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
Cevher Deha TÜRKÖZER
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 For this purpose, the connecting beams have been replaced by an equivalent layered medium and unit forces have been applied in the directions of the degrees of freedom to find the displacements of the system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
Cevher Deha TÜRKÖZER
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
4.3.4.1.(4). The Relative Vertical Displacement due to
the Shearing Effect in the Laminae ......................... 38
4.3.4.1.(5). The Relative Vertical Displacement due to
the Flexibility of the Connections ........................... 39
4.3.4.1.(6). The Relative Vertical Displacement due to
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
Table 6.3. Maximum displacement (m) in the X direction of point G
for damped case in Example 2 ............................................................ 89
Table 6.4. Comparison of the natural frequencies (Hz) obtained from the
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
Table 6.7. Comparison of the natural frequencies (Hz) obtained from the
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
Table 6.9. Maximum displacement (m) in the Y direction of point G
for damped case in Example 6 .......................................................... 107
Table 6.10. Maximum displacement (m) in the Y direction of point G
for undamped case in Example 6 ...................................................... 109
Table 6.11. Maximum displacement (m) in the Y direction of point G
for damped case in Example 6 .......................................................... 109
Table 6.12. Comparison of the natural frequencies (Hz) obtained from the
present program and SAP2000 in Example 7 .................................... 113
Table 6.13. Maximum displacement (m) in the X direction of point G
for undamped case in Example 8 ...................................................... 118
Table 6.14. Maximum displacement (m) in the X direction of point G
for damped case in Example 8 .......................................................... 118
IX
Table 6.15. Comparison of the natural frequencies (Hz) obtained from the
present program and SAP2000 in Example 9 .................................... 122
Table 6.16. Maximum displacement (m) in the X direction of point G
for undamped case in Example 10 .................................................... 127
Table 6.17. Maximum displacement (m) in the X direction of point G
for damped case in Example 10 ........................................................ 127
Table 6.18. Comparison of the natural frequencies (Hz) obtained from the
present program and SAP2000 in Example 11 .................................. 131
Table 6.19. Maximum displacement (m) in the X direction of point G
for undamped case in Example 12 .................................................... 136
Table 6.20. Maximum displacement (m) in the X direction of point G
for damped case in Example 12 ........................................................ 136
Table 6.21. Comparison of the natural frequencies (Hz) obtained from the
present program and SAP2000 in Example 13 .................................. 141
Table 6.22. Maximum displacement (m) in the X direction of point G
for undamped case in Example 14 .................................................... 146
Table 6.23. Maximum displacement (m) in the X direction of point G
for damped case in Example 14 ........................................................ 146
Table 6.24. Comparison of the natural frequencies (Hz) obtained from the
present program and SAP2000 for unstiffened case in Example 15... 150
Table 6.25. Comparison of the natural frequencies (Hz) obtained from the
present program and SAP2000 for stiffened case in Example 15 ...... 151
Table 6.26. Maximum displacement (m) in the X direction of point G
for unstiffened case in Example 16 ................................................... 159
Table 6.27. Maximum displacement (m) in the X direction of point G
for stiffened case in Example 16 ....................................................... 159
Table 6.28. Comparison of the natural frequencies (Hz) obtained from the
present program and SAP2000 for unstiffened case in Example 17... 163
Table 6.29. Comparison of the natural frequencies (Hz) obtained from the
present program and SAP2000 for stiffened case in Example 17 ...... 164
X
Table 6.30. Maximum displacement (m) in the X direction of point G
for unstiffened case in Example 18 ................................................... 172
Table 6.31. Maximum displacement (m) in the X direction of point G
for stiffened case in Example 18 ....................................................... 172
Table 6.32. Comparison of the natural frequencies (Hz) obtained from the
present program and SAP2000 for unstiffened case in Example 19... 176
Table 6.33. Comparison of the natural frequencies (Hz) obtained from the
present program and SAP2000 for stiffened case in Example 19 ...... 177
Table 6.34. Maximum displacement (m) in the X direction of point G
for unstiffened case in Example 20 ................................................... 185
Table 6.35. Maximum displacement (m) in the X direction of point G
for stiffened case in Example 20 ....................................................... 185
Table 6.36. Comparison of the natural frequencies (Hz) obtained from the
present program and SAP2000 in Example 21 .................................. 189
Table 6.37. Maximum displacement (m) in the X direction of the point O
for undamped case in Example 22 .................................................... 194
Table 6.38. Maximum displacement (m) in the X direction of the point O
for damped case in Example 22 ........................................................ 194
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
directions found by both the present program and SAP2000
in Example 1 ...................................................................................... 86
Figure 6.5. Comparison of seventh and eighth mode shapes in X and Y
directions found by both the present program and SAP2000
in Example 1 ...................................................................................... 87
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
Figure 6.9. Time-varying displacements in X direction at the top of the
shear wall for damped case with 5 % damping ratio in Example 2 ...... 90
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
directions found by both the present program and SAP2000
in Example 3 ...................................................................................... 94
Figure 6.13. Comparison of fourth, fifth and sixth mode shapes in X and Y
directions found by both the present program and SAP2000
in Example 3 ...................................................................................... 95
Figure 6.14. Comparison of seventh and eighth mode shapes in X and Y
directions found by both the present program and SAP2000
in Example 3 ...................................................................................... 96
Figure 6.15. Cross-sectional view of the structure and applied
dynamic load in Example 4 ................................................................ 97
Figure 6.16. Rectangular pulse force in Example 4................................................. 97
Figure 6.17. Time-varying displacements in X direction at the top of the
shear wall for undamped case in Example 4 ....................................... 99
Figure 6.18. Time-varying displacements in X direction at the top of the
shear wall for damped case with 5 % damping ratio in Example 4 ...... 99
Figure 6.19. Cross-sectional view of the structure in Example 5 ........................... 100
Figure 6.20. Frame model of the structure in Example 5 and its 3-D view ............ 101
Figure 6.21. Comparison of first, second and third mode shapes in X and Y
directions found by both the present program and SAP2000
in Example 5 .................................................................................... 103
Figure 6.22. Comparison of fourth, fifth and sixth mode shapes in X, Y and Teta
directions found by both the present program and SAP2000
in Example 5 .................................................................................... 104
Figure 6.23. Comparison of seventh and eighth mode shapes in X, Y and Teta
directions found by both the present program and SAP2000
in Example 5 .................................................................................... 105
Figure 6.24. Cross-sectional view of the structure and applied
dynamic load in Example 6 .............................................................. 106
Figure 6.25. Rectangular pulse force in Example 6............................................... 107
XIV
Figure 6.26. Time-varying displacements in Y direction at the top of the
shear wall for undamped case in Example 6 ..................................... 108
Figure 6.27. Time-varying displacements in Y direction at the top of the
shear wall for damped case with 5 % damping ratio in Example 6 .... 108
Figure 6.28. Sine pulse force in Example 6 .......................................................... 109
Figure 6.29. Time-varying displacements in Y direction at the top of the
shear wall for undamped case in Example 6 ..................................... 110
Figure 6.30. Time-varying displacements in Y direction at the top of the
shear wall for damped case with 5 % damping ratio in Example 6 .... 110
Figure 6.31. Cross-sectional view of the structure in Example 7 ........................... 111
Figure 6.32. Frame model of the structure in Example 7 and its 3-D view ............ 112
Figure 6.33. Comparison of first, second and third mode shapes in X and Y
directions found by both the present program and SAP2000
in Example 7 .................................................................................... 114
Figure 6.34. Comparison of fourth, fifth and sixth mode shapes in X, Y and Teta
directions found by both the present program and SAP2000
in Example 7 .................................................................................... 115
Figure 6.35. Comparison of seventh and eighth mode shapes in X, Y and Teta
directions found by both the present program and SAP2000
in Example 7 .................................................................................... 116
Figure 6.36. Cross-sectional view of the structure and applied
dynamic load in Example 8 .............................................................. 117
Figure 6.37. Triangular pulse force in Example 8 ................................................. 118
Figure 6.38. Time-varying displacements in X direction at the top of the
shear wall for undamped case in Example 8 ..................................... 119
Figure 6.39. Time-varying displacements in X direction at the top of the
shear wall for damped case with 6 % damping ratio in Example 8 .... 119
Figure 6.40. Cross-sectional view of the structure in Example 9 ........................... 120
Figure 6.41. Frame model of the structure in Example 9 and its 3-D view ............ 121
XV
Figure 6.42. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by t both he present program and SAP2000
in Example 9 .................................................................................... 123
Figure 6.43. Comparison of fourth, fifth and sixth mode shapes in X and Y
directions found by both the present program and SAP2000
in Example 9 .................................................................................... 124
Figure 6.44. Comparison of seventh and eighth mode shapes in X and Y
directions found by both the present program and SAP2000
in Example 9 .................................................................................... 125
Figure 6.45. Cross-sectional view of the structure and applied
dynamic load in Example 10 ............................................................ 126
Figure 6.46. Triangular pulse force in Example 10 ............................................... 126
Figure 6.47. Time-varying displacements in X direction at the top of the
shear wall for undamped case in Example 10 ................................... 128
Figure 6.48. Time-varying displacements in X direction at the top of the
shear wall for damped case with 7 % damping ratio in Example 10 .. 128
Figure 6.49. Cross-sectional view of the structure in Example 11 ......................... 129
Figure 6.50. Frame model of the structure in Example 11 and its 3-D view .......... 130
Figure 6.51. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by both the present program and SAP2000
in Example 11 .................................................................................. 132
Figure 6.52. Comparison of fourth, fifth and sixth mode shapes in X, Y and Teta
directions found by both the present program and SAP2000
in Example 11 .................................................................................. 133
Figure 6.53. Comparison of seventh and eighth mode shapes in X and Y
directions found by both the present program and SAP2000
in Example 11 .................................................................................. 134
Figure 6.54. Cross-sectional view of the structure and applied
dynamic load in Example 12 ............................................................ 135
Figure 6.55. Rectangular pulse force in Example 12 ............................................. 136
XVI
Figure 6.56. Time-varying displacements in X direction at the top of the
shear wall for undamped case in Example 12 ................................... 137
Figure 6.57. Time-varying displacements in X direction at the top of the
shear wall for damped case with 7 % damping ratio in Example 12 .. 137
Figure 6.58. Non-planar non-symmetrical structure in Example 13 ...................... 138
Figure 6.59. Cross-sectional view of the structure in Example 13 ......................... 139
Figure 6.60. Frame model of the structure in Example 13 and its 3-D view .......... 140
Figure 6.61. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by both the present program and SAP2000
in Example 13 .................................................................................. 142
Figure 6.62. Comparison of fourth, fifth and sixth mode shapes in X, Y and Teta
directions found by both the present program and SAP2000
in Example 13 .................................................................................. 143
Figure 6.63. Comparison of seventh and eighth mode shapes in X and Y
directions found by both the present program and SAP2000
in Example 13 .................................................................................. 144
Figure 6.64. Cross-sectional view of the structure and applied
dynamic load in Example 14 ............................................................ 145
Figure 6.65. Sine pulse force in Example 14 ........................................................ 146
Figure 6.66. Time-varying displacements in X direction at the top of the
shear wall for undamped case in Example 14 ................................... 147
Figure 6.67. Time-varying displacements in X direction at the top of the
shear wall for damped case with 6 % damping ratio in Example 14 .. 147
Figure 6.68. Cross-sectional view of the structure in Example 15 ......................... 148
Figure 6.69. Frame model of the structure in Example 15 and its 3-D view .......... 149
Figure 6.70. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by both the present program and
SAP2000 for stiffened case in Example 15 ....................................... 152
Figure 6.71. Comparison of fourth, fifth and sixth mode shapes in X and Y
directions found by both the present program and
SAP2000 for stiffened case in Example 15 ....................................... 153
XVII
Figure 6.72. Comparison of seventh and eighth mode shapes in X and Y
directions found by both the present program and
SAP2000 for stiffened case in Example 15 ....................................... 154
Figure 6.73. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by the present program (CCM) for
stiffened and unstiffened cases in Example 15 .................................. 155
Figure 6.74. Comparison of fourth, fifth and sixth mode shapes in X and Y
directions found by the present program (CCM) for
stiffened and unstiffened cases in Example 15 .................................. 156
Figure 6.75. Comparison of seventh and eighth mode shapes in X and Y
directions found by the present program (CCM) for
stiffened and unstiffened cases in Example 15 .................................. 157
Figure 6.76. Cross-sectional view of the structure and applied
dynamic load in Example 16 ............................................................ 158
Figure 6.77. Rectangular pulse force in Example 16 ............................................. 159
Figure 6.78. Time-varying displacements in X direction at the top of the
shear wall for undamped case in Example 16 ................................... 160
Figure 6.79. Time-varying displacements in X direction at the top of the
shear wall for damped case with 5 % damping ratio in Example 16 .. 160
Figure 6.80. Cross-sectional view of the structure in Example 17 ......................... 161
Figure 6.81. Frame model of the structure in Example 17 and its 3-D view .......... 162
Figure 6.82. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by both the present program and
SAP2000 for stiffened case in Example 17 ....................................... 165
Figure 6.83. Comparison of fourth, fifth and sixth mode shapes in X and Y
directions found by both the present program and
SAP2000 for stiffened case in Example 17 ....................................... 166
Figure 6.84. Comparison of seventh and eighth mode shapes in X and Y
directions found by both the present program and
SAP2000 for stiffened case in Example 17 ....................................... 167
XVIII
Figure 6.85. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by the present program (CCM) for
stiffened and unstiffened cases in Example 17 .................................. 168
Figure 6.86. Comparison of fourth, fifth and sixth mode shapes in X and Y
directions found by the present program (CCM) for
stiffened and unstiffened cases in Example 17 .................................. 169
Figure 6.87. Comparison of seventh and eighth mode shapes in X and Y
directions found by the present program (CCM) for
stiffened and unstiffened cases in Example 17 .................................. 170
Figure 6.88. Cross-sectional view of the structure and applied
dynamic load in Example 18 ............................................................ 171
Figure 6.89. Rectangular pulse force in Example 18 ............................................. 172
Figure 6.90. Time-varying displacements in X direction at the top of the
shear wall for undamped case in Example 18 ................................... 173
Figure 6.91. Time-varying displacements in X direction at the top of the
shear wall for damped case with 5 % damping ratio in Example 18 .. 173
Figure 6.92. Cross-sectional view of the structure in Example 19 ......................... 174
Figure 6.93. Frame model of the structure in Example 19 and its 3-D view .......... 175
Figure 6.94. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by both the present program and
SAP2000 for stiffened case in Example 19 ....................................... 178
Figure 6.95. Comparison of fourth, fifth and sixth mode shapes in X and Y
directions found by both the present program and
SAP2000 for stiffened case in Example 19 ....................................... 179
Figure 6.96. Comparison of seventh and eighth mode shapes in X and Y
directions found by both the present program and
SAP2000 for stiffened case in Example 19 ....................................... 180
Figure 6.97. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by the present program (CCM) for
stiffened and unstiffened cases in Example 19 .................................. 181
XIX
Figure 6.98. Comparison of fourth, fifth and sixth mode shapes in X and Y
directions found by the present program (CCM) for
stiffened and unstiffened cases in Example 19 .................................. 182
Figure 6.99. Comparison of seventh and eighth mode shapes in X and Y
directions found by the present program (CCM) for
stiffened and unstiffened cases in Example 19 .................................. 183
Figure 6.100. Cross-sectional view of the structure and applied
dynamic load in Example 20 ........................................................... 184
Figure 6.101. Triangular pulse force in Example 20 .............................................. 185
Figure 6.102. Time-varying displacements in X direction at the top of the
shear wall for undamped case in Example 20 .................................. 186
Figure 6.103. Time-varying displacements in X direction at the top of the
shear wall for damped case with 5 % damping ratio in Example 20 . 186
Figure 6.104. Cross-sectional view of the 1st region of the structure
in Example 21 ................................................................................. 187
Figure 6.105. Cross-sectional view of the 2nd region of the structure
in Example 21 ................................................................................. 188
Figure 6.106. Frame model of the structure in Example 21 and its 3-D view ......... 188
Figure 6.107. Comparison of first, second and third mode shapes in X and Y
directions found by both the present program and SAP2000
in Example 21 ................................................................................. 190
Figure 6.108. Comparison of fourth, fifth and sixth mode shapes in X and Y
directions found by both the present program and SAP2000
in Example 21 ................................................................................. 191
Figure 6.109. Comparison of seventh and eighth mode shapes in X and Y
directions found by both the present program and SAP2000
in Example 21 ................................................................................. 192
Figure 6.110. Cross-sectional view of the structure and applied
dynamic load in Example 22 ........................................................... 193
Figure 6.111. Rectangular pulse force in Example 22............................................ 194
XX
Figure 6.112. Time-varying displacements in X direction at the top of the
shear wall for undamped case in Example 22 .................................. 195
Figure 6.113. Time-varying displacements in X direction at the top of the
shear wall for damped case with 5 % damping ratio
in Example 22 ................................................................................. 195
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
1. INTRODUCTION Cevher Deha TÜRKÖZER
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-
1. INTRODUCTION Cevher Deha TÜRKÖZER
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
1. INTRODUCTION Cevher Deha TÜRKÖZER
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.
2. PREVIOUS STUDIES Cevher Deha TÜRKÖZER
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.
2. PREVIOUS STUDIES Cevher Deha TÜRKÖZER
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.
2. PREVIOUS STUDIES Cevher Deha TÜRKÖZER
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.
2. PREVIOUS STUDIES Cevher Deha TÜRKÖZER
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.
3. METHODS OF ANALYSIS Cevher Deha TÜRKÖZER
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
3. METHODS OF ANALYSIS Cevher Deha TÜRKÖZER
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.
3. METHODS OF ANALYSIS Cevher Deha TÜRKÖZER
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).
3. METHODS OF ANALYSIS Cevher Deha TÜRKÖZER
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
3. METHODS OF ANALYSIS Cevher Deha TÜRKÖZER
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
3. METHODS OF ANALYSIS Cevher Deha TÜRKÖZER
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
3. METHODS OF ANALYSIS Cevher Deha TÜRKÖZER
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
3. METHODS OF ANALYSIS Cevher Deha TÜRKÖZER
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
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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.
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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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
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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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.
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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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
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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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
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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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
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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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
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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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′
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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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)
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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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
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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)
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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
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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,
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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)
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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)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
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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∑∑ ∫∫=
−
=
++=+
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
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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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
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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( )
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
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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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
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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)
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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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)
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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( ) 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
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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( ) 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.
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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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
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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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)
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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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)
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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)
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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)
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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)
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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)
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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 .
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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)
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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)
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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)
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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)
Reorganizing (4.124),
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)
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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)
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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)
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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)
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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)
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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)
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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)
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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
−
−−−
Reorganizing (4.166),
( )[ ]
( ) ( )[ ]( ) ( )[ ] ( ) ( )[ ]
( ) ( )[ ] ( ) ( )( ) ( ) ( )[ ] 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 =
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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
−
−−−
Reorganizing (4.170),
( )[ ]
( ) ( )[ ]( ) ( )[ ]
( ) ( )[ ]( ) ( )( ) ( ) ( )[ ] 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)
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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,
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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)
4. FREE VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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)
5. FORCED VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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)
5. FORCED VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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)
5. FORCED VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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)
5. FORCED VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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.
5. FORCED VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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:
5. FORCED VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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.
5. FORCED VIBRATION ANALYSIS OF NON-PLANAR COUPLED SHEAR WALLS Cevher Deha TÜRKÖZER
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.
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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.
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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)
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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.
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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.
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
85
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
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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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
Mode 4 in 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=1,02500 Hz SBY NF=1,01237 Hz
Mode 5 in X 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=1,53440 Hz CCM DF=1,57316 Hz
Mode 6 in X 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=2,52580 Hz
CCM NF=2,59400 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
87
Figure 6.5. Comparison of seventh and eighth mode shapes in X and Y directions
found by both the present program and SAP2000 in Example 1
Mode 7 in Y Direction
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
Mode 8 in X Direction
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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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.
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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
calculated by the computer program, were concentrated at the center of the whole
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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
92
Figure 6.11. Frame model of the structure in Example 3 and its 3-D view
In this example, the model of the structure was chosen symmetrical with
respect to X and Y axes and the free vibration analysis was carried out. Table 6.4
compares the natural frequencies corresponding to each mode found by the program
prepared in the present work and the SAP2000 structural analysis program,
expressing the percentage differences.
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
93
Table 6.4. Comparison of the natural frequencies (Hz) obtained from the present program and SAP2000 in Example 3
Mode
Present Study (CCM)
SAP2000 (Frame Method)
% difference
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
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. 6.12-14 present the mode shapes of the shear wall, found by the present
program and the SAP2000 structural analysis program, both in the same figure.
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
94
Figure 6.12. Comparison of first, second and third mode shapes in X and Y directions
found by both the present program and SAP2000 in Example 3
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=0,97582 Hz
CCM NF=0,97409 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=1,35782 Hz
CCM NF=1,36735 Hz
Mode 3 in X 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=5,18059 Hz CCM NF=5,20893 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
95
Figure 6.13. Comparison of fourth, fifth and sixth mode shapes in X and Y directions
found by both the present program and SAP2000 in Example 3
Mode 4 in 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=6,00303 Hz
CCM NF=6,00116 Hz
Mode 5 in X 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=11,75796 Hz
SBY NF=11,82911 Hz
Mode 6 in 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=16,51685 Hz CCM DF=16,55209 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
96
Figure 6.14. Comparison of seventh and eighth mode shapes in X and Y directions
found by both the present program and SAP2000 in Example 3
Mode 7 in X 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=21,06216 Hz
CCM NF=21,25545 Hz
Mode 8 in 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=31,74091 Hz
CCM NF=31,93361 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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.
Figure 6.15. Cross-sectional view of the structure and applied dynamic load in
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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
98
At the top of the shear wall, the maximum displacement in the X direction of
point O, the mass center, was calculated by the computer program prepared in the
present study and compared with those of the SAP2000 structural analysis program
in Tables 6.5-6 for both damped and undamped cases.
Table 6.5. Maximum displacement (m) in the X direction of point O for undamped case in Example 4
SAP2000
(Frame Method) Present Study
(CCM) % difference
0.066510 0.065713 1.21
The results of both methods for 5 % damping ratio are given in Table 6.6.
Table 6.6. Maximum displacement (m) in the X direction of point O for damped case in Example 4
SAP2000
(Frame Method) Present Study
(CCM) % difference
0.058340 0.057648 1.20
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
are presented in Figs. 6.17-18.
It is observed that the results obtained in the present work (CCM) coincide
with those of the SAP2000 structural analysis program, perfectly.
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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
Figure 6.18. Time-varying displacements in X direction at the top of the shear wall
for damped case with 5 % damping ratio in Example 4
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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.
Figure 6.19. Cross-sectional view of the structure in Example 5
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.
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 (0.125; 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)
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
101
The structure was solved both by the present method using the CCM and by
the SAP2000 structural analysis program using the frame method for which the
model and its 3-D view are given in Fig. 6.20.
Figure 6.20. Frame model of the structure in Example 5 and its 3-D view
Table 6.7 compares the natural frequencies corresponding to each mode
found by the program prepared in the present work and the SAP2000 structural
analysis program, expressing the percentage differences.
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
102
Table 6.7. Comparison of the natural frequencies (Hz) obtained from the present program and SAP2000 in Example 5
Mode Present Study
(CCM) SAP2000
(Frame Method) % difference Natural Frequencies Natural Frequencies
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
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, Y and Teta directions were compared by normalizing with
respect to the displacements at the top of the structure.
Figs. 6.21-23 present the mode shapes of the shear wall, found by the present
program and the SAP2000 structural analysis program, both in the same figure.
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
103
Figure 6.21. Comparison of first, second and third mode shapes in X and Y directions
found by both the present program and SAP2000 in Example 5
Mode 1 in Y Direction
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
)
Mode 2 in X Direction
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
Mode 3 in 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=4,14449 Hz CCM NF=4,14919 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
104
Figure 6.22. Comparison of fourth, fifth and sixth mode shapes in X, Y and Teta
directions found by both the present program and SAP2000 in Example 5
Mode 4 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=6,86626 Hz SBY NF=6,88659 Hz
Mode 5 in 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=11,33655 Hz
CCM DF=11,36339 Hz
Mode 6 in X 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=16,62251 Hz
CCM NF=16,69150 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
105
Figure 6.23. Comparison of seventh and eighth mode shapes in X, Y and Teta
directions found by both the present program and SAP2000 in Example 5
Mode 7 in 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=21,76464 Hz CCM NF=21,86704 Hz
Mode 7 in Teta Direction
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
Mode 8 in X Direction
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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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.
Figure 6.24. Cross-sectional view of the structure and applied dynamic load in
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)
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
107
100
Figure 6.25. Rectangular pulse force in Example 6
At the top of the shear wall, the maximum displacement in the Y direction of
point G, the mass center, was calculated by the computer program prepared in the
present study and compared with those of the SAP2000 structure analysis program in
Tables 6.8-9 for both damped and undamped cases.
Table 6.8. Maximum displacement (m) in the Y direction of point G for undamped case in Example 6
SAP2000
(Frame Method) Present Study
(CCM) % difference
0.121225 0.122312 0.90
The results of both methods for 5 % damping ratio are given in Table 6.9.
Table 6.9. Maximum displacement (m) in the Y direction of point G for damped case in Example 6
SAP2000
(Frame Method) Present Study
(CCM) % difference
0.110000 0.110792 0.72
The responses for both damped and undamped systems to rectangular force
were determined and the time-varying displacements in the Y direction of point G
are presented in Figs. 6.26-27.
It is observed that the results obtained in the present work (CCM) coincide
with those of the SAP2000 structural analysis program, perfectly.
P(lb)
t(s) 5
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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
for undamped case in Example 6
-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 6.27. Time-varying displacements in Y direction at the top of the shear wall
for damped case with 5 % damping ratio in Example 6
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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
(Frame Method) Present Study
(CCM) % difference
0.123144 0.124157 0.82
The results of both methods for 5 % damping ratio are given in Table 6.11.
Table 6.11. Maximum displacement (m) in the Y direction of point G for damped case in Example 6
SAP2000
(Frame Method) Present Study
(CCM) % difference
0.117102 0.118075 0.83
The responses for both damped and undamped systems to sine force were
determined and the time-varying displacements in the Y direction of point G are
presented in Figs. 6.29-30.
t (s)
P(lb)
150
5
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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
Figure 6.29. Time-varying displacements in Y direction at the top of the shear wall
for undamped case in Example 6
-0,150
-0,100
-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
Figure 6.30. Time-varying displacements in Y direction at the top of the shear wall
for damped case with 5 % damping ratio in Example 6
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
111
Example 7:
In this example, a non-planar non-symmetrical coupled shear wall with the
cross-section shown in Fig. 6.31 was considered and the free vibration analysis was
carried out. The total height of the shear wall is 48 m, the storey height is 3 m, the
thickness is 0.4 m, the height of the connecting beams is 0.5 m and the elasticity and
shear moduli are E = 2.85×106 kN/m2 and G = 1055556 kN/m2, respectively.
Figure 6.31. Cross-sectional view of the structure in Example 7
According to the lumped mass idealization, the lumped masses, which were
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).
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. 6.32.
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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
112
Figure 6.32. Frame model of the structure in Example 7 and its 3-D view
Table 6.12 compares the first twenty natural frequencies corresponding to
each mode found by the program prepared in the present work and the SAP2000
structural analysis program, expressing the percentage differences.
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
113
Table 6.12. Comparison of the natural frequencies (Hz) obtained from the present program and SAP2000 in Example 7
Mode
Present Study (CCM)
SAP2000 (Frame Method)
% difference
Natural Frequencies Natural Frequencies
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
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, Y and Teta directions were compared by normalizing with
respect to the displacements at the top of the structure.
Figs. 6.33-35 present the mode shapes of the shear wall, found by the present
program and the SAP2000 structural analysis program, both in the same figure.
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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
Mode 1 in X Direction
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
Mode 2 in Y Direction
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
Mode 3 in 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=4,14449 Hz CCM NF=4,14919 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
115
Figure 6.34. Comparison of fourth, fifth and sixth mode shapes in X, Y and Teta
directions found by both the present program and SAP2000 in Example 7
Mode 4 in Teta 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=6,86626 Hz
SBY NF=6,88659 Hz
Mode 5 in 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=11,33655 Hz
CCM DF=11,36338 Hz
Mode 6 in X 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=16,62251 Hz
CCM NF=16,69150 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
116
Figure 6.35. Comparison of seventh and eighth mode shapes in X, Y and Teta
directions found by both the present program and SAP2000 in Example 7
Mode 7 in Y Direction
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
Mode 7 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=21,76464 Hz
CCM DF=21,86704 Hz
Mode 8 in X 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=30,65600 Hz
CCM NF=30,88525 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
117
Example 8:
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 sixteen storey shear wall were taken as in Example 7.
The forced vibration analysis of the shear wall was carried out by the computer
program prepared in the present work and the SAP2000 structural analysis program
for damped and undamped cases.
The dynamic load, P(t), was applied at the top of the structure in the global Y
direction in the plane of the shear wall as in Fig. 6.36 and the triangular pulse force
was chosen as in Fig. 6.37.
Figure 6.36. Cross-sectional view of the structure and applied dynamic load in
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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
118
Figure 6.37. Triangular pulse force in Example 8
At the top of the shear wall, the maximum displacement in the X direction of
point G, the mass center, was calculated by the computer program prepared in the
present study and compared with those of the SAP2000 structural analysis program
in Tables 6.13-14 for both damped and undamped cases.
Table 6.13. Maximum displacement (m) in the X direction of point G for undamped case in Example 8
SAP2000
(Frame Method) Present Study
(CCM) % difference
0.036393 0.036112 0.77
The results of both methods for 6 % damping ratio are given in Table 6.14.
Table 6.14. Maximum displacement (m) in the X direction of point G for damped case in Example 8
SAP2000
(Frame Method) Present Study
(CCM) % difference
0.036130 0.035939 0.53
The responses for both damped and undamped systems to triangular force
were determined and the time-varying displacements in the X direction of point G
are presented in Figs. 6.38-39.
It is observed that the results obtained in the present work (CCM) coincide
with those of the SAP2000 structural analysis program, perfectly.
P(lb)
t (s)
5 2,5
150
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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
Figure 6.38. Time-varying displacements in X direction at the top of the shear wall
for undamped case in Example 8
-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
Figure 6.39. Time-varying displacements in X direction at the top of the shear wall
for damped case with 6 % damping ratio in Example 8
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
120
Example 9:
In this example, a non-planar non-symmetrical coupled shear wall with the
cross-section shown in Fig. 6.40 was considered and the free vibration analysis was
carried out. The total height of the shear wall is 50 m, the storey height is 2.5 m and
the thicknesses of the piers and the connecting beams are shown in Fig. 6.40. The
height of the connecting beams is 0.35 m and the elasticity and shear moduli are
=E 2.85×106 kN/m2 and =G 1055556 kN/m2, respectively.
Figure 6.40. Cross-sectional view of the structure in Example 9
According to the lumped mass idealization, the lumped masses, which were
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
(1.111; -1.045).
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. 6.41.
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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
121
Figure 6.41. Frame model of the structure in Example 9 and its 3-D view
Table 6.15 compares the first twenty natural frequencies corresponding to
each mode found by the program prepared in the present work and the SAP2000
structural analysis program, expressing the percentage differences.
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
122
Table 6.15. Comparison of the natural frequencies (Hz) obtained from the present program and SAP2000 in Example 9
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
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, Y and Teta directions were compared by normalizing with
respect to the displacements at the top of the structure.
Figs. 6.42-44 present the mode shapes of the shear wall, found by the present
program and the SAP2000 structural analysis program, both in the same figure.
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
123
Figure 6.42. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by both the present program and SAP2000 in Example 9
Mode 3 in Teta Direction
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
Mode 1 in X Direction
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
Mode 2 in Y Direction
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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
124
Figure 6.43. Comparison of fourth, fifth and sixth mode shapes in X and Y directions
found by both the present program and SAP2000 in Example 9
Mode 4 in Y Direction
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
Mode 5 in Y Direction
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
Mode 6 in X Direction
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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
125
Figure 6.44. Comparison of seventh and eighth mode shapes in X and Y directions
found by both the present program and SAP2000 in Example 9
Mode 7 in Y Direction
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
Mode 8 in X Direction
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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
126
Example 10:
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 twenty storey shear wall were taken as in Example 9.
The forced vibration analysis of the shear wall was carried out by the computer
program prepared in the present work and the SAP2000 structural analysis program
for damped and undamped cases.
The dynamic load, P(t), was applied at the top of the structure in the global X
direction in the plane of the connection beam as in Figure 6.45 and the triangular
pulse force was chosen as in Fig. 6.46.
Figure 6.45. Cross-sectional view of the structure and applied dynamic load in
Example 10
Figure 6.46. Triangular pulse force in 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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
127
At the top of the shear wall, the maximum displacement in the X direction of
point G, the mass center, was calculated by the computer program prepared in the
present study and compared with those of the SAP2000 structural analysis program
in Tables 6.16-17 for both damped and undamped cases.
Table 6.16. Maximum displacement (m) in the X direction of point G for undamped case in Example 10
SAP2000
(Frame Method) Present Study
(CCM) % difference
0.183850 0.182404 0.79
The results of both methods for 7 % damping ratio are given in Table 6.17.
Table 6.17. Maximum displacement (m) in the X direction of point G for damped case in Example 10
SAP2000
(Frame Method) Present Study
(CCM) % difference
0.172310 0.171008 0.76
The responses for both damped and undamped systems to triangular force
were determined and the time-varying displacements in the X direction of point G
are presented in Figures 6.47-48.
It is observed that the results obtained in the present work (CCM) coincide
with those of the SAP2000 structural analysis program, perfectly.
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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
Figure 6.47. Time-varying displacements in X direction at the top of the shear wall
for undamped case in Example 10
-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
Figure 6.48. Time-varying displacements in X direction at the top of the shear wall
for damped case with 7 % damping ratio in Example 10
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
129
Example 11:
In this example, a non-planar non-symmetrical coupled shear wall with the
cross-section shown in Fig. 6.49 was considered and the free vibration analysis was
carried out. The total height of the shear wall is 48 m, the storey height is 3 m and the
thicknesses of the piers and the connecting beams are shown in Fig. 6.49. The height
of the connecting beams is 0.4 m and the elasticity and shear moduli are
=E 2.85×106 kN/m2 and =G 1055556 kN/m2, respectively.
Figure 6.49. Cross-sectional view of the structure in Example 11
According to the lumped mass idealization, the lumped masses, which were
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)
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
130
The structure was solved both by the present method using the CCM and by
the SAP2000 structural analysis program using the frame method for which the
model and its 3-D view are given in Fig. 6.50.
Figure 6.50. Frame model of the structure in Example 11 and its 3-D view
Table 6.18 compares the first twenty natural frequencies corresponding to
each mode found by the program prepared in the present work and the SAP2000
structural analysis program, expressing the percentage differences.
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
131
Table 6.18. Comparison of the natural frequencies (Hz) obtained from the present program and SAP2000 in Example 11
Mode Present Study
(CCM) SAP2000
(Frame Method) % difference
Natural Frequencies Natural Frequencies
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
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, Y and Teta directions were compared by normalizing with
respect to the displacements at the top of the structure.
Figs. 6.51-53 present the mode shapes of the shear wall, found by the present
program and the SAP2000 structural analysis program, both in the same figure.
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
132
Figure 6.51. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by both the present program and SAP2000 in Example 11
Mode 1 in X Direction
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
Mode 2 in Y Direction
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
Mode 3 in Teta 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=2,46396 Hz CCM NF=2,47098 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
133
Figure 6.52. Comparison of fourth, fifth and sixth mode shapes in X, Y and Teta
directions found by both the present program and SAP2000 in Example 11
Mode 4 in 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=4,11839 Hz SBY NF=4,12617 Hz
Mode 5 in X 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=6,47507 Hz
CCM DF=6,51577 Hz
Mode 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=11,45595 Hz
CCM NF=11,50762 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
134
Figure 6.53. Comparison of seventh and eighth mode shapes in X and Y directions
found by both the present program and SAP2000 in Example 11
Mode 7 in Y Direction
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
Mode 8 in X 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=20,10512 Hz
CCM NF=20,48279 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
135
Example 12:
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 sixteen storey shear wall were taken as in Example 11.
The forced vibration analysis of the shear wall was carried out by the computer
program prepared in the present work and the SAP2000 structural analysis program
for damped and undamped cases.
The dynamic load, P(t), was applied at the top of the structure in the global X
direction in the plane of the connection beam as in Fig. 6.54 and the rectangular
pulse force was chosen as in Fig. 6.55.
Figure 6.54. Cross-sectional view of the structure and applied dynamic load in
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)
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
136
100
Figure 6.55. Rectangular pulse force in Example 12
At the top of the shear wall, the maximum displacement in the X direction of
point G, the mass center, was calculated by the computer program prepared in the
present study and compared with those of the SAP2000 structural analysis program
in Tables 6.19-20 for both damped and undamped cases.
Table 6.19. Maximum displacement (m) in the X direction of point G for undamped case in Example 12
SAP2000
(Frame Method) Present Study
(CCM) % difference
0.045550 0.045408 0.31
The results of both methods for 7 % damping ratio are given in Table 6.20.
Table 6.20. Maximum displacement (m) in the X direction of point G for damped case in Example 12
SAP2000
(Frame Method) Present Study
(CCM) % difference
0.040480 0.040359 0.30
The responses for both damped and undamped systems to rectangular force
were determined and the time-varying displacements in the X direction of point G
are presented in Figures 6.56-57.
It is observed that the results obtained in the present work (CCM) coincide
with those of the SAP2000 structural analysis program, perfectly.
P(lb)
t(s)
5
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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
Figure 6.56. Time-varying displacements in X direction at the top of the shear wall
for undamped case in Example 12
-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
Figure 6.57. Time-varying displacements in X direction at the top of the shear wall
for damped case with 7 % damping ratio in Example 12
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
138
Example 13:
In this example, a non-planar non-symmetrical coupled shear wall with the
cross-section shown in Fig. 6.58 was considered and the free vibration analysis was
carried out. The total height of the shear wall is 48 m, the storey height is 3 m and the
thicknesses of the piers and the connecting beams are shown in Fig. 6.59. The height
of the connecting beams is 0.4 m and the elasticity and shear moduli are
=E 2.85×106 kN/m2 and =G 1055556 kN/m2, respectively.
Figure 6.58. Non-planar non-symmetrical structure in Example 13
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 (-0.016; 0.018).
H =
48
m
h =
3 m
X
Y
connecting beams
Z
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
139
Figure 6.59. Cross-sectional view of the structure in Example 13
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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
140
The structure was solved both by the present method using the CCM and by
the SAP2000 structural analysis program using the frame method for which the
model and its 3-D view are given in Fig. 6.60.
Figure 6.60. Frame model of the structure in Example 13 and its 3-D view
Table 6.21 compares the first twenty natural frequencies corresponding to
each mode found by the program prepared in the present work and the SAP2000
structural analysis program, expressing the percentage differences.
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
141
Table 6.21. Comparison of the natural frequencies (Hz) obtained from the present program and SAP2000 in Example 13
Mode
Present Study (CCM)
SAP2000 (Frame Method)
% difference
Natural Frequencies Natural Frequencies
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
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, Y and Teta directions were compared by normalizing with
respect to the displacements at the top of the structure.
Figs. 6.61-63 present the mode shapes of the shear wall, found by the present
program and the SAP2000 structural analysis program, both in the same figure.
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
142
Figure 6.61. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by both the present program and SAP2000 in Example 13
Mode 1 in X Direction
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
Mode 2 in Y Direction
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
Mode 3 in Teta 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=3,20475 Hz
CCM NF=3,20768 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
143
Figure 6.62. Comparison of fourth, fifth and sixth mode shapes in X, Y and Teta
directions found by both the present program and SAP2000 in Example 13
Mode 4 in 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=4,12314 Hz SBY NF=4,13771 Hz
Mode 5 in X 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=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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
144
Figure 6.63. Comparison of seventh and eighth mode shapes in X and Y directions
found by both the present program and SAP2000 in Example 13
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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
145
Example 14:
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 sixteen storey shear wall were taken as in Example 13.
The forced vibration analysis of the shear wall was carried out by the computer
program prepared in the present work and the SAP2000 structural analysis program
for damped and undamped cases.
The dynamic load, P(t), was applied at the top of the structure in the global X
direction in the plane of the connection beam as in Fig. 6.64 and the sine pulse force
was chosen as in Fig. 6.65.
Figure 6.64. Cross-sectional view of the structure and applied dynamic load in
Example 14
H =
48
m
h =
3 m
X
Y Z
P(t)
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
146
Figure 6.65. Sine pulse force in Example 14
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 6.22-23 for both damped and undamped cases.
Table 6.22. Maximum displacement (m) in the X direction of point G for undamped case in Example 14
SAP2000
(Frame Method) Present Study
(CCM) % difference
0.025920 0.026004 0.32
The results of both methods for 6 % damping ratio are given in Table 6.23.
Table 6.23. Maximum displacement (m) in the X direction of point G for damped case in Example 14
SAP2000
(Frame Method) Present Study
(CCM) % difference
0.024480 0.024562 0.33
The responses for both damped and undamped systems to sine force were
determined and the time-varying displacements in the X direction of point G are
presented in Figs. 6.66-67.
It is observed that the results obtained in the present work (CCM) coincide
with those of the SAP2000 structural analysis program, perfectly.
P(lb)
t (s)
150
5
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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
)
Present studySAP2000
Figure 6.66. Time-varying displacements in X direction at the top of the shear wall
for undamped case in Example 14
-0,030
-0,020
-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
)
Present studySAP2000
Figure 6.67. Time-varying displacements in X direction at the top of the shear wall
for damped case with 6 % damping ratio in Example 14
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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.
Figure 6.68. Cross-sectional view of the structure in Example 15
According to the lumped mass idealization, the lumped masses, which were
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
(0.125; 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)
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
149
The structure was solved both by the present method using the CCM and by
the SAP2000 structural analysis program using the frame method for which the
model and its 3-D view are given in Fig. 6.69.
Figure 6.69. Frame model of the structure in Example 15 and its 3-D view
Tables 6.24-25 compare the natural frequencies corresponding to each mode
found by the program prepared in the present work and the SAP2000 structural
analysis program for unstiffened and stiffened cases, expressing the percentage
differences.
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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
(Frame Method) % difference Natural Frequencies Natural Frequencies
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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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
(Frame Method) % difference Natural Frequencies Natural Frequencies
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.
Mode shapes in X, Y and Teta directions were compared by normalizing with
respect to the displacements at the top of the structure.
Figs. 6.70-72 present the mode shapes of the shear wall for stiffened case
found by the present program and the SAP2000 structural analysis program and Figs.
6.73-75 compare the mode shapes of the shear wall for stiffened and unstiffened
cases.
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
152
Figure 6.70. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by both the present program and SAP2000 for stiffened case in Example 15
Mode 3 in Teta Direction
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
Mode 1 in X Direction
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
Mode 2 in Y Direction
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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
153
Figure 6.71. Comparison of fourth, fifth and sixth mode shapes in X and Y directions
found by both the present program and SAP2000 for stiffened case in Example 15
Mode 4 in X 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=7,44742 Hz CCM NF=7,48094 Hz
Mode 6 in 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=20,44814 Hz
CCM NF=20,61359 Hz
Mode 5 in 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=11,23488 Hz
CCM DF=11,24119 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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
Mode 7 in Y Direction
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
Mode 8 in 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=32,29070 Hz
CCM NF=32,47404 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
155
Figure 6.73. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by the present program (CCM) for stiffened and unstiffened cases in Example 15
Mode 1 in X Direction
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
Mode 2 in Y Direction
0 6
12
18
24
30
36
42
48
0,0 0,5 1,0
Hei
ght (
m)
Unstiffened caseNF=1,65427 Hz
Stiffened case NF=2,04419 Hz
Mode 3 in Teta Direction
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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
156
Figure 6.74. Comparison of fourth, fifth and sixth mode shapes in X and Y directions
found by the present program (CCM) for stiffened and unstiffened cases in Example 15
Mode 4 in X Direction
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
Mode 5 in Y Direction
0 6
12
18
24
30
36
42
48
-1,0 -0,5 0,0 0,5 1,0
Hei
ght (
m)
Unstiffened caseNF=11,36339 Hz
Stiffened case DF=11,24119 Hz
Mode 6 in X Direction
0
6
12
18
24
30
36
42
48
-1,5 -0,5 0,5 1,5
Hei
ght (
m)
Unstiffened case NF=16,69150 Hz
Stiffened case NF=20,61359 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
157
Figure 6.75. Comparison of seventh and eighth mode shapes in X and Y directions
found by the present program (CCM) for stiffened and unstiffened cases in Example 15
Mode 7 in Y Direction
0
6
12
18
24
30
36
42
48
-1,5 -0,5 0,5 1,5
Hei
ght (
m)
Unstiffened case NF=21,86704 Hz
Stiffened case DF=21,54959 Hz
Mode 8 in X Direction
0
6
12
18
24
30
36
42
48
-1,5 -0,5 0,5 1,5
Hei
ght (
m)
Unstiffened case NF=30,88525 Hz
Stiffened case NF=32,47404 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
158
Example 16:
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 15.
The forced vibration analysis of the shear wall was carried out by the computer
program prepared in the present work and the SAP2000 structural analysis program
for damped and undamped cases.
The dynamic load, P(t), was applied at the top of the structure in the global X
direction in the plane of the connection beam as in Fig. 6.76 and the rectangular
pulse force was chosen as in Fig. 6.77.
Figure 6.76. Cross-sectional view of the structure and applied dynamic load in
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)
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
159
100
Figure 6.77. Rectangular pulse force in Example 16
At the top of the shear wall, the maximum displacement in the X direction of
point G, the mass center, was calculated by the computer program prepared in the
present study and compared with those of the SAP2000 structural analysis program
in Tables 6.26-27 for both unstiffened and stiffened cases. Damping ratio was chosen
5 % for this example.
Table 6.26. Maximum displacement (m) in the X direction of point G for unstiffened case in Example 16
Unstiffened case SAP2000
(Frame Method) Present Study
(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
(Frame Method) Present Study
(CCM) % difference
Undamped 0.021600 0.021297 1.43
Damped 0.017840 0.017650 1.06
The responses for both damped and undamped systems to rectangular force
were determined for stiffened and unstiffened cases and the time-varying
displacements in the X direction of point G are presented in Figs. 6.78-79.
P(lb)
t(s)
5
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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
Figure 6.78. Time-varying displacements in X direction at the top of the shear wall
for undamped case in Example 16
-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
Figure 6.79. Time-varying displacements in X direction at the top of the shear wall
for damped case with 5 % damping ratio in Example 16
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
161
Example 17:
In this example, the free vibration analysis of a non-planar non-symmetrical
coupled shear wall with and without stiffening beam was carried out using the
present program and the SAP2000 structural analysis program. Two stiffening beam
of 3.0 m height were placed at the mid-height and at the top of the wall which had 16
stories.
The geometrical properties and the cross-sectional view of the structure are
given in Fig. 6.80. The total height of the shear wall is 48 m, the storey height is 3 m,
the height of the connecting beams is 0.5 m and the elasticity and shear moduli are
E= 2.85×106 kN/m2 and G = 1055556 kN/m2, respectively.
Figure 6.80. Cross-sectional view of the structure in Example 17
According to the lumped mass idealization, the lumped masses, which were
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
(-0.545; 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)
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
162
The structure was solved both by the present method using the CCM and by
the SAP2000 structural analysis program using the frame method for which the
model and its 3-D view are given in Fig. 6.81.
Figure 6.81. Frame model of the structure in Example 17 and its 3-D view
Tables 6.28-29 compare the first twenty natural frequencies corresponding to
each mode found by the program prepared in the present work and the SAP2000
structural analysis program for unstiffened and stiffened cases, expressing the
percentage differences.
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
163
Table 6.28. Comparison of the natural frequencies (Hz) obtained from the present program and SAP2000 for unstiffened case in Example 17
Unstiffened case
Mode
Present Study (CCM)
SAP2000 (Frame Method) % difference
Natural Frequencies Natural Frequencies
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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
164
Table 6.29. Comparison of the natural frequencies (Hz) obtained from the present program and SAP2000 for stiffened case in Example 17
Stiffened case
Mode
Present Study (CCM)
SAP2000 (Frame Method) % difference
Natural Frequencies Natural Frequencies
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
In each method, after obtaining the natural frequencies for unstiffened and
stiffened cases of the non-planar coupled shear wall, the mode shape vectors of the
system were found.
Mode shapes in X, Y and Teta directions were compared by normalizing with
respect to the displacements at the top of the structure.
Figs. 6.82-84 present the mode shapes of the shear wall for stiffened case
found by the present program and the SAP2000 structural analysis program and Figs.
6.85-87 compare the mode shapes of the shear wall for stiffened and unstiffened
cases.
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
165
Figure 6.82. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by both the present program and SAP2000 for stiffened case in Example 17
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=0,47228 Hz
CCM NF=0,47361 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=0,50174 Hz
CCM NF=0,50691 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=2,15674 Hz
CCM NF=2,15597 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
166
Figure 6.83. Comparison of fourth, fifth and sixth mode shapes in X and Y directions
found by both the present program and SAP2000 for stiffened case in Example 17
Mode 4 in 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=2,97419 Hz SBY NF=2,98694 Hz
Mode 5 in X 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=6,34009 Hz
CCM NF=6,21172 Hz
Mode 6 in Y Direction
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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
167
Figure 6.84. Comparison of seventh and eighth mode shapes in X and Y directions
found by both the present program and SAP2000 for stiffened case in Example 17
Mode 7 in 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=10,24592 Hz
CCM DF=10,24086 Hz
Mode 8 in 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=15,97689 Hz
CCM NF=16,27233 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
168
Figure 6.85. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by the present program (CCM) for stiffened and unstiffened cases in Example 17
Mode 1 in Y Direction
0
6
12
18
24
30
36
42
48
0,0 0,5 1,0
Hei
ght (
m)
Unstiffened case NF=0,43177 Hz
Stiffened case NF=0,47361 Hz
Mode 2 in X Direction
0 6
12
18
24
30
36
42
48
0,0 0,5 1,0
Hei
ght(m
)
Unstiffened caseNF=0,49023 Hz
Stiffened case NF=0,50691 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)
Unstiffened case NF=2,04893 Hz Stiffened case NF=2,15597 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
169
Figure 6.86. Comparison of fourth, fifth and sixth mode shapes in X and Y directions
found by the present program (CCM) for stiffened and unstiffened cases in Example 17
Mode 4 in Y Direction
0
6
12
18
24
30
36
42
48
-1,0 -0,5 0,0 0,5 1,0
Hei
ght (
m)
Unstiffened caseNF=3,04620 Hz Stiffened case NF=2,98694 Hz
Mode 5 in X Direction
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=5,25108 Hz Stiffened case NF=6,21172 Hz
Mode 6 in Y Direction
0
6
12
18
24
30
36
42
48
-1,0 -0,5 0,0 0,5 1,0
Hei
ght (
m)
Unstiffened caseNF=8,49342 Hz Stiffened case NF=8,39412 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
170
Figure 6.87. Comparison of seventh and eighth mode shapes in X and Y directions
found by the present program (CCM) for stiffened and unstiffened cases in Example 17
Mode 7 in X Direction
0
6
12
18
24
30
36
42
48
-1,5 -0,5 0,5 1,5
Hei
ght (
m)
Unstiffened case NF=9,94970 Hz
Stiffened case NF=10,24086 Hz
Mode 8 in Y Direction
0
6
12
18
24
30
36
42
48
-1,5 -0,5 0,5 1,5
Hei
ght (
m)
Unstiffened case NF=16,18170 Hz
Stiffened case NF=16,27233 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
171
Example 18:
In this example, the forced vibration analysis of a non-planar non-
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.
The dynamic load, P(t), was applied at the top of the structure in the global X
direction in the plane of the connection beam as in Fig. 6.88 and the rectangular
pulse force was chosen as in Fig. 6.89.
Figure 6.88. Cross-sectional view of the structure and applied dynamic load in
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)
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
172
11
100
Figure 6.89. Rectangular pulse force in Example 18
At the top of the shear wall, the maximum displacement in the X direction of
point G, the mass center, was calculated by the computer program prepared in the
present study and compared with those of the SAP2000 structural analysis program
in Tables 6.30-31 for both unstiffened and stiffened cases. Damping ratio was chosen
5 % for this example.
Table 6.30. Maximum displacement (m) in the X direction of point G for unstiffened case in Example 18
Unstiffened case SAP2000
(Frame Method) Present Study
(CCM) % difference
Undamped 0.083100 0.083607 0.61
Damped 0.076500 0.076998 0.65
Table 6.31. Maximum displacement (m) in the X direction of point G for stiffened case in Example 18
Stiffened case SAP2000
(Frame Method) Present Study
(CCM) % difference
Undamped 0.054360 0.053835 0.97
Damped 0.050310 0.049506 1.60
The responses for both damped and undamped systems to rectangular force
were determined for stiffened and unstiffened cases and the time-varying
displacements in the X direction of point G are presented in Figs. 6.90-91.
P(lb)
t(s)
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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
)
Unstiffened caseStiffened case
Figure 6.90. Time-varying displacements in X direction at the top of the shear wall
for undamped case in Example 18
-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
)
Unstiffened caseStiffened case
Figure 6.91. Time-varying displacements in X direction at the top of the shear wall
for damped case with 5 % damping ratio in Example 18
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
174
Example 19:
In this example, the free vibration analysis of a non-planar non-symmetrical
coupled shear wall with and without stiffening beam was carried out using the
present program and the SAP2000 structural analysis program. The geometric and
material properties of the sixteen storey shear wall were taken as in Example 11. A
stiffening beam of 3.0 m height was placed at the height of 30 m on the tenth storey.
The total height of the shear wall is 48 m, the storey height is 3 m and the
thicknesses of the piers and the connecting beams are shown in Fig. 6.92. The height
of the connecting beams is 0.4 m and the elasticity and shear moduli are
=E 2.85×106 kN/m2 and =G 1055556 kN/m2, respectively.
Figure 6.92. Cross-sectional view of the structure in Example 19
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)
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
175
According to the lumped mass idealization, the lumped masses, which were
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
(-0.795; 0.322).
The structure was solved both by the present method using the CCM and by
the SAP2000 structural analysis program using the frame method for which the
model and its 3-D view are given in Fig. 6.93.
Figure 6.93. Frame model of the structure in Example 19 and its 3-D view
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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
structural analysis program, expressing the percentage differences.
Table 6.32. Comparison of the natural frequencies (Hz) obtained from the present program and SAP2000 for unstiffened case in Example 19
Unstiffened case
Mode
Present Study (CCM)
SAP2000 (Frame Method)
% difference
Natural Frequencies Natural Frequencies
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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
177
Table 6.33. Comparison of the natural frequencies (Hz) obtained from the present program and SAP2000 for stiffened case in Example 19
Stiffened case
Mode
Present Study (CCM)
SAP2000 (Frame Method) % difference
Natural Frequencies Natural Frequencies
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
In each method, after obtaining the natural frequencies for unstiffened and
stiffened cases of the non-planar coupled shear wall, the mode shape vectors of the
system were found.
Mode shapes in X and Y directions were compared by normalizing with
respect to the displacements at the top of the structure.
Figs 6.94-96 present the mode shapes of the shear wall for stiffened case
found by the present program and the SAP2000 structural analysis program and Figs.
6.97-99 compare the mode shapes of the shear wall for stiffened and unstiffened
cases.
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
178
Figure 6.94. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by both the present program and SAP2000 for stiffened case in Example 19
Mode 1 in X Direction
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
Mode 2 in Y Direction
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
Mode 3 in Teta 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=2,58037 Hz
CCM NF=2,60238 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
179
Figure 6.95. Comparison of fourth, fifth and sixth mode shapes in X and Y directions
found by both the present program and SAP2000 for stiffened case in Example 19
Mode 4 in 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=4,11147 Hz SBY NF=4,11919 Hz
Mode 6 in Y Direction
0
6
12
18
24
30
36
42
48
-1,5 -0,5 0,5 1,5
Hei
ght (
m)
SAP 2000 NF=11,43606 Hz
CCM NF=11,48704 Hz
Mode 5 in X 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=6,57733 Hz
CCM NF=6,63493 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
180
Figure 6.96. Comparison of seventh and eighth mode shapes in X and Y directions
found by both the present program and SAP2000 for stiffened case in Example 19
Mode 7 in 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=12,77066 Hz
CCM NF=12,96880 Hz
Mode 8 in 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=20,05863 Hz
CCM NF=20,42581 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
181
Figure 6.97. Comparison of first, second and third mode shapes in X, Y and Teta
directions found by the present program (CCM) for stiffened and unstiffened cases in Example 19
Mode 1 in X Direction
0
6
12
18
24
30
36
42
48
0,0 0,5 1,0
Hei
ght (
m)
Unstiffened case NF=0,50138 Hz
Stiffened case NF=0,62562 Hz
Mode 2 in Y Direction
0 6
12
18
24
30
36
42
48
0,0 0,5 1,0
Hei
ght(m
)
Unstiffened caseNF=0,66135 Hz
Stiffened case NF=0,66052 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)
Unstiffened case NF=2,47098 Hz Stiffened case NF=2,60238 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
182
Figure 6.98. Comparison of fourth, fifth and sixth mode shapes in X and Y directions
found by the present program (CCM) for stiffened and unstiffened cases in Example 19
Mode 4 in Y Direction
0
6
12
18
24
30
36
42
48
-1,0 -0,5 0,0 0,5 1,0
Hei
ght (
m)
Unstiffened caseNF=4,12617 Hz Stiffened case NF=4,11919 Hz
Mode 5 in X Direction
0
6
12
18
24
30
36
42
48
-1,0 -0,5 0,0 0,5 1,0
Hei
ght (
m)
Unstiffened caseNF=6,51577 Hz Stiffened case NF=6,63493 Hz
Mode 6 in Y Direction
0
6
12
18
24
30
36
42
48
-1,5 -0,5 0,5 1,5
Hei
ght (
m)
Unstiffened case NF=11,50762 Hz
Stiffened case NF=11,48704 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
183
Figure 6.99. Comparison of seventh and eighth mode shapes in X and Y directions
found by the present program (CCM) for stiffened and unstiffened cases in Example 19
Mode 7 in X Direction
0
6
12
18
24
30
36
42
48
-1,5 -0,5 0,5 1,5
Hei
ght (
m)
Unstiffened case NF=12,51956 Hz
Stiffened case NF=12,96880 Hz
Mode 8 in Y Direction
0
6
12
18
24
30
36
42
48
-1,5 -0,5 0,5 1,5
Hei
ght (
m)
Unstiffened case NF=20,48279 Hz
Stiffened case NF=20,42581 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
184
Example 20:
In this example, the forced vibration analysis of a non-planar non-
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 19. The forced vibration analysis of the shear wall was carried out by the
computer program prepared in the present work and the SAP2000 structural analysis
program for damped and undamped cases.
The dynamic load, P(t), was applied at the top of the structure in the global X
direction in the plane of the connection beam as in Fig. 6.100 and the triangular pulse
force was chosen as in Fig. 6.101.
Figure 6.100. Cross-sectional view of the structure and applied dynamic load in
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)
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
185
Figure 6.101. Triangular pulse force in Example 20
At the top of the shear wall, the maximum displacement in the X direction of
point G, the mass center, was calculated by the computer program prepared in the
present study and compared with those of the SAP2000 structural analysis program
in Tables 6.34-35 for both damped and undamped cases. Damping ratio was chosen 5
% for this example.
Table 6.34. Maximum displacement (m) in the X direction of point G for unstiffened case in Example 20
Unstiffened case SAP2000
(Frame Method) Present Study
(CCM) % difference
Undamped 0.035710 0.035658 0.14
Damped 0.035270 0.035192 0.22
Table 6.35. Maximum displacement (m) in the X direction of point G for stiffened case in Example 20
Stiffened case SAP2000
(Frame Method) Present Study
(CCM) % difference
Undamped 0.024550 0.023820 3.06
Damped 0.023570 0.022860 3.01
The responses for both damped and undamped systems to rectangular force
were determined and the time-varying displacements in the X direction of point G
are presented in Figs. 6.102-103.
P(lb)
t (s)
5 2,5
150
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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
)
Unstiffened caseStiffened case
Figure 6.102. Time-varying displacements in X direction at the top of the shear wall
for undamped case in Example 20
-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
Figure 6.103. Time-varying displacements in X direction at the top of the shear wall
for damped case with 5 % damping ratio in Example 20
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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
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.
According to the lumped mass idealization, the lumped masses, which were
calculated by computer program, were concentrated at the center of the whole cross-
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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
188
Figure 6.105. Cross-sectional view of the 2nd region of the structure in Example 21
Figure 6.106. Frame model of the structure in Example 21 and its 3-D view
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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
189
Table 6.36. Comparison of the natural frequencies (Hz) obtained from the present program and SAP2000 in Example 21
Mode
Present Study (CCM)
SAP2000 (Frame Method)
% difference
Natural Frequencies Natural Frequencies
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
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. 6.107-109 present the mode shapes of the shear wall, found by the
present program and the SAP2000 structural analysis program, both in the same
figure.
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
190
Figure 6.107. Comparison of first, second and third mode shapes in X and Y
directions found by both the present program and SAP2000 in Example 21
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=1,07366 Hz
CCM NF=1,07270 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=1,60347 Hz
CCM NF=1,61807 Hz
Mode 3 in 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=4,01831 Hz CCM NF=4,01622 Hz
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
191
Figure 6.108. Comparison of fourth, fifth and sixth mode shapes in X and Y
directions found by both the present program and SAP2000 in Example 21
Mode 4 in X 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=5,85470 Hz CCM NF=5,88496 Hz
Mode 5 in X 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=10,63805 Hz
CCM DF=10,68451 Hz
Mode 6 in Y Direction
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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
192
Figure 6.109. Comparison of seventh and eighth mode shapes in X and Y directions
found by both the present program and SAP2000 in Example 21
Mode 7 in 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=19,19478 Hz
CCM DF=19,27886 Hz
Mode 8 in Y Direction
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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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 21. The forced vibration analysis of the shear wall was carried out by the
computer program prepared in the present work and the SAP2000 structural analysis
program for damped and undamped cases.
The dynamic load, P(t), was applied at the top of the structure in the global X
direction in the plane of the connection beam as in Fig. 6.110 and the rectangular
pulse force was chosen as in Fig. 6.111.
Figure 6.110. Cross-sectional view of the structure and applied dynamic load 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
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
194
150
Figure 6.111. Rectangular pulse force in Example 22
At the top of the shear wall, the maximum displacement in the X direction of
the point O was calculated by the computer program prepared in the present study
and compared with those of the SAP2000 structural analysis program in Tables 6.37-
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
(Frame Method) Present Study
(CCM) % difference
0.036393 0.036112 0.77
Table 6.38. Maximum displacement (m) in the X direction of the point O for damped case in Example 22
SAP2000
(Frame Method) Present Study
(CCM) % difference
0.036130 0.035939 0.53
The responses for both damped and undamped systems to rectangular force
were determined and the time-varying displacements in the X direction of the point
O are presented in Figs. 6.112-113.
It is observed that the results obtained in the present work (CCM) coincide
with those of SAP2000 structural analysis program perfectly.
P(lb)
t(s)
6
6. NUMERICAL RESULTS Cevher Deha TÜRKÖZER
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
Figure 6.112. Time-varying displacements in X direction at the top of the shear wall
for undamped case in Example 22
-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
Figure 6.113. Time-varying displacements in X direction at the top of the shear wall
for damped case with 5 % damping ratio in Example 22
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
7. CONCLUSIONS Cevher Deha TÜRKÖZER
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
computer program prepared in the present work and the SAP2000 structural analysis
program. In the free vibration analyses, the natural frequencies and mode shape
vectors, thus obtained, are compared. In the forced vibration analyses, the 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.
7. CONCLUSIONS Cevher Deha TÜRKÖZER
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
259
:************** 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)
261
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
264
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)
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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
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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)
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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
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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)
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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
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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)
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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
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@ ,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)
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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)
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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)+
314
- 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
317
@ ,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
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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: #
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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
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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
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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
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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
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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