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Families of Shape Functions, Numerical Integration, Physical and Master Element, Concept and Mapping,
Element Stiffness and Load Vector Calculations
PM Mohite
Department of Aerospace Engineering Indian Institute of Technology Kanpur
TEQIP School on Computational Methods in Engineering Application
Knowledge Incubation for TEQIP
Single variable or multivariable problems
The phenomenon is represented through partial differential equations
Finite element formulation involves same steps as one dimensional case
Boundary of a two dimensional domain Ω, in general, is a curve
Two dimensional shapes like triangle and rectangle/quadrilaterals are used to approximate the geometry.
Two Dimensional Problems:
• Poisson equation
gradient operator
These equation represent
• Heat conduction
• Electrostatics
• Stream function
• Magnetic statics
• Torsion of non-circular sections, Transverse deflection of elastic membranes
fuk in
yj
xi
ˆˆ
Sample Boundary Value Problems:
it becomes Poisson’s equation.
Finite element discretization:
• The number, shape and type (linear, quadratic,……….) of elements should be such that the geometry of the domain is accurately represented. • Density of elements should be such that the regions of large gradients of the solution are adequately modeled.
11 12 21 22 , in u u u u
k k k k f x yx x y y x x
12 21 11 22with 0andk k k k k
Sample Boundary Value Problems: Heat Conduction
Weighted residual form:
We know that,
Therefore, Similarly,
x
wqw
x
qwq
xx
xx
dAwdAw
y
uk
x
uk
yy
uk
x
uk
x22221211
x
wqwq
xwq
xxxx
Heat Conduction Equation: Weak Formulation
y
wqwq
ywq
yyyy
Now using the Divergence Theorem, and where nx and ny are the direction cosines or the components of the unit normal vector n
dsnwqdAwq
xxxx
dsnwqdAwq yyy
y
Heat Conduction Equation: Weak Formulation
ˆ ˆ ˆ ˆcos sinx yn i n j i j n
writing the weak form over an element
Looking at boundary term:
• Primary variable – essential BC and
• coefficient of weight function in boundary term form the natural BC.
dAwfy
uk
x
uk
y
w
y
uk
x
uk
x
w
e
222112110
dsy
uk
x
ukn
y
uk
x
uknw
e yx
22212211
uw
Heat Conduction Equation: Weak Formulation
qn - is the secondary variable.
qn - projection of the vector along the unit normal n
- positive outward from the surface as the as we more counter-clock wise along the boundary
Thus,
k u
on e
11 12 21 220e e
n
w u u w u uk k k k wf dA wq ds
x x y y x y
y
uk
x
ukn
y
uk
x
uknq yxn 22211211
Heat Conduction Equation: Weak Formulation
Or where, - bilinear from, is not symmetric. It is symmetric only when - linear from Quadratic Potential:
wFuwB ,
eedswqdAfwwF n
uwB , 2112 kk
e
dAy
uk
x
uk
y
w
y
uk
x
uk
x
wuwB 22211211,
Heat Conduction Equation: Weak Formulation
F w
,I u B u u F u
Approximation of over as
- are the nodal values of at node.
- are the Lagrange interpolation functions with the property
Substitute in the weak form and to get equation
replace w by to give
u e
1
, , ,
NODEee
h j jj
u x y u x y u x y
eju
ej ,e
i j j ijx y
1
, ,
NODEe e
h j j
j
u x y u x y
ei
11 12 21 22
1
0e
NODEj j j ji i
j
k k k k dAx x y y x y
ee
dsqdAf nii
Heat Conduction Equation: Finite Element Formulation
, 1,2, ,i j NODE
This leads to: where, and In matrix notation, is symmetric only when
1
NODEe e e eij j i i
j
k u f Q
dAy
kx
kyy
kx
kx
ke
jiijjie
ij
22211211
dsqQdAff e
in
e
i
e
i
e
iee
,
eeee Qfuk
2112 kk ek
Heat Conduction Equation: Finite Element Formulation
The approximate solution over an element, for the assurance of convergence to the actual solution as the number of elements is increased and their size is decreased, must satisfy: 1) It must be continuous over the element and differentiable as required by the
weak form. - It ensures a nonzero coefficient matrix 2) It must not allow a strain to appear if the nodal displacements are compatible with a rigid-body displacements.
yxuh ,
Interpolation Functions: Requirements
3) If the nodal displacements are compatible with a uniform strain in the elements, then the interpolation functions must yield this strain for the nodal displacements. In other fields: If the nodal variables are compatible with uniform states of the variable and any of its derivatives up to the highest in the appropriate quadratic functional , then these uniform states must be preserved in the element as it shrinks to zero size. This condition is called completeness. - This requirement is necessary in order to capture all possible states of the actual solution
Interpolation Functions: Requirements
I
Example: If a linear polynomial without the constant term is used to represent the temperature distribution in a one-dimensional system, the approximate solution can never be able to represent a uniform state of temperature in the element.
Interpolation Functions: Requirements
4) The interpolation functions should be chosen so that the strain remains finite at the boundary. Note, however, that the strain can be indeterminate at a boundary. For example, if the strains involve only the first derivatives, then the displacement field must be continuous across a boundary. (strain energy must be finite) In other fields: The dependent variable and all its derivatives up to a one less than the highest order derivative in the appropriate quadratic functional should be continuous at the element interfaces. (The quadratic potential must be finite) This requirement is called compatibility condition. Elements satisfying these requirements are called conforming elements.
Interpolation Functions: Requirements
Degree of Compatibility achieved by interpolation functions at the element: • continuity is achieved when the field variable only and none of its derivatives maintain continuity at an interelement interface. • continuity is achieved when the field variable and its first derivatives maintain continuity at an interelement interface. • continuity is achieved when the field variable, its first derivatives and second derivatives all maintain continuity at an interelement interface.
Interpolation Functions: Requirements
0C
1C
2C
Continuity Requirement for completeness and compatibility: For assurance of finite element convergence for a functional having as the highest order of a derivative – • continuity – requirement for completeness • continuity – requirement for compatibility
Interpolation Functions: Requirements
p
pC
1pC
Lagrange Family: • Interpolation functions derived using the dependent unknown only and not its derivatives, at nodes. • - continuity
Hermite Family: • Interpolation functions derived using the dependent unknown and its derivatives as well, at nodes. • continuity – continuity of the first derivative of the dependent unknown. Cubic interpolation functions • continuity – continuity of the first and second derivatives of the dependent unknown. Quintic interpolation functions
Interpolation Functions: Types of Families
0C
1C
2C
must be at least linear in x and y.
Let
Three linearly independent terms, linear in both x and y.
to be expressed terms of nodal values of . Hence, we must have three nodes.
This is possible for a triangle with its vertices as nodes.
ycxccyxuh 321,
ic
e
i
Linear Interpolation Functions:
hu
• Consider a polynomial
• Four linear independent terms, linear in x, y with a bilinear term in x and y.
• This requires an element with four nodes.
• Two possible geometries:
- A triangle with the three nodes at its vertices and the fourth node at its centre or centroid.
- A rectangle with the nodes at the vertices
xycycxccyxuh 4321,
Linear Interpolation Functions:
• A triangle with the fourth node at its centre does not provide a single valued variation of primary variable at interelement boundaries.
• This results in incompatible variations of primary variable there. Therefore, this is not admissible.
• Thus, the only possible element with assumed polynomial approximation is rectangle with the nodes at the four vertices
Linear Interpolation Functions:
• Consider a polynomial with five constants
• This is an incomplete quadratic polynomial
• Note that the terms x2 and y2 can not be varied independently.
• The polynomial requires an element with five nodes.
• The possible geometry is a rectangle with a node at each vertex and the fifth node at its centre.
• This again does not give single valued variation of primary variable.
22
54321, yxcxycycxccyxuh
Quadratic Interpolation Functions:
• Consider a polynomial with six constants
• This is a complete quadratic polynomial
• The polynomial requires an element with six nodes.
• The possible geometry is a triangle with a node at each vertex and a node at the midpoint of each side.
2
6
2
54321, ycxcxycycxccyxuh
Quadratic Interpolation Functions:
A triangular element with three nodes at vertices
Approximate solution over an element is of the form
We need three conditions to find the unknowns Ci ‘s.
Using nodal values of solution we get three conditions
ycxccyxuh 321,
Interpolation Functions: Linear Triangular Element
1 2
3
Thus, we have
Or in a matrix form
unknowns Ci ‘s are found as
where,
Interpolation Functions: Linear Triangular Element
33321333
23221222
13121111
,
,
,
ycxccyxuu
ycxccyxuu
ycxccyxuu
h
h
h
3
2
1
3
2
1
33
22
11
3
2
1
1
1
1
c
c
c
A
c
c
c
yx
yx
yx
u
u
u
3
2
1
3
2
1
1
c
c
c
u
u
u
A
321
321
321
1
2
1
eAA
3212 eA
Solving for Ci ‘s:
- determinant of
- area of the triangle
where,
Interpolation Functions: Linear Triangular Element
3322113
3322112
3322111
2
1
2
1
2
1
uuuA
c
uuuA
c
uuuA
c
e
e
e
order natural ain permute ,, and kjikji
xx
yy
yxyx
kji
kji
jkkji
eA2
eA
A
Substituting values of Ci ‘s:
where,
With the properties:
Interpolation Functions: Linear Rectangular Element
3
1
332211332211332211
,
2
1,
i
e
i
e
i
e
h
yxu
yuuuxuuuuuuA
yxu
3,2,1 2
1, i
Ayx e
i
e
i
e
i
e
e
i
,, ij
e
j
e
j
e
i yx
3
1
,1i
e
i
,03
1
i
e
i
x
0
3
0
i
e
i
y
Lagrange Interpolation Functions: Linear Triangular Element
Linear interpolation functions for the three-noded triangular element
1 2
3
1
ψ1
1 2
3
1
ψ2
1 2
3
1
ψ3
Lagrange Interpolation Functions: Triangular Geometries to Avoid
• The nodes are almost in-line.
• The resulting coefficient matrix is nearly singular or non-invertible.
L
a L>>a
L
h L>>h
Lagrange Interpolation Functions: Linear Triangular Element
Shapes that should be avoided in the meshing.
A rectangular element with four nodes at the corner and of sides a and b.
Approximate solution over an element is of the form
We need four conditions to find the unknowns Ci ‘s.
yxcycxccyxuh 4321,
Interpolation Functions: Linear Rectangular Element
1 2
4 3
y
x
x
y
b
a
Using the nodal solution values, we get four conditions: Gives the values of constants as Putting these back and re-arranging,
110,0 cuuh 2120, cacuauh
abcbcaccubauh 43213, bccubuh 314,0
ab
uuuuc
b
uuc
a
uucuc 2143
414
312
211 ,,,
b
y
a
x
b
yu
b
y
a
xu
b
y
a
x
a
xu
ab
yx
b
y
a
xuyxuh 4321 1,
Interpolation Functions: Linear Rectangular Element
44332221, uuuuyxuh
b
x
a
x
b
y
a
x ee 1,11 21
b
y
a
x
b
y
a
x ee
1, 43
b
yy
a
xxyx iii
111,1e
i
,,e
i ijjj yx
4
1
1i
e
i
where, And can be given as With the properties:
Interpolation Functions: Linear Rectangular Element
Lagrange Interpolation Functions: Linear Rectangular Element
Linear interpolation functions for the four-noded rectangular element
ψ4
1
2
3
4 1
1
1
2
3
4
ψ2
1
2
3
1 4
ψ1
1 1
2
3
4
ψ3
b
y
a
x
b
y
a
x
b
x
a
x
b
y
a
x eeee
1,,1,11 4321
• Can also be obtained from tensor product of 1D linear interpolation functions associated with sides 1-2 and 2-3
32
411
1
b
y
b
y
a
xa
x
Interpolation Functions: Linear Rectangular Element
a
a
x1
x
a
x
b
b
y1
y
b
y
Another approach: Derivation of We know that and And zero on the lines Therefore, is of the form: Putting first condition that Hence, is In a similar way, the other interpolation functions can be derived.
4,3,2for 0,1 iyx ii
e 1, 111 yxe
e
1
byax and
yxe ,1
ybxacyxe 11 ,
0,0at 1,1 yxyxeab
c1
1
b
y
a
xybxa
abyxe 11
1,1
Interpolation Functions: Linear Rectangular Element
e
1
• The interpolation functions of higher degree can be systematically developed with the help of Pascal’s Triangle.
• It contain the terms of polynomials of various degrees in the two coordinates x and y.
• x and y denote the local coordinate (and not the global coordinates of the problem).
• Shape of the triangle need not be equilateral triangle.
• order triangular element has n nodes given by the relation
• A complete polynomial of degree is given by
thp
212
1 ppn
n
i
n
j
jj
sr
i uyxayxu1 1
),( psr with
Interpolation Functions: Higher Order Element
thp
Interpolation Functions: Pascal’s Triangle
1
x y
2x 2y
3y
4y
5y
6y
7y
3x
4x
5x
6x
7x
xy
yx2 2xy
yx3 22 yx3xy
yx4 23 yx32 yx
4xy
yx5 24 yx 33 yx42 yx
5xy
yx6 25 yx 34 yx 43 yx52 yx 6xy
Polynomial Degree Element
We choose the terms
in the polynomial
Interpolation Functions: Pascal’s Triangle
1
x y
2x 2y
3y
4y
5y
6y
7y
3x
4x
5x
6x
7x
xy
yx2 2xy
yx3 22 yx3xy
yx4 23 yx32 yx
4xy
yx5 24 yx 33 yx42 yx
5xy
yx6 25 yx 34 yx 43 yx52 yx 6xy
0 1
1, cyxuh
Polynomial Degree Element
Interpolation Functions: Pascal’s Triangle
1
x y
2x 2y
3y
4y
5y
6y
7y
3x
4x
5x
6x
7x
xy
yx2 2xy
yx3 22 yx3xy
yx4 23 yx32 yx
4xy
yx5 24 yx 33 yx42 yx
5xy
yx6 25 yx 34 yx 43 yx52 yx 6xy
1 3
ycxccyxuh 321,
Polynomial Degree Element
Interpolation Functions: Pascal’s Triangle
1
x y
2x 2y
3y
4y
5y
6y
7y
3x
4x
5x
6x
7x
xy
yx2 2xy
yx3 22 yx3xy
yx4 23 yx32 yx
4xy
yx5 24 yx 33 yx42 yx
5xy
yx6 25 yx 34 yx 43 yx52 yx 6xy
2
6
2
6
2
54
321,
ycxcxyc
ycxccyxuh
Polynomial Degree Element
We choose the terms
in the polynomial
Interpolation Functions: Pascal’s Triangle
1
x y
2x 2y
3y
4y
5y
6y
7y
3x
4x
5x
6x
7x
xy
yx2 2xy
yx3 22 yx3xy
yx4 23 yx32 yx
4xy
yx5 24 yx 33 yx42 yx
5xy
yx6 25 yx 34 yx 43 yx52 yx 6xy
0
1
2
3
4
5
6
7
1
3
6
10
15
21
28
212
1 ppn
Lagrange Interpolation Functions: Quadratic Triangular Element
Geometric variation of Lagrangian interpolation functions
P is an arbitrary point inside a linear triangular element Let Ai, Aj and Ak be the area of triangles formed by point P with an edge of the triangle. For example, Ak is the area formed by edge i-j and point P. Define: Natural /triangular/ barycentric coordinate and A
AL
A
AL
A
AL k
k
j
ji
i ,,
Interpolation Functions: Using Area Co-ordinates
P i
j
k
(i)
(j)
(k)
iL
kji AAAA
• For P at node i: • For P at node j: • For P at node k: Connection with triangular element:
Interpolation Functions: Using Area Co-ordinates
0,0,1 kji LLL
0,1,0 kji LLL
1,0,0 kji LLL
P 1
2
3
(1)
(2)
(3)
0,0,1 321 LLL
0,0,1 312 LLL
0,0,1 213 LLL3L
2L
1L
• We needed to invert coefficient matrix when we derived interpolation functions. This can be avoided in this approach. Consider a quadratic triangular element We express natural co-ordinate Li at any point by using superscript to identify the nodes. Example: Point P moves to node 6.
Interpolation Functions: Using Area Co-ordinates
P 1
2
3
(1)
(2)
(3) 4
5
6
,2
116
1 A
AL
P 1
2
3
(1) (3) 4
5
6
,0026
2 AA
AL
2
136
3 A
AL
• Point P moves to node at 4. • Point P moves to node at 5.
Interpolation Functions: Using Area Co-ordinates
,2
114
1 A
AL
,2
124
2 A
AL 0
034
3 AA
AL
P
1
2
3
(1) (2)
4 5
6
,0015
1 AA
AL
,2
125
2 A
AL
2
135
3 A
AL
P 1
2
3 (2)
(3) 4
5
6
• For a cubic triangular element Example: Point P moves to node 10. Example: Point P moves to node 4.
Interpolation Functions: Using Area Co-ordinates
,3
1110
1 A
AL ,
3
1210
2 A
AL
3
1310
3 A
AL
P 1
2
3
(1)
(2)
(3) 4
5 6
7
8 9
10
P 1
2
3
(1)
(2)
(3) 4
5 6
7
8 9
10
P
1
2
3
(1)
(2)
4
5 6
7
8 9
10 ,
3
214
1 A
AL ,
3
124
2 A
AL 0
034
3 AA
AL
To get the interpolation functions, we first transform the Lagrangian interpolation formula from Cartesian to natural coordinates. where j = one of the three natural coordinates n (=p) = order of interpolation polynomial r = free index from 1 up to the total number of nodes Thus, for each node:
1
1 for 1
r
jnL
rj
j jr
i
nL iL L nL
i
Interpolation Functions: Using Area Co-ordinates
1 for 0
r
j jr
L L nL
1 2 3r r r rN L L L L L L
Example: For quadratic triangular element (n=p=2) For N1, put r=1 and considering j=1 first Thus, and . Therefore,
Interpolation Functions: Using Area Co-ordinates
1
12 2 1 2r
jnL L
1
12 21 1
1 11 1
1 11 1
2 1 2 1
2 1 1 2 2 1 2 1
1 2
L
i i
L i L iL L
i i
L LL L
2 31 10, 0L L L L 1 1 12 1N L L
Example: For quadratic triangular element (n=p=2) For N4, put r=4 and considering j=1 first Thus, Now, considering j=2
Interpolation Functions: Using Area Co-ordinates
4
1
12 2 1
2
r
jnL L
4
12 11 1 1
1 141 1
2 1 2 1 2 1 12
1
L
i i
L i L i LL L L
i i
4
2
12 2 1
2
r
jnL L
2 242L L L
Example: For quadratic triangular element (n=p=2) For N4, considering j=3 Thus, For remaining nodes,
Interpolation Functions: Using Area Co-ordinates
4
32 2 0 0r
jnL L 3 41L L
4 1 24N L L
2 2 2 3 3 3
5 2 3 6 1 3
2 1 , 2 1 ,
4 , 4
N L L N L L
N L L N L L
Polynomial Degree = 1 1
Interpolation Functions: Pascal’s Triangle
x y
2x 2y
3y
4y
5y
6y
7y
3x
4x
5x
6x
7x
xy
yx2 2xy
yx3 22 yx 3xy
yx4 23 yx32 yx
4xy
yx5 24 yx 33 yx42 yx 5xy
yx6 25 yx34 yx 43 yx 52 yx 6xy
xycycxccyxuh 4321,
Polynomial Degree = 2
1
Interpolation Functions: Pascal’s Triangle
x y
2x 2y
3y
4y
5y
6y
7y
3x
4x
5x
6x
7x
xy
yx2 2xy
yx3 22 yx 3xy
yx4 23 yx32 yx
4xy
yx5 24 yx 33 yx42 yx 5xy
yx6 25 yx34 yx 43 yx 52 yx 6xy
22
9
2
8
2
7
2
6
2
54
321,
yxcxycyxc
ycxcxyc
ycxccyxuh
Polynomial terms Polynomial Degree = 3
can be chosen 1
Interpolation Functions: Pascal’s Triangle
x y
2x 2y
3y
4y
5y
6y
7y
3x
4x
5x
6x
7x
xy
yx2 2xy
yx3 22 yx 3xy
yx4 23 yx32 yx
4xy
yx5 24 yx 33 yx42 yx 5xy
yx6 25 yx34 yx 43 yx 52 yx 6xy
32
16
3
15
3
14
33
13
23
12
3
11
3
10
22
9
2
8
2
7
2
6
2
54
321,
yxcxycycyxc
yxcyxcxc
yxcxycyxc
ycxcxyc
ycxccyxuh
Polynomial terms Polynomial Degree Element
can be chosen 1
Interpolation Functions: Pascal’s Triangle
x y
2x 2y
3y
4y
5y
6y
7y
3x
4x
5x
6x
7x
xy
yx2 2xy
yx3 22 yx 3xy
yx4 23 yx32 yx
4xy
yx5 24 yx 33 yx42 yx 5xy
yx6 25 yx34 yx 43 yx 52 yx 6xy
1
2
3
4
2
No. of Nodes 1n p
Polynomial Degree Element
Rectangular array
Interpolation Functions: Pascal’s Triangle in Rectangular Array Form
1
2
3
2
No. of Nodes 1n p
333233
232222
32
321
yxyxxyy
yxyxxyy
yxyxxyy
xxx
0
The Lagrange interpolation functions associated with rectangular elements can be obtained from corresponding one-dimensional Lagrange interpolation functions by taking tensor product.
T
b
b
b
b
b
a yy
byy
ayy
aa
xx
aaa
axx
aa
axa
x
2
2
4
2
2
21
2
21
21
963
852
741
2
2
2
)(
)(
)(
)(
)(
)(
)(
)(2
)(2
Interpolation Functions: Higher Order Rectangular Elements
114
1,11
4
1
,114
1,11
4
1
41
21
2 2 2 2 2 21 2 3
2 2 2 2 2 24 5 6
2 2 2 2 2 27 8 9
1 1 1, 1 , ,
4 2 4
1 1 11 , 1 1 , 1 ,
2 4 2
1 1 1, 1 ,
4 2 4
Interpolation Functions: Master Rectangular Elements
1,1 1,1
1,1 1,1
1 2
34
1,1 1,1
1,1 1,1
1 2 3
4 56
7 8 9
Tensor product of 1D functions:
Lagrange Interpolation Functions: Quadratic Rectangular Element
Geometric variation of Lagrangian interpolation functions
Lagrange Interpolation Functions: Quadratic Rectangular Element
Geometric variation of
Lagrange interpolation
functions:
• The shape function presented earlier are for the interpolation of primary variables.
• The Hermite cubic interpolation functions based on the interpolation of can be generated.
Interpolation for
Variable u
derivative
2
, , ,u u u
ux y x y
2 21
1 216
i i i i i
Hermite Interpolation Functions: Master Rectangular Element
u
2 21
2 216
i i i i
Derivative
Derivative
where, i=1,….4 are the nodes of the element.
2 2
1
11 1
16i i i i i
2 21
2 116
i i i i i
Hermite Interpolation Functions: Master Rectangular Element
u
2u
Polynomial Degree Element
Terms inside this
cone are not
included
1
Interpolation Functions: Pascal’s Triangle
2
3
4
x y
2x 2y
3y
4y
5y
6y
7y
3x
4x
5x
6x
7x
xy
yx2 2xy
yx3 22 yx 3xy
yx4 23 yx32 yx
4xy
yx5 24 yx 33 yx42 yx 5xy
yx6 25 yx34 yx 43 yx 52 yx 6xy
Polynomial Degree = 2
1
Interpolation Functions: Pascal’s Triangle
2
8
2
7
2
6
2
54
321,
xycyxc
ycxcxyc
ycxccyxuh
x y
2x 2y
3y
4y
5y
6y
7y
3x
4x
5x
6x
7x
xy
yx2 2xy
yx3 22 yx 3xy
yx4 23 yx32 yx
4xy
yx5 24 yx 33 yx42 yx 5xy
yx6 25 yx34 yx 43 yx 52 yx 6xy
Polynomial Degree = 3
1
Interpolation Functions: Pascal’s Triangle
3
12
3
11
3
10
3
9
2
8
2
7
2
6
2
54
321,
xycycyxc
xcxycyxc
ycxcxyc
ycxccyxuh
x y
2x 2y
3y
4y
5y
6y
7y
3x
4x
5x
6x
7x
xy
yx2 2xy
yx3 22 yx 3xy
yx4 23 yx32 yx
4xy
yx5 24 yx 33 yx42 yx 5xy
yx6 25 yx34 yx 43 yx 52 yx 6xy
Polynomial terms Polynomial Degree Element
can be chosen 1
Interpolation Functions: Pascal’s Triangle
2
3
4
x y
2x 2y
3y
4y
5y
6y
7y
3x
4x
5x
6x
7x
xy
yx2 2xy
yx3 22 yx 3xy
yx4 23 yx32 yx
4xy
yx5 24 yx 33 yx42 yx 5xy
yx6 25 yx34 yx 43 yx 52 yx 6xy
Derivation of : vanishes on the lines
Interpolation Functions: Derivation of Serendipity Functions
1,1 1,1
1,1 1,1
1 2 3
4 5
6 7 8
1 0 1 0
1 0
1 0
1 0
1 0
1 0
1 0
1
1
1 0, 1 0,
1 0
Therefore, is of the from
It has a value of 1 at
vanishes along the lines
Let
1
1 , 1 1 1C
1
11 1 1
4
2 1 0,1 0,1 0
2 1 1 1c
2 1 at 0,-1 12c
Interpolation Functions: Derivation of Serendipity Functions
1, 1 14
c
22
11 1
2
13 4
214 4
215 2
1 1 1
1 1
1 1
16 4
217 2
18 4
1 1 1
1 1
1 1 1
Similarly, Note: The serendipity interpolation functions are different than Lagrange interpolation functions as internal nodes are not present.
Interpolation Functions: Derivation of Serendipity Functions
• The set of interpolation functions of polynomial degree p should be in the set of interpolation functions of polynomial degree (p+1). • The number of interpolation functions which do not vanish at vertices and sides should be smallest possible. • Example: 1D Hierarchic interpolation functions
Interpolation Functions: Hierarchic Functions
• Nodal Interpolation functions:
- Four nodal interpolation functions
- Exactly same as the functions for four noded quadrilateral element
Hierarchic Shape Functions: Rectangular Elements
114
1,11
4
1
,114
1,11
4
1
41
21
• Side Nodes Interpolation functions:
- There are interpolation functions associated with the sides of finite elements with the order of approximation
- For Side 1
- For Side 2
- For Side 3
- For Side 4
Hierarchic Shape Functions: Rectangular Elements
piii ,,2 12
11
14 p
2p
piii ,,2 12
12
piii ,,2 12
13
piii ,,2 12
14
• Internal Nodes Interpolation functions:
- There are interpolation functions associated with the internal nodes of finite elements with the order of approximation
and so on.
Hierarchic Shape Functions: Rectangular Elements
0 0
2 2 3 21 2
0 0
2 3 4 23 4
0 0
3 3 2 45 6
, ,
, ,
, ,
2 3 2p p 4p
where, polynomial of degree j is:
- is Legendre polynomial.
Using the properties of Legendre polynomials
Hierarchic Shape Functions: Rectangular Elements
1
1
2 1, 2,3,
2j j
jP t dt j
2
1
2 2 1j j jP P
j
jP
Legendre polynomial is given by Bonnet Recursion formula:
with
and
Hierarchic Shape Functions: Rectangular Elements
1 11 2 1 , 1,2,n n nn P n P nP n
1
1
2 for
2 1
0 for i j
i jP P d i
i j
0 11,P P
• Nodal Interpolation functions:
- Three nodal interpolation functions -
- Exactly same as the functions for three noded triangular element
• Side Modes:
- There are side modes and vanish on the other two sides
Sides 1:
where,
Hierarchic Shape Functions: Rectangular Elements
1 2 3, ,L L L
3 1p
1
2 1 2 1 , 2, ,i iL L L L i p
211 , 2, ,
4j j j p
For example:
• Internal Modes:
- There are internal modes
Mode 1:
Other modes:
and so on.
Hierarchic Shape Functions: Rectangular Elements
1 2 2p p
0
1 2 31 L L L
22 3 4
76, 10 , 5 1
8
0 0
1 2 3 1 2 1 1 2 3 1 32 3, 2 1L L L P L L L L L P L
• Non rectangular domains cannot be represented using rectangular elements. However, it can be represented more accurately by quadrilateral elements.
• Interpolation functions are easily available for a rectangular element
• And it is easy to evaluate integral over rectangular geometric
• Therefore, we transform the finite element integral statement over quadrilaterals to a rectangular.
• Similarly, the finite element integral statement over a triangular element is transformed to an isosceles triangle.
• Caution: Transformation is for numerical integration purpose only ! The actual domain is not mapped to another domain.
Mapping:
• Then numerical integration schemes like Gauss-Legendre are used to evaluate the integrals.
• These schemes require that the integral be evaluated on a specific domain or with respect to a specific coordinate system.
• In general it is a coordinate system such that
• Transformation between physical element and master element
where, interpolation functions over master element
, 1,1
e
1 1
ˆ ˆ, , ,
m me e e ej j j j
j j
x x y y
e
j
Mapping:
In general, the dependent variables(s) are approximated by expressions of the form
The degree of approximations used for the geometry and the dependent variables(s) are different.
Depending upon it, the finite element formulations are classified as:
- Sub-parametric
- Iso-parametric
- Super parametric
1
, ,
ne ej i
j
u x y u x y
Mapping:
a) super- parametric : The approximation used for the geometry is of
higher order than used for the dependent variable (s).
b) iso-parametric : Equal degree of approximation for geometry and
dependent variable(s).
- This is extensively used in h- version of finite element programs.
nm
nm
Mapping:
c) Sub-parametric : Approximation order used for geometry is lower than that used for primary variable (s).
• Coordinate transformations is only for the purpose of numerical integration.
• When an element is transformed to its master element for the purpose of numerical integration, the integrand must also be expressed in terms of the coordinates of the master element.
nm
,
Mapping:
For example, consider the stiffness coefficient Here, we need to relate with and using the transformation. • can be expressed in terms of local coordinate and . Hence, the chain rule for partial differentiation gives In matrix form,
, ( , ) ( , )e
e ee ej je e ei i
ij i jk a x y b x y c x y dxdyx x y y
,e ei i
x y
e
i
e
i
),( yxe
i
e e ei i i
e e ei i i
x y
x y
x y
x y
e ei i
eeii
x y
x
x y
Jacobian Matrix
Mapping :
We need to relate with
Hence, we need the inverse of the Jacobian matrix.
must be non singular.
Now, using the transformation, we can write
,e ei i
x y
ˆ ˆand
e ei i
1
ˆ
ˆ
eeii
e ei i
xJ
y
J
1 1
1 1
ˆ ˆ, ,
ˆ ˆ,
em m ej i
j j
j j
em m ej i
j j
j j
x yx y
x yx y
Mapping :
1 1
1 1
ˆ ˆ
ˆ ˆ
m me ei i
j j
j j
m me ei i
j j
j j
x y
J
x y
1 11 2
2 2
1 2
ˆˆ ˆ
ˆˆ ˆ
m
m
m m
x y
x y
x y
A necessary and sufficient condition for to exist is that the determinant , called the Jacobian, be non negative at every point in . Thus, the function must be continuous, differentiable and invertible.
1J J
,
det 0x y x y
J J
),(),,( yxyx
Mapping :
• In case of higher order triangular and rectangular elements the placement of edge and interior nodes is restricted.
• It can be shown that the side nodes should be placed at a distance greater than a quarter of the length of the side form either corner node.
)21(4
)21(4
by
ax
1 2 2 1 2 2 1 0J b a
Mapping :
0,0 1,0
0,1
1 2
3
0
0
1
4
56
x
y
1 2
3
4
56 ,a b
0.25
0.25
Mapping the elemental area dA from x,y to We have Now, the derivatives of the shape functions with respect to x,y can be given as
ˆ
ˆˆ ˆ ˆ ˆ
dA dx dy dx dy k
x x y yd i d j d i d j k
x y y yd d
J d d
1 *
ˆ ˆ
ˆ ˆ
e eei ii
e e ei i i
xJ J
y
Mapping : d d
Now, using this we can write the element calculations over master element as or
11 12 11 12
ˆ
21 22 21 22
ˆ ˆˆ ˆˆ
ˆ ˆ
e
j je i iij i j
j ji i
i i i ii j
k a b c dxdyx x y y
a J J J J
b J J J J c J d d
Mapping :
ˆ
,eijk F d d
Linear Master Element Linear Physical Element
• The affine mapping essentially stretches, translates, and rotates the triangle.
• Straight or planar faces of the reference cell are therefore mapped onto straight or planar faces in the physical coordinate system.
• Preferred when all the edges of an element are straight.
Physical to Master Element Mapping: Linear Triangular Element
0,0 1,0
0,1
1 2
3
0
0
1
x
y
1
2
3
3
1
3
1
ˆ ,ˆj
e
jj
j
e
jj yyxx
For three noded triangular element the transformation of co-ordinates is given by
• where, are the interpolation functions over the master three noded triangular element.
• Master triangle is an isosceles triangle with unit length and right angle
3 3
1 1
ˆ ˆ, , ,i i i i
i i
x x y y
i
1
2
3
ˆ 1
ˆ
ˆ
Numerical Integration: Triangular Element
The Jacobian matrix, [J] for the linear triangular element is
Inverse transformation from the element to is:
where, A the area of the element
e
1 3 1 1 3 1
1 1 2 1 2 1
1
2
1
2
x x y y y y x xA
x x y y y y x xA
e
Mapping: Triangular Element
1313
1212
yyxx
yyxxJ
Linear Master Element Linear Physical Element
Physical to Master Element Mapping: Linear Rectangular Element
4
1
4
1
ˆ ,ˆj
e
jj
j
e
jj yyxx
1,1 1,1
1,1 1,1
1 2
34
x
y
1
2
3
4
Physical to Master Element Mapping: Quadratic Triangular Element
Quadratic Master Element Quadratic Physical Element
• The straight faces of the reference triangle are mapped onto curved faces of parabolic shape in the physical coordinate system.
0,0 1,0
0,1
1 2
3
0
0
1
4
56
x
y
1
2
3
4
5
6
6
1
6
1
ˆ ,ˆj
e
jj
j
e
jj yyxx
Quadratic Master Element Quadratic Physical Element
Physical to Master Element Mapping: Quadratic Rectangular Element
8
1
8
1
ˆ ,ˆj
e
jj
j
e
jj yyxx
x
y
1
2
3
4
56
78
1,1 1,1
1,1 1,1
1 2
34
67
8
5
Curved Quadratic Physical Element with
Side 2 is the only curved side.
Curve given in parametric
form so that
are the coordinates of node i
Physical to Master Element Mapping: Quadratic Rectangular Element
,i iX Y
x
y
1
2
3
4
2
2
x x
y y
2 2,x x y y
2 2 2 2 2 3 2 31 , 1 , 1 , 1x X y Y x X y Y
We have
The first four terms are the linear mapping terms.
The fifth term is the product of two functions:
The bracketed term represents the difference between
and the x-coordinates of the chord that connects and
The other is the linear blending function , which is unity along side 2 and zero along side 4.
Physical to Master Element Mapping: Quadratic Rectangular Element
1 2 3 4
2 2 3
1 1 1 11 1 1 1 1 1 1 1
4 4 4 4
1 1 1
2 2 2
x X X X X
x X X
2x
2 2,X Y 3 3,X Y
1 2
Therefore, we simplify as
Similarly,
In the general case all sides may be curved. We write the curved sides in the parametric form:
Mapping functions are:
Physical to Master Element Mapping: Quadratic Rectangular Element
1 4 2
1 1 11 1 1 1
4 4 2x X X x
, , 1 1 1,2,3,4i ix x y y i
1 4 2
1 1 11 1 1 1
4 4 2y Y Y y
1 2 3 4
1 2 3 4
1 1 1 11 1 1 1 1 1 1 1
4 4 4 4
1 1 1 1
2 2 2 2
x X X X X
x x x x
Mapping functions for y coordinate are:
Inverse mapping cannot be given explicitly.
Newton-Raphson method or similar procedure is used.
One can see the quadratic iso-parametric mapping as a special application of the blending function method.
Physical to Master Element Mapping: Quadratic Rectangular Element
1 2 3 4
1 2 3 4
1 1 1 11 1 1 1 1 1 1 1
4 4 4 4
1 1 1 1
2 2 2 2
y Y Y Y Y
y y y y
1 1 1
1ˆ 1 1 1
, , ,
M
j j
j
F d d F d d w d
ji
M
i
N
j
ji wwF
1 1
,
M,N - Number of quaddrature points in and directions. - Gauss points, - corresponding weights. The Gauss point locations are given by the tensor product of one dimensional Gauss points.
,i i
ji ww ,
M N
Numerical Integration:
ˆ
,eijk F d d
Numerical Integration: Rectangular Element
Selection of integration order
and location of the Gauss
points:
1 1
1ˆ 0 0
, , ,
NINT
j j j
j
F d d F d d w
NINT – Number of quadrature points - Gauss points, - corresponding weight.
,i i
iw
Numerical Integration:
ˆ
,eijk F d d
Line integration: Area integration: Transformation of the co-ordinates to natural co-ordinates: where, are the global coordinates of ith node.
Numerical Integration: Using Area Co-ordinates
1 2
! !
1 !
bm k
a
m kL L ds b a
m k
1 2 2 2! ! !
2 !
m k l
A
Am k l
L L L dAm k l
1 1
,
n n
i i i i
i i
x x L y y L
,i ix y
Quadratic Master Element Quadratic Physical Element
• This is also an Iso-parametric mapping of quadratic element.
Physical to Master Element Mapping: Quadratic Rectangular Element
8
1
8
1
ˆ ,ˆj
e
jj
j
e
jj yyxx
1,1 1,1
1,1 1,1
1 2 3
4 56
7 8 9
x
y
1
2
3
4
56
78