FEM FOR 2D SOLIDS

(1)

Finite Element Method

(2)

CONTENTS

 INTRODUCTION

 _{LINEAR TRIANGULAR ELEMENTS}

– Field variable interpolation

– Shape functions construction

– Using area coordinates

– Strain matrix

– Element matrices

 _{LINEAR RECTANGULAR ELEMENTS}

– Shape functions construction

– Strain matrix

– Gauss integration

(3)

INTRODUCTION



2D solid elements are applicable for the analysis

of plane strain and plane stress problems.



A 2D solid element can have a

triangular

,

rectangular

or

quadrilateral

shape with

straight

or

curved

edges.



A 2D solid element can deform only in the plane

of the 2D solid.



At any point, there are

two components

in the

x

and

y

directions for the displacement as well as

(5)

INTRODUCTION



For plane strain problems, the thickness of the

element is

unit

, but for plane stress problems, the

actual thickness

must be used.



In this course, it is assumed that the element has a

uniform

thickness

h

.



Formulating 2D elements with a given variation of

(6)

2D solids

–

plane stress and plane strain

(7)

LINEAR TRIANGULAR

ELEMENTS



Less accurate than quadrilateral elements



Used by most mesh generators for complex

geometry



A linear triangular element:

y, v

2 (x2, y2) (u2, v2) 3 (x3, y3)

(u3, v3)

A

fsx

(8)

Field variable interpolation

( , )

h

e

x y



x y

U

N

d

3 node at nts displaceme 2 node at nts displaceme 1 node at nts displaceme 3 3 2 2 1 1                              v u v u v u e d 3 1 2 3 1 2 Node 2

Node 1 Node 3

0 0 0 0 0 0 N N N N N N    _ _   N where x, u y, v

1 (x1, y1) (u1, v1)

2 (x2, y2) (u2, v2) 3 (x3, y3)

(u3, v3)

A

fsx

fsy

(9)

Shape functions construction

1 1 1 1

N

 

a

b x c y



2 2 2 2

N  a b x c y

3 3 3 3

N

 

a

b x c y



i i i i

N  a b x c y

Assume,

i= 1, 2, 3



1



T

i

i i

i a

N x y b

c

   

 _{ } 

   

p



(10)

Delta function property:

1 for

( , )

0 for

i j j

i j N x y

i j  

  _



1 1 1

1 2 2

1 3 3

( , ) 1

( , ) 0

N x y N x y N x y

  

Therefore, 1 1 1 1 1 1 1 1

1 2 2 1 1 2 1 2 1 3 3 1 1 3 1 3

( , ) 1

( , ) 0

N x y a b x c y

   

Solving, 2 3 3 2 2 3 3 2

1 , 1 , 1

2 _e 2 _e 2 _e

x y x y y y x x

a b c

A A A

  

(11)

1 1

2 2 2 3 3 2 2 3 1 3 2 1

3 3

1

1 1 1

1 [( ) ( ) ( ) ]

2 2 2

1

e

x y

A x y x y x y y y x x x y

x y

 P       

Area of triangle Moment matrix

Substitute a₁, b₁and c₁back into N₁ = a₁ + b₁x + c₁y:

1 2 3 2 3 2 2

1

[( )( ) ( )( )]

2 _e

N y y x x x x y y

A

(12)

Shape functions construction

Similarly,

2 1 1

2 2 2

2 3 3

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

N x y N x y N x y

  

2 3 1 1 3 3 1 1 3

3 1 3 1 3 3

1

[( ) ( ) ( ) ]

2 1

[( )( ) ( )( )]

2

e

N x y x y y y x x x y

A

y y x x x x y y

A

     

     

3 1 1

3 2 2

3 3 3

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

N x y

  

3 1 2 1 1 1 2 2 1

1 2 1 2 1 1

1

[( ) ( ) ( ) ]

2 1

[( )( ) ( )( )]

2

e

N x y x y y y x x x y

A

y y x x x x y y

A

     

(13)

i i i i

N

 

a

b x c y



1

( )

2 1

( )

2 1

( )

2

i j k k j

e

i j k

e

i k j

e

a x y x y

A

b y y

A

c x x

A

 

where

i

j k

i= 1, 2, 3

J, k determined from cyclic

permutation

i = 1, 2

(14)

Using area coordinates



Alternative method of constructing shape

functions

i,1

j, 2

k, 3

x y

P

A1

1 2 2 2 3 3 2 2 3 3 2

3 3

1

1 1

1 [( ) ( ) ( ) ]

2 2

1

x y

A x y x y x y y y x x x y

x y

      

1 1

e

A L

A

  2-3-P:

Similarly,  3-1-P A₂

 1-2-P A₃

2 2

e

A L

A 

3 3

e

A L

A

(15)

Using area coordinates

1 2 3

1

L



L



L



Partitions of unity:

3 2 2 3

2 2

1 2 3

1

e e e e

A

L

A









Delta function property: e.g. L₁ = 0 at if P at nodes 2 or 3

Therefore,

1 1

,

2 2

,

3 3

N



L

N



L

N



L

( , )

h

x y



x y

(16)

16

Strain matrix

xx yy xy u x v y u v y x               

LU





where 0 0 x y y x  _            _     _ _  _ _      L e e

Bd

LNd

LU





0 0 x y y x  _           _{ } _        _ _     

B LN N

1 2 3

1 1 2 2 3 3

0 0 0

a a a

b b b

b a b a b a

           B 

(17)

Element matrices

0

d ( d ) d d

e e e

h

T T T

e

V A A

V z A h A







 





k B cB B cB B cB

Constant matrix

 T

e



hA

e

k

B cB

0

d d d d

e e e

h

T T T

e

V A A

V x A h A









 





(18)

Element matrices

1 1 1 2 1 3

2 1 2 2 2 3

3 1 3 2 3 3

0 0 0

d

0 0 0

e

A

N N N N N N

h A

N N N N N N

          _ _          



m

For elements with uniform density and thickness,

A p n m p n m A L L L n p

A m 2 )! 2 ( ! ! ! d 3 2

1  _ _ _



(19)

Element matrices

                     2 0 2 . 1 0 2 0 1 0 2 1 0 1 0 2 0 1 0 1 0 2 12 sy hA e  m x, u y, v

1 (x1, y1)

(u1, v1)

2 (x2, y2)

(u2, v2)

3 (x3, y3)

(u3, v3)

A fsx fsy l f f l sy sx

e [ ] ₂ ₃ d

T



_      N f                _ y x e f f f l 0 0 2 1 3 2 f

(20)

LINEAR RECTANGULAR

ELEMENTS



Non-constant strain matrix



More accurate representation of stress and strain

(21)

x, u y, v

1 (x1, y1) (u1, v1)

2 (x2, y2) (u2, v2)

3 (x₃, y₃) (u₃, v₃)

2a

fsy

fsx

4 (x4, y4) (u4, v4)

2b





Consider a rectangular element

1 1 2 2 3 3 4 4

displacements at node 1

e

u v u u u u u u

   _   _     

        

   _   _     

     _

(22)

Shape functions construction

x, u y, v

1 (x1, y1)

(u1, v1)

2 (x2, y2)

(u2, v2)

3 (x3, y3)

(u3, v3)

2a

fsy

fsx

4 (x4, y4)

(u4, v4)

2b





1 (1, 1) (u1, v1)

2 (1, 1) (u2, v2)

3 (1, +1) (u3, v3)

2a

4 (1, +1) (u4, v4)

2b





, y b

a

x 

 



( , ) ( , )

h

e

x y  x y

U N d 1 2 3 4

3

1 2 4

Node 2 Node 3

Node 1 Node 4

0

0 0 0

0

0 0 0

N

N N N

N

N N N

 

  

 

N

where

(23)

) 1 )( 1 ( ) 1 )( 1 ( ) 1 )( 1 ( ) 1 )( 1 ( 4 1 4 4 1 3 4 1 2 4 1 1                     N N N N 1 1

3 at node 1 4 ₁

1

1 3 _{at node 2} 4

1 1

1 3 _{at node 3} 4

1 1

1 3 _{at node 4} 4

1

(1 )(1 ) 0

(1 )(1 ) 1

(1 )(1 ) 0

N N N N                                         Delta function property 4

1 2 3 4

1

1_[(1 ₎₍₁ _{) (1} ₎₍₁ _{) (1} ₎₍₁ _{) (1} ₎₍₁ _)]

i i

N N N N N

                        



Partition of

)

1

)(

1

(

4

1





j j

j

N





1 (1, 1) (u1, v1)

2 (1, 1) (u2, v2)

3 (1, +1) (u3, v3)

2a

4 (1, +1) (u4, v4)

2b



(24)

Strain matrix

                                    a b a b a b a b b b b b a a a a                 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 LN B

(25)

Element matrices

, y b

a

x 

 

  dxdy = ab dd

Therefore,

 d d

d 1 T

1 1 1 T

cB B

k h A abh

A

e



 

 

 

  

 

 d d d d 1 d d

1 1 1

0 N N N N N N

N N

m T T

A T

A h T

V

e



V



x A



h A

 

abh



 



 

(26)

Element matrices

l f f l sy sx

e [ ] ₂ ₃ d

T



_      N f x, u y, v

1 (x1, y1)

(u₁, v₁)

2 (x2, y2)

(u₂, v₂) 3 (x3, y3)

(u3, v3)

2a

fsy

fsx

4 (x₄, y₄) (u₄, v₄)

2b



 For uniformly distributed load,

(27)

Gauss integration

 For evaluation of integrals in k_e and m_e (in practice)

In 1 direction: ( )d ( )

1 1

1 j j

m

j

f w f

I









 

 



m gauss points gives exact solution of

polynomial integrand of n = 2m - 1

1 1 1 1

1 1

( , )d d ( , )

y

x n

n

i j i j

i j

I   f     w w f  

 

 



 



 

(28)

Gauss integration

m j wj Accuracy n

1 0 2 1

2 -1/3, 1/3 1, 1 3

3 -0.6, 0, 0.6 5/9, 8/9, 5/9 5 4 -0.861136, -0.339981,

0.339981, 0.861136

0.347855, 0.652145, 0.652145, 0.347855

7 5 -0.906180, -0.538469, 0,

0.538469, 0.906180

0.236927, 0.478629, 0.568889, 0.478629, 0.236927

9

6 -0.932470, -0.661209, -0.238619, 0.238619, 0.661209, 0.932470

0.171324, 0.360762, 0.467914, 0.467914, 0.360762, 0.171324

(29)

(30)

Evaluation of

m

e E.g. ) 1 )( 1 ( 4 ) 1 )( 1 ( ) 1 )( 1 ( 16 3 1 3 1 1 1 1 1 1 1 1 1 j i j i j i j i j i ij hab d d hab d d N N hab m



















        



 

        9 4 ) 1 1 1 )( 1 1 1 ( 4 3 1 3 1 33 hab hab

m 



       



(31)

LINEAR QUADRILATERAL

ELEMENTS



Rectangular elements have limited application



Quadrilateral elements with unparallel edges are

more useful



Irregular shape requires coordinate mapping

(32)

Coordinate mapping

2 (x2, y2)

y

x 1 (1, 1) 2 (1, 1)

3 (1, +1) 4 (1, +1)

 

3 (x₃, y₃) 4 (x4, y4)

1 (x₁, y₁)

Physical coordinates Natural coordinates

( , ) ( , )

h

e

 



 

U N d (Interpolation of displacements)

( , )

 



( , )

 

_e

(33)

Coordinate mapping

( , )

 

 ( , )

 

_e

X N x

where x

y        X _, 1 1 2 2 3 3 4 4

coordinate at node 1

e x y x y x y x y    _   _                  _   _           _ x ) 1 )( 1 ( ) 1 )( 1 ( ) 1 )( 1 ( ) 1 )( 1 ( 4 1 4 4 1 3 4 1 2 4 1 1                     N N N N i i i x N

x ( , )

4 1  



  y N

y ( , )

4

 



(34)

Coordinate mapping

Substitute 1 into _i _i

i

x N

x ( , )

4 1  



 

2 (x2, y2)

y

x 1 (1, 1) 2 (1, 1)

3 (1, +1) 4 (1, +1)



3 (x3, y3)

4 (x4, y4)

1 (x1, y1)

3 2 1 2 2 1 3 2 1 2 2 1 ) 1 ( ) 1 ( ) 1 ( ) 1 ( y y y x x x             or ) ( ) ( ) ( ) ( 2 3 2 1 3 2 2 1 2 3 2 1 3 2 2 1 y y y y y x x x x x          

Eliminating , { ( )} ( )

) ( ) ( 3 2 2 1 3 2 2 1 2 3 2 3 y y x x x y y x x

y    

(35)

Strain matrix

                              y y N x x N N y y N x x N N i i i i i

i i _i

i i N _N x N N y      _ _  _  _ _ _   _ _ _     _ _  _  __ _ __   J or x y x y         _ _         _ _    J

where (Jacobian matrix)

1 1

3

1 2 4

2 2

x y N

N N N

x y                _ _ _ _               J

(36)

Strain matrix

1

i i

i _i

N N

x

N _N

y









 



 

 

 _  _

 

  

 __ _

 

 _  _

  _ _

J

Therefore,

N LN

B

  

 

  

 

  



  

  

x y

y x

0

Replace differentials of N_i w.r.t. x and y

with differentials of N_i w.r.t.  and 

(37)

Element matrices

Murnaghan (1951) : dA=det |J | dd

1 1

T

1 1

det

d d

e

h

 

 

 



 

k

B cB

J

(38)

Remarks

 Shape functions used for interpolating the coordinates are

the same as the shape functions used for interpolation of the displacement field. Therefore, the element is called an

isoparametric element.

 _{Note that the shape functions for coordinate interpolation}

and displacement interpolation do not have to be the same.

 _{Using the different shape functions for coordinate}

interpolation and displacement interpolation, respectively,

will lead to the development of so-called subparametric or

(39)

HIGHER ORDER ELEMENTS

 Higher order triangular elements

i (I,J,K) (0,0,p)

L₁

L₃ L2

(0,p1,1) (0,1,p1)

(1,0,p1) (2,0,p2)

n_d= (p+1)(p+2)/2

I

  

J

K

p

Node i,

Argyris, 1968 :

1 2 3

( ) ( ) ( )

I J K

i I J K

N



l L l

L l

L

0 1 ( 1)

( )( ) ( )

( )

( )( ) ( )

L L L L L L

l_ L_           

(40)

HIGHER ORDER ELEMENTS

 Higher order triangular elements (Cont’d)

x, u y, v

1

2 3

4 5 6

1 2 2 (2 1 1) 1

N  N  N  L  L

4 5 6 4 1 2

N  N  N  L L

x, u y, v

1

2 3

4 5

6 7 8

9 ₁₀

1 2 3 1 1 1

1

(3 1)(3 2) 2

N N N  L  L  L

4 9 1 2 1

9

(3 1) 2

N N  L L L 

10 27 1 2 3

N  L L L

Cubic element

(41)

 Higher order rectangular elements

0





(0,m) (n,m)

i(n,m)

Lagrange type:

1 1

( ) ( )

D D n m

i I J I J

N



N N



l



l



0 1 1 1

( )( ) ( )( ) ( )

( )

( )( ) ( )( ) ( )

n k k n

k

k k k k k k k n

l                 _  _  

    



    

(42)

42

 Higher order rectangular elements (Cont’d)

1 2 3 4   5 6 7 8 9 1 1

1 1 1

1 1

2 2 1

1 1

3 2 2

1 1

4 1 2

1

( ) ( ) (1 ) (1 )

4 1

( ) ( ) (1 ) (1 )

4 1

( ) ( ) (1 )(1 )

4 1

( ) ( ) (1 )(1 )

4

D D

N N N

                                          1 1 5 3 1

1 1 6 2 3 1 1 7 3 2 1 1 8 1 1

1 1 2 2

9 3 3

1

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

1

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

1

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

1

( ) ( ) (1 )(1 ) 2

( ) ( ) (1 )(1 )

D D

N N N

N N N N N N

                                                    

(43)

HIGHER ORDER ELEMENTS

Serendipity type:

1 2

3

4 



5

6 7

8 0





=1

=1

1 4

2 1

2

2 1

2

(1 )(1 )( 1) 1, 2, 3, 4 (1 )(1 ) 5, 7

(1 )(1 ) 6, 8

j j j j j

j j

N j

          

  

     

   

(44)

1 2

3 4





5 6

7 8 9

10

11

12

2 2

1 32

2 9

32

1 3 2

9 32

(1 )(1 )(9 9 10)

for corner nodes 1, 2, 3, 4

(1 )(1 )(1 9 )

for side nodes 7, 8, 11, 12 where 1 and

(1 )(1 )(1

j j j

j j

N

j

N

j

N

          

 

  

    



   

    

   

1 3

9 )

for side nodes 5, 6, 9, 10 where and 1

j

j j

j

 

 

    

(45)

ELEMENT WITH CURVED

EDGES

4 3

8

7

6 1

4 2 5

3 6

1

2 3

4 5 6

3 4

6 7

(46)

COMMENTS (GAUSS

INTEGRATION)

 When the Gauss integration scheme is used, one has to

decide how many Gauss points should be used.

 _{Theoretically, for a one-dimensional integral, using} m

points can give the exact solution for the integral of a polynomial integrand of up to an order of (2m1).

 _{As a general rule of thumb, more points should be used for}

(47)

COMMENTS (GAUSS

INTEGRATION)

 _{Using a smaller number of Gauss points tends to}

counteract the over-stiff behaviour associated with the

displacement-based method.

 _{Displacement in an element is assumed using shape}

(48)

COMMENTS ON GAUSS

INTEGRATION

 Two Gauss points for linear elements, and two or three

points for quadratic elements in each direction should be sufficient for most cases.

 _{Most of the explicit FEM codes based on explicit}

(49)

CASE STUDY

(50)

CASE STUDY

Elastic Properties of Polysilicon Young’s Modulus, E 169GPa

Poisson’s ratio,  0.262 Density,  2300kgm-3

10N/m

(51)

CASE STUDY

(52)

CASE STUDY

(53)

CASE STUDY

(54)

CASE STUDY

Analysis no. 4: Von Mises stress distribution using 24 eight-nodal,

(55)

CASE STUDY

(56)

CASE STUDY

Analysis no.

Number / type of elements

Total number of nodes in

model

Maximum Von Mises Stress (GPa) 1 24 bilinear,

quadrilateral 41 0.0139

2 96 bilinear,

3 144 bilinear,

4 24 quadratic,

quadrilateral 105 0.0191 5 192 linear,

FEM FOR 2D SOLIDS

Finite Element Method

 INTRODUCTION

CONTENTS

 LINEAR QUADRILATERAL ELEMENTS

INTRODUCTION

of the 2D solid.

In this course, it is assumed that the element has a

Field variable interpolation

Shape functions construction

Element matrices

Shape functions construction

Gauss integration

m j wj Accuracy n

Evaluation of

Coordinate mapping

HIGHER ORDER ELEMENTS

HIGHER ORDER ELEMENTS

INTEGRATION)

COMMENTS (GAUSS

CASE STUDY