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Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 1MAE 323 Lecture 3 Shape Functions and Meshing
Part I: Shape Function Basics
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 2MAE 323 Lecture 3 Shape Functions and Meshing
•In the previous lecture, we saw a bar or truss element which could be used to
solve truss problems in structural mechanics. We constructed shape functions, Ni
by solving the governing differential equations.
•However, in FEM we need a more general procedure for generating such
functions. It turns out that we can use any trial functions, as long as they satisfy
the boundary conditions and certain continuity requirements
•This is generally done by polynomial interpolation, which we will discuss now
We’ll start with the linear shape functions for the truss element:
1 2
1
( ) 1n
i i
i
x xu x u N u u
L L=
= − +
∑
Lagrange Interpolation and Natural Coordinates
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 3MAE 323 Lecture 3 Shape Functions and Meshing
•These shape functions form a linear piecewise polynomial field which interpolate
the points x=0 and x=L
•For now, we’ll define interpolate to mean that the polynomial takes on a value
u(xi) at x=xi (node i) equal to the ith coefficient, ui
•There is an automatic interpolation procedure which will give such functions given
the interpolating points, xi. It is called Lagrange Interpolation. So, if we have k+1
data points, u0,…, uk the Lagrange formula, L may be stated as:
0
( ) : ( )k
j j
j
L x u l x=
=∑where:
0,
( ) :k
ij
i i j j i
x xl x
x x= ≠
−=
−∏
•Note that k equals the degree of the polynomial sought
Lagrange Interpolation and Natural Coordinates (Cont.)
(1)
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 4MAE 323 Lecture 3 Shape Functions and Meshing
•So, if we want a linear polynomial that interpolates two points p0=(x0,u0) and
p1=(x1,u1), equation (1) would yield:
Lagrange Interpolation and Natural Coordinates (Cont.)
010 1
0 1 1 0
( )x xx x
L x u ux x x x
−−= +
− − (2)
0 1 0 1( )x L x L x x
L x u u u uL L L L
− − = + = +
−
OR:
0 1( ) 1
x xL x u u
L L
= − +
0 1x
L
y
x0,u0 x1,u1
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 5MAE 323 Lecture 3 Shape Functions and Meshing
•If we want a bi-linear interpolation field over a square domain (as might be used
for plane stress elements, for example), it is straightforward to extend equation (1)
to two dimensions:
Lagrange Interpolation and Natural Coordinates (Cont.)
•Similarly, for a cubic domain:
0 0
( , ) ( ) ( )n m
ij i j
i j
L x y u l x l y= =
=∑∑
0 0 0
( , , ) ( ) ( ) ( )n m o
ijk i j k
i j k
L x y z u l x l y l z= = =
=∑∑∑
•Equation (3a) is equivalent to:
(3a)
(4)
0 0
( ) : ( ) * ( )n m
i i j j
i j
L x u l x u l x= =
= ∑ ∑ (3b)
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 6MAE 323 Lecture 3 Shape Functions and Meshing
•Equation (3b) represents a tensor or outer product. Thus, the extension to two (or
more) dimensions can be accomplished by interpolating in one spatial dimension,
then another, and finding the outer product of the two interpolations
•Although this is very convenient, some problems arise as can be seen when we try
to construct a bi-quadratic interpolation over a square domain
•Before we continue, some remarks are
in order. The element coordinate
system (which we will use to
interpolate) is placed in the centroid of
the square. This will be convenient
when dealing with more arbitrary
shapes. Also, equations (3a) and (3b)
suggest that we need to interpolate nine
points (three in each direction for 2nd
degree polynomial)! This means we
need nine nodes
x,u
y,v
a
b
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 7MAE 323 Lecture 3 Shape Functions and Meshing
•We’ll start by interpolating a curve on the x-axis (y=0) utilizing (1)
Lagrange Interpolation and Natural Coordinates (Cont.)
•The three points used to interpolate this curve are {-a,u1},{0,u2},{a,u3}
1
2
2
2
3
2
( )
2
( )
(
((
)
))
2
u x a x
a
u a x a x
a
u x x
a
x
a
=
− +
− + +−
+
N
x,u
y,v
a
u1
u2 u3
-a
Remember: We don’t really have a curve to interpolate. We’re just
interpolating three points symbolically as if they pass thru a 2nd
degree curve. This gives us the form of a 2nd degree interpolant to re-
use in particular circumstances
x1 x2 x3
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 8MAE 323 Lecture 3 Shape Functions and Meshing
•Now interpolate along the y-axis (x=0)
Lagrange Interpolation and Natural Coordinates (Cont.)
•The three points used to interpolate this curve are {-b,v1},{0,v2},{b,v3}
x,u
y,vb
-b
v1
v2
v3
1
2
2
2
3
2
( )
2
( )
(
((
)
))
2
v y b y
b
v b y b y
b
v y y
b
y
b
=
− +
− + +−
+
N
y1
y2
y3
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 9MAE 323 Lecture 3 Shape Functions and Meshing
•Next, use equation (3b) to find the bi-quadratic interpolant over the entire
domain
Lagrange Interpolation and Natural Coordinates (Cont.)
( , ) ( ) ( )x y y x= ⊗N N N
1
2 2
2
2 2
3
2 2
4
2 2
5
2 2
6
2 2
7
2 2
8
2 2
9
( ) ( )
4
( )( ) ( )
2
( ) ( )
4
( )( )( )
2
( )( )( )( )
( )( )( )
2
( ) (
( , )
)
4
( )( ) ( )
2
c x a x y b y
a b
c a x a x y b y
a b
c x a x y b y
a b
c x a x b y b y
a b
c a x a x b y b y
a b
c x a x b y b y
a b
c x a x y b y
a b
c a x a x y b y
a b
y
c
x
− + − +
− + + − +−
+ − +
− + − + +−
− + + − + +
+ − + +−
− +
+ + +
=
+
−−
N
2 2
( ) ( )
4
x a x y b y
a b
+ +
1 2 3
4 5 6
7 8 9
We now number the nodes to match the coefficients
in the outer product. However, note that the actual
numbering is unimportant. What IS important to
match each basis function coefficient to it’s proper
spatial location
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 10MAE 323 Lecture 3 Shape Functions and Meshing
•We now have a basis that could serve nicely as a rectangular continuum element (maybe
plane stress or plane strain). The shape functions, N form the basis for both the u and v
displacement fields (and all their derivatives)
Lagrange Interpolation and Natural Coordinates (Cont.)
9
1
( , ) ( , )i i
i
u x y N x y u=
=∑9
1
( , ) ( , )i i
i
v x y N x y v=
=∑•The shape functions just calculated are capable of interpolating all three domains
shown below , by just changing values of a and b and and multiplying the stiffness
matrix by a rotation
yx
yx
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 11MAE 323 Lecture 3 Shape Functions and Meshing
Lagrange Interpolation and Natural Coordinates (Cont.)
An Isoparametric Rectangular Lagrange Element
•But what if the element were distorted as below?
•We must first explain why we might want to support such shapes. The reason is
that the element domains must discretize arbitrary geometries (such discretizations
are carried out by some meshing technique). The closer they can conform to
surface contours, the fewer elements we will need for a given degree of accuracy*
•It turns out, there are some fairly straightforward ways to handle this. The idea is
to first find a bi-variate polynomial that interpolates the nodal locations, and then
re-use that same polynomial to interpolate the displacement field
xy
xy
*This statement provides the basis for what is known as the p-method of mesh
refinement
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 12MAE 323 Lecture 3 Shape Functions and Meshing
•Elaborating a bit further: first parameterize the element coordinates so that the element
corners are always at (±1,±1).
Lagrange Interpolation and Natural Coordinates (Cont.)
An Isoparametric Rectangular Lagrange Element (Cont.)
•As before, the origin (0,0) is at the centroid (or barycenter) of the domain. Next,
the nodal points themselves are interpolated in the local, or natural coordinate
system, such that the local-to-global coordinate transformation may be given as:
rs
(1,1)
(1,-1)
(-1,1)
(-1,-1)
(1,0)
(-1,0)
(0,1)
(0,-1)
(0,0)
x
y
1
1
n
i i
i
n
i i
i
x N x
y N y
=
=
=
=
∑
∑
The r,s system is the
natural (parametric)
coordinate system
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 13MAE 323 Lecture 3 Shape Functions and Meshing
•The shape functions are obtained by using the shape functions from before for a rectangular
domain, setting a and b to 1, and replace x and y with r and s
Lagrange Interpolation and Natural Coordinates (Cont.)
An Isoparametric Rectangular Lagrange Element (Cont.)
•The s,r system is the natural (parametric) coordinate system (by
definition, the centroid lies at (0,0)1
( 1 ) ( 1 )4
1( 1 )(1 )( 1 )
2
1(1 )( 1 )
4
1( 1 ) ( 1 )(1 )
2
( 1 )(1 )( 1 )(1 )
1(1 )( 1 )(1 )
2
1( 1 ) (1 )
4
1( 1 )(1 ) (1 )
2
1(1 ) (1 )
4
( , )
r r s s
r r s s
r r s s
r r s s
r r s s
r r s s
r rs s
r r s s
r
r
s
s
r s
− + − +
− − + + − +
+ − +
− − + − + +
− + + − + +
− + − + +
− +
=
+
−
+
− + + +
+
N
The parametric
shape functions are
shown at right
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 14MAE 323 Lecture 3 Shape Functions and Meshing
•Note that the local-to-global mapping has the same form as the displacement solution:
Lagrange Interpolation and Natural Coordinates (Cont.)
An Isoparametric Rectangular Lagrange Element (Cont.)
•Element formulations with this property (where the solution interpolation has the
same form as the parametric coordinate mapping) are said to be isoparametric.
The bi-quadratic element formulation just shown is known as a Lagrangian
isoparametric rectangular element.
•One question that arises with this type of element is: Where to place the mid-side
and center nodes? The common answer is: at the midpoints and centroid,
respectively. This will work, of course, but the node at the center is not strictly
necessary and having to calculate the centroid beforehand is inconvenient.
1
1
n
i i
i
n
i i
i
u N u
v N v
=
=
=
=
∑
∑
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 15MAE 323 Lecture 3 Shape Functions and Meshing
•Another problem with the Lagrange element we just described is that convergence issues
arise when connecting such elements to elements of lower polynomial degree.
•Consider the following bi-linear Lagrange rectangle (constructed the same way we
constructed the bi-quadratic rectangle)
Serendipity Elements
r
s
(1,-1)
(1,1)(-1,1)
(-1,-1)1 2
3 4 1(1 )(1 )
4
1(1 )(1 )
4
1(1 )(1 )
4
1(1 )
( ,
(1 )4
)
r s
r s
r s
r s
r s
− −
+ −
− +
+ +
=
N
•What happens if we connect this element to the bi-quadratic Lagrange element?
The strains may be
incompatible along this
edge
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 16MAE 323 Lecture 3 Shape Functions and Meshing
•So, we’d really like the flexibility of creating an element like the one below, which can be
linear along any edge you choose, and quadratic elsewhere (and without a node at the
centroid)
Serendipity Elements
Linear along
this edge
Quadratic
along these
three
•And, since we don’t really need the center node, we’d like to get rid of it as well.
•Such elements can be obtained, and usually go by the name transition elements. The
techniques to derive them are general and produce a class of elements called Serendipity
Elements. These are some of the most popular and common finite elements (most
continuum elements in the ANSYS library fall under this category). There are many ways to
construct them*, but we will focus on just one
*see, for example:
R. D. Cook, D. S. Malkus, and M. E. Plesha, Concepts and Applications of Finite Element Analysis,
3rd ed. New York, NY, USA: John Wiley & Sons, 1989.
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 17MAE 323 Lecture 3 Shape Functions and Meshing
•The method we’ll use can be found in C.A. Felippa’s course on finite elements at the
University of Colorado*
•The method is based on the observation that most isoparametric shape functions are given
as products of fairly simple polynomials given in the natural coordinates. Thus a particular
shape function at node i may be given as:
Serendipity Elements
where Lj…Lm = 0 are the homogeneous equation of lines or curves expressed as linear
functions in the natural coordinates and ci is a normalization coefficient (not to be confused
with the shape function coefficient)
* http://www.colorado.edu/engineering/cas/courses.d/IFEM.d/IFEM.Ch18.d/IFEM.Ch18.pdf
1 2...i i mN c L L L=
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 18MAE 323 Lecture 3 Shape Functions and Meshing
•The rules for determining Ni are as follow:
Serendipity Elements
1. Select the Lj as the minimal number of lines or curves linear in the natural
coordinates that cross all nodes except the ith node. Primary choices in
2D are element sides and medians
2. Set the coefficient ci so that Ni has the value 1 at the ith node
3. Check that Ni vanishes over all element sides that do not contain node i
4. Check the polynomial order* over each side that contains node i. If the
degree is n, there must be exactly n+1 nodes on the side for compatibility
to hold.
5. If (3) and (4) are satisfied, check that the sum of the shape functions is
identically one at an arbitrary point within the element domain
*Recall that by “polynomial order”, we usually mean degree + 1
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 19MAE 323 Lecture 3 Shape Functions and Meshing
•To begin, let’s try to find the shape functions for an eight-node, quadratic rectangle
Serendipity Elements
A Quadratic Serendipity Rectangle
rs
1
2
3
4
5
67
8
Note: We are now switching to a more
conventional node numbering scheme.
The one we used previously was simply
convenient as it mirrored terms in the
outer product of two Lagrangian
functions
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 20MAE 323 Lecture 3 Shape Functions and Meshing
•Step 1: Select the Lj as the minimal number of lines or curves linear in the natural
coordinates that cross all nodes except the ith node.
•So, for node 1, we would have the following:
Serendipity Elements
A Quadratic Serendipity Rectangle
rs
1
2
3
4
5
67
8
L4-3: s=1L2-3: r=1
L5-8: r+s=-1
1 1 4 3 2 3 5 8N c L L L− − −=
4 3
2 3
5 8
1
1
1
L s
L r
L r s
−
−
−
= −
= −
= + +
1 1( 1)( 1)( 1)N c s r r s= − − + +
Now, substitute back in:
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 21MAE 323 Lecture 3 Shape Functions and Meshing
•Step 2: Set the coefficient ci so that Ni has the value 1 at the ith node
•At node 1, (r,s)=(-1,-1). Plugging those values in for N1 yields:
Serendipity Elements
A Quadratic Serendipity Rectangle
rs
1
2
3
4
5
67
8
L4-3: s=1L2-3: r=1
L5-8: r+s=-1
•Therefore, c1=-1/4
1 1( 1, 1) ( 4)N c− − = −
1
1( 1)( 1)( 1)
4N s r r s= − − − + +
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 22MAE 323 Lecture 3 Shape Functions and Meshing
•Step 3: Check that Ni vanishes over all element sides that do not contain node i
•For node 1, the relevant element sides are at s=1 and r=1. It is immediately clear that N1=0
along these sides.
Serendipity Elements
A Quadratic Serendipity Rectangle
rs
1
2
3
4
5
67
8
L4-3: s=1L2-3: r=1
L5-8: r+s=-1
•Step 4: Check the polynomial order over each side that contains node i the degree
is n, there must be exactly n+1 nodes on the side for compatibility to hold.
•The two sides that contain node 1
are at s=-1 and r=-1. Plugging
these into the expression for N1
reveals that the polynomial is
quadratic in r on s=-1, and
quadratic r on r=-1. There are
correspondingly three nodes on
each of these sides
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 23MAE 323 Lecture 3 Shape Functions and Meshing
•We must wait until we have calculated all the other shape functions for this element before
performing the final check in step 5. So, let’s move on to node 5:
Serendipity Elements
A Quadratic Serendipity Rectangle
1
rs
2
3
4
5
67
8
L4-3: s=1L2-3: r=1
L1-4: r=-1
5 5 4 3 2 3 1 4N c L L L− − −=
4 3
2 3
1 4
1
1
1
L s
L r
L r
−
−
−
= −
= −
= +
•Plugging back in:
2
5 5 ( 1)( 1)N c s r= − −At (0,-1), N5=2. So c5=1/2
2
5
1( 1)( 1)
2N s r= − −
Steps 3 reveals that the N5=0 at all
points other than node 5. Also,
since no line contains node 5 in N5,
step 4 is irrelevant
•Step 2:
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 24MAE 323 Lecture 3 Shape Functions and Meshing
•For node 2:
Serendipity Elements
A Quadratic Serendipity Rectangle
2
1
rs
3
4
5
67
8
L4-3: s=1
L5-6: s=r-1
L1-4: r=-1
2 2 1 4 4 3 5 6N c L L L− − −=
4 3
1 4
5 6
1
1
1
L s
L r
L s r
−
−
−
= −
= +
= − +
•Plugging back in:
2 2( 1)( 1)( 1)N c s r s r= − + − +
At (1,-1), N2=4. So c4=1/4
2
1( 1)( 1)( 1)
4N s r s r= − + − +
Steps 3 and 4 yield the same
results as for N1
•Step 2:
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 25MAE 323 Lecture 3 Shape Functions and Meshing
•The calculations for the five remaining nodes will be left as an exercise for the student. He
(she) should be able to verify the following:
Serendipity Elements
A Quadratic Serendipity Rectangle
2
1
rs
3
4
5
67
8
2
2
2
2
1( 1)( 1)( 1)
4
1( 1)( 1)( 1)
4
1( 1)( 1)( 1)
4
1( 1)( 1)( 1)
4( , )
1( 1)( 1)
2
1( 1)( 1)
2
1( 1)(1 )
2
1( 1)( 1)
2
r s r s
r s s r
r s s r
r s s r
r s
r s
r s
r s
r s
− − − + + + − − + + + + −
− − + − −
= − − + −
− − +
− −
N•We can now
implement the final
check, Step 5.
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 26MAE 323 Lecture 3 Shape Functions and Meshing
•Step 5: If steps 3 and 4 are satisfied, check that the sum of the shape functions is identically
one at an arbitrary point within the element domain
•We will arbitrarily check the point, r=1/2, s=1/2
Serendipity Elements
A Quadratic Serendipity Rectangle
1
8
3
16
0
3
161 1
, 32 2
16
9
16
9
16
3
16
−
−
−
=
N
8
1
1 1 24 8, 1
2 2 16 16i
i
N=
= − =
∑
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 27MAE 323 Lecture 3 Shape Functions and Meshing
•Using the five steps on slide 17, plus the concept of “mapping”, where a low-order
configuration is used as a parent element while selectively adding higher order terms*, one
can solve for virtually any hybrid combination of polynomial degrees. For example, the
transition element of slide 15 may be constructed by starting with the bi-linear element (slide
14) and augmenting with higher order terms while obeying steps 3 and 4.
Serendipity Elements
A linear-quadratic transition element
1 2
34
5
6
7
r
s
*Doing this in a systematic way requires more assumptions
2
2
2
1( 1) ( 1)
4
1( 1)( 1)( 1)
4
1( 1)( 1)( 1)
4
1( , ) ( 1) ( 1)
4
1( 1)( 1)
2
1( 1)( 1)
2
1( 1)( 1)
2
r r s
r s s r
r s s r
r s r r s
r s
r s
r s
− − −
+ − − + + − − +
= − +
− − − + − − − +
N
•Note that every
term except
those for nodes
1 and 4 are the
same as the
eight-node
serendipity
rectangle
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 28MAE 323 Lecture 3 Shape Functions and Meshing
Part II: Meshing
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 29MAE 323 Lecture 3 Shape Functions and Meshing
Constructing a Finite Element Mesh
Structured (or “Regular”) Grids
•In the numerical approximation of
differential equations, a domain must first
be discretized. Such discretizations are
made of cells, which form the support of
finite elements. The earliest and simplest
way of doing this was to generate a
structured grid
•Definition: A structured grid is one in
which each cell can be addressed by the
index (I,j) in two dimensions, or (I,j,k) in
three dimensions, and each vertex has
coordinates (i⋅dx,j⋅dy) in two dimensions,
or (i⋅dx,j⋅dy,k⋅dz) in three
•Such grids are necessary in the Finite
Difference Method (above), and can be
useful in the Finite Element Method
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 30MAE 323 Lecture 3 Shape Functions and Meshing
Constructing a Finite Element Mesh
Structured (or “Regular”) Grids
•A severe limitation of the structured grid is the
requirement that it be representable as a unique
coordinate mapping (as shown at the right)*
•Most modern meshing algorithms allow a user
to either define the coordinate system and
mapping explicitly, or to extract both from the
topology of the manifold to be meshed. The last
option is not trivial, and to my knowledge there
are no algorithms for doing this in a systematic
and robust way. However, this operation
becomes almost trivial if the manifold is 2m-
sided, where m is the number of spatial
dimensions
*the picture is from a CFD mesh, described on http://www.cgl-
erlangen.com/downloads/Manual/ch09s16s01.html
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 31MAE 323 Lecture 3 Shape Functions and Meshing
Constructing a Finite Element Mesh
Structured (or “Regular”) Grids
•However, most geometric entities encountered
by the analyst will not meet this requirement.
So, if a structured grid is required, the domain
must be sub-divided into 2m-sided polytopes by
the user (manually). Again, this is not a trivial
task in most cases.
•For example, most automatic mesh generators
would not be able to generate a structured grid
out of the domain at the right
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 32MAE 323 Lecture 3 Shape Functions and Meshing
Constructing a Finite Element Mesh
Structured (or “Regular”) Grids
•A common solution would be to “split” the
polygon into quadrilaterals as shown at the right.
•Most meshers could now create five structured
grids, which share nodes between them as
shown below
Four quadrilaterals
Five edges shared
between
neighbors
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 33MAE 323 Lecture 3 Shape Functions and Meshing
Constructing a Finite Element Mesh
Structured (or “Regular”) Grids
•Most finite element applications do not make use of
the nodal coordinate mapping produced (or implied) by
structured grids, and so the mapping itself is commonly
not calculated (when this is the case, such grids are
referred to by the shape of their elements). Instead,
nodal connectivity data is stored for each element.
•In the quadrilateral grid shown to the right, node
numbers are marked with circles, while element
numbers are marked in squares
•The element data are stored according the element
number and nodal connectivity as shown. For example,
element 1 and 2 would be stored as:
1
1 2 3 4 5
6 7 8 9 10
2 3 4
i j
kl
Element: 1
Nodes: 1,2,7,6
Element: 2
Nodes: 2,3,8,7
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 34MAE 323 Lecture 3 Shape Functions and Meshing
Constructing a Finite Element Mesh
Unstructured Grids
•Thus, it can immediately be seen by simply looking at
the element connectivity data structures for elements
one and two on the previous slide, that they share
nodes 2 and 7.
•The element connectivity data structure is very
flexible, and can accommodate any kind of element.
•For example, consider a domain meshed by quadratic
triangular elements. The data structures for element 1
and 2 may be given by:1 21
2
3
4
5
6
7
8
9
Element: 1
Nodes: 1,2,3,4,5,6
Element: 2
Nodes: 2,7,3,5,8,9
•Triangular elements have the advantage that they do not require a domain to
have 2m sides . They may thus be used to create unstructured meshes over
arbitrary domains in an automated fashion
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 35MAE 323 Lecture 3 Shape Functions and Meshing
Constructing a Finite Element Mesh
Unstructured Grids
•Let’ s take another look at the two domains we have
already used as examples
•Unstructured
quadratic triangle
mesh. No manual
intervention
•Structured
quadratic
rectangle mesh.
Area split into four
sub-domains
•Unstructured
quadratic triangle
mesh. No manual
intervention
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 36MAE 323 Lecture 3 Shape Functions and Meshing
Constructing a Finite Element Mesh
Unstructured Grids
•These examples illustrate that an unstructured grid may be created
automatically with fairly standard algorithms (described next) over most
geometric domains
•In two dimensions, the unstructured element type is triangular. In three
dimensions, it is a tetrahedron. This is because these shapes form what is
known as a simplex in their respective spaces. For our purposes, we only
need to know that a simplex is the simplest possible polytope of dimension
m which can be used to tile a space of the same dimension
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 37MAE 323 Lecture 3 Shape Functions and Meshing
Constructing a Finite Element Mesh
Unstructured Grids - Delaunay Triangulation
•To understand how unstructured grids are created, we’ll start by looking at
the algorithm which plays some role in most unstructured meshing
programs: the Delaunay Algorithm
•We’ll begin by considering an arbitrary collection of points (nodes) in two
dimensions
•Note that there isn’t a unique
triangular mesh for these points
•But this one… …looks “better” than this one…
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 38MAE 323 Lecture 3 Shape Functions and Meshing
Constructing a Finite Element Mesh
Unstructured Grids - Delaunay Triangulation
•So, before we continue, we should probably give a definition of a “good”
mesh. What are the goals when constructing meshes? The following
definition comes from a paper by Bern, Eppstein, and Gilbert*:
•The mesh must conform to the boundary of the region, which may
consist of more than one connected component
•The mesh must be fine enough to produce an acceptable
approximation to the original problem at all points of interest
•The number of elements in the mesh should be as small as possible
given the two previous requirements
•The individual elements must be “well-shaped”. The most important
restriction is on minimum angle:
•No small angles: For most problems, elements with small
angles lead to ill-condtioned linear systems. Angles close to 180
degrees pose further problems
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 39MAE 323 Lecture 3 Shape Functions and Meshing
Constructing a Finite Element Mesh
Unstructured Grids - Delaunay Triangulation
•The Delaunay Criterion states
that any node under consideration
as a tetrahedron or triangle vertex
must not be contained within the
circumsphere* of any other node
which is already being used as a
vertex
•Numerically, this criterion is
easily checked by making sure that
the sum of opposite angles is less
than 180°
•Note that in this case, β+δ>180.
So this triangulation fails*Note the difference
between “contained
within” and “lying on”
AB
CD
α β
γδ
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 40MAE 323 Lecture 3 Shape Functions and Meshing
Constructing a Finite Element Mesh
Unstructured Grids - Delaunay Triangulation
•Once the criterion fails, the
adjacent edge is “flipped”. That is
to say the other two vertices of
the four points are used as the
adjacent edge
•The Delaunay Criterion is checked
again.
•It this case, it passes because
α+γ<180A
B
CD
αβ
γδ
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 41MAE 323 Lecture 3 Shape Functions and Meshing
Constructing a Finite Element Mesh
Unstructured Grids - Delaunay Triangulation
•Thus the Delaunay algorithm proceeds by arbitrarily drawing a
circumsphere around three points (in two dimensions. Four points in three
dimensions). It then arbitrarily chooses a neighboring point and draws
another circumphere and checks the Delaunay Criterion. It either passes or
fails, in which case the adjacent edge is flipped. It then chooses another
point and draws a circumsphere around it, and so on until the entire set of
points is meshed. For a given collection of points, the Delaunay Algorithm
guarantees the best mesh according to the “well-shaped” criteria stated
earlier (it maximizes the minimum angle)
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 42MAE 323 Lecture 3 Shape Functions and Meshing
Constructing a Finite Element Mesh
Unstructured Grids - Delaunay Triangulation
•But that’s not the whole story. Recall the first requirement from slide 42.
We MUST have nodes on the boundary. Even in the interior of the domain,
we usually don’t simply create points randomly.
•So, the Delaunay algorithm is often combined with some judicious point
insertion algorithm. Of these, there are several different varieties. Three
very common ones are: Methods based on Delaunay Refinement, Octree
(in three dimensions – Quadtree in two) and Advancing Front.
Advancing Front Quadtree
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 43MAE 323 Lecture 3 Shape Functions and Meshing
Constructing a Finite Element Mesh
Unstructured Grids - Delaunay Triangulation
•Many commercial finite element codes offer several different algorithms for
flexibility in meshing difficult geometry.
•Historically, the advancing front algorithms have tended to produce higher quality
meshes, but at the expense of requiring absolutely error-free geometry
•The quadtree and Deluanay-based algorithms (and other similar ones) are
currently enjoying a resurgence due to problems relating to CAD import. These
algorithms are capable of producing meshes on broken or incomplete geometry
•For more detailed discussion, see the references below
http://www.cs.berkeley.edu/~jrs/mesh/present.html
For a list of meshing algorithms:
For a more thorough discussion of meshing generally:
http://morden.csee.usf.edu/dragon/kpalbrec/mesh.html
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 44MAE 323 Lecture 3 Shape Functions and Meshing
Review
•Elements represent the support (domain) over which shape functions are defined
•Higher order shape functions (polynomials of degree greater than 1) require mid-side
nodes. This is because nodes represent locations corresponding to shape function
coefficients. The number of midside nodes required along an element edge is equal to n-
1, where n is the degree of the polynomial (recall that a 2nd degree polynomial has three
terms and thus requires three coefficients)
•The number of degrees of freedom (DoF’s) an element has is equal to the number of
nodes times the spatial dimension. This is illustrated below for a 2-dimensional linear
element
rs
4 nodes times 2 DoF’s
per node. This
element has 8 DoF’s
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 45MAE 323 Lecture 3 Shape Functions and Meshing
Part III:
Element Types in
Workbench
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 46MAE 323 Lecture 3 Shape Functions and Meshing
2D Continuum Element Types
rs rs
6-node triangle
Quadratic shape
functions
8-node quadrilateral
Quadratic shape
functions
4-node quadrilateral
Linear shape
functions
3-node triangle
Linear shape
functions
Structural
Parent Shape MAPDL Type
3-node triangular Plane182
4-node quad. Plane182
6-node triangular Plane183
8-node quad. Plane183
Thermal
Parent Shape MAPDL Type
3-node triangular Plane55
4-node quad. Plane55
6-node triangular Plane77
8-node quad. Plane77
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 47MAE 323 Lecture 3 Shape Functions and Meshing
r
s
t
r
s
t
3D Continuum Element Types
20-node Hex.
Quadratic shape
functions
10-node tet.
Quadratic
shape
functions
8-node hex.
Linear shape
functions
4-node tet.
Linear shape
functions
Parent Shape MAPDL Type
4-node tetrahedral solid185
8-node hexahedral solid185
10-node tetrahedral solid186
20-node hexahedral solid187
Structural Thermal
Parent Shape MAPDL Type
4-node tetrahedral Solid70
8-node hexahedral Solid70
10-node tetrahedral Solid87
20-node hexahedral Solid90
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 48MAE 323 Lecture 3 Shape Functions and Meshing
3D Reduced-Continuum Element Types
2-node beam 3-node beam 8-node quad. shell
Structural Thermal
t
rs
12
3
4
2
3
4
rst
1
t
rs
1
2
3
Parent Shape MAPDL Type
2-node beam beam188
3-node beam beam189
3-node triang. Shell shell181
4-node quad. Shell shell181
6-node trang. Shell shell281
8-node quad. Shell shell281
Parent Shape MAPDL Type
2-node beam link33
3-node beam link33
3-node triang. Shell shell131
4-node quad. Shell shell131
6-node trang. Shell shell132
8-node quad. Shell shell132
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 49MAE 323 Lecture 3 Shape Functions and Meshing
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 50MAE 323 Lecture 3 Shape Functions and Meshing
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 51MAE 323 Lecture 3 Shape Functions and Meshing
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 52MAE 323 Lecture 3 Shape Functions and Meshing
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 53MAE 323 Lecture 3 Shape Functions and Meshing
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 54MAE 323 Lecture 3 Shape Functions and Meshing
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 55MAE 323 Lecture 3 Shape Functions and Meshing
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 56MAE 323 Lecture 3 Shape Functions and Meshing
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 57MAE 323 Lecture 3 Shape Functions and Meshing
Part IV:
Lab 3
Shape Functions and MeshingMAE 323: Lecture 3
2011 Alex Grishin 58MAE 323 Lecture 3 Shape Functions and Meshing
Tet mesh: 1
element thru
thickness
Tet mesh with
edge
refinement
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