Dynamic Analysis of Non-Planar Coupled Shear Walls
Dynamic Analysis of Non-Planar Coupled Shear Walls
Dynamic Analysis of Non-Planar Coupled Shear Walls
Ph.D. THESIS
ADANA, 2008
UKUROVA NVERSTES
FEN BLMLER ENSTTS
mza:........................................
mza:..............................................
DANIMAN
YE
YE
mza:...........................................
mza:......................................
YE
YE
Not: Bu tezde kullanlan zgn ve baka kaynaktan yaplan bildirilerin, izelge, ekil ve
fotoraflarn kaynak gsterilmeden kullanm, 5846 sayl Fikir ve Sanat Eserleri
Kanunundaki hkmlere tabidir.
Dedicated to my parents,
Cevher Trkzer and Sdka Trkzer
ABSTRACT
Ph.D. THESIS
DYNAMIC ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS
Z
DOKTORA TEZ
DZLEMSEL OLMAYAN
BOLUKLU DEPREM PERDELERNN DNAMK ANALZ
II
ACKNOWLEDGEMENT
First of all I would like to thank Prof. Dr. Orhan AKSOAN for his
supervision and encouragement that enabled me to complete my research
satisfactorily. It has been a great pleasure for me to work with him.
I would also like to thank Asst. Prof. Dr. Engin Emsen for his beneficial
discussions, valuable comments and encouragement and his close friendship.
Finally, I wish to thank my family for their support, care, encouragement and
belief in my success in my education life.
III
CONTENTS
PAGE NUMBER
ABSTRACT ..............................................................................................................I
Z ........................................................................................................................... II
ACKNOWLEDGEMENT ...................................................................................... III
CONTENTS ........................................................................................................... IV
LIST OF TABLES............................................................................................... VIII
LIST OF FIGURES ................................................................................................ XI
NOMENCLATURE ............................................................................................ XXI
1. INTRODUCTION ................................................................................................ 1
2. PREVIOUS STUDIES.......................................................................................... 5
3. METHODS OF ANALYSIS ............................................................................... 10
3.1. Introduction ................................................................................................. 10
3.2. Continuous Connection Method (CCM) ...................................................... 11
3.3. Comparison of the Results of the Present Method for Non-planar Coupled
Shear Walls with those of the Frame Method............................................... 13
3.4. Stiffening of Coupled Shear Walls............................................................... 16
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED
SHEAR WALLS ............................................................................................... 19
4.1. Introduction ................................................................................................. 19
4.2. Determination of Mass Matrix ..................................................................... 20
4.3. Determination of Stiffness Matrix................................................................ 21
4.3.1. Introduction ....................................................................................... 21
4.3.2. General Information........................................................................... 22
4.3.2.1. Transformation of the Local Displacements .......................... 24
4.3.2.2. Transformation of the Moments of Inertia ............................. 25
4.3.2.3. The Assumptions of the Continuous Connection Method ...... 28
4.3.3. The Relation between the Axial Force Ti and the Shear Force
per Unit Length qi .......................................................................... 30
4.3.4. Formulation ....................................................................................... 31
4.3.4.1. Compatibility Equations ....................................................... 31
IV
VI
App. 3. Computation Procedure for the Sectorial Area in an Open Section............ 247
App. 4. List of Input Data File of a Computer Program Prepared
in Fortran Language for the Dynamic Analysis of
Non-Planar Coupled Shear Walls Using CCM ......................................... 256
App. 5. List of the Computer Program Prepared in Fortran Language for
the Dynamic Analysis of Non-Planar Coupled Shear Walls
Using CCM .............................................................................................. 260
VII
LIST OF TABLES
Table 6.1.
PAGE NUMBER
Table 6.2.
Table 6.3.
Table 6.4.
Table 6.5.
Table 6.6.
Table 6.7.
Table 6.8.
Table 6.9.
VIII
Table 6.15. Comparison of the natural frequencies (Hz) obtained from the
present program and SAP2000 in Example 9 .................................... 122
Table 6.16. Maximum displacement (m) in the X direction of point G
for undamped case in Example 10 .................................................... 127
Table 6.17. Maximum displacement (m) in the X direction of point G
for damped case in Example 10 ........................................................ 127
Table 6.18. Comparison of the natural frequencies (Hz) obtained from the
present program and SAP2000 in Example 11 .................................. 131
Table 6.19. Maximum displacement (m) in the X direction of point G
for undamped case in Example 12 .................................................... 136
Table 6.20. Maximum displacement (m) in the X direction of point G
for damped case in Example 12 ........................................................ 136
Table 6.21. Comparison of the natural frequencies (Hz) obtained from the
present program and SAP2000 in Example 13 .................................. 141
Table 6.22. Maximum displacement (m) in the X direction of point G
for undamped case in Example 14 .................................................... 146
Table 6.23. Maximum displacement (m) in the X direction of point G
for damped case in Example 14 ........................................................ 146
Table 6.24. Comparison of the natural frequencies (Hz) obtained from the
present program and SAP2000 for unstiffened case in Example 15... 150
Table 6.25. Comparison of the natural frequencies (Hz) obtained from the
present program and SAP2000 for stiffened case in Example 15 ...... 151
Table 6.26. Maximum displacement (m) in the X direction of point G
for unstiffened case in Example 16 ................................................... 159
Table 6.27. Maximum displacement (m) in the X direction of point G
for stiffened case in Example 16 ....................................................... 159
Table 6.28. Comparison of the natural frequencies (Hz) obtained from the
present program and SAP2000 for unstiffened case in Example 17... 163
Table 6.29. Comparison of the natural frequencies (Hz) obtained from the
present program and SAP2000 for stiffened case in Example 17 ...... 164
IX
LIST OF FIGURES
PAGE NUMBER
XI
XII
Figure 6.11. Frame model of the structure in Example 3 and its 3-D view .............. 92
Figure 6.12. Comparison of first, second and third mode shapes in X and Y
directions found by both the present program and SAP2000
in Example 3 ...................................................................................... 94
Figure 6.13. Comparison of fourth, fifth and sixth mode shapes in X and Y
directions found by both the present program and SAP2000
in Example 3 ...................................................................................... 95
Figure 6.14. Comparison of seventh and eighth mode shapes in X and Y
directions found by both the present program and SAP2000
in Example 3 ...................................................................................... 96
Figure 6.15. Cross-sectional view of the structure and applied
dynamic load in Example 4 ................................................................ 97
Figure 6.16. Rectangular pulse force in Example 4................................................. 97
Figure 6.17. Time-varying displacements in X direction at the top of the
shear wall for undamped case in Example 4 ....................................... 99
Figure 6.18. Time-varying displacements in X direction at the top of the
shear wall for damped case with 5 % damping ratio in Example 4 ...... 99
Figure 6.19. Cross-sectional view of the structure in Example 5 ........................... 100
Figure 6.20. Frame model of the structure in Example 5 and its 3-D view ............ 101
Figure 6.21. Comparison of first, second and third mode shapes in X and Y
directions found by both the present program and SAP2000
in Example 5 .................................................................................... 103
Figure 6.22. Comparison of fourth, fifth and sixth mode shapes in X, Y and Teta
directions found by both the present program and SAP2000
in Example 5 .................................................................................... 104
Figure 6.23. Comparison of seventh and eighth mode shapes in X, Y and Teta
directions found by both the present program and SAP2000
in Example 5 .................................................................................... 105
Figure 6.24. Cross-sectional view of the structure and applied
dynamic load in Example 6 .............................................................. 106
Figure 6.25. Rectangular pulse force in Example 6............................................... 107
XIII
XIV
Figure 6.42. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by t both he present program and SAP2000
in Example 9 .................................................................................... 123
Figure 6.43. Comparison of fourth, fifth and sixth mode shapes in X and Y
directions found by both the present program and SAP2000
in Example 9 .................................................................................... 124
Figure 6.44. Comparison of seventh and eighth mode shapes in X and Y
directions found by both the present program and SAP2000
in Example 9 .................................................................................... 125
Figure 6.45. Cross-sectional view of the structure and applied
dynamic load in Example 10 ............................................................ 126
Figure 6.46. Triangular pulse force in Example 10 ............................................... 126
Figure 6.47. Time-varying displacements in X direction at the top of the
shear wall for undamped case in Example 10 ................................... 128
Figure 6.48. Time-varying displacements in X direction at the top of the
shear wall for damped case with 7 % damping ratio in Example 10 .. 128
Figure 6.49. Cross-sectional view of the structure in Example 11 ......................... 129
Figure 6.50. Frame model of the structure in Example 11 and its 3-D view .......... 130
Figure 6.51. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by both the present program and SAP2000
in Example 11 .................................................................................. 132
Figure 6.52. Comparison of fourth, fifth and sixth mode shapes in X, Y and Teta
directions found by both the present program and SAP2000
in Example 11 .................................................................................. 133
Figure 6.53. Comparison of seventh and eighth mode shapes in X and Y
directions found by both the present program and SAP2000
in Example 11 .................................................................................. 134
Figure 6.54. Cross-sectional view of the structure and applied
dynamic load in Example 12 ............................................................ 135
Figure 6.55. Rectangular pulse force in Example 12 ............................................. 136
XV
XVI
XVII
Figure 6.85. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by the present program (CCM) for
stiffened and unstiffened cases in Example 17 .................................. 168
Figure 6.86. Comparison of fourth, fifth and sixth mode shapes in X and Y
directions found by the present program (CCM) for
stiffened and unstiffened cases in Example 17 .................................. 169
Figure 6.87. Comparison of seventh and eighth mode shapes in X and Y
directions found by the present program (CCM) for
stiffened and unstiffened cases in Example 17 .................................. 170
Figure 6.88. Cross-sectional view of the structure and applied
dynamic load in Example 18 ............................................................ 171
Figure 6.89. Rectangular pulse force in Example 18 ............................................. 172
Figure 6.90. Time-varying displacements in X direction at the top of the
shear wall for undamped case in Example 18 ................................... 173
Figure 6.91. Time-varying displacements in X direction at the top of the
shear wall for damped case with 5 % damping ratio in Example 18 .. 173
Figure 6.92. Cross-sectional view of the structure in Example 19 ......................... 174
Figure 6.93. Frame model of the structure in Example 19 and its 3-D view .......... 175
Figure 6.94. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by both the present program and
SAP2000 for stiffened case in Example 19 ....................................... 178
Figure 6.95. Comparison of fourth, fifth and sixth mode shapes in X and Y
directions found by both the present program and
SAP2000 for stiffened case in Example 19 ....................................... 179
Figure 6.96. Comparison of seventh and eighth mode shapes in X and Y
directions found by both the present program and
SAP2000 for stiffened case in Example 19 ....................................... 180
Figure 6.97. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by the present program (CCM) for
stiffened and unstiffened cases in Example 19 .................................. 181
XVIII
Figure 6.98. Comparison of fourth, fifth and sixth mode shapes in X and Y
directions found by the present program (CCM) for
stiffened and unstiffened cases in Example 19 .................................. 182
Figure 6.99. Comparison of seventh and eighth mode shapes in X and Y
directions found by the present program (CCM) for
stiffened and unstiffened cases in Example 19 .................................. 183
Figure 6.100. Cross-sectional view of the structure and applied
dynamic load in Example 20 ........................................................... 184
Figure 6.101. Triangular pulse force in Example 20 .............................................. 185
Figure 6.102. Time-varying displacements in X direction at the top of the
shear wall for undamped case in Example 20 .................................. 186
Figure 6.103. Time-varying displacements in X direction at the top of the
shear wall for damped case with 5 % damping ratio in Example 20 . 186
Figure 6.104. Cross-sectional view of the 1st region of the structure
in Example 21 ................................................................................. 187
Figure 6.105. Cross-sectional view of the 2nd region of the structure
in Example 21 ................................................................................. 188
Figure 6.106. Frame model of the structure in Example 21 and its 3-D view ......... 188
Figure 6.107. Comparison of first, second and third mode shapes in X and Y
directions found by both the present program and SAP2000
in Example 21 ................................................................................. 190
Figure 6.108. Comparison of fourth, fifth and sixth mode shapes in X and Y
directions found by both the present program and SAP2000
in Example 21 ................................................................................. 191
Figure 6.109. Comparison of seventh and eighth mode shapes in X and Y
directions found by both the present program and SAP2000
in Example 21 ................................................................................. 192
Figure 6.110. Cross-sectional view of the structure and applied
dynamic load in Example 22 ........................................................... 193
Figure 6.111. Rectangular pulse force in Example 22............................................ 194
XIX
XX
NOMENCLATURE
ai
(ax, ay)
A ji
A ci
A*ci
bi
bk
(bx, by)
: bimoment,
BE
Bi
Bi
di
C
~
C
: damping matrix,
Dji
D ji
dPX , dPY
: elasticity modulus,
: shear modulus,
XXI
x g ji , y g ji
x g ji , y g ji
Hp
hi
hs
: region number,
Ici
I x ji , I y ji
I xy ji
I x ji , I y ji
I xy ji
I wji
I wi
I wi
: a geometrical constant,
IqXi , IqYi
: pier number,
Jji
Ji
XXII
Jo
: stiffness matrix,
~
K
K1i, K2i,
K3i, K4i
K ci
msv
: mass matrix,
~
M
M EXi , M EYi
M Eti
Mmp
Mtop
M ti
M ti
Mtot
Msv
Mw
: degree of freedom,
Ni
O(X,Y,Z)
: load vector,
XXIII
~
P
~
Pef
Q x ji , Q y ji
qi
Rji
ri
: a geometrical constant
si
: eigenvector,
S ji
x s ji , y s ji
Sx , Sy
S wx , S wy
Sw
Ti
tk
ux , uy , uz
us
ui
u ji
XXIV
vi
v ji
Vi
&
X
&&
X
&
Y
&&
Y
d1i
d2i
d3i
d4i
: the relative
vertical displacement
due to
the shear
d6i
qi
XXV
qji
qz
F
w
: modal matrix,
ws
: sectorial area,
wj
f ji
: angle between the global axes and the principal axes of jth pier
: circular frequency,
in region i,
g ci
XXVI
1. INTRODUCTION
1. INTRODUCTION
1. INTRODUCTION
analyses. Practically, the design of non-planar coupled shear walls requires special
consideration of dynamic behaviour in case of the wind and seismic loads. It is well
known that the deformation of a coupled shear wall subjected to lateral loading is not
confined to its plane. In other words, either applied loading is not confined to the
plane of the wall or the cross-sections of the piers are not planar. In non-planar
coupled shear walls, both the flexural and torsional behaviours under the dynamic
loading have to be taken into account in the analysis. The bending analysis of the
structure is rather simple. However, its torsional analysis is rather difficult and needs
to be explained in detail.
This thesis considers the dynamic analysis of non-planar coupled shear walls
resting on rigid foundations. In this study, continuous connection method (CCM) and
Vlasovs theory of thin-walled beams are employed to find the stiffness matrix.
In the CCM, the discrete connecting beams between the piers are replaced by
an equivalent continuous system of laminae (Rosman, 1964). Based on this method,
a host of investigations have been made about the static and dynamic analyses of
planar coupled shear walls. However, the studies about non-planar coupled shear
walls are far from being adequate.
In the theory of flexure, the assumption of plane sections remaining plane
during bending, is usually referred to as the Bernoulli-Navier hypothesis. Torsion
was considered to be completely determined by St.Venants theory in 1850s. The
crucial point in St.Venants theory, is that the cross-section is free to deform out of
its plane during torsion. This is called free warping of the cross-section. St.
Venants theory was applied to uniform, as well as non-uniform, sections.
A general theory of non-uniform torsion came in 1905, when Timoshenko
considered the effect of restraining the warping of a beam at its ends. The coupling
of flexure and torsion was explained in 1921, when Maillart introduced the concept
of shear center and showed that the transverse loads and supports must act through
this center, if no torsion is to result. A comprehensive theory of combined torsion
and flexure of open thin-walled bars was developed by Vlasov in 1940s. However,
his work had not become generally known, until his book was translated into English
in 1961. This theory is generally called Vlasovs theory. Vlasov's theory for thin-
1. INTRODUCTION
walled open-section beams points out the fact that the sections cannot warp freely as
assumed in the St. Venant solutions. This theory is an approximate theory developed
for engineering purposes and it is based on certain simplifying assumptions.
When thin-walled structures are twisted, there is a so-called warping of
the cross-section and the Bernoulli-Navier hypothesis is violated. Warping is defined
as the out-of-plane distortion of the cross-section of a beam in the direction of the
longitudinal axis. The warping of shear walls is greatly restrained by the floor slabs
and the foundations. As a result of this interaction, two types of warping stresses,
namely direct stresses in the longitudinal direction and shear stresses in the tangential
direction of the piers, are introduced. A classical analysis of warping torsion requires
the prior evaluation of the shear center location, the principal sectorial area diagram,
the warping moment of inertia and the torsion constant.
When the height restrictions prevent connecting beams from fulfilling their
tasks of reducing the maximum total shear wall bending moments and the maximum
lateral displacements at the top, beams with high moments of inertia, called
Stiffening Beams, are placed at certain heights to make up for this deficiency.
Stiffening of coupled shear walls decreases the lateral displacements, thus, rendering
an increase in the height of the building possible. Hence, assigning some stories of
the building as storages, service areas and the like and placing high beams on those
floors seems to be a logical solution.
The crucial assumption made in applying CCM to planar shear walls is that
the connecting and stiffening beams are assumed to be infinitely rigid in the axial
direction. It is well known that this assumption is equivalent to the rigid diaphragm
model for floors and has been used widely for a long time. The extension of this
assumption to non-planar shear walls shows itself as a straight forward application of
the rigid diaphragm model.
In this thesis, the top, the bottom and the heights at which there are changes
of storey height, beam height and/or connection stiffness will be called ends and
the section between any two consecutive ends will be called a region. Hence,
every region of the shear wall, in the vertical direction, has no stiffening beam and
1. INTRODUCTION
change in cross-sectional shape in it and all properties of the piers and the connecting
beams including the storey heights are exactly same.
In the present research, the general dynamic analysis of a non-planar nonsymmetrical stiffened coupled shear wall with a stepwise variable cross-section,
resting on a rigid foundation, is studied. In the dynamic analysis of non-planar
coupled shear wall, the system stiffness matrix is found by using Vlasovs theory of
thin-walled beams and the CCM. In the free vibration analysis, the natural
frequencies and mode shape vectors of non-planar shear wall are determined. The
behavior of the coupled shear wall under the dynamic loading is investigated in the
forced vibration analysis. Then, using those analytical results, a computer program in
Fortran language has been prepared and various examples have been solved. The
results are verified via comparison with those obtained by the SAP2000 structural
analysis program.
2. PREVIOUS STUDIES
2. PREVIOUS STUDIES
Continuous Connection Method was originated by Chitty in 1940s, and
developed by Beck (1962), Rosman (1964) and Coull (1966). It has been extended
by Glck (1970) and Tso and Biswas (1973) to deal with three dimensional
shear/core wall assemblies.
Vlasov (1961) presented a comprehensive theory of combined torsion and
flexure of open thin-walled bars. The theory stipulated that, Bernoulli-Navier
hypothesis of the bending of beams was not applicable to thin-walled beams, because
of the distortion (warping) of the section.
Beck (1962) presented an approximate method of analysis where a
continuous system replaces the discontinuous frame system which allows a simple
calculation in high multistory buildings.
Clough et al. (1964) presented a method based on the wide-column frame
analogy for the analysis of planar coupled shear wall structures. Both vertical and
lateral loading of structures with an arbitrary system of shear walls were considered
in that study.
Rosman (1964) used the CCM to solve a two bay symmetrical coupled shear
wall on rigid foundation. In his analysis, he kept the number of unknown functions to
a minimum, to solve the static problem.
Coull and Smith (1966) proposed a method based on the continuous
connection method.
Soane (1967) analyzed high multi-bay shear walls in complex buildings using
analog computers.
Coull and Puri (1968) found the solution for a shear wall with variations in
the cross-sectional area and compared their analytical results with those of
experimental works.
Glck (1970) presented a study on the three dimensional application of the
continuous connection method to solve the problem of a structure consisting of
coupled, prismatic or non-prismatic shear walls and frames arranged asymmetrically
in the horizontal plane.
2. PREVIOUS STUDIES
2. PREVIOUS STUDIES
Macleod and Hosny (1977) employed a method for analyzing shear wall
cores and proposed two types of elements that comprise a generalized column
element and a solid wall element for modeling planar wall units.
Choo and Coull (1984) investigated the effect of two stiffening beams, one at
the top and the other at the bottom, on the behaviour of a single bay coupled shear
wall on elastic foundation subject to lateral loads. That analysis was carried out
employing the CCM, which ended up in a closed form solution. The authors
investigated the effects of the stiffening beams on the forces and displacements
evoked in the shear wall for various types of soils.
Chan and Kuang (1988) studied a single bay coupled shear wall with a
stiffening beam at any height resting on elastic foundation, employing the CCM.
They proposed that the stiffening beam be placed at 0.2-0.5 of the total height
depending on their results. Working on the same problem (1989), they presented
graphs to express the effect of the height and the stiffness of the stiffening beam on
the structural behaviour of the shear wall.
Coull and Bensmail (1991) investigated single bay constant cross-section
coupled shear walls with two stiffening beams at arbitrary heights resting on elastic
foundation using the CCM. The authors gave a closed form solution and presented
the graphics of various quantities.
Smer and Akar (1992) carried out a study on an open mono-symmetric core
wall coupled with connecting beams considering cases where shear deformation of
the walls is significant.
Kwan (1993), in his article, investigated the erroneous results in the widecolumn frame analogy when shear deformation of the walls is significant. He
suggested the use of beam elements with vertical rigid arms for the coupling beams
together with solid wall elements with no rotational degree of freedom to eliminate
the underestimation of the beam end rotation due to shear strain in the walls.
Aksogan et al. (1993) employed the CCM to investigate shear walls with any
number of stiffening beams on elastic foundation and prepared a general purpose
computer program to implement their analysis.
2. PREVIOUS STUDIES
Li and Choo (1994) carried out a continuous-discrete method for the free
vibration analysis of a single bay coupled shear wall without stiffening beam. In this
work, the structure is considered as a discrete system of lumped masses. Then, using
the CCM and loading each lumped mass with a unit force the flexibility matrix
determined. The stiffness matrix was determined by taking the inverse of the
flexibility matrix and the free vibration analysis is carried out. The accuracy of the
method was illustrated with two examples.
Arslan and Aksogan (1995) analyzed coupled shear walls on elastic
foundation having a stepwise varying cross-section and any number of stiffening
beams and implemented the analysis, for the case with any number of regions along
the height, by a computer program.
Li and Choo (1996) investigated a single bay coupled shear wall on elastic
foundation with two or three stiffening beams.
Arviddson (1997) proposed a method for the analysis of three dimensional
structures consisting of non-planar coupled shear walls. The theory of analysis is
based on the CCM by taking into account both flexural and torsional behaviour of
such walls. To simplify the analysis, the author ignored the effect of St. Venants
torsion in his work.
Arslan (1999) studied the dynamic analysis of stiffened coupled shear walls
with variable cross-sections on flexible foundations, considering the effects of shear
deformations in the walls and beams.
Mendis (2001) published an article on an open section thin-walled beam with
connecting beams for estimating the longitudinal stresses in the piers. The study
employed the CCM and expressed the relevance of warping stresses in a torsionally
loaded concrete core. The cases with and without connecting beams were analyzed to
study the effect of their presence.
Bike (2002) carried out a study on the static and dynamic analyses of multibay planar coupled shear walls on elastic foundation, with finite number of stiffening
beams based on the CCM. The author prepared two computer programs in the
Mathematica programming language for the static and dynamic analyses, separately.
2. PREVIOUS STUDIES
Aksogan, Arslan and Akavc (2003) published an article about the stiffening
of coupled shear walls on elastic foundation with flexible connections and stepwise
changes in width.
Arslan, Aksogan and Choo (2004) presented an article about free vibrations
of flexibly connected elastically supported stiffened coupled shear walls with
stepwise changes in width.
Bikce, Aksogan and Arslan (2004) published an article about stiffened multibay coupled shear walls on elastic foundation.
Resatolu (2005) presented a study on the static analysis of non-planar
coupled shear walls, based on Vlasovs theory of thin-walled beams and the CCM
having flexible connections and rigid foundation, with the properties of the
connecting beams and the rotational stiffnesses of their end connections varying from
region to region in the vertical direction.
Emsen (2006) carried out a study on the static analysis of non-planar coupled
shear walls with any number of stiffening beams and stepwise cross-sectional
changes using Vlasovs theory of thin walled beams and the CCM.
3. METHODS OF ANALYSIS
3. METHODS OF ANALYSIS
3.1. Introduction
Shear walls are used to resist the lateral loads that arise from the effect of
winds and earthquakes. However, these walls are pierced by windows, doors and
service ducts. These features turn a simple shear wall into a coupled pair, which can
be considered as two smaller walls, coupled together by a system of lintel beams.
This thesis considers the dynamic analysis of non-planar coupled shear walls
resting on rigid foundations. The analysis deals with coupled shear walls with a finite
number of stiffening beams, the properties of which vary from region to region along
the height. In this study, continuous connection method (CCM) and Vlasovs theory
of thin-walled beams (see Appendix 1) are employed to find the stiffness matrix.
The CCM was developed by assuming that the discrete system of
connections, in the form of individual coupling beams or floor slabs, could be
replaced by continuous laminae. In the dynamic analysis of shear walls with the
CCM, following an approach similar to the one used by Tso and Biswas (1973) both
the flexural and the torsional behaviour (see Appendix 1) are taken into account. The
deformation of a coupled shear wall subjected to lateral loading is not always
confined to one plane. Thus, the present analysis is a three dimensional analysis of
coupled shear walls.
While the discrete structure is formulated as a continuous medium, the
continuously distributed mass of the structure is discretized to a system of
lumped masses for finding the corresponding stiffness matrix. After obtaining
the standard frequency equation of the discrete system, the circular frequencies
are determined in a straightforward manner and used to find the modes of vibration.
The forced vibration analysis of the structure is resolved by uncoupling the
system of differential equations obtained, using mode superposition technique, which
renders the mass and stiffness matrices diagonal. The Newmark method, one of the
numerous numerical methods available in the literature, is employed to carry out the
timehistory analysis.
10
3. METHODS OF ANALYSIS
(b)
qz
st
1 pier
laminae
H
11
2nd pier
3. METHODS OF ANALYSIS
In this thesis, the CCM is used for the analysis of non-planar coupled shear
walls on rigid foundations subjected to lateral loads. The warping of the piers due to
twist has been considered, as well as their bending, in the analysis. The compatibility
equation has been written at the mid-points of the connecting beams. The differential
equations for the compatibility of the displacements and the equilibrium of the forces
are used in the analysis. After the application of the loading and boundary conditions,
mathematical solutions are obtained for the determination of the shears and moments
in the beams and the walls.
The basic assumptions of the CCM for non-planar coupled shear walls can be
summarized as follows:
1.
The geometric and material properties are constant throughout each region i
It is assumed that, halves of the moments of inertia and the rotational spring
constants of the connecting beams in each region, are assigned to the ends of it.
4.
The discrete set of connecting beams with bending stiffness EIci and
Vlasovs theory for thin-walled beams of open section is valid for each pier.
6.
The outline of a transverse section of the coupled shear wall at a floor level
remains unchanged in plan (due to the rigid diaphragm assumption for floors).
Moreover, the parts of the shear wall between floor levels are also assumed to satisfy
this condition. Depending on the foregoing assumption, the axis of each connecting
beam remains straight in plan and does not change its length. Furthermore, the slope
and curvature at the ends of a connecting beam in the vertical plane are also assumed
to be equal. Consequently, it can be proved in a straightforward manner that,
depending on assumption 2 and that there are no vertical external forces on the
connecting beams, their mid-points are points of contraflexure.
12
3. METHODS OF ANALYSIS
7.
The discrete shear forces in the connecting beams in region i are replaced by
an equivalent continuous shear flow function qi, per unit length in the vertical
direction along the mid-points of the connecting laminae.
8.
9.
When there are changes in the cross-section along the height, GJiqi term is
sections, the warping deformations of the two regions are assumed to be independent
of each other.
11.
12.
3.3. Comparison of the Results of the Present Method for Non-planar Coupled
Shear Walls with Those of the Frame Method
The frame analysis, which is utilized for frame systems composed of beams
and columns, cannot be used for the analysis of multi-storey shear walls. An
idealized frame structure can be utilized to simulate the behaviour of the shear walls.
The basic difference between the frame analysis for walls and the method for frame
systems is that the effect of the finite widths of the members, in particular those of
the walls, cannot be neglected in the former. Therefore, the term wide-columnframe analogy is generally used to describe the aforementioned method to refer to
the application of the latter method to the former case.
The wide-column-frame analogy is popular in design offices for the analysis
of shear/core wall buildings. This analogy was developed first by Clough et al.
(1964) and Macleod (1967) for the analysis of plane coupled shear wall structures.
Basically, the method treats the walls and lintel beams as discrete frame members
with the finite width of the walls allowed for by horizontal rigid arms incorporated in
the beam elements. It is also commonly known as the frame method (see Fig. 3.2).
13
3. METHODS OF ANALYSIS
Figure 3.2. Equivalent frame model for planar coupled shear walls
As mentioned previously, open section shear walls resist torsion mainly by
the out-of-plane bending or warping of the walls. These open sections subject to
interaction effects of surrounding beams or slabs resist torsion to a great extent by
the continuously distributed membrane or St.Venant shear stresses. The analogous
frame in this thesis is similar to the equivalent frame method.
The frame method was extended to analyze the three dimensional coupled
shear/core wall structures by treating the non-planar shear/core walls as assemblies
of two dimensional planar wall units individually as discrete column members
residing at the centroidal axis of the wall units as shown in Fig. 3.3(a)-(b)
respectively. In the extended method, in order to allow connection between adjacent
planar wall units to form the non-planar walls, the nodes are placed along the vertical
wall joints instead of column members at the centers of the wall units and to the ends
of the coupling beams. In this frame model, so that the wall cross-section can
undergo warping deformations, the ends of the rigid arms on the connection lines of
the shear wall units are donated with hinges rotating freely about lines normal to the
shear wall units. The connecting beams are rigidly connected to the rigid arms such
14
3. METHODS OF ANALYSIS
that the rotations of the beams at the beam-wall joints are equal to those of the rigid
arms. This method was developed in the 1970s by Macleod (Fig. 3.3).
(a)
rigid arms
beam member
(b)
column
member
15
3. METHODS OF ANALYSIS
continuity of the connection between the wall units. This modification was observed
to improve the results by various comparisons with the CCM.
rigid arms
beam member
column
member
storey
levels
16
3. METHODS OF ANALYSIS
shear walls is realized by placing high connecting beams at the levels of whole or
partial stories used as storage or service areas. Such beams can be steel trusses or
reinforced concrete beams of very high bending stiffness. The number and levels of
these high beams, to improve the structural behaviour of the buildings, is up to the
design engineer (Fig. 3.5).
hs1
Stiffening beams
hs2
17
3. METHODS OF ANALYSIS
program, which can carry out algebraic computations, to write computer programs
for multi-region coupled shear walls. All of the aforementioned works dealt with
planar coupled shear walls having one row of openings under statical loading.
Not many authors have worked on the dynamics of coupled shear walls. Coull
and Mukherjee (1973) worked on unstiffened shear walls with one row of openings
and Li and Choo (1996) worked on the free vibration of coupled shear walls, having
one row of openings, with two stiffening beams, one at the top and one at the bottom.
Aksogan et al. (1999) carried out the free vibration analysis of coupled shear walls
with one row of openings and any number of stiffening beams having different
thicknesses in different regions. Bikce, Aksogan and Arslan (2004) carried out the
dynamic analysis of stiffened planar coupled shear walls on elastic foundation having
any number of rows of openings.
All of the above analyses on stiffened coupled shear walls concern
themselves with planar coupled shear walls. Recently, Resatoglu (2005) and Emsen
(2006) carried out the static analysis of non-planar coupled shear walls in their
doctoral theses. No study has been made, to the knowledge of the author, concerning
the dynamic analysis of stiffened non-planar coupled shear walls, so far.
In this thesis, the dynamic analysis of non-planar coupled shear walls with
any number of stiffening beams, having flexible beam-wall connections and resting
on rigid foundations is carried out. Furthermore, the change of wall cross-section and
the heights of the stories and connecting beams from region to region along the
height are taken into consideration. A computer program has been prepared in
Fortran Language to implement both the free and forced vibration analyses of nonplanar coupled shear walls.
18
19
M1
(1)
M2
h1
z2
M3
M4
z3
h2
(2)
stiffening
beams
(3)
zi-1
(i-1)
zi
H
hi-1
(i)
hi
zi+1
(i+1)
zn-1
z
(n-1)
hn-1
zn
MN-1
(n)
hn
MN
Figure 4.1. Non-planar coupled shear wall and its lumped mass model
20
21
x s1i
xs2i
x g 2i
x g 1i
Y 2i
S2i
Y1i
f2 i
S1i
y s 1i
G2i
X1i
f1i
y g1i
connecting beam
G1i
1st pier
X 2i
y g 2i
2nd pier
ci
Figure 4.2. Plan of non-planar coupled shear wall in region i
A non-planar coupled shear wall and its plan for one region are given in Figs.
4.2-3 with global axes OX, OY and OZ, the origin being at the mid-point of the clear
span in the base plane. The X axis is along the longitudinal direction of the
connecting beams. The Z axis is the vertical axis and the Y axis is in the horizontal
plane perpendicular to the X axis. The plan of the non-planar coupled shear wall in
region i is seen in Fig. 4.2. Throughout the analysis of the general problem in this
thesis, the cross-sectional form of the non-planar coupled shear wall will be assumed
to have this shape.
22
Z
z1
(1)
h1
z2
(2)
h2
z3
stiffening beams
(3)
zi-1
(i-1)
zi
hi-1
(i)
hi
zi+1
(i+1)
zn-1
(n-1)
hn-1
zn
(n)
hn
subscripts 1 and 2 express the left and the right piers, respectively. The subscript i
(i = 1,2,..., n ) , refers to the number of region. Similarly, the shear centers of the piers
are located at x s1i , ys1i and x s 2 i , ys 2 i , respectively. The coordinates referred to the
23
( j = 1,2) , (i = 1,2,..., n )
(4.1)
where uji and vji are the lateral displacements of the shear centers with reference to
the principal axes of pier j (j=1,2) and q ji is the rotation about Z ji axis.
The equilibrium equations of the whole structure are expressed in the (X, Y,
Z) system of axes. The transformation of the geometric and physical properties of the
shear walls into this system is possible by combined rotation and translation. The
rotation and translation matrices can be expressed, respectively, as follows:
Cosf ji
R ji = - Sin f ji
0
Sin f ji
Cosf ji
0
0
0
1
1 0 - ys ji
Tji = 0 1 x s ji
0 0
1
( j = 1,2) , (i = 1,2,..., n )
(4.2)
( j = 1,2) , (i = 1,2,..., n )
(4.3)
in which x s ji and ys ji are the coordinates of the shear center of pier j relative to the
global axes.
The displacement vector of the shear wall in the local axes passing through
the shear center are related to the corresponding displacements u i , v i and qi in the
global axes, which are given in compact notation as
24
ui
Di = vi
qi
(i = 1,2,..., n )
(4.4)
where u i and v i are the displacements at the mid-points of the connecting laminae
in the X and Y directions of the global system of axes, and qi is the rotation of the
rigid diaphragm about the Z axis, in region i. Thus, the transformation of the
displacement vector can be expressed as
D ji = R ji Tji Di
( j = 1,2) , (i = 1,2,..., n )
(4.5)
( j = 1,2) , (i = 1,2,..., n )
(4.6)
Sin f ji
Cosf ji
0
- y s ji u i
x s ji v i
1 qi
( j = 1,2) , (i = 1,2,..., n )
(4.7)
v ji = v i + x s ji qi
( j = 1,2) , (i = 1,2,..., n )
(4.8)
q ji = qi
( j = 1,2) , (i = 1,2,..., n )
(4.9)
25
values with respect to another system of axes Oxy parallel to the first system can
be given as follows (see Fig. 4.4):
y
a
O
b
(4.10)
Iy = I y + a 2 A
(4.11)
Ixy = I xy + ab A
(4.12)
Consider two sets of axes Oxy and O x y , the latter being inclined at an
angle f with the former (see Fig. 4.5).
dA
x
f
O
( )
26
x Cosf Sin f x
=
y - Sin f Cosf y
(4.13)
In addition to this, applying the rotation matrix in (4.13), to the moments of inertia of
a cross-section, the moments of inertia with respect to the system of axes O x y can
be found in a straightforward manner. It should be noted that, to calculate the
moments of inertia with respect to the system of axes O x y in terms of the moments
of inertia with respect to the system of axes Oxy, starting from their definitions:
Ix = y dA = (- xSin f + yCosf) dA
(4.14)
I y = x dA = (xCosf + ySin f ) dA
(4.15)
(4.16)
Using the definitions of the moments of inertia with respect to the system of axes
Oxy, the relations in (4.14-16) can be rewritten as follows:
I x = I x Cos 2f + I ySin 2f - 2I xySin fCosf
(4.17)
(4.18)
(4.19)
The system of axes for which the product of inertia I xy of a planar area is zero
is called principal system of axes. As seen from expression (4.19), I xy becomes
zero for f0 defined by
tg 2f0 = -
2 I xy
Ix - Iy
(4.20)
27
I1, 2 =
Ix + I y
2
I -I
x y + I 2 xy
2
(4.21)
M1i
M2i
q y1i
q y2i
V1i
V2i
Li
28
M1i=
EIci
4q y1i + 2q y2 i + 6y i + M12i
Li
(4.22)
M1i=
6EIci
q yi
Li
(4.23)
M2i=
6EIci
q yi
Li
(4.24)
Since M1i and M2i are equal, they can be called by the same name as Mi. Considering
the free body diagram of the connecting beam in Fig. 4.6, the summation of the
moments about the left end being set equal to zero, yields
V2i=
2M i
Li
(4.25)
29
(4.26)
M mp i =
2M i Li
- Mi = 0
Li
2
(4.27)
meaning that the bending moment at the mid-points of the connecting beams vanish.
4.3.3. The Relation between the Axial Force Ti and the Shear Force per Unit
Length qi
Fig. 4.7 shows an element of infinitesimal length dz in region i of a coupled
shear wall during the application of the continuous connection method. The shear
force per unit length of the continuous medium is exposed by the vertical cut along
the line through the mid-points. The axial forces evoked in one of the piers by these
shear forces and the shear forces on that pier satisfy the vertical force equilibrium
equation, i.e.,
Ti+dTi
dz
G2i
Ti+dTi
qi
G1i
Ti
Ti
30
(4.28)
(4.29)
which gives the relation between the axial force and the shear force per unit length at
any height, when divided by dz, as follows:
dTi
= -q i
dz
(4.30)
4.3.4. Formulation
4.3.4.1. Compatibility Equations
While obtaining the compatibility equations in a non-planar coupled shear
wall analysis, it is assumed that all rows of connecting laminae will be cut through
the mid-points, which are the points of zero moment. The compatibility of the
relative vertical displacements, on the two sides of the mid-points of the connecting
laminae, occurs as a result of six actions.
4.3.4.1.(1). The relative vertical displacements due to the deflections and
rotations of the piers
The first contribution to the total relative displacement at the mid-point of a
lamina is due to the deflection and rotation of the piers. These are caused by the
bending of the piers in the principal directions and their twists about the shear
centers. In the non-planar system of coupled shear walls given in Fig. 4.8, x g ji and
y g ji (j=1,2, i=1,2,,n), which are the distances of the centroids from point O,
31
G2i
S1i
f1i
X1i
G1i
f2 i
S2i
yg 2 i
y g1i
X
x g 1i
x g2i
32
Y 2i
Y1i
X 2i
G2i
X1i
G1i
u 2 i x g 2 i
- u 1 i x g 1 i
Y 2i
Y1i
X 2i
G2i
X1i
G1i
- v 1 i y g 1 i
v 2 i y g 2 i
S2i
S1i
- q1 i w1i
q2i w 2i
c) Twisting about the shear centers
Figure 4.9. Relative displacements at the mid-point of a lamina due to the deflection
and rotation of the piers in region i
33
) (
(4.31)
The first two terms in equation (4.31), represent the contributions of the bending of
the piers about the principal axes and the last represents the contribution of the
twisting of the piers. w1i and w2i are the sectorial areas at points on the left and the
right side of the cut for piers 1 and 2, respectively, as shown in Fig. 4.10.
S2i
S1i
w1i
+
w2i
Figure 4.10. The principal sectorial area diagrams of the piers in region i
A detailed exposition of thin-walled beam theory and the definition of
sectorial area with formulas were made in Chapter 3. Substitution of equations (4.79) into (4.31), yields
d1i = y s1i x g1i qi - y s2i x g 2i q - x s1i y g1i qi + x s2 i y g 2 i qi
- x g1i ui + x g 2 i u i - y g1i vi + y g 2 i vi + qi w1i - qi w2i
Rearranging the equation (4.32)
34
(4.32)
(i = 1,2,..., n )
d1i = u i a i + vi b i + qi (wi + d i )
(4.33)
where i is the number of a particular region and the new constants are defined as
follows:
wi = w1i - w2 i
(i = 1,2,..., n )
(4.34)
a i = x g 2 i - x g 1i
(i = 1,2,..., n )
(4.35)
b i = y g 2 i - y g 1i
(i = 1,2,..., n )
(4.36)
(i = 1,2,..., n )
(4.37)
V1
(1)
(i-1)
z1
q1
1st pier
qi-1
2nd pier
Vi
(i)
zi
qi
Ti
Ti
Figure 4.11. Vertical forces on the piers
35
zi
i -1
t =1
zt
z t +1
t =1
(i = 1,2,..., n )
Ti = q i dz + ( q t dz) + Vt
(4.38)
The axial force, Ti, produces axial deformation in the piers and, in turn,
causes additional relative vertical displacement along the cut as shown in Fig. 4.12.
Ti
Ti
d*2i
d*2*i
ci/2
ci/2
Figure 4.12. Relative displacements due to the axial forces in the piers
The normal strain being defined as
s
T
=
, the total axial elongation of pier 1 can
E EA
be written as
z
1 n 1 t
1 1 i
Ti dz
d =-
Tt dz -
E t =i +1 A1t zt +1
E A1i zi+1
*
2i
(i = 1,2,..., n )
(4.39)
(i = 1,2,..., n )
(4.40)
1 n 1 t
1 1 i
Ti dz
d =-
Tt dz -
E t =i +1 A 2 t zt +1
E A 2i zi+1
**
2i
36
d 2i = -
z
z
1 1
1
1 n 1
1 t
T
dz
+
+
t
Ti dz
E t =i+1 A1t A 2 t zt +1
E A1i A2i zi+1
(i = 1,2,..., n )
(4.41)
PL3
. Hence, the shear force in each lamina of dz height
deflection equal to d =
3EI
which is
Pi = q i dz
(4.42)
Pi
d*3i
d*3*i
ci/2
Pi
ci/2
37
d*3i = -
(i = 1,2,..., n )
(4.43)
(i = 1,2,..., n )
(4.44)
and
q i dz (c i / 2)
3EIci
dz
hi
d*3*i = -
where hi is the storey height and ci is the clear span of the laminae in region i.
Therefore, the total relative displacement caused by the shear forces in the laminae,
which is shown in Fig. 4.13, can be obtained by adding expressions (4.43) and (4.44)
as
h i ci
qi
12EIci
(i = 1,2,..., n )
(4.45)
4.3.4.1.(4). The Relative Vertical Displacement due to the Shearing Effect in the
Laminae
A point load P at the end of a cantilever beam of length L, causes an end
PL
The same shear force Pi (see Fig. 4.14) as mentioned before in Section
4.3.1.3, causes relative displacements in the lamina as
qi (ci / 2) dz qi (ci / 2) dz
q ch
=- i i *i
*
*
Ac
Ac
GAci
G i dz
G i dz
hi
hi
38
(i = 1,2,..., n )
(4.46)
A *ci =
A ci
(i = 1,2,..., n )
mc
(4.47)
in which the constant m c takes the value 1.2 for rectangular cross-sections.
Therefore, the total relative vertical displacement due to the shearing effect in
the lamina can be written as follows:
d 4i = -
1.2 q i c i h i
GA ci
(i = 1,2,..., n )
(4.48)
Pi
d*4i
d*4*i
ci/2
Pi
ci/2
Figure 4.14. Relative vertical displacement due to the shear deformation in a lamina
4.3.4.1.(5). The Relative Vertical Displacement due to the Flexibility of the
Connections
In the rigid case, a lamina takes a form tangent to line 1, which is
perpendicular to the deformed axis as shown in Fig. 4.15. When the connections are
flexible, the lamina will take a form such that it is tangent to line 2.
The relative vertical displacement due to the rotational spring (see assumption
2, 4) at the left end of the lamina can be written as
39
d*5i = -a i
ci
2
(4.49)
K ci dz
hi
ai ), where K ci is the
rotational spring constant of a connecting beam. From this relation it is clear that
( ai =
Mi h i
). Therefore, substituting this expression into (4.49)
K ci dz
d*5i = -
M i h i ci
K ci dz 2
(4.50)
can be obtained. Mi in equation (4.50) is the moment at the end-point of the lamina.
It is equal to (Mi = Pi ci / 2 ) due to the shear force, Pi , at the mid-point of the
lamina. It is also known that ( Pi = q i dz ). Therefore, the total relative vertical
displacement due to the flexibility of the connections can be written as
qi h ici
qhc
qhc
- i i i =- i i i
4K ci
4 K ci
2K ci
(i = 1,2,..., n )
2
ai
(1)
(4.51)
d*5i
ci/2
40
(i = 1,2,..., n )
(4.52)
On the other hand, applying (4.33) for region i-1 at the common boundary ( z = z i ),
results in
d1( i-1)
z =z i
= u( i-1)
a (i-1) + v( i-1)
z =z i
z=zi
z=zi
(w(i-1) + d (i-1) )
(4.53)
The difference of (4.52) and (4.53) expressions gives the additional relative vertical
displacement at the mid-point of the lamina at the boundary i as
i = d1i z=z - d1(i -1)
i
(i = 1,2,..., n )
z=zi
(4.54)
For any number of changes in cross-section, (4.54) expression is computed for all of
the boundaries of change below the level of concern and added to the relative
displacement at that level. This additional quantity, which is constant for a region,
can be expressed as follows:
d6i =
t =
t =i +1
[d
n
t =i +1
1t z = z
t
41
(i = 1,2,..., n )
(4.55)
ys 2 i
(i = 1,2,..., n )
d1i + d 2i + d3i + d 4 i + d 5i + d 6i = 0
(4.56)
Therefore, the compatibility equation for the vertical displacement in region i can be
written as
u iai + vibi + q i(wi + d i )
1
E
z
1
1 1
zj
1
1
+
T
dz
Ti dz
+
j
A2 j zj +1
E A1i A2 i zi +1
j =i +1 A1 j
3
2
Ti 1.2 hi ci
hc
hc
+ i i + i i
E GAci
12 EI ci 2 K ci
E + d 6 i = 0 (i = 1, 2,..., n ) (5.57)
(i = 1,2,..., n )
(4.58)
(i = 1,2,..., n )
(4.59)
where
3
2
1 . 2 hi c i
hi c i
hi c i
g ci =
+
+
GA c
12 EI c i 2 K c i
i
42
d t
(Tt )dz = 0
dz zt +1
(4.60)
d
(Ti )dz = dz (Ti ) - dzi+1 (Ti ) = Ti
dz
dz
dz zi +1
(4.61)
are employed.
4.3.4.2. Equilibrium Equations
4.3.4.2.(1). Bending Moment Equilibrium Equations
The coordinate system and positive directions of internal bending moments
acting on the different components of the coupled shear wall are adapted as shown
vectorially in Fig. 4.16.
E I y 2 i u 2i
E Iy1i u1i
G2i
G1i
EIx1i v1i
f1i
f2i
bi
EIx2i v2i
X
O
ai
Figure 4.16. Internal bending moments acting on the components of the coupled
shear wall
These internal moments, along with the couple produced by the axial force,
Ti, balance the external bending moments M EXi and M EYi . For the equilibrium of
the moments about X and Y axes, the following relationships can be derived using
Vlasovs theory of thin walled beams:
43
(4.62)
(4.63)
E I y1i u1i cosf1i - I x1i v1i sinf1i + I y2i u2i cosf2i - I x 2i v2i sinf2i + Ti a i - MEYi = 0
E I y1i u1i sinf1i + I x1i v1i cosf1i + I y2i u2i sinf2i + I x 2i v2i cosf2i + Ti bi - MEXi = 0
On substituting the local displacement expressions (4.7-9) and the moments of inertia
expressions (4.17-20) into equations (4.62-63), the bending moment components are
found, in terms of the global displacements at point O, as
M EYi = E I Yi ui + E I XYi vi - E IqYi qi + Ti a i
(i = 1,2,..., n )
(4.64)
(i = 1,2,..., n )
(4.65)
where
I Yi = I y1i + I y2 i
(4.66)
I X i = I x1i + I x 2 i
(4.67)
I XYi = I xy1i + I xy 2i
(4.68)
(4.69)
(4.70)
Terms I y ji and I x ji are the second moments of area of the cross-sections, and I xy ji is
the product of inertia of pier j (j=1,2, i=1,2,,n) about axes parallel to the global
axes and passing through the centroids. I y ji and I x ji are the second moments of area
of the cross-sections of pier j (j=1,2, i=1,2,,n) about its respective principal axes.
44
E I y2i u2i
E Iy1i u1i
- E Iw2i q2i
- E Iw1i q1i
EIx 2i v2i
S2i f 2i
E Ix1i v1i
S1i f1i
y s 2i
O
y s1i
x s 2i
x s1i
45
(H-z)
M yq 2 i
V1
q1
M xq 2 i
Bq 2 i
Vi
M yq1i
qi
Ti
M xq1i
B q1i
Ti
Figure 4.18. 3-D view of the additional internal bending moments and bimoments
due to shear forces in connecting medium
In Fig. 4.17, the internal bimoment expressions - E Iwji qji (j=1,2, i=1,2,,n),
are caused by the non-uniform warping of the cross-sections of the piers. It must be
mentioned that the computation of the internal bimoments created at point O by the
bending moments of the wall, are carried out after transferring the bending moments
to the shear centers of the piers.
Let Bi be the resultant bimoment about point O, which is due to the
resistance offered by the piers. It can be written as (see Fig. 4.17)
Bi = -E I w1i q1i - E I w2i q2i
( )
( )
v (- x )- E I
+ E I y1i u 1i y s1i + E I y 2 i u 2i y s 2 i
+ E I x1i
46
1i
s1i
x 2i
( )
v2i x s 2 i
(4.71)
(i = 1,2,..., n )
Bi = E I qYi u i - E I qX i vi - E I wi qi
(4.72)
(4.73)
M yq 2 i
M yq1i
Bq 2 i
M xq 2 i
B q1i
S2i f 2i
M xq1i
S1i f1i
y s2i
O
y s1i
x s 2i
x s1i
Figure 4.19. Cross-sectional view of the additional internal bending moments and
bimoments due to the shear forces in the connecting medium
47
(4.74)
(4.75)
( )
( )
( )
(4.76)
These additional bending moments and bimoments acting on an element are shown
in Fig. 4.18. From the consideration of equilibrium of moment about Y1i and X1i
axes for pier 1, the following relations can be obtained, respectively:
( )
M xq + Ti y g1i = 0
1i
(4.77)
48
M yq - Ti - x g1i = 0
(4.78)
M xq = -Ti y g1i
(4.79)
M yq = -Ti x g1i
(4.80)
1i
Thus,
1i
1i
(4.81)
M yq = Ti x g 2i
(4.82)
2i
2i
(4.83)
BE i = Bi + Bi
(i = 1,2,..., n )
(4.84)
(i = 1,2,..., n )
(4.85)
49
E I x 2 i v2i
E I x1i v1i
S1i
y s1i
E I y 2 i u 2i
S2i
E I y1i u1i
f1i
y s2i
f 2i
- E I w2 i q2i + GJ 2 i q2 i
O
X
x s1i
x s 2i
50
( )
( )
v ( - x ) - E I
s1i
1i
x 2i
( )
v2i x s2 i
(4.86)
On substituting the expressions (4.7-9), (4.13) and (4.17-20) into (4.86), the resultant
resisting twisting moment of the piers for all regions in the structure are found as
(i = 1,2,..., n )
(4.87)
(i = 1,2,..., n )
J i = J1i + J 2 i
(4.88)
Q y 2i
Q x 2i
Ti+dTi
Q y 2i
G2i
dz
Q y1i
G1i
Q x1i
Q y1i
Q x1i
Q x 2i
M tq 2 i
M tq 2 i
qi
M tq1i
M tq1i
Ti
Ti
Figure 4.21. 3-D view of the additional internal shear forces and twisting moments
due to the shear flow in the connecting medium
51
Q y2 i
Q y1i
S1i
y s1i
Qx2i
S2i
Q x1i
f1i
M tq2 i
y s2i
M tq1i
f 2i
x s1i
x s 2i
Figure 4.22. Cross-sectional view with the additional internal twisting moments and
shear forces due to the shear flow q i
All quantities are related to the shear centers of the piers, S1i and S2i. In order
to determine the twisting moment due to the shear flow q i , the relationship between
the force components in the piers and q i may be determined from the free body
diagram in Fig. 4.21.
Let M t i be the resultant twisting moment due to the additional twisting
moments and shear forces about point O as shown in Fig. 4.22.
The shear flow q i in the laminae produces bimoments on the piers. For pier
1, the bimoment dB z1i is caused by the external force q i dz and is equal to the
product of this force and the principal sectorial area, w1i , of the point of its
application. Therefore, it can be written in the following form:
dB z1i = q i dz w1i
(i = 1,2,..., n )
(4.89)
(i = 1,2,..., n )
(4.90)
52
These bimoments produce the flexural twisting moments M tq1i and M tq2 i , related to
the shear centers S1i and S2i, respectively. Using the relation obtained in (3.60),
M tq1i =
M tq 2i =
dB z1i
dz
dB z 2i
dz
= q i w1i
(i = 1,2,..., n )
(4.91)
= -q i w2 i
(i = 1,2,..., n )
(4.92)
The resultant twisting moment, M t i , due to these additional torques and shear
forces about point O is, then, found as
( )
( )
( )
(4.93)
These additional torques and shear forces acting on an element are shown in Fig.
4.21. From the consideration of equilibrium of moments about Y1i and X1i axes for
pier 1, the following relations can be obtained, respectively:
Q x1i dz + q i dz x g1i = 0
(4.94)
Q y1i dz + q i dz y g1i = 0
(4.95)
Thus,
Q x 1i = - q i x g1i
(i = 1,2,..., n )
(4.96)
Q y1i = - q i yg 1i
(i = 1,2,..., n )
(4.97)
53
(i = 1,2,..., n )
(4.98)
Q y 2 i = q i yg 2 i
(i = 1,2,..., n )
(4.99)
(i = 1,2,..., n )
(4.100)
M Eti = M ti + M t i
(i = 1,2,..., n )
(4.101)
(i = 1,2,..., n )
(4.102)
Finally, using the compatibility equation (4.57) and the four equilibrium
equations (4.64), (4.65), (4.85), and (4.102), the 4n unknowns of the problem,
namely u, v, q , and T, can be found under the unit loadings M EX i , M EYi , BE i , and
M Eti .
54
ui =
E IYi
(4.103)
vi =
(4.104)
(
T (a I
XYi
(4.105)
55
(4.106)
(
(
I I +I I
a I - bi I Yi
IXi I Yi - I 2XYi
I Xi IYi - I2XYi
) (
(
(
I I +I I
a I - bi IXYi
ui =
M EXi IXYi
M EYi IXi
E
+
) (
(4.107)
(4.108)
D i = ( I Xi I Yi - I 2XYi
K1i =
K 2i =
(I
Xi
(4.109)
(4.110)
(4.111)
Di
(I
K 3i =
K 4i = -
XYi
IqYi + I Yi IqXi
(4.112)
Di
(a I
i
Xi
- bi I XYi
(4.113)
Di
(a I
i
XYi
- bi I Yi
(4.114)
Di
56
vi =
M I
M I
1
- E qi K 2 i - Ti K 4i - EYi XYi + EXi Yi
E
Di
Di
ui =
M I
M I
1
E qi K1i - Ti K 3i - EXi XYi + EYi Xi
E
Di
Di
(i = 1,2,..., n )
(4.115)
(i = 1,2,..., n )
(4.116)
vi =
M I
M I
1
- E qiK 2 i - TiK 4i - EYi XYi + EXi Yi
E
Di
Di
u i =
M I
M I
1
E qiK 1i - Ti K 3i - EXi XYi + EYi Xi
E
Di
Di
(4.117)
(4.118)
M Eti = E IqYi
M I
M I
1
E qiK1i - Ti K 3i - EXi XYi + EYi Xi
E
Di
Di
-E IqXi
M I
M I
1
- E qiK 2i - TiK 4i - EYi XYi + EXi Yi
E
Di
Di
(4.119)
Reorganizing (4.119),
M Et i = - Ti (wi + d i ) + Ti I qXi K 4i - Ti I qYi K 3i
+ E I qXi qi K 2 i + E I qy qi K1i - E I wi qi
+ MEYi
(I
Xi
) - M ( I
EXi
57
XYi
I qYi + I Yi I qXi
Di
) + G J q
i
(4.120)
(4.121)
(4.122)
(4.123)
(i = 1,2,..., n )
(4.124)
a i I XY - b i I Y
i
i
ri = wi + d i - I qXi D
ri = wi + d i +
) +I
qYi
a i I X - b i I XY
i
i
I X I qY + I XY I qX
I XY IqY + I Y I qX
i
i
i
i
i
i
ri = wi + d i + a i i i
- bi
Di
Di
and
58
(4.125)
(4.126)
Di
(4.127)
ri = wi + di + a i K1i - bi K 2i
(4.128)
qi =
1
E ri
Ti
- Tig ci
- M EXi K 3i - M EXi K 4i +
Ai
(i = 1,2,..., n )
(4.129)
Reorganizing (4.124),
(4.130)
Derivatives of equations (4.129) and (4.130) with respect to z twice and once,
respectively, are as follows:
qi=
1
E ri
T
i K 3i - MEX
i K 4i + i - Ti g ci
- MEX
Ai
G J i qi = -Tiri + MEY
i K1i - MEX
i K 2i - METi
E Iwi qi-
(4.131)
(4.132)
Iwi
1
ri
T
i K 3i - MEX
i K 4i + i - Ti g ci
- MEX
Ai
1
T
-G J i
- M EXi K 3i - M EXi K 4i + i - Tig ci
E ri
Ai
i K1i - MEX
i K 2i - MEti
= -Tiri + MEY
(4.133)
Simplifying equation (4.133), the single fourth order governing differential equation
for the axial force, Ti, is obtained as follows:
59
(g
ci wi
)T - AI
G J i g ci
G Ji
2
Ti =
+ ri Ti+
E
A
E
i
i
i I wi K 3i + K1i ri - MEX
i I wi K 4i - K 2i ri +
- MEY
wi
GJ i
M EYi K 3i + M EXi K 4i + MEt i ri
E
(4.134)
Equation (4.134) can be reorganized and put into the following form:
i I wi K 3i + K1i ri - MEX
i Iwi K 4i - K 2i ri +
- MEY
GJ i
M EYi K 3i + M EXi K 4i + MEti ri
E
(i = 1,2,..., n )
(4.135)
where
b1i = g ci Iwi
(4.136)
b2i =
I wi G J i g ci
2
+
+ ri
Ai
E
(4.137)
b 3i =
G Ji
E Ai
(4.138)
60
z - z'
= (z - z ')
z - z'
=0
and
z - z'
=1
for
z > z'
and
z - z'
=0
for
z z'
(4.139)
Thus, in the general form, the external effects M EX i , M EYi and M Eti for any
unit loading is found, using the following expression for the particular case:
M EXi = H p - z
M EYi = H p - z
M Eti =
(4.140)
(4.141)
(- d PY ) +(d PX )
(4.142)
if
if
Hp > z
Hp z
;
;
H p - z = (H p - z )
1
(4.143)
Hp - z = 0
M EXi = (H p - z )
i = -1
M EX
i = 0
M EX
M EYi = (H p - z )
i = -1
M EY
i = 0
M EY
(4.144)
Hence, for the part above the unit load, M EX i , M EYi and M Eti are equal to zero and
61
(k = 1,2,... )
(4.145)
t =1
BEi = 0
(4.146)
b 3i E
[GJ (K
i
3i
M EYi + K 4i M EX i
)]
(i = 1,2,..., n )
(4.147)
in which
b - b 2 - 4b b
2i
1i 3i
a1i = 2 i
2 b1i
(i = 1,2,..., n )
(4.148)
b + b 2 - 4b b
2i
1i 3i
a 2 i = 2i
2 b1i
(i = 1,2,..., n )
(4.149)
62
1 n 1
1 t
Tt dz
-
+
E t =i A1t A 2 t zt +1
Vi
E
3
2
1.2 ci
ci
ci
+
+
+ d6i = 0 (i = 1,2,..., n )
GAsi 12EIsi 2K sbi
(4.150)
1 n 1
1 t
T
dz
+
t
E t =i A1t A 2 t zt +1
3
2
Ti z=z 1.2h c
h i ci
h i ci
i i
i
+
+
+
i
GA c i 12EIc i 2K cb i
GA s i 12EIs i 2K sb i
(4.152)
12EIc i 2K cb i
GA c i
Si =
3
2
1.2 ci
c
c
+ i + i
GA s i 12EIs i 2K sb i
(4.153)
(4.154)
63
(4.155)
un z = 0 = vn z = 0 = qn z = 0 = 0
(4.156)
Applying equation (4.57) at ( z = 0 ) and using (4.156), the first boundary condition is
Tn z = 0 = 0
2-
(4.157)
the twisting moment expression (4.102) and using qn z = 0 = 0 , the second boundary
condition at the base ( z = 0 ) is found as
T
- MEYn K 3n - MEXn K 4n + n - g cn Tn - Tn K 3n
M Etn - IqYn
+
Dn
rn
An
M I - MEYn I XYn K 2n
T
- MEYn K 3n - MEXn K 4 n + n - g cn Tn - Tn K 4 n
+ IqXn EXn Yn
rn
An
Dn
Iw
T
+ n - MEYn K 3n - MEXn K 4n + n - g cn Tn + (wn + d n )Tn = 0
(4.158)
rn
An
3-
From the equilibrium of the vertical forces in each pier in the uppermost
T1 = q1dz + V1
(4.159)
64
T1
z =H
= - S1 T1 z =H
(4.160)
(4.161)
G11
V1
q1
T1
Figure 4.23. The free-body diagram of a part of the shear wall at the top
4-
expression (4.85) and applying it at the top ( z = H ), the fourth boundary condition is
obtained as
T
- M EY1 K 31 - M EX1 K 41 + 1 - g c1 T1 - T1 K 31
BE1 - IqY1
+
r1
A1
D1
T
- M EY1 K 31 - M EX1 K 41 + 1 - g c1 T1 - T1 K 41
+ I qX1
D1
r1
A1
Iw
T
+ 1 - M EY1 K 31 - M EX1 K 41 + 1 - g c1 T1 + (w1 + d1 ) T1 = 0
(4.162)
r1
A1
65
From the vertical force equilibrium of one piece of the stiffening beam in
T(i -1)
z =z i
+ Vi = Ti
(4.163)
z = zi
T(i -1)
z = zi
- Si Ti z = z = Ti z = z
i
(i = 2,3,..., n )
(4.164)
T(i-1)
G1i
Vi
Ti
Figure 4.24. The vertical forces acting on one piece of the stiffening beam at the
height z = zi
6-
Applying the compatibility equation (4.57) for two neighbouring regions (i)
and (i1) at height z = z i , considering (4.55) for the case of cross-sectional changes,
the following equation is obtained as the sixth type boundary condition:
1.2h (i-1)c (i-1) h (i -1) c3(i-1) h (i -1) c 2(i-1)
T(i-1) z=z
+
+
i
GA
12
EI
2
K
c ( i -1 )
c ( i -1 )
cb ( i -1)
2
1.2h i c i h i c i 3
hc
= Ti z=z
+
+ i i
i
GA ci 12EIci 2K cb i
66
(i = 2,3,..., n )
(4.165)
Since the total twisting moments of the two neighbouring regions (i) and
+ E I wi qi+ (wi + d i ) Ti
(4.166)
Reorganizing (4.166),
[M
Et( i -1)
] [
]
v( ) - E I v] + [- G J ( )q( ) + G J q ]
q( ) - E I q] + [(w( ) + d ( ) ) T( ) - (w + d ) T]
+ E I qX(i-1)
+ E I w(i-1)
i -1
qXi
i -1
wi
i -1
i -1
i -1
i -1
i i
i -1
=0
(4.167)
(i = 2,3,..., n )
(4.168)
where
C1i =
g ci K 1i I qYi
ri
g ci K 2i IqXi
ri
C 2i = IqYi K 3i - I qXi K 4i -
I wi g c i
I qYi K1i
A i ri
ri
-
I qXi K 2i
A i ri
I wi
A i ri
+ wi + d i
C 3i = i i
Di
Di
+ M Eti
- i
ri
ri
67
(4.169)
(i = 2,3,..., n )
Since the total bimoments of the two neighbouring regions (i) and (i1), at
(4.170)
Reorganizing (4.170),
[B
E( i -1 )
][
]
v] + [E I
w( i -1)
q(i-1) - E Iwi qi
(4.171)
Expressing the other unknown functions in terms of Ti and its derivatives, after some
necessary rearrangements, the eighth type boundary condition is obtained as
C1(i -1) T(i-1) - C1i Ti+ C 2 (i -1) T(i-1) - C 2i Ti + C 4( i-1) - C 4i = 0
(4.172)
where
I XY IqY M X + I Yi I qXi M X I XYi I qXi M Y + I Xi I qYi M Y
-
C 4i = i i
D
D
i
i
+ B Ei
- i
ri
ri
(4.173)
68
qi =
1
Eri
Ti
T
dz dz + G1i z i + G 2i
M
K
M
K
+
g
EX
3
i
EX
4
i
i
c
i
i
i
A
i
(i = 1,2,, n )
(4.174)
3-4.
Due to the complete fixity, there is no rotation or warping at the base. Hence,
q n z =0 = 0
(4.175)
qn z =0 = 0
(4.176)
From the continuity of q and q for two neighbouring regions (i) and (i1) at
qi z=z = qi -1 z=z
(i = 2,3,, n )
(4.177)
(i = 2,3,, n )
(4.178)
69
ui =
E
K
T
K
q
+
i
1
i
i
3
i
dz dz + R 1i z i + R 2i
E
Di
D i
(i = 1,2,, n )
vi =
(4.179)
E
K
T
K
q
+
i
2
i
i
4
i
dz dz + N1i z i + N 2i
E
Di
D i
(i = 1,2,, n )
(4.180)
Substituting from (4.129) into (4.179) and (4.180), ui and vi can be expressed
in terms of the variable z only. In the resulting expressions of ui and vi, there are 4n
integration constants Ri and Ni to be determined from the boundary conditions of the
problem. These boundary conditions come from the equivalence of the horizontal
displacements and the respective slopes for every pair of neighbouring regions at
their common boundary (z = zi)
u i z =z = u i-1 z=z
i
v i z=z = vi -1 z =z
i
ui z =z = ui-1 z=z
i
vi z=z = vi -1 z =z
i
(i = 2,3,, n )
(4.181)
(i = 2,3,, n )
(4.182)
(i = 2,3,, n )
(4.183)
(i = 2,3,, n )
(4.184)
and the vanishing of the horizontal displacements and the respective slopes at the
bottom, i.e.,
un
z =0
= v n z =0 = 0
(4.185)
un
z =0
= vn z =0 = 0
(4.186)
70
K=F
-1
(4.187)
K - w2 M = 0
(4.188)
where is the circular frequency, M is the mass matrix and K is the stiffness matrix
of the structure. The respective eigenvectors, si, are found by substituting each and
every circular frequency, i, in the following equation at a time:
(K - w M )s = 0
2
71
i = 1,2,..., m
(4.189)
P = P(t)
(vary with time)
P (static)
Inertial forces
(a) Static loading
72
.
m N1
m12
m 22
m 32
.
m N2
m13
m 23
m 33
.
m N3
.
.
.
.
k N1 k N 2 k N 3
.
.
.
.
.
.
c12
c 22
c 32
.
c N2
c13
c 23
c 33
.
c N3
k1N
k 2 N
k 3N
.
k NN
x1
x
2
x3 =
.
x N
.
.
.
.
.
c1N
c 2 N
c3N
.
c NN
x& 1
x&
2
x& 3
.
x& N
P1
P
2
P3
.
PN
(5.1)
It is observed that every equation in the set of Eq. (5.1) involves entities
belonging to each and every node. Despite the fact that the solution of this equation
set is possible, it gets rather difficult for shear walls with a high degree of freedom.
Employing the mode superposition method, the solution can be rendered simpler as
~
m
11
.
.
.
~
m
22
.
~
m 33
.
.
.
.
~
m
NN
~
k11
.
+ .
.
.
&x&1 ~c11
&x& .
2
&x& 3 + .
. .
&x& N .
.
~
k 22
.
.
.
~
k 33
.
.
.
.
.
.
.
.
.
73
.
~c
22
.
.
.
~c
33
.
~
k NN
x1
x
2
x3 =
.
x N
.
.
.
P1
P
2
P3
.
PN
.
.
.
.
~c
NN
x& 1
x&
2
x& 3
.
x& N
(5.2)
F = F1 F 2 F 3 LF i LF N
(5.3)
where, F i is the eigenvector pertaining to the ith eigenvalue and the dimension of it is
equal to Nx1, in which N represents the degree of freedom and Ni shows the number
of lumped masses between the ends of section i.
n
N = ( N i + 1) + 1
(5.4)
i =1
Obtaining the modal matrix as mentioned above, the real displacement vector
Y is described in terms of X modal displacement vector as follows:
Y=F X
(5.5)
Using equation (5.5) and its first and second time derivatives
Y& = F X&
(5.6)
Y&& = F X&&
the equation of motion of the multi-degree of freedom system
&& + C Y
& +K Y =P
MY
(5.7)
74
(5.8)
(5.9)
~
M = FT M F =
~
m
22
~
m
NN
(5.10)
c11
~
~
C = FT C F =
~c
22
~c
NN
(5.11)
~
k 11
~
K = FT K F =
~
k 22
~
k NN
75
(5.12)
(5.13)
(5.14)
After necessary computations and finding X modal displacement vector, the real
displacement vector of the system, Y , is expressed in the following form:
Y = F X = F1 F 2 F 3 L F N
X1
X
2
X3
:
X N
(5.15)
76
(5.16)
(5.17)
in which, the parameters a and d define the variation of acceleration over a time step
and determine the stability and accuracy characteristics of the method. According to
the assumptions of linear acceleration and constant average acceleration, the
parameters a and d are taken as 1/6 and 1/2, respectively. The solution of this
method is based on following steps:
1)
The starting value is assumed for the equation obtained in (5.14). The
displacement vector and the velocity vector are taken to be zero for t = 0 , i.e.,
X0 = 0
&0 =0
X
(5.18)
The starting value of the acceleration vector is obtained by substituting the values in
(5.18) into equation (5.14) as follows:
~ &
~ &&
~
~
MX
0 + C X 0 + K X 0 = P( t )
(5.19)
77
~ -1 ~
~ &
~
&& 0 = M
X
P( t ) - C X
0 - K X0
2)
(5.20)
The time increment value Dt and the parameters a and d are determined:
d = 0.50
a 0.25 (0.5 + d)
3)
(5.21)
The constants which will be used in the Newmark method are calculated:
1.0
a Dt 2
d
a1 =
a Dt
1.0
a2 =
a Dt
1.0
- 1.0
a3 =
2a
d
a 4 = - 1.0
a
Dt d
a5 =
- 2.0
2.0 a
a 6 = Dt (1.0 - d)
a0 =
(5.22)
a 7 = a Dt
4)
5)
(5.23)
All computations determined in the previous steps are repeated for every Dt
78
ys 2 i
a)
~
Pef
~
~
~
& t +a X
&& t + C
& t +a X
&& t
= P t + Dt + M a 0 X t + a 2 X
a1 Xt + a 4 X
3
5
b)
t + Dt
(5.24)
c)
(5.25)
t + Dt
The acceleration
&& t + Dt = a (X t + Dt - X ) - a X
& t -a X
&& t
X
0
t
2
3
(5.26)
(5.27)
The displacement
Y t + Dt = F X t + Dt
(5.28)
the velocity,
& t + Dt = F X
& t + Dt
Y
(5.29)
(5.30)
79
6. NUMERICAL RESULTS
6. NUMERICAL RESULTS
6.1. Introduction
In this thesis, the dynamic analysis of non-planar stiffened shear walls has
been studied in detail. The CCM and Vlasovs theory of thin-walled beams has been
employed to find the structure stiffness matrix. For this purpose, the connecting
beams have been replaced by an equivalent layered medium and unit forces have
been applied in the directions of the degrees of freedom to find the displacements of
the system corresponding to each of them. In the dynamic analysis of shear walls, the
lumped mass is concentrated at the center of the whole cross-sectional area of the
structure. Following the free vibration analysis, the uncoupled stiffness, damping and
mass matrices have been found employing the mode superposition method. A timehistory analysis has been carried out using Newmark numerical integration method to
find the system displacement vector for every time step. A computer program has
been prepared in Fortran language to analyze free vibration and forced vibration of
non-planar stiffened shear walls.
In the literature, there is no analytical work about the dynamic analysis of
non-planar coupled shear walls. In order to verify the present computer program,
examples were solved both by the present method (CCM) and by the SAP2000
structural analysis program.
To illustrate the application of the present theory, various examples have been
solved. In examples 1-14 non-planar coupled shear walls with constant cross-sections
were considered. Examples 15-20 were selected to study the effect of stiffened
beams on the dynamic analysis of shear walls. Examples 21-22 are chosen from
multi-region structures with different geometric properties in each region.
In the modeling of coupled shear walls as frames for the application of
SAP2000 computer program, the moments of inertia of the connecting and stiffening
beams were considered to be at the storey levels. When the heights of the stiffening
beams were equal to the storey height, the moment of inertia was considered to be at
the level of the upper storey.
80
6. NUMERICAL RESULTS
81
6. NUMERICAL RESULTS
H = 48 in
6 in
(-4.805, 5.195)
(4.805, 5.195)
5.195 in
0.39 in
0.39 in
(0, 1.687)
0.39 in
2
(-4.805, 0)
(0,0)
(2, 0)
(-2, 0)
2.805 in
0.39 in
2 in
2 in
2
(4.805, 0)
2.805 in
Figure 6.1. Geometrical properties of the model originally considered by Tso and
Biswas
82
6. NUMERICAL RESULTS
Figure 6.2. Frame model of the structure in Example 1 and its 3-D view
In this example, the model of the structure was chosen symmetrical with
respect to Y axis and the free vibration analysis was carried out. Table 6.1 compares
the natural frequencies corresponding to each mode found by the program prepared
in the present work and the SAP2000 structural analysis program, expressing the
percentage differences.
83
6. NUMERICAL RESULTS
Table 6.1. Comparison of the natural frequencies (Hz) obtained from the present
program and SAP2000 in Example 1
Mode
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Present Study
(CCM)
SAP2000
(Frame Method)
Natural Frequencies
Natural Frequencies
0,16427
0,16902
0,71226
1,01237
1,57316
2,59400
2,79376
3,81979
5,20122
5,39394
6,57889
7,62432
8,76270
12,7231
16,7636
19,8694
0,16636
0,16823
0,69767
1,02500
1,53440
2,52580
2,82740
3,70460
5,00040
5,45530
6,25420
7,18320
8,85340
12,83700
16,88600
19,99000
% difference
1,26
0,47
2,09
1,23
2,53
2,70
1,19
3,11
4,02
1,12
5,19
6,14
1,02
0,89
0,72
0,60
84
6. NUMERICAL RESULTS
48
42
Height (m)
36
Mode 1 in Y Direction
30
24
SAP 2000
NF=0,16636 Hz
18
CCM
NF=0,16427 Hz
12
6
0
0,0
0,5
1,0
48
42
Height(m)
36
Mode 2 in X Direction
30
SAP 2000
NF=0,16823 Hz
24
18
CCM
NF=0,16902 Hz
12
6
0
0,0
0,5
1,0
48
42
Height (m)
36
30
24
SAP 2000
NF=0,69767 Hz
18
CCM
NF=0,71226 Hz
12
6
0
-0,5
0,0
0,5
1,0
Figure 6.3. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by both the present program and SAP2000 in Example 1
85
6. NUMERICAL RESULTS
48
42
36
Mode 4 in Y Direction
Height (m)
30
SAP 2000
NF=1,02500 Hz
24
18
SBY
NF=1,01237 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
36
Mode 5 in X Direction
Height (m)
30
24
SAP 2000
NF=1,53440 Hz
18
CCM
DF=1,57316 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
36
Mode 6 in X Direction
Height (m)
30
SAP 2000
NF=2,52580 Hz
24
18
CCM
NF=2,59400 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
Figure 6.4. Comparison of fourth, fifth and sixth mode shapes in X and Y directions
found by both the present program and SAP2000 in Example 1
86
6. NUMERICAL RESULTS
48
42
36
Mode 7 in Y Direction
Height (m)
30
SAP 2000
NF=2,82740 Hz
24
18
CCM
DF=2,78376 Hz
12
6
0
-1,0
0,0
1,0
48
42
36
Mode 8 in X Direction
Height (m)
30
SAP 2000
NF=3,70460 Hz
24
18
CCM
NF=3,81979 Hz
12
6
0
-1,0
0,0
1,0
Figure 6.5. Comparison of seventh and eighth mode shapes in X and Y directions
found by both the present program and SAP2000 in Example 1
87
6. NUMERICAL RESULTS
Example 2:
In this example, the forced vibration analysis of a non-planar coupled shear
wall on a rigid foundation was considered. The geometric and material properties of
the eight storey shear wall were taken as in Example 1. The forced vibration analysis
of the shear wall was carried out by the computer program prepared in the present
work and the SAP2000 structural analysis program for damped and undamped cases.
The dynamic load, P(t), was applied at the top, in the global X direction in the
plane of the connection beam as in Fig. 6.6 and the triangular pulse force was chosen
as in Fig. 6.7.
(-4.805, 5.195)
(4.805, 5.195)
0.39 in
0.39 in
(0, 1.687)
P(t)
0.39 in
1
2
(-4.805, 0)
(0,0)
(-2, 0)
0.39 in
1
(2, 0)
2
(4.805, 0)
Figure 6.6. Cross-sectional view of the structure and applied dynamic load in
Example 2
P(lb)
100
t (s)
4
88
6. NUMERICAL RESULTS
At the top of the shear wall, the maximum displacement in the X direction of
point G, the mass center, was calculated by the computer program prepared in the
present study and compared with those of the SAP2000 structural analysis program
in Tables 6.2-3 for both damped and undamped cases.
Table 6.2. Maximum displacement (m) in the X direction of point G for undamped
case in Example 2
SAP2000
(Frame Method)
Present Study
(CCM)
% difference
0.040238
0.039819
1.04
The results of both methods for 5 % damping ratio are given in Table 6.3.
Table 6.3. Maximum displacement (m) in the X direction of point G for damped case
in Example 2
SAP2000
(Frame Method)
Present Study
(CCM)
% difference
0.038404
0.038003
1.05
The responses for both damped and undamped systems to triangular force
were determined and the time-varying displacements in the X direction of point G
are presented in Figs. 6.8-9.
It is observed that the results obtained in the present study (CCM) coincide
with those of the SAP2000 structural analysis program, perfectly.
89
6. NUMERICAL RESULTS
0,500
0,400
Present study
Top Displacement (m)
0,300
Sap2000
0,200
0,100
0,000
0
10 12 14 16 18 20 22 24
-0,100
-0,200
-0,300
Time (s)
Figure 6.8. Time-varying displacements in X direction at the top of the shear wall for
undamped case in Example 2
0,500
0,400
Present study
0,300
Sap2000
0,200
0,100
0,000
0
10 12 14
16 18 20 22 24
-0,100
-0,200
Time (s)
Figure 6.9. Time-varying displacements in X direction at the top of the shear wall for
damped case with 5 % damping ratio in Example 2
90
6. NUMERICAL RESULTS
Example 3:
In this example, a non-planar coupled shear wall system on a rigid foundation
was solved to show the agreement in the results of free vibration analysis. The total
height of the shear wall is 24 m, the storey height is 3 m, the thickness is 0.3 m, the
height of the connecting beams is 0.5 m and the elasticity and shear moduli of the
structure are E = 2.85106 kN/m2 and G = 1055556 kN/m2, respectively. Fig. 6.10.
(-3, 2)
2m
0.3 m
0.3 m
(-3, 0)
0.3 m
2
1
(0,0)
2m
0.3 m
1
(-1, 0)
2
1
(3, 0)
(1, 0)
3
0.3 m
(3, 2)
0.3 m
(-3, -2)
2m
1m
1m
(3, -2)
2m
91
6. NUMERICAL RESULTS
Figure 6.11. Frame model of the structure in Example 3 and its 3-D view
In this example, the model of the structure was chosen symmetrical with
respect to X and Y axes and the free vibration analysis was carried out. Table 6.4
compares the natural frequencies corresponding to each mode found by the program
prepared in the present work and the SAP2000 structural analysis program,
expressing the percentage differences.
92
6. NUMERICAL RESULTS
Table 6.4. Comparison of the natural frequencies (Hz) obtained from the present
program and SAP2000 in Example 3
Mode
Present Study
(CCM)
SAP2000
(Frame Method)
% difference
0,97409
1,36735
5,20893
6,00116
11,82911
16,55209
21,25545
31,93361
33,45567
47,76713
51,83542
62,34009
73,52751
75,21096
99,04972
117,37420
0,97582
1,35782
5,18059
6,00303
11,75796
16,51685
21,06219
31,74091
32,97118
46,71418
51,22450
60,43418
70,80442
73,73767
96,23732
113,25790
0,18
0,70
0,55
0,03
0,61
0,21
0,92
0,61
1,47
2,25
1,19
3,15
3,85
2,00
2,92
3,63
93
6. NUMERICAL RESULTS
48
42
Height (m)
36
Mode 1 in Y Direction
30
24
SAP 2000
NF=0,97582 Hz
18
CCM
NF=0,97409 Hz
12
6
0
0,0
0,5
1,0
48
42
Height(m)
36
Mode 2 in X Direction
30
SAP 2000
NF=1,35782 Hz
24
18
CCM
NF=1,36735 Hz
12
6
0
0,0
0,5
1,0
48
42
36
Mode 3 in X Direction
Height (m)
30
SAP 2000
NF=5,18059 Hz
24
18
CCM
NF=5,20893 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
Figure 6.12. Comparison of first, second and third mode shapes in X and Y directions
found by both the present program and SAP2000 in Example 3
94
6. NUMERICAL RESULTS
48
42
36
Mode 4 in Y Direction
Height (m)
30
24
SAP 2000
NF=6,00303 Hz
18
CCM
NF=6,00116 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
Height (m)
36
Mode 5 in X Direction
30
SAP 2000
NF=11,75796 Hz
24
18
SBY
NF=11,82911 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
Height (m)
36
Mode 6 in Y Direction
30
SAP 2000
NF=16,51685 Hz
24
18
CCM
DF=16,55209 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
Figure 6.13. Comparison of fourth, fifth and sixth mode shapes in X and Y directions
found by both the present program and SAP2000 in Example 3
95
6. NUMERICAL RESULTS
48
42
Height (m)
36
Mode 7 in X Direction
30
SAP 2000
NF=21,06216 Hz
24
18
CCM
NF=21,25545 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
Height (m)
36
Mode 8 in Y Direction
30
SAP 2000
NF=31,74091 Hz
24
18
CCM
NF=31,93361 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
Figure 6.14. Comparison of seventh and eighth mode shapes in X and Y directions
found by both the present program and SAP2000 in Example 3
96
6. NUMERICAL RESULTS
Example 4:
In this example, the forced vibration analysis of a non-planar symmetrical
coupled shear wall on a rigid foundation was considered. The geometric and material
properties of the eight storey shear wall were taken as in Example 3. The forced
vibration analysis of the shear wall was carried out by the computer program
prepared in the present work and the SAP2000 structural analysis program for
damped and undamped cases.
The dynamic load, P(t), was applied at the top, in the global X direction in the
plane of the connecting beam as in Fig. 6.15 and the rectangular pulse force was
chosen as in Fig. 6.16.
(-3, 2)
2m
0.3 m
0.3 m
P(t)
(-3, 0)
0.3 m
2
1
(0,0)
2m
0.3 m
2
1
(-1, 0)
(3, 0)
(1, 0)
3
0.3 m
(3, 2)
0.3 m
(-3, -2)
2m
1m
1m
(3, -2)
2m
Figure 6.15. Cross-sectional view of the structure and applied dynamic load in
Example 4
P(lb)
150
t(s)
5
Figure 6.16. Rectangular pulse force in Example 4
97
6. NUMERICAL RESULTS
At the top of the shear wall, the maximum displacement in the X direction of
point O, the mass center, was calculated by the computer program prepared in the
present study and compared with those of the SAP2000 structural analysis program
in Tables 6.5-6 for both damped and undamped cases.
Table 6.5. Maximum displacement (m) in the X direction of point O for undamped
case in Example 4
SAP2000
(Frame Method)
Present Study
(CCM)
% difference
0.066510
0.065713
1.21
The results of both methods for 5 % damping ratio are given in Table 6.6.
Table 6.6. Maximum displacement (m) in the X direction of point O for damped case
in Example 4
SAP2000
(Frame Method)
Present Study
(CCM)
% difference
0.058340
0.057648
1.20
98
6. NUMERICAL RESULTS
0,090
0,070
Present study
SAP2000
0,050
0,030
0,010
-0,010 0
10
-0,030
-0,050
Time (s)
Figure 6.17. Time-varying displacements in X direction at the top of the shear wall
for undamped case in Example 4
0,080
0,060
Present study
0,040
SAP2000
0,020
0,000
0
10
-0,020
-0,040
Time (s)
Figure 6.18. Time-varying displacements in X direction at the top of the shear wall
for damped case with 5 % damping ratio in Example 4
99
6. NUMERICAL RESULTS
Example 5:
In this example, a non-planar non-symmetrical coupled shear wall with the
cross-section shown in Fig. 6.19 was considered and the free vibration analysis was
carried out. The total height of the shear wall is 24 m, the storey height is 3 m, the
thickness is 0.3 m, the height of the connecting beams is 0.5 m and the elasticity and
shear moduli are E=2.85106 kN/m2 and G=1055556 kN/m2, respectively.
3 (4, 3)
3m
0.3 m
G
(-5, 0)
(-3, 0)
2
1
(0, 0)
1
(-1, 0)
0.3 m
1
(1, 0)
2
(4, 0)
2m
(0,125, 0,208)
0.3 m
3
0.3 m
4
(-3, -2)
2m
2m
1m
1m
3m
100
6. NUMERICAL RESULTS
The structure was solved both by the present method using the CCM and by
the SAP2000 structural analysis program using the frame method for which the
model and its 3-D view are given in Fig. 6.20.
Figure 6.20. Frame model of the structure in Example 5 and its 3-D view
Table 6.7 compares the natural frequencies corresponding to each mode
found by the program prepared in the present work and the SAP2000 structural
analysis program, expressing the percentage differences.
101
6. NUMERICAL RESULTS
Table 6.7. Comparison of the natural frequencies (Hz) obtained from the present
program and SAP2000 in Example 5
Mode
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Present Study
SAP2000
(CCM)
(Frame Method)
Natural Frequencies Natural Frequencies
0,69367
0,69248
1,65427
1,64679
4,14919
4,14449
6,88659
6,86626
11,36339
11,33655
16,69150
16,62251
21,86704
21,76464
30,88525
30,65600
35,45502
35,16467
49,27674
48,65858
51,41270
50,74482
67,68475
66,43667
70,87069
69,48448
80,19234
78,38068
92,87637
90,32802
109,78064
106,11409
% difference
0,17
0,45
0,11
0,30
0,24
0,42
0,47
0,75
0,83
1,27
1,32
1,88
2,00
2,31
2,82
3,46
102
6. NUMERICAL RESULTS
48
42
Height (m)
36
Mode 1 in Y Direction
30
24
SAP 2000
NF=0,69248 Hz
18
CCM
NF=0,69367 Hz
12
6
0
0,0
0,5
1,0
48
42
Height (m)
36
Mode 2 in X Direction
30
SAP 2000
NF=1,64679 Hz
24
18
CCM
NF=1,65427 Hz
12
6
0
0,0
0,5
1,0
48
42
Height (m)
36
Mode 3 in Y Direction
30
SAP 2000
NF=4,14449 Hz
24
18
CCM
NF=4,14919 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
Figure 6.21. Comparison of first, second and third mode shapes in X and Y directions
found by both the present program and SAP2000 in Example 5
103
6. NUMERICAL RESULTS
48
42
Height (m)
36
30
24
SAP 2000
NF=6,86626 Hz
18
SBY
NF=6,88659 Hz
12
6
0
-0,5
0,0
0,5
1,0
48
42
Height (m)
36
Mode 5 in Y Direction
30
24
SAP 2000
NF=11,33655 Hz
18
CCM
DF=11,36339 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
36
Mode 6 in X Direction
Height (m)
30
SAP 2000
NF=16,62251 Hz
24
18
CCM
NF=16,69150 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
Figure 6.22. Comparison of fourth, fifth and sixth mode shapes in X, Y and Teta
directions found by both the present program and SAP2000 in Example 5
104
6. NUMERICAL RESULTS
48
42
Height (m)
36
Mode 7 in Y Direction
30
SAP 2000
NF=21,76464 Hz
24
18
CCM
NF=21,86704 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
Height (m)
36
30
SAP 2000
NF=21,76464 Hz
24
18
CCM
DF=21,86704 Hz
12
6
-2,5
-1,5
0
-0,5
0,5
1,5
2,5
48
42
36
Mode 8 in X Direction
Height (m)
30
SAP 2000
NF=30,65600 Hz
24
18
CCM
NF=30,88525 Hz
12
6
0
-1,0
0,0
1,0
Figure 6.23. Comparison of seventh and eighth mode shapes in X, Y and Teta
directions found by both the present program and SAP2000 in Example 5
105
6. NUMERICAL RESULTS
Example 6:
In this example, the forced vibration analysis of a non-planar nonsymmetrical coupled shear wall on a rigid foundation was considered. The geometric
and material properties of the eight storey shear wall were taken as in Example 5.
The forced vibration analysis of the shear wall was carried out by the computer
program prepared in the present work and the SAP2000 structural analysis program
for damped and undamped cases.
The dynamic load, P(t), was applied at the top of the structure in the global Y
direction in the plane of the shear wall as in Fig. 6.24 and the rectangular pulse force
was chosen as in Fig. 6.25.
3 (4, 3)
3m
0.3 m
G
(-3, 0)
2
2
(0,125, 0,208)
0.3 m
(0, 0)
0.3 m
1
(-1, 0)
(4, 0)
(1, 0)
P(t)
2m
(-5, 0)
0.3 m
4
(-3, -2)
2m
2m
1m
1m
3m
Figure 6.24. Cross-sectional view of the structure and applied dynamic load in
Example 6
106
6. NUMERICAL RESULTS
P(lb)
100
t(s)
5
Figure 6.25. Rectangular pulse force in Example 6
At the top of the shear wall, the maximum displacement in the Y direction of
point G, the mass center, was calculated by the computer program prepared in the
present study and compared with those of the SAP2000 structure analysis program in
Tables 6.8-9 for both damped and undamped cases.
Table 6.8. Maximum displacement (m) in the Y direction of point G for undamped
case in Example 6
SAP2000
(Frame Method)
Present Study
(CCM)
% difference
0.121225
0.122312
0.90
The results of both methods for 5 % damping ratio are given in Table 6.9.
Table 6.9. Maximum displacement (m) in the Y direction of point G for damped case
in Example 6
SAP2000
(Frame Method)
Present Study
(CCM)
% difference
0.110000
0.110792
0.72
The responses for both damped and undamped systems to rectangular force
were determined and the time-varying displacements in the Y direction of point G
are presented in Figs. 6.26-27.
It is observed that the results obtained in the present work (CCM) coincide
with those of the SAP2000 structural analysis program, perfectly.
107
6. NUMERICAL RESULTS
0,150
0,100
Present study
0,050
SAP2000
0,000
0
10
-0,050
-0,100
-0,150
Time (s)
Figure 6.26. Time-varying displacements in Y direction at the top of the shear wall
for undamped case in Example 6
0,150
0,100
Top Displacement (m)
Present study
SAP2000
0,050
0,000
0
10
-0,050
-0,100
Time (s)
Figure 6.27. Time-varying displacements in Y direction at the top of the shear wall
for damped case with 5 % damping ratio in Example 6
108
6. NUMERICAL RESULTS
When the dynamic load applied at the top of the shear wall was chosen as the
sine pulse force in Fig. 6.28 for the same example, the maximum displacement in the
Y direction of point G was calculated and compared with those of the SAP2000
structural analysis program in Tables 6.10-11 for both damped and undamped cases.
P(lb)
150
5
t (s)
Present Study
(CCM)
% difference
0.123144
0.124157
0.82
The results of both methods for 5 % damping ratio are given in Table 6.11.
Table 6.11. Maximum displacement (m) in the Y direction of point G for damped
case in Example 6
SAP2000
(Frame Method)
Present Study
(CCM)
% difference
0.117102
0.118075
0.83
The responses for both damped and undamped systems to sine force were
determined and the time-varying displacements in the Y direction of point G are
presented in Figs. 6.29-30.
109
6. NUMERICAL RESULTS
0,150
Present study
0,100
SAP2000
0,050
0,000
0
10
-0,050
-0,100
-0,150
Time (s)
Figure 6.29. Time-varying displacements in Y direction at the top of the shear wall
for undamped case in Example 6
0,150
0,100
Top Displacemnt (m)
Present study
SAP2000
0,050
0,000
0
10
-0,050
-0,100
-0,150
Time (s)
Figure 6.30. Time-varying displacements in Y direction at the top of the shear wall
for damped case with 5 % damping ratio in Example 6
110
6. NUMERICAL RESULTS
Example 7:
In this example, a non-planar non-symmetrical coupled shear wall with the
cross-section shown in Fig. 6.31 was considered and the free vibration analysis was
carried out. The total height of the shear wall is 48 m, the storey height is 3 m, the
thickness is 0.4 m, the height of the connecting beams is 0.5 m and the elasticity and
shear moduli are E = 2.85106 kN/m2 and G = 1055556 kN/m2, respectively.
2.0 m
Y
0.4 m
4
3 (5.5, 5)
(-3.5, 5)
5.0 m
0.4 m
0.4 m
5.0 m
(-5.5, 5)
(-0.45, 1.75)
0.4 m
2
(-5.5, 0)
4.0 m
1
(-1.5, 0)
1.5 m
(0, 0)
1.5 m
0.4 m
1
(1.5, 0)
2
(5.5, 0)
4.0 m
111
6. NUMERICAL RESULTS
Figure 6.32. Frame model of the structure in Example 7 and its 3-D view
Table 6.12 compares the first twenty natural frequencies corresponding to
each mode found by the program prepared in the present work and the SAP2000
structural analysis program, expressing the percentage differences.
112
6. NUMERICAL RESULTS
Table 6.12. Comparison of the natural frequencies (Hz) obtained from the present
program and SAP2000 in Example 7
Mode
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Present Study
(CCM)
SAP2000
(Frame Method)
Natural Frequencies
Natural Frequencies
0,4109880
0,4931529
1,8800960
2,9823973
4,715429
8,2914745
8,8708069
14,381514
16,169147
21,22184
26,609686
29,368472
38,776526
39,572676
49,372204
55,006839
61,01724
72,831525
73,465525
86,298492
0,409867
0,492887
1,877845
2,975817
4,709624
8,248660
8,851927
14,331390
16,015483
21,110277
26,203663
29,148453
38,379274
38,681877
48,702600
53,280684
59,950797
69,774177
71,852793
83,978085
% difference
0,27
0,05
0,12
0,22
0,12
0,52
0,21
0,35
0,96
0,53
1,55
0,75
1,04
2,30
1,37
3,24
1,78
4,38
2,24
2,76
113
6. NUMERICAL RESULTS
48
42
Height (m)
36
Mode 1 in X Direction
30
24
SAP 2000
NF=0,69248 Hz
18
CCM
NF=0,69367 Hz
12
6
0
0,0
0,5
1,0
48
42
Height (m)
36
Mode 2 in Y Direction
30
SAP 2000
NF=1,64679 Hz
24
18
CCM
NF=1,65426 Hz
12
6
0
0,0
0,5
1,0
48
42
36
Mode 3 in Y Direction
Height (m)
30
SAP 2000
NF=4,14449 Hz
24
18
CCM
NF=4,14919 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
Figure 6.33. Comparison of first, second and third mode shapes in X and Y directions
found by both the present program and SAP2000 in Example 7
114
6. NUMERICAL RESULTS
48
42
36
Height (m)
30
24
SAP 2000
NF=6,86626 Hz
18
SBY
NF=6,88659 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
Height (m)
36
Mode 5 in Y Direction
30
24
SAP 2000
NF=11,33655 Hz
18
CCM
DF=11,36338 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
Height (m)
36
Mode 6 in X Direction
30
SAP 2000
NF=16,62251 Hz
24
18
CCM
NF=16,69150 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
Figure 6.34. Comparison of fourth, fifth and sixth mode shapes in X, Y and Teta
directions found by both the present program and SAP2000 in Example 7
115
6. NUMERICAL RESULTS
48
42
36
Mode 7 in Y Direction
Height (m)
30
SAP 2000
NF=21,76464 Hz
24
18
CCM
NF=21,86704 Hz
12
6
0
-1,5
-0,5
0,5
1,5
48
42
36
Height (m)
30
SAP 2000
NF=21,76464 Hz
24
18
CCM
DF=21,86704 Hz
12
6
0
-1,5
-0,5
1,5
0,5
48
42
36
Mode 8 in X Direction
Height (m)
30
SAP 2000
NF=30,65600 Hz
24
18
CCM
NF=30,88525 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
Figure 6.35. Comparison of seventh and eighth mode shapes in X, Y and Teta
directions found by both the present program and SAP2000 in Example 7
116
6. NUMERICAL RESULTS
Example 8:
In this example, the forced vibration analysis of a non-planar nonsymmetrical coupled shear wall on a rigid foundation was considered. The geometric
and material properties of the sixteen storey shear wall were taken as in Example 7.
The forced vibration analysis of the shear wall was carried out by the computer
program prepared in the present work and the SAP2000 structural analysis program
for damped and undamped cases.
The dynamic load, P(t), was applied at the top of the structure in the global Y
direction in the plane of the shear wall as in Fig. 6.36 and the triangular pulse force
was chosen as in Fig. 6.37.
2.0 m
Y
0.4 m
4
(-5.5, 5)
3 (5.5, 5)
(-3.5, 5)
5.0 m
0.4 m
0.4 m
(-0.45, 1.75)
P(t)
(-5.5, 0)
G
O
0.4 m
(0, 0)
1
(-1.5, 0)
4.0 m
1.5 m
5.0 m
0.4 m
1
(1.5, 0)
1.5 m
2
(5.5, 0)
4.0 m
Figure 6.36. Cross-sectional view of the structure and applied dynamic load in
Example 8
117
6. NUMERICAL RESULTS
P(lb)
150
t (s)
2,5
5
Figure 6.37. Triangular pulse force in Example 8
At the top of the shear wall, the maximum displacement in the X direction of
point G, the mass center, was calculated by the computer program prepared in the
present study and compared with those of the SAP2000 structural analysis program
in Tables 6.13-14 for both damped and undamped cases.
Table 6.13. Maximum displacement (m) in the X direction of point G for undamped
case in Example 8
SAP2000
(Frame Method)
Present Study
(CCM)
% difference
0.036393
0.036112
0.77
The results of both methods for 6 % damping ratio are given in Table 6.14.
Table 6.14. Maximum displacement (m) in the X direction of point G for damped
case in Example 8
SAP2000
(Frame Method)
Present Study
(CCM)
% difference
0.036130
0.035939
0.53
The responses for both damped and undamped systems to triangular force
were determined and the time-varying displacements in the X direction of point G
are presented in Figs. 6.38-39.
It is observed that the results obtained in the present work (CCM) coincide
with those of the SAP2000 structural analysis program, perfectly.
118
6. NUMERICAL RESULTS
0,040
0,030
0,020
Present study
SAP2000
0,010
0,000
0
10
12
14
-0,010
Time (s)
Figure 6.38. Time-varying displacements in X direction at the top of the shear wall
for undamped case in Example 8
0,040
0,030
0,020
Present study
SAP2000
0,010
0,000
0
10
12
14
-0,010
Time (s)
Figure 6.39. Time-varying displacements in X direction at the top of the shear wall
for damped case with 6 % damping ratio in Example 8
119
6. NUMERICAL RESULTS
Example 9:
In this example, a non-planar non-symmetrical coupled shear wall with the
cross-section shown in Fig. 6.40 was considered and the free vibration analysis was
carried out. The total height of the shear wall is 50 m, the storey height is 2.5 m and
the thicknesses of the piers and the connecting beams are shown in Fig. 6.40. The
height of the connecting beams is 0.35 m and the elasticity and shear moduli are
E = 2.85106 kN/m2 and G = 1055556 kN/m2, respectively.
(4, 2)
0.3 m
4
(7, 2)
2m
3
2
0.2 m
1
0.3 m
O
1
3m
1
(1, 0)
(-1, 0)
(4, 0)
0.3 m
(1,111, -1.045)
0.2 m
1.5 m
(-3, -3)
0.2 m
3m
(-8, -4.5)
(0, 0)
4.5 m
(-3, 0)
2
(4, -4.5)
(-5, -4.5)
2m
2m
1m 1m
3m
3m
120
6. NUMERICAL RESULTS
Figure 6.41. Frame model of the structure in Example 9 and its 3-D view
Table 6.15 compares the first twenty natural frequencies corresponding to
each mode found by the program prepared in the present work and the SAP2000
structural analysis program, expressing the percentage differences.
121
6. NUMERICAL RESULTS
Table 6.15. Comparison of the natural frequencies (Hz) obtained from the present
program and SAP2000 in Example 9
Mode
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Present Study
(CCM)
SAP2000
(Frame Method)
Natural Frequencies
Natural Frequencies
0,2728988
0,5955637
1,4747386
3,2946904
3,9478691
7,6107878
8,9650465
12,483401
17,374156
18,547388
25,791174
28,549694
34,197754
42,45922
43,743252
54,389935
59,075664
66,078435
78,358690
78,710916
0,272051
0,594873
1,473539
3,288973
3,945944
7,604040
8,931684
12,463691
17,256600
18,500908
25,696652
28,241979
34,024317
41,788411
43,447923
53,915274
57,783123
65,349682
76,080273
77,634855
% difference
0,31
0,12
0,08
0,17
0,05
0,09
0,37
0,16
0,68
0,25
0,37
1,09
0,51
1,61
0,68
0,88
2,24
1,12
2,99
1,39
122
6. NUMERICAL RESULTS
50
45
40
Height (m)
35
Mode 1 in X Direction
30
SAP 2000
NF=0,27205 Hz
25
20
CCM
NF=0,27289 Hz
15
10
5
0
0,0
0,5
1,0
50
45
40
35
Mode 2 in Y Direction
Height(m)
30
25
SAP 2000
NF=0,59487 Hz
20
CCM
NF=0,59556 Hz
15
10
5
0
0,0
0,5
1,0
50
45
40
35
Height (m)
30
SAP 2000
NF=1,47354 Hz
25
20
CCM
NF=1,47474 Hz
15
10
5
0
-1,0
-0,5
0,0
0,5
1,0
Figure 6.42. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by both the present program and SAP2000 in Example 9
123
6. NUMERICAL RESULTS
50
45
40
35
Mode 4 in Y Direction
Height (m)
30
SAP 2000
NF=3,28897 Hz
25
20
SBY
NF=3,29469 Hz
15
10
5
0
-1,0
-0,5
0,0
0,5
1,0
50
45
40
35
Mode 5 in Y Direction
Height (m)
30
SAP 2000
NF=3,94594 Hz
25
20
CCM
DF=3,94787 Hz
15
10
5
0
-1,0
-0,5
0,0
0,5
1,0
50
45
40
Height (m)
35
Mode 6 in X Direction
30
SAP 2000
NF=7,60404 Hz
25
20
CCM
NF=7,61078 Hz
15
10
5
0
-1,0
1,0
0,0
Figure 6.43. Comparison of fourth, fifth and sixth mode shapes in X and Y directions
found by both the present program and SAP2000 in Example 9
124
6. NUMERICAL RESULTS
50
45
40
Height (m)
35
Mode 7 in Y Direction
30
SAP 2000
NF=8,93168 Hz
25
20
CCM
DF=8,96504 Hz
15
10
5
0
-1,0
0,0
1,0
50
45
40
Height (m)
35
Mode 8 in X Direction
30
SAP 2000
NF=12,46369 Hz
25
20
CCM
NF=12,48340 Hz
15
10
5
0
-1,0
-0,5
0,0
0,5
1,0
Figure 6.44. Comparison of seventh and eighth mode shapes in X and Y directions
found by both the present program and SAP2000 in Example 9
125
6. NUMERICAL RESULTS
Example 10:
In this example, the forced vibration analysis of a non-planar nonsymmetrical coupled shear wall on a rigid foundation was considered. The geometric
and material properties of the twenty storey shear wall were taken as in Example 9.
The forced vibration analysis of the shear wall was carried out by the computer
program prepared in the present work and the SAP2000 structural analysis program
for damped and undamped cases.
The dynamic load, P(t), was applied at the top of the structure in the global X
direction in the plane of the connection beam as in Figure 6.45 and the triangular
pulse force was chosen as in Fig. 6.46.
Y
(4, 2)
0.3 m
4
(7, 2)
2m
P(t)
0.2 m
1
0.3 m
O
1
3m
1
(1, 0)
(-1, 0)
(4, 0)
0.3 m
(1,111, -1.045)
0.2 m
1.5 m
(-3, -3)
5
(-8, -4.5)
0.2 m
4
4
(4, -4.5)
(-5, -4.5)
3m
(0, 0)
4.5 m
(-3, 0)
2
2m
2m
1m 1m
3m
3m
Figure 6.45. Cross-sectional view of the structure and applied dynamic load in
Example 10
P(lb)
150
t (s)
2,5
5
Figure 6.46. Triangular pulse force in Example 10
126
6. NUMERICAL RESULTS
At the top of the shear wall, the maximum displacement in the X direction of
point G, the mass center, was calculated by the computer program prepared in the
present study and compared with those of the SAP2000 structural analysis program
in Tables 6.16-17 for both damped and undamped cases.
Table 6.16. Maximum displacement (m) in the X direction of point G for undamped
case in Example 10
SAP2000
(Frame Method)
Present Study
(CCM)
% difference
0.183850
0.182404
0.79
The results of both methods for 7 % damping ratio are given in Table 6.17.
Table 6.17. Maximum displacement (m) in the X direction of point G for damped
case in Example 10
SAP2000
(Frame Method)
Present Study
(CCM)
% difference
0.172310
0.171008
0.76
The responses for both damped and undamped systems to triangular force
were determined and the time-varying displacements in the X direction of point G
are presented in Figures 6.47-48.
It is observed that the results obtained in the present work (CCM) coincide
with those of the SAP2000 structural analysis program, perfectly.
127
6. NUMERICAL RESULTS
0,200
0,150
Present study
SAP2000
0,100
0,050
0,000
0
10
12
14
-0,050
-0,100
Time (s)
Figure 6.47. Time-varying displacements in X direction at the top of the shear wall
for undamped case in Example 10
0,200
0,150
Present study
SAP2000
0,100
0,050
0,000
0
10
12
14
-0,050
-0,100
Time (s)
Figure 6.48. Time-varying displacements in X direction at the top of the shear wall
for damped case with 7 % damping ratio in Example 10
128
6. NUMERICAL RESULTS
Example 11:
In this example, a non-planar non-symmetrical coupled shear wall with the
cross-section shown in Fig. 6.49 was considered and the free vibration analysis was
carried out. The total height of the shear wall is 48 m, the storey height is 3 m and the
thicknesses of the piers and the connecting beams are shown in Fig. 6.49. The height
of the connecting beams is 0.4 m and the elasticity and shear moduli are
E = 2.85106 kN/m2 and G = 1055556 kN/m2, respectively.
2m
2m
2m
2m
3m
Y
0.2 m
(-7, 4)
10
11
10
0.4 m
3
(-5, 4)
(-1, 4)
(1, 4)
0.4 m
G
0.2 m
8
(-7, 0)
3m
(4, 4)
0.4 m
0.2 m
4m
0.4 m
4
(-3, 4)
9
(-5, 0)
(-3, 0)
(-0.795, 0.322)
0.4 m
1
0.4 m
O
1
(-1, 0)
(1, 0)
(-7, -3)
0.2 m
0.4 m
0.4 m
0.4 m
6
(-3, -3)
4m
(4, 0)
0.4 m
0.2 m
7
(-1, -3)
2m
6
(4, -3)
(1, -3)
2m
3m
129
6. NUMERICAL RESULTS
The structure was solved both by the present method using the CCM and by
the SAP2000 structural analysis program using the frame method for which the
model and its 3-D view are given in Fig. 6.50.
Figure 6.50. Frame model of the structure in Example 11 and its 3-D view
Table 6.18 compares the first twenty natural frequencies corresponding to
each mode found by the program prepared in the present work and the SAP2000
structural analysis program, expressing the percentage differences.
130
6. NUMERICAL RESULTS
Table 6.18. Comparison of the natural frequencies (Hz) obtained from the present
program and SAP2000 in Example 11
Mode
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Present Study
(CCM)
SAP2000
(Frame Method)
Natural Frequencies
Natural Frequencies
0,5013758
0,6613498
2,4709828
4,126175
6,5157725
11,50762
12,519564
20,482792
22,458122
30,369575
36,969828
42,140664
54,984537
55,732593
71,036909
76,429148
87,855184
101,190190
105,832480
120,366150
0,500474
0,661167
2,463962
4,118391
6,475072
11,455949
12,375324
20,105120
22,271148
29,549525
36,474049
40,571794
52,995514
53,893310
66,594367
74,306353
81,065347
95,995947
97,418722
110,846184
% difference
0,18
0,03
0,28
0,19
0,63
0,45
1,17
1,88
0,84
2,78
1,36
3,87
3,75
3,41
6,67
2,86
8,38
5,41
8,64
8,58
131
6. NUMERICAL RESULTS
48
42
Height (m)
36
Mode 1 in X Direction
30
24
SAP 2000
NF=0,50047 Hz
18
CCM
NF=0,50137 Hz
12
6
0
0,0
0,5
1,0
48
42
Height(m)
36
Mode 2 in Y Direction
30
SAP 2000
NF=0,66117 Hz
24
18
CCM
NF=0,66135 Hz
12
6
0
0,0
0,5
1,0
48
42
36
Height (m)
30
24
SAP 2000
NF=2,46396 Hz
18
CCM
NF=2,47098 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
Figure 6.51. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by both the present program and SAP2000 in Example 11
132
6. NUMERICAL RESULTS
48
42
36
Mode 4 in Y Direction
Height
(m)
30
SAP 2000
NF=4,11839 Hz
24
18
SBY
NF=4,12617 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
36
Mode 5 in X Direction
Height
(m)
30
SAP 2000
NF=6,47507 Hz
24
18
CCM
DF=6,51577 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
Height (m)
36
30
SAP 2000
NF=11,45595 Hz
24
CCM
NF=11,50762 Hz
18
12
6
0
-1,5
-0,5
0,5
1,5
Figure 6.52. Comparison of fourth, fifth and sixth mode shapes in X, Y and Teta
directions found by both the present program and SAP2000 in Example 11
133
6. NUMERICAL RESULTS
48
42
36
Mode 7 in Y Direction
Height (m)
30
SAP 2000
NF=12,37512 Hz
24
18
CCM
DF=12,51956 Hz
12
6
0
-1,5
-0,5
1,5
0,5
48
42
36
Mode 8 in X Direction
Height (m)
30
SAP 2000
NF=20,10512 Hz
24
18
CCM
NF=20,48279 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
Figure 6.53. Comparison of seventh and eighth mode shapes in X and Y directions
found by both the present program and SAP2000 in Example 11
134
6. NUMERICAL RESULTS
Example 12:
In this example, the forced vibration analysis of a non-planar nonsymmetrical coupled shear wall on a rigid foundation was considered. The geometric
and material properties of the sixteen storey shear wall were taken as in Example 11.
The forced vibration analysis of the shear wall was carried out by the computer
program prepared in the present work and the SAP2000 structural analysis program
for damped and undamped cases.
The dynamic load, P(t), was applied at the top of the structure in the global X
direction in the plane of the connection beam as in Fig. 6.54 and the rectangular
pulse force was chosen as in Fig. 6.55.
2m
2m
2m
2m
3m
Y
0.2 m
(-7, 4)
10
11
10
0.4 m
3
(-5, 4)
4m
(4, 4)
(-1, 4)
(1, 4)
0.4 m
0.2 m
0.4 m
0.2 m
8
(-7, 0)
3m
0.4 m
4
(-3, 4)
P(t)
0.4 m
2
(-3, 0)
(-5, 0)
(-0.795, 0.322)
0.4 m
O
1
(-1, 0)
(1, 0)
(-7, -3)
0.2 m
0.4 m
0.4 m
0.4 m
6
(-3, -3)
4m
(4, 0)
0.4 m
0.2 m
7
(-1, -3)
2m
6
(4, -3)
(1, -3)
2m
3m
Figure 6.54. Cross-sectional view of the structure and applied dynamic load in
Example 12
135
6. NUMERICAL RESULTS
P(lb)
100
t(s)
5
Figure 6.55. Rectangular pulse force in Example 12
At the top of the shear wall, the maximum displacement in the X direction of
point G, the mass center, was calculated by the computer program prepared in the
present study and compared with those of the SAP2000 structural analysis program
in Tables 6.19-20 for both damped and undamped cases.
Table 6.19. Maximum displacement (m) in the X direction of point G for undamped
case in Example 12
SAP2000
(Frame Method)
Present Study
(CCM)
% difference
0.045550
0.045408
0.31
The results of both methods for 7 % damping ratio are given in Table 6.20.
Table 6.20. Maximum displacement (m) in the X direction of point G for damped
case in Example 12
SAP2000
(Frame Method)
Present Study
(CCM)
% difference
0.040480
0.040359
0.30
The responses for both damped and undamped systems to rectangular force
were determined and the time-varying displacements in the X direction of point G
are presented in Figures 6.56-57.
It is observed that the results obtained in the present work (CCM) coincide
with those of the SAP2000 structural analysis program, perfectly.
136
6. NUMERICAL RESULTS
0,060
0,040
Present study
0,020
SAP2000
0,000
0
10
-0,020
-0,040
-0,060
Time (s)
Figure 6.56. Time-varying displacements in X direction at the top of the shear wall
for undamped case in Example 12
0,060
0,040
Top Displacement (m)
Present study
SAP2000
0,020
0,000
0
10
-0,020
-0,040
Time (s)
Figure 6.57. Time-varying displacements in X direction at the top of the shear wall
for damped case with 7 % damping ratio in Example 12
137
6. NUMERICAL RESULTS
Example 13:
In this example, a non-planar non-symmetrical coupled shear wall with the
cross-section shown in Fig. 6.58 was considered and the free vibration analysis was
carried out. The total height of the shear wall is 48 m, the storey height is 3 m and the
thicknesses of the piers and the connecting beams are shown in Fig. 6.59. The height
of the connecting beams is 0.4 m and the elasticity and shear moduli are
E = 2.85106 kN/m2 and G = 1055556 kN/m2, respectively.
h=3m
H = 48 m
connecting
beams
Y
X
138
2m
(-5, 4)
(-3, 4)
0.3 m
7
2m
1.5 m
(-1, 4)
4m
0.3 m
0.3 m
0.3 m
139
(-7, 0)
(-1, 0)
1
(-3, 0)
(4, 2)
0.3 m
O
(0, 0)
0.3 m
(1, 0)
3m
(8, 2)
9
(4, 0)
(-0.016, 0.018)
10
0.3 m
0.3 m
(6, 2)
0.3 m
2m
(2.5, 4)
4
2m
0.3 m
(1, 4)
1.5 m
0.3 m
5m
0.3 m
(-7, 4)
4
2m
6. NUMERICAL RESULTS
2m
2m
6. NUMERICAL RESULTS
5
0.3 m
(-7, -3)
(-3, -3)
10
0.3 m
(-5, -3)
2m
11
(1, -3)
2m
4m
(4, -3)
(2.5, -3)
1.5 m
0.3 m
8
(8, -3)
1.5 m
4m
10
6. NUMERICAL RESULTS
The structure was solved both by the present method using the CCM and by
the SAP2000 structural analysis program using the frame method for which the
model and its 3-D view are given in Fig. 6.60.
Figure 6.60. Frame model of the structure in Example 13 and its 3-D view
Table 6.21 compares the first twenty natural frequencies corresponding to
each mode found by the program prepared in the present work and the SAP2000
structural analysis program, expressing the percentage differences.
140
6. NUMERICAL RESULTS
Table 6.21. Comparison of the natural frequencies (Hz) obtained from the present
program and SAP2000 in Example 13
Mode
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Present Study
(CCM)
SAP2000
(Frame Method)
Natural Frequencies
Natural Frequencies
0,5693957
0,7100410
3,2076784
4,1377091
8,7021153
11,453579
16,857499
22,314747
27,669057
36,710061
41,091277
54,580514
57,06977
75,518967
75,853501
96,291975
100,415440
119,119440
128,070530
143,520020
0,569495
0,709684
3,20475
4,123140
8,671171
11,362686
16,732066
21,995505
27,322003
35,882920
40,312723
52,798368
55,544476
72,420642
72,846471
91,756689
94,567838
112,044569
118,607331
133,046100
% difference
0,02
0,05
0,09
0,35
0,36
0,80
0,75
1,45
1,27
2,31
1,93
3,38
2,75
4,28
4,13
4,94
6,18
6,31
7,98
7,87
141
6. NUMERICAL RESULTS
48
42
36
Mode 1 in X Direction
Height (m)
30
24
SAP 2000
NF=0,56949 Hz
18
CCM
NF=0,56939 Hz
12
6
0
0,0
0,5
1,0
48
42
36
Mode 2 in Y Direction
Height (m)
30
SAP 2000
NF=0,70968 Hz
24
18
CCM
NF=0,71004 Hz
12
6
0
0,0
1,0
0,5
48
42
36
Height (m)
30
24
SAP 2000
NF=3,20475 Hz
18
CCM
NF=3,20768 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
Figure 6.61. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by both the present program and SAP2000 in Example 13
142
6. NUMERICAL RESULTS
48
42
Height (m)
36
Mode 4 in Y Direction
30
24
SAP 2000
NF=4,12314 Hz
18
SBY
NF=4,13771 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
Height (m)
36
Mode 5 in X Direction
30
24
SAP 2000
NF=8,67117 Hz
18
CCM
DF=8,70211 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
Height (m)
36
30
SAP 2000
NF=11,36269 Hz
24
CCM
NF=11,45358 Hz
18
12
6
0
-1,5
-0,5
0,5
1,5
Figure 6.62. Comparison of fourth, fifth and sixth mode shapes in X, Y and Teta
directions found by both the present program and SAP2000 in Example 13
143
6. NUMERICAL RESULTS
48
42
36
a) Mod 7 on X Direction
Height (m)
30
SAP 2000
NF=16,73207 Hz
24
18
CCM
DF=16,85750 Hz
12
6
0
-1,5
-0,5
1,5
0,5
48
42
36
a) Mod 8 on Y Direction
Height (m)
30
SAP 2000
NF=21,99550 Hz
24
18
CCM
NF=22,31475 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
Figure 6.63. Comparison of seventh and eighth mode shapes in X and Y directions
found by both the present program and SAP2000 in Example 13
144
6. NUMERICAL RESULTS
Example 14:
In this example, the forced vibration analysis of a non-planar nonsymmetrical coupled shear wall on a rigid foundation was considered. The geometric
and material properties of the sixteen storey shear wall were taken as in Example 13.
The forced vibration analysis of the shear wall was carried out by the computer
program prepared in the present work and the SAP2000 structural analysis program
for damped and undamped cases.
The dynamic load, P(t), was applied at the top of the structure in the global X
direction in the plane of the connection beam as in Fig. 6.64 and the sine pulse force
was chosen as in Fig. 6.65.
h=3m
H = 48 m
P(t)
Y
X
Figure 6.64. Cross-sectional view of the structure and applied dynamic load in
Example 14
145
6. NUMERICAL RESULTS
P(lb)
150
5
t (s)
Present Study
(CCM)
% difference
0.025920
0.026004
0.32
The results of both methods for 6 % damping ratio are given in Table 6.23.
Table 6.23. Maximum displacement (m) in the X direction of point G for damped
case in Example 14
SAP2000
(Frame Method)
Present Study
(CCM)
% difference
0.024480
0.024562
0.33
The responses for both damped and undamped systems to sine force were
determined and the time-varying displacements in the X direction of point G are
presented in Figs. 6.66-67.
It is observed that the results obtained in the present work (CCM) coincide
with those of the SAP2000 structural analysis program, perfectly.
146
6. NUMERICAL RESULTS
0,030
0,020
Top Displacement (m)
Present study
SAP2000
0,010
0,000
0
10
12
-0,010
-0,020
-0,030
Time (s)
Figure 6.66. Time-varying displacements in X direction at the top of the shear wall
for undamped case in Example 14
0,030
0,020
Top Displacement (m)
Present study
SAP2000
0,010
0,000
0
10
12
-0,010
-0,020
-0,030
Time (s)
Figure 6.67. Time-varying displacements in X direction at the top of the shear wall
for damped case with 6 % damping ratio in Example 14
147
6. NUMERICAL RESULTS
Example 15:
In this example, the free vibration analysis of a non-planar non-symmetrical
coupled shear wall with and without stiffening beam was carried out using the
present program and the SAP2000 structural analysis program. Two stiffening beam
of 3.0 m height were placed at the mid-height and at the top of the wall which had 8
stories.
The geometrical properties and the cross-sectional view of the structure are
given in Fig. 6.68. The total height of the shear wall is 24 m, the storey height is 3 m,
the thickness is 0.3 m, the height of the connecting beams is 0.5 m and the elasticity
and shear moduli are E=2.85106 kN/m2 and G=1055556 kN/m2, respectively.
3 (4, 3)
3m
0.3 m
G
(-5, 0)
(-3, 0)
2
1
(0, 0)
1
(-1, 0)
0.3 m
1
(1, 0)
2
(4, 0)
2m
(0,125, 0,208)
0.3 m
3
0.3 m
4
(-3, -2)
2m
2m
1m
1m
3m
148
6. NUMERICAL RESULTS
The structure was solved both by the present method using the CCM and by
the SAP2000 structural analysis program using the frame method for which the
model and its 3-D view are given in Fig. 6.69.
Figure 6.69. Frame model of the structure in Example 15 and its 3-D view
Tables 6.24-25 compare the natural frequencies corresponding to each mode
found by the program prepared in the present work and the SAP2000 structural
analysis program for unstiffened and stiffened cases, expressing the percentage
differences.
149
6. NUMERICAL RESULTS
Table 6.24. Comparison of the natural frequencies (Hz) obtained from the present
program and SAP2000 for unstiffened case in Example 15
Unstiffened case
Mode
Present Study
(CCM)
Natural Frequencies
SAP2000
(Frame Method)
Natural Frequencies
% difference
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
0,69367
1,65427
4,14919
6,88659
11,36339
16,69150
21,86704
30,88525
35,45502
49,27674
51,41271
67,68475
70,87070
80,19235
92,87637
109,78064
0,69248
1,64679
4,14449
6,86626
11,33655
16,62251
21,76464
30,65600
35,16467
48,65858
50,74482
66,43667
69,48448
78,38068
90,32802
106,11409
0,17
0,45
0,11
0,30
0,24
0,42
0,47
0,75
0,83
1,27
1,32
1,88
2,00
2,31
2,82
3,46
150
6. NUMERICAL RESULTS
Table 6.25. Comparison of the natural frequencies (Hz) obtained from the present
program and SAP2000 for stiffened case in Example 15
Stiffened case
Mode
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Present Study
(CCM)
Natural Frequencies
0,68010
2,04419
4,04453
7,48094
11,24119
20,61359
21,54959
32,47404
35,04346
50,28789
51,83913
67,01399
71,54452
79,22520
94,90283
108,83989
SAP2000
(Frame Method)
Natural Frequencies
0,67889
2,02707
4,04126
7,44742
11,23489
20,44814
21,51302
32,29070
35,03757
50,16020
51,47142
66,36363
70,78988
77,74788
92,79546
105,56731
% difference
0,18
0,85
0,08
0,45
0,06
0,81
0,17
0,57
0,02
0,26
0,71
0,98
1,07
1,90
2,27
3,10
In each method, after obtaining the natural frequencies for unstiffened and
stiffened cases of the non-planar coupled shear wall, the mode shape vectors of the
system were found.
Mode shapes in X, Y and Teta directions were compared by normalizing with
respect to the displacements at the top of the structure.
Figs. 6.70-72 present the mode shapes of the shear wall for stiffened case
found by the present program and the SAP2000 structural analysis program and Figs.
6.73-75 compare the mode shapes of the shear wall for stiffened and unstiffened
cases.
151
6. NUMERICAL RESULTS
48
42
Height (m)
36
Mode 1 in X Direction
30
24
SAP 2000
NF=0,67889 Hz
18
CCM
NF=0,68010 Hz
12
6
0
0,0
0,5
1,0
48
42
Height (m)
36
Mode 2 in Y Direction
30
SAP 2000
NF=2,02706 Hz
24
18
CCM
NF=2,04419 Hz
12
6
0
0,0
0,5
1,0
48
42
36
Height (m)
30
24
SAP 2000
NF=4,04126 Hz
18
CCM
NF=4,04454 Hz
12
6
0
-2,0
-1,0
0,0
1,0
2,0
Figure 6.70. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by both the present program and SAP2000 for
stiffened case in Example 15
152
6. NUMERICAL RESULTS
48
42
36
Mode 4 in X Direction
Height
(m)
30
24
SAP 2000
NF=7,44742 Hz
18
CCM
NF=7,48094 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
36
Mode 5 in Y Direction
Height
(m)
30
24
SAP 2000
NF=11,23488 Hz
18
CCM
DF=11,24119 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
36
Mode 6 in X Direction
Height
(m)
30
SAP 2000
NF=20,44814 Hz
24
18
CCM
NF=20,61359 Hz
12
6
0
-1,5
-0,5
0,5
1,5
Figure 6.71. Comparison of fourth, fifth and sixth mode shapes in X and Y directions
found by both the present program and SAP2000 for stiffened case in
Example 15
153
6. NUMERICAL RESULTS
48
42
36
Mode 7 in Y Direction
Height (m)
30
SAP 2000
NF=21,51302 Hz
24
18
CCM
DF=21,54959 Hz
12
6
0
-1,5
-0,5
1,5
0,5
48
42
36
Mode 8 in X Direction
Height (m)
30
SAP 2000
NF=32,29070 Hz
24
18
CCM
NF=32,47404 Hz
12
6
0
-1,5
-0,5
0,5
1,5
Figure 6.72. Comparison of seventh and eighth mode shapes in X and Y directions
found by both the present program and SAP2000 for stiffened case in
Example 15
154
6. NUMERICAL RESULTS
48
42
Height (m)
36
Mode 1 in X Direction
30
24
Unstiffened case
NF=0,69367 Hz
18
Stiffened case
NF=0,68010 Hz
12
6
0
0,0
0,5
1,0
48
42
Height (m)
36
Mode 2 in Y Direction
30
Unstiffened case
NF=1,65427 Hz
24
18
Stiffened case
NF=2,04419 Hz
12
6
0
0,0
0,5
1,0
48
42
36
Height (m)
30
24
Unstiffened case
NF=4,14919 Hz
18
Stiffened case
NF=4,04453 Hz
12
6
0
-2,0
-1,0
0,0
1,0
2,0
Figure 6.73. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by the present program (CCM) for stiffened and
unstiffened cases in Example 15
155
6. NUMERICAL RESULTS
48
42
36
Mode 4 in X Direction
Height (m)
30
24
Unstiffened case
NF=6,88659 Hz
18
Stiffened case
NF=7,48094 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
36
Mode 5 in Y Direction
Height (m)
30
Unstiffened case
NF=11,36339 Hz
24
18
Stiffened case
DF=11,24119 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
36
Mode 6 in X Direction
Height (m)
30
Unstiffened case
NF=16,69150 Hz
24
18
Stiffened case
NF=20,61359 Hz
12
6
0
-1,5
-0,5
0,5
1,5
Figure 6.74. Comparison of fourth, fifth and sixth mode shapes in X and Y directions
found by the present program (CCM) for stiffened and unstiffened
cases in Example 15
156
6. NUMERICAL RESULTS
48
42
36
Mode 7 in Y Direction
Height (m)
30
Unstiffened case
NF=21,86704 Hz
24
18
Stiffened case
DF=21,54959 Hz
12
6
0
-1,5
-0,5
1,5
0,5
48
42
36
Mode 8 in X Direction
Height (m)
30
Unstiffened case
NF=30,88525 Hz
24
18
Stiffened case
NF=32,47404 Hz
12
6
0
-1,5
-0,5
0,5
1,5
Figure 6.75. Comparison of seventh and eighth mode shapes in X and Y directions
found by the present program (CCM) for stiffened and unstiffened cases
in Example 15
157
6. NUMERICAL RESULTS
Example 16:
In this example, the forced vibration analysis of a non-planar nonsymmetrical coupled shear wall on a rigid foundation was considered. The geometric
and material properties of the eight storey shear wall were taken as in Example 15.
The forced vibration analysis of the shear wall was carried out by the computer
program prepared in the present work and the SAP2000 structural analysis program
for damped and undamped cases.
The dynamic load, P(t), was applied at the top of the structure in the global X
direction in the plane of the connection beam as in Fig. 6.76 and the rectangular
pulse force was chosen as in Fig. 6.77.
3 (4, 3)
3m
0.3 m
G
P(t)
(-3, 0)
(-5, 0)
3
2
2
(0,125, 0,208)
0.3 m
(0, 0)
0.3 m
1
(-1, 0)
(4, 0)
2m
(1, 0)
3
0.3 m
4
(-3, -2)
2m
2m
1m
1m
3m
Figure 6.76. Cross-sectional view of the structure and applied dynamic load in
Example 16
158
6. NUMERICAL RESULTS
P(lb)
100
t(s)
5
Figure 6.77. Rectangular pulse force in Example 16
At the top of the shear wall, the maximum displacement in the X direction of
point G, the mass center, was calculated by the computer program prepared in the
present study and compared with those of the SAP2000 structural analysis program
in Tables 6.26-27 for both unstiffened and stiffened cases. Damping ratio was chosen
5 % for this example.
Table 6.26. Maximum displacement (m) in the X direction of point G for unstiffened
case in Example 16
Present Study
(CCM)
0.034714
% difference
Undamped
SAP2000
(Frame Method)
0.034650
Damped
0.028530
0.028194
1.17
Unstiffened case
0.18
Table 6.27. Maximum displacement (m) in the X direction of point G for stiffened
case in Example 16
Present Study
(CCM)
0.021297
% difference
Undamped
SAP2000
(Frame Method)
0.021600
Damped
0.017840
0.017650
1.06
Stiffened case
1.43
The responses for both damped and undamped systems to rectangular force
were determined for stiffened and unstiffened cases and the time-varying
displacements in the X direction of point G are presented in Figs. 6.78-79.
159
6. NUMERICAL RESULTS
0,040
0,030
Unstiffened case
0,020
Stiffened case
0,010
0,000
0
10
-0,010
-0,020
-0,030
Time (s)
Figure 6.78. Time-varying displacements in X direction at the top of the shear wall
for undamped case in Example 16
0,040
0,030
0,020
Unstiffened case
Stiffened case
0,010
0,000
0
10
-0,010
-0,020
Time (s)
Figure 6.79. Time-varying displacements in X direction at the top of the shear wall
for damped case with 5 % damping ratio in Example 16
160
6. NUMERICAL RESULTS
Example 17:
In this example, the free vibration analysis of a non-planar non-symmetrical
coupled shear wall with and without stiffening beam was carried out using the
present program and the SAP2000 structural analysis program. Two stiffening beam
of 3.0 m height were placed at the mid-height and at the top of the wall which had 16
stories.
The geometrical properties and the cross-sectional view of the structure are
given in Fig. 6.80. The total height of the shear wall is 48 m, the storey height is 3 m,
the height of the connecting beams is 0.5 m and the elasticity and shear moduli are
E= 2.85106 kN/m2 and G = 1055556 kN/m2, respectively.
2.0 m
3.0 m
Y
0.4 m
4
0.4 m
(-5.5, 4) 3
(2.5, 5)
4 (-2.5, 4)
3 (4.5, 5)
1.0 m
4.0 m
5.0 m
5
(-2.5, 3)
(-0,545, 2,091)
0.4 m
0.4 m
0.4 m
1
(-5.5, 0)
(0, 0)
X
1
1.5 m
1.5 m
2
(4.5, 0)
(1.5, 0)
(-1.5, 0)
4.0 m
0.4 m
3.0 m
161
6. NUMERICAL RESULTS
The structure was solved both by the present method using the CCM and by
the SAP2000 structural analysis program using the frame method for which the
model and its 3-D view are given in Fig. 6.81.
Figure 6.81. Frame model of the structure in Example 17 and its 3-D view
Tables 6.28-29 compare the first twenty natural frequencies corresponding to
each mode found by the program prepared in the present work and the SAP2000
structural analysis program for unstiffened and stiffened cases, expressing the
percentage differences.
162
6. NUMERICAL RESULTS
Table 6.28. Comparison of the natural frequencies (Hz) obtained from the present
program and SAP2000 for unstiffened case in Example 17
Unstiffened case
Mode
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Present Study
(CCM)
SAP2000
(Frame Method)
Natural Frequencies
Natural Frequencies
0,43177
0,49023
2,04893
3,04620
5,25108
8,49342
9,94970
16,18170
16,57520
23,91606
27,28633
33,12682
40,58494
43,76342
55,74229
56,41818
68,90739
74,70329
82,98062
95,29495
0,42973
0,48973
2,05602
3,03307
5,25612
8,40696
9,91554
16,05605
16,26648
23,61614
26,47472
32,52422
38,82465
42,67564
53,05564
53,92346
66,04236
68,85061
78,70490
85,81492
163
% difference
0,48
0,10
0,35
0,43
0,10
1,03
0,35
0,78
1,90
1,27
3,07
1,85
4,53
2,55
5,06
4,63
4,34
8,50
5,43
11,05
6. NUMERICAL RESULTS
Table 6.29. Comparison of the natural frequencies (Hz) obtained from the present
program and SAP2000 for stiffened case in Example 17
Stiffened case
Mode
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Present Study
(CCM)
SAP2000
(Frame Method)
Natural Frequencies
Natural Frequencies
0,47361
0,50691
2,15597
2,98694
6,21172
8,39412
10,24086
16,27233
16,67436
24,09009
26,98125
34,38680
39,88931
43,62386
55,81717
55,86423
68,41781
73,46209
83,48164
94,28620
0,47228
0,50174
2,15674
2,97419
6,34009
8,31093
10,24592
15,97689
16,59881
23,86781
26,20980
34,13438
38,23439
42,76440
52,65419
54,37579
66,08382
67,98693
79,98379
85,37953
% difference
0,28
1,03
0,04
0,43
2,03
1,00
0,05
1,85
0,46
0,93
2,94
0,74
4,33
2,01
6,01
2,74
3,53
8,05
4,37
10,43
In each method, after obtaining the natural frequencies for unstiffened and
stiffened cases of the non-planar coupled shear wall, the mode shape vectors of the
system were found.
Mode shapes in X, Y and Teta directions were compared by normalizing with
respect to the displacements at the top of the structure.
Figs. 6.82-84 present the mode shapes of the shear wall for stiffened case
found by the present program and the SAP2000 structural analysis program and Figs.
6.85-87 compare the mode shapes of the shear wall for stiffened and unstiffened
cases.
164
6. NUMERICAL RESULTS
48
42
Height (m)
36
Mode 1 in Y Direction
30
24
SAP 2000
NF=0,47228 Hz
18
CCM
NF=0,47361 Hz
12
6
0
0,0
0,5
1,0
48
42
Height(m)
36
Mode 2 in X Direction
30
SAP 2000
NF=0,50174 Hz
24
18
CCM
NF=0,50691 Hz
12
6
0
0,0
0,5
1,0
48
42
36
Height (m)
30
24
SAP 2000
NF=2,15674 Hz
18
CCM
NF=2,15597 Hz
12
6
0
-0,5
0,0
0,5
1,0
Figure 6.82. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by both the present program and SAP2000 for
stiffened case in Example 17
165
6. NUMERICAL RESULTS
48
42
Height (m)
36
Mode 4 in Y Direction
30
24
SAP 2000
NF=2,97419 Hz
18
SBY
NF=2,98694 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
Height (m)
36
Mode 5 in X Direction
30
SAP 2000
NF=6,34009 Hz
24
18
CCM
NF=6,21172 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
Height (m)
36
Mode 6 in Y Direction
30
SAP 2000
NF=8,31093 Hz
24
18
CCM
NF=8,39412 Hz
12
6
0
-1,5
-0,5
0,5
1,5
Figure 6.83. Comparison of fourth, fifth and sixth mode shapes in X and Y directions
found by both the present program and SAP2000 for stiffened case in
Example 17
166
6. NUMERICAL RESULTS
48
42
Height (m)
36
Mode 7 in X Direction
30
SAP 2000
NF=10,24592 Hz
24
18
CCM
DF=10,24086 Hz
12
6
0
-1,5
-0,5
0,5
1,5
48
42
Height (m)
36
Mode 8 in Y Direction
30
SAP 2000
NF=15,97689 Hz
24
18
CCM
NF=16,27233 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
Figure 6.84. Comparison of seventh and eighth mode shapes in X and Y directions
found by both the present program and SAP2000 for stiffened case in
Example 17
167
6. NUMERICAL RESULTS
48
42
36
Mode 1 in Y Direction
Height (m)
30
24
Unstiffened case
NF=0,43177 Hz
18
Stiffened case
NF=0,47361 Hz
12
6
0
0,0
0,5
1,0
48
42
36
Mode 2 in X Direction
Height(m)
30
Unstiffened case
NF=0,49023 Hz
24
18
Stiffened case
NF=0,50691 Hz
12
6
0
0,0
0,5
1,0
48
42
Height (m)
36
30
Unstiffened case
NF=2,04893 Hz
24
18
Stiffened case
NF=2,15597 Hz
12
6
0
-0,5
0,0
0,5
1,0
Figure 6.85. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by the present program (CCM) for stiffened and
unstiffened cases in Example 17
168
6. NUMERICAL RESULTS
48
42
36
Mode 4 in Y Direction
Height (m)
30
24
Unstiffened case
NF=3,04620 Hz
18
Stiffened case
NF=2,98694 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
36
Mode 5 in X Direction
Height (m)
30
24
Unstiffened case
NF=5,25108 Hz
18
Stiffened case
NF=6,21172 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
36
Mode 6 in Y Direction
Height (m)
30
Unstiffened case
NF=8,49342 Hz
24
18
Stiffened case
NF=8,39412 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
Figure 6.86. Comparison of fourth, fifth and sixth mode shapes in X and Y directions
found by the present program (CCM) for stiffened and unstiffened
cases in Example 17
169
6. NUMERICAL RESULTS
48
42
36
Mode 7 in X Direction
Height (m)
30
Unstiffened case
NF=9,94970 Hz
24
18
Stiffened case
NF=10,24086 Hz
12
6
0
-1,5
-0,5
1,5
0,5
48
42
36
Mode 8 in Y Direction
Height (m)
30
Unstiffened case
NF=16,18170 Hz
24
Stiffened case
NF=16,27233 Hz
18
12
6
0
-1,5
-0,5
0,5
1,5
Figure 6.87. Comparison of seventh and eighth mode shapes in X and Y directions
found by the present program (CCM) for stiffened and unstiffened cases
in Example 17
170
6. NUMERICAL RESULTS
Example 18:
In this example, the forced vibration analysis of a non-planar nonsymmetrical stiffened coupled shear wall on a rigid foundation was considered. The
geometric and material properties of the sixteen storey shear wall were taken as in
Example 17. The forced vibration analysis of the shear wall was carried out by the
computer program prepared in the present work and the SAP2000 structural analysis
program for damped and undamped cases.
The dynamic load, P(t), was applied at the top of the structure in the global X
direction in the plane of the connection beam as in Fig. 6.88 and the rectangular
pulse force was chosen as in Fig. 6.89.
2.0 m
3.0 m
Y
0.4 m
4
0.4 m
(-5.5, 4) 3
(2.5, 5)
4 (-2.5, 4)
3 (4.5, 5)
1.0 m
5.0 m
4.0 m
5
(-2.5, 3)
(-0,545, 2,091)
0.4 m
0.4 m
P(t)
2
0.4 m
1
(-5.5, 0)
(0, 0)
X
1
(1.5, 0)
(-1.5, 0)
4.0 m
0.4 m
1.5 m
1.5 m
2
(4.5, 0)
3.0 m
Figure 6.88. Cross-sectional view of the structure and applied dynamic load in
Example 18
171
6. NUMERICAL RESULTS
P(lb)
100
t(s)
11
Figure 6.89. Rectangular pulse force in Example 18
At the top of the shear wall, the maximum displacement in the X direction of
point G, the mass center, was calculated by the computer program prepared in the
present study and compared with those of the SAP2000 structural analysis program
in Tables 6.30-31 for both unstiffened and stiffened cases. Damping ratio was chosen
5 % for this example.
Table 6.30. Maximum displacement (m) in the X direction of point G for unstiffened
case in Example 18
Present Study
(CCM)
0.083607
% difference
Undamped
SAP2000
(Frame Method)
0.083100
Damped
0.076500
0.076998
0.65
Unstiffened case
0.61
Table 6.31. Maximum displacement (m) in the X direction of point G for stiffened
case in Example 18
Present Study
(CCM)
0.053835
% difference
Undamped
SAP2000
(Frame Method)
0.054360
Damped
0.050310
0.049506
1.60
Stiffened case
0.97
The responses for both damped and undamped systems to rectangular force
were determined for stiffened and unstiffened cases and the time-varying
displacements in the X direction of point G are presented in Figs. 6.90-91.
172
6. NUMERICAL RESULTS
0,100
0,080
Unstiffened case
0,060
Stiffened case
0,040
0,020
0,000
-0,020
10
12
14
16
18
20
22
-0,040
-0,060
-0,080
Time (s)
Figure 6.90. Time-varying displacements in X direction at the top of the shear wall
for undamped case in Example 18
0,100
0,080
0,060
Unstiffened case
Stiffened case
0,040
0,020
0,000
0
10
12
14
16
18
20
22
-0,020
-0,040
-0,060
Time (s)
Figure 6.91. Time-varying displacements in X direction at the top of the shear wall
for damped case with 5 % damping ratio in Example 18
173
6. NUMERICAL RESULTS
Example 19:
In this example, the free vibration analysis of a non-planar non-symmetrical
coupled shear wall with and without stiffening beam was carried out using the
present program and the SAP2000 structural analysis program. The geometric and
material properties of the sixteen storey shear wall were taken as in Example 11. A
stiffening beam of 3.0 m height was placed at the height of 30 m on the tenth storey.
The total height of the shear wall is 48 m, the storey height is 3 m and the
thicknesses of the piers and the connecting beams are shown in Fig. 6.92. The height
of the connecting beams is 0.4 m and the elasticity and shear moduli are
E = 2.85106 kN/m2 and G = 1055556 kN/m2, respectively.
2m
2m
2m
2m
3m
Y
0.2 m
(-7, 4)
10
11
10
0.4 m
3
(-5, 4)
(-1, 4)
(1, 4)
0.4 m
G
0.2 m
8
(-7, 0)
3m
(4, 4)
0.4 m
0.2 m
4m
0.4 m
4
(-3, 4)
9
(-5, 0)
(-3, 0)
(-0.795, 0.322)
0.4 m
1
0.4 m
O
1
(-1, 0)
(1, 0)
(-7, -3)
0.2 m
0.4 m
0.4 m
(-3, -3)
4m
0.4 m
0.2 m
7
2
(4, 0)
2m
0.4 m
6
(-1, -3)
(1, -3)
2m
(4, -3)
3m
174
6. NUMERICAL RESULTS
According to the lumped mass idealization, the lumped masses, which were
calculated by computer program, were concentrated at the center of the whole crosssectional area of the structure. The coordinates of the mass center were calculated as
(-0.795; 0.322).
The structure was solved both by the present method using the CCM and by
the SAP2000 structural analysis program using the frame method for which the
model and its 3-D view are given in Fig. 6.93.
Figure 6.93. Frame model of the structure in Example 19 and its 3-D view
175
6. NUMERICAL RESULTS
Mode
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Present Study
(CCM)
SAP2000
(Frame Method)
Natural Frequencies
Natural Frequencies
0,50138
0,66135
2,47098
4,12617
6,51577
11,50762
12,51956
20,48279
22,45812
30,36957
36,96983
42,14066
54,98454
55,73259
71,03691
76,42915
87,85518
101,19019
105,83248
124,366150
0,50047
0,66117
2,46396
4,11839
6,47507
11,45595
12,37532
20,10512
22,27115
29,54952
36,47405
40,57179
52,99551
53,89331
66,59437
74,30635
81,06535
95,99595
97,41872
110,846184
176
% difference
0,18
0,03
0,29
0,19
0,63
0,45
1,17
1,88
0,84
2,78
1,36
3,87
3,75
3,41
6,67
2,86
8,38
5,41
8,64
12,20
6. NUMERICAL RESULTS
Table 6.33. Comparison of the natural frequencies (Hz) obtained from the present
program and SAP2000 for stiffened case in Example 19
Stiffened case
Mode
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Present Study
(CCM)
SAP2000
(Frame Method)
Natural Frequencies
Natural Frequencies
0,62562
0,66052
2,60238
4,11919
6,63493
11,48704
12,96880
20,42581
22,44974
30,66376
36,84898
42,27349
54,90843
55,57030
71,31118
76,23393
87,42965
100,76738
105,59907
124,11197
0,62284
0,66030
2,58037
4,11147
6,57733
11,43606
12,77066
20,05863
22,26720
29,81654
36,37067
40,73950
52,91929
53,85963
66,97627
74,21348
80,90873
96,05297
97,18234
110,98024
% difference
0,45
0,03
0,85
0,19
0,88
0,45
1,55
1,83
0,82
2,84
1,32
3,77
3,76
3,18
6,47
2,72
8,06
4,91
8,66
11,83
In each method, after obtaining the natural frequencies for unstiffened and
stiffened cases of the non-planar coupled shear wall, the mode shape vectors of the
system were found.
Mode shapes in X and Y directions were compared by normalizing with
respect to the displacements at the top of the structure.
Figs 6.94-96 present the mode shapes of the shear wall for stiffened case
found by the present program and the SAP2000 structural analysis program and Figs.
6.97-99 compare the mode shapes of the shear wall for stiffened and unstiffened
cases.
177
6. NUMERICAL RESULTS
48
Height (m)
42
36
Mode 1 in X Direction
30
24
SAP 2000
NF=0,62284 Hz
18
CCM
NF=0,62562 Hz
12
6
0
0,0
0,5
1,0
48
Height (m)
42
36
Mode 2 in Y Direction
30
SAP 2000
NF=0,66030 Hz
24
18
CCM
NF=0,66052 Hz
12
6
0
0,0
1,0
0,5
48
42
36
Height (m)
30
24
SAP 2000
NF=2,58037 Hz
18
CCM
NF=2,60238 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
Figure 6.94. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by both the present program and SAP2000 for
stiffened case in Example 19
178
6. NUMERICAL RESULTS
48
42
36
Mode 4 in Y Direction
Height (m)
30
SAP 2000
NF=4,11147 Hz
24
18
SBY
NF=4,11919 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
36
Mode 5 in X Direction
Height (m)
30
SAP 2000
NF=6,57733 Hz
24
18
CCM
NF=6,63493 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
36
Mode 6 in Y Direction
Height (m)
30
SAP 2000
NF=11,43606 Hz
24
18
CCM
NF=11,48704 Hz
12
6
0
-1,5
-0,5
0,5
1,5
Figure 6.95. Comparison of fourth, fifth and sixth mode shapes in X and Y directions
found by both the present program and SAP2000 for stiffened case in
Example 19
179
6. NUMERICAL RESULTS
48
42
36
Mode 7 in X Direction
Height (m)
30
SAP 2000
NF=12,77066 Hz
24
18
CCM
NF=12,96880 Hz
12
6
0
-1,5
-0,5
0,5
1,5
48
42
36
Mode 8 in Y Direction
Height (m)
30
SAP 2000
NF=20,05863 Hz
24
18
CCM
NF=20,42581 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
Figure 6.96. Comparison of seventh and eighth mode shapes in X and Y directions
found by both the present program and SAP2000 for stiffened case in
Example 19
180
6. NUMERICAL RESULTS
48
42
Height (m)
36
Mode 1 in X Direction
30
24
Unstiffened case
NF=0,50138 Hz
18
Stiffened case
NF=0,62562 Hz
12
6
0
0,0
0,5
1,0
48
42
Height(m)
36
Mode 2 in Y Direction
30
Unstiffened case
NF=0,66135 Hz
24
18
Stiffened case
NF=0,66052 Hz
12
6
0
0,0
0,5
1,0
48
42
Height (m)
36
30
Unstiffened case
NF=2,47098 Hz
24
18
Stiffened case
NF=2,60238 Hz
12
6
0
-0,5
0,0
0,5
1,0
Figure 6.97. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by the present program (CCM) for stiffened and
unstiffened cases in Example 19
181
6. NUMERICAL RESULTS
48
42
36
Mode 4 in Y Direction
Height (m)
30
24
Unstiffened case
NF=4,12617 Hz
18
Stiffened case
NF=4,11919 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
36
Mode 5 in X Direction
Height (m)
30
24
Unstiffened case
NF=6,51577 Hz
18
Stiffened case
NF=6,63493 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
36
Mode 6 in Y Direction
Height (m)
30
Unstiffened case
NF=11,50762 Hz
24
18
Stiffened case
NF=11,48704 Hz
12
6
0
-1,5
-0,5
0,5
1,5
Figure 6.98. Comparison of fourth, fifth and sixth mode shapes in X and Y directions
found by the present program (CCM) for stiffened and unstiffened
cases in Example 19
182
6. NUMERICAL RESULTS
48
42
Height (m)
36
Mode 7 in X Direction
30
Unstiffened case
NF=12,51956 Hz
24
18
Stiffened case
NF=12,96880 Hz
12
6
0
-1,5
-0,5
0,5
1,5
48
42
36
Mode 8 in Y Direction
Height (m)
30
Unstiffened case
NF=20,48279 Hz
24
18
Stiffened case
NF=20,42581 Hz
12
6
0
-1,5
-0,5
0,5
1,5
Figure 6.99. Comparison of seventh and eighth mode shapes in X and Y directions
found by the present program (CCM) for stiffened and unstiffened cases
in Example 19
183
6. NUMERICAL RESULTS
Example 20:
In this example, the forced vibration analysis of a non-planar nonsymmetrical stiffened coupled shear wall on a rigid foundation was considered. The
geometric and material properties of the sixteen storey shear wall were taken as in
Example 19. The forced vibration analysis of the shear wall was carried out by the
computer program prepared in the present work and the SAP2000 structural analysis
program for damped and undamped cases.
The dynamic load, P(t), was applied at the top of the structure in the global X
direction in the plane of the connection beam as in Fig. 6.100 and the triangular pulse
force was chosen as in Fig. 6.101.
2m
2m
2m
2m
3m
Y
0.2 m
(-7, 4)
10
11
10
0.4 m
3
(-5, 4)
4m
(4, 4)
(-1, 4)
(1, 4)
0.4 m
0.2 m
0.4 m
0.2 m
8
(-7, 0)
3m
0.4 m
4
(-3, 4)
P(t)
0.4 m
2
(-3, 0)
(-5, 0)
(-0.795, 0.322)
0.4 m
O
1
(-1, 0)
(1, 0)
(-7, -3)
0.2 m
0.4 m
0.4 m
0.4 m
6
(-3, -3)
4m
(4, 0)
0.4 m
0.2 m
7
(-1, -3)
2m
6
(4, -3)
(1, -3)
2m
3m
Figure 6.100. Cross-sectional view of the structure and applied dynamic load in
Example 20
184
6. NUMERICAL RESULTS
P(lb)
150
t (s)
2,5
5
Figure 6.101. Triangular pulse force in Example 20
At the top of the shear wall, the maximum displacement in the X direction of
point G, the mass center, was calculated by the computer program prepared in the
present study and compared with those of the SAP2000 structural analysis program
in Tables 6.34-35 for both damped and undamped cases. Damping ratio was chosen 5
% for this example.
Table 6.34. Maximum displacement (m) in the X direction of point G for unstiffened
case in Example 20
Present Study
(CCM)
0.035658
% difference
Undamped
SAP2000
(Frame Method)
0.035710
Damped
0.035270
0.035192
0.22
Unstiffened case
0.14
Table 6.35. Maximum displacement (m) in the X direction of point G for stiffened
case in Example 20
Present Study
(CCM)
0.023820
% difference
Undamped
SAP2000
(Frame Method)
0.024550
Damped
0.023570
0.022860
3.01
Stiffened case
3.06
The responses for both damped and undamped systems to rectangular force
were determined and the time-varying displacements in the X direction of point G
are presented in Figs. 6.102-103.
185
6. NUMERICAL RESULTS
0,040
0,030
0,020
Unstiffened case
Stiffened case
0,010
0,000
0
10
-0,010
-0,020
Time (s)
Figure 6.102. Time-varying displacements in X direction at the top of the shear wall
for undamped case in Example 20
0,040
0,030
0,020
Unstiffened case
Stiffened case
0,010
0,000
0
10
-0,010
Time (s)
Figure 6.103. Time-varying displacements in X direction at the top of the shear wall
for damped case with 5 % damping ratio in Example 20
186
6. NUMERICAL RESULTS
Example 21:
The computer program prepared in this study has been made such that the
change in the cross-section of the coupled shear wall along the height can also be
taken into account.
In this example, the coupled shear wall considered in Example 3 was solved
once more, when the stories above the fourth are of a different cross-section than the
ones below as shown in Figs. 6.104-105. The solution was carried out both by the
present method and the SAP2000 structural analysis program, for which the model
and its 3-D view are given in Fig. 6.106.
The total height of the shear wall is 24 m, the storey height is 3 m, the
thickness is 0.3 m, the height of the connecting beams is 0.5 m and the elasticity and
shear moduli of the structure are E = 2.85106 kN/m2 and G = 1055556 kN/m2 ,
respectively.
According to the lumped mass idealization, the lumped masses, which were
calculated by computer program, were concentrated at the center of the whole crosssectional area of the structure. The mass center was located on the point O.
(3, 2)
Y
2m
0.3 m
2
(-3, 0)
0.3 m
(0,0)
2
1
(3, 0)
(1, 0)
(-1, 0)
0.3 m
2m
0.3 m
(-3, -2)
2m
1m
1m
2m
Figure 6.104. Cross-sectional view of the 1st region of the structure in Example 21
187
6. NUMERICAL RESULTS
(-3, 2)
2m
0.3 m
0.3 m
(-3, 0)
0.3 m
2
1
(0,0)
2m
0.3 m
1
(-1, 0)
2
1
(3, 0)
(1, 0)
3
0.3 m
(3, 2)
0.3 m
(-3, -2)
2m
1m
1m
(3, -2)
2m
Figure 6.105. Cross-sectional view of the 2nd region of the structure in Example 21
Figure 6.106. Frame model of the structure in Example 21 and its 3-D view
188
6. NUMERICAL RESULTS
Table 6.36. Comparison of the natural frequencies (Hz) obtained from the present
program and SAP2000 in Example 21
Mode
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Present Study
(CCM)
SAP2000
(Frame Method)
Natural Frequencies
Natural Frequencies
1,07270
1,61807
4,01622
5,88496
10,68451
14,19554
19,27886
25,67233
31,10287
38,98559
46,79324
53,47172
65,79977
69,71574
91,63290
107,58726
1,07366
1,60347
4,01831
5,85470
10,63805
14,14731
19,19478
25,44843
30,91459
38,45470
46,35447
52,34941
64,26048
67,71944
88,45925
103,86768
% difference
0,09
0,91
0,05
0,52
0,44
0,34
0,44
0,88
0,61
1,38
0,95
2,14
2,40
2,95
3,59
3,58
189
6. NUMERICAL RESULTS
48
42
Height (m)
36
Mode 1 in Y Direction
30
24
SAP 2000
NF=1,07366 Hz
18
CCM
NF=1,07270 Hz
12
6
0
0,0
0,5
1,0
48
42
Height (m)
36
Mode 2 in X Direction
30
SAP 2000
NF=1,60347 Hz
24
18
CCM
NF=1,61807 Hz
12
6
0
0,0
1,0
0,5
48
42
36
Mode 3 in Y Direction
Height (m)
30
SAP 2000
NF=4,01831 Hz
24
18
CCM
NF=4,01622 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
Figure 6.107. Comparison of first, second and third mode shapes in X and Y
directions found by both the present program and SAP2000 in
Example 21
190
6. NUMERICAL RESULTS
48
42
36
Mode 4 in X Direction
Height (m)
30
24
SAP 2000
NF=5,85470 Hz
18
CCM
NF=5,88496 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
36
Mode 5 in X Direction
Height (m)
30
24
SAP 2000
NF=10,63805 Hz
18
CCM
DF=10,68451 Hz
12
6
0
-1,0
-0,5
0,0
0,5
1,0
48
42
36
Mode 6 in Y Direction
Height (m)
30
SAP 2000
NF=14,14731 Hz
24
18
CCM
NF=14,19554 Hz
12
6
0
-1,5
-0,5
0,5
1,5
Figure 6.108. Comparison of fourth, fifth and sixth mode shapes in X and Y
directions found by both the present program and SAP2000 in
Example 21
191
6. NUMERICAL RESULTS
48
42
36
Mode 7 in X Direction
Height (m)
30
SAP 2000
NF=19,19478 Hz
24
18
CCM
DF=19,27886 Hz
12
6
0
-1,5
-0,5
0,5
1,5
48
42
36
Mode 8 in Y Direction
Height (m)
30
SAP 2000
NF=25,44843 Hz
24
18
CCM
NF=25,67233 Hz
12
6
0
-1,5
-0,5
0,5
1,5
Figure 6.109. Comparison of seventh and eighth mode shapes in X and Y directions
found by both the present program and SAP2000 in Example 21
192
6. NUMERICAL RESULTS
Example 22:
In this example, the forced vibration analysis of the change in the crosssection of the coupled shear wall on a rigid foundation was considered. The
geometric and material properties of the eight storey shear wall were taken as in
Example 21. The forced vibration analysis of the shear wall was carried out by the
computer program prepared in the present work and the SAP2000 structural analysis
program for damped and undamped cases.
The dynamic load, P(t), was applied at the top of the structure in the global X
direction in the plane of the connection beam as in Fig. 6.110 and the rectangular
pulse force was chosen as in Fig. 6.111.
Z
H = z1 = 24 m
z2 = 12 m
2nd region
h2 = 3 m
1st region
h1 = 3 m
P(t)
Figure 6.110. Cross-sectional view of the structure and applied dynamic load in
Example 22
193
6. NUMERICAL RESULTS
P(lb)
150
t(s)
6
Figure 6.111. Rectangular pulse force in Example 22
At the top of the shear wall, the maximum displacement in the X direction of
the point O was calculated by the computer program prepared in the present study
and compared with those of the SAP2000 structural analysis program in Tables 6.3738 for both damped and undamped cases. Damping ratio was chosen 5 % for this
example.
Table 6.37. Maximum displacement (m) in the X direction of the point O for
undamped case in Example 22
SAP2000
(Frame Method)
Present Study
(CCM)
% difference
0.036393
0.036112
0.77
Table 6.38. Maximum displacement (m) in the X direction of the point O for damped
case in Example 22
SAP2000
(Frame Method)
Present Study
(CCM)
% difference
0.036130
0.035939
0.53
The responses for both damped and undamped systems to rectangular force
were determined and the time-varying displacements in the X direction of the point
O are presented in Figs. 6.112-113.
It is observed that the results obtained in the present work (CCM) coincide
with those of SAP2000 structural analysis program perfectly.
194
6. NUMERICAL RESULTS
Present study
0,080
SAP2000
0,060
0,040
0,020
0,000
0
10
12
-0,020
-0,040
-0,060
Time (s)
Figure 6.112. Time-varying displacements in X direction at the top of the shear wall
for undamped case in Example 22
0,080
0,060
Present study
SAP2000
0,040
0,020
0,000
0
10
12
-0,020
-0,040
Time (s)
Figure 6.113. Time-varying displacements in X direction at the top of the shear wall
for damped case with 5 % damping ratio in Example 22
195
7. CONCLUSIONS
7. CONCLUSIONS
In this study, the dynamic analysis of non-planar coupled shear walls with
any number of stiffening beams, having flexible beam-wall connections and resting
on rigid foundations is carried out. In this study, continuous connection method
(CCM) and Vlasovs theory of thin-walled beams are employed to find the structure
stiffness matrix. A computer program has been prepared in Fortran Language to
implement both the free and forced vibration analyses of non-planar coupled shear
walls and various examples have been solved by applying it to different structures.
In the first and second examples, the model of non-planar coupled shear wall
has been taken same as the example considered by Tso and Biswas in 1973. In the
first example, the free vibration analysis is carried out both by the computer program
prepared in the present work and the SAP2000 structural analysis program. The
natural frequencies and the mode shape vectors, thus obtained, have been compared.
In the second example, the forced vibration analysis is carried out and the timevarying displacements have been plotted in Fig.6.8-9 for damped and undamped
cases. It is seen that the results of the present study coincide with those of SAP2000
perfectly.
In the third and fourth examples, the model of the structure is chosen
symmetrical with respect to X and Y axes. Both piers are assigned as star sections, in
which the sectorial areas are equal to zero at all points of the cross-section. The free
and forced vibration analyses are carried out both by the computer program prepared
in the present work and the SAP2000 structural analysis program. In the fourth
example, the dynamic load is applied at the top of structure in the global X direction
and the rectangular pulse force is chosen. Then, the time-varying displacements have
been plotted in Fig.6.17-18 for damped and undamped cases. The results obtained are
compared with those of SAP2000 and a perfect match is observed.
In the fifth and sixth examples, the non-symmetrical non-planar coupled shear
wall structures, in which both piers have zero sectorial area values, are considered. In
the fifth example, the free vibration analysis is carried out both by the computer
program and SAP2000. The natural frequencies and mode shape vectors, thus
196
7. CONCLUSIONS
obtained, are compared. In the sixth example, the dynamic load is applied at the top
of the structure in the global Y direction and the forced vibration analysis is carried
out. Both the rectangular and sine pulse forces are chosen in this example. The timevarying displacements have been plotted in Figs. 6.26-29 for the damped and
undamped cases. It is observed that the results obtained in the present work coincide
with those of the SAP2000 structural analysis program perfectly.
In the examples, from seventh to fourteenth, the non-symmetrical non-planar
shear wall structures, in which both sections have non-zero sectorial area values, are
considered. The free and forced vibration analyses are carried out both by the
computer program prepared in the present work and the SAP2000 structural analysis
program. In the free vibration analyses, the natural frequencies and mode shape
vectors, thus obtained, are compared. In the forced vibration analyses, the timevarying displacements are plotted for the damped and undamped cases. The results
obtained are compared with those of SAP2000 and a perfect match is observed.
The examples, from fifteenth to twentieth, are selected to study the effect of
the stiffening beams on the dynamic behaviour of non-planar coupled shear walls.
In the fifteenth and sixteenth examples, the dynamic analysis of the nonplanar non-symmetrical coupled shear wall having 2 stiffening beams is considered.
In the fifteenth example, the free vibration analysis is carried out. The natural
frequencies and mode shape vectors, thus obtained, are compared with and without
stiffening beams using the present program and the SAP2000 structural analysis
program. In the sixteenth example, the forced vibration analysis is carried out and the
time-varying displacements have been plotted in Figs. 6.77-78 for stiffened and
unstiffened cases.
In the seventeenth and eighteenth examples, a non-planar non-symmetrical
stiffened coupled shear wall structure is considered. Two stiffening beams of 3.0 m
height are placed, one at the mid-height and the other at the top of the wall which has
16 stories. The free and forced vibration analyses are carried out both by the present
program and the SAP2000 structural analysis program. In the eighteenth example,
the forced vibration analysis is carried out and the time-varying displacements have
been plotted in Figs. 6.77-78 for stiffened and unstiffened cases.
197
7. CONCLUSIONS
198
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201
202
CURRICULUM VITAE
I was born in Adana on May 20, 1982. I graduated from Private Adana
College in 1998. In the same year, I started my BSc. study at the Karadeniz
Technical University and graduated from the Department of Civil Engineering in
2002. Then, I started MSc. programme in Structural Mechanics at the ukurova
University and graduated in 2004. Then, I started my PhD. study as a student of my
supervisor Prof. Dr. Orhan Aksoan at the ukurova University in 2004. I have, also,
been working as a Control Engineer in General Directorate of Highways in
5.Division Directorate since 2005.
203
APPENDICES
204
205
L
b
b + 2c
0.1
L
206
(A1.1)
Thin-walled beams are characterized by the fact that their three dimensions (t,
b+2c, L) are all of different orders of magnitude as depicted by (A1.1). The thickness
of a beam is small compared with any characteristic dimension of the cross-section,
and the cross-sectional dimensions are small compared with the length of the beam.
A1.2.2. Coordinate System
In order to show the development of Vlasovs basic theory, the general form
of an open tube and an open bar can be considered as shown in Fig. A1.2 and in Fig.
A1.3, respectively, and they are, generally, called thin-walled open section beams.
An axis in the middle surface parallel to the beam axis is called the generator and the
intersection of the middle surface with a plane perpendicular to the generator is
called the contour line. An orthogonal system of coordinates (z, s) is chosen, the first
being in the direction of the longitudinal generators and the second tangent to a
contour line. The coordinate z starts from one end and the coordinate s from any
generator. P(z, s) is an arbitrary point on the middle surface and its displacements in
the coordinate directions are called u z and u s .
s
O
Contour line
P(z,s)
us
Generator
uz
207
z
O
s
Contour line
J=
1 n
3
bk t k
3 k =1
(A1.2)
where bk is the width of section k and tk is its thickness and n is the number of parts
in the cross-section (n=3 in Fig. A1.4). It must be noted that J is not the polar
moment of inertia when the section is non-circular as in Fig. A1.4. The twisting
rigidity of a section is given by GJ. G is known as the shear modulus of the material.
208
t
Mt
ds h s dw s
=
2
2
A
P(sp)
ds
hs
dA =
A
s
209
Tangent Line
The sectorial area of a point on a section is computed using the median line of
the outline of the section. ws is the sectorial area for a point, P, with coordinate sp
and is equal to twice the area swept by the line connecting R, an arbitrary center of
rotation (initial pole), to the points on the contour line, starting from O and ending at
P, as shown in Fig. A1.5. Its mathematical definition is:
sp
ws = h s ds
(A1.3)
where
ws : sectorial area,
h s : distance from point R to the line tangent to the contour line at point A,
s
The parameter ws is used in the warping of the section. The sectorial area
corresponding to the shear center (principal pole) as the pole, is called the principal
sectorial area, and its plot for the section is the principal sectorial area diagram. The
process of calculating the sectorial areas of a section is laborious and time
consuming. Therefore, to speed up these calculations, a program in Fortran language
is written in this thesis. The program is given in Appendix 5 and a worked example is
given in Appendix 3.
The sign of the increment of sectorial area is positive if the sliding radius,
210
The following points constitute simple rules for the determination of the shear
center location for some typical cross-sections:
1. The shear center always falls on a cross-sectional axis of symmetry (two
sections on the left in Fig. A1.6).
2. If the cross-section contains two perpendicular axes of symmetry, or a
symmetry center, then the shear center is located at the intersection of the
symmetry axes and the symmetry center, respectively (two sections on the
right in Fig. A1.6).
S G
211
for locating the origin of the principal linear axes called the centroid of the section or
simply G and the third gives the direction of the principal linear axes. These values
are not sufficient for the analysis of a thin walled beam. There are some other
properties of sections which are defined on the basis of sectorial area and named as
sectorial properties of a section. For the analysis of internal stresses, these properties
must be known to determine the location of the shear center through the use of the
sectorial area. Fig. A1.7 shows the procedure for the determination of the location of
the shear center based on the sectorial area. The moment of the shear stresses in the
cross-section with respect to a certain point D is:
s2
M D = t t h s ds = t t h s ds
A
(A1.4)
s1
y
ds
t
hs
s1
D
x
t
s2
(A1.5)
If t is eliminated using
t=
Vx S y
Iy t
Vy S x
(A1.6)
Iy t
212
where
Vx and Vy
Sx and Sy
axes,
Ix , Iy
axes,
t
Substituting equation (A1.6) into equation (A1.4), moment about point D is found as
follows:
Vy
MD =
Ix
S
A
dws
dw s
V
dA
dA + x S y
dA
dA
Iy A
(A1.7)
Using integration by parts technique, the first integral in (A1.7) can be evaluated in
the following form:
dws
dA = Sx ws
dA
s2
s1
dSx
ws dA
dA
A
(A1.8)
where the limits s1 and s2 signify the end points of the cross-section (see Fig. A1.7).
It is known that the statical moment about x axis is
S x = y dA
(A1.9)
Sx ws s 2 = 0
(A1.10)
213
and
dS x
=y
dA
(A1.11)
dw s
dA = y w s dA
dA
A
(A1.12)
MD = -
Vy
Ix
Vx
y w dA - I x w dA
s
(A1.13)
y A
If the point D coincides with the shear center, the moment MD is zero regardless of
the values Vx and Vy. Therefore, the entities called the sectorial statical moments of a
section about axes x and y can be written, respectively, as
S wx = y ws dA
(A1.14)
S wy = x w s dA
(A1.15)
and
The principal pole (shear center) of a section can be defined by equating the
expressions (A1.14) and (A1.15) to zero. The line segment connecting the shear
center and the point on the contour line at which the sectorial area is zero is called
the principal radius. Hence, to locate the principal radius, the sectorial statical
moment of the section is set equal to zero (see Appendix 3),
214
S w = ws dA = 0
(A1.16)
The method of finding the position of the shear center of a section resembles
that of finding the principal radius. Let us assume an arbitrarily placed initial pole
and an initial radius. The theoretical formulas for calculating the location of the shear
center and evaluating the coordinates for the shear center are given in the following
equations (see Appendix 2):
a x = bx +
S
1
ws ydA = b x + wx
Ix
Ix A
(A1.17)
a y = by -
S
1
ws xdA = b y - wy
Iy A
Iy
(A1.18)
where ax and ay are linear coordinates of a principal pole, bx and by are linear
coordinates of an arbitrarily placed pole (see Fig. A1.8).
S(ax,ay)
R(bx,by)
y
Figure A1.8. Arbitrarily placed pole R and principal pole S
215
These formulas are valid only if axes x and y are identical with the principal
axes of a section, in other words, only if Ixy=0. In the case where the given axes x and
y do not coincide with the principal axes of a section (Ixy 0), but pass through the
center of gravity, the following equations must be used:
a x = bx +
I ySwx - I xySwy
I x I y - I2xy
(A1.19)
a y = by -
I xSwy - IxySwx
I x I y - I 2xy
(A1.20)
The derivation of the formulas for the position of a shear center (A1.19 and A1.20)
and the location of principal radius are given in Appendix 2.
A1.2.6. Warping Moment of Inertia (Sectorial Moment of Inertia)
Warping moment of inertia expresses the warping torsional resistance of a
section, or in other words, the capacity of the section to resist warping torsion. It is
analogous to the moment of inertia in bending. The warping moment of inertia is
derived from the sectorial area distribution, and can be written as
Iw = ws2 dA
(A1.21)
Similar to flexural rigidity EI, EIw is the warping rigidity of a section. The integral
expression in (A1.21) shows up in section A1.5 in Vlasovs theory.
All the star sections (including all angle-sections and T-sections) shown in
Fig. A1.9, have their shear centers at the intersection of their branches. Since all
shear force increments pass through the shear centers, the sectorial areas are equal to
zero at all points of the cross-sections. Therefore, the warping moments of inertia are
all equal to zero for star sections.
216
S
S
S
Figure A1.9. Some star sections for which ws = I w = 0
A1.3. Physical Meanings of St.Venant Twist and Flexural Twist
A1.3.1. St. Venant Twist
A thin walled beam of open section is liable to warp when subject to a
twisting moment. For two beams of identical material and cross-sectional shape, the
warping effect is much greater for the open section. The fact is that the open section
does not have high warping resistance because of the continuity of its contour line
and also does not allow the St. Venant torsional stresses to circulate around the
contour and thereby not develop a high internal torsional resistance.
The theory of the twisting moment acting on a thick walled beam was
developed by St.Venant and is a generalization of the problem of the twisting of a
circular shaft as shown in Fig. A1.10. St. Venants theory is perfectly valid for a
circular cross-section and remains practically valid for any thick walled beam.
z
qz
L
g
x
The twisting of a circular shaft does not produce any longitudinal stresses, but
only shear stresses. St.Venants theory is based on the hypothesis that the twisting of
a shaft of any cross-section produces only shear stresses. The warping effect of a
twisting moment can be seen clearly in Fig. A1.11(a) and (b). (Although the
displacements due to warping are not very large in magnitude and do not have a
major significance, they gain importance when warping is prevented). It is apparent
in Fig. A1.10 that the deformation is equal to g L where g is the shear strain.
Msv (St.Venant Twist)
q
(b) Plan
(a) Elevation
218
The presence of longitudinal stresses infers that part of the work done by the
twisting moment is used up in developing them, and only the remainder will develop
shear stresses associated with the St.Venant twist. These longitudinal stresses can
become more important to the safety of a structure than shears caused by a St.Venant
twist.
D
rigidly fixed
(a) Elevation
219
M tot = M w + M sv
(A1.22)
where
Mtot : Twisting moment acting on a thin-walled beam
M w : Flexural twisting moment (warping torque)
Msv : St.Venant (pure) twisting moment
D
2D
q
+
q
(a-) Elevation
(c-) due to M w
P.e=-M
P
h
A bimoment can be either positive or negative and has a direction in the same
manner as a bending moment. It is a vector not a scalar quantity. In this study, a
bimoment is assumed to be positive when the direction of one moment seen from the
plane of the other moment is clockwise (as in Fig. A1.14). Bimoment has units of
force times the square of the length (lb-in2, kip-ft2, kN-m2, t-m2, etc.). The magnitude
of a bimoment is given by the product of the distance of these parallel planes and the
moment on one of them.
Bz = M h
(A1.23)
221
P
2
P
2
P
2
P
2
P
2
P
2
P
2
P
2
=
P
2
(a) Vertical load at corner
P
2
P
2
P
2
P
2
P
2
P
2
P
2
+
Warping
stresses
222
2D
2D
M=P.e
2P
P
-M=P.e
Bz
223
determine the internal stresses and strains as simple as the ones in a thick walled
beam due to bending moments and shear forces.
A1.4. St. Venants Theory (Uniform Torsion)
A1.4.1. Torsion of a Bar with Circular Cross-Section
A cylinder of a circular cross-section being twisted by couples applied at the
end planes was given before in Fig. A1.10. The behaviour of a circular cylinder
under torsion is such that all cross-sections normal to the axis remain plane after
deformation. The shearing stress in a cylider of circular cross-section under torsion is
known from the strength of materials. It is well known that these stresses satisfy the
boundary conditions, therefore, they represent the exact solution for a circular
cylinder. The exact solution for that problem was first formulated by St. Venant and
is generally called St. Venants theory. St. Venant assumed that the projection of any
deformed cross-section on the x-y plane (see Fig. A1.10) rotates as a rigid body, the
angle of twist per unit length being constant.
The displacement of any point P due to rotation is shown in Fig. A1.18. The
line SP which is equal to r (radial distance), rotates through a small angle qz about
S, which is called the center of twist and its displacement in the horizontal plane is
zero. Since the angle of rotation qz is small, arc PP* is assumed to be a straight line
normal to SP. The x and y components of the displacement of P, then, are given by
uz= 0
ux=- r qz sinb = - y qz
(A1.24a)
uy= r qz cosb = x qz
224
t
P* tyz
txz
qz
(A1.24b)
uy = q'z x z
These displacements produce only shear strains in the bar so that by generalized
Hookes law the only non-zero stresses in the bar are the shear stresses, i.e.,
e z = 0, e x = 0, e y = 0
g xz = -q 'z y, g yz = q 'z x , g xy = 0
txz = Gg xz = -Gq'z y
(A1.25)
t yz = Gg yz = Gq'z x
From Fig. A1.18, these shear stresses can be used to relate the internal torque, Msv, to
the twist per unit length as follows:
225
Msv= Gq 'z x 2 + y 2 dA
A
Msv= Gq 'z J o
(A1.26)
J o = x 2 + y 2 dA = r 2 dA
A
(A1.27)
Figure A1.19. (a) Torsion of a circular bar with no out-of-plane warping (b) Torsion
of a non-circular bar with warping
226
P*
tzx
uy
y
Msv
qz
R
ux
P
y
b
x
(A1.28)
u y = q 'z x
where f(x,y) represents the warping function (describing the z displacement of each
section independent of position along the axis). These displacements produce strains
227
in the bar so that by generalized Hookes law the only non-zero stresses in the bar are
the shear stresses. Within the elastic limit, shear strain is proportional to the shear
t
stress g = .
G
e z = 0, e x = 0, e y = 0
f
f
g xz = - q'z - y , g yz = q'z + x , g xy = 0
x
y
t xz
= Gq - y
x
(A1.29)
'
z
t yz = Gq 'z + x
y
The internal torque Msv is a constant since the only loading is a pair of
twisting moments at the ends of the bar. From Fig. A1.20, these shear stresses can be
used to relate the internal torque, Msv, to the twist per unit length as follows:
Msv= (xt yz - yt xz )dA
(A1.30)
Substituting the shear stress components from equation (A1.29) into (A1.30), the
following expression is found:
M sv = GJqz
(A1.31)
228
rigid plane
warping
deformation
Mz
229
A
+
(+) :
Tensile stress
(-) :
Compressive stress
Mz
z
230
The first assumption depicts that the shape of the outline of a cross-section
remains unchanged under loading. It means that, under external loading, the profile
of a section may be translated or rotated from its initial position, but the relative
position of points on the profile will remain unchanged in x-y plane not along
longitudinal axis z as shown in Fig. A1.23.
Position after
loading
Unloaded crosssection
ds
qz
Tangent line
x
hs
R
y
Figure A1.23. Displacement of a section in non-uniform torsion
The transverse displacement us in the direction of the tangent to the contour line is
given by:
u s (z , s ) = q z h s
(A1.32)
where
u s ( z, s) : displacement of point A measured along curve s of the profile of a section
qz
of a beam. It states that the contour of the section remains perpendicular to the
231
longitudinal axis after deformation, meaning that shear deflections are equal to zero.
Hence , according to Vlasovs second assumption:
g=
u z (z, s ) u s (z, s )
+
=0
s
z
(A1.33)
(A1.34)
u z (z, s ) = - h s ds
0
dq z
dz
(A1.35)
where u z (z, s ) is the longitudinal displacement along the generator and qz is the
relative angle of torsion (also referred to as torsional warping). The product hs.ds is
equal to twice the area of the triangle whose base and height are equal to ds and hs,
respectively, and is usually given the symbol dws . Hence,
u z (z, s) = - dws
0
dq z
dz
(A1.36)
u z ( z , s) = -
dq z
ws
dz
(A1.37)
232
e(z, s ) =
u z (z, s )
z
(A1.38)
e(z, s ) = -q 'z' ws
(A1.39)
Considering the in plane rigidity of the section contour and considering the shearing
strain to be zero (i.e. the section will remain orthogonal after deformation), equation
(A1.39) can be obtained as for longitudinal deformation.
In this study, the warping stress expression is given by Vlasovs theory of
thin-walled beams of open section. Vlasovs first theorem is as follows:
Theorem 1 : The stress in a longitudinal fibre of a thin walled beam due to a
bimoment is equal to the product of this bimoment and the principal sectorial area
divided by the principal sectorial moment of inertia of the cross-section.
Vlasovs theory, which concerns beams consisting of thin plates, considers only the
normal stresses in the direction of the generator of the middle surface and the shear
stresses in the direction of the tangent to the contour line. Using the physical
relations between the stresses and strains in the beam, normal stresses s( z, s ) , and
shear stresses t(z, s ) can be found. Equation (A1.39) does not determine the strain
explicitly yet, since the function qz is still unknown. When the beam is deformed,
internal elastic forces arise in it. These forces represent normal and shear stresses in
233
the cross-section. According to Vlasovs theory, the shear stresses, which are
directed along the normal to the contour line, are assumed to vanish.
In Vlasovs theory it is assumed that the normal stresses are constant over the
thickness of the beam wall and that the shear stresses in the thickness direction vary
according to a linear law as shown in Fig. A1.24(c). The shear stresses lead to a
force, t t , per unit length of the cross-section acting along the tangent to the contour
as shown in Fig. A1.24(a). On the other hand, the total torsional moment Msv in the
whole cross-section can be assumed to be the sum of the torsional moments msv (z, s )
per unit length of the cross-section (along the contour line) as shown in Fig.
A1.24(b).
ds
ds
t2
t1
1
t1 t ds
msv ds
t1
t2
A
t1 = t(z, s )
(b) due to St. Venant twist
(M )
(t1+t2)
1
(t1-t2)
234
The state of stress in the cross-sectional plane can be expressed by the normal
stresses s( z, s ) , the average shear stresses t(z, s ) and the torsional moments. The
shear and normal stresses are considered as functions of the two variables z and s.
The torsional moment which depends on the difference of the shear stresses at the
extreme points of the wall are shown in Fig. A1.24(b). The torsional moment ( M sv )
in the whole cross-section is a function of the variable z only.
By using the physical relations between the stresses and strains in the beam, it
is possible to find the stresses (s , t ) and the moments Msv from the strains.
According to Hookes law
s( z , s ) = E e( z , s )
(A1.40)
Substituting equation (A1.39) into equation (A1.40), the normal stress is found as
follows:
s( z, s ) = - E q 'z' w s
(A1.41)
The difference between Vlasovs and St. Venants theories is the inclusion of
sectorial properties in the former, as it is seen from equation (A1.41). Multiplying
both sides of this equation by ws t and integrating over the whole contour line yields
B z = s(z, s ) ws t ds = - Eq
0
''
z
2
s
t ds
(A1.42)
The integral on the right was defined as the warping moment of inertia, I w , which
expresses the warping torsional resistance of the cross-section (see section A1.2.6).
Hence,
Bz = - Eq'z' Iw
(A1.43)
235
q'z' = -
Bz
EIw
(A1.44)
s(z, s ) =
Bz
ws
Iw
(A1.45)
236
s ( z, s ) +
s(z, s)
dz
z
t(z, s) +
t(z, s)
ds
s
dz
t (z , s)
ds
t( z, s)
s( z, s)
s(z, s)
t(z, s )
dz ds + t
dz ds = 0
z
s
(A1.46)
Hence,
s(z, s) t(z, s )
=0
+
s
z
(A1.47)
t(z, s) = Eq
'''
z
w ds
(A1.48)
237
1
t(z, s ) = Eq 'z'' ws dA
t
A
(A1.49)
The integral on the right side of (A1.49) was defined, in section A1.2.5.2, as the
sectorial or warping statical moment, Sw , of the cross-section.
Hence,
t(z, s ) = Eq'z''
Sw
t
(A1.50)
Equation (A1.50), infers the shear stresses in s direction which appear in the
restrained torsion of a thin-walled beam and are distributed uniformly across the wall
as shown in Fig. A1.24(a).
The flexural torsional moment (flexural twist) carried by membrane shear
stresses as shown in Fig. A1.24(a) can be obtained as follows:
(A1.51)
Substituting the stresses t(z, s ) , found as in equation (A1.48) into (A1.51), and
remembering the definition of sectorial warping moment of inertia (see equation
(A1.21)),
M w = - EI wqz
(A1.52)
q'z'' = -
Mw
EIw
(A1.53)
238
By substituting the value found in (A1.53) into equation (A1.50), the shear stress
component parallel to the z axis can be obtained expressing the mathematical
definition of Vlasovs second theorem
1 M
t(z, s ) = - w Sw
t Iw
(A1.54)
(A1.55)
The sum of the moments ( M sv ) and ( M w ) gives the total torsional moment, that is
denoted by ( M tot ) . Hence,
M tot = M sv + M w = - EI wq 'z'' + GJq 'z
(A1.56)
239
(A1.57)
(A1.58)
or
n
B z = Pk wk
(A1.59)
k =1
(A1.60)
240
The validity of this theorem becomes apparent from the definition of a bimoment as
a pair of parallel bending moments as shown in Fig.A1.26.
M
M
Bz=M.e
M
241
APPENDIX 2. Derivation of the Formulas for the Position of the Shear Center
It was mentioned in Appendix 1 that there are some properties of crosssections of thin-walled beams which are defined on the basis of sectorial area. For
the analysis of internal stresses in thin-walled beams, these properties must be known
to determine the location of the shear center through the use of the sectorial area. The
procedure for the determination of the location of the shear center formulas based on
the sectorial area is as follows:
The cross-section of a thin-walled beam is shown in Fig. A2.1. The
coordinate axes x-y pass through the center of gravity of the cross-section, but they
do not coincide with the principal axes of the cross-section.
Tangent line
S(ax,ay
)
R(bx,by)
dwS
2
dwR
2
f
R
hS
hR
242
(A2.1)
dw R = -ds h R
(A2.2)
(A2.3)
dwS = ds (SR - h R ) = ds SR + dw R
(A2.4)
and
The distance SR can be given by the coordinates of the points S and R (see Fig.
A2.2) as follows:
SR = R R + R S = (b y - a y )Cosf + (a x - b x )Sin f
(A2.5)
ax-bx
O
by-ay
dy
ds
f
R
dx
R
R
243
Cosf =
dx
ds
dy
ds
(A2.6)
dx
dy
+ (a x - b x )
ds
ds
(A2.7)
, Sin f =
Hence,
SR = ( b y - a y )
(A2.8)
(A2.9)
(A2.10)
S wx = y ws dA =0
(A2.11)
and
Hence, multiplying equation (A2.9) by x and y and substituting into (A2.10) and
(A2.11), respectively,
244
x dA + (a y - b y ) x 2 dA - (a x - b x ) xy dA + C x dA = 0
(A2.12)
y dA + (a y - b y ) xy dA - (a x - b x ) y 2 dA + C y dA = 0
(A2.13)
and
Since the axes x-y pass through the center of gravity of the cross-section,
x dA = y dA = 0
(A2.14)
Hence, the coordinates of the shear center are independent of the choice of the initial
radius of the system, wR .
With definitions
x dA = I
2
(A2.15)
(A2.16)
y dA = I
2
xy dA = I
(A2.17)
xy
x dA = SRwx
(A2.18)
y dA = SRwy
(A2.19)
245
(a
- b y )I y - (a x - b x ) I xy + SRwx = 0
(A2.20)
(a
- b y )I xy - (a x - b x ) I x + SRwy = 0
246
-x3
-x2
Contour line
S3
S2
S1
y2=y3
y1
ys
O1
yo
G
x
y4
S4
twice
the
area
of
the
hatched
247
triangle
RO1S4.
In
other
words
From S2 to S3, the angle O1 R S 2 decreases and the radius RO1 rotates in the
clockwise direction. Hence, this additional area (twice the area of the triangle S3RS2)
is negative. Therefore, wRO1S 3 = wRO 1S 2 - ( y3 - ys ) x 2 - x 3 .
The method of computation of the sectorial area diagram of a section can best
be explained by a numerical example. In Fig. A3.2, such an example is given for the
calculation of the sectorial area diagram and the resulting warping moment of inertia.
In Fig. A3.3 the contour of the section is given, with the origin at the centroid (to be
calculated) of the section.
9.25 cm
B
S ax,ay
tf
a
X
G(0,0)
19 cm
tw
tf
tf : 1.0 cm
tw : 0.5 cm
5.25 cm
248
9 cm
2.3043 cm
6.6957 cm
10.5652 cm
18 cm
7.4348 cm
3
C
2.3043 cm
2.6957 cm
5 cm
249
Thus,
Coordinates of node A : (6.696, 7.435)
Coordinates of node B : (-2.304, 7.435)
Coordinates of node C : (-2.304, -10.565)
Coordinates of node D : (2.696, -10.565)
Step:2 Calculate the values of IX, IY and IXY for the set of orthogonal axes X-Y
with the origin at the centre of gravity.
IX = 1321.8188 cm4
IY = 162.7237 cm4
IXY = 169.044 cm 4
IXY0, in other words, given axes X and Y do not coincide with the principal
axes of the section. Therefore, equations (A1.19) and (A1.20) can be used for the
determination of the shear center.
Generally, the direction of the principal axes as shown in Fig. A3.2, can be
determined from the well-known formula
tan 2a =
- 2I xy
Ix - Iy
a = -8.1305 o
Step:3 Choose any trial position for the pole (R) and an initial radius (RO1) for
the sectorial area.
The location of the trial pole and the initial radius can be chosen arbitrarily.
For simplicity, the trial pole is chosen at corner C and the initial radius along web
CB.
250
and
9
Step:5 Find the coordinates (ax,ay) of the principal pole (shear center) S from
expressions (A1.19) and (A1.20) since IXY0.
Substituting the values found from the previous step into expressions (A1.19)
and (A1.20) in Appendix 1, ax and ay can be obtained as
a x = -2.304 +
a y = -10.565 -
Step:6 Assume an arbitrary direction for the initial radius from the principal
pole (shear center) and find the sectorial area values.
For finding the initial sectorial area diagram, the initial radius is chosen from
the shear center S to point D (see Fig. A3.3)
251
-64.122
-29.898
initial radius
-70.986
-70.986
252
253
15.951
-18.273
S
O
Principal radius
-25.137 -
45.849
254
Iw = (45.849 - 14.197 s ) 1 ds
0
18
255
256
FIN
OMG
TT1
XAPP
YAPP
Not
: This example data file is given for a non-planar coupled shear wall with one
region. For the case of multi-region walls, region number (i) varies in the respective
range.
TITLE
BSAY
GSAY
KSAY
PSAY
PN
EN
: Total number of elements for a pier (equal to the number of the wall units)
XI
YJ
ELEM
CI
CJ
PT
: Shear modulus
: Elasticity modulus
C(i)
BKT(i)
HKAT(i)
KCB(i)
KSB(i)
: Rotational spring constant at the ends of the stiffening beam on the boundary i
HKIR(i)
HGKIR(i)
Z(i)
KSI
: Damping ratio
ETSUR
257
YGEN
DT
ASUR
: Duration of analysis
LTIP
JI (joint # )
IAP
FIN
OMG
TT1
XAPP
YAPP
Not: The coding of the elements and joints of the example structures in this thesis has
been shown on their cross-sectional views. The analyses have been carried out after
the preparation of the data files according to this coding.
List of Input Data File for Example 17 (for the stiffened case)
EXAMPLE 17
2.,2.,16.,2.
:************************ REGION 1 ****************************************************
:************** LOCATION OF THE FIRST PIER *****************************************
54
1 -1.50 0.00
2 -5.50 0.00
3 -5.50 4.00
4 -2.50 4.00
5 -2.50 3.00
1 1 2 0.4
2 2 3 0.4
3 3 4 0.4
4 4 5 0.4
:************** LOCATION OF THE SECOND PIER ****************************************
43
1 1.50 0.00
2 4.50 0.00
3 4.50 5.00
4 2.50 5.00
1 1 2 0.4
2 2 3 0.4
3 3 4 0.4
:************************ REGION 2 ****************************************************
:************** LOCATION OF THE FIRST PIER *****************************************
54
1 -1.50 0.00
2 -5.50 0.00
3 -5.50 4.00
4 -2.50 4.00
5 -2.50 3.00
1 1 2 0.4
2 2 3 0.4
3 3 4 0.4
4 4 5 0.4
:************** LOCATION OF THE SECOND PIER ****************************************
43
1 1.50 0.00
2 4.50 0.00
3 4.50 5.00
4 2.50 5.00
1 1 2 0.4
2 2 3 0.4
3 3 4 0.4
:******************************************************************************************
1055555.,2850000.
G, Elas
258
259
APPENDIX 5. List of the Computer Program Prepared in Fortran Language for the
Dynamic Analysis of Non-planar Coupled Shear Walls Using CCM
c *********************************************************************
c
c
DYNAMIC ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS
c
WITH STIFFENING BEAMS AND STEPWISE CROSS-SECTINAL CHANGES
c
c *********************************************************************
PARAMETER (N=65)
IMPLICIT REAL*8 (A-H,K-Z)
c
REAL*8
c(N),ec(N),BKTHICK(N),eBKTHICK(N),a(N),ea(N),b(N),eb(N)
,EJ(N),eEJ(N),EIx(N),eEIx(N),EIxy(N),eEIxy(N),EIy(N)
,eEIy(N),Delta(N),eDelta(N),EIxc(N),eEIxc(N),EIyc(N)
,eEIyc(N),K1(N),eK1(N),K2(N),eK2(N),K3(N),eK3(N),K4(N)
,eK4(N),ALAN(N),eALAN(N),d(N),ed(N),w(N),ew(N),r(N),er(N)
,EIw(N),eEIw(N),EIOw(N),eEIOw(N)
,EKS(N),EKSozel(N),dMx(N),dMy(N),dMt(N),dBt(N)
,TOPMOMX(N),TOPMOMY(N),STOPMOMX(N),STOPMOMY(N)
,TOPMOMT(N),STOPMOMT(N),EKSz(N),EKSozelz(N),STz(N)
,TOPMOMB(N),STOPMOMB(N),FLEX(N,N),STIFF(N,N)
,EEV(N),EKUT(N),MASSMAT(N,N),EGNVEC(N,N),EGNVAL(N)
,DUMM(N),DUMV(N,N),CFREQ(N),NFREQ(N),EKUTT(N)
@
@
REAL*8 AMAS1(N,N),AMAS2(N,N),STIF1(N,N),EGNVEC2(N,N),TEGNVEC(N,N)
,GSSTF1(N,N),GSSTF(N,N),GSMSS1(N,N),GSMSS(N,N),GSMS(N,N)
,GSSN(N,N),GSST(N,N),YVEK(N),EFFK(N,N)
REAL*8 yenX(N),XN(N),XNN(N),yenEX(N),EXNN(N),YN1(N),YN2(N)
,YN3(N),YN4(N),EYV(N),XGZ(N),GYVEK(N)
260
INTEGER ibsay,BSAY,GSAY,KSAY,PSAY,PN,EN
CHARACTER*200 CIKTI,GIRDI
WRITE(*,*)
'GIRDI DOSYASININ ADI'
READ(*,10)
GIRDI
WRITE(*,*)
'CIKTI DOSYASININ ADI'
READ(*,10)
CIKTI
OPEN(5,FILE=GIRDI,FORM='FORMATTED')
OPEN(6,FILE=CIKTI,FORM='FORMATTED')
10
20
22
23
25
FORMAT(A200)
FORMAT(A50)
FORMAT(1X,30F10.3)
FORMAT(1X,30F13.3)
FORMAT(1X,30F13.4)
READ(5,20) BASLIK
WRITE(6,20) BASLIK
READ(5,*)
BSAY,GSAY,KSAY,PSAY
WRITE(6,*) '
BSAY
GSAY
KSAY
PSAY
WRITE(6,*) '-----------------------------------------------WRITE(6,22) BSAY,GSAY,KSAY,PSAY
c
c
c
'
'
ibsay=BSAY+1
910
920
922
923
925
c
c
c
C
C
FORMAT(A8)
FORMAT(A50)
FORMAT(1X,30F10.3)
FORMAT(1X,30F10.2)
FORMAT(1X,30F13.4)
WRITE(6,*) '
WRITE(6,*) '************* SECTION PREPERTIES ****************
WRITE(6,*) '
PERDELERN GEOMETRK ZELLKLER DATA DOSYASINDAN OKUNARAK
W DYAGRAMI VE IW BURULMA ATALET HESAPLANACAK
DO 432 JJ=1,BSAY
READ(5,20) BILGI
TOPALAN=0.0
DO 431 II=1,PSAY
READ(5,20) BILGI
c
c
c
c
READ(5,*)
ePN,eEN
WRITE(6,*) '
'
WRITE(6,*) '
PN
EN
(Number of point on section)'
WRITE(6,*) '--------------------------'
WRITE(6,922) PN,EN
eePN=ePN+1
eeEN=eEN+1
c
c
c
c
WRITE(6,*) '
'
WRITE(6,*) '
POINT
X
Y
(point coordinates)'
WRITE(6,*) '------------------------------------'
DO I=1,ePN
READ(5,*)
P,eDX(P),eDY(P)
WRITE(6,922) P,eDX(P),eDY(P)
ENDDO
261
'
'
'
eDX(eePN)=0.0
eDY(eePN)=0.0
c
c
c
WRITE(6,*) '
'
WRITE(6,*) '
ELEMENT
I
J
THICKNESS
'
WRITE(6,*) '-----------------------------------------------------'
DO I=1,eEN
READ(5,*)
E,eEDI(E),eEDJ(E),eTHICK(E)
WRITE(6,923) E,eEDI(E),eEDJ(E),eTHICK(E)
ENDDO
eEDI(eeEN) = 1.0
eEDJ(eeEN) = eePN
eTHICK(eeEN)= 0.0
PNN(II)=eePN
ENN(II)=eeEN
CALL
c
c
c
SECPREP(II,eePN,eeEN,P,eDX,eDY,eTHICK,eEDI,eEDJ
,eCSAREA,EX,EY,sIx,sIy,sIxy,eJp,CX,CY)
WRITE(6,*) '
WRITE(6,*) '************ SECTORIAL PREPERTIES *****************
WRITE(6,*) '
'
'
'
eSx=CX(II)
eSy=CY(II)
eSCN=0.0
TOPALAN=TOPALAN+eCSAREA(II)
CALL CAIw(II,eCSAREA,eePN,eeEN,eSCN,eSx,eSy,P,eDX,eDY
,eEDI,eEDJ,eTHICK,ePRSECAREA,ePRSA,eSIW)
c
c
c
c
431
WRITE(6,*)
WRITE(6,*)
WRITE(6,*)
WRITE(6,*)
CONTINUE
'
'****************************************************
'****************************************************
'
eEXG(JJ)=(eCSAREA(1)*EX(1)+eCSAREA(2)*EX(2))/TOPALAN
eEYG(JJ)=(eCSAREA(1)*EY(1)+eCSAREA(2)*EY(2))/TOPALAN
c
c
c
c
c
c
c
c
c
c
c
c
c
c
ea(JJ)
= EX(2)-EX(1)
WRITE(6,*) '
'
WRITE(6,*) 'a=
'
WRITE(6,*) ea(JJ)
eb(JJ)
= EY(2)-EY(1)
WRITE(6,*) '
'
WRITE(6,*) 'b=
'
WRITE(6,*) eb(JJ)
eEJ(JJ)
= EJp(1)+EJp(2)
WRITE(6,*) '
'
WRITE(6,*) 'J=
'
WRITE(6,*) eEJ(JJ)
eEIx(JJ)
= SIx(1)+SIx(2)
WRITE(6,*) '
'
WRITE(6,*) 'Ix=
'
WRITE(6,*) eEIx(JJ)
eEIxy(JJ) = SIxy(1)+SIxy(2)
WRITE(6,*) '
'
WRITE(6,*) 'Ixy=
'
262
'
'
'
'
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
WRITE(6,*) eEIxy(JJ)
eEIy(JJ)
= SIy(1)+SIy(2)
WRITE(6,*) '
'
WRITE(6,*) 'Iy=
'
WRITE(6,*) eEIy(JJ)
eDelta(JJ) = (eEIx(JJ)*eEIy(JJ) - eEIxy(JJ)**2)
WRITE(6,*) '
'
WRITE(6,*) 'Delta=
'
WRITE(6,*) eDelta(JJ)
eEIxc(JJ) = CX(1)*SIx(1)+CX(2)*SIx(2)-CY(1)*SIxy(1)-CY(2)*SIxy(2)
WRITE(6,*) '
'
WRITE(6,*) 'Ixc=
'
WRITE(6,*) eEIxc(JJ)
eEIyc(JJ) = CY(1)*SIy(1)+CY(2)*SIy(2)-CX(1)*SIxy(1)-CX(2)*SIxy(2)
WRITE(6,*) '
'
WRITE(6,*) 'Iyc=
'
WRITE(6,*) eEIyc(JJ)
eK1(JJ)
= (eEIxc(JJ)*eEIxy(JJ) + eEIx(JJ)*eEIyc(JJ))/eDelta(JJ)
WRITE(6,*) '
'
WRITE(6,*) 'K1=
'
WRITE(6,*) eK1(JJ)
eK2(JJ)
= (eEIxc(JJ)*eEIy(JJ) + eEIxy(JJ)*eEIyc(JJ))/eDelta(JJ)
WRITE(6,*) '
'
WRITE(6,*) 'K2=
'
WRITE(6,*) eK2(JJ)
eK3(JJ)
=
(ea(JJ)*eEIx(JJ) - eb(JJ)*eEIxy(JJ))/eDelta(JJ)
WRITE(6,*) '
'
WRITE(6,*) 'K3=
'
WRITE(6,*) eK3(JJ)
eK4(JJ)
=
(eb(JJ)*eEIy(JJ) - ea(JJ)*eEIxy(JJ))/eDelta(JJ)
WRITE(6,*) '
'
WRITE(6,*) 'K4=
'
WRITE(6,*) eK4(JJ)
eALAN(JJ) = 1/((1/eCSAREA(1))+(1/eCSAREA(2))
@
+ea(JJ)*eK3(JJ)+eb(JJ)*eK4(JJ))
WRITE(6,*) '
'
WRITE(6,*) 'Alan=
'
WRITE(6,*) eALAN(JJ)
ed(JJ)
= CX(2)*EY(2)-CY(2)*EX(2)+CY(1)*EX(1)-CX(1)*EY(1)
WRITE(6,*) '
'
WRITE(6,*) 'd=
'
WRITE(6,*) ed(JJ)
ew(JJ)
= ePRSA(1,PNN(1))-ePRSA(2,PNN(2))
WRITE(6,*) '
'
WRITE(6,*) 'w=
'
WRITE(6,*) ew(JJ)
er(JJ)
= ew(JJ) + ed(JJ) + ea(JJ)*eK1(JJ) - eb(JJ)*eK2(JJ)
WRITE(6,*) '
'
WRITE(6,*) 'r=
'
WRITE(6,*) er(JJ)
eEIw(JJ)
= eSIw(1)+eSIw(2)+CX(1)**2*SIx(1)+CX(2)**2*SIx(2)
@
+CY(1)**2*SIy(1)+CY(2)**2*SIy(2)
@
-2*CX(1)*CY(1)*SIxy(1)
@
-2*CX(2)*CY(2)*SIxy(2)
WRITE(6,*) '
'
WRITE(6,*) 'Iw=
'
WRITE(6,*) eEIw(JJ)
eEIOw(JJ) = eEIw(JJ) - eEIyc(JJ)*eK1(JJ) - eEIxc(JJ)*eK2(JJ)
WRITE(6,*) '
'
WRITE(6,*) 'IOw=
'
WRITE(6,*) eEIOw(JJ)
eTALAN(JJ)=TOPALAN
432
CONTINUE
263
READ(5,20)
c
c
c
c
BILGI
READ(5,*)
G,Elas
WRITE(6,*) '
WRITE(6,*) '
G
Elas
WRITE(6,*) '------------------------WRITE(6,23) G,Elas
'
'
'
c
c
c
c
c
READ(5,*)
Met,Bet
WRITE(6,*) '
WRITE(6,*) '
Met
Bet
WRITE(6,*) '---------------------------------WRITE(6,23) Met,Bet
c
c
c
c
c
READ(5,*)
Px,Wx,dy
WRITE(6,*) '
WRITE(6,*) '
Px
Wx
dy
WRITE(6,*) '------------------------------WRITE(6,22) Px,Wx,dy
'
'
'
c
c
c
c
c
READ(5,*)
Py,Wy,dx
WRITE(6,*) '
WRITE(6,*) '
Py
Wy
dx
WRITE(6,*) '------------------------------WRITE(6,22) Py,Wy,dx
'
'
'
READ(5,20)
c
c
c
c
c
c
c
c
c
c
c
c
c
c
'
'
'
BILGI
WRITE(6,*) '
c
BAG.KR. KALINLII
WRITE(6,*) '------------------------------------------DO I=1,bsay
READ(5,*)
ec(I),eBKTHICK(I)
WRITE(6,22) ec(I),eBKTHICK(I)
ENDDO
ec(ibsay)=ec(bsay)
eBKTHICK(ibsay)=eBKTHICK(bsay)
'
'
WRITE(6,*) '
WRITE(6,*) '-----Kat Ykseklii-----WRITE(6,*) '------------------------DO 1 I=1,bsay
READ(5,*)
eHKAT(I)
WRITE(6,22) eHKAT(I)
1 CONTINUE
eHKAT(ibsay)=eHKAT(bsay)
'
'
'
WRITE(6,*) '
WRITE(6,*) '-----Balant Rij Katsays-Ccb----WRITE(6,*) '-----------------------------------DO 2 I=1,bsay
READ(5,*) eCCB(I)
WRITE(6,22) eCCB(I)
2 CONTINUE
eCCB(ibsay)=eCCB(bsay)
'
'
'
WRITE(6,*) '
WRITE(6,*) '-----Balant Rij Katsays-Csb----WRITE(6,*) '-----------------------------------DO 3 I=1,bsay
READ(5,*) eCSB(I)
'
'
'
264
WRITE(6,22) eCSB(I)
3 CONTINUE
eCSB(ibsay)=eCSB(bsay)
c
c
c
WRITE(6,*) '
WRITE(6,*) '-----Kiri Ykseklii-----WRITE(6,*) '--------------------------DO 4 I=1,bsay
READ(5,*)
eHKIR(I)
WRITE(6,22) eHKIR(I)
4 CONTINUE
eHKIR(ibsay)=eHKIR(bsay)
READ(5,20)
c
c
c
c
c
c
c
c
c
c
c
BILGI
WRITE(6,*) '
WRITE(6,*) '-G Kiri Ykseklii----Yerden Mesafe---WRITE(6,*) '-----------------------------------------DO 5 I=1,ibsay
READ(5,*)
eHGKIR(I),eZ(I)
WRITE(6,22) eHGKIR(I),eZ(I)
5 CONTINUE
READ(5,*)
HP
WRITE(6,*) ' HP (Ykn uyguland ykseklik) '
WRITE(6,*) '-----------'
WRITE(6,22) HP
H
= eZ(1)
WRITE(6,*) '
WRITE(6,*) 'H=
WRITE(6,*) H
'
'
ea(ibsay)=ea(bsay)
eb(ibsay)=eb(bsay)
eEJ(ibsay)=eEJ(bsay)
eEIx(ibsay)=eEIx(bsay)
eEIxy(ibsay)=eEIxy(bsay)
eEIy(ibsay)=eEIy(bsay)
eDelta(ibsay)=eDelta(bsay)
eEIxc(ibsay)=eEIxc(bsay)
eEIyc(ibsay)=eEIyc(bsay)
eK1(ibsay)=eK1(bsay)
eK2(ibsay)=eK2(bsay)
eK3(ibsay)=eK3(bsay)
eK4(ibsay)=eK4(bsay)
eALAN(ibsay)=eALAN(bsay)
ed(ibsay)=ed(bsay)
ew(ibsay)=ew(bsay)
er(ibsay)=er(bsay)
eEIw(ibsay)=eEIw(bsay)
eEIOw(ibsay)=eEIOw(bsay)
'
'
'
265
'
'
'
DO 6 I=1,ibsay
eAc(I)=eBKTHICK(I)*eHKIR(I)
eAs(I)=eBKTHICK(I)*eHGKIR(I)
eEIc(I)=(eBKTHICK(I)*eHKIR(I)**3.)/12.
eEIs(I)=(eBKTHICK(I)*eHGKIR(I)**3.)/12.
eGAMA(I)=Elas*((ec(I)**2*ehkat(I))/(2.*eCcb(I))+
(ec(I)*ehkat(I)*1.2)/(eAc(I)*G)+
(ec(I)**3*ehkat(I))/(12.*Elas*eEIc(I)))
eGAMAstf(I)=Elas*((ec(I)**2)/(2.*eCsb(I))+
(ec(I)*1.2)/(eAs(I)*G)+
(ec(I)**3)/(12.*Elas*eEIs(I)))
eS(I)=eGAMA(I)/eGAMAstf(I)
eBETA1(I) = eEIOw(I)*eGAMA(I)
eBETA2(I) = eEIOw(I)/eALAN(I)+(eEJ(I)*G*eGAMA(I))/Elas+er(I)**2
eBETA3(I) = (eEJ(I)*G)/(Elas*eALAN(I))
eALFA1(I)=Sqrt((eBETA2(I)-Sqrt(eBETA2(I)**2
-4*eBETA1(I)*eBETA3(I)))/eBETA1(I))/Sqrt(2.)
eALFA2(I)=Sqrt((eBETA2(I)+Sqrt(eBETA2(I)**2
*
-4*eBETA1(I)*eBETA3(I)))/eBETA1(I))/Sqrt(2.)
*
6 CONTINUE
TOL=0.00001
iksay=KSAY+1
DO I=1,ibsay
IF(eZ(I).EQ.H) THEN
KAT(iksay)=H
GOTO 160
ENDIF
DO 170 in=1,iksay
KAT(iksay-in)=KAT(iksay-in+1)-eHKAT(I-1)
IF(KAT(iksay-in).LT.TOL) THEN
KAT(iksay-in)=0.d0
GOTO 160
ENDIF
IF(ABS(eZ(I)-KAT(iksay-in)).LT.TOL) THEN
iksay=iksay-in
GOTO 160
ENDIF
170 CONTINUE
160 ENDDO
iksay=KSAY+1
DO 1000 JJ=iksay,2,-1
HP=KAT(jj)
C
Px=0
Py=0
Met=0
dx=eEXG(1)
dy=eEYG(1)
266
DO 800 IH=1,3
IF(IH.EQ.1) THEN
Px=1
Py=0
Met=0
dy=0
ENDIF
IF(IH.EQ.2) THEN
Px=0
Py=1
Met=0
dy=0
ENDIF
IF(IH.EQ.3) THEN
Px=0
Py=0
Met=1
dy=0
ENDIF
C
DO I=1,60
a(I)=0.d0
b(I)=0.d0
EJ(I)=0.d0
EIx(I)=0.d0
EIxy(I)=0.d0
EIy(I)=0.d0
DELTA(I)=0.d0
EIxc(I)=0.d0
EIyc(I)=0.d0
K1(I)=0.d0
K2(I)=0.d0
K3(I)=0.d0
K4(I)=0.d0
ALAN(I)=0.d0
d(I)=0.d0
w(I)=0.d0
r(I)=0.d0
EIw(I)=0.d0
EIOw(I)=0.d0
c(I)=0.d0
BKTHICK(I)=0.d0
HKAT(I)=0.d0
CCB(I)=0.d0
HKIR(I)=0.d0
HGKIR(I)=0.d0
Ac(I)=0.d0
EIc(I)=0.d0
TALAN(I)=0.d0
GAMA(I)=0.d0
BETA1(I)=0.d0
BETA2(I)=0.d0
BETA3(I)=0.d0
ALFA1(I)=0.d0
ALFA2(I)=0.d0
Z(I)=0.d0
As(I)=0.d0
CSB(I)=0.d0
EIs(I)=0.d0
GAMAstf(I)=0.d0
267
S(I)=0.d0
=0.d0
=0.d0
=0.d0
=0.d0
My(I)
=0.d0
Myz(I)
=0.d0
Myzz(I) =0.d0
Myzzz(I) =0.d0
Mt(I)
=0.d0
Mtz(I)
=0.d0
Mtzz(I) =0.d0
Bt(I)
=0.d0
C1(I)=0.d0
C2(I)=0.d0
DD1(I)=0.d0
DD2(I)=0.d0
SDD1(I)=0.d0
SDD2(I)=0.d0
Tozel(I)=0.d0
Tozelz(I)=0.d0
Tozelzz(I)=0.d0
Tozelzzz(I)=0.d0
Mx(I)
Mxz(I)
Mxzz(I)
Mxzzz(I)
STozel(I)=0.d0
STozelz(I)=0.d0
STozelzz(I)=0.d0
STozelzzz(I)=0.d0
EKSozel(I)=0.d0
EKSozelz(I)=0.d0
EKSozel(I)=0.d0
EKSozelz(I)=0.d0
EKS(I)=0.d0
EKSz(I)=0.d0
ST(I)=0.d0
STz(I)=0.d0
STOPMOMX(I)=0.d0
STOPMOMY(I)=0.d0
STOPMOMT(I)=0.d0
STOPMOMB(I)=0.d0
dMx(I) =0.d0
dMy(I) =0.d0
dMt(I) =0.d0
dBt(I) =0.d0
TOPMOMX(I)=0.d0
TOPMOMY(I)=0.d0
TOPMOMT(I)=0.d0
TOPMOMB(I)=0.d0
ENDDO
C
IF(JJ.EQ.(iksay-1))THEN
Program Kontrol Satiri
iksay=KSAY+1
ENDIF
ibadd=0
ibsay2=bsay+1
do ii=1,ibsay2
if(ABS(eZ(ii)-HP).lt.TOL) THEN
ibsay=bsay+1
DO I=1,ibsay
a(I)=ea(I)
b(I)=eb(I)
EJ(I)=eEJ(I)
EIx(I)=eEIx(I)
EIxy(I)=eEIxy(I)
EIy(I)=eEIy(I)
268
DELTA(I)=eDELTA(I)
EIxc(I)=eEIxc(I)
EIyc(I)=eEIyc(I)
K1(I)=eK1(I)
K2(I)=eK2(I)
K3(I)=eK3(I)
K4(I)=eK4(I)
ALAN(I)=eALAN(I)
d(I)=ed(I)
w(I)=ew(I)
r(I)=er(I)
EIw(I)=eEIw(I)
EIOw(I)=eEIOw(I)
c(I)=ec(I)
BKTHICK(I)=eBKTHICK(I)
HKAT(I)=eHKAT(I)
CCB(I)=eCCB(I)
HKIR(I)=eHKIR(I)
HGKIR(I)=eHGKIR(I)
Ac(I)=eAc(I)
TALAN(I)=eTALAN(I)
EIc(I)=eEIc(I)
GAMA(I)=eGAMA(I)
BETA1(I)=eBETA1(I)
BETA2(I)=eBETA2(I)
BETA3(I)=eBETA3(I)
ALFA1(I)=eALFA1(I)
ALFA2(I)=eALFA2(I)
Z(I)=eZ(I)
As(I)=eAs(I)
CSB(I)=eCSB(I)
EIs(I)=eEIs(I)
GAMAstf(I)=eGAMAstf(I)
S(I)=eS(I)
ENDDO
endif
if((eZ(ii)-HP).GT.TOL.and.(HP-eZ(ii+1)).GT.TOL) THEN
ibsay=bsay+2
ibadd=ii
DO I=1,ibsay
IF(I.LE.ibadd) THEN
a(I)=ea(I)
b(I)=eb(I)
EJ(I)=eEJ(I)
EIx(I)=eEIx(I)
EIxy(I)=eEIxy(I)
EIy(I)=eEIy(I)
DELTA(I)=eDELTA(I)
EIxc(I)=eEIxc(I)
EIyc(I)=eEIyc(I)
K1(I)=eK1(I)
K2(I)=eK2(I)
K3(I)=eK3(I)
K4(I)=eK4(I)
ALAN(I)=eALAN(I)
d(I)=ed(I)
w(I)=ew(I)
r(I)=er(I)
EIw(I)=eEIw(I)
EIOw(I)=eEIOw(I)
c(I)=ec(I)
BKTHICK(I)=eBKTHICK(I)
HKAT(I)=eHKAT(I)
CCB(I)=eCCB(I)
HKIR(I)=eHKIR(I)
269
HGKIR(I)=eHGKIR(I)
Ac(I)=eAc(I)
TALAN(I)=eTALAN(I)
EIc(I)=eEIc(I)
GAMA(I)=eGAMA(I)
BETA1(I)=eBETA1(I)
BETA2(I)=eBETA2(I)
BETA3(I)=eBETA3(I)
ALFA1(I)=eALFA1(I)
ALFA2(I)=eALFA2(I)
ELSE
a(I)=ea(I-1)
b(I)=eb(I-1)
EJ(I)=eEJ(I-1)
EIx(I)=eEIx(I-1)
EIxy(I)=eEIxy(I-1)
EIy(I)=eEIy(I-1)
DELTA(I)=eDELTA(I-1)
EIxc(I)=eEIxc(I-1)
EIyc(I)=eEIyc(I-1)
K1(I)=eK1(I-1)
K2(I)=eK2(I-1)
K3(I)=eK3(I-1)
K4(I)=eK4(I-1)
ALAN(I)=eALAN(I-1)
d(I)=ed(I-1)
w(I)=ew(I-1)
r(I)=er(I-1)
EIw(I)=eEIw(I-1)
EIOw(I)=eEIOw(I-1)
c(I)=ec(I-1)
BKTHICK(I)=eBKTHICK(I-1)
HKAT(I)=eHKAT(I-1)
CCB(I)=eCCB(I-1)
HKIR(I)=eHKIR(I-1)
HGKIR(I)=eHGKIR(I-1)
Ac(I)=eAc(I-1)
TALAN(I)=eTALAN(I-1)
EIc(I)=eEIc(I-1)
GAMA(I)=eGAMA(I-1)
BETA1(I)=eBETA1(I-1)
BETA2(I)=eBETA2(I-1)
BETA3(I)=eBETA3(I-1)
ALFA1(I)=eALFA1(I-1)
ALFA2(I)=eALFA2(I-1)
ENDIF
IF(I.LE.ibadd)THEN
Z(I)=eZ(I)
As(I)=eAs(I)
CSB(I)=eCSB(I)
EIs(I)=eEIs(I)
GAMAstf(I)=eGAMAstf(I)
S(I)=eS(I)
ENDIF
IF(I.EQ.(ibadd+1))THEN
Z(I)=HP
As(I)=0
CSB(I)=0
EIs(I)=0
GAMAstf(I)=Elas*((c(I)**2)/(2.*Csb(I))+
(c(I)*1.2)/(As(I)*G)+
(c(I)**3)/(12.*Elas*EIs(I)))
S(I)=GAMA(I)/GAMAstf(I)
ENDIF
IF(I.GT.(ibadd+1))THEN
Z(I)=eZ(I-1)
270
As(I)=eAs(I-1)
CSB(I)=eCSB(I-1)
EIs(I)=eEIs(I-1)
GAMAstf(I)=eGAMAstf(I-1)
S(I)=eS(I-1)
ENDIF
ENDDO
endif
enddo
DO I=1,ibsay
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
WRITE(6,*) a(I)
WRITE(6,*) b(I)
WRITE(6,*) EJ(I)
WRITE(6,*) EIx(I)
WRITE(6,*) EIxy(I)
WRITE(6,*) EIy(I)
WRITE(6,*) Delta(I)
WRITE(6,*) EIxc(I)
WRITE(6,*) EIyc(I)
WRITE(6,*) K1(I)
WRITE(6,*) K2(I)
WRITE(6,*) K3(I)
WRITE(6,*) K4(I)
WRITE(6,*) ALAN(I)
WRITE(6,*) d(I)
WRITE(6,*) w(I)
WRITE(6,*) r(I)
WRITE(6,*) EIw(I)
WRITE(6,*) EIOw(I)
WRITE(6,*) CCB(I)
WRITE(6,*) '-----------------------------------------WRITE(6,22) c(I),BKTHICK(I)
ENDDO
ETOL=-0.00001
DO I=1,ibsay
My(I)
Myz(I)
Myzz(I)
Myzzz(I)
Mt(I)
Mtz(I)
Mtzz(I)
=-(dy*Px)+dx*Py-dy*Wx*(HP-Z(I))+dx*Wy*(HP-Z(I))+Met
=dy*Wx - dx*Wy
=0
ELSE
271
'
Mx(I)
Mxz(I)
Mxzz(I)
Mxzzz(I)
=0
=0
=0
=0
My(I)
Myz(I)
Myzz(I)
Myzzz(I)
=0
=0
=0
=0
Mt(I)
Mtz(I)
Mtzz(I)
=0
=0
=0
ENDIF
Bt(I)
= Bet
WRITE(6,*) ' My(',I,') My(',I,')
WRITE(6,22) Mx(I),My(I),Mt(I)
Mt(',I,')
'
ENDDO
DO 7 I=1,ibsay
C1(I)=
-
d(I)+EIyc(I)*K3(I)-EIxc(I)*K4(I)+EIw(I)/
(Alan(I)*r(I))-(EIyc(I)*K1(I))/(Alan(I)*r(I))(EIxc(I)*K2(I))/(Alan(I)*r(I))+w(I)
C2(I)= -(EIw(I)/r(I))*GAMA(I)+(GAMA(I)*EIyc(I)*K1(I))/r(I)
+(GAMA(I)*EIxc(I)*K2(I))/r(I)
DD1(I)=
DD2(I)=
-
((EIxy(I)*EIyc(I)*Mxz(I)+EIxc(I)*EIy(I)*Mxz(I)
-EIxc(I)*EIxy(I)*Myz(I)-EIx(I)*EIyc(I)*Myz(I))/Delta(I)((EIw(I)-EIyc(I)*K1(I)-EIxc(I)*K2(I))*(K4(I)*Mxz(I)
+K3(I)*Myz(I)))/r(I))+Mt(I)
(EIxy(I)*EIyc(I)*Mx(I)+EIxc(I)*EIy(I)*Mx(I)
-EIxc(I)*EIxy(I)*My(I)-EIx(I)*EIyc(I)*My(I))/Delta(I)((EIw(I)-EIyc(I)*K1(I)-EIxc(I)*K2(I))*(K4(I)*Mx(I)
+K3(I)*My(I)))/r(I)
SDD1(I)= ((EIxy(I)*EIyc(I)*Mxz(I+1)+EIxc(I)*EIy(I)*Mxz(I+1)
-EIxc(I)*EIxy(I)*Myz(I+1)-EIx(I)*EIyc(I)*Myz(I+1))/Delta(I)((EIw(I)-EIyc(I)*K1(I)-EIxc(I)*K2(I))*(K4(I)*Mxz(I+1)
+K3(I)*Myz(I+1)))/r(I))+Mt(I+1)
SDD2(I)= (EIxy(I)*EIyc(I)*Mx(I+1)+EIxc(I)*EIy(I)*Mx(I+1)
-EIxc(I)*EIxy(I)*My(I+1)-EIx(I)*EIyc(I)*My(I+1))/Delta(I)((EIw(I)-EIyc(I)*K1(I)-EIxc(I)*K2(I))*(K4(I)*Mx(I+1)
+K3(I)*My(I+1)))/r(I)
7 CONTINUE
-
C
c
WRITE(6,*)'
WRITE(6,*)'
Tozel(I)
DO 55 I=1,ibsay
-
Tozelz(I)
Tozelzz(I)
'
Tozelzzz(I)'
Tozel(I)=(G*EJ(I)*K4(I)*Mx(I)-Elas*EIOw(I)*K4(I)*Mxzz(I)+
G*EJ(I)*K3(I)*My(I)-Elas*EIOw(I)*K3(I)*Myzz(I)+
Elas*Mtz(I)*r(I)+Elas*K2(I)*Mxzz(I)*r(I)-
272
Elas*K1(I)*Myzz(I)*r(I))/(BETA3(I)*Elas)
Tozelz(I)=(G*EJ(I)*K4(I)*Mxz(I)-Elas*EIOw(I)*K4(I)*Mxzzz(I)+
G*EJ(I)*K3(I)*Myz(I)-Elas*EIOw(I)*K3(I)*Myzzz(I)+
Elas*Mtzz(I)*r(I)+Elas*K2(I)*Mxzzz(I)*r(I)Elas*K1(I)*Myzzz(I)*r(I))/(BETA3(I)*Elas)
Tozelzz(I)=(G*EJ(I)*K4(I)*Mxzz(I)+G*EJ(I)*K3(I)*Myzz(I))/
(BETA3(I)*Elas)
Tozelzzz(I)= 0.
C
WRITE(6,*) Tozel(I),Tozelz(I),Tozelzz(I),Tozelzzz(I)
CONTINUE
55
C
C
WRITE(6,*)'
WRITE(6,*)'
STozel(I)
DO 56 I=1,ibsay
STozelz(I)
STozelzz(I)
'
STozelzzz(I)'
STozel(I)=(G*EJ(I)*K4(I)*Mx(I+1)-Elas*EIOw(I)*K4(I)*Mxzz(I+1)+
G*EJ(I)*K3(I)*My(I+1)-Elas*EIOw(I)*K3(I)*Myzz(I+1)+
Elas*Mtz(I+1)*r(I)+Elas*K2(I)*Mxzz(I+1)*r(I)Elas*K1(I)*Myzz(I+1)*r(I))/(BETA3(I)*Elas)
STozelz(I)=(G*EJ(I)*K4(I)*Mxz(I+1)-Elas*EIOw(I)*K4(I)*Mxzzz(I+1)+
G*EJ(I)*K3(I)*Myz(I+1)-Elas*EIOw(I)*K3(I)*Myzzz(I+1)+
Elas*Mtzz(I+1)*r(I)+Elas*K2(I)*Mxzzz(I+1)*r(I)Elas*K1(I)*Myzzz(I+1)*r(I))/(BETA3(I)*Elas)
STozelzz(I)=(G*EJ(I)*K4(I)*Mxzz(I+1)+G*EJ(I)*K3(I)*Myzz(I+1))/
(BETA3(I)*Elas)
STozelzzz(I)= 0.
C
56
WRITE(6,*) STozel(I),STozelz(I),STozelzz(I),STozelzzz(I)
CONTINUE
TOL=0.00001
273
P2(I)=0.d0
V(I)=0.d0
G1(I)=0.d0
G2(I)=0.d0
TETA(I)=0.d0
ENDDO
IF(JJ.EQ.(iksay-1))THEN
Program Kontrol Satiri
iksay=KSAY+1
ENDIF
IL=1
IK=0
DO I=1,ibsay
IF(I.EQ.1) THEN
@
@
@
@
CALL COEF1A(Tozel(I),Tozelz(I),ALFA1(I),ALFA2(I)
,Z(I),S(I),REK1,REKB1)
CALL COEF1B(Tozel(I),Tozelzz(I),ALFA1(I),ALFA2(I)
,GAMA(I),Z(I),EIw(I),ALAN(I),r(I),d(I),w(I)
,EIxc(I),EIyc(I),K1(I),K2(I),K3(I),K4(I),Bet
,REK2,REKB2)
ENDIF
IF((I.GT.1).AND.(I.LT.ibsay)) THEN
CALL COEF2A(ibsay,ALFA1,ALFA2,Z(I),
Tozelz,STozelz,GAMA,I,REK1,REKB1)
CALL COEF2B(ibsay,ALFA1,ALFA2,Z(I),S(I),
Tozel,Tozelz,STozel,I,REK2,REKB2)
CALL COEF2C(ibsay,Tozelz,Tozelzzz,STozelz,STozelzzz,
ALFA1,ALFA2,Z(I),C1,C2,DD1,SDD1,I,REK3,REKB3)
CALL COEF2D(ibsay,Tozel,Tozelzz,STozel,STozelzz,
ALFA1,ALFA2,Z(I),C1,C2,DD2,SDD2,I,REK4,REKB4)
ENDIF
IF(I.EQ.ibsay) THEN
@
@
@
@
CALL COEF3A(Tozelz(I),ALFA1(I),ALFA2(I),Z(I),REK1,REKB1)
CALL COEF3B(Tozelz(I),Tozelzzz(I),ALFA1(I),ALFA2(I),
Z(I),GAMA(I),Mxz(I),Myz(I),Mt(I),
EIw(I),ALAN(I),r(I),w(I),EIyc(I),EIxc(I),EIx(I),
EIy(I),EIxy(I),K1(I),K2(I),K3(I),K4(I),d(I),
Delta(I),REK2,REKB2)
ENDIF
IF(I.EQ.1) THEN
DO J=1,4
ZK(I,J)=ZK(I,J)+REK1(J)
ZB(I)=REKB1
ZK(I+1,J)=ZK(I+1,J)+REK2(J)
ZB(I+1)=REKB2
ENDDO
ENDIF
IF((I.GT.1).AND.(I.LT.ibsay)) THEN
274
DO J=1,8
ZK(I+IL-3,J+IK-4)=ZK(I+IL-3,J+IK-4)+REK1(J)
ZB(I+IL-3)=REKB1
ZK(I+IL-2,J+IK-4)=ZK(I+IL-2,J+IK-4)+REK2(J)
ZB(I+IL-2)=REKB2
ZK(I+IL-1,J+IK-4)=ZK(I+IL-1,J+IK-4)+REK3(J)
ZB(I+IL-1)=REKB3
ZK(I+IL,J+IK-4)=ZK(I+IL,J+IK-4)+REK4(J)
ZB(I+IL)=REKB4
ENDDO
ENDIF
IK=IK+4
IF(I.EQ.ibsay) THEN
DO J=1,4
ZK(I+IL-3,J+IK-8)=ZK(I+IL-3,J+IK-8)+REK1(J)
ZB(I+IL-3)=REKB1
ZK(I+IL-2,J+IK-8)=ZK(I+IL-2,J+IK-8)+REK2(J)
ZB(I+IL-2)=REKB2
ENDDO
ENDIF
IL=IL+3
do j=1,8
REK1(j)=0.d0
REK2(j)=0.d0
REK3(j)=0.d0
REK4(j)=0.d0
enddo
REKB1=0
REKB2=0
REKB3=0
REKB4=0
ENDDO
in=4*(ibsay-1)
C
C
C
C
C
C
C
C
C
C
C
WRITE(6,*) '
'
write(6,*) 'Coefficient Matrix'
DO I =1,n
WRITE(6,22) (zk(I,J),J=1,N)
enddo
77 FORMAT(1X,40F22.10)
WRITE(6,*) '
'
write(6,*) 'Right Hand Side Vector'
WRITE(6,77) (zb(J),J=1,N)
CALL GAUSS(ZK,in,ZB,SONUC1)
WRITE(6,*) '
'
write(6,*) 'Result Vector'
WRITE(6,77) (SONUC1(J),J=1,N)
im=0
DO I=1,ibsay
if(I.EQ.ibsay)THEN
im=im-4
endif
D1(I)=D1(I)+SONUC1(I+im)
D2(I)=D2(I)+SONUC1(I+1+im)
D3(I)=D3(I)+SONUC1(I+2+im)
D4(I)=D4(I)+SONUC1(I+3+im)
im=im+3
ENDDO
IF(JJ.EQ.(iksay-1))THEN
Program Kontrol Satiri
iksay=KSAY+1
ENDIF
275
WRITE(6,*)'
WRITE(6,*)'
WRITE(6,*)'
c
'
'
'
c
c
c
c
c
c
c
write(6,*) '
'
write(6,*) '--KAT SEViYELERi--'
write(6,*) '
'
do in=1,iksay
im=iksay-in+1
write(6,*) KAT(im)
enddo
WRITE(6,*)'
WRITE(6,*)'
WRITE(6,*)'
'
'
'
c**************************************
iksay=KSAY+1
im=0
DO I=1,ibsay
DO 911 in=1,iksay
im=iksay-in+1
IF((HP-KAT(im)).GT.ETOL) THEN
EKSozel(im)=(2*BETA2(I)*G*EJ(I)*(K3(I)*Wx + K4(I)*Wy)+
BETA3(I)*(-2*Elas*(EIOw(I)*K3(I)*Wx-dy*r(I)*Wx+
K1(I)*r(I)*Wx+EIOw(I)*K4(I)*Wy+dx*r(I)*WyK2(I)*r(I)*Wy)+G*EJ(I)*(HP-KAT(im))*(2*K3(I)*Px+
2*K4(I)*Py+HP*K3(I)*Wx+HP*K4(I)*Wy-(K3(I)*Wx+
K4(I)*Wy)*KAT(im))))/(2.*BETA3(I)**2*Elas)
EKSozelz(im)= (G*EJ(I)*(-(K3(I)*Wx)-K4(I)*Wy)*(HP-KAT(im))G*EJ(I)*(2*K3(I)*Px+2*K4(I)*Py+HP*K3(I)*Wx+HP*K4(I)*Wy(K3(I)*Wx+K4(I)*Wy)*KAT(im)))/(2.*BETA3(I)*Elas)
ELSE
EKSozel(im)=0.
EKSozelz(im)=0.
ENDIF
@
@
@
@
EKS(im)= D1(I)*SINH(ALFA1(I)*KAT(im))
+D2(I)*COSH(ALFA1(I)*KAT(im))
+D3(I)*SINH(ALFA2(I)*KAT(im))
+D4(I)*COSH(ALFA2(I)*KAT(im))
+EKSozel(im)
@
@
@
@
EKSz(im)= D1(I)*ALFA1(I)*COSH(ALFA1(I)*KAT(im))
+D2(I)*ALFA1(I)*SINH(ALFA1(I)*KAT(im))
+D3(I)*ALFA2(I)*COSH(ALFA2(I)*KAT(im))
+D4(I)*ALFA2(I)*SINH(ALFA2(I)*KAT(im))
+EKSozelz(im)
276
IF((I.GT.1).AND.(I.LT.ibsay)) THEN
@
@
@
@
ST(I)= D1(I-1)*SINH(ALFA1(I-1)*Z(I))
+D2(I-1)*COSH(ALFA1(I-1)*Z(I))
+D3(I-1)*SINH(ALFA2(I-1)*Z(I))
+D4(I-1)*COSH(ALFA2(I-1)*Z(I))
+STozel(I-1)
@
@
@
@
STz(I)= D1(I-1)*ALFA1(I-1)*COSH(ALFA1(I-1)*Z(I))
+D2(I-1)*ALFA1(I-1)*SINH(ALFA1(I-1)*Z(I))
+D3(I-1)*ALFA2(I-1)*COSH(ALFA2(I-1)*Z(I))
+D4(I-1)*ALFA2(I-1)*SINH(ALFA2(I-1)*Z(I))
+STozelz(I-1)
STOPMOMX(I)= (Mx(I)-ST(I)*b(I-1))
STOPMOMY(I)=-(My(I)-ST(I)*a(I-1))
STOPMOMT(I)= (Mt(I)+(w(I-1)+d(I-1))*STz(I))
STOPMOMB(I)= (Bt(I)+(w(I-1)+d(I-1))*ST(I))
ENDIF
IF((HP-KAT(im)).GT.ETOL) THEN
dMx(im) =Py*(HP - KAT(im)) + (Wy*(HP - KAT(im))**2)/2.
dMy(im) =Px*(HP - KAT(im)) + (Wx*(HP - KAT(im))**2)/2.
dMt(im) =-(dy*Px)+dx*Py-dy*Wx*(HP-KAT(im))+dx*Wy*(HP-KAT(im))+Met
dMt(im) =-(dy*Px)+dx*Py-dy*Wx*(HP-KAT(im))+dx*Wy*(HP-KAT(im))+Met
ELSE
dMx(im)
dMy(im)
dMt(im)
ENDIF
=0.
=0.
=0.
dBt(im)
= Bet
TOPMOMX(im)= (dMx(im)-EKS(im)*b(I))
TOPMOMY(im)=-(dMy(im)-EKS(im)*a(I))
TOPMOMT(im)= (dMt(im)+(w(I)+d(I))*EKSz(im))
TOPMOMB(im)= (dBt(im)+(w(I)+d(I))*EKS(im))
IF(ABS(Z(I+1)-KAT(im)).LT.TOL) THEN
iksay=im
GOTO 811
ENDIF
IF(KAT(im).LT.TOL) THEN
GOTO 811
ENDIF
911 CONTINUE
811 ENDDO
iksay=KSAY+1
244 FORMAT(1X,30F20.12)
iksay=ksay+1
WRITE(6,*)'
'
WRITE(6,*)'********************************************'
WRITE(6,*)'
'
WRITE(6,*)' Kat Hizalarnda PERDE EKSENEL KUVVET (T) '
WRITE(6,*)'------------------------------------------- '
277
WRITE(6,*)'
do I=1,iksay
J=iksay-I+1
write(6,244) EKS(J)
enddo
'
c
c
c
c
c
c
c
c
WRITE(6,*)'
'
WRITE(6,*)'
'
WRITE(6,*)' Blge snrlar hemen zerinde PERDE EKS KUV (ST)'
WRITE(6,*)'--------------------------------------------------'
WRITE(6,*)'
'
do I=1,ibsay
write(6,244) ST(I)
enddo
c
c
c
c
c
c
c
c
c
c
WRITE(6,*)'
'
WRITE(6,*)'********************************************'
WRITE(6,*)'
'
WRITE(6,*)'Glob-X etrafnda TOPLAM PERDE ELME MOMENT'
WRITE(6,*)'--------------------------------------------'
WRITE(6,*)'
'
do I=1,iksay
J=iksay-I+1
write(6,244) TOPMOMX(J)
enddo
c
c
c
c
c
c
c
c
c
WRITE(6,*)'
'
WRITE(6,*)'
'
WRITE(6,*)'
Blge snrlar hemen zerinde
'
WRITE(6,*)'Glob-X etrafnda TOPLAM PERDE ELME MOMENT'
WRITE(6,*)'--------------------------------------------'
WRITE(6,*)'
'
do I=1,ibsay
write(6,244) STOPMOMX(I)
enddo
c
c
c
c
c
c
c
c
c
c
WRITE(6,*)'
'
WRITE(6,*)'********************************************'
WRITE(6,*)'
'
WRITE(6,*)'Glob-Y etrafnda TOPLAM PERDE ELME MOMENT'
WRITE(6,*)'--------------------------------------------'
WRITE(6,*)'
'
do I=1,iksay
J=iksay-I+1
write(6,244) TOPMOMY(J)
enddo
c
c
c
c
c
c
c
c
c
WRITE(6,*)'
'
WRITE(6,*)'
'
WRITE(6,*)'
Blge snrlar hemen zerinde
'
WRITE(6,*)'Glob-Y etrafnda TOPLAM PERDE ELME MOMENT'
WRITE(6,*)'--------------------------------------------'
WRITE(6,*)'
'
do I=1,ibsay
write(6,244) STOPMOMY(I)
enddo
c
c
c
c
c
c
WRITE(6,*)'
'
WRITE(6,*)'********************************************'
WRITE(6,*)'
'
WRITE(6,*)' Glob-Z etrafnda TOPLAM PERDE BURULMA MOMENT (Mt)'
WRITE(6,*)'---------------------------------------------------'
WRITE(6,*)'
'
do I=1,iksay
J=iksay-I+1
278
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
write(6,244) TOPMOMT(J)
enddo
WRITE(6,*)'
'
WRITE(6,*)'
'
WRITE(6,*)'
Blge snrlar hemen zerinde
'
WRITE(6,*)' Glob-Z etrafnda TOPLAM PERDE BURULMA MOMENT (Mt)'
WRITE(6,*)'---------------------------------------------------'
WRITE(6,*)'
'
do I=1,ibsay
write(6,244) STOPMOMT(I)
enddo
WRITE(6,*)'
'
WRITE(6,*)'********************************************'
WRITE(6,*)'
'
WRITE(6,*)'
'
WRITE(6,*)' TOPLAM PERDE BMOMENT (Bt)'
WRITE(6,*)'-----------------------------'
WRITE(6,*)'
'
do I=1,iksay
J=iksay-I+1
write(6,244) TOPMOMB(J)
enddo
c
c
c
c
c
c
WRITE(6,*)'
'
WRITE(6,*)'
'
WRITE(6,*)' Blge snrlar hemen zerinde '
WRITE(6,*)'
TOPLAM PERDE BMOMENT (Bt)
'
WRITE(6,*)'---------------------------------'
WRITE(6,*)'
'
do I=1,ibsay
write(6,244) STOPMOMB(I)
enddo
WRITE(6,*)'
'
WRITE(6,*)'********************************************'
WRITE(6,*)'
'
c
c
c
c
c*******************************************
c
@
@
@
CALL COEF1Ua(ibsay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,Z(I),GAMA,G,EJ,K1,K2,K3,K4,HP,DELTA
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan,EIx,EIxy
,Elas,D1,D2,D3,D4,REK1,REKB1,I)
@
@
@
CALL COEF2Ua(ibsay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,Z(I),GAMA,G,EJ,K1,K2,K3,K4,HP,DELTA
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan,EIx,EIxy
,Elas,D1,D2,D3,D4,REK2,REKB2,I)
279
ENDIF
IF(I.EQ.ibsay) THEN
@
@
@
CALL COEF3Ua(ibsay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,Z(I),GAMA,G,EJ,K1,K2,K3,K4,HP,DELTA
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan,EIx,EIxy
,Elas,D1,D2,D3,D4,REK1,REKB1,I)
@
@
@
CALL COEF4Ua(ibsay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,Z(I),GAMA,G,EJ,K1,K2,K3,K4,HP,DELTA
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan,EIx,EIxy
,Elas,D1,D2,D3,D4,REK2,REKB2,I)
ENDIF
else
IF((hpp.LT.TOL).AND.(hpp.GT.ETOL)) THEN
@
@
@
CALL COEF3Ub(ibsay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,Z(I),GAMA,G,EJ,K1,K2,K3,K4,HP,DELTA
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan,EIx,EIxy
,Elas,D1,D2,D3,D4,REK1,REKB1,I)
@
@
@
CALL COEF4Ub(ibsay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,Z(I),GAMA,G,EJ,K1,K2,K3,K4,HP,DELTA
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan,EIx,EIxy
,Elas,D1,D2,D3,D4,REK2,REKB2,I)
else
@
@
@
CALL COEF1Ub(ibsay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,Z(I),GAMA,G,EJ,K1,K2,K3,K4,HP,DELTA
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan,EIx,EIxy
,Elas,D1,D2,D3,D4,REK1,REKB1,I)
@
@
@
CALL COEF2Ub(ibsay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,Z(I),GAMA,G,EJ,K1,K2,K3,K4,HP,DELTA
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan,EIx,EIxy
,Elas,D1,D2,D3,D4,REK2,REKB2,I)
ENDIF
endif
IF(I.LT.ibsay) THEN
DO J=1,4
YK(I+iL-1,J+iK)=YK(I+iL-1,J+iK)+REK1(J)
YB(I+iL-1)=REKB1
YK(I+iL,J+iK)=YK(I+iL,J+iK)+REK2(J)
YB(I+iL)=REKB2
ENDDO
ENDIF
iK=iK+2
IF(I.EQ.ibsay) THEN
DO J=1,2
YK(I+iL-1,J+iK-2)=YK(I+iL-1,J+iK-2)+REK1(J)
YB(I+iL-1)=REKB1
YK(I+iL,J+iK-2)=YK(I+iL,J+iK-2)+REK2(J)
YB(I+iL)=REKB2
ENDDO
ENDIF
280
iL=iL+1
do j=1,4
rek1(j)=0.d0
rek2(j)=0.d0
rek3(j)=0.d0
rek4(j)=0.d0
enddo
REKB1=0
REKB2=0
ENDDO
in=2*(ibsay-1)
c
c
c
c
c
c
im=0
DO I=1,ibsay
F1(I)=F1(I)+SONUC3(I+im)
F2(I)=F2(I)+SONUC3(I+1+im)
im=im+1
ENDDO
F1(ibsay)=F1(ibsay-1)
F2(ibsay)=F2(ibsay-1)
C
U(Z) FONKSIYONU
iksay=KSAY+1
DO I=1,ibsay
DO 91 in=1,iksay
im=iksay-in+1
HPP=HP-Z(I)
if(hpp.GT.ETOL) then
@
@
@
@
@
@
@
@
CALL UFONKa(ibsay,iksay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,GAMA,K1,K2,K3,K4,Elas,G,EJ,DELTA
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan,HP
,EIx,EIxy,D1,D2,D3,D4,U,KAT,I,im
,F1,F2)
else
CALL UFONKb(ibsay,iksay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,GAMA,K1,K2,K3,K4,Elas,G,EJ,DELTA
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan,HP
,EIx,EIxy,D1,D2,D3,D4,U,KAT,I,im
,F1,F2)
endif
281
IF(ABS(Z(I+1)-KAT(im)).LT.TOL) THEN
iksay=im
GOTO 81
ENDIF
IF(KAT(im).LT.TOL) THEN
GOTO 81
ENDIF
91 CONTINUE
81 ENDDO
iksay=KSAY+1
WRITE(6,*) '
'
WRITE(6,*) 'U= '
do I=1,iksay
J=iksay-I+1
write(6,24) U(J)
enddo
in=2*(ibsay-1)
do j=1,in
YB(j)=0.d0
enddo
do I =1,in
do J =1,in
yk(I,J)=0.d0
enddo
enddo
do j=1,4
rek1(j)=0.d0
rek2(j)=0.d0
rek3(j)=0.d0
rek4(j)=0.d0
enddo
WRITE(6,*)'
'
WRITE(6,*)'********************************************'
c
@
@
@
CALL COEF1Va(ibsay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,Z(I),GAMA,G,EJ,K1,K2,K3,K4,HP,DELTA
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan,EIy,EIxy
,Elas,D1,D2,D3,D4,REK1,REKB1,I)
@
@
@
CALL COEF2Va(ibsay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,Z(I),GAMA,G,EJ,K1,K2,K3,K4,HP,DELTA
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan,EIy,EIxy
,Elas,D1,D2,D3,D4,REK2,REKB2,I)
282
ENDIF
IF(I.EQ.ibsay) THEN
@
@
@
CALL COEF3Va(ibsay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,Z(I),GAMA,G,EJ,K1,K2,K3,K4,HP,DELTA
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan,EIy,EIxy
,Elas,D1,D2,D3,D4,REK1,REKB1,I)
@
@
@
CALL COEF4Va(ibsay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,Z(I),GAMA,G,EJ,K1,K2,K3,K4,HP,DELTA
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan,EIy,EIxy
,Elas,D1,D2,D3,D4,REK2,REKB2,I)
ENDIF
else
IF((hpp.LT.TOL).AND.(hpp.GT.ETOL)) THEN
@
@
@
CALL COEF3Vb(ibsay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,Z(I),GAMA,G,EJ,K1,K2,K3,K4,HP,DELTA
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan,EIy,EIxy
,Elas,D1,D2,D3,D4,REK1,REKB1,I)
@
@
@
CALL COEF4Vb(ibsay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,Z(I),GAMA,G,EJ,K1,K2,K3,K4,HP,DELTA
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan,EIy,EIxy
,Elas,D1,D2,D3,D4,REK2,REKB2,I)
else
@
@
@
CALL COEF1Vb(ibsay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,Z(I),GAMA,G,EJ,K1,K2,K3,K4,HP,DELTA
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan,EIy,EIxy
,Elas,D1,D2,D3,D4,REK1,REKB1,I)
@
@
@
CALL COEF2Vb(ibsay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,Z(I),GAMA,G,EJ,K1,K2,K3,K4,HP,DELTA
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan,EIy,EIxy
,Elas,D1,D2,D3,D4,REK2,REKB2,I)
ENDIF
endif
IF(I.LT.ibsay) THEN
DO J=1,4
YK(I+iL-1,J+iK)=YK(I+iL-1,J+iK)+REK1(J)
YB(I+iL-1)=REKB1
YK(I+iL,J+iK)=YK(I+iL,J+iK)+REK2(J)
YB(I+iL)=REKB2
ENDDO
ENDIF
iK=iK+2
IF(I.EQ.ibsay) THEN
DO J=1,2
YK(I+iL-1,J+iK-2)=YK(I+iL-1,J+iK-2)+REK1(J)
YB(I+iL-1)=REKB1
YK(I+iL,J+iK-2)=YK(I+iL,J+iK-2)+REK2(J)
YB(I+iL)=REKB2
ENDDO
283
ENDIF
iL=iL+1
do j=1,4
rek1(j)=0.d0
rek2(j)=0.d0
rek3(j)=0.d0
rek4(j)=0.d0
enddo
ENDDO
in=2*(ibsay-1)
c
c
c
c
c
c
im=0
DO I=1,ibsay
P1(I)=P1(I)+SONUC4(I+im)
P2(I)=P2(I)+SONUC4(I+1+im)
im=im+1
ENDDO
P1(ibsay)=P1(ibsay-1)
P2(ibsay)=P2(ibsay-1)
C
V(Z) FONKSIYONU
iksay=KSAY+1
DO I=1,ibsay
DO 92 in=1,iksay
im=iksay-in+1
HPP=HP-Z(I)
if(hpp.GT.ETOL) then
@
@
@
@
CALL VFONKa(ibsay,iksay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,GAMA,K1,K2,K3,K4,Elas,G,EJ,DELTA
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan,HP
,EIy,EIxy,D1,D2,D3,D4,V,KAT,I,im
,P1,P2)
else
@
@
@
@
CALL VFONKb(ibsay,iksay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,GAMA,K1,K2,K3,K4,Elas,G,EJ,DELTA
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan,HP
,EIy,EIxy,D1,D2,D3,D4,V,KAT,I,im
,P1,P2)
284
endif
IF(ABS(Z(I+1)-KAT(im)).LT.TOL) THEN
iksay=im
GOTO 82
ENDIF
IF(KAT(im).LT.TOL) THEN
GOTO 82
ENDIF
92 CONTINUE
82 ENDDO
iksay=KSAY+1
WRITE(6,*) '
'
WRITE(6,*) 'V= '
do I=1,iksay
J=iksay-I+1
write(6,24) V(J)
enddo
in=2*(ibsay-1)
do j=1,in
YB(j)=0.d0
enddo
do I =1,in
do J =1,in
yk(I,J)=0.d0
enddo
enddo
do j=1,4
rek1(j)=0.d0
rek2(j)=0.d0
rek3(j)=0.d0
rek4(j)=0.d0
enddo
WRITE(6,*)'
'
WRITE(6,*)'********************************************'
c
@
@
@
CALL COEF1Ya(ibsay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,Z(I),GAMA,G,EJ,K1,K2,K3,K4,HP
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan
,Elas,D1,D2,D3,D4,REK1,REKB1,I)
CALL COEF2Ya(ibsay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,Z(I),GAMA,G,EJ,K1,K2,K3,K4,HP
285
@
@
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan
,Elas,D1,D2,D3,D4,REK2,REKB2,I)
ENDIF
IF(I.EQ.ibsay) THEN
@
@
@
CALL COEF3Ya(ibsay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,Z(I),GAMA,G,EJ,K1,K2,K3,K4,HP
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan
,Elas,D1,D2,D3,D4,REK1,REKB1,I)
@
@
@
CALL COEF4Ya(ibsay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,Z(I),GAMA,G,EJ,K1,K2,K3,K4,HP
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan
,Elas,D1,D2,D3,D4,REK2,REKB2,I)
ENDIF
else
IF((hpp.LT.TOL).AND.(hpp.GT.ETOL)) THEN
@
@
@
CALL COEF3Yb(ibsay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,Z(I),GAMA,G,EJ,K1,K2,K3,K4,HP
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan
,Elas,D1,D2,D3,D4,REK1,REKB1,I)
@
@
@
CALL COEF4Yb(ibsay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,Z(I),GAMA,G,EJ,K1,K2,K3,K4,HP
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan
,Elas,D1,D2,D3,D4,REK2,REKB2,I)
else
@
@
@
CALL COEF1Yb(ibsay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,Z(I),GAMA,G,EJ,K1,K2,K3,K4,HP
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan
,Elas,D1,D2,D3,D4,REK1,REKB1,I)
@
@
@
CALL COEF2Yb(ibsay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,Z(I),GAMA,G,EJ,K1,K2,K3,K4,HP
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan
,Elas,D1,D2,D3,D4,REK2,REKB2,I)
ENDIF
endif
IF(I.LT.ibsay) THEN
DO J=1,4
YK(I+iL-1,J+iK)=YK(I+iL-1,J+iK)+REK1(J)
YB(I+iL-1)=REKB1
YK(I+iL,J+iK)=YK(I+iL,J+iK)+REK2(J)
YB(I+iL)=REKB2
ENDDO
ENDIF
iK=iK+2
IF(I.EQ.ibsay) THEN
DO J=1,2
YK(I+iL-1,J+iK-2)=YK(I+iL-1,J+iK-2)+REK1(J)
YB(I+iL-1)=REKB1
YK(I+iL,J+iK-2)=YK(I+iL,J+iK-2)+REK2(J)
286
YB(I+iL)=REKB2
ENDDO
ENDIF
iL=iL+1
do j=1,4
rek1(j)=0.d0
rek2(j)=0.d0
rek3(j)=0.d0
rek4(j)=0.d0
enddo
ENDDO
in=2*(ibsay-1)
c
c
c
c
c
c
im=0
DO I=1,ibsay
G1(I)=G1(I)+SONUC2(I+im)
G2(I)=G2(I)+SONUC2(I+1+im)
im=im+1
ENDDO
G1(ibsay)=G1(ibsay-1)
G2(ibsay)=G2(ibsay-1)
C
TETA(Z) FONKSIYONU
iksay=KSAY+1
DO I=1,ibsay
DO 90 in=1,iksay
im=iksay-in+1
HPP=HP-Z(I)
if(hpp.GT.ETOL) then
@
@
@
CALL TETAFONKa(ibsay,iksay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,GAMA,K1,K2,K3,K4,Elas,G,EJ
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan,HP
,D1,D2,D3,D4,TETA,KAT,I,im,G1,G2)
else
@
@
@
CALL TETAFONKb(ibsay,iksay,ALFA1,ALFA2,BETA1,BETA2,BETA3
,GAMA,K1,K2,K3,K4,Elas,G,EJ
,Px,Py,Wx,Wy,dx,dy,r,EIOw,Alan,HP
,D1,D2,D3,D4,TETA,KAT,I,im,G1,G2)
287
endif
IF(ABS(Z(I+1)-KAT(im)).LT.TOL) THEN
iksay=im
GOTO 80
ENDIF
IF(KAT(im).LT.TOL) THEN
GOTO 80
ENDIF
90 CONTINUE
80 ENDDO
iksay=KSAY+1
24 FORMAT(1X,30F20.12)
iksay=ksay+1
WRITE(6,*) '
'
WRITE(6,*) 'TETA='
do I=1,iksay
J=iksay-I+1
write(6,24) TETA(J)
enddo
C *********************************************************************
C
c
c
c
c
c
c
c
c
c
C
C
C
iksay=KSAY+1
IKSA=3*IKSAY
IF(JJ.EQ.1)THEN
DO NM=1,IKSA
U(NM)=0.00000000000000000001
V(NM)=0.00000000000000000001
TETA(NM)=0.00000000000000000001
ENDDO
ENDIF
**********************
** ESNEKLIK MATRISI **
**********************
288
c
c
c
c
c
c
KOL=iksay-JJ+IH
iksay=KSAY+1
DO NN=1,iksay
M=iksay-NN+1
FLEX(NN,KOL)=FLEX(NN,KOL)+U(M)
enddo
KOL=3*(iksay-JJ)+IH
iksay=KSAY+1
DO NN=1,ksay
M=iksay-NN+1
FLEX(3*NN-2,KOL)=FLEX(3*NN-2,KOL)+UG(M)
ENDDO
iksay=KSAY+1
DO NN=1,ksay
M=iksay-NN+1
FLEX(3*NN-1,KOL)=FLEX(3*NN-1,KOL)+VG(M)
ENDDO
iksay=KSAY+1
DO NN=1,ksay
M=iksay-NN+1
FLEX(3*NN,KOL)=FLEX(3*NN,KOL)+TETA(M)
ENDDO
800 CONTINUE
1000
CONTINUE
iksay=KSAY+1
write(6,*) '
'
write(6,*) '--KAT SEViYELERi--'
write(6,*) '
'
do in=1,iksay
im=iksay-in+1
write(6,*) KAT(im)
enddo
write(6,*) '
'
write(6,*) '--ESNEKLK MATRS--'
IKSA=3*KSAY
DO I=1,IKSA
WRITE(6,24) (FLEX(I,J),J=1,IKSA)
ENDDO
C
C
C
iksay=ksay+1
IKSA=3*KSAY
*****************************************
**RIJITLIK MATRISI BULUNUYOR [K]=[F]**(-1)**
*****************************************
CALL INVMATRIS(FLEX,STIFF,IKSA)
write(6,*) '
'
write(6,*) '--RIJITLIK MATRISI--'
IKSA=3*KSAY
DO I =1,IKSA
WRITE(6,24) (STIFF(I,J),J=1,IKSA)
enddo
289
C
C
C
GAM=2.4
******************************
********* KUTLE MATRISI **********
******************************
IKSAY=KSAY+1
DO I=1,IBSAY
DO 89 M=1,KSAY
IF((KAT(IKSAY-M+1)-Z(I)).LT.TOL)THEN
IF((Z(I)-H).LT.TOL)THEN
IF(HGKIR(I).LT.TOL)THEN
EEV(M)= TALAN(I)*(HKAT(I)/2)+c(I)*BKTHICK(I)*HKIR(I)
EKUT(M)=EEV(M)*GAM
else
EEV(M)= TALAN(I)*(HKAT(I)/2)+c(I)*BKTHICK(I)*HGKIR(I)
EKUT(M)=EEV(M)*2.4
ENDIF
ENDIF
IF((Z(I).GT.TOL).AND.(H-(Z(I)).GT.TOL)) THEN
IF(HGKIR(I).LT.TOL)THEN
EEV(M)=((TALAN(I)+TALAN(I-1))/2)*((HKAT(I)+HKAT(I-1))/2)
+c(I)*BKTHICK(I)*HKIR(I)
EKUT(M)=EEV(M)*2.4
else
EEV(M)=((TALAN(I)+TALAN(I-1))/2)*((HKAT(I)+HKAT(I-1))/2)
+c(I)*BKTHICK(I)*HGKIR(I)
EKUT(M)=EEV(M)*2.4
ENDIF
ENDIF
ENDIF
IF(((Z(I)-KAT(IKSAY-M+1)).GT.TOL).AND.
((KAT(IKSAY-M+1)-Z(I+1)).GT.TOL)) THEN
EEV(M)=TALAN(I)*HKAT(I)+c(I)*BKTHICK(I)*HKIR(I)
EKUT(M)=EEV(M)*2.4
ENDIF
89
CONTINUE
ENDDO
iksay=KSAY+1
write(6,*) '
'
write(6,*) '--KUTLE VEKTORU--'
do M=1,ksay
write(6,*) EKUT(M)
enddo
DO NN=1,ksay
MASSMAT(3*NN-2,3*NN-2)=MASSMAT(3*NN-2,3*NN-2)+EKUT(NN)
ENDDO
DO NN=1,KSAY
MASSMAT(3*NN-1,3*NN-1)=MASSMAT(3*NN-1,3*NN-1)+EKUT(NN)
ENDDO
*****
*****
DO NN=1,KSAY
EKUTT(NN)=0.0000000000000000001
MASSMAT(3*NN,3*NN)=MASSMAT(3*NN,3*NN)+EKUTT(NN)
ENDDO
write(6,*) '
'
write(6,*) '--KUTLE MATRiSi--'
IKSA=3*KSAY
DO I=1,IKSA
290
WRITE(6,24) (MASSMAT(I,J),J=1,IKSA)
ENDDO
write(6,*) '
'
888
DO 888 I=1,IKSA
DO 888 J=1,IKSA
STIF1(I,J)=STIFF(I,J)
AMAS1(I,J)=MASSMAT(I,J)
AMAS2(I,J)=AMAS1(I,J)
CONTINUE
DO I =1,IKSA
WRITE(6,22) (STIF1(I,J),J=1,IKSA)
enddo
DO I =1,IKSA
WRITE(6,22) (AMAS1(I,J),J=1,IKSA)
enddo
iksay=ksay+1
IKSA=3*KSAY
CALL JACK3(IKSA,STIFF,MASSMAT,EGNVAL,EGNVEC)
write(6,*) '
'
write(6,*) '--OZVEKTORLER MATRISI--'
IKSA=3*KSAY
DO I =1,IKSA
WRITE(6,24) (EGNVEC(I,J),J=1,IKSA)
enddo
DO J=1,IKSA
DO I=1,IKSA
EGNVEC2(I,J)=(1/EGNVEC(1,J))*EGNVEC(I,J)
ENDDO
ENDDO
C
C
C
C
C
c
write(6,*) '
'
write(6,*) '--NORMALIZE EDILMIS OZVEKTORLER MATRISI--'
DO I =1,IKSA
WRITE(6,22) (EGNVEC2(I,J),J=1,IKSA)
enddo
**********************************************
DO J=1,8
WRITE(6,*) 'MOD',J,' ICIN X SEKIL VECTORU'
DO I=1,KSAY
WRITE(6,*) EGNVEC(3*I-2,J)
ENDDO
ENDDO
write(6,*) '
'
DO J=1,8
WRITE(6,*) 'MOD',J,' ICIN Y SEKIL VEKTORU'
DO I=1,KSAY
WRITE(6,*) EGNVEC(3*I-1,J)
ENDDO
ENDDO
write(6,*) '
'
DO J=1,8
WRITE(6,*) 'MOD',J,' ICIN TETA SEKIL VEKTORU'
DO I=1,KSAY
WRITE(6,*) EGNVEC(3*I,J)
ENDDO
ENDDO
291
c **************************************************
write(6,*) '
'
write(6,*) '--OZDEGERLER VEKTORU--'
IKSA=3*KSAY
DO I =1,IKSA
WRITE(6,24) EGNVAL(I)
enddo
c
c
c
c
c 144
c
c
c
145
DO 144 KLM=1,IKSA
DUMM(KLM)=EGNVAL(IKSA+1-KLM)
DO 144 KKLM=1,IKSA
DUMV(KKLM,KLM)=EGNVEC(KKLM,IKSA+1-KLM)
CONTINUE
PI=4*ATAN(1.0)
DO 145 KLM1=1,IKSA
EGNVAL(KLM1)=DUMM(KLM1)
CFREQ(KLM1)=SQRT(EGNVAL(KLM1))
NFREQ(KLM1)=CFREQ(KLM1)/(2*PI)
DO 145 KLM3=1,IKSA
EGNVEC(KLM3,KLM1)=DUMV(KLM3,KLM1)
CONTINUE
WRITE(6,*) '
EIGENVALUE
CIRCULAR FREQ.
NATURAL FR
@EQ.'
WRITE(6,*) ' ---------------------------------------------------@----'
DO 146 KLM2=1,IKSA
WRITE(6,34) EGNVAL(KLM2),CFREQ(KLM2),NFREQ(KLM2)
146 CONTINUE
34 FORMAT(E18.8,1X,',',E18.8,1X,',',E18.8)
WRITE(6,*) ' ====================================================
@===='
WRITE(6,*) ' '
C
C
C
c
****************************************************************
**
Serbest Titreim analizi yapld,
*********
**
Zorlanm Titreim Analizi balyor
*********
****************************************************************
READ(5,*)KSI
READ(5,*)ETSUR
READ(5,*)YGEN
READ(5,*)DT
READ(5,*)ASUR
DO 151 I1=1,IKSA
DO 151 I2=1,IKSA
TEGNVEC(I1,I2)=EGNVEC(I2,I1)
151 CONTINUE
CALL MTRXML(STIF1,IKSA,IKSA,EGNVEC,IKSA,GSSTF1,N)
CALL MTRXML(TEGNVEC,IKSA,IKSA,GSSTF1,IKSA,GSSTF,N)
C
C
CALL MTRXML(SONUM,NEV,NEV,EGNVEC,NEV,GSSNM1,NW)
CALL MTRXML(TEGNVEC,NEV,NEV,GSSNM1,NEV,GSSNM,NW)
CALL MTRXML(AMAS1,IKSA,IKSA,EGNVEC,IKSA,GSMSS1,N)
CALL MTRXML(TEGNVEC,IKSA,IKSA,GSMSS1,IKSA,GSMSS,N)
292
DO 153 I=1,IKSA
DO 153 J=1,IKSA
GSMS(I,J)=0.0
GSSN(I,J)=0.0
GSST(I,J)=0.0
153 CONTINUE
DO 152 I1=1,IKSA
GSMS(I1,I1)=GSMSS(I1,I1)
GSSN(I1,I1)=2*GSMS(I1,I1)*KSI*CFREQ(I1)
GSST(I1,I1)=GSSTF(I1,I1)
152 CONTINUE
C
WRITE(6,*)'=======================================================
C
@===================================================='
C
DO 154 I=1,NEV
C
WRITE(6,*) ' '
C 154 WRITE(6,*) (GSSN(I,IJ),IJ=1,NEV)
WRITE(*,*)'
'
WRITE(*,*) 'TIME-HISTORY ANALZ=>1'
WRITE(*,*) '
SPEKTRUM ANALZ=>2'
READ(5,*)ISEC
IF(ISEC.EQ.1) THEN
WRITE(*,*)'
'
WRITE(*,*)'YUK TIPINI GIRIN'
WRITE(*,*)'
'
WRITE(*,*)'HARMONIK YUKLEME=>1'
WRITE(*,*)'
UCGEN YUKLEME=>2'
WRITE(*,*)'
ADIM YUKLEME=>3'
WRITE(*,*)'
'
READ(5,*)LTIP
WRITE(*,*)'
'
WRITE(*,*)'ENTER THE JOINT NUMBER TO BE ANALIZED'
WRITE(*,*)'
'
READ(5,*)JI
WRITE(*,*)'
'
WRITE(*,*)'ENTER THE JOINT NUMBER, THE FORCE ACTING ON'
WRITE(*,*)'
'
READ(5,*)I31
WRITE(*,*)'
'
WRITE(*,*)'ENTER THE LOAD VALUE AT t=0'
WRITE(*,*)'
'
READ(5,*)FIN
WRITE(*,*)'
'
SIGMA=0.5
AFA=0.25*(0.5+SIGMA)**2
A0=1./(AFA*DT**2)
A1=SIGMA/(AFA*DT)
A2=1./(AFA*DT)
A3=1./(2*AFA)-1.
A4=SIGMA/AFA-1.
A5=DT/2.*(SIGMA/AFA-2.)
A6=DT*(1.0-SIGMA)
A7=DT*SIGMA
293
*****************************************
DO 1618 IA1=1,IKSA
YVEK(IA1)=0.0
yenX(IA1)=0.0
yenEX(IA1)=0.0
XN(IA1)=0.0
XNN(IA1)=0.0
EXNN(IA1)=0.0
DO 1618 IB1=1,IKSA
1618 EFFK(IA1,IB1)=GSST(IA1,IB1)+A0*GSMS(IA1,IB1)+A1*GSSN(IA1,IB1)
DMAX=yenX(JI)
YVEK(I31)=FIN
CALL MTRXML1(TEGNVEC,IKSA,IKSA,YVEK,IKSA,XNN,N)
DO 165 LP=1,IKSA
165 EXNN(LP)=XNN(LP)
READ(5,*)OMG
READ(5,*)TT1
READ(5,*)PXC
READ(5,*)PYC
IF(I31.EQ.1)THEN
MOMENTKOLU=DY-PYC
ELSE
MOMENTKOLU=PXC-DX
ENDIF
ZD=ASUR/DT
IZD=ZD
DO 1628 IZ=1,IZD
ZZ=IZ*DT
IF(LTIP.EQ.1)THEN
YUK=0.0
IF(ZZ.LE.ETSUR)YUK=YGEN*SIN(OMG*ZZ)
ELSEIF(LTIP.EQ.2)THEN
IF(ZZ.LE.TT1)THEN
YUK=ZZ/TT1*YGEN
ELSEIF(ZZ.GT.TT1.AND.ZZ.LE.ETSUR)THEN
YUK=YGEN-YGEN*(ZZ-TT1)/(ETSUR-TT1)
ELSEIF(ZZ.GT.ETSUR)THEN
YUK=0.0
ENDIF
ELSE
YUK=0.0
IF(ZZ.LE.ETSUR)YUK=YGEN
ENDIF
YVEK(I31)=YUK
YVEK(3)=YUK*MOMENTKOLU
163
DO 163 IR=1,IKSA
YN1(IR)=A0*yenX(IR)+A2*XN(IR)+A3*XNN(IR)
YN2(IR)=A1*yenX(IR)+A4*XN(IR)+A5*XNN(IR)
CALL MTRXML1(GSMS,IKSA,IKSA,YN1,IKSA,YN3,N)
CALL MTRXML1(GSSN,IKSA,IKSA,YN2,IKSA,YN4,N)
CALL MTRXML1(TEGNVEC,IKSA,IKSA,YVEK,IKSA,GYVEK,N)
294
DO 164 IR=1,IKSA
164 EYV(IR)=GYVEK(IR)+YN3(IR)+YN4(IR)
CALL GAUSS2(EFFK,IKSA,EYV,yenX)
DO 166 I9=1,IKSA
XNN(I9)=A0*(yenX(I9)-yenEX(I9))-A2*XN(I9)-A3*XNN(I9)
yenEX(I9)=yenX(I9)
XN(I9)=XN(I9)+A6*EXNN(I9)+A7*XNN(I9)
166 EXNN(I9)=XNN(I9)
CALL MTRXML1(EGNVEC,IKSA,IKSA,yenX,IKSA,XGZ,N)
IF(XGZ(JI).GT.DMAX) DMAX=XGZ(JI)
WRITE(6,*)XGZ(JI)
1628 CONTINUE
WRITE(6,*)'*****************TIME-HISTORY ANALIZI******************
@**********'
WRITE(6,*)'
'
WRITE(6,*)'MAX DEFLECTION IS',DMAX
WRITE(6,*)'
'
WRITE(6,*)'XG= ',eEXG(1)
WRITE(6,*)'YG= ',eEYG(1)
c
ELSE
ENDIF
GO TO 99
99
STOP
END
c ******************************************************************
c *********
A L T
P R O G R A M L A R
***************
c ******************************************************************
SUBROUTINE COEF1A(ETozel,ETozelz,EALFA1,EALFA2
,EZ,ES,REK1,REKB1)
295
SUBROUTINE COEF1B(ETozel,ETozelzz,EALFA1,EALFA2
,EGAMA,EZ,EEIw,EALAN,Er,Ed,Ew,EEIxc,EEIyc
,EK1,EK2,EK3,EK4,EBet,REK2,REKB2)
@
@
296
SUBROUTINE COEF2B(ibsay,EALFA1,EALFA2,EZ,ES,
ETozel,ETozelz,ESTozel,J,REK2,REKB2)
REK3(2)= EALFA1(J-1)*(EC1(J-1)+EALFA1(J-1)**2*EC2(J-1))*
Sinh(EALFA1(J-1)*EZ)
REK3(3)= EALFA2(J-1)*(EC1(J-1)+EALFA2(J-1)**2*EC2(J-1))*
Cosh(EALFA2(J-1)*EZ)
REK3(4)= EALFA2(J-1)*(EC1(J-1)+EALFA2(J-1)**2*EC2(J-1))*
Sinh(EALFA2(J-1)*EZ)
REK3(5)= -(EALFA1(J)*(EC1(J)+EALFA1(J)**2*EC2(J))*
Cosh(EALFA1(J)*EZ))
REK3(6)= -(EALFA1(J)*(EC1(J)+EALFA1(J)**2*EC2(J))*
Sinh(EALFA1(J)*EZ))
REK3(7)= -(EALFA2(J)*(EC1(J)+EALFA2(J)**2*EC2(J))*
Cosh(EALFA2(J)*EZ))
REK3(8)= -(EALFA2(J)*(EC1(J)+EALFA2(J)**2*EC2(J))*
Sinh(EALFA2(J)*EZ))
@
@
@
REKB3=-(-EDD1(J) + EDD1(J-1)
-(EC1(J)*ETozelz(J))+EC1(J-1)*ETozelz(J-1)
-EC2(J)*ETozelzzz(J)+EC2(J-1)*ETozelzzz(J-1))
RETURN
END
SUBROUTINE COEF2D(ibsay,ETozel,ETozelzz,ESTozel,ESTozelzz,
EALFA1,EALFA2,EZ,EC1,EC2,EDD2,ESDD2,J,REK4,REKB4)
297
REK4(2)= (EC1(J-1)+EALFA1(J-1)**2*EC2(J-1))*
Cosh(EALFA1(J-1)*EZ)
REK4(3)= (EC1(J-1)+EALFA2(J-1)**2*EC2(J-1))*
Sinh(EALFA2(J-1)*EZ)
REK4(4)= (EC1(J-1)+EALFA2(J-1)**2*EC2(J-1))*
Cosh(EALFA2(J-1)*EZ)
REK4(5)= -((EC1(J)+EALFA1(J)**2*EC2(J))*
Sinh(EALFA1(J)*EZ))
REK4(6)= -((EC1(J)+EALFA1(J)**2*EC2(J))*
Cosh(EALFA1(J)*EZ))
REK4(7)= -((EC1(J)+EALFA2(J)**2*EC2(J))*
Sinh(EALFA2(J)*EZ))
REK4(8)= -((EC1(J)+EALFA2(J)**2*EC2(J))*
Cosh(EALFA2(J)*EZ))
REKB4=-(-EDD2(J) + ESDD2(J-1)
-(EC1(J)*ETozel(J))+EC1(J-1)*ESTozel(J-1)
-EC2(J)*ETozelzz(J)+EC2(J-1)*ESTozelzz(J-1))
@
@
RETURN
END
SUBROUTINE COEF3A(ETozelz,EALFA1,EALFA2,EZ,REK1,REKB1)
implicit real*8 (A-H,K-Z)
dimension REK1(8)
REK1(1)=EALFA1*Cosh(EALFA1*EZ)
REK1(2)=EALFA1*Sinh(EALFA1*EZ)
REK1(3)=EALFA2*Cosh(EALFA2*EZ)
REK1(4)=EALFA2*Sinh(EALFA2*EZ)
REKB1=-ETozelz
RETURN
END
SUBROUTINE COEF3B(ETozelz,ETozelzzz,EALFA1,EALFA2,
EZ,EGAMA,EMxz,EMyz,EMt,
EEIw,EALAN,Er,Ew,EEIyc,EEIxc,EEIx,EEIy,EEIxy,
EK1,EK2,EK3,EK4,Ed,EDelta,REK2,REKB2)
@
@
@
298
EEIxc*EK2-EALAN*(EALFA1**2*EGAMA*(EEIyc*EK1+EEIxc*EK2)+
Er*(Ed+EEIyc*EK3-EEIxc*EK4+Ew))))/(EALAN*Er))
REK2(2)=0.
REK2(3)= -((EALFA2*((-1+EALAN*EALFA2**2*EGAMA)*EEIw+EEIyc*EK1+
EEIxc*EK2-EALAN*(EALFA2**2*EGAMA*(EEIyc*EK1+EEIxc*EK2)+
Er*(Ed+EEIyc*EK3-EEIxc*EK4+Ew))))/(EALAN*Er))
REK2(4)=0.
REKB2=
-((EDelta*(EEIw-EEIyc*EK1-EEIxc*EK2)*ETozelz+
EALAN*((EEIxc*EEIy*EMxz+EEIxy*EEIyc*EMxz-EEIxc*EEIxy*EMyzEEIx*EEIyc*EMyz)*Er+EDelta*(Er*(EMt+Ed*ETozelz)EEIw*(EK4*EMxz+EK3*EMyz+EGAMA*ETozelzzz)+
EEIyc*(EK1*EK4*EMxz+EK1*EK3*EMyz+EK3*Er*ETozelz+
EGAMA*EK1*ETozelzzz)+
EEIxc*(EK2*EK4*EMxz+EK2*EK3*EMyz-EK4*Er*ETozelz+
EGAMA*EK2*ETozelzzz)+Er*ETozelz*Ew)))/
(EALAN*EDelta*Er))
RETURN
END
@
@
@
SUBROUTINE COEF1Ya(ibsay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EZ,EGAMA,EG,EEJ,EK1,EK2,EK3,EK4,EHP
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan
,EElas,ED1,ED2,ED3,ED4,REK1,REKB1,J)
@
@
@
@
REK1(1)=1
REK1(2)=0
REK1(3)=-1
REK1(4)=0
REKB1= -((EALFA1(J)**2*EALFA2(J)**2*EZ**2*
(-4*EBETA3(J)*(EAlan(J)*EBETA3(J)*EElas-EEJ(J)*EG)*
(EK3(J)*(EPx+EHP*EWx)+EK4(J)*(EPy+EHP*EWy))+
2*EBETA3(J)*(EAlan(J)*EBETA3(J)*EElas-EEJ(J)*EG)*
(EK3(J)*EWx+EK4(J)*EWy)*EZ)+
2*EALFA1(J)**2*EALFA2(J)**2*EZ*
(6*EAlan(J)*EBETA3(J)*
(2*EEJ(J)*EG*EGAMA(J)*(EK3(J)*EWx+EK4(J)*EWy)+
EBETA3(J)*EElas*EHP*
(2*EK3(J)*EPx+2*EK4(J)*EPy+EHP*EK3(J)*EWx+
EHP*EK4(J)*EWy))6*(2*EBETA2(J)*EEJ(J)*EG*(EK3(J)*EWx+EK4(J)*EWy)+
EBETA3(J)*(EEJ(J)*EG*EHP*
(2*EK3(J)*EPx+2*EK4(J)*EPy+EHP*EK3(J)*EWx+
EHP*EK4(J)*EWy)-2*EElas*
(EEIOw(J)*EK3(J)*EWx-Edy*Er(J)*EWx+EK1(J)*Er(J)*EWx+
EEIOw(J)*EK4(J)*EWy+Edx*Er(J)*EWy-EK2(J)*Er(J)*EWy)))
-4*EBETA3(J)*(EAlan(J)*EBETA3(J)*EElas-EEJ(J)*EG)*
(EK3(J)*(EPx+EHP*EWx)+EK4(J)*(EPy+EHP*EWy))*EZ+
EBETA3(J)*(EAlan(J)*EBETA3(J)*EElas-EEJ(J)*EG)*
(EK3(J)*EWx+EK4(J)*EWy)*EZ**2)+
24*EALFA1(J)*EALFA2(J)**2*EBETA3(J)**2*ED2(J)*EElas*
(-1+EAlan(J)*EALFA1(J)**2*EGAMA(J))*Sinh(EALFA1(J)*EZ)+
24*EBETA3(J)**2*EElas*
(EALFA1(J)*EALFA2(J)**2*ED1(J)*
(-1+EAlan(J)*EALFA1(J)**2*EGAMA(J))*Cosh(EALFA1(J)*EZ)+
EALFA1(J)**2*EALFA2(J)*ED3(J)*
(-1+EAlan(J)*EALFA2(J)**2*EGAMA(J))*Cosh(EALFA2(J)*EZ)+
EALFA1(J)**2*EALFA2(J)*ED4(J)*
(-1+EAlan(J)*EALFA2(J)**2*EGAMA(J))*Sinh(EALFA2(J)*EZ)))/
(24.*EAlan(J)*EALFA1(J)**2*EALFA2(J)**2*
299
EBETA3(J)**2*EElas**2*Er(J))
-(EALFA1(J-1)**2*EALFA2(J-1)**2*EZ**2*
(-4*EBETA3(J-1)*(EAlan(J-1)*
EBETA3(J-1)*EElas-EEJ(J-1)*EG)*
(EK3(J-1)*(EPx+EHP*EWx)+EK4(J-1)*(EPy+EHP*EWy))+
2*EBETA3(J-1)*(EAlan(J-1)*EBETA3(J-1)*EElas-EEJ(J-1)*EG)*
(EK3(J-1)*EWx+EK4(J-1)*EWy)*EZ)+
2*EALFA1(J-1)**2*EALFA2(J-1)**2*EZ*
(6*EAlan(J-1)*EBETA3(J-1)*
(2*EEJ(J-1)*EG*EGAMA(J-1)*(EK3(J-1)*EWx+EK4(J-1)*EWy)+
EBETA3(J-1)*EElas*EHP*
(2*EK3(J-1)*EPx+2*EK4(J-1)*EPy+EHP*EK3(J-1)*EWx+
EHP*EK4(J-1)*EWy))6*(2*EBETA2(J-1)*EEJ(J-1)*EG*(EK3(J-1)*EWx+EK4(J-1)*EWy)+
EBETA3(J-1)*(EEJ(J-1)*EG*EHP*
(2*EK3(J-1)*EPx+2*EK4(J-1)*EPy+EHP*EK3(J-1)*EWx+
EHP*EK4(J-1)*EWy)-2*EElas*
(EEIOw(J-1)*EK3(J-1)*EWx-Edy*Er(J-1)*
EWx+EK1(J-1)*Er(J-1)*EWx+
EEIOw(J-1)*EK4(J-1)*EWy+Edx*Er(J-1)*
EWy-EK2(J-1)*Er(J-1)*EWy)))
-4*EBETA3(J-1)*(EAlan(J-1)*
EBETA3(J-1)*EElas-EEJ(J-1)*EG)*
(EK3(J-1)*(EPx+EHP*EWx)+EK4(J-1)*(EPy+EHP*EWy))*EZ+
EBETA3(J-1)*(EAlan(J-1)*EBETA3(J-1)*EElas-EEJ(J-1)*EG)*
(EK3(J-1)*EWx+EK4(J-1)*EWy)*EZ**2)+
24*EALFA1(J-1)*EALFA2(J-1)**2*
EBETA3(J-1)**2*ED2(J-1)*EElas*
(-1+EAlan(J-1)*EALFA1(J-1)**2*
EGAMA(J-1))*Sinh(EALFA1(J-1)*EZ)+
24*EBETA3(J-1)**2*EElas*
(EALFA1(J-1)*EALFA2(J-1)**2*ED1(J-1)*
(-1+EAlan(J-1)*EALFA1(J-1)**2*
EGAMA(J-1))*Cosh(EALFA1(J-1)*EZ)
+EALFA1(J-1)**2*EALFA2(J-1)*ED3(J-1)*
(-1+EAlan(J-1)*EALFA2(J-1)**2*
EGAMA(J-1))*Cosh(EALFA2(J-1)*EZ)
+EALFA1(J-1)**2*EALFA2(J-1)*ED4(J-1)*
(-1+EAlan(J-1)*EALFA2(J-1)**2*
EGAMA(J-1))*Sinh(EALFA2(J-1)*EZ)
))/(24.*EAlan(J-1)*EALFA1(J-1)**2*EALFA2(J-1)**2*
EBETA3(J-1)**2*EElas**2*Er(J-1)))
return
end
@
@
@
SUBROUTINE COEF2Ya(ibsay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EZ,EGAMA,EG,EEJ,EK1,EK2,EK3,EK4,EHP
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan
,EElas,ED1,ED2,ED3,ED4,REK2,REKB2,J)
@
@
@
@
REK2(1)=EZ
REK2(2)=1
REK2(3)=-EZ
REK2(4)=-1
REKB2= -((EALFA1(J)**2*EALFA2(J)**2*EZ**2*
(6*EAlan(J)*EBETA3(J)*
(2*EEJ(J)*EG*EGAMA(J)*(EK3(J)*EWx+EK4(J)*EWy)+
EBETA3(J)*EElas*EHP*
300
(2*EK3(J)*EPx+2*EK4(J)*EPy+EHP*EK3(J)*EWx+
EHP*EK4(J)*EWy))6*(2*EBETA2(J)*EEJ(J)*EG*(EK3(J)*EWx+EK4(J)*EWy)+
EBETA3(J)*(EEJ(J)*EG*EHP*
(2*EK3(J)*EPx+2*EK4(J)*EPy+EHP*EK3(J)*EWx+
EHP*EK4(J)*EWy)-2*EElas*
(EEIOw(J)*EK3(J)*EWx-Edy*Er(J)*EWx+EK1(J)*Er(J)*EWx+
EEIOw(J)*EK4(J)*EWy+Edx*Er(J)*EWy-EK2(J)*Er(J)*EWy)))
-4*EBETA3(J)*(EAlan(J)*EBETA3(J)*EElas-EEJ(J)*EG)*
(EK3(J)*(EPx+EHP*EWx)+EK4(J)*(EPy+EHP*EWy))*EZ+
EBETA3(J)*(EAlan(J)*EBETA3(J)*EElas-EEJ(J)*EG)*
(EK3(J)*EWx+EK4(J)*EWy)*EZ**2)+
24*EALFA2(J)**2*EBETA3(J)**2*ED2(J)*EElas*
(-1+EAlan(J)*EALFA1(J)**2*EGAMA(J))*Cosh(EALFA1(J)*EZ)+
24*EBETA3(J)**2*EElas*
(EALFA1(J)**2*ED4(J)*(-1+EAlan(J)*EALFA2(J)**2*EGAMA(J))*
Cosh(EALFA2(J)*EZ)+
EALFA2(J)**2*ED1(J)*(-1+EAlan(J)*EALFA1(J)**2*EGAMA(J))*
Sinh(EALFA1(J)*EZ)+
EALFA1(J)**2*ED3(J)*(-1+EAlan(J)*EALFA2(J)**2*EGAMA(J))*
Sinh(EALFA2(J)*EZ)))/
(24.*EAlan(J)*EALFA1(J)**2*EALFA2(J)**2*
EBETA3(J)**2*EElas**2*Er(J))
-(EALFA1(J-1)**2*EALFA2(J-1)**2*EZ**2*
(6*EAlan(J-1)*EBETA3(J-1)*
(2*EEJ(J-1)*EG*EGAMA(J-1)*(EK3(J-1)*EWx+EK4(J-1)*EWy)+
EBETA3(J-1)*EElas*EHP*
(2*EK3(J-1)*EPx+2*EK4(J-1)*EPy+EHP*EK3(J-1)*EWx+
EHP*EK4(J-1)*EWy))6*(2*EBETA2(J-1)*EEJ(J-1)*EG*(EK3(J-1)*EWx+EK4(J-1)*EWy)+
EBETA3(J-1)*(EEJ(J-1)*EG*EHP*
(2*EK3(J-1)*EPx+2*EK4(J-1)*EPy+EHP*EK3(J-1)*EWx+
EHP*EK4(J-1)*EWy)-2*EElas*
(EEIOw(J-1)*EK3(J-1)*EWx-Edy*Er(J-1)*
EWx+EK1(J-1)*Er(J-1)*EWx+
EEIOw(J-1)*EK4(J-1)*EWy+Edx*Er(J-1)*
EWy-EK2(J-1)*Er(J-1)*EWy)))
-4*EBETA3(J-1)*(EAlan(J-1)*
EBETA3(J-1)*EElas-EEJ(J-1)*EG)*
(EK3(J-1)*(EPx+EHP*EWx)+EK4(J-1)*(EPy+EHP*EWy))*EZ+
EBETA3(J-1)*(EAlan(J-1)*EBETA3(J-1)*EElas-EEJ(J-1)*EG)*
(EK3(J-1)*EWx+EK4(J-1)*EWy)*EZ**2)+
24*EALFA2(J-1)**2*EBETA3(J-1)**2*ED2(J-1)*EElas*
(-1+EAlan(J-1)*EALFA1(J-1)**2*
EGAMA(J-1))*Cosh(EALFA1(J-1)*EZ)+
24*EBETA3(J-1)**2*EElas*
(EALFA1(J-1)**2*ED4(J-1)*(-1+EAlan(J-1)*EALFA2(J-1)**2*
EGAMA(J-1))*Cosh(EALFA2(J-1)*EZ)+
EALFA2(J-1)**2*ED1(J-1)*(-1+EAlan(J-1)*EALFA1(J-1)**2*
EGAMA(J-1))*Sinh(EALFA1(J-1)*EZ)+
EALFA1(J-1)**2*ED3(J-1)*(-1+EAlan(J-1)*EALFA2(J-1)**2*
EGAMA(J-1))*Sinh(EALFA2(J-1)*EZ)))/
(24.*EAlan(J-1)*EALFA1(J-1)**2*EALFA2(J-1)**2*
EBETA3(J-1)**2*EElas**2*Er(J-1)))
return
end
@
@
@
SUBROUTINE COEF3Ya(ibsay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EZ,EGAMA,EG,EEJ,EK1,EK2,EK3,EK4,EHP
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan
,EElas,ED1,ED2,ED3,ED4,REK1,REKB1,J)
implicit real*8 (A-H,K-Z)
dimension REK1(4),EBETA1(ibsay),EBETA2(ibsay),EBETA3(ibsay)
301
@
@
@
@
,EALFA1(ibsay),EALFA2(ibsay),EGAMA(ibsay)
,ED1(ibsay),ED2(ibsay),ED3(ibsay),ED4(ibsay)
,EEJ(ibsay),EK1(ibsay),EK2(ibsay),EK3(ibsay)
,EK4(ibsay),Er(ibsay),EEIOw(ibsay),EAlan(ibsay)
REK1(1)=1
REK1(2)=0
REKB1= -(-((EALFA1(J)*EALFA2(J)**2*ED1(J)*
(-1+EAlan(J)*EALFA1(J)**2*EGAMA(J))+
EALFA1(J)**2*EALFA2(J)*ED3(J)*
(-1+EAlan(J)*EALFA2(J)**2*EGAMA(J)))/
(EAlan(J)*EALFA1(J)**2*EALFA2(J)**2*EElas*Er(J))))
return
end
@
@
@
SUBROUTINE COEF4Ya(ibsay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EZ,EGAMA,EG,EEJ,EK1,EK2,EK3,EK4,EHP
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan
,EElas,ED1,ED2,ED3,ED4,REK2,REKB2,J)
@
@
@
@
REK2(1)=0
REK2(2)=1
REKB2= -((-24*EALFA2(J)**2*EBETA3(J)**2*ED2(J)*EElas*
(-1+EAlan(J)*EALFA1(J)**2*EGAMA(J))24*EALFA1(J)**2*EBETA3(J)**2*ED4(J)*EElas*
(-1+EAlan(J)*EALFA2(J)**2*EGAMA(J)))/
(24.*EAlan(J)*EALFA1(J)**2*EALFA2(J)**2*
EBETA3(J)**2*EElas**2*Er(J)))
return
end
@
@
@
SUBROUTINE COEF1Yb(ibsay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EZ,EGAMA,EG,EEJ,EK1,EK2,EK3,EK4,EHP
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan
,EElas,ED1,ED2,ED3,ED4,REK1,REKB1,J)
@
@
@
@
REK1(1)=1
REK1(2)=0
REK1(3)=-1
REK1(4)=0
REKB1= -(-((EALFA1(J)*EALFA2(J)**2*ED1(J)*(1 - EAlan(J)*
EALFA1(J)**2*EGAMA(J))*
Cosh(EALFA1(J)*EZ) +
EALFA1(J)*EALFA2(J)**2*ED2(J)*(1 EAlan(J)*EALFA1(J)**2*EGAMA(J))*
Sinh(EALFA1(J)*EZ) EALFA1(J)**2*(-1 + EAlan(J)*EALFA2(J)**2*EGAMA(J))*
(EALFA2(J)*ED3(J)*Cosh(EALFA2(J)*EZ) +
EALFA2(J)*ED4(J)*Sinh(EALFA2(J)*EZ)))/
302
(EAlan(J)*EALFA1(J)**2*EALFA2(J)**2*EElas*Er(J))) +
(EALFA1(J-1)*EALFA2(J-1)**2*ED1(J-1)*(1 - EAlan(J-1)*
EALFA1(J-1)**2*EGAMA(J-1))*
Cosh(EALFA1(J-1)*EZ) +
EALFA1(J-1)*EALFA2(J-1)**2*ED2(J-1)*
(1 - EAlan(J-1)*EALFA1(J-1)**2*EGAMA(J-1))
*Sinh(EALFA1(J-1)*EZ) EALFA1(J-1)**2*(-1 + EAlan(J-1)*EALFA2(J-1)**2*EGAMA(J-1))*
(EALFA2(J-1)*ED3(J-1)*Cosh(EALFA2(J-1)*EZ) +
EALFA2(J-1)*ED4(J-1)*Sinh(EALFA2(J-1)*EZ)))/
(EAlan(J-1)*EALFA1(J-1)**2*EALFA2(J-1)**2*EElas*Er(J-1)))
return
end
@
@
@
SUBROUTINE COEF2Yb(ibsay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EZ,EGAMA,EG,EEJ,EK1,EK2,EK3,EK4,EHP
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan
,EElas,ED1,ED2,ED3,ED4,REK2,REKB2,J)
@
@
@
@
REK2(1)=EZ
REK2(2)=1
REK2(3)=-EZ
REK2(4)=-1
REKB2= -(-((EALFA2(J)**2*ED2(J)*(1 - EAlan(J)*
EALFA1(J)**2*EGAMA(J))*
Cosh(EALFA1(J)*EZ) +
EALFA2(J)**2*ED1(J)*(1 - EAlan(J)*EALFA1(J)**2*EGAMA(J))*
Sinh(EALFA1(J)*EZ) EALFA1(J)**2*(-1 + EAlan(J)*EALFA2(J)**2*EGAMA(J))*
(ED4(J)*Cosh(EALFA2(J)*EZ) + ED3(J)*Sinh(EALFA2(J)*EZ)))/
(EAlan(J)*EALFA1(J)**2*EALFA2(J)**2*EElas*Er(J))) +
(EALFA2(J-1)**2*ED2(J-1)*(1 - EAlan(J-1)*EALFA1(J-1)**2*
EGAMA(J-1))*Cosh(EALFA1(J-1)*EZ) +
EALFA2(J-1)**2*ED1(J-1)*(1 - EAlan(J-1)*
EALFA1(J-1)**2*EGAMA(J-1))*
Sinh(EALFA1(J-1)*EZ) EALFA1(J-1)**2*(-1 + EAlan(J-1)*EALFA2(J-1)**2*EGAMA(J-1))*
(ED4(J-1)*Cosh(EALFA2(J-1)*EZ) + ED3(J-1)*
Sinh(EALFA2(J-1)*EZ)))/
(EAlan(J-1)*EALFA1(J-1)**2*EALFA2(J-1)**2*EElas*Er(J-1)))
return
end
@
@
@
SUBROUTINE COEF3Yb(ibsay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EZ,EGAMA,EG,EEJ,EK1,EK2,EK3,EK4,EHP
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan
,EElas,ED1,ED2,ED3,ED4,REK1,REKB1,J)
@
@
@
@
303
REK1(1)=1
REK1(2)=0
REK1(3)=-1
REK1(4)=0
REKB1= -((EALFA1(J)**2*EALFA2(J)**2*EZ**2*
(-4*EBETA3(J)*(EAlan(J)*EBETA3(J)*EElas-EEJ(J)*EG)*
(EK3(J)*(EPx+EHP*EWx)+EK4(J)*(EPy+EHP*EWy))+
2*EBETA3(J)*(EAlan(J)*EBETA3(J)*EElas-EEJ(J)*EG)*
(EK3(J)*EWx+EK4(J)*EWy)*EZ)+
2*EALFA1(J)**2*EALFA2(J)**2*EZ*
(6*EAlan(J)*EBETA3(J)*
(2*EEJ(J)*EG*EGAMA(J)*(EK3(J)*EWx+EK4(J)*EWy)+
EBETA3(J)*EElas*EHP*
(2*EK3(J)*EPx+2*EK4(J)*EPy+EHP*EK3(J)*EWx+
EHP*EK4(J)*EWy))6*(2*EBETA2(J)*EEJ(J)*EG*(EK3(J)*EWx+EK4(J)*EWy)+
EBETA3(J)*(EEJ(J)*EG*EHP*
(2*EK3(J)*EPx+2*EK4(J)*EPy+EHP*EK3(J)*EWx+
EHP*EK4(J)*EWy)-2*EElas*
(EEIOw(J)*EK3(J)*EWx-Edy*Er(J)*EWx+EK1(J)*Er(J)*EWx+
EEIOw(J)*EK4(J)*EWy+Edx*Er(J)*EWy-EK2(J)*Er(J)*EWy)))
-4*EBETA3(J)*(EAlan(J)*EBETA3(J)*EElas-EEJ(J)*EG)*
(EK3(J)*(EPx+EHP*EWx)+EK4(J)*(EPy+EHP*EWy))*EZ+
EBETA3(J)*(EAlan(J)*EBETA3(J)*EElas-EEJ(J)*EG)*
(EK3(J)*EWx+EK4(J)*EWy)*EZ**2)+
24*EALFA1(J)*EALFA2(J)**2*EBETA3(J)**2*ED2(J)*EElas*
(-1+EAlan(J)*EALFA1(J)**2*EGAMA(J))*Sinh(EALFA1(J)*EZ)+
24*EBETA3(J)**2*EElas*
(EALFA1(J)*EALFA2(J)**2*ED1(J)*
(-1+EAlan(J)*EALFA1(J)**2*EGAMA(J))*Cosh(EALFA1(J)*EZ)+
EALFA1(J)**2*EALFA2(J)*ED3(J)*
(-1+EAlan(J)*EALFA2(J)**2*EGAMA(J))*Cosh(EALFA2(J)*EZ)+
EALFA1(J)**2*EALFA2(J)*ED4(J)*
(-1+EAlan(J)*EALFA2(J)**2*EGAMA(J))*Sinh(EALFA2(J)*EZ)))/
(24.*EAlan(J)*EALFA1(J)**2*EALFA2(J)**2*
EBETA3(J)**2*EElas**2*Er(J))+
(EALFA1(J-1)*EALFA2(J-1)**2*ED1(J-1)*(1 - EAlan(J-1)*
EALFA1(J-1)**2*EGAMA(J-1))*
Cosh(EALFA1(J-1)*EZ) +
EALFA1(J-1)*EALFA2(J-1)**2*ED2(J-1)*
(1 - EAlan(J-1)*EALFA1(J-1)**2*EGAMA(J-1))
*Sinh(EALFA1(J-1)*EZ) EALFA1(J-1)**2*(-1 + EAlan(J-1)*EALFA2(J-1)**2*EGAMA(J-1))*
(EALFA2(J-1)*ED3(J-1)*Cosh(EALFA2(J-1)*EZ) +
EALFA2(J-1)*ED4(J-1)*Sinh(EALFA2(J-1)*EZ)))/
(EAlan(J-1)*EALFA1(J-1)**2*EALFA2(J-1)**2*EElas*Er(J-1)))
return
end
@
@
@
SUBROUTINE COEF4Yb(ibsay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EZ,EGAMA,EG,EEJ,EK1,EK2,EK3,EK4,EHP
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan
,EElas,ED1,ED2,ED3,ED4,REK2,REKB2,J)
@
@
@
@
304
REK2(4)=-1
REKB2= -((EALFA1(J)**2*EALFA2(J)**2*EZ**2*
(6*EAlan(J)*EBETA3(J)*
(2*EEJ(J)*EG*EGAMA(J)*(EK3(J)*EWx+EK4(J)*EWy)+
EBETA3(J)*EElas*EHP*
(2*EK3(J)*EPx+2*EK4(J)*EPy+EHP*EK3(J)*EWx+
EHP*EK4(J)*EWy))6*(2*EBETA2(J)*EEJ(J)*EG*(EK3(J)*EWx+EK4(J)*EWy)+
EBETA3(J)*(EEJ(J)*EG*EHP*
(2*EK3(J)*EPx+2*EK4(J)*EPy+EHP*EK3(J)*EWx+
EHP*EK4(J)*EWy)-2*EElas*
(EEIOw(J)*EK3(J)*EWx-Edy*Er(J)*EWx+EK1(J)*Er(J)*EWx+
EEIOw(J)*EK4(J)*EWy+Edx*Er(J)*EWy-EK2(J)*Er(J)*EWy)))
-4*EBETA3(J)*(EAlan(J)*EBETA3(J)*EElas-EEJ(J)*EG)*
(EK3(J)*(EPx+EHP*EWx)+EK4(J)*(EPy+EHP*EWy))*EZ+
EBETA3(J)*(EAlan(J)*EBETA3(J)*EElas-EEJ(J)*EG)*
(EK3(J)*EWx+EK4(J)*EWy)*EZ**2)+
24*EALFA2(J)**2*EBETA3(J)**2*ED2(J)*EElas*
(-1+EAlan(J)*EALFA1(J)**2*EGAMA(J))*Cosh(EALFA1(J)*EZ)+
24*EBETA3(J)**2*EElas*
(EALFA1(J)**2*ED4(J)*(-1+EAlan(J)*EALFA2(J)**2*EGAMA(J))*
Cosh(EALFA2(J)*EZ)+
EALFA2(J)**2*ED1(J)*(-1+EAlan(J)*EALFA1(J)**2*EGAMA(J))*
Sinh(EALFA1(J)*EZ)+
EALFA1(J)**2*ED3(J)*(-1+EAlan(J)*EALFA2(J)**2*EGAMA(J))*
Sinh(EALFA2(J)*EZ)))/
(24.*EAlan(J)*EALFA1(J)**2*EALFA2(J)**2*
EBETA3(J)**2*EElas**2*Er(J))
+
(EALFA2(J-1)**2*ED2(J-1)*(1 - EAlan(J-1)*EALFA1(J-1)**2*
EGAMA(J-1))*Cosh(EALFA1(J-1)*EZ) +
EALFA2(J-1)**2*ED1(J-1)*(1 - EAlan(J-1)*
EALFA1(J-1)**2*EGAMA(J-1))*
Sinh(EALFA1(J-1)*EZ) EALFA1(J-1)**2*(-1 + EAlan(J-1)*EALFA2(J-1)**2*EGAMA(J-1))*
(ED4(J-1)*Cosh(EALFA2(J-1)*EZ) + ED3(J-1)*
Sinh(EALFA2(J-1)*EZ)))/
(EAlan(J-1)*EALFA1(J-1)**2*EALFA2(J-1)**2*EElas*Er(J-1)))
return
end
@
@
@
SUBROUTINE TETAFONKa(ibsay,iksay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EGAMA,EK1,EK2,EK3,EK4,EElas,EG,EEJ
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan,EHP
,ED1,ED2,ED3,ED4,ETETA,ETET,J,jm,EG1,EG2)
@
@
@
@
@
ETET(jm)*EG1(J)+EG2(J)-(EALFA1(J)**2*
EALFA2(J)**2*ETET(jm)**2*
(6*EAlan(J)*EBETA3(J)*
(2*EEJ(J)*EG*EGAMA(J)*(EK3(J)*EWx+EK4(J)*EWy)+
EBETA3(J)*EElas*EHP*
(2*EK3(J)*EPx+2*EK4(J)*EPy+EHP*EK3(J)*EWx+
305
EHP*EK4(J)*EWy))6*(2*EBETA2(J)*EEJ(J)*EG*(EK3(J)*EWx+EK4(J)*EWy)+
EBETA3(J)*(EEJ(J)*EG*EHP*
(2*EK3(J)*EPx+2*EK4(J)*EPy+EHP*EK3(J)*EWx+
EHP*EK4(J)*EWy)-2*EElas*
(EEIOw(J)*EK3(J)*EWx-Edy*Er(J)*EWx+EK1(J)*Er(J)*EWx+
EEIOw(J)*EK4(J)*EWy+Edx*Er(J)*EWy-EK2(J)*Er(J)*EWy)))
-4*EBETA3(J)*(EAlan(J)*EBETA3(J)*EElas-EEJ(J)*EG)*
(EK3(J)*(EPx+EHP*EWx)+EK4(J)*(EPy+EHP*EWy))*ETET(jm)+
EBETA3(J)*(EAlan(J)*EBETA3(J)*EElas-EEJ(J)*EG)*
(EK3(J)*EWx+EK4(J)*EWy)*ETET(jm)**2)+
24*EALFA2(J)**2*EBETA3(J)**2*ED2(J)*EElas*
(-1+EAlan(J)*EALFA1(J)**2*EGAMA(J))*
Cosh(EALFA1(J)*ETET(jm))+
24*EBETA3(J)**2*EElas*
(EALFA1(J)**2*ED4(J)*(-1+EAlan(J)*EALFA2(J)**2*EGAMA(J))*
Cosh(EALFA2(J)*ETET(jm))+
EALFA2(J)**2*ED1(J)*(-1+EAlan(J)*EALFA1(J)**2*EGAMA(J))*
Sinh(EALFA1(J)*ETET(jm))+
EALFA1(J)**2*ED3(J)*(-1+EAlan(J)*EALFA2(J)**2*EGAMA(J))*
Sinh(EALFA2(J)*ETET(jm))))/
(24.*EAlan(J)*EALFA1(J)**2*EALFA2(J)**2*
EBETA3(J)**2*EElas**2*Er(J))
return
end
@
@
@
SUBROUTINE TETAFONKb(ibsay,iksay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EGAMA,EK1,EK2,EK3,EK4,EElas,EG,EEJ
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan,EHP
,ED1,ED2,ED3,ED4,ETETA,ETET,J,jm,EG1,EG2)
@
@
@
@
@
@
@
@
SUBROUTINE COEF1Ua(ibsay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EZ,EGAMA,EG,EEJ,EK1,EK2,EK3,EK4,EHP,EDELTA
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan,EEIx,EEIxy
,EElas,ED1,ED2,ED3,ED4,REK1,REKB1,J)
306
@
@
,EK4(ibsay),Er(ibsay),EEIOw(ibsay),EAlan(ibsay)
,EDELTA(ibsay),EEIx(ibsay),EEIxy(ibsay)
REK1(1)=1
REK1(2)=0
REK1(3)=-1
REK1(4)=0
REKB1= -((EALFA1(J)**2*EALFA2(J)**2*EZ**2*
(EDELTA(J)*(-4*EBETA3(J)*
(EAlan(J)*EBETA3(J)*EElas*EK1(J) - EEJ(J)*EG*EK1(J)+
EAlan(J)*EEJ(J)*EG*EK3(J)*Er(J))*
(EK3(J)*(EPx + EHP*EWx) + EK4(J)*(EPy + EHP*EWy))+
2*EBETA3(J)*
(EAlan(J)*EBETA3(J)*EElas*EK1(J) - EEJ(J)*EG*EK1(J)+
EAlan(J)*EEJ(J)*EG*EK3(J)*Er(J))*
(EK3(J)*EWx + EK4(J)*EWy)*EZ)
+ EAlan(J)*EBETA3(J)**2*EElas*Er(J)*
(EEIx(J)*(4*(EPx + EHP*EWx) - 2*EWx*EZ) +
EEIxy(J)*(-4*(EPy + EHP*EWy) + 2*EWy*EZ))) +
2*EALFA1(J)**2*EALFA2(J)**2*EZ*
(EDELTA(J)*(12*EAlan(J)*EBETA2(J)*EEJ(J)*EG*EK3(J)*Er(J)*
(EK3(J)*EWx + EK4(J)*EWy) +
6*EAlan(J)*EBETA3(J)**2*EElas*EHP*EK1(J)*
(2*EK3(J)*EPx + 2*EK4(J)*EPy + EHP*EK3(J)*EWx +
EHP*EK4(J)*EWy) 12*EAlan(J)*EBETA3(J)*EElas*EK3(J)*Er(J)*
(Er(J)*(-(Edy*EWx) + EK1(J)*EWx + Edx*EWy EK2(J)*EWy) + EEIOw(J)*(EK3(J)*EWx + EK4(J)*EWy))+
6*EAlan(J)*EBETA3(J)*EEJ(J)*EG*
(2*EHP*EK3(J)*(EK3(J)*EPx + EK4(J)*EPy)*Er(J) +
2*EGAMA(J)*EK1(J)*(EK3(J)*EWx + EK4(J)*EWy) +
EHP**2*EK3(J)*Er(J)*(EK3(J)*EWx + EK4(J)*EWy)) 6*EK1(J)*(2*EBETA2(J)*EEJ(J)*EG*(EK3(J)*EWx + EK4(J)*EWy)+
EBETA3(J)*
(EEJ(J)*EG*EHP*
(2*EK3(J)*EPx + 2*EK4(J)*EPy + EHP*EK3(J)*EWx +
EHP*EK4(J)*EWy) 2*EElas*
(EEIOw(J)*EK3(J)*EWx - Edy*Er(J)*EWx +
EK1(J)*Er(J)*EWx + EEIOw(J)*EK4(J)*EWy +
Edx*Er(J)*EWy - EK2(J)*Er(J)*EWy))) 4*EBETA3(J)*
(EAlan(J)*EBETA3(J)*EElas*EK1(J) - EEJ(J)*EG*EK1(J)+
EAlan(J)*EEJ(J)*EG*EK3(J)*Er(J))*
(EK3(J)*(EPx + EHP*EWx) + EK4(J)*(EPy + EHP*EWy))*EZ +
EBETA3(J)*(EAlan(J)*EBETA3(J)*EElas*EK1(J)-EEJ(J)*EG*EK1(J)+
EAlan(J)*EEJ(J)*EG*EK3(J)*Er(J))*(EK3(J)*EWx+EK4(J)*EWy)*
EZ**2) +
EAlan(J)*EBETA3(J)**2*EElas*Er(J)*
(EEIx(J)*(-6*EHP*(2*EPx + EHP*EWx) +
4*(EPx + EHP*EWx)*EZ - EWx*EZ**2) +
EEIxy(J)*(6*EHP*(2*EPy + EHP*EWy) 4*(EPy + EHP*EWy)*EZ + EWy*EZ**2))) 24*EDELTA(J)*EBETA3(J)**2*EElas*
(EALFA1(J)*EALFA2(J)**2*ED1(J)*
(EK1(J) - EAlan(J)*EALFA1(J)**2*EGAMA(J)*EK1(J) EAlan(J)*EK3(J)*Er(J))*Cosh(EALFA1(J)*EZ) +
EALFA1(J)*EALFA2(J)**2*ED2(J)*
(EK1(J) - EAlan(J)*EALFA1(J)**2*EGAMA(J)*EK1(J) EAlan(J)*EK3(J)*Er(J))*Sinh(EALFA1(J)*EZ) EALFA1(J)**2*((-1 + EAlan(J)*EALFA2(J)**2*EGAMA(J))*EK1(J)+
EAlan(J)*EK3(J)*Er(J))*
(EALFA2(J)*ED3(J)*Cosh(EALFA2(J)*EZ) +
EALFA2(J)*ED4(J)*Sinh(EALFA2(J)*EZ))))/
(24.*EDELTA(J)*EAlan(J)*EALFA1(J)**2*EALFA2(J)**2*EBETA3(J)**2*
EElas**2*Er(J)) + (-(EALFA1(J-1)**2*EALFA2(J-1)**2*EZ**2*
307
(EDELTA(J-1)*(-4*EBETA3(J-1)*
(EAlan(J-1)*EBETA3(J-1)*EElas*
EK1(J-1) - EEJ(J-1)*EG*EK1(J-1) +
EAlan(J-1)*EEJ(J-1)*EG*EK3(J-1)*Er(J-1))*
(EK3(J-1)*(EPx + EHP*EWx) + EK4(J-1)*(EPy + EHP*EWy))+
2*EBETA3(J-1)*
(EAlan(J-1)*EBETA3(J-1)*EElas*
EK1(J-1) - EEJ(J-1)*EG*EK1(J-1) +
EAlan(J-1)*EEJ(J-1)*EG*EK3(J-1)*Er(J-1))*
(EK3(J-1)*EWx + EK4(J-1)*EWy)*EZ
) + EAlan(J-1)*EBETA3(J-1)**2*EElas*Er(J-1)*
(EEIx(J-1)*(4*(EPx + EHP*EWx) - 2*EWx*EZ) +
EEIxy(J-1)*(-4*(EPy + EHP*EWy) + 2*EWy*EZ)))) 2*EALFA1(J-1)**2*EALFA2(J-1)**2*EZ*
(EDELTA(J-1)*(12*EAlan(J-1)*EBETA2(J-1)*
EEJ(J-1)*EG*EK3(J-1)*Er(J-1)*
(EK3(J-1)*EWx + EK4(J-1)*EWy) +
6*EAlan(J-1)*EBETA3(J-1)**2*EElas*EHP*EK1(J-1)*
(2*EK3(J-1)*EPx + 2*EK4(J-1)*EPy + EHP*EK3(J-1)*EWx+
EHP*EK4(J-1)*EWy) 12*EAlan(J-1)*EBETA3(J-1)*EElas*EK3(J-1)*Er(J-1)*
(Er(J-1)*(-(Edy*EWx) + EK1(J-1)*EWx + Edx*EWy EK2(J-1)*EWy) + EEIOw(J-1)*
(EK3(J-1)*EWx + EK4(J-1)*EWy)) +
6*EAlan(J-1)*EBETA3(J-1)*EEJ(J-1)*EG*
(2*EHP*EK3(J-1)*(EK3(J-1)*EPx + EK4(J-1)*EPy)*Er(J-1) +
2*EGAMA(J-1)*EK1(J-1)*(EK3(J-1)*EWx + EK4(J-1)*EWy)+
EHP**2*EK3(J-1)*Er(J-1)*(EK3(J-1)*EWx + EK4(J-1)*EWy))6*EK1(J-1)*(2*EBETA2(J-1)*EEJ(J-1)*EG*
(EK3(J-1)*EWx + EK4(J-1)*EWy) +
EBETA3(J-1)*
(EEJ(J-1)*EG*EHP*
(2*EK3(J-1)*EPx + 2*EK4(J-1)*EPy + EHP*EK3(J-1)*EWx+
EHP*EK4(J-1)*EWy) 2*EElas*
(EEIOw(J-1)*EK3(J-1)*EWx - Edy*Er(J-1)*EWx +
EK1(J-1)*Er(J-1)*EWx + EEIOw(J-1)*EK4(J-1)*EWy+
Edx*Er(J-1)*EWy - EK2(J-1)*Er(J-1)*EWy))) 4*EBETA3(J-1)*
(EAlan(J-1)*EBETA3(J-1)*EElas*
EK1(J-1) - EEJ(J-1)*EG*EK1(J-1) +
EAlan(J-1)*EEJ(J-1)*EG*EK3(J-1)*Er(J-1))*
(EK3(J-1)*(EPx + EHP*EWx) + EK4(J-1)*(EPy + EHP*EWy))*EZ+
EBETA3(J-1)*(EAlan(J-1)*EBETA3(J-1)*EElas*EK1(J-1) EEJ(J-1)*EG*EK1(J-1) + EAlan(J-1)*
EEJ(J-1)*EG*EK3(J-1)*Er(J-1))*
(EK3(J-1)*EWx + EK4(J-1)*EWy)*EZ**2) +
EAlan(J-1)*EBETA3(J-1)**2*EElas*Er(J-1)*
(EEIx(J-1)*(-6*EHP*(2*EPx + EHP*EWx) +
4*(EPx + EHP*EWx)*EZ - EWx*EZ**2) +
EEIxy(J-1)*(6*EHP*(2*EPy + EHP*EWy) 4*(EPy + EHP*EWy)*EZ + EWy*EZ**2))) +
24*EDELTA(J-1)*EBETA3(J-1)**2*EElas*
(EALFA1(J-1)*EALFA2(J-1)**2*ED1(J-1)*
(EK1(J-1) -EAlan(J-1)*EALFA1(J-1)**2*EGAMA(J-1)*EK1(J-1)EAlan(J-1)*EK3(J-1)*Er(J-1))*Cosh(EALFA1(J-1)*EZ) +
EALFA1(J-1)*EALFA2(J-1)**2*ED2(J-1)*
(EK1(J-1) -EAlan(J-1)*EALFA1(J-1)**2*EGAMA(J-1)*EK1(J-1)EAlan(J-1)*EK3(J-1)*Er(J-1))*Sinh(EALFA1(J-1)*EZ) EALFA1(J-1)**2*((-1 + EAlan(J-1)*EALFA2(J-1)**2*EGAMA(J-1))*
EK1(J-1) + EAlan(J-1)*EK3(J-1)*Er(J-1))*
(EALFA2(J-1)*ED3(J-1)*Cosh(EALFA2(J-1)*EZ) +
EALFA2(J-1)*ED4(J-1)*Sinh(EALFA2(J-1)*EZ))))/
(24.*EDELTA(J-1)*EAlan(J-1)*EALFA1(J-1)**2*
EALFA2(J-1)**2*EBETA3(J-1)**2*
EElas**2*Er(J-1)))
308
return
end
@
@
@
SUBROUTINE COEF2Ua(ibsay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EZ,EGAMA,EG,EEJ,EK1,EK2,EK3,EK4,EHP,EDELTA
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan,EEIx,EEIxy
,EElas,ED1,ED2,ED3,ED4,REK2,REKB2,J)
@
@
@
@
@
REK2(1)=EZ
REK2(2)=1
REK2(3)=-EZ
REK2(4)=-1
REKB2= -((EALFA1(J)**2*EALFA2(J)**2*EZ**2*
(EDELTA(J)*(12*EAlan(J)*EBETA2(J)*EEJ(J)*EG*EK3(J)*Er(J)*
(EK3(J)*EWx + EK4(J)*EWy) +
6*EAlan(J)*EBETA3(J)**2*EElas*EHP*EK1(J)*
(2*EK3(J)*EPx + 2*EK4(J)*EPy + EHP*EK3(J)*EWx +
EHP*EK4(J)*EWy) 12*EAlan(J)*EBETA3(J)*EElas*EK3(J)*Er(J)*
(Er(J)*(-(Edy*EWx) + EK1(J)*EWx + Edx*EWy EK2(J)*EWy) + EEIOw(J)*(EK3(J)*EWx + EK4(J)*EWy))+
6*EAlan(J)*EBETA3(J)*EEJ(J)*EG*
(2*EHP*EK3(J)*(EK3(J)*EPx + EK4(J)*EPy)*Er(J) +
2*EGAMA(J)*EK1(J)*(EK3(J)*EWx + EK4(J)*EWy) +
EHP**2*EK3(J)*Er(J)*(EK3(J)*EWx + EK4(J)*EWy)) 6*EK1(J)*(2*EBETA2(J)*EEJ(J)*EG*(EK3(J)*EWx + EK4(J)*EWy)+
EBETA3(J)*
(EEJ(J)*EG*EHP*
(2*EK3(J)*EPx + 2*EK4(J)*EPy + EHP*EK3(J)*EWx+
EHP*EK4(J)*EWy) 2*EElas*
(EEIOw(J)*EK3(J)*EWx - Edy*Er(J)*EWx +
EK1(J)*Er(J)*EWx + EEIOw(J)*EK4(J)*EWy+
Edx*Er(J)*EWy - EK2(J)*Er(J)*EWy))) 4*EBETA3(J)*
(EAlan(J)*EBETA3(J)*EElas*EK1(J) - EEJ(J)*EG*EK1(J)+
EAlan(J)*EEJ(J)*EG*EK3(J)*Er(J))*
(EK3(J)*(EPx + EHP*EWx) + EK4(J)*(EPy + EHP*EWy))*EZ+
EBETA3(J)*(EAlan(J)*EBETA3(J)*
EElas*EK1(J) - EEJ(J)*EG*EK1(J) +
EAlan(J)*EEJ(J)*EG*EK3(J)*Er(J))*
(EK3(J)*EWx + EK4(J)*EWy)*
EZ**2) +
EAlan(J)*EBETA3(J)**2*EElas*Er(J)*
(EEIx(J)*(-6*EHP*(2*EPx + EHP*EWx) +
4*(EPx + EHP*EWx)*EZ - EWx*EZ**2) +
EEIxy(J)*(6*EHP*(2*EPy + EHP*EWy) 4*(EPy + EHP*EWy)*EZ + EWy*EZ**2))) 24*EDELTA(J)*EBETA3(J)**2*EElas*
(EALFA2(J)**2*ED2(J)*
(EK1(J) - EAlan(J)*EALFA1(J)**2*EGAMA(J)*EK1(J) EAlan(J)*EK3(J)*Er(J))*Cosh(EALFA1(J)*EZ) +
EALFA2(J)**2*ED1(J)*
(EK1(J) - EAlan(J)*EALFA1(J)**2*EGAMA(J)*EK1(J) EAlan(J)*EK3(J)*Er(J))*Sinh(EALFA1(J)*EZ) EALFA1(J)**2*((-1 + EAlan(J)*EALFA2(J)**2*EGAMA(J))*EK1(J)+
EAlan(J)*EK3(J)*Er(J))*
309
(ED4(J)*Cosh(EALFA2(J)*EZ) + ED3(J)*Sinh(EALFA2(J)*EZ))))/
(24.*EDELTA(J)*EAlan(J)*EALFA1(J)**2*EALFA2(J)**2*EBETA3(J)**2*
EElas**2*Er(J)) + (-(EALFA1(J-1)**2*EALFA2(J-1)**2*EZ**2*
(EDELTA(J-1)*(12*EAlan(J-1)*EBETA2(J-1)*
EEJ(J-1)*EG*EK3(J-1)*Er(J-1)*
(EK3(J-1)*EWx + EK4(J-1)*EWy) +
6*EAlan(J-1)*EBETA3(J-1)**2*EElas*EHP*EK1(J-1)*
(2*EK3(J-1)*EPx + 2*EK4(J-1)*EPy + EHP*EK3(J-1)*EWx+
EHP*EK4(J-1)*EWy) 12*EAlan(J-1)*EBETA3(J-1)*EElas*EK3(J-1)*Er(J-1)*
(Er(J-1)*(-(Edy*EWx) + EK1(J-1)*EWx + Edx*EWy EK2(J-1)*EWy) + EEIOw(J-1)*
(EK3(J-1)*EWx + EK4(J-1)*EWy)) +
6*EAlan(J-1)*EBETA3(J-1)*EEJ(J-1)*EG*
(2*EHP*EK3(J-1)*(EK3(J-1)*EPx +
EK4(J-1)*EPy)*Er(J-1) +
2*EGAMA(J-1)*EK1(J-1)*(EK3(J-1)*EWx + EK4(J-1)*EWy)+
EHP**2*EK3(J-1)*Er(J-1)*(EK3(J-1)*EWx + EK4(J-1)*EWy))6*EK1(J-1)*(2*EBETA2(J-1)*EEJ(J-1)*EG*
(EK3(J-1)*EWx + EK4(J-1)*EWy) +
EBETA3(J-1)*
(EEJ(J-1)*EG*EHP*
(2*EK3(J-1)*EPx + 2*EK4(J-1)*EPy + EHP*EK3(J-1)*EWx+
EHP*EK4(J-1)*EWy) 2*EElas*
(EEIOw(J-1)*EK3(J-1)*EWx - Edy*Er(J-1)*EWx +
EK1(J-1)*Er(J-1)*EWx + EEIOw(J-1)*EK4(J-1)*EWy+
Edx*Er(J-1)*EWy - EK2(J-1)*Er(J-1)*EWy))) 4*EBETA3(J-1)*
(EAlan(J-1)*EBETA3(J-1)*EElas*EK1(J-1)-EEJ(J-1)*EG*EK1(J-1)+
EAlan(J-1)*EEJ(J-1)*EG*EK3(J-1)*Er(J-1))*
(EK3(J-1)*(EPx + EHP*EWx) + EK4(J-1)*(EPy + EHP*EWy))*EZ
+ EBETA3(J-1)*
(EAlan(J-1)*EBETA3(J-1)*EElas*EK1(J-1) EEJ(J-1)*EG*EK1(J-1) +
EAlan(J-1)*EEJ(J-1)*EG*EK3(J-1)*
Er(J-1))*(EK3(J-1)*EWx + EK4(J-1)*EWy)*
EZ**2) +
EAlan(J-1)*EBETA3(J-1)**2*EElas*Er(J-1)*
(EEIx(J-1)*(-6*EHP*(2*EPx + EHP*EWx) +
4*(EPx + EHP*EWx)*EZ - EWx*EZ**2) +
EEIxy(J-1)*(6*EHP*(2*EPy + EHP*EWy) 4*(EPy + EHP*EWy)*EZ + EWy*EZ**2)))) +
24*EDELTA(J-1)*EBETA3(J-1)**2*EElas*
(EALFA2(J-1)**2*ED2(J-1)*
(EK1(J-1) - EAlan(J-1)*EALFA1(J-1)**2*EGAMA(J-1)*EK1(J-1)EAlan(J-1)*EK3(J-1)*Er(J-1))*Cosh(EALFA1(J-1)*EZ) +
EALFA2(J-1)**2*ED1(J-1)*
(EK1(J-1) - EAlan(J-1)*EALFA1(J-1)**2*EGAMA(J-1)*EK1(J-1)EAlan(J-1)*EK3(J-1)*Er(J-1))*Sinh(EALFA1(J-1)*EZ) EALFA1(J-1)**2*((-1 + EAlan(J-1)*EALFA2(J-1)**2*EGAMA(J-1))*
EK1(J-1) + EAlan(J-1)*EK3(J-1)*Er(J-1))*
(ED4(J-1)*Cosh(EALFA2(J-1)*EZ) +
ED3(J-1)*Sinh(EALFA2(J-1)*EZ))))/
(24.*EDELTA(J-1)*EAlan(J-1)*EALFA1(J-1)**2*
EALFA2(J-1)**2*EBETA3(J-1)**2*
EElas**2*Er(J-1)))
return
end
@
@
@
SUBROUTINE COEF3Ua(ibsay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EZ,EGAMA,EG,EEJ,EK1,EK2,EK3,EK4,EHP,EDELTA
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan,EEIx,EEIxy
,EElas,ED1,ED2,ED3,ED4,REK1,REKB1,J)
310
@
@
@
@
@
REK1(1)=1
REK1(2)=0
REKB1= -((EALFA1(J)*EALFA2(J)**2*ED1(J)*
(EK1(J) - EAlan(J)*EALFA1(J)**2*EGAMA(J)*
EK1(J)-EAlan(J)*EK3(J)*Er(J)) EALFA1(J)**2*EALFA2(J)*ED3(J)*
((-1 + EAlan(J)*EALFA2(J)**2*EGAMA(J))*
EK1(J)+EAlan(J)*EK3(J)*Er(J)))/
(EAlan(J)*EALFA1(J)**2*EALFA2(J)**2*EElas*Er(J)))
return
end
@
@
@
SUBROUTINE COEF4Ua(ibsay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EZ,EGAMA,EG,EEJ,EK1,EK2,EK3,EK4,EHP,EDELTA
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan,EEIx,EEIxy
,EElas,ED1,ED2,ED3,ED4,REK2,REKB2,J)
@
@
@
@
@
REK2(1)=0
REK2(2)=1
REKB2= -((EALFA2(J)**2*ED2(J)*(EK1(J) - EAlan(J)*
EALFA1(J)**2*EGAMA(J)*EK1(J) EAlan(J)*EK3(J)*Er(J)) EALFA1(J)**2*ED4(J)*((-1 + EAlan(J)*
EALFA2(J)**2*EGAMA(J))*EK1(J)+
EAlan(J)*EK3(J)*Er(J)))/
(EAlan(J)*EALFA1(J)**2*EALFA2(J)**2*EElas*Er(J)))
return
end
@
@
@
SUBROUTINE COEF1Ub(ibsay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EZ,EGAMA,EG,EEJ,EK1,EK2,EK3,EK4,EHP,EDELTA
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan,EEIx,EEIxy
,EElas,ED1,ED2,ED3,ED4,REK1,REKB1,J)
@
@
@
@
@
311
REKB1=-((-(EALFA1(J)*EALFA2(J)**2*ED1(J)*
(EK1(J) - EAlan(J)*EALFA1(J)**2*EGAMA(J)*EK1(J) EAlan(J)*EK3(J)*Er(J))*Cosh(EALFA1(J)*EZ)) EALFA1(J)*EALFA2(J)**2*ED2(J)*
(EK1(J) - EAlan(J)*EALFA1(J)**2*EGAMA(J)*EK1(J) EAlan(J)*EK3(J)*Er(J))*Sinh(EALFA1(J)*EZ) +
EALFA1(J)**2*((-1 + EAlan(J)*EALFA2(J)**2*EGAMA(J))*
EK1(J) + EAlan(J)*EK3(J)*Er(J))*
(EALFA2(J)*ED3(J)*Cosh(EALFA2(J)*EZ) +
EALFA2(J)*ED4(J)*Sinh(EALFA2(J)*EZ)))/
(EAlan(J)*EALFA1(J)**2*EALFA2(J)**2*EElas*Er(J)) +
(EALFA1(J-1)*EALFA2(J-1)**2*ED1(J-1)*
(EK1(J-1) - EAlan(J-1)*EALFA1(J-1)**2*EGAMA(J-1)*
EK1(J-1) - EAlan(J-1)*EK3(J-1)*Er(J-1))*
Cosh(EALFA1(J-1)*EZ) +
EALFA1(J-1)*EALFA2(J-1)**2*ED2(J-1)*
(EK1(J-1) - EAlan(J-1)*EALFA1(J-1)**2*EGAMA(J-1)*
EK1(J-1) - EAlan(J-1)*EK3(J-1)*Er(J-1))*
Sinh(EALFA1(J-1)*EZ) EALFA1(J-1)**2*
((-1 + EAlan(J-1)*EALFA2(J-1)**2*EGAMA(J-1))*
EK1(J-1) + EAlan(J-1)*EK3(J-1)*Er(J-1))*
(EALFA2(J-1)*ED3(J-1)*Cosh(EALFA2(J-1)*EZ) +
EALFA2(J-1)*ED4(J-1)*Sinh(EALFA2(J-1)*EZ)))/
(EAlan(J-1)*EALFA1(J-1)**2*EALFA2(J-1)**2*EElas*Er(J-1)))
return
end
@
@
@
SUBROUTINE COEF2Ub(ibsay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EZ,EGAMA,EG,EEJ,EK1,EK2,EK3,EK4,EHP,EDELTA
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan,EEIx,EEIxy
,EElas,ED1,ED2,ED3,ED4,REK2,REKB2,J)
@
@
@
@
@
REK2(1)=EZ
REK2(2)=1
REK2(3)=-EZ
REK2(4)=-1
REKB2= -((-(EALFA2(J)**2*ED2(J)*
(EK1(J) - EAlan(J)*EALFA1(J)**2*EGAMA(J)*EK1(J) EAlan(J)*EK3(J)*Er(J))*Cosh(EALFA1(J)*EZ)) EALFA2(J)**2*ED1(J)*
(EK1(J) - EAlan(J)*EALFA1(J)**2*EGAMA(J)*EK1(J) EAlan(J)*EK3(J)*Er(J))*Sinh(EALFA1(J)*EZ) +
EALFA1(J)**2*((-1 + EAlan(J)*EALFA2(J)**2*EGAMA(J))*
EK1(J) + EAlan(J)*EK3(J)*Er(J))*
(ED4(J)*Cosh(EALFA2(J)*EZ) +
ED3(J)*Sinh(EALFA2(J)*EZ)))/
(EAlan(J)*EALFA1(J)**2*EALFA2(J)**2*EElas*Er(J)) +
- (EALFA2(J-1)**2*ED2(J-1)*
(EK1(J-1) - EAlan(J-1)*EALFA1(J-1)**2*EGAMA(J-1)*
EK1(J-1) - EAlan(J-1)*EK3(J-1)*Er(J-1))*
Cosh(EALFA1(J-1)*EZ) +
EALFA2(J-1)**2*ED1(J-1)*
(EK1(J-1) - EAlan(J-1)*EALFA1(J-1)**2*EGAMA(J-1)*
EK1(J-1) - EAlan(J-1)*EK3(J-1)*Er(J-1))*
Sinh(EALFA1(J-1)*EZ) EALFA1(J-1)**2*
312
((-1 + EAlan(J-1)*EALFA2(J-1)**2*EGAMA(J-1))*
EK1(J-1) + EAlan(J-1)*EK3(J-1)*Er(J-1))*
(ED4(J-1)*Cosh(EALFA2(J-1)*EZ) +
ED3(J-1)*Sinh(EALFA2(J-1)*EZ)))/
(EAlan(J-1)*EALFA1(J-1)**2*EALFA2(J-1)**2*EElas*Er(J-1)))
return
end
@
@
@
SUBROUTINE COEF3Ub(ibsay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EZ,EGAMA,EG,EEJ,EK1,EK2,EK3,EK4,EHP,EDELTA
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan,EEIx,EEIxy
,EElas,ED1,ED2,ED3,ED4,REK1,REKB1,J)
@
@
@
@
@
REK1(1)=1
REK1(2)=0
REK1(3)=-1
REK1(4)=0
REKB1= -((EALFA1(J)**2*EALFA2(J)**2*EZ**2*
(EDELTA(J)*(-4*EBETA3(J)*
(EAlan(J)*EBETA3(J)*EElas*EK1(J) - EEJ(J)*EG*EK1(J)+
EAlan(J)*EEJ(J)*EG*EK3(J)*Er(J))*
(EK3(J)*(EPx + EHP*EWx) + EK4(J)*(EPy + EHP*EWy))+
2*EBETA3(J)*
(EAlan(J)*EBETA3(J)*EElas*EK1(J) - EEJ(J)*EG*EK1(J)+
EAlan(J)*EEJ(J)*EG*EK3(J)*Er(J))*
(EK3(J)*EWx + EK4(J)*EWy)*EZ)
+ EAlan(J)*EBETA3(J)**2*EElas*Er(J)*
(EEIx(J)*(4*(EPx + EHP*EWx) - 2*EWx*EZ) +
EEIxy(J)*(-4*(EPy + EHP*EWy) + 2*EWy*EZ))) +
2*EALFA1(J)**2*EALFA2(J)**2*EZ*
(EDELTA(J)*(12*EAlan(J)*EBETA2(J)*EEJ(J)*EG*EK3(J)*Er(J)*
(EK3(J)*EWx + EK4(J)*EWy) +
6*EAlan(J)*EBETA3(J)**2*EElas*EHP*EK1(J)*
(2*EK3(J)*EPx + 2*EK4(J)*EPy + EHP*EK3(J)*EWx +
EHP*EK4(J)*EWy) 12*EAlan(J)*EBETA3(J)*EElas*EK3(J)*Er(J)*
(Er(J)*(-(Edy*EWx) + EK1(J)*EWx + Edx*EWy EK2(J)*EWy) + EEIOw(J)*(EK3(J)*EWx + EK4(J)*EWy))+
6*EAlan(J)*EBETA3(J)*EEJ(J)*EG*
(2*EHP*EK3(J)*(EK3(J)*EPx + EK4(J)*EPy)*Er(J) +
2*EGAMA(J)*EK1(J)*(EK3(J)*EWx + EK4(J)*EWy) +
EHP**2*EK3(J)*Er(J)*(EK3(J)*EWx + EK4(J)*EWy)) 6*EK1(J)*(2*EBETA2(J)*EEJ(J)*EG*(EK3(J)*EWx + EK4(J)*EWy)+
EBETA3(J)*
(EEJ(J)*EG*EHP*
(2*EK3(J)*EPx + 2*EK4(J)*EPy + EHP*EK3(J)*EWx +
EHP*EK4(J)*EWy) 2*EElas*
(EEIOw(J)*EK3(J)*EWx - Edy*Er(J)*EWx +
EK1(J)*Er(J)*EWx + EEIOw(J)*EK4(J)*EWy +
Edx*Er(J)*EWy - EK2(J)*Er(J)*EWy))) 4*EBETA3(J)*
(EAlan(J)*EBETA3(J)*EElas*EK1(J) - EEJ(J)*EG*EK1(J)+
EAlan(J)*EEJ(J)*EG*EK3(J)*Er(J))*
(EK3(J)*(EPx + EHP*EWx) + EK4(J)*(EPy + EHP*EWy))*EZ +
EBETA3(J)*(EAlan(J)*EBETA3(J)*EElas*EK1(J)-EEJ(J)*EG*EK1(J)+
313
EAlan(J)*EEJ(J)*EG*EK3(J)*Er(J))*(EK3(J)*EWx+EK4(J)*EWy)*
EZ**2) +
EAlan(J)*EBETA3(J)**2*EElas*Er(J)*
(EEIx(J)*(-6*EHP*(2*EPx + EHP*EWx) +
4*(EPx + EHP*EWx)*EZ - EWx*EZ**2) +
EEIxy(J)*(6*EHP*(2*EPy + EHP*EWy) 4*(EPy + EHP*EWy)*EZ + EWy*EZ**2))) 24*EDELTA(J)*EBETA3(J)**2*EElas*
(EALFA1(J)*EALFA2(J)**2*ED1(J)*
(EK1(J) - EAlan(J)*EALFA1(J)**2*EGAMA(J)*EK1(J) EAlan(J)*EK3(J)*Er(J))*Cosh(EALFA1(J)*EZ) +
EALFA1(J)*EALFA2(J)**2*ED2(J)*
(EK1(J) - EAlan(J)*EALFA1(J)**2*EGAMA(J)*EK1(J) EAlan(J)*EK3(J)*Er(J))*Sinh(EALFA1(J)*EZ) EALFA1(J)**2*((-1 + EAlan(J)*EALFA2(J)**2*EGAMA(J))*EK1(J)+
EAlan(J)*EK3(J)*Er(J))*
(EALFA2(J)*ED3(J)*Cosh(EALFA2(J)*EZ) +
EALFA2(J)*ED4(J)*Sinh(EALFA2(J)*EZ))))/
(24.*EDELTA(J)*EAlan(J)*EALFA1(J)**2*EALFA2(J)**2*EBETA3(J)**2*
EElas**2*Er(J)) +
(EALFA1(J-1)*EALFA2(J-1)**2*ED1(J-1)*
(EK1(J-1) - EAlan(J-1)*EALFA1(J-1)**2*EGAMA(J-1)*
EK1(J-1) - EAlan(J-1)*EK3(J-1)*Er(J-1))*
Cosh(EALFA1(J-1)*EZ) +
EALFA1(J-1)*EALFA2(J-1)**2*ED2(J-1)*
(EK1(J-1) - EAlan(J-1)*EALFA1(J-1)**2*EGAMA(J-1)*
EK1(J-1) - EAlan(J-1)*EK3(J-1)*Er(J-1))*
Sinh(EALFA1(J-1)*EZ) EALFA1(J-1)**2*
((-1 + EAlan(J-1)*EALFA2(J-1)**2*EGAMA(J-1))*
EK1(J-1) + EAlan(J-1)*EK3(J-1)*Er(J-1))*
(EALFA2(J-1)*ED3(J-1)*Cosh(EALFA2(J-1)*EZ) +
EALFA2(J-1)*ED4(J-1)*Sinh(EALFA2(J-1)*EZ)))/
(EAlan(J-1)*EALFA1(J-1)**2*EALFA2(J-1)**2*EElas*Er(J-1)))
return
end
@
@
@
SUBROUTINE COEF4Ub(ibsay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EZ,EGAMA,EG,EEJ,EK1,EK2,EK3,EK4,EHP,EDELTA
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan,EEIx,EEIxy
,EElas,ED1,ED2,ED3,ED4,REK2,REKB2,J)
@
@
@
@
@
REK2(1)=EZ
REK2(2)=1
REK2(3)=-EZ
REK2(4)=-1
REKB2= -((EALFA1(J)**2*EALFA2(J)**2*EZ**2*
(EDELTA(J)*(12*EAlan(J)*EBETA2(J)*EEJ(J)*EG*EK3(J)*Er(J)*
(EK3(J)*EWx + EK4(J)*EWy) +
6*EAlan(J)*EBETA3(J)**2*EElas*EHP*EK1(J)*
(2*EK3(J)*EPx + 2*EK4(J)*EPy + EHP*EK3(J)*EWx +
EHP*EK4(J)*EWy) 12*EAlan(J)*EBETA3(J)*EElas*EK3(J)*Er(J)*
(Er(J)*(-(Edy*EWx) + EK1(J)*EWx + Edx*EWy EK2(J)*EWy) + EEIOw(J)*(EK3(J)*EWx + EK4(J)*EWy))+
314
6*EAlan(J)*EBETA3(J)*EEJ(J)*EG*
(2*EHP*EK3(J)*(EK3(J)*EPx + EK4(J)*EPy)*Er(J) +
2*EGAMA(J)*EK1(J)*(EK3(J)*EWx + EK4(J)*EWy) +
EHP**2*EK3(J)*Er(J)*(EK3(J)*EWx + EK4(J)*EWy)) 6*EK1(J)*(2*EBETA2(J)*EEJ(J)*EG*(EK3(J)*EWx + EK4(J)*EWy)+
EBETA3(J)*
(EEJ(J)*EG*EHP*
(2*EK3(J)*EPx + 2*EK4(J)*EPy + EHP*EK3(J)*EWx+
EHP*EK4(J)*EWy) 2*EElas*
(EEIOw(J)*EK3(J)*EWx - Edy*Er(J)*EWx +
EK1(J)*Er(J)*EWx + EEIOw(J)*EK4(J)*EWy+
Edx*Er(J)*EWy - EK2(J)*Er(J)*EWy))) 4*EBETA3(J)*
(EAlan(J)*EBETA3(J)*EElas*EK1(J) - EEJ(J)*EG*EK1(J)+
EAlan(J)*EEJ(J)*EG*EK3(J)*Er(J))*
(EK3(J)*(EPx + EHP*EWx) + EK4(J)*(EPy + EHP*EWy))*EZ+
EBETA3(J)*(EAlan(J)*EBETA3(J)*
EElas*EK1(J) - EEJ(J)*EG*EK1(J) +
EAlan(J)*EEJ(J)*EG*EK3(J)*Er(J))*
(EK3(J)*EWx + EK4(J)*EWy)*
EZ**2) +
EAlan(J)*EBETA3(J)**2*EElas*Er(J)*
(EEIx(J)*(-6*EHP*(2*EPx + EHP*EWx) +
4*(EPx + EHP*EWx)*EZ - EWx*EZ**2) +
EEIxy(J)*(6*EHP*(2*EPy + EHP*EWy) 4*(EPy + EHP*EWy)*EZ + EWy*EZ**2))) 24*EDELTA(J)*EBETA3(J)**2*EElas*
(EALFA2(J)**2*ED2(J)*
(EK1(J) - EAlan(J)*EALFA1(J)**2*EGAMA(J)*EK1(J) EAlan(J)*EK3(J)*Er(J))*Cosh(EALFA1(J)*EZ) +
EALFA2(J)**2*ED1(J)*
(EK1(J) - EAlan(J)*EALFA1(J)**2*EGAMA(J)*EK1(J) EAlan(J)*EK3(J)*Er(J))*Sinh(EALFA1(J)*EZ) EALFA1(J)**2*((-1 + EAlan(J)*EALFA2(J)**2*EGAMA(J))*EK1(J)+
EAlan(J)*EK3(J)*Er(J))*
(ED4(J)*Cosh(EALFA2(J)*EZ) + ED3(J)*Sinh(EALFA2(J)*EZ))))/
(24.*EDELTA(J)*EAlan(J)*EALFA1(J)**2*EALFA2(J)**2*EBETA3(J)**2*
EElas**2*Er(J)) + (EALFA2(J-1)**2*ED2(J-1)*
(EK1(J-1) - EAlan(J-1)*EALFA1(J-1)**2*EGAMA(J-1)*
EK1(J-1) - EAlan(J-1)*EK3(J-1)*Er(J-1))*
Cosh(EALFA1(J-1)*EZ) +
EALFA2(J-1)**2*ED1(J-1)*
(EK1(J-1) - EAlan(J-1)*EALFA1(J-1)**2*EGAMA(J-1)*
EK1(J-1) - EAlan(J-1)*EK3(J-1)*Er(J-1))*
Sinh(EALFA1(J-1)*EZ) EALFA1(J-1)**2*
((-1 + EAlan(J-1)*EALFA2(J-1)**2*EGAMA(J-1))*
EK1(J-1) + EAlan(J-1)*EK3(J-1)*Er(J-1))*
(ED4(J-1)*Cosh(EALFA2(J-1)*EZ) +
ED3(J-1)*Sinh(EALFA2(J-1)*EZ)))/
(EAlan(J-1)*EALFA1(J-1)**2*EALFA2(J-1)**2*EElas*Er(J-1)))
return
end
@
@
@
@
SUBROUTINE UFONKa(ibsay,iksay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EGAMA,EK1,EK2,EK3,EK4,EElas,EG,EEJ,EDELTA
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan,EHP
,EEIx,EEIxy,ED1,ED2,ED3,ED4,EU,EKAT,J,jm
,EF1,EF2)
implicit real*8 (A-H,K-Z)
315
dimension EU(iksay),EKAT(iksay),EF1(ibsay),EF2(ibsay)
,ED1(ibsay),ED2(ibsay),ED3(ibsay),ED4(ibsay)
,EALFA1(ibsay),EALFA2(ibsay),EGAMA(ibsay)
,EBETA1(ibsay),EBETA2(ibsay),EBETA3(ibsay)
,EEJ(ibsay),EK1(ibsay),EK2(ibsay),EK3(ibsay)
,EK4(ibsay),Er(ibsay),EEIOw(ibsay),EAlan(ibsay)
,EDELTA(ibsay),EEIx(ibsay),EEIxy(ibsay)
@
@
@
@
@
@
EU(jm)=
-
return
end
SUBROUTINE UFONKb(ibsay,iksay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EGAMA,EK1,EK2,EK3,EK4,EElas,EG,EEJ,EDELTA
316
@
@
@
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan,EHP
,EEIx,EEIxy,ED1,ED2,ED3,ED4,EU,EKAT,J,jm
,EF1,EF2)
@
@
@
@
@
@
return
end
@
@
@
SUBROUTINE COEF1Va(ibsay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EZ,EGAMA,EG,EEJ,EK1,EK2,EK3,EK4,EHP,EDELTA
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan,EEIy,EEIxy
,EElas,ED1,ED2,ED3,ED4,REK1,REKB1,J)
@
@
@
@
@
REK1(1)=1
REK1(2)=0
REK1(3)=-1
REK1(4)=0
REKB1= -((-(EALFA1(J)**2*EALFA2(J)**2*EZ**2*
(EDELTA(J)*(4*EBETA3(J)*
(-(EAlan(J)*EBETA3(J)*EElas*EK2(J)) + EEJ(J)*EG*EK2(J)+
EAlan(J)*EEJ(J)*EG*EK4(J)*Er(J))*
(EK3(J)*(EPx + EHP*EWx) + EK4(J)*(EPy + EHP*EWy))+
2*EBETA3(J)*
(EAlan(J)*EBETA3(J)*EElas*EK2(J) - EEJ(J)*EG*EK2(J)EAlan(J)*EEJ(J)*EG*EK4(J)*Er(J))*(EK3(J)*EWx + EK4(J)*EWy)*EZ
) + EAlan(J)*EBETA3(J)**2*EElas*Er(J)*
(EEIxy(J)*(4*(EPx + EHP*EWx) - 2*EWx*EZ) +
EEIy(J)*(-4*(EPy + EHP*EWy) + 2*EWy*EZ)))) 2*EALFA1(J)**2*EALFA2(J)**2*EZ*
(EDELTA(J)*(-12*EAlan(J)*EBETA2(J)*EEJ(J)*EG*EK4(J)*Er(J)*
(EK3(J)*EWx + EK4(J)*EWy) +
6*EAlan(J)*EBETA3(J)**2*EElas*EHP*EK2(J)*
(2*EK3(J)*EPx + 2*EK4(J)*EPy + EHP*EK3(J)*EWx +
EHP*EK4(J)*EWy) +
12*EAlan(J)*EBETA3(J)*EElas*EK4(J)*Er(J)*
(Er(J)*(-(Edy*EWx) + EK1(J)*EWx + Edx*EWy EK2(J)*EWy) + EEIOw(J)*(EK3(J)*EWx + EK4(J)*EWy))6*EAlan(J)*EBETA3(J)*EEJ(J)*EG*
317
318
(EK3(J-1)*EWx + EK4(J-1)*EWy) +
EBETA3(J-1)*
(EEJ(J-1)*EG*EHP*
(2*EK3(J-1)*EPx + 2*EK4(J-1)*EPy + EHP*EK3(J-1)*EWx +
EHP*EK4(J-1)*EWy) 2*EElas*
(EEIOw(J-1)*EK3(J-1)*EWx - Edy*Er(J-1)*EWx +
EK1(J-1)*Er(J-1)*EWx + EEIOw(J-1)*EK4(J-1)*EWy +
Edx*Er(J-1)*EWy - EK2(J-1)*Er(J-1)*EWy))) +
4*EBETA3(J-1)*
(-(EAlan(J-1)*EBETA3(J-1)*EElas*
EK2(J-1)) + EEJ(J-1)*EG*EK2(J-1) +
EAlan(J-1)*EEJ(J-1)*EG*EK4(J-1)*Er(J-1))*
(EK3(J-1)*(EPx + EHP*EWx) + EK4(J-1)*(EPy + EHP*EWy))*EZ+
EBETA3(J-1)*(EAlan(J-1)*EBETA3(J-1)*EElas*EK2(J-1) EEJ(J-1)*EG*EK2(J-1) - EAlan(J-1)*
EEJ(J-1)*EG*EK4(J-1)*Er(J-1))*
(EK3(J-1)*EWx + EK4(J-1)*EWy)*EZ**2) +
EAlan(J-1)*EBETA3(J-1)**2*EElas*Er(J-1)*
(EEIxy(J-1)*(-6*EHP*(2*EPx + EHP*EWx) +
4*(EPx + EHP*EWx)*EZ - EWx*EZ**2) +
EEIy(J-1)*(6*EHP*(2*EPy + EHP*EWy) 4*(EPy + EHP*EWy)*EZ + EWy*EZ**2))) +
24*EDELTA(J-1)*EBETA3(J-1)**2*EElas*
(EALFA1(J-1)*EALFA2(J-1)**2*ED1(J-1)*
((-1 + EAlan(J-1)*EALFA1(J-1)**2*EGAMA(J-1))*EK2(J-1) EAlan(J-1)*EK4(J-1)*Er(J-1))*Cosh(EALFA1(J-1)*EZ) +
EALFA1(J-1)*EALFA2(J-1)**2*ED2(J-1)*
((-1 + EAlan(J-1)*EALFA1(J-1)**2*EGAMA(J-1))*EK2(J-1) EAlan(J-1)*EK4(J-1)*Er(J-1))*Sinh(EALFA1(J-1)*EZ) +
EALFA1(J-1)**2*((-1 + EAlan(J-1)*
EALFA2(J-1)**2*EGAMA(J-1))*
EK2(J-1) - EAlan(J-1)*EK4(J-1)*Er(J-1))*
(EALFA2(J-1)*ED3(J-1)*Cosh(EALFA2(J-1)*EZ) +
EALFA2(J-1)*ED4(J-1)*Sinh(EALFA2(J-1)*EZ))))/
(24.*EDELTA(J-1)*EAlan(J-1)*EALFA1(J-1)**2*
EALFA2(J-1)**2*EBETA3(J-1)**2*
EElas**2*Er(J-1)))
return
end
@
@
@
SUBROUTINE COEF2Va(ibsay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EZ,EGAMA,EG,EEJ,EK1,EK2,EK3,EK4,EHP,EDELTA
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan,EEIy,EEIxy
,EElas,ED1,ED2,ED3,ED4,REK2,REKB2,J)
@
@
@
@
@
REK2(1)=EZ
REK2(2)=1
REK2(3)=-EZ
REK2(4)=-1
REKB2= -((-(EALFA1(J)**2*EALFA2(J)**2*EZ**2*
(EDELTA(J)*(-12*EAlan(J)*EBETA2(J)*EEJ(J)*EG*EK4(J)*Er(J)*
(EK3(J)*EWx + EK4(J)*EWy) +
6*EAlan(J)*EBETA3(J)**2*EElas*EHP*EK2(J)*
(2*EK3(J)*EPx + 2*EK4(J)*EPy + EHP*EK3(J)*EWx +
EHP*EK4(J)*EWy) +
12*EAlan(J)*EBETA3(J)*EElas*EK4(J)*Er(J)*
319
320
EEJ(J-1)*EG*EK2(J-1) +
EAlan(J-1)*EEJ(J-1)*EG*EK4(J-1)*Er(J-1))*
(EK3(J-1)*(EPx + EHP*EWx) + EK4(J-1)*(EPy + EHP*EWy))*EZ+
EBETA3(J-1)*(EAlan(J-1)*EBETA3(J-1)*EElas*EK2(J-1) EEJ(J-1)*EG*EK2(J-1) - EAlan(J-1)*
EEJ(J-1)*EG*EK4(J-1)*Er(J-1))*
(EK3(J-1)*EWx + EK4(J-1)*EWy)*EZ**2) +
EAlan(J-1)*EBETA3(J-1)**2*EElas*Er(J-1)*
(EEIxy(J-1)*(-6*EHP*(2*EPx + EHP*EWx) +
4*(EPx + EHP*EWx)*EZ - EWx*EZ**2) +
EEIy(J-1)*(6*EHP*(2*EPy + EHP*EWy) 4*(EPy + EHP*EWy)*EZ + EWy*EZ**2))) +
24*EDELTA(J-1)*EBETA3(J-1)**2*EElas*
(EALFA2(J-1)**2*ED2(J-1)*
((-1 + EAlan(J-1)*EALFA1(J-1)**2*EGAMA(J-1))*EK2(J-1)EAlan(J-1)*EK4(J-1)*Er(J-1))*Cosh(EALFA1(J-1)*EZ)+
EALFA2(J-1)**2*ED1(J-1)*
((-1 + EAlan(J-1)*EALFA1(J-1)**2*EGAMA(J-1))*EK2(J-1)EAlan(J-1)*EK4(J-1)*Er(J-1))*Sinh(EALFA1(J-1)*EZ) +
EALFA1(J-1)**2*((-1 + EAlan(J-1)*EALFA2(J-1)**2*EGAMA(J-1))*
EK2(J-1) - EAlan(J-1)*EK4(J-1)*Er(J-1))*
(ED4(J-1)*Cosh(EALFA2(J-1)*EZ) +
ED3(J-1)*Sinh(EALFA2(J-1)*EZ))))/
(24.*EDELTA(J-1)*EAlan(J-1)*EALFA1(J-1)**2*
EALFA2(J-1)**2*EBETA3(J-1)**2*
EElas**2*Er(J-1)))
return
end
@
@
@
SUBROUTINE COEF3Va(ibsay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EZ,EGAMA,EG,EEJ,EK1,EK2,EK3,EK4,EHP,EDELTA
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan,EEIy,EEIxy
,EElas,ED1,ED2,ED3,ED4,REK1,REKB1,J)
@
@
@
@
@
REK1(1)=1
REK1(2)=0
REKB1= -((EALFA1(J)*EALFA2(J)**2*ED1(J)*
((-1 + EAlan(J)*EALFA1(J)**2*EGAMA(J))*
EK2(J) - EAlan(J)*EK4(J)*Er(J)) +
EALFA1(J)**2*EALFA2(J)*ED3(J)*
((-1 + EAlan(J)*EALFA2(J)**2*EGAMA(J))*
EK2(J) - EAlan(J)*EK4(J)*Er(J)))/
(EAlan(J)*EALFA1(J)**2*EALFA2(J)**2*EElas*Er(J)))
return
end
@
@
@
SUBROUTINE COEF4Va(ibsay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EZ,EGAMA,EG,EEJ,EK1,EK2,EK3,EK4,EHP,EDELTA
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan,EEIy,EEIxy
,EElas,ED1,ED2,ED3,ED4,REK2,REKB2,J)
implicit real*8 (A-H,K-Z)
321
dimension REK2(4),EBETA1(ibsay),EBETA2(ibsay),EBETA3(ibsay)
,EALFA1(ibsay),EALFA2(ibsay),EGAMA(ibsay)
,ED1(ibsay),ED2(ibsay),ED3(ibsay),ED4(ibsay)
,EEJ(ibsay),EK1(ibsay),EK2(ibsay),EK3(ibsay)
,EK4(ibsay),Er(ibsay),EEIOw(ibsay),EAlan(ibsay)
,EDELTA(ibsay),EEIy(ibsay),EEIxy(ibsay)
@
@
@
@
@
REK2(1)=0
REK2(2)=1
REKB2= -((EALFA2(J)**2*ED2(J)*((-1 + EAlan(J)*EALFA1(J)**2*
EGAMA(J))*EK2(J) - EAlan(J)*EK4(J)*Er(J)) +
EALFA1(J)**2*ED4(J)*((-1 + EAlan(J)*
EALFA2(J)**2*EGAMA(J))*EK2(J) EAlan(J)*EK4(J)*Er(J)))/
(EAlan(J)*EALFA1(J)**2*EALFA2(J)**2*EElas*Er(J)))
return
end
@
@
@
SUBROUTINE COEF1Vb(ibsay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EZ,EGAMA,EG,EEJ,EK1,EK2,EK3,EK4,EHP,EDELTA
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan,EEIy,EEIxy
,EElas,ED1,ED2,ED3,ED4,REK1,REKB1,J)
@
@
@
@
@
REK1(1)=1
REK1(2)=0
REK1(3)=-1
REK1(4)=0
REKB1= -((-(EALFA1(J)*EALFA2(J)**2*ED1(J)*
((-1 + EAlan(J)*EALFA1(J)**2*EGAMA(J))*EK2(J) EAlan(J)*EK4(J)*Er(J))*
Cosh(EALFA1(J)*EZ)) EALFA1(J)*EALFA2(J)**2*ED2(J)*
((-1 + EAlan(J)*EALFA1(J)**2*EGAMA(J))*EK2(J) EAlan(J)*EK4(J)*Er(J))*
Sinh(EALFA1(J)*EZ) EALFA1(J)**2*((-1 + EAlan(J)*EALFA2(J)**2*EGAMA(J))*EK2(J) EAlan(J)*EK4(J)*Er(J))*
(EALFA2(J)*ED3(J)*Cosh(EALFA2(J)*EZ) +
EALFA2(J)*ED4(J)*Sinh(EALFA2(J)*EZ)))/
(EAlan(J)*EALFA1(J)**2*EALFA2(J)**2*EElas*Er(J)) +
- (EALFA1(J-1)*EALFA2(J-1)**2*ED1(J-1)*
((-1 + EAlan(J-1)*EALFA1(J-1)**2*EGAMA(J-1))*EK2(J-1) EAlan(J-1)*EK4(J-1)*Er(J-1))*Cosh(EALFA1(J-1)*EZ) +
EALFA1(J-1)*EALFA2(J-1)**2*ED2(J-1)*
((-1 + EAlan(J-1)*EALFA1(J-1)**2*EGAMA(J-1))*EK2(J-1) EAlan(J-1)*EK4(J-1)*Er(J-1))*Sinh(EALFA1(J-1)*EZ) +
EALFA1(J-1)**2*((-1 + EAlan(J-1)*
EALFA2(J-1)**2*EGAMA(J-1))*EK2(J-1) EAlan(J-1)*EK4(J-1)*Er(J-1))*
(EALFA2(J-1)*ED3(J-1)*Cosh(EALFA2(J-1)*EZ) +
EALFA2(J-1)*ED4(J-1)*Sinh(EALFA2(J-1)*EZ)))/
(EAlan(J-1)*EALFA1(J-1)**2*EALFA2(J-1)**2*EElas*Er(J-1)))
return
end
322
@
@
@
SUBROUTINE COEF2Vb(ibsay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EZ,EGAMA,EG,EEJ,EK1,EK2,EK3,EK4,EHP,EDELTA
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan,EEIy,EEIxy
,EElas,ED1,ED2,ED3,ED4,REK2,REKB2,J)
@
@
@
@
@
REK2(1)=EZ
REK2(2)=1
REK2(3)=-EZ
REK2(4)=-1
REKB2=-((-(EALFA2(J)**2*ED2(J)*
((-1 + EAlan(J)*EALFA1(J)**2*EGAMA(J))*EK2(J) EAlan(J)*EK4(J)*Er(J))*Cosh(EALFA1(J)*EZ)) EALFA2(J)**2*ED1(J)*((-1 + EAlan(J)
*EALFA1(J)**2*EGAMA(J))*EK2(J) EAlan(J)*EK4(J)*Er(J))*Sinh(EALFA1(J)*EZ) EALFA1(J)**2*((-1 + EAlan(J)*
EALFA2(J)**2*EGAMA(J))*EK2(J) EAlan(J)*EK4(J)*Er(J))*
(ED4(J)*Cosh(EALFA2(J)*EZ) + ED3(J)*Sinh(EALFA2(J)*EZ)))/
(EAlan(J)*EALFA1(J)**2*EALFA2(J)**2*EElas*Er(J)) +
(EALFA2(J-1)**2*ED2(J-1)*((-1 + EAlan(J-1)*
EALFA1(J-1)**2*EGAMA(J-1))*EK2(J-1) EAlan(J-1)*EK4(J-1)*Er(J-1))*Cosh(EALFA1(J-1)*EZ) +
EALFA2(J-1)**2*ED1(J-1)*
((-1 + EAlan(J-1)*EALFA1(J-1)**2*EGAMA(J-1))*EK2(J-1) EAlan(J-1)*EK4(J-1)*Er(J-1))*Sinh(EALFA1(J-1)*EZ) +
EALFA1(J-1)**2*((-1 + EAlan(J-1)*EALFA2(J-1)**2*
EGAMA(J-1))*EK2(J-1) EAlan(J-1)*EK4(J-1)*Er(J-1))*
(ED4(J-1)*Cosh(EALFA2(J-1)*EZ) +
ED3(J-1)*Sinh(EALFA2(J-1)*EZ)))/
(EAlan(J-1)*EALFA1(J-1)**2*EALFA2(J-1)**2*EElas*Er(J-1)))
return
end
@
@
@
SUBROUTINE COEF3Vb(ibsay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EZ,EGAMA,EG,EEJ,EK1,EK2,EK3,EK4,EHP,EDELTA
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan,EEIy,EEIxy
,EElas,ED1,ED2,ED3,ED4,REK1,REKB1,J)
@
@
@
@
@
REK1(1)=1
REK1(2)=0
REK1(3)=-1
REK1(4)=0
REKB1= -((-(EALFA1(J)**2*EALFA2(J)**2*EZ**2*
(EDELTA(J)*(4*EBETA3(J)*
(-(EAlan(J)*EBETA3(J)*EElas*EK2(J)) + EEJ(J)*EG*EK2(J)+
EAlan(J)*EEJ(J)*EG*EK4(J)*Er(J))*
323
324
return
end
@
@
@
SUBROUTINE COEF4Vb(ibsay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EZ,EGAMA,EG,EEJ,EK1,EK2,EK3,EK4,EHP,EDELTA
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan,EEIy,EEIxy
,EElas,ED1,ED2,ED3,ED4,REK2,REKB2,J)
@
@
@
@
@
REK2(1)=EZ
REK2(2)=1
REK2(3)=-EZ
REK2(4)=-1
REKB2= -((-(EALFA1(J)**2*EALFA2(J)**2*EZ**2*
(EDELTA(J)*(-12*EAlan(J)*EBETA2(J)*EEJ(J)*EG*EK4(J)*Er(J)*
(EK3(J)*EWx + EK4(J)*EWy) +
6*EAlan(J)*EBETA3(J)**2*EElas*EHP*EK2(J)*
(2*EK3(J)*EPx + 2*EK4(J)*EPy + EHP*EK3(J)*EWx +
EHP*EK4(J)*EWy) +
12*EAlan(J)*EBETA3(J)*EElas*EK4(J)*Er(J)*
(Er(J)*(-(Edy*EWx) + EK1(J)*EWx + Edx*EWy EK2(J)*EWy) + EEIOw(J)*(EK3(J)*EWx + EK4(J)*EWy)) 6*EAlan(J)*EBETA3(J)*EEJ(J)*EG*
(2*EHP*EK4(J)*(EK3(J)*EPx + EK4(J)*EPy)*Er(J) 2*EGAMA(J)*EK2(J)*(EK3(J)*EWx + EK4(J)*EWy) +
EHP**2*EK4(J)*Er(J)*(EK3(J)*EWx + EK4(J)*EWy)) 6*EK2(J)*(2*EBETA2(J)*EEJ(J)*EG*(EK3(J)*EWx + EK4(J)*EWy)+
EBETA3(J)*
(EEJ(J)*EG*EHP*
(2*EK3(J)*EPx + 2*EK4(J)*EPy + EHP*EK3(J)*EWx+
EHP*EK4(J)*EWy) 2*EElas*
(EEIOw(J)*EK3(J)*EWx - Edy*Er(J)*EWx +
EK1(J)*Er(J)*EWx + EEIOw(J)*EK4(J)*EWy +
Edx*Er(J)*EWy - EK2(J)*Er(J)*EWy))) +
4*EBETA3(J)*
(-(EAlan(J)*EBETA3(J)*EElas*EK2(J)) + EEJ(J)*EG*EK2(J)+
EAlan(J)*EEJ(J)*EG*EK4(J)*Er(J))*
(EK3(J)*(EPx + EHP*EWx) + EK4(J)*(EPy + EHP*EWy))*EZ
+ EBETA3(J)*
(EAlan(J)*EBETA3(J)*EElas*EK2(J) - EEJ(J)*EG*EK2(J)EAlan(J)*EEJ(J)*EG*EK4(J)*Er(J))*(EK3(J)*EWx + EK4(J)*EWy)*
EZ**2) +
EAlan(J)*EBETA3(J)**2*EElas*Er(J)*
(EEIxy(J)*(-6*EHP*(2*EPx + EHP*EWx) +
4*(EPx + EHP*EWx)*EZ - EWx*EZ**2) +
EEIy(J)*(6*EHP*(2*EPy + EHP*EWy) 4*(EPy + EHP*EWy)*EZ + EWy*EZ**2)))) 24*EDELTA(J)*EBETA3(J)**2*EElas*
(EALFA2(J)**2*ED2(J)*
((-1 + EAlan(J)*EALFA1(J)**2*EGAMA(J))*EK2(J) EAlan(J)*EK4(J)*Er(J))*Cosh(EALFA1(J)*EZ) +
EALFA2(J)**2*ED1(J)*
((-1 + EAlan(J)*EALFA1(J)**2*EGAMA(J))*EK2(J) EAlan(J)*EK4(J)*Er(J))*Sinh(EALFA1(J)*EZ) +
EALFA1(J)**2*((-1 + EAlan(J)*EALFA2(J)**2*EGAMA(J))*EK2(J)EAlan(J)*EK4(J)*Er(J))*
(ED4(J)*Cosh(EALFA2(J)*EZ) + ED3(J)*Sinh(EALFA2(J)*EZ))))/
325
(24.*EDELTA(J)*EAlan(J)*EALFA1(J)**2*EALFA2(J)**2*EBETA3(J)**2*
EElas**2*Er(J)) + (EALFA2(J-1)**2*ED2(J-1)*((-1 + EAlan(J-1)*
EALFA1(J-1)**2*EGAMA(J-1))*EK2(J-1) EAlan(J-1)*EK4(J-1)*Er(J-1))*Cosh(EALFA1(J-1)*EZ) +
EALFA2(J-1)**2*ED1(J-1)*
((-1 + EAlan(J-1)*EALFA1(J-1)**2*EGAMA(J-1))*EK2(J-1) EAlan(J-1)*EK4(J-1)*Er(J-1))*Sinh(EALFA1(J-1)*EZ) +
EALFA1(J-1)**2*((-1 + EAlan(J-1)*EALFA2(J-1)**2*
EGAMA(J-1))*EK2(J-1) EAlan(J-1)*EK4(J-1)*Er(J-1))*
(ED4(J-1)*Cosh(EALFA2(J-1)*EZ) +
ED3(J-1)*Sinh(EALFA2(J-1)*EZ)))/
(EAlan(J-1)*EALFA1(J-1)**2*EALFA2(J-1)**2*EElas*Er(J-1)))
return
end
@
@
@
@
SUBROUTINE VFONKa(ibsay,iksay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EGAMA,EK1,EK2,EK3,EK4,EElas,EG,EEJ,EDELTA
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan,EHP
,EEIy,EEIxy,ED1,ED2,ED3,ED4,EV,EKAT,J,jm
,EP1,EP2)
@
@
@
@
@
@
326
EAlan(J)*EBETA3(J)**2*EElas*Er(J)*
(EEIxy(J)*(-6*EHP*(2*EPx + EHP*EWx) +
4*(EPx + EHP*EWx)*EKAT(jm) - EWx*EKAT(jm)**2) +
EEIy(J)*(6*EHP*(2*EPy + EHP*EWy) 4*(EPy + EHP*EWy)*EKAT(jm) + EWy*EKAT(jm)**2))) +
24*EDELTA(J)*EBETA3(J)**2*EElas*
(EALFA2(J)**2*ED2(J)*
((-1 + EAlan(J)*EALFA1(J)**2*EGAMA(J))*EK2(J) EAlan(J)*EK4(J)*Er(J))*Cosh(EALFA1(J)*EKAT(jm)) +
EALFA2(J)**2*ED1(J)*
((-1 + EAlan(J)*EALFA1(J)**2*EGAMA(J))*EK2(J) EAlan(J)*EK4(J)*Er(J))*Sinh(EALFA1(J)*EKAT(jm)) +
EALFA1(J)**2*((-1 + EAlan(J)*
EALFA2(J)**2*EGAMA(J))*EK2(J)EAlan(J)*EK4(J)*Er(J))*
(ED4(J)*Cosh(EALFA2(J)*EKAT(jm)) +
ED3(J)*Sinh(EALFA2(J)*EKAT(jm)))))/
(24.*EDELTA(J)*EAlan(J)*EALFA1(J)**2*
EALFA2(J)**2*EBETA3(J)**2*
EElas**2*Er(J))
return
end
@
@
@
@
SUBROUTINE VFONKb(ibsay,iksay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3
,EGAMA,EK1,EK2,EK3,EK4,EElas,EG,EEJ,EDELTA
,EPx,EPy,EWx,EWy,Edx,Edy,Er,EEIOw,EAlan,EHP
,EEIy,EEIxy,ED1,ED2,ED3,ED4,EV,EKAT,J,jm
,EP1,EP2)
@
@
@
@
@
@
return
end
C
C
C
C
C
#################################################################
#################################################################
#
GAUSS-ELIMINASYON ILE [AA]{XB}={B.X}
#
#
DONUSTURULUYOR VE SONDEN YERINE KOYMA
#
#
ILE {XB} VEKTORU HESAPLANIYOR
#
327
C
C
#################################################################
#################################################################
SUBROUTINE GAUSS(SS,N,EVT,EV2)
IMPLICIT REAL*8(A-H,O-Z)
PARAMETER(NE=65)
REAL*8 SS(NE,NE),EVT(NE),EV2(NE),AA(NE,NE+1)
M=N+1
C *** ARTTIRILMIS MATRIS OLUSTURULUYOR ***
DO 3 I=1,N
DO 3 J=1,N
AA(I,J)=SS(I,J)
3
AA(I,M)=EVT(I)
L=N-1
DO 12 K=1,L
JJ=K
BIG=DABS(AA(K,K))
KP1=K+1
DO 7 I=KP1,N
AB=DABS(AA(I,K))
IF(BIG-AB)6,7,7
6
BIG=AB
JJ=I
7
CONTINUE
IF(JJ-K)8,10,8
8
DO 9 J=K,M
TEMP=AA(JJ,J)
AA(JJ,J)=AA(K,J)
9
AA(K,J)=TEMP
10
DO 11 I=KP1,N
QUOT=AA(I,K)/AA(K,K)
DO 11 J=KP1,M
11
AA(I,J)=AA(I,J)-QUOT*AA(K,J)
DO 12 I=KP1,N
12
AA(I,K)=0.
C *** SONDAN YERINE KOYMA ***
EV2(N)=AA(N,M)/AA(N,N)
DO 14 NN=1,L
SUM=0.
I=N-NN
IP1=I+1
DO 13 J=IP1,N
13
SUM=SUM+AA(I,J)*EV2(J)
14
EV2(I)=(AA(I,M)-SUM)/AA(I,I)
RETURN
END
C
C
C
C
#########################################################################
# MATRIX INVERSION USING GAUSS-JORDAN REDUCTION AND PARTIAL PIVOTING. #
# MATRIX B IS THE MATRIX TO BE INVERTED AND A IS THE INVERTED MATRIX. #
#########################################################################
SUBROUTINE INVMATRIS(B,A,N)
IMPLICIT REAL*8(A-H,O-Z)
PARAMETER(NWW=65)
DIMENSION B(NWW,NWW),A(NWW,NWW),INTER(NWW,2)
DO 2 I=1,N
DO 2 J=1,N
2
A(I,J)=B(I,J)
C CYCLE PIVOT ROW NUMBER FROM 1 TO N
DO 12 K=1,N
JJ=K
IF(K.EQ.N)GO TO 6
KP1=K+1
BIG=DABS(A(K,K))
C SEARCH FOR LARGEST PIVOT ELEMENT
DO 5 I=KP1,N
328
AB=DABS(A(I,K))
IF(BIG-AB)4,5,5
4
BIG=AB
JJ=I
5
CONTINUE
C MAKE DECISION ON NECESSITY OF ROW INTERCHANGE AND STORE THE NUMBER
C OF THE ROWS INTERCHANGED DURING KTH REDUCTION. IF NO INTERCHANGE,
C BOTH NUMBERS STORED EQUAL K.
6
INTER(K,1)=K
INTER(K,2)=JJ
IF(JJ-K)7,9,7
7
DO 8 J=1,N
TEMP=A(JJ,J)
A(JJ,J)=A(K,J)
8
A(K,J)=TEMP
C CALCULATE ELEMENTS OF REDUCED MATRIX
C FIRST CALCULATE NEW ELEMENTS OF PIVOT ROW
9
DO 10 J=1,N
IF(J.EQ.K)GO TO 10
A(K,J)=A(K,J)/A(K,K)
10
CONTINUE
C CALCULATE ELEMENT REPLACING PIVOT ELEMENT
A(K,K)=1./A(K,K)
C CALCULATE NEW ELEMENTS NOT IN PIVOT ROW OR COLUMN
DO 11 I=1,N
IF(I.EQ.K)GO TO 11
DO 110 J=1,N
IF(J.EQ.K)GO TO 110
A(I,J)=A(I,J)-A(K,J)*A(I,K)
110
CONTINUE
11
CONTINUE
C CALCULATE NEW ELEMENT FOR PIVOT COLUMN--EXCEPT PIVOT ELEMENT
DO 120 I=1,N
IF(I.EQ.K)GO TO 120
A(I,K)=-A(I,K)*A(K,K)
120
CONTINUE
12
CONTINUE
C REARRANGE COLUMNS OF FINAL MATRIX OBTAINED
DO 13 L=1,N
K=N-L+1
KROW=INTER(K,1)
IROW=INTER(K,2)
IF(KROW.EQ.IROW)GO TO 13
DO 130 I=1,N
TEMP=A(I,IROW)
A(I,IROW)=A(I,KROW)
A(I,KROW)=TEMP
130
CONTINUE
13
CONTINUE
RETURN
END
c *****************************************************************************
c
--------------------------------------------------c
CALCULATION OF SECTION PREPERTIES OF A BEAM ELEMENT
c
--------------------------------------------------c
This program calculates to sectional preperties of a beam element
c which has an arbitrary cross section. Cross-section area, location of the
c centroid, angle of principal radius, bending and polar moments of inertia,
c max and min moments of inertia and location of the shear center are obtained
c during this calculation.
c
c ******************************************************************************
329
SUBROUTINE SECPREP(JJ,PN,EN,P,DX,DY,THICK,EDI,EDJ,
sSUMAREA,sXC,sYC,sTAMX,sTAMY,sTAMXY,
sTCONS,sXGSHR,sYGSHR)
PARAMETER (N=65)
IMPLICIT REAL*8 (A-H,K-Z)
I and J are variables
DIMENSION DX(N),DY(N),THICK(N),EDI(N),EDJ(N)
DIMENSION W(N),WO(N),DIS(N),EKIX(N),EKIY(N),EKJX(N),EKJY(N)
@
,ELAREA(N),EKXCC(N),EKYCC(N),AME1(N),AME2(N),TETA(N)
@
,SN2TE(N),CS2TE(N),WPRM(N),RR(N),SCAR1(N),SECAREA11(N)
@
,Cix(N),Ciy(N),Cjx(N),Cjy(N)
@
,AR(N),AR1(N),AR2(N),XCC(N),SPR(N),XC1(N),XC2(N)
DIMENSION sSUMAREA(N),sXC(N),sYC(N),sTAMX(N),sTAMY(N)
,sTAMXY(N),sTCONS(N),sXGSHR(N),sYGSHR(N)
10
15
22
23
25
26
FORMAT(A8)
FORMAT(A50)
FORMAT(1X,30F10.3)
FORMAT(1X,30F10.2)
FORMAT(1X,30F13.4)
FORMAT(//7X,'EL',6X,'LENGTH',9X,'CROSS SEC',6X,'MOM.IN.-LOC.X'
@
,3X,'MOM.IN.-LOC.Y',4X,'ANG-GLOB.X',/6X,4('-'),4X,9('-')
@
,7X,9('-'),6X,12('-'),4X,12('-'),3X,12('-'))
27 FORMAT(5X,I4,4X,E12.5,4X,E12.5,4X,E12.5,4X,E12.5,4X,E12.5)
28 FORMAT(1X,30F10.4)
TOL1=0.9999999
TOL2=1.0000001
ARG=1.0
PI=4.0*DATAN(ARG)
E=1.0E-4
C
DX(EDI(I))
DY(EDI(I))
DX(EDJ(I))
DY(EDJ(I))
c
c
c
c
WRITE(6,*)
WRITE(6,*)
WRITE(6,*)
WRITE(6,*)
'
'--------------------' LENGTH OF ELEMENTS
'---------------------
'
'
'
'
DO I=1,EN
DIS(I)=((EKJX(I)-EKIX(I))**2+(EKJY(I)-EKIY(I))**2)**(0.5)
WRITE(6,22) DIS(I)
ENDDO
C-----------------------------------------------------------C
CALCULATING THE CROSS-SECTION AREA OF THE EACH ELEMENT
C-----------------------------------------------------------C
330
C
C
fark=EDJ(I)-EDI(I)
IF((fark.GT.TOL1).AND.(fark.LT.TOL2)) THEN
ELAREA(I)= DIS(I) * THICK(I)
else
ELAREA(I)= (DIS(I)-THICK(I)/2) * THICK(I)
ENDIF
ENDDO
C
C
C
C
CALCULATING THE MOMENTS OF INERTIA OF THE EACH ELEMENT IN OWN LOCAL AXES
DO I=1,EN
AME1(I)=(DIS(I)*(THICK(I)**3))/12.0
AME2(I)=((DIS(I)**3)*THICK(I))/12.0
ENDDO
C
AXIS
103
105
104
106
107
108
109
110
c
c
c
c
c
C
CALCULATING THE ANGLE BETWEEN LOCAL AXES OF THE EACH ELEMENT AND GLOBAL X
DO I=1,EN
IF(DABS(EKIX(I)-EKJX(I)).LE.E) GOTO 103
IF(DABS(EKIY(I)-EKJY(I)).LE.E) GOTO 104
GOTO 107
IF(EKIY(I).LT.EKJY(I)) GOTO 105
TETA(I)=PI/2.0
GOTO 110
TETA(I)=-PI/2.0
GOTO 110
IF(EKIX(I).LT.EKJX(I)) GOTO 106
TETA(I)=PI
GOTO 110
TETA(I)=0.0
GOTO 110
SLOPE=(EKIY(I)-EKJY(I))/(EKIX(I)-EKJX(I))
IF(EKIX(I).GT.EKJX(I)) GOTO 108
GOTO 109
TETA(I)=PI-DATAN(SLOPE)
GOTO 110
TETA(I)= -DATAN(SLOPE)
SN2TE(I)=DSIN(2.*TETA(I))
CS2TE(I)=DCOS(2.*TETA(I))
ENDDO
WRITE(6,26)
DO I=1,EN
TETA(I)=TETA(I)*180/PI
WRITE(6,27) I,DIS(I),ELAREA(I),AME1(I),AME2(I),TETA(I)
ENDDO
WRITE(6,*)
WRITE(6,*)
WRITE(6,*)
WRITE(6,*)
'
'
'----------------------------------------'
' TOTAL CROSS-SECTION AREA OF THE SECTION'
'----------------------------------------'
331
ENDDO
WRITE(6,25) SUMAREA
sSUMAREA(JJ)=SUMAREA
c
c
c
c
c
c
WRITE(6,*) '
WRITE(6,*) '
Coordinates of the Centroid
WRITE(6,*) '-------------------------------WRITE(6,*) '
Xc
Yc
WRITE(6,*) '-------------------------------WRITE(6,25) XC,YC
sXC(JJ)=XC
sYC(JJ)=YC
'
'
'
'
'
C
C
c
c
c
c
c
c
c
DO I=1,EN
GAMXe =(AME1(I)+AME2(I))/2.0+(AME1(I)-AME2(I))*CS2TE(I)/2.0
GAMYe =(AME1(I)+AME2(I))/2.0-(AME1(I)-AME2(I))*CS2TE(I)/2.0
GAMXYe=(AME1(I)-AME2(I))*SN2TE(I)/2.0
DDX=(XC-EKXCC(I))
DDY=(YC-EKYCC(I))
AMX =GAMXe+ELAREA(I)*DDY**2
AMY =GAMYe+ELAREA(I)*DDX**2
AMXY =GAMXYe+ELAREA(I)*DDX*DDY
TAMX =TAMX+AMX
TAMY =TAMY+AMY
TAMXY=TAMXY+AMXY
ENDDO
WRITE(6,*) '
'
WRITE(6,*) ' Moments of inertia of the section acording to '
WRITE(6,*) ' x-y axes locating on centroid of the section '
WRITE(6,*) '------------------------------------------------ '
WRITE(6,*) '
Ixc
Iyc
Ixyc
'
WRITE(6,*) '------------------------------------------------ '
WRITE(6,25) TAMX,TAMY,TAMXY
sTAMX(JJ)=TAMX
sTAMY(JJ)=TAMY
sTAMXY(JJ)=TAMXY
DETERMINATION OF MAX AND MIN MOMENTS OF INERTIA
DSRT=DSQRT((TAMX-TAMY)**2.0+4.0*TAMXY**2.0)
AMMAX=(TAMX+TAMY)/2.0+DSRT/2.0
AMMIN=(TAMX+TAMY)/2.0-DSRT/2.0
IF(DABS(TAMX-TAMY).LE.E) GOTO 51
ALFA=(DATAN(2.0*TAMXY/(TAMY-TAMX)))/2.0
IF(TAMX.LT.TAMY.AND.TAMXY.GE.0.0) GOTO 134
IF(TAMX.LT.TAMY.AND.TAMXY.LT.0.0) ALFA=PI/2.0+ALFA
332
GOTO 50
134 ALFA=PI/2.0-ALFA
GOTO 50
51 ALFA=PI/4.0
IF(TAMXY.GE.0.0) ALFA=-PI/4.0
50 DALFA=ALFA*180.0/PI
c
c
c
c
c
c
WRITE(6,*) '
'
WRITE(6,*) '
MAX AND MIN MOMENTS OF INERTIA
'
WRITE(6,*) '----------------------------------------------- '
WRITE(6,*) '
Imaxc
Iminc
ALFA
'
WRITE(6,*) '----------------------------------------------- '
WRITE(6,25) AMMAX,AMMIN,DALFA
C-----------------------------------------------------C
DETERMINATION OF TORSIONAL CONSTANT (J)
C-----------------------------------------------------TCONS=0.0
DO I=1,EN
TCONS=TCONS+(1./3.)*(DIS(I)*THICK(I)**3)
ENDDO
c
c
c
c
c
c
WRITE(6,*) '
WRITE(6,*) ' TORSIONAL CONSTANT
WRITE(6,*) '-------------------WRITE(6,*) '
J
WRITE(6,*) '-------------------WRITE(6,25) TCONS
sTCONS(JJ)=TCONS
'
'
'
'
'
C-----------------------------------------------------C
DETERMINATION OF THE SHEAR CENTER
C-----------------------------------------------------C
C
C
C
C
C
c
c
WRITE(6,*) '
'
WRITE(6,*) '
'
WRITE(6,*) '
DETERMINATION OF X AND Y COORDINATES OF NODES
'
WRITE(6,*) '
'
WRITE(6,*) '
Nix
Niy
Njx
Njy
'
WRITE(6,*) '---------------------------------------------------- '
DO I=1,EN
Centroid'te bulunan, global eksen takmna paralel
x,y eksen takmna gre dm noktas koordinatlar
Cix(I)=EKIX(I)-XC
Ciy(I)=EKIY(I)-YC
Cjx(I)=EKJX(I)-XC
Cjy(I)=EKJY(I)-YC
WRITE(6,28) CPRMix(I),CPRMiy(I),CPRMjx(I),CPRMjy(I)
ENDDO
C
C
C
WRITE(6,*) '
'
DO I=1,EN
iki vektrn vektrel arpm
SCAR1(I)=((EKIX(I)-XC)*(EKJY(I)-YC)-(EKJX(I)-XC)*(EKIY(I)-YC))
WRITE(6,22) SCAR1(I)
ENDDO
WRITE(6,*) '
'
WRITE(6,*) '------------------------------------------ '
C
C
C
C
333
C
C
WRITE(6,*) '
W PRIME (the centroid is the pole)
'
WRITE(6,*) '------------------------------------------ '
SECAREA11(1)=0.0
PNF=PN-1
DO I=1,PNF
fark=EDJ(I)-EDI(I)
IF((fark.GT.TOL1).AND.(fark.LT.TOL2)) THEN
SECAREA11(I+1)=SECAREA11(I)+SCAR1(I)
else
SECAREA11(I+1)=SECAREA11(I+1-fark)+SCAR1(I)
ENDIF
ENDDO
WRITE(6,*) '
'
DO I=1,PN
WRITE(6,22) SECAREA11(I)
ENDDO
c
C
C-----------------------------------------------------C
CALCULATION OF COORDINATES OF THE SHEAR CENTER
C-----------------------------------------------------SWXga=0.0
SWYga=0.0
DO I=1,EN
SWXga=SWXga+(1./6.)*DIS(I)*THICK(I)
*(SECAREA11(EDI(I))*(2*Ciy(I)+Cjy(I))
+ SECAREA11(EDJ(I))*(Ciy(I)+2*Cjy(I)))
@
@
SWYga=SWYga+(1./6.)*DIS(I)*THICK(I)
*(SECAREA11(EDI(I))*(2*Cix(I)+Cjx(I))
+ SECAREA11(EDJ(I))*(Cix(I)+2*Cjx(I)))
@
@
c
c
c
c
ENDDO
WRITE(6,*) '
'
WRITE(6,*) '
SWXga
SWYga
'
WRITE(6,*) '------------------------------'
WRITE(6,22) SWXga,SWYga
XSHEAR= (TAMY*SWXga-TAMXY*SWYga)/(TAMX*TAMY-TAMXY**2)
YSHEAR=-(TAMX*SWYga-TAMXY*SWXga)/(TAMX*TAMY-TAMXY**2)
XGSHR=XC+XSHEAR
YGSHR=YC+YSHEAR
c
c
c
c
c
c
WRITE(6,*) '
'
WRITE(6,*) ' COORDINATES OF THE SHEAR CENTER (IN LOCAL AXES)
'
WRITE(6,*) '-----------------------------------------------------'
WRITE(6,*) '
Xsl
Ysl
'
WRITE(6,*) '-----------------------------------------------------'
WRITE(6,25) XSHEAR,YSHEAR
c
c
c
c
c
c
WRITE(6,*) '
'
WRITE(6,*) ' COORDINATES OF THE SHEAR CENTER (IN GLOBAL AXES)
'
WRITE(6,*) '-----------------------------------------------------'
WRITE(6,*) '
Xsg
Ysg
'
WRITE(6,*) '-----------------------------------------------------'
WRITE(6,25) XGSHR,YGSHR
sXGSHR(JJ)=XGSHR
sYGSHR(JJ)=YGSHR
RETURN
END
334
c *********************************************************************
c
c
ARPILMA ATALET MOMENT HESABI (Iw)
c
c *********************************************************************
SUBROUTINE CAIw(JJ,CSAREA,PN,EN,SCN,Sx,Sy,P,DX,DY,
EDI,EDJ,THICK,PRSECAREA,PRSA,SIW)
PARAMETER (N=65)
IMPLICIT REAL*8 (A-H,K-Z)
I ve J dedisken olarak kaldi
dimension DX(N),DY(N),THICK(N),DIS(N),PRSECAREA(N)
,EDI(N),EDJ(N),EKIX(N),EKIY(N),EKJX(N),EKJY(N)
,SECAREA1(N),PRSA(N,N),SCAR(N),SIW(N),CSAREA(N)
@
@
910
920
922
923
925
FORMAT(A8)
FORMAT(A50)
FORMAT(1X,30F10.3)
FORMAT(1X,30F10.2)
FORMAT(1X,30F13.4)
TOL1=0.9999999
TOL2=1.0000001
c
C
C
c
c
c
DO I=1,EN
EKIX(I)= DX(EDI(I))
EKIY(I)= DY(EDI(I))
EKJX(I)= DX(EDJ(I))
EKJY(I)= DY(EDJ(I))
ENDDO
WRITE(6,*) '
'
DO I=1,EN
iki vektrn vektrel arpm
SCAR(I)=((EKIX(I)-Sx)*(EKJY(I)-Sy)-(EKJX(I)-Sx)*(EKIY(I)-Sy))
WRITE(6,922) SCAR(I)
ENDDO
WRITE(6,*) '
'
WRITE(6,*) ' W PRIME (for initial radius) '
WRITE(6,*) '------------------------------- '
SECAREA1(1)=0.0
1. YNTEM
PNF=PN-1
DO I=1,PNF
fark=EDJ(I)-EDI(I)
IF((fark.GT.TOL1).AND.(fark.LT.TOL2)) THEN
SECAREA1(I+1)=SECAREA1(I)+SCAR(I)
else
SECAREA1(I+1)=SECAREA1(I+1-fark)+SCAR(I)
ENDIF
ENDDO
c
c
C
WRITE(6,*) '
'
DO I=1,PN
WRITE(6,922) SECAREA1(I)
ENDDO
2. YNTEM
335
C
C
C
C
C
C
C
DO I=2,PN
SECAREA1(I)=SECAREA1(EDI(I-1))+SCAR(I-1)
ENDDO
WRITE(6,*) '
'
DO I=1,PN
WRITE(6,922) SECAREA1(I)
ENDDO
c
c
c
WRITE(6,*) '
'
WRITE(6,*) ' LENGTH OF ELEMENTS '
WRITE(6,*) '--------------------- '
c
c
c
c
DO I=1,EN
DIS(I)=((DX(EDJ(I))-DX(EDI(I)))**2
@
+(DY(EDJ(I))-DY(EDI(I)))**2)**(0.5)
WRITE(6,922) DIS(I)
ENDDO
WRITE(6,*) '
'
WRITE(6,*) ' SWS '
WRITE(6,*) '-------'
SWS=0.0
DO I=1,EN-1
SWS=SWS+(0.5)*THICK(I)*DIS(I)*(SECAREA1(EDI(I))+SECAREA1(EDJ(I)))
ENDDO
WRITE(6,922) SWS
R=SWS/CSAREA(JJ)
c
c
c
c
c
c
c
WRITE(6,*) '
'
WRITE(6,*) ' R
'
WRITE(6,*) '------'
WRITE(6,922) R
WRITE(6,*) '
'
WRITE(6,*) ' W PRI '
WRITE(6,*) '---------'
DO I=1,PN
PRSECAREA(I)=SECAREA1(I)-R
WRITE(6,922) PRSECAREA(I)
ENDDO
DO I=1,PN
PRSA(JJ,I)=PRSECAREA(I)
ENDDO
SSIW=0.0
DO I=1,EN-1
SSIW=SSIW+(1./3.)*DIS(I)*THICK(I)*((PRSECAREA(EDI(I)))**2.+
@
PRSECAREA(EDI(I))*PRSECAREA(EDJ(I))+
@
(PRSECAREA(EDJ(I)))**2.)
ENDDO
c
c
c
WRITE(6,*) '
'
WRITE(6,*) '
Iw
'
WRITE(6,*) '---------'
WRITE(6,922) SSIW
SIW(JJ)=SSIW
336
RETURN
END
C
c
C
C
C
C
C
#############################################################
#
PROGRAM MAIN
#
#
JACK3.FOR
#
#
JACOBI METODU ILE SERBEST TITRESIM ANALIZI YAPILIYOR.
#
#
SISTEM MATRISLERI KARE OLARAK SAKLANIYOR.
#
#
OZVEKTORLER KUTLEYE GORE NORMALIZE EDILIYOR.
#
#############################################################
SUBROUTINE JACK3(NTD,SRM,SKM,EGNVAL,EGNVEC)
IMPLICIT REAL*8 (A-H,O-Z)
PARAMETER(NQ=65)
DIMENSION SRM(NQ,NQ),SKM(NQ,NQ)
DIMENSION EGNVEC(NQ,NQ),EGNVAL(NQ),NORD(NQ),EGNVECS(NQ,NQ)
,EGNVALS(NQ)
##### J A C O B I
M E T O D U #######
CALL AXLBX(NTD,SRM,SKM,EGNVAL,EGNVEC,NQ)
DO 315 J=1,NTD
315
NORD(J)=J
C *** Ozdegerler siraya konuyor ***
DO 310 I=1,NTD
II=NORD(I)
I1=II
C1=EGNVAL(II)
J1=I
DO 300 J=I,NTD
IJ=NORD(J)
IF(C1.LE.EGNVAL(IJ)) GO TO 300
C1=EGNVAL(IJ)
I1=IJ
J1=J
300
CONTINUE
IF(I1.EQ.II) GO TO 310
NORD(I)=I1
NORD(J1)=II
310
CONTINUE
C *********************************************************************
PI=4.0D0*DATAN(1.0D0)
DO 230 II=1,NTD
DO 230 IA=1,NTD
NVEC=NORD(II)
c
WRITE(6,695)
c695
FORMAT('#######################################################',
c
$'##########')
c
WRITE(*,700)II,EGNVAL(NVEC)
EGNVALS(II)=EGNVAL(NVEC)
C700
FORMAT('OZDEGER(',I2,')=',E11.5)
C
WRITE(6,232)II
C232
FORMAT(/,T20,'OZVEKTOR(',I2,')')
230
EGNVECS(IA,II)=EGNVEC(IA,NVEC)
C230
WRITE(*,540)(EGNVEC(I,NVEC),I=1,NTD)
C540
FORMAT(4E15.5)
233
C
C
DO 233 I=1,NTD
DO 233 J=1,NTD
EGNVAL(I)=EGNVALS(I)
EGNVEC(J,I)=EGNVECS(J,I)
RETURN
END
#################################################################
#
OZDEGER PROBLEMINI COZMEK ICIN ALTPROGRAM:
#
337
C
C
C
C
#
[A]{X} = LAMBDA.[B]{X}
#
#
PROGRAM SADECE POSITIVE-DEFINITE [B] MATRISI ICIN COZUM
#
#
YAPAR V, VT, W AND IH'NIN DIMENSION'LARI AYNI OLMALIDIR.
#
#################################################################
SUBROUTINE AXLBX(N,A,B,XX,X,NQ)
IMPLICIT REAL*8 (A-H,O-Z)
PARAMETER(NEQ=65)
DIMENSION A(NQ,NQ),B(NQ,NQ),XX(NQ),X(NQ,NQ)
DIMENSION V(NEQ,NEQ),VT(NEQ,NEQ),W(NEQ,NEQ),IH(NEQ)
10
DO 10 I=1,N
DO 10 J=1,N
B(J,I)=B(I,J)
20
30
DO 30 I=1,N
IF(B(I,I)) 20,30,30
WRITE(6,80)
STOP
CONTINUE
40
DO 40 I=1,N
DO 40 J=1,N
VT(I,J)=V(J,I)
50
DO 50 I=1,N
B(I,I)=1.0/DSQRT(B(I,I))
C
C
60
DO 60 J=1,N
XX(J)=A(J,J)
C
C
338
C
C
C
C
C
C
C
C
20
30
40
60
70
75
80
90
100
110
120
130
140
150
160
170
180
190
200
210
#################################################################
#
AMAC : [Q] MATRISINI DIAGONAL HALE GETIRMEK
#
#
DEGISKENLERIN TANIMI
#
#
N
: REEL,SIMETRIK [Q] MATRISININ MERTEBESI (N > 2)
#
#
[Q] : DIAGONAL HALE GETIRILECEK MATRIS
#
#
[V] : OZVEKTOR MATRISI
#
#
M
: UYGULANAN ROTASYON SAYISI
#
#################################################################
SUBROUTINE JACOBI (N,Q,V,X,IH,NQ)
IMPLICIT REAL*8 (A-H,O-Z)
DIMENSION Q(NQ,NQ),V(NQ,NQ),X(NQ),IH(NQ)
EPSI=1.0D-08
DO 40 I=1,N
DO 40 J=1,N
IF(I-J) 30,20,30
V(I,J)=1.0
GO TO 40
V(I,J)=0.0
CONTINUE
M=0
MI=N-1
DO 70 I=1,MI
X(I)=0.0
MJ=I+1
DO 70 J=MJ,N
IF(X(I)-DABS(Q(I,J))) 60,60,70
X(I)=DABS(Q(I,J))
IH(I)=J
CONTINUE
DO 100 I=1,MI
IF(I-1) 90,90,80
IF(XMAX-X(I)) 90,100,100
XMAX=X(I)
IP=I
JP=IH(I)
CONTINUE
IF(XMAX-EPSI) 500,500,110
M=M+1
IF(Q(IP,IP)-Q(JP,JP)) 120,130,130
TANG=-2.0*Q(IP,JP)/(DABS(Q(IP,IP)-Q(JP,JP))+DSQRT((Q(IP,IP)
&
-Q(JP,JP))**2+4.0*Q(IP,JP)**2))
GO TO 140
TANG=2.0*Q(IP,JP)/(DABS(Q(IP,IP)-Q(JP,JP))+DSQRT((Q(IP,IP)
&
-Q(JP,JP))**2+4.0*Q(IP,JP)**2))
COSN=1.0/DSQRT(1.0+TANG**2)
SINE=TANG*COSN
QII=Q(IP,IP)
Q(IP,IP)=COSN**2*(QII+TANG*(2.*Q(IP,JP)+TANG*Q(JP,JP)))
Q(JP,JP)=COSN**2*(Q(JP,JP)-TANG*(2.*Q(IP,JP)-TANG*QII))
Q(IP,JP)=0.0
IF(Q(IP,IP)-Q(JP,JP)) 150,190,190
TEMP=Q(IP,IP)
Q(IP,IP)=Q(JP,JP)
Q(JP,JP)=TEMP
IF(SINE) 160,170,170
TEMP=COSN
GO TO 180
TEMP=-COSN
COSN=DABS(SINE)
SINE=TEMP
DO 260 I=1,MI
IF(I-IP) 210,260,200
IF(I-JP) 210,260,210
IF(IH(I)-IP) 220,230,220
339
220
230
240
250
260
270
280
290
300
320
330
340
350
380
390
400
410
430
450
500
C
C
C
10
IF(IH(I)-JP) 260,230,260
K=IH(I)
TEMP=Q(I,K)
Q(I,K)=0.0
MJ=I+1
X(I)=0.0
DO 250 J=MJ,N
IF(X(I)-DABS(Q(I,J))) 240,240,250
X(I)=DABS(Q(I,J))
IH(I)=J
CONTINUE
Q(I,K)=TEMP
CONTINUE
X(IP)=0.0
X(JP)=0.0
DO 430 I=1,N
IF(I-IP) 270,430,320
TEMP=Q(I,IP)
Q(I,IP)=COSN*TEMP+SINE*Q(I,JP)
IF(X(I)-DABS(Q(I,IP))) 280,290,290
X(I)=DABS(Q(I,IP))
IH(I)=IP
Q(I,JP)=-SINE*TEMP+COSN*Q(I,JP)
IF(X(I)-DABS(Q(I,JP))) 300,430,430
X(I)=DABS(Q(I,JP))
IH(I)=JP
GO TO 430
IF(I-JP) 330,430,380
TEMP=Q(IP,I)
Q(IP,I)=COSN*TEMP+SINE*Q(I,JP)
IF(X(IP)-DABS(Q(IP,I))) 340,350,350
X(IP)=DABS(Q(IP,I))
IH(IP)=I
Q(I,JP)=-SINE*TEMP+COSN*Q(I,JP)
IF(X(I)-DABS(Q(I,JP))) 300,430,430
TEMP=Q(IP,I)
Q(IP,I)=COSN*TEMP+SINE*Q(JP,I)
IF(X(IP)-DABS(Q(IP,I))) 390,400,400
X(IP)=DABS(Q(IP,I))
IH(IP)=I
Q(JP,I)=-SINE*TEMP+COSN*Q(JP,I)
IF(X(JP)-DABS(Q(JP,I))) 410,430,430
X(JP)=DABS(Q(JP,I))
IH(JP)=I
CONTINUE
DO 450 I=1,N
TEMP=V(I,IP)
V(I,IP)=COSN*TEMP+SINE*V(I,JP)
V(I,JP)=-SINE*TEMP+COSN*V(I,JP)
GO TO 75
RETURN
END
#################################################################
#
[C]=[A][B] MATRIS CARPIMI OLUSTURULUYOR
#
#################################################################
SUBROUTINE MTRXML(A,N,M,B,L,C,NQ)
IMPLICIT REAL*8 (A-H,O-Z)
DIMENSION A(NQ,NQ),B(NQ,NQ),C(NQ,NQ)
DO 10 I=1,N
DO 10 J=1,L
C(I,J)=0.0
DO 10 K=1,M
C(I,J)=C(I,J)+A(I,K)*B(K,J)
RETURN
END
340
C
C
C
C
10
C
C
C
C
10
#################################################################
# A MATRISI ILE B VEKTORUNU CARPARAK C VEKTORUNU OLUSTURULUYOR #
#
[C]=[A][B] MATRIS CARPIMI OLUSTURULUYOR
#
#################################################################
SUBROUTINE MTRXML1(A,N,M,B,L,C,NQ)
IMPLICIT REAL*8 (A-H,O-Z)
DIMENSION A(NQ,NQ),B(NQ),C(NQ)
DO 10 I=1,N
DO 10 J=1,L
C(I)=0.0
DO 10 K=1,M
C(I)=C(I)+A(I,K)*B(K)
RETURN
END
#################################################################
#
A MATRISI ILE B VEKTORUNU CARPARAK C SATISI OLUSTURULUYOR #
#
[C]=[A][B] MATRIS CARPIMI OLUSTURULUYOR
#
#################################################################
SUBROUTINE MTRXML2(A,N,B,L,C,NQ)
IMPLICIT REAL*8 (A-H,O-Z)
DIMENSION A(NQ),B(NQ)
C=0.0
DO 10 I=1,N
C=C+A(I)*B(I)
RETURN
END
C
C
C
C
C
C
C
#################################################################
#################################################################
#
GAUSS-ELIMINASYON ILE [AA]{XB}={B.X}
#
#
DONUSTURULUYOR VE SONDEN YERINE KOYMA
#
#
ILE {XB} VEKTORU HESAPLANIYOR
#
#################################################################
#################################################################
SUBROUTINE GAUSS2(SS,N,EVT,EV2)
IMPLICIT REAL*8(A-H,O-Z)
PARAMETER(NE=65)
REAL*8 SS(NE,NE),EVT(NE),EV2(NE),AA(NE,NE)
M=N+1
C *** ARTTIRILMIS MATRIS OLUSTURULUYOR ***
DO 3 I=1,N
DO 3 J=1,N
AA(I,J)=SS(I,J)
3
AA(I,M)=EVT(I)
L=N-1
DO 12 K=1,L
JJ=K
BIG=DABS(AA(K,K))
KP1=K+1
DO 7 I=KP1,N
AB=DABS(AA(I,K))
IF(BIG-AB)6,7,7
6
BIG=AB
JJ=I
7
CONTINUE
IF(JJ-K)8,10,8
8
DO 9 J=K,M
TEMP=AA(JJ,J)
AA(JJ,J)=AA(K,J)
9
AA(K,J)=TEMP
10
DO 11 I=KP1,N
QUOT=AA(I,K)/AA(K,K)
DO 11 J=KP1,M
11
AA(I,J)=AA(I,J)-QUOT*AA(K,J)
DO 12 I=KP1,N
341
12
AA(I,K)=0.
C *** SONDAN YERINE KOYMA ***
EV2(N)=AA(N,M)/AA(N,N)
DO 14 NN=1,L
SUM=0.
I=N-NN
IP1=I+1
DO 13 J=IP1,N
13
SUM=SUM+AA(I,J)*EV2(J)
14
EV2(I)=(AA(I,M)-SUM)/AA(I,I)
RETURN
END
342