Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini

8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini

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FINITE ELEMENT ANALYSIS FOR MINIMAL SHAPE

Vinicius Arcaro, Katalin Klinka, Dario Gasparini

This text describes a mathematical model for minimizingpath length, surface area and volume using a line element,

a triangle element and a tetrahedron element respectively.

The elements can also be used together through the unifying

concept of minimizing shape volume. The square of the

relative change in the elements volume is used to define

an isovolumetric element. A quasi-Newton method is used,

which avoids the evaluation of the Hessian matrix as

required in a Newton method. The source and executable

computer codes of the algorithm are available from the

first author's website.

Keywords: element, line, membrane, minimization, nonlinear,

optimization, tetrahedron, triangle.

1 Introduction

The geometrical shape minimization problem can be

formulated as an unconstrained nonlinear programming

problem, where the objective function is the final total

volume and the displacements of the nodal points are the

unknowns. The shape volume is discretized into line,triangle and tetrahedron elements. The elements are

interconnected at their nodal points. It is important to

emphasize that by construction, the final element state is

uniquely defined in terms of its nodal displacements.

The following conventions apply unless otherwise specified

or made clear by the context. A Greek letter expresses a

scalar. A lower case letter represents a column vector.

2 Line element definition

Figure 1 shows the geometry of the element. The nodes are

labeled 1 and 2. The nodal displacements transform the

element from its initial state to its final state.
http://www.arcaro.org/http://civil.case.edu/people/dag6http://civil.case.edu/people/dag6http://civil.case.edu/people/dag6http://www.arcaro.org/


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2x

v

1x

1u

2u

2x

v1x

Figure 1

2 1v v u u= +

( )1

T 2v v v=

The previous expression can be written as:

( ) ( ) ( )1

T 2T T 2 1 2 1 2 1v v v 2v u u u u u u = + +

2.1 Element final volume

Supposing that the line element has a fixed cross sectional

area , its final volume can be written as:

v =

2.2 Gradient of the element final volume

The nodal displacements vectors are numbered according to

its node numbers. Their individual components are numbered

as follows:


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1 4

1 2

2 5

3 6

u u

u u , u u

u u

= =

The gradient of the element final volume with respect tothe nodal displacements can be written as:

v1

vv

= +

3 Triangle element definition

Figure 2 shows the geometry of the element. The nodes are

labeled 1, 2 and 3 while traversing the sides in

counterclockwise fashion. Each side is labeled with the

number of its opposite node. The nodal displacements

transform the element from its initial state to its final

state.

3x

2v

1

Figure 2

1 1 3v v u= + 2u

2 2 1v v u= + 3u

3 3 2v v u= + 1u

1 2w v v=

1x

2x

3x

1x

1u

2u

3u

v

2x

3

v

2v

1v

3v


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1 2

w v v=

The previous expression can be written as:

1 1 2 2 3 3

1 2 2 3 3 1

w wv u v u v u

u u u u u u

= ++ + +

+ + +

+

The vectors w and w are orthogonal to the element in the

initial state and final state respectively. Notice that

these vectors point toward the observer.


Supposing that the triangle element has a fixed thickness ,its final volume can be written as:

w2

=


The derivatives of the element final volume with respect to

the nodal displacements can be written as:

31 2

1 2 3j j j j

i i i i

wd w ww w w

du 2 w u u u

= + +



as follows:

1 4

1 2 32 5

3 6

u u

u u , u u , u u

u u u

= = =

7

8

9

u

The gradient of the element final volume can be written as:


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1

2

1 2

w v

w v2 w

w v w v

=

4 Tetrahedron element definition

Figure 3 shows the geometry of the element. The base nodes

are labeled 1, 2 and 3 while traversing the sides in

counterclockwise fashion looking from the apex, which is

labeled 4. The nodal displacements transform the element

from its initial state to its final state.

4x

Figure 3

1 1 1v v u= + 4u

2 2 2v v u= + 4u

3 3 3v v u= + 4u


1x

2x

3x

4x

3v

2v

1v

3u

2u

1u

4u

2x

1x

3x

1v3

v2

v


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The final volume of the tetrahedron element can be written

as:

( ) ( )T

1 21v v v

6 = 3

Note that if the apex moves below the base, the volume

becomes negative.

1 1 1 4

2 2 2 4

3 3 3 4

v v u u

v v u u

v v u u

= +

= + = +

( ) ( ) ( )T

1 1 4 2 2 4 3 3 41v u u v u u v u u

6 = + + +

i i 4d u u=

+

+

10

11

12

u

u

( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

T1 2 3

T T T1 2 3 2 3 1 3 1 2

T T T1 2 3 2 3 1 3 1 2

T1 2 3

6 v v v

d v v d v v d v v

v d d v d d v d d

d d d

= +

+ + +

+ + +

+




as follows:

1 4 7

1 2 3 4

2 5 8

3 6 9

u u u

u u , u u , u u , u u

u u u

= = = =

The gradient of the element final volume can be written as:


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3 2

1 3

2 1

3 2 1 3 2 1

v v

v v1

6 v v

v v v v v v

=

5 Isovolumetric element definition

An element that does not change its volume is desirable for

many problems types. The square of the relative change in

the elements volume can be used to define an isovolumetric

element. Ideally, the volume change should be zero for the

isovolumetric element in its final state.


Considering 0 and as the initial and final volumesrespectively, an isovolumetric element can be defined

through the following volume function, where is a penaltyparameter.

2

0 0

02

=


Using the chain rule, the gradient of the element final

volume with respect to its nodal displacements can be

written as:

0

0

=

5.3 Severe cancellation

Inaccuracy often results from severe cancellation that

occurs when nearly equal values are subtracted. Notice that

inaccuracy can result from the difference between the final

and initial volumes because they can be arbitrarily close


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for the isovolumetric element. However, severe cancellation

can usually be eliminated by algebraic reformulation.

0

1

=

20

2

=

=

The expression for , already reformulated in an attempt toavoid severe cancellation, can be written for the line,

triangle and tetrahedron elements.

Line element

1v v

v=

( )2 11

z uv

= u

T T2v z z z

v z

+ =

+ + 1

Triangle element

w w w=

( )

( )

( )

1 2 1

2 3 2

3 1 2

w z

v u u

v u u

u v v u

=

+ +

+ +

+ + + 1

T T2w z z z

w z 1

+ =

+ +

Tetrahedron element


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i i

i

1v v

v=

( )i ii1

z u

v

= 4u

+

+

+

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

T1 2 3

T1 2 2 3

T2 3 3 1

T3 1 1 2

T1 2 3

v v v

v v z z

v v z z

v v z z

z z z

=

+ +

+ +

+ +

+

6 Final shape volume

Considering u as the vector of unknown displacements, the

final shape volume function and its gradient can bewritten as follows:

( ) ( )elements

u u =

( ) ( )elements

u u =

In order to find the local minimum points of a nonlinear

multivariate function, the general strategy that can be

used is: Choose a starting point and move in a given

direction such that the function decreases. Find the

minimum point in this direction and use it as a new

starting point. Continue this way until a local minimum

point is reached. In the quasi-Newton method, a descent

direction or a direction such that the function decreases

can be defined with the help of the gradient vector. Theadvantages of this approach are: It is not necessary to

derive an expression for the Hessian matrix and it is not

necessary to solve any system of equations. The computer

code uses the limited memory BFGS to tackle large scale

problems as described by [3]. It also employs a line search

procedure with safeguards as described by [1].


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7 Examples

The primary colors red, green and blue are used for the

line, triangle and tetrahedron elements respectively. The

secondary colors cyan, magenta and yellow are used for the

isovolumetric line, triangle and tetrahedron elementsrespectively.

Example 1: An initially flat circular surface with

thickness = 2, radius r = 1 and the boundary displaced

according to the following hyperbolic paraboloid equation,

where h = 1/2. Figure 4 shows the meshes for the initial

and final surfaces.

Figure 4

2 2

x yz hr r =

Example 2: A cylinder with thickness = 1, radius r = 1 and

height h = 1. Figure 5 shows the meshes for the initial and

final surfaces.


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Figure 5

The final surface is symmetrical about the Z axis. The

following analytical solution for the cross-section of thesurface in the YZ plane is described by [2].

( )z

y z c coshc

=

Note that y(0) = c. The value c is a solution of the

following equation. In this example, c = 0.8483379.

2r 2c hcosh

h h 2c

=

Table 1 shows the relative error for y(0) with different

initial meshes.

Table 1

Elements y(0) Error

384 0.8470554 0.15 %

1536 0.8480256 0.04 %

6144 0.8482543 0.01 %

Example 3: A frustum cone with thickness = 2, upper radius= 0.5, lower radius = 1 and height = 0.9. Figure 6 shows

the meshes for the initial and final surfaces.


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Figure 6

Example 4: On the top, Figure 7 shows an initial path

through the corner points of a rectangle with horizontaldimension equal to 4 and vertical dimension equal to 2. The

line elements have area = 1. On the bottom, Figure 7 shows

the final path.

Figure 7


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The general problem of connecting n points by the shortest

path length is called Steiner problem [2]. Its solution

contains straight lines intersecting at 120. The number ofintersections is between zero and (n - 2).

Example 5: An initially flat square surface with thickness= 1, side = 1 and two opposite corners displaced by +1/2

while the two other opposite corners displaced by -1/2. The

edges have line elements with area = 5. Figure 8 shows the

meshes for the initial and final surfaces.

Note that minimizing the total volume results in opposite

effects on the lengths of the free edges. It tends to

decrease the surface area defined by the triangle elements

and consequently tends to increase the lengths of the free

edges. It tends to decrease the path lengths defined by the

line elements and consequently tends to decrease the

lengths of the free edges. In this example, the free edges

are curved due to relatively small value for the areas of

the line elements.

Figure 8

Example 6: An initially flat square surface with thickness

= 1, side = 1 and two opposite corners displaced by +1/2


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while the two other opposite corners displaced by -1/2. The

edges have line elements with area = 500. Figure 9 shows

the meshes for the initial and final surfaces.

Note that minimizing the total volume results in opposite

effects on the lengths of the free edges. It tends to

decrease the surface area defined by the triangle elements

and consequently tends to increase the lengths of the free

edges. It tends to decrease the path lengths defined by the

line elements and consequently tends to decrease the

lengths of the free edges. In this example, the free edges

are straight due to relatively big value for the areas of

the line elements.

Figure 9

Example 7: A straight prismoid with height = 3. The bottom

and top regular triangles are inscribed in a circle ofradius = 1. It is composed by 3 line elements and 9

isovolumetric line elements. The line elements have area =

1. The penalty parameter = 1.0E+05. The top triangle

rotates 150 degrees clockwise relatively to the bottom

triangle. Figure 10 shows the initial and final shapes.

Appendix 1 presents analytical expressions for this type of

prismoid.


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Figure 10

Example 8: A straight prismoid with height = 2. The bottom

and top regular pentagons are inscribed in a circle of

radius = 0.75 and 0.5 respectively. It is composed by 5

line elements and 15 isovolumetric line elements. The lineelements have area = 1. The penalty parameter = 1.0E+05.

The top pentagon rotates 126 degrees clockwise relatively

to the bottom pentagon. Figure 11 shows the initial and

final shapes. Appendix 1 presents analytical expressions

for this type of prismoid.


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Figure 11

Example 9: A straight prismoid with height = 6.651155. The

projection of the bottom and top closed polygonal chains is

a regular 12-sided polygon inscribed in a circle of radius

= 1. The parameter delta = 0.4277998. The Z coordinates of

the bottom points alternate between +/- delta while the Zcoordinates of the corresponding top points alternate

between the height of the prismoid -/+ delta. The

corresponding bottom and top points are connected by

isovolumetric elements only when its distance is the

smallest between the two possible values. Isovolumetric

elements connect successive top points, and also successive

bottom points, of the polygonal chain. This prevents the

collapse of the prismoid into a line. The prismoid is

composed by 12 line elements and 54 isovolumetric line

elements. The line elements have area = 1. The penalty

parameter = 1.0E+05. The top polygonal chain rotates 120degrees clockwise relatively to the bottom polygonal chain.

Figure 12 shows the initial and final shapes. Appendix 2

presents analytical expressions for this type of prismoid.


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Figure 12

Example 10: A circular prismoid with axis on a

circumference of radius = 10. The section is defined by a

regular triangle inscribed in a circle of radius = 1. It is

composed by 30 line elements and 60 isovolumetric line

elements. The line elements have area = 1. The penaltyparameter = 1.0E+05. Figure 13 shows the initial and final

shapes.


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Figure 13

Example 11: A square with side = 1 composed by 8

isovolumetric triangle elements with thickness = 10. The

squares perimeter has line elements with area = 1. The

penalty parameter = 1.0E+05. The triangle elements preserve

the squares area while the line elements minimize itsperimeter. The square turns into an octagon. Figure 14

shows the initial and final areas.


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Figure 14

Example 12: A cube with side = 2 composed by 24

isovolumetric tetrahedron elements. The cubes surface has

triangle elements with thickness = 1. The penalty parameter

= 1.0E+05. The tetrahedron elements preserve the cubes

volume while the triangle elements minimize its surface.

The cube turns into a 24 faces polyhedron. Figure 15 shows

the initial and final volumes.

Figure 15


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8 Appendix 1

Figure 16 shows a straight prismoid. The bottom and top

regular polygons are inscribed in circles of different

radius. The sum of the lengths of the red lines is

minimized by rotating the top polygon counterclockwise withrespect to the bottom polygon, while the lengths of the

black lines remain constant.

ip

ip

Figure 16

8.1 Geometry

For n-sided regular polygons, the coordinates of thevertices can be written as:

2

n

=

( )

( )ir cos i

p r sin i

0

=

( )( )i

r cos i

p r sin i

h

+ = +

ip

i i 1b p

+=


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i i id p p

+= 1

i iv p= ip

8.2 Final height

The square of the norm of vector vi can be written as:

i i iv p p=

2i 2 2v r r 2rr cos h

2

= + +

Since this norm is constant, it leads to the following

expression that relates the initial and final heights.

( )2 20h h 2rr 1 cos =

8.3 Rotation angle

The square of the norm of vector di can be written as:

i i ib d v 0+ =

( ) ( )2 2 2i i i 2d v b 2r 1 cos 2rr cos cos= +

Due to symmetry, minimizing the sum of the norms of all

vectors is equivalent to minimizing the square of the norm

of only one vector.

2i

d sin0 tan

cos 1

= =

Notice that this expression is valid when the vectors

connect the corresponding bottom and top points in any

symmetric way. Table 2 shows the rotation angle in degrees

and the values for the relation between the initial and

final heights for some n-sided regular polygons.

( )2 2 10

2h h

2 2 cosrr

= + 2


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Table 2

n 2 20h h

rr

3 150.0 3.7320514 135.0 3.414214

5 126.0 3.175571

6 120.0 3.000000

7 115.7 2.867767

8 112.5 2.765367

9 110.0 2.684040

90.0 2.000000

8.4 Minimum initial height

The minimum initial height is given by:

( )1

22

0h rr 2 2 cos 2rr h + 0

9 Appendix 2

Figure 17 shows a straight prismoid. The projection of the

bottom and top closed polygonal chains is a regular polygon

inscribed in a circle of radius = r. The Z coordinates ofthe bottom points alternate between while the Zcoordinates of the corresponding top points alternate

between the height of the prismoid . The correspondingbottom and top points are connected by elements only when

its distance is the smallest between the two possible

values. The sum of the lengths of the red lines is

minimized by rotating the top polygonal chain

counterclockwise with respect to the bottom polygonal

chain, while the lengths of the black lines remain

constant.


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Figure 17

9.1 Geometry

For n-sided regular polygons, the coordinates of the

vertices can be written as:

2

n

=

( )

( )( )

i

i

r cos i

p r sin i

1

=

( )

( )

( )

i

i

r cos i

p r sin i

h 1

+

= +

ip

i i 1b p

+=

i i id p p += 1

i iv p= ip

An expression for can be written by setting the anglebetween the vectors (pi+1 pi) and (pi+2 pi) equal to .

ip

ip

i 2+p

i 2+p


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( )2

2

1 cos 11 cos

r 4 cos

+ = 2

9.2 Final height

The square of the norm of vector vi can be written as:

i i iv p p=

( ) ( )2 ii 2 2 2

v 2r 1 cos h 4 1 4 h = + +

Notice that there are two different values for this norm.

The smallest value, associated with i even, is given by:

( )2

i 2 2 2v 2r 1 cos h 4 4 h = + +

Since this norm is constant, it leads to the following

expression that relates the initial and final heights.

( ) ( ) ( )2 2 2

0h 2 h 2 2r 1 cos =

9.3 Rotation angle

The square of the norm of vector di can be written as:

i i ib d v 0+ =

( ) ( )

2 2 2i i i

2 2

d v b

2r 1 cos 2r cos cos 4h 8

= + +

+ 2

Due to symmetry, minimizing the sum of the norms of all

vectors di is equivalent to minimizing the square of the

norm of only one vector.

( )

2i

2d

2r 0

= =


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Notice that this expression is valid when the vectors

connect the corresponding bottom and top points in any

symmetric way.

( ) ( )

( ) ( )1

2 2 2

0

cos 1 sin sin cos

2 sin

h 2 2r 1 cos

= +

sintan

cos 1

=

The rotation angle is in the following interval.

< <

9.4 Minimum initial height

The minimum initial height, which is associated with = 180degrees, is given by:

0h 2r 2 h 2 +

9.5 Special case

The initial height can be set in order that a 2n-sided

prismoid with n vertical elements will rotate the same as

its associated n-sided prismoid with n vertical elements.

( )

( )

sin 2tan

cos 2 1

=

( )

( )

22

0h 2 cos

2 2

r r 1 cos r

= + +

1 sin

( )

h 2

r 1 cos r

=

Table 3 shows the initial and final heights for = 60degrees.


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Table 3

n0h

r

h

r

6 3.808383 2.828427 150.08 4.603421 4.060207 135.0

10 5.595633 5.236068 126.0

12 6.651155 6.386283 120.0

14 7.730797 7.522434 115.7

10 Appendix 3

1 1 1

1 2 33 2

2 2 2

3 1

1 2 3

2 1

3 3 3

1 2 3

v v v

x x x 0 a av v v

v a x a 0 ax x x

a a 0v v v

x x x

= =

1 1 2 2 3 3

1 2 2 3 3 1

w w

v u v u v u

u u u u u u

= +

+ + +

+ + +

+

1 1 1

j j jj j1 2 33 2

j j2 2 2

3 1j j j

1 2 3 j j

2 1

3 3 3

j j j

1 2 3

w w w

u u u0 v v

w w wv 0 v

u u uv v 0

w w w

u u u

=

11 References

[1] Gill, P. E. and Murray, W. Newton type methods for

unconstrained and linearly constrained optimization,

Mathematical Programming 7, 311-350, 1974.

[2] Isenberg, C., The Science of Soap Films and Soap

Bubbles, Dover Publications, 1992.


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[3] Nocedal, J. and Wright, S. J., Numerical Optimization,

Springer-Verlag, 1999.

Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini

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