chapter 10site.iugaza.edu.ps/marafa/files/fem-chapter-10-2017-18m2.pdf · 2017-12-14 · the term...

Finite Element Method

Chapter 10

Isoparametric Formulation

Basic Principle of Isoparametric Elements

Definition:

The term isoparametric (same parameters) is derived from the use of

the same shape (interpolation) functions N to define the element’s

geometric shape as are used to define the displacements within the

element.

The basic principle of isoparametric elements is that the interpolation

functions for the displacements are also used to represent the

geometry of the element.

Alternatively:

4 4

1 1

4 4

1 1

,

,

i i i i

i i

i i i i

i i

u N u v N v

x N x y N y


In this formulation, displacements are expressed in terms of the

natural (local) coordinates and then differentiated with respect to

global coordinates. Accordingly, a transformation matrix [J], called

Jacobian, is produced.

If the geometric interpolation functions are of lower order than the

displacement shape functions, the element is called

subparametric. If the reverse is true, the element is referred to as

superparametric.

The isoparametric formulation is generally applicable to 1-, 2-

and 3- dimensional stress analysis. The isoparametric family

includes elements for plane, solid, plate, and shell problems. Also,

it is applicable for nonstructural problems.


The isoparametric formulation makes it possible to generate

elements that are nonrectangular and have curved sides. So it

can facilitate an accurate representation of irregular elements.

Numerous commercial computer programs have adopted this

formulation for their various libraries of elements.

Two-Noded Bar Isoparametric Element

Basic Principle of Isoparametric Elements in 2D

As shown in the figure, the local (natural) coordinate system (,) for

the two elements have their origins at the centroids of the elements,

with (,) varying form –1 to 1. The natural coordinate system needs

not to be orthogonal and neither has to be parallel to the x-y axes.

The coordinate transformation will map the point (,) in the master

element to x(,) and y(,) in the slave element.

s

t

1 11

1

s

t

y

x

Examples

s

t

1

1

y

x

s

t

Isoparametric Formulation of Two-

Noded Bar Element Stiffness Matrix

Isoparametric Formulation of the Bar

Element Stiffness Matrix

Step 1: Select the Element Type

We consider the bar element to have two degrees of freedom s.

For the special case when the s and x axes are parallel to each other,

the s and x coordinates can be related by

where xc is the global coordinate of the element centroid.

Using the global coordinates x1 and x2 in

1 2

2 1

2

2

x xs x

x x

2c

Lx x s

Step 2: Select Displacement Functions

We begin by relating the natural coordinate to the global coordinate by

This linear shape functions map the s coordinate of any point in the element

to the x coordinate

1 2

1

1 2 1 2

2

1 1

11 1

2

1 1,

2 2

s

x s x s x

x s sx N N N N

x


The displacement function within the bar is now defined by the same

shape functions

Step 3: Define the Strain/Displacement and Stress/Strain Relationships

1

1 2 1 2

2

1 1,

2 2

u s su N N N N

u

Using Chain R le

2

1 2

u

x

x

u

x

u u x

s x s

u x Lu xJ

s sx s

u u u

x J s L s


2 1

1

2

2

2 1 1

since

1 1

x

x

x

u uu

s

uu u

ux L s L L

B d

BL L

E E B d

Step 4 Derive the Element Stiffness Matrix

0

1

10

1 1

1 1

[ ] [ ][ ]

J:Jacobian Matrix

2

[ ] [ ] [ ][ ] [ ] [ ][ ]

1 1[ ]

2

1 1

L

T

L

T T

k B D B Adx

f x dx f s J ds

x LJ

s

Lk B D B A J d

EAk

s B D B A ds

L

Body Forces

1

0 1

1

1

[ ] { }

[ ] { } [ ] { }

1

12{ }

1 12 2

2

T

b b

V

L

T T

b b b

bb b

f N X dV

f A N X dx A N X J ds

s

ALXLf A X ds

s

Surface Forces

[ ] { }T

s x

S

f N T dS

Assuming the cross section is constant and the traction is uniform

over the perimeter and along the length of the element, we obtain

1

0 1

1

1

[ ] { } [ ] { } [ ] { }

1

12{ }

1 12 2

2

L

T T T

s s x x x

S

b x x

f f N T dS N T dx N T J ds

s

L Lf T ds T

s

Isoparametric Formulation of the

Three-Noded linear strain bar

Three-Noded linear strain isoparametric bar

Three-noded linear strain isoparametric bar

1

1 2 3 2

3

2

1 2 3

1 1, , 1

2 2

2

1 2

x

x

x N N N x

x

s s s sN N N s

u

x

u u x x LJ

s x s s

u u u

x J s L s

Three-noded linear strain bar isoparametric element

13

, 2

1

3

1

2

3

2 1 2 12

2 2

2 2 1 2 1 4

2 1 2 1 4

i s i

i

uu s s

N u s us

u

uu u s s s

ux L s L L L

u

s s sB

L L L

1

1[ ] [ ] [ ][ ]

2

Tk B D B A J ds

LJ

The Stiffness Matrix of three-noded linear strain bar

The Stiffness Matrix of three-noded linear strain bar

The Stiffness Matrix of Three-Noded linear strain bar

Numerical Integration

Gauss Quadrature

Numerical Integration

Gauss Quadrature

• Gauss quadrature implements a strategy of

positioning any two points on a curve to define a

straight line that would balance the positive and

negative errors.

• Hence, the area evaluated under this straight line

provides an improved estimate of the integral.

Two points Gauss-Legendre

Formula• Assume that the two Integration points are xo and x1 such that:

• c0 and c1 are constants, the function arguments x0 and x1 are

unknowns…….(4 unknowns)

1

0 0 1 1

1

( ) ( )I f x dx c f x c f x

Two points Gauss-Legendre

Formula

• Thus, four unknowns

to be evaluated require

four conditions.

• If this integration is

exact for a constant,

1st order, 2nd order, and

3rd order functions:

1

1

3

1100

1

1

2

1100

1

1

1100

1

1

1100

0)()(

3

2)()(

0)()(

21)()(

dxxxfcxfc

dxxxfcxfc

dxxxfcxfc

dxxfcxfc

Two points Gauss-Legendre Formula

• Solving these 4 equations, we can determine c0, c1, x0 and x1.

3

1

3

1ffI

0 1

0

1

The weighting factors are: 1

The Integration points are:

10.5773503

3

10.5773503

3

c c

x

x

Two points Gauss-Legendre Formula

• Since we used limits for the previous integration from –1 to 1 and the actual limits are usually from a to b, then we need first to transform both the function and the integration from the x-system to the xd-system

1ax

1bx

d2

abdx

2

ab

2

abx

f(x)

x

a b

f(xo)

f(x1)

xo x1

-1 1

f()

Higher-Points Gauss-Legendre Formula

1

11

( ) ( )n

i i

i

I f d W f

Weight Integration point

1

11

1 1 2 2

( ) ( )

( ) ( ) ( )

n

i i

i

n n

I f d W f

I W f W f W f

Multiple Points Gauss-LegendrePoints Weighting factor Function argument Exact for

2 1.0 -0.577350269=−1/ 3 up to 3rd

1.0 0.577350269= 1/ 3 degree

3 0.5555556=5/9 -0.774596669=−1/ 0.6 up to 5th

0.8888889=8/9 0.0 degree0.5555556=5/9 0.774596669= 1/ 0.6

4 0.3478548 -0.861136312 up to 7th

0.6521452 -0.339981044 degree0.6521452 0.3399810440.3478548 0.861136312

6 0.1713245 -0.932469514 up to 11th

0.3607616 -0.661209386 degree0.4679139 -0.2386191860.4679139 0.2386191860.3607616 0.6612093860.1713245 0.932469514

Example 1

Evaluate the stiffness matrix for the three-noded bar element using

two-point Gaussian quadrature.

Three–noded bar with two Gauss points

We have for the stiffness matrix

Using two-point Gaussian quadrature, we evaluate the stiffness matrix at the two

points shown :

Example 1

Multiple Integration

dydxyxfI

d

cy

b

ax

),(

• Double integral:

AA

ddJfdydxyxf ),(),(

In General transformation to natural

coordinate:

1 1

1 1( , )I f d d

1 1

1 11 1

( , ) ( , )n n

ij i j

i j

I f d d W f

Where Wij =Wi Wj

2 21 1

1 11 1

( , ) ( , )

1 1 1 1 1 1 1 1( , ) ( , ) ( , ) ( , )

3 3 3 3 3 3 3 3

ij i j

i j

I f d d W f

f f f f

For n=2 Wij =Wi Wj=1

1

1

1 1

3

1,

3

1

3

1,

3

1

3

1,

3

1

3

1,

3

1

Shape function for Isoparametric 6-Nodes triangular Elements

1

2

3

4

5

6

2 2 1 1

2 1

2 1

4 (1 )

4

4 (1 )

N

N

N

N

N

N

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

ts

tsf 1~),(

3

1,

3

1

2

1fI

s

t

1

11/3

1/3

Gauss Points for Isoparametric triangular Elements

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

1/2

21

3

Gauss Points for Isoparametric triangular Elements

Isoparametric Formulation of the Plane

Quadrilateral Element Stiffness Matrix


In other words, we look for shape functions that map the regular

shape element in isoparametric coordinates to the quadrilateral

in the x-y coordinates whose size and shape are determined by

the eight nodal coordinates x1, y1, x2, y2, ….., x4, y4.

8765

4321

),(

),(

aaaay

aaaax


1 2 3 4

5 6 7 8

1

2

8

( , )

( , )

( , ) 1 0 0 0 0

( , ) 0 0 0 0 1

a a a ax

y a a a a

a

ax

y

a

1 1 1 1

2 2 2 2

3 3 3 3

4 4 4 4

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 11

1 1 1 1 1 1 1 14

1 1 1 1 1 1 1 1

x a a x

x a a x

x a a x

x a a x


])1)(1()1)(1()1)(1()1)(1[(4

1

1111

1111

1111

1111

4

111),(

4321

4

3

2

1

4

3

2

1

xxxx

x

x

x

x

a

a

a

a

x

i

i

i

i

i

i

yN

xN

y

x

y

x

y

x

y

x

NNNN

NNNN

y

x4

1

4

1

4

4

3

3

2

2

1

1

4321

4321

0000

0000

),(

),(


)1)(1(4

1,)1)(1(

4

1

)1)(1(4

1,)1)(1(

4

1

43

21

NN

NN

These shape functions are seen to map the (,) coordinates of

any point in the rectangular element in the above master element

to x and y coordinates in the quadrilateral (slave) element.

For example, consider the coordinates of node 1, where:

=-1,=-1 using the above equation, we get x=x1 , y=y1

Shape Function for 4-Nodes quadrilateral Elements

),,,2,1(,11

niNn

i

i

where n = the number of shape functions associated with

number of nodes



nodesotherallat

inodeatN i

0

1


i

i

i

i

i

i

vN

uN

v

u

v

u

v

u

v

u

NNNN

NNNN

v

u4

1

4

1

4

4

3

3

2

2

1

1

4321

4321

0000

0000

),(

),(

][][ dNv

u

where u and v are displacements parallel to the global x and y

coordinates

Step 3: Define the Strain/Displacement and Stress/Strain

Relationships

y

y

fx

x

ff

y

y

fx

x

ff

ff x y

x

f x y f

y

Using Chain Rule


1

N fN f

x xJ J

N N f f

y y

NN x y

x

N x y N

y

Can be

computed

We want to compute

these for the B matrix

This is known as the

Jacobian matrix (J) for the

mapping (,) → (x,y)

Define the Strain/Displacement and Stress/Strain Relationships

i

i

ii

i

i

i

i

ii

i

i

yN

xN

yN

xN

yx

yx

J4

1

4

1

4

1

4

1][

1 1[ ]

where

y y

Jx xJ

x y x yJ

1

1

1

1

N N

xJ

N N

y

N y y N

x

N x x NJ

y

N y N y N

x J

N x N x N

y J

Since:


{ }

( ) ( )0

1 ( ) ( )0

( ) ( ) ( ) ( )

{ } [ ] [

x

y

xy

x

y

xy

u

x

v

y

u v

y x

y y

ux x

vJ

x x y y

uD

v

][ ][ ]D N d


(3 8) (3 2) (2 8)

{ } [ ][ ][ ]

{ } [ ][ ]

( ) ( )0

1 ( ) ( )[ ] 0

( ) ( ) ( ) ( )

[ ] [ ] [ ]

D N d

B d

y y

x xD

J

x x y y

B D N


Derive the Element Stiffness Matrix and Equations

1 1

1 1

[ ] [ ] [ ][ ]

( , ) ( , )

[ ] [ ] [ ][ ]

T

A

A A

T

k B D B t dx dy

f x y dx dy f J d d

k B D B t J d d

The shape function are:

)1(4

1,)1(

4

1,)1(

4

1,)1(

4

1

)1(4

1,)1(

4

1,)1(

4

1,)1(

4

1

4321

4321

NNNN

and

NNNN

)1)(1(4

1,)1)(1(

4

1

)1)(1(4

1,)1)(1(

4

1

43

21

NN

NN


Their derivatives:

i

i

ii

i

i

i

i

ii

i

i

yN

xN

yN

xN

yx

yx

J4

1

4

1

4

1

4

1][


i

i

ii

i

i

i

i

ii

i

i

yNy

JxNx

J

yNy

JxNx

J

4

1

22

4

1

21

4

1

12

4

1

11

,

,

44

33

22

11

,4,3,2,1

,4,3,2,1

2221

1211][

yx

yx

yx

yx

NNNN

NNNN

JJ

JJJ


22 1 2 3 4

12 1 2 3 4

11 1 2 3 4

21 1 2 3 4

11 1 1 1

4

11 1 1 1

4

11 1 1 1

4

11 1 1 1

4

J y y y y

J y y y y

J x x x x

J x x x x


11 12

21 22

11 22 12 21

[ ]J J

JJ J

J J J J J

1

11 12 2

1 2 3 4

21 22 3

4

0 1 1

1 0 11

1 0 18

1 1 0

y

J J yJ x x x x

J J y

y

Explicit formulation for |J| for 4 node Element


][][][ NDB

22 12

1 2 3 4

11 21

1 2 3 4

11 21 22 12

( ) ( )0

0 0 0 01 ( ) ( )[ ] 0

0 0 0 0

( ) ( ) ( ) ( )

J J

N N N NB J J

N N N NJ

J J J J

,12,22,21,11

,21,11

,12,22

0

0

][

iiii

ii

ii

i

NJNJNJNJ

NJNJ

NJNJ

B


1 2 3 4

1B B B B B

J


22 , 12 ,

11 , 21 ,

11 , 21 , 22 , 12 ,

0

[ ] 0

i i

i i i

i i i i

J N J N

B J N J N

J N J N J N J N



22 1 2 3 4

12 1 2 3 4

11 1 2 3 4

21 1 2 3 4

11 1 1 1

4

11 1 1 1

4

11 1 1 1

4

11 1 1 1

4

J y y y y

J y y y y

J x x x x

J x x x x

In the our text Book



A

T

V

Tb dydxtXNdVXNf }{][}{][}{

The element body force matrix

1 1

1 1{ } [ ] { }

(8 1) (8 2) (2 1)

T

bf N X t J d d

b

b

Y

XX}{


and Xb and Yb are the weight densities (body weight/unit volume)

in the x and y directions, respectively.

L

T

S

Ts dxtTNdSTNf }{][}{][}{

1

1{ } [ ] { }

(4 1) (4 2) (2 1)

T

sf N T t J d

1

143

43

4

4

3

3

200

00

dL

tp

p

NN

NN

f

f

f

fT

s

s

s

s

Note that for one-dimensional transformation | J | = L / 2. Also,

p, p are the pressure distributions in and , respectively.

The element surface force matrix


We assumed surface loading at edge with overall length

L (Since N1 = 0 and N2 = 0 along edge =1, and hence,

no nodal forces exist at nodes 1 and 2,

1 13

4

12

1

1

Problem: Consider the following isoparamteric map

y

34

1 2(3,1) (6,1)

(6,6)(3,6)

xISOPARAMETRIC COORDINATES

GLOBAL COORDINATES

1 1 2 2 3 3 4 4

1 1 2 2 3 3 4 4

u N u N u N u N u

v N v N v N v N v

Displacement interpolation

Shape functions in isoparametric coordinate system

)1)(1(4

1,)1)(1(

4

1

)1)(1(4

1,)1)(1(

4

1

43

21

NN

NN

1 1 2 2 3 3 4 4

1 1 2 2 3 3 4 4

( , ) ( , ) ( , ) ( , )

( , ) ( , ) ( , ) ( , )

1 1(1 )(1 ) 3 (1 )(1 ) 6

4 4

1 1 3(3 )(1 )(1 ) 6 (1 )(1 ) 3

4 4 2

1 1(1 )(1 ) 1 (1 )(1 ) 1

4 4

1 1(1 )(1 ) 6

4 4

x N x N x N x N x

y N y N y N y N y

x

y

7 5

(1 )(1 ) 62

The isoparamtric mapping

The Jacobian matrix

NOTE: The diagonal terms are due to stretching of the sides along the

x-and y-directions. The off-diagonal terms are zero because the

element does not shear.

30

2

50

2

x y

Jx y

3(3 )

2

7 5

2

x

y

since

1 3 / 2 0 2 / 3 01 15

and0 5 / 2 0 2 / 515 / 4 4

J J

Hence, if I were to compute the columns of the B1 matrix along the

positive x-direction

Hence

22 1, 12 1,

1 11 1, 21 1,

11 1, 21 1, 22 1, 12 1,

0

[ ] 0

J N J N

B J N J N

J N J N J N J N

1

5 15 1 00 082 4

3 13 1[ ] 0 0 0

2 4 8

3 1 5 1 3 1 5 10 0

2 4 2 4 8 8

B

Example1

For the four-noded linear plane element shown with a uniform surface

traction along side 2–3, evaluate the force matrix by using the energy

equivalent nodal forces. Let the thickness of the element be h=0.1 in.

Example1

Shape functions N2 and N3 must be used,

as we are evaluating the surface traction

along side 2–3 (at s = 1).

Example1

evaluated along s = 1

Example1

Flowchart to evaluate k by four-point Gaussian quadrature

Evaluate the stiffness matrix for the

quadrilateral element shown in Figure

using the four-point Gaussian quadrature

rule.

Let E = 30×106 psi, = 0.25 and h=1 in.

Example 2

we evaluate the k matrix. Using the four-point rule, the four points are:

Solution

1 1

2 2

3 3

4 4

, 0.5773, 0.5773

, 0.5773,0.5773

, 0.5773, 0.5773

, 0.5773,0.5773

1 2 3 4 1.0W W W W

Example 2

Example 2

22 1, 12 1,

1 11 1, 21 1,

11 1, 21 1, 22 1, 12 1,

0

[ ] 0

J N J N

B J N J N

J N J N J N J N

22 1 2 3 4

22

12 1

1

1 21

,

1,

11 1 1 1

4

12 0.5773 1 2 1 0.5773 4 1 0.5773 4 1 0.5773 1.0

4

with similar computations used to obtain , . Also,

1 1(1 ) (1 0.577

and

3) 0.39434 4

1(1

4

J J J

J y y y y

J

N

N

1

) (1 0.5773) 0.39434

Example 2

Example 3 For the rectangular element shown previous

Example, assume plane stress conditions

Let E = 30×106 psi, = 0.3 and displacements:

u1 = 0 , v1 = 0

u2 = 0.001 , v2 = 0.0015

u3 = 0.003 , v3 = 0.0016

u4 = 0 , v4= 0

Evaluate the stresses at s=0 , t=0

Solution

Example 3

22 , 12 ,

11 , 21 ,

11 , 21 , 22 , 12 ,

22 1 2 3 4

22

12 11 21

0

[ ] 0

11 1 1 1

4

12 0 1 2 1 0

Simili

4 1 0 4 1 0 14

0,arly 1, 0

i i

i i i

i i i i

J N J N

B J N J N

J N J N J N J N

J y y y y

J

J J J

1, 2, 3, 4,

1, 2, 3, 4,

1 1 1 1 1 1(1 ) (1 0) ,Similarly

Similarly

and4 4 4 4 4 4

1 1 1 1 1 1(1 ) (1 0) , and

4 4 4 4 4 4

N N N N

N N N N

Example 3

Higher-Order Shape Functions

In general, higher-order element shape functions can

be developed by adding additional nodes to the sides

of the linear element.

These elements result in higher-order strain variations

within each element, and convergence to the exact

solution thus occurs at a faster rate using fewer

elements.

Another advantage of the use of higher-order

elements is that curved boundaries of irregularly

shaped bodies can be approximated more closely than

by the use of simple straight-sided linear elements.

or, in compact index notation, we express

where i is the number of the shape function

Shape function of a quadratic isoparametric

element

Shape function of a cubic isoparametric element

HW: 10.6, 10.8, 10.15 and 10.21

chapter 10site.iugaza.edu.ps/marafa/files/fem-chapter-10-2017-18m2.pdf · 2017-12-14 · the term...

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