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1
Finite Element Method
FEM FOR 2D SOLIDS
for readers of all backgrounds
G. R. Liu and S. S. Quek
CHAPTER 7:
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Finite Element Method by G. R. Liu and S. S. Quek2
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 ofme
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Finite Element Method by G. R. Liu and S. S. Quek3
CONTENTS
LINEAR QUADRILATERAL ELEMENTS
Coordinate mapping
Strain matrix
Element matrices
Remarks
HIGHER ORDER ELEMENTS
COMMENTS (GAUSS INTEGRATION)
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Finite Element Method by G. R. Liu and S. S. Quek4
INTRODUCTION
2D solid elements are applicable for the analysisof plane strain and plane stress problems.
A 2D solid element can have a triangular,rectangular or quadrilateral shape with straight orcurved 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 asforces.
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Finite Element Method by G. R. Liu and S. S. Quek5
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 ofthickness is also straightforward, as the procedure
is the same as that for a uniform element.
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Finite Element Method by G. R. Liu and S. S. Quek6
2D solids plane stress and plane strain
Plane stress Plane strain
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LINEAR TRIANGULAR
ELEMENTS Less accurate than quadrilateral elements
Used by most mesh generators for complex
geometry A linear triangular element:
x, u
y, v
1 (x1, y1)
(u1, v1)
2 (x2, y2)
(u2, v2)
3 (x3, y3)
(u3, v3)
sxsy
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Field variable interpolation
( , ) ( , )he
x y x yU N d
3nodeatntsdisplaceme
2nodeatntsdisplaceme
1nodeatntsdisplaceme
3
3
2
2
1
1
v
u
v
u
v
u
ed
31 2
31 2
Node 2Node 1 Node 3
00 000 0
NN N
NN N
N
where
x, u
y, v
1 (x1, y1)
(u1, v1)
2 (x2, y2)
(u2, v2)
3 (x3, y3)
(u3, v3)
Asx
sy
(Shape functions)
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Shape functions construction
1 1 1 1N a b x c y
2 2 2 2N a b x c y
3 3 3 3
N 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
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Shape 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
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Shape 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
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, b1 and c1 back intoN1= a1+ b1x + c1y:
1 2 3 2 3 2 2
1[( )( ) ( )( )]
2 eN y y x x x x y y
A
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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
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
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Shape functions construction
i i i iN 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 yA
b y yA
c x xA
where
i
jk
i= 1, 2, 3
J, kdetermined from cyclic
permutationi = 1, 2
j = 2, 3k= 3, 1
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Using area coordinates
Alternative method of constructing shape
functions
i,1
, 2
k, 3
x
P
A1
1 2 2 2 3 3 2 2 3 3 2
3 3
1
1 11 [( ) ( ) ( ) ]2 2
1
x y
A x y x y x y y y x x x y
x y
1
1
e
AL
A
2-3-P:
Similarly, 3-1-P A2
1-2-P A3
22
e
ALA
3
3
e
AL
A
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Using area coordinates
1 2 31L L L Partitions of unity:
3 2 2 32 2
1 2 3 1e 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
( , ) ( , )h
ex y x yU N d
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Strain matrix
xx
yy
xy
ux
v
y
u vy x
LU where
0
0
x
y
y x
L
ee BdLNdLU
0
0
x
y
y x
B LN N
1 2 3
1 2 3
1 1 2 2 3 3
0 0 0
0 0 0
a a a
b b b
b a b a b a
B
(constant strain element)
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Element matrices
0d ( d ) d d
e e e
hT T T
e
V A A
V z A h A k B cB B cB B cB
Constant matrix
Te ehAk B cB
0d d d d
e e e
hT T T
eV A AV x A h A
m N N N N N N
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Element 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 0 d0 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 AN 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):
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19
Element matrices
2
02.
102
0102
10102
010102
12
sy
hAe
m
x, u
, v
1 (x1, y1)
(u1, v1)
2 (x2, y2)
(u2, v
2)
3 (x3, y3)
(u3, v3)
Asx
sy
l
f
f
lsy
sx
e d][32
T
Nf
y
x
y
x
e
f
f
f
fl
0
0
2
132fUniform distributed load:
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20
LINEAR RECTANGULAR
ELEMENTS
Non-constant strain matrix More accurate representation of stress and strain
Regular shape makes formulation easy
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Shape functions construction
x, u
y, v
1 (x1,y1)(u1, v1)
2 (x2,y2)(u2, v2)
3 (x3,y3)(u3, v3)
2a
sy
sx
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
u
u
u
u
u
d
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Shape functions construction
x, u
y, v
1 (x1,y1)
(u1, v1)
2 (x2,y2)
(u2, v2)
3 (x3,y3)
(u3, v3)
2a
sy
sx
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
, byax
( , ) ( , )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
Nwhere
(Interpolation)
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Finite Element Method by G. R. Liu and S. S. Quek23
Shape functions construction
)1)(1(
)1)(1(
)1)(1(
)1)(1(
4
14
4
13
4
12
4
11
N
N
NN
1 13 4at node 11
113 4at node 2
1
113 4at node 31
11
3 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 4
1
14
14
[(1 )(1 ) (1 )(1 ) (1 )(1 ) (1 )(1 )]
[2(1 ) 2(1 )] 1
i
i
N N N N N
Partition of
unity
)1)(1(4
1 jjjN
1 (1, 1)(u1, v1)
2 (1, 1)(u2, v2)
3 (1, +1)
(u3, v3)
2a
4 (1, +1)(u4, v4)
2b
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Finite Element Method by G. R. Liu and S. S. Quek24
Strain matrix
abababab
bbbb
aaaa
11111111
1111
1111
0000
0000
LNB
Note: No longer a constant matrix!
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Finite Element Method by G. R. Liu and S. S. Quek25
Element matrices
, byax dxdy = ab dd
Therefore,
ddd T1
1
1
1
TcBBcBBk habAh
A
e
dddddd1
1
1
10
NNNNNNNNmTT
A
T
A
hT
V
e abhAhAxV
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Finite Element Method by G. R. Liu and S. S. Quek26
Element matrices
lf
f
lsy
sx
e d][32
T
Nf
x, u
, v
1 (x1,y1)
(u1, v1)
2 (x2,y2)
(u2, v2)
3 (x3,y3)
(u3, v3)
2a
sy
sx
4 (x4,y4)
(u4, v4)
2b
For uniformly distributed load,
0
0
0
0
y
x
y
x
e
f
f
f
f
bf
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Finite Element Method by G. R. Liu and S. S. Quek27
Gauss 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 ofn = 2m - 1
1 1
1 11 1
( , )d d ( , )yxnn
i j i j
i j
I f ww f
In 2 directions:
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Finite Element Method by G. R. Liu and S. S. Quek28
Gauss integrationm j wj Accuracy n1 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
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Finite Element Method by G. R. Liu and S. S. Quek29
Evaluation of me
404.
204
0204
10204
010204
2010204
02010204
9
sy
habe
m
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Finite Element Method by G. R. Liu and S. S. Quek30
Evaluation of me
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(43
1
3
133
habhab
m
Note: In practice, Gauss integration is often used
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Finite Element Method by G. R. Liu and S. S. Quek31
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
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Finite Element Method by G. R. Liu and S. S. Quek32
Coordinate mapping
2 (x2,y2)
x1 (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)
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Finite Element Method by G. R. Liu and S. S. Quek33
Coordinate 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
xy
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
ii
i
xNx ),(4
1
ii
i
yNy ),(4
1
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Finite Element Method by G. R. Liu and S. S. Quek34
Coordinate mapping
Substitute 1 into iii
xNx ),(4
1
2 (x2,y2)
x1 (1, 1) 2 (1, 1)
3 (1, +1)4 (1, +1)
3 (x3,y3)4 (x4,y4)
1 (x1,y1)
321
221
321
221
)1()1(
)1()1(
yyy
xxx
or
)()(
)()(
232
1322
1
232
1322
1
yyyyy
xxxxx
Eliminating , )()}({)(
)(322
1322
1
23
23 yyxxxyy
xxy
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Finite Element Method by G. R. Liu and S. S. Quek35
Strain matrix
y
y
Nx
x
NN
y
y
Nx
x
NN
iii
iii i i
ii
N N
x
NN
y
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 ,
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Finite Element Method by G. R. Liu and S. S. Quek36
Strain matrix
1
ii
i i
NN
x
N N
y
JTherefore,
NLNB
xy
y
x
00
Replace differentials ofNiw.r.t.x andy
with differentials ofNiw.r.t. and
(Relationship between differentials of shape
functions w.r.t. physical coordinates and
differentials w.r.t. natural coordinates)
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Finite Element Method by G. R. Liu and S. S. Quek37
Element matrices
Murnaghan (1951) : dA=det |J | dd
1 1T
1 1 det 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
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Finite Element Method by G. R. Liu and S. S. Quek38
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
superparametric elements.
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Finite Element Method by G. R. Liu and S. S. Quek39
HIGHER ORDER ELEMENTS
Higher order triangular elements
i (I,J,K)
(p,0,0) (0,p,0)
(0,0,p)
(p1,1,0)
L1
L3
L2
(0,p1,1)
(0,1,p1)(1,0,p1)
(2,0,p2)
nd= (p+1)(p+2)/2
I J K p Node i,
Argyris, 1968 :
1 2 3( ) ( ) ( )I J Ki I J K N l L l L l L
0 1 ( 1)
0 1 ( 1)
( )( ) ( )( )
( )( ) ( )I I I
L L L L L Ll L
L L L L L L
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Finite Element Method by G. R. Liu and S. S. Quek40
HIGHER ORDER ELEMENTS
Higher order triangular elements (Contd)
x, u
, v
1
2
3
4
56
1 2 2 1 1(2 1)N N N L L
4 5 6 1 24N N N L L
x, u
, v
1
2
3
45
6
78
9
10
1 2 3 1 1 1
1(3 1)(3 2)
2N N N L L L
4 9 1 2 1
9(3 1)
2N N L L L
10 1 2 327N L L L
Cubic element
Quadratic element
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Finite Element Method by G. R. Liu and S. S. Quek41
HIGHER ORDER ELEMENTS
Higher order rectangular elements
(0,0)
0
(n,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
0 1 1 1
( )( ) ( )( ) ( )( )
( )( ) ( )( ) ( )
n k k nk
k k k k k k k n
l
[Zienkiewicz et al., 2000]
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Finite Element Method by G. R. Liu and S. S. Quek42
HIGHER ORDER ELEMENTS
Higher order rectangular elements (Contd)
1 2
34
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 )
41
( ) ( ) (1 ) (1 )4
1( ) ( ) (1 )(1 )
4
1
( ) ( ) (1 )(1 )4
D D
D D
D D
D D
N N N
N N N
N N N
N N N
1 1
5 3 1
1 1
6 2 3
1 17 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
D D
D D
D D
D D
N N N
N N N
N N N
N N N
N N N
(nine node quadratic element)
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Finite Element Method by G. R. Liu and S. S. Quek43
HIGHER ORDER ELEMENTS
Higher order rectangular elements (Contd)
Serendipity type:
1 2
34
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)
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Finite Element Method by G. R. Liu and S. S. Quek44
HIGHER ORDER ELEMENTS
Higher order rectangular elements (Contd)
1 2
34
5 6
7
8
910
11
12
2 21
32
29
32
13
29
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 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)
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Finite Element Method by G. R. Liu and S. S. Quek45
ELEMENT WITH CURVED
EDGES
4
2
3
8
1
5
7
6
1
4 2
5
3
6
1
2
3
4
56
1 2
34
5
6
7
8
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Finite Element Method by G. R. Liu and S. S. Quek46
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.
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Finite Element Method by G. R. Liu and S. S. Quek47
COMMENTS (GAUSS
INTEGRATION) Using a smaller number of Gauss points tends to
counteract the over-stiff behaviour associated with thedisplacement-based method.
Displacement in an element is assumed using shapefunctions. This implies that the deformation of the elementis 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 softerthan lower order ones. This is because using higher orderelements gives fewer constraint to the elements.
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Finite Element Method by G. R. Liu and S. S. Quek48
COMMENTS ON GAUSSINTEGRATION
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 explicitformulation tend to use one-point integration to achieve the
best performance in saving CPU time.
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Finite Element Method by G. R. Liu and S. S. Quek49
CASE STUDY
Side drive micro-motor
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Finite Element Method by G. R. Liu and S. S. Quek50
CASE STUDY
Elastic Properties of Polysilicon
Youngs Modulus, E 169GPa
Poissons ratio, 0.262
Density, 2300kgm-3
10N/m
10N/m
10N/m
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Finite Element Method by G. R. Liu and S. S. Quek51
CASE STUDY
Analysis no. 1: Von Mises stress
distribution using 24 bilinear
quadrilateral elements (41 nodes)
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Finite Element Method by G. R. Liu and S. S. Quek52
CASE STUDY
Analysis no. 2: Von Mises stress
distribution using 96 bilinear
quadrilateral elements (129 nodes)
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Finite Element Method by G. R. Liu and S. S. Quek53
CASE STUDY
Analysis no. 3: Von Mises stress
distribution using 144 bilinear
quadrilateral elements (185 nodes)
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Finite Element Method by G. R. Liu and S. S. Quek54
CASE STUDY
Analysis no. 4: Von Mises stress
distribution using 24 eight-nodal,
quadratic elements (105 nodes)
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Finite Element Method by G. R. Liu and S. S. Quek55
CASE STUDY
Analysis no. 5: Von Mises stress
distribution using 192 three-nodal,
triangular elements (129 nodes)
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CASE STUDYAnalysis
no.
Number / type of
elements
Total number
of nodes inmodel
Maximum
Von MisesStress (GPa)
124 bilinear,
quadrilateral41 0.0139
2 96 bilinear,quadrilateral
129 0.0180
3144 bilinear,
quadrilateral185 0.0197
424 quadratic,
quadrilateral105 0.0191
5 192 linear,
triangular129 0.0167
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