-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
1/27
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/ -
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
2/27
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:
-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
3/27
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
-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
4/27
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.
3.1 Element final volume
Supposing that the triangle element has a fixed thickness ,its final volume can be written as:
w2
=
3.2 Gradient of the element final volume
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
= + +
The nodal displacements vectors are numbered according to
its node numbers. Their individual components are numbered
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:
-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
5/27
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
4.1 Element final volume
1x
2x
3x
4x
3v
2v
1v
3u
2u
1u
4u
2x
1x
3x
1v3
v2
v
-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
6/27
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
= +
+ + +
+ + +
+
4.2 Gradient of the element final volume
The nodal displacements vectors are numbered according to
its node numbers. Their individual components are numbered
as follows:
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:
-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
7/27
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.
5.1 Element final volume
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
=
5.2 Gradient of the element final volume
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
-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
8/27
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
-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
9/27
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].
-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
10/27
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.
-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
11/27
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.
-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
12/27
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
-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
13/27
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
-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
14/27
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.
-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
15/27
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.
-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
16/27
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.
-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
17/27
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.
-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
18/27
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.
-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
19/27
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
-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
20/27
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
+=
-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
21/27
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
-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
22/27
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.
-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
23/27
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
-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
24/27
( )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
= =
-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
25/27
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.
-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
26/27
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.
-
8/9/2019 Finite Element Analysis for Minimal Shape by Arcaro, Klinka, Gasparini
27/27
[3] Nocedal, J. and Wright, S. J., Numerical Optimization,
Springer-Verlag, 1999.