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THE APPLICATION OF THE MODULAR MODELING METHOD TO
FINITE ELEMENTS ANALYSIS

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THE APPLICATION OF MODULAR MODELING METHODS IN FINITE
ELEMENT ANALYSIS

BY

Xin Li

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

MASTER OF SCIENCE
Department of Mechanical Engineering

2000.05

ABSTRACT
The Application of Modular Modeling Methods in Finite Element Analysis
By

Xin Li

Structural modeling analysis is important to engineering design and
development. An important issue is the interconnectivity problem for multi degree
of freedom subsystems. The traditional finite element method requires re—
meshing subsystems for mesh compatibility at interconnections. As the number
of subsystems increase, the computational burden of re-meshing increases
rapidly, sometimes exponentially.

The Modular Modeling method is applied here to develop a new finite element
analysis approach. Modular Modeling through the use of input 1 output ports on
the boundaries of interconnected parts is introduced for this purpose. Using input
I output ports and the Modular Modeling method to connect parts in system
design does not require expensive remodeling and re-meshing. This approach
uses the original FEA model and combines parts with the input I output ports
among their boundaries. As the number of available alternate parts in a structural
system increases, the modular F EA method minimizes the time and cost for the
system integration.

A two-dimensional example of FEA elastic strain-stress problems is used to
illustrate this technique. This example shows it is not necessary to re-mesh
components to assemble a structural system. Computation error is evaluated to

show that this modular finite element method produces accurate system models
for engineering design.

Copyright by
Xin Li

2000

TABLE OF CONTENTS

List of Figures ....................................................................................................... v
Introduction ........................................................................................................... 1
Multi-Component Modular Modeling for Finite Element Analysis .......................... 4
2.1. FEA Modular Modeling and the related DOF analysis ............................... 4
2.2. Visual Component Boundary Port — FEA Input/Output Connector ........... 7
2.3. FEA Modular Modeling Procedures and Formulations ............................... 9
2.3.1. Sub-system Equation Combination .................................................... 10
2.3.2. Applying Boundary Displacements Constraints ................................. 12
2.3.3. Applying Boundary Forces Constraints ............................................. 21
2.3.4. Combine the Supplemental Boundary Constraint ............................... 24
2.4. Result and Discussion ............................................................................. 27
Two-Dimensional Example for Elastic Strain-Stress Problem ............................ 28
3.1. F EA static stress-strain analysis by modular modeling ...................... . ...... 2 8
3.1.1 2-D Elastic Plane A: ............................................................................ 29
3.1.2. 2-D Elastic Plane B: ........................................................................... 30
3.1.3. Using Modular Modeling To Integrate 2-D Elastic Components ......... 31
3.2. Error Estimation and Analysis .................................................................. 32
3.2.1. Standard Reference Models for Accuracy Comparison ..................... 32
3.2.2. Modular Modeling Method Accuracy Comparison .............................. 33
Conclusions ........................................................................................................ 37
4.1 Contributions ............................................................................................. 37
4.2 Future work ................................................................................................ 38
References ......................................................................................................... 39

Appendix I. .........42
Appendix II..... ......46
Appendix III.. .49
Appendix IV.... ...51
Appendix VI...... ...67

LIST OF FIGURES

Figure 1.1. The modular modeling using input / output Ports (joint) ............................... 3
Figure 2.1. The modular modeling using input / output Ports (connectors) ...................... 5
Figure 2.2. The modular modeling method for the FEA application ................................ 7
Figure 2.3. The modular modeling input / output Ports along the FEA boundary ............. 8
Figure 2.4. At least 6 constraint are needed for 2-D FEA modular modeling ................. 11
Figure 2.5. Nodal displacements along the boundary elements ....................................... 13
Figure 2.6. Nodal compatible situation along the boundary elements ............................. 14
Figure 2.7. Nodal displacements in the boundary elements ............................................ 15

Figure 2.8. Linear transformation on the boundary elements (case 1: interpolation) ....... 16

Figure 2.9. Linear transformation on the boundary elements (case 2: extrapolation) ....... 18

Figure 2.10. Linear transformation on the boundary elements (case 3: equality) ............. 18
Figure 2.11. The power conservation along the Boundary ............................................. 23
Figure 2.12. The “solvable system” and related equations. ............................................ 24
Figure 3.1. Two-Dimensional Example for Elastic Strain-Stress Problem ..................... 28
Figure 3.2 Plane A meshing details and nodal distribution ............................................ 29
Figure 3.3. Plane B meshing details and nodal distribution ........................................... 30
Figure 3.4. Re-meshed reference model with higher or lower meshing resolution .......... 32
Figure 3.5(a). Displacement contour lines for reference model (7x7). ........................... 33
Figure 3.5(b). Displacement contour lines for the Joint Modular Model ........................ 33
Figure 3 .6. Displacement difference between the Joint Modular Model and reference
model (7x7) ............................................................................................................. 34
Figure 3.7(a). Displacement contour lines for reference model (3x3). ........................... 35
Figure 3.7(b). Displacement contour lines for the Joint Modular Model ........................ 35
Figure 3.8. Displacement difl‘erence between the Joint Modular Model and reference
model (3x3) ............................................................................................................. 35
Figure 3.9. The difference of nodal displacements along the boundary ........................... 36
Figure 4.1. Comparing the Modular Model and other models ........................................ 37

INTRODUCTION

During the design of complex systems, reliable predictions of detailed
stress states for coupled components are often critical to correctly predict the
performance of those systems. The finite element analysis (F EA) yields accurate,
detailed stress predictions for individual system components with the resolution
of mesh distribution. When more precision is needed, a higher precision mesh
needs to be generated. In engineering practice, system components can be so
complex that the finite element method is mandatory for the design phase. It is

computationally intensive and often requires extensive re-modeling effort.

Coupling system components into assemblies is a time intensive task
frequently requiring reformulation for finite element analysis models. For
example, the Chrysler large car models used for mechanical geometry studies
contain representations of over 5500 interconnected physical subsystems
[Computers in Engineering: “Chrysler designs paperless cars”, 1998]. To reduce
the length and cost of the design process for a complex system, efficient
modeling techniques that guarantee reliable predictions of detailed stress states
are needed. These techniques must be not only computationally efficient, but
also dramatically reduce or eliminate the necessity of re-modeling. According to
[Computers in Engineering: “Chrysler designs paperless cars”, 1998], Chrysler

engineers solved design issues by developing a computer model instead of

physical prototypes, reducing the cycle time from 39 to 31 months and saving the

company more than $75 million.

Finite element analysis of system of components typically requires
compatible node geometry at each component connection. When constructing
the two-dimensional FEA models, it is a mandatory requirement to keep mesh
nodally compatible [Zienkiewicz and Taylor, 1989]. At present, the methods
available to ensure FEA nodal compatibility include the global llocal analysis
method, the coupling analysis method, and the interface element method, [M.
Aminpour, 1995; V. S. Kothnur, 1999; J. B. Ransom, 1993]. These methods have
drawbacks that prevent them from being widely accepted as standard design
tools, [2. Zhao, 1998]. The common limitation of these methods is that they can
only be applied to mesh discretizations with a one-to-one nodal correspondence
across boundaries between structural subsystems. Methods that do not require
nodal compatibility on the component boundaries have much more modeling

flexibility, and eliminate the need for re-meshing.

A new finite element analysis approach is introduced in this paper which
employs the fixed input / output structure "modular modeling method" [8. Byam
and C. Radcliffe, 1999]. In Modular Modeling, the mathematical model that
describes each component remains the same independent of the system model
in which it is used. Fixed input / output structure means the input and output ports

are standardized, therefore the internal equations of mathematical subsystem

models of engineering systems have the same modularity as the engineering
system. Fixed input I output structure modular modeling is a power-based,
physically intuitive, top-down methodology systematic modeling methodology that
eliminates equation reformulation from large model development across multiple

energy domain [8. Byam and C. Radcliffe, 1999]

This paper demonstrates the modular modeling method in FEA. The work
presented here developed an analytical framework with the Modular Modeling
analysis method, a solution strategy, and a demonstration of this solution on a
F EA example. Finally, the accuracy and convergence of a Modular FEA Model
will be evaluated by comparison with various conventional FEA solutions to a 2-D

plate problem (Fig, 1.1).

   

Ub=S-Ua Inlemal boundary
Ea'+Eb"S=U constrains

Figure 1.1. The modular modeling using input / output Ports (joint)

MULTI-COMPONENT MODULAR MODELING FOR FINITE ELEMENT
ANALYSIS

2.1. FEA Modular Modeling and the related DOF analysis

Modular modeling is a new modeling method designed to eliminate the model
reformulation and enhance model performance verification. It is defined by a
standard input-output structure, in which connection model physical interaction
between components via power transfer. Power Transfer between components
requires the product of two signed variables such as force times velocity,
pressure times flow rate, and voltage times current. The two physical variables
form a component input-output pair. For each type of physical interaction there is
a standard input-output definition for modular modeling components. [3 Byam
and C. Radcliffe, 1999]

Modular modeling can be applied to F EA model components. In the
discussion below, a basic 2-D elastic stress-strain FEA problem will demonstrate
the basic approach for connecting different FEA component models together,
even when their nodes are geometrically incompatible. The following figure
shows the basic concept for modular modeling in FEA. In this approach, a single,
multi-node input/output port is defined to connect two F EA models. The power,
force and displacement transfers through this connection port have standard
definitions [8. Byam and C. Radcliffe, 1999].

 

 

 

 

 

 

leattomtch,U

 

Figure 2.1. The modular modeling using input / output Ports (connectors)

The analysis of system degrees of freedom (DOF) is the foundation of all
other analysis. We can not begin any static or dynamic analysis before we decide
the system DOF. In FEA modular modeling, the connection of two independent
FEA components produces a modular model with a unique solution if it is the
zero DOF. In another words, modular connectors provide enough constraints for
whole system to generate a non-singular system equation. For example, if we
combine two independent FEA components, each with 6, unconstrained DOF,
we need to find at least 2x6=12 constraints for displacement to generate a
system model with stable, unique solutions for different force inputs. These
constraints should provide both external boundary conditions and internal
boundary conditions. In Figure 1.1, a constraint set would fix one component to
the ground (external constraint) and the other component is fixed on the first one
(internal constraint). This combination of constraints will remove all rigid body
degrees of freedom and generate system equations with a unique solution.

Supplemental boundary constraint equations are critical to control the FEA
modular model DOF. In the first step of the proposed analysis, we combine
system equations and their degrees of freedom. For example, in 2-D stress-strain
analysis, the three degree of freedom unconstrained components have a singular
stiffness matrix, whose rank is N—3 where N is the dimension of the stiffness

matrix. When we combine two three degree of freedom components together, the

combined stiffness matrix has a rank of N+M—6, Where N and M are the
dimensions of the two component stiffness matrices. External boundary
conditions combined with connection constraints must increase the combined
system rank to N+M to generate a model with unique solutions to external force
inputs. This idea can be easily understood if we consider this with a physical
model. In two-dimensional space, a 3—DOF subject can be held fixed if 3
boundary conditions are provided. When we connect two 3-DOF subjects
together, an additional 3 boundary conditions prevent relative motion between
the two components. When the 3 boundary conditions on one component are
combined with 3 internal constraints, the resulting stiffness matrix is non-singular
and yields unique solutions to input forcing.

 

2.2. Visual Component Boundary Port — FEA Input/Output Connector

The modular modeling method generates non-singular equations by using the
original component system equations and the modular input/output equation.
Actually, this is one of the advantages of the modular modeling method. With this
method, a great deal of effort can be saved because no remodeling and re-
verification is needed [8. Byam and C. Radcliffe, 1999]. When applying this idea
into F EA, the benefit is clear: an FEA model can be used directly in system re-
mesh and re-verification of component models are not necessary.

2w: 2f*u=0

I ————— i ~~~~~~ / I "I
K [: 3:. a: a ‘2 :> mitt]

J/f‘w.‘ ——————————— ’TIN/ U//f u\\f

Constrain
m n ... ..-...

Figure 2.2. The modular modeling method for the FEA application

 

 

 

 

 

 

Definition of an l/O port on each component connection will provide proper
constraint equations to make the combined system equation non-singular. The
component system equations will be applied directly to generate the combined
system equations. The Modular Modeling Method (Fig. 2.2) allows the
development of standard integration and computer automation methods.
Development of IIO port constraint equations begins with the basic properties of
FEA model. Force and displacement transferred across a component connection
port must satisfy basic energy and work conservation requirements [Yang, T.Y.
1986,]. The total work transferred through the IIO port must sum to zero.

6WA+6WB =0 2, 6W =‘5W3 .......................................... (21)

A
in which :
6W4 = ZfaN ' uaN = fzai-t ° ”2am +f2al 'uzai +fzaiz-i '"2a2-1 +fzaz 'uzaz + +fdll 'uaN
5W3 = Zfau 'uaM = bel-l 'uzbi—i +f2bl '“2bi +f2bZ-l '“2b2-1 +f2b2 'uzbz +"'+me '“Bm

In the upper equations, the indices a1, a2,... aN, and b1,b2,... bN refer to the
nodal number along the connection forming the IIO port.

 

 

 

 

 

 

 

 

FEA m FEA m B
A
System Equation ‘
for Modular A +
comm“ Constroin Equation
--— System ( for IIO Port
Equation
\ System Equation
for Modular B
It \ :10 port room F. u /
Comoction
pocflon

 

Figure 2.3. The modular modeling input I output Ports along the FEA boundary

Theoretical solution is provided here for a two-dimensional, straight-line
boundary and three degree of freedom elastic stress-strain analyses. Future
work will be needed for three-dimensional, curve-surface boundary, different type
of elements and six degree of freedom finite element analysis.

2.3. F EA Modular Modeling Procedures and Formulations

New formulations for the modular modeling need to be developed for the two-
dimensional F EA models. The procedures and formulations for one-dimensional
FEA modular modeling have been demonstrated [8. Byam and C. Radcliffe,
1999]. The two dimensional problems are different and need updated procedures
and formulations. Two-dimensional F EA modular modeling does not require
mesh nodal compatibility along the connection boundary. To make F EA modular
modeling, we must develop new procedures and formulations for making the
F EA modular model work under the situation of mesh nodal non-compatibility
along the connection boundary. Methods to connect modular F EA models
without compatible nodal point geometry are given below.

In this paper, four steps develop a two-dimensional FEA modular modeling
formulation, which is demonstrated in section §2.3.1, §2.3.2, §2.3.3 and §2.3.4.
The combined system equation of FEA models is provided by step one. Step
two applies the boundary displacement constraint equations to transform the
system equations. Step three puts the system equations into standard linear
algebraic equation form by moving the unknown force into the unknown variable
vector. In step four, the boundary force constraint equation will supplement the
system equation, and make the system stiffness matrix square. The system
model equations are now complete, non-singular and have a unique solution
given a set of independent inputs.

2.3.1. Sub-system Equation Combination

The first step is to get the combined system equation. Consider 2 FEA
components: A and B. A has N nodes and B has M nodes and their stress-strain
equation is following. To avoid confusion, The displacement and force on subject
A and B are represented by uA, us and fit, f3. They have same definitions and

 

 

 

 

 

 

units.
K A -UA = FA 0r
_ku kn km - pun — rf1 q
kn kzz km “2 = f2 (2 2)
Lkm km kNN A J'N _ A LfN 3».
KB -UB = FB or
_ku kn km - -ul - -f1 —
kn kn km “2 = f: (2 3)
_km km kMM_B _“MJB LfM_B

 

 

 

 

 

 

In both equations, KA and K3, are singular. For two dimentional problems the
subsystem models allow “free body" motion with two orthogonal deflection and
one rotation. 80 the ranks of Kit and K3 are N—3 and M—3. For every individual

equation could be solved if boundary condition such as following provided

11. 0 vll 0
u;= u1, = 0; and u;= vb = O ........................... (2.4)
uc A 0 VC 3 0

Any 3 U value equal to zero could eliminate the rank of K by 3, and make it
non-singular. Now we combine both components together without any re-mesh
work. From this equation, we can see that it contains all of the old elements and
nodes.

10

K -u =f ................................................ (2.5)

syn sys sys
in which
K A E 0 u A f A
Km = . ; (um): ; (1'8”):
0 5 KB uB fB
Obviously, the rank of K3,, could be available from following equation:
Rank(Km)=Rank(KA)+Rank(KB)=(N-3)+(M-3)=N+M—6 ........ (2.6)
From equation 2.6 we know that the combined system needs at least 6
internal and external boundary condition to become solvable.

      
 

FEA Modular A FEA Modular 8

Figure 2.4. At least 6 constraint are needed for 2-D FEA modular modeling

11

2.3.2. Applying Boundary Displacements Constraints

The boundary displacement constraint needs to be defined at first, like most
regular FEA procedures. This procedure will make the system equation (2.5) a
full-rank linear equation. Therefore, this is the crucial step to make the system
solvable. According to the previous analysis, the internal (boundary between 2
components) displacement constraint equation must be more than rank of 3.
Assume the following equation:

(“21)an =Smxn .(u8)mxm u............................... (27)

is the boundary displacements constraint equation, we need to make sure that:

Rank(smxn)23 (2.3)

The simulation result with the FEA model is a linear approximation for real
situation [Zienkiewicz, O. C. and Taylor, R. L., 1989]. Suppose we are using
linear elements, the boundary displacements constraint analysis should be based
on every nodal value on both components boundaries linear approximation. The
finite element method provides a displacement value within every element only
according to the nodal value in itself. Therefore the boundary constraint equation
could be found by the local boundary elements interpolation. Figure 2.4
demonstrates the mean of the local boundary elements nodal displacement
relationship.

12

 

 

   

FEA Modular B

 

 

2-Dimentional
Displacement

Figure 2.5. Nodal displacements along the boundary elements

The boundary displacement in Figure 2.4 at node “i” was defined as “u2i-1”
and “u2i”. If the boundary nodes on modular A and B is a1, a2,...aN and b1,
b2,... bM, then the boundary displacement constraint equation could be written

as.
(“bl “b2 ace “WK =.'So(llal “02 can uaM);ooooaaoaaaoaoooooo(2.9)
specially, the 2 - D equation situation :
r
(um—i u2bl "2122—1 "21:2 ' ° ' uZW—l ”2w )4 :
Smxzrv ' (”201-1 “2m u2a2-l "2a2 ' ° ' uZaN—l uzwv)

In most cases the rank of equation (2.9) will be more than 3, otherwise at
least one component has only one node on the boundary. So, if we can develop
boundary displacement constraint relationships, equation (2.9) will have enough
constraints as required. For both subjects (Fig. 2.5), the nodal displacement
value along the boundary should satisfy the boundary interpolation [Langhe,

K. De, 1995].

13

2-Dimentional
Displacement

 

Case 3

Case 1

 

Figure 2.6. Nodal compatible situation along the boundary elements

Figure 2.5 shows the position of node “b1” has only 3 possibilities relative to
the corresponding elements. Case 1 is located on the edge of corresponding
boundary element. Case 2 is located on the straight extended line of the edge of
corresponding boundary element. Case 3 is coincided with one node of the
corresponding boundary element.

Independent of which component has more boundary nodes than another,
both sides of the port can be used to develop boundary displacement constraint
equations. In other words, equation 2.9 has 2 equivalent forms and they are
equivalent constraints. Here we present results for nodes on module 8 along the
boundary.

CASE 1. From the definition of finite element, any internal point value can be
represented by the linear transformation of nodal values [Segerlind, L. J., 1984].
For the boundary nodes in another component, we also use similar linear
interpolation approach to figure out their values [Langhe, K. 09., 1995 and
Kothnur, V.S. , Yu Xie, 1999]. In this paper, the discussion was focused in the
basic theoretical solution for two-dimensional, straight-line boundary and three
degree of freedom elastic stress-strain problem. So we can apply the similar
approach from reference [Langhe, KDe., 1995].

14

U082) l

U(2a2-1)

'0
.-
.0

  
 

Boundary

 

2-Dimentional
Displacement

Figure 2.7. Nodal displacements in the boundary elements

As it shows in figure 2.6, we can get the shape function of an arbitrary point in
the element which has the coordinates of (x, y) [Yang, T.Y. 1986]:

“21“: y) = u2al +clx+c2y

u2i—1(x’y)=uzal-l+c3x+c4y .............................. (2.10)

p The c1, c2, c3, c4 is determined by the shape of element and can be derived
from the following equation:

 

 

(“ZaI-l I
(C, I fb, -b, o b, o -b,. o ‘ um
c2 = 1 a, -aj 0 -ak 0 a}. 0 . “2a2-1 ...... (2.“)
c3 ajbk -akbj 0 b; -bk 0 bit 0 -b1 “zaz
Kg) _ 0 a, —a1 0 -a,, 0 a, _ uh},

 

 

 

 

( “243 )

From equation (2.10) and (2.11), we get a general interpolation form for any
point located in the boundary element. Because the values of c1, c2, 03, c4 are
determined linearly only by properties of the corresponding boundary element,
we could conclude that the u2b1-1 and u2b1 in (2.10) are linear functions of
u2a1, u2a1-1, u2a2, and u2a2-1, u2a3, u2a3—1 correspondingly.

15

The transformation matrix 81, 82 in the upper equation is decided by the
corresponding boundary element and the x, y coordinate value of node “b1”.
After we figure out all of the equations (2.12) for all boundary nodes on
component B, we combine them and get the equation (2.9).

r
u2b1-l =81 '(uzai-i u2a2-l “us—1)

u2bl=sz.(um “202 “203)7 (2.12)

In this paper, we are discussing the straight-line boundary problem. So all
boundary node “b1”, “b2”, ...... “bM” is locate on the edge of boundary elements.
In this situation, the equation (2.12) could be simplified and decided only by
nodal value on boundary edge [Segerlind, L. J., 1984].

     

 

 

 

 

 

 

M2“) UQa1-1)'Na1(x y) “um Uaflmam'”

: +UQa2-1)‘Na2(x,y) +UQa2)‘Na2(x,y) i

I : I

IUWJ) :Ule-l) ' . :U b1 Euaaz)

: E :urzaz-n 1UP“) :9 ’ :

.1? J1 a2I : at! :m ia2

«0—0- 8 *——_"' 3

.1 t 82 : 81 f 82 s:

Figure 2.8. Linear transformation on the boundary elements (case 1:
interpolation)

Figure 2.7 shows the linear interpolation of node “b1”. Suppose the distance
from node “b1” and node “at” is $1, and from node “b1” and node “a1” is $2,
which could be derived from the their coordinate values, we can get the linear
interpolation equation from the shape function [Langhe, K.De., 1995]:

 

 

_ _ 32 S1
u2b1-l — Na1(s)'u2a1—1 +Na2(5)'“2a2-1 " ' ' _' ' 'uzai-i 1' '“2a2-1
sl +s2 13-1-32
u2b1 =Na1(s)'u2a1+Na2(3)'u2a2 = ”u2a1+ 'uzaz (2-13)
31 +s2 sl +s2

16

The distance from node “b1” and node “at”, which is $1, and from node “b1”
and node “a1”, which is 52, could be derived from following coordinate

 

 

computation.
St = Jam —xal)2 +0511" yal)2
32 =J(be-xa2)2+(ybl —ya2)2 .............................. (2.14)

It is necessary to mention that the s1 and s2 values are always positive in

case1.

CASE 2. In this case, node “b1” is located on the straight extended line of the
edge of corresponded boundary element. This happens because node “b1” is the
terminal point on the boundary, and it does not contact with the corresponded
component. In the following figure (2.8), we can see that equation (2.13) is still
valid. The only thing that needs to be modified is that we should to re-define st
and s2 according to the situation of “b1 ” location. If the “b1” location comes with
the sequence of “a1, a2, b1”, the s1 and 52 should be:

 

SI = ‘J(xbl -xal)2 + (ybl " yal)2
32 = “/09,l "xa2)2 + 0’»: - ya)? .............................. (2_15)

 

If the “b1” location comes with the sequence of “b1, a1, a2”, the st and 32
should be:

 

31 = -J(xbl —xal)2 +(ybl "' yal)2
52 zfixbl ‘1‘”): +0,“ _ yd)? .............................. (2.16)

 

17

“UQi-I)
U(2a1-1)'Na1(x,y)
+U(2a2-1)'Na2(x ,y)

     

uaaz)

 

I

I

:Ule-I) :UQbI) :UQOI)
82: lbl _ b1! :81 A2 L

81 a 8 : SI s2

 

 

p—-—----—

I,
d
on

I

Figure 2.9. Linear transformation on the boundary elements (case 2:
extrapolation)

CASE 3. In this case, node “b1” is located on the exact same nodal point of
corresponded element. Clearly, this is a “nodal compatibility” situation. It is easy
to conclude that if “b1” is located on same nodal point of “a1”

 

 

 

_ _ _ 52 SI
u2b1-l -uZal-l ‘1'“2ai-i +o'“2«2-i " '“2a1-1 '1' ' 'uzaz-i
31 +32 31 52
“2b: =“2a1 =1’“2ai +0'“2a2 = '“2a1 1' '"2a2
S, +32 S1 +32

If “D1” is located on same nodal point of “a2”

 

- _ _ 52 31
u2b1-l —u2a2-l ‘0'“2a1-I +1'“2a2-1 "' 'uzai +—'_'u2a2
1 52 31 +32
3 s
— — . . -—2 e 1 '
u2b1 -u202 ‘0 “2111'” “2a " uza 1' “2a2

.t',+s2 ' sl+s2

U0“) ”90 0 111-1 u 1.1

I U(2b1-1)=UQa2-1) ll ung’Jafi'; )
r UleFUQaZ) :
: I
' I
: ' i :Uaazi
I I ' '
I : : :

a1; ‘2: ”1 min ; a2

 

 

 

 

”32:0 :7 -H--sl=0 82 t

31 ._ g E
i _. E

II

Figure 2.10. Linear transformation on the boundary elements (case 3: equality)

18

Therefore, equation (2.13) is valid too for this case. When the condition of
“s1=0 or 51:0” was represented in (2.13), the output gets the simplified result as
equation (2.17) or (2.18). Overall, we could conclude that in the 2-dimentional
elastic problem, the displacement constraint equation along the straight-line
boundary is equation (2.13). Based on different nodal location, the interpolation

parameter s1 and 82 should be computed according to equations (2.14), (2.15),
(2.16), (2.17) or (2.18).

From these equations, we could get all interpolation functions for “b1,

b2,...bM” in figure 2.4. In equation (2.19), sb11 and sb12 represent the s1 and s2
value for “b1”, and Lb1 represent s1+s2 value for “b1”,

bl bl bl bl
.. -2— . ...L. . — ...2_. . _I_ .
u2b1-l " L "201-! ‘1' uZaZ-l . u2b1 " “2a: 1' L “zaz
In La La b1
b2 b2 b2 b2
u -—2—-u +‘—-u u -—2--u +—‘—-u
2112-1 "' L 2a2—i 203-1’ 2b2 ‘ 2a2 L 2a3
b2 Ln Lbz b2
saw by saw an
.— ._2_ . I_ . — -2— . _I_ . ......
quM—l - L “um-1H '1' uZaN—l ’ “21w " L uZaUV-I) 1' “2m (219)
an an m m

If we define displacements by node as following:
u2bl-l “2122-1 “zen-1
u = u = ’ ...... u =
bl ( um } b2 [ “252 ] bM [ “25M )

Equation group (2.19) could be gotten if all boundary nodal coordinate values

are available. Also, the equations of (2.19) could be rewritten as the boundary
displacement constraint equation (2.20).

19

“bl {Cbl’al
“b2 = 0
um I 0

 

Cbl’aZ 0
Cb2‘a2 Cb2‘a3
0 0

CW‘aUV-I) CW'aN J

 

 

 

(2.20)

In upper constraint equation, the constraint elements was defined as:

Cbl'al =

i=1 or 2;

 

 

(31: 0 I
L“ s” I Cbl'a2=
0 _2_.
I L»)
decided by

Ii
Lu
0

 

I

aY position

\
0

bl
EL.

Lu)

 

Or the equation (2.20) could be write as:

I “251-: \

 

 

If}: 0
bl b]
0 it.
bl
0 0
0 0
0 0
0 0
\

 

— 0 o
Lbl bl
o 3‘— o
02 L“ b2
3.2_ 0 §I_
L112 b2 2
o ‘2 o
L,2
o o o
o o 0

20

”ME."

 

of

r” If": C

O

 

(if:
___ Lax
0 Si—
I
N
0 o
o o
o o
o o
i 0
Lb”
0 ii
I“)

 

 

\

“201-1
“201
“zaz-i
u2a2
“203-1

“243

“um-D4
“zerN-i)

uZaN-l

 

 

I“2a~)

2.3.3. Applying Boundary Forces Constraints

Boundary force constraints analysis is required to solve the system equation.
From the description of §2.3.1, we know that after boundary displacement
constraint equations convert the system equation to full rank, the boundary force
constraint equations and initial conditions are also required to solve the system.
In fact, the boundary force constraints equation is very important to enable the
system equation to get the final solution. The corresponding mathematical

Km'“1xm =fm; n) m. After applying the

description is, if we need to solve
system boundary displacement constraints equation, the equality of the rank of
stiffness matrix K' and f ' can only guarantee the solution exist, but it can not
guarantee that the solution is unique. It is absolutely necessary to have the force

constraint equations for the system equation.

The physical sense of this statement is: both the boundary displacement and
force constraint equations are required to ensure that the total power (for
dynamic model) or work (for static model) flow through the boundary is
conserved. Modular modeling ports provide the power constraint. Power is
conserved across modular ports because modular ports are power (or work)
transfer mechanisms. The power at the connected modular modeling ports
always sum to zero. For the elastic stress-strain problem, we need not make
dynamic analysis work. Therefore, the work at the connected modular modeling
ports always sum to zero.

The work along the boundary defined as: (WV = 23W, = 2:6, -u,.) =1"r ~u

Because work across the connection of component A and component B will
be

mA+mB=o z, MA=_MB ....................................... (221)
-. (rT-u),+(rT-u),, =rI-u,+r;-u,,=o

21

from Equation (2.20), we get
(ff -u,,)+(f; -C.uA)=0=> (f;w +f; -C)-uA =0
Because the power conservation is valid for any displacement under elastic
range, we can say that the equation (2.22) is valid for any of those displacement.
Therefore, we could conclude

r,’ +r; -C=0 and r,+c-r,, =0

Especially, for 2-dimentional case:

r T
I f 2a1 \ (“201—1 \ f bel I (u2b1-1 \
f 2a1-1 “201 f 2b1-1 “2121
f 2a2-1 “2a2-1 f 21,2-1 “21:2-1

.0

6wx=2fm'uuv= fzaz ' “zaz ma=2fsu'ueu= f2b2 ' “21:2

fZaN-l uZaN-l beM -1 “nu-1

 

 

 

 

 

 

 

 

(fzalv) (“2w ) (fun) (“21:14)

The equation (2.1) should be the beginning step of the constraint equation for
nodal forces. From equation (2.1), we get

( f2al NT (“201—1 \ f f2“ NT r“zen-1 \
f 241-1 “m f zen-1 “2b:
f 2a 2-1 “2a 2-1 f 2b2-1 “21:2-1
mx+ma=2fm'um+£fm'um = f2a2 ' “2.22 '1‘ fun ' “2b: :0
fZaN-l uZaN—l fun-1 quM-l
(fun; (“zany (fsz 1 (“2w )

 

 

 

 

 

 

 

 

When we place equation (2.20) into the above equation,

22

 

 

 

 

 

 

 

 

I f 241 )7 r“zen-1 \ I f 251 IT (“w-1 \
f2a1-1 “2411 f 2b1-l “2a:
f2a2-l u2a2-l f 2b2-1 “2.:2-1
5W4 + We = fzaz ' “2n '1' f2b2 ' 8' “zaz = 0 """"" (2°22)
f2aN—l uZaN-l f 2m -1 “zen-1
If“) Iuw) Ifwl Iuw)

Is“ s“ \
A 0 —‘—- 0 O 0 0 0 0 0
La La
s“ s“
0 J— 0 l 0 0 O O 0 0
Lb1 1%]
Std Sta
0 0 J— 0 —I— 0 0 O 0 0
Ln [12
in which 8: s” s”
0 0 O J— 0 ' 0 0 0 0

 

 

 

 

 

All work is done by elastic
force and displacement

   

2-Dimentional
Displacement
FEA Modular 8

Figure 2.11. The power conservation along the Boundary

‘ Figure 2.10 shows the sum of work transferred along the boundary equal
ZERO under the situation of the elastic range.

23

2.3.4. Combine the Supplemental Boundary Constraint

From the boundary constraint equations (2.20) and (2.23), we get the total 2M
constraint equations (if M<N) or 2N constraint equations (if M>N). From figure
2.12, we can see that when the system DOF is 0, there must be M>3 and N>3,
and the system equation is mathematically non-singular.

    

UFS'UP lntemal boundary

External U2i-1=0; 3431310 constrains

boundary U2i=0;
conditions: U2k-1=0; \T

Figure 2.12. The “solvable system” and related equations.

» For any solvable system like figure 2.12, the system equation, boundary
condition and boundary constraint equation should be combined through the
following procedures.

Step 1. Use the FEA models to get the combined system equation (2.5), the
boundary displacement constraint equation (2.20), and the boundary force
constraint equation (2.23). All of these constraint equations could come out from

nodal coordinate values and HG port shape. If KA is pxp, KB is qxq matrix, then

K A 5 0 uA “,4 f4
=KAB° —_- .............................. (2.23)
0 KB “3 “3 f3
KAB is a (p+q) x(p+q) matrix.

24

‘ Step 2. Apply the external boundary conditions and boundary displacement
constraint equation (2.20) to transform system equation. This step will replace

[u2H u2i u2k_,]=[0 0 O]; and
[UZbI-l Um UZbZ-l Um U2bM-l Ule 37

=S' [Um-t U2al UZaZ-l Um U2aN-l UZeNK
the displacement of some boundary nodes. For example in figure 2.11 the

external boundary conditions and boundary displacement constraint equation is

Using upper equations, we can get rid of u2i-1; u2i; u2k-1; and u2b1-1; u2b1;
u2b2-1; u2b2; U2bM-1 and U2bM-1 from UA and U8, and get following
equation (assume N>M):

uA u’A KA E 0 u, uA fA

u
g!
cI-l
ll
’3
a
II

(2.24)

In which TU is a (p+q-3-M) x(p+q) transformation matrix, and K'AB is a (p+q-
3-M) x(p+q) matrix that is transformed from KAB.

Step 3. Move the unknown forces (external and internal boundary forces) to
the left side, and leave the right side of the equation as a totally known force
vector. For example, if 1], fk, fl, fm in [FA FB]T is unknown, we can simply replace
it by “0". This will make the system equation become a standard linear algebra
equation, in which TF is a (2M+3) x(p+q) matrix and K’AB is a (p+q+M) x(p+q)

matrix.
{u’A \ f u" 1
“’4 {A f’A
K’u' = (K'AB 5 Tr)’ “’3 :10”. “’8 = (2.25)
“’3 f8 f’B
Kf’A+B ) \f’A‘l-B )

 

 

 

 

25

Step 4. Apply the boundary force constraint equation. Because all
boundary forces in equation(2.23) is unknown forces that have been moved
to left side, we can directly apply the boundary force constraint equation into

the system equation.

(u’A \ (u’AI (PAN
fital f2M K’ T ..
f f AB ' F
2.02 + 2:112 ’SF=Q =9 . . “’8 =K"’AB. “,8 = f’A °-- (2.26)
0 : S’
r r ’
2aN 2»! (rant) (rho) KO)

 

 

 

 

 

 

Because the boundary force could provide M constraint, the system equation
matrix becomes a square matrix with full rank, which SP is a (2M+3) x(M) matrix,

and K’”AB is a (p+q+M) x(p+q+M) matrix. Therefore, the final integrated system

is
(u’A I (PA)
K’AB T;
“’3 = f’B ..................... (2.27)
o s 8',
If”) I“)

 

 

 

 

By solving equation (2.27), the system internal forces f’A+B and displacement
distribution u’A and u’B are available. In following chapter, I am going to
demonstrate this approach with a example. Through this example, the system
modeling and integrating details could be shown clearly.

26

2.4. Result and Discussion

Modular modeling system FEA equations can be developed from the
subsystem equations, external boundary conditions, and internal constraint
equations. If enough constraint equations are available and the DOF is zero, the
modular modeling FEA equations can be solvable. The general form of the
combined system equation (2.27) has been developed here for the two-
dimensional elastic problem.

Modular modeling FEA equations could be realized automatically with a
proper algorithm. The reason is that all of parameters used in combined systems
come from individual FEA model and connection description, such as nodal
coordinates and boundary conditions. This point is valuable for the modular
modeling method because this means that the system could be integrated by the
user's selection of parts and connection type.

Solution of the modular modeling formulation can be shown to be equivalent
to solving the associated Lagrangian formulation [Appendix I, Radcliffe and Diaz,
2000]. Although the modular modeling formulation is derived based on only work
constraint at the boundary, its solution is identical to the solution of the
Lagrangian formulation which analysis minimum potential energy for the system.
It is gratifying that the solution for the modular modeling method coincides with
the traditional energy-based Lagrangian approach.

27

TWO-DIMENSIONAL EXAMPLE FOR ELASTIC STRAIN-STRESS PROBLEM

3.1. FEA static stress-strain analysis by modular modeling

. To demonstrate the last chapter work about integrating the modular modeling
FEA systems, a simple example for a 2-D elastic plane force-displacement
problem will be solved by the approach and procedures in the chapter 2. After
the combined system has been solved, the result was compared with same
object but derived from FEM using two different mesh densities.

In figure 3.1, there is a combined plane willbe a subject of load. The two
triangular components have FEA models: plane A has a FEA stiffness matrix KA,
plane B has a FEA stiffness matrix K3,

3] /’ $1 Plane A

 

  

 

   

 

 

 

 

 

 

Load

I
/,’ y Plane B

Load
Figure 3.1. Two-Dimensional Example for Elastic Strain-Stress Problem

From figure 3.1, the subject has 3 external boundary conditions so it is O
DOF, using the approach in last chapter and original equation (3.1) and (3.2),
applying the external and internal boundary constraint (port) equations. The
system equations are singular.

28

3.1.1 2-D Elastic Plane A:

We need to get the individual plane meshing and system equation. From the
following procedures, it could be solved like any traditional examples [L.
Segerlind, 1984]. The related system properties are: Young’s modulus: E =
15x106 (N/cm2); Poisson’s ratio: u = 1/4; Plane thickness: t = 0.1cm.

Finite element summary:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

eleme i j k

(6 (8 (1) 1 3 2

(2) 3 4 5

(2 (4 (3) 2 3 5

(3 (4) 2 5 6

2 Y (5) 4 7 8

(1 f ' (6) 5 4 a

,x (7) 5 8 9

¢ 1 (8) 6 5 9

(9) 6 9 10

Nodes’ coordinator values:
Nod 1 2 3 4 5 6 7 8 9 10
e

X 0 2.0 0 0 2.0 4.0 0 2.0 4.0 6.0
Y 0 2.0 2.0 4.0 4.0 4.0 6.0 6.0 6.0 6.0

 

 

 

 

 

 

 

 

 

 

 

Figure 3.2 Plane A meshing details and nodal distribution

 

‘ By the traditional method [L. Segerlind, 1984], we get all element stiffness
matrixes in plane A: [K( 1)], [K(2)], [K(3)], [K(4)], [K(5)], [K(6)] and combine them
together as following. See the Appendix II for the detail of computer software.

29

 

 

50

1

X
00000000003000004603
00000000000200006480
00000000506M00685ߣ6
00000000056500622586
000000506fi0044522400
00000005650064253600
0000006600005H4£0000
000000420000fl5£60000
00300000665fl00006fl20
OOOZOOOOMfiflSOOOOfiéOS
006%506640IM66006M5000
oossoswsumwsoossosoo
OOOOGB5£6£004£500000
00006425M600£8050000
£8665fl££500000000000
64%6a566050000000000
305266006M2000000000
02%54Aw600660300000000
08202600000000000000
30038600000000000000

=
m

3.1.2. 2-D Elastic Plane B

~ The meshing and system equation for plane B could be done by similar way.

It has same static properties as plane A: Young’s modulus: E = 15x106 (N/cm2);

Poisson’s ratio: u = 1/4; Plane thickness: t = 0.1cm.

Finite element summary:

 

 

 

 

 

eleme

nt

0)

 

 

m
m
IA\

 

 

 

Loa

 

 

 

Nodes’ coordinator values:

 

6.0

6.0

 

6.0
3.0

 

3.0
3.0

 

6.0

 

3.0

 

1

 

 

Node

 

 

 

 

Figure 3.3. Plane B meshing details and nodal distribution

30

By the traditional method [L. Segerlind, 1984], we could get all element
stiffness matrixes in plain A: [K(1)], [K(2)], [K(3)], [K(4)], and combine them
together as following. See the Appendix III for the detail of computer software.

can

-8 -2 0
-3 -3 0

(DO

rods
and:
N

”'ro

{301

II

coco
m¢$oo

I

or

I

d
om

x105

I
N
I
CO
01
.5.
.5
o
o
I
0:
I
m
(00600000

I
O)
I
0|
0
00
N
N
01
I
d
or
I
01
ONOOOOOO

I
(D
I
I
..I.
a)
I
U!
N
N
U'l
I
03 CO
I

I
[0
doc:
I
01
cl:
(1,01
MS
“I
I
care

 

 

OOOONOOO
00000000
0
CO
GOD
I
N
I
a:
O
000

3.1.3. Using Modular Modeling To Integrate 2-D Elastic Components

We are using this example to demonstrate the details of the modular
modeling FEA approach. We need to follow the steps one by one according to
the description in previous chapter. The final solvable system equation is too
large to be shown here. For detailed computing procedure and result, please
review Appendix V.

31

3.2. Error Estimation and Analysis

To validate the result from §3.1, we need to compare it with a traditional
model. Both higher and lower meshing resolution models are needed for the
comparison to verify the modular modeling result. For a linear problem, higher
resolution meshes yield a more accurate result. The modular model mesh has
more resolution than then low resolution reference model but less resolution than
the higher resolution model. We expect the modular model to be more accurate
than the lower resolution mesh model and less accurate than the higher
resolution mesh model. Comparing these models will provide useful information
about the detail in the model and component connections at the boundary.

3.2.1. Standard Reference Models for Accuracy Comparison

Two standard reference FEA models (Fig. 3.4) will be used to evaluate the
accuracy of the 13 element modular modeling example (Fig. 3.1). The higher
resolution standard reference model has 72 elements while the lower resolution
standard reference model has only eight elements. We use the same coordinate
systems, load and FEA equations to ensure the accuracy of the formulations are

directly comparable.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

/l 43 44 45 46 47 4s 49 /l

(61) (63) (65) (67) (69 (71) 7 a 9
36 (62) (64) (66) (68) (70) (72) 42

(49) (51) (53) (55) (57) (59) (5) (6) (7) (a)

29 (50) (52) (54) (56) (58) (60) 35

(37) (39) (41) (43) (45) (47)

22 (33) (40) (42) (44) (46) (48) 28

(25) (27) (29) (31) (33) (35)

15 (26) (28) (30) (32) (34) (36) 21

(13) (15) (17) (19) (21) (23) (1) (2) (3) (4)

s (14) (16) (18) (20) (22) (24) 14

(1) (3) (5) (7) (9) (ll

(2) (4) (6) (8) (10) (12)
1 2 3 4 5 6 7 1 2
Load Load

Figure 3.4. Re-meshed reference model with higher or lower meshing resolution

32

3.2.2. Modular Modeling Method Accuracy Comparison

Displacement contour lines were used for comparing the overall system
accuracy. Displacement means the total displacement at every node. If a node
has “a” displacement on X direction and “b” displacement on “Y” direction, its

total displacement is (Ja’ +b2 ). Because we are focusing on the linear case and

the load is same, the linear static elastic problem should have the same and
repeatable result. Therefore this comparing is reliable.

Predicted displacement for the modular model were compared to the
displacements predicted by the higher resolution standard reference model.
EXCEL was used to compute and plot total model displacements (Fig. 3.5). The
3—‘D displacement difference surface (Fig. 3.6) shows the largest model
differences are less than 10%. In particular, nOte that the modular model results
are almost universally smaller displacement than the higher resolution reference
model. This indicates that the modular model is stiffer than the higher resolution

reference model.

 

 

3031...... . 9);;ng U,
’ 47'

...-W

l- W SE
\
0!. ’\\m
“‘
I..,
'r
"u I I“:’,. ,1.
. 9., . .~.
“.‘_‘ h. '
. "I.
"'2’”,
' to
:33: .. %

 

 

 

 

 

 

 

 

Load

Figure 3.5(a). Displacement contour Figure 3.5(b). Displacement contour
lines for reference model (7x7). lines for the Joint Modular Model

33

 

Nodal Dleplecem ant Difference

   

j; _ ' .1500); 40.00%
45.00% ‘ .,

.0056 45.0056
.0056 «10.00%

 

 

 

Figure 3.6. Displacement difference between the Joint Modular Model and
reference model (7x7)

Predicted displacement for the modular model were next compared to the
displacements predicted by the lower resolution standard reference model.
EXCEL was again used to compute and plot total model displacements (Fig.
3.7). The 3-D displacement difference surface (Fig. 3.8) shows that in this case,
the largest model differences approach 25%. In particular, note that the modular
model results are almost universally larger displacement than the lower
resolution reference model. In contrast to the comparison with the higher
resolution reference model, this results indicates that the modular model is less
stiff than the lower resolution reference model.

 

 

 

 

 

 

 

 

Figure 3.7(a). Displacement contour Figure 3.7(b). Displacement contour
lines for reference model (3x3). lines for the Joint Modular Model

 

Nodal Displacement Difference

n20.00%-25.00%
I15.00%-20.00%
l10.00%-15.00%
I5.00%-10.00%
D0.00%-5.00%

s7 n-5.oo%-o.oo%
I-10.00%--5.00%
[moose-40.00%

 

 

 

 

Figure 3.8. Displacement difference between the Joint Modular Model and
reference model (3x3)

Comparison of the three models (fig. 3.9) indicates that nodal displacements
along the boundary of plane A and B uniformly converge to the higher resolution
standard model result as the number of elements in the model increase.
Because the number of elements in the modular model lies between the lower
and higher resolution reference models, its displacements are bracketed by the

35

two reference model displacements. As expected, the model stiffness uniformly
decreases as the number of elements is increased, indicating the modular model
accuracy is comparable to traditional FEM models with the same mesh

 

 

 

 

 

 

 

 

 

resolution.
Bet-day III-placement
Load
1 0.0014
‘3‘ 0.0012 z'
\9\ 0.001
\4
\\5 ; 0.0000
\
\ 0.0000
\5 1
%] \ 0 0004
\ 7

 

 

 

 

 

 

 

 

 

 

 

Figure 3.9. The difference of nodal displacements along the boundary

36

CONCLUSIONS
4.1 Contributions

Application of the modular modeling method in FEA has been discussed; its
advantages and efficiency have been shown in examples and numerical results.
Most importantly, the 2-dimensional boundary constraint equation that is used to
connect FEA modules has been developed and verified in examples. Modular
modeling method is shown to use the original FEA data, removing the need for
re-meshing and verification, so it saves cost and time.

From a general description and example, we believe that the equation (2.5),
(2.20) and (2.23) is the proper way to get the combined system equation and
boundary constraint equations. For actual engineering problem, just like the
example, we can sequentially use the equations (2.5), (2.24), (2.25), (2.26) and
(2.27) to get the solution. For 2-Dimentional elastic stress-strain problem, if the
system have 0 DOF, we can conclude that the developed system equation of
FEA modular model is non-singular and could be solved with the boundary
constraint equations, external boundary conditions and loads.

The modular model has good computational accuracy and quality that is close
to the re-meshed higher resolution model and is much better than the relatively
low-resolution model. The modular modeling method was shown to provide
accuracy comparable to traditional FEA models with comparable mesh
resolution. The ability to achieve this accuracy without remeshing is a significant
improvement to FEA modeling technology.

Result Preclslon and Quality improving

 

 

 

 

 

\
Low Meshing Resolution Modular Model High Meshing Resolution
Reference Model Assembly Reference Model

 

 

 

Figure 4.1. Comparing the Modular Model and other models

37

4.2 Future work

The equation developed in this paper can be only applied to a simple 2-D
problem, the accuracy of this approach will be decided by the equality of
boundary equation to make this approach practically useful in industrial
applications, we must develop the following topics:

9 Convergence of the stiffness matrix

0 All kinds of meshes

o The 3 dimensional flat boundary

0 The 3 dimensional arbitrary boundaries ‘

o The Boundary Constraint Equation development automation,

'. Apply modular modeling method into the nonlinear FEA problem.

There are 2 ways to develop the boundary constraint equation. From
equations (2.20) and (2.23), we can see that component B was treated as the
subject of interpolation from component A. It is easy to understand, the contrary
way could also provide similar constraint equations. Those equations should also
be enough to make the system solvable. But the problem is, “which one is ‘
better?” I believe that this is a more complex and practical problem and needs
research through great number of actual case, cause the meshing and constraint
have so many kind of types that it is very hard to simply identify them.

Theoretically, the boundary interface input / output equation between
components will be only decided by their shape function. Therefore, the
automatically formulation should be practically realized by proper programming.
This feature should be one of the critical block in the future modular modeling
FEA software.

38

REFERENCES

Byam, B. P. and Radcliffe, C. J., 1999, “Modular Modeling of Engineering
Systems Using Fixed Input-Output Structure”, 1999 IMECE, Proceedings of The
Symposium of Systematic Modeling, ASME, New York, November

“Computers in Engineering: Chrysler designs paperless cars”, 1998, Automotive
Engineering lntemational, Volume 106, Number 6, (June): Page 48

Segerlind, L. J., 1984, Applied Finite Element Analysis, 2nd Edition, John W&S,
New York

Zenkiewicz, O. C. and Taylor, R. L., 1989, The Finite Element Method, 4th
Edition, McGraw-Hill, New York

Yang, T.Y. 1986, Finite Element Structural Analysis, Prentice-Hall, New Jersey

Meirovitch, L., 1967, Analytical Methods in Viberations, MacMillan Publishing,
New York

Rosenberg, R. C. and Kamopp, D. C., 1983, introduction to Physical System
Dynamics, McGraw-Hill, New York

Langhe, K.De., Finite Elements with boundary interpolation applied to the solid
mechanics field problem, Computers 81 Structures, Vol.54, No.5, pp.851-857,
1995.

Aminpour, M.A., Ransom, J.B., and McCleary, S.L., “A Coupled Analysis Method
for Structures with Independently Modelled Finite Element Subdomains,’
lntemational Journal for Numerical Methods in Engineering, Vol38, pp. 3695-
3718, 1995.

Kothnur, V.S. , Yu Xie , “T wo-dimensional linear elasticity by the boundary node
method,” lntemational Journal of Solids and Structures, Vol.36, No.8, pp.1129-
1147, Mar. 1999

39

Ransom, J.B., McCleary, S.L., and Aminpour, MA, “A New Interface Element for
Connecting Independently Modeled Substructures,” AIAA Paper Number 93-
1503, 1993.

Belytschko, T., Lu, Y.Y., Tabbara, M., “Element-Free Galerkin Methods for Static
and Dynamic Fracture," lntemational Journal of Solid Structure, Vol.32,
No.17/18. PP 2547-2570, 1995

Hesthaven, J.S., Gottlieb, 0., “Stable Special Methods for conservation laws on
triangles with unstructured grids.” Computer Methods in Applied Mechanics and
Engineering, Vol.175, 361 -381 , 1999

Zhao, Zhiye, “A Simple Error Indicator for Adaptive Boundary Element Method.”
Computers and Structures, Vol.68, 433-443, 1998

Besseling, J.F., Gong, 0.6., “Numerical Simulation of Spatial Mechanisms and
Manipulators with Flexible Links,” Finite Elements in Analysis and Design, v.8 n1-
3. 139121-128, 1994

Elkaranshawy, H.A., Dokainish, M.A. “Corotational finite element analysis of
planar flexible multibody systems,” Computers and Structures, Vol.54, n5, pp881-
890, 1995

40

IV.

VI.

APPENDIX

Lagrangian FEM Modular Modeling notes

The Mathematica® program for plant A. system stiffness matrix
computation.

Filename: Plane_a.nb

The Mathematica® program for plant B. system stiffness matrix
computation.

Filename: Plane_b.nb

The Mathematica® program for re-meshed plant. Work as the reference
for error analysis.

Filename: Plane_all.nb

The Mathematica® program for combined plant. System integration.
Filename: Plane_a+b_connect_6.nb

The MS Excel data table for the result of FEA computation and error

analysis.

41

TO: Mr. Xin Li

From: Clark Radcliffe

Subject: Lagrangian FEM Modular Modeling
Date: April 7, 1999

Per discussion with A. Diaz 4/6/00 at your MS defense, I have outlined below the
relationship between the Modular Modeling Method and the Lagrangian
approach to the joining of two bodies with constraints.

The Modular FEM Modeling Problem (Xin Li MS thesis,4/6/00) is formulated as
K, 0 . u, = 1’, (1a)
0 K, n, f,

‘,=S-fi,and (1b)
W,+W,=0 (10)

where u, =[2‘]. u, =[2’], f, =[g‘], f, =[;"]
A B A B

and the work done on the plates A and B by the connection are W, = f, -1’i , and

W, =f',-1‘1, respectively. Substitution of the connection forces and
displacements into (1c) yields the relationship between internal joint forces

A

fl=$flg (my

subject to the constraints

All variables denoted ”bar" variables (7) refer to 'extemal” quantities not
involved in a joint while all “hat“ variables (it) refer to 'intemal" joint variables.
Partitioning the equations and rewriting them

 

 

 

 

 

 

K, E, o 0 'fi,‘ 1,,
K5 K4 _" 3 .‘I‘ = 5‘ (2a)
0 O K B K a “a fa
I. O O E; KBJ 1.6 B .I ..f3 ..
subject to constraints Ii , = 90, (2b)
and r, = -s’r, (20)

Question: Is the solution of this problem equivalent to the solution of the
Lagrangian formulation?

The Lagrangian Problem (A. Diaz, 4/6/00)
Given the potential energy function

42

H =-;—u:K,u, Jig, +-;—u;K,u, 41$, (3a)

subject to displacement constraints
11, = S «11, (3b)

The Lagrangian is
L=II+(fi, -S-0,)T J.
(4)

=%u:K,u, 4111?, +%u;K,u, -fi;f, +(1’i, -3-1’i,)T VI
The solution of (3a) and (3b) is a stationary point of (4) that minimizes the
potential energy (3a).

Taking the derivative of L with respect to each of the independent variables
u,,u,,}».

 

 

 

 

3f =0=K,fi,+f(_,fi,—f, (4a)
Bu,
3L Ar-- “ a “r
.. =0=K,u,+K,u,—S/1 (4b)
3“)
3_L =0=K,fi,+fi,fi,—f, (4c)
an,
a]. =o=f<';i,+1‘i,0,—2 (4d)
Bun
8L . .,
-a—A—=0=u,-Su, (46)
Putting (4a)-(4e) into matrix form yields,
F— :— ‘ -_- ---
If, K, o o o u, f,
K’,’ K, o o —s’ fi, 0
o o Ki, 0"fin=fa (5)
0 0 K; K, I fin 0
_o —S o I 0_-/1-_0.

 

 

 

 

 

 

Note that (5) is a symmetric formulation.

11 can be shown that solutions to (4a-e) rewritten as (5) are also solutions to (2a-
c). First note that the Lagrange formulation equation (4e) is equivalent to
Modular Modeling constraint (2b).

0:0, —Sfi, 4:: 0, =50, (6)

43

Let the Lagrange multiplier, 2. = —f,, the intemal boundary force on plate B.
Then, the Lagrange formulation equation (4d) is equivalent to the fourth equation
in the Modular Modeling matrix equation (2a).

0: K ,u,+K,fi,+/1¢=IK,u,+K,fi,=f, for/1: (7)

The Lagrange formulation equation (4b) is equivalent to the second equation in
the Modular Modeling matrix equation (2a). Use the force constraint (2c) from

the Modular Modeling formulation to redefine the internal force f, appearing in
the second equation in the matrix equation (2a) to write

o= K (u,+K,u,—s’2eKTu,+K,u, =f, forf, =-s’r, =S’A (8)

The Lagrange formulation equations (4a,c) are identical to the first and third
equations in the Modular Modeling matrix equation (2a).

0 = K5, +‘K’,a, —f, :1 Km, +K‘,0, =1”, (9)
o = K5, +K,a, —f, :1 Km, +K,a, = f, (10)

The conclusion is that the two formulations have identical solutions with the
recognition that the Lagrange multiplier it = —f,, the joint internal force.

General Form of the Displacement Constraint

A more general form of the displacement constraint is the linear relationship
SAuA =8,“, (11)

with this relationship, the Lagrangian formulation becomes
1
r1 = 5.15191, -..51, +—;-u;1<,u, —u,T,f, (12a)

subject to displacement constraints
S,u,=S,u, (12b)

Note that in this case, I have not partitioned the displacements u, and u,. The
body forces 1‘, and I, represent all external forces applied to the two bodies and

are simply a superset of the previous external forces f, and i, that also allow
external tractions to be applied at the position of joint connections. The
Lagrangian is

L=H+(S,u, —S, -u,)T J1

13
=%U:K,u, -u:fA +%u;KBuB "uirafa +(Saua _SA ~u,)T '2' ( )

whose stationary value occurs at the solution to

 

 

31L =0=K,u, —f, -s§2 (14a)
A

3‘1" =0=K,u, -f, -s;2 (14b)
B
%=0=S,u, —S,u, . (14c)

Putting (14a)-(14c) into matrix form yields,

' K, 0 —S§ u, f,
0 K, Sf, 11, = f, (15)
-S, S, 0 ,1 0

Note that (15) is a symmetric formulation with u, and 11, defined as in (1c).

With this definition, the joint constraint matrices in (15) can be used to define the
original Modular Modeling formulation by defining,

0 0
S,=[0 5,] (16a)
0 0
and S,=[0 1] (16b)

45

APPENDIX II

1
II 8 15000000; u I 73 I: a 0.1.3
'ayeten propertlea: Young'a modulus (II/4:119); Poieeon'a ratio; Plane thickneaa («)3 '3

II- {10 3: 2: 2: ‘1 5r 5: 6: ‘}}W8 {3: ‘0 3: 5: 7p ‘1 B, 5: 9},

“I {2' 5: 5: 6r .1 a! 9: 9! 10},

final: for plane A: nodee diatrihution, pick up the 13:] 3k: of In element: by:
“In-filial]: :lnafldifnllr haullnll'r

'3

X:- {0, 2, 0, 0, 2, 4, 0, 2, A, 6}3 Y:- {6, 4, 4, 2, 2, 2, 0, 0, 0, 0};
up: value tor plane A nodee, pick up the In a 1nd of In node by: '3
'anx[[n]]3 m-![[n]] '3

”fulfil: ”IMIYIBL Bltmeyllk, 9];Array[c:l., 9];Array[c:j, 9]3Array[¢k, 913

9013143] 'I(Y[[(W[[n]])ll)-(Y[[(NKIIDI])]]): {nu 9}]:
DOIDSIDII(YIUNKIIIIIJHH'(Yllflullnnnl): {no 9}]:
I”[33:01]3(YIIWIIIIIIDII)-(Y[[(N~T[[n]])l])n (n: 91]:
DOICi-lnl-(IIHflKIIDIIHI)-(3[[(U[[n]l)]])a (n. 91]:
Dolcilln]I(XIIINIIIBIIHJ):(IIHNRIIDIIHIL (n. 9}]:
no[c1:[n]a(X[[(m[[n]])]])-(1[[(u1[[n]])]]), (n: 9}]:

'Detlne 31,83,319 c1,cj,c1: for all eleuente, cowute every value by: " 3
'31 [n] “3 [n] Jilin] 3 3:1 in] “Hill ~11nt I It in] .n In] -!:I in] ' 3
'c1[n]-n[n] -xj [n] 3 c: [n] .11 [n] -n [n] 3 Ch [n] :13 [n] -n [n] " 3

't - “bl-[{IIHNKIIBII)llorliflmlfnllll]I.
{n.9}]3nlatrlot:[t:,vlot:aolned -> True, Plotnabel->'e ']';

31 ('0 1: .).-I
C1(", 1: 'I"! (C1[1]),

Do[Pr:l.nt['Bi(', 1, -)--, (31(1)), -
(EH11): ' “('I 1: 'I"! (”(11): '

 

' Cj(', 1: ')"1 (C3 [1])!

1 llHNIIInIIHI
1 IIIOWHDJDII

1
no[a- -xDet[(
’ 1 xrumrnnm

1
Arraypl, 9]; Do[nn[n] a fi[

-pr1ne[ae[n])-, (n, 9} ]; Dn-

Arraymel-ent, 9] t
not“ - (Trauma-[(881111)] -DD) 2

31 [n] 0
0 (:1 [n] 0
(:1. [n] 31 [:1] c3 [n]

 

l-ua

' “('1 1, 'I": (“[1])10 {1, 9} 13

YIHNIIIDJIHI
Y[[(W[[nll)ll

)13 'Prlnt[h]', {n a 9}];
YIHNKIIDIIHI

a: [n] 0 urn] 0
cjfn] 0 Chill] 3
Ella] Chin] 3Mn]

133;.)

m 2 Hana] 3 x a 1: 015m; Keleuenfln] :- nationallaeflt] 3
I'l’rint: [xeluent [n] ] ', (n, 9}]

46

3

 

(000000000000000000001

00000000000000000000

00.000000000000000000

00000000000000000000

00000000000000000000

00000000000000000000

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 O 0 0 O 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 O 0 0 0 0 0 0 0 0 O
0 0 0 0 O 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
(0 O 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 O 0)

 

8.11 I

'31-1-11kfll.21-13k[21.25-1-1dkf3l.21-1-13k[4].zk-1-13k[s],2x-1jk[6],-,

Do[Print['elenent: ', e]3 ijk[1] - (2xfl1[[e]])-l3

Arrayfljk, 6] 3

1:139] . (”WING“): 13kI3I-(3XNJIIOII)-13 13k“) ' (3XWIIOII):

1211:1518 (3XM[[O]])-1r 11k”) 8 (”WING”): Mann-80101311140]:
DO [mini-llliiklilo ilklillll 8M11II13RI1]: ilklllll «tutti. 5]]: {1. 5}].

(j! 6}]! {.0 9}],

'rill the element K into the integral 8(xall), baeed on every corrdinated nodee';

447

 

000)

0
0
0

0
0
0
0
0
0

0 0 0 0
0 0 0 0

0
0

0
0

0
0
0 0

~3 -3
~2 ~8

3
0

2
22

0 0 0
0 0 0 0
0 0 0 0
0 0 0 0

3

-5

~6

5 ~16 -5

22

2
0

0
0
0

-5 -6 0 0 —5 ~16 2
~3 —3
—2 —8

22

5

-3 —2 ~16

5
0
—5
-6

0
5

5

-5
-6

0
0
0

5 22
~3 ~2

~5
~3 —8

-3 ~8

0

0 -3 ~3
~2 ~8

0
-5

5 ~16
S 22

22

0

0

-5

~5 ~16

0
0
-6

0
0
0 0
0 0

0
~5
~5 ~16

~5

44

0 0 -6

-6 0 0

10 ~16

10 44

~5
~5 ~6

5 ~16

0
0
0

0
5
0
0

0 0
-5

5
—6

-5
22

0
0
-3
-3

0 0 0
0 0 0

5
22

-5
-6

0 0 ~16
0 0

—3 ~2

~3 -8

—5 ~16 2 0

5

~5

5 ~8

5 11 -2
~8 -2 22

11

0
0
0

0
0
~6

0 0
0 0

0
0

0
-5

5 ~8 —3 0 0
-3

22 -2

0
0
—6

0 0

0 -3 ~3 S

—5 -8

~5 ~16 0 0 ~3
0 0 0

0 0 0 0

0 ~5 ~16

5

0

5 -8 ~3

22
-2
~3

-2 22

-3

0 0

~2 -3

5
-8
-3

5 0

0

0
0

3)

3

0

I

48

APPENDIX Ill

1
n 8 150000003 u 8 T; t n 0.1;

'syeten propertiea: Young'a noduluem/cm’); Poisson's ratio; Plane thickneea(an)3 '3

III-{1, 2, 2, 4);”: {4, 4, 5, 6km: {2, 5, 3, 5};
'neah for plane A: nodea diatribution, pic): up the 1333):; of in element by: '3
~1n=x1[[n]]: :ln-Milni]: manila]?!

13(01 3' ‘3 3' 6' 6}; ’3 {0' o, o. -3, '3’ -6};
-r,! value for plane A nodea, pic): up the n a m3 of tn node by: '3
annual]: tn-Yflnii '3

marl”. 4]: multi. 4]; Arreyfnk. 4]: Muriel. 4]; Arr-tried. 4]: Arreylck. 4]:
0018113] . (YIIMJIIDIIHI) - (Y[[(NK[[n]])l]). tn. 4} ]3

0013113] - (Yllflmllnnnll - (Yllffiiiniinl). (n. 4} I}

001315an 1* (Y[[(N1[[n]])]])-'-(YIIWJIInIIHII. {no 4} l:

bowl-in] - (xmmrtnnm) - (HUN-7111111)”). (n. 4} i:
DOlczllnl-(1[[(NI[[n]])i])-(311013111111th (n. A} 13

Miami] - (111(m11n11111)-<xrr(u:11n11>11). (n. 4} ]:

'Detine bi,bj,3k,¢i,¢j,¢k tor all elemente, coupute every value by: '3
“31131-15 In] 41513] 3 3:1 [11] .“Mill -Y=|- [n] 3 33! [n] .n [n] -¥:l In] " 3

“C1 In] .13: In] 41 in] 3 Cd [DI-111111415111] 3 Chin] =le [111-11111] ' r

'1: ' TIblOHXlHl-fllnllnlflll("011311)”).
(31.4)]3LiatPlot[t,PlotJoined -> True, Plothabel->'e ']'3

Do[Pr1nt["B:l(', i, ')I", (Bi[:l.]), " 330', :l, ')s", (Bj[:1.]),
' “('1 10 '1"! (“111)! ' Ci(', 1' '1"! (Ci-[1]):
" CM": 1: ')='a (C3111): " CH“: 1. ')-'- (035111)]: {1. 4}]

1 1 xmmrnnm YIHNIIMD]!
Do[A=-2—xnet[[1 111014111411)“ YIHWIInllllll]:'Printlhi'. {n41}:
1 xmmrnum trumrrnnm

1 Bi [n] 0 33 [n] 0 h):[n] 0
Arr-”[33, 9]; Do[nn[n] - — [ 0 (:1 [n] 0 c: [n] 0 Ck[n] 3
3“ c1 [n] 31 [n] c: [n] 3:1 [n] C):[n] Bk[n]

 

l u 0
'Print[nn[n]]', {n, 4”,». n [u 1 0 ]
1‘“: 0 0 ‘—’“—
2

Array[xelenent, 4] 3
Dom - (Tram-Ouufnill JD) 3
m - KR.BB[n] 3 R s t *Atm; Kelenent [n] s Aationalireflt] 3
'Print [xelenent [n] ] ', (n, 4}]

49

r0 0 11 o o 11 o o 1) o 0 01

0 0 1) 0 0 11 o o 1) o o o

o 0 1) o 11 0 11 0 0 11 o 0

o 0 1) 0 0 <1 0 0 <1 0 o o

0 o 11 0 0 1) 0 11 o o 11 o

0 o 11 o 0 11 o 0 11 o o 0
M11'000000000000’

o o 11 o 0 11 0 o 11 0 0 o

o 0 11 o 0 r) o 0 1) 0 0 0

o 0 11 0 o 1) 0 0 11 o 0 o

o o 11 o 11 o 11 0 0 11 o 0

(0 0 1) o 0 1) 0 0 11 0 0 1n

 

 

mulijkr ‘13
'2i-l-ijk[l],21-ijk[2],Zj-l-ijk[3],2j-l-ijk[4],2k-1-ijk[5],2k-ijkl61,'3
no[Print['ele1aent: ', e]3 1:1):[11 e (2xfl1[[e]]) -13
111312] - (3x11111011): 111513] . (3x11311011) -1:1:lk[4] - (2x11311011):
1:1):[5] :- (2xux[[e]]) -13 ijk[6] - (2 xuxflellh Rteap-xelenent[e]3
DO [DOIMIIIIUIKli]. 1111(3)” -h11[[1:lk[1]. 1111:1311] +Kt-lvlli. 3]]. {1. 61].
{:l. 6}]. {0. All:
'Pill the element K into the integral x(xall), baaed on every corrdinated nodee';

m1 / 100000
( 8 0 —8 -2 o o o 2 o o o 01
o 3 -3 -3 o 0 3 o o o o o
~8 —3 22 5 —8 -2 ~6 —5 0 5 0 o
-2 —3 5 22 -3 -3 —5 -16 5 o o o
o 0 -8 -3 11 5 o 0 -3 -2 o o
0 0 —2 —3 11 0 0 —3 -s o o
o 3 -6 -5 o o 22 5 -16 -5 o 2
2 0 —5 ~16 o 0 5 22 -5 ~6 3 o
o o o 5 -3 —3 —16 -5 22 5 -3 —2
o 0 5 o -2 —8 -5 -6 5 22 -3 -8
0 o o 0 o 0 o 3 -3 —3 3 0
( o 0 o 0 o o 2 o —2 -8 o s)

 

 

50

APPENDIX IV

1
II I 150000003 u s -‘-3 t a 0.13

Isyaten propertiee: Young'a modulus (II/Inf); Poieeon'e ratio3 Plane thickneee(cn)3 '3

I! a {l, l, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 6, 8, 9, 9, 10, 10, ll, ll, l2, l2, l3, l3, 15, 15, 16,
l6, 17, 17, 16, 18, l9, 19, 20, 20, 22, 22, 23, 23, 24, 24, 25, 25, 26, 26, 27, 27, 29, 29,
30, 30, 31, 31, 32, 32, 33, 33, 34, 34, 36, 36, 37, 37, 38, 36, 39, 39, 40, 40, 41, 4l}3

NJ- {0, 9, 9, 10, 10, ll, ll, l2, l2, l3, l3, 14, 15, l6, l6, l7, 17, 10, 10,
l9, 19, 20, 20, 21, 22, 23, 23, 24, 24, 25, 25, 26, 26, 27, 27, 26, 29,

30, 30, 31, 31, 32, 32, 33, 33, 34, 34, 35, 36, 37, 37, 36, 36, 39, 39,

40, 40, 41, 41, 42, 43, 44, 44, 45, 45, 46, 46, 47, 47, 40, 40, 49}3
IR- {9, 2, 10, 3, ll, 4, l2, 5, l3, 6, l4, 7, 16, 9, 17, 10, 10, ll, l9, 12,

20, 13, 21, 14, 23, 16, 24, 17, 25, 16, 26, 19, 27, 20, 20, 21, 30,

33, 31' 24. 33, 25, 33, 26, 34, 27, 35, 26, 37, 30, 30, 31, 39, 32, 40,

33, 41, 34, 42, 35, 44, 37, 45, 30, 46, 39, 47, 40, 46, 41, 49, 42}3
'neeh for plane A: nodea diatribution, pic): up the i3j3k3 o! in element by: '3
'13-'1118111 :InIl-‘Illnllr anIIRIInll'i

r. {0, l, 2, 3, 4, 5, 6, 0, l, 2, 3, 4, 5, 6, 0, l, 2, 3, 4, 5, 6, 0, l,
2. 3, 4. 5. 5. 0. 1. 2' 3. 4. 5. 5. 0. 10 3. 3: 4. 5: 6. 0. 1. 2. 3. 4, 5. 6}:
12(6, 6, 6, 6, 6, 6, 6, 5, 5, 5, 5, 5, 5, 5, 4, 4, 4, 4, 4, 4, 4, 3, 3, 3, 3,
3, 3, 3, 2, 2. 2, 2, 2, 2, 2, l, l, l, l, l, l, l, 0, 0, 0, 0, 0, 0, 0}3
'1,! value for plane A nodea, pick up the In a m: of In node by: '3
“In-1118]]: “I11!!!” '3

Arr-win. virus-earl”. “ism-313k. 72]:

mulch 72lrmnr1C3. wishful“. 72]:

00131111] I (![[(N-T[fn]])]]) - (YIIOIRHDIIHI). in. '72} I:
00131111] I (YllfmffnllHl) -(![[(N1[[nll)]]). (n. 72} II
not-11:11-(11101110111111)42110111121111”). in. 72} 13
hotel!!!) I (311(lelnlllill - (Xllflwllnlllll). (n. 72} l:
micflnl I (31101111101111) - (811(Mllnlllll). (n. 73} l:
001C311!) I (11101111101111)«11101111141111». (n. 72} l:
'Detine Pi,bj,bk,ci,c5,c1: for all elenente, oeauute every value by: '3
'31 [n] I” [n] 415111]: 3:1 in] Infra] 41 In]: Skin] 1'11an -Y:l [nl'r
“(31181-13nt 41 [n] 3 6:1 in] .n in] 41: In] 1 Chin] II! [n] 4113] " I

Do[Print['bi(', 1, “)I', (BlIil), " Bj(", 1, I')-",
(3311]): " “('11 1. “I": (3311]): " 01": 1o '1": (C1111):
" C32”": 1. ')"o (321111): " “I“: 1: '1": (93111)]- {11 731]

1 1 xmmrtnlim rtumrnnm
no[1-;xn-e[ 1 xtumrrnnm 11101710111111 ]:-mn6111-.1n.721]
1 xmmrnnm rmmtnnm

51

1 mm o Bj[n] o bkln] o
m-ytu.721;no[u1n1-——l o c11n1 0 cm] 0 Cklnl . 121.721]:
3‘ c1111] 8:1[n] c1 [n] :3 [n] Ch[n] lkln]
1 u 0
”9,, " 1: 1. o
‘“'“' o 11 3:1

 

16(130000 4000000 0
4000000 16000000 0
0 0 W0

maylxele-ant, 72] 3
Don! I (Tran-poeel (”[n] ) ] .w) 3
m a Kmbblnl 3 x a tom-m3 xeluant[n] a Rationallaew] 3
'Print [blunt [n] ] ', (n, 72]]

0 : 0
all I [ee 8 ..] 3
0 ‘ ° 90x90
mayfljh, 6])
'21-1-1jh[1] ,21-1jkt2] ,21-1-13313] ,23-1-131114] ,2h-1-13kl5] ,2k-1jkl6] , . 3
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017, 018, 019, 020, 021, 022, 023, 024, 025, 026, 027, 028, 029, 030, 031, 032, 033,
034, 035, 036, 037, 038, 039, 040, 041, 042, 043, 044, 045, 046, 047, 048, 049, 050,
051, 052, 053, 054, 055, 056, 057, 058, 059, 060, 061, 062, 063, 064, 065, 066,
067, 068, 069, 070, 071, 072, 073, 074, 075, 076, 077, 078, 079, 080, 081, 082,
083, 084, 085, 086, 087, 088, 089, 090, 091, 092, 093, 094, 095, 096, 097, 098)]

{{n 40.12» 100.185» 100.111 -10,U2-:0.
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260252991266536944617225 10173328405 199981 188065579673797672572561775087656937209 }}

57923800939607003274446190180321706822582242227616209840059619855745503238513840000

U79 —0

 

 

U804

 

 

U824-

 

U83 -1

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U87 -:

 

U88 -3

 

 

U90»-

 

U91 -:

 

U92 -:

 

U93 -:

 

 

U95 -:-

 

U964-

 

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0115: I
{01, 02, 03, 04, 05, 06, 07, 06, 09, 010, 011, 012, 013, 014, 015, 016, 017, 010, 019, 020,

021, 022, 023, 024, 025, 026, 027, 026, 029, 030, 031, 032, 033, 034, 035, 036, 037, 036,
039, 040, 041, 042, 043, 044, 045, 046, 047, 040, 049, 050, 051, 052, 053, 054, 055,
056, 057, 050, 059, 060, 061, 062, 063, 064, 065, 066, 067, 066, 069, 070,
071, 072, 073, 074, 075, 076, 077, 076, 079, 060, 061, 062, 063, 034, 035,
036, 067, 066, 069, 090, 091, 092, 093, 094, 095, 096, 097, 096} I. 9:

'Oive 0(n) value to liat of 011et. 011at[[l,n]]-0n'3

'double check 12 011st 1a correct.'3 011et[[1, 3]]

 

m . ram-[\fmn-qu, (2x (7:: (3-1) +1))]])'+ (an-1:111, (2x (7:: (1-1) +1) -1)]])’ .

15. 71. 11. 71]:
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56

am

0 0000397902
000055708 0.000663627
0.000776718 0000833543
000087507 0.00089985
0000890793 0000878236
0000816166 0000759257
0000740533 0.000549315

blatantourrlot [I'D]

7

0000636563
0000770334
0000872305
0000915 15 1
0.000893 197
0.00081 1204
0000709462

0000793737
0000857346
0.00091493 1
0000942044
0000931388
0000892521
0000864815

 

 

 

 

 

1 2 3 4

- ContourGraphics -

57

0.0009 1301 3
0000936392
0000965958
0000984314
0000985405

000097927
0000992764

000101909
000101069
000102364
000103857
000104844
000106492
000110837

0.001 12839
000107818

0.0010854
0.001 10172
0.001 1 1826
0.001 14791
000123026

APPENDIX V

-3
-8
-5
-6

-3
-2
-16

-5
-16

5
22
-5
-6

0
-3
-2
22

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22

-3
-a

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-8

-3
-3

22
-2
-8

5

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-5
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-16

-3
-3

-5
44

-16

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-16

0 0 -5 -16

44

10
- 16

-16

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-8

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11

-3
-3

5 11
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3 3
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-s
-15

5223
2..

-16

 

4 1000003

 

01

-5
-16

-3 22 5 -8 -2 -6
-3 22 -3 -3 -5

-3
-3
-16

-3 11 5
11

-16

-3 -3 -16
-2 -8

-3
-8

-3
-2

31

58

 

 

3"-

£2
£3
£4
£50
£60
£70
£60
£90
£100
£11
£12
£13
£140
£150
£160
£170
£160
£19
£20
91
92
930
940
950
960

96
990
9100
911

 

 

1 013 1

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59

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or
m o o o o o o o o o o o o o o o o o o o o o o o o o o o o 1 o 01
o o o o o o o o o o o o o o o o o o o o o o o o o o o o 1. o o o
o o o o o o o o o o o o o o o o o o o o o o o o o 1 o o o o o o
o o o o o o o o o o o o o o o o o o o o o o o o 1 o o o o o o o
o o o o o o o o o o o o o o o o o o o o o o o 1. o o o o o o o o
o o o o o o o o o o o o o o o o o o o o o o 1. o o o o o o o o o
o o o o o o o o o o o o o o o o o o o 1. o o o o o o o o o o o 1
o o o o o o o o o o o o o o o o o o 1 o o o o o o o o o o o 1 o
o o o o o o o 0 o o o o o o o o o 1 o o o o o o o o o o o o o o
o o o o o o o o o o o o o o o o 1 o o 0 o o o o o o o o o o o o
o o o o o o o o o o o o o o o 1 o o o o o o o o o o o o o o o o
o o o o o o o o o o o o o o 1. o o o o o o o o o o o o o o o o o
o o o o o o o o o o o o o 1. o o o o o o o o o o o o o o o o o o
o o o o o o o o o o o o 1 o o o o o o o o o o o o o o o o o o o
o o o o o o o o o o o 1 o o o o o o o o o o o o o o o 1-2 0 o o o
o o o o o o o o o o 1 o o o o o o o o o o o o o o o ...—2 o o o o o
o o o o o o o o o 1.. o o o o o o o o o o o o o o o o o o o o o o
o o o o o o o o 1. o o o o o o o o o o o o o o o o o o o o o o o
o o o o o o o 1. o o o o o o o o o o o o o o o o o o o o o o o o
o o o o o o 1 o o o o o o o o o o o o o o o o o o o o o o o o o
o o o o o 1. o o o o o o o o o o o o o o o o o o o o o o o o o o
o o o o 1. o o o o o o o o o o o . o o o o o o o o o o o o o o o o
o o o 1. o o o o o o o o o o o o o o o o o o o o o o o 1-2 0 o o o
o o 1 o o o o o o o o o o o o o o o o o o o o o o o 1-2 0 o o o o
o 1 o o o o o o o o o o o o o o o o o o o 1. o o o o o o o o o o
1. o o o o o o o o o o o o o o o o o o o 1. o o o o o o o o o o o

 

1'01I

60

3

 

1000000000000000000000001

00000000000000000000000
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01000000000000000000000
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00000000000000100000000
00000000000000010000000
00000000000000001000000
00000000000000000100000
00000000000000000010000

00000000000000000001000

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0
10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1]

 

‘1'02 I

1'0 :1 201.1023 feat Tranntomtionnatrixby : T0.0all '

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62

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0
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5
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-16

-3
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22

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5
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5
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10 -16 -5
44 -6

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5

22

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11
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s
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63

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0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

 

n.6[xnr]

218493540493073842176000000000000000000000000000000000000000000000000000000000.
00000000000000000000000000000000000000000000000000000000000

801"[{m.0' II 2020], {03, 04, 05, 06, 07, 08, 09, 010, 011,
012, 014, 015, 016, 017, 018, 019, 020, V3, V4, V5, V6, V9, V10,
£1, £2, £3, £4, £11, £12, £13, £19, £20, g1, g2, g7, 98, 911, 912]]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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{{034 -0.00015326539217269, 044 -0.00015326539217269. 054
-0.00015326539217269, U64 -0.00071475324324991, 074 -0.0001812188769314,
084 -0.00024856457649176903,094 -0.0001937977674698, 0104
-0.000285645697424133, 0114 —0.000285645697424133, 0124
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0164 -0.00029906009398439, 0174 -0.000238025082672836.

0184 -0.00038117719721275, 0194 -0.000285645697424133.

0204 —0.000285645697424133, V34 -0.00006369624013694,V44
-0.00035315173762, V54 -0.0001938874971248, V64 -0.0005959370274947.

V94 -0.00018502470134319, V104 -0.00047104487235471, £14
0.95001142635067, £24 109.62907398655632, £34 -4.141778764472281, £44
-21.631811496107005,£114 12.70620981649933, £124 -4.932619555920048, £134
98.61541943476531, £194 16.670395461247498, £204 19.353827219991384. 914
—0.95001142635067, 924 -109.62907398655632, 974 8.423994255565799, 984
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m.o;uz-03013.0;v1.o,v2.0,v7.(034.1111)”;
V! a (04 +012) [2; V11 1:019) V12 .0202
“Boundary Conditions and 131.91.00.11“: Con-train Ignition";

mm: a {01, 02, 93, 04, 05, 06, 07, no. 09,

010, 011, 012, 013, 014, ms, 016, m7, 018, 019, 020};
mm: a {V1, V2, V3, V4, V5, V6, v7, V8, V9, V10. V11, V12};
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um

{O . 000000000000000000, 0 . 000826648034066665.
0.000731311222584203, 0.000847252275786886.
0 . 000832209342095123, 0 . 000932414037579913, 0 . 000684510360745850.
0 . 000727484384202640, O . 000891287339200160, 0 . 001191523504541550}

 

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001087953 IOWOOOOOO, 0010963540372081521 ,0.001 191523504541550}

66

APPENDIX VI
The Result of FEA computation and Displacement Difference Analysis

Reference Model Result: (From re-meshed 7x7 modal. 72 Elements)

 

  
   
  
  
   
 

0 0.0003 0.00059 0.00079 0.00091 0.00102 0.001128
0.00042 0.00046 0.00077 0.00086 0.00094 0.00101 0.001078
0.00078 0.00083 0.00087 0.00091 0.00097 0.00102 0.001085
0.00088 0.0009 0.00092 0.00094 0.00098 0.00104 0.001102
0.00089 0.00088 0.00089 0.00093 0.00099 0.00105 0.001118
0.00082 0.00076 0.00081 0.00089 0.00098 0.00106
0.00074 0.00065 0.00071 0.00086 0.00099 0.00111

 

 

Modular Model Result (interpolation value in yellow area).

 

Boundary
Displacement:

 
    
 
 

, 0 0.00026 0.00052 0.00077 0.00086 0.00095 0.001031
0.00037 0.00041 0.00067 0.00081 0.00088 0.00094 0.001009
0.00073 0.00078 0.00083 0.00084 0.00089 0.00094 0.000986
0.00079 0.00081 0.00083 0.00088 0.00091 0.00094 0.000964
0.00085 0.00084 0.00083 0.00088 0.00093 0.00099 0.00104
0.00077 0.00077 0.00078 0.00085 0.00091 0.00106 0.001116
0.00068 0.00071 0.00073 0.00_0_81 0.00089 0.00104 0.001192

 
    
  
   
 
 

570.1%
45.59%

 

 
   
   
  

—0— Joint
Modular
Model

-- O— -Re-meshed
Reference
Model (7X7)

- - i - - Reference
3: I 10.00%-15.00% Model (3X3)
, 3‘ I 5.00%-10.00%

0.00°/o-5.00%

 

 

 

 

 

67

 

The Result of FEA computation and Displacement Difference Analysis (2)
Reference Model Result: (From 3x3 (8 elements) model).

0 0.00024 0.00049 0.00073 0.00084 0.00095 0.00106
0.00026 0.00027 0.00059 0.00076 0.00085 0.00093 0.00102
0.00051 0.0006 0.00055 0.00079 0.00086 0.00092 0.00098
0.00077 0.00078 0.0008 0.00082 0.00086 0.0009 0.00094
0.00073 0.00076 0.00079 0.00081 0.00091 0.00093 0.00099

‘ 0.0007 0.00073 0.00077 0.0008 0.00088 0.00099 0.00103
0.00055, 0.00071 0.00075 0.00079. 0.00089 0.00098 0.00108

Modular Model Result (interpolation value in yellow area).

0 0.00026 0.00052 000077000086 0.00095 0.00103
0.00037 0.00041 0.00067 0.00081 0.00088 0.00094 0.00101
0.00073 0.00078 0.00083 0.00084 0.00089 0.00094 0.00099
0.00079 0.00081 0.00083 0.00088 0.00091 0.00094 0.00096
0.00085 0.00084 0.00083 0.00088 0.00093 0.00099 0.00104
0.00077 0.00077 0.00078 0.00085 0.00091 0.00106 0.00112
0.00068 0.00071 0.00073 0.00081 0.00089 0.00104 0.00119

Displacement difference: (result-reference)/MAX(reference) (%)

0.00% 1.16% 2.31% 13.47% 1.52% -0.43% +2.38%
9.66% 12.13% 6.73% 3.95% 2.31% 0.68%» 095%»
19.32% 15.24% 24.25% 4.42% 3.11% 1.79% 0.48%
2.11% 2.17% 2.22% 4.90% 3.90% 12:90%311.91%
10.22% 7.13% 4.03% 5.97% 2.19% 5.00% 4.66%
6.16% 3.59% 1.03% 3.62% 2.80% 6.18% 7.42%
2.09% 0.06% 4.97% 1.27% 0.27% 5.22% 10.17%

 

Nodal Displeament Difference

I -15.00%--10.00%

 

 

 

68

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