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DECOMPOSITION
AND CONSECUTIVE DYNAMIC CONDENSATION METHODS
FOR STATIC AND DYNAMIC ANALYSIS OF SINGLE LAYER
LATTICE PLATES

By

Vera Vladimirovna Galishnikova

A DISSERTATION
Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of
DOCTOR OF PHILOSOPHY
Department of Civil and Environmental Engineering

2004

ABSTRACT

DECOMPOSITION
AND CONSECUTIVE DYNAMIC CONDENSATION METHODS
FOR STATIC AND DYNAMIC ANALYSIS OF SINGLE LAYER
LATTICE PLATES

BY

Vera Vladimirovna Galishnikova

Approximate yet accurate methods for analyzing large lattice structures are very
efficient for the preliminary structural analysis and design, and parametric optimization
of large lattice structures.

Two classes of new and effective approximate methods for static and dynamic
analysis of large lattice single layer plates using decomposition and consecutive dynamic
condensation techniques are developed in this work. These developments are extensions
of the decomposition method proposed by Pshenichnov and the dynamic condensation
method proposed by Ignatiev.

Simple and accurate approximate analytical formulas for the displacements, force
responses, and first eigenvalue of the boundary value problems of thin plates with elastic
supports are obtained using the decomposition method. Static and dynamic problems of
latticed plates with elastic supports are efficiently solved using continuum modeling. The
developed analytical dependencies are used to obtain optimal lattice geometries for a
class of plate problems. Shear deformations and joint flexibility are not considered.

The decomposition method also is used with a finite difference formulation that is
able to model the original discrete lattice plate. This alternate method has similar

accuracy to that based on a continuum modeling for simple, regular lattices.

While the decomposition method is effective and accurate for static analysis and
for estimating fundamental frequency of lattice plates, it is intractable for estimating
higher frequencies and mode shapes.

An energy form of the consecutive dynamic condensation method is developed in
this work. It is demonstrated that the combination of static condensation with the energy
form of consecutive dynamic condensation yield accurate estimates of most frequencies
and mode shapes of lattice plates. This technique is computationally efficient due to the
resulting block diagonal equations and is suitable for implementation on parallel

computers.

TO MY FAMILY

iv

ACKNOWLEDGEMENTS

I would like to express my sincere gratitude to Professor Ronald S. Harichandran,
my advisor, for his encouragement and significant help as well as for his friendly attitude
over the years of my study.

I wish to acknowledge Professors Ronald C. Averill, Rigoberto Burguefio, and
Amit H. Varma, the members of advisory committee, for their interest in my work and
suggestions.

I would like to express my deepest appreciation to Dr. Thomas L. Maleck for
providing financial support that made it possible for me to finish this work. I extend my
gratitude to Dr. Salvatore Castronovo who was a kind and generous host for me during
one of my semesters at MSU.

I would like to thank all my friends at MSU, and especially Professors Emeriti
William E. Saul, William C. Taylor, and Frank Hatfield for their constant friendly
support and encouragement.

I wish to appreciate my collaborators, colleagues and friends at the Volgograd
State University of Architecture and Civil Engineering and Saratov State Technical
University (Russia).

Last, but not least, I want to thank my family for sharing all the challenge with
me: my father Vladimir Aleksandrovich Ignatiev for inspiration and guidance, my
husband Evgeny Lebed for his understanding and faith, and professional help with the
software development, and my sons Aleksander and Ilia for being the main source of

cheer for me.

TABLE OF CONTENTS

LIST OF TABLES .................................................................................. ix
LIST OF FIGURES ................................................................................ xii
CHAPTER 1
INTRODUCTION AND LITERATURE REVIEW ............................................. 1
1.1 Introduction ........................................................................ 1
1.2 Motivation ........................................................................... 2
1.3 Background ........................................................................ 3
1.3.1 Substructuring Methods .................................................... 4
1.3.2 Order Reduction MethodsS
1.3.3 Continuum Modeling ....................................................... 8
CHAPTER 2
DECOMPOSITION METHOD
FOR SOLVING PROBLEMS OF STRUCTURAL MECHANICS ...................... 10
2.1 Main Concept of the Decomposition Method .............................. 10
2.2 Bending of Rectangular Plate with Non-Symmetric
Elastic Supports .................................................................. 12
2.2.1 Problem Statement12
2.2.2 Decomposition of the Problem ........................................... 15
2.2.3 Numerical Results ............................................................ 21
2.3 Symmetric Bending Problem of Rectangular Plates
with Elastic Supports ........................................................... 23
2.3.1 Problem Statement ....................................................... 23
2.3.2 Decomposition of the Problem ............................................. 24
2.3.3 Solution techniques for the interconnection equation ................. 27
2.3.4 Solution of the Problem
for the Case ofaNon-uniform Load......... 36
2.4 Bending of Rectangular Plate with One Free Edge

and Three Elastically Supported Edges ....................................... 39
2.4.1 Problem Statement ......................................................... 39
2.4.2 Decomposition of the problem ........................................... 40
2.4.3 Numerical results ................................................................... 44

vi

2.5

CHAPTER 3

Transverse Free Vibration of a Plate with Elastic Supports ........... 50

STATIC AND DYNAMIC ANALYSIS OF SINGLE LAYER LATTICE PLATES:
CONTINUUM MODELING AND SOLUTION

BY THE DECOMPOSITION METHOD ........................................................ 55
3.1 Calculation Model of Lattice Plate .......................................... 55
3.2 Solution of Bending Problem for Lattice Plate
with Elastic Supports ........................................................... 59
3.3 Comparative Analysis for Different Types of Lattices .................. 65
3.4 Transverse Free Vibration of Lattice Plate
with Elastic Supports .............................................................. 72
CHAPTER 4

STATIC AND DYNAMIC ANALYSIS OF SINGLE LAYER LATTICE PLATES
BY THE DECOMPOSITION METHOD

BASED ON FINITE DIFFERENCE DISCRETIZATION ................................. 89
4.1 Notations and the Main Operators of the
Finite Difference Formulation ................................................ 90
4.2 Constitutive Equations for Regular Rectangular Grid
Stated in Finite Difference Form ............................................ 93
4.2.1 Method of Virtual Work ................................................ 94
4.2.2 Displacement Method and Mixed Method ........................... 98
4.3 Governing Finite Difference Bending Equation
for Lattice Plate ................................................................ 102
4.4 Boundary Conditions ......................................................... 105
4.5 Solution of the Bending of Lattice Plates
with Orthogonal Grids ........................................................ 107
4.6 Free Vibration Problem of Lattice Plate
with an Orthogonal Grid ..................................................... 119
CHAPTER 5

DYNAMIC ANALYSIS OF LATTICE PLATES
BY CONSECUTIVE DYNAMIC CONDENSATION .................................... 124

vii

5.1 Problem Statement ........................................................... 124
5.2 Frequency - Dynamic Condensation Method .......................... 126
5.2.1 Condensation based on the Displacement Method ................ 126
5.2.2 Condensation based on the Method of Forces ...................... 128
5.3 Consecutive Dynamic Condensation Method .......................... 130
5.4 Energy Form of the Consecutive Dynamic
Condensation Method ....................................................... 132
5.4.1 Condensation Using the Smallest Natural Frequency
of Partial Systems ...................................................... 133
5.4.2 Condensation Using the Reduced Spectrum of Eigenvalues
and Eigenvectors of Partial Systems ................................. 135
5.4.3 Combined Static and Consecutive Dynamic Condensation ...... 140
CHAPTER 6
CONCLUSIONS AND RECOMMENDATIONS .......................................... 160
APPENDICES .................................................................................... 164
APPENDIX A
ARBITRARY FUNCTIONS FOR THE PROBLEM
PRESENTED IN SECTION 2.4.2 .................................................. 165
APPENDIX B
PROGRAM “PLAS” FOR STATIC ANALYSIS
OF LATTICE PLATES ................................................................ 169
APPENDIX C
FORMULAE FOR THE MAIN DIFFERENCES AND SUMS ................ 179
APPENDIX D
ELEMENTS OF MATRICES OF COEFFICIENTS
FOR THE PROBLEM PRESENTED IN SECTION 4.5 ....................... 180
APPENDIX E

EXAMPLES OF DYNAMIC ANALYSIS USING
THE COMBINED STATIC AND CONSECUTIVE DYNAMIC
CONDENSATION METHOD ........................................................ 181

REFERENCES .................................................................................. 190

viii

TABLE 2.1.

TABLE 2.2.

TABLE 2.3.

TABLE 2.4.

TABLE 2.5.

TABLE 2.6.

TABLE 2.7.

TABLE 2.8.

TABLE 2.9.

TABLE 2.10.

TABLE 2.11.

TABLE 2.12.

TABLE 2.13.

TABLE 3.1.

TABLE 3.2.

TABLE 3.3.

TABLE 3.4.

LIST OF TABLES

Accuracy of results obtained using Bubnov-Galerkin

condition (2.30) .................................................................... 22
Accuracy of results obtained using Bubnov-Galerkin

conditions (2.46) ................................................................... 28
Calculation results for k]: k2 = 0 (fixed supports)

obtained using (2.47) ............................................................. 29
Calculation results for k; =k2 =1 (pinned supports)

obtained using (2.47) .............................................................. 3]

Calculation results for k; = k2 = 0 (fixed supports)

obtained using (2.48) .............................................................. 33

Calculation results for k; = k2 = 1 (pinned supports)

obtained using (2.48) ............................................................. 34
Calculation results for rectangular plate under non-uniform load

for k1 =k2 =1 (pinned supports) obtained using (2.47) ........................ 37
Maximum deflection of the free edge for k] =1 (pinned supports)... .......45
Deflection at the middle of the plate for k, =1 (pinned supports) ........... 46
Maximum bending moment for k, =1 (pinned supports) ..................... 47
Maximum deflection of the free edge for k; =0 (fixed supports) ............ 48
Deflection at the middle of the plate for k; =0 (fixed supports) ............. 49
Expressions for P for different types of supports ............................. 54
Results for k1 = k2 = 1 (pinned supports) ...................................... 64
Results for k1 = k2 = 0 (fixed supports) ........................................ 64

Main types of support conditions and corresponding expressions
for the square of the first frequency of free vibration ........................ 76

Results for lattice plate with pinned supports
and governing angle of (p = 450 .................................................. 77

TABLE 3.5.

TABLE 3.6.

TABLE 3.7.

TABLE 3.8.

TABLE 3.9.

TABLE 3.10.

TABLE 3.11.

TABLE 3.12.

TABLE 3.13.

TABLE 3.14.

TABLE 3.15.

TABLE 4.1

TABLE 4.2

TABLE 4.3

TABLE 4.4

TABLE 4.5.

Results for lattice plate with fixed supports

and governing angle of (p = 450 .................................................. 78
Results for lattice plate with pinned supports

and governing angle of (p = 300 .................................................. 79
Results for lattice plate with fixed supports

and governing angle of (p = 300 .................................................. 80
Results for lattice plate with pinned supports

and governing angle of (p = 600 .................................................. 81
Results for lattice plate with fixed supports

and governing angle of (p = 600 .................................................. 82
Results for lattice plate with supports of Type 1

obtained by (3.41) ................................................................. 83
Results for lattice plate with supports of Type 2

obtained by (3.42) ................................................................. 84
Results for lattice plate with supports of Type 3

obtained by (3.43) ................................................................. 85
Results for lattice plate with supports of Type 4

obtained by (3.44) ................................................................. 86
Results for lattice plate with supports of Type 5

obtained by (3.45) ................................................................. 87
Results for lattice plate with supports of Type 6

obtained by (3.46) ................................................................. 88
Lower-order central difference operators ...................................... 92

Deflections at the middle of the plate for the different types of grid

for k; =1 (pinned support) ....................................................... 117
Bending moments for the plate with the 16x16 grid

for 1:, =1 (pinned support) and IT, =0 (fixed support) ....................... 117
Deflections in the middle of the plate with the 16x16 grid

for different values of support rigidity 1T1 .................................... 1 18
Dimensionless values of the first free vibration frequency ................ 123

TABLE 5.1.

TABLE 5.2

TABLE 5.3

TABLE 5.4.

TABLE 5.5.

TABLE 5.6.

TABLE 5.7.

TABLE 5.8.

TABLE 5.9.

TABLE E. 1.

TABLE E.2.

Results for Test Problem 1 ...................................................... 135
Arrangement of the primary degrees of freedom

for the Test Problem 2 ........................................................... 138
Results for Test Problem 2 ...................................................... l 39
Variants of condensation for Test Problem 3 ................................. 142
Results for the Test Problem 3 for the case with 22 primary d.o.f. . . . 145
Results for Test Problem 4 (condensation to 11 primary d.o.f.). . . . . l 50
Results for Test Problem 4 (condensation to 8 primary d.o.f.) ............. 153
Results for Test Problem 4 (condensation to 3 primary d.o.f.). . . . . . . . 1 54
Results for Test Problem 5 ....................................................... 157
Results for Problem 1 ............................................................. 183
Results for Problem 2 ............................................................. 187

xi

Figure 1.1.
Figure 2.1.
Figure 2.2.

Figure 2.3.

Figure 2.4.

Figure 2.5.

Figure 2.6.

Figure 2.7.

Figure 2.8.

Figure 2.9.

Figure 2.10.

Figure 2.11.

Figure 2.12.

Figure 2.13.

Figure 2.14.

Figure 2.14.

LIST OF FIGURES

Examples of latticed structures ................................................
Rectangular plate with non-symmetric elastic supports. . . . . . . . . . . ..

Rectangular plate with symmetric elastic supports.............

Maximum deflections at the middle of the plate obtained by (2.47)

for k1 = k2 = 0 (fixed supports) ................................................

Maximum bending moments on the edge of the plate

obtained by (2.47) for k1= k2 = 0 (fixed supports)............................

Maximum deflections at the middle of the plate obtained by (2.47)

for k1 = k2 = 1 (pinned supports) ..............................................

Maximum bending moments at the middle of the plate

obtained by (2.47) for k. = k; =1 (pinned supports)..........................

Maximum deflections at the middle of the plate obtained by (2.48)

for k; = k2 = 0 (fixed supports) ................................................

Maximum bending moments at the middle of the plate

obtained by (2.48).for k1 = k2 = 0 (fixed supports) .........................

Maximum bending moments on the edge of the plate

obtained by (2.48) for k1= k2 = 0 (fixed supports)............................

Maximum deflections at the middle of the plate obtained by (2.48)

for k1 = k2 = l (pinned supports) ...............................................

Maximum bending moments at the middle of the plate

obtained by (2.48) for k1 = k2 = 1 Qainned supports) .......................

Deflections at the middle of the plate under non-uniform load

for k1 =k2 =1 (pinned supports) obtained using (2.47) .....................

Bending moments at the middle of the plate under non-uniform load

for k1 =k2 =1 (pinned supports) obtained using (2.47) .....................
Rectangular plate with one free edge ........................................

Maximum deflection of the free edge for kl =1 (pinned supports). .

xii

....1

13

.23

...30

.30

....31

.32

...33

...34

.34

...35

...36

...38

...38

...39

..45

Figure 2.15. Deflection at the middle of the plate for k; =1 (pinned supports) ................. 46

Figure 2.16. Maximum bending moment for k; =1 (pinned supports) ........................ 47
Figure 2.17. Maximum deflection of the free edge for k1 =0 (fixed supports)... .........48
Figure 2.18. Deflection at the middle of the plate for k1 =0 (fixed supports) ................ 49
Figure 3.1 Lattice plate with elastic supports............... ....56
Figure 3.2 Internal forces and moments in a rod ............................................ 57
Figure 3.3 Distributed internal forces and moments in the continuum model. . . . . . ....57
Figure 3.4 Types ofgrids for lattice plates ..60
Figure 3.5. Plate with an orthogonal lattice .................................................... 64
Figure 3.6. Maximum deflection at the middle of the plate

for kl = R; = 0 (fixed supports) .................................................. 68
Figure 3.7. Maximum moment at the middle ofthe plate for k1 = k2 = O. . . . . ........68
Figure 3.8. Maximum deflection at the middle of the plate

for k. = k2 = 1 (pinned supports) ................................................. 69
Figure 3.9. Maximum moment at the middle of the plate

for k1 = k2 = 1 (pinned supports) ................................................. 69
Figure 3.10. Maximum deflection at the middle of the plate for k1 = k2 = 0.5............70
Figure 3.11. Maximum moment at the middle of the plate for k; = k; = 0.5 ............... 70
Figure 3.12. Maximum deflection at the middle of the plate

for different combinations of support rigidities ................................ 71
Figure 3.13 Lattice plate with non-symmetric elastic supports ............................. 72
Figure 3.14. First frequency of free vibration of lattice plate with pinned supports

and governing angle of (p = 450 .................................................. 77
Figure 3.15. First frequency of free vibration of lattice plate with fixed supports

and governing angle of (p = 450 .................................................. 78
Figure 3.16. First frequency of free vibration of lattice plate

with pinned supports and governing angle of (p = 300 ........................ 79

xiii

Figure 3.17.

Figure 3.18.

Figure 3.19.

Figure 3.20.

Figure 3.21.

Figure 3.22.

Figure 3.23.

Figure 3.24.

Figure 3.25.

Figure 4.1

Figure 4.2

First frequency of free vibration of lattice plate with fixed supports

and governing angle of (p = 300 ................................................... 80
First frequency of free vibration of lattice plate with pinned supports

and governing angle of (p = 600 .................................................. 81
First fi'equency of free vibration of lattice plate with fixed supports

and governing angle of (p = 600 ................................................... 82
First frequency of free vibration of lattice plate

with supports of Type 1 obtained by (3.41) .................................... 83
First frequency of free vibration of lattice plate

with supports of Type 2 obtained by (3.42) .................................... 84
First frequency of free vibration of lattice plate

with supports of Type 3 obtained by (3.43) .................................... 85
First frequency of free vibration of lattice plate

with supports of Type 4 obtained by (3.44) .................................... 86
First frequency of free vibration for lattice plate

with supports of Type 5 obtained by (3.45) .................................... 87
First frequency of free vibration of lattice plate

with supports of Type 6 obtained by (3.46) .................................... 88
Lattice plate with regular rectangular grid ..................................... 89
The original beam and auxiliary load cases for Example 1 .................. 95

Figure 4.3 Auxiliary load states for the system of orthogonal beams for Example 2 ...... 97

Figure 4.4

Figure 4.5

Figure 4.7

Figure 4.8.

Figure 5.1.

The original beam and the main systems of the displacement
method and the mixed method for Example 3 .................................. 99

Boundary conditions for the beam in Example 4 ............................. 105

Deflections in the middle of the plate with the grid 16x16
for different values of support rigidity I?! .................................... I 18

Lattice plate with an orthogonal grid and concentrated nodal masses. . . .1 19

Simply supported beam with equidistant masses
for test problems 1 and 2 ........................................................ 134

xiv

Figure 5.2.

Figure 5.3.

Figure 5.4.

Figure 5.5.

Figure 5.6.

Figure 5.7.

Figure 5.8.

Figure 5.9.

Figure 5.10.
Figure 5.11.
Figure 5.12.

Figure 5.13.

Figure 5.14.

Figure E.l.

Figure E.2.
Figure E.3.

Figure E.4.

Figure E.5.

Figure E.6.

Location of the primary d.o.f. and secondary d.o.f. blocks

for the case with 22 primary d.o.f. for Test Problem 3 ...................... 144
Graphs of errors for Test Problem 3 for the case

with 22 primary d.o.f. .......................................................... 147
Graphs of errors for the Test Problem 3 for the case

with 14 primary d.o.f. .......................................................... 147
Graphs of errors for Test Problem 3 for the case

with 10 primary d.o.f. .......................................................... 148
Graphs of errors for Test Problem3 for the case

with 6 primary d.o.f. ........................................................... 148
Location of the primary d.o.f. and secondary d.o.f. blocks

for Test Problem 4 ............................................................... l 49
Results for Test Problem 4 (condensation to 11 primary d.o.f.) ........... 150

Mode shapes for Test Problem 4 (condensation to 11 primary d.o.f.). ...151

Results for Test Problem 4 (condensation to 8 primary d.o.f.) ............. 153
Graphs of errors for Test Problem 4 .......................................... 154
Rectangular lattice plate with a 16x16 orthogonal grid ..................... 155
Condensation schemes for Test Problem 5: (a) to 29 primary nodes;

(b) to 49 primary nodes; and (c) to 81 primary nodes........................156
Graphs of errors for Test Problem 5 ............................................ 158
(a) The frame for Problem 1; (b) Arrangement of the primary d.o.f.

and secondary d.o.f. blocks for Problem 1 .................................... 182
Results for Problem 1 ............................................................ 184
Graphs of errors for Problem 1 ................................................. 184

Location of condensation nodes for Problem 2: (a) condensation to
63 d.o.f.; (b) condensation to 75 d.o.f.; (c) condensation to 99 d.o.f. l 86

Results for Problem 2 ............................................................ 189

Graphs of errors for Problem 2 ................................................. 189

XV

CHAPTER 1

INTRODUCTION AND LITERATURE REVIEW

1.1 Introduction

A latticed structure consists of a very large number of elements or cells
interconnected to form a periodic (repetitive) array. Such structures are used extensively
in different areas of engineering. In civil construction their potential for freedom of form
over long spans makes them architecturally attractive. From the engineering point of

view they have advantages such as lightness, high rigidity and rapid erection.

 

 

 

 

 

 

 

 

 

\

 

\
XXXVI

\‘VV

 

 

\

 

 

   

Figure 1.1. Examples of latticed structures

A review of early studies in this area is available in “Lattice Structure: State-of-
the Art Report” (1976) and a comprehensive bibliography is given by Sherman (1972).
The development of large space structures (LSS) has provided new impetus for research.
Latticed structures having dimensions of the order of 102-103 m are the dominant form
for LSS, due to their low weight and high stiffness as well as ease of transporting and

assembling in space.

1.2 Motivation

The finite element method can be used to solve most problems involving lattice
structures. Nevertheless, the large-scale algebraic systems that result in many cases pose
significant challenges. Linear static finite element analysis of a large lattice structure may
involve the solution of thousands of linear simultaneous algebraic equations. Adding
complex boundary conditions or geometrical or physical nonlinearity complicates the
problem. Conventional finite element analysis for such problems is usually used at the
conclusive stage of design, to confirm the strength, stability and stiffness of the structure.
However, numerical analysis does not provide the analytical dependence between force
and deformation characteristics, which is desirable for the prediction of the overall
structural behavior and optimal design. Further, because the calculation process is
cumbersome, it is difficult to use it during the iterative preliminary stages of the design

and for parametric studies of lattice structures:

Therefore, the development of new approximate numerical and analytical
methods convenient for preliminary design and parametric studies is desirable. These

methods provide practical solutions for global structural behavior. They are efficient

during preliminary design to develop the optimal lattice arrangement and estimate the

initial cross section of members.

Another area of application for such methods is the dynamic and stability analysis
of large lattice structures. Computation of the responses due to dynamic loadings requires
the solution of thousands of coupled differential equations. The computational difficulties
of solving such equations often limit the use of full-scale dynamic analysis in design.
Stability analysis of large lattice structures also represents a complicated problem.
Accurate approximate methods for conducting dynamic and stability analyses of such

structures can be very useful in preliminary design.

1.3 Background

Approximate methods that have been proposed for simplified structural analysis
can be divided into three main types: substructure synthesis, order reduction methods and
continuum modeling. The first type involves methods of decomposing or breaking up a
large complex problem into a set of subproblems of lower dimensionality, the union of
which is equivalent to the original problem. This approach includes substructuring
methods, methods of domain decomposition and equation decomposition. The second
type includes techniques for reducing the degrees of freedom in complex structural
systems, and has been referred to as reduction methods or condensation methods. The
third type includes methods that substitute the actual lattice structure by a continuum
model with equivalent properties. These three types of methods are briefly explained

below.

1.3.1 Substructuring Methods

The main concept in substructure synthesis methods is to represent a complex
structure as a set of substructures, each representing an aggregate of basic finite elements.
In this approach, each substructure is defined in a convenient system of selected
coordinates and analyzed independently with boundaries common to the other
substructures. This analysis provides a more compact and tractable stiffness matrix of the
substructure and its nodal loading matrix. The substructure for which such matrices have

been defined is sometimes called a superelement.

The system of equations written for the boundary nodes of the superelements
expresses the equilibrium conditions of the entire structure as an aggregate of
superelements. This system of equations contains far fewer unknowns than if the entire
system is modeled by standard finite elements. Computation of displacements of the

substructure nodes referred to as the “forward step,” comprises the first major step.

In the so-called “backward step,” each substructure is analyzed for the given
loading and the boundary displacements found in the forward step. These calculations
offer no particular difficulty since the substructures are invariably described by relatively
small systems of equations. From the viewpoint of the classic method of displacements,
each substructure (superelement) in such an approach represents a complex element of

the main system of the displacement method (Przemieniecki 1963, 1968).

The conventional form of substructure analysis has several drawbacks. These
include computations in several stages, storage of the stiffness matrices of substructures
at all levels, and limitations of the static condensation procedure which prohibits

blockwise elimination of the boundary nodes in the Gauss method.

During the past decades the concept of substructuring was extended to the
dynamic response of structures. Hurty (1960) proposed the component mode synthesis
method. Craig and Bampton (1968) further developed this method. Substructure synthesis
also includes such methods as branch-mode analysis (Gladwell, 1964), and component
mode substitution (Bajan et al., 1969). A general review of substructuring methods

developed in 1960s and 1970s is provided by Nelson (1979).

Further research in this area yielded solution procedures for extracting
eigenvalues and eigenvectors from linear dynamic systems using the finite element
method. Eigensolution techniques that provide only a partial eigensolution are efficient
because they extract only a subset (normally the lowest) of the eigenvalues and
eigenvectors required for the analysis of systems. These techniques include the subspace
iteration method, the Lanczos method, the conjugate gradient method, the Ritz vector

method, the substructure synthesis method, condensation techniques, etc.
1.3.2 Order Reduction Methods

Techniques for reducing the degrees of freedom (d.o.f.) in complex structural
systems are referred to as reduction methods or condensation methods. The basic concept
in reduction methods is to condense a large system (of algebraic and/or differential
equations) to a similar much smaller system of substitute equations. In dynamic
problems, the full set of equations of complex systems is reduced by selecting a set of
master d.o.f. and eliminating all other (slave) d.o.f. from the primary governing equation.
Guyan (1965) first proposed a consistent method of reducing both the stiffness and mass

matrices. Methods of static and dynamic condensation based on reducing the order of the

characteristic matrix by exchanging all secondary (auxiliary) d.o.f. have gained

considerable application in recent years.

The method of static condensation (Guyan 1965) is one of the most convenient
and simple methods of reducing the unknowns in the substructuring method of solving
dynamic problems. The slave coordinates are those in which, at low frequency, the inertia
forces are considered negligible compared to the static forces. This technique can greatly
reduce the computational effort necessary to calculate the system eigenpairs. ‘However it
has some shortcomings. The major drawback is the error arising from the assumption that
the inertial forces in the secondary nodes are negligible. Another drawback is that the

accuracy of the result depends on the selection of the condensed nodes (master d.o.f.).

Dynamic condensation by modal synthesis of substructures has been discussed in
several works (Hurty 1965, Bathe and Wilson, 1972). Displacements of the secondary
(slave) nodes of the substructure are represented as the sum of their static displacements
caused by displacements of the primary nodes and displacements of firmly fixed primary
nodes in the substructure represented in term of natural modes of vibrations. The dynamic
condensation techniques can yield solutions of very high accuracy depending on the
number of modes used. The frequencies of the first few modes barely differ from those
calculated by the static condensation method. Therefore, the use of dynamic condensation

methods for determining only the lower frequencies is not recommended.

Meirovitch and Hale (1981) demonstrated that the component mode synthesis is
essentially a different form of the Rayleigh-Ritz method. Based on this, Meirovitch and
Kwak (1991) proposed the construction of an approximate eigensolution from the space

of admissible functions, and not necessarily from the component modes. They also

proposed choosing the trial vectors from the space of quazi-comparison functions, a new
class of functions with high convergence characteristics. These functions represent linear
combinations of admissible functions that act like comparison functions. A comparison
function satisfies the boundary conditions but not necessarily the differential equation.
Quazi-comparison functions obtained can satisfy natural boundary conditions to any
degree of accuracy, and the eigensolutions obtained exhibit superior convergence
characteristics compared to those based on admissible functions.

Jonsson et el. (1995) proposed a recursive substructuring of finite elements for
repetitive structures. In each recursive step the problem is transformed in to a new
problem involving half the number of identical substructures. The computational work
involved in factorization grows only logarithmically as opposed to linear growth in
conventional methods.

Farhatt and Geradin (1994) developed a Hybrid Craig-Bampton method involving
the original CB method for assembling substructures, hybrid variational formulation and
finite element procedure for incompatible substructures. This method can be used as an
interface reduction method.

Archer and Graham (2001) present the variation of the component mode
technique for the dynamic substructuring of large-scale structural systems. The principal
innovation of the proposed method is that the resulting matrix of the reduced
substructures remains diagonal. The reduction is accomplished by transforming the
degrees of freedom in the substructure using boundary shapes and internal shapes. Then
diagonalization of the mass matrix takes place. To recover the accuracy, lost in the

diagonalization, additional pseudo-rigid—body-mode shapes are included.

In recent years reduction methods have been extensively developed covering a
wide range of problems. Newer techniques using reduction methods in conjunction with

substructuring and operator splitting techniques also have been proposed.
1.3.3 Continuum Modeling

In the continuum approach, the actual lattice structure is substituted by a
continuum model with equivalent properties derived from those of the discrete members.
The behavior of a discrete structure can be determined by studying that of the continuous
one. Large sets of algebraic equations used in numerical methods are replaced by a small
number of partial differential equations that can be solved analytically or numerically. In
many cases continuum modeling provides practical solution methods for global
structural behavior, and can be used efficiently in preliminary design and parametric
studies. It has been successfully applied to study the vibration and buckling of latticed

structures.

Existing continuum modeling methods differ by how the appropriate relationships
between the geometric and material properties of the original lattice structure and its
continuum model are determined. Most of them fall into one of several main categories.
One group of methods uses the relation between force or deformation characteristics of a
repeating cell of a lattice structure and those of the continuum model (Wright 1965,
Pshenichnov 1982, Necib and Sun 1989). Displacement equations for a lattice cell can be
written in terms of finite difference operators and transformed to differential operators
(Renton 1970, Kollar and Hegedus 1985). A second category includes methods using
energy equivalencies between the lattice and continuum models (Noor, Andersen and

Green 1978, Noor 1988, Dow and Huyer 1987, 1989, and Lee 1990, 1991, 1994, 1998.

Methods of the third category are based on the finite element model of a repeating cell.
The model is subjected to static loading (Sun and Kim 1985, Sun, Kim and Bogdanoff
1988) or wave propagation (Abrate 1991) and the equivalent properties of the model are
determined from the results of these studies. The method suggested by Nayfeh and Hefzy
(1981) combines decomposition of the structural member array and an analytical

geometry approach.

The detailed survey of the application of continuum modeling and an extensive
bibliography in this area are available in the reviews by Noor and Mikulas (1988), and

Abrate (1985, 1988, and 1991).

CHAPTER 2

DECOMPOSITION METHOD
FOR SOLVING PROBLEMS OF STRUCTURAL MECHANICS

2.1 Main Concept of the Decomposition Method

The decomposition method for solving differential equations and boundary value
problems was proposed by G.I. Pshenichnov (1985). Unlike the domain decomposition
methods in which the structure is decomposed into substructures, this method
decomposes the governing equation and boundary conditions into subproblems. The main
concept in this method is to replace the task of solving the complex boundary value
problem by the analysis of simpler auxiliary problems stated in terms of additional
unknown functions. The form of these fimctions and their relationship to the field

equation and the boundary conditions is the key to this method.

Assume that the solution y = {y1 (x), ..., ym (x)} of the boundary value problem

Li(y)=f,-(x),i=1,...,m,er (2.1)

lj(y)=(0j(X),j:l,...,r,xer (22)

is to be found, where L,- and 1i are the operators of the equations and the boundary
conditions, respectively, f,-(x) and ¢i(x) are given functions, and x = {x1,...,x,,}. The
domain F consists of pieces of the whole boundary of the domain 0, and may include

some regions inside this domain. Domain 0 may be multiply connected, and the solution

may be multivalued. Let the operators of the system be represented in the form

10

[1

Li = kZLik (2'3)
=1

where some of the terms of the operators Lik may not occur in Li.
Introducing the notation
Lik (y) = ft" (x) (2.4)

it follows from equations (2.1), (2.3), and (2.4) that

h
fi(x) = Xfik (X), i=1,...,m (2.5)
k=l

The following h auxiliary problems for yk = {ylk (x),..., y,1f,(x)} are now added:

L,k(y")=f,."(x), er,i=l,...,m,k=1,...,h (2.6)
Ij<yk>=<oj(x). xerjker. j=i1k,....irk (2.7)

The conditions (2.7) are chosen so that conditions (2.2) are satisfied by at least one

solution yk at each fixed point on the contour F.

The solutions of the problem described by (2.1) and (2.2) coincide with the
solutions common to all the problems characterized by (2.6) and (2.7). Boundary
conditions will be satisfied as a consequence of the selections in (2.7). Furthermore, (2.1)
will be satisfied as well, since (2.3), (2.5), and (2.6) hold. On the other hand, a solution y
of the problem described by (2.1) and (2.2) is a solution of each of the h problems
characterized by (2.6) and (2.7) as well. Consequently, the task of solving the boundary

value problem may be replaced by that of finding solutions of the auxiliary problems

11

(2.6), containing m x h unknown functions f k (x) with the addition of m x h conditions

(2.7) on the solutions.

The merit of this method is the flexibility in the decomposition of the original
problem, which provides wide latitude for choosing the auxiliary problems that facilitate
the construction of the desired solution. As a result, simple and highly accurate
approximate analytical formulas for displacements, force responses, and eigenvalues of
the boundary value problems can be obtained in many cases, where other methods must

usually resort to numerical solutions.

In this work the decomposition method is developed for the bending and free

vibration problems of thin isotropic and single-layer lattice plates with elastic supports.

2.2 Bending of Rectangular Plate with Non-Symmetric Elastic Supports

The application of the decomposition method is first illustrated with a non-
symmetric problem of bending of the rectangular thin isotropic plate shown in Figure 2.1

subjected to a uniform transverse load.

2.2.1 Problem Statement

In the Cartesian coordinate system the differential equation governing the bending

problem has the well-known form

64w 64w 64w q(x)
+ = ——

 

(2.8)

12

where w is the transverse deflection, q is the uniformly distributed transverse load

intensity, and D is the plate’s flexural rigidity.

 

 

 

 

 

 

 

\ «r -__ 7‘ ""_'1l >
rl \ y
a I r3 r.. I
’l \
r
L_._ 2 J

 

 

><

Figure 2.1. Rectangular plate with non-symmetric elastic supports

Assuming flexible elastic supports along the edges of the plate, (2.8) must be

solved under the following boundary conditions:

w=0, Mlz—rlfl (x=0)
6x
6w
w=0, M = — =a
1 r26x (x )
w=0, M2=—r3—al (y=0) (2.9)
6x
6w
w=0, M =r — =b
2 46x 0’ )

where r,- is the stiffiiess per unit length of the distributed rotational springs along the

corresponding support (0 Sr,- <oo). Using the moment-curvature relationships

13

sz—D(62w/6x2)and Myz—D(62w/6y2), the boundary conditions may be

written as:

2
w=0, Da—wmrlialw (x=0)
6x2 6x
62w 6w
w=0, D—+r2—=O (x=a)
ax2 6x
62w aw
w=0, D—-——r —=o ( =0) (2.10)
ayz 35,y y
62w

w=0, D——2+r4§!=0 (y=b)
ay 6y

For a rigid support, the stiffness coefficient is infinity, and the equations above are
inconvenient for obtaining the analytical dependencies. It is expedient to introduce the

following expressions for the dimensionless stiffness coefficients of the elastic supports:

k1=___1_, k,=__1__
1+r1a/D l+r2a/D

(2.11)
1 1

k3 =———, k4 :—
l+r3b/D l+r4b/D

Since0_<_ r,- <oo, it follows that OS ki 51, and the extreme cases ki =0and kt =1

correspond to rigid and hinged supports at the edges of the plate. Furthermore, the

following notations are introduced to reduce the whole problem to non-dimensional form:

a=x/a, ,B=y/b, i=b/a 21
D (2.12)

14

where v = dimensionless deflection function.

The dimensionless differential equation corresponding to (2.8) is

4 4 4
a:+—22———62"2 +-—12——a: :1 (2.13)
60: 2 50: 6,6 ,1 an

 

and the dimensionless boundary conditions corresponding to (2.10) are

62v
2

 

v=0, k1 (1—k1)§—v—=O (a=0)
6a

6a

62v
630:2

 

v=0, k2 +(l—k2)3v—=O (a=l)
6a

82v

V=0, k3 -
aaz

 

(1—k3)-§—Y—=0 (,B=0) (2.14)
a

2
V=O, k4av+(l—k4)fl=0 (fl=1)
aaz 6a

 

2.2.2 Decomposition of the Problem

The boundary value problem characterized by (2.13) and (2.14) is solved by the

decomposition method. The dimensionless deflection v(a, ,6) is represented in the three

forms v] , v2 , and V3 . Three auxiliary problems, two of which are boundary value
problems, are introduced to determine these forms.
The first auxiliary problem (boundary value problem) is to find the solution to the

differential equation

64V] _

a 4 fl(a9fl) (215)
a

 

15

subject to the conditions

 

62v] av,
v1=0, k1——(1—k1)—=0 (a=0)
aa2 6a
(2.16)
62v] avl
v1=0, k2 +(l—k2)—=O (0:1)
aa2 0a

The second auxiliary problem (boundary value problem) is to find the solution to

the differential equation

64V2
—,— = new) (2.17)
5V3

subject to the conditions

 

azvz avz
V2=O, k3——2—(l—k3)——=O (5:0)
6a 50'
(2.18)
OZVZ 6V2
122:0, k4 2+(1—k4)—=O (,6=1)
aa ad

The third auxiliary problem is to find the solution to the interconnection

differential equation

2 64V3

?W=‘fl(aafl)— %f2(a,fl)+ 1 (2°19)
a

These problems include two unknown functions f1 (a, ,6) and f2(a,,6).

Assuming that

v = v1 2 v2 2 v3 (2.20)

16

and separately summing the left- and right-hand terms of the equations in the three
auxiliary problems, yields the lefi- and right-hand terms of the original problem in (2.13).
Similarly, summation of the boundary conditions of the auxiliary problems results in the
boundary conditions of the original problem (2.14). Thus, the solution of the boundary
value problem (2.13) and (2.14) can be replaced by the solution of auxiliary problems

(2.15), (2.17), and (2.19) satisfying conditions (2.20).

If the condition (2.20) is satisfied exactly, the three forms of the solution

v1, v2 , and v3 coincide, and the exact solution of the original boundary value problem is

obtained. In this work an approximate solution is sought by representing of the unknown
functions f1(a, ,6) and f2 (a, ,6) as power series. These functions do not depend upon
boundary conditions, and therefore can be represented by the same system of basic
functions. Since the fourth derivative of v3 is approximated by these functions, retaining

only a few terms of the power series expansion can still yield accurate results for v.

Assmne that

fl(a9fl) =Wl<18>
f2(a,fl)=Il/2(a)

(2.21)

where (#166) and I/lz (a) are arbitrary functions.

Consider the first auxiliary problem (2.15) and (2.16). The integration of (2.15)

yields

4 3 2
a a a
V1=Wl(fl)[§+C3?+Cz7+Cla+C0] (2.22)

17

which C ,- , i = 0.3 , are the integration constants depending on the values of A (the aspect

ratio of the plate) and k,- (the dimensionless stiffness coefficients of the elastic supports).

These constants are found by satisfying boundary conditions (2.16):

 

 

 

 

C020
C] :_1i k11+5k ) =_1_.kl(1+5k2)
23(1+k)1+k 2)— 2—(1 k k2) 12 R12
C2=_11_. 1—k1)(1+5k) :_1_.(1—k1)(1+5k2) (2.23)
3“(1+k1)(1+k2) 2(1 kl k 2) 12 R12

2(1+2k,)R,,+ +5k2)(1- klk k)

(1
6R,2(1+k2)

 

C3:—

where R12 =3(l+k])Ll+k2)—2(l-kl k2)=l+3(kl +k2)+5k1k2.

The expression for the first form of the deflection function can now be written as

 

 

 

 

V'=W17(2m{3a4_1e12(1+k2) [211,20 + 2k )+ (1 + 5k )(1- k k )]a3
(2.24)
+ 3(1+ 5k2)[(1—k1)r12 + 2klal}
R12
A similar procedure for the second auxiliary problem (2.17) and (2.18) yields
v2 =l’2752a—){3fi — R34(1+k4) [2R34(1+2k )+ (1+5k4)(1—k3k4)],83
(2.25)

 

+3(1;5k4I[(1-k3),62 +2k3fl1}

34

where R34 =3(l+k3)(l+k4)—2(l—k3 k4)=l+3(k3 +k4)+5k3 k4

Satisfying condition v1: v2 from (2.20) yields the arbitrary functions ([11 (,6) and 012(61):

 

 

111168) p{3fl R34(1+k4) [2R34(1+2k4)+(1+5k4)(1 k3k4)],6
+3(1+5k4)[(1_k3)fl2 +2k3 fl]} (226)
R34

or MA) = pram)

 

 

:: 4 — 2 _ 3
(112(a) p {3a R,2(1+k2) [2R12 (1+2k2)+ (1+5k2)(1 klk2)]a
+3(1+5k2)[(1‘-k1)a2 +2k1a1} (227)
R12

0" I1112(0) = PC0201)

where p is an unknown constant. Substituting wlw) and 1/12 (a) into (2.24) and (2.25)

yields the following expression for the dimensionless deflection function:

 

v: v1: v2 =%{3a4 — R,2(lik2) [2R12 (1+2k2)+(1+5k2)(1—k1k2)]a3

 

+ 3(1+5k2I[(1—k,)az +2k1al}

R12
(2.28)

x 4— 2 T 3
{3p R34(1+k4) [2R34(1+2k4)+(1+5k4)(1 k3k4)],6

 

+3(1+5k4)

 

34

[(1—k3),62 +2k3fll}

Equation (2.28) is an approximate solution to the original problem with the

unknown constant to be determined by solving the third auxiliary problem.

19

The approximate solution of the third auxiliary problem (2.19) is found by

assuming v3 = v] =v2 = v. The following the discrepancy function based on (2.13),
(2.15), and (2.17) is proposed:

64173

(D __
(05,3): fl(a;6)+ :Zaaz 6,32

+fi-f2(a,/3)— 1 (2.29)

If the solution is exact, the function (D(a, ,6) is identically zero. In the approximate

solution that is sought, the arbitrary constant p is determined by minimizing the

discrepancy function <D(a,,8) using the Bubnov-Galerkin method. The vanishing

condition for the discrepancy function can be written as:

1 1
H¢(a,fl)¢2(a)¢1(fl)dadfl = o (2.30)
0 0

Using the notations in (2.26), (2.27), and (2.28) yields the expression

J; —¢'(’6)+ 72. 272% (a “)9”! (,B)+—/:—4(01(fl)](02(a)g01(/3)dad,6=0

where the primes denote derivatives with respect to the arguments of the functions.
Performing the integrations and necessary transformations yields the following

expression for the arbitrary constant

49 R12 R34 612 G34

p =
4 7
7R12 012 F34 + EH12 H34 + 11R34G34F12

 

(2.31)

where

R12 =1+3(k1+k2)+5k1k2, R34 =1+3(k3 +k4)+5k3k4

20

612 =1+8(k1+k2)+55k,k2 , G34 =1+8(k3 +k4)+55k3k4
11112 =1+13(k1 +k2)+58(k,2 +k22)+154k,k2 +625(k,2k2 +k,k§) + 2125k12k22
H34 =1+13(k3 + k4) + 58(k32 + k3) +154k3k4 + 625(kg7‘k4 + k3k3) + 2125k§k§
Fm =1+15(k1+k2)+60(k12 +k22)+208k1k2 +765(k12k2 +k,k§)+ 2575k3k22
F34 =1+15(k3 + k4) + 60(k,2 + k3) + 208k3k4 + 765 (1.321., + 1:31:42) + 2575 k32k}

After the value of p is obtained from (2.31), the dimensionless deflection can be

computed using (2.28). The actual deflection and bending moments at any point of the

plate are calculated using the formulas

 

4
w=1‘i—v
D
2 2
M1 =-qa2 -a—‘:—+Vi—:— (2.32)
6a 6,6
2 2
M2 =-qa2 V a :+§—:
6a 6,8

2.2.3 Numerical Results

Test results were obtained from the decomposition method for an isotropic square
plate with different combinations of boundary conditions under the uniformly distributed
load and compared with accurate solutions (Timoshenko and Voinowsky-Kn'eger 1959).

The calculation results and the associated errors are presented in Table 2.1.

The error in the deflections computed by the decomposition method do not exceed

5%, but the error in the computed bending moments is significant and reaches 20%. This

21

is because the bending moment is related to the second derivative of the displacement
which is prone to greater error. The number of terms in the series expansion should be

increased to improve the results for the bending moment.

TABLE 2.1. Accuracy of results obtained using Bubnov-Galerkin condition (2.30)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

v(0, 0)x1o3 Mow, 0)x1 112
Support
Type Exact Exact
value D" E M) value M° (1’ 0) 51%)
3:; 1.32 1.26 -4.8 2.77 2.31 -19.0
1' _ '1
: : 4.14 4.04 -2.0 5.17 4.79 -9.0
///Z//ll(l
1| 1.61 1.57 -2.5 3.26 2.83 -15.9
3,777,”)I
- _ —l
a : 2.82 2.80 -0.7 4.31 3.90 -10.4
i g 1.98 1.92 -3.12 2.77 2.44 -13.5
— — _|
a : 2.40 2.30 -4.3 3.02 2.81 -7.5
Notations: W fixed support, 2:— pinned support

22

2.3 Symmetric Bending Problem of Rectangular Plates
with Elastic Supports

In this section the decomposition method is used to obtain the bending response of
the thin isotropic rectangular plate with elastic supports shown in Figure 2.2 under an
arbitrarily distributed load. Collocation methods are introduced for minimizing the

discrepancy function.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

a _ , 1
( '- r r 1 \ ’ l k k \
2 2 . »
ll ii i x y_. i 2 2 \
11 ’ 1 f
a ’l r i y 1 ’ll k ’ ,
l—,———‘—:,—d—,-4 LL: 1, , —,J
v X v ‘ “
Figure 2.2. Rectangular plate with symmetric elastic supports
2.3.1 Problem Statement
The differential equation for the bending problem is
04w 64w 64w Z
— + 2 — — (2.33)

 

+ _

where Z is the arbitrary distributed transverse load.

23

Assuming symmetrical elastic supports along the opposite edges of the plate, (2.33) must

be solved under the following boundary conditions:

2
w=O, Dflirléfi=0 (x=ia)
6x2 6x
(2.34)

2
w=0, Da_;vir2§_v::0 (y=ib)
ay 6y

Using the notations (2.11) and (2.12), the problem is reduced to the non-dimensional
form:

64v + 2 04v _1_ 64v
aa“ 22 6012632 24 ap“

 

= (Kan/3) (2.35)

 

where q(a, ,6) = = the dimensionless load function. The dimensionless boundary

'2 max l
conditions corresponding to (2.34) are

 

2

v=O, kla:i(l—kl)§v—=O (a=i1) (2.36a)
6a a
2

v=0, kzi;i(l—k2)-a—v—=O (,Bzil) (2.36b)
ap afl

2.3.2 Decomposition of the Problem

The boundary value problem characterized by (2.35) and (2.36) is solved by the
decomposition method using three auxiliary problems. The first auxiliary problem

(boundary value problem) is to find the solution to the differential equation

 

a4
‘2 =f1(a,fl) (2.37)
Ga

24

subject to the conditions (2.36a), where v = vl .

The second auxiliary problem (boundary value problem) is to find the solution to

the differential equation

64v

,2 =f2(a,fl) (238)
613

 

subject to the conditions (2.36b), where v = v2.

The third auxilia1y problem (solution of the differential equation) is

64
cum/3); 2, 2V3 , +ma,/3)+—17f2(a,/3)—q(a.m=0 (2.39)
2 aa 65 2

 

These problems include two unknown functions f1 (01,,6) and f2 (0:, ,6). An approximate
solution is sought by retaining the first two terms of the power series expansion of

f1(a, ,8) and f2 (a, ,6). Due to symmetry in the coordinates a and ,6, it is assumed that

f1 (M) = f1(fl)+ a2f3(fl), f2(a,fl) = f2(a) + fl2f4(a) (2.40)

where f,(,B), f2(a), f3(fl), f4(a) are arbitrary functions.

Solutions of the boundary value problems (2.37) and (2.38) are then obtained as

v, 7171i 4 -2(1+2k1)a2 +1+4k1]f1(,6)
(2.41)

1
+§6_O_[a5 —3(1+4k1)a2 + 2(1+6k1)]f3(fl)

25

V2 =-le[fl4 -2(1+2k2)fl2 +l+4k2]f2(a)

1
763% —3(1+4k2>/32 + 2(1 + 6k2)]f4(a)

Satisfying conditions (2.20) yields

1
—w1(l)(a)[p1w1(2)(fl)+psi/I?) (5)]

V‘:V2=24

l
+355W§1)(a)lp2wl2)(fl)+p4W§2)(fl)i

where

Wink!) =a4 —2(1+2k1)a2 +1+4kl
film) =a6 —3(1+4k1)a2 +2(l+6k1)
91/906) = ,84 —2(1+2k2),62 +1+4k2

w§2)(fl)=fl6-3<1+4k2)/32 +2<1+6k2)

and pi = arbitrary constants.

(2.42)

(2.43)

(2.44)

Equation (2.43) is an approximate solution to the original problem with four constants to

be determined by solving the third auxiliary problem. Satisfying conditions (2.20) and

using (2.39) and (2.43) yields

1 N N 1
(Wm/3) = 121 [wfzhflHl—fiz—wl‘” (aw?) (fl)+Fv/1m(a)]

1 H N 1
+p2[a2w§2’(fl)+——2w§” (aw/1‘2) (fl)+—-—4-w§l)(a)]
1801 15,1

26

2 1 l H 2 H 15
+ p3 [1115 )(fl) +1315”: ’ (am ) (,6) +174—fl2wf”(a)]

1 .. .. 1
+ p4[a2w§2)<fl) + WW9) (aw/52’ (fl) + 1.132121%]— q(mfl) = 0 (2.45)

The primes denote derivatives with respect to the arguments of the functions.

2.3.3 Solution techniques for the interconnection equation

The third auxiliary problem (interconnection equation) can be solved by different

methods. For the problem at hand, three different approaches were used to evaluate the

arbitrary constants pi: the Bubnov-Galerkin method and two forms of the collocation

method. The load was assumed to be uniformly distributed: q(a, ,6) = l .

Bubnov-Galerkin Method. In the first approach using the Bubnov-Galerkin

method, the vanishing conditions for the discrepancy function can be written as:

l l
I I¢<a,fl>wf”(a)wf2’(mda dfl = o
00

11
l l¢(a,fl)V/§l)(a)w§2)(fl)da dfl = 0
oo

(2.46)
l l

II¢(a,fl)1I/f”(a)w§2)(fl)da dfl = 0
00

1 l
1 I¢(a,fl)v1§1)(a)w1(2)(fl) da cw = 0
00

Substituting the discrepancy function (2.45) into (2.46) and performing the
integrations yields a system of four linear algebraic equations which can be solved to

obtain the four unknowns p], 1);, p3, and p4. The dimensionless deflection at any point of

27

the plate is then obtained from (2.43) and bending moments are computed using the
moment curvature relationships. Results obtained from the decomposition method for a
square plate with three different types of boundary conditions, and the associated errors
compared with accurate solutions (Timoshenko and Voinowsky-Krieger 1959) are given
in Table 2.2. The example demonstrates the accuracy of the technique for the case of

square plates.

TABLE 2.2. Accuracy of results obtained using Bubnov-Galerkin conditions (2.46)

 

 

 

 

 

 

 

 

 

 

 

 

Support Type v (0, 0) a (%) Mo(0, 0) a (%) M0 (1, 0) a PM
Pimef‘ “EM“s 0.00406 0 0.0474 1.04 0 0
(k1- k2 " 1)
Fixedfuflpms 0.00126 0 0.0228 1.2 0.0512 03
(k1- k2 - 0)
Mixed supports
(R1: 0’ k2: 1) 0.00192 0.1 0.0246 -0.6 0.0668 1.3

 

However, the solution using this approach loses the accuracy for aspect ratios
other than 1. Also, the Bubnov-Galerkin method for solving the third auxiliary problem is
cumbersome and does not lend itself to automation. An alternate method is to obtain the
equations for determining the unknown constants by equating the discrepancy function or

its derivatives to zero at several collocation points.

Collocation Method with Single Point. In the second approach the arbitrary

constants are determined by minimizing <D(a,,6) in the middle section of the plate.
Accordingly, the function and some of its lower derivatives with respect to a and ,6 are

set to zero at the center of the plate.

2 2 4
¢(0,0) = 0, 13:40.0) = 0, a—gwp) = 0, f3—3‘:(0,0) = 0 (247)
0a 00 0a

28

Since all odd derivatives are zero at the center of the plate owing to symmetry, these are
not used in (2.47).
Using (2.39) and (2.47) results in a system of four linear algebraic equations, the

solution of which yield the constants pi. The values of these constants are then

substituted into (2.43) to compute the dimensionless deflection function. The real
deflections and moments can be calculated using (2.32). Results obtained from this
technique are compared with accurate solutions (Timoshenko and Voinowsky-Krieger
1959) in Tables 2.3 and 2.4, and Figures (2.3) through (2.6). It can be seen that this
approach provides reasonably good results for square plates and for long plates with high

aspect ratios, but yields large discrepancies for intermediate aspect ratios.

TABLE 2.3. Calculation results for k1= k2 = 0 (fixed supports) obtained using (2.47)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

b v(0, 0) x 103 -Mo(1, 0)x102
A a Exact Value [3:333 a (%) Exact Value 03:33: a (%)
1.0 20.16 20.29 0.64 20.52 20.42 -0.49
1.1 24.00 23.74 -1.08 23.24 22.80 -1.89
1.2 27.52 26.75 -2.80 25.56 24.79 -3.01
1.3 30.56 29.30 -4.12 27.48 26.43 -3.82
1.4 33.12 31.42 -5.13 29.04 27.75 -4.44
1.5 35.20 33.16 -5.80 30.28 28.81 -4.85
1.6 36.80 34.59 -6.00 31.20 29.65 -4.97
1.7 38.08 35.75 -6.12 31.96 30.32 -5.13
1.8 39.20 36.70 -6.38 32.48 30.86 -4.99
1.9 39.84 37.48 -5.92 32.88 31.28 -4.87
2.0 40.64 38.12 -6.20 33.16 31.63 -4.61
00 41.60 41.67 0.17 33.32 33.33 0.03

 

29

 

 

 

 

 

 

 

 

1
1 401
1 ...41
1| .’o__,.—4—
1 8 35 ,«-"°’
2 ,o»
X ..r' 1
A /
c 301 ,0” 1
1 5 "
1 > 25 -< .I'
1
1
1 20 I 1 1 1 1 , , T
i 1 1,1 1,2 1,3 1,4 1,5 1,6 1,7 1,8 1,9 2
1 AspectRatio
1
1
1

1—9—- Exact Solution — 9-- :Decompositionw Method

Figure 2.3. Maximum deflections at the middle of the plate obtained by (2.47)

for k1= k2 = 0 (fixed supports)

E

 

”w” “_ Th1
1
|
l
l
l
l

M(‘l;0)x100

 

 

 

 

 

1,4 1,5 1,6 1,7 1,8 1,9 2

Aspect Ratio

1,2 1,3

1 1,1

 

1 1—0— Exact Solution — 0- Decomposition Method

Figure 2.4. Maximum bending moments on the edge of the plate obtained by (2.47)

for k1= kz- - 0 (fixed supports)

30

TABLE 2.4 Calculation results for k1=k2 =1 (pinned supports) obtained using (2.47)

 

 

 

 

 

 

 

 

 

 

 

 

 

b v(0, 0)x102 Mo(0, 0)x102
A _ Z Exact Value 03:33:: a (%) Exact Value [3:31:15 3 (%)
1.0 6.50 6.41 -1.38 19.16 18.97 -0.99
1.2 9.02 8.78 -2.66 25,08 26.24 4.62
1.4 11.28 10.88 -3.55 30.20 32.60 7.95
1.6 13.28 12.66 -4.67 34.48 37.88 9.86
1.8 14.90 14.12 -5.23 37.92 42.10 11.0
2.0 16.16 15.31 -5.26 40.80 45.40 11.3
3.0 19.52 18.62 -4.61 47.56 53.28 11.9
4.0 20.48 19.83 -3.17 49.40 54.77 10.9
5.0 20.80 20.33 —2.26 49.84 54.53 9.41
00 20.80 20.80 0 50.00 50.57 0.01

 

 

 

 

 

 

 

 

v(0;0)x100

 

 

 

 

1,2 1,4

Exact-Solution -___:-_ DEcompositioflfl Me_thod 1

1

1,6

1,8

2

2,2 2,4

Aspect Ratio

 

 

2,6 2,8

 

Figure 2.5. Maximum deflections at the middle of the plate obtained by (2.47)

for k1: k2 = l (pinned supports)

31

 

 

m0; 0)x100

 

 

 

 

! 1 1,2 1,4 1,6 1,8 2 2,2 2,4 2,6 2,8 3
I Aspect Ratio
1

 

1 1;'-°-_:':§>§2£§2_'L'tj¢3 :1: 99911323319214.2992}

 

 

Figure 2.6. Maximum bending moments at the middle of the plate obtained by (2.47)

for k1= k2 = l (pinned supports)

Collocation Method with Multiple Points. The third approach is to obtain the
equations for determining the unknown constants by equating the discrepancy function to
zero at several collocation points. For the given symmetrical problem, only a quarter of
the plate needs to be analyzed. For four equidistant collocation points, the linear algebraic

equations can be solved to obtain p1, p2, p3, and p4:
<1)(0,0)-_- 0, <1>(0,0.5)= 0, c1>(0.5,0) = 0, <1>(0.5,0.5)= 0 (2.48)

Tables 2.5 and 2.6 show the results calculated for plates with fixed and pinned
supports, respectively, for different aspect ratios. These results are also illustrated in
Figures 2.7 through 2.11. The collocation method was used to obtain the results shown in
these tables. The comparison of deflections and bending moments with exact solutions

show the high accuracy of this method.

32

TABLE 2.5. Calculation results for k1= k2 = 0 (fixed supports) obtained using (2.48)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

b v(0, 0) x 103 Mo(0, 0) Mo(1, 0)
’1 = " Exact Decomp. Exact Decomp. Exact Decomp.

a Value Method ‘3 (%) Value Method 8 (%) Value Method 8 ( A)
1.0 20.16 20.11 -0.25 9.24 9.14 -1.06 20.52 20.13 -1.88
1.1 24.00 23.96 -0.14 10.56 10.64 0.71 23.24 22.98 -1.11
1.2 27.52 27.41 -0.41 11.96 11.93 -0.23 25.56 25.41 -0.57
1.3 30.56 30.37 -0.62 13.08 13.02 -0.45 27.48 27.43 -0.20
1.4 33.12 32.86 -0.79 13.96 13.91 -0.34 29.04 29.05 0.03
1.5 35.20 34.91 -0.82 14.72 14.53 -1.28 30.28 30.34 0.19
1.6, 36.80 36.58 -0.61 15.24 15.20 -0.24 31.20 31.34 0.46
1.7 38.08 37.92 -0.42 15.68 15.65 -0.17 31.96 32.12 0.49
1.8 39.20 38.99 -0.54 16.04 16.00 -0.22 32.48 32.71 0.70
1.9 39.84 39.84 0.00 16.28 16.28 -0.03 32.88 33.15 0.83
2.0 40.64 40.51 -0.32 16.48 16.48 0.02 33.16 33.49 0.98
00 41.60 41.67 0.16 - - - 33.32 33.33 0.04

17"" " ‘7 1
1 40 - 1
1
1
o 35 ‘ i
O 1
° 1
i 1
c 30 ~ 1
a 1
1 3': 1
1 25 - ,
1 1
1 20 , . . . . e . . .
1 1,1 1,2 1,3 1,4 1,5 1,6 1,7 1,8 1,9 2
i Aspect Ratio
1 [—O—-_Exact Solution : j -_- - Decomppsition Method _1

Figure 2.7. Maximum deflections at the middle of the plate obtained by (2.48)

for k1= k2 = 0 (fixed supports)

33

 

17
16 -
15
14 «
13 ~«
1 12 ~
1 11 4
1 10 «
1

1

1

 

M(O; 0)x100

 

 

 

9 ‘ . 2 . , . .
1 1,1 1,2 1,3 1,4 1,5 1,6 1,7 1,8 1,9 2
Aspect Ratio

}’_.__ i A ‘E'xé'd Quipn: T—Tfoec'caThprsRiSn’Meihpdfi

#A,_;-___~_____a~_ii_-i__,______._~_.—__._._f J

Figure 2.8. Maximum bending moments at the middle of the plate obtained by (2.48)

for k1= k2 = 0 (fixed supports)

 

 
 
 

30

m1; 0)x100
N N
O) on

24~

 

 

 

 

1 1,1 1,2 1,3 1,4 1,5 1,6 1.7 1.8 1.9 2 {
AspectRatio ‘

{ 1_.; $381 9E1'fi074T: Decbumpositioanethod 1

 

Figure 2.9. Maximum bending moments on the edge of the plate obtained by (2.48)

for k1= k2 = 0 (fixed supports)

34

TABLE 2.6. Calculation results for k1= k2 = 1 (pinned supports) obtained using (2.48)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

v(0, 0) x 102 Mo(0, 0)x102
,1 = —
Decomp. o Decomp. o
Exact Value Method 5M) Exact Value Method 3M)
1.0 6.50 6.41 -1.45 19.16 18.96 -1.07
1.2 9.02 8.91 -1.27 25,08 24.76 -1.28
1.4 11.28 11.15 -1.13 30.20 29.79 -1.34
1.6 13.28 13.06 -1.63 . 34.48 33.97 -1.49
1.8 14.90 14.64 -1.76 37.92 37.34 -1.53
2.0 16.16 15.91 -1.54 40.68 40.03 -1.60
3.0 19.52 19.33 -0.98 47.56 47.08 -1.02
4.0 20.48 20.45 -0.17 49.40 49.31 -0.19
5.0 20.80 20.82 0.11 49.84 50.04 0.41
00 20.80 20.83 0.16 50.00 50.00 0.00
1 20
1 181
1
1e .
O
O
i 14—
0
g 12 4
>
’ 10 ~
1 8~
1 6 ‘ - , w .
i 1 1,2 1,4 1,6 1,8 2 2,2 2,4 2,6 2,8 3
1 Aspect Ratio
1 __,.,..,__,_-_ L__.-_. _,
1 + _ , Exact 30191102 :- -: (9.999111919311199 1!?!th E

Figure 2.10. Maximum deflections at the middle of the plate obtained by (2.48)

for k1= k2 = l (pinned supports)

35

 

 

M (0; 0)x100

 

 

 

 

1 1 1 . . . 1 1
; 1 1,2 1,4 1,6 1,8 2 2,2 2,4 2,6 2,8 3 1
1 Aspect Ratio

1 .-.. _ __2--.-2-22W22,_...__.2--.__,__

1 1—+— Exact Solution - -o— - Decomposition Method :

1 , _._ _. _ _ _._ _ _._ __ _ __. _ _ . ._ . _ _ .__ __ .

Figure 2.11. Maximum bending moments at the middle of the plate

obtained by (2.48) for k1= k2 = l (pinned supports)

2.3.4 Solution of the Problem for the Case of a Non-uniform Load

For the rectangular plate with pinned supports the boundary value problem has the

form (2.35) and (2.36). Assume the load distribution function as

mnfl
2 (2.49)

 

q(a,,6) = cosfg—O—l— cos

n, m =1,3,5...

The approximate solution of the problem has the form (2.43) with respect to (2.44). The

arbitrary constants p, are to be determined from the conditions (2.47). Using the

. rm m7: . .
notations 2,, :7, 2m :7, the values of the load function and its low even

derivatives at the central point of the plate are determined as

36

62 62 a4
3—‘12—10,O)=-4%, $<0,0)=—4%., a—‘g10,0)=—4i, (2.50)
a a

9(0,0) =1,

Using the expression (2.35) for discrepancy function and applying conditions
(2.47) yields the system of four linear algebraic equations with four unknown constants.
The resulting values of constants are then substituted into (2.43) to obtain the equation of

the dimensionless deflection function.

The non-dimensional values of deflections and bending moments for the plates
with different aspect ratios obtained from this technique are presented in the Table 2.7
and Figures 2.12 and 2.13. To estimate the accuracy, the solution by Timoshenko method

was obtained. The comparison shows a reasonable accuracy of the proposed technique.

TABLE 2.7. Calculation results for rectangular plate under non-uniform load

for k1=k2 =1 (pinned supports) obtained using (2.47)

 

 

 

 

 

 

 

 

 

 

 

 

b v(0, 0)x1o2 Mo(0, 0)x102

A - 3 Exact Value 2:;33' c(%) Exact Value 3:32: c(%)
1.0 4.11 4.16 1.2 10.1 10.2 1.0
1.2 5.72 5.68 -0.7 18.3 18.3 0.0
1.4 7.20 7.09 -1.5 23.1 22.8 -1.2
1.6 8.49 8.28 -2.5 27.2 26.8 -1.8
1.8 9.59 9.32 -2.8 30.8 30.1 -2.0
2.0 10.5 10.2 -2.8 33.7 32.9 -2.1
3.0 13.3 12.8 -3.3 42.7 41.9 -1.9
4.0 14.5 14.1 -3.2 46.6 46.1 -1.2
5.0 15.2 14.7 -3.1 48.7 48.4 -0.7
10.0 16.1 15.7 -2.6 51.6 52.0 0.6

 

 

 

 

 

 

 

 

 

37

 

_.—.—"
'_'—’-'—
.._—.-
_."

v (o, O)x100
\

 

 

 

41:.T1..4,..~.44.
11,522,533,544.555.566,577,588,5 99,510

Aspect Ratio

 

1+ Exact Solution - -o—- Decomposition Method 1

1
1

Figure 2.12. Deflections at the middle of the plate under non-uniform load

for k1=k2 =1 (pinned supports) obtained using (2.47)

 

(D

1'“ ..‘V..__-.7 * A a.-_ ..~-!..._.__..__-

Figure 2.13. Bending moments at the middle of the plate under non-uniform load

 

 

 

11,22,33.44,

5 5 5 5

55.
5

fi

6

6.
5

7

Aspect Ratio

 

1;0—Exact Solution — 4- Decomposition Method 1

for k1=k2 =1 (pinned supports) obtained using (2.47)

38

2.4 Bending of Rectangular Plate with One Free Edge
and Three Elastically Supported Edges

The transverse bending of a thin isotropic rectangular plate with one free edge and
elastic supports along the three other edges is investigated (Figure 2.14). The load is

assumed to be unifome distributed.

 

 

 

 

 

 

 

 

 

 

 

 

b 1
= * r — > 15 ' * J>
1'l y k1 1”
(1" 1
1’ 1
a . I] 1 ; kl
<1. (.5
1'
i a. .l 1 k1
O
V" 11

 

 

Figure 2.14. Rectangular plate with one free edge

2.4.1 Problem Statement

The problem is stated in the non-dimensional form with respect to notations

(2.10). Assume that the edges a = O, 0: =1, and fl = 0 are supported elastically, and the

edge fl = 1 is free. In this case the boundary value problem can be written as

 

 

4 4 4
a:+—25——‘:"2+—-IZ‘9:=1 (2.51)
6a /1 6a 6,6 1. 6,6

39

 

2
v=O, k] a vi(1—k1)2v—=O (a=0, (1:1) (2.52a)
62v 6v
v = 11 —3-(1-k1)—— = 0 (a = 0) (2.52b)
an (M

_ 3 3
4 aa 63 ,1 6,6
(2.52c)

1 62v 62v
M =———+#— =0 (13:1)
” [/12 6,62 65:2]

2.4.2 Decomposition of the problem

The boundary value problem characterized by (2.51) and (2.52) is solved by the
decomposition method using three auxiliary problems. The first auxiliary problem
(boundary value problem) is to find the solution to the differential equation

64V] _

6614

 

f1 (on/3) (2.53)

subject to the conditions (2.52a), where v = vl :

2
‘3 v1i(1—k1)—avl=0 (01:0. a=1> (254)
2 6a

 

V1=k1

The second auxiliary problem (boundary value problem) is to find the solution to

the differential equation

4
a ”f =f2(a,fl) (2.55)
641

 

subject to the conditions (2.52b) and (2.52c):

4O

2
v2 =1:1 a v2 —(1—k )a—Vl: —0 (fl=0) (2.56a)

 

 

 

 

 

6,62 afl
(2—#)62Vl 6V2 1 53v2
V4“ 2 +7 3 =
4 aa afl 2 6,6
(2.56b)
M 1 azvz 62v, 0 (13 1)
:— ~—— + = =
fl 12 6,62 ’1 aa2

Note that boundary conditions (2.56b) now include the derivatives of the
deflection function in both directions. For this case it is more convenient to use the
“weaker” form of the boundary conditions. To obtain it, the summation of work done by

tractions along the edge ,8 =1 is set equal to zero:

 

_(—2 av a v163v '
A”) 2 flIaaZ 1V 1—3‘—32 J'Vl da =0
0
(2.560)
1 62 v 6v 62 v 6v
UM— _ ____2_ lJ‘a_a Id fill!_ l _lda :0 (flzl)

226,620 W526

The third auxiliary problem (solution of the interconnection differential equation) is

4
mm ,8)=— ‘3

l
12W +f1(0!,,3)+Ff2(a,13)—1=0 (257)

Note that solutions of the auxiliary problems must satisfy the condition
v=vl =v2 = v3 (2.58)

The approximating ftmclions are taken in the following forms:

f1(a,fl)= 11(4), f2(a,fl)= flf2(a)+ fl2f3(a)+ 413/400 (2.59)

41

where f,.(t), i=1,...4,a1ea1bin‘aryfimctionsoftheargmnents a and ,6.

The solution of the first boundary value problem has the form
v]: 32.114 —2a3 —2(—1 + k,)/(1+ k1)a2 + 2k1a/(1 + k1)]fl (,6) (2.60)

Consider the second auxiliary problem. Integration of the differential equation

(2.55) yields

3 2
V2 = f2(0!)[-1—.55 +'flL-Cs +£2—C2 +flC1+Co]

120 6
+f(a)[L,B6 +giC +gi—C +,BC +C (261)
3 360 6 7 2 6 5 4 '
+f(a)i67+fl—3C +fl—2C +,6C +C
4 840 6 11 2 10 9 8

where C ,- , i = 1,K 11, are the integration constants depending on the values of A (the aspect

ratio of the plate) and k1 (the dimensionless stiffness coefficient of the elastic supports).
These constants are found by satisfying the boundary conditions (2.56a) and (2.56c). The

expressions for C ,- are obtained using the Maple computer algebra system, and are given in
Appendix A.

Satisfying condition v1: v2 from (2.58) yields:
2 2 2
f1(16)= p] 101‘ )(fl)+p2 111$ )(fl)+p3 1/45, )(fl)

f2(a) = f3 (00 = f4 (a) = iwf'ka) (2.62)

42

where

Wi1)(a)=a4 —2a3 -2(-1+k1)/(1+k1)0‘2 +2[Ha/(”l”)

3 2
2 1
1,1/f)(fl)=1—2—Ofl5+g€C3+gz—C2+3CI+CO

1 6 ,33 132
— +—C +—C + C +C
360fl 7 2 6 ,3 5 4

1421113): 6

3 2
W(2)(fl)=8—1— 1313 7+g6—C11+%C10+13C9+C8

The expression for the deflection function can now be written as

V] :3;

wf')(a)(p w12)(/3) + pz 14)” (16) + 123 142’ (m)

(2.63)

(2.64)

where p1, p2 , and 123 are arbitrary constants that are to be determined from the third

auxiliary problem with respect to the conditions in (2.58).

Substituting the expression (2.64) into (2.57) and using (2.63) yields

d>(a,13)=P1[W1(+2)(fl)W11)/12(a)V/1(2)(fl)]

+p21/11//§Z)(16)+w1"/12(a) 1111(2) W]

b

 

+1231! 1412113141 712.11% 01))w2’(fl)]

L

241241119) ( )(fl+fl +16 )

43

(2.65)

An overdefined point collocation method is used for the solution of this problem, with 90
collocation points spaced equidistantly over the surface of the plate. For each point the following
condition is applied:

1 l
”42(a,,6)W,. da dfl = o
0 0 (2.66)

W1 = 5(a -a,-,,6—,6,~), i =1,2,K K,K N
where 5 is Dirac delta-function, and (a, , A) are coordinates of the collocation points.

The constants p1 , p2 and p3 are obtained as the least-square solution of the resulting

overdefined system of linear algebraic equations. The dimensionless deflection function is
determined using (2.64), and the real deflections and bending moments are calculated

using the expressions in (2.32).

2.4.3 Numerical results

Test results were obtained from the decomposition method for an isotropic
rectangular plate with different aspect ratios under the uniformly distributed load and
compared with available accurate solutions (Timoshenko and Voinowsky-Krieger
1959) and finite element solutions. The finite element analysis was performed using
the LIRA software (Kiev, 2000). A refined mesh of quadrilateral finite elements was
used to achieve convergent solutions. All dimensional values of deflections obtained
by the finite element method were converted into non-dimensional form. Calculation

results are presented in Tables 2.8 through 2.12 and in Figures 2.14 through 2.18.

44

TABLE 2.8. Maximum deflection of the free edge for k; =1 (pinned supports)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

v(o.5; 1)x1o3
21 - b Decomp
— _ Timoshenko , ' o e (%)
a solution FEM 5V“) “fin“ 5”” DMI FEM
0.5 7.1 7.09 -O.14 717 1.02 1.11
0.67 9.68 9.68 0.00 9.79 1.13 1.14
0.71 10.23 10.28 0.49 10.40 1.65 1.11
0.77 10.92 10.91 -0.09 11.03 0.98 1.03
0.83 11.58 11.56 -0.17 11.67 0.75 0.93
0.91 12.32 12.21 -0.89 12.30 -0.13 0.80
1.0 12.86 12.85 -0.08 12.93 0.51 0.59
1.1 13.41 13.39 -0.15 13.46 0.38 0.51
1.2 13.84 13.82 -0.14 13.88 0.25 0.40
1.3 14.17 14.15 -0.15 14.20 0.19 0.34
1.4 14.42 14.40 -O.13 14.45 0.19 0.33
1.5 14.62 14.59 -0.18 14.65 0.18 0.37
2.0 15.07 15.05 -O.14 15.24 1.12 1.25
l 16
1
1 151
i 141
' g 131
O
i 12 1
lb: 11 .
9; 1o-
1 91 r
1 8 . 1
* 7 . . a . . , . . . . . . . . . ‘
l 0,5 0,6 0.7 0.8 0,9 1 1.1 1.2 1.3 1.4 1,5 1,6 1,7 1,8 1.9 2
1 Aspect ratio
'1 1-*— ~111119951318974311111011 4:? 175111 "1-1' 9.93559981131714611199 ,

 

 

Figure 2.14. Maximum deflection of the free edge for k, =1 (pinned supports)

45

TABLE 2.9. Deflection at the middle of the plate for k; =1 (pinned supports)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

140.5; 0.5)x1o3
x1 = P— Decomp. 61%)
a FEM ”33°“ DM vs. FEM
0.5 3.81 3.87 1.57
0.67 5.43 5.53 1.77
0.71 5.82 5.96 2.43
0.77 6.31 6.42 1.74
0.83 6.81 6.92 1.70
0.91 7.35 7.47 1.64
1.0 7.94 8.06 1.57
1.1 8.51 8.64 1.52
1.2 9.02 9.15 1.50
1.3 9.47 9.61 1.49
1.4 9.87 10.02 1.50
1.5 10.22 10.38 1.53
2.0 11.50 11.69 1.65

 

 

 

 

 

 

 

      

 

 

 

1 12
3 ,,,,,, F
1 10«
; o
l 8 91
1 2 8~
In
1 o' 7 "
1 to 6*
5 2 5«
. >
1 4
1 31
1 2 r r f 1 r r 1 1 I . r . r 7
1 0,5 0.6 0,7 0,8 0,9 1 1,1 1,2 1,3 1,4 1,5 1,6 1,7 1.8 1,9 2
1 AspectRatio
i f_._" Femffiecbfib‘pdsitisnhemod

 

‘L- __ ,A_ _H

l
_4

 

 

 

Figure 2.15. Deflection at the middle of the plate for k, =1 (pinned supports)

46

 

TABLE 2.10. Maximum bending moment for k; =1 (pinned supports)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

. M (0.5; 1)x102
A = 2 Tlmoshenko Decomp. £(°/)
a solution megaod o
0.5 6.0 6.25 4.16
0.67 8.3 8.53 2.75
0.71 8.8 9.06 2.94
0.77 9.4 9.61 2.20
0.83 10.0 10.16 1.63
0.91 10.7 10.72 0.18
1.0 11.2 11.26 0.54
1.1 11.7 11.73 0.23
1.2 12.1 12.09 -0.10
1.3 12.4 12.37 -0.26
1.4 12.6 12.59 -0.11
1.5 12.8 12.76 -0.31
2.0 13.2 13.27 0.53

 

 

 

15-

10~

M (0.5; 1)x100

1

 

 

 

5

Aspect Ratio

 

 

0.5 0,7 0,9 1,1 1,3 1,5 1,7 1.9 2,1 2,3 2,5 2,7 2,9

 

1+. 7; Ti‘moshenko Method - - - Decomposition M51568",

Figure 2.16. Maximum bending moment for k; =1 (pinned supports)

47

TABLE 2.11. Maximum deflection of the free edge for k, =0 (fixed supports)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

140.5; 1)x1o‘

2:—

Decomp. £(%)

FEM method DM vs FEM

0.6 22.24 22.10 0.63
0.7 24.88 24.65 0.92
0.8 26.44 26.16 1.06
0.9 27.27 26.97 1.10
1.0 27.66 27.33 1.19
1.25 27.78 27.34 1.58
1.5 27.55 27.06 1.78

1777777— —_W7__ —_4771

1 c . 1

1 Q. _________ A- ..... fi'

1 ."'A
o E
O 1
O
O 1
i 1
lb: 1

1 e 1

1 > 1

1

1

1

1 0,6 0.7 0.8 0,9 1 1,1 1,2 1,3 1,4 1,5 1

1 Aspect Ratio '

1 1:+hf;FE'7L-t'999219351102119539.

Figure 2.17. Maximum deflection of the free edge for k; =0 (fixed supports)

48

TABLE 2.12. Deflection at the middle of the plate for k; =0 (fixed supports)

 

 

 

 

 

 

 

 

 

v(0.5; 0.5)x104
x1 = 2 Decomp. 61%)
a FEM "1°33.“ 8 (%) 0M vs FEM
0.6 10.89 10.90 15.5 0.16
0.7 13.34 13.27 16.5 -0.48
0.8 15.48 15.35 17.0 -0.85
0.9 17.35 17.19 17.8 -0.95
1.0 18.96 18.82 18.2 -0.76
1.25 22.01 22.14 17.7 0.59
1.5 25.32 24.53 15.4 -3.12

 

 

 

 

 

 

 

 

 

 

 

 

 

1+ M: {EM +39:- 98903900714300

1 o 1
o 1
1 o
O
1 E
1 It!
O
1 m“
9.
1 >
1 1o. . . . . . . . . .
1 0,6 0,7 0,8 0,9 1 1,1 1,2 1,3 1,4 1,5;
1 AspectRatlo 5
1
1
1
1

Figure 2.18. Deflection at the middle of the plate for k; =0 (fixed supports)

Percentage errors are reported in the tables for comparisons between the
decomposition method and the Timoshenko solution, the FEM and the Timoshenko

solution, and the decomposition method and the FEM. The results clearly demonstrate

49

that the collocation form of the decomposition method yields accurate analytical
dependencies for displacements and force responses. The proposed technique is easy for
computer implementation. Depending on the point where the solution is sought one can

use a different distribution and density of collocation points.

2.5 Transverse Free Vibration of a Plate with Elastic Supports

Consider an isotropic elastic rectangular plate of constant thickness shown in

Figure 2.1. The differential equation governing the transverse free vibration of the plate is

64w+ 64w +64w_pha)2
6x4 6x26y2 ay“ D

 

(2.67)

where a) = circular natural frequency of free vibration, w = transverse displacement, p =

mass density, h = plate thickness, and D = flexural rigidity of the plate.

Using the dimensionless stiffness coefficients of elastic supports (2.11); the

boundary conditions can be written in the form:

62w 6w
w=0, klay—(l—k1)-éx—=O (x=0)

62w 6w
W=O, kzagxi+(l—k2)a=0 (x=a)
(2.68)

82w 6w
w=0, k3bgf—(l_k3)3y_=0 (y=0)

2
w=O, k4b9—§-+(1-k4)§3=0 (y=b)
6y 5y

50

The homogeneous equation (2.67) and the homogeneous boundary conditions

(2.68) represent an eigenvalue problem, and the lowest value of a) which provides a non-

trivial solution is the fundamental natural frequency of the plate.

Using the decomposition method, the solution w is sought in terms of three

components w, w, and W3 that constitute the unknowns in three auxiliary problems.

The first problem is

64w

6 4' = f1(x,y)
x

 

subject to the boundary conditions

 

 

2

W1=0, klaawl-(l—k1)§fl=0 (x=0)
6x2 6x
52W] 6W]

w1=O, kza +(1—k2)—=0 (x=a)
6x2 fix

The second auxiliary problem is

64W2 _

4 f2(x,y)

 

dy

subject to the boundary conditions

 

 

2

W2:o, k3ba‘22—(1-k3>9-Wl=o 0:0)
ay ay
2

W2:0, k4ba 22+(1—k4)-a—-w-2—=0 (y=b)
6y 0y

The third auxiliary problem is

51

(2.69)

(2.70)

(2.71)

(2.72)

a4 2
CI>(a..6’)22 “’3 —””“’ W3+f1(X,y)+f2(x,y)=0 (2.73)
axzay2 D

 

where CD(a, ,8) is the discrepancy function.

To find the approximate solution to the original problem, it is assumed that
W1: W2 = W3 (2-74)

and the functions f1(x, y) and f2(x, y) are

fl(x9y) =fl(y)

(2.75)
f2(x,y) = f2(x)
Integrating (2.69) four times and satisfying the boundary conditions (2.70) yields
f1(y) 4 2 3
w =——— 3x ———————- 2R l+2k + l—k k 1+5k ax
1 72 R12(1+k2) 1 12( 2) ( 12)( 2)]
(2.76)

+§£1—+—5£2—Z[(1 —k1)a2 x2 +2k1a3 x1}
R12

where R12 =1 + 3(k1 + k2) + 5k1k2. Similarly, the solution to the second auxiliary problem

(2.71) and (2.72) is

 

f2(X) {1V4 2 -[2R34(1+2k4)+(l—k3k4)(1+5k4)]by3

72 " R34(1+ k4)
(2.77)

+flg—Sk‘Q10—k3w2 y2 +2k3b3 121}
34

where R34 = 1 + 3(k3 + k4) + 5k3k4.

To satisfy (2.74), fi(y) and f2(x) must be proportional to the quantities in braces in

(2.76) and (2.77), respectively, and hence

52

w] = W2 = imxwo) (2.78)

 

72
where
(0(X)= 3x4 --———£—— '[2R12(1+ 21(2) ‘1‘ (l — k1k2)(1+ 51(2) ]ax3
R12(1+ k2)
(2.79)
+Mkl -k1)a2 x2 + 2k1a3 X]
R12
(My) = 3y4 - —-—-2--—- ° [21334 (1 + 2k4 ) + (1 - k3k4)(1+ 51(4)]by3
R34 (1 + 1‘4)
(2.80)
+M10—k3wz y2 + 2k3 b3 y]
R34
Setting W3 = w] and using the Bubnov-Galerkin method with (2.73) yields:
ba 64W1 ,0th
”2 2 2— D w1+f](y)+f2(x)w1dxdy=0 (2.81)
00 5x 5y

Integrating (2.81) yields the expression for square of the first natural frequency of the

plate

,0th = 504 Ran + 288 H12 H34 +504 R34G34
D a4 1’12 azb2 F12F34 b4 F34

 

, (2.82)

where R12,R34 , G12 , G34 , H12 , H34 , F12 , and F34 have the same values as in (2.31).
Formula (2.82) is symmetric with respect to coefficients k, and the side lengths a and b,

and yields the following dimensionless parameter proportional to the lowest natural

frequency:

P = (0521183 (2.83)
D

53

Expressions for P obtained for the six possible combinations of fixed (k = O) and pinned
(k = 1) supports for the plate are given in Table 2.13. Comparison of the results with
exact solutions (Umansky 1973) shows that the error in (2.83) does not exceed 0.5%, and

confirms the high accuracy of the decomposition method.

TABLE 2.13. Expressions for P for different types of supports

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2 .
P = wb 1/ph D Maxumum Error
Boundary /
Conditions
Decomposition Method Exact Solution 51%) Location
b
I“ _ ‘1 2 2 0 1 All
a] 1 9880 + 4 ) 9.87(l + 2 ) . values
_ B _
a: 1 9.881/1 + 2.3022 + 2.45214 9.87\/ 1 + 2.3312 + 2.4424 0-1 0
b
— — —|
a? I 154511 +1.08/12 + 24 15.4211 + 1.1222 + 24 0-5 l
b 1’
a; '—*—* 2
a 2 ; 9.881fi+ 2.4322 + 5.1724 9.871/1 + 2.4922 + 5.1424 0-2 0.57
//////////1////4 2 4
ai , 22.451/1+0.542 +0472 22.37% + 0.57/12 +0.47% 0.4 1.21
’47////////7/”///
b
3m 22.451/1+ 0.5722 + .14 22.371/1+ 0.6112 + 24 0-4 1
Notations: x1 = b/ a , W fixed support, —— - — pinned support

54

CHAPTER 3

STATIC AND DYNAMIC ANALYSIS OF SINGLE LAYER LATTICE PLATES:
CONTINUUM MODELING
AND SOLUTION BY THE DECOMPOSITION METHOD

3.1 Calculation Model of Single Layer Lattice Plate

In order to compute the responses of single layer lattice plates, a continuum model
based on the theory of lattice plates and shells by Pshenichnov (1993) is used.
Constitutive equations of the model that are based on the lattice structure and material are
obtained by relating force and deformation characteristics of the rods that constitute the
lattice plate to those of the continuum model. In this work all members of the lattice plate
are assumed to lie in a single plane. While a continuum model can be developed for
lattice plates constructed of 3-D trusses, such a model would need to include shear
deformations. Flexibility of joints in the lattice plate is neglected. Joint flexibility can be
approximately accounted for in a continuum model by a shear deformable plate. This is

beyond the scope of this work.

Figure 3.1 illustrates a single layer lattice plate with n families of rods. The
position of the axes of the it]? family of rods (1 _<_ i S n) is characterized by the angle (01

measured from the x—axis to the y—axis.

A rod’s deformation is assumed to be equal to that of the mid-surface in the

continuum model. Using the transformation relations for the components of deformation

55

in the theory of elasticity, the following expressions are obtained for the components of

deformation of the axis of the i ’h family of rods:
* 2 2 . t
K,- = K101 +K25i + K12 sm 2gp, , 6,- = s,- c,- (K2 —K1)+ K12 cosZgoi (3.1)
._ - _ _ 2 2 __ 2 2
where s,- -srn(o,-, c,- —cosqo,-, K] —-6 w/ax , K2 ——6 w/ay ,

and K12 = —62w/(6xay) are bending and twisting curvatures of the plate’s mid-surface,

and K; and 6; are the curvature and twist angle of the i (h family of rods.

 

 

 

 

 

T"n

x l 1

Figure 3.1 Lattice plate with elastic supports

 

56

The positive directions of internal forces and bending and twisting moments in a rod are

shown in Fig. 3.2. Their dependence on the deformation components is assumed to be
It It
Mi = —Ei1iKia Ti = _GiJigi , Q1: ’ViMi (3-2)

a . . . .
where V ,- = c,- a + s,— a 18 a l1near differential operator, E ,- = Young’s modulus,

G = shear modulus, 1,. = moment of inertia, and J ,- = torsion moment of inertia.

 

Figure 3.2 Internal forces and moments in a rod

m2
ql t, In]

Figure 3.3 Distributed internal forces and moments in the continuum model

Assuming that the rod’s forces and moments are distributed continuously across
the continuum model’s cross section, the following expressions are obtained for the

forces and moments in the continuum model shown in Figure 3.3:

57

€11: i(QiCi)/aia (12 = fXQiSfl/ai
i=1 i=1

n ’7
m1: 2(Mici2 +Tisici)/aia m2 = 2(Mi512 —TiSiC,° )/ai (3-3)

i=1 (:1

n n
’1=-Z(MiSiCi—Ti0i2)/aia ’2 =‘Z(Mi3ici +TiSi2)/ai

i=1 i=1

where a,- = distance between the axes of the ith family of rods. By substituting (3.2) into

(3.3) and taking (3.1) into account, the following constitutive equations are obtained for

the continuum model:

n l
m] : ZFCiVi(Ei1i ciVi —GiJi SiAi)w’
i=1 i

n 1
m2 = Z—a—Sivi(Ei1iSiVi "GiJi CiAi)W

.21 .

I I (3.4)
n l '
’1=‘Z—CiVi(Ei1iSiVi-GiJiCiAiWs

i=lai

n l
’2 = “z";sivi(Eili CiVi +GiJi SiAi)W
i=1 i

6

where A,- = s,- 6— -—c,- a is a linear differential operator orthogonal to the axes of the
x

i (h family of rods.
Equilibrium equations for an element of the plate has the form

afl+§ql+2:0’ __—_—q2:0, at—z—fafl—q120 (3.5)
(3x 5y 5x 5y 5y 5x

Using (3.4) in the last two equations of (3.5) yields

58

n l
q] = —Z_V12(Ei1i ciVi +GiJi SiAi)w’

i=lai

(3.6)
n 1 2
‘12 = “'22—‘71“ (5111' SiVi *GiJi ciAi)W
i=1 1'
while the first equation in (3.5) yields
L(w) — z = 0 (3.7)
where the linear differential operator L has the form
_ " 1 2 2 2
L(W)— Z—V1(Ei1ivi +GiJi Ai)W (3.8)

i=1“:

3.2 Solution of Bending Problem for Lattice Plate with Elastic Supports

Consider a plate with the lattice type shown in Figure 3.4(a). It consists of four
families of rods (n = 4) and the rods of the first and second families are identical. For this

specific case

¢1=¢, ¢2=-(o, (03:7r/2, go4=0
“1:612:61, a=Za3s=2a4c (39)

E111: E212 = E1, G1J1= 02-12 = GJ
With respect to (3.9), the constitutive equations can be written as

m1=fl11K1+fllzK2a m2=fl12K1+522Kza
(3.10)
t1=1331K12, t2:,341K12

where the coefficients ,6 depend upon geometrical and physical characteristics of the

lattice:

59

,6“: —2(E1c4 + E4I4c+GJs202)/a

fl12=—2s2c2(E1—GJ)/a

fizz: —2(EIS4 +E313s+GJs2c2)/a (3,11)
fl31=(EIsin2 2(0 + ZGJc2c052rp + 204.14 c)/a

,841=(EIsin2 2p - ZGJ 32c052q) + ZG3J3 s)/a

Note that the parameters without subscripts refer to the first two families of rods.

i=3 9:

 

 

 

 

 

 

 

 

 

 

x l
(a) Typel

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4 | , .1) ~ .,
d M d d \ d
x x l x l

(c) Type 3 ((1) Type 4 (e) Type 5

Figure 3.4 Types of grids for lattice plates

60

Using (3.7) and (3.8) yields the bending equation for the continuum model as

4 4 4
016—43+Dz—i—w—2—+D3§¥=LZ (3.12)
6x 6x 5y 5y 2E1

where

D, = c4 + 75262 + g4c

D2 = 6c232 +y(s4 —4c2s2 +c4)+73s+74c

D3 = s4 +76232 +g3s (3.13)

Bending equations for plates with other types of lattice geometries shown in
Figures 3.4(b) to 3.4(e) may be obtained from (3.12).and (3.13) by considering the terms

of coefficients corresponding to the family of rods that are not present to be zero.

Since the plates are assumed to have elastic supports, (3.12) must be solved under

the following boundary conditions:
w=O, M1 zirlgfl (x=iA)
6x
w=0, M2 =ir29w- (y=:tB)
5y

Here r; and r2 are the stiffness per unit length of the distributed rotational springs along

the supports. Using (3.10) and (3.11), these boundary conditions may be expressed as

62w aw
w=0, —ir—=0 xziA 3.14a
flit 6x2 lax ( ) ( )

61

2
w=0, fizzé—g’irzia‘iw (y=iB) (3.14b)
ay 6y
The following notations are introduced to reduce the problem to non-dimensional

form:

a=X/A, fl2y/B, II=B/A, ”IZDZ/Dl’ 772=D3/D1

2E1 D
9 fZ(a9fl)=Z/p2 a V=W——4——1—
aA pZ

 

p2 = male

where fz (a,fl)= dimensionless load function, and V: dimensionless deflection
function.
Non-dimensional forms of the stiffness coefficients of the elastic supports are

takenas k1=fl11/(fl11+r1A) ande =fl22/(fl22 +rzB),Wh€l‘€ 0 5 [61,2 S1.

The dimensionless form of (3. 12) is

 

 

4 4 4
av+m 6v +5730.)

 

—— = fz(a,fl) (3.15)
aa“ 22 6a2 6,62 .14 6,64
to be solved under the boundary conditions
2
v=0, k1a ”:(1—k1)—a—"—=o (a=i1) (3.16a)
6a2 6a
2
v=O, kZ—Q—g-ia—kflialw (5::1) (3.16b)
an (M

The boundary value problem characterized by (3.15) and (3.16) is solved by the
decomposition method as described in Section 2.3. The interconnection equation of the

problem is solved by the multiple-point collocation method. The dimensionless

62

displacement is computed using (2.43), from which the actual displacement is recovered.
The internal forces in the rods are finally computed using (3.1) and (3.2). The

displacement and internal forces in the rods are given by

4
w=_A__a£Z_v (3.17)
2191151
2
z- .Lflx’.’ (3,13)

r g! 2D] I

AzaPz

 

Ti:7i 2D1 91' (3-19)
Q,- =— fl3+3}?— M,- (3.20)
A 6a B 6,6

Numerical example

Consider the rectangular lattice plate shown in Fig. 3.5 which is uniformly loaded
at the joints with the loads P), = l N. The lattice consists of two orthogonal families of
rods with the following characteristics: I; = 0.1 m, 12 = 0.15 m, E11 = E12 = 105 N.m2. For
simplicity and without any loss of generalization, torsional rigidities were assumed to be
zero, i.e. GJkJ = GJm = 0. Deflections and bending moments are calculated at the four
locations indicated in Fig. 6 for pinned and fixed supports. Table 3.1 shows the results
obtained using the finite element method (FEM) and the decomposition method (DM). In
the FEM, each rod was taken as a separate finite element.

The example demonstrates that the continuum model, together with the
decomposition method, yields an accuracy of within 2% for displacements and bending

moments, which is adequate for preliminary design and optimization purposes.

63

 

 

 

I I «
III/IIIII/IIIII
I II

 

 

  

 

 

 

 

.—

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

E
2
°1
0.]
/ I /_n _ __ -« n;-
fig __64 x (_1_.1 m- g _ _l
Figure 3.5. Plate with an orthogonal lattice
TABLE 3.1. Results for k1= k2 = l (pinned supports)
3 --1
Location w (x 10 m) M (x 10 N.m)
"Mb" FEM DM aw.) FEM DM s(%)
O 2.29 2.30 0.1 4.96 5.01 1.2
1 1.93 1.96 0.2 4.41 4.46 1.1
2 1.15 1.16 0.8 2.77 2.83 1.0
3 0 0 — O O —
TABLE 3.2. Results for k. = k; = 0 (fixed supports)
lLocation w (x 103 m) M (x 10‘1 N.m)
Numb" FEM DM 8 (%) FEM DM 5 (%)
0 0.555 0.560 0.9 1.40 1.42 1.4
1 0.407 0.415 1.9 1.98 1.99 0.5
2 0.148 0.149 0.7 -0.265 -O.266 1.0
3 0 0 - -3.12 -3. 125 0.4

 

 

 

 

 

 

 

 

64

 

 

3.3 Comparative Analysis for Different Types of Lattices

In this section the bending problem of a lattice plate is analyzed for the different
types of lattices shown in Figure 3.4, and different values of support rigidities. The rod’s

material volume per unit area of the middle surface of the plate is fixed, i.e.,

F
fi+fl+~3~+Iji=t7=const (3-21)
(11 02 a3 a4

where P} = the cross-section area of a rod in the i'h family. The rods of all four families

are assumed to be made of the same material and to have the same cross-section
F] = F2 = F3 = F4 = F (3.22)

With respect to (3.22) and notations (3.9), the condition (3.21) can be written as

2—F(61+63s+§4c) (3.23)
a

a =
Coefficients 61, 63, and 64 take the values of 1 if the corresponding family of
rods is present or 0 if the corresponding family of rods is absent. The coefficient of

torsional rigidity of the 1"” family of rods is defined as 7,- = 6,- 7 , where 7 is determined

by (3.13). The formulae for the coefficients 771 and 772 of the bending equation become

6615in2 (pc05(p+61y(cos—1(19-6sin2 $cosgo+§3tggp+ 54) (3 24)
771= '
61c053 (0+64 +617 sin2 (pcosgo

 

 

- 3 - 2
2 : sm ¢Ig¢+63tggo+§1ysm (pcosgo (3.25)

dlcos3 ¢2+64 +617 sin2 (pcosw

65

Taking the angle (a as the controlling parameter, using (3.23) the other variable a

can be expressed in terms of e2 since the material volume is kept constant:

a = 2(61+53S+54C)F/U
or (3.26)

2F

a=a0(61+53s+64c), where a0=—:—
0

Substituting this value into (3.17) and (3.18) and assuming a uniform transverse load, the

equations for deflection and bending moments can be written as

 

 

4
w: wo m (3.25)

E]
M,- = — M? .42an (3.26)

where

W0 =V (61+53S-I-64C) (327)

2121

.1 6 +6 s+6 c

M194.- g1-(1 3 4) (3.28)

2121

The dimensionless displacement function is determined by (2.43) and (2.44), and the
curvature K; is found from (3.1). The dimensionless coefficient D1 is given in (3.13).

The a0 = F / 6 ratio is constant for the particular problem and is known from the

problem definition.

The technique described above was implemented into the PLAST computer

program for analyzing rectangular lattice plates with different types of lattices and

66

different values of support rigidities. PLAST is written in the C programming language
and can be used on personal computers. The listing of the program is given in Appendix

B. Some of the results obtained using the program are shown in graphical form below.

Figures 3.6 through 3.11 represent the dependencies of maximum values wgnax

and M ,0 max on the angle (0 for the values k, =0, 1, and 0.5. The three curves on the

graphs correspond to three different types of lattices shown in Figure 3.4. The graphs
identify the optimal lattice type and governing angle value for each type of support
condition. Figures 3.8 and 3.9 show that for pinned supports the optimal lattice type that
minimizes moments and deflections is rhombic (Type 2) and the optimal value of the
angle is (p = 450. For fixed supports (Figures 3.6 and 3.7) the rhombic lattice with a 450
angle yields smaller moments but at the same time gives larger deflections than lattices
with the governing angle close to O0 or 90°. For partially restrained supports (Figures 3.10

and 3.11) the rhombic lattice is also optimal.

Figure 3.12 shows examples of similar dependencies of wgm on angle (0 for

different combinations of support rigidities.

67

 

 

 

 

 

 

 

 

 

 

 

g 21
2
j; 20 4 ’
°.
3 l
x 19 4
N
E
3 1e 4
17 ‘ 1 Y T i T r f fl r 1 r r f r
0 51015202530354045505560657075808590q
Governing Angle ;
f —°—Q __Mjétfigéfiié 1— 3- Laws? Ups 2 - .‘ -" La'itiééLTvai-égi ‘
Figure 3.6. Maximum deflection at the middle of the plate
for k1= k2 = 0 (fixed supports)
26 ~
c, 24-
O
3 224
S; 20 -
g; 18 «
§1G« .___.._._' .___._._._ 1
14 i k‘x... .r.’ I“
12- ..4f'.'."'......
0 51015202530354045505560657075808590
Governing Angle

 

1+: i 1442123: 1'.- $949332? 511*"? L‘éttiézifié

Figure 3.7.

 

Maximum moment at the middle of the plate

for k1= k2 = 0 (fixed supports)

68

 

W max (0.0) x 1000

M max (0.0) x100

80

 

 

 

78~
764
741
72* ,-a——**'——t—-__
’,e—- e.‘

70'l ’r ‘k

I \\
681 ax; . ¢ + ¢ . N
66‘ F.~.‘- ...r.-—'-
644 "~--—-.--—.-._..—--—-"
62+
60 r r T T r r Y 1 7 t T

 

 

 

Figure 3.8.

18

0 51015202530

35 40 45 50 55 60 65 70 75 80 85 90 '
Governing Angle

_+Léttis?1i9s 1“- :ffittiéefibe , -

Maximum deflection at the middle of the plate

for k1= k2 = l (pinned supports)

 

17-

16‘

15~

14

 

13

 

 

 

Figure 3.9.

0 5101520253035

T T T i T r ‘T i T

40 45 50 55 60 65 70 75 80 85 90 i
Governing Angle ‘

 

—_°— + 3 "12341251212“ ' - *3 93103096 2* - 1‘7 Patties We 3 5

Maximum moment at the middle of the plate

for k1= k2 = l (pinned supports)

69

 

 

 

W max (0,0) x 1000

 

 

 

46 l T ‘ i T i T i l’ l T T T T 1
0 5 10152025 30 354045 50 556065 70 75 80 85 90
Governing Angle

 

i+£afiicéfib€ 1 :7?- 1:5; _1— +5— aWs

Figure 3.10. Maximum deflection at the middle of the plate

 

 

 

for kl=k2=0.5

14

8 e- t-
r“— ‘~o—-__ _._-r” ~‘t.

313‘ r,’ ‘— \\‘
O
O
X o..,_. ...,.-
212‘ F.~'."—-I— _.._._. I—--"'"'r
E

11 T T i l T T T

 

0 51015202530354045505560657075808590
Governing Angle

 

lat-+— Latticefijl'ype 1 — O— - Lattice TypeZ *3— + -'Laittice Type 3 1

Figure 3.11. Maximum moment at the middle of the plate

for k1= k2 = 0.5

70

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1 k1 =0. k2 = 0.5
1 0,05 «
1 52
9..
X
N
E 1
1 g 1
1 '1
I 1 . ‘ 1 . 1
5 15 25 35 45 55 65 75 85 1
1
1 Governing Angle
1
7—0; 1511151111213: 5314-va? 9 31.111... fvpé? '
oEéFT‘C'TT'T'TTF
1 0,05 1
' 8
1 92 0.041 1
' x
G
l E 0.03 4 1
1 3 1
1 0.02
l 0.01 . . . . . . .
1 5 15 25 35 45 55 65 75 85 1
1 Governing Angle 1
1 % .--, _._ -_ ~_._i. 4*-.. q— 1
1 1—o—Lattice Type1 —----Lattice Type2 ——.—-Lattice Type3 l *
1 no _ ,_ +2 2___ fl
‘1 1
1 a: l
1 e -
1 g 1
l E '1
3
1
1
1 0,02 . . . 1
1 5 15 25 35 45 55 65 75 85

1 Governing Angle
1 1

T;._ Lattice Type 1 - - .- - - Lattice‘Type 23— a- - Lattice Type 3 1 1

 

 

 

Figure 3.12. Maximum deflection at the middle of the plate
for different combinations of support rigidities

71

3.4 Transverse Free Vibration of Lattice Plate with Elastic Supports

Consider the plate shown in Figure 3.13 with non-symmetric elastic supports. The
lattice consists of four families of rods (n = 4), as shown in Figure 3.4(a), and the cross-
section of the rods in all the families are identical. The continuum model described in

Section 3.1 is used for this problem.

 

 

 

 

 

Figure 3.13 Lattice plate with non-symmetric elastic supports

The differential equation governing the transverse free vibration of the plate can
be obtained from the bending equation (3.12) by substituting the external load Z by the

inertia forces due to its movement:

 

(3.29)

where p,- is the density of material of the 1'” family of rods. For the case of steady

harmonic free vibration at the frequency a)

72

2
n O O
Z=wzw2p'F’=20) pF

i=1 01'

(1+sin(o+cosgo) (3.30)

 

Substituting this value into the right-hand side of (3.12), the governing equation becomes

4 4 4
Dl§T+D2—iw—2+D3-a—hw—=—£2-p—(l+singo+cos¢) (3.31)
6x 6x 6y2 6y Er2 cosgo

where r is the radius of gyration of the rods, and the coefficients D1, D2 , and D3 are

defined by (3.13). The boundary conditions for the problem using the non-dimensional

coefficients of support rigidity are

62W

w=k1A—————(1—k1—aw=,(x=0) (3.32a)
6x2 6x
62w aw

w: sz-—--(1— k2)— =0, (x=A) (3.32b)
6x2 6x
2

W: k3Ba——2-—(1—k3)§—w= 0, (y=o) (3.320)
6y2 6y
2

w: k4Ba—2——(1-k4)§—w=0, (y=B) (3.32d)
6y 6y

The three auxiliary problems introduced in the decomposition method have the

form:

 

=fibd) 03»

subject to the boundary conditions (3.32a,b);

73

 

2. D2 4 =f2(x,y) (3.34)

subject to the boundary conditions (3.32c,d); and

64W3 _ (02,0

3. D3
6x2 ayz Erzcosq)

(1+sin<o+cosco>=—f1(x.y)-f2(x,y) (3.35)

 

The approximating functions are taken in the form

f1(x,y) = W10) f2(x,J’)= W2(x)

From the solution of the first two boundary value problems and satisfying

condition w] = wz

 

 

W1=W2 =11x1(y)1u2(x) (3.36)
where
1 y(y-B) ‘ 2
= 1 3k 3k 5k k -
WU) 2402X(1+3k3+3k4+5k3k4)x{y ("L 3+ 4+ 3 4)
yB(1+k3 +5k4 +5k3k4)—282(k3 +5k3k4)]
(3.37)
W2(X) 1 x(x—A) [x2(1+3k1+3k2 +5klk2)-

: X X
24 D, (1+ 3k] + 3k2 + Sklkz)
xA(1+k1 +512 +5klk2)—2A2(kl 4.5/11112)]

The Bubnov-Galerkin method is used for solving the third auxiliary problem,

assuming that W3 a W] :

BA 54W1 (02p 1
H D3a—26—3+w1(y)+1/12(x)—w1E—Z——(1+sm(p+cos¢) wldxdy=0 (3.38)
00 x y r cosp

74

The expression for (02 is determined from (3.38). For k,- =1 (pinned support):

(02 __ 108 x (86801 A“ +86703 A282 +868DZB‘)Er2 cosgo

 

___ 3.39
961 ,0A"B4 (1+sin(p+c08(p) ( )
For k,- = 0 (fixed support):
4 2 2 4 2
(02 =72(7D,A +2D3A B +7023 )Er cosgo (3.40)

 

,oA‘tB4 (1+ singo+ comp)

Expressions for 02 determined for six different types of support combinations

are summarized in Table (3.3).

Numerical examples

Consider the rectangular lattice plate shown in Figure 3.13. The lattice consists of
four families of rods (Type 1 in Figure 3.4). The rods are standard steel tubes. The values
of the first frequency of free vibration a) are calculated using the decomposition method
and compared with the solutions obtained by the FEM using the LIRA software (Kiev

2000)

Tables 3.4 -— 3.9 and Figures 3.14 — 3.19 demonstrate the results obtained for

lattice plates with pinned and fixed supports and the values of the governing angle of the

lattice (o = 450, 300, and 600. The longer side of the plate was taken as b=5 m, and the

step of the lattice along side b was taken 1 m.

Tables 3.10 - 3.15 and Figures 3.20 -— 3.25 show the results obtained using

expressions (3.41) — (3.46) for lattice plates with different combinations of supports and

75

different aspect ratios. The results presented correspond to the case when b=10 m, and

(0:450.

The results demonstrate that the decomposition method yields sufficient accuracy

for preliminary design applications. The analytical dependencies obtained can be used for

optimization purposes.

TABLE 3.3. Main types of support conditions and corresponding expressions for

the square of the first frequency of free vibration

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

TYPO 0f 2
supports Expression for a)
1 3’4“”? (0, _2 (63D,A‘ +361), A282 +1331),B4)Er2 cow (3 41)
g,,,,,7,,;| 19 p A484 (1+ sin go + cos (a) 0
1 //////1
I‘ 1
2 1 , __ 72 (13301.44 +36D3A282 +63DZB4)Erzcosg/) 342
2 fl 9 a) "T9— A‘B4(1 ' ) (' )
2////////l p + sm ¢ + cos (0
3 g— “ ‘1 a), _ 648 (133D,A‘ +72133 A232 +133D,B4)Er2 cosqp (3 43)
3,,,,,,, I 361 pA‘B4(1+singo+cosgo) '
4 g : w, __ 216 (266D,A‘ +3061), A282 +6511),13“)Er2 cosgo (3 44)
2 l 589 pA“B4 (1 +singo+ cosgo) '
5 g 2 w, __73 (421),,44 +511)3 A232 +2171),13“)Er2 cosgo (3 45)
2 1, 2 31 pA"B4 (1+sin(p+cos¢)) '
a r 1' 7 2 216 (6SIDIA‘ +3061)3 A282 +2661),13“)15r2 cosgo
6 2 6 w = 1 . . (3.46)
/:——— —/ 589 pA B (1+sm(p+cosgo)
Notations: W fixed support, ——-—: pinned support

76

 

TABLE 3.4. Results for lattice plate with pinned supports

and governing angle of q) = 450

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

b Mass First frequency of free vibration a), {—1—]

,1 = — (metric sec
0 ton) FEM DM 1.1%)

1 1.147 85.64 87.17 1.8
1.2 1.370 72.42 72.77 0.5
1.4 1.593 64.44 64.36 -0.1
1.6 1.816 59.26 59.04 -0.4
1.8 2.039 55.70 55.46 -0.4
2 2.262 53.16 52.94 -0.4
3 3.378 47.14 47.02 -0.3
4 4.493 45.04 44.94 -0.2
5 5.608 44.08 43.95 -0.3

 

 

 

 

 

 

 

 

 

o
o
g ,
x 1
‘1? 1
I 1
> 1
o
r: 1
o
a 1
1 g 1
1 fl 1
1 40 1 1 1 1
1 1,0 2,0 3,0 4,0 5,0 11
Z Aspect Ratio 1
1 _'
1 1

 

1+: 75M -- 9 #,:1D¢cqm£°§iti:>2-Mét'76d 1

Figure 3.14. First frequency of free vibration of lattice plate with pinned supports

and governing angle of (p = 450

77

TABLE 3.5. Results for lattice plate with fixed supports

and governing angle of (p = 450

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

b Mass First frequency of free vibration a), [L]
,1 = — (metric sec
a ton) FEM DM s(%)
1 1.147 163.5 164.2 0.4
1.2 1.370 139.2 138.2 -O.7
1.4 1.593 126.1 124.4 -1.3
1.6 1.816 118.4 116.4 ~1.6
1.8 2.039 113.4 111.4 -1.7
2 2.262 109.7 108.1 -1.4
3 3.378 103.7 101.1 -2.5
4 4.493 101.7 98.86 -2.8
5 5.608 100.8 97.77 -3.0
1_,7,-77,_____ __,___ 77,
1 170
1 g 160
1 E 150 4
1.7 140 1
E. 130 « 1
6' 1
§ 120 ‘ ‘1
6' 110 ‘ 1
IE 100 . — - — _ :QT _____ i? 1
1 so 1 - 1
1 1,0 2,0 3,0 4,0 5,0 1
1 Aspect Ratio
1
1

11+ " 755111 7 #3 :Qfiznpsvsifiemeth‘éd 'I

7. 7 7 7 7, ,7,, 77. 1 , 7777 77—.7— 7 77 7

Figure 3.15. First frequency of free vibration of lattice plate with fixed supports

and governing angle of (p = 450

78

TABLE 3.6. Results for lattice plate with pinned supports

and governing angle of «p = 300

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

b Mass First frequency of free vibration a), (L)
A = — (metric sec
a ton) FEM DM s(%)
1.04 0.845 80.83 77.38 -4.3
1.27 1.026 71.07 68.26 -4.0
1.38 1.116 67.57 65.44 -3.2
1.50 1.207 64.84 63.30 -2.4
1.61 1.297 62.91 61.61 -2.1
1.85 1.478 59.74 59.17 -1.0
2.08 1.658 57.99 57.55 -0.8
3.00 2.381 53.98 54.37 0.7
3.93 3.104 52.34 53.16 1.6
4.96 3.916 51.48 52.48 1.9
90
O
8
K
74‘
E.
3
5
3
5
IL

 

 

 

 

 

1,00 2,00 3,00 4,00 1
Aspect Ratio

+411": — @gaasfién

Figure 3.16. First frequency of free vibration of lattice plate

with pinned supports and governing angle of (p = 300

79

TABLE 3.7. Results for lattice plate with fixed supports

and governing angle of (p = 300

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

b Mass First frequency of free vibration w,(—1—)
l = — (metric sec
a ton) FEM DM £(%)
1.04 0.845 147.2 147.5 0.2
1.27 1.026 132.9 133.9 0.8
1.38 1.116 128.8 130.1 1.0
1.50 1.207 125.8 127.3 1.2
1.61 1.297 123.5 125.3 1.5
1.85 1.478 120.4 122.4 1.7
2.08 1.658 118.4 120.7 1.9
3.00 2.381 115.0 117.5 2.2
3.93 3.104 113.9 116.4 2.2
4.96 3.916 113.4 115.8 2.1
150
o 145
8
P 140
X
1; 135 4
5
5. 130
g 125
g 120 ~
“- 115
1 110 r 1 '
1 1,00 2,00 3,00 4,00 1
1 Aspect Ratio

 

——e_—FEM ‘- é — -03compositm1Method

1 _._ ___

Figure 3.17. First frequency of free vibration of lattice plate with fixed supports

and governing angle of (p = 300

80

TABLE 3.8. Results for lattice plate with pinned supports

and governing angle of q) = 600 .

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

b Mass First frequency of free vibration a), [-1—1
A = — (metric sec

a ton) FEM DM aw.)
1.04 0.435 81.61 82.94 1.6
1.38 0.569 61.07 59.88 -1.9
1.73 0.703 51.3 49.69 -3.1
2.08 0.838 45.9 44.35 -3.4
2.42 0.972 42.6 41.22 -3.2
2.77 1.106 40.46 39.27 -2.9
3.11 1.241 38.98 37.89 -2.8
3.46 1.375 37.91 37.01 -2.4
3.81 1.510 37.13 36.28 -2.3
4.16 1.644 36.53 35.76 -2.1
4.50 1.778 36.06 35.35 -2.0
4.85 1.913 35.69 34.91 -2.2
5.20 2.047 35.39 34.65 -2.1

 

 

 

 

 

 

Frequency (Hz) x 1000

 

 

 

 

 

 

1,00 2,00 3,00 4,00
Aspect Ratio

1+ FEM - é‘ibééqrfieésirfiéfibbsthod

Figure 3.18. First frequency of free vibration of lattice plate with pinned supports

and governing angle of (p = 600

81

TABLE 3.9. Results for lattice plate with fixed supports

and governing angle of q) = 600

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

b Mass First frequency of free vibration w,(—1—)
A=— (metric sec
a ton) FEM DM £(%)
1.04 0.435 135.96 142.35 4.7
1.38 0.569 103.72 108.56 4.7
1.73 0.703 90.41 94.42 4.4
2.08 0.838 84.15 87.59 4.1
2.42 0.972 80.82 83.74 3.6
2.77 1.106 78.89 81.58 3.4
3.11 1.241 77.87 79.99 3.0
3.46 1.375 76.86 78.85 2.6
3.81 1.510 76.29 77.99 2.2
4.16 1.644 75.88 77.35 1.9
4.50 1.778 75.57 76.86 1.7
4.85 1.913 75.34 76.45 1.5
5.20 2.047 75.15 76.14 1.3
1 150
1 8 140
‘ O
1 ; 130 1
1 §120«
; 110 «
1 §1oo1
1 g 90
s E 801
1 70 1 . .
1 1,00 2,00 3,00 4,00 5,00

1 Aspect Ratio

 

1 18+ _f‘ ._ FEM - a - 0600099911190 1??thon

 

 

Figure 3.19. First frequency of free vibration of lattice plate with fixed supports

and governing angle of q) = 600

82

TABLE 3.10. Results for lattice plate with supports of Type 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

obtained by (3.41)
b ”I“ frequency of free vibration a), (L)
A = _ sec

a FEM DM 61%)
1.0 33.92 34.24 0.9
1.1 31.79 31.94 0.5
1.2 30.25 30.31 0.2
1.3 29.11 29.09 .0.1
1.4 28.25 28.18 -02
1-5 27.57 27.47 -0.4
1.6 27.05 26.91 -05
1.7 26.62 26.46 -0.6
1.8 26.28 26.10 -0]
1.9 26.0 25,08 -03
2.0 25.76 25.54 -09
2.5 25.02 24.72 -12
3.0 24.66 24.29 -1 _5

 

 

 

 

 

 

 

 

 

1 ;

__“1

1,5

2,0

 

_i;—'—:___ - FEM -_ 9- P

ecompo

Aspect Ratio

2,5

91.192 Méthw i

 

Figure 3.20. First frequency of free vibration of lattice plate

with supports of Type 1 obtained by (3.41)

83

TABLE 3.11. Results for lattice plate with supports of Type 2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

obtained by (3.42)
First frequency of free vibration to, (i)
2, = _ sec

FEM DM s(%)
1.0 33.92 34.24 0.9
1.1 30.29 30.31 0.1
1.2 27.62 27.46 -0.6
1.3 25.61 25.34 -1.1
1.4 24.07 23.73 -1.4
1.5 22.87 22.50 -1.6
1.6 21.92 21.53 -1.8
1.7 21.16 20.75 -1.9
1.8 20.55 20.13 -2.0
1.9 20.04 19.63 -2.0
2.0 19.63 19.21 -2.1
2.5 18.33 17.92 -2.2
3.0 17.70 17.29 -2.3

 

 

 

 

 

Frequency (Hz) x 1000

 

 

 

 

 

 

Aspect Ratio

 

+ FENi - 43 j Decomposition Method

Figure 3.21. First frequency of free vibration of lattice plate

with supports of Type 2 obtained by (3.42)

84

TABLE 3.12. Results for lattice plate with supports of Type 3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  

 

 

 

 

obtained by (3.43)
1
b First frequency of free vibration to, [—j
,1 = _ sec
a FEM DM £(%)

1.0 29.02 28.87 -0.5

1.1 26.53 26.24 -1.1

1.2 24.69 24.32 -1.5

1.3 23.30 22.89 -1.8

1.4 22.23 21.80 -1.9

1.5 21.38 20.95 -2.0

1.6 20.71 20.28 -2.1

1.7 20.16 19.74 -2.1

1.8 19.72 19.30 -2.1

1.9 19.35 18.93 -2.2

2.0 19.04 18.62 -2.2

2.5 18.05 17.65 -2.2

3.0 17.54 17.14 -2.3

30
O
8
31 25 1
N
E
g
g 20 —
.1 ————————— 4
15 . 1 Y I ‘
1,0 1.5 2,0 2,5 3,0 1
Aspect Ratio

:8 —°—' 7 E14? 3179ifl°§1§®00d }

Figure 3.22. First frequency of free vibration of lattice plate

with supports of Type 3 obtained by (3.43)

85

TABLE 3.13. Results for lattice plate with supports of Type 4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

obtained by (3.44)
First frequency of free vibration (o, [—1-)
’1 = _ SCC
FEM DM s(%)

1.0 25.47 26.44 3.8
1.1 23.82 24.58 3.2
1.2 22.58 23.21 2.8
1.3 21.63 22.17 2.5
1.4 20.89 21.36 2.2
1.5 20.29 20.72 2.1
1.6 19.81 20.20 2.0
1.7 19.42 19.78 1.9
1.8 19.09 19.43 1.8
1.9 18.82 19.14 1.7
2.0 18.58 18.88 1.6
2.5 17.82 18.06 1.3
3.0 17.41 17.60 1.1
30

O

8

x 25

’N‘

15

>4

8

g 20 —

6'

2

I].
15

1,0 2,0 3,0
Aspect Ratio 1

+ FEM - a - Decomposition‘flejbcfii

 

 

Figure 3.23. First frequency of free vibration of lattice plate
with supports of Type 4 obtained by (3.44)

86

TABLE 3.14. Results for lattice plate with supports of Type 5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

obtained by (3.45)
b First frequency of free vibration a), (L)
A = _ sec
a FEM DM 31%)
1.0 30.94 31.06 0.4
1.1 29.56 29.59 0.1
1.2 28.55 28.52 -0.1
1.3 27.79 27.72 -0.3
1.4 27.20 27.09 -0.4
1.5 26.73 26.60 -0.5
1.6 26.36 26.20 -0.6
1.7 26.06 25.88 -0.7
1.8 25.81 25.61 -0.8
1.9 25.60 25.38 -0.9
2.0 25.43 25.18 -1.0
2.5 24.86 24.54 -1.3
3.0 24.56 24.18 -1.5
1 35
O
8
X
i?
5
8‘
5
3
g

 

 

 

 

1,0

1,5

 

 

Aspect Ratio

 

;F_EM -: ¥_gecomposifl_on 11191th1

17. __777 _—,7,7

 

Figure 3.24. First frequency of free vibration for lattice plate
with supports of Type 5 obtained by (3.45)

87

TABLE 3.15. Results for lattice plate with supports of Type 6

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

obtained by (3.46)
b First frequency of free vibration a), {—l—j
,1 = _ sec

0 FEM DM g(%)
1.0 30.94 31.87 3.0
1.1 27.00 27.68 2.5
1.2 24.03 24.56 2.2
1.3 21.78 22.17 1.8
1.4 20.02 20.32 1.5
1.5 18.62 18.86 1.3
1.6 17.49 17.68 1.1
1.7 16.58 16.71 0.8
1.8 15.83 15.92 0.6
1.9 15.20 15.26 0.4
2.0 14.68 14.72 0.3
2.5 13.02 13.02 0.0
3.0 12.19 12.19 0.0

 

 

 

 

 

 

 

 

 

 

 

 

O
' O
i 2
1 x
17
E.
>
O
c
0
3
a
2 3
u.
10 1 1
1,0 2,0
AspectRatio

 

1 27—7 ___— __..,7_7 7__-

1 tar—FEM -, *3 -. DeéemeositiorLMethod ,

Figure 3.25. First frequency of free vibration of lattice plate

with supports of Type 6 obtained by (3.46)

88

CHAPTER 4

STATIC AND DYNAMIC ANALYSIS OF SINGLE LAYER LATTICE PLATES
BY THE DECOMPOSITION METHOD
BASED ON FINITE DIFFERENCE DISCRETIZATION

Lattice plates with a regular rectangular grid as shown in Figure 4.1 are
investigated in this chapter. The governing equations for bending and free vibrations
stated in the finite difference formulation are obtained. The decomposition method is then

used to obtain solutions.

 

 

 

 

 

 

 

 

 

Figure 4.1 Lattice plate with regular rectangular grid

89

4.1 Notations and the Main Operators of the Finite Difference

Formulation

This chapter follows the notations introduced by A. Markov (1911) and used in

the works of Bleich and Melan (1936), and Ignatiev (1979).

Assume that (bx = f (xi) is a function for the discrete argument x defined in the

interval [a, b]. The discrete argument x takes the values
xi=x0+hi, i=0,i1,d:2,...in (4.1)
where x0 is a fixed number, and h > O is the step size. Without any loss of generality it

is assumed that x0 = O and h = 1. All functions (bx introduced in this work are assumed

to be single-valued, real, and bounded.

The main operator of the finite difference calculus is the difference operator of the

first order A (the forward difference operator) defined as
4.- =Af(x,-)=f(x.- +h)—f(x,-) (4.2)

or, in short notation

Ai=Afi=fi+l—fia 11:1 (43)

The backward difference operator is defined as

7574-41—11-], h=1 (4.4)

The forward and backward shift operators, E and E ‘1 , are defined as

Ef(x,-)= f(x,-+1), E"‘f(x.-)= f(x,-_1) (4.5)

or in short notation:

9O

13:1,“, E'1=1,-_1 (4.6)

Higher-order shift operators can be written as

E"f=1,-+k, E"‘I=1,-_k (4.7)
The following identities exist for the difference and shift operators:
E=1+A, E'lzl—V, Af,-=Vfi+1, Vf,=Af,._1 (4.8)

The higher-order differences may be expressed in terms of the following recurrent

dependencies:

__1 _1 _
A?=A" (AL-H" fin—A" 1f,-

(4-9)
V? = V"“(Vf,-) = V"“f.- — V"“f.--1
These differences can be expressed in terms of the values of the function as
n
A? = Z(-1)’C(n,r)f[i+(n—r)1
r=0 (4.10)

V? = A’f—n

(r)

r!

 

n . . . . .
where C(n,r)= are the b1nom1a1 coeffiments, n 1S a whole number, and nm 18 a

factorial polynomial.

The central difference operators defined in the domain of equidistantly located

points i = 0, i1, i 2,K , i n are used to obtain the constitutive equations for a regular

lattice plate. The central difference operator can be expressed in terms of the forward

difference operator as

91

* n _ n—1 n—1 n _ n-2 n—2 n—2
A,- -1Ai+(—l)" —A,- 1(4) —Ai_1 —2A,. +731.+1 (4.11)

The formulae and graphical templates for some lower-order central difference operators

are given in Table 4.1.

TABLE 4.1 Lower-order central difference operators

 

 

 

 

 

Notation Graphical template and analytical formulation
,1, —c 4.: 3 c 0—
Ai l' 1 l
fi‘fiA
_G 3 © 3 O—
'1‘ A2- 1-1 1' i+1
I

fi—l "Zfi +fi+l

 

 

—o——o——© c 4.)—
A3- i-2 i-l i i+1 i+2
z

- fi—Z + 3fi—1 - 3f: + fi+1

 

 

* —c + 3 c o—

A‘} i—2 H i 1+1 1+2
.fi—Z _4fi—l + 6f: ‘ 4fi+l + fi+2

 

 

 

 

Using the central difference operators yields easier formulations that maintain the
symmetric structure of the main equations. Only this type of operator is used in this work,

and henceforth the asterisk is omitted for simplicity.

In finite difference calculus, the function G,- is called the sum of f,- if its first

difference in the given domain is equal to f1:

92

If AG,‘ = fi’ then SAG,‘ = Gi = Sf}

and

a+nh a+nh a+nh
S AGi =|G,-|a = S f,- (4.12)
a a
The operator of summation, S, is the inverse of the difference operator. Therefore the
summation formulae can be obtained by inverting the difference formulae. For the

general case of a factorial polynomial

Aim =t(i-1)(”1) (4.13)
_ (i+1)(t+l)
i0) ___—(+1 +C (4.14)

The main differences and sums used in this work are given in Appendix C.

Note that the operator of finite difference summation is related to the operator of

algebraic summation as follows:

ind

(4.15)

echtr
n
a M?"

4.2 Constitutive Equations for Regular Rectangular Grid Stated in Finite

Difference Form

The finite difference equations obtained from a system of algebraic equations
yield a banded coefficient matrix. The first and the last equations of these systems usually

serve as the boundary conditions.

93

Different methods are used to obtain the band structure of the coefficient matrix.
Three different approaches using the principle of virtual work, the displacement method

and the mixed method are illustrated in this chapter.

4.2.1 Method of Virtual Work

In the first approach, auxiliary states of the system are introduced when localized
self-equilibrating groups of forces are applied to any possible statically determinate
system obtained from the original system. The virtual-work equation is then constructed

for the displacements of the original system.

Example 1. Consider the prismatic beam shown in Figure 4.2(a), loaded at equidistant
points. The virtual state shown in Figure 4.2(c) that results in a localized bending moment
diagram is used to obtain the main bending equation for the beam in finite differences.

Equating the virtual work of external and internal forces gives

[2
‘yi—1+2yi—yi+l = EEO/[1'4 +4Mi +Mi+l)

Using the finite difference notation this may be expressed as

2
2 1 2
A. .=_-———A +6~M~ 4.16
I (M) 6 E I ( )1 ( I) ( )
From the equilibrium condition for node i, it follows that

Mi_Mi—1_Mi+l_Mi

 

=8-

01'

43114.14 (m

94

The set of equations for all node numbers i yields the full system of algebraic equations

 

 

 

 

 

 

 

for the original problem.
(a)
P0 P1 Pi I)n-l Pn
., 1 1 1 m 1 1 r,
. - z 7 A 7
Q1, ..-, ; ' 'i ' /'*:2v ;
l T WW flflflflfl
L = n l >1
y 11
(b) M0
M01 1 111M 11111111 T111111M
1'1 2 1/7- 1111 1 _/fl‘ \\~~— - 1111 1111
M 1-1 1 Mi“
1 2
131 1 ’51

Figure 4.2 The original beam and auxiliary load cases for Example 1

Equations (4.16) and (4.17) can be transformed to yield the finite difference

equation with respect to the unknown displacements y ,- :

I3

41003—157142 +6118) (4.18)

95

Further, (4.16) and (4.17) can also be manipulated to yield expressions for the bending

moment in the i ”7 nodal section and the shear force between nodes i - 1 and i :

E1 131 E] P]

 

Mi=7z—(—yi—l+2yi_yi+l)+_6_='l_§_A2yi+ '6 (4-19)
M. -M-
Q]. ___—___I 1-2 =21?” +1413. (4.20)
l 13 6

The limits of the expressions (4.16), (4.17), (4.19), and (4.20)

 

 

2 2
A .
lim y, =d yx, lim [———1—(A2+6)iM,-]=—Mx
l—)dx [2 dxz l—nix 6E1
P- AZM- d2M
lim —'—=qx, lim ' = x
l—~>dxl l—>dx [2 dxz

A4 . 4 .
lim y, = d yx, lim _1_(A2 +6)i[fl] =qx
1—>dx 14 dx4 l—>dx 6E1 l

limM—i—i=0, lim flA3yi+lAPi =Ely'"
I—>dx I l—->dx [3 6

 

demonstrate that these relations are the difference analogues of the corresponding
differential identities. Note that the inverse transformation from the differential identities
to those written in finite difference form does not necessarily yield the original form of

the latter.

Example 2. Consider the plate with the regular rectangular grid shown in Figure 4.1
loaded with arbitrary nodal forces. This plate can be treated as a system of orthogonal

beams in the two directions i and j. Assume that the elements of the grid have zero

96

torsional rigidity (GJ k = 0) . The auxiliary load stateij in the form of the two groups of

self-equilibrating forces shown in Fig. 4.3 is used to obtain constitutive equations.

 

 

 

 

 

 

 

Figure 4.3 Auxiliary load states for the system of orthogonal beams for Example 2

The first group of forces is applied to the j ”7 beam in the statically determinate system

obtained from the original system as shown in Fig.4.3 (a):

LS’UD) = LAP)

The second group of forces is applied to the i ”7 beam in the statically determinate system
obtained from the original system as shown in Fig.4.3 (b):
(2) _
Li,- (1’) — L j (P)
The operators in these expressions have the following form

14(1)): _li—l +2i _1i+1
Lj(P)=—1j_1+2j—1j+1

97

Two equations similar to (4.16) are obtained using the virtual work method:

A? (2,]. )+ 6—212—11-(A2 +6),- (Mé): o (4.21)
A7;(Z,-j )+ ggW + 611.1114,” 1: 0 (4.22)

where M J and M ill-I are nodal bending moments in the ith and j ”7 beams, respectively.

The third equation is obtained from the equilibrium condition of the node if of

the system:

2 2
A' 1 A ( 11

The three equations (4.21), (4.22), and (4.23) form the full system of constitutive finite

difference equations resulting from the method of virtual work for the original system.

4.2.2 Displacement Method and Mixed Method

Example 3. Consider the prismatic beam shown in Figure 4.4(a) under the distributed
load. The main system of the displacement method shown in Fig. 4.4(b) is formed by
introducing nodal constraints in the form of clamps and rollers. The main system of the
mixed method shown in Figure 4.4 (c) is formed by introducing nodal vertical supports

(rollers), hinges and nodal moments as redundant.

For the triplet of nodes i-l, i, and i +1 of the main system shown in Figure 4.4 (b),
reactions in the 1‘h node due to unit displacements of all restraints obtained from the

slope-deflection diagrams are

98

12E] 6E1
Ri =73— i—l _Zyi +yi+1)+ ‘1—2—(¢i—1 —¢i+1)—Pi = 0

6E1 2E1
Mi = 7701—1‘yi+l)+l—'(¢i—1+4¢i—(0i+1)+0 = 0

 

 

 

 

 

. -9“)
(a) - + 1 1 1 + i r ‘
s I 1
i r l l I 1 i l ‘i l l i i 1 J ‘i j. >
4.1.1.; __,1__ x
L g1
Y1 ’
H "T ”T T “ ~ 2
, ’ f 1 T” t 2-; .
(b) 1 1 1 1 1 I 1 .
(C)

 

 

 

 

Figure 4.4 The original beam and the main systems of the displacement method

and the mixed method for Example 3

99

The system of finite difference equations of the displacement method can be written in

the operator form as

Llllyi)+ L12(¢i)= 141(4)

(4.24)
Lzllyi)+ L22(¢i)= L2(q)

where

L11 = 12151014 *21' +1:+1)/13
L21: L12 =6EI(1i-l 4140/12

L22 = 2EI(1i—1 + 4i + 1140/1
If q(x) = q = constant, then L1 (q) = ql / 2, L2 (q) = 0.

For the main system of the mixed method (Figure 4.4 c) reactions in the ith node

due to the unit displacements and unit moments in the triplet of nodes i-l, i, and i +1 are
Rt = 0+—1_(Mi—1_2Mi+Mi+l)—Pi =0
l
1 l ( )
(Di =;(yi—1—2yi+yi+l)+’6fl’ Mi—l +4Mi +Mi+1 =0

This system can be written in the operator form as

Lillyi)+L12(Mi)=L1(q)

(4.25)
inlyi)+ L22(Mi)= L2(¢1)

where

100

L21=(1i—1—2i+1i+1)/1

L22 =50: _11+4 +1141)

Generalization of this approach on the regular system of orthogonal beams in
Example 2 yields systems of finite difference equations similar to (4.24) and (4.25). The

system of the displacement method has the form

L11(Zy'1+ L121¢1JW1+ L131¢2v1= L1(Pij1
L21(Z,-j1+ 2221¢QY)1+0 = o (4.26)

L31(Z,-j1+0+ L331¢1f11=0

where

L11=21AAE+BA§1 A=6E1/113, 3:62:12/13, le-l

6E1 6E1
L12 = L21 ='T](li-1"1i+l)' L13 - L31=—2—2(1j—1—1j+l)
1 ’2 (4.27)
2E1 2E1
L22- _ 1 _1— (A2+ 61, L33: [2 2 ———(A2+ 61

For the main system of the mixed method the finite difference equations have the form

L11(Mj )+0+Ll3(zij=) 0
o + L22 (Mg-1 )+ L23 (zij) = 0 (4.28)

L31(M,§-)+ L32<M,§’) + L33 (2.)) = L<P.—,~>

where

101

L” = (A2 +6),-/A112, L = -1

L12=1/21=O

(4.283)
1/22 = (42 +6)1/Bl12a 123 = L32 = Ai/lz,

2
L13 = L31=Ai/ll’ L33 =0

Equations (4.28) written in the full form are identical to equations (4.21), (4.22), and

(4.23) obtained earlier using the virtual work method.

4.3 Governing Finite Difference Bending Equation for Lattice Plate

The systems of equations in (4.26) and (4.28) include three unknowns. Using
operator transformations, both systems can be reduced to one equation in one unknown.

System (4.28) can be transformed to one equation with respect to any of the three

unknownle M” orZ'

(L11L23 L32 + L22 L13 L3111M1)" 1: L22 L13 L (P17) (4298)
(L11L23L32 + L22L13L3111My1'11: L11L23LlPij) (42913)
(L11L23L32 + L22L13L31X2111= L11L22L(Pif) (4296)

Similarly the system of equations (4.26) is reduced to the equation

£12111: 1L11L23L33 + Li3L22 + Li2L33 1121] 1: L22L33(Pij) (4-30)

Substituting the corresponding expressions for operators L into equations (4.29c) and

(4.30) and performing transformations yields the same equation

102

{AM-(AZ+61j1+B[A‘1-(A2+61i1 1(zy1=[(A2+61i(A2+61j-](P,j1 (4.31)

where A =65]l /113 and B: 6E12/13 . Equation (4.31) is a finite difference analogue of a
differential bending equation of the thin isotropic plate.
The system of equations resulting from the mixed method is more general since it

allows the governing equation to be obtained both in terms of internal forces and

displacements.

Accounting for torsion and shear, the operators (4.27) of the finite difference

equation (4.26) become

LH=(A1Al-2+A2A3), L=—1
L12: L2]: BIL!" L13: L31: 32L} (4.32)

L22=(D,A?—F2A§+G,1, L33=(DZA3.-F,A}+Gz1

where

 

 

-5111. -2222.
’71— 2 ’ 2— 2
l1 12
12E] 12E]
A]: 1 1 , A2: 2 1
[l3 1+12771 I; 1+12772
l l

Bz—Al, B=—Al
1211 2222

D _2E11(1-57711 D _2E1211-67721
1 11 1+12m’ 2 12 14.12772

 

 

(4.33)

103

where 71 and 172 are the engineering shear strains. If shear deformations are neglected

i.e., 71 = )72 = 0 , then the coefficients in (4.33) become

 

 

 

 

771-772-0
12E] 12E] 1
A1: 31 2: 32, B1=—A111a 32-—4212
’1 ’2
2 2E] 2
D1=2£L=A1le Dz = 2 = 21—2 (4-34)
11 6 12 6
GI GI
01-4111, G2=A212, 171-7“: F2: 1”
1 2

Substituting (4.34) into (4.30) and simplifying yields

F A F A 2 F
—6 22 2 A? 21 1 A? +—:—2§ F] [L——3 AfAji
[1A1 12 A2 1112 6 A2

2
+F ll-fl 21.221441}? 12+}? 121212212. +A A4(A2+61
2 6 A] 1 j 12 21 1 j l 1 j

2 2
+A241142+6). 11%) 3— 1’2”: 141-125-141

[12122 Al A2

 

 

 

2 2
+614??? 21,? A3 -6112 413 -6212“ 21,? 1411212 +61, (212 +61j1111’i-1 (4.35)

 

 

 

which is the finite difference analogue of the corresponding differential equation.

104

4.4 Boundary Conditions

Example 4. In this example the boundary conditions for the cantilever beam with the

left end fixed and the right end free (Figure 4.5) are obtained.

Pl P2 P3 Pn-] Pn

 

 

UK
1"

\\ \\\

 

Figure 4.5 Boundary conditions for the beam in Example 4

The system of equations resulting from the displacement method for this case has the

following form

 

12EI 6E1 1
R1 - _(2y1 Y21—72—(02 -P1 =0
1n0del
6E1 2E1
M1=‘7 2)+—l-(4<01+(021=0
1251 6E1 ‘
Ki: —( J’i— l+2yi Yi+l)+—12_(—¢i—l +(”HQ—Pi :0
6E] 2E1 1 interior nodei
Miz- —(— yi— l+yi+l)+‘T'(¢i— 1+4¢i+¢i+l): O

 

105

12E1 6E1 1
Rn =l_3(yn —yn—l)+‘IT(—¢n +¢n—l)_Pn = 0
1n0de n
6E1 2E1
Mn=—*l"§— n—Yn—1)+“T(2(Pn +(I’rz—1)=0

 

The general case equations for an interior node i apply to all nodes except for
nodes 1 and n. Therefore, the two equations corresponding to nodes 1 and n are the
boundary conditions for the system. When changing from the system of two equations to
one equation with one unknown function, only one boundary equation containing one

unknown function for each end of the beam needs to be obtained.

For the left support, the first equation of the system for node 1 yields

12 12E1
P- 4 +
¢2=6E111——(Y1 y211

Substituting it into the second equation for node 1 yields

 

12 1 6E1
(P1:

—P —— 4 +

M, 1 1 y. M]

Substituting these expressions for (p1 and (02 into the first and the second equations for
node 2 produces two expressions for (03, and equating these yields one boundary

equation with one unknown fimction:

3

12E1 (7101 +1921

9Y1 - 453/2 + Y3 -

For the right end of the beam the equations for Rn and R "_1 yield

106

_ I2 ‘P__6EI _12EI( _ )
(on—1‘65” n l3 (on ’73—— yn yn—l

12 6E] 12E1
Con—2 = Pn

"6E? ‘75—¢n_7(yn-2yn—l+yn—2 )1

Substituting these expressions into the equations for M n and Mn_1 and equating the

resulting expressions for (p n yields the boundary equation

6713310)” — 2yn—1 + yn—2 1 = 6Pn + Pn—l

Boundary conditions for lattice plates can be constructed similarly since at every edge of
the boundary either a displacement or a force has a fixed value. Using the operator
transformation allows the original system of finite difference equations to be reduced to
one equation similar to the corresponding differential equation for a continuous structure.
This transformation allows the same methods and algorithms that are used for the

solution of differential equations, such as the decomposition method, to be used.

4.5 Solution of the Bending of Lattice Plates with Orthogonal Grids

Consider the rectangular lattice plate shown in Figure 4.1 formed by two
orthogonal families of rods that are parallel to the edges of the plate. The plate has elastic
supports along the edges. The external load is applied at the nodes of the grid. Assuming
that the torsional rigidities of the elements of the grid are negligible, the bending equation

has the form (4.31). This equation can be written in the non-dimensional form
114111114121 +611): A2 411A? +6111 (1711- 1=1Ai +6111?)- +6,-1(E-,-) (4.36)

107

where

6E1 13 6E] 13 ~ 1 ~ P113
A] : Ii, A2 : 2 i, WU. : Wij’ Pi]. : 1
E10 [1 E10 [2 W0 E10 W0

 

 

 

(4.37)

A] and A2 are dimensionless coefficients, 1'17 ,1- is the dimensionless displacement of the

node ij , and 15,]- is the dimensionless load. The subscript 0 denotes a reference value of

the corresponding function.

According to the decomposition method the original problem (4.36) can be

replaced by the following three auxiliary problems:

1. 6A, A? (11,1 1=f1(i,j) (4.38)
2. 6A2 51.1%,,” 1: f2 (i, j) (4.39)
3. 1A1(A’1A2j1+ 4.1/1% Aj- 1] (115.” 1: (A? +6,1(A2j +6 ,- P},)- f1 (i, j)— f2 (i, j) (4.40)
where f1(i, j) and f2 (i, j) are unknown functions of discrete arguments. Solutions to
these auxiliary problems must be subject to condition

1.1 = M)!” (4.41)

Equations (4.38) and (4.39) can be regarded as the bending equations of the non-

interconnected beams in the directions i and j . The beams have elastic supports in the

form of rotational springs. Boundary conditions for the first auxiliary problem for the

beam parallel to the x -axis (i-direction) can be written as

108

j
174,]. (5 + 4?, )— $95,, (4 + 0.5/2', )+ .74., = Al [13L]. (4 — 0.5?1 )+ 13', j ] (4.42a)
l
at i=0
Wlnj=0
w,,L,.(5+4k,)— w,_L,(4+o.5k,)w WM}: _1—1[P_L,(4— O.5k)+F,,_2J-] (4.42b)
at i=n

where [:1 =(1+rlll /3E1,)_l is the dimensionless stiffness coefficients of the elastic

supports. Boundary conditions for the second auxiliary problem have a similar form with

corresponding subscripts and superscripts.

An approximate solution to the problem is found by approximating the functions

of discrete argument f] (i, j) and f2 (1', j). For the case of symmetric load it is assumed

that

m

Maj):Morgan-((42)“(U-ml”)1‘3—1]

(4.43)
f2(z,1)=W1(I)-W2(I)((1(2)+J(')—m1(‘))——2——1]
where wl(j), l/lz(j), {/11(i), and (112(1) are arbitrary functions, and im, and j(’) are

factorial polynomials: i0) =i(i—1)(i—-2)...(i—t+1), j(’) = j(j—l)(j—2)...(j—t+l).
The first auxiliary problem can be considered as one-dimensional and can be written in

the following form

6A A,- 4(w, 4): DL+Dzim+D3i<2> (4.44)

109

O I 4 C 4 I
Where D1 =W1(J)+l//2(J)a Dz =("—1)—2V/2(1)a D3 =——2W2(J)o
n n

Using formula (4.14), the solution to the first auxiliary problem is

-(4) ' (5) - (6) -(3) -(2)
74,5. =—1—— Dll———+D2 (”2) +D3-(fi2-l— +CL’——+C2’——+C3i(‘)+C4 (4.45)
6A1 24 120 360 6 2

 

with the arbitrary constants C,- to be obtained from the boundary conditions (4.42).

Comparing equation (4.44) of the first auxiliary problem with the known bending

equation of the beam loaded at equidistantly located points
4 ~I 2 “‘1
4141' (W11): (AI + 6:“ )(Pi )

implies that

—;-[Dl + 02 i0) + D3 im]: (A? + 6,)(1?)

Solving for 25,} yields

~] 1 1 .1 .2
p, 7K0, _ED,)+D,.<>+D,.< >]

From the first condition of (4.42) it follows that C4 = O. The other three conditions yield

the following system of equations:

Cl +C2(— 0.51?l —1)+C3(3E,)C, = 36171[ 01(5—051?l )+ D2(6—E1)+D3G+—;—§LH

C—+C —+C =—
1 6 2 2 3"

"(3) "(2) 1[ ”(4) D "(5) no]

110

 

 

(n—l)(3) ~ ~ ("—2)“) ("_3)(3>
C,( 6 (5+4k,)—(4+o.5k,) 6 + 6

 

+ C2 9551236 + 41?1 )— (4 + 0.51?1 ) (’1 ‘22)(2) + (" ‘23)(3)(

 

+ C3 (n —1)(5 +412", )— (4+ 0.5?1 )(n—2)+(n —3)(

L.

 

 

= _1_ (D1 [5 ‘005E1_(5+4;1)(n—1)(4)+(4+ O5;J(n-2)(4) _ (n —3)(4) (
36A] 4 4 4

 

 

+ D2 B4 - 0.5;] )(n —1)+(n + 2)_ (5 + 4E1)“ — 0(5) + (4 + 0.5/:1)“ _ 2)(5) — (n __ 3)(5) :|

 

 

 

20 20 20
' ~ .. ,, _ (6)
+ D3 ((4 — 0.5/cI )((n —1)(2) —%j + (n + 2)(2) — g — (5 + 4k1)( 61))
+ (4 + 0.521 )(n - 2)(6) __ (n - ”(6)“
6O 60

For convenience this system can be written in the matrix form

[B,,](C’ }= 361717[Gn](D1 ( (4.46)

In this equation (C1(=[Cl C2 C3]T, (Dl}=[D1 D2 D3]T, and [8"] and

[6”] are the matrices of coefficients of C ,- and D,- correspondingly. The elements of

these matrices are given in Appendix D. From (4.46)

{c’}=[B.r‘[c.1{D'}—‘—

36Al

111

and substituting it into (4.45) yields the following expression for the dimensionless

deflection function

{0112-}

where

 

{m ‘ ({a.}+tc.1T[B;']’{b.-})

36A]

 

20 60

1(4) (p.215) (i+2)(6) T
{ai}=(: 4 J

T
{b,-}=(fl .(2) W]

’_
6 2
Similarly, the solution for the second auxiliary problem is

"1’ =1D”} {f1}

where

(DI/1:151 52 531T

~ . . ~ 4 . ~ 4
D1=W1(1)+W2(l), Dz =(m-1)—2 W20): D3 =——*¢//2(i)
m

{1,1 ,—6—‘,({a} [cm] [3.211% {41)

(4.47)

(4.48)

(4.49)

(4.50)

Expressions (4.47) and (4.49) with respect to (4.44) and (4.50) can be written in the form

112

{4711' = (01(1) W1U)+ (02(1) W20)

(4.51)
17g” =¢1(f) W1(i)+¢2U)V/2(i)
where
wW=MHfi11WW=MmK01
¢1(j)=[d1]{fj 1 (02(1): [dl,m]{fj}
(4.52)

M14 ooLkmkp $441_%(

14.1:[1 1,1-1) "3,1

m

Accounting for the symmetry of the system, the arbitrary functions 1/11(1) and 1/12 (i) can

be assumed as follows:

W10): a] ¢1(i)+ “2 (02(1): W20): 0'3 ¢1(i)+a4 (02(1)

~.~. .~.~. (453)
W1(f)=a1¢1(1)+az¢2(l)a W2(J)=a3<01(1)+a4(02(1)
Satisfying 172,-],- 2 174,51 from (4.41) implies that
671 =61], [2'2 =02 , 53 =a3 , 54 =(14 (4.54)

Coefficients a are determined from the third auxiliary problem. Introducing the

discrepancy function (13(1', j), the third auxiliary problem can be written as

113

 

¢(i+f)=141 (4142111 42 (4141)](‘7’1111‘1141 + 511(5) + 61 )1 (131')
(4.55)

+f1(i+j)+f2(i+j)= 0

Assuming if} = P = constant and performing algebraic manipulations yields

«no-.1):41.414111111141111»41.411111114111111]
+111 [42111111114141 (11441121114111 (1)]
+111 [1,131,113,1011,131,113,111]
+a1lA1A‘3-1111-1A11111-h141411111411101
1101111411 (11411111111 (11111.41
112111-1401 111111-1111

+ a4 1' (02(i)f(f+m)— (02(j)f(i+")1- 36” = 0 (4'56)

where 11,11): ((2) 11(1) _ 111-(0)32 -1 and 10,111) = (1(2) 1 1(1) _ 111-(0),";‘2 _1

The third auxiliary problem is solved by setting (D(i, j) and some of its even lower

differences to zero at the center of the plate:

<1>(i,j) = 01 A? [$031)] = 0+ 421' [$031)] = 0’ Aw! [W’ 1.)] = 0 (4 57)
i=n/2, j=m/2 .

114

The arbitrary constants a,- are obtained by solving the resulting system of linear
algebraic equations. The values of a,- are then substituted into the expression (4.53).

Using (4.53), (4.54) and (4.41), the equation for the deflection function can be written as
1111- =[a](¢1(i) 11(1)] (458)
where

[01:16“ 02 a3 “4]

”(01(1) (P10)—
(02(1) (01(1)
(01(1) (P20)
(01(1) (P201

(4.59)

(1112-) 1pm}

 

 

The value of bending moment in any rod of the lattice can be found using the expressions

My = T(— Wi—l + 2W1— Wi+1)+ %
(4.60)
E] P1-

11
My =—l—(—Wj_l+2Wj—Wj+l)+—6—-

Using notation (4.37), the equation for the dimensionless bending moment in any j’h

beam in the i ”1 direction can be written as

My = ”(WI—Lj _Zwi,j + Wi+l,j)+“A—l—Plj (4.61)

where

115

 

~ 12 , 613
M,.’- =M,-- _1_ , A1: 0 (4.62)
1 1 11201511 Elowo

In equation (4.61), 13,-]! is the nodal load applied to the beams in the first direction (1' ).

Using (4.44) and (4.53)

251’ 31310411001» (4.63)

where

{D(i,j)}=[D1(i.j) 02131) 030:1) D4(i,j)lT

 

D1 (111) = «21(1). Dz (1,1) = (110)0(1. 11), 03(1) = 12(1) (464)
,(1) 1(2)
D(i,n)=1+niz+(n—1)5n—2—— 4’12

Numerical examples

The square lattice plate shown in Figure 4.1 (L1 = L2) unifome loaded at the
joints with unit loads was analyzed for six types of grids differing by the number of rods
in the two directions (m=n= 8, 16, 24, 32, 64, and 128). The following characteristics of

the lattice were assumed: 11 =12 =1, E11 = E12 = E], GJ1= (U; = 0.

Table 4.2 shows the maximum dimensionless values of deflections obtained for
the plate with pinned support for six different types of grid using the decomposition
method (DM) in finite difference formulation and differential formulation based on the
continuum model. The results are compared with those obtained using the finite element

method (FEM). The LIRA software (Kiev 2000) was used for the finite element analysis.

116

Maximum bending moments for the plate with a 16x16 grid are shown in Table

4.3 for the cases of pinned and fixed support. Table 4.4 demonstrates the

maximum

deflections of the same plate with varying coefficients of support rigidities. The results

presented in Table 4.4 are illustrated in Figure 4.7.

Table 4.2 Deflections at the middle of the plate for the different types of grid

~

k 1 =1 (pinned support)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

172 (flflj x E1 for the type of grid
2 2 pl}
888 16x16 24X24 32x32 64x64 128x128
FEM 0.06403 0.13085 0.19636 0.26213 0.52487 1.05003
DM
(finite difference 0.06289 0.13076 0.19737 0.26352 0.52997 1.07959
model)
s(%) -1.78 -0.07 0.51 0.53 0.97 2.82
(continugrs model) 0.06768 0.13530 0.20170 0.26381 0.52539 1.05108
s(%) 5.7 3.4 2.72 0.64 0.1 0.1
Table 4.3 Bending moments for the plate with the 161116 grid
for k1 =1 (pinned support) and k1 =0 (fixed support)
~ 1 n m 1 ~ 1 m 1
My- —,— x — -M,+j O,—— x —
FEM DM 50%.) FEM DIVI £(%)
Pinned supports
~ 1.2305 1.2329 0.20 - - -
kl =1
Fixed supports
1: -0 0.4083 0.4059 -0.58 0.9109 0.9152 0.47
1-

 

 

 

 

 

 

 

 

 

117

 

Table 4.4

Deflections in the middle of the plate with the 16x16 grid
for different values of support rigidity k1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

172(n/2, m/2)x[ E1 ]

k1

FEM Decomposition Method s(%)
1.0 0.13085 0.13076 -0.06
0.99 0.11028 0.11055 0.25
0.95 0.07268 0.07287 0.25
0.8 0.04162 0.04162 0.0
0.6 0.03283 0.03273 -0.3
0.17 0.02772 0.02753 -0.6

0 0.02663 0.02642 -0.8
0.14
1 0,12 » ++— — — + —
1 A 0.1 + — + i
1 N
' E 0.08 + ——» w- —
N

g E 0.06 +4
i 3 0.04 + ~+ - -- —
1 o 02 F 2 fl I .._._2.___ __ ‘
‘1 o I Y I ‘ I
g 0 0,1 0,2 0,3 0,4 0.5 0,6 0.7 0,8 0.9 1
1 Coefficient of Rigidity of Elastic Support
1 T-Fi'EM-a-992991_Eé§116514_eiho?
Figure 4.7 Deflections in the middle of the plate with the grid 1611116

for different values of support rigidity [Cl

118

 

Results obtained demonstrate sufficient accuracy of the decomposition method
stated in finite differences for bending problems of lattice plates with different types of
grids and support rigidities. The finite difference formulation of the method provides
better results than the differential formulation for sparse grids. However, for dense grids

the results have similar accuracy.

4.6 Free Vibration Problem of Lattice Plate with an Orthogonal Grid

Consider a free vibration problem of a rectangular lattice plate with an orthogonal

grid shown in Figure 4.8 with concentrated masses at the nodes.

 

  
 
 

  

 

111/mull
r ' VII,”

11 ' L
MI,” 1/ 1,
00 A 10 1'31: i0 110/

 

 

 

xi

 

Figure 4.8. Lattice plate with an orthogonal grid and concentrated nodal masses

119

The governing equation of the problem has the form (Ignatiev, 1981)
[AIM-(A214 6,- )+ A2 Art-(133+ 61- )(Vvij-H: [(13% + 6,-)+ (A2]. +6j)](fii,-j 602 175,1) (4.65)

where {iii}- = mij lg / E10 , mg is a point mass at the node i, j , and a) is the frequency of

free vibration. The problem is subject to the boundary conditions (4.42).

The decomposition method is used to solve the problem, introducing the three
auxiliary problems. The first two boundary value problems have the form (4.38) and
(4.39). The third auxiliary problem stated in the form of discrepancy function is written

$031): [A1 (4143* A2 (AiAlfllfi’H‘ [(412+ 6:‘)(Ai+61')l(’fiy'wzwy)
+f1(i,j)+f2(i,j)=0

The solutions for the first two auxiliary problems are known from Section 4.5,

and are given by (4.51), (4.52), and (4.53). Substituting expressions for f1(i, j) and

f2(i, j) from (4.43), and 17.3,]- from (4.51) into (4.65) and performing algebraic

manipulations yields
. ._ 4 . 2 . 2 . 4 .
(MAJ)-0‘1[AlAj<PI(1)Ai<PI(J)+AzAr<P1(1)Aj<P1(I)]
+ 0‘2[A1AL}"P2(")A%(P1(J)+ AzAi<P2 (053% (1)]

+a3[A,At-cpz(i)A%cp2(J-)+A2A1<p1(i)A§-cpz(j)]

120

+ 0£4 [AIA‘}<P2(i)Ai<P2 (j)+ AzAilpz (1)4th (1')]
+a1[<p1<j)+<p.<j)-a.,-maximal-)1

+a21<92(i)-¢1(j)f(i,n)-fiiyw2<p2(i)¢1(j)l (4.67)

+0t31<p2(i)-<m(i)f(j,m)-r71g- w2¢1(i)<p2 (1')]
+a4[-<pz(i)f(i,"1)-@20)f(i,n)-fiig- (oz cpz(z-)<p2(j)]=o

The third auxiliary problem is solved using the collocation method by setting

<D(i, j) to zero at the following four collocation points:

(a?) (3%), 13%), (as?)

The resulting system of four linear algebraic equations can be written in matrix form as

follows
[A — (028][61] = o (4.68)
where
[a]=[a] a2 a3 (14] (4.69)

and A, and B are the matrices of coefficients. The members of the matrix A are

determined as

01k = AIA?<P1(i)Ag'<P1(j)+Az A?¢1(i)AL} <P1(j)+<P|(j)+<P1(j)

121

 

02k = AiAl<P2(i)Azj<P1(j)+ AzAi<P2(i)A‘}<P1(j)+ <P2(i)- (p1(j)f(i,n)
03k = AIAl¢2(i)Azj<Pz(f)+ 142414)] (1743432 (J)+ (92(1)- <P1(i)f(1m) (4-70)
04k = A1A3<P2(i)4?}<l>2(f)+ AzAilp1(i)A§<P2(j)+ <P2(J')‘ (P1(i)f(f,m)

and the members of matrix B are

blk = fiy¢1(i)¢1(j)

b2]. = fingfikm (i)

(4.71)
b3k = rig-(01 (W12 U)
b4k = fig¢2(i)¢2(f)
where the subscript k = 1...4 denote the four collocation points.
For the general case [a]¢ 0, therefore
lA—sz‘ = 0 (4-72)

Multiplying this equation from the right by B", (4.72) can be written as the eigenvalue

problem
|C~2n=0 «J»

where ,1 = 602 , C = A B-1 , and I is the identity matrix.

Solving (4.73) yields the four eigenvalues from which the smallest is chosen.

122

 

Results for the systems of orthogonal beams shown in Figure 4.8 with pinned
support and six different grid sizes are presented in Table 4.5. The problem was solved in
the dimensionless form using the following assumptions: [1 =12 =1, E11 2 E12 2 El ,

and mi]- = m . The results obtained by the decomposition method are compared with the

exact solutions obtained by Ignatiev (1979).

TABLE 4.5. Dimensionless values of the first free vibration frequency

 

 

 

 

 

 

51 X 103
Grid size 8x8 16x16 24x24 32x32 64x64 128x128
Decomposition
method 211.045 54.5811 24.3659 13.6513 3.43240 0.854986
Exact solution 218.086 54.5223 24.2322 13.6306 3.40764 0.851912
s(%) -3.22 0.11 0.55 0.15 0.72 0.36

 

 

 

 

 

 

 

The results demonstrate that the decomposition method yields good accuracy for
the first frequency. To obtain higher frequencies, the other forms of approximation
functions must be used. However, the higher forms of vibration are hard to predict, and
formal approximation may not be exact. For this reason, the number of sequence
members retained in the approximating function must be at least four times greater than
the number of frequencies to be determined. The resulting procedure is hard to implement
by means of the decomposition method. Therefore, other approximate methods have to be
developed for determining larger spectrum of natural frequencies for dynamic problems.
The consecutive dynamic condensation method for solving the eigenproblems of lattice

plates is presented in the next chapter.

123

 

CHAPTER 5

DYNAMIC ANALYSIS OF LATTICE PLATES
BY CONSECUTIVE DYNAMIC CONDENSATION

Free vibration problems of complex structures involve the solution of large

systems of algebraic equations ( (N 5103 —105). Since only a limited number of

eigenvalues and corresponding eigenvectors are required in practice the approximate
methods of static and dynamic condensation can be used. The work presented in this
chapter is based on the frequency dynamic condensation method (Grinenko, Mokeev
1988), and the consecutive dynamic condensation method proposed by Ignatiev (1992).

A brief overview of these methods is given in Sections 5.2 and 5.3 of this chapter.

The energy form of the consecutive dynamic condensation method is developed
in this chapter and three different techniques of using it to solve dynamic problems are
described: condensation using the smallest natural frequency of subsystems;
condensation using the eigenvectors of subsystems; and condensation using preliminary

static condensation in the form of the displacement method.

5.1 Problem Statement

The problem of free vibration or stability of a system described by a discrete

analytical model is characterized by the eigenvalue problem

124

[C—ll]{z}=0 (5.1)

In scalar form, the above equation represents a system of homogeneous linear
equations. A non-trivial solution of this system is possible when the determinant of the

matrix of coefficients is equal to zero:
det[ C — 2.1 ]= o (5.2)

The matrix of coefficients [C —/11 ] in (5.1) is called the characteristic matrix of the

given matrix C, and (5.2) is the characteristic equation of C with respect to ,1 .The roots

of this equation are the eigenvalues 21(1' 2 1,2,... N) that form the full matrix spectrum.

When the free vibration problem is solved by the displacement method

(neglecting damping), the matrix C and the eigenvalue )1. in (5.1) are determined as
C=M“K, 1:602 (5.3)

and (5.1) has the form
[K — 2M]{z} = o (5.4)

where M and K are the mass and stiffness matrices, respectively; (0 is the frequency of
free vibration of the system, and {z} is the vector of nodal displacements.

In the force method, the matrix C has the form

C = M6 (5.5)

and equation (5.1) becomes

[6M — u]{z} = 0 (5.6)

125

where M is the mass matrix, 5 is the flexibility matrix, A =1/a)2 is the eigenvalue, and I

is the identity matrix.

5.2 Frequency - Dynamic Condensation Method

Similar to other reduction methods, this method divides the full set of degrees-of-
freedom (d.o.f.) into primary (master) d.o.f. and secondary (auxiliary, or slave) d.o.f. The
secondary d.o.f. are eliminated on the basis of the initial matrix equation, and the
condensation of mass and stiffness (or flexibility) matrices is achieved by equating some
frequencies of the condensed system with those of the original system.

5.2.1 Condensation based on the Displacement Method

Classifying all the d.o.f. of the system under consideration as primary and

secondary, equation (5.4) can be written as

krr krs Mrr Mrs Zr
— A = o (5.7)
kSI‘ kSS MS? MSS ZS

Subscripts r and s are used to denote the elements of matrices related to the primary and

secondary d.o.f. correspondingly.

Eliminating the secondary unknowns zs , (5.7) is condensed to one equation with

the primary unknowns:

[(Kn--1Mrr)+D§i)(/1)](zr)=0 (5.8)

where

126

D5901): _(Kss _AMrs )(Kss _1Mss)—1(Ksr _1Msr) (5'9)

Equation (5.8) can be approximated as

[(Krr _AMrr)+(K§r€) _1Mr(':))](zr)=0 (5-10)

where K )5 ) and M ,(j ) are found from the assumption that the eigenvalues of the original

equation (5.4) are equal to those of the condensed equation written in the forms (5.8) and
(5.10). Using the two eigenvalues from (5 .4) and equating (5.8) and (5.10), the following

system of equations is obtained:

K29) _Ale Mg) = D5?) (11,13)

 

 

(5.11)
16:) - 22... M5? = 05:122..)
where 21,“ and 22,” are the limiting eigenvalues (Amax and 1min) of (5.4).
From system (5.11)
1
MS) = [D£i)(ll,rs )_ D£:)(/12,rs )]
(12,13 — 11 rs)
1 (5.12)
KS) = [12,“ D£:)(’11,rs)_ ’11,rs 05:)(1 2,rs )]
(12,“ T 2'1 rs)

The limiting eigenvalues 21,” and 22,”, and the natural frequencies

corresponding to them (basic frequencies, or condensation range) are obtained from
available approximate solutions. The choice of the basic frequencies significantly affects
the accuracy of approximation of dynamic conversion matrices (5.10). This is a serious

drawback of this method.

127

5.2.2 Condensation based on the Method of Forces

After separating the full set of d.o.f. into r primary and s secondary ones, (5.6) can

be written in the form

5 6 M O I O 2
rr rs r _ l r r = O (5.13)
5sr 55s 0 Ms 0 Is zs
where I, and I S are identity matrices of orders r and s. Lumped mass system is used
here yielding the diagonal mass matrix.

The relation between the primary 2, and secondary zs unknowns is obtained

from the second equation of (5.13):
z, = —(6,, Ms — 11,11 5,, M, 2, (5.14)
Substituting (5.14) into the first equation of (5.13) yields
[(6” M —,)2.1)+ DfiS)(/1)]{} {}z, = (5.15)
in which
1))? (,1) = -6,, M, (as, M, - 1 1,)“1 as, M, (5.16)

is the dynamic transformation matrix of auxiliary displacements. Matrix Dfif) can be

approximated in the condensation range by
5§S)(4)=6" M1” (5.17)

where M 5“ ) is the matrix of mass transport from auxiliary nodes 5 into primary nodes r.

128

 

Coefficients of this matrix are obtained from the condition that matrices D901) and
59(2) coincide in the minimal condensation range:
DEW mm )= 13590 max,1.) (5.18)

where Amax, ,5: 1/ (aim is the upper value of the condensation range calculated

approximately by any known method.

From (5.17) and (5.18)
M1” =5: 51% max,rs) (5.19)

Accounting for (5.18), equation (5.15) can be written as

[arr Mr +029)“ max,rs )_ Mr ]{zr}: 0 (5'20)

01'
16,, (M, + M55) )— 11, ]{z, } = o (5.21)

The eigenvalues and the corresponding eigenvectors for the chosen condensation range

are found from the characteristic equation
[a,, (M, + M5.” )— M, ]= o (5.22)

The drawback of this approach is the need to obtain the initial value of A max by

some method. It also involves conversion of high order matrices when the total number
of d.o.f. is high, and the number of selected primary d.o.f. is low.
An advantage of the frequency-dynamic condensation method is the possibility of

stepwise elimination of secondary d.o.f. and evaluation of correctness of their assignment

129

as secondary ones based on comparing the approximation errors of the dynamic

conversion matrix in the condensation range.

5.3 Consecutive Dynamic Condensation Method

The main idea of the consecutive dynamic condensation method is to represent
the entire characteristic matrix of the system (5.1) in block form. Thus, the secondary
d.o.f. of the system are subdivided into several groups and each group is treated
separately. For each group the corresponding partial eigenvalues are found by solving

equation (5.13). Selecting lmax from these eigenvalues, dynamic transformation of the

given group of masses to the primary group is performed using (5.19). This iterative
procedure is repeated with each submatrix. Afier dynamic transformation of all groups of

masses, the final equation for the reduced system is obtained as

= o (5.23)

 

I
6,,(M, + 2M9) J — 2.1,

5:]

 

Thus the solution of the large-scale problem is substituted by the iterative
procedure of finding the eigenvectors and eigenvalues of small matrices that requires
minimum computer capacity. The method is particularly well-suited for implementation

on parallel computers.

For the case when the original system is reduced to a system with a single d.o.f.,
(5.13) can be considered as a matrix system of two linear algebraic equations. Solution of

this system yields

130

l
lmaxz'2—(Mr 6" +Ms 683)

(5.24)

 

1
+\[Z(Mr 6rr +(”36.85)’[MrA’Is(5rr63s _6rs53r)

Substituting (5.24) into (5.19) and (5.16) and performing algebraic manipulations, the

equation of mass transport from node s to node r is obtained in the following form:

arsasr Ms Mr

 

 

M)” = (5.25)
arr (653 M s T ’1 max,rs )
The coefficient of mass condensation to the main node is introduced as
(S)
kr,s = Mr = 67S 6S? Mr (526)
MS 671653 Ms _ ’1 max,rs)
The final equation for the reduced system becomes

N
5,, M, 1+ 2km — 2 = 0 (5.27)

s=1

 

 

where N is the total number of d.o.f. of the system.
From (5.27) the approximate equation for the maximum eigenvalue of the system is

obtained that can be used for the following refined analysis:

N
2,0, = 6,, M, 1+ 2km (5.28)
3:]
Test analyses of natural frequencies of beams and rectangular plates demonstrated
the high accuracy of the results obtained by (5.28). However, the technique described

provides accurate results only for the upper part of the eigenvalue spectrum (Ignatiev,

1992).

131

5.4 Energy Form of the Consecutive Dynamic Condensation Method

The energy form of the consecutive frequency-dynamic condensation method is

based on the following assumptions:

0 Equality of eigenvectors and eigenvalues of the reduced system and the
corresponding part of the eigenvector and eigenvalue spectrum of the original

system

0 Equality of kinetic energies of the reduced system and the original system

The equation for the kinetic energy of the original system has the form:
Uorig :gzmrwzzg +é—stw2 z? (5.29)

After mass transfer to the condensation nodes, kinetic energy of the reduced system

becomes

Ureduced_ ézmrw222+g Zm(s)w 2 22 (5.30)

where my) are the condensation additives resulting from transformation of all masses
ms to masses m,.

Equating (5.29) and (5.30) yields

st zs 2=mez, 2 (5.31)

or, in matrix form

{2 Mlmsl }= 2r}[m(s)]{z} (5.32)

132

 

where {2,} and {Z5} are the subvectors of the total vector of nodal displacements
corresponding to the primary and secondary d.o.f.; and [ms] and [mfg] are diagonal
matrices of orders 5 and r, respectively.

5.4.1 Condensation Using the Smallest Natural Frequency of Partial Systems

For the first form of vibration corresponding to the fundamental (smallest) natural

frequency cumin , equation (5.14) for the secondary nodal displacements can be written as

}=[D‘i’(4 4m...)l{zr} (5.331

where
[01:111.1]=—a..M.—111-*16..11M.1 (5.34)
Substituting (5.33) into (5.32) yields the dependence
121 1111111 115...] 1m. 111191....1112 14411691141 15.35)

from which the matrix of condensation additives is found:

[m1 5’] [01WmalTlmsllDr/S’amwl 15.36)

The right side of (5.36) results in a square matrix of order r x r. In order to reduce

it to the diagonal form corresponding to the structure of [mg )] , all the rows are added

and the result are written in the diagonal form. For every partial system 2mm, is found

solving (5.27).

133

Performing similar operations with all t groups of masses, the final equation for

the reduced system becomes

= 0 (5.37)

 

t
6,,[M, + 2M9] — 21,

3:1

 

Test Problem 1

Consider the prismatic simply supported beam shown in Figure 5.1. The beam

carries 12 equidistantly located point masses m = 1.

 

 

1=(N+l)a

Figure 5.1. Simply supported beam with equidistant masses

for test problems 1 and 2

The free vibration problem of the beam was analyzed by the energy form of the
consecutive dynamic condensation method using the fundamental frequency of partial
systems. Condensation was performed to two, four, and six primary d.o.f., by blocks of
two secondary d.o.f. The results are presented in Table 5.1. The values of the natural
frequencies obtained by the proposed technique are compared with accurate solutions

calculated using the formula by Ignatiev (1979).

134

TABLE 5.1. Results for Test Problem 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Arrangement 6:122:51,” Frequency wk
of the primary d.o.f. k Exact ”0139399 e (%)
solution technique
4 + 4% q, 1 0.13328 0.13409 0.61
n = 2, s = 10 2 0.00833 0.01121 25.7
. 1 0.13328 0.13341 0.09
,3." ” T‘ T” T ’ “a”: "" 2 0.00833 0.00897 7.2
n = 4, 5:8 3 0.00165 0.00206 20.1
4 0.00052 0.00092 13.8
1 0.13328 0.13375 0.35
2 0.00833 0.00891 6.5
i“ #91 e—e—e me—o-—& ‘5 3 0.00165 0.00182 9.3
T n = 6' s = 6 4 0.00052 0.00063 17.4
5 0.00021 0.00031 32.2
6 0.00010 0.00017 41.1

 

 

 

 

 

 

 

 

The results indicate that only the largest eigenvalue of the original system is
determined with sufficient accuracy. The reason for this is that the condensation
coefficient at each step accounts only for the largest eigenvalue of the characteristic

matrix of the partial system.

5.4.2 Condensation Using the Reduced Spectrum of Eigenvalues and Eigenvectors

of Partial Systems

In order to improve the accuracy of the computed natural frequencies, a different
technique for determining the matrix of condensation coefficients is proposed in this
section. As before, the set of d.o.f. of the original system is subdivided into primary and

secondary d.o.f. The displacement vector is divided into two subvectors

135

{z}= [2, 2.1 (5.38)

The procedure is based on the method of forces. The main equation of the original
system has the matrix form of (5.13) and the dependence between the primary and

secondary unknown displacements z, and zs has the form (5. 14).

Let us write (5.14) in the form
{ZS}: [Asr llzr} (539)

where [A5,] is the transformation matrix from secondary to primary unknowns.

Equation (5.39) is applicable for any k’h mode of vibration:
[zs,k ] = [Asr “Zn/c] (5'40)

To construct the matrix [A s, 1 for the partial system the first r of n eigenvectors

of the partial system are used (k = r). Since the matrix [2,1,] is square and nonsingular,

[A3, ] can be determined as

[Asr l: [Zs,k] [Zr,k l—1 (541)

Using the notation kr.s for the coefficient of mass condensation introduced earlier in

(5.26), expression (5.32) can be written as

{251T [112.1123 } = {Zr 1T [kns "611211 (5.42)

Substituting (5.41) into (5.42) yields the matrix of the coefficients of

condensation

136

[km ] = [As 17 [ms 114., l [1/ m1 (5.43)

After determining the matrices of condensation coefficients for all partial systems, the

matrix of condensed masses can be found as
[A7,]=[mrlllr +2195] (544)
The equation of the general type for the reduced system can be written as
[6,,M, - 2 1, ]{z, }= o (5.45)
and the characteristic equation for the reduced system is
[6,,M, — 2 1, = 0 (5.46)

After solving (5.46) and (5.45), the reduced spectrum of eigenvalues and eigenvectors of

the system is obtained. The eigenvectors corresponding to the secondary d.o.f. of the
system can be determined by the backward transformation. Matrix [A5, 1 is constructed

for each partial system and can be considered as the matrix of vibration modes of the ith

partial system. The matrix of vibration modes for the whole system can be written as
[B]=[I, 4,4917, (5.47)

where t is the number of partial systems. The matrix of eigenvectors can be determined as

{z}= (:3 = [8114} (5.48)

137

Test Problem 2

Consider the prismatic simply supported beam shown in Figure 5 .1. The free

vibration problem of the beam is analyzed by the energy form of the consecutive dynamic

condensation method using the reduced spectrum of eigenvalues and eigenvectors of

partial systems. The total number of d.o.f. of the system is twelve. Condensation was

carried out using two, four, and six primary d.o.f. The schemes of the arrangements of

condensation nodes are shown in Table 5.2. The total number s of the secondary d.o.f.

was divided into b blocks. The number of blocks (b) and the number of secondary

unknowns in each block (8,) varied.

TABLE 5.2 Arrangement of the primary degrees of freedom for the Test Problem 2

 

 

 

 

 

 

 

 

 

 

 

Scheme Arrangement n/s Block size

N0 of the primary d.o.f. ratio bxs.
‘ 10x1

n = 2, s = 10 2x5

T—c- ‘—+—++¢v+—.—o—o——. o “8*“: 8X1

II "‘ ““' 1/2 4x2
n = 4, 5:8 2x4

, 6x1

T H-..fl+w+jfi(

Ill .1. .4 1 3x2
n = 6, s = 6 2x3

 

 

 

 

Results are presented in Table 5.3. The eigenvalues obtained for the reduced

system are compared with accurate values calculated using the equation obtained by

Ignatiev (1979) for the case N=12:

(0k -—
ma

138

2 _ 48EI(S, 4 kn

 

TABLE 5.3 Results for Test Problem 2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Order Frequency a)
scam "/7 Of frequency bxsb Exact Proposedk 0
ratio k solution technique 8 (A)
10x1 0.13362 0.26
1 5X2 0.13328 0.13471 1.07
l 1,5 2X5 0.13193 -1.01
10x1 0.00903 8.4
2 5X2 0.00833 0.00904 8.5
2X5 0.01134 36.1
8X1 0.13356 0.21
1 4X2 0.13328 0.13342 0.10
2X4 0.13343 0.11
8X1 0.00849 1.92
2 4X2 0.00833 0.00862 3.48
II 112 2X4 0.0085 2.04
8X1 0.00169 2.37
3 4X2 0.00165 0.00179 7.82
2X4 0.00173 4.62
8X1 0.00062 16.13
4 4X2 0.00052 0.00059 11.86
2X4 0.00058 10.34
6X1 0.13345 0.13
1 3X2 0.13328 0.13347 0.14
2X3 0.13343 0.11
6X1 0.0084 0.84
2 3X2 0.00833 0.00839 0.72
2X3 0.00838 0.60
6X1 0.00167 1.20
3 3X2 0.00165 0.00175 5.71
Ill 1 2X3 0.00168 1.79
6X1 0.00056 7.14
4 3X2 0.00052 0.00056 7.14
2X3 0.00057 8.77
6X1 0.00023 8.70
5 3X2 0.00021 0.00023 8.70
2X3 0.00028 25.00
6X1 0.00012 16.67
6 3X2 0.00010 0.00009 -11.11
2X3 0.00019 47.37

 

 

 

 

 

 

 

139

 

 

The following observations are maid:
- The proposed technique provides good accuracy for approximately 25% of the
spectrum of eigenvalues
o The number of blocks does not affect the results significantly
0 Increasing the number of condensation nodes improves accuracy. The

recommended ratio of primary to secondary nodes is ‘/2 < n/s <1

5.4.3 Combined Static and Consecutive Dynamic Condensation

The eigenproblem stated in the form of the displacement method has the form
(5.4). The technique proposed in this section combines the two methods presented earlier
— static condensation and the energy form of consecutive dynamic condensation. After
classifying all the d.o.f. of the system under consideration as primary and secondary, the
main equation (5.4) transforms to (5.7). The system is then subdivided into a set of partial

systems, and the secondary d.o.f. are represented in a block form.

In the first step Guyan’s transformation is performed for the partial systems. The

vector of displacements of the partial system is written as

{H'} =[A,]{z.} (5.49)

where

M

Ir
1 (5.50)
_' Ks; Ksr

is the matrix of static condensation.

140

The statically condensed stiffness and mass matrices of the partial system are

obtained through

[K:r]= [Ar]T [K ][Ar] = Krr "Krs Ks_sl Ksr
(5.51)
[M:r]=[Ar]T[M][Ar] =Mrr—MrsMs-31Msr

The eigenvalues and eigenvectors of the partial system are then obtained from the

solution of the characteristic equation
[ [Kg ]- ,1[M:, H {z} = 0 (5.52)

In the second step, refined condensation is performed based on the energy form of
the consecutive dynamic condensation method using the reduced spectrum of eigenvalues

and eigenvectors of partial systems described in Section 5.4.2.

Condensed stiffness and mass matrices of the partial system are obtained through

the transformation

[29]: [BIT [KirllBl
(5.53)
[Mg-4431f [M:.][Bl

Performing condensation of all 1 blocks of secondary d.o.f., the final stiffness and mass

matrices of the system are obtained by the summations

[Err]: :[Er'ir 1‘ (t-1)[Krr] (5-54)

141

[M..]=§[M£.]—(z-1)[M..]

(5.55)

The eigenvalues and eigenvectors of the reduced system are obtained from the

characteristic equation

Test Problem 3

[ [g,,]_.[12,,]]{.}=o

(5.56)

Consider a prismatic single-span beam with fixed ends loaded with a uniformly

distributed mass of intensity m = 1. All geometrical and physical characteristics of the

cross section are taken equal to unity. The beam is divided along the length into 72 linear

finite elements. The length of each element is assumed to be a =1.

TABLE 5.4. Variants of condensation for Test Problem 3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Number of
Total number of Number of Number of
secondary d.o.f.
dfi'f primary d.o.f n/s, "lo blogks in one block
n 85
2 68
6 4.4
4 34
2 66
1 0 7.6 3 44
6 22
2 64
142
14 1 1 4 32
8 16
2 60
22 18 3 4 30
' 6 20
1 2 1 0

 

 

 

 

 

142

 

The problem was solved by the combined static and consecutive dynamic
condensation method. Different variants of condensation were performed by varying the

number of primary d.o.f. and the secondary d.o.f. blocks as shown in Table 5.4.

Figure 5.2 illustrates the location of the primary d.o.f. and secondary d.o.f. blocks
for the case with 22 primary d.o.f. Results for this case are presented in Table 5.5. The
eigenvalues obtained by both methods are compared with the results obtained by solving
the full system of equations. Figure 5.3 illustrates the distribution of errors for the
eigenvalues obtained by different techniques. Similar graphs for the other cases listed in

Table 5.4 are shown in Figures 5.4 —— 5.6.

Test Problem 4

Consider a beam described in Test Problem 3 with three equidistant intermediate
roller supports added (Figure 5.7). Each span consists of 18 finite elements. The total

number of d.o.f. of the structure is 139.

Condensation was performed using eleven, eight, and three primary d.o.f.
Arrangement schemes of the primary d.o.f. and blocks of the secondary d.o.f. are shown

in Figure 5.7(a), (b), and (c).

Results are presented in Tables 5.6, 5.7, and 5.8 and illustrated in Figures 5.8, 5.9,

5.10, and 5.11.

143

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econ :88 I I

 

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144

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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145

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

00... 0000.. 00.0. 000... 0.0. 0000... 00... 0000.. 00.00 0.05.. .0000 00
00. .0 0000.0 00.00 0000.. 00.0 0500.0 00.0 0005.0 00.00 0000.. 0055.0 ..0
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00.0 0000.0 .00 5500.0 00.0 0000.0 .00 0000.0 00.00 0000.0 0000.0 0..
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00x0 208 00x0 0...". 0.0.0 0050.. m in

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0050.0 00300000000 0.0.050 0>30000000 0.3 .0 000. 3.000 03030 0.3 .0 0030.00 m. m...

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146

Error, %

~-—-(-_>-- 4 blocks x 30 d.o.f. +2 blocks x 60 d.o.f. 1

Error, “lo

204 ------------------------------- — ————————————————————
10« ------------------------------------------------------
o ._—--.“. .. .. ._.; ‘

! :0;8 _b—locks ;1—t5 d.o.f.

 

 

 

12 3 4 5 6 7 8 910111213141516171819202122
Eigenvalue Number

 

131 Static condensation +12 blocks x 10 d.o.f ”— _ —6 Hike-{2070.0}. ' ;

l

 

Figure 5.3. Graphs of errors for Test Problem 3 for the case
with 22 primary d.o.f.

 

 

 

 

 

OIJIIIIII
12345678910111213141

Eigenvalue Number

 

+4 bloeks x 32 d.o.f. +{bl0ks 06400.13 *7 .

 

Figure 5.4. Graphs of errors for the Test Problem 3 for the case
with 14 primary d.o.f.

147

Error, %

Error, %

 

 

 

 

0§—-Fl a o

 

 

1 2 3 4 5 6 7 8 9 10‘

 

Figure 5.5. Graphs of errors for Test Problem 3 for the case

 

 

 

 

 

with 10 primary d.o.f.
3________-___-_.---_
2 . _______________________________________________________
1 .. _______________________________________________________
oh I I .

1 2 3 4 5 6

Eigenvalue Number

 

,' + 1 blocks x 34 d.o.f. +2 blocks x 68 d.o.f ‘

Figure 5.6. Graphs of errors for Test Problem3 for the case
with 6 primary d.o.f.

148

 

3030.... 300,—. 00.. 00.00... .30... 000000000 000 3.0.0 E080:— 0... ..0 0030004

.50 2:00.

 

 

 

 

 

 

 

 

0000 0.0.0 - I 0000 0.0000000 - o
E 0 0
a
0 0 0 . .0.
E 0
a
0 0 .
.x _\ _\ .\
.7 a... ._ a... a... ._ a: ._
E o o
a
0 N . A0.

149

TABLE 5.6. Results for Test Problem 4 (condensation to 11 primary d.o.f.)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

g Eigenvalues obtained by

3: s?L‘ii'§§‘.?§.§."° fifgifggmfiggf' °°"§Zf.32§§£¥§§mi°

% N = 142 to 11 primary d.o.f.

g, 24,)th3 Mxloz’ e(%) kkx103 8(%)

E M I ¢i I
1 1.2628 1.2763 1,07 1.26287 0.01 0.12
2 2.2645 2.3066 1,86 2.26451 0.00 0.02
3 3.7804 3.8886 2,86 3.78043 0.00 0.2
4 4.7683 4.9242 3,27 4.76842 0.00 0.06
5 17.4851 22.1025 26,41 17.48739 0.01 0.17
6 23.7822 32.4956 36,64 23.78883 0.03 0.27
7 31.6421 49.2494 55,65 31.66137 0.06 0.78
8 36.2342 64.0146 76,67 36.26416 0.08 1.65
9 83.8759 151.1493 80,21 84.6864 0.97 11.02
10 103.539 230.7529 122,87 105.76222 2.15 25.6
11 126.4732 330.8788 161,62 137.7662 8.93 102

 

 

 

 

‘ 350
300 ~
250 4
200 .
150 .

 

 

Eigenvalue x 1000

 

 

 

1 2 3 4 5 6 7 8 9 10 11 ,
Eigenvalue Number

i + Exact solution
i —I-- Static condensation to 11 d.o.f.
L - 10 — Dynamic condensation to 11 d.o.f.

 

Figure 5.8. Results for Test Problem 4 (condensation to 11 primary d.o.f.)

150

3.0.0 0.3.00:— 3 00 00m000000000v v .00—00.5 000a. 00.0 000000 0002 .06 0.0—ME

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*

 

 

 

 

152

TABLE 5.7. Results for Test Problem 4 (condensation to 8 primary d.o.f.)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

A
C
J 1

G _‘ Eigenvalues obtained by
3
E '- Solution of the Consecutive dynamic
E g full system St??? cgnmcgensgtgtzn condensation
g; N= 142 P 'V ' " to11 primary d.o.f.
‘” A... x103 m 103 e (%) 1... x103 8 (%)
1 1.2628 1.28703 1.92 1 .26288 0.01
2 2.2645 2.33339 3.04 2.26453 0.00
3 3.7804 3.92298 3.77 3.78049 0.00
4 4.7683 4.92419 3.27 4.76843 0.00
5 17.4851 28.65426 63.88 17.5231 0.22
6 23.7822 40.88108 71.90 23.87258 0.38
7 31.6421 56.21311 77.65 31.83166 0.60
8 36.2342 64.01454 76.67 36.4974 0.73
[_ _ __ _ _ - __ _ ._ _ _ _._ s
70 i
8 60 I
O i
P 50
X
0 40 - '
2
g 30 -
5 20
.9
m

 

 

 

 

 

O

 

1 2 3 4 5 6 7 8

EigenvalueNumber

 

§.+ Exact solution .
—I— Static condensation \
[ :v-QLConsecutwidynamic condensation

Figure 5.10. Results for Test Problem 4 (condensation to 8 primary d.o.f.)

153

TABLE 5.8. Results for Test Problem 4 (condensation to 3 primary d.o.f.)

 

 

 

 

 

 

 

0 a‘ Eigenvalues obtained by
321.133.:3“ sgtgcpcgggggaggn “02:32:33?“
g, g N = 142 ' ' ‘ to 3 primary d.o.f.
x0103 2...);103 0%) 9.0.103 s(%)
1 .2628 1 ,153033 -8,69 1 ,263967 0,09
2.2645 1 ,690078 -25,37 2,275782 0,50
3.7804 4,000915 5,83 3,861551 2,15

 

 

 

 

 

 

 

 

 

 

 

 

 

\
3 4 5 6 7 8 91011l
Eigenvalue Number ‘

\

 

+Dynamic condensation to 11 d.o.f.” _ if t
+ Dynamic condensation to 8 d.o.f.
_ _-_jA - Dynamic condensation to 3 d.o.f.

0 __. -, mgifi, , i

 

Figure 5.11. Graphs of errors for Test Problem 4

154

Test Problem 5

Consider a free vibration problem of a rectangular lattice plate with a 16x16
orthogonal grid shown in Figure 5.12. Concentrated masses are applied at the nodes. The

problem was solved in the non-dimensional form using the following

assumptionszll =12 =1, E11 = E12 = E1, and mij = m. The transformation to the

dimensional form can be done using the expression (0:071/EI / ml3 wherea? is

dimensionless frequency.

The problem was solved by the combined static and consecutive dynamic
condensation method using three condensation schemes shown in Figure 5.13. The
dimensionless values of natural frequencies are presented in Table 5.6. The results
obtained by the proposed method are compared with the exact solutions obtained by

Ignatiev (1979).

   

 
 
  
  

”I, l
I”II/I/III’”II
[W l/

12

 

Zv

Figure 5.12. Rectangular lattice plate with a 16x16 orthogonal grid

155

 

 

 

 

 

 

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aw ) \W
( cw (

156

 

Figure 5.13. Condensation schemes for Test Problem 5: (a) to 29 primary nodes;
(b) to 49 primary nodes; and (c) to 81 primary nodes

TABLE 5.6. Results for Test Problem 5

 

Dimensionless frequency values obtained by

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

% f Combined static and consecutive dynamic condensation method
g g Exact with condensation to

a 5 SOIUtiON 29 nodes 49 nodes 81 nodes

(751x102 e (%) 651x102 e (%) (751x102 s (%)

1 5.452232 5.465317 0.24 5.456594 0.08 5.450051 -0.04
2 21.80858 21.83475 0.12 21.79985 -0.04 21.9089 0.46
3 49.06569 48.96265 -0.21 49.0755 0.02 49.18345 0.24
4 87.20919 87.40977 0.23 87.23535 0.03 87.64523 0.5
5 136.1962 136.4958 0.22 136.1825 -0.01 137.6398 1.06
6 195.9207 191.9827 -2.01 196.0383 0.06 196.7044 0.4
7 266.1501 269.1309 1.12 266.2299 0.03 266.6824 0.2
8 346.4102 359.2966 3.72 346.3755 -0.01 346.7912 0.11
9 435.7914 462.7233 6.18 436.0528 0.06 439.2777 0.8
10 532.6364 554.581 4.12 533.0093 0.07 526.0317 -1.24
11 634.0843 654.1214 3.16 633.8307 -0.04 640.1716 0.96
12 735.5039 774.0443 5.24 733.8858 -0.22 744.6241 1 .24
13 830.0138 897.577 8.14 831.1759 0.14 835.4089 0.65
14 908.5588 1001.959 10.28 908.2863 -0.03 910.1034 0.17
15 961.1921 1006.656 4.73 963.4989 0.24 974.9371 1.43
16 979.7959 992.2393 1.27 980.9717 0.12 985.2828 0.56

 

 

 

 

 

 

 

 

157

 

 

 

 

 

'4T Vii? V.7 ‘ \
123456789101112131415161

Number of frequency

 

i —O— Condensation to 29 nodes + Condensation to 49 nodes
i - 1A — Condensation to 81 nodes

Figure 5.14. Graphs of errors for Test Problem 5

Examples of dynamic analysis of a frame and an isotropic plate using the
proposed method are given in Appendix E. Based on the obtained results the following

conclusions can be made:
0 The proposed technique allows for solution of a broad range of problems

0 A preliminary static condensation used in combination with the energy form of
the consecutive dynamic condensation method provides better results than both

methods used alone

0 The proposed technique allows to determine approximately 70% of the reduced

spectrum of eigenvalues with sufficient accuracy for different types of problems

0 Block form of condensation provides a reduction of computation effort and

improves the efficiency of proposed technique

158

o The number of blocks does not significantly affect the accuracy; therefore it is
possible to use lesser number of blocks with larger number of secondary d.o.f.

when appropriate

0 The number of main d.o.f. and their location play the main role in accuracy of

results

159

CHAPTER 6

CONCLUSIONS AND RECOMMENDATIONS

Two classes of new and effective approximate methods for static and dynamic
analysis of large lattice structures using decomposition and consecutive dynamic
condensation techniques are developed in this work. The main results of the work are

discussed below.

A decomposition method proposed by Pshenichnov is developed for solving
bending and free vibration problems of thin isotropic plates with elastic supports. Simple
and accurate approximate analytical formulas for the displacements, force responses, and
eigenvalues of these boundary value problems are obtained. A comparison of results with
well-known solutions for rigid and hinged supports, demonstrates the high accuracy of
this method. The merit of this method is the flexibility in the decomposition of the
original problem, which provides wide latitude for choosing the auxiliary problems that

facilitate the construction of the desired solution.

An effective technique for solving bending and free vibration problems of lattice
plates with different combinations of supports is developed based on continuum
modeling of lattice plates. The continuum model developed by Pshenichnov is used in
this work. It is demonstrated that the continuum model together with the decomposition
method yields an accuracy of within 2% for displacements and bending moments, which

is adequate for preliminary design and optimization purposes.

160

It is demonstrated that the developed analytical dependencies can be used to
obtain optimal lattice geometries for a class of plate problems. The proposed technique
was implemented into the PLAST computer program for analyzing rectangular lattice
plates with different types of lattices and different values of support rigidities. PLAST is

written in the C programming language and can be used on personal computers.

Analytical formulas for calculating the fundamental frequency are obtained for
lattice plates with different combinations of support rigidities. The results obtained for
the test problems demonstrate that the decomposition method yields sufficient accuracy
for the fundamental frequency. The analytical dependencies obtained can be used for
optimization purposes. However, the method is intractable for estimating higher

frequencies and mode shapes.

The decomposition method is generalized to include bending and free vibration
equations derived from finite difference formulations. This approach allows the original
discrete models of lattice plates to be used. Simple approximate analytical solutions are
obtained for bending and free vibration problems of lattice plates in the form of systems
of orthogonal beams with different types of supports. The results demonstrate that the
decomposition method stated in finite difference form for bending problems of lattice
plates with different types of grids and support rigidities is sufficiently accurate. The
finite difference formulation of the method provides better results than the differential
formulation for sparse grids. However, for dense grids both formulations yield similar

accuracy.

An energy form of the consecutive dynamic condensation method proposed by

Ignatiev is developed. It is demonstrated that preliminary static condensation used in

161

combination with the energy form of the consecutive dynamic condensation method
provides better results than both methods used alone. The proposed technique is capable
of determining approximately 70% of the reduced spectrum of eigenvalues with
sufficient accuracy for different types of problems. The block form of condensation
yields a reduction of computational effort and improves the efficiency of the proposed
technique. The number of blocks does not significantly affect the accuracy; therefore it is
possible to use fewer blocks with a large number of secondary d.o.f. when appropriate.
The proposed technique is computationally efficient due to the resulting block diagonal

equations and is suitable for implementation on parallel computers.

Based on the results, the following directions for future research are

recommended:

0 Inclusion of shear deformation and joint flexibility in the continuum and finite

difference formulation of the decomposition method

0 Application of the decomposition method to nonlinear static and dynamic

problems

0 Development of a continuous model for composite laminated plates that will
make it possible to use the decomposition method for static and dynamic analysis

of laminated structural elements

0 Application of the decomposition method to stability problems

162

Extension of the proposed technique to broader class of lattice structures
including static, dynamic and stability problems of multi-layer lattice plates and

large lattice shells

Use of the finite difference formulation of the method to account for more
complicated types of grids, for example grids with openings, grids with double

regularity, etc.

Development of computer programs for static and dynamic analysis of lattice

plates and shells

Application of the energy form of consecutive dynamic condensation method to
the analysis of complex structural systems such as lattice shells, thin-walled

cellular structures, complex frames, etc.

Development of computer programs based on the consecutive dynamic

condensation method for the analysis of complex structural systems

163

 

APPENDICES

164

APPENDIX A

ARBITRARY FUNCTIONS FOR THE PROBLEM
PRESENTED IN SECTION 2.4.2

CI = 0.03333333333 (0.3808425 107 1.2 k14 + 0.1105425 107 1.4 k,4 + 0.2652250 107 k]4
+ 416160. 1.4 1:13 + 0.1030000107 k13 + 0.1456380107 12 k13 + 65178.14 kl2

+ 151500. k,2 + 220980. x2 k]2 + 10000. k1 + 15660. 7.2 k1 + 4896. 1.4 kl + 250.
4153.1.4 +435.1.2)kl / (79866. 7.4 k,2 +0.1838550107 82 k,5

+ 0.2353905 107 1.4 k14 + 25. + 809760. 1.2 1:13 + 0.2541630 107 1.2 k14 + 5355. 34 k1

+ 611694. 1.4 1113 + 87850. 1:13 + 0331627510" 14 k15 + 14150. k,2 + 975. kl
+ 210. 12 + 114240. 1.2 kl2 + 7770. 32 kl+162225.k14 — 265225. k15 + 1531‘)

C2 = - 0.03333333333 (_1. + k1) (0.3808425 107 1.2 k14 + 0.1105425 107 x4 k14
+ 0.2652250 107 k14 + 416160. A4 kf + 0.1030000 107 1:13 + 0.1456380 107 1.2 kl3

+ 65178. at“ 112 + 151500.152 + 220980. 1.2 k12 + 10000. k1 + 15660. A2 k1
+ 4896. 1‘ k1 + 250+ 153.1.4 + 435. 7.2) / (79866. 1.4 k12 + 0.1838550 107 1.2 kls

+ 0.2353905 107 1.4 1:14 + 25. + 809760. 7.2 k13 + 0.2541630 107 32 k14 + 5355. 14 kl

+ 611694. 7.4 k13 + 87850. k13 + 0.3316275 107 x4 k,5 4141501]2 + 975. k1
+ 210. 7.2 + 11424031.2 kl2 + 7770. 7.2 k1 + 162225. kl4 — 265225.115 +1533“)

165

C3 = — 0.05000000000 (250. + 418200. 1.2 kl2 + 29675. 1.2 k1 + 878500. 1:13
+ 141500. k12 + 0.2749400 107 1.2 kf + 0.1622250 107 k14 + 0.1574370 107 1.4 kf

+ 220014. A4 kl2 + 825. 1.2 + 9750. k1 + 0.5527125 107 7.4 kl5 — 0.2652250 107 kl5
+ 15453. 1.4 k1 — 218875. 1.2 kls + 0.5397075 107 7.4 k14 + 459. 7.4

+ 0.7139175 107 7.2 1:14) / (79866.1.4 kl2 + 0.1838550 107 7.2 k,5

+ 0.2353905 107 1.4 k14 + 25. + 809760. 7.2 kl3 + 0.2541630 107 7.2 k14 + 5355. 7.4 k1
+ 611694. 1.4 1:13 + 87850. 1:13 + 0.3316275 107 7.4 k]5 + 14150.1:12 + 975. k1

+ 210. 78 4.11424032 kl2 + 7770.).2 k1+162225.kl4 — 2652251]5 +1531“)

C5 = 005000000000 (0.1400800 107 22 k14 + 368475. 1.4 k14 + 0.1326125 107 [(14
+ 138720. 1.4 1:13 + 515000. k13 + 535680. 1.2 kf + 21726. 1.4 k,2 + 75750. kl2

+ 81280.1.2 k,2 + 5000. kl + 5760.1.2 k1 + 1632. 1.4 k1+125'+ 51. 1.4 + 160. 7.2) kI
/ (79866. 7.4 kl2 + 0.1838550 107 1.2 kl’ + 0.2353905 107 1.4 kl4 + 25.

+ 809760. )3 1:13 + 0.2541630 107 1.2 kl4 + 5355. 7.4 k1 + 611694. 7.4 kl3 + 87850. kl3

+ 0.3316275 107 1.4 kl“ +14150.k12 + 975./(1+ 210.12 411424012 1,2
+ 7770.12 11 + 162225. kl4 — 265225.k15 + 1531‘)

166

C6 = — 005000000000 (—1. + kl) (0.1400800 107 12 k14 + 368475. 7.4 k14
+ 0.1326125 107 kl‘1 4138720.).4 1:13 + 515000. k13 + 535680. 1.2 kl3

+ 2172614 1:,2 + 75750.1:12 + 81280.1.2 k12 + 5000.1:l + 5760. 7.2 kl + 1632. 7.4 k1
+ 125. +51. 7.4+ 160. 1.2) / (79866. x4k12+0.1838550 10713195

+ 0.2353905 107 14 k14 + 25. + 809760. 78 kf + 0.2541630 107 1.2 1:14 + 5355. 7.4 k1

+ 611694. 1.4 1:13 + 87850. 1:13 + 03316275107 70‘ kls 414150.152 + 975. k1
+ 210.1.2 +114240.1.2 kl2 + 7770.1.2 kl+162225.kl4 - 265225. kl5 +1531“)

C7 = — 003333333333 (250. + 10251. 7.4 k1 - 0.1707225 107 x2 kl5
+ 0.3316275 107 1.4 k15 + 0.1809300 107 1.2 k13 + 141500.k12 + 570. 7.2 + 9750.1l

+ 282540. 1.2 k12 + 20325. 1.2 k1 + 0.4337490 107 7.2 k14 + 0.3459330 107 i4 k"
+ 0.1027854 107 1.4 k13 + 145044. 1.4 kl2 + 878500. k13 + 306. 1.4

+ 0.1622250107 kl4 -O.2652250107 kls) / (79866. 7.4 kl2

+ 0.1838550 107 1.2 kl5 + 0.2353905 107 1.4 k14 + 25+ 809760. A2 kl3
+ 0.2541630 107 1.2 k14 + 5355. 1.4 k1 + 611694. 1.4 1:13 + 87850. 1:13

+ 0.3316275 107 1.4 1,5 414150.ch2 + 975.1:l + 210.1.2 + 114240. 7.2 k,2
+ 7770.12 kl+162225.k14 — 265225.k15 4.15314)

167

C9 = 002857142857 (0.1488350 107 7.2 k14 + 368475. 14 k14 + 0.1856575 107 kl4
+ 13872014 1:13 + 721000. 1:13 + 569160. 7.2 1:13 + 21726.14 kl2 + 106050. k12

+ 86360.1.2 kl2 + 7000. kl + 6120. 1.2 kl+ 1632.1.4 kI + 175. + 51. x4 + 170. 7(2) kl
/ (79866. 1.4 k,2 + 0.1838550 107 7.2 kl5 + 0.2353905 107 1.4 k14 + 25.

+ 809760. 1.2 1:13 + 0.2541630 107 7.?- k14 + 5355. 1.4 k1+ 611694. 1.4 1:13 + 87850. k13

+ 0.3316275107 1.4 kl5 +14150.k12 + 975~k1+ 210. 7.2 + 11424012 k,2
+ 7770.1.2 k1 + 162225114 — 265225.195 4.15324)

C10 = — 0.02857142857 (-1. + k1 ) (0.1488350 107 22 k14 + 368475. 7.4 kl4
+ 0.1856575 107 k14 + 13872014 1:13 + 721000. 1:13 + 569160. 32 kl3

+ 21726. 1.4 kl2 + 106050.k12 + 86360. 1.2 kl2 + 7000. k1 + 6120. 1.2 k1 + 1632. x4 11
+ 175.+51.x4+170.1.2)/(79866.14k12+0.1838550 1071.2195

+ 0.2353905 107 1.4 k14 + 25. + 809760. 1.2 1:13 + 0.2541630 107 1.2 k14 + 5355. 1.4 kl

+ 611694. 1.4 1:13 + 87850. k13 + 0.3316275107 1.4 1:15 +14150.kl2 + 975. kl
+ 210. 7.2 4114240.).2 k,2 + 7770.1.2 kl+162225.kl4 - 265225.k15 +153.>.4)

Cll = — 0007142857143 (875. + 49420. 13 kl — 0.8579900 107 1.2 k15
+ 0.7737975 107 1.4 k]5 + 0.5677875 107 k,4 - 0.9282875 107 k.5 + 495250. k,2

+ 1400. 12 + 0.3074750 107 k13 + 675920. 1.2 kl2 + 0.8975960 107 1.2 k14
+ 0.8440245 107 7.4 k14 + 34125. k1 + 0.2537046 107 1.4 1:13 + 25551. 1.4 k1 + 765. x4

+ 0.4189360 107 1.2 1:13 + 360162. 7.4 klz) / (79866. 7.4 kl2 + 0.1838550 107 1.2 k,5

+ 0.2353905 107 1.4 k14 + 25. + 809760. 12 1:13 + 0.2541630 107 1.2 k14 + 5355. 1.4 k1
+ 61169414 k" + 87850. 11,3 + 0.3316275107 1.4 kl5 414150.112 + 975. k1

+ 210.12 +114240.1.’- k,2 + 7770. 7.2 kl+162225.k14 — 265225.k15 +1531“)

168

APPENDIX 8

PROGRAM “PLAS” FOR ANALYSIS OF LATTICE PLATES

/***************************************~k*~k***************/

PLAS

Analysis of lattice plates with elastic support

#include <graphics.h>
#include <stdlib.h>
#include <math.h>
#include <stdio.h>
#include <bios.h>
#include <conio.h>
#include <io.h>

int errorcode, graphdriver, graphmode;

char work,str[6][30];
union inkey char ch[2]; int ii; jj, dd;

int i,ij,j,ji,k,ki,l,li,m,mi,n,ni,e[5][5];

double
dl,ee,fl,f2,f3,f4,ga,k1,k2,k3,k4,12,l4,lm,lml,lm
2,nu, pi,t1,t2,zl,22,23,z4;

int dl,d2,d3,d4,e0;

double

1mm{21],a[5] [5],X[9],y[9],V[9] [9],W[9] [9],f[2] [7];

double

aa,a0,a1,a2,a3,a4,a5,bb,b0,bl,b2,b3,b4,b5,cc,c0,cl,c2
c3,c4,el,e2,rr,51,52,s3,s4,xl,x2,x3,x4,x5,x6,yl,
312.113,
y4,y5,y6,ul,u2,u3,u4,vl,v2,v3,v4;

double O[5] [5],g[5].qqrq0,p[5] [5],b[2] [5],C[4] [4];

void opred(),tire(),ramka(),risl();

169

/* nepemeHHme rpaQMKM */

int

il,iZ,i3,i4,i5,i6,jl,j2,j3,j4,j5,j6,iOy,ij,lx,ly,
lxl,ly1,xx{2][9],yy[2][9],ZZ[9][2];

FILE *fd; FILE *fp;

main()
{
clrscr(); /* quCTKa aKpaHa */
if ( access("plas.dn",0)==—l )
printf( "\n HeT @afina mcxonnux naHme — plas.dn\n" );

exit(l);
fd=fopen("plas.dn","r");
fp=fopen("plas.pr","w");

printf("\n Pacqu nnaCTMHKM Ha ynpyPOM OCHOBaHMM\n");

printf("\n Been mcxonawx naHHux\n");
fscanf(fd,"%lf%lf%lf%d",&lml,&lm2,&dl,&ij);
fscanf(fd,"%lf%lf%lf",&tl,&t2,&ga);
fscanf(fd,"%d%d%d%d",&dl,&d2,&d3,&d4);
fscanf(fd,"%lf%lf%lf%lf",&fl,&f2,&f3,&f4);
fscanf(fd,"%lf%lf%lf%lf",&kl,&k2,&k3,&k4);

tire();
fprintf(fp,"Pacqu nnaCTMHKM Ha ynpyPOM OCHOBaHMM ");
tire();
fprintf(fp,"\n M c x o n H u e n a H H m e");
tire();
fprintf(fp,"\n lml lm2 dl
ij");
fprintf(fp,"\n%l8.2f%15.2f%15.2f%12d\n",lml,lm2,dl,ij);
fprintf(fp,"\n t1 t2 ga");
fprintf(fp,"\n%18.2f%15.2f%15.2f\n",tl,t2,ga);
fprintf(fp,"\n dl d2 d3
d4");
fprintf(fp,"\n%15d%15d%15d%15d\n",dl,d2,d3,d4);
fprintf(fp,"\n fl f2 f3
f4");
fprintf(fp,"\n%l8.2f%15.2f%15.2f%15.2f\n",fl,f2,f3,f4);
fprintf(fp,"\n kl k2 k3
k4");
fprintf(fp,"\n%l8.2f%15.2f%15.2f%15.2f\n",kl,k2,k3,k4);
tire();
n=O;

170

if (lml<l II lml>lOO) n=l;
if (lm2<l ll lm2>lOO) n=l;
if (dl<O.l II dl>lO) n=l;
if (tl<O ll tl>5) n=1;
if (t2<O || t2>5) n=l;
if (dl<0 ll dl>l) n=l;
if (d2<0 ll d2>1) n=l;
if (d3<O || d3>l) n=l;
if (d4<O II d4>l) n=1;
if (fl<—9O ll fl>90) n=l;
if (f2<-9O ll f2>90) n=l;
if (f3<-9O ll f3>90) n=l;
if (f4<-9O ll f4>90) n=l;
if (kl<O ll k1>l) n=l;
if (k2<O ll k2>l) n=l;
if (k3<0 II k3>l) n=l;
if (k4<0 ll k4>l) n=l;
m=(lm2—lml)/dl+l; /* quTgMK uMKna no lm */
if (m>20)
printf("\n gmcno maroa no /nHM6na/ npeemmaeT 20 — N =
%d\n",m);
exit(l);
if (n==l)

printf("\n OnMH M3 mcxonumx napamerpoa MMeeT HenonyCTMMoe
3Haqeame !\n");

exit(l);
pi=3.14159265359; ee=2.7l828183; nu=O.3;
fl=fl*pi/180; f2=f2*pi/l80;
f3=f3*pi/l80; f4=f4*pi/l80;

printf("\n Hagano paooru nporpaMMH\n");

sl=sin(fl); cl=cos(fl);
a0=dl+sl*d3+c1*d4;
rr=d1*(cl*cl*cl*cl)+d4*cl+dl*ga*(sl*sl)*(cl*cl);

/* Koeoomumearm npm HeMBBeCTme CMCTeMu ypaBHeHMfi */

lm=lml; 12=lm*lm; l4=lm*lm*lm*lm;
ul=l+2*kl; u2=l+4*kl; u3=1+6*kl;
vl=l+2*k2; v2=l+4*k2; v3=l+6*k2;

a[1][l]=l+4*k2+2*u1*u1*tl/(3*12)+u2*t2/l4;

a[l][2]=u2*vl*tl/(15*12)+2*u3*t2/(15*14);

a[1][3]=2*V3+u1*V2*t1/12;

a[l][4]=u2*v2*tl/(lO*lZ);

a[l][O]=1;
a[2][l]=(-4)*(Vl*tl/12+ul*t2/l4);

171

a[2][2]=2*(l+4*k2-u2*t2/(5*l4));
ai2][3]=(-6)*(V2*t1/12);
a[2][4]=4*v3;
a[2][O]=O;
a[3][l]=(—4)*(1+2*k2+ul*tl/12);
a[3][2]=(—2)*u2*tl/(5*12);
a[3][3]=(-6)*(l+4*k2-5*u2*t2/l4);
a[3][4]=4*u3*t2/l4;
a[3][O]=O;
a[4][1]=6*t2/12;
a[4][2]=(-2)*V1*t1;
a[4][3]=0,
a[4][4]=(-3)*v2*tl;
a[4][0]=0;
/* enmnquue KOGQQMUMGHTH npm MMHOan */
for (l=l; l<=4; l++)
for (m=l; m<=4; m++)
k=l+m; eO=-l;
for (n=1; n<=k—l; m++) eO=eO*(-l);

e[l][m]=e0;
/* qumcneame rnaBHoro onpenenmrena */
for (m=l; m<=4; m++)
for (n=0; m<=4; m++)
pimiin1=a[m][n];
Oimiini=a[m][n];
Opredi); g[O]=qq;

/* qumcneame BTopocreneHHux onpenenmrenefi */
for (i=1; i<=4; i++)
for (m=l; m<=4; m++)
for (n=1; m<=4; m++)
pim][n]=0[m][n];
for (m=1; m<=4; m++)
pimiiil=a[m][0];
opred(); g[i]=qq;
/* BHaquMH HeMBBeCTme */
if (g[O]<0.000l)
printf("\n PnaBHmfi onpenenMTenb paBeH Hynm !\n");

exit(l);
21=g[1]/g[0]: 22=g[2]/g[0];
z3=g[3]/g[0]; z4=g[4]/g[01;

/*********************~k***********************************/

/* qumcneame npormooa M MOMeHToe */
X[0]=0; Y[0]=0;
for (k=l; k<=8; k++)
x[k]=x[k-l]+0.125; y[k]=y[k-1]+O.125;

172

for (j=O; j<=8; j++)
for (i=0; i<=8; i++)
/* KOOpflMHaTH */

xl=x[i]; y1=y[j];

x2=x1*x[i]; y2=y1*y[j];
x3=x2*x[i]; y3=y2*y[j]i
x4=x3*X[i]; y4=y3*y[j];
x5=x4*x[i]; y5=y4*Y[j];

x6=x5*x[i]; y6=y5*y[j];

/* @ynxumm "on" */
ul=1+2*kl; u2=1+4*kl; u3=1+6*kl;
vl=1+2*k2; v2=l+4*k2; v3=l+6*k2;
f[l][l]=y4—2*u1*y2+u2;

f[O][l ] =x4-2*vl*x2+v2;

f[l][2 ] =y6— 3*u2*y2+2*u3;
f[O ][2]=x6— 3*v2*x2+2*v3;

f[l][3 ] =4*y3- 4*ul*y1;

f[O][3]= 4*x3- 4*vl*xl;

f[l][4 ]= —6*y5- 6*u2*yl;
f[O ][4 ]= 6*x5-6*V2*x1;

f[l][S ]=12*y2-4*ul;

f[O][5]=12*x2—4*v1;

f[l][6]=30*y4-6*u2;

f[O][6]=30*x4-6*v2;

el=l; e2=l;

e1=el/24; e2=e2/360;
bl=el*f[l][l]*(zl*f[0][l]+z3*f[0][2]);
b2=e2*f[l][2]*(22*f[O][l]+z4*f[0][2])°

v[j][i]=bl+b2;
al=el*f[l][5]*(z l*f[0][l]+z3*f[0]
a2=e2*f[l][6]*(22*f[0][1]+z4*f[0]
a3=el*f[l][1]*(zl*f[0 ][5]+z3*f[0]
a4=e2*f[l][2]*(22*f[O][5 ]+z4*f[0 ]

w[j][i]=al+a2+nu*(a3+a4);

[2]);
[2]);
[6]);
[6]);

/***************************************~k~k**~k~k~k******~k*/

printf("\n\n Bueon peeyanaToe Ha neqarb\n");

f1=fl*l80/pi; f2=f2*180/pi;
f3=f3*180/pi; f4=f4*180/pi;
tire();
fprintf(fp,"\n 21 22 23
24");

fprintf(fp,"\n%l8.4f%15.4f%15.4f%15.4f\n",21,22,23,24);

173

tire();

tire();
fprintf(fp,"\nKoopnMHaTm Toqu nnaCTMHKM no X M no Y");
tire();
fprintf(fp," O l 2 3 4");
fprintf(fp," 5 6 7 8");
tire();
fprintf(fp,"\n %8.3f",x[O]);
for (k=l; k<=8; k++) fprintf(fp,"%8.3f",x[k]);
fprintf(fp,"\n %8.3f",y[0]);
for (k=l; k<=8; k++) fprintf(fp,"%8.3f",y[k]);
tire();
fprintf(fp,"\nfipormom nnaCTMHKM no TquaM");
tire();
fprintf(fp," O l 2 3 4");
fprintf(fp," 5 6 7 8");
tire();

for (i=0; i<=8; i++)
fprintf(fp,"\n %8.3f",v[i][0]);
for (k=l; k<=8; k++) fprintf(fp,"%8.3f”,v[i][k]);

tire();
fprintf(fp,"\nM3PMoawmme MOMGHTH nnaCTMHKM no TOQKaM");
tire();
fprintf(fp," O l 2 3 4");
fprintf(fp," 5 6 7 8");
tire();
for (i=0; i<=8; i++)
fprintf(fp,"\n %8.3f",w[i][0]);
for (k=1; k<=8; k++) fprintf(fp,"%8.3f”,w[i][k]);

tire();
fprintf(fp,"\n Konen peayanaToe \n");

else
printf("\n\n Ppaonqecxmfi BmBon \n");

174

/**************************~k*************************~k~k*/

/* HonPOTOBKa PpaQquCKOPO nonH*/

/* KoopnMHaTm rpaomqecxoro nona MOHMTOpa */
/* il=0; jl=0; 12:639; j2=349; */
i1=0; jl=O; i2=639; j2=479;

/* KoopnMHaTm nonn maoopaxenma */
13=il+20; j3=jl+25; 14:12-20; j4=j2-55;

/* paamepa nona KpMBofi */
lx=i4-i3; ly=j4-j3;
/* KOOpflMHaTH KoopnMHaTme ocen */
iOy=lx/2+i3; ij=ly/2+j3;
/* machaoMpoaanme meoopaxennn */
al=lx; bl=ly; el=al/bl;
if (lm>el)
lx=(lX/l6)*l6; ly=lx/lm; ly=(ly/l6)*16;
lxl=lx/l6; ly1=ly/l6;
else
ly=(ly/l6)*l6; lx=ly*lm; lx=(lx/16)*l6;
lxl=lx/l6; lyl=ly/l6;
al= fabs( w[O ][O]) bl=fabs(v[0][0]);

for (i=1; i<= 8; i++)
a2=fabs(w[0][i]); a3=fabs(w[i][O
b2=fabs(v[0][i]): b3=fabs(v[i][0

if (al<a2) al=a2;
if (al<a3) al=a3;
if (b1<b2) b1=b2;
if (b1<b3) bl=b3;
if (lxl>lyl) el=lxl; else el=lyl;
if (al>bl) e0=3*el/al;
else e0=3*e1/bl;
for (i=0; i<= 8; i++)
YY[0][i l]=W[0][i]*eO;
yyiliiii=W[i][0]*e0;
XX[O][i]=V[O ][i]*e0:
Xil ][i]=V[i ][0 ]*e0;
clrscr(); /* OQMCTKa eraHa */

/* orkpmrne Ppaonqecxoro pexmma */
errorcode = registerbgidriver(EGAVGA_driver);
if (errorcode < O)

printf("Graphics error: %s\n", grapherrormsg(errorcode));
exit(l);

175

detectgraph( &graphdriver, &graphmode);
initgraph( &graphdriver, &graphmode, "");

setfillstyle(SOLID_FILL, EGA_CYAN);
bar(i1,jl,i2,j2);
setfillstyle(SOLID_FILL, EGA_WHITE);
bar(i1+8,j1+8,12-8,j2-38);
setfillstyle(SOLID_FILL, EGA_WHITE);
bar(il+90,j2—30,12—90,j2-8);
setcolor(EGA_BLACK);
i5=i1+10; j5=j2—8; i6=il+50; j6=j2-30;
ramka();
outtextxy(iZ-60, j2-23, "PLAST");
outtextxy(il+20, j2-23, "Esc");

i5=iOy—lx/2; j5=ij-ly/2;i6=i0y+lx/2;
j6=ij+ly/2;

ramka();

line(i5,j0x,i6,j0x); line(iOy,j5,iOy,j6);

/* noacnenna 1< rpaQMKaM */
sprintf(str[1],"B/A=%4.2f",lm);
outtextxy(il+460, j2-23, str[l]);

setcolor(EGA_RED); line(i1+120,j2-
20,il+150,j2-20);
sprintf(str[3],"Mmax=%5.3f",al);
outtextxy(il+l60, j2-23, str[3]);

i5=i0y; i6=i0y;
for (i=0; i<=8; i++)
j5=j0Xi j6=j0X-yy[0][i];
line(i5,j5,i6,j6);
zz[i][O]=i6; zz[i][l]=j6;
i5=15+lx1; i6=i6+lxl;
risl();
j5=j0x; j6=ij;
for (i=0; i<=8; i++)
i5=iOy; i6=iOy+yy[1][i];
line(i5,j5,i6,j6);
zz[i][O]=i6; zz[i][1]=j6;
j5=j5+lyl; j6=j6+lyl;
risl();

setcolor(EGA_BLUE); line(i1+270,j2-
20,i1+300,j2-20);
sprintf(str[2],"Wmax=%5.3f",bl);
outtextxy(il+310, j2-23, str[2]);
i5=i0y; i6=i0y;
for (i=0; i<=8; i++)

176

j5=j0x; j6=ij—xx[0][i];
line<i5,j5,i6,j6);

zz[i][O]=i6; zz[i][l]=j6;
i5=i5-lxl; i6=i6-lxl;
risl();
j5=j0x; j6=j0x;
for (i=0; i<=8; i++)
i5=i0y; i6=i0y+xx[l][i];
line(i5,j5,i6,j6);
ZZ[i][O]=i6: 22[i][l]=j6;
j5=j5—lyl; j6=j6—lyl;
risl();
work=1;
while (work: ) /* noxa He HaxaTa Esc */

jj. ii= bioskey(0);
switch (jj. ch[O])

case OxlB: /* Knaamma Esc - Bmxon */
work=0;
break;
default: break;
closegraph(); /* saxpurme rpaomqecxoro pexmma */
clrscr(); /* OQMCTKa ekpana */

printf("\n Paoora nporpaMMH 3axonqena\n");
} /* Koneu — main() */

/*******************************************************~k*/

void opred()

qq=0:
for (1:0; l<=l; l++)
for (m=l; m<=4; m++) b[l][m]=0;

for (k=l; k<=4; k++)
b[0][k]=p[1][k];

switch(k)
case 1: li=2; mi=3; ni=4; break;
case 2: li=l; mi=3; ni=4; break;
case 3: 11:1; mi=2; ni=4; break;
case 4: li=l; mi=2; mi=3; break;
default: exit(l);
for (n: n<= 3; m++)
Cin][1]=p[n+l][li];
C[n1[2]-WP[n+1][ i];
c[n][3]— =p[n+l][ni];
a2=C[1][1]*C[2][2]*C[3][3]-C[1][1]*C[2][3]*C[3][2]:

177

a3=C[l][2]*C[2][3]*C[3][l];

a1=a2+a3;
b2=C[1][2]*C[2][1]*C[3][3]-C[l][3]*C[2][l]*C[3][2];
b3=C[1][3]*C[2][2]*C[3][1]:

b1=b2+b3;
b[1][k]=(a1-bl)*b[0][k];

for (k=1; k<=4; k++)
qq=qq+b[1][k]*e[l][k];
return;
void risl()
for (i=1; i<=8; i++)

i5=zz[i-l][0]; j5=zz[i-1][1];
i6=zz[i][0]; j6=zz[i][1];
line(i5,j5,i6,j6);
return;
void tire() int iv;
fprintf(fp,"\n ");
for (iv=1; iv<=74; iv++)fprintf(fp,"—");
fprintf(fp,"\n");
return;

void ramka()
rectangle(i5,j5,i6,j6);

return;
/~k******~k**~k***********~k******~k***************~k***********/

/* K o H e u n p o P p a M M m */

178

APPENDIX C

FORMULAE FOR THE MAIN DIFFERENCES AND SUMS

 

Differences of the second order

Differences of the 4‘" order

 

A2i(2) = 2

A4i(4) = 24

 

7- (3) = 6(i _ ”(1)

Mi ')—(5 =i—120( 2)“)

 

1321(4) = 12(i —1)(2)

A4i5)( =i—-360( 2)(2)

 

A2i(5) = 20(i—1)(3)

 

A2i(6) = 30(i -1)(4)

 

 

Sums

 

57(3) = M51) +C

 

 

 

Si = i + C
4
Si = (i +;)(2) + C Sim = (i +506) + C

 

 

 

 

5,45) =£:1)_‘6_’+C

 

179

 

APPENDIX D

ELEMENTS OF MATRICES OF COEFFICIENTS
FOR THE PROBLEM PRESENTED IN SECTION 4.5

Elements of matrix C ,- :

 

 

cll =1 c12 = —0.5i€, -1 cl3 = 3%,
C 29: C —& C -n
21 6 22 _ 2 23 _

_ (3) ~ ~ _ (3) _ (3)
c3l=(" 61) (5+4k,)—(4+0.5k,)(" 62) +(" 63)

_ (2) ~ ~ _ (2) _ (3)
c32=@_21_(5+4k1)-(4+0.5k1)(" 2) +(" 23)

C33 =(n-1X5+4l:,)—(4+0.5/:, Xn—2)+(n—3)

Elements of matrix 0,:

 

 

d11=5"0-5E1 d12=6—E1 d13=l+l~1
3 6
(4) (5) (6)
d21=L (122-1L 6123:1—
4 20 6O

2. ~ _ (4) ~ _ (4) _ (4)
d3l=5—0.5k,—(5+4k,)(n 1) +(4+0.5k1)(" 42) —(" 43)

(s)

d32=(4— 0.5k n—l))+(n+2 )—(5+ 4k)——+ (4+o.5/.:1)(‘20)"‘)(52 "(“2231)“
~ _ (6)
d33=(—4 0.5513(11(_1)(2 Lyjnflypgjsflkl (n 01) +

60
~ _ (6) _ (6)
+(4+0.5k1)(n 6(2)) - (n 6(3))

180

 

APPENDIX E

EXAMPLES OF DYNAMIC ANALYSIS USING THE COMBINED STATIC
AND CONSECUTIVE DYNAMIC CONDENSATION METHOD

Problem 1. Dynamic Analysis of a Two-Story One-Bay Frame with Fixed
Supports and Rigid Joints.

Consider the concrete frame shown in Figure E.1 (a). The beams have the
following characteristics: dimensions of the cross-section 0.3x0.6 m, the area of the
cross-section A = 0.18 m2, modulus of elasticity E = 210 GPa, moment of inertia I =
0.0054 m4, mass per unit length m = 0.45 (kN*sec2)/m. The columns have the following
characteristics: dimensions of the cross-section 0.3x0.3 m, the area of the cross-section A
= 0.09 m2, modulus of elasticity E = 210 GPa, moment of inertia I = 0.000675 m4, mass
per unit length m = 0.225 (kN*sec2)/m. The frame is divided into 24 linear finite
elements .with equal lengths of beam and column elements: a = b = 1 m. The
condensation was performed using ten main d.o.f. concentrated in the four main joints of
the frame (Figure B] b). The secondary nodes were structurally subdivided into six
groups. Two types of blocks of the secondary unknowns were used. The number of
unknowns in the first and the fourth blocks corresponding to the beam groups of nodes
was ten, and the number of unknowns in the second, third, fifth, and sixths blocks
corresponding to the column groups was four. The problem was solved by the combined
static and consecutive dynamic condensation method. Results presented in Table E.1 are
compared to the results obtained by solving the original problem with the full set of d.o.f.

Results are illustrated in Figures [3.2 and E3.

181

 

(a)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

   

 

 

 

A

E

M

V

E

M

7777 ////
6m
I
A

(b) C “ _
, ll
E

M

II

1v 3

A ('3‘

E

M

II

..D

x

M

\| \
////
6xa=6m

Figure E]. (a) The frame for Problem 1; (b) Arrangement of the primary d.o.f.

and secondary d.o.f. blocks for Problem 1

182

 

TABLE E.l. Results for Problem 1

 

Eigenvalues obtained by

 

 

 

 

 

 

 

 

 

 

 

 

 

° 2:

:; Siam" 3...... 3:31:32?" .. 1o ...f..::.:::r:::::::::.:...

a; N = 46 ' ' ' d.o.f.

2... x 105 2... x105 8 (%) 2.. x105 8 (%)

1 0.94334 095033 0.79 0.94334 0.00
2 4.15933 532339 23.07 4.15935 0.01
3 5.39297 771121 42.99 5.39397 0.02
4 3.19999 3.40341 2.54 3.20214 0.03
5 37.927 54.71233 44.23 33.3213 1.04
3 39.5599 33.532 33.13 40.3213 1.93
7 43.4939 30.935 74.13 53.0437 20.54
3 51.9747 123.302 147.32 93.7232 33.10
9 39.0507 219.7353 213.22 135.351 93.45
10 93.5372 353.323 232.12 131.334 34.03

 

 

 

 

 

 

183

 

 

400
350

Eigenvalue
A a N N O)
0 (II 0 01 O
O O O O O

 

  
 
 
 

________________________________________________________

A.

————————————————————————————————————————————————————————

L

 

 

 

50 ------------------------------------------
0
1 2 3 4 5 6 7 8 9 10
Number of eigenvalue
—-0— Full System —8— Static condensation to 10 d.o.f.

k

— a — Proposed Technique

Figure E.2. Results for Problem 1

 

 

300.00 ‘

250.00 _____________________________________

   
   

200.00 2 ______________________________________

150.00 2 ......................................

Error, %

100.00 2

50.00 2

 

 

 

0.00 i
1 2 3 4 5 6 7 8 9 10 i

Eigenvalue Number

+ Static condensation to 10 d.o.f. —I— Proposed Technique

 

Figure E.3. Graphs of errors for Problem 1

184

 

Problem 2. Dynamic Analysis of an Isotropic Plate with Fixed Supports.

Consider a thin rectangular isotropic plate with fixed supports. The plate has the
following parameters: length A = 3 m, width B = 2.8 In, thickness t = 0.1 m, Poisson’s
ratio v = 0.3, Young’s modulus E = 210 GPa. The plate is loaded by the uniformly distributed
mass of intensity 771 = 10 (kN*s2)/m. The plate is divided into a mesh of rectangular finite

elements with dimensions a = 0.3m, b = 0.2 m (Figure ED.4 a).

The plate was divided into four substructures by assigning the main nodes along
the centerlines of the plate. Additional main nodes were assigned concentrated in the
middle of each block. The condensation was carried out using 63, 75, and 99 primary

d.o.f. Three variants of the arrangement of condensation nodes are shown in Figure 13.4

(a), (b), and (C)-

Dynamic analysis of the plate was performed by the combined static and
consecutive dynamic condensation method. Results are presented in Table E2 and
illustrated in Figures 135 and E6. The eigenvalues obtained by proposed technique are
compared with those obtained solving the full system of equations. Alternative

calculations were made using the static condensation to 99 primary d.o.f.

185

(a)

=28m

be

 

 

 

10xa=3m g

(b)

 

(C)

 

Figure E.4. Location of condensation nodes for Problem 2: (a) condensation to 63

d.o.f.; (b) condensation to 75 d.o.f.; (c) condensation to 99 d.o.f.

186

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

mod- ommwdo and mom F .5 EN 00860 mndw case. 5 wmwmdo —. _.
no.0 oooodm 3.0 momodm CNN oommdm and ammo. 5 88.5 3.
no.0 Emmdm Sac mm Stow omd oomodm mmdw F Fomdm omomdm m
5.0. 836m 36 mnmmdm 54.? 00 5.8 8.? N3 ...ov _.Nmm.mm a
oo.o vam. _.m med 0 Em. Fm Ed oomodm om.m ommmdm vam. 5 n
vod mmomém mod ovpmém 5. T oomm.v~ mm. S. cm ENN ommnéw @
cod $3.2. 5.0 88.9. 56 83.9. and 83.8 9392. m
vod 300.3. mod mo 5.3. 5.0 003.3. cod? mommdr N393. v
5.0 m vas 3.0 $.va «v.0. oomvfi cad 8me m Fmvs m
5.0.. owmmd cod 53.0 Nod oommd mud owmmd mmmmd N
mod omooé mod- mmmwé n _. .o oonmé 8.0 m Ewe whom; —.

so: saw?“ so: can __& .3: can ...a 13: van ...a can 3.x BE

...oé taste 8 ..6... basic ms c6... 352...: 3 3n .1. 2 m

o. coauocoocoo 5.3 3 o. :WMMmRNEcwm 033m “flaw” x. m

oofioE cosmos—Eon 3:356 25:32.3 new 233 ooEnEoo no cozaom m

3 uofifino 33.—«Emma m.

 

 

 

N Bozo...— ..o.— 3153— .N.H ”ma—fiat

187

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

m a .o 88.8 8.0 88.8 ow.» 88.8 8.8 8 F .2 F 88 P .88 8
88 85.8. Rd 88.82 88 88.82 8.8 88.88 88. Ft 8
88 88th 88 888: 2. : 08:2 8.8 88.58 88ng 8
SN «v8.2: 82:. 8882 8.9 0838 8.8 88.88 88.82 8
84 28.88 31 88.89 S... 8882 8.? 88.8m 88.82 8
Ed $8.89 8.? 8883 88 88.? 8.8 88.08 8882 2
8.? 8 3.8? 84 S 8.82 8.8 8881 8.8 8882 88.89 8
8.o 88.89 8.o 88.82 8.8 838: 8.8 88. 8? 88.8? 8
a F .o 33.8? 8.0 88.»? NS. 88. F 2 8.8 88.83 8889 3
mg. 85.8 2.0 «889 8. S 88. F 3 9.8 8288 98.8 2
m e .o 88.8 88 88.8 88 88.8 8.8 08 F .m: 88.8 2
3.: 3:: 3.: .39 ..a 3.: .31; 83 .31; .29:

ed... 52. 8 ...6... £2: 8 ..6... 53:. 8 .8 u z m .w...
#66 EuEtn w e

2 :ozamcoucoo 5:5 mm 2 33.8528 £83 53!? =2 m. m

.352: cozamcoucoo 382.? 2533250 new ”55» ooEnEoo 15 “_o cow—go...” x. m.

 

 

an .3:an 3:32.35

 

 

 

82553 N .5333 .8 333m .3 mama?

188

350 ------------------------------------------------------------

300 --------------------------------------------------------- -—
g 250 ------------------------------------------------------------
E

200 -------------------------------------------------------- .-.
a M
m 150- ---------------------------------------- -------

100. ............................... _--/ — "— .................

A
50 H """"""""""" 4” é """"""""""""""""""""""""
0 _- — Y T r r 7 T f r I T f r 7 7 T
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
Number of eigenvalue

+ Exact solution + Static condensation to 99 d.o.f.

_.fi—

 

 

 

Dynamic condensation to 63 d.o.f. -—-~---~+f——- Dynamic condensation to 75 d.o.f.

+ Dynamic condensation to 99 d.o.f.

 

Figure E.S. Results for Problem 2

 

 

Error, %

 

 

 

 

T

2 3 4 5 6 7 8 910111213141516171819202122

Eigenvalue Number

 

 

E—I— Dynamic condensation to 63 d. of - -A- Dynamic condensation to 75 d. of

+ Dynamic condensation to 99 d o.fm _ 7 . I

 

 

Figure E.6. Graphs of errors for Problem 2

189

REFERENCES

190

 

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