fem Plane

8/8/2019 fem Plane

1/14

M. Vable Notes for finite element method: Intro to FEM 2-D

8-1

FEM in two-dimension

Strain energy density: Uo

1

2---

xx

xx

yy

yy

xy

xy+ +[ ]=

Define: { }

xx

yy

xy

= { }

xx

yy

xy

= Uo

1

2--- { }

T{ }=

Generalized Hookes law

Plane stress (All stresses with subscript z are zero)

Plane strain (All strains with subscript z are zero)

xx

E

xx

yy+[ ]

1 2

( )-------------------------------=

yy

E

yy

xx+[ ]

1 2

( )-------------------------------=

xy

E2 1 +( )--------------------

xy=

xx

E1 ( )

xx

yy+[ ]

1 +( ) 1 2( )------------------------------------------------=

yy

E1 ( )

yy

xx+[ ]

1 +( ) 1 2( )------------------------------------------------=

xy E2 1 +( )-------------------- xy=

Plane Stress Plane Strain

E[ ]E

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

1 0 1 0

0 01 ( )

2----------------

= E[ ]E

1 2( ) 1 +( )--------------------------------------

1 ( ) 0 1 ( ) 0

0 01 2( )

2--------------------

=

Plane StrainPlane Stress

{ } E[ ] { }= E

[ ] E

[ ]

T

=E E 1 2

( ) 1 ( )

Uo

1

2--- { }

T{ } 1

2--- { }

TE[ ]

T{ } 1

2--- { }

TE[ ] { }= = =

8/8/2019 fem Plane

2/14


8-2

Strain-Displacement

xx x

u=

yy yv

= xy y

ux

v+=

{ }

x 0

0y

y

x

u

v

=

Displacements approximating:

u x( ) ui

e( )fi

x y,( )

i 1=

n

= v x( ) vie( )

fi

x y,( )

i 1=

n

=

Define the nodal displacement vector as: d{ }

u1

e( )

v1

e( )

u2e( )

v2

e( )

un

e( )

vne( )

=

u

v f1 0 f2 0 fn 0

0 f1

0 f2

0 fn

d{ }=

8/8/2019 fem Plane

3/14

{ }

x

0

0y

y

x

f1

0 f2

0 fn

0

0 f1 0 f2 0 fnd{ } B[ ] d{ }= =

B[ ]

x

f1

0x

f2

0 x

fn

0

0y

f1 0

y

f2 0 y

fn

y

f1

x

f1

y

f2

x

f2

y

fn

x

fn

=


8-3

Matrix [B] is called strain-displacement matrix

Uo

e( ) 12--- d{ }

TB{ }

TE[ ] B{ } d{ }=

Strain Energy:U e( ) Uoe( ) Vd

V

=

Ue( ) 1

2--- d{ }

TB{ }

TE[ ] B{ } d{ } Vd

V

1

2--- d{ }

TK

e( )[ ] d{ }= =

Element stiffness matrix: Ke( )

[ ] B[ ]T

E[ ] B[ ] Vd

V

=

Variation in potential energy: e( )

d{ }T

Ke( )

[ ] d{ } Re

{ }( )=

8/8/2019 fem Plane

4/14


8-4

Overview of approximate methods

Some Jargon

1-D Heat Conduction:

T

x L=

T

o

=

kxd

dT

x 0=

0=

x2

2

d

d T0= 0 x L Differential Equation

Natural Boundary Condition

Essential Boundary Condition

L

x

Beam Bending

V

xd

dEI

x2

2

d

d v

P= =

v 0( ) 0=

x2

2

d

d EI

x2

2

d

d v

py= 0 x L

xd

dv

x 0=

0=

M EI

x2

2

d

d v0= =

Differential Equation



L

P

py

8/8/2019 fem Plane

5/14


8-5

Approximation of boundary value problem

Deu ge=

Dn

u gn

=

in

on e

on n

Differential Equation



e n

Lu f=

u cj

j

1=

n

=j set of approximating functions

set ofj is complete and independent.

ed

cj

Lj

j 1=

n

f= Error in Differential Equation

Error in Natural Boundary Condition

Error in Essential Boundary Condition

en cjDnjj 1=

n

gn=

ee cjDejj 1=

n

ge=

8/8/2019 fem Plane

6/14


8-6

Commonality and Differences in

Approximate Methods

Commonalities

Produce a set of algebraic equations in the unknown constants cj. Choose i to set one (or two) of the errors ed, ee, or en to zero Minimize the remaining error(s).

Differences

Which error is set to zero

Domain Methods: ee =0 or en =0Boundary Methods: ed=0

Error Minimizing Process

Independence ofi

No i can be obtained from a linear combination of other is inthe set.

If the set of functions i

are not independent then the equations inthe matrix will not be independent and the matrix will be singular.

Completeness ofi

In a series sequence no term should be skipped. If a set is not complete then the solution may not converge forsome problems.

8/8/2019 fem Plane

7/14


8-7

Error Minimization

Weighted Residue

id( )e

dxd yd

i

e( )ee

sd

e

in( )e

nsd

n

+ + 0=

FEM-Stiffness version: ee = 0

i

d( )e

dxd yd

i

n( )e

nsd

n

+ 0=

FEM-Flexibility version: en = 0

i

d( )e

dxd yd

i

e( )e

esd

e

+ 0=

BEM: ed = 0

i

n( )e

nsd

i

e( )e

esd

+ 0=

FEM: Discretization process is on domain of the entire body

BEM: Discretization process is on the boundary of the body

In FEM stiffness matrix the equilibrium equation on stresses (differen-

tial equations) and boundary conditions on stresses (natural boundary

conditions) are approximately satisfied.

In FEM flexibility matrix the compatibility equation on stresses (dif-

ferential equations) and boundary conditions on displacements (essen-

tial boundary conditions) are approximately satisfied.

8/8/2019 fem Plane

8/14


8-8

Constant Strain Triangle (CST)

Displacements are linear in x and y, resulting in constant strains.u a

0a

1x a

2y+ += v b

0b

1x b

2y+ +=

xx x

ua

1= =

yy yv

b2

= = xy y

ux

v+ a

2b

1+= =

1

32

y

x

u x y,( ) Ni

x y,( )ui

e( )

i 1=

3

=

v x y,( ) Ni

x y,( )vi

e( )

i 1=

3

=d

e( ){ }

u1

e

v1

e( )

u2

e( )

v2

e( )

u3

e( )

v3

e( )

=u1

e( )

u2e( )

u3e( )

v1e( )

v3e( )

v2e( )

xx

yy

xy

1

2A-------

y23

0 y31

0 y12

0

0 x32

x13

x21

x32

y23

x13

y31

x21

y12

d e( ){ } B[ ] d e( ){ }= =

xij

xi

xj

= yij

yi

yj

=

Ke( )

[ ] B[ ]T

E[ ] B[ ] Vd

V

B[ ]T

E[ ] B[ ]tA= =

1

32

1

32

1

321

1 1

N1

N2

N3

8/8/2019 fem Plane

9/14


8-9

Pascals Triangle

Used for determining a complete set of polynomial terms in twodimensions.

1

x

x2

x3

x4

x5

x

y2

y

y3

y4

y5

xy

x2y xy2

x3y x2y2 xy3

x4y x3y2 x2y3 xy4

Greater the degree of freedom, less stiff will be element. Interpolation functions are easier to develop with areacoordinates.

8/8/2019 fem Plane

10/14


8-10

Natural Coordinates

Coordinates which vary between 0 and 1 or -1 and 1. Natural coordinates and non-dimensional coordinates.

1-d Coordinatesx

x1x2

Node 1 Node 2

Node 1 Node 2

=1

Node 1 Node 2

=1=1

Possibility 1 Possibility 2

L1 x( )x x

2

x1

x2

-----------------

= L2

x( )x x

1

x2

x1

-----------------

=

L1 ( ) 1 ( )=

L2

( ) =

L1 ( ) 1 ( ) 2=

L2

( ) 1 +( ) 2=2-D Triangular elements (Area Coordinates)

AI

I

J

K

AJ

AK

LI

AI

A------=

LI=0

LI=1

LJ

AJ

A------=

LK

AK

A--------=

LI

LJ

LK

+ + 1=

LK

LJ

8/8/2019 fem Plane

11/14


8-11

Linear Strain Triangle

I

J

K

1

2

3

4

5

6

N1

LI

LI

1 2( )

1( ) 1 2( )-------------------------------- 2L

IL

I1

2---

= =

N2

LIL

J

1 2( ) 1 2( )------------------------------ 4L

IL

J= =

Homework Problem: Write cubic interpolation functions using area coor-

dinates for nodes 1,2 and 10.

I

J

K

1 2

3

4

5

67

8

9

10

8/8/2019 fem Plane

12/14


8-12

Bi-Linear Quadrilateral

u a0

a1x a

2y a

3xy+ + += v b

0b

1x b

2y b

3xy+ + +=

xx xu a

1a

3y+= = yy y

v b2

b3x+= =

xy y

ux

v+ a

2b

1+( ) a

3x b

3y+ += =

1 2

3 4

x

y

a a

b

b

u1e( )

v1

e( )

u2e( )

v2

e( )

u3

e( )v3

e( )v4

e( )

u4

e( )

u x y,( ) Ni

x y,( )ui

e( )

i 1=

4

=

v x y,( ) Ni

x y,( )vi

e( )

i 1=

4

=

Interpolation functions in natural coordinates

x a= y b=

N1

1

4--- 1 ( ) 1 ( )= N

21

4--- 1 +( ) 1 ( )=

N3

1

4--- 1 ( ) 1 +( )= N

41

4--- 1 +( ) 1 +( )=

x

Ni

Nix

1a---

Ni= =

y

Ni

Niy

1b---

Ni= =

8/8/2019 fem Plane

13/14


8-13

Other Quadrilaterals

Complete quadratic

1 2 3

4x

y

a a

b

b

5 6

7 8 9

The stiffness (row and column) related to node 5 is known atelement level and as rows and columns of other elements do notadd to it.

Quadratic element often used in practice:

1 2 3

4x

y

a a

b

b

5

6 7 8

When internal nodes are eliminated care has to be exercised toensure the mesh from such elements will converge.

8/8/2019 fem Plane

14/14


8-14

Mechanical Loads

There are three types of mechanical loads

1. Concentrated Forces or Moments

The loads must be applied at nodes when making the mesh.

Theoretically the stresses are infinite at the point of application, hence

in the neighborhood of concentrated load a large stress gradient can be

anticipated.

2. Tractions

Forces that act on the bounding surfaces.

Sx xxnx xyny+= Sy yxnx yyny+=nx and ny are the direction cosines of the unit normal

Tractions has units of force per unit area and are distributed forces.

Usually the tractions are specified in local normal and tangential coor-

dinates.

These distributed forces must be converted to nodal forces.

In two dimensions the bounding surface is a curve. Distributed forces can

be converted to nodal forces as was done in 1-d axial and bending prob-

lems. (work equivalency) Rj p x( )fj x( ) xd

0

L

=

p

L

pL/2 pL/2 pL/6 pL/62pL/3

Linear Lagrange Quadratic Lagrange

3. Body Forces

Forces that act at each and every point on the body.

Gravity, magnetic, inertial are some examples.

These forces must be converted to nodal forces.

(See section 3.9)

fem Plane

Documents