

12

11,1f

f1=1

f1=0

What is the lowest order Polynomial F1(,) that satisfiesF1(-1,-1) =1 & F1(1,-1)=F1(1,1)=F1(-1,1)=0?

Bi-Linear Surface

(1,)(1,1)

Construction of Lagrange Shape Functions

f2(-1,)f1(-1)

PP

PPPP ffNF

114

1

,11,, 2111

PP

P

Construction of Lagrange Shape Functions

114

1,11,, 2111 ffNF

For Every Point ()

(1,1)

1 (-1,-1)F1(,)

Construction of Lagrange Shape Functions

(1,)(1,1)

2 (1,-1)

12

11,1f

Bi-Linear Surface

114

1,11,, 2122 ffNF

12

1,12f

F2(,)

Construction of Lagrange Shape Functions

3 (1,1)

12

11,1f

Bi-Linear Surface

114

1,11,, 2133 ffNF

12

1,12f

F3(,)

12

1,12f

Construction of Lagrange Shape Functions

4 (-1,1)

Bi-Linear Surface

114

1,11,, 2144 ffNF

F4(,)

12

11,1f

In Summary

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

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

114

14N

114

13N

114

12N

114

11N

Shape Functions in a Direct Way

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

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

Assume Bi-Linear

Variation

dcbaN i ,

Complete Polynomial

i=1,2,3,4

With conditions

jiif

jiifN jji 0

1, j=1,2,3,4

Shape Functions in a Direct Way

dcbaN ,1

Each, i, provides 4 Equations with 4 unknowns

e.g. i=1

jiif

jiifN jji 0

1,

11,11 dcbaN

01,11 dcbaN

01,11 dcbaN

01,11 dcbaN 11

4

11N

Field Variables in Discrete Form

114

14N

114

13N

114

12N

114

11N

Geometry

44332211 xNxNxNxNx

44332211 yNyNyNyNy

Displacement

74533211 qNqNqNqNu

84634221 qNqNqNqNv

q8

q7

q4

q3

q1

q2

vu

q6

q5

Field Variables in Discrete Form

4

4

1

1

4321

4321

0000

0000

y

x

y

x

NNNN

NNNN

y

x

Geometry

8

1

4321

4321

0000

0000

q

q

NNNN

NNNN

v

u

Displacement

Intrinsic Coordinate System

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

4 (-1,1)

Map Element

Parent

3 (1,1)

4

1

2

3

JDefine Jacobian

Transformation

The Jacobian from (x,y) to (,) may be obtained as:

Consider any function f(x,y)

defined on area of element

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

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

y

y

fx

x

ff

y

y

fx

x

ff

Transformation

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

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

y

y

fx

x

ff

y

y

fx

x

ff

y

fx

f

yx

yx

f

f

y

fx

f

f

f

J

Jacobian of Transformation

44332211 xNxNxNxNx

44332211 yNyNyNyNy

44

33

22

11

4

1

xN

xN

xN

xN

xNx

ii

i

4

1ii

i xNx

4

1ii

i yNy

4

1ii

i yNy

yx

yx

J

Jacobian of Transformation

iiii

iN

14

111

4

1

iiiiiN

14

111

4

1

Jacobian of Transformation

4321

4

1

4

111

11114

1

14

1

xxxx

xxNx

J ii

iiii

i

4321

4

1

4

121

11114

1

14

1

xxxx

xxNx

J ii

iiii

i

Jacobian of Transformation

4321

4

1

4

121

11114

1

14

1

yyyy

yyNy

J ii

iiii

i

4321

4

1

4

122

11114

1

14

1

yyyy

yyNy

J ii

iiii

i

In Summary

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

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

4

1

2

3

Without Proof

2221

1211

JJ

JJyx

yx

J

ddJdxdydA det

Jacobian of Transformation

4321

4

1

4

111

11114

1

14

1

xxxx

xxNx

J ii

iiii

i

4321

4

1

4

121

11114

1

14

1

xxxx

xxNx

J ii

iiii

i

Jacobian of Transformation

4321

4

1

4

121

11114

1

14

1

yyyy

yyNy

J ii

iiii

i

4321

4

1

4

122

11114

1

14

1

yyyy

yyNy

J ii

iiii

i

In Summary

Or the inverse

y

fx

f

JJ

JJf

f

2221

1211

f

f

JJ

JJ

Jf

f

JJ

JJ

y

fx

f

1121

1222

1

2221

1211

det

1

In Summary

114

14N

114

13N

114

12N

114

11N

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

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

44332211 xNxNxNxNx

44332211 yNyNyNyNy 74533211 qNqNqNqNu

84634221 qNqNqNqNv

Strain Tensor from Nodal Values of Displacements

x

v

y

uy

vx

u

ε

Strain Tensor from Nodal Values of Displacements

Thus we need to evaluate

and

y

ux

u

y

vx

v

Strain Tensor from Nodal Values of Displacements

Recall that

Consequently,

f

f

JJ

JJ

Jy

fx

f

1121

1222

det

1

f=u

u

u

JJ

JJ

Jy

ux

u

1121

1222

det

1

f=v

v

v

JJ

JJ

Jy

vx

v

1121

1222

det

1

Strain Tensor from Nodal Values of Displacements

v

v

u

u

JJJJ

JJ

JJ

J

x

v

y

uy

vx

u

12221121

1121

1222

00

00

det

1ε

A

Strain Tensor from Nodal Values of Displacements

Next we need to evaluate

Nqu

8

1

4321

4321

0000

0000),(

q

q

NNNN

NNNN

v

u

Tvvuu

Strain Tensor from Nodal Values of Displacements

74

53

32

11

4

112 q

Nq

Nq

Nq

Nq

Nu

ii

i

4

112

ii

i qNu

4

12

ii

i qNv

4

12

ii

i qNv

Nqu

8

1

4321

4321

0000

0000),(

q

q

NNNN

NNNN

v

u

Strain Tensor from Nodal Values of Displacements

8

2

1

10101010

10101010

01010101

01010101

4

1

q

q

q

v

v

u

u

8

2

1

q

q

q

v

v

u

u

G

Strain Tensor from Nodal Values of Displacements

= B q

Both A and G are linear functions of and

= AG q

Stresses

xy

y

x

xy

y

xE

2

100

01

01

1 2

Dεσ = B q

DBqσ

Element Stiffness Matrix ke

eV

Te dVU σDε

2

1

= B qe= D B qe

e

A

TTe

el

TTe

e

dAt

tdA

U

e

qDBBq

qDBBq

2

1

2

1

ke

tdAdV

4

1

2

3

8x8 matrix

Element Stiffness Matrix ke

dJdttdAdV det

1

1

1

1

det dJdt

dAt

T

A

Te

DBB

DBBk

Furthermore

and

Numerical

Integration

Integration of Stiffness Matrix

1

1

1

1

det dJdt Te DBBk

B (3x8)

D (3x3)

BT(8x3)

ke (8x8)

Integration of Stiffness Matrix

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

0

2

4

6

8

10

0 0.5 1 1.5 2 2.5 3

x2

Numerical Integration

Integrals

Indefinite Definite

Cxdxx 32

3

1

2

0

2

3

8dxx

Numerical Integration

Definite integrals can be computed numerically

b

a iii xfwdxxf

Objective:

• Determine points xi

• Determine coefficients wi

0

2

4

6

8

10

0 0.5 1 1.5 2 2.5 3

x2

Numerical Integration

Depending on choice of wi and xi

Midpoint Rule

Trapezoidal Rule

Simpson's

Gaussian Quadratures

etc

Numerical Integration – Upper & Lower Bounds

Lower Sum

L(f;xi)

Upper Sum

U(f;xi)

a=x1 x2 x3 x4 x5 x6 x7=b

a=x1 x2 x3 x4 x5 x6 x7=b

Numerical Integration

ib

a iiii xfUxfwdxxfxfL ;;

It can be shown that

ii

b

a iiii

ixfUxfwdxxfxfL ;lim;lim

Mid-Point Rule

a=x1 x2 x3 x4 x5 x6 x7=b

212

1xx

322

1xx

542

1xx 762

1xx

1

111 2

1n

iiiii

b

a iii xxfxxxfwdxxf

76676 2

1xxfxxA

12121 2

1xxfxxA

Mid Point Rule

1

111 2

1n

iiiii

b

a iii xxfxxxfwdxxf

Simple to comprehend and implement

Large number of intervals is required for accuracy

a=x1

…

b=xn

Trapezoid Rule

a=x1 x2 x3 x4 x5 x6 x7=b

1

1112

1 n

iiiii

b

a iii xfxfxxxfwdxxf

67676 2

1xfxfxxA

12121 2

1xfxfxxA

Trapezoid Rule

Simple to comprehend and implement

Exact for polynomials f(x) = ax+b

Large number of intervals is required for accuracy

(Less than midpoint rule)

1

1112

1 n

iiiii

b

a iii xfxfxxxfwdxxf

Simpson’s Rule

hafhafafh

dxxfha

a

243

2

Step 1

ha

a

dxxf2

h ha a+2h

h

xi xi+1xi+1/2

h

Simpson’s Rule

h

xi xi+1xi+1/2

h

i

x

x

Ihafhafafh

dxxfi

i

243

1

Step 2

1

0

n

ii

b

a

Idxxf

Ii

Ii+1…

Quadratures

nn

b

a iii xfwxfwxfwxfwdxxf 2211

Objective

Where do such formulae come from?

Theory of Interpolation….

Let

n

iii xfxlxpxf

1

Recall Shape Functions

li(x): cardinal functions

Quadratures

i

n

ii

b

a

i

n

ii

b

a

b

a

wxfdxxlxfdxxpdxxf

11

It will give correct values for the integral of

every polynomial of degree n-1

Mid-Point Rule is an example of a

quadrature with n=1

Example

Establish coefficients of quadrature for the interval

[-2,2] and nodes –1,0,1

-2 -1 0 1 2

f(x)

ni1 1

n

ijj ji

ji xx

xxxl

Cardinal Functions:

Example

xxxx

xx

xxxl

ijj ji

j

23

11 2

1

11

1

01

0

110

1

10

1 23

12

xxx

xx

xxxl

ijj ji

j

xxxx

xx

xxxl

ijj ji

j

23

13 2

1

01

0

11

1

Cardinal Functions – Lagrange Polynomials:

Example

12

1011

2

1 222 fxxfxfxxxf

ii

iii

i wxfdxxlxfdxxpdxxf

3

1

2

2

3

1

2

2

2

2

xxxl 21 2

1 122 xxl xxxl 2

3 2

1

with

2

2

dxxlw ii

Example

2

2

22

2

22

2

11 3

8

2

1

2

1dxxxdxxxdxxlw

3

41

2

2

22

2

22

dxxdxxlw

3

8

2

1

2

1 2

2

22

2

22

2

33

dxxxdxxxdxxlw

Example

13

80

3

41

3

82

2

fffdxxf

-2 -1 0 1 2

f(x)=x2

3

161

3

80

3

41

3

82

2

2

dxx

Quadrature

3

16

3 2

232

2

2

x

dxx

Exact

Quadratures

So far the placement of nodes has been arbitrary

a=x1 x2 x3 x4 x5 x6 x7=b

They should belong to the interval of integration

Quadrature is accurate for polynomials of degree n-1 (n is the number of nodes)

Gaussian Quadrature

Karl Friedriech Gauss discovered that by a

special placement of nodes the accuracy of the

numerical integration could be greatly increased

Gaussian Quadrature

Theorem on Gaussian nodes

Let q be a polynomial of degree n such that

1-n0,1,...,k 0 dxxxq kb

a

Let x1,x2,…,xn be the roots of q(x). Then

nn

b

a iii xfwxfwxfwxfwdxxf 2211

with xi’s as nodes is exact for all polynomials of

degree 2n-1.

Gaussian Quadrature 2-point

-1 1

f()

311

f()W1=1

312

f()W2=1

3113111

1

ffdxxf

Gauss Points and Weights

Compare

Let f(x)=x2

MidPointMidPoint

1

1

2 32dxx

Use 2-points

X1=-1, X2=0, X3=1

215.0)5.0(1

1

222

dxx

X=-0.5 X=0.5

Compare

Let f(x)=x2

TrapezoidalTrapezoidal

1

1

2 32dxx

Use 2-points

X1=-1, X2=0, X3=1

1)1()0(12

1)0()1(1

2

1 221

1

222

dxx

Compare

Let f(x)=x2

GaussGauss

1

1

2 32dxx

Use 2-points

3231131122

1

1

2

dxx

311 311

1,31 111 wx 1,31 222 wx

2-Dimensional Integration

Gaussian Quadrature

1

1

1

1

, ddf

n

j

n

ijiij fww

1 1

,

1

1 1

, dfwn

iii

2-D Integration 2-point formula

311 312

311

312

11 w

12 w

111 ww 112 ww

122 ww121 ww

2-D Integration 2-point formula

311 312

311

311

1

1

1

1

, ddf

2222212112121111 ,,,, fwwfwwfwwfww

Integration of Stiffness Matrix

1

1

1

1

det dJdt

dAt

T

A

Te

DBB

DBBk

Stiffness Matrix

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

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

Integration over

Integration of Stiffness Matrix

12221121

1121

1222

00

00

det

1

JJJJ

JJ

JJ

JA

(3x4)

10101010

10101010

01010101

01010101

4

1G

(4x8)= B q= AG q

B (3x8)

Integration of Stiffness Matrix

1

1

1

1

det dJdt Te DBBk

B (3x8)

D (3x3)

BT(8x3)

ke (8x8)

Integration of Stiffness Matrix

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

Integration of Stiffness Matrix

311 312

311

311

2222212112121111 ,,,, gwwgwwgwwgww

1

1

1

1

, ddgt

11 w

12 w

22211211 ,,,, gggg

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

Modeling Issues: Degenerate Quadrilaterals

Three nodes collinear

1

2

3

4

xx

xx

1

2

3

4 x

xx

x

Less accurate

Integration Bias

Modeling Issues: Degenerate Quadrilaterals

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

Convergence Considerations

For monotonic convergence of solution

Elements (mesh) must be compatible

Elements must be complete

Requirements

Mesh Compatibility

OK

NO NO!

Mesh compatibility - Refinement

Acceptable Transition

However…Compatibility of displacements OK

Stresses?

Convergence Considerations

We will revisit the issue…