The Finite Element Method A Practical Course FEM FOR 2D SOLIDS CHAPTER 5:
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Transcript of The Finite Element Method A Practical Course FEM FOR 2D SOLIDS CHAPTER 5:
The FThe Finite Element Methodinite Element Method A Practical CourseA Practical Course
FEM FOR 2D SOLIDS
CHAPTER 5:
CONTENTSCONTENTS 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 me
CONTENTSCONTENTS
LINEAR QUADRILATERAL ELEMENTS– Coordinate mapping– Strain matrix– Element matrices– Remarks
HIGHER ORDER ELEMENTS COMMENTS (GAUSS INTEGRATION) CASE STUDY
INTRODUCTIONINTRODUCTION
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.
2D solid element can deform only in the plane of the 2D solid.
At any point, there are two components in x and y directions for the displacement as well as forces.
INTRODUCTIONINTRODUCTION
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 2-D elements with given variation of thickness is also straightforward, as the procedure is the same as that for a uniform element.
22D solids – plane stress and plane strainD solids – plane stress and plane strain
fx
fy
x
y
x
y fx
z
Plane stress Plane strain
LINEAR TRIANGULAR ELEMENTSLINEAR TRIANGULAR ELEMENTS
Less accurate than quadrilateral elementsUsed by most mesh generators for complex
geometryLinear triangular element
x, u
y, v
1 (x1, y1) (u1, v1)
2 (x2, y2) (u2, v2)
3 (x3, y3) (u3, v3)
A fsx
fsy
Triangular elements Nodes
Field variable interpolationField variable interpolation
( , ) ( , )hex y x yU N d
3 nodeat ntsdisplaceme
2 nodeat ntsdisplaceme
1 nodeat ntsdisplaceme
3
3
2
2
1
1
v
u
v
u
v
u
ed
31 2
31 2
Node 2Node 1 Node 3
00 0
00 0
NN N
NN N
N
where
(Shape functions)
x, u
y, v
1 (x1, y1) (u1, v1)
2 (x2, y2) (u2, v2)
3 (x3, y3) (u3, v3)
A fsx
fsy
Shape functions constructionShape functions construction
1 1 1 1N a b x c y
2 2 2 2N a b x c y
3 3 3 3N a b x c y
i i i iN a b x c y
Assume,
i= 1, 2, 3
1 T
T
i
i i
i
a
N x y b
c
p
p
or
Shape functions constructionShape functions construction
Delta function property:
1 for ( , )
0 for i j j
i jN x y
i j
1 1 1
1 2 2
1 3 3
( , ) 1
( , ) 0
( , ) 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
( , ) 0
N x y a b x c y
N x y a b x c y
N x y a b x c y
Solving, 2 3 3 2 2 3 3 21 1 1, ,
2 2 2e e e
x y x y y y x xa b c
A A A
Shape functions constructionShape functions construction
1 1
2 2 2 3 3 2 2 3 1 3 2 1
3 3
11 1 1
1 [( ) ( ) ( ) ]2 2 2
1e
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, b1 and c1 back into N1 = a1 + b1x + c1y:
1 2 3 2 3 2 2
1[( )( ) ( )( )]
2 e
N y y x x x x y yA
Shape functions constructionShape 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
e
N x y x y y y x x x yA
y y x x x x y yA
3 1 1
3 2 2
3 3 3
( , ) 0
( , ) 0
( , ) 1
N x y
N x y
N x y
3 1 2 1 1 1 2 2 1
1 2 1 2 1 1
1[( ) ( ) ( ) ]
2
1[( )( ) ( )( )]
2
e
e
N x y x y y y x x x yA
y y x x x x y yA
Shape functions constructionShape functions construction
i i i iN a b x c y
1( )
2
1( )
2
1( )
2
i j k k je
i j ke
i k je
a x y x yA
b y yA
c x xA
where
i
jk
i= 1, 2, 3
J, k determined from cyclic permutation
i = 1, 2
j = 2, 3k = 3, 1
Using area Using area coordinatescoordinatesAlternative 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
11 1
1 [( ) ( ) ( ) ]2 2
1
x y
A x y x y x y y y x x x y
x y
11
e
AL
A
2-3-P:
Similarly, 3-1-P A2
1-2-P A3
22
e
AL
A
33
e
AL
A
Using area coordinatesUsing area coordinates
1 2 3 1L L L Partitions of unity:
3 1 2 31 21 2 3 1
e e e e
A A A AA AL L L
A A A 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
( , ) ( , )hex y x yU N d
Strain matrixStrain matrix
xx
yy
xy
u
xv
y
u v
y x
LU where
0
0
x
y
y x
L
ee BdLNdLU
0
0
x
y
y x
B LN N1 2 3
1 2 3
1 1 2 2 3 3
0 0 0
0 0 0
b b b
c c c
c b c b c b
B
(constant strain element)
Element matricesElement matrices
0d ( d ) d d
e e e
hT T Te
V A A
V z A h A k B cB B cB B cB
Constant matrix T
e ehAk B cB
0d d d d
e e e
hT T Te
V A A
V z A h A m N N N N N N
Element matricesElement matrices
1 1 1 2 1 3
1 1 1 2 1 3
2 1 2 2 2 3
2 1 2 2 2 3
3 1 3 2 3 3
3 1 3 2 3 3
0 0 0
0 0 0
0 0 0d
0 0 0
0 0 0
0 0 0
e
e
A
N N N N N N
N N N N N N
N N N N N Nh A
N N N N N N
N N N N N N
N N N N N N
m
For elements with uniform density and thickness,
Apnm
pnmALLL pn
A
m 2)!2(
!!!d321
Eisenberg and Malvern (1973):
Element Element matricesmatrices
2
02.
102
0102
10102
010102
12
sy
hAe
m
T
2 3 d
sxe l
sy
fl
f
f N
y
x
y
xe
f
f
f
fl
0
0
2
132fUniform distributed load:
x, u
y, v
1 (x1, y1) (u1, v1)
2 (x2, y2) (u2, v2)
3 (x3, y3) (u3, v3)
A fsx
fsy
LINEAR RECTANGULAR LINEAR RECTANGULAR ELEMENTSELEMENTS
Non-constant strain matrixMore accurate representation of stress and
strainRegular shape makes formulation easy
Shape functions Shape functions constructionconstruction
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
displacements at node 2
displacements at node 3
displacements at node 4
e
u
v
u
v
u
v
u
v
d
Shape functions constructionShape 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)
2
4 ( 1, +1) (u4, v4)
2
2 1 2 1( ) / 2 ( ) / 2,
x x x y y y
a b
( , ) ( , )hex y x yU N d 31 2 4
31 2 4
Node 2 Node 3Node 1 Node 4
00 0 0
00 0 0
NN N N
NN N N
N
where
Shape functions constructionShape functions construction
)1)(1(
)1)(1(
)1)(1(
)1)(1(
41
4
41
3
41
2
41
1
N
N
N
N
113 4at node 1 1
113 4at node 2 1
113 4at node 3 1
113 4at node 4 1
(1 )(1 ) 0
(1 )(1 ) 0
(1 )(1 ) 1
(1 )(1 ) 0
N
N
N
N
Delta function property
4
1 2 3 41
14
14
[(1 )(1 ) (1 )(1 ) (1 )(1 ) (1 )(1 )]
[2(1 ) 2(1 )] 1
ii
N N N N N
Partition of unity
)1)(1(41 jjjN
1 ( 1, 1) (u1, v1)
2 (1, 1) (u2, v2)
3 (1, +1) (u3, v3)
2
4 ( 1, +1) (u4, v4)
2
Strain matrixStrain matrix
1 1 1 1
1 1 1 1
1 1 1 1 1 1 1 1
0 0 0 01
0 0 0 04
a a a a
b b b b
a a a ab b b b
B LN
Note: No longer a constant matrix!
Element matricesElement matrices
, byax dxdy = ab dd
Therefore,
ddd T1
1
1
1
T cBBcBBk habAhA
e
1 1
0 1 1d d d d d d
hT T T Te
V A A
V z A h A abh
m N N N N N N N N
Element matricesElement matrices
T
2 3 d
sxe l
sy
fl
f
f N
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,
0
0
0
0
y
x
y
x
e
ff
f
f
bf
Gauss integrationGauss integration
For evaluation of integrals in ke and me (in practice)
In 1 direction: )()d(1
1
1 jj
m
j
fwfI
m gauss points gives exact solution of polynomial integrand of n = 2m - 1
1 1
1 11 1
( , )d d ( , )yx
nn
i j i ji j
I f w w f
In 2 directions:
Gauss integrationGauss integration
m Gauss Point j Gauss Weight wj Accuracy order 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
11
Evaluation of Evaluation of mmee
4
04.
204
0204
10204
010204
2010204
02010204
9
sy
habe
m
Evaluation of Evaluation of mmee
E.g.
)1)(1(4
)1)(1()1)(1(16
31
31
1
1
1
1
1
1
1
1
jiji
jiji
jiij
hab
ddhab
ddNNhabm
94)111)(111(
4 31
31
33
habhabm
Note: In practice, gauss integration is often used
LINEAR QUADRILATERAL LINEAR QUADRILATERAL ELEMENTSELEMENTS
Rectangular elements have limited applicationQuadrilateral elements with unparallel edges are
more useful Irregular shape requires coordinate mapping
before using Gauss integration
Coordinate mappingCoordinate 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
( , ) ( , )he U N d (Interpolation of displacements)
( , ) ( , ) e X N x (Interpolation of coordinates)
Coordinate mappingCoordinate mapping
( , ) ( , ) e X N x
wherex
y
X ,
1
1
2
2
3
3
4
4
coordinate at node 1
coordinate at node 2
coordinate at node 3
coordinate at node 4
e
x
y
x
y
x
y
x
y
x
)1)(1(
)1)(1(
)1)(1(
)1)(1(
41
4
41
3
41
2
41
1
N
N
N
N
iii
xNx ),(4
1
iii
yNy ),(4
1
Coordinate mappingCoordinate mapping
Substitute 1 into iii
xNx ),(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)
1 12 32 2
1 12 32 2
(1 ) (1 )
(1 ) (1 )
x x x
y y y
or
)()(
)()(
2321
3221
2321
3221
yyyyy
xxxxx
Eliminating ,3 2 1 1
2 3 2 32 23 2
( ){ ( )} ( )
( )
y yy x x x y y
x x
Strain matrixStrain matrix
y
y
Nx
x
NN
y
y
Nx
x
NN
iii
iii i i
ii
N N
xNNy
Jor
x y
x y
Jwhere (Jacobian matrix)
1 131 2 4
2 2
3 331 2 4
4 4
x yNN N N
x y
x yNN N N
x y
JSince ( , ) ( , ) e X N x ,
Strain matrixStrain matrix
1
ii
i i
NN
xN Ny
JTherefore,
NLNB
xy
y
x
0
0
Replace differentials of Ni w.r.t. x and y with differentials of Ni w.r.t. and
(Relationship between differentials of shape functions w.r.t. physical coordinates and differentials w.r.t. natural coordinates)
Element matricesElement matrices
Murnaghan (1951) : dA=det |J | dd
1 1 T
1 1det d de h
k B cB J
dddet
dddd
1
1
1
1
0
JNN
NNNNNNm
T
T
A
T
A
hT
V
e
h
AhAxV
RemarksRemarks
Shape functions used for interpolating the coordinates are the same as the shape functions used for the interpolation of the displacement field. Therefore, the element is called 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 superparametric elements.
HIGHER ORDER ELEMENTSHIGHER ORDER ELEMENTS
Higher order triangular elements
i (I,J,K)
(p,0,0) (0,p,0)
(0,0,p)
(p 1,1,0)
L1
L3
L2
(0,p 1,1)
(0,1,p 1) (1,0,p 1)
(2,0,p 2)
nd = (p+1)(p+2)/2
I J K p Node i,
Argyris, 1968 :
1 2 3( ) ( ) ( )I J Ki I J KN l L l L l L
0 1 ( 1)
0 1 ( 1)
( )( ) ( )( )
( )( ) ( )
L L L L L Ll L
L L L L L L
HIGHER ORDER ELEMENTSHIGHER ORDER ELEMENTS
Higher order triangular elements (Cont’d)
x , u
y , v
1
2
3
4
5 6
1 1 1( 2 1 )N L L
4 1 24N L L
x , u
y , v
1
2
3
4 5
6
7 8
9 1 0
1 1 1 1
1( 3 1 ) ( 3 2 )
2N L L L
4 1 2 1
9( 3 1 )
2N L L L
1 0 1 2 32 7N L L L
Cubic element
Quadratic element
HIGHER ORDER ELEMENTSHIGHER ORDER ELEMENTS
Higher order rectangular elements
(0,0)
0
(n,0)
(0,m) (n,m)
i(I,J)
Lagrange type:
1 1 ( ) ( )D D n mi I J I JN N N l l
0 1 1 1
0 1 1 1
( )( ) ( )( ) ( )( )
( )( ) ( )( ) ( )n k k nk
k k k k k k k n
l
(Zienkiewicz et al., 2000)
HIGHER ORDER ELEMENTSHIGHER ORDER ELEMENTS
Higher order rectangular elements(Cont’d)
1 2
3 4
5
6
7
8 9
I=0 I=1 I=2
J=0
J=1
J=2
1 11 0 0
1 12 2 0
1 13 2 2
1 14 0 2
1( ) ( ) (1 ) (1 )
41
( ) ( ) (1 ) (1 )4
1( ) ( ) (1 )(1 )
41
( ) ( ) (1 )(1 )4
D D
D D
D D
D D
N N N
N N N
N N N
N N N
1 15 1 0
1 16 2 1
1 17 1 2
1 18 0 1
1 1 2 29 1 1
1( ) ( ) (1 )(1 )(1 )
21
( ) ( ) (1 )(1 )(1 )21
( ) ( ) (1 )(1 )(1 )2
1( ) ( ) (1 )(1 )
2
( ) ( ) (1 )(1 )
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
(9 node quadratic element)
HIGHER ORDER ELEMENTSHIGHER ORDER ELEMENTS
Higher order rectangular elements(Cont’d)
Serendipity type:
1 2
3 4
5
6
7
8 0
=1
= 1
14
212
212
(1 )(1 )( 1) 1, 2, 3, 4
(1 )(1 ) 5, 7
(1 )(1 ) 6,8
j j j j j
j j
j j
N j
N j
N j
(eight node quadratic element)
HIGHER ORDER ELEMENTSHIGHER ORDER ELEMENTS
Higher order rectangular elements(Cont’d)
1 2
3 4
5 6
7
8
9 10
11
12
2 2132
2932
13
2932
(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 j
j j
j j
N
j
N
j
N
13
9 )
for side nodes 5, 6, 9, 10 where and 1
j
j jj
(twelve node cubic element)
ELEMENT WITH CURVED ELEMENT WITH CURVED EDGESEDGES
4
2
3
8
1 5
7
6
1 4 2
5
3
6
1
2
3
4
5 6
1 2
3 4
5
6
7
8
COMMENTS (GAUSS INTEGRATION)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
higher order of elements.
COMMENTS (GAUSS COMMENTS (GAUSS INTEGRATION)INTEGRATION)
Using 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 be. It is often observed that higher order elements are usually softer than lower order ones. This is because using higher order elements gives less constraint to the elements.
COMMENTS (GAUSS COMMENTS (GAUSS INTEGRATION)INTEGRATION)
Two gauss points for linear elements, and two or three points for quadratic elements in each direction should be sufficient for many 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 STUDYCASE STUDY
Side drive micro-motor
CASE STUDYCASE STUDY
Elastic Properties of Polysilicon
Young’s Modulus, E 169GPa
Poisson’s ratio, 0.262
Density, 2300kgm-3
10N/m
10N/m
10N/m
CASE STUDYCASE STUDY
Analysis no. 1: Von Mises stress distribution using 24 bilinear
quadrilateral elements (41 nodes)
CASE STUDYCASE STUDY
Analysis no. 2: Von Mises stress distribution using 96 bilinear
quadrilateral elements (129 nodes)
CASE STUDYCASE STUDY
Analysis no. 3: Von Mises stress distribution using 144 bilinear
quadrilateral elements (185 nodes)
CASE STUDYCASE STUDY
Analysis no. 4: Von Mises stress distribution using 24 eight-nodal, quadratic elements (105 nodes)
CASE STUDYCASE STUDY
Analysis no. 5: Von Mises stress distribution using 192 three-nodal,
triangular elements (129 nodes)
CASE STUDYCASE STUDYAnalysis
no.Number / type of
elements
Total number of nodes in
model
Maximum Von Mises
Stress (GPa)
124 bilinear,
quadrilateral 41 0.0139
296 bilinear, quadrilateral
129 0.0180
3144 bilinear, quadrilateral
185 0.0197
424 quadratic, quadrilateral
105 0.0191
5 192 linear, triangular
129 0.0167