Part I: Shape Function Basics - PADT, Inc. · PDF filePart I: Shape Function Basics. MAE 323:...

Shape Functions and MeshingMAE 323: Lecture 3

2011 Alex Grishin 1MAE 323 Lecture 3 Shape Functions and Meshing

Part I: Shape Function Basics



•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



•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)



•So, if we want a linear polynomial that interpolates two points p0=(x0,u0) and

p1=(x1,u1), equation (1) would yield:


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



•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:


•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)



•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



•We’ll start by interpolating a curve on the x-axis (y=0) utilizing (1)


•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



•Now interpolate along the y-axis (x=0)


•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



•Next, use equation (3b) to find the bi-quadratic interpolant over the entire

domain


( , ) ( ) ( )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



•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)


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




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



•Elaborating a bit further: first parameterize the element coordinates so that the element

corners are always at (±1,±1).


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



•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



•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



•Note that the local-to-global mapping has the same form as the displacement solution:



•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

=

=

=

=

∑

∑



•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



•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)


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.



•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:


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=



•The rules for determining Ni are as follow:


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



•To begin, let’s try to find the shape functions for an eight-node, quadratic rectangle


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



•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:



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:



•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:



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= − − − + +



•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.



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



•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:



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:



•For node 2:



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:



•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:



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.



•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



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=

= − =

∑



•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.


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



Part II: 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





•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





•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





•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





•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




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




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




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




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…





•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





•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





•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





•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)





•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





•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



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



Part III:

Element Types in

Workbench



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



Thermal








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


4-node tetrahedral solid185

8-node hexahedral solid185

10-node tetrahedral solid186

20-node hexahedral solid187

Structural Thermal


4-node tetrahedral Solid70

8-node hexahedral Solid70

10-node tetrahedral Solid87

20-node hexahedral Solid90



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


2-node beam beam188

3-node beam beam189

3-node triang. Shell shell181

4-node quad. Shell shell181

6-node trang. Shell shell281



2-node beam link33

3-node beam link33

3-node triang. Shell shell131


6-node trang. Shell shell132




Part IV:

Lab 3



Tet mesh: 1

element thru

thickness

Tet mesh with

edge

refinement



Tet mesh w/

sphere of

influence

refinement

Hex mesh: 1

element thru

thicknesss

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