dynamic analysis of non-planar coupled shear walls

ÇUKUROVA UNIVERSITY

INSTITUTE OF BASIC AND APPLIED SCIENCES

Ph.D. THESIS Cevher Deha TÜRKÖZER DYNAMIC ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS

DEPARTMENT OF CIVIL ENGINEERING ADANA, 2008

ÇUKUROVA ÜNİVERSİTESİ FEN BİLİMLERİ ENSTİTÜSÜ

DYNAMIC ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS

Cevher Deha TÜRKÖZER

DOKTORA TEZİ

İNŞAAT MÜHENDİSLİĞİ BÖLÜMÜ

Bu tez 07/10/2008 tarihinde aşağıdaki jüri üyeleri tarafından oybirliği ile kabul edilmiştir. İmza:....................................... İmza:........................................ İmza:..............................................

Prof. Dr. Orhan AKSOĞAN Prof. Dr. Vebil YILDIRIM Doç. Dr. H. Murat ARSLAN

DANIŞMAN ÜYE ÜYE İmza:........................................... İmza:......................................

Yard. Doç. Dr. S. Seren GÜVEN Yard. Doç. Dr. Murat BİKÇE

ÜYE ÜYE

Bu tez Enstitümüz İnşaat Mühendisliği Anabilim Dalında hazırlanmıştır.

Kod No: Prof. Dr. Aziz ERTUNÇ Enstitü Müdürü Not: Bu tezde kullanılan özgün ve başka kaynaktan yapılan bildirişlerin, çizelge, şekil ve

fotoğrafların kaynak gösterilmeden kullanımı, 5846 sayılı Fikir ve Sanat Eserleri Kanunundaki hükümlere tabidir.

Dedicated to my parents,

Cevher Türközer and Sıdıka Türközer

I

ABSTRACT

Ph.D. THESIS

DYNAMIC ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS


DEPARTMENT OF CIVIL ENGINEERING INSTITUTE OF BASIC AND APPLIED SCIENCES

UNIVERSITY OF ÇUKUROVA

Supervisor: Prof. Dr. Orhan AKSOĞAN

Year: 2008, Pages: 342

Jury : Prof. Dr. Orhan AKSOĞAN : Prof. Dr. Vebil YILDIRIM : Assoc. Prof. Dr. H. Murat ARSLAN : Asst. Prof. Dr. Seren GÜVEN : Asst. Prof. Dr. Murat BİKÇE

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 has been carried out. 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 as well. In the analysis, Vlasov’s theory of thin-walled beams and Continuous Connection Method (CCM) have been employed to find the stiffness matrix of the system. For this purpose, the connecting beams have been replaced by an equivalent layered medium and unit forces have been applied in the directions of the degrees of freedom to find the displacements of the system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. The warping of the cross-sections of the piers due to their twist, as well as their bending, has been considered in obtaining the displacements. The system mass matrix has been found in the form of lumped masses at the heights where the unit forces have been applied. Following the free vibration analysis, uncoupled stiffness, damping and mass matrices have been found employing the mode superposition method. A time-history analysis has been carried out using Newmark numerical integration method to find the system displacement vector for every time step. Finally, a computer program has been prepared in Fortran language and various examples have been solved. The results have been verified via comparisons with those of the SAP2000 structural analysis program.

Key Words: Continuous Connection Method, Non-planar Coupled Shear Wall,

Dynamic Analysis, Vlasov’s Theory, Newmark Method.

II

ÖZ

DOKTORA TEZİ

DÜZLEMSEL OLMAYAN BOŞLUKLU DEPREM PERDELERİNİN DİNAMİK ANALİZİ


ÇUKUROVA ÜNİVERSİTESİ FEN BİLİMLERİ ENSTİTÜSÜ

İNŞAAT MÜHENDİSLİĞİ ANABİLİM DALI

Danışman: Prof. Dr. Orhan AKSOĞAN

Yıl: 2008, Sayfa: 342

Jüri : Prof. Dr. Orhan AKSOĞAN : Prof. Dr. Vebil YILDIRIM : Doç. Dr. H. Murat ARSLAN : Yard. Doç. Dr. Seren GÜVEN : Yard. Doç. Dr. Murat BİKÇE

Bu tezde, rijit temeller üzerine oturan, bağlantı kirişi-duvar birleşim noktalarında esneklik olan ve istenen sayıda güçlendirici kirişe sahip düzlemsel olmayan boşluklu deprem perdelerinin dinamik analizi yapılmıştır. Perde kesiti ile kat ve bağlantı kirişi yüksekliklerinin perde yüksekliği boyunca bölgeden bölgeye değişimleri de dikkate alınmıştır. Analizde, sistem rijitlik matrisini bulmak için Vlasov’un ince cidarlı kiriş teorisi ve sürekli bağlantı yöntemi (SBY) kullanılmıştır. Bu amaçla bağlantı kirişleri sürekli bir ortama dönüştürülmüş ve serbestlik dereceleri doğrultusunda birim yükler uygulanarak her birim yük için bina boyunca yerdeğiştirmeler hesaplanmıştır. Yerdeğiştirme ifadeleri elde edilirken duvarların eğilmelerine ek olarak burulmadan dolayı kesitlerin çarpılması da göz önüne alınmıştır. Sistem kütle matrisi toplanmış kütle kabulüne göre yükseklik boyunca istenilen sayıda kütle alınarak elde edilmiştir. Serbest titreşim analizinden sonra mod-süperpozisyon yöntemi ile girişimsiz rijitlik, sönüm ve kütle matrisleri oluşturulmuştur. Zaman tanım alanında (Time-history) analiz, Newmark sayısal integrasyon yöntemi kullanılarak her zaman değerine karşı gelen sistem yerdeğiştirme değerleri hesaplanarak yapılmıştır. Son olarak, yapılan analize göre Fortran dilinde genel amaçlı bir bilgisayar programı hazırlanıp çeşitli örnekler çözülmüştür. Elde edilen sonuçlar SAP2000 yapı analizi programı kullanılarak bulunan sonuçlarla karşılaştırılarak doğrulanmıştır. Anahtar Kelimeler: Sürekli Bağlantı Yöntemi, Düzlemsel Olmayan Boşluklu Deprem Perdeleri, Dinamik Analiz, Vlasov Teorisi, Newmark Yöntemi.

III

ACKNOWLEDGEMENT

First of all I would like to thank Prof. Dr. Orhan AKSOĞAN 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.

IV

CONTENTS PAGE NUMBER

ABSTRACT ..............................................................................................................I

ÖZ ........................................................................................................................... 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

V

4.3.4.1.(1). The Relative Vertical Displacements due to

the Deflections and Rotations of the Piers .............. 31

4.3.4.1.(2). The Relative Vertical Displacement due to the Axial

Deformation of the Piers......................................... 35

4.3.4.1.(3). The Relative Vertical Displacement due to

the Bending of the Laminae ................................... 37


the Shearing Effect in the Laminae ......................... 38


the Flexibility of the Connections ........................... 39


the Change in Cross-Section .................................. 41

4.3.4.2. Equilibrium Equations ............................................................. 43

4.3.4.2.(1). Bending moment equilibrium equations ................. 43

4.3.4.2.(2). Bimoment equilibrium equation ............................. 45

4.3.4.2.(3). Twisting moment equilibrium equation .................. 50

4.3.4.3. Method of Solution .................................................................. 55

4.3.4.4. Determination of the shear forces in the stiffening beams ......... 62

4.3.4.5. Boundary Conditions ............................................................... 64

4.3.4.6. Determination of the rotation function ( θ ) ............................... 69

4.3.4.7. Determination of the lateral displacement

functions (u and v) .................................................................. 69

4.4. Determination of Eigenvalues and Eigenvectors .......................................... 71

5. FORCED VIBRATION ANALYSIS OF NON-PLANAR COUPLED

SHEAR WALLS ............................................................................................... 72

5.1. Introduction ................................................................................................. 72

5.2. Mode Superposition Method........................................................................ 73

5.2.1. Introduction ....................................................................................... 73

5.2.2. Determination of Uncoupled Equation Set ......................................... 74

5.3. Time-History Analysis ................................................................................ 76

5.3.1. Newmark Method .............................................................................. 77

VI

6. NUMERICAL RESULTS ................................................................................... 80

6.1. Introduction ................................................................................................. 80

6.2. Numerical Applications ............................................................................... 81

Example 1 ................................................................................................... 81

Example 2 ................................................................................................... 88

Example 3 ................................................................................................... 91

Example 4 ................................................................................................... 97

Example 5 ................................................................................................. 100

Example 6 ................................................................................................. 106

Example 7 ................................................................................................. 111

Example 8 ................................................................................................. 117

Example 9 ................................................................................................. 120

Example 10 ............................................................................................... 126

Example 11 ............................................................................................... 129

Example 12 ............................................................................................... 135

Example 13 ............................................................................................... 138

Example 14 ............................................................................................... 145

Example 15 ............................................................................................... 148

Example 16 ............................................................................................... 158

Example 17 ............................................................................................... 161

Example 18 ............................................................................................... 171

Example 19 ............................................................................................... 174

Example 20 ............................................................................................... 184

Example 21 ............................................................................................... 187

Example 22 ............................................................................................... 193

7. CONCLUSIONS .............................................................................................. 196

REFERENCES ..................................................................................................... 199

CURRICULUM VITAE ....................................................................................... 203

APPENDICES ...................................................................................................... 204

App. 1. Torsional Behaviour and Theory of Open Section Thin-Walled Beams .... 205

App. 2. Derivation of the Formulas for the Position of the Shear Center ............... 242

VII

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

VIII

LIST OF TABLES PAGE NUMBER

Table 6.1. Comparison of the natural frequencies (Hz) obtained from the

present program and SAP2000 in Example 1 ...................................... 84

Table 6.2. Maximum displacement (m) in the X direction of point G

for undamped case in Example 2 ........................................................ 89


for damped case in Example 2 ............................................................ 89


present program and SAP2000 in Example 3 ...................................... 93

Table 6.5. Maximum displacement (m) in the X direction of point O

for undamped case in Example 4 ........................................................ 98

Table 6.6. Maximum displacement (m) in the X direction of point O

for damped case in Example 4 ............................................................ 98


present program and SAP2000 in Example 5 .................................... 102

Table 6.8. Maximum displacement (m) in the Y direction of point G

for undamped case in Example 6 ...................................................... 107


for damped case in Example 6 .......................................................... 107











IX




for undamped case in Example 10 .................................................... 127


for damped case in Example 10 ........................................................ 127


present program and SAP2000 in Example 11 .................................. 131












present program and SAP2000 for unstiffened case in Example 15... 150


present program and SAP2000 for stiffened case in Example 15 ...... 151


for unstiffened case in Example 16 ................................................... 159


for stiffened case in Example 16 ....................................................... 159





X















Table 6.37. Maximum displacement (m) in the X direction of the point O


Table 6.38. Maximum displacement (m) in the X direction of the point O


XI

LIST OF FIGURES PAGE NUMBER

Figure 3.1. Discrete and substitute systems .......................................................... 11

Figure 3.2. Equivalent frame model for planar coupled shear walls ...................... 14

Figure 3.3. Conventional wide-column-frame analogy for the analysis of

three dimensional coupled wall structures .......................................... 15

Figure 3.4. Arrangement of wall and beam elements ............................................ 16

Figure 3.5. Stiffened non-planar coupled shear wall ............................................. 17

Figure 4.1. Non-planar coupled shear wall and lumped mass model ..................... 20

Figure 4.2. Plan of non-planar coupled shear wall in region i ............................... 22

Figure 4.3. Non-planar coupled shear wall ........................................................... 23

Figure 4.4. Translation of axes ............................................................................. 26

Figure 4.5. Rotation of axes ................................................................................. 26

Figure 4.5. General displacement of a connecting beam ....................................... 28

Figure 4.7. Free body diagram of a differential element ....................................... 30

Figure 4.8. Cross-section of a non-planar coupled shear wall ............................... 32

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

Figure 4.10. The principal sectorial area diagrams of the piers in region i .............. 34

Figure 4.11. Vertical forces on the piers ................................................................. 35

Figure 4.12. Relative displacements due to the axial forces in the piers .................. 36

Figure 4.13. Relative displacement of a lamina due to bending .............................. 37

Figure 4.14. Relative vertical displacement due to the shear deformation

in a lamina ......................................................................................... 39

Figure 4.15. Elastic connection condition for the lamina in region i ....................... 40

Figure 4.16. Internal bending moments acting on the components

of the coupled shear wall .................................................................... 43

Figure 4.17. Internal bimoments and bending moments .......................................... 45

Figure 4.18. 3-D view of the additional internal bending moments and

bimoments due to shear forces in connecting medium ........................ 46

XII

Figure 4.19. Cross-sectional view of the additional internal bending moments

and bimoments due to the shear forces in the connecting medium ...... 47

Figure 4.20. Internal twisting moments and shear forces ........................................ 50

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

Figure 4.22. Cross-sectional view with the additional internal twisting moments

and shear forces due to the shear flow iq ........................................... 52

Figure 4.23. The free-body diagram of a part of the shear wall at the top................ 65

Figure 4.24. The vertical forces acting on one piece of the stiffening beam

at the height z = zi .............................................................................. 66

Figure 5.1. Basic difference between static and dynamic loads ............................. 72

Figure 6.1. Geometrical properties of the model originally considered

by Tso and Biswas ............................................................................. 82

Figure 6.2. Frame model of the structure in Example 1 and its 3-D view .............. 83

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

Figure 6.4. Comparison of fourth, fifth and sixth mode shapes in X and Y


in Example 1 ...................................................................................... 86

Figure 6.5. Comparison of seventh and eighth mode shapes in X and Y


in Example 1 ...................................................................................... 87

Figure 6.6. Cross-sectional view of the structure and applied

dynamic load in Example 2 ................................................................ 88

Figure 6.7. Triangular pulse force in Example 2 ................................................... 88

Figure 6.8. Time-varying displacements in X direction at the top of the

shear wall for undamped case in Example 2 ....................................... 90


shear wall for damped case with 5 % damping ratio in Example 2 ...... 90

Figure 6.10. Cross-sectional view of the structure in Example 3 ............................. 91

XIII

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


in Example 3 ...................................................................................... 94



in Example 3 ...................................................................................... 95



in Example 3 ...................................................................................... 96


dynamic load in Example 4 ................................................................ 97

Figure 6.16. Rectangular pulse force in Example 4................................................. 97


shear wall for undamped case in Example 4 ....................................... 99


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



in Example 5 .................................................................................... 103

Figure 6.22. Comparison of fourth, fifth and sixth mode shapes in X, Y and Teta


in Example 5 .................................................................................... 104

Figure 6.23. Comparison of seventh and eighth mode shapes in X, Y and Teta


in Example 5 .................................................................................... 105


dynamic load in Example 6 .............................................................. 106

Figure 6.25. Rectangular pulse force in Example 6............................................... 107

XIV

Figure 6.26. Time-varying displacements in Y direction at the top of the

shear wall for undamped case in Example 6 ..................................... 108


shear wall for damped case with 5 % damping ratio in Example 6 .... 108

Figure 6.28. Sine pulse force in Example 6 .......................................................... 109









in Example 7 .................................................................................... 114



in Example 7 .................................................................................... 115



in Example 7 .................................................................................... 116


dynamic load in Example 8 .............................................................. 117

Figure 6.37. Triangular pulse force in Example 8 ................................................. 118







XV


directions found by t both he present program and SAP2000

in Example 9 .................................................................................... 123



in Example 9 .................................................................................... 124



in Example 9 .................................................................................... 125


dynamic load in Example 10 ............................................................ 126

Figure 6.46. Triangular pulse force in Example 10 ............................................... 126


shear wall for undamped case in Example 10 ................................... 128


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



in Example 11 .................................................................................. 132



in Example 11 .................................................................................. 133



in Example 11 .................................................................................. 134



Figure 6.55. Rectangular pulse force in Example 12 ............................................. 136

XVI





Figure 6.58. Non-planar non-symmetrical structure in Example 13 ...................... 138





in Example 13 .................................................................................. 142



in Example 13 .................................................................................. 143



in Example 13 .................................................................................. 144



Figure 6.65. Sine pulse force in Example 14 ........................................................ 146








directions found by both the present program and

SAP2000 for stiffened case in Example 15 ....................................... 152




XVII





directions found by the present program (CCM) for

stiffened and unstiffened cases in Example 15 .................................. 155

























XVIII































XIX








dynamic load in Example 20 ........................................................... 184

Figure 6.101. Triangular pulse force in Example 20 .............................................. 185


shear wall for undamped case in Example 20 .................................. 186


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



in Example 21 ................................................................................. 190



in Example 21 ................................................................................. 191



in Example 21 ................................................................................. 192


dynamic load in Example 22 ........................................................... 193

Figure 6.111. Rectangular pulse force in Example 22............................................ 194

XX


shear wall for undamped case in Example 22 .................................. 195


shear wall for damped case with 5 % damping ratio

in Example 22 ................................................................................. 195

XXI

NOMENCLATURE

ai : distance between the centroids of the piers in X direction in

region i,

(ax, ay) : linear coordinates of principal pole,

jiA : cross sectional area of the jth pier in region i,

icA : cross sectional area of connecting beams in region i,

*ci

A : effective cross-sectional area in shear for rectangular sections

in region i,

bi : distance between the centroids of the piers in Y direction in

region i,

bk : width of kth unit in a thin-walled beam,

(bx, by) : linear coordinates of an arbitrarily placed pole,

B : bimoment,

EB : external bimoment value,

iB : resultant bimoment about point O, due to the resistance

offered by the piers in region i,

iB : resultant bimoment, about the vertical axis through O, due to

the component bending moments and bimoments in region i,

di : a geometric property, defined in (4.37),

C : damping matrix,

C~ : uncoupled damping matrix,

Dji : displacement vector of shear wall in local axes in region i,

jiD : displacement vector of shear wall in principal axes in region i,

dPX , dPY : moment arms of the components of the unit force,

e : distance of bimoment from shear center,

E : elasticity modulus,

G : shear modulus,

XXII

jiji gg y,x : coordinates of the centroid of jth pier in region i, referring to

global axes, X and Y, respectively,

jiji gg y,x : coordinates of centroid of the jth pier in region i, referring to

principal axes, X and Y , respectively,

H : total height of shear wall,

Hp : height of the unit load,

hi : storey height in region i,

hs : distance of a pole from the tangent line at a point on the

contour line,

i : region number,

icI : moment of inertia of connecting beams in region i,

jiji yx I , I : moments of inertia of pier j w.r.t. global X and Y axes in

region i, respectively,

jixyI : product of inertia of pier j w.r.t. global X and Y axes in region

i,

jiji yx I , I : moments of inertia of pier j w.r.t. principal X and Y axes in

region i, respectively,

jixyI : product of inertia of pier j w.r.t. principal X and Y axes in

region i,

jiIω : sectorial moment of inertia of pier j in region i,

iIω : sum of the sectorial moments of inertia of the two piers at

point O in region i,

iIω : a geometrical constant,

iXIθ ,iYIθ : sectional properties, defined in (4.69-70),

j : pier number,

Jji : St. Venant torsional constant (moment of inertia) of pier j in

region i,

Ji : sum of the St. Venant torsional constants of the two piers in

region i,

XXIII

Jo : polar moment of inertia of a circular cross-section,

K : stiffness matrix,

K~ : uncoupled stiffness matrix,

K1i, K2i,

K3i, K4i : geometrical quantities related to moments of inertia,

icK rotational stiffness of connecting beam-wall connections in

region i,

msv : total torsional moment of the shear stress distribution per unit

length along the contour line, w.r.t. any point,

M : bending moment in an arbitrary plane,

M : mass matrix,

M~ : uncoupled mass matrix,

ii EYEX M,M : external bending moments in region i about the respective

axes due to the unit loading,

iEtM : external twisting moment in region i due to the unit loading

above the cross-section considered,

Mmp : bending moment at the mid-points of the connecting beams,

Mtop : an external bimoment acting at the top

itM : resultant torque about point O, due to resistance offered by

piers in region i,

itM : resultant torque, about the vertical axis through O, due to

component shears and torques in region i,

Mtot : total twisting moment in a thin-walled beam,

Msv : St. Venant twisting moment,

ωM : flexural twisting moment,

n : number of regions in vertical direction,

N : degree of freedom,

Ni : number of lumped masses,

O(X,Y,Z) : orthogonal system of global axes,

P : load vector,

XXIV

P~ : modal force vector,

efP~ : effective load vector,

jiji y x Q , Q : shear forces developed in a region due to shear flow, qi, in the

laminae,

qi : shear flow in laminae per unit length in region i,

Rji : rotation matrix in region i, defined in (4.2),

ri : a geometrical constant

si : eigenvector,

S : shear center (principal pole),

jiS : shear center of jth pier in region i,

jiji ss y,x : coordinates of the shear center of the jth pier, referring to

global axes, X and Y, respectively,

Sx , Sy : statical moments of an area with respect to principal X and Y

axes, respectively,

yx S,S ωω : sectorial statical moments of a section about global axes X

and Y, respectively,

ωS : sectorial statical moment of a section from its center of twist,

Ti : axial force in region i,

t : thickness of a thin-walled beam,

tk : thickness of kth unit in a thin-walled beam,

ux , uy , uz : displacement components in the directions of orthogonal

global system of axes O(X,Y,Z),

us : displacement component in the tangential direction to the

contour line,

ui : horizontal global displacement of point O in X direction in

region i,

jiu : horizontal principal displacement of jth pier in jiX direction in

region i,

XXV

vi : horizontal global displacement of point O in Y direction in

region i,

jiv : horizontal principal displacement of jth pier in jiY direction in

region i,

Vi : shear force in ith stiffening beam,

z : spatial coordinate measured along the height of the structure,

X : modal displacement vector,

X& : modal velocity vector,

X&& : modal acceleration vector,

Y : real displacement vector,

Y& : real velocity vector,

Y&& real acceleration vector,

β1i , β2i , β3i : geometrical constants, defined in (4.136-138),

δ1i : the relative vertical displacement due to the bending of the

piers in X and Y directions and due to the warping of the

piers,

δ2i : the relative vertical displacement due to the axial deformation

of the piers caused by the induced axial forces arising from

the shear flow qi in the connecting medium,

δ3i : the relative vertical displacement due to the bending of the

connecting beams,

δ4i : the relative vertical displacement due to the shear

deformations in the laminae,

δ5i : the relative vertical displacement due to the relative rotation

of the ends of the connecting beams w.r.t. the piers,

δ6i the relative vertical displacement due to the change in cross-

section

θi : global rotational displacement of the rigid diaphragm in

region i,

XXVI

θji principal rotational displacement of jth pier about jiZ direction

in region i,

zθ′ : angle of twist per unit length,

Φ : modal matrix, ω : circular frequency,

sω : sectorial area,

ωj : sectorial area of pier j at point O,

jiφ : angle between the global axes and the principal axes of jth pier

in region i,

icγ : a constant property of connecting beams, defined in (4.59),

Δi : a geometrical constant, defined in (4.110).

1. INTRODUCTION Cevher Deha TÜRKÖZER

1

1. INTRODUCTION

High-rise buildings have become an important part of urban design since

1880s. Especially, the rapid growth of urban population has obliged people to use

existing areas in the most economic way and also aim to be close to each other to

avoid a continuous urban sprawl. For this purpose architects and engineers have to

participate in the stages of the project in order to come up with an economical

building. However, the higher the buildings get, the more increased are the lateral

loads, in addition to the vertical loads. For this reason, especially in earthquake

regions, shear walls are preferred instead of columns.

In modern tall buildings made of reinforced concrete, the lateral loads

induced by wind and earthquake are often resisted by specially arranged shear walls.

The most elementary shape in which a shear wall is employed in a tall building is a

planar shear wall without openings. The behaviour of such a shear wall is essentially

similar to a deep, slender cantilever beam. Shear wall components may be planar, are

usually located at the sides of the building or in the form of a core which houses

staircases or elevator shafts. When one or more rows of openings divide the shear

wall into solid walls connected by lintel beams, the resulting structural system is

called a coupled shear wall. ‘‘Pierced shear wall’’ and ‘‘shear wall with openings’’

are other commonly used terminologies for such structural elements in civil

engineering practice. Weakening of shear walls in tall buildings by doors, windows

and corridor openings is one of the most frequently encountered problems of

structural engineering. When the coupling action between the walls separated by

openings becomes important, some of the external moment is resisted by the couple

formed by the axial forces in the walls due to the increase in the stiffness of the

coupled system by the connecting beams. In planar coupled shear wall analyses, the

lateral loads are applied in such a way that the deformation of the shear wall is

confined within its own plane.

Studies considering in-plane, out-of-plane and torsional deformations in the

investigation of coupled shear walls are called non-planar coupled shear wall


2

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

Vlasov’s 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.Venant’s theory in 1850s. The

crucial point in St.Venant’s 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.

Venant’s 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 Vlasov’s theory. Vlasov's theory for thin-


3

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


4

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 non-

symmetrical 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 Vlasov’s 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 Cevher Deha TÜRKÖZER

5

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 Glück (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.

Glück (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.


6

Macleod (1970) published an article regarding different aspects of the

interactions of shear walls and frames. In that study, it was stated that the finite

element method would give, best results for shear walls.

Heidebrecht and Swift (1971) proposed a method to analyze coupled

asymmetric shear walls, considering the warping of the piers in shear walls and the

coupling of piers by floor beams and slabs.

Smith and Taranath (1972) carried out a study on the open section shear cores

subjected to torsional loads. Experimental results for a model core structure with and

without floor slabs are shown to compare loosely with the theoretical results.

Coull and Subedi (1972) treated the case of shear walls with two rows of non-

symmetrical and three rows of symmetrical openings by the CCM. They compared

their results with those of their experimental work on models.

Mukherjee and Coull (1973) used the CCM for the free vibration analysis of a

single bay coupled shear walls. In this study, the free vibration shapes and the

frequencies of the coupled shear wall without any cross-sectional change and

stiffened beam were determined using Galerkin Method. The examples considered

were compared with the analytical and experimental results.

Tso and Biswas (1973) presented the analysis of a coupled non-planar wall

structure based on the CCM and Vlasov’s theory. In their analysis, rotation is taken

as the main unknown. They plotted their analytical and experimental results for

comparison.

Coull (1974) worked on non-symmetrical single bay shear walls on elastic

foundation with a stiffening beam at the top, employing the CCM, and found closed

form solutions. An example was solved at the end of that paper and the improvement

of the structural behaviour of the wall due to the presence of the stiffening beam was

mentioned.

Tso and Biswas (1974) carried out a study on the three dimensional analysis

of shear wall buildings subject to lateral loads. To simplify the analysis, they ignored

the effect of St. Venant’s torsion in their work.


7

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.

Sümer and Aşkar (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 wide-

column 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.


8

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. Venant’s

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.

Bikçe (2002) carried out a study on the static and dynamic analyses of multi-

bay 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.


9

Aksogan, Arslan and Akavcı (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 multi-

bay coupled shear walls on elastic foundation.

Resatoğlu (2005) presented a study on the static analysis of non-planar

coupled shear walls, based on Vlasov’s 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 Vlasov’s theory of thin walled beams and the CCM.

3. METHODS OF ANALYSIS Cevher Deha TÜRKÖZER

10

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 Vlasov’s 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

time–history analysis.


11

3.2. Continuous Connection Method (CCM)

This method allows a broad look at the behaviour of planar and non-planar

coupled shear wall structures and, simultaneously, gives a good understanding of the

relative influences of the piers and the connecting beams in resisting lateral forces.

The CCM was first originated by Chitty in 1940s and subsequently developed

by Beck (1962), but probably the most comprehensive treatment has been carried out

by Rosman (1964). The importance of torsional analysis in the design of non-planar

coupled shear walls using the CCM was first studied by Glück (1970). Then it was

extended by Tso and Biswas (1973).

In its most basic form the theory assumes that the elastic structural properties

of the coupled wall system remain constant over the height and the points of

inflection of all the beams are at mid-span. In this method, the individual connecting

beams are replaced by continuous laminae. Under the effect of lateral loads, the walls

induce shear forces in the laminae as shown in Fig. 3.1(a)-(b).

Figure 3.1. Discrete and substitute systems

zq

z

(a) (b)

H

laminae

1st pier 2nd pier


12

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

along the height.

2. The rotational stiffness constants at the ends of a connecting beam are

assumed to be equal and are modeled as equivalent linear rotational springs.

3. 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

rotational stiffness Kci in region i are replaced by an equivalent continuous

connecting medium of flexural rigidity EIci/hi and rotational stiffness Kci/hi per unit

length in the vertical direction.

5. Vlasov’s 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.


13

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. The torsional stiffness of the connecting beams is neglected.

9. When there are changes in the cross-section along the height, GJiθi′ term is

neglected in the torsional equilibrium equation.

10. At the common boundaries of two neighbouring regions with different cross-

sections, the warping deformations of the two regions are assumed to be independent

of each other.

11. The walls and beams are assumed to be linearly elastic.

12. Bernoulli-Navier hypothesis is assumed to be valid for the connecting beams.

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-column-

frame 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).


14

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


15

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).

Figure 3.3. Conventional wide-column-frame analogy for the analysis of three

dimensional coupled wall structures In this thesis, the non-planar coupled shear wall structures are analyzed using

Macleod’s (1977) frame method, also. The shear wall structures consist of

interconnected planar walls. The central wall panel has a row of openings and is thus

in effect a pair of coupled shear walls. The coupled non-planar wall structure is

therefore actually composed of planar wall units and a row of coupling beams. In

Macleod’s method, the planar wall units are modeled as column members in wide

column analogy.

Fig. 3.4 shows the arrangement of the wall and beam elements used in the

analysis for comparison. In this thesis, as a modification to Macleod’s method,

additional rigid beam members are placed between the storey levels to improve the

(a)

(b) hinges placed along vertical wall joints

column member

rigid arms beam member


16

continuity of the connection between the wall units. This modification was observed

to improve the results by various comparisons with the CCM.

Figure 3.4. Arrangement of wall and beam elements

In this thesis, the shear deformation of the walls is automatically neglected

due to the second assumption in Vlasov’s theory. This causes over stiffness in the

walls. For comparison purposes the shear deformation is neglected in SAP2000

applications, also.

3.4. Stiffening of Coupled Shear Walls

Studies carried on shear walls taking the rows of openings in them into

consideration have shown that economical design of shear walls and the buildings in

which they reside limits the number of stories to 30-40. The design of even higher

buildings necessitates the stiffening of coupled shear walls to meet the general rules

of design (like the horizontal displacement of the highest point of the building not

being greater than 1/500 of the height of it). Such coupled shear walls are called

“stiffened coupled shear walls”.

Stiffening of coupled shear walls causing a decrease in the top displacement

of buildings, the heights of buildings can be increased. The stiffening of coupled

column member

rigid arms hinges placed along vertical wall joints

storey levels

beam member

h


17

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).

Figure 3.5. Stiffened non-planar coupled shear wall

Besides other methods of modeling, stiffened coupled shear walls have been

treated also by the continuous connection method analytically, putting a stiffening

beam at the top (Coull, 1974) and two stiffening beams, one at the top and one at the

bottom (Choo and Coull, 1984), with a single region along the height. Later, putting

the stiffening beam at an intermediate level, the number of regions increased to two

and the variation of the level has been studied for its effect on the behaviour of the

building (Chan and Kuang, 1988). Coull and Bensmail (1988) increased the number

of intermediate stiffening beams to two and gave analytical solution for problems

with three regions. Both the tediousness of the analysis and the limit on the number

of regions forced some authors (Aksogan et al., 1993) to surrender to Mathematica

Y

Z

H

X O

h s1

Stiffening beams

h s2


18

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 non-

planar coupled shear walls.

4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER

19

4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR

WALLS

4.1. Introduction

It is, generally, of extreme relevance to know the free vibration characteristic

of a structure in assessing the behaviour of it due to seismic effects.

All of the dynamic analyses in the literature on stiffened coupled shear walls

concern themselves with planar coupled shear walls. No study has been made

concerning the dynamic analysis of stiffened non-planar coupled shear walls, so far.

Li and Choo (1984) carried out a research on the effect of stiffeners on the dynamic

behaviour of planar coupled shear walls. In that work the authors handled the free

vibration of a coupled shear wall divided into two sections by one intermediate

stiffening beam.

In this study, the analysis considers non-planar 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 Vlasov’s

theory of thin-walled beams are employed to find the structure stiffness matrix. The

structure mass matrix is found with the lumped mass idealization. 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. Then, a computer

program has been prepared in Fortran Language to analyze free vibration of non-

planar coupled shear walls. The structure is solved both by the present method using

CCM and by the SAP2000 structural analysis program using the frame method. It is

observed that the results obtained by the present method coincide with those of

SAP2000 structural analysis program perfectly.


20

4.2. Determination of Mass Matrix

In the dynamic analysis of non-planar coupled shear walls in conjunction with

CCM, the structure is considered as a discrete system of lumped masses at the

selected levels along the height of the structure (see Fig. 4.1). Lumped masses are

concentrated at the center of the whole cross-sectional area of the structure. Since

each point has three degrees of freedom, in X, Y and Teta directions, the dimension

of mass matrix is equal to 3nx3n, where n represents the number of masses.

However, the mass matrix elements associated with the rotational degrees of freedom

will be zero because of the assumption that the mass is lumped at nodes which have

no rotational inertia. Thus, the lumped-mass matrix is a diagonal matrix which has

zero diagonal elements for the rotational degrees of freedom.

Figure 4.1. Non-planar coupled shear wall and its lumped mass model

(i-1)

(i)

(i+1)

(n-1)

(n)

(1)

(2)

(3)

Z

X

Y

z1

z2

z3

zi-1

zi

zi+1

zn-1

zn

z

H

O

h1

h2

hi-1

hi

hn-1

hn

stiffening beams

M1

M2

M3

M4

MN

MN-1


21

The mass matrix of the coupled shear wall was found as a diagonal matrix

employing a lumped mass approach. To explain this procedure, the top, bottom and

each height at which there is a stiffening beam and/or change of wall thickness will

be called ‘‘ends’’ and the section between any two consecutive ends will be called a

‘‘region’’. Equidistant masses of a suitable number were placed in each region. The

masses in a region were found by dividing the total mass of the region by this

number and half of that was assigned to the ends of the region. Completing this

procedure for each region and adding to each end the additional mass due to the

pertinent stiffening beam, the mass matrix was found as a diagonal matrix.

Forming the mass matrix as described in the previous paragraph, the CCM is

used to determine the stiffness matrix.

4.3. Determination of Stiffness Matrix

4.3.1. Introduction

After the determination of the mass matrix, the second step is the

determination of the stiffness matrix of the structure for the degrees of freedom

chosen during the determination of the mass matrix. This procedure is carried out by

applying two horizontal unit forces in the directions of X and Y axes and one unit

moment about Z axis at every height with a lumped mass. For every one of these

loadings, a solution is carried out making use of CCM and writing down the

compatibility equation for the vertical displacements at the midpoints of the

connecting beams. Then, employing the equilibrium equations, the corresponding

displacements are obtained. The displacements of the points where the lumped

masses are located are determined by using the rigid floor diaphragm assumption.

Thus, each unit loading gives one column of the flexibility matrix as the

displacements at the points where the lumped masses are. Hence, the analysis for the

three loading cases for one floor will suffice to introduce the complete solution

procedure for the flexibility matrix. The stiffness matrix of the structure will be

determined by taking the inverse of the flexibility matrix.


22

4.3.2. General Information

Based on Vlasov’s theory of thin-walled beams and the continuous

connection method (CCM), an approximate procedure is presented for the analysis of

non-planar coupled shear walls subjected to lateral loads which produce combined

bending and torsional deformations.

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.

Y

i1sy

ci

i2gx i1gx

i2gy

i1gy

i2sx i1sx

1iX

i1φconnecting beam

O

S1i

G1i

S2i

G2i i2φ

2nd pier 1st pier X

1iY 2iX

2iY


23

Figure 4.3. Non-planar coupled shear wall

Referring to the axes OX and OY, the location of the centroids of the piers

are taken to be ( )i1i1 gg y,x and ( )

i2i2 gg y,x , respectively. Throughout this thesis, the

subscripts 1 and 2 express the left and the right piers, respectively. The subscript i

( )n1,2,...,i = , refers to the number of region. Similarly, the shear centers of the piers

are located at ( )i1i1 ss y,x and ( )

i2i2 ss y,x , respectively. The coordinates referred to the

principal axes of pier j ( )1,2j = which are represented by ( )jiji Y,X , making an angle

jiφ with the respective global axes are shown in Fig. 4.2. The jiZ axis is parallel to

the global Z axis.

(i-1)

(i)

(i+1)

(n-1)

(n)

(1)

(2)

(3)

Z

X

Y

z1

z2

z3

zi-1

zi

zi+1

zn-1

zn

z

H

O

h1

h2

hi-1

hi

hn-1

hn

stiffening beams


24

4.3.2.1. Transformation of the Local Displacements

The displacement vector of the shear center of each pier in its local system of

axes can be written as

=jiD

θ ji

ji

ji

vu

( )1,2j = , ( )n1,2,...,i = (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 jiθ is the rotation about jiZ 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:

=jiR

φφ−φφ

1000CosSin0SinCos

jiji

jiji

( )1,2j = , ( )n1,2,...,i = (4.2)

−=

100x10y01

Tji

ji

s

s

ji ( )1,2j = , ( )n1,2,...,i = (4.3)

in which jisx and

jisy 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 iu , iv and iθ in the

global axes, which are given in compact notation as


25

θ=

i

i

i

i vu

D ( )n1,2,...,i = (4.4)

where iu and iv are the displacements at the mid-points of the connecting laminae

in the X and Y directions of the global system of axes, and iθ 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

ijijiji DTRD = ( )1,2j = , ( )n1,2,...,i = (4.5)

Substituting the expressions (4.1-4) into (4.5),

θ

φφ−

−φφ=

θ i

i

i

sjiji

sjiji

ji

ji

ji

vu

100xCosSinySinCos

vu

ji

ji

( )1,2j = , ( )n1,2,...,i = (4.6)

From these geometric relationships, the local displacements jiu , jiv and jiθ

(j=1,2) parallel to the global axes at the shear centers of piers 1 and 2 may be

expressed in terms of the global displacements iu , iv and iθ as follows:

isiji jiyuu θ−= ( )1,2j = , ( )n1,2,...,i = (4.7)

isiji ji

xvv θ+= ( )1,2j = , ( )n1,2,...,i = (4.8) iji θ=θ ( )1,2j = , ( )n1,2,...,i = (4.9)

4.3.2.2. Transformation of the Moments of Inertia

Let the moments of inertia values of an area A with respect to an Oxy system

of axes be Ix, Iy and Ixy. If these three values are known, the moments of inertia


26

values with respect to another system of axes O′x′y′ parallel to the first system can

be given as follows (see Fig. 4.4):

Figure 4.4. Translation of axes

AbII 2xx +=′ (4.10)

AaII 2

yy +=′ (4.11) AabII xyxy +=′ (4.12)

Consider two sets of axes Oxy and yxO , the latter being inclined at an

angle φ with the former (see Fig. 4.5).

Figure 4.5. Rotation of axes

Any point in the plane can be described by the coordinates ( )y,x or by ( )y,x . These

coordinates are related by a rotation matrix with the transformation

φ φ

dA x

y

x y

y

x O

A x

y

x O b

a

y

O′

G

y′

x′


27

yx =

φφ−φφ

yx

CosSinSinCos

(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 yxO 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 yxO in terms of the moments

of inertia with respect to the system of axes Oxy, starting from their definitions:

( ) dAyCosxSindAyI2

A

2

Ax ∫∫∫∫ φ+φ−== (4.14)

( ) dAySinxCosdAxI2

A

2

Ay ∫∫∫∫ φ+φ== (4.15)

( ) ( )dAyCosxSinySinxCosdAxyI

A

2

Axy φ+φ−⋅φ+φ== ∫∫∫∫ (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:

φφ−φ+φ= CosSinI2SinICosII xy2

y2

xx (4.17) φφ+φ+φ= CosSinI2CosISinII xy

2y

2xy (4.18)

( ) ( )φ−φ+φφ−= 22

xyyxxy SinCosICosSinIII (4.19)

The system of axes for which the product of inertia xyI of a planar area is zero

is called “principal system of axes”. As seen from expression (4.19), xyI becomes

zero for 0φ defined by

yx

xy0 II

I22tg

−−=φ (4.20)


28

The axes which make angles φ0 and φ0 + π/2 with the x axis are the principal axes.

The moments of inertia computed with respect to the principal axes are the “principal

moments of inertia” which are given by the following formula:

xy2

2yxyx

2,1 I2

II2

III +

−±

+= (4.21)

Every symmetry axis of a planar area is a principal axis of it.

4.3.2.3. The Assumptions of the Continuous Connection Method

The assumptions in the foregoing analysis were given in Chapter 4. In view

of assumption 6, the overall cross-section of the structure cannot be deformed in its

own plane. The movements of any point on the contour can then be expressed in

terms of the global generalized displacement variables, which consist of two

horizontal displacements, ui and vi of point O and a rotation, θ i (see Fig. 4.1).

In order to understand assumption 6 mentioned in Chapter 4, the general

displacement of a connecting beam and the moments induced at the ends are shown

in Fig. 4.6.

Figure 4.6. General displacement of a connecting beam

V1i V2i

M1i M2i

i2yθ

Li

i1yθ 1 2


29

The floor slabs surrounded by the coupled shear wall system are assumed to

be rigid in their own planes and flexible in planes normal to it, so that the cross-

sections of the piers remain undistorted in plan. Because of this assumption, actual

rotations i1yθ and

i2yθ of the connecting beam shown in Fig. 4.6 can be assumed to

be equal to each other. The end moments M1i and M2i may be considered as being

caused by a combination of moments due to the fixed end moments developed by the

transverse loading on the member, the relative normal translation and the actual

rotations, i1yθ and

i2yθ , of its ends.

Using the slope deflection method, the general form of the left end moment

can be written as follows:

M1i= ( )ii2i1 12iyy

i

ci M624L

EI+ψ+θ+θ (4.22)

Since ii2i1 yyy θ=θ=θ , 0i =ψ and assuming there is no transverse loading, which is

a logical assumption for this analysis, then, M1i and M2i can be rewritten,

respectively, as follows:

M1i=iy

i

ci

LEI6

θ (4.23)

M2i=iy

i

ci

LEI6

θ (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=i

i

LM2 (4.25)


30

From the equilibrium of the vertical forces

i2i1i VVV == (4.26)

Taking moment about the mid-point of the member yields

0M2L

LM2M i

i

i

iimp =−×= (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.,

Figure 4.7. Free body diagram of a differential element

qi

Ti+dTi dz

Ti

G1i

G2i

Ti+dTi

Ti


31

0dzqTdTT iiii =+−+ (4.28)

Simplification and rearrangement of this equation yields

dzqdT ii −= (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:

ii q

dzdT

−= (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, jigx and

jigy (j=1,2, i=1,2,…,n), which are the distances of the centroids from point O,


32

measured along the principal directions of the piers, will be considered for

expressing the contribution in the heading to the compatibility equation.

Figure 4.8. Cross-section of a non-planar coupled shear wall

The vertical displacement due to bending can be obtained as the product of

the slope at the section considered and the distance of point O from the respective

neutral axis. In addition, vertical displacement arises, also, due to the twisting of the

piers, and is equal to the value of the twist at the section considered, times the

sectorial area, iω , at point O.

Graphically, the vertical displacements to the left and right of the cut due to

bending and twisting are shown all together in Fig. 4.9. Here, the downwards

displacements of the end on the left of the cut and the upwards displacements of the

end on the right of it will be taken to be positive in expressing the relative vertical

displacements. Here and in the rest of the formulation, a prime will show

differentiation with respect to the variable z.

Y

i2gxi1gx

i2gy

i1gy

i1φ

O

S1i

G1i

S2i G2i i2φ

X

2iX

2iY

1iY

1iX


33

a) Bending about 1iY and 2iY axes

b) Bending about 1iX and 2iX axes

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

i2i2 ωθ′

i1i1 ωθ′− S1i

S2i

i1gi1 xu ′−

2iY

1iX 2iX

1iY

i2gi2 xu ′

G1i

G2i

2iY

1iX 2iX

1iY

i2gi2 yv ′ i1gi1 yv ′−

G1i

G2i


34

Let i1δ denote the total relative vertical displacement due to the deflections

and rotations of the piers, which can be given as

( ) ( ) ( )i1i1i2i2gi1gi2gi1gi2i1 i1i2i1i2 yvyvxuxu ωθ′−ωθ′−′−′+′−′=δ (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. i1ω and i2ω 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.

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 (4.7-

9) into (4.31), yields

i2ii1iigigigig

igsigsgsigsi1

vyvyuxux

yxyxxyxy

i2i1i2i1

i2i2i1i1i2i2i1i1

ωθ′−ωθ′+′+′−′+′−

θ′+θ′−θ′−θ′=δ

Rearranging the equation (4.32)

O +

+

ω1i

+

+ −

ω2i

+ −

− −

− −

S1i S2i

(4.32)


35

)d(bvau iiiiiiii1 +ωθ′+′+′=δ ( )n1,2,...,i = (4.33)

where i is the number of a particular region and the new constants are defined as

follows:

i2i1i ω−ω=ω ( )n1,2,...,i = (4.34)

i1i2 ggi xxa −= ( )n1,2,...,i = (4.35)

i1i2 ggi yyb −= ( )n1,2,...,i = (4.36)

i1i1i1i1i2i2i2i2 gsgsgsgsi yxxyxyyxd −+−= ( )n1,2,...,i = (4.37)

4.3.4.1.(2). The Relative Vertical Displacement due to the Axial Deformation of

the Piers

The axial force in each pier shown in Fig. 4.11 is found by writing down the

vertical force equilibrium equation for the part of one pier above any horizontal

cross-section as

Figure 4.11. Vertical forces on the piers

1st pier

Ti z

z1

zi

V1

q1

Vi

qi

qi-1

(1)

(i-1)

(i)

2nd pier

Ti


36

∑∑ ∫∫=

−

=

++=+

i

1tt

1i

1t

z

zt

z

zii V)dzq(dzqT

t

1t

i

( )n1,2,...,i = (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.

Figure 4.12. Relative displacements due to the axial forces in the piers

The normal strain being defined as ,EAT

E=

σ the total axial elongation of pier 1 can

be written as

∫∫∑++

−−=δ

+=

i

1i

t

1t

z

zi

i1

z

zt

n

1it t1

*i2 dzT

A1

E1dzT

A1

E1 ( )n1,2,...,i = (4.39)

Similarly, the total axial shortening of pier 2 is

∫∫∑++

−−=δ

+=

i

1i

t

1t

z

zi

i2

z

zt

n

1it t2

**i2 dzT

A1

E1dzT

A1

E1 ( )n1,2,...,i = (4.40)

*i2δ

**i2δ

ci/2 ci/2

Ti Ti


37

From (4.39-40), the total relative displacement due to the change of length in the

piers can be written as

dzTA1

A1

E1dzT

A1

A1

E1 z

zi

i2i1

n

1it

z

zt

t2t1i2

1i

t

1t

∫∑ ∫++

+−

+−=δ

+=

( )n1,2,...,i = (4.41)

4.3.4.1.(3). The Relative Vertical Displacement due to the Bending of the

Laminae

A point load P at the end of a cantilever beam of length L, causes an end

deflection equal to

=δ

EI3PL3

. Hence, the shear force in each lamina of dz height

which is

dzqP ii = (4.42)

causes relative displacements (see Fig. 4.13):

Figure 4.13. Relative displacement of a lamina due to bending

Pi

*i3δ

**i3δ Pi

ci/2 ci/2


38

( )

dzhEI3

2/cdzq

i

c

3ii*

i3i

−=δ ( )n1,2,...,i = (4.43)

and

( )

dzhEI3

2/cdzq

i

c

3ii**

i3i

−=δ ( )n1,2,...,i = (4.44)

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

ic

3ii**

i3*

i3i3 qEI12ch

i

−=δ+δ=δ ( )n1,2,...,i = (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

deflection due to the shear deformation of the beam, equal to

=δ

GAPL .

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

*c

iii

i

*c

ii

i

*c

ii**i4

*i4i4

iiiGA

hcq

dzhA

G

dz)2/c(q

dzhA

G

dz)2/c(q−=−−=δ+δ=δ ( )n1,2,...,i = (4.46)

where effective cross-sectional shear area *ci

A is given as


39

c

c*c

i

i

AA

µ= ( )n1,2,...,i = (4.47)

in which the constant cµ 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:

ic

iiii4 GA

hcq2.1−=δ ( )n1,2,...,i = (4.48)

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

ci/2

Pi

*i4δ

**i4δ

ci/2

Pi


40

2ci

i*

i5 ×α−=δ (4.49)

The moment value in that spring is equal to ( ii

ci h

dzKM i α×= ), where

icK is the

rotational spring constant of a connecting beam. From this relation it is clear that

(dzKhM

ic

iii =α ). Therefore, substituting this expression into (4.49)

2c

dzKhM i

c

ii*i5

i

×−=δ (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 = 2/cP ii × ) due to the shear force, iP , at the mid-point of the

lamina. It is also known that ( dzqP ii ×= ). Therefore, the total relative vertical

displacement due to the flexibility of the connections can be written as

iii c

2iii

c

2iii

c

2iii**

i5*

i5i5 K2chq

K4chq

K4chq

−=−−=δ+δ=δ ( )n1,2,...,i = (4.51)

Figure 4.15. Elastic connection condition for the lamina in region i

ci/2

*i5δ

1

2 αi (1)


41

4.3.4.1.(6). The Relative Vertical Displacement due to the Change in Cross-

section

When there are stepwise changes in the cross-sectional form, an additional

relative vertical displacement will take place at the boundaries of cross-sectional

change due to the difference between the bending and the warping of neighbouring

regions. To find this additional displacement, (4.33) expression is written at the level

of the boundary ( izz = ) and after simplifications

)d(bvau iizziizziizzizzi1iiii

+ωθ′+′+′=δ====

( )n1,2,...,i = (4.52)

On the other hand, applying (4.33) for region i-1 at the common boundary ( izz = ),

results in

)d(bvau )1i()1i(zz)1i()1i(zz)1i()1i(zz)1i(zz)1i(1iiii

−−=−−=−−=−=− +ωθ′+′+′=δ (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

ii zz)1i(1zzi1i =−=δ−δ=∇ ( )n1,2,...,i = (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:

[ ]∑∑+=

=−=+=

δ−δ=∇=δn

1itzz)1t(1zzt1

n

1itti6

tt ( )n1,2,...,i = (4.55)


42

Under lateral loads the two ends of a lamina experience vertical

displacements at the cuts consisting of contributions ( )i6i5i4i3i2i1 and,,,, δδδδδδ as

explained before. Since, the two piers are connected by the laminae, the

compatibility condition depicts that the relative displacements must vanish, i.e.,

0i6i5i4i3i2i1 =δ+δ+δ+δ+δ+δ ( )n1,2,...,i = (4.56)

Therefore, the compatibility equation for the vertical displacement in region i can be

written as

( )

),...,2,1(0212

2.1

111111

6

23

211 21 11

niEKch

EIch

GAch

ET

dzTAAE

dzTAAE

dbvau

ic

ii

c

ii

c

iii

z

zi

ii

n

ij

z

zj

jj

iiiiiii

iii

i

j

j

==+

++

′+

+−

+−

+′+′+′

∫∑ ∫++

+=

δ

ωθ

Differentiating (4.57) with respect to z ends up in

( ) 0ETT

A1

A1

E1dbvau

ici

ii2i1

iiiiiii =γ′′

+

+−+ωθ′′+′′+′′ ( )n1,2,...,i = (4.58)

where

icγ = E

Kch

EIch

GAch

iii c

ii

c

ii

c

ii

++

2122.1 23

( )n1,2,...,i = (4.59)

and well-known properties

(5.57)

i2sy


43

( ) 0dzTdzd t

1t

z

zt =∫

+

(4.60)

( ) ( ) ( ) ii1i

i

z

zi TT

dzdzT

dzdzdzT

dzd

1i

=−= +∫+

(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.

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 iEXM and

iEYM . For the equilibrium of

the moments about X and Y axes, the following relationships can be derived using

Vlasov’s theory of thin walled beams:

i1x vIEi1

′′

ai

O

i1φ

i2φ

bi

i1y uIEi1

′′

i2x vIEi2

′′

X

Y

i2y uIEi2

′′

G1i

G2i


44

( ) 0MaTsinvIcosuIsinvIcosuIEii2i2i1i1 EYiii2i2xi2i2yi1i1xi1i1y =−+φ′′−φ′′+φ′′−φ′′ (4.62)

( ) 0MbTcosvIsinuIcosvIsinuIEii2i2i1i1 EXiii2i2xi2i2yi1i1xi1i1y =−+φ′′+φ′′+φ′′+φ′′ (4.63)

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

iiiYiXYiYEY aTIEvIEuIEMiiii

+θ′′−′′+′′= θ ( )n1,2,...,i = (4.64)

iiiXiXiXYEX bTIEvIEuIEMiiii

+θ′′+′′+′′= θ ( )n1,2,...,i = (4.65)

where

i2i1i yyY III += (4.66)

i2i1i xxX III += (4.67)

i2i1i xyxyXY III += (4.68)

i2i2i1i1i2i2i1i1i xysxysxsxsX IyIyIxIxI −−+=θ (4.69)

i2i2i1i1i2i2i1i1i xysxysysysY IxIxIyIyI −−+=θ (4.70)

Terms jiyI and

jixI are the second moments of area of the cross-sections, and jixyI 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. jiyI and

jixI 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.


45

4.3.4.2.(2). Bimoment Equilibrium Equation

When non-planar coupled shear walls are rigidly constrained at the base, the

cross-sections of the piers do not warp uniformly along the height and internal

stresses evoke in the wall due to the bimoments. The internal force resultants shown

in Fig. 4.16 are caused by biaxial bending. These bending moments form bimoments

with the resisting moments at the bottom of the wall and show up in the bimoment

equilibrium equation. This point is explained in Section 3.6.2 in detail.

In order to obtain the bimoment equilibrium equation, the coupled shear wall

will be cut through by a horizontal plane such that an upper part is isolated from the

lower part of the structure. The bimoment equilibrium equation is written by

equating to zero, the bimoment evoked at point O by the internal and external forces

and moments. The internal bimoments in the structure consist of two parts; one

contributed by the individual piers as shown in Fig. 4.17 and the other due to the

resistance of the connecting laminae and stiffening beams as shown in Fig. 4.18.

Figure 4.17. Internal bimoments and bending moments

Y

i2sx i1sx

i2sy

i1sy

i1φ

O

S1i

S2i i2φ

X

i1y uIEi1

′′

i1x vIEi1

′′i2x vIE

i2′′

i2y uIEi2

′′

i1i1IE θ′′− ω

i2i2IE θ′′− ω


46

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 jijiIE θ′′− ω (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 iB 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)

( ) ( )( ) ( )i2i2i1i1

i2i2i1i1

i2i1

si2xsi1x

si2ysi1y

i2i1i

xvIExvIE

yuIEyuIE

IEIEB

′′−−′′+

′′+′′+

θ ′′−θ ′′−= ωω

(H-z)

qi

Vi

q1

V1

Ti

Ti i1yqM

i1xqM i1qB

i2yqM

i2xqMi2qB

(4.71)


47

On substituting the expressions (4.7-9), (4.13) and (4.17-20) into (4.71), the resultant

resisting bimoment of the piers for all regions in the structure are found as

iiXiYi iiiIEvIEuIEB θ ′′−′′−′′= ωθθ ( )n1,2,...,i = (4.72)

in which ii YX IandI θθ are defined in (4.69-70) and

i2i2i2i1i1i1i2i2i1i1

i2i2i1i1i2i1i

xyssxyssy2sy

2s

x2sx

2s

Iyx2Iyx2IyIy

IxIxIII

−−++

+++= ωωω

In addition, there are bending moments i1xqM ,

i1yqM , i2xqM ,

i2yqM and

bimoments i1qB and

i2qB , due to the summation of the shear forces in the laminae, as

shown in Fig. 4.18-19.

Figure 4.19. Cross-sectional view of the additional internal bending moments and bimoments due to the shear forces in the connecting medium

Y

i2sx i1sx

i2sy

i1sy

i1φ

O

S1i

S2i i2φ

X

i1yqM

i1xqMi2xqM

i2yqM

i1qB

i2qB

(4.73)


48

All quantities are related to the shear centers of the piers, S1i and S2i. In order

to determine the bimoment due to the shear forces in the laminae, the relationship

between the force components in the piers, may be determined from the free body

diagram in Fig. 4.18.

Let iB be the resultant bimoment due to the additional bending moments and

bimoments about the vertical axis through point O, as shown in Fig. 4.19.

The shear forces in the laminae produce bimoments on the piers. For pier 1,

the bimoment i1qB is caused by the Ti (instead of summation of the shear forces in

the connecting medium, see equation (4.38)), and is equal to the product of this force

and the principal sectorial area i1ω of the point of its application. This property is

explained in Section 3.6.1 in detail. Therefore, it can be written in the following form

i1iq TBi1

ω−= (4.74)

and analogously for pier 2

i2iq TBi2

ω= (4.75)

The resultant bimoment, iB , due to these additional bending moments and

bimoments about point O is, then, found as

( ) ( ) ( ) ( )i2i2i2i2i1i1i1i1i2i1 syqsxqsyqsxqqqi yMxMyMxMBBB +−+−++= (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 i1Y and i1X

axes for pier 1, the following relations can be obtained, respectively:

( ) 0yTMi1i1

gixq=+ (4.77)


49

( ) 0xTM i1i1

giyq=−− (4.78)

Thus,

i1i1

gixqyTM −= (4.79)

i1

i1giyq

xTM −= (4.80)

Similar consideration for the other pier gives

i2i2

gixqyTM = (4.81)

i2

i2giyq

xTM = (4.82)

Substituting (4.74-75) and (4.79-82) in (4.76), using (4.13,34,37) and simplifying,

the resultant bimoment, about the vertical axis through point O, due to the

component bending moments and bimoments is found as

( ) iiii TdB +ω−= (4.83)

Equating the external bimoment, iEB , to the internal resisting bimoments, the

bimoment equilibrium equation of the isolated part of the coupled shear wall above

an arbitrary horizontal plane is established. Finally, the foregoing equilibrium

equations for all regions of the structure can be written as follows:

iiE BBBi

+= ( )n1,2,...,i = (4.84)

( ) iiiiiXiYE TdIEvIEuIEBiiii

+ω−θ′′−′′−′′= ωθθ ( )n1,2,...,i = (4.85)


50

4.3.4.2.(3). Twisting Moment Equilibrium Equation

In order to obtain the twisting moment equilibrium equation, the coupled

shear wall will be cut through by a horizontal plane such that an upper free body

diagram is isolated from the rest of the structure.

The internal twisting moments (torque) in the structure consist of two parts;

one contributed by the individual piers as shown in Fig. 4.20 and the other due to the

resistance of the connecting laminae, in other words, the differential effect of the

shear flow iq in the connecting medium as shown in Fig. 4.21.

Figure 4.20. Internal twisting moments and shear forces

In Fig. 4.20, expressions jijiJG θ′ (j=1,2, i=1,2,…,n) are the St. Venant twisting

moments. Expressions jijiIE θ ′′′− ω (j=1,2, i=1,2,…,n) are the additional twisting

moments due to the non-uniform warping of the piers along the height. Furthermore,

the total twisting moment due to the shear forces in the cross-sections of the piers

about point O must also be considered.

Let it

M be the resultant torque about the vertical axis through point O, which

is due to the resistance offered by the piers. It can be written as (see Fig. 4.20)

Y

i2sx i1sx

i2sy

i1sy

i1φ

O

S1i

S2i i2φ

X

i1x vIEi1

′′′

i1y uIEi1

′′′

i2y uIEi2

′′′

i2x vIEi2

′′′

i2i2i2 GJIEi2

θ′+θ ′′′− ω

i1i1i1 GJIEi1

θ′+θ ′′′− ω


51

( ) ( )( ) ( )i2i2i1i1

i2i2i1i1

i2i1i

si2xsi1x

si2ysi1y

i2i1i2i2i1i1t

xvIExvIE

yuIEyuIE

IEIEGJGJM

′′′−−′′′+

′′′+′′′+

θ ′′′−θ ′′′−θ′+θ′= ωω

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

iiiiXiYt iiiiIEJGvIEuIEM θ ′′′−θ′+′′′−′′′= ωθθ ( )n1,2,...,i = (4.87)

where ii YX I,I θθ and

iIω are as defined in (4.69,70,73), respectively, and

i2i1i JJJ += ( )n1,2,...,i = (4.88)

In addition, there are shear forces i1xQ ,

i1yQ , i2xQ ,

i2xQ , and twisting

moments i1tqM and

i2tqM , developed in the section due to the shear force, iq , in the

laminae, as shown in Fig. 4.21-22.

Figure 4.21. 3-D view of the additional internal shear forces and twisting moments

due to the shear flow in the connecting medium

qi

Ti+dTi

dz

G1i

G2i

Ti+dTi

Ti

i1yQ

i1xQ

i1tqM i1yQ

i1xQ

i1tqM Ti

i2yQ

i2xQ

i2yQ

i2xQ

i2tqM

i2tqM

(4.86)


52

Figure 4.22. Cross-sectional view with the additional internal twisting moments and

shear forces due to the shear flow iq

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 iq , the relationship between

the force components in the piers and iq may be determined from the free body

diagram in Fig. 4.21.

Let it

M 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 iq in the laminae produces bimoments on the piers. For pier

1, the bimoment i1zdB is caused by the external force dzq i and is equal to the

product of this force and the principal sectorial area, i1ω , of the point of its

application. Therefore, it can be written in the following form:

i1iz dzqdBi1

ω= ( )n1,2,...,i = (4.89)

and analogously for pier 2

i2iz dzqdBi2

ω−= ( )n1,2,...,i = (4.90)

Y

i2sx i1sx

i2sy

i1sy

i1φ

O

S1i

S2i i2φ

X

i1yQ

i1xQ

i2yQ

i1tqM

i2xQ

i2tqM


53

These bimoments produce the flexural twisting moments i1tqM and

i2tqM , related to

the shear centers S1i and S2i, respectively. Using the relation obtained in (3.60),

i1iz

tq qdz

dBM i1

i1ω== ( )n1,2,...,i = (4.91)

i2iz

tq qdz

dBM i2

i2ω−== ( )n1,2,...,i = (4.92)

The resultant twisting moment, it

M , due to these additional torques and shear

forces about point O is, then, found as

( ) ( ) ( ) ( )i2i2i2i2i1i1i1i1i2i1isysxsysxtqtqt xQyQxQyQMMM +−−−−+= (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 i1Y and i1X axes for

pier 1, the following relations can be obtained, respectively:

0xdzqdzQ i1i1gix =+ (4.94)

0ydzqdzQi1i1 giy =+ (4.95)

Thus,

i1i1gix xqQ −= ( )n1,2,...,i = (4.96)

i1i1 giy yqQ −= ( )n1,2,...,i = (4.97)


54

Similar consideration for the other pier gives

i2i2gix xqQ = ( )n1,2,...,i = (4.98)

i2i2 giy yqQ = ( )n1,2,...,i = (4.99)

Substituting (4.91-92) and (4.96-99) in (4.93), using (4.13,34,37) and simplifying,

the resultant twisting moment, about the vertical axis through point O, due to the

component twisting moments and shear forces is found as

( ) iiit qdMi

+ω= ( )n1,2,...,i = (4.100)

Equating the external twisting moment, iEtM , to the internal resisting

moments, with opposite senses, the equilibrium equation of the isolated part of the

coupled shear wall above an arbitrary horizontal plane is established. Finally, the

foregoing equilibrium equation for all regions of the structure can be written as

follows:

ii ttEti MMM += ( )n1,2,...,i = (4.101)

( ) iiiiiiiXiYEt TdIEJGvIEuIEMiiii

′+ω−θ ′′′−θ′+′′′−′′′= ωθθ ( )n1,2,...,i = (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, θ , and T, can be found under the unit loadings iEXM ,

iEYM , iEB , and

iEtM .


55

4.3.4.3. Method of Solution

For any given unit loading on the shear wall system, the values of the applied

moments, bimoment and torque, iEXM ,

iEYM , iEB , and

iEtM , at any height of the

structure can be found from the static equilibrium equations. The deformation of the

structure is then found by obtaining the solution of the system of differential

equations (4.58), (4.64), (4.65), (4.85), and (4.102). It is more convenient to reduce

the 4n unknowns to n unknowns, Ti, and solve the n equations employing the

appropriate boundary conditions.

Equating the iu ′′ expressions obtained from (4.64) and (4.65)

i

iii

i

iii

XY

iiiXiXEX

Y

iiiYiXYEYi IE

bTIEvIEMIE

aTIEvIEMu

−θ′′−′′−=

−θ′′+′′−=′′ θθ (4.103)

Similar procedure for iv ′′ yields

i

iii

i

iii

X

iiiXiXYEX

XY

iiiYiYEYi IE

bTIEuIEMIE

aTIEuIEMv

−θ′′−′′−=

−θ′′+′′−=′′ θθ (4.104)

Simultaneous solution of (4.103) and (4.104) yields

( ) ( )( )

iiiiii

iiiiiii

YEXXYEYYiXYii

YXYXYi2XYYXi

IMIMIbIaT

IIIIEIIIEv

+−−

++θ′′−=−′′ θθ

( ) ( )( )

iiiiii

iiiiiii

XEYXYEXXYiXii

XXYYXi2XYYXi

IMIMIbIaT

IIIIEIIIEu

−+−

++θ′′−=−′′− θθ

Dividing both equations by ( )2XYYX iii

III −

(4.105)

(4.106)


56

( )( )

( )( )

( ) ( )

−+

−−

−

−+

−

+θ′′−

=′′

θθ

2XYYX

YEX

2XYYX

XYEY

2XYYX

YiXYii2

XYYX

XYYXYi

i

iii

ii

iii

ii

iii

ii

iii

iiii

III

IM

III

IM

III

IbIaT

III

IIIIE

E1v (4.107)

( )( )

( )( )

( ) ( )

−+

−−

−

−−

−

+θ′′

=′′

θθ

2XYYX

XEY

2XYYX

XYEX

2XYYX

XYiXii2

XYYX

XXYYXi

i

iii

ii

iii

ii

iii

ii

iii

iiii

III

IM

III

IM

III

IbIaT

III

IIIIE

E1u (4.108)

Here and in the sequel the following definitions will be employed:

i4ii3ii2i1i

KbKaA1

A1

A1

++

+= (4.109)

( )2XYYXi iii

III −=∆ (4.110)

( )i

XXYYXi1

iiiiIIII

K∆

+= θθ (4.111)

( )i

XYYXYi2

iiiiIIII

K∆

+= θθ (4.112)

( )i

XYiXii3

iiIbIa

K∆

−= (4.113)

( )i

YiXYii4

iiIbIa

K∆

−−= (4.114)


57

Thus,

∆

+∆

−−θ′′−=′′i

YEX

i

XYEYi4ii2ii

iiiiIMIM

KTKEE1v ( )n1,2,...,i = (4.115)

∆

+∆

−−θ′′=′′i

XEY

i

XYEXi3ii1ii

iiiiIMIM

KTKEE1u ( )n1,2,...,i = (4.116)

Differentiation of equations (4.115) and (4.116) with respect to z yields

∆

′+

∆

′−′−θ ′′′−=′′′

i

YEX

i

XYEYi4ii2ii

iiiiIMIM

KTKEE1v (4.117)

∆

′+

∆

′−′−θ ′′′=′′′

i

XEY

i

XYEXi3ii1ii

iiiiIMIM

KTKEE1u (4.118)

Substituting (4.117) and (4.118) into (4.102)

( ) iiiiii

i

YEX

i

XYEYi4ii2iX

i

XEY

i

XYEXi3ii1iYEt

TdIEJG

IMIMKTKE

E1IE

IMIMKTKE

E1IEM

i

iiii

i

iiii

ii

′+ω−θ ′′′−θ′+

∆

′+

∆

′−′−θ ′′′−−

∆

′+

∆

′−′−θ ′′′=

ω

θ

θ

Reorganizing (4.119),

( ) i3Yii4XiiiiEt KITKITdTMiii θθ ′−′++ω′−=

ii1iyi2iX iiIEKIEKIE θ ′′′−θ ′′′+θ ′′′+ ωθθ

( ) ( )ii

i

XYYXYEX

i

XXYYXEY JG

IIIIM

IIIIM iiii

i

iiii

iθ′+

∆

+′−

∆

+′+ θθθθ (4.120)

(4.119)


58

Simplifying (4.120) further,

[ ]i3Yi4XiiiEt KIKIdTMiii θθ +−+ω′−=

[ ]i1Yi2Xi KIKIIEiii θθω −−θ ′′′−

iii2EXi1EY JGKMKMii

θ′+′−′+ (4.121)

Employing the definitions

i3Yi4Xiii KIKIdrii θθ +−+ω= (4.122)

i1Yi2X KIKIIIiiii θθωω −−= (4.123)

equation (4.121) takes the following form:

iiii2EXi1EYiiEt JGIEKMKMrTM iiiiθ′+θ ′′′−′−′+′−= ω ( )n1,2,...,i = (4.124)

Substituting (4.111-114) in (4.122) and rearranging, i.e.,

( ) ( )

∆

−+

∆

−−−+ω= θθ

i

XYiXiY

i

YiXYiXiii

ii

i

ii

i

IbIaI

IbIaIdr (4.125)

i

YXYiYXi

i

XYiXXYiiii

iiiiiiiiIIbIIaIIbIIa

dr∆

−+

∆

−++ω= θθθθ (4.126)

( ) ( )

∆

+−

∆

+++ω= θθθθ

i

XYYXYi

i

XXYYXiiii

iiiiiiiiIIII

bIIII

adr (4.127)

and


59

i2ii1iiii KbKadr −++ω= (4.128)

Upon substitution of expressions (4.115-116) into (4.58),

γ′′−+−−=θ′′

iii cii

ii4EXi3EX

ii T

ATKMKM

rE1 ( )n1,2,...,i = (4.129)


iiii Eti2EXi1EYiiiii MKMKMrTJGIE −′−′+′−=θ′−θ ′′′ω (4.130)

Derivatives of equations (4.129) and (4.130) with respect to z twice and once,

respectively, are as follows:

γ′′′′−

′′+′′−′′−=θ ′′′′

iii cii

ii4EXi3EX

ii T

ATKMKM

rE1 (4.131)

iiii ETi2EXi1EYiiiii MKMKMrTJGIE ′−′′−′′+′′−=θ′′−θ ′′′′ω (4.132)

Elimination of iθ terms from equations (4.129), (4.131) and (4.132) yields

iii

iii

iiii

Eti2EXi1EYii

cii

ii4EXi3EX

ii

cii

ii4EXi3EX

i

MKMKMrT

TATKMKM

rE1JG

TATKMKM

r1I

′−′′−′′+′′−=

γ′′−+−−−

γ′′′′−

′′+′′−′′−ω

Simplifying equation (4.133), the single fourth order governing differential equation

for the axial force, Ti, is obtained as follows:

(4.133)


60

( )( ) ( )

( ) iEti4EXi3EYi

ii2i4EXii1i3EY

ii

ii

2i

ci

iic

rMKMKME

GJ

rKKIMrKKIM

TAEJGTr

EJG

AITI

iii

iiii

iiii

′++

+−′′−+′′−

=

+′′

+

γ+−′′′′γ

ωω

ωω

Equation (4.134) can be reorganized and put into the following form:

( ) ( ) ( )( ) ( )

( )( )n,...,2,1i

rMKMKME

GJ

rKKIMrKKIM

TTT

iEti4EXi3EYi

ii2i4EXii1i3EY

ii3ii2ii1

iii

iiii

=

′++

+−′′−+′′−

=β+′′β−′′′′β

ωω

where

iiIci1 ωγ=β (4.136)

2i

ci

ii2 r

EJG

AI ii +

γ+=β ω (4.137)

i

ii3 AE

JG=β (4.138)

Thus, the governing differential equation of the analysis of non-planar stiffened

coupled shear walls is found as (4.135). In this equation, in the solution for a unit

loading, iEXM and

iEYM are the external bending moments and iEtM is the external

twisting moment about the respective global axes for the particular unit loading.

Equation (4.135) is written for each region separately. However, in this context when

the unit load is applied at an internal point of a region, it divides that region into two

(4.134)

(4.135)


61

new regions. The system of Macaulay’s brackets should be understood, here and in

the sequel, as

( )nn zzzz '' −=− and 1' 0 =− zz for 'zz >

0' =− nzz and 0' 0 =− zz for 'zz ≤ (4.139)

Thus, in the general form, the external effects iEXM ,

iEYM and iEtM for any

unit loading is found, using the following expression for the particular case:

1

zHM pEXi−= (4.140)

1

zHM pEYi−= (4.141)

( ) ( ) 1

PXPYEt ddMi

+−= (4.142)

Employing the Macaulay’s brackets,

if zH p > ; ( )zHzH pp −=−1

if zH p ≤ ; 01

=− zH p

in accordance with the system of Macaulay’s brackets, we can rewrite iEXM and

iEYM for the part beneath the unit load as follows:

( )zHM pEXi−= 1−=′

iEXM 0=′′iEXM

( )zHM pEYi−= 1−=′

iEYM 0=′′iEYM

Hence, for the part above the unit load, iEXM ,

iEYM and iEtM are equal to zero and

(4.143)

(4.144)


62

( )∑=

×=k

1tbmtopE tti

dMB ( )1,2,...k = (4.145)

0BiE =′ (4.146)

Using these expressions in (4.135) and solving the resulting differential

equation yields:

[ ] [ ] [ ] [ ]zCoshDzSinhDzCoshDzSinhDT i2i4i2i3i1i2i1i1i α+α+α+α=

( )[ ]ii EXiEYii

i

MKMKGJE 43

3

1++

β

( )n1,2,...,i = (4.147)

in which

βββ−β−β

=αi1

i3i12

i2i2i1 2

4 ( )n1,2,...,i = (4.148)

βββ−β+β

=αi1

i3i12

i2i2i2 2

4 ( )n1,2,...,i = (4.149)

Employing the boundary conditions to determine the integration constants, Ti

can be obtained in a straightforward manner.

4.3.4.4. Determination of the Shear Forces in the Stiffening Beams

The compatibility equation for any region was obtained as in (4.57).

Similarly, for the stiffening beam at the upper boundary of that region


63

( )

)n,...,2,1i(0K2c

EI12c

GAc2.1

EV

dzTA1

A1

E1

dbvau

i6sb

2i

s

3i

s

ii

n

it

z

zt

t2t1

iiiiiii

iii

t

1t

==δ+

++−

+−

+ωθ′+′+′

∑ ∫=

+

Applying equation (4.57) for the uppermost position (z = zi)

( )

)n,...,2,1i(0K2ch

EI12ch

GAch2.1

E

T

dzTA1

A1

E1

dbvau

i6cb

2ii

c

3ii

c

iizzi

n

it

z

zt

t2t1

iiiiiii

iii

i

t

1t

==δ+

++

′+

+−

+ωθ′+′+′

=

=∑ ∫

+

Subtracting (4.151) from (4.150) and reorganizing

iiiiiii

zzicb

2ii

c

3ii

c

iii1

sb

2i

s

3i

s

i TK2ch

EI12ch

GAch2.1V

K2c

EI12c

GAc2.1

=′

++−=

++ (4.152)

Employing a new constant defined as

++

++

=

iii

iii

sb

2i

s

3i

s

i

cb

2ii

c

3ii

c

ii

i

K2c

EI12c

GAc2.1

K2ch

EI12ch

GAch2.1

S (4.153)

the shear forces in the stiffening beams are found as

izziii TSV=

′−= (4.154)

(4.150)

(4.151)


64

4.3.4.5. Boundary Conditions

To determine the integration constants in the single fourth order differential

equation (4.147) the following 4n boundary conditions are employed:

1- The structure being rigidly fixed at the base ( 0z = )

0vu0zn0zn0zn =θ==

=== (4.155)

0vu0zn0zn0zn =θ′=′=′

=== (4.156)

Applying equation (4.57) at ( 0z = ) and using (4.156), the first boundary condition is

0T0zn =′

= (4.157)

2- Substituting nnn andv,u θ ′′′′′′′′′ from (4.117-118) and the derivative of (4.129) in

the twisting moment expression (4.102) and using 00zn =θ′

=, the second boundary

condition at the base ( 0z = ) is found as

( )

( )

( ) 0TdTATKMKM

rI

KTTATKMKM

rKIMIM

I

KTTATKMKM

rKIMIM

IM

nnnncn

nn4EXn3EY

n

n4nncn

nn4EXn3EY

n

n2

n

XYEYYEXX

n3nncn

nn4EXn3EY

n

n1

n

XYEXXEYYEt

nnn

n

nnn

nnnn

n

nnn

nnnn

nn

=′+ω+

′′′γ−

′+′−′−+

′−

′′′γ−

′+′−′−−

∆

′−′+

′−

′′′γ−

′+′−′−+

∆

′−′−

ω

θ

θ

3- From the equilibrium of the vertical forces in each pier in the uppermost

region of the shear wall (see Fig. 4.23)

1

H

z 11 VdzqT += ∫ (4.159)

(4.158)


65

Applying this equation for the uppermost possition (z = H), the first term on the right

drops out. Considering expression (4.154), the third boundary condition is found as

follows:

Hz11Hz1 TST== ′−= (4.160)

As a special case, when there is no stiffening beam at the top

0T Hz1 == (4.161)

Figure 4.23. The free-body diagram of a part of the shear wall at the top

4- Substituting 111 andv,u θ′′′′′′ from (4.115-116) and (4.129) in the bimoment

expression (4.85) and applying it at the top ( Hz = ), the fourth boundary condition is

obtained as

( )

( )

( ) 0TdTATKMKM

rI

KTTATKMKM

rKIMIM

I

KTTATKMKM

rKIMIM

IB

1111c1

141EX31EY

1

4111c1

141EX31EY

1

21

1

XYEYYEXX

3111c1

141EX31EY

1

11

1

XYEXXEYYE

111

1

111

1111

1

111

1111

11

=+ω+

′′γ−+−−+

−

′′γ−+−−−

∆−

+

−

′′γ−+−−+

∆

−−

ω

θ

θ

q1

V1

T1

G11

(4.162)


66

5- From the vertical force equilibrium of one piece of the stiffening beam in

either of the piers at height z = zi (see Fig. 4.24)

ii zziizz)1i( TVT ==− =+ (4.163)

Substituting (4.154) in (4.163), the fifth type boundary condition is obtained as

follows:

iii zzizziizz)1i( TTST

===− =′− ( )n2,3,...,i = (4.164)

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 (i−1) at height izz = , considering (4.55) for the case of cross-sectional changes,

the following equation is obtained as the sixth type boundary condition:

( )( ) ( )

( )

( ) ( )

( )

( ) ( )

( )

( )n2,3,...,iK2ch

EI12ch

GAch2.1T

K2ch

EI12ch

GAch2.1

T

iiii

1i1i1ii

cb

2ii

c

3ii

c

iizzi

cb

21i1i

c

31i1i

c

1i1i

zz1i

=

++′=

++′

=

−−−−−−

=−

−−−

Vi

Ti

G1i

T(i-1)

(4.165)


67

7- Since the total twisting moments of the two neighbouring regions (i) and

(i−1), at height izz = balance each other

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

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

( ) iiii

iiiXiYEt

1i1i1i1i

1i1i1iX1iYEt

TdIE

JGvIEuIEM

TdIE

JGvIEuIEM

i

iii

1i

1i1i1i

′+ω+θ ′′′+

θ′−′′′+′′′−=

′+ω+θ ′′′+

θ′−′′′+′′′−

ω

θθ

−−−−ω

−−−θ−θ

−

−−−


( )[ ]

( ) ( )[ ]( ) ( )[ ] ( ) ( )[ ]

( ) ( )[ ] ( ) ( )( ) ( ) ( )[ ] 0TdTdIEIE

JGJGvIEvIE

uIEuIEMM

iii1i1i1ii1i

ii1i1iiX1iX

iY1iYEtEt

i1i

i1i

i1ii1i

=′+ω−′+ω+θ ′′′−θ ′′′+

θ′+θ′−+′′′−′′′+

′′′+′′′−+−

−−−ω−ω

−−θ−θ

θ−θ

−

−

−−

Applying assumption 9 mentioned in Chapter 4 and expressing all unknown

functions in terms of Ti and its derivatives, after some rearrangements, the seventh

type boundary condition is found as

( ) ( ) 0CCTCTCTCTC i3)1i(3ii21i)1i(2ii11i)1i(1 =−+′−′+′′′−′′′ −−−−− ( )n2,3,...,i = (4.168)

where

i

iii

iiiiiiii

iii

ii

iiiiii

Eti

Yi3Xi4

i

i2Xi1Y

i

YYXYXXY

i

XXYXYXYi3

iiiiii

i2X

ii

i1Yi4Xi3Yi2

i

c

i

Xi2c

i

Yi1ci1

Mr

MKMKr

KIKII

MIIMIIMIIMIIC

drA

IrA

KIrAKI

KIKIC

rI

rIK

rIK

C

+

′+′

−−−

∆

′+′−

∆

′+′=

+ω++−−−=

γ−

γ+

γ=

θθω

θθθθ

ωθθθθ

ωθθ

(4.169)

(4.166)

(4.167)

( )n2,3,...,i =


68

8- Since the total bimoments of the two neighbouring regions (i) and (i−1), at

height izz = balance each other

( ) ( ) ( ) ( ) ( )

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

( ) iiiiiXiYE

1i1i1i1i

1iX1iYE

TdIEvIEuIEB

TdIE

vIEuIEB

iiii

1i

1i1i1i

+ω+θ′′+′′+′′−=

+ω+θ′′+

′′+′′−

ωθθ

−−−−ω

−θ−θ

−

−−−


( )[ ]

( ) ( )[ ]( ) ( )[ ]

( ) ( )[ ]( ) ( )( ) ( ) ( )[ ] 0TdTd

IEIEvIEvIE

uIEuIEBB

iii1i1i1i

i1iiX1iX

iY1iYEE

i1ii1i

i1ii1i

=+ω−+ω+

θ′′−θ′′+′′−′′+

′′+′′−+−

−−−

ω−ωθ−θ

θ−θ

−−

−−

Expressing the other unknown functions in terms of Ti and its derivatives, after some

necessary rearrangements, the eighth type boundary condition is obtained as

( ) ( ) 0CCTCTCTCTC i4)1i(4ii21i)1i(2ii11i)1i(1 =−+−+′′−′′ −−−−− (4.172)

where

i

iii

iiiiiiii

Ei

Yi3Xi4

i

i2Xi1Y

i

YYXYXXY

i

XXYXYXYi4

Br

MKMKr

KIKII

MIIMIIMIIMIIC

+

+

−−−

∆

+−

∆

+=

θθω

θθθθ

To determine the integration constants D1i to D4i, the boundary conditions at

the top, bottom and between each pair of consecutive regions are used. Substituting

them in expression (4.147), the general solution for Ti ( )n,1,2,i …= can be found.

(4.170)

(4.171)

(4.173)


69

4.3.4.6. Determination of the Rotation Function ( θ )

The rotation expressions in n regions can be found by substituting the Ti

expressions into equation (4.129) and integrating two times with respect to z as

i2ii1cii

ii4EXi3EX

ii GzGdzdzT

ATKMKM

Er1

iii++

γ′′−+−−=θ ∫ ∫

( )n,1,2,i …= (4.174)

To determine the integration constants, n2 boundary conditions are needed.

The boundary conditions for the unknowns in θ i function are as follows:

1-2. Due to the complete fixity, there is no rotation or warping at the base. Hence,

00zn =θ

= (4.175)

00zn =θ′

= (4.176)

3-4. From the continuity of θ and θ′ for two neighbouring regions (i) and (i−1) at

height izz = , the following conditions are obtained:

ii zz1izzi =−=

θ=θ ( )n,2,3,i …= (4.177)

ii zz1izzi =−=θ′=θ′ ( )n,2,3,i …= (4.178)

4.3.4.7. Determination of the Lateral Displacement Functions (u and v)

Integrating (4.115) and (4.116) twice with respect to z,


70

i2ii1i

XEY

i

XYEXi3ii1ii RzRdzdz

IMIMKTKE

E1u iiii ++

∆

+∆

−−θ′′= ∫ ∫

( )n,1,2,i …= (4.179)

i2ii1i

YEX

i

XYEYi4ii2ii NzNdzdz

IMIMKTKE

E1v iiii ++

∆

+∆

−−θ ′′−= ∫ ∫

( )n,1,2,i …= (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)

ii zz1izzi uu

=−== ( )n,2,3,i …= (4.181)

ii zz1izzi vv=−=

= ( )n,2,3,i …= (4.182)

ii zz1izzi uu

=−=′=′ ( )n,2,3,i …= (4.183)

ii zz1izzi vv

=−=′=′ ( )n,2,3,i …= (4.184)

and the vanishing of the horizontal displacements and the respective slopes at the

bottom, i.e.,

0vu0zn0zn ==

== (4.185)

0vu0zn0zn =′=′

== (4.186)


71

Having determined the displacements for unit loadings at each and every one

of the levels of lumped masses, the flexibility matrix of the structure can be found.

Each unit loading gives one column of the flexibility matrix as the displacements at

the points where the lumped masses are. Then, the stiffness matrix of the structure

will be determined by taking the inverse of the flexibility matrix:

1−= FK (4.187)

4.4. Determination of Eigenvalues and Eigenvectors

Having formed the mass matrix and the stiffness matrix as described in the

previous section, the circular frequencies are determined from the following standard

frequency equation for the lumped mass system:

02 =− MK ω (4.188)

where ω 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, ωi, in the following equation at a time:

( ) 02 =− ii sMK ω mi ,...,2,1= (4.189)

5. FORCED VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER

72

5. FORCED VIBRATION ANALYSIS OF NON-PLANAR COUPLED

SHEAR WALLS

5.1. Introduction

Forced vibration analysis is concerned with the behaviour of the structure

under loads which vary with time. The static and dynamic methods of analysis are

fundamentally different in character. A structural-dynamic problem differs from its

static-loading counterpart in two important respects. The first difference to be noted,

by definition, is the time-varying nature of the dynamic problem. Because both

loading and response vary with time, it is evident that a dynamic problem does not

have a single solution, as a static problem does, instead the analyst must establish a

succession of solutions corresponding to all times of interest in the response history.

Thus, a dynamic analysis is clearly more complex and time-consuming than a static

analysis. The second and more fundamental distinction between static and dynamic

problems is illustrated in Fig. 5.1. If a simple beam is subjected to a static load P, as

shown in Fig. 5.1a, inertial moments and shears and deflected shape depend only

upon this load. On the other hand, if the load P(t) is applied dynamically, as shown in

Fig. 5.1b, the resulting displacements of the beam depend not only upon this load but

also upon inertial forces which oppose the accelerations producing them.

(a) Static loading (b) Dynamic loading

Figure 5.1. Basic difference between static and dynamic loads

P (static)

Inertial forces

P = P(t) (vary with time)


73

5.2. Mode Superposition Method

5.2.1. Introduction

In the forced vibration analysis of multi-degree of freedom structures, the

solution of the equation set gets harder as the degree of freedom increases and,

especially, as the time increment decreases or the number of steps increases, the

amount of computation increases, also, and a high computer capacity is needed. The

equation of motion pertaining to the coupled shear wall, for which the mass and

stiffness matrices were determined in the previous section, can be written as follows:

+

N

3

2

1

NN3N2N1N

N3333231

N2232221

N1131211

N

3

2

1

NN3N2N1N

N3333231

N2232221

N1131211

x.

xxx

c.ccc.....

c.cccc.cccc.ccc

x.

xxx

m.mmm.....

m.mmmm.mmmm.mmm

&

&

&

&

&&

&&

&&

&&

=

+

N

3

2

1

N

3

2

1

NN3N2N1N

N3333231

N2232221

N1131211

P

PPP

x

xxx

kkkk

kkkkkkkkkkkk

...

........

(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

+

N

3

2

1

NN

33

22

11

N

3

2

1

NN

33

22

11

x.

xxx

c~...........c~.....c~.....c~

x.

xxx

m~...........m~.....m~.....m~

&

&

&

&

&&

&&

&&

&&

=

+

N

3

2

1

N

3

2

1

NN

33

22

11

P

PPP

x

xxx

k

kk

k

..~....

.....

..~..

...~.

....~

(5.2)


74

Since, every equation involves entities belonging only to one node in equation set

(5.2), the solution gets easier.

5.2.2. Determination of Uncoupled Equation Set

In the first step of this method, the mode-shape matrix is found by writing

eigenvectors pertaining to each eigenvalue in the related column of the matrix in the

following form:

[ ]Ni321 ΦΦΦΦΦ=Φ LL (5.3)

where, iΦ 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.

( ) 11NNn

1ii ++= ∑

=

(5.4)

Obtaining the modal matrix as mentioned above, the real displacement vector

Y is described in terms of X modal displacement vector as follows:

XY Φ= (5.5)

Using equation (5.5) and its first and second time derivatives XY && Φ=

XY &&&& Φ=

the equation of motion of the multi-degree of freedom system

PYKYCYM =++ &&& (5.7)

(5.6)


75

can be written as

PXKXCXM =Φ+Φ+Φ &&& (5.8)

If equation (5.8) is multiplied by the transpose of the mode-shape matrix TΦ , it

takes the following form:

PXKXCXM TTTT Φ=ΦΦ+ΦΦ+ΦΦ &&& (5.9)

Using the orthogonality properties of the eigenvectors, uncoupled mass matrix,

=ΦΦ=

NN

22

11

T

m~

m~m~

MM~ (5.10)

uncoupled damping matrix,

=ΦΦ=

NN

22

11

T

c~

c~c~

CC~ (5.11)

uncoupled stiffness matrix,

=ΦΦ=

NN

22

11

T

k~

k~k~

KK~ (5.12)


76

and modal forces,

=Φ=

N

2

1

T

P~

P~P~

PP~ (5.13)

are obtained. Finally, uncoupled equation of motion can be obtained as

)(~~~~ tPXKXCXM =++ &&& (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:

[ ]

ΦΦΦΦ=Φ=

N

N

X

XXX

XY:

3

2

1

321 L (5.15)

5.3. Time-History Analysis

The computation of the displacements or member end forces of the structure

under time dependent loads is called the time-history analysis. The uncoupled

equations of motion (5.14) were obtained in matrix form before. When equation set

(5.14) is examined, it is, obviously, seen to be a set of second order differential

equations and, especially, for non-planar coupled shear walls with many degrees of

freedom, the solution gets considerably hard and time consuming. Different

numerical solution methods have been developed to render the solution easier and to

reduce the computation time. One of these methods is the “Newmark” method

which has been used in the present work.


77

5.3.1. Newmark Method

The Newmark integration method is based on the assumptions of linear

acceleration and constant average acceleration. In the Newmark formulation, the

basic integration equations for the final velocity and displacement are expressed as

follows:

( )[ ] txx1xx tttttt ∆δ+δ−+= ∆+∆+ &&&&&& (5.16)

( )[ ] 2ttttttt txx5.0txxx ∆α+α−+∆+= ∆+∆+ &&&&& (5.17)

in which, the parameters α and δ 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 α and δ 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 0=t , i.e.,

0X0X

0

0

=

=&

(5.18)

The starting value of the acceleration vector is obtained by substituting the values in

(5.18) into equation (5.14) as follows:

)t(P~XK~XC~XM~ 000 =++ &&& (5.19)

Arranging this equation, the starting value of the acceleration vector is

determined:


78

[ ]001

0 XK~XC~)t(P~M~X −−=− &&& (5.20)

2) The time increment value ∆t and the parameters α and δ are determined:

( )25.025.0

50.0

δ+≥α

=δ (5.21)

3) The constants which will be used in the Newmark method are calculated:

( )ta

0.1ta

0.20.2ta

0.1a

0.12

0.1a

t0.1a

ta

t0.1a

7

6

5

4

3

2

1

20

∆α=δ−∆=

−

αδ∆

=

−αδ

=

−α

=

∆α=

∆αδ

=

∆α=

(5.22)

4) The system effective stiffness matrix is formed:

C~aM~aK~K~ 10ef ++= (5.23)

5) All computations determined in the previous steps are repeated for every ∆t

time increment value.


79

a) The effective load vector is determined for t+∆t:

( ) ( )t5t4t1t3t2t0ttttef XaXaXaC~XaXaXaM~P~P~ &&&&&& ++++++= ∆+∆+ (5.24)

b) The displacement vector of the structure is determined for t+∆t:

ttefttef P~YK~

∆+∆+ = (5.25)

c) The acceleration

( ) t3t2ttt0tt XaXaXXaX &&&&& −−−= ∆+∆+ (5.26)

and the velocity,

tt7t6ttt XaXaXX ∆+∆+ ++= &&&&&& (5.27)

are calculated for t+∆t.

d) The displacement

tttt XY ∆+∆+ Φ= (5.28)

the velocity,

tttt XY ∆+∆+ Φ= && (5.29)

and the acceleration

tttt XY ∆+∆+ Φ= &&&& (5.30)

vectors are determined for the structure.

i2sy

6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER

80

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 Vlasov’s 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 time-

history 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.


81

6.2. Numerical Applications

Example 1:

In the first example, the free vibration analysis of a non-planar coupled shear

wall on a rigid foundation was considered. The example of non-planar coupled shear

wall has been taken same as the onee considered by Tso and Biswas in 1973. This

example was solved by the computer program prepared in the present work and the

SAP2000 structural analysis program and the natural frequencies of the shear wall,

thus obtained, were compared. The geometrical and material properties of the

example shear wall model are same as those given by the authors in the original

model, which they chose suitable for experimental convenience.

The geometrical properties and the cross-sectional view of the model are

given in Fig. 6.1. The non-planar coupled shear wall system rests on a rigid

foundation. The total height is 48 in, the storey height is 6 in, the thickness is 0.39 in,

the height of the connecting beams is 1.5 in and the elasticity and shear moduli of the

model are psi1040.0E 6×= and psi10148.0G 6×= , respectively.

The cross-section having been formed by two angles, each angle has its shear

center at the intersection of its sides. The sectorial areas are equal to zero at all points

of the cross-section. Hence, the warping moment of inertia, for this example, is

identically equal to zero for each pier.

According to the lumped mass idealization, the lumped masses, which were

calculated by the computer program, were concentrated at the center of the whole

cross-sectional area of the structure. The coordinates of the mass center were

calculated as (0.000, 1.687).

The model structure was solved both by the present method using the CCM

and by the SAP2000 structural analysis program using the frame method (also called

wide column analogy) for which the model and its 3-D view are given in Fig. 6.2.


82

Figure 6.1. Geometrical properties of the model originally considered by Tso and

Biswas

Y

H =

48

in

X

6 in

O

Z 5.

195

in

O X

Y

2.805 in 2.805 in 2 in 2 in

2

3

(0,0)

(-2, 0) (-4.805, 0)

(-4.805, 5.195) (4.805, 5.195)

(2, 0) (4.805, 0)

2

1 1

0.39 in

0.39 in 0.39 in

0.39 in

2 1 1 2

3

G (0, 1.687)


83

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.


84

Table 6.1. Comparison of the natural frequencies (Hz) obtained from the present program and SAP2000 in Example 1

Mode Present Study

(CCM) SAP2000

(Frame Method) % difference Natural Frequencies Natural Frequencies

1 0,16427 0,16636 1,26 2 0,16902 0,16823 0,47 3 0,71226 0,69767 2,09 4 1,01237 1,02500 1,23 5 1,57316 1,53440 2,53 6 2,59400 2,52580 2,70 7 2,79376 2,82740 1,19 8 3,81979 3,70460 3,11 9 5,20122 5,00040 4,02 10 5,39394 5,45530 1,12 11 6,57889 6,25420 5,19 12 7,62432 7,18320 6,14 13 8,76270 8,85340 1,02 14 12,7231 12,83700 0,89 15 16,7636 16,88600 0,72 16 19,8694 19,99000 0,60

In each method, after obtaining the natural frequencies 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. 6.3-5 present the mode shapes of the shear wall, found by the present

program and the SAP2000 structural analysis program, both in the same figure.


85


directions found by both the present program and SAP2000 in Example 1

Mode 1 in Y Direction

0

6

12

18

24

30

36

42

48

0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=0,16636 Hz CCM NF=0,16427 Hz

Mode 2 in X Direction

0

6

12

18

24

30

36

42

48

0,0 0,5 1,0

Hei

ght(m

)

SAP 2000NF=0,16823 Hz

CCM NF=0,16902 Hz

Mode 3 in Teta Direction

0

6

12

18

24

30

36

42

48

-0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=0,69767 Hz

CCM NF=0,71226 Hz


86

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


0

6 12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=1,02500 Hz SBY NF=1,01237 Hz


0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=1,53440 Hz CCM DF=1,57316 Hz


0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=2,52580 Hz

CCM NF=2,59400 Hz


87

Figure 6.5. Comparison of seventh and eighth mode shapes in X and Y directions



0

6

12

18

24

30

36

42

48

-1,0 0,0 1,0

Hei

ght (

m)

SAP 2000 NF=2,82740 Hz

CCM DF=2,78376 Hz


0

6

12

18

24

30

36

42

48

-1,0 0,0 1,0

Hei

ght (

m)

SAP 2000 NF=3,70460 Hz

CCM NF=3,81979 Hz


88

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.

Figure 6.6. Cross-sectional view of the structure and applied dynamic load in

Example 2

Figure 6.7. Triangular pulse force in Example 2

O X

Y

2

3

(0,0)

(-2, 0) (-4.805, 0)

(-4.805, 5.195) (4.805, 5.195)

(2, 0) (4.805, 0)

2

1 1

0.39 in

0.39 in 0.39 in

0.39 in

P(t) 2 1 1 2

3

G (0, 1.687)

100

P(lb)

t (s)

8 4


89

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.


90

-0,300

-0,200

-0,100

0,000

0,100

0,200

0,300

0,400

0,500

0 2 4 6 8 10 12 14 16 18 20 22 24

Time (s)

Top

Dis

plac

emen

t (m

)

Present study

Sap2000

Figure 6.8. Time-varying displacements in X direction at the top of the shear wall for

undamped case in Example 2

-0,200

-0,100

0,000

0,100

0,200

0,300

0,400

0,500

0 2 4 6 8 10 12 14 16 18 20 22 24

Time (s)

Top

Dis

plac

emen

t (m

) Present study

Sap2000

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


91

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 = 2.85×106 kN/m2 and G = 1055556 kN/m2, respectively. Fig. 6.10.

Figure 6.10. Cross-sectional view of the structure in Example 3

Both T-section piers in this system have their shear centers at the intersection

of their branches. Therefore, the sectorial areas are equal to zero at all points of the

cross-section. The warping moments of inertia are all equal to zero, also.

According to lumped mass idealization, the lumped masses which were


cross-sectional area of the structure. The mass center was located at point O.

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. 6.11.

2 m

Y

2 m

O X

2 m

2

3

(0,0) (-1, 0)

(-3, 0)

(-3, 2) (3, 2)

(1, 0)

(3, 0)

2

1 1

0.3 m

0.3 m 0.3 m

0.3 m

(-3, -2) (3, -2)

3 3

0.3 m 0.3 m

1 m 1 m 2 m

2

4

1 1

2

3

4


92


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.


93


Mode

Present Study (CCM)


% difference

Natural Frequencies Natural Frequencies

1 0,97409 0,97582 0,18 2 1,36735 1,35782 0,70 3 5,20893 5,18059 0,55 4 6,00116 6,00303 0,03 5 11,82911 11,75796 0,61 6 16,55209 16,51685 0,21 7 21,25545 21,06219 0,92 8 31,93361 31,74091 0,61 9 33,45567 32,97118 1,47

10 47,76713 46,71418 2,25 11 51,83542 51,22450 1,19 12 62,34009 60,43418 3,15 13 73,52751 70,80442 3,85 14 75,21096 73,73767 2,00 15 99,04972 96,23732 2,92 16 117,37420 113,25790 3,63








94

Figure 6.12. Comparison of first, second and third mode shapes in X and Y directions



0

6

12

18

24

30

36

42

48

0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=0,97582 Hz

CCM NF=0,97409 Hz


0

6

12

18

24

30

36

42

48

0,0 0,5 1,0

Hei

ght(m

)

SAP 2000NF=1,35782 Hz

CCM NF=1,36735 Hz


0

6 12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=5,18059 Hz CCM NF=5,20893 Hz


95




0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=6,00303 Hz

CCM NF=6,00116 Hz


0

6 12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=11,75796 Hz

SBY NF=11,82911 Hz


0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=16,51685 Hz CCM DF=16,55209 Hz


96




0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=21,06216 Hz

CCM NF=21,25545 Hz


0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=31,74091 Hz

CCM NF=31,93361 Hz


97

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.


Example 4

150

Figure 6.16. Rectangular pulse force in Example 4

2 m

Y

2 m

O X

2 m

2

3

(0,0) (-1, 0)

(-3, 0)

(-3, 2) (3, 2)

(1, 0)

(3, 0)

2

1 1

0.3 m

0.3 m 0.3 m

0.3 m

(-3, -2) (3, -2)

3 3

0.3 m 0.3 m

1 m 1 m 2 m

2

4

1 1

2

3

4

P(t)

t(s)

P(lb)

5


98


point O, the mass center, was calculated by the computer program prepared in the



Table 6.5. Maximum displacement (m) in the X direction of point O for undamped case in Example 4

SAP2000


(CCM) % difference

0.066510 0.065713 1.21


Table 6.6. Maximum displacement (m) in the X direction of point O for damped case in Example 4

SAP2000


(CCM) % difference

0.058340 0.057648 1.20

The responses of both damped and undamped systems to rectangular force

were determined and the time-varying displacements in the X direction of point O


It is observed that the results obtained in the present work (CCM) coincide



99

-0,050

-0,030

-0,010

0,010

0,030

0,050

0,070

0,090

0 2 4 6 8 10

Time (s)

Top

Dis

plac

emen

t (m

)

Present studySAP2000

Figure 6.17. Time-varying displacements in X direction at the top of the shear wall

for undamped case in Example 4

-0,040

-0,020

0,000

0,020

0,040

0,060

0,080

0 2 4 6 8 10

Time (s)

Top

Dis

plac

emen

t (m

)

Present study

SAP2000


for damped case with 5 % damping ratio in Example 4


100

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=2.85×106 kN/m2 and G=1055556 kN/m2, respectively.


Both T and L section piers in this system have their shear centers at the

intersection of their branches. Therefore, the sectorial areas are equal to zero at all

points of the cross-section. The warping moments of inertia are all equal to zero,

also.




calculated as (0.125; 0.208).

X

Y

3 m

2 m 2 m 1 m 1 m 3 m

2 m

O

(-1, 0)

(-5, 0)

(-3, -2)

(-3, 0)

(0, 0)

(1, 0)

(4, 3)

(4, 0)

1 2

3

2

1 1

4

2

3

0.3 m 0.3 m

1

2

0.3 m

0.3 m

3

G (0,125, 0,208)


101





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.


102


Mode Present Study

(CCM) SAP2000


1 0,69367 0,69248 0,17 2 1,65427 1,64679 0,45 3 4,14919 4,14449 0,11 4 6,88659 6,86626 0,30 5 11,36339 11,33655 0,24 6 16,69150 16,62251 0,42 7 21,86704 21,76464 0,47 8 30,88525 30,65600 0,75 9 35,45502 35,16467 0,83

10 49,27674 48,65858 1,27 11 51,41270 50,74482 1,32 12 67,68475 66,43667 1,88 13 70,87069 69,48448 2,00 14 80,19234 78,38068 2,31 15 92,87637 90,32802 2,82 16 109,78064 106,11409 3,46



Mode shapes in X, Y and Teta directions were compared by normalizing with





103

Figure 6.21. Comparison of first, second and third mode shapes in X and Y directions



0

6

12

18

24

30

36

42

48

0,0 0,5 1,0

SAP 2000 NF=0,69248 Hz CCM NF=0,69367 Hz H

eigh

t (m

)


0 6

12 18 24 30 36 42 48

0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=1,64679 Hz CCM NF=1,65427 Hz


0 6

12 18 24 30 36 42 48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=4,14449 Hz CCM NF=4,14919 Hz


104




0

6

12

18

24

30

36

42

48

-0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=6,86626 Hz SBY NF=6,88659 Hz


0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=11,33655 Hz

CCM DF=11,36339 Hz


0 6

12 18 24

30 36 42 48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=16,62251 Hz

CCM NF=16,69150 Hz


105




0 6

12 18 24 30 36 42 48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=21,76464 Hz CCM NF=21,86704 Hz


0 6

12 18 24 30 36 42 48

-2,5 -1,5 -0,5 0,5 1,5 2,5

Hei

ght (

m)

SAP 2000 NF=21,76464 Hz

CCM DF=21,86704 Hz


0

6

12

18

24

30

36

42

48

-1,0 0,0 1,0

Hei

ght (

m)

SAP 2000 NF=30,65600 Hz

CCM NF=30,88525 Hz


106

Example 6:

In this example, the forced vibration analysis of a non-planar non-

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 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.


Example 6

X

Y

3 m

2 m 2 m 1 m 1 m 3 m

2 m

O

(-1, 0)

(-5, 0)

(-3, -2)

(-3, 0)

(0, 0)

(1, 0)

(4, 3)

(4, 0) 1 2

3

2

1 1

4

2

3

0.3 m 0.3 m

1

2

0.3 m

0.3 m

3

P(t)

G (0,125, 0,208)


107

100


At the top of the shear wall, the maximum displacement in the Y direction of


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


(CCM) % difference

0.121225 0.122312 0.90


Table 6.9. Maximum displacement (m) in the Y direction of point G for damped case in Example 6

SAP2000


(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




P(lb)

t(s) 5


108

-0,150

-0,100

-0,050

0,000

0,050

0,100

0,150

0 2 4 6 8 10

Time (s)

Top

Dis

plac

emen

t (m

) Present study

SAP2000

Figure 6.26. Time-varying displacements in Y direction at the top of the shear wall


-0,100

-0,050

0,000

0,050

0,100

0,150

0 2 4 6 8 10

Time (s)

Top

Dis

plac

emen

t (m

)

Present study

SAP2000




109

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.

Figure 6.28. Sine pulse force in Example 6

Table 6.10. Maximum displacement (m) in the Y direction of point G for undamped case in Example 6

SAP2000


(CCM) % difference

0.123144 0.124157 0.82


Table 6.11. Maximum displacement (m) in the Y direction of point G for damped case in Example 6

SAP2000


(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.

t (s)

P(lb)

150

5


110

-0,150

-0,100

-0,050

0,000

0,050

0,100

0,150

0 2 4 6 8 10

Time (s)

Top

Dis

plac

emen

t (m

)

Present study

SAP2000



-0,150

-0,100

-0,050

0,000

0,050

0,100

0,150

0 2 4 6 8 10

Time (s)

Top

Dis

plac

emnt

(m)

Present study

SAP2000




111

Example 7:



carried out. The total height of the shear wall is 48 m, the storey height is 3 m, the


shear moduli are E = 2.85×106 kN/m2 and G = 1055556 kN/m2, respectively.



calculated by computer program, were concentrated at the center of the whole cross-

sectional area of the structure. The coordinates of the mass center were calculated as

(-0.450; 1.750).




5.0

m

5.0

m

2.0 m

O

Y

X (-1.5, 0) (0, 0)

(-5.5, 0)

(-5.5, 5)

(-3.5, 5)

(1.5, 0) (5.5, 0)

(5.5, 5)

1 2

3 4

1 2

3

0.4 m 0.4 m

0.4 m 0.4 m

1

2

3

1

2

0.4 m

4.0 m 4.0 m 1.5 m 1.5 m

(-0.45, 1.75)

G


112


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.


113


Mode

Present Study (CCM)


% difference


1 0,4109880 0,409867 0,27 2 0,4931529 0,492887 0,05 3 1,8800960 1,877845 0,12 4 2,9823973 2,975817 0,22 5 4,715429 4,709624 0,12 6 8,2914745 8,248660 0,52 7 8,8708069 8,851927 0,21 8 14,381514 14,331390 0,35 9 16,169147 16,015483 0,96

10 21,22184 21,110277 0,53 11 26,609686 26,203663 1,55 12 29,368472 29,148453 0,75 13 38,776526 38,379274 1,04 14 39,572676 38,681877 2,30 15 49,372204 48,702600 1,37 16 55,006839 53,280684 3,24 17 61,01724 59,950797 1,78 18 72,831525 69,774177 4,38 19 73,465525 71,852793 2,24 20 86,298492 83,978085 2,76








114

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


0

6

12

18

24

30

36

42

48

0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=0,69248 Hz

CCM NF=0,69367 Hz


0

6

12

18

24

30

36

42

48

0,0 0,5 1,0

Hei

ght (

m)

SAP 2000NF=1,64679 Hz

CCM NF=1,65426 Hz


0

6 12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=4,14449 Hz CCM NF=4,14919 Hz


115




0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=6,86626 Hz

SBY NF=6,88659 Hz


0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=11,33655 Hz

CCM DF=11,36338 Hz


0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=16,62251 Hz

CCM NF=16,69150 Hz


116




0

6

12

18

24

30

36

42

48

-1,5 -0,5 0,5 1,5

Hei

ght (

m)

SAP 2000 NF=21,76464 Hz

CCM NF=21,86704 Hz


0

6

12

18

24

30

36

42

48

-1,5 -0,5 0,5 1,5

Hei

ght (

m)

SAP 2000 NF=21,76464 Hz

CCM DF=21,86704 Hz


0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=30,65600 Hz

CCM NF=30,88525 Hz


117

Example 8:



and material properties of the sixteen storey shear wall were taken as in Example 7.




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



Example 8

5.0

m

5.0

m

2.0 m

O

Y

X (-1.5, 0)

(0, 0) (-5.5, 0)

(-5.5, 5)

(-3.5, 5)

(1.5, 0) (5.5, 0)

(5.5, 5)

1 2

3 4

1 2

3

0.4 m 0.4 m

0.4 m 0.4 m

1

2

3

1

2

0.4 m

P(t)

4.0 m 4.0 m 1.5 m 1.5 m

(-0.45, 1.75) G


118







SAP2000


(CCM) % difference

0.036393 0.036112 0.77



SAP2000


(CCM) % difference

0.036130 0.035939 0.53






P(lb)

t (s)

5 2,5

150


119

-0,010

0,000

0,010

0,020

0,030

0,040

0 2 4 6 8 10 12 14

Time (s)

Top

Dis

plac

emen

t (m

)

Present study

SAP2000



-0,010

0,000

0,010

0,020

0,030

0,040

0 2 4 6 8 10 12 14

Time (s)

Top

Dis

plac

emen

t (m

)

Present study

SAP2000




120

Example 9:



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 2.85×106 kN/m2 and =G 1055556 kN/m2, respectively.



calculated by computer program, were concentrated at the center of whole cross-


(1.111; -1.045).




0.3 m

1

X

Y

1 m 3 m 3 m 1 m 2 m 2 m

1.5

m

3 m

O

3 m

1 (-1, 0)

(0, 0)

(-3, 0)

(-3, -3)

(-5, -4.5) (-8, -4.5)

(1, 0)

(4, 0)

(4, -4.5)

(4, 2)

(7, 2)

0.3 m

0.2 m

2

4

1

2

3

4

3

0.2 m

0.2 m

0.3 m

1

2

3

4 5

2

3 4

5

4.5

m

2 m

(1,111, -1.045)

G


121






122


Mode Present Study

(CCM) SAP2000

(Frame Method) % difference

Natural Frequencies Natural Frequencies 1 0,2728988 0,272051 0,31 2 0,5955637 0,594873 0,12 3 1,4747386 1,473539 0,08 4 3,2946904 3,288973 0,17 5 3,9478691 3,945944 0,05 6 7,6107878 7,604040 0,09 7 8,9650465 8,931684 0,37 8 12,483401 12,463691 0,16 9 17,374156 17,256600 0,68 10 18,547388 18,500908 0,25 11 25,791174 25,696652 0,37 12 28,549694 28,241979 1,09 13 34,197754 34,024317 0,51 14 42,45922 41,788411 1,61 15 43,743252 43,447923 0,68 16 54,389935 53,915274 0,88 17 59,075664 57,783123 2,24 18 66,078435 65,349682 1,12 19 78,358690 76,080273 2,99 20 78,710916 77,634855 1,39








123




0

5

10

15

20

25

30

35

40

45

50

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=1,47354 Hz

CCM NF=1,47474 Hz


0

5

10

15

20

25

30

35

40

45

50

0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=0,27205 Hz

CCM NF=0,27289 Hz


0

5

10

15

20

25

30

35

40

45

50

0,0 0,5 1,0

Hei

ght(m

)

SAP 2000NF=0,59487 Hz

CCM NF=0,59556 Hz


124




0 5

10

15

20

25

30

35

40

45

50

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=3,28897 Hz SBY NF=3,29469 Hz


0

5

10

15

20

25

30

35

40

45

50

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=3,94594 Hz

CCM DF=3,94787 Hz


0

5

10

15

20

25

30

35

40

45

50

-1,0 0,0 1,0

Hei

ght (

m)

SAP 2000 NF=7,60404 Hz

CCM NF=7,61078 Hz


125




0

5

10

15

20

25

30

35

40

45

50

-1,0 0,0 1,0

Hei

ght (

m)

SAP 2000 NF=8,93168 Hz

CCM DF=8,96504 Hz


0

5

10

152025

3035

40

4550

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=12,46369 Hz

CCM NF=12,48340 Hz


126

Example 10:



and material properties of the twenty storey shear wall were taken as in Example 9.




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.


Example 10


P(t)

0.3 m

1

X

Y

1 m 3 m 3 m 1 m 2 m 2 m

1.5

m

3 m

O

3 m

1 (-1, 0)

(0, 0)

(-3, 0)

(-3, -3)

(-5, -4.5) (-8, -4.5)

(1, 0)

(4, 0)

(4, -4.5)

(4, 2)

(7, 2)

0.3 m

0.2 m

2

4

1

2

3

4

3

0.2 m

0.2 m

0.3 m

1

2

3

4 5

2

3 4

5

4.5

m

2 m

(1,111, -1.045)

G

P(lb)

t (s)

5 2,5

150


127






SAP2000


(CCM) % difference

0.183850 0.182404 0.79



SAP2000


(CCM) % difference

0.172310 0.171008 0.76



are presented in Figures 6.47-48.




128

-0,100

-0,050

0,000

0,050

0,100

0,150

0,200

0 2 4 6 8 10 12 14

Time (s)

Top

Dis

plac

emen

t (m

)

Present study

SAP2000



-0,100

-0,050

0,000

0,050

0,100

0,150

0,200

0 2 4 6 8 10 12 14

Time (s)

Top

Dis

plac

emen

t (m

)

Present study

SAP2000




129

Example 11:



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




calculated by computer program, were concentrated at the center of whole cross-

sectional area of the structure. The coordinates of the mass center were calculated

as (-0.795; 0.322).

4 m

3

m

2 m 2 m 2 m 3 m 2 m

4 m 2 m 2 m 3 m

O 1

(-1, 0)

2

3 4 4 11

10

8

7 5 6 6

2

3

5

9

(-3, 0)

(-3, 4) (-1, 4)

(-3, -3) (-1, -3) (-7, -3)

(-7, 0) (-5, 0)

(-5, 4)

(-7, 4)

(4, 0)

(4, 4)

(1, 4)

(4, -3) (1, -3)

0.2 m

1

2

3

5

1

3

5

8

10

6

0.2 m

0.2 m

0.2 m

4

9

7

2

4

Y

X

0.2 m 0.4 m

0.4 m

0.4 m

0.4 m

0.4 m

0.4 m

0.4 m

0.4 m

0.4 m

0.4 m

1 (1, 0)

G (-0.795, 0.322)


130









131


Mode Present Study

(CCM) SAP2000

(Frame Method) % difference


1 0,5013758 0,500474 0,18 2 0,6613498 0,661167 0,03 3 2,4709828 2,463962 0,28 4 4,126175 4,118391 0,19 5 6,5157725 6,475072 0,63 6 11,50762 11,455949 0,45 7 12,519564 12,375324 1,17 8 20,482792 20,105120 1,88 9 22,458122 22,271148 0,84 10 30,369575 29,549525 2,78 11 36,969828 36,474049 1,36 12 42,140664 40,571794 3,87 13 54,984537 52,995514 3,75 14 55,732593 53,893310 3,41 15 71,036909 66,594367 6,67 16 76,429148 74,306353 2,86 17 87,855184 81,065347 8,38 18 101,190190 95,995947 5,41 19 105,832480 97,418722 8,64 20 120,366150 110,846184 8,58








132




0

6

12

18

24

30

36

42

48

0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=0,50047 Hz

CCM NF=0,50137 Hz


0

6

12

18

24

30

36

42

48

0,0 0,5 1,0

Hei

ght(m

)

SAP 2000NF=0,66117 Hz

CCM NF=0,66135 Hz


0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=2,46396 Hz CCM NF=2,47098 Hz


133




0

6 12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght

(m)

SAP 2000 NF=4,11839 Hz SBY NF=4,12617 Hz


0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght

(m)

SAP 2000 NF=6,47507 Hz

CCM DF=6,51577 Hz


0

6

12

18

24

30

36

42

48

-1,5 -0,5 0,5 1,5

Hei

ght (

m)

SAP 2000 NF=11,45595 Hz

CCM NF=11,50762 Hz


134




0

6

12

18

24

30

36

42

48

-1,5 -0,5 0,5 1,5

Hei

ght (

m)

SAP 2000 NF=12,37512 Hz

CCM DF=12,51956 Hz


0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=20,10512 Hz

CCM NF=20,48279 Hz


135

Example 12:








direction in the plane of the connection beam as in Fig. 6.54 and the rectangular



Example 12

4 m

3

m

2 m 2 m 2 m 3 m 2 m

4 m 2 m 2 m 3 m

O 1

(-1, 0)

2

3 4 4 11

10

8

7 5 6 6

2

3

5

9

(-3, 0)

(-3, 4) (-1, 4)

(-3, -3) (-1, -3) (-7, -3)

(-7, 0) (-5, 0)

(-5, 4)

(-7, 4)

(4, 0)

(4, 4)

(1, 4)

(4, -3) (1, -3)

0.2 m

1

2

3

5

1

3

5

8

10

6

0.2 m

0.2 m

0.2 m

4

9

7

2

4

Y

X

0.2 m 0.4 m

0.4 m

0.4 m

0.4 m

0.4 m

0.4 m

0.4 m

0.4 m

0.4 m

0.4 m

P(t) 1

(1, 0)

G (-0.795, 0.322)


136

100







SAP2000


(CCM) % difference

0.045550 0.045408 0.31



SAP2000


(CCM) % difference

0.040480 0.040359 0.30



are presented in Figures 6.56-57.



P(lb)

t(s)

5


137

-0,060

-0,040

-0,020

0,000

0,020

0,040

0,060

0 2 4 6 8 10

Time (s)

Top

Dis

plac

emen

t (m

) Present study

SAP2000



-0,040

-0,020

0,000

0,020

0,040

0,060

0 2 4 6 8 10

Time (s)

Top

Dis

plac

emen

t (m

)

Present study

SAP2000




138

Example 13:



carried out. The total height of the shear wall is 48 m, the storey height is 3 m and the




Figure 6.58. Non-planar non-symmetrical structure in Example 13




calculated as (-0.016; 0.018).

H =

48

m

h =

3 m

X

Y

connecting beams

Z


139


Y

X

2 m 2 m 2 m 1.5 m 2 m 1.5 m 2 m 2 m

2 m 4 m 1.5 m 2 m 1.5 m 4 m

4 m

3

m

2 m

5

m 1

(-1, 0)

1

0.3 m

1

(1, 0)

(-3, 0)

2

7

9

3

4

6 10 11

(-7, 0)

(-7, 4)

(-7, -3) (-5, -3)

(-3, -3)

5

(-5, 4) (-3, 4) (-1, 4)

8

(1, -3)

2

3

4 5

6 8

9 10

7

(4, 0)

(4, 2)

(2.5, 4) (1, 4)

(4, -3)

(2.5, -3) (8, -3)

(8, 2) (6, 2)

0.3 m

1 2

6 3

8 5 5

4

9 10 6

7 4

3

2

7

8

9

0.3 m

0.3 m 0.3 m

0.3 m

0.3 m 0.3 m

0.3 m

0.3 m 0.3 m

0.3 m

0.3 m

0.3 m O (0, 0)

0.3 m

G (-0.016, 0.018)

6. NU

MER

ICA

L RESU

LTS

Cevher D

eha TÜR

KÖ

ZER

139


140









141


Mode

Present Study (CCM)


% difference


1 0,5693957 0,569495 0,02 2 0,7100410 0,709684 0,05 3 3,2076784 3,20475 0,09 4 4,1377091 4,123140 0,35 5 8,7021153 8,671171 0,36 6 11,453579 11,362686 0,80 7 16,857499 16,732066 0,75 8 22,314747 21,995505 1,45 9 27,669057 27,322003 1,27 10 36,710061 35,882920 2,31 11 41,091277 40,312723 1,93 12 54,580514 52,798368 3,38 13 57,06977 55,544476 2,75 14 75,518967 72,420642 4,28 15 75,853501 72,846471 4,13 16 96,291975 91,756689 4,94 17 100,415440 94,567838 6,18 18 119,119440 112,044569 6,31 19 128,070530 118,607331 7,98 20 143,520020 133,046100 7,87








142




0

6

12

18

24

30

36

42

48

0,0 0,5 1,0

Hei

ght (

m) SAP 2000

NF=0,56949 Hz CCM NF=0,56939 Hz


0

6

12

18

24

30

36

42

48

0,0 0,5 1,0

Hei

ght (

m)

SAP 2000NF=0,70968 Hz

CCM NF=0,71004 Hz


0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=3,20475 Hz

CCM NF=3,20768 Hz


143




0 6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=4,12314 Hz SBY NF=4,13771 Hz


0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=8,67117 Hz

CCM DF=8,70211 Hz

Mod 6 in Teta Direction

0

6

12

18

24

30

36

42

48

-1,5 -0,5 0,5 1,5

Hei

ght (

m)

SAP 2000 NF=11,36269 Hz

CCM NF=11,45358 Hz


144



a) Mod 7 on X Direction

0

6

12

18

24

30

36

42

48

-1,5 -0,5 0,5 1,5

Hei

ght (

m)

SAP 2000 NF=16,73207 Hz

CCM DF=16,85750 Hz

a) Mod 8 on Y Direction

0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=21,99550 Hz

CCM NF=22,31475 Hz


145

Example 14:








direction in the plane of the connection beam as in Fig. 6.64 and the sine pulse force



Example 14

H =

48

m

h =

3 m

X

Y Z

P(t)


146

Figure 6.65. Sine pulse force in Example 14






SAP2000


(CCM) % difference

0.025920 0.026004 0.32



SAP2000


(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.



P(lb)

t (s)

150

5


147

-0,030

-0,020

-0,010

0,000

0,010

0,020

0,030

0 2 4 6 8 10 12

Time (s)

Top

Dis

plac

emen

t (m

)




-0,030

-0,020

-0,010

0,000

0,010

0,020

0,030

0 2 4 6 8 10 12

Time (s)

Top

Dis

plac

emen

t (m

)





148

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=2.85×106 kN/m2 and G=1055556 kN/m2, respectively.





(0.125; 0.208).

X

Y

3 m

2 m 2 m 1 m 1 m 3 m

2 m

O

(-1, 0)

(-5, 0)

(-3, -2)

(-3, 0)

(0, 0)

(1, 0)

(4, 3)

(4, 0)

1 2

3

2

1 1

4

2

3

0.3 m 0.3 m

1

2

0.3 m

0.3 m

3

G (0,125, 0,208)


149





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.


150

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) SAP2000


1 0,69367 0,69248 0,17 2 1,65427 1,64679 0,45 3 4,14919 4,14449 0,11 4 6,88659 6,86626 0,30 5 11,36339 11,33655 0,24 6 16,69150 16,62251 0,42 7 21,86704 21,76464 0,47 8 30,88525 30,65600 0,75 9 35,45502 35,16467 0,83 10 49,27674 48,65858 1,27 11 51,41271 50,74482 1,32 12 67,68475 66,43667 1,88 13 70,87070 69,48448 2,00 14 80,19235 78,38068 2,31 15 92,87637 90,32802 2,82 16 109,78064 106,11409 3,46


151

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 Present Study

(CCM) SAP2000


1 0,68010 0,67889 0,18 2 2,04419 2,02707 0,85 3 4,04453 4,04126 0,08 4 7,48094 7,44742 0,45 5 11,24119 11,23489 0,06 6 20,61359 20,44814 0,81 7 21,54959 21,51302 0,17 8 32,47404 32,29070 0,57 9 35,04346 35,03757 0,02 10 50,28789 50,16020 0,26 11 51,83913 51,47142 0,71 12 67,01399 66,36363 0,98 13 71,54452 70,78988 1,07 14 79,22520 77,74788 1,90 15 94,90283 92,79546 2,27 16 108,83989 105,56731 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.



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.


152


directions found by both the present program and SAP2000 for stiffened case in Example 15


0

6

12

18

24

30

36

42

48

-2,0 -1,0 0,0 1,0 2,0

Hei

ght (

m)

SAP 2000 NF=4,04126 Hz

CCM NF=4,04454 Hz


0

6

12

18

24

30

36

42

48

0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=0,67889 Hz

CCM NF=0,68010 Hz


0

6

12

18

24

30

36

42

48

0,0 0,5 1,0

Hei

ght (

m)

SAP 2000NF=2,02706 Hz

CCM NF=2,04419 Hz


153


found by both the present program and SAP2000 for stiffened case in Example 15


0 6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght

(m)

SAP 2000 NF=7,44742 Hz CCM NF=7,48094 Hz


0

6

12

18

24

30

36

42

48

-1,5 -0,5 0,5 1,5

Hei

ght

(m)

SAP 2000 NF=20,44814 Hz

CCM NF=20,61359 Hz


0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght

(m)

SAP 2000 NF=11,23488 Hz

CCM DF=11,24119 Hz


154

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


0

6

12

18

24

30

36

42

48

-1,5 -0,5 0,5 1,5

Hei

ght (

m)

SAP 2000 NF=21,51302 Hz

CCM DF=21,54959 Hz


0

6

12

18

24

30

36

42

48

-1,5 -0,5 0,5 1,5

Hei

ght (

m)

SAP 2000 NF=32,29070 Hz

CCM NF=32,47404 Hz


155


directions found by the present program (CCM) for stiffened and unstiffened cases in Example 15


0

6

12

18

24

30

36

42

48

0,0 0,5 1,0

Hei

ght (

m)

Unstiffened case NF=0,69367 Hz

Stiffened case NF=0,68010 Hz


0 6

12

18

24

30

36

42

48

0,0 0,5 1,0

Hei

ght (

m)

Unstiffened caseNF=1,65427 Hz



0

6

12

18

24

30

36

42

48

-2,0 -1,0 0,0 1,0 2,0

Hei

ght (

m)

Unstiffened caseNF=4,14919 Hz Stiffened case NF=4,04453 Hz


156


found by the present program (CCM) for stiffened and unstiffened cases in Example 15


0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

Unstiffened case NF=6,88659 Hz Stiffened case NF=7,48094 Hz


0 6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)


Stiffened case DF=11,24119 Hz


0

6

12

18

24

30

36

42

48

-1,5 -0,5 0,5 1,5

Hei

ght (

m)




157




0

6

12

18

24

30

36

42

48

-1,5 -0,5 0,5 1,5

Hei

ght (

m)


Stiffened case DF=21,54959 Hz


0

6

12

18

24

30

36

42

48

-1,5 -0,5 0,5 1,5

Hei

ght (

m)




158

Example 16:



and material properties of the eight storey shear wall were taken as in Example 15.








Example 16

X

Y

3 m

2 m 2 m 1 m 1 m 3 m

2 m

O

(-1, 0)

(-5, 0)

(-3, -2)

(-3, 0)

(0, 0)

(1, 0)

(4, 3)

(4, 0) 1 2

3

2

1 1

4

2

3

0.3 m 0.3 m

1

2

0.3 m

0.3 m

3 P(t)

G (0,125, 0,208)


159

100





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

Unstiffened case SAP2000


(CCM) % difference

Undamped 0.034650 0.034714 0.18

Damped 0.028530 0.028194 1.17

Table 6.27. Maximum displacement (m) in the X direction of point G for stiffened case in Example 16

Stiffened case SAP2000


(CCM) % difference

Undamped 0.021600 0.021297 1.43

Damped 0.017840 0.017650 1.06


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.

P(lb)

t(s)

5


160

-0,030

-0,020

-0,010

0,000

0,010

0,020

0,030

0,040

0 2 4 6 8 10

Time (s)

Top

Dis

plac

emen

t (m

) Unstiffened caseStiffened case



-0,020

-0,010

0,000

0,010

0,020

0,030

0,040

0 2 4 6 8 10

Time (s)

Top

Dis

plac

emen

t (m

)

Unstiffened caseStiffened case




161

Example 17:



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= 2.85×106 kN/m2 and G = 1055556 kN/m2, respectively.





(-0.545; 2.091).

1

0.4 m

1

0.4 m

4.0 m

O

Y

(-1.5, 0) (0, 0)

(-5.5, 0) (1.5, 0) (4.5, 0)

(-5.5, 4) (-2.5, 4)

(-2.5, 3)

(4.5, 5)

(2.5, 5)

X

1.5 m 1.5 m 3.0 m

5.0

m

4.0

m

3.0 m 2.0 m

1.0 m

2

3 4

5

2

3 4

1 2

3 4

1

2

3

0.4 m

0.4 m 0.4 m

0.4 m

G (-0,545, 2,091)


162





Tables 6.28-29 compare the first twenty natural frequencies corresponding to


structural analysis program for unstiffened and stiffened cases, expressing the

percentage differences.


163


Unstiffened case

Mode

Present Study (CCM)

SAP2000 (Frame Method) % difference


1 0,43177 0,42973 0,48 2 0,49023 0,48973 0,10 3 2,04893 2,05602 0,35 4 3,04620 3,03307 0,43 5 5,25108 5,25612 0,10 6 8,49342 8,40696 1,03 7 9,94970 9,91554 0,35 8 16,18170 16,05605 0,78 9 16,57520 16,26648 1,90

10 23,91606 23,61614 1,27 11 27,28633 26,47472 3,07 12 33,12682 32,52422 1,85 13 40,58494 38,82465 4,53 14 43,76342 42,67564 2,55 15 55,74229 53,05564 5,06 16 56,41818 53,92346 4,63 17 68,90739 66,04236 4,34 18 74,70329 68,85061 8,50 19 82,98062 78,70490 5,43 20 95,29495 85,81492 11,05


164


Stiffened case

Mode

Present Study (CCM)



1 0,47361 0,47228 0,28 2 0,50691 0,50174 1,03 3 2,15597 2,15674 0,04 4 2,98694 2,97419 0,43 5 6,21172 6,34009 2,03 6 8,39412 8,31093 1,00 7 10,24086 10,24592 0,05 8 16,27233 15,97689 1,85 9 16,67436 16,59881 0,46

10 24,09009 23,86781 0,93 11 26,98125 26,20980 2,94 12 34,38680 34,13438 0,74 13 39,88931 38,23439 4,33 14 43,62386 42,76440 2,01 15 55,81717 52,65419 6,01 16 55,86423 54,37579 2,74 17 68,41781 66,08382 3,53 18 73,46209 67,98693 8,05 19 83,48164 79,98379 4,37 20 94,28620 85,37953 10,43



system were found.



Figs. 6.82-84 present the mode shapes of the shear wall for stiffened case



cases.


165




0

6

12

18

24

30

36

42

48

0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=0,47228 Hz

CCM NF=0,47361 Hz


0

6

12

18

24

30

36

42

48

0,0 0,5 1,0

Hei

ght(m

)

SAP 2000NF=0,50174 Hz

CCM NF=0,50691 Hz


0

6 12

18

24

30

36

42

48

-0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=2,15674 Hz

CCM NF=2,15597 Hz


166




0 6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=2,97419 Hz SBY NF=2,98694 Hz


0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=6,34009 Hz

CCM NF=6,21172 Hz


0

6

12

18

24

30

36

42

48

-1,5 -0,5 0,5 1,5

Hei

ght (

m)

SAP 2000 NF=8,31093 Hz CCM NF=8,39412 Hz


167




0

6

12

18

24

30

36

42

48

-1,5 -0,5 0,5 1,5

Hei

ght (

m)

SAP 2000 NF=10,24592 Hz

CCM DF=10,24086 Hz


0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=15,97689 Hz

CCM NF=16,27233 Hz


168




0

6

12

18

24

30

36

42

48

0,0 0,5 1,0

Hei

ght (

m)




0 6

12

18

24

30

36

42

48

0,0 0,5 1,0

Hei

ght(m

)




0

6

12

18

24

30

36

42

48

-0,5 0,0 0,5 1,0

Hei

ght (

m)



169




0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)



0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)



0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)



170




0

6

12

18

24

30

36

42

48

-1,5 -0,5 0,5 1,5

Hei

ght (

m)




0

6

12

18

24

30

36

42

48

-1,5 -0,5 0,5 1,5

Hei

ght (

m)




171

Example 18:


symmetrical 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.





Example 18

1

0.4 m

1

0.4 m

4.0 m

O

Y

(-1.5, 0) (0, 0)

(-5.5, 0) (1.5, 0) (4.5, 0)

(-5.5, 4) (-2.5, 4)

(-2.5, 3)

(4.5, 5)

(2.5, 5)

X

1.5 m 1.5 m 3.0 m

5.0

m

4.0

m

3.0 m 2.0 m

1.0 m

2

3 4

5

2

3 4

1 2

3 4

1

2

3

0.4 m

0.4 m 0.4 m

0.4 m

P(t)

G (-0,545, 2,091)


172

11

100





in Tables 6.30-31 for both unstiffened and stiffened cases. Damping ratio was chosen

5 % for this example.




(CCM) % difference

Undamped 0.083100 0.083607 0.61

Damped 0.076500 0.076998 0.65




(CCM) % difference

Undamped 0.054360 0.053835 0.97

Damped 0.050310 0.049506 1.60


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.

P(lb)

t(s)


173

-0,080

-0,060

-0,040

-0,020

0,000

0,020

0,040

0,060

0,080

0,100

0 2 4 6 8 10 12 14 16 18 20 22

Time (s)

Top

Dis

plac

emen

t (m

)




-0,060

-0,040

-0,020

0,000

0,020

0,040

0,060

0,080

0,100

0 2 4 6 8 10 12 14 16 18 20 22

Time (s)

Top

Dis

plac

emen

t (m

)





174

Example 19:



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





4 m

3

m

2 m 2 m 2 m 3 m 2 m

4 m 2 m 2 m 3 m

O 1

(-1, 0)

2

3 4 4 11

10

8

7 5 6 6

2

3

5

9

(-3, 0)

(-3, 4) (-1, 4)

(-3, -3) (-1, -3) (-7, -3)

(-7, 0) (-5, 0)

(-5, 4)

(-7, 4)

(4, 0)

(4, 4)

(1, 4)

(4, -3) (1, -3)

0.2 m

1

2

3

5

1

3

5

8

10

6

0.2 m

0.2 m

0.2 m

4

9

7

2

4

Y

X

0.2 m 0.4 m

0.4 m

0.4 m

0.4 m

0.4 m

0.4 m

0.4 m

0.4 m

0.4 m

0.4 m

1 (1, 0)

G (-0.795, 0.322)


175




(-0.795; 0.322).






176

Tables 6.32-33 compare the first twenty natural frequencies corresponding to

each mode found by the program prepared in the present work, and the SAP2000



Unstiffened case

Mode

Present Study (CCM)


% difference


1 0,50138 0,50047 0,18 2 0,66135 0,66117 0,03 3 2,47098 2,46396 0,29 4 4,12617 4,11839 0,19 5 6,51577 6,47507 0,63 6 11,50762 11,45595 0,45 7 12,51956 12,37532 1,17 8 20,48279 20,10512 1,88 9 22,45812 22,27115 0,84

10 30,36957 29,54952 2,78 11 36,96983 36,47405 1,36 12 42,14066 40,57179 3,87 13 54,98454 52,99551 3,75 14 55,73259 53,89331 3,41 15 71,03691 66,59437 6,67 16 76,42915 74,30635 2,86 17 87,85518 81,06535 8,38 18 101,19019 95,99595 5,41 19 105,83248 97,41872 8,64 20 124,366150 110,846184 12,20


177


Stiffened case

Mode

Present Study (CCM)



1 0,62562 0,62284 0,45 2 0,66052 0,66030 0,03 3 2,60238 2,58037 0,85 4 4,11919 4,11147 0,19 5 6,63493 6,57733 0,88 6 11,48704 11,43606 0,45 7 12,96880 12,77066 1,55 8 20,42581 20,05863 1,83 9 22,44974 22,26720 0,82 10 30,66376 29,81654 2,84 11 36,84898 36,37067 1,32 12 42,27349 40,73950 3,77 13 54,90843 52,91929 3,76 14 55,57030 53,85963 3,18 15 71,31118 66,97627 6,47 16 76,23393 74,21348 2,72 17 87,42965 80,90873 8,06 18 100,76738 96,05297 4,91 19 105,59907 97,18234 8,66 20 124,11197 110,98024 11,83



system were found.



Figs 6.94-96 present the mode shapes of the shear wall for stiffened case



cases.


178




0

6

12

18

24

30

36

42

48

0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=0,62284 Hz

CCM NF=0,62562 Hz


0

6

12

18

24

30

36

42

48

0,0 0,5 1,0

Hei

ght (

m)

SAP 2000NF=0,66030 Hz

CCM NF=0,66052 Hz


0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=2,58037 Hz

CCM NF=2,60238 Hz


179




0 6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=4,11147 Hz SBY NF=4,11919 Hz


0

6

12

18

24

30

36

42

48

-1,5 -0,5 0,5 1,5

Hei

ght (

m)

SAP 2000 NF=11,43606 Hz

CCM NF=11,48704 Hz


0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=6,57733 Hz

CCM NF=6,63493 Hz


180




0

6

12

18

24

30

36

42

48

-1,5 -0,5 0,5 1,5

Hei

ght (

m)

SAP 2000 NF=12,77066 Hz

CCM NF=12,96880 Hz


0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=20,05863 Hz

CCM NF=20,42581 Hz


181




0

6

12

18

24

30

36

42

48

0,0 0,5 1,0

Hei

ght (

m)




0 6

12

18

24

30

36

42

48

0,0 0,5 1,0

Hei

ght(m

)




0

6 12

18

24

30

36

42

48

-0,5 0,0 0,5 1,0

Hei

ght (

m)



182




0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)



0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)



0

6

12

18

24

30

36

42

48

-1,5 -0,5 0,5 1,5

Hei

ght (

m)




183




0

6

12

18

24

30

36

42

48

-1,5 -0,5 0,5 1,5

Hei

ght (

m)




0

6

12

18

24

30

36

42

48

-1,5 -0,5 0,5 1,5

Hei

ght (

m)




184

Example 20:


symmetrical 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





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.


Example 20

4 m

3

m

2 m 2 m 2 m 3 m 2 m

4 m 2 m 2 m 3 m

O 1

(-1, 0)

2

3 4 4 11

10

8

7 5 6 6

2

3

5

9

(-3, 0)

(-3, 4) (-1, 4)

(-3, -3) (-1, -3) (-7, -3)

(-7, 0) (-5, 0)

(-5, 4)

(-7, 4)

(4, 0)

(4, 4)

(1, 4)

(4, -3) (1, -3)

0.2 m

1

2

3

5

1

3

5

8

10

6

0.2 m

0.2 m

0.2 m

4

9

7

2

4

Y

X

0.2 m 0.4 m

0.4 m

0.4 m

0.4 m

0.4 m

0.4 m

0.4 m

0.4 m

0.4 m

0.4 m

P(t) 1

(1, 0)

G (-0.795, 0.322)


185





in Tables 6.34-35 for both damped and undamped cases. Damping ratio was chosen 5

% for this example.




(CCM) % difference

Undamped 0.035710 0.035658 0.14

Damped 0.035270 0.035192 0.22




(CCM) % difference

Undamped 0.024550 0.023820 3.06

Damped 0.023570 0.022860 3.01




P(lb)

t (s)

5 2,5

150


186

-0,020

-0,010

0,000

0,010

0,020

0,030

0,040

0 1 2 3 4 5 6 7 8 9 10

Time (s)

Top

Dis

plac

emen

t (m

)




-0,010

0,000

0,010

0,020

0,030

0,040

0 2 4 6 8 10

Time (s)

Top

Dis

plac

emen

t (m

)





187

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


shear moduli of the structure are E = 2.85×106 kN/m2 and G = 1055556 kN/m2,

respectively.



sectional area of the structure. The mass center was located on the point O.

Figure 6.104. Cross-sectional view of the 1st region of the structure in Example 21

O X

Y

2 m

2

3

(0,0) (-1, 0)

(-3, 0)

(-3, -2)

(3, 2)

(1, 0)

(3, 0)

2

1 1

0.3 m

0.3 m 0.3 m

0.3 m

2 m

1 m 1 m 2 m

2

1 1

2

3

2 m


188

Figure 6.105. Cross-sectional view of the 2nd region of the structure in Example 21


2 m

Y

2 m

O X

2 m

2

3

(0,0) (-1, 0)

(-3, 0)

(-3, 2) (3, 2)

(1, 0)

(3, 0)

2

1 1

0.3 m

0.3 m 0.3 m

0.3 m

(-3, -2) (3, -2)

3 3

0.3 m 0.3 m

1 m 1 m 2 m

2

4

1 1

2

3

4


189


Mode

Present Study (CCM)


% difference


1 1,07270 1,07366 0,09 2 1,61807 1,60347 0,91 3 4,01622 4,01831 0,05 4 5,88496 5,85470 0,52 5 10,68451 10,63805 0,44 6 14,19554 14,14731 0,34 7 19,27886 19,19478 0,44 8 25,67233 25,44843 0,88 9 31,10287 30,91459 0,61 10 38,98559 38,45470 1,38 11 46,79324 46,35447 0,95 12 53,47172 52,34941 2,14 13 65,79977 64,26048 2,40 14 69,71574 67,71944 2,95 15 91,63290 88,45925 3,59 16 107,58726 103,86768 3,58





Figs. 6.107-109 present the mode shapes of the shear wall, found by the

present program and the SAP2000 structural analysis program, both in the same

figure.


190




0

6

12

18

24

30

36

42

48

0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=1,07366 Hz

CCM NF=1,07270 Hz


0

6

12

18

24

30

36

42

48

0,0 0,5 1,0

Hei

ght (

m)

SAP 2000NF=1,60347 Hz

CCM NF=1,61807 Hz


0

6 12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=4,01831 Hz CCM NF=4,01622 Hz


191




0 6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=5,85470 Hz CCM NF=5,88496 Hz


0

6

12

18

24

30

36

42

48

-1,0 -0,5 0,0 0,5 1,0

Hei

ght (

m)

SAP 2000 NF=10,63805 Hz

CCM DF=10,68451 Hz


0

6

12

18

24

30

36

42

48

-1,5 -0,5 0,5 1,5

Hei

ght (

m)

SAP 2000 NF=14,14731 Hz

CCM NF=14,19554 Hz


192




0

6

12

18

24

30

36

42

48

-1,5 -0,5 0,5 1,5

Hei

ght (

m)

SAP 2000 NF=19,19478 Hz

CCM DF=19,27886 Hz


0

6

12

18

24

30

36

42

48

-1,5 -0,5 0,5 1,5

Hei

ght (

m)

SAP 2000 NF=25,44843 Hz

CCM NF=25,67233 Hz


193

Example 22:

In this example, the forced vibration analysis of the change in the cross-

section 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 22

Y

P(t)

X O

H =

z1 =

24

m

z 2 =

12

m

Z

1st region

2nd region

h 2 =

3 m

h 1

= 3

m


194

150



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 6.37-

38 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


(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


(CCM) % difference

0.036130 0.035939 0.53


were determined and the time-varying displacements in the X direction of the point

O are presented in Figs. 6.112-113.


with those of SAP2000 structural analysis program perfectly.

P(lb)

t(s)

6


195

-0,060

-0,040

-0,020

0,000

0,020

0,040

0,060

0,080

0 2 4 6 8 10 12

Time (s)

Top

Dis

plac

emen

t (m

)Present study

SAP2000



-0,040

-0,020

0,000

0,020

0,040

0,060

0,080

0 2 4 6 8 10 12

Time (s)

Top

Dis

plac

emen

t (m

) Present study

SAP2000



7. CONCLUSIONS Cevher Deha TÜRKÖZER

196

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 Vlasov’s 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 time-

varying 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


197

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 time-

varying 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


program. In the free vibration analyses, the natural frequencies and mode shape

vectors, thus obtained, are compared. In the forced vibration analyses, the time-

varying 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 non-

planar 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.


198

In the nineteenth and twentieth examples, a stiffening beam of 3.0 m height is

placed at the height of 30 m on the tenth storey. In the nineteenth example, free

vibration analysis is carried out. Then, the natural frequencies and mode shape

vectors are obtained for the stiffened and unstiffened cases. The results obtained are

compared with those of SAP2000 and a good agreement is observed. In the twentieth

example, the forced vibration analysis is carried out and the time-varying

displacements have been plotted in Figs. 6.101-102 for the stiffened and unstiffened

cases.

The computer program prepared in this study has been made such that

changes in the cross-section of the coupled shear wall along the height can also be

taken into account. In the twenty-first and twenty-second examples, the coupled

shear wall considered in the third example is solved once more, with the stories

above the fourth being of a different cross-section than the ones below. The results

obtained are compared with those of SAP2000 and a perfect match is observed.

The method proposed in this thesis has two main advantages, which are that

the data preparation is much easier compared to the equivalent frame method for

non-planar coupled shear walls and the computation time needed is much shorter

compared to other methods. Hence, the method presented in this thesis is very useful

for predesign and dimensioning purposes while determining the geometry of non-

planar coupled shear wall structures.

199

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203

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 Çukurova

University and graduated in 2004. Then, I started my PhD. study as a student of my

supervisor Prof. Dr. Orhan Aksoğan at the Çukurova University in 2004. I have, also,

been working as a Control Engineer in General Directorate of Highways in

5.Division Directorate since 2005.

204

APPENDICES

205

APPENDIX 1. TORSIONAL BEHAVIOUR AND THEORY OF OPEN

SECTION THIN- WALLED BEAMS

A1.1. Introduction

The behaviour of a one dimensional structure can be investigated for two

cases, torsion and flexure, as stated in Chapter 1. The bending analysis of a beam is

rather simple. However, its torsional analysis is rather complex and needs to be

studied in detail.

The elementary theory of extension or compression, flexure and pure torsion

is included in the general engineering theory of solid beams, which is based on the

application of St.Venant’s principle. St.Venant’s theory is based on the hypothesis

that the twisting of any cross-section will produce only shear stresses. St.Venant’s

theory is suitable only for thick walled beams. The theory of flexure is based on the

assumption of plane sections remaining plane during bending, usually referred to as

the Bernoulli-Navier hypothesis.

The high-rise buildings with wide spans and reduced self weight encouraged

the application of thin-walled beams in design. It is known that, the behaviour of a

thin-walled beam in bending and axial loading is very similar to that of a solid beam.

However, under torsional loading, relative axial displacement of the beam

complicates the behaviour. Thin-walled beams must be given special consideration in

their analysis and design. The main feature of thin-walled beams is that they can

undergo longitudinal extension as a result of torsion. Consequently, longitudinal

normal stresses proportional to strains are created, which lead to an internal

equilibrium of the longitudinal forces in each cross-section. These stresses, which

arise as a result of the relative warping of the section and which are not examined in

the theory of pure torsion, can attain very large values in thin-walled beams. When a

thin-walled beam with an open section is subjected to simultaneous bending and

torsion, it is necessary to calculate some new geometrical properties of the cross-

section called sectorial properties. The main principles of the theory of thin-walled

206

beams were developed by Vlasov (1961) and, therefore, it is generally called

Vlasov’s theory.

Warping is defined as the out-of-plane distortion of the cross-section of a

beam in the direction of the longitudinal axis, which violates the Bernoulli-Navier

hypothesis. Structural engineers, generally, are not familiar with the concept of

warping behaviour and its methods of analysis. Engineers should be able to

appreciate whether an open section is liable to twist and warp. The aim of this

chapter is to establish an understanding of the influence of certain structural

parameters related to warping.

A1.2. Cross-Sectional Properties of an Open Section Thin-Walled Beam

A1.2.1. Dimensional Properties

A thin-walled member is one of the basic elements in structural engineering

which is made from thin plates joined along their edges. Thin-walled structures are

the most modern and optimal structures designed for minimum weight and maximum

stiffness. They are used extensively in long span bridges and other structures where

weight and cost are prime considerations. They are not always made of steel. For

example, a box girder can be constructed as a reinforced concrete structure, known as

thin-walled tube. To classify a given structure (Fig. A1.1) as thin-walled beam, the

proportions must satisfy the relations:

Figure A1.1. Geometrical dimensions of a thin walled beam

1.0L

c2b,1.0c2b

t≤

+≤

+ (A1.1)

b

t L c

207

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 Vlasov’s 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 zu and su .

Figure A1.2. General form of an open tube

Contour line

Generator su

zu

O

z

s

P(z,s)

208

Figure A1.3. General form of an open bar

A1.2.3. Torsion Constant

St. Venant’s theory is based on the hypothesis that the twisting of a shaft of

any cross-section will produce only shear stresses. This theory will be explained in

detail in Section (A1.4). When an open section is subjected to a torque, it twists,

producing shear stresses of the type as shown in Fig. A1.4. The stresses are

distributed linearly across the thickness of the wall, acting in opposite directions. As

the effective lever arm of these stresses is equal to only two-thirds of the wall

thickness, the torsional resistance of these stresses is very low for an open section.

The torsion constant for this twisting action can be written as

∑=

=n

1k

3kk tb

31J (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.

Contour line

s

z

O

209

Figure A1.4. Stress distribution in a twisted open section beam

A1.2.4. Sectorial Area

In addition to the familiar flexural properties of cross-sections, such as

centroid, static moments, moments of inertia, etc., there are some other properties

which are used in the analysis of thin-walled beams. For this purpose, a new

coordinate called the sectorial area is introduced. The diagram of an element of the

sectorial area of the cross-section of a beam is shown in Fig. A1.5.

Figure A1.5 Computation of sectorial area

Mt τ

τ

2d

2hds

dA ss ω=

⋅=

R

A A′

h s

O

ds

s

P(sp)

Tangent Line

210

The sectorial area of a point on a section is computed using the median line of

the outline of the section. sω 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. It’s mathematical definition is:

dshps

0ss ∫=ω (A1.3)

where

sω : sectorial area,

sh : distance from point R to the line tangent to the contour line at point A,

s : coordinate of a point on the contour line.

The parameter sω 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 it’s 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, →

RA , rotates in the counter-clockwise direction (as in Fig. A1.5) and negative if

clockwise.

A1.2.5. Shear Center

A1.2.5.1 Introduction

There is a special point in the cross-section of a thin walled beam, called the

shear center of the cross-section, S, which is defined to be the point where a shear

force, when applied to the cross-section, will produce bending only (no twisting).

211

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).

Figure A1.6. Shear centers of typical cross-sectional shapes due to symmetry

Shear center is also the point, to which warping properties of a section are

related, in the way that the bending properties of a section are related to the centroid

(through which the neutral axis passes). If the cross-section contains no axis of

symmetry or only one axis of symmetry, the determination of the exact location of

shear center requires a more detailed analysis.

A1.2.5.2. Determination of the Shear Center Using Sectorial Area

In the calculation of the shear and bending stresses in a section, linear

coordinates (x,y) of a section are used for finding the five flexural properties which

are Sx, Sy, Ixy, Ix and Iy. The position of the origin and the direction of the principal

axes are defined by the conditions, Sx=Sy=0 and Ixy=0. The first two of these are used

GS GS GS GS

212

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:

dshtdshtM2

1

s

sss

AD ∫∫ τ=τ= (A1.4)

Since dshd ss =ω

sA

D dtM ωτ= ∫ (A1.5)

If τ is eliminated using

tI

SVtI

SV

y

xy

y

yx +=τ (A1.6)

Figure A1.7. Determination of the shear center

s2

y

x D

s1

τ ds

hs

t

213

where

Vx and Vy : components of the shear force along the principal x, y axes,

Sx and Sy : statical moment of the cross-section with respect to the principal x, y

axes,

yx I,I : moment of inertia of the cross-section with respect to the principal x, y

axes,

t : thickness of the cross-section where the shearing stress is to be

determined.

Substituting equation (A1.6) into equation (A1.4), moment about point D is found as

follows:

dAdAd

SIVdA

dAd

SIV

M s

Ay

y

xs

Ax

x

yD

ω+

ω= ∫∫ (A1.7)

Using integration by parts technique, the first integral in (A1.7) can be evaluated in

the following form:

dAdAdSSdA

dAdS s

A

xs

ssxs

Ax

2

1ω−ω=

ω∫∫ (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

∫=A

x dAyS (A1.9)

and Sx is zero for points 1 and 2. Hence,

2

1

s

ssxS ω = 0 (A1.10)

214

and

ydAdSx = (A1.11)

Thus, the integral under consideration becomes

dAydAdAd

SA

ss

Ax ∫∫ ω=

ω (A1.12)

The second integral appearing in equation (A1.7) is transformed in a similar way.

Finally, the moment of the shear stresses about point D is found as

dAxIVdAy

IV

MA

sy

x

As

x

yD ∫∫ ω−ω−= (A1.13)

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

dAySA

sx ∫ ω=ω (A1.14)

and

dAxSA

sy ∫ ω=ω (A1.15)

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),

215

0dASA

s =ω= ∫ω (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):

x

xx

As

xxx I

SbydAI1ba ω+=ω+= ∫ (A1.17)

y

yy

As

yyy I

SbxdA

I1ba ω−=ω−= ∫ (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).

Figure A1.8. Arbitrarily placed pole R and principal pole S

y

S(ax,ay)

R(bx,by)

x G

216

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:

2xyyx

yxyxyxx III

SISIba

−−

+= ωω (A1.19)

2xyyx

xxyyxyy III

SISIba

−−

−= ωω (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

dAIA

2s∫ω=ω (A1.21)

Similar to flexural rigidity EI, ωEI is the warping rigidity of a section. The integral

expression in (A1.21) shows up in section A1.5 in Vlasov’s 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.

217

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. Venant’s theory is perfectly valid for a

circular cross-section and remains practically valid for any thick walled beam.

S

S

S

Figure A1.9. Some star sections for which 0Is ==ω ω

Figure A1.10. Twisting of a circular shaft (no warping)

z

x

zθ′

L

γ

218

The twisting of a circular shaft does not produce any longitudinal stresses, but

only shear stresses. St.Venant’s 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 γ L where γ is the shear strain.

A1.3.2. Flexural Twist and Bimoment

The same beam is shown with one end rigidly fixed (restrained) in Fig.

A1.12. The warping of the bottom surface of this beam is restrained. When a thin-

walled beam is constrained against warping, a distribution of longitudinal stresses

develop to eliminate the warping displacements at the constrained section. These

normal stresses vary along the member. If an imaginary elemental beam is separated

as shown in Fig. A1.12(b), its deflected shape represents a cantilever deformed by

forces acting between this cantilever and the remaining portion of the beam. This

deflection will cause an internal bending moment in the cantilever, which in turn will

cause longitudinal stresses.

Figure A1.11. Twisting of a rectangular beam (warping not restrained)

(a) Elevation (b) Plan

Msv (St.Venant Twist)

θ

219

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.

Considering an I-section beam loaded by a twisting moment at its

unrestrained end as shown in Fig. A1.13(a). The final distortion of this beam takes

place as the summation of the distortions shown in Fig. A1.13(b,c). The first

distortion in Fig. A1.13(b) is due to the action of the St. Venant twist and the second

is caused by a twisting moment resulting from the bending of the rectangular

components of the beam. The second distortion caused by a bending twist (flexural

twist) or flexural torsional moment is denoted by ωM . Because of the considerable

rigidity of flanges, the twisting moments producing deflection ( ∆ ) is greater than the

component producing a rotation (angle θ). It can be concluded from Fig. A1.13 that a

twisting moment acting on a thin walled beam is composed of flexural and pure

twisting moments. Therefore, the total twisting moment acting on a thin-walled beam

can be written as follows:

(a) Elevation (b) Imaginary elemental beam

Figure A1.12. Twisting of a rectangular beam (warping restrained)

∆δ

rigidly fixed M

220

svtot MMM += ω (A1.22)

where

Mtot : Twisting moment acting on a thin-walled beam

ωM : Flexural twisting moment (warping torque)

Msv : St.Venant (pure) twisting moment

It is clear that flexural twist always causes some bending moments in a

structure. This assumption is too general since it is known that a flexural twist evokes

a pair of bending moments. Such a pair of bending moments is called a ‘Bimoment’

and represented by zB . A bimoment is a mathematical function introduced by

Vlasov(1961). The general definition of bimoment is a pair of equal but opposite

bending moments acting in two parallel planes as shown in Fig. A1.14.

(a-) Elevation (b-) due to Msv (c-) due to ωM

Figure A1.13. A twisting moment and resulting distortions

= θθ

θ

+

∆2 ∆ ∆

Figure A1.14. Representation of a bimoment

P

P

h

= P.e=-M

M P

P

e h

Bz

221

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.

hMBz = (A1.23)

As mentioned earlier, a flexural twist acting on a thin-walled beam is

accompanied by a bimoment. An example of bimoment caused by a twisting moment

is shown in Fig. A1.13(a). The flexural twist at the top of this column produces

bending moments in both flanges. These bending moments are equal and opposite in

the opposite flanges.

Now let us consider an I-shaped shear wall which is loaded at one corner by a

load P as an external force parallel to the longitudinal axis as shown in Fig. A1.15. In

the cases of the first three loadings on the right side, which represent the axial

loading and the bending moments about x and y axes, the Bernoulli-Navier

hypothesis is valid. In the last loading case, it is obvious that the cross-section does

not remain plane. The flanges which are bending in opposite directions, produce

significant warping stresses as shown in Fig. A1.15(e).

As defined by St. Venant, uniform torsion component of shear stresses is not

dependent on s, the coordinate variable along the contour line. Warping stresses,

however, vary with s, since they result from a non-uniform distribution of axial

stresses over the contour.

222

+

Figure A1.15. A thin-walled I-shaped cantilever column subjected to an eccentric

load

The resistance offered by the interconnecting web is not strong enough. The

bending action of the flanges can be thought of as being brought about by equal and

opposite horizontal forces parallel to the flanges. The compatibility condition

between the web and flanges results in a rotation of the cross-section as shown in

Fig. A1.16.

+

2P

2P2

P 2P

2P

2P

2P2

P

+

(a) Vertical load at corner

(b) Axial loads (c) Bending about x-axis

(d) Bending about y-axis (e) Bimoment load

P

2P

2P

2P

2P 2

P 2P

2P

2P

=

Warping stresses

223

In Fig. A1.17, the two external forces P are producing equal and opposite

moments acting on the opposite flanges of a beam. The deformation of this column is

practically same as the deformation caused by the twisting moment previously shown

in Fig. A1.13(a). In this case, there is no external twist. Therefore, total twisting

moment acting on a thin-walled beam is equal to zero (Mtot = ωM + Msv = 0) or ( ωM

= -Msv ≠ 0). Although there is no external twist, shearing stresses exist.

Hence, an internal flexural twist is created by an external bimoment. The

relation between bimoment and flexural twist will be explained in detail in section

A1.5 in which the Vlasov’s theorems will be given. These theorems enable us to

e

P(external load)

P -M=P.e

M=P.e h

2P

± Bz

Figure A1.17. Bimoment creating a twist

θ

(a) Displacement of flanges due to bimoment load

(b) Rotation with geometric compatibility between flanges and web

Figure A1.16. Plan section of an I-shaped column

2 ∆

=

2 ∆

224

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. Venant’s 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. Venant’s 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 ρ (radial distance), rotates through a small angle zθ about

S, which is called the center of twist and its displacement in the horizontal plane is

zero. Since the angle of rotation zθ 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=- ρ zθ sinβ = - y zθ

uy= ρ zθ cosβ = x zθ

(A1.24a)

225

If normal cross-sections remain plane after deformation, one might assume

that the deformation in the longitudinal direction is zero. If the cross-section is at a

distance z from the origin, the angle of rotation is given by ( zθ = 'zθ z), where '

zθ is

the angle of twist per unit length along the z-direction. Then, the displacements are

uz = 0

ux = - 'zθ y z (A1.24b)

uy = 'zθ x z

These displacements produce only shear strains in the bar so that by generalized

Hooke’s law the only non-zero stresses in the bar are the shear stresses, i.e.,

0,0,0 yxz =ε=ε=ε

0,x,y xy'zyz

'zxz =γθ=γθ−=γ

yGG 'zxzxz θ−=γ=τ

xGG 'zyzyz θ=γ=τ

From Fig. A1.18, these shear stresses can be used to relate the internal torque, Msv, to

the twist per unit length as follows:

(A1.25)

Figure A1.18. Shear stresses in uniform torsion

τyz

S x

ρ y

τxz P

β

τ

zθ

P*

226

Msv= ( )dAyxA

xzyz∫ τ−τ

Msv= ( )dAyxGA

22'z ∫ +θ

Msv= o'z JGθ (A1.26)

Jo is the polar moment of inertia of the cross-section given by

( ) dAdAyxJA

2

A

22o ∫∫ ρ=+= (A1.27)

A1.4.2. Torsion of a Bar with Non-Circular Cross-Section

In the previous section it was assumed that each planar cross-section of a

circular bar remained planar and merely rotated about the central axis of the bar.

When a torque is applied to a non-circular cross-section, the cross-section again

rotates about the central axis but there is also a significant distortion of the cross-

section in z-direction, i.e. the cross-sections both rotate and warp as shown in Fig.

A1.19(b).

Figure A1.19. (a) Torsion of a circular bar with no out-of-plane warping (b) Torsion of a non-circular bar with warping

227

For a circular cross-section, the displacements are assumed to be due to a pure

rotation of each cross-section with no displacement in the axial direction. On the

other hand, for a non-circular cross-section there are axial displacements. However,

since the warping of each cross-section is assumed to be the same, all points on a line

parallel to the z-axis move through the same distance in the z-direction. The

displacement of any point P due to rotation is shown in Fig. A1.20.

Figure A1.20. Displacements of a non-circular bar due to torsion

Incorporating the warping action into our previous description of the

displacements of the bar mentioned in equations (A1.24.a and b), the following

relations can be written:

( )y,xfu 'zz θ=

yu 'zx θ−= (A1.28)

xu 'zy θ=

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

Msv

R

x

y

P

τzy

τzx

X

Y

P*

R

y

x

r

P

ux

uy

β

θz

228

in the bar so that by generalized Hooke’s law the only non-zero stresses in the bar are

the shear stresses. Within the elastic limit, shear strain is proportional to the shear

stress

τ

=γG

.

0,0,0 yxz =ε=ε=ε

0,xyf,y

xf

xy'zyz

'zxz =γ

+

∂∂

θ=γ

−

∂∂

θ−=γ

−

∂∂

θ=τ yxfG '

zxz

+

∂∂

θ=τ xyfG '

zyz

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= ( )dAyxA

xzyz∫ τ−τ (A1.30)

Substituting the shear stress components from equation (A1.29) into (A1.30), the

following expression is found:

zsv GJM θ′= (A1.31)

where J is a torsional constant and G is the shear modulus of the material.

(A1.29)

229

A1.5. Vlasov’s Theory (Non-Uniform Torsion)

The cross-sections of shear walls used for bracing tall buildings are

frequently open and are characterized by the fact that their three dimensions, height,

width and thickness, are all of different orders of magnitude. When such a system is

subjected to torsion, it suffers warping displacements and may develop axial stresses

due to the restraint at the foundation.

When the sections are free to warp, a beam responds in uniform torsion

(St.Venant torsion). This behaviour was given in section A1.4. On the contrary, if

warping is prevented, due to the complex distribution of longitudinal stresses, shear

stresses in the cross-section can be related to two different modes of torsional

behaviour. One is due to the uniform torsional behaviour of the structure, therefore,

varying linearly over the thickness of the beam and does not change with sectorial

areas. The other is due to the warping torsion, coupled with transverse bending and

axial loading of the beam, uniform over the thickness of the beam and changes with

s.

As an example, consider an I-section bar (see Fig.A1.21). The end B is forced

to rotate and warp, whereas, the end A is a fixed rigid plane which will not let the

end section to rotate or warp. Thus, while z = 0 cross-section has no warping, z = L

cross-section can have an arbitrary warping displacement function. In between these

two ends (0<z<L), this function varies continuously. This is a non-uniform torsion.

In such a deformation zθ′ value is not constant, i.e., )z(θ function is not linear.

Figure A1.21. Non-uniform torsion

Mz

A x

y

z

rigid plane

warping deformation

B

230

An important point about non-uniform torsion is that, the variation of the

warping displacement function uz from section to section causes normal stresses to

appear in the cross-sections. The normal stress distribution at end A is shown in Fig.

A1.22.

Figure A1.22. Normal stress distribution in the cross-section of the bar

Since there are no other cross-sectional normal forces, the normal stresses due

to non-uniform warping in the cross-section must be self-balancing.

Vlasov’s theory explains the difference of behaviour between thin-walled and

thick-walled beams under the same loading. Furthermore, it not only explains but

renders it possible to calculate stresses and distortions of a thin-walled beam which

cannot be explained by the classic thick-walled beam theory. Only open section thin-

walled beams are considered and analyzed in this study. Vlasov introduced two new

types of effects which are called Flexural Twist ( ωM ) and Bimoment (Bz).

Furthermore, additional properties of a section called sectorial properties are

introduced. According to Vlasov’s theory of thin-walled beams, the following

assumptions are made:

Assumption 1 : The cross-section is completely rigid in its own plane.

Assumption 2 : The shear strain of the middle surface is negligible (s direction is

perpendicular to z direction after deformation).

z

Mz

A

+ _

_ +

(+) : Tensile stress

(−) : Compressive stress B

231

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.

The transverse displacement us in the direction of the tangent to the contour line is

given by:

szs h)s,z(u ⋅θ= (A1.32)

where

)s,z(u s : displacement of point A measured along curve s of the profile of a section

zθ : angle of rotation of the profile at a distance z from the base

The other assumption is similar to that made in the normal theory of bending

of a beam. It states that the contour of the section remains perpendicular to the

Figure A1.23. Displacement of a section in non-uniform torsion

Tangent line

y

ds

Unloaded cross-section

Position after loading

R

hs

x zθ

A

232

longitudinal axis after deformation, meaning that shear deflections are equal to zero.

Hence , according to Vlasov’s second assumption:

( ) ( ) 0z

s,zus

s,zu sz =∂

∂+

∂∂

=γ (A1.33)

Substituting equation (A1.32) into equation (A1.33),

( ) 0dzdh

ss,zu z

sz =

θ⋅+

∂∂ (A1.34)

Integrating this equation with respect to s, the longitudinal displacement of point A

along z-axis, is found as

( )dzddshs,zu z

s

0sz

θ−= ∫ (A1.35)

where ( )s,zu z is the longitudinal displacement along the generator and zθ′ 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 dωs . Hence,

dzdd)s,z(u z

s

0sz

θω−= ∫ (A1.36)

After some necessary arrangements, equation (A1.36) can be written as

sz

z dzd)s,z(u ωθ

−= (A1.37)

233

where sω is as defined in section A1.2.4. Equation (A1.37) determines the

longitudinal displacement that does not obey the law of plane sections and arises as a

result of torsion. This is the sectorial warping of the section. Warping is defined as

the out-of-plane distortion of the cross-section of a beam in the direction of the

longitudinal axis. This warping is given by the law of sectorial areas. Since the

displacement ( )s,zu z changes along the distance z, the longitudinal strain of a point

measured along z can be written as follows:

( ) ( )z

s,zus,z z

∂∂

=ε (A1.38)

( ) s''zs,z ωθ−=ε (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 Vlasov’s theory of

thin-walled beams of open section. Vlasov’s 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.

Vlasov’s 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(σ , and

shear stresses ( )s,zτ can be found. Equation (A1.39) does not determine the strain

explicitly yet, since the function zθ is still unknown. When the beam is deformed,

internal elastic forces arise in it. These forces represent normal and shear stresses in

234

the cross-section. According to Vlasov’s theory, the shear stresses, which are

directed along the normal to the contour line, are assumed to vanish.

In Vlasov’s 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, tτ , 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 ( )s,z

per unit length of the cross-section (along the contour line) as shown in Fig.

A1.24(b).

Figure A1.24. Effect of shear stresses in an open section thin-walled beam

1

2

A

τ1

τ1

t

)s,z(1 τ=τ

(a) due to flexural twist (Mω)

ds A

τ1 t ds 1

2 A

A

τ2

τ2

t

(b) due to St. Venant twist (M )

msv ds

ds

1

2 A

A

(τ1+τ2)

t

(c) due to total twist (Mtot)

(τ1-τ2)

235

The state of stress in the cross-sectional plane can be expressed by the normal

stresses )s,z(σ , the average shear stresses ( )s,zτ 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 ( svM )

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 ( )τσ , and the moments Msv from the strains.

According to Hooke’s law

)s,z(E)s,z( ε=σ (A1.40)

Substituting equation (A1.39) into equation (A1.40), the normal stress is found as

follows:

s''zE)s,z( ωθ−=σ (A1.41)

The difference between Vlasov’s and St. Venant’s 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 tsω and integrating over the whole contour line yields

( ) dstEdsts,zBs

0

2s

''zs

s

0z ∫∫ ωθ−=ωσ= (A1.42)

The integral on the right was defined as the warping moment of inertia, ωI , which

expresses the warping torsional resistance of the cross-section (see section A1.2.6).

Hence,

ωθ−= IEB ''zz (A1.43)

236

Dividing both sides by ωEI

ω

−=θEIBz''

z (A1.44)

Substituting expression (A1.44) into (A1.41), longitudinal normal stresses can be

obtained as

( ) sz

IBs,z ω=σ

ω

(A1.45)

Equation (A1.45) is the mathematical definition of the first theorem. It determines

the normal stresses which arise because the cross-sections do not remain plane under

torsion.

Vlasov’s second theorem is as follows:

Theorem 2 : A shear stress in a fibre of a thin walled beam caused by a flexural

twist is equal to the product of this flexural twist and the sectorial statical moment of

this point divided by the wall thickness (at this point) and the principal sectorial

moment of inertia.

Normal stresses described by equation (A1.45) were determined by using the

relation between stress and strain for the elastic beam. However, the analogous of

elastic equation cannot be used to determine the average shear stresses ( )s,zτ shown

in Fig. A1.24(b), since the shearing strains caused by the shear stresses are taken to

be zero in Vlasov’s theory.

To determine the shear stresses, the condition of equilibrium of the vertical

forces acting on an infinitesimal beam element with the sides dz and ds as shown in

Fig. A1.25, is set equal to zero,

237

0dsdzs

)s,z(tdsdzz

)s,z(t =∂

τ∂+

∂σ∂ (A1.46)

Hence,

0s

)s,z(z

)s,z( =∂

τ∂+∂

σ∂ (A1.47)

Differentiating the normal stress expression (A1.41) with respect to z, substituting it

into equation (A1.47) and integrating

dsE)s,z(s

0s

'''z ∫ωθ=τ (A1.48)

The value of shear stress at a point A in Fig. A1.23 can be obtained by

integrating the shear stress expression (A1.48) from a free edge (s=0) to s. Denoting

(t ds) by (dA), the shear stress is found as

Figure A1.25. Free body diagram of an element (t dz ds) of a thin-walled beam

ds

dz

t

)s,z(σ

)s,z(τ

)s,z(τ dss

)s,z()s,z(∂

τ∂+τ

dzz

)s,z()s,z(∂

σ∂+σ

238

dAEt1)s,z(

As

'''z ∫ ωθ=τ (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, ωS , of the cross-section.

Hence,

( )t

SEs,z '''z

ωθ=τ (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:

( ) ( ) s

s

0s

s

0

dts,zdshts,zM ωτ=τ= ∫∫ω (A1.51)

Substituting the stresses ( )s,zτ , found as in equation (A1.48) into (A1.51), and

remembering the definition of sectorial warping moment of inertia (see equation

(A1.21)),

zEIM θ ′′′−= ωω (A1.52)

Rearranging equation (A1.51)

ω

ω−=θEIM'''

z (A1.53)

239

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 Vlasov’s second theorem

( )

−=τ ω

ω

ω SI

Mt1s,z (A1.54)

Therefore, from relations obtained in equation (A1.44) and (A1.53),

'zBM =ω (A1.55)

The sum of the moments ( )M sv and ( ωM ) gives the total torsional moment, that is

denoted by ( )M tot . Hence,

'z

'''zsvtot GJEIMMM θ+θ−=+= ωω (A1.56)

A1.6. Internal Bimoment and Flexural Twist due to External Loading

In the previous sections, it was mentioned that it would not be simple to find

the relationship between the external loading and the internal forces which were in

our case bimoment and flexural twist. This relationship is given by a differential

equation which will be studied in the following sections. Before dealing with that

differential equation, it is useful in practice to know two theorems which will enable

us to recognize the presence of an internal bimoment.

A1.6.1. Bimoment caused by a Force Parallel to the Longitudinal Axis of a Thin-

walled Beam

The relationship between an internal bimoment and an external force is given

by the first theorem as follows:

240

Theorem 1 : A bimoment caused by an external force parallel to the longitudinal

axis of a beam is equal to the product of this force and the principal sectorial co-

ordinate of the point of its application.

This theorem can be expressed mathematically as shown below:

pz PB ω×= (A1.57)

This equation, actually, is the other mathematical definition of bimoment given in

equation (A1.42) which is

( ) dAs,zB sA

z ωσ= ∫ (A1.58)

or

k

n

1kkz PB ω×= ∑

=

(A1.59)

A1.6.2. Bimoment caused by an External Bending Moment Acting in a Plane

Parallel to the Longitudinal Axis of a Beam

The relation between an external bending moment and its internal bimoment

is defined by the second theorem as follows:

Theorem 2 : A bimoment caused by a bending moment acting in a plane parallel to

the longitudinal axis of a beam is equal to the product of the value of this bending

moment and the distance of its plane from the shear center of that beam.

This theorem can be expressed mathematically by the formula:

eMBz ×= (A1.60)

241

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.

The two theorems formulated by equations (A1.57) and (A1.60) lead to the

important general conclusion that, whenever dealing with a thin-walled beam, the

presence of a bimoment and a flexural twist should be expected. That is, in a thin-

walled beam torsional stresses, i.e. bimoment, flexural twist and St.Venant twist, can

be present even when no external twist moments are present.

Figure A1.26. Bending moment creating a bimoment

shear center axis

e

e

e

M

M

M

M

M Bz=M.e

242

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 cross-

sections 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.

Figure A2.1. Cross-section of a thin-walled beam

The point S in this figure is located at the principal pole (shear center) of the cross-

section. The point R represents an arbitrarily chosen pole (trial pole) of the sectorial

diagram. The increments of the sectorial areas in these two systems over a distance

ds are given, respectively, by the following equations:

S(ax,ay)

R(bx,by)

R′

φ

hS

hR

2d Rω 2

d Sω

y

x G

Tangent line

243

SS hdsd =ω (hatched in Fig. A2.1) (A2.1)

RR hdsd −=ω (A2.2)

The distance between point S and R′ is,

RS ′ = SR hh + (A2.3)

and

( ) RRS dRSdshRSdsd ω+′=−′=ω (A2.4)

The distance RS ′ can be given by the coordinates of the points S and R (see Fig.

A2.2) as follows:

φ−+φ−=′′+′′′=′ Sin)ba(Cos)ab(SRRRRS xxyy (A2.5)

Figure A2.2. Geometric and trigonometric relations in a thin-walled beam

The trigonometrical function of an angle φ can be replaced by the derivatives as

follows:

φ

R

R′

S

R ′′

ax-bx

by-ay

φ

φ

O

dx

dy ds

φ

244

dsdySin,

dsdxCos =φ=φ (A2.6)

Hence,

dsdy)ba(

dsdx)ab(RS xxyy −+−=′ (A2.7)

Substituting (A2.7) into (A2.4),

( ) ( ) RxxyyS ddybadxbad ω+−−−=ω (A2.8)

Integration of equation (A2.8) yields,

( ) ( ) Cybaxba xxyyRS +−−−+ω=ω (A2.9)

where C is a constant of integration, depending on the choice of initial radii of both

systems of sectorial areas. It is known from Appendix 1 that, the principal pole

(shear center) of a section can be defined by the two conditions that the sectorial

statical moments about x and y axes, must be equal to zero, i.e.,

dAxSA

sy ∫ ω=ω =0 (A2.10)

and

dAySA

sx ∫ ω=ω =0 (A2.11)

Hence, multiplying equation (A2.9) by x and y and substituting into (A2.10) and

(A2.11), respectively,

245

( ) ( ) 0dAxCdAxybadAxbadAxAA

xxA

2yy

AR =+−−−+ω ∫∫∫∫ (A2.12)

and

( ) ( ) 0dAyCdAybadAxybadAyAA

2xx

Ayy

AR =+−−−+ω ∫∫∫∫ (A2.13)

Since the axes x-y pass through the center of gravity of the cross-section,

0dAydAxA A

==∫ ∫ (A2.14)

Hence, the coordinates of the shear center are independent of the choice of the initial

radius of the system, Rω .

With definitions

yA

2 IdAx =∫ (A2.15)

xA

2 IdAy =∫ (A2.16)

xyA

IdAxy =∫ (A2.17)

xRA

R SdAx ω=ω∫ (A2.18)

yRA

R SdAy ω=ω∫ (A2.19)

(A2.12-13) are put into the following forms:

246

( ) ( ) 0SIbaIba xRxyxxyyy =+−−− ω

(A2.20)

( ) ( ) 0SIbaIba yRxxxxyyy =+−−− ω

The simultaneous solution of these two equations for ax and ay yield

expressions (A1.19) and (A1.20) in Appendix 1. These expressions are used when

the given axes x and y do not coincide with the principal axes of the section (Ixy≠0),

but pass through the center of gravity. If the axes x and y coincide with the principal

axes of a cross-section (Ixy=0), the formulas are simplified to the forms (A1.17) and

(A1.18) in Appendix 1.

247

APPENDIX 3. Computation Procedure for the Sectorial Area in an Open

Section

The sectorial area of the cross-section of a thin-walled beam refers to the

center line (contour line) of its outline. This property and the other related sectorial

properties were defined in Appendix 1. In Fig. A3.1, an example is given for the

calculation of the sectorial area of a section for an arbitrary chosen center of rotation

(pole) R and an initial (trial) radius 1RO→

.

Figure A3.1. Part of the sectorial area diagram for a section

The angle 41 SRO∧

is measured from 1RO→

in the clockwise direction. Thus

the sectorial area is negative. The absolute value of the sectorial area 41SROω is equal

to twice the area of the hatched triangle RO1S4. In other words

x

R

y

O1

S1

S2 S3

S4

Contour line

h

yo

-x3

y1 y2=y3

ys

y4

G

-x2

248

=ω41SRO ( )04 yyh +×− . The value of the sectorial area at point S1 is positive and is

equal to ( )01SRO yyh11

−×=ω . The value of the sectorial area at point S2 can be

obtained by adding, algebraically, twice the area of triangle RS1S2 to 11SROω , because

the radius swept from pole R moving from point S1 to S2, rotates in an anti-clockwise

direction. The sign of this additional area is also positive. In other words,

( )12SROSRO yyh1121

−×+ω=ω or ( )02SRO yyh21

−×=ω .

From S2 to S3, the angle 21 SRO∧

decreases and the radius 1RO→

rotates in the

clockwise direction. Hence, this additional area (twice the area of the triangle S3RS2)

is negative. Therefore, ( ) 32s3SROSRO xxyy2131

−×−−ω=ω .

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.

Figure A3.2. Plan of an arbitrary section

Y Y

5.25 cm

X

9.25 cm

19 cm

tf

tw

tf

( )yx a,aS

G(0,0)

X

α

tf : 1.0 cm tw : 0.5 cm

A B

C D

249

The computation procedure for the sectorial properties is given below:

Step:1 Find the position of the centre of gravity of the section.

The area of this section is,

A= ( ) 2cm23195.075.475.81 =×++×

The position of the centre of gravity is,

x = ( ) cm2.304323/625.275.4625.475.8 =×+×

y = ( ) cm 10.565223/9195.011875.8 =××+××

Figure A3.3. Contour of the section

5 cm

x

D C

A B 1

18 c

m

9 cm

2.3043 cm 6.6957 cm

7.43

48 c

m

10.5

652

cm

Y

X

y

2.3043 cm 2.6957 cm

2

3

G

250

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 = 4cm 169.044

IXY≠0, 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

yx

xy

III2

2tan−

−=α

-8.1305=α 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.

251

Step:4 Calculate the values of the sectorial linear statical moments.

The sectorial area BCs )(ω can be given as -18s. The sectorial statical

moments are

( ) ( ) 59

0CBX cm5417ds435.7s18S −=−= ∫ω

and

( ) ( ) 59

0CBY cm2696ds696.6s18S −=−= ∫ω

Step:5 Find the coordinates (ax,ay) of the principal pole (shear center) S from

expressions (A1.19) and (A1.20) since IXY≠0.

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

( )( )2x 169.044 1321.82177

2696169.0441775417304.2a−×

−×−×−+−= = −4.5870 cm

( ) ( )( )2y 03.1681310177

541703.16826961310565.10a−×

−×−−×−−= = 3.6321 cm

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)

252

The sectorial areas of points A, B, C and D are:

( ) 2SDD cm0.0=ω

( ) ( ) 2SDC cm70.9863.632110.565250.0 −=+×−=ω

( ) ( ) 2SDB cm29.898 2.30434.58701870.986 −=−×+−=ω

( ) ( ) 2SDA cm64.1223.63217.4348929.898 −=−×−−=ω

Figure A3.3. Initial sectorial area diagram ω′

Step:7 Find the principal radius.

The principal radius is found using the following formula:

( ) ( ) dAA1

SDSDO ∫ ω=ω

This equation gives the actual sectorial area value for the end (D) of the second

initial radius. The integral on the right side of equation (A3.2), which is equal to the

sectorial statical moment of the section is found as follows:

-70.986

−

-64.122 -29.898

S

C

initial radius

D

−

-70.986

−

253

( ) 4SD cm-1054.527S =ω

The sectorial area of the point on the section contour line through which the actual

principal radius SO passes, is

( ) 2SDO cm-45.849

231054.527-

==ω

Step:8 Find the principal sectorial area diagram of the section.

The principal sectorial area values for the points of the section can be found

either by re-calculating them from the principal radius or from the following

formula:

( ) ( ) ( )OSDSSDSSO ω−ω=ω

Here, the latter method will be employed. Hence,

( ) 2OD cm45.849)45.849(0.0 =−−=ω

( ) 2OC cm25.137)45.849(70.986 −=−−−=ω

( ) 2OB cm15.951)45.849(29.898 =−−−=ω

( ) 2OA cm18.273)45.849(64.122 −=−−−=ω

254

The principal sectorial area diagram is given in Fig. A3.4.

Figure A3.4. Principal sectorial area diagram

The equations for the parts of the principal sectorial area diagram can be

written as follows:

Between points D and C:

( ) ( ) ( ) ( )s

DCSODSOC

SODSOs ×ω−ω

+ω=ω

( ) s197.1445.849s5

45.84925.13745.849SOs −=×−−

+=ω

Between points C and B:

( ) ( ) ( ) ( )s

CBSOCSOB

SOCSOs ×ω−ω

+ω=ω

( ) s2826.225.137SOs +−=ω

+

+ 45.849

-25.137

15.951

-18.273

-

S

O Principal radius

+ -

255

Between points B and A:

( ) ( ) ( ) ( )s

BASOBSOA

SOBSOs ×ω−ω

+ω=ω

( ) s803.315.951SOs −=ω

Step:9 Calculate the principal sectorial moment of inertia (warping moment of

inertia) ωI from the formula.

( )[ ] dAIA

2SOs∫ ω=ω

( )

( )

( ) 629

0

218

0

25

0

cm4982.518ds1s803.315.951

ds5.0s2826.225.137

ds1s197.1445.849I

=×−+

×+−+

×−=

∫

∫

∫ω

256

APPENDIX 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

TITLE

BSAY, GSAY, KSAY, PSAY

:=============== REGION (i) ===============

:************ location of the first pier**************

PN(1), EN(1)

XI(1), YJ(1)

ELEM, CI(1), CJ(1), PT(1)

:********** location of the second pier**************

PN(2), EN(2)

XI(2), YJ(2)

ELEM, CI(2), CJ(2), PT(2)

: ========================================

G, E

:***********repeating by number of region***********

C(i), CBT(i)

HKAT(i)

KCB(i)

KSB(i)

HKIR(i)

:************* ends of the regions ***************

HGKIR(1), Z(1)

HGKIR(i), Z(i)

HGKIR(n), Z(n)

:========= DYNAMIC PARAMETERS =========

KSI

ETSUR

YGEN

DT

ASUR

LTIP

JI

IAP

257

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 : Explanatory information row for the coupled shear wall structure

BSAY : Total number of the regions

GSAY : Total number of the stiffening beams

KSAY : Total number of the stories

PSAY : Total number of the piers (equal to 2 for this thesis)

PN : Total number of joints for a pier (equal to the number of intersections of

the wall units)

EN : Total number of elements for a pier (equal to the number of the wall units)

XI : Global X coordinate of a joint in a pier

YJ : Global Y coordinate of a joint in a pier

ELEM : Element number of a wall unit in a pier

CI : Number of the first joint of an element in a pier

CJ : Number of the second joint of an element in a pier

PT : Thickness of an element in a pier

G : Shear modulus

E : Elasticity modulus

C(i) : Span length of the connecting beams in region i

BKT(i) : Thickness of the connecting beams in region i

HKAT(i) : Storey height in region i

KCB(i) : Rotational spring constant at the ends of the connecting beams in region i

KSB(i) : Rotational spring constant at the ends of the stiffening beam on the boundary i

HKIR(i) : Heights of the connecting beams in region i

HGKIR(i) : Height of the stiffening beam on the boundary i

Z(i) : Distance of the boundary i from the ground level

KSI : Damping ratio

ETSUR : Duration of dynamic loading

258

YGEN : Magnitude of the dynamic loading

DT : Time increment value

ASUR : Duration of analysis

LTIP : Type of the dynamic loading

JI (joint # ) : The joint where the analysis is performed

IAP : The joint where the dynamic loading is applied

FIN : Load value for t=0

OMG : The angular frequency of dynamic load

TT1 : Characteristic time in the value of dynamic load (s)

XAPP : The location of the dynamic load on the X axis

YAPP : The location of the dynamic load on the Y axis

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 ***************************************** 5 4 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 **************************************** 4 3 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 ***************************************** 5 4 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 **************************************** 4 3 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

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:************** REPETS BY NUMBER OF REGION ******************************************* 3.,0.4 3.,0.4 3.0 3.0 1.e50 1.e50 1.e50 1.e50 0.5 0.5 :************** ENDS OF THE REGIONS ************************************************** 3.0,48.0 3.0,24.0 0.0,0.0 :========= DYNAMIC PARAMETERS ========= 0.0 11.0 100 0.005 22 3 1 1 100 1.256637061 11 0.0 0.0

260

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 I and J remained as variable REAL*8 Teta(N),U(N),V(N),T(N),ST(N),KYUK(N),KAT(N) REAL*8 HKAT(N),eHKAT(N),HKIR(N),eHKIR(N),CCB(N),eCCB(N),CSB(N) @ ,eCSB(N),HGKIR(N),eHGKIR(N),Z(N),eZ(N) @ ,Ac(N),eAc(N),As(N),eAs(N),EIc(N),eEIc(N),EIs(N),eEIs(N) @ ,GAMA(N),eGAMA(N),GAMAstf(N),eGAMAstf(N) @ ,BETA1(N),eBETA1(N),BETA2(N),eBETA2(N),BETA3(N),eBETA3(N) @ ,ALFA1(N),eALFA1(N),ALFA2(N),eALFA2(N) @ ,S(N),eS(N),Tozel(N),Tozelz(N),Tozelzz(N),Tozelzzz(N) @ ,STozel(N),STozelz(N),STozelzz(N),STozelzzz(N) @ ,Mx(N),Mxz(N),Mxzz(N),Mxzzz(N) @ ,My(N),Myz(N),Myzz(N),Myzzz(N) @ ,Mt(N),Mtz(N),Mtzz(N),Mtzzz(N),Bt(N) @ ,ZK(N,N),ZB(N),D1(N),D2(N),D3(N),D4(N),SDD1(N),SDD2(N) @ ,REK1(N),REK2(N),REK3(N),REK4(N),C1(N),C2(N),DD1(N),DD2(N) @ ,YK(N,N),YB(N),SONUC1(N),SONUC2(N),SONUC3(N),SONUC4(N) @ ,G1(N),G2(N),F1(N),F2(N),P1(N),P2(N),eSIW(N),ePRSA(N,N) @ ,eDX(N),eDY(N),eEDI(N),eEDJ(N),eTHICK(N),ePRSECAREA(N) @ ,CX(N),CY(N),EX(N),EY(N),eCSAREA(N),eTALAN(N),TALAN(N) @ ,SIx(N),SIxy(N),SIy(N),EJp(N),PNN(N),ENN(N),UG(N),VG(N) @ ,eEXG(N),eEYG(N) 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)

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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 FORMAT(A200) 20 FORMAT(A50) 22 FORMAT(1X,30F10.3) 23 FORMAT(1X,30F13.3) 25 FORMAT(1X,30F13.4) READ(5,20) BASLIK WRITE(6,20) BASLIK READ(5,*) BSAY,GSAY,KSAY,PSAY c WRITE(6,*) ' BSAY GSAY KSAY PSAY ' c WRITE(6,*) '------------------------------------------------ ' c WRITE(6,22) BSAY,GSAY,KSAY,PSAY ibsay=BSAY+1 910 FORMAT(A8) 920 FORMAT(A50) 922 FORMAT(1X,30F10.3) 923 FORMAT(1X,30F10.2) 925 FORMAT(1X,30F13.4) c WRITE(6,*) ' ' c WRITE(6,*) '************* SECTION PREPERTIES **************** ' c WRITE(6,*) ' ' C PERDELERİN GEOMETRİK ÖZELLİKLERİ DATA DOSYASINDAN OKUNARAK C W DİYAGRAMI VE IW BURULMA ATALETİ HESAPLANACAK DO 432 JJ=1,BSAY READ(5,20) BILGI TOPALAN=0.0 DO 431 II=1,PSAY READ(5,20) BILGI READ(5,*) ePN,eEN c WRITE(6,*) ' ' c WRITE(6,*) ' PN EN (Number of point on section)' c WRITE(6,*) '--------------------------- ' c WRITE(6,922) PN,EN eePN=ePN+1 eeEN=eEN+1 c WRITE(6,*) ' ' c WRITE(6,*) ' POINT X Y (point coordinates)' c WRITE(6,*) '------------------------------------- ' DO I=1,ePN READ(5,*) P,eDX(P),eDY(P) c WRITE(6,922) P,eDX(P),eDY(P) ENDDO

262

eDX(eePN)=0.0 eDY(eePN)=0.0 c WRITE(6,*) ' ' c WRITE(6,*) ' ELEMENT I J THICKNESS ' c WRITE(6,*) '-----------------------------------------------------' DO I=1,eEN READ(5,*) E,eEDI(E),eEDJ(E),eTHICK(E) c 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 SECPREP(II,eePN,eeEN,P,eDX,eDY,eTHICK,eEDI,eEDJ - ,eCSAREA,EX,EY,sIx,sIy,sIxy,eJp,CX,CY) c WRITE(6,*) ' ' c WRITE(6,*) '************ SECTORIAL PREPERTIES ***************** ' c 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 WRITE(6,*) ' ' c WRITE(6,*) '**************************************************** ' c WRITE(6,*) '**************************************************** ' c WRITE(6,*) ' ' 431 CONTINUE eEXG(JJ)=(eCSAREA(1)*EX(1)+eCSAREA(2)*EX(2))/TOPALAN eEYG(JJ)=(eCSAREA(1)*EY(1)+eCSAREA(2)*EY(2))/TOPALAN ea(JJ) = EX(2)-EX(1) c WRITE(6,*) ' ' c WRITE(6,*) 'a= ' c WRITE(6,*) ea(JJ) eb(JJ) = EY(2)-EY(1) c WRITE(6,*) ' ' c WRITE(6,*) 'b= ' c WRITE(6,*) eb(JJ) eEJ(JJ) = EJp(1)+EJp(2) c WRITE(6,*) ' ' c WRITE(6,*) 'J= ' c WRITE(6,*) eEJ(JJ) eEIx(JJ) = SIx(1)+SIx(2) c WRITE(6,*) ' ' c WRITE(6,*) 'Ix= ' c WRITE(6,*) eEIx(JJ) eEIxy(JJ) = SIxy(1)+SIxy(2) c WRITE(6,*) ' ' c WRITE(6,*) 'Ixy= '

263

c WRITE(6,*) eEIxy(JJ) eEIy(JJ) = SIy(1)+SIy(2) c WRITE(6,*) ' ' c WRITE(6,*) 'Iy= ' c WRITE(6,*) eEIy(JJ) eDelta(JJ) = (eEIx(JJ)*eEIy(JJ) - eEIxy(JJ)**2) c WRITE(6,*) ' ' c WRITE(6,*) 'Delta= ' c WRITE(6,*) eDelta(JJ) eEIxc(JJ) = CX(1)*SIx(1)+CX(2)*SIx(2)-CY(1)*SIxy(1)-CY(2)*SIxy(2) c WRITE(6,*) ' ' c WRITE(6,*) 'Ixc= ' c WRITE(6,*) eEIxc(JJ) eEIyc(JJ) = CY(1)*SIy(1)+CY(2)*SIy(2)-CX(1)*SIxy(1)-CX(2)*SIxy(2) c WRITE(6,*) ' ' c WRITE(6,*) 'Iyc= ' c WRITE(6,*) eEIyc(JJ) eK1(JJ) = (eEIxc(JJ)*eEIxy(JJ) + eEIx(JJ)*eEIyc(JJ))/eDelta(JJ) c WRITE(6,*) ' ' c WRITE(6,*) 'K1= ' c WRITE(6,*) eK1(JJ) eK2(JJ) = (eEIxc(JJ)*eEIy(JJ) + eEIxy(JJ)*eEIyc(JJ))/eDelta(JJ) c WRITE(6,*) ' ' c WRITE(6,*) 'K2= ' c WRITE(6,*) eK2(JJ) eK3(JJ) = (ea(JJ)*eEIx(JJ) - eb(JJ)*eEIxy(JJ))/eDelta(JJ) c WRITE(6,*) ' ' c WRITE(6,*) 'K3= ' c WRITE(6,*) eK3(JJ) eK4(JJ) = (eb(JJ)*eEIy(JJ) - ea(JJ)*eEIxy(JJ))/eDelta(JJ) c WRITE(6,*) ' ' c WRITE(6,*) 'K4= ' c WRITE(6,*) eK4(JJ) eALAN(JJ) = 1/((1/eCSAREA(1))+(1/eCSAREA(2)) @ +ea(JJ)*eK3(JJ)+eb(JJ)*eK4(JJ)) c WRITE(6,*) ' ' c WRITE(6,*) 'Alan= ' c WRITE(6,*) eALAN(JJ) ed(JJ) = CX(2)*EY(2)-CY(2)*EX(2)+CY(1)*EX(1)-CX(1)*EY(1) c WRITE(6,*) ' ' c WRITE(6,*) 'd= ' c WRITE(6,*) ed(JJ) ew(JJ) = ePRSA(1,PNN(1))-ePRSA(2,PNN(2)) c WRITE(6,*) ' ' c WRITE(6,*) 'w= ' c WRITE(6,*) ew(JJ) er(JJ) = ew(JJ) + ed(JJ) + ea(JJ)*eK1(JJ) - eb(JJ)*eK2(JJ) c WRITE(6,*) ' ' c WRITE(6,*) 'r= ' c 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) c WRITE(6,*) ' ' c WRITE(6,*) 'Iw= ' c WRITE(6,*) eEIw(JJ) eEIOw(JJ) = eEIw(JJ) - eEIyc(JJ)*eK1(JJ) - eEIxc(JJ)*eK2(JJ) c WRITE(6,*) ' ' c WRITE(6,*) 'IOw= ' c WRITE(6,*) eEIOw(JJ) eTALAN(JJ)=TOPALAN 432 CONTINUE

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READ(5,20) BILGI READ(5,*) G,Elas c WRITE(6,*) ' ' c WRITE(6,*) ' G Elas ' c WRITE(6,*) '------------------------- ' c WRITE(6,23) G,Elas c READ(5,*) Met,Bet c WRITE(6,*) ' ' c WRITE(6,*) ' Met Bet ' c WRITE(6,*) '---------------------------------- ' c WRITE(6,23) Met,Bet c READ(5,*) Px,Wx,dy c WRITE(6,*) ' ' c WRITE(6,*) ' Px Wx dy ' c WRITE(6,*) '------------------------------- ' c WRITE(6,22) Px,Wx,dy c READ(5,*) Py,Wy,dx c WRITE(6,*) ' ' c WRITE(6,*) ' Py Wy dx ' c WRITE(6,*) '------------------------------- ' c WRITE(6,22) Py,Wy,dx READ(5,20) BILGI c WRITE(6,*) ' c BAG.KİR. KALINLIĞI ' c WRITE(6,*) '------------------------------------------- ' DO I=1,bsay READ(5,*) ec(I),eBKTHICK(I) c WRITE(6,22) ec(I),eBKTHICK(I) ENDDO ec(ibsay)=ec(bsay) eBKTHICK(ibsay)=eBKTHICK(bsay) c WRITE(6,*) ' ' c WRITE(6,*) '-----Kat Yüksekliği------ ' c WRITE(6,*) '------------------------- ' DO 1 I=1,bsay READ(5,*) eHKAT(I) c WRITE(6,22) eHKAT(I) 1 CONTINUE eHKAT(ibsay)=eHKAT(bsay) c WRITE(6,*) ' ' c WRITE(6,*) '-----Bağlantı Rij Katsayısı-Ccb----- ' c WRITE(6,*) '------------------------------------ ' DO 2 I=1,bsay READ(5,*) eCCB(I) c WRITE(6,22) eCCB(I) 2 CONTINUE eCCB(ibsay)=eCCB(bsay) c WRITE(6,*) ' ' c WRITE(6,*) '-----Bağlantı Rij Katsayısı-Csb----- ' c WRITE(6,*) '------------------------------------ ' DO 3 I=1,bsay READ(5,*) eCSB(I)

265

c WRITE(6,22) eCSB(I) 3 CONTINUE eCSB(ibsay)=eCSB(bsay) c WRITE(6,*) ' ' c WRITE(6,*) '-----Kiriş Yüksekliği------ ' c WRITE(6,*) '--------------------------- ' DO 4 I=1,bsay READ(5,*) eHKIR(I) c WRITE(6,22) eHKIR(I) 4 CONTINUE eHKIR(ibsay)=eHKIR(bsay) READ(5,20) BILGI c WRITE(6,*) ' ' c WRITE(6,*) '-Güç Kiriş Yüksekliği----Yerden Mesafe---- ' c WRITE(6,*) '------------------------------------------ ' DO 5 I=1,ibsay READ(5,*) eHGKIR(I),eZ(I) c WRITE(6,22) eHGKIR(I),eZ(I) 5 CONTINUE c READ(5,*) HP c WRITE(6,*) ' HP (Yükün uygulandığı yükseklik) ' c WRITE(6,*) '-----------' c WRITE(6,22) HP H = eZ(1) c WRITE(6,*) ' ' c WRITE(6,*) 'H= ' c 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) c Kesit değişimi durumunda "GJTetaz" ihmal ediliyor IF(bsay.gt.1) then DO I=1,bsay IF(eEIOw(I).ne.eEIOw(I+1)) then DO J=1,ibsay eEJ(J)=0.0000000001 ENDDO ENDIF ENDDO ENDIF

266

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)

267

DO 800 IH=1,3 IF(IH.EQ.1) THEN Px=1 Py=0 Met=0 c dy=0 ENDIF IF(IH.EQ.2) THEN Px=0 Py=1 Met=0 c dy=0 ENDIF IF(IH.EQ.3) THEN Px=0 Py=0 Met=1 c 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

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S(I)=0.d0 Mx(I) =0.d0 Mxz(I) =0.d0 Mxzz(I) =0.d0 Mxzzz(I) =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 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 IF(JJ.EQ.(iksay-1))THEN C 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)

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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)

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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)

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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 WRITE(6,*) a(I) c WRITE(6,*) b(I) c WRITE(6,*) EJ(I) c WRITE(6,*) EIx(I) c WRITE(6,*) EIxy(I) c WRITE(6,*) EIy(I) c WRITE(6,*) Delta(I) c WRITE(6,*) EIxc(I) c WRITE(6,*) EIyc(I) c WRITE(6,*) K1(I) c WRITE(6,*) K2(I) c WRITE(6,*) K3(I) c WRITE(6,*) K4(I) c WRITE(6,*) ALAN(I) c WRITE(6,*) d(I) c WRITE(6,*) w(I) c WRITE(6,*) r(I) c WRITE(6,*) EIw(I) c WRITE(6,*) EIOw(I) c WRITE(6,*) CCB(I) c WRITE(6,*) '------------------------------------------ ' c WRITE(6,22) c(I),BKTHICK(I) ENDDO ETOL=-0.00001 DO I=1,ibsay c Dis Kuvvetlerin Momenti (Bolge Sinirlarinda) HPP=HP-Z(I) IF(HPP.GT.ETOL) THEN Mx(I) =Py*(HP - Z(I)) + (Wy*(HP - Z(I))**2)/2. Mxz(I) =-Py - Wy*(HP - Z(I)) Mxzz(I) =Wy Mxzzz(I) =0 My(I) =Px*(HP - Z(I)) + (Wx*(HP - Z(I))**2)/2. Myz(I) =-Px - Wx*(HP - Z(I)) Myzz(I) =Wx Myzzz(I) =0 Mt(I) =-(dy*Px)+dx*Py-dy*Wx*(HP-Z(I))+dx*Wy*(HP-Z(I))+Met Mtz(I) =dy*Wx - dx*Wy Mtzz(I) =0 ELSE

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Mx(I) =0 Mxz(I) =0 Mxzz(I) =0 Mxzzz(I) =0 My(I) =0 Myz(I) =0 Myzz(I) =0 Myzzz(I) =0 Mt(I) =0 Mtz(I) =0 Mtzz(I) =0 ENDIF Bt(I) = Bet WRITE(6,*) ' My(',I,') My(',I,') Mt(',I,') ' WRITE(6,22) Mx(I),My(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)= ((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) DD2(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 WRITE(6,*)' ' c WRITE(6,*)' Tozel(I) Tozelz(I) Tozelzz(I) Tozelzzz(I)' DO 55 I=1,ibsay 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)-

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- 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) 55 CONTINUE C WRITE(6,*)' ' C WRITE(6,*)' STozel(I) STozelz(I) STozelzz(I) STozelzzz(I)' DO 56 I=1,ibsay 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 WRITE(6,*) STozel(I),STozelz(I),STozelzz(I),STozelzzz(I) 56 CONTINUE TOL=0.00001 c INTEGRASYON SABITLERININ ELDE EDILMESI DO I=1,60 DO J=1,60 ZK(I,J)=0.d0 YK(I,J)=0.d0 ZB(I)=0.d0 YB(I)=0.d0 SONUC1(I)=0.d0 SONUC2(I)=0.d0 ENDDO ENDDO DO I=1,60 D1(I)=0.d0 D2(I)=0.d0 D3(I)=0.d0 D4(I)=0.d0 F1(I)=0.d0 F2(I)=0.d0 U(I)=0.d0 P1(I)=0.d0

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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 C 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

275

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 WRITE(6,*) ' ' C write(6,*) 'Coefficient Matrix' C DO I =1,n C WRITE(6,22) (zk(I,J),J=1,N) C enddo 77 FORMAT(1X,40F22.10) C WRITE(6,*) ' ' C write(6,*) 'Right Hand Side Vector' C WRITE(6,77) (zb(J),J=1,N) CALL GAUSS(ZK,in,ZB,SONUC1) C WRITE(6,*) ' ' C write(6,*) 'Result Vector' C 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 C Program Kontrol Satiri iksay=KSAY+1 ENDIF

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WRITE(6,*)' ' WRITE(6,*)' ' WRITE(6,*)' ' c KAT DEPLASMANLARININ BULUNMASI c write(6,*) ' ' c write(6,*) '--KAT SEViYELERi--' c write(6,*) ' ' c do in=1,iksay c im=iksay-in+1 c write(6,*) KAT(im) c 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)*Wy- - K2(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)

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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) =0. dMy(im) =0. dMt(im) =0. ENDIF 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 Hizalarında PERDE EKSENEL KUVVETİ (T) ' WRITE(6,*)'------------------------------------------- '

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WRITE(6,*)' ' do I=1,iksay J=iksay-I+1 write(6,244) EKS(J) enddo c WRITE(6,*)' ' c WRITE(6,*)' ' c WRITE(6,*)' Bölge sınırları hemen üzerinde PERDE EKS KUV (ST)' c WRITE(6,*)'--------------------------------------------------' c WRITE(6,*)' ' c do I=1,ibsay c write(6,244) ST(I) c enddo c WRITE(6,*)' ' c WRITE(6,*)'********************************************' c WRITE(6,*)' ' c WRITE(6,*)'Glob-X etrafında TOPLAM PERDE EĞİLME MOMENTİ' c WRITE(6,*)'--------------------------------------------' c WRITE(6,*)' ' c do I=1,iksay c J=iksay-I+1 c write(6,244) TOPMOMX(J) c enddo c WRITE(6,*)' ' c WRITE(6,*)' ' c WRITE(6,*)' Bölge sınırları hemen üzerinde ' c WRITE(6,*)'Glob-X etrafında TOPLAM PERDE EĞİLME MOMENTİ' c WRITE(6,*)'--------------------------------------------' c WRITE(6,*)' ' c do I=1,ibsay c write(6,244) STOPMOMX(I) c enddo c WRITE(6,*)' ' c WRITE(6,*)'********************************************' c WRITE(6,*)' ' c WRITE(6,*)'Glob-Y etrafında TOPLAM PERDE EĞİLME MOMENTİ' c WRITE(6,*)'--------------------------------------------' c WRITE(6,*)' ' c do I=1,iksay c J=iksay-I+1 c write(6,244) TOPMOMY(J) c enddo c WRITE(6,*)' ' c WRITE(6,*)' ' c WRITE(6,*)' Bölge sınırları hemen üzerinde ' c WRITE(6,*)'Glob-Y etrafında TOPLAM PERDE EĞİLME MOMENTİ' c WRITE(6,*)'--------------------------------------------' c WRITE(6,*)' ' c do I=1,ibsay c write(6,244) STOPMOMY(I) c enddo c WRITE(6,*)' ' c WRITE(6,*)'********************************************' c WRITE(6,*)' ' c WRITE(6,*)' Glob-Z etrafında TOPLAM PERDE BURULMA MOMENTİ (Mt)' c WRITE(6,*)'---------------------------------------------------' c WRITE(6,*)' ' do I=1,iksay J=iksay-I+1

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c write(6,244) TOPMOMT(J) enddo c WRITE(6,*)' ' c WRITE(6,*)' ' c WRITE(6,*)' Bölge sınırları hemen üzerinde ' c WRITE(6,*)' Glob-Z etrafında TOPLAM PERDE BURULMA MOMENTİ (Mt)' c WRITE(6,*)'---------------------------------------------------' c WRITE(6,*)' ' do I=1,ibsay c write(6,244) STOPMOMT(I) enddo c WRITE(6,*)' ' c WRITE(6,*)'********************************************' c WRITE(6,*)' ' c WRITE(6,*)' ' c WRITE(6,*)' TOPLAM PERDE BİMOMENTİ (Bt)' c WRITE(6,*)'-----------------------------' c WRITE(6,*)' ' do I=1,iksay J=iksay-I+1 c write(6,244) TOPMOMB(J) enddo c WRITE(6,*)' ' c WRITE(6,*)' ' c WRITE(6,*)' Bölge sınırları hemen üzerinde ' c WRITE(6,*)' TOPLAM PERDE BİMOMENTİ (Bt) ' c WRITE(6,*)'---------------------------------' c WRITE(6,*)' ' do I=1,ibsay c write(6,244) STOPMOMB(I) enddo c WRITE(6,*)' ' c WRITE(6,*)'********************************************' c WRITE(6,*)' ' c******************************************* c U(Z) DEPLASMAN FONKSIYONUNUN ELDE EDILMESI iL=0 iK=0 DO I=2,ibsay HPP=HP-Z(I) if(hpp.gt.TOL) then IF(I.LT.ibsay) THEN 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)

280

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

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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 write(6,*) 'Coefficient Matrix' c DO I =1,in c WRITE(6,233) (yk(I,J),J=1,in) c enddo 234 FORMAT(1X,40F22.10) c write(6,*) 'Right Hand Side Vector' c WRITE(6,234) (yb(J),J=1,in) CALL GAUSS(YK,in,YB,SONUC3) c write(6,*) 'Result Vector' c WRITE(6,22) (SONUC3(J),J=1,in) 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

282

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 V(Z) DEPLASMAN FONKSIYONUNUM ELDE EDILMESI iL=0 iK=0 DO I=2,ibsay HPP=HP-Z(I) if(hpp.gt.TOL) then IF(I.LT.ibsay) THEN 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)

283

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

284

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 write(6,*) 'Coefficient Matrix' c DO I =1,in c WRITE(6,233) (yk(I,J),J=1,in) c enddo 235 FORMAT(1X,40F22.10) c write(6,*) 'Right Hand Side Vector' c WRITE(6,235) (yb(J),J=1,in) CALL GAUSS(YK,in,YB,SONUC4) c write(6,*) 'Result Vector' c WRITE(6,22) (SONUC4(J),J=1,in) 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)

285

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 Teta(Z) DEPLASMAN FONKSIYONUNUM ELDE EDILMESI iL=0 iK=0 DO I=2,ibsay HPP=HP-Z(I) if(hpp.gt.TOL) then IF(I.LT.ibsay) THEN 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

286

@ ,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)

287

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 write(6,*) 'Coefficient Matrix' c DO I =1,in c WRITE(6,233) (yk(I,J),J=1,in) c enddo 233 FORMAT(1X,40F22.10) c write(6,*) 'Right Hand Side Vector' c WRITE(6,233) (yb(J),J=1,in) CALL GAUSS(YK,in,YB,SONUC2) c write(6,*) 'Result Vector' c WRITE(6,22) (SONUC2(J),J=1,in) 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)

288

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 KUTLE MERKEZİ DEPLASMANLARININ HESAPLANMASI do I=1,iksay J=iksay-I+1 UG(J)=U(J)-TETA(J)*dy VG(J)=V(J)+TETA(J)*dx enddo WRITE(6,*) ' ' WRITE(6,*) 'UG=' do I=1,iksay J=iksay-I+1 write(6,24) UG(J) enddo WRITE(6,*) ' ' WRITE(6,*) 'VG=' do I=1,iksay J=iksay-I+1 write(6,24) VG(J) enddo c iksay=KSAY+1 c IKSA=3*IKSAY c IF(JJ.EQ.1)THEN c DO NM=1,IKSA c U(NM)=0.00000000000000000001 c V(NM)=0.00000000000000000001 c TETA(NM)=0.00000000000000000001 c ENDDO c ENDIF C ********************** C ** ESNEKLIK MATRISI ** C **********************

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c KOL=iksay-JJ+IH c iksay=KSAY+1 c DO NN=1,iksay c M=iksay-NN+1 c FLEX(NN,KOL)=FLEX(NN,KOL)+U(M) c 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,*) '--ESNEKLİK MATRİSİ--' IKSA=3*KSAY DO I=1,IKSA WRITE(6,24) (FLEX(I,J),J=1,IKSA) ENDDO iksay=ksay+1 IKSA=3*KSAY C ***************************************** C **RIJITLIK MATRISI BULUNUYOR [K]=[F]**(-1)** C ***************************************** 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

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GAM=2.4 C ****************************** C ********* KUTLE MATRISI ********** C ****************************** 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 C ***** kütlenin dönel ataleti ihmal ediliyor ***** 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

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WRITE(6,24) (MASSMAT(I,J),J=1,IKSA) ENDDO write(6,*) ' ' 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) 888 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 write(6,*) ' ' C write(6,*) '--NORMALIZE EDILMIS OZVEKTORLER MATRISI--' C DO I =1,IKSA C WRITE(6,22) (EGNVEC2(I,J),J=1,IKSA) C enddo c ********************************************** 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

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c ************************************************** write(6,*) ' ' write(6,*) '--OZDEGERLER VEKTORU--' IKSA=3*KSAY DO I =1,IKSA WRITE(6,24) EGNVAL(I) enddo c DO 144 KLM=1,IKSA c DUMM(KLM)=EGNVAL(IKSA+1-KLM) c DO 144 KKLM=1,IKSA c DUMV(KKLM,KLM)=EGNVEC(KKLM,IKSA+1-KLM) c 144 CONTINUE PI=4*ATAN(1.0) DO 145 KLM1=1,IKSA c EGNVAL(KLM1)=DUMM(KLM1) CFREQ(KLM1)=SQRT(EGNVAL(KLM1)) NFREQ(KLM1)=CFREQ(KLM1)/(2*PI) c DO 145 KLM3=1,IKSA c EGNVEC(KLM3,KLM1)=DUMV(KLM3,KLM1) 145 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 ** Serbest Titreşim analizi yapıldı, ********* C ** Zorlanmış Titreşim Analizi başlıyor ********* c **************************************************************** 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 CALL MTRXML(SONUM,NEV,NEV,EGNVEC,NEV,GSSNM1,NW) C 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)

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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 ANAL˜Z˜=>1' C WRITE(*,*) ' SPEKTRUM ANAL˜Z˜=>2' c READ(5,*)ISEC c 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

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C ***************************************** 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 DO 163 IR=1,IKSA YN1(IR)=A0*yenX(IR)+A2*XN(IR)+A3*XNN(IR) 163 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)

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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 c WRITE(*,*)'ENTER THE JOINT NUMBER TO BE ANALIZED' c 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) implicit real*8 (A-H,K-Z) dimension REK1(8) c real*8 EALFA1,EALFA2,EZ,ES REK1(1)=EALFA1*ES*Cosh(EALFA1*EZ) + Sinh(EALFA1*EZ) REK1(2)=Cosh(EALFA1*EZ) + EALFA1*ES*Sinh(EALFA1*EZ) REK1(3)=EALFA2*ES*Cosh(EALFA2*EZ) + Sinh(EALFA2*EZ) REK1(4)=Cosh(EALFA2*EZ) + EALFA2*ES*Sinh(EALFA2*EZ) REKB1=-ETozel -ES*ETozelz c WRITE(6,*) REK1(1),REK1(2) RETURN END

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SUBROUTINE COEF1B(ETozel,ETozelzz,EALFA1,EALFA2 @ ,EGAMA,EZ,EEIw,EALAN,Er,Ed,Ew,EEIxc,EEIyc @ ,EK1,EK2,EK3,EK4,EBet,REK2,REKB2) implicit real*8 (A-H,K-Z) dimension REK2(8) REK2(1)=(((-1 + EALAN*EALFA1**2*EGAMA)*EEIw + EEIyc*EK1+ - EEIxc*EK2 - EALAN* - (EALFA1**2*EGAMA*(EEIyc*EK1 + EEIxc*EK2)+ - Er*(Ed + EEIyc*EK3 - EEIxc*EK4 + Ew)))* - Sinh(EALFA1*EZ))/(EALAN*Er) REK2(2)=(((-1 + EALAN*EALFA1**2*EGAMA)*EEIw + EEIyc*EK1+ - EEIxc*EK2 - EALAN* - (EALFA1**2*EGAMA*(EEIyc*EK1 + EEIxc*EK2)+ - Er*(Ed + EEIyc*EK3 - EEIxc*EK4 + Ew)))* - Cosh(EALFA1*EZ))/(EALAN*Er) REK2(3)=(((-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)))* - Sinh(EALFA2*EZ))/(EALAN*Er) REK2(4)=(((-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)))* - Cosh(EALFA2*EZ))/(EALAN*Er) REKB2= -(EBet+(-Ed - EEIyc*EK3 + EEIxc*EK4)*ETozel+ - ((-EEIw + EEIyc*EK1 + EEIxc*EK2)*ETozel)/(EALAN*Er)+ - (EGAMA*(EEIw - EEIyc*EK1 - EEIxc*EK2)*ETozelzz)/Er- - ETozel*Ew) RETURN END SUBROUTINE COEF2A(ibsay,EALFA1,EALFA2,EZ, @ ETozelz,ESTozelz,EGAMA,J,REK1,REKB1) implicit real*8 (A-H,K-Z) dimension REK1(8),EALFA1(ibsay),EALFA2(ibsay), @ ETozelz(ibsay),ESTozelz(ibsay),EGAMA(ibsay) REK1(1)=EALFA1(J-1)*EGAMA(J-1)*Cosh(EALFA1(J-1)*EZ) REK1(2)=EALFA1(J-1)*EGAMA(J-1)*Sinh(EALFA1(J-1)*EZ) REK1(3)=EALFA2(J-1)*EGAMA(J-1)*Cosh(EALFA2(J-1)*EZ) REK1(4)=EALFA2(J-1)*EGAMA(J-1)*Sinh(EALFA2(J-1)*EZ) REK1(5)=-(EALFA1(J)*EGAMA(J)*Cosh(EALFA1(J)*EZ)) REK1(6)=-(EALFA1(J)*EGAMA(J)*Sinh(EALFA1(J)*EZ)) REK1(7)=-(EALFA2(J)*EGAMA(J)*Cosh(EALFA2(J)*EZ)) REK1(8)=-(EALFA2(J)*EGAMA(J)*Sinh(EALFA2(J)*EZ)) REKB1=-(-(EGAMA(J)*ETozelz(J))+EGAMA(J-1)*ETozelz(J-1)) RETURN END

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SUBROUTINE COEF2B(ibsay,EALFA1,EALFA2,EZ,ES, @ ETozel,ETozelz,ESTozel,J,REK2,REKB2) implicit real*8 (A-H,K-Z) dimension REK2(8),EALFA1(ibsay),EALFA2(ibsay), @ ETozel(ibsay),ETozelz(ibsay),ESTozel(ibsay) REK2(1)=Sinh(EALFA1(J-1)*EZ) REK2(2)=Cosh(EALFA1(J-1)*EZ) REK2(3)=Sinh(EALFA2(J-1)*EZ) REK2(4)=Cosh(EALFA2(J-1)*EZ) REK2(5)=-(EALFA1(J)*ES*Cosh(EALFA1(J)*EZ))-Sinh(EALFA1(J)*EZ) REK2(6)=-(EALFA1(J)*ES*Sinh(EALFA1(J)*EZ))-Cosh(EALFA1(J)*EZ) REK2(7)=-(EALFA2(J)*ES*Cosh(EALFA2(J)*EZ))-Sinh(EALFA2(J)*EZ) REK2(8)=-(EALFA2(J)*ES*Sinh(EALFA2(J)*EZ))-Cosh(EALFA2(J)*EZ) REKB2=-(-ETozel(J)+ESTozel(J-1)-ES*ETozelz(J)) RETURN END SUBROUTINE COEF2C(ibsay,ETozelz,ETozelzzz,ESTozelz,ESTozelzzz, @ EALFA1,EALFA2,EZ,EC1,EC2,EDD1,ESDD1,J,REK3,REKB3) implicit real*8 (A-H,K-Z) dimension REK3(8),EALFA1(ibsay),EALFA2(ibsay),EDD1(ibsay), @ EC1(ibsay),EC2(ibsay),ETozelz(ibsay),ESTozelz(ibsay), @ ESDD1(ibsay),ETozelzzz(ibsay),ESTozelzzz(ibsay) REK3(1)= EALFA1(J-1)*(EC1(J-1)+EALFA1(J-1)**2*EC2(J-1))* @ Cosh(EALFA1(J-1)*EZ) 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)

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implicit real*8 (A-H,K-Z) dimension REK4(8),EALFA1(ibsay),EALFA2(ibsay),EDD2(ibsay), @ EC1(ibsay),EC2(ibsay),ETozel(ibsay),ESTozel(ibsay), @ ESDD2(ibsay),ETozelzz(ibsay),ESTozelzz(ibsay) REK4(1)= (EC1(J-1)+EALFA1(J-1)**2*EC2(J-1))* @ Sinh(EALFA1(J-1)*EZ) 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) implicit real*8 (A-H,K-Z) dimension REK2(8) REK2(1)= -((EALFA1*((-1+EALAN*EALFA1**2*EGAMA)*EEIw+EEIyc*EK1+

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- 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*EMyz- - EEIx*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) implicit real*8 (A-H,K-Z) dimension REK1(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) 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*

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- 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) implicit real*8 (A-H,K-Z) 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) 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*

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- (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)

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@ ,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) implicit real*8 (A-H,K-Z) 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) 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) implicit real*8 (A-H,K-Z) dimension REK1(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) 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)))/

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- (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) implicit real*8 (A-H,K-Z) 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) 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) implicit real*8 (A-H,K-Z) dimension REK1(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)

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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) implicit real*8 (A-H,K-Z) 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) REK2(1)=EZ REK2(2)=1 REK2(3)=-EZ

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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) implicit real*8 (A-H,K-Z) dimension ETETA(iksay),ETET(iksay),EG1(ibsay),EG2(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) ETETA(jm)= - - 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+

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- 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) implicit real*8 (A-H,K-Z) dimension ETETA(iksay),ETET(iksay),EG1(ibsay),EG2(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) ETETA(jm)= ETET(jm)*EG1(J) + EG2(J) + - (EALFA2(J)**2*ED2(J)*(1 - EAlan(J)*EALFA1(J)**2*EGAMA(J))* - Cosh(EALFA1(J)*ETET(jm)) + - EALFA2(J)**2*ED1(J)*(1 - EAlan(J)*EALFA1(J)**2*EGAMA(J))* - Sinh(EALFA1(J)*ETET(jm)) - - EALFA1(J)**2*(-1 + EAlan(J)*EALFA2(J)**2*EGAMA(J))* - (ED4(J)*Cosh(EALFA2(J)*ETET(jm)) + - ED3(J)*Sinh(EALFA2(J)*ETET(jm))))/ - (EAlan(J)*EALFA1(J)**2*EALFA2(J)**2*EElas*Er(J)) return end 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) implicit real*8 (A-H,K-Z) dimension REK1(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)

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@ ,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*

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- (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)))

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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) implicit real*8 (A-H,K-Z) 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),EEIx(ibsay),EEIxy(ibsay) 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))*

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- (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)

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implicit real*8 (A-H,K-Z) dimension REK1(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),EEIx(ibsay),EEIxy(ibsay) 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) implicit real*8 (A-H,K-Z) 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),EEIx(ibsay),EEIxy(ibsay) 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) implicit real*8 (A-H,K-Z) dimension REK1(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),EEIx(ibsay),EEIxy(ibsay) REK1(1)=1 REK1(2)=0 REK1(3)=-1 REK1(4)=0

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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) implicit real*8 (A-H,K-Z) 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),EEIx(ibsay),EEIxy(ibsay) 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*

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- ((-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) implicit real*8 (A-H,K-Z) dimension REK1(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),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)+

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- 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) implicit real*8 (A-H,K-Z) 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),EEIx(ibsay),EEIxy(ibsay) 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))+

315

- 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)

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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)= EKAT(jm)*EF1(J) + EF2(J) + (-(EALFA1(J)**2* - EALFA2(J)**2*EKAT(jm)**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))*EKAT(jm) - + 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)* - EKAT(jm)**2) + - EAlan(J)*EBETA3(J)**2*EElas*Er(J)* - (EEIx(J)*(-6*EHP*(2*EPx + EHP*EWx) + - 4*(EPx + EHP*EWx)*EKAT(jm) - EWx*EKAT(jm)**2) + - EEIxy(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)* - (EK1(J) - EAlan(J)*EALFA1(J)**2*EGAMA(J)*EK1(J) - - EAlan(J)*EK3(J)*Er(J))*Cosh(EALFA1(J)*EKAT(jm)) + - 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)*EKAT(jm)) - - EALFA1(J)**2*((-1 + EAlan(J)*EALFA2(J)**2*EGAMA(J))*EK1(J)+ - EAlan(J)*EK3(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 UFONKb(ibsay,iksay,EALFA1,EALFA2,EBETA1,EBETA2,EBETA3 @ ,EGAMA,EK1,EK2,EK3,EK4,EElas,EG,EEJ,EDELTA

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@ ,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) 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)= EKAT(jm)*EF1(J)+EF2(J)+(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)*EKAT(jm)) + - 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)*EKAT(jm)) - - EALFA1(J)**2*((-1 + EAlan(J)*EALFA2(J)**2*EGAMA(J))*EK1(J) + - EAlan(J)*EK3(J)*Er(J))*(ED4(J)*Cosh(EALFA2(J)*EKAT(jm)) + - ED3(J)*Sinh(EALFA2(J)*EKAT(jm))))/ - (EAlan(J)*EALFA1(J)**2*EALFA2(J)**2*EElas*Er(J)) 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) implicit real*8 (A-H,K-Z) dimension REK1(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) 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*

318

- (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* - (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))))/ - (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)*(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))+ - 2*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) - + EAlan(J-1)*EBETA3(J-1)**2*EElas*Er(J-1)* - (EEIxy(J-1)*(4*(EPx + EHP*EWx) - 2*EWx*EZ) + - EEIy(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*EK4(J-1)*Er(J-1)* - (EK3(J-1)*EWx + EK4(J-1)*EWy) + - 6*EAlan(J-1)*EBETA3(J-1)**2*EElas*EHP*EK2(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*EK4(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*EK4(J-1)*(EK3(J-1)*EPx + EK4(J-1)*EPy)*Er(J-1)- - 2*EGAMA(J-1)*EK2(J-1)*(EK3(J-1)*EWx + EK4(J-1)*EWy)+ - EHP**2*EK4(J-1)*Er(J-1)*(EK3(J-1)*EWx + EK4(J-1)*EWy))- - 6*EK2(J-1)*(2*EBETA2(J-1)*EEJ(J-1)*EG*

319

- (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) implicit real*8 (A-H,K-Z) 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)=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)*

320

- (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))))/ - (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*EK4(J-1)*Er(J-1)* - (EK3(J-1)*EWx + EK4(J-1)*EWy) + - 6*EAlan(J-1)*EBETA3(J-1)**2*EElas*EHP*EK2(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*EK4(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*EK4(J-1)*(EK3(J-1)*EPx + EK4(J-1)*EPy)*Er(J-1)- - 2*EGAMA(J-1)*EK2(J-1)*(EK3(J-1)*EWx + EK4(J-1)*EWy) + - EHP**2*EK4(J-1)*Er(J-1)*(EK3(J-1)*EWx + EK4(J-1)*EWy))- - 6*EK2(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*EK2(J-1)) +

321

- 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) implicit real*8 (A-H,K-Z) dimension REK1(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) 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)

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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) implicit real*8 (A-H,K-Z) dimension REK1(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) 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

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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) implicit real*8 (A-H,K-Z) 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)=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) implicit real*8 (A-H,K-Z) dimension REK1(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) 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))*

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- (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* - (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* - (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))))/ - (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)* - ((-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)))

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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) implicit real*8 (A-H,K-Z) 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)=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))))/

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- (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) implicit real*8 (A-H,K-Z) dimension EV(iksay),EKAT(iksay),EP1(ibsay),EP2(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),EEIy(ibsay),EEIxy(ibsay) EV(jm)= EKAT(jm)*EP1(J) + EP2(J) + (EALFA1(J)**2* - EALFA2(J)**2*EKAT(jm)**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))*EKAT(jm)+ - 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)* - EKAT(jm)**2) +

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- 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) implicit real*8 (A-H,K-Z) dimension EV(iksay),EKAT(iksay),EP1(ibsay),EP2(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),EEIy(ibsay),EEIxy(ibsay) EV(jm)= EKAT(jm)*EP1(J) + EP2(J) + (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))))/ - (EAlan(J)*EALFA1(J)**2*EALFA2(J)**2*EElas*Er(J)) return end C ################################################################# C ################################################################# C # GAUSS-ELIMINASYON ILE [AA]{XB}={B.X} # C # DONUSTURULUYOR VE SONDEN YERINE KOYMA # C # ILE {XB} VEKTORU HESAPLANIYOR #

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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 # MATRIX INVERSION USING GAUSS-JORDAN REDUCTION AND PARTIAL PIVOTING. # C # MATRIX B IS THE MATRIX TO BE INVERTED AND A IS THE INVERTED MATRIX. # C ######################################################################### 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

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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 ******************************************************************************

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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) c 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 FORMAT(A8) 15 FORMAT(A50) 22 FORMAT(1X,30F10.3) 23 FORMAT(1X,30F10.2) 25 FORMAT(1X,30F13.4) 26 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 CALCULATING THE END COORDINATES OF THE EACH ELEMENT 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 C CALCULATING THE LENGTH OF THE ELEMENTS c WRITE(6,*) ' ' c WRITE(6,*) '--------------------- ' c WRITE(6,*) ' LENGTH OF ELEMENTS ' c WRITE(6,*) '--------------------- ' DO I=1,EN DIS(I)=((EKJX(I)-EKIX(I))**2+(EKJY(I)-EKIY(I))**2)**(0.5) c WRITE(6,22) DIS(I) ENDDO C------------------------------------------------------------ C CALCULATING THE CROSS-SECTION AREA OF THE EACH ELEMENT C------------------------------------------------------------ C KÖŞE NOKTALARINDA FAZLA ALANI IHMAL ICIN YAKLAŞIK YÖNTEM DO I=1,EN

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C fark=EDJ(I)-EDI(I) C IF((fark.GT.TOL1).AND.(fark.LT.TOL2)) THEN ELAREA(I)= DIS(I) * THICK(I) C else C ELAREA(I)= (DIS(I)-THICK(I)/2) * THICK(I) C ENDIF ENDDO C CALCULATING THE CENTROID OF THE EACH ELEMENT DO I=1,EN EKXCC(I)=(EKJX(I)+EKIX(I))/2 EKYCC(I)=(EKJY(I)+EKIY(I))/2 ENDDO 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 CALCULATING THE ANGLE BETWEEN LOCAL AXES OF THE EACH ELEMENT AND GLOBAL X AXIS 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 107 103 IF(EKIY(I).LT.EKJY(I)) GOTO 105 TETA(I)=PI/2.0 GOTO 110 105 TETA(I)=-PI/2.0 GOTO 110 104 IF(EKIX(I).LT.EKJX(I)) GOTO 106 TETA(I)=PI GOTO 110 106 TETA(I)=0.0 GOTO 110 107 SLOPE=(EKIY(I)-EKJY(I))/(EKIX(I)-EKJX(I)) IF(EKIX(I).GT.EKJX(I)) GOTO 108 GOTO 109 108 TETA(I)=PI-DATAN(SLOPE) GOTO 110 109 TETA(I)= -DATAN(SLOPE) 110 SN2TE(I)=DSIN(2.*TETA(I)) CS2TE(I)=DCOS(2.*TETA(I)) ENDDO c WRITE(6,26) DO I=1,EN TETA(I)=TETA(I)*180/PI c WRITE(6,27) I,DIS(I),ELAREA(I),AME1(I),AME2(I),TETA(I) ENDDO c WRITE(6,*) ' ' c WRITE(6,*) '----------------------------------------' c WRITE(6,*) ' TOTAL CROSS-SECTION AREA OF THE SECTION' c WRITE(6,*) '----------------------------------------' C CALCULATING THE TOTAL CROSS-SECTION AREA OF THE SECTION SUMAREA = 0.0 DO I=1,EN SUMAREA = SUMAREA + ELAREA(I)

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ENDDO c WRITE(6,25) SUMAREA sSUMAREA(JJ)=SUMAREA C CALCULATING COORDINATES OF THE CENTROID OF THE SECTION SUMXSA=0.0 SUMYSA=0.0 DO I=1,EN XSBAR=ELAREA(I)*EKXCC(I) YSBAR=ELAREA(I)*EKYCC(I) SUMXSA=SUMXSA+XSBAR SUMYSA=SUMYSA+YSBAR ENDDO XC=SUMXSA/SUMAREA YC=SUMYSA/SUMAREA c WRITE(6,*) ' ' c WRITE(6,*) ' Coordinates of the Centroid ' c WRITE(6,*) '-------------------------------- ' c WRITE(6,*) ' Xc Yc ' c WRITE(6,*) '-------------------------------- ' c WRITE(6,25) XC,YC sXC(JJ)=XC sYC(JJ)=YC C DETERMINATION OF MOMENT OF INERTIA OF THE SECTION ABOUT AXES C PARALLEL TO THE GLOBAL AXES AT THE CENTROID TAMX=0.0 TAMY=0.0 TAMXY=0.0 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 c WRITE(6,*) ' ' c WRITE(6,*) ' Moments of inertia of the section acording to ' c WRITE(6,*) ' x-y axes locating on centroid of the section ' c WRITE(6,*) '------------------------------------------------ ' c WRITE(6,*) ' Ixc Iyc Ixyc ' c WRITE(6,*) '------------------------------------------------ ' c WRITE(6,25) TAMX,TAMY,TAMXY sTAMX(JJ)=TAMX sTAMY(JJ)=TAMY sTAMXY(JJ)=TAMXY C 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

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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 WRITE(6,*) ' ' c WRITE(6,*) ' MAX AND MIN MOMENTS OF INERTIA ' c WRITE(6,*) '----------------------------------------------- ' c WRITE(6,*) ' Imaxc Iminc ALFA ' c WRITE(6,*) '----------------------------------------------- ' c 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 WRITE(6,*) ' ' c WRITE(6,*) ' TORSIONAL CONSTANT ' c WRITE(6,*) '-------------------- ' c WRITE(6,*) ' J ' c WRITE(6,*) '-------------------- ' c WRITE(6,25) TCONS sTCONS(JJ)=TCONS C------------------------------------------------------ C DETERMINATION OF THE SHEAR CENTER C------------------------------------------------------ C WRITE(6,*) ' ' C WRITE(6,*) ' ' C WRITE(6,*) ' DETERMINATION OF X AND Y COORDINATES OF NODES ' C WRITE(6,*) ' ' C WRITE(6,*) ' Nix Niy Njx Njy ' C WRITE(6,*) '---------------------------------------------------- ' DO I=1,EN c Centroid'te bulunan, global eksen takımına paralel c x,y eksen takımına göre düğüm noktası koordinatları Cix(I)=EKIX(I)-XC Ciy(I)=EKIY(I)-YC Cjx(I)=EKJX(I)-XC Cjy(I)=EKJY(I)-YC C WRITE(6,28) CPRMix(I),CPRMiy(I),CPRMjx(I),CPRMjy(I) ENDDO C DETERMINATION OF SECTORIAL AREA DIAGRAM WITH AN ARBITRARY ORIGIN C AND A POLE (IN THIS PROGRAM, THE ORIGIN IS THE I NODE OF THE ELEMENT C AND THE CENTROID IS THE POLE) C WRITE(6,*) ' ' DO I=1,EN C iki vektörün vektörel çarpımı SCAR1(I)=((EKIX(I)-XC)*(EKJY(I)-YC)-(EKJX(I)-XC)*(EKIY(I)-YC)) C WRITE(6,22) SCAR1(I) ENDDO C WRITE(6,*) ' ' C WRITE(6,*) '------------------------------------------ '

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C WRITE(6,*) ' W PRIME (the centroid is the pole) ' C 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 c WRITE(6,*) ' ' DO I=1,PN C WRITE(6,22) SECAREA11(I) ENDDO 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))) ENDDO c WRITE(6,*) ' ' c WRITE(6,*) ' SWXga SWYga ' c WRITE(6,*) '------------------------------' c 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 WRITE(6,*) ' ' c WRITE(6,*) ' COORDINATES OF THE SHEAR CENTER (IN LOCAL AXES) ' c WRITE(6,*) '-----------------------------------------------------' c WRITE(6,*) ' Xsl Ysl ' c WRITE(6,*) '-----------------------------------------------------' c WRITE(6,25) XSHEAR,YSHEAR c WRITE(6,*) ' ' c WRITE(6,*) ' COORDINATES OF THE SHEAR CENTER (IN GLOBAL AXES) ' c WRITE(6,*) '-----------------------------------------------------' c WRITE(6,*) ' Xsg Ysg ' c WRITE(6,*) '-----------------------------------------------------' c WRITE(6,25) XGSHR,YGSHR sXGSHR(JJ)=XGSHR sYGSHR(JJ)=YGSHR RETURN END

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c ********************************************************************* c c ÇARPILMA ATALET MOMENTİ 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) c 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 FORMAT(A8) 920 FORMAT(A50) 922 FORMAT(1X,30F10.3) 923 FORMAT(1X,30F10.2) 925 FORMAT(1X,30F13.4) TOL1=0.9999999 TOL2=1.0000001 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 c WRITE(6,*) ' ' DO I=1,EN C iki vektörün vektörel çarpımı SCAR(I)=((EKIX(I)-Sx)*(EKJY(I)-Sy)-(EKJX(I)-Sx)*(EKIY(I)-Sy)) C WRITE(6,922) SCAR(I) ENDDO c WRITE(6,*) ' ' c WRITE(6,*) ' W PRIME (for initial radius) ' c WRITE(6,*) '------------------------------- ' SECAREA1(1)=0.0 C 1. YÖNTEM 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 WRITE(6,*) ' ' DO I=1,PN c WRITE(6,922) SECAREA1(I) ENDDO C 2. YÖNTEM

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C DO I=2,PN C SECAREA1(I)=SECAREA1(EDI(I-1))+SCAR(I-1) C ENDDO C WRITE(6,*) ' ' C DO I=1,PN C WRITE(6,922) SECAREA1(I) C ENDDO c WRITE(6,*) ' ' c WRITE(6,*) ' LENGTH OF ELEMENTS ' c WRITE(6,*) '--------------------- ' DO I=1,EN DIS(I)=((DX(EDJ(I))-DX(EDI(I)))**2 @ +(DY(EDJ(I))-DY(EDI(I)))**2)**(0.5) c WRITE(6,922) DIS(I) ENDDO c WRITE(6,*) ' ' c WRITE(6,*) ' SWS ' c 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 c WRITE(6,922) SWS R=SWS/CSAREA(JJ) c WRITE(6,*) ' ' c WRITE(6,*) ' R ' c WRITE(6,*) '------' c WRITE(6,922) R c WRITE(6,*) ' ' c WRITE(6,*) ' W PRI ' c WRITE(6,*) '---------' DO I=1,PN PRSECAREA(I)=SECAREA1(I)-R c 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 WRITE(6,*) ' ' c WRITE(6,*) ' Iw ' c WRITE(6,*) '---------' c WRITE(6,922) SSIW SIW(JJ)=SSIW

337

RETURN END C ############################################################# c # PROGRAM MAIN # C # JACK3.FOR # C # JACOBI METODU ILE SERBEST TITRESIM ANALIZI YAPILIYOR. # C # SISTEM MATRISLERI KARE OLARAK SAKLANIYOR. # C # OZVEKTORLER KUTLEYE GORE NORMALIZE EDILIYOR. # C ############################################################# 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) C ##### 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) DO 233 I=1,NTD DO 233 J=1,NTD EGNVAL(I)=EGNVALS(I) 233 EGNVEC(J,I)=EGNVECS(J,I) RETURN END C ################################################################# C # OZDEGER PROBLEMINI COZMEK ICIN ALTPROGRAM: #

338

C # [A]{X} = LAMBDA.[B]{X} # C # PROGRAM SADECE POSITIVE-DEFINITE [B] MATRISI ICIN COZUM # C # YAPAR V, VT, W AND IH'NIN DIMENSION'LARI AYNI OLMALIDIR. # C ################################################################# 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) C [B] MATRISI DIAGONAL HALE GETIRILIYOR CALL JACOBI (N,B,V,XX,IH,NEQ) C DIAGONAL [B] SIMETRIK HALE GETIRILIYOR DO 10 I=1,N DO 10 J=1,N 10 B(J,I)=B(I,J) C [B]'NIN POSITIVE-DEFINITE DURUMU KONTROL EDILIYOR DO 30 I=1,N IF(B(I,I)) 20,30,30 20 WRITE(6,80) STOP 30 CONTINUE C [B]'NIN OZVEKTORLERI ARRAY V(I,J)'DE SAKLANIYOR DO 40 I=1,N DO 40 J=1,N 40 VT(I,J)=V(J,I) C [F]=[VT][A][V] BULUNUYOR VE [A] MATRISI OLARAK SAKLANIYOR CALL MTRXML (VT,N,N,A,N,W,NQ) CALL MTRXML (W,N,N,V,N,A,NQ) DO 50 I=1,N 50 B(I,I)=1.0/DSQRT(B(I,I)) C [Q]=[B][A][B] BULUNUYOR VE [A] MATRISI OLARAK SAKLANIYOR CALL MTRXML (B,N,N,A,N,W,NQ) CALL MTRXML (W,N,N,B,N,A,NQ) C [Q]{Z}=LAMDA{Z} OLUSTURULUYOR VE [Q] DIAGONAL HALE GETIRILIYOR C ( OZDEGERLER HESAPLANIYOR ) CALL JACOBI (N,A,VT,XX,IH,NEQ) C OZDEGERLER DIAG [A] OLARAK GERI DONUYOR DO 60 J=1,N 60 XX(J)=A(J,J) C ASAGIDAKI ILISKIDEN OZVEKTORLER HESAPLANIYOR, C {X}=[V][GI]{Z}=[V][B][VT] CALL MTRXML (V,N,N,B,N,W,NQ) CALL MTRXML (W,N,N,VT,N,X,NQ) 80 FORMAT (/,'*** [GLM] MATRISI POSITIVE-DEFINITE DEGIL ***') RETURN END

339

C ################################################################# C # AMAC : [Q] MATRISINI DIAGONAL HALE GETIRMEK # C # DEGISKENLERIN TANIMI # C # N : REEL,SIMETRIK [Q] MATRISININ MERTEBESI (N > 2) # C # [Q] : DIAGONAL HALE GETIRILECEK MATRIS # C # [V] : OZVEKTOR MATRISI # C # M : UYGULANAN ROTASYON SAYISI # C ################################################################# 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 20 V(I,J)=1.0 GO TO 40 30 V(I,J)=0.0 40 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 60 X(I)=DABS(Q(I,J)) IH(I)=J 70 CONTINUE 75 DO 100 I=1,MI IF(I-1) 90,90,80 80 IF(XMAX-X(I)) 90,100,100 90 XMAX=X(I) IP=I JP=IH(I) 100 CONTINUE IF(XMAX-EPSI) 500,500,110 110 M=M+1 IF(Q(IP,IP)-Q(JP,JP)) 120,130,130 120 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 140 130 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)) 140 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)) 150,190,190 150 TEMP=Q(IP,IP) Q(IP,IP)=Q(JP,JP) Q(JP,JP)=TEMP IF(SINE) 160,170,170 160 TEMP=COSN GO TO 180 170 TEMP=-COSN 180 COSN=DABS(SINE) SINE=TEMP 190 DO 260 I=1,MI IF(I-IP) 210,260,200 200 IF(I-JP) 210,260,210 210 IF(IH(I)-IP) 220,230,220

340

220 IF(IH(I)-JP) 260,230,260 230 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))) 240,240,250 240 X(I)=DABS(Q(I,J)) IH(I)=J 250 CONTINUE Q(I,K)=TEMP 260 CONTINUE X(IP)=0.0 X(JP)=0.0 DO 430 I=1,N IF(I-IP) 270,430,320 270 TEMP=Q(I,IP) Q(I,IP)=COSN*TEMP+SINE*Q(I,JP) IF(X(I)-DABS(Q(I,IP))) 280,290,290 280 X(I)=DABS(Q(I,IP)) IH(I)=IP 290 Q(I,JP)=-SINE*TEMP+COSN*Q(I,JP) IF(X(I)-DABS(Q(I,JP))) 300,430,430 300 X(I)=DABS(Q(I,JP)) IH(I)=JP GO TO 430 320 IF(I-JP) 330,430,380 330 TEMP=Q(IP,I) Q(IP,I)=COSN*TEMP+SINE*Q(I,JP) IF(X(IP)-DABS(Q(IP,I))) 340,350,350 340 X(IP)=DABS(Q(IP,I)) IH(IP)=I 350 Q(I,JP)=-SINE*TEMP+COSN*Q(I,JP) IF(X(I)-DABS(Q(I,JP))) 300,430,430 380 TEMP=Q(IP,I) Q(IP,I)=COSN*TEMP+SINE*Q(JP,I) IF(X(IP)-DABS(Q(IP,I))) 390,400,400 390 X(IP)=DABS(Q(IP,I)) IH(IP)=I 400 Q(JP,I)=-SINE*TEMP+COSN*Q(JP,I) IF(X(JP)-DABS(Q(JP,I))) 410,430,430 410 X(JP)=DABS(Q(JP,I)) IH(JP)=I 430 CONTINUE DO 450 I=1,N TEMP=V(I,IP) V(I,IP)=COSN*TEMP+SINE*V(I,JP) 450 V(I,JP)=-SINE*TEMP+COSN*V(I,JP) GO TO 75 500 RETURN END C ################################################################# C # [C]=[A][B] MATRIS CARPIMI OLUSTURULUYOR # C ################################################################# 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 10 C(I,J)=C(I,J)+A(I,K)*B(K,J) RETURN END

341

C ################################################################# C # A MATRISI ILE B VEKTORUNU CARPARAK C VEKTORUNU OLUSTURULUYOR # C # [C]=[A][B] MATRIS CARPIMI OLUSTURULUYOR # C ################################################################# 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 10 C(I)=C(I)+A(I,K)*B(K) RETURN END C ################################################################# C # A MATRISI ILE B VEKTORUNU CARPARAK C SATISI OLUSTURULUYOR # C # [C]=[A][B] MATRIS CARPIMI OLUSTURULUYOR # C ################################################################# 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 10 C=C+A(I)*B(I) RETURN END C ################################################################# C ################################################################# C # GAUSS-ELIMINASYON ILE [AA]{XB}={B.X} # C # DONUSTURULUYOR VE SONDEN YERINE KOYMA # C # ILE {XB} VEKTORU HESAPLANIYOR # C ################################################################# C ################################################################# 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

342

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

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