FEM FOR 2D SOLIDS
FEM FOR 2D SOLIDS
1 Finite Element Method 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
- – Element matrices
- – Gauss integration
- – Evaluation of m
e CONTENTS
LINEAR QUADRILATERAL ELEMENTS
- – Coordinate mapping
- – Strain matrix
- – Element matrices
- – Remarks
HIGHER ORDER ELEMENTS
COMMENTS (GAUSS INTEGRATION)
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 forces.
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 thickness is also straightforward, as the procedure is the same as that for a uniform element.
Plane stress Plane strain
2D solids
- – plane stress and plane strain
Less accurate than quadrilateral elements
Used by most mesh generators for complex geometry
A linear triangular element: , y
3 (x
3 3 ) y, v
(u , v )
3
3
f
sy f sxA , y 2 (x
2 2 ) (u , v )
2
2 , y
1 (x )
1
1 (u , v )
1
1 x, u Field variable interpolation ( , ) ( , ) h e
U x y x y N d 3 node at nts displaceme 2 node at nts displaceme
2 Node 2 Node 1 Node 3 N N N N N N
A f sx f sy
3 )
3 , v
3 ) (u
3 , y
2 ) 3 (x
2 , v
2 ) (u
2 , y
1 ) (u 1 , v 1 ) 2 (x
1 , y
1 (x
N where x, u y, v
1
1 node at nts displaceme
3
2
1
3
v u v u v u e d
1
1
2
2
3
3
(Shape functions) Shape functions construction Assume,
N a b x c y
1
1
1
i
1 N a b x c y
= 1, 2, 3 i i i i a
N a b x c y
i
2
2
2
2
T p
N
1 x y b or i i T c
N a b x c y p i
3
3
3
3 Shape functions construction Delta function property:
N x y ( , ) 1
1
1
1 i j
1 for N x y N x y
( , ) 0 ( , )
1
2
2 i j j
i j
0 for N x y
( , ) 0
1
3
3 N x y ( , ) a b x c y
1 Therefore,
1
1
1 1 1 1 1 1
N x y ( , ) a b x c y
1
2
2 1 1 2 1 2
N x y a b x c y ( , )
1
3
3 1 1 3 1 3
x y x y y y x x
2 3 3 2
2
3
3
2 Solving, a b c
, ,
1
1
1 A A A
2
2
2 e e e
N y y x x x x y y A
:
1 , b
1 and c
1 back into N
1 = a
1
1 x + c
1 y
1
P Area of triangle Moment matrix
2
3
2
3
2
2
1 [( )( ) ( )( )]
2 e
Substitute a
1 e x y A x y x y x y y y x x x y x y
Shape functions construction
2
1
1
2 2 2 3 3 2
2
3
1
3
1
2
3
3
1
1
1
1 1 [( ) ( ) ( ) ]
2
2
- + b
1 N x y N x y x y y y x x x y
( , ) 0 [( ) ( ) ( ) ]
2
1
1 2 3 1 1 3
3
1
1
3
2 A
e N x y ( , ) 1
2
2
2
1 N x y
( , ) 0 [( y y )( x x ) ( x x )( y y )]
2
3
3
3
1
3
1
3
3 A
2 e
1 N x y
( , ) 0 N [( x y x y ) ( y y x ) ( x x y ) ]
3
1
1 3 1 2 1 1
1
2
2
1 A
2 e
N x y ( , ) 0
3
2
2
1 y y x x x x y y
[( )( ) ( )( )]
1
2
1
2
1
3
3
3
1 N x y ( , ) 1
2 A
e Shape functions construction N a b x c y i i i i
1 a x y x y
( ) i where
= 1, 2, 3 i j k k j
2 A
e J
, k determined from cyclic
1 b ( y y ) i j k permutation
2 A e i
= 1, 2 i
1 c ( x x ) i k j
2 A e k j j
= 2, 3 k
= 3, 1 Using area coordinates
Alternative method of constructing shape functions i,
A
1
1
eA L A 2-3-P:
Similarly, 3-1-P A
2 1-2-P
3
2
2
2 e
A L A
3
3
e A L A
1 x y A x y x y x y y y x x x y x y
2
1 j ,
2
2 k ,
3 x y
P A
1
1
2 2 2 3 3 2
3
1
1 [( ) ( ) ( ) ]3
2
3
3
1
1
Using area coordinates
1
A A A A A A L L L
A A A A
( , ) ( , ) h e
N L N L N L
3 , ,
3
2
2
1
1
1 = 0 at if P at nodes 2 or 3 Therefore,
Delta function property: e.g. L
1 e e e e
2
3
2
1
2
2
3
2
2
3
1 L L L Partitions of unity:
3
U x y x y N d Strain matrix xx yy xy u x v y u v y x
2
a a a b b b b a b a b a
3
3
2
2
1
1
3
2
1
3
1
B LN N
Bd LNd LU x y
L e e
LU
where
x y y x
B (constant strain element) Element matrices 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 e
hA k B cB d d d d e e e h T T T e
V A A V x A h A
m N N N N N N Element matrices
1
N N N N N N
N N N N N N
N N N N N N h2
3
3
3
1
3
2
3
3 d e e A
A
N N N N N N
N N N N N NN N N N N N
1
m
For elements with uniform density and thickness, A
p n m
p n m A L L L p n A m2 )! 2 ( ! ! ! d
3
2
1
3
3
1
3
1
2
1
3
1
1
1
2
1
2
3
1
2
2
2
3
2
1
2
2
2
Eisenberg and Malvern (1973): Element matrices
] d [
2
) 3 (x
3 , y
3
) (u
3
, v
3
)
A f sx f sy l f f l sy sx e
3
2
N f
y x y x e f f f f l
2
1
3
2 f
Uniform distributed load:
, v
) (u
2
2 2 .
1
2
1
2
1
1
2
1
1
12 sy hA e
2
m x, u y, v
1 (x
1 , y
1
) (u
1
, v
1
) 2 (x
2 , y
2 T
LINEAR RECTANGULAR ELEMENTS
Non-constant strain matrix
More accurate representation of stress and strain
Regular shape makes formulation easy Shape functions construction x, u y, v
1 (x
2
4
) (u
4 , v 4 )
2b
Consider a rectangular element
1
1
2
4
3
3
4
4 displacements at node 1 displacements at node 2 displacements at node 3 displacements at node 4 e u v u u u u u u
, y
4 (x
1
2
, y
1
) (u
1 , v 1 )
2 (x
2 , y 2 )
(u
, v
) 2a f sy f sx
2
) 3 (x
3 , y 3 )
(u
3
, v
3
d Shape functions construction x, u y, v
1 (x
U x y x y N d
4 (1, +1) (u
4
, v
4
)
2b
, b y a x
( , ) ( , ) h e
3
3 (1, +1) (u
1
2
4
3
1
2
4 Node 2 Node 3 Node 1 Node 4
N N N N N N N N
3 , v 3 ) 2a
2 , v 2 )
1 , y
3 , v
1 ) (u
1 , v
1 ) 2 (x
2 , y
2 ) (u
2 , v
2 ) 3 (x
3 , y
3 ) (u
3 )
2 (1, 1) (u
2a f sy f sx
4 (x
4 , y
4 ) (u
4 , v
4 )
2b
1 (1, 1) (u
1 , v 1 )
N where (Interpolation) Shape functions construction ) 1 )(
1 ( ) 1 )( 1 (
3
4 [(1 )(1 ) (1 )(1 ) (1 )(1 ) (1 )(1 )] [2(1 ) 2(1 )] 1 i i
1
4
1
1
4
2
1
4
Delta function property
N N N N N
1 (1 )(1 ) N N N N
3 (1, +1) (u
)
4
, v
4
4 (1, +1) (u
3 , v 3 ) 2a
2 , v 2 )
Partition of unity ) 1 )(
2 (1, 1) (u
1 , v 1 )
1 (1, 1) (u
N
1 j j j
4
1 (
1 (1 )(1 ) (1 )(1 ) (1 )(1 )
) 1 )( 1 ( )
4
1
1
4
2
1
3
1
1
4
4
1
4
1 )( 1 (
N N N N
1
4 at node 4
1
3
1
1
1
4 at node 3
3
1
3
1
4 at node 2
3
1
1
1
4 at node 1
2b Strain matrix
1 LN B
1
Note: No longer a constant matrix!
1
1
1
1
1
1
1
1
1
1
1
1
1
a b a b a b a b b b b b a a a a
1 Element matrices , b y a x
dxdy = ab d d Therefore,
d d d d d d
1
1
1
B cB cB B k A h ab h
d d d T
1 T
A e
1
1
1 N N N N N N N N m T T
A
T A h TV e A abh h A x
V
1
Element matrices l f f l sy sx e
N f x, u y, v
] d [
2 T
1 (x
2a f sy f sx
y x y x e f f f f b f
For uniformly distributed load,
4 , v 4 ) 2b
4 ) (u
4 , y
4 (x
3 )
1 , y
3 , v
3 , y 3 ) (u
3
2 , v
2 , y 2 ) (u
1 ) 2 (x
1 , v
1 ) (u
2 ) 3 (x Gauss integration
For evaluation of integrals in k and m (in practice) e e m
1 In 1 direction: I f w f ( )d ( ) j j
1 j
1 m gauss points gives exact solution of polynomial integrand of n = 2m - 1 n n y x
1
1 I f w w f
In 2 directions: ( , )d d ( , )
i j i j
1
1
i j
1
1 Gauss integration m j w j
Accuracy n
1
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,
11
- 0.238619, 0.238619, 0.661209, 0.932470 0.171324, 0.360762, 0.467914, 0.467914, 0.360762, 0.171324
4
9 sy hab e
4
2
1
2
4
2
1
2
2
1
4
2
1
4
2
4
2
4 4 .
m Evaluation of m e
1
1 m hab N N d d
ij i j
1
1
1
1 hab
d d ( 1 )( 1 ) ( 1 )( 1 ) i j i j
1
1
16 hab
1
1 ( 1 )( 1 ) i j i j
3
3
4 hab
hab
1
1 E.g. m ( 1 1 1 )( 1 1 1 ) 4
33
3
3
4
9 Note: In practice, Gauss integration is often used LINEAR QUADRILATERAL ELEMENTS
Rectangular elements have limited application
Quadrilateral elements with unparallel edges are more useful
Irregular shape requires coordinate mapping before using Gauss integration Coordinate mapping 2 (x
2 , y
2 ) y x
1 (1, 1) 2 (1, 1) 3 (1, +1) 4 (1, +1)
3 (x
3 , y
3 ) 4 (x
4 , y
4 ) 1 (x
1 , y 1 ) Physical coordinates Natural coordinates ( , ) ( , ) h e
U N d (Interpolation of displacements)
( , ) ( , ) e
X N x (Interpolation of coordinates) Coordinate mapping
x 1 coordinate at node 1
y
1
X N x ( , ) ( , ) e
x
2 coordinate at node 2
x
y
2 x
X e where ,
3 y
x coordinate at node 3
y
3
4 coordinate at node 4
x1
N y
( 1 )( 1 )
4
1
4
1 N
( 1 )( 1 )
2
4
4
1 N
x N x ( 1 )( 1 ) ( , )
3 i i
4 i
1
1 N
( 1 )( 1 )
4
4
4
y N ( , ) y i i
i
1 Coordinate mapping Substitute 1 into
i i
iN x x
) , (
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
) (
3
2
2
1
3
2
2
1
2
3
2
3 y y x x x y y x x y
2
2
4
1 , y 1 )
1
2 (x
2 , y 2 ) y x
1 (1, 1) 2 (1, 1) 3 (1, +1) 4 (1, +1)
3 (x
3 , y 3 )
4 (x
4 , y 4 )
1 (x
3
) ( ) ( ) ( ) (
2
1
2
2
1
3
2
1
2
2
1 ) ) 1 (
1 ( ) ) 1 ( 1 ( y y y x x x
or
Strain matrix
4
4
2
2
3
3
3
1
2
4
1
4 x y N N N N x y N x y
N N N x y
J Since ( , ) ( , ) e
2
3
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
1
J or x y x y
J where (Jacobian matrix)
1
X N x , Strain matrix
1 i i i i
N N x N
N y
J Therefore,
N LN B
x y y x
Replace differentials of N i w.r.t. x and y with differentials of N i w.r.t. and (Relationship between differentials of shape functions w.r.t. physical coordinates and differentials w.r.t. natural coordinates) Element matrices Murnaghan (1951) : dA=det |J | d d
1
1 T
T A T A h T
V
A h A x
V e h
N N N N N N m T
1
1
1
1
d d det d d d d
k B cB J
1 det d d e h
1 J N N
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, subparametric will lead to the development of so-called or superparametric elements.
HIGHER ORDER ELEMENTS
2
I L L L L L L l L L L L L L L
I I
( )( ) ( ) ( ) ( )( ) ( )
1 ( 1) 1 ( 1)
I J K N l L l L l L
I J K i
3 ( ) ( ) ( )
Higher order triangular elements i (I,J,K)
(p,0,0) (0,p,0)
I J K p Node i, Argyris, 1968 :
= (p+1)(p+2)/2
n d
(1,0,p1) (2,0,p2)
(0,p1,1) (0,1,p1)
2
3 L
1 L
L
(0,0,p) (p1,1,0)
1
HIGHER ORDER ELEMENTS
1
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
9 (3 1)
4
2 N N L L L
10
1
2
3
27 N L L L
Cubic element Quadratic element
5
Higher order triangular elements (Cont’d) x, u y, v
1
2
2
3
4
5
6
1
2
1
2
1 (2 1) N N N L L
4
5
6
1
2 N 4 N N L L x, u
y, v
1
3
HIGHER ORDER ELEMENTS
1
[Zienkiewicz et al., 2000]
( )( ) ( )( ) ( ) n k k n k k k k k k k k n
l
1 ( )( ) ( )( ) ( ) ( )
1
1
1
Higher order rectangular elements
(0,0)
I J N N N l l
I J
D D n m i
1 ( ) ( )
1
Lagrange type:
(n,0) (0,m) (n,m) i (n,m)
1
HIGHER ORDER ELEMENTS
1
1
1
2
3
7
1
3
1
2
6
1
1
1
3
5
8
1
1
1 ( ) ( ) (1 )(1 )(1 )
(nine node quadratic element)
D D D D D D D D D D N N N N N N N N N N N N N N N
1 ( ) ( ) (1 )(1 ) 2 ( ) ( ) (1 )(1 )
2
1 ( ) ( ) (1 )(1 )(1 )
2
2
1
1 ( ) ( ) (1 )(1 )(1 )
3
3
9
2
2
1
1
Higher order rectangular elements (Cont’d)
1
1
1
1
1
1
1
1
9
2
8
7
6
5
4
3
2
2
1
2
1 ( ) ( ) (1 )(1 ) D D D D D D D D N N N N N N N N N N N N
4
1 ( ) ( ) (1 )(1 )
4
1 ( ) ( ) (1 ) (1 )
4
1 ( ) ( ) (1 ) (1 )
1
1
4
1
1
2
2
3
1
HIGHER ORDER ELEMENTS
Higher order rectangular elements (Cont’d) Serendipity type:
=1
4
3
7
1
N j
(1 )(1 )( 1) 1, 2, 3, 4 j j j j j
4
2
1
6 N j
(1 )(1 ) 5, 7
j j
2
8
2
1 N (1 )(1 ) j 6, 8
j 2 j
2
1
5 =1 (eight node quadratic element)
HIGHER ORDER ELEMENTS
Higher order rectangular elements (Cont’d)
2
2
1 N (1 )(1 )(9 9 10)
j j j
32
4
10
9
3 j for corner nodes 1, 2, 3, 4
2
9 N (1 )(1 )(1 9 ) j j j
32
8
11
1 j for side nodes 7, 8, 11, 12 where 1 and j j
3
7
12
2
9 N (1 )(1 )(1 9 ) j j j
32
1 j for side nodes 5, 6, 9, 10 where and
1 j j
3
2
1
6
5 (twelve node cubic element) ELEMENT WITH CURVED EDGES
4
2
7
6
5
4
3
2
1
6
5
4
3
1
2
6
3
5
2
4
1
6
7
5
1
8
3
8
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 (2m1).
As a general rule of thumb, more points should be used for a higher order of elements.
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 functions. This implies that the deformation of the element is somehow prescribed in a fashion of the shape function.
This prescription gives a constraint to the element. The so- constrained element behaves stiffer than it should. It is often observed that higher order elements are usually softer than lower order ones. This is because using higher order elements gives fewer constraint to the elements.
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 formulation tend to use one-point integration to achieve the best performance in saving CPU time.
CASE STUDY
Side drive micro-motor
CASE STUDY
Elastic Properties of Polysilicon Young ’s Modulus, E 169GPa
Poisson ’s ratio, 0.262 Density, 2300kgm
10N/m
- 3
10N/m
10N/m
CASE STUDY
Analysis no. 1: Von Mises stress distribution using 24 bilinear quadrilateral elements (41 nodes)
CASE STUDY
Analysis no. 2: Von Mises stress distribution using 96 bilinear quadrilateral elements (129 nodes)
CASE STUDY
Analysis no. 3: Von Mises stress distribution using 144 bilinear quadrilateral elements (185 nodes)
CASE STUDY
Analysis no. 4: Von Mises stress distribution using 24 eight-nodal, quadratic elements (105 nodes)
CASE STUDY
Analysis no. 5: Von Mises stress distribution using 192 three-nodal, triangular elements (129 nodes)
CASE STUDY
Analysis no.