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Introduction to the Finite Element MethodLecture 18
P.S. Koutsourelakis
[email protected] Hollister Hall
November 22 2010
Last Updated: November 22, 2010
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Higher-order isoparametric elements
In many problems there might be a need to increase the numberof shape functions (without increasing the number of elements)and/or represent more accurately the boundary of the domain.
We discuss some higher order isoparametric elements thatachieve this goal.
There is no free lunch! Every time you increase the number ofshape functions you will also increase the number of unknownsand the size of the system you need to solve.
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Serendipity Elements
e
(xe1 , ye1 ) (xe2 , y
e2 )
(xe3 , ye3 )
(xe4 , ye4 )
(xe5 , ye5 )
(xe6, ye
6)
(xe7 , ye7 )
(xe8 , ye8 )
x
y
(1,1) (1,1)
(1, 1)(1, 1)
(0,1)
(1, 0)
(0, 1)
(1, 0)
x = x()
= (x)
11 22
334
4
55
66
77
88
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Serendipity Elements
Serendipity elements consist of additional nodes along the
boundaryObserve the numbering!
Let the mapping be defined as follows:
x(, ) =
8a=1 Na(, )x
ea
y(, ) = 8a=1 Na(, )yea x() =
8
a=1 Na()xea
To obtain the functions Na we first assume a form:x(, ) = b0 + b1 + b2 + b3 + b4
2 + b52 + b6
2 + b72
y(, ) = c0 + c1 + c2 + c3 + c42 + c5
2 + c62 + c7
2
where the parameters b and c should be determined by satisfying:
x(a, a) = xea
y(a, a) = yea
for a= 1, . . . , 8
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Serendipity Elements
We obtain the following shape functions:
N1 =14
(1 )(1 ) 12
(N8 + N5)N2 =
14
(1 + )(1 ) 12
(N5 + N6)
N3 =14
(1 + )(1 + ) 12
(N6 + N7)
N4 =14
(1 )(1 + ) 12
(N7 + N8)
N5 =
1
2 (1
2
)(1
)N6 =12
(1 + )(1 2)N7 =
12
(1 2)(1 + )
N8 =12
(1 )(1 2)
Note that N1 arises from the linearshape function of the 4-nodequad (which is equal to 1/2 at nodes 5 and 8) by subtracting1/2 N5 and 1/2 N8 in order to become zero at nodes 5 and 8.
Similarly for N2, N3, N4.
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Serendipity Elements
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Serendipity Elements
To find the local stiffness matrix ke:
ke =
e
(Be)T D Be d
=
11
11
(Be(, ))T D Be(, ) j(, ) dd
The strain displacement matrix Be:
31
=B
e316
de
161=
Be
dex,1dey,1.
.dex,16dey,16
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S di i El
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Serendipity Elements
Lets break Be in two parts:
=
uxxuyyuxy
+uyx
= 1
j(, )
y, y, 0 0
0 0 x, x,x, x, y, y,
uxuxuyuy
or:
= Be1
uxuxuy
uy
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S di it El t
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Serendipity Elements
and the second part:
uxuxuyuy
=
N1, 0 . . . . . . N9, 0
N1, 0 . . . . . . N9, 0
0 N1, . . . . . . 0 N9,0 N1, . . . . . . 0 N9,
de
or:
uxuxuyuy
= B
e2 d
e
Hence:B
e316
= Be134
Be2
416
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S di it El t
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Serendipity Elements
When is the Jacobian determinant not zero?
acceptable position for node 5L/4L/4
12
3
4
5
6
7
8
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S di it El t
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Serendipity Elements
Example: Find local force vector
q3q4
1 2
34
5
6
7
8
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Lagrange Elements
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Lagrange Elements
(xe2 , ye2 )
x
y
x = x()
= (x)
11 22
334
4
55
66
77
88 99
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Lagrange Elements
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Lagrange Elements
Lagrange elementsconsist of additional nodes in the interior ofthe element
Observe the numbering!
Lagrange polynomials. Suppose one is given npairs of values
(i, i = (i)). The the function can be approximated as follows:
L11 + L22 + . . . + Lnn
where:L1 =
(2)(3)...(n)(21)(31)...(n1)
L2 =(1)(2)...(n)
(12)(33)...(n3)
. . .
Ln =(1
)(2
)...(n1
)(1n)(2n)...(n1n)
Note: Li(j) = i,j
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Lagrange Elements
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Lagrange Elements
For example, for n = 2, 1 = 1, 2 = +1:
L(2)1 =
(2)(21)
= 12
(1 )
L(2)2 =
(1)(12)
= 12
(1 + )
For n = 3 and 3 = 0:
L(3)1 =
(2)(3)(21)(31)
= 12(1 )
L(3)2 =
(1)(3)(12)(32)
= 12(1 + )
L(3)3 = (1
)(2
)(13)(23) = 12 (1 2)
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Lagrange Elements
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Lagrange Elements
1 2
34
5
6
7
89
To generate the shape functions for plane isoparametric elements, it
suffices to multiply the Lagrange polynomials w.r.t. to and of
appropriate order. For example, for the 9 node element:
N1(, ) = L(3)1 () L
(3)1 () =
14
(1 )(1 )
N6(, ) = L(3)
2 () L(3)
3 () =1
4 (1 + )(1
2
)
or :
N9(, ) = L(3)2 () L
(3)2 () = (1
2)(1 2) bubble function
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Lagrange Elements
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Lagrange Elements
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Lagrange Elements
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Lagrange Elements
Naturally, we could derive the same shape functions if we follow the
procedure discussed for serendipity elements and use interpolating
functions of the form:
b0 + b1 + b2 + b3 + b42 + b5
2 + b62
+ b72
+ b82
2
Conversely, we can use the Lagrange polynomials to re-derive the
shape functions of the isoparametric elements we have seen before.
For example, the 4-node quadrilateral:
N1(, ) =1
4(1 )(1 ) = L
(2)1 ()L
(2)1 ()
Or the shape functions of lateral nodes of the 8-node serendipity
element, e.g.:
N5(, ) = L(3)3 ()L
(2)1 () =
1
2(1 2)(1 )
One has to be more careful for the shape functions of the corner nodes!
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Isoparametric Quadrilateral Elements with Variable
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Isoparametric Quadrilateral Elements with Variable
Number of nodes
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Isoparametric Quadrilateral Elements with Variable
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Isoparametric Quadrilateral Elements with Variable
Number of nodes
5-node quadrilateral:
x
y
x = x()
= (x)
11 22
33 44
55
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Isoparametric Quadrilateral Elements with Variable
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Isoparametric Quadrilateral Elements with Variable
Number of nodes
The shape function of lateral nodecan be found from Lagrange
polynomials:
N5 = L(3)3 () L
(2)1 () =
1
2(1 2)(1 )
To find the shape functions of the corner nodeswe use the onesfrom the 4-node quadrilateral and subtract 1/2N5 IF NEEDED sothat they become 0 at all other nodes. Hence:
N1 =14
(1 )(1 ) 12
N5N2 =
1
4
(1 + )(1 ) 1
2
N5
and:N3 =
14
(1 + )(1 + )
N4 =14
(1 )(1 + )
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Summary of Shape functions for Isoparametric
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Summary of Shape functions for Isoparametric
Quadrilateral Elements
i = 5 i = 6 i = 7 i = 8 i = 9N1 =
14
(1 )(1 ) 12
N5 12
N814
N9N1 =
14
(1 )(1 ) 12
N5 12
N614
N9N1 =
14
(1 )(1 ) 12
N6 12
N714
N9N1 =
14
(1 )(1 ) 12
N7 12
N814
N9
N5 = 12 (1 2)(1 ) 12 N9N5 =
12
(1 + )(1 2) 12
N9N5 =
12
(1 2)(1 + ) 12
N9N5 =
12
(1 )(1 2) 12
N9
N9 = (1 2)(1 2)
The shape functions in columns i = 5 9 are activated only if therespective node appears in the element and they are added tothe appropriate shape function in the first column.
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