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- Permanent Link:
- http://ufdc.ufl.edu/UF00090209/00001
## Material Information- Title:
- Finite element model for composite masonry walls
- Series Title:
- Finite element model for composite masonry walls
- Creator:
- Boulton, George X.,
- Place of Publication:
- Gainesville FL
- Publisher:
- University of Florida
- Publication Date:
- 1984
## Subjects- Subjects / Keywords:
- Bricks ( jstor )
Coordinate systems ( jstor ) Degrees of freedom ( jstor ) Integers ( jstor ) Matrices ( jstor ) Stiffness ( jstor ) Stiffness matrix ( jstor ) Structural deflection ( jstor ) Subroutines ( jstor ) Terminology ( jstor )
## Record Information- Source Institution:
- University of Florida
- Holding Location:
- University of Florida
- Rights Management:
- Copyright George X. Boulton. Permission granted to the University of Florida to digitize, archive and distribute this item for non-profit research and educational purposes. Any reuse of this item in excess of fair use or other copyright exemptions requires permission of the copyright holder.
- Resource Identifier:
- 000487104 ( alephbibnum )
11902053 ( oclc )
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FINITE ELEMENT MODEL FOR COMPOSITE MASONRY WALLS By GEORGE X. BOULTON A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY UNIVERSITY OF FLORIDA 1984 This dissertation is dedicated to Almighty God in thanksgiving for all of His blessings. ACKNOWLEDGMENTS I would like to thank Dr. James H. Schaub for his active personal interest and support which encouraged me to attend the University of Florida, and for his willingness to serve on my supervisory committee. Dr. Morris W. Self deserves special thanks for serving as chairman of my supervisory committee, being my graduate advisor, and affording me the opportunity to work on this project. Special thanks also go to Dr. John M. Lybas for providing much invaluable aid and guidance during the development of this model. Further gratitude is extended to Professor William W. Coons, Dr. Fernando E. Fagundo, and Dr. Harry J. Reitz, Jr. for also serving on my supervisory committee. Each individual has greatly contributed to the value of my graduate studies and is an asset to their profession and to the University of Florida. Appreciation is expressed to Dr. Clifford 0. Hays. From his classes and notes, the author acquired the technical background necessary to undertake this effort. Thanks also go to Dr. Mang Tia for his technical assistance and recommendations. Randall Brown and Kevin Toye were valuable sources of advice, wisdom, and friendship throughout the author's graduate studies and deserve special mention. The assistance received from Krai Soongswang is also gratefully acknowledged. The unselfish sacrifice, constant support, and endless love provided by the author's parents have been a tremendous source of inspiration and must not go unnoticed. The encouragement of family, friends, and former teachers is also appreciated. Finally, the author extends heartfelt thanks to Ms. Cindy Zimmerman and Ms. Joanne Stevens for typing the manuscript and helping with its preparation. TABLE OF CONTENTS Page ACKNOWLEDGMENTS......... ...........................................iii LIST OF TABLES............... .....................................viii LIST OF FIGURES.. ..................................................x ABSTRACT. .......... .............................. .............. xvii CHAPTER ONE INTRODUCTION.. ................................. ........ .. 1 1.1 Background ......... ... ....... ....... ...... ......... 1.2 Objective............................................6 TWO DEVELOPMENT OF THE FINITE ELEMENT MODEL....................7 2.1 Matrix Analysis of Structures by the Direct Stiffness Method ........................... ...........7 2.2 Construction of the Element Stiffness Matrix..........9 2.3 Construction of the Element Index Matrix..............12 2.4 Construction of the Structure Stiffness Matrix........15 2.5 Construction of the Structure Force Matrix............ 17 2.6 Solving For the Structure Displacement Matrix.........18 2.7 Solving For the Element Displacement Matrix...........20 2.8 Solving For the Element Force Matrix...................21 THREE SPECIAL CONSIDERATIONS ................................. 22 3.1 Different Materials................................22 3.2 Different Types of Elements...........................22 3.3 Shear Deformation .................................25 3.4 Moment Magnification .................................. 36 3.5 Material Nonlinearity..............................44 3.6 Equation Solving Techniques............o ............50 3.6.1 Gauss Elimination...............................50 3.6.2 Static Condensation............................56 3.7 Solution Convergence..................................63 FOUR USING THE MODEL TO ANALYZE COMPOSITE MASONRY WALLS.........68 4.1 Structural Idealization of Wall....................... 68 4.2 Types of Elements....................................71 4.2.1 Brick Element................................. 71 4.2.2 Collar Joint Element..........................71 4.2.3 Concrete Block Element.........................74 4.3 Experimental Determination of Material Properties.....76 4.3.1 Brick..........................................76 4.3.2 Collar Joint................................... 83 4.3.3 Concrete Block................................83 4.4 Load Application......................................86 4.5 Solution Procedure....................................97 FIVE NUMERICAL EXAMPLES........................................ 102 5.1 General Comments.....................................102 5.2 Material Property Data................................103 5.2.1 P-A Curves....................................103 5.2.2 M-0 Curves .................................... 113 5.2.3 P-M Interaction Diagrams......................130 5.3 Example Number 1 Finite Element Analysis of a Test Wall........ ................ ..... .......137 5.4 Illustrative Examples................................141 5.4.1 Example Number 2................ ..........141 5.4.2 Example Number 3.............................145 5.4.3 Example Number 4............................. 145 5.4.4 Example Number 5............................145 SIX RESULTS OF ANALYSIS.......................................150 6.1 Wall Failure ....................................... 150 6.2 Lateral Wall Deflection Versus Height................150 6.3 Brick Wythe Vertical Deflection Versus Height........150 6.4 Block Wythe Vertical Deflection Versus Height........ 160 6.5 Plate Load Versus End Rotation........................160 6.6 Wall Wythe Vertical Load Versus Height...............160 6.7 Brick Wythe Moment Versus Height.....................185 6.8 Block Wythe Moment Versus Height....................185 6.9 Collar Joint Shear Stress Versus Height............. 185 SEVEN CONCLUSIONS AND RECOMMENDATIONS............................200 APPENDIX A COMPUTER PROGRAM..........................................202 A.1 Introduction........................................202 A.2 Detailed Program Flowchart..........................204 A.3 Program Nomenclature................................204 A.4 Listing of Program and Subroutines..................227 B SUBROUTINES...............................................282 B.1 Subroutine NULL.....................................282 B.2 Subroutine EQUAL....................................282 B.3 Subroutine ADD......................................282 B.4 Subroutine MULT.....................................282 B.5 Subroutine SMULT....................................282 B.6 B.7 B.8 B.9 B.10 B.11 B.12 B. 13 B.14 B.15 B.16 B.17 B.18 B.19 B.20 B.21 B.22 B. 23 B.24 B.25 B.26 B.27 B.28 B.29 B. 30 B.31 B. 32 B. 33 B.34 B. 35 B. 36 Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine C USER'S MANUAL.................... ....................... 368 C.1 General Information..................................68 C.2 Data Input Guide...................................70 D DATA FILES FOR NUMERICAL EXAMPLES.......................... 86 D.1 D.2 D.3 D.4 D.5 Example Number 1....................................386 Example Number 2.................................... 91 Example Number 3..................................... 396 Example Number 4 ....................................401 Example Number 5....................................406 REFERENCES..................................... .................... 410 BIOGRAPHICAL SKETCH.............................................. 412 BMULT.................................... 292 INSERT........................................292 BNSERT........ ........................... 292 EXTRAK.................................. 298 PULROW................................. 303 PULMAT....................... ...........303 GAUSS1.................. .................03 STACON..................................10 TITLE...................................310 READ ................................... 310 COORD ...................................310 CURVES.... ..... ............. ... ...... 18 PRINT.............................. ..... 318 WRITE................. .. ..... ..... .. 318 BRPDMT. ................................... 318 CJPDMT................... .............. 327 BLPDMT..................... .... ...... ... 327 STIFAC.................................. 336 BRESM.................................... 336 CJESM ..................................336 BLESM.......................... .......... 36 INDXBR..................................36 INDXCJ............ .................... ....347 INDXBL................................... 47 FORCES.................................3.47 APPLYF ....................347 PLATEK........... ........................47 DISPLA................................... 356 CHKTOL .................................356 CHKFAI..................................356 WYTHE.... ... ..........................356 LIST OF TABLES Page Variables Used in Constructing Brick Element Stiffness Matrix................................... ....... 80 Variables Used in Constructing Block Element Stiffness Matrix ........................................87 Summary of Wall Failure For Examples Number 1 Through 5................... .... ................ ...... 151 Variables Used in Detailed Program Flowchart...............205 Program Nomenclature.....................................215 Nomenclature For Subroutine NULL..........................283 Table 4.1 4.2 6.1 A.1 A.2 B.1 B.2 B.3 B.4 B.5 B.6 B.7 B.8 B.9 B.10 B.11 B.12 B.13 B.14 B.15 B.16 For For For For For For For For For For For For For For For Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine EQUAL ................. ...... 285 ADD........................... 287 MULT.......................... 289 SMULT........................ 291 BMULT....................... 294 INSERT ............ .......... 296 BNSERT....................... 299 EXTRAK........................ 01 PULROW............. ........... 304 PULMAT ....................... 306 GAUSS1......................... 3C8 STACON........................311 TITLE........................ 314 READ................... ..... 16 COORD................ ......... 319 Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature B.17 B.18 B.19 B.20 B.21 B.22 B.23 B.24 B.25 B.26 B.27 B.28 B.29 B. 30 B.531 B. 32 B. 33 B.34 B.35 B. 36 C.1 Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature Nomenclature For For For For For For For For For For For For For For For For For For For For Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Data Input Guide.........................................375 CURVES........................ 321 PRINT........................ 323 WRITE......................... 325 BRPDMT........................ 328 CJPDMT......................... 331 BLPDMT........................ 333 STIFAC....................... 337 BRESM......................... 339 CJESM....................... 341 BLESM.........................343 INDXBR............. ...........345 INDXCJ........................348 INDXBL...................... 350 FORCES...................... 352 APPLYF............... ........ 354 PLATEK..........................357 DISPLA....................... 359 CHKTOL ...................... 361 CHKFAI....................... 363 WYTHE........................... 365 LIST OF FIGURES Figure Page 1.1 Typical Composite Masonry Wall Section...................... 2 1.2 Typical Composite Masonry Wall Loadbearing Detail...........2 1.3 Stress and Strain Distribution For a Composite Prism Under Vertical Load...................................4 2.1 Development of Element Stiffness Matrix....................10 2.2 Basic Element Stiffness Matrix.............................13 2.3 Structure and Element Degrees of Freedom For a Frame........14 2.4 Construction of the Structure Force Matrix.................19 3.1 Structure and Element Degrees of Freedom For a Frame With Different Materials and Element Types............24 3.2 Element Stiffness Matrices For Elements 1 and 3 in Figure 3.1 ................................................. 26 3.3 Development of Stiffness Matrix For Element 2 of Figure 3.1............................................... 27 3.4 Element Stiffness Matrix For Element 2 in Figure 3.1.......28 3.5 Shear Deformation and Its Importance......................29 3.6 Application of Displacements to Establish Stiffness Coefficients Considering Shear Deformation.................31 3.7 Element Stiffness Matrix Considering Shear Deformation.....37 3.8 Moment Magnification.......................................38 3.9 Element Used in Derivation of Moment Magnification Terms..................................................... .40 3.10 Element Stiffness Matrix Considering Moment Magnification ............................................. 45 3.11 Element Stiffness Matrix Considering Shear Deformation and Moment Magnification.................................. 46 3.12 Nonlinear Load Deformation Curve............................48 3.13 Advantage of Symmetry and Bandedness of Structure Stiffness Matrix...........................................55 3.14 Comparison of the Equation Solving Operations of Standard Gauss Elimination Versus Modified Gauss Elimination.....................................................57 3.15 Computational Savings of Modified Gauss Elimination Over Standard Gauss Elimination............................ 58 3.16 Comparison of Equation Solving Operations of Modified Gauss Elimination Versus Modified Static Condensation......64 3.17 Computational Savings of Modified Static Condensation Over Modified Gauss Elimination............................65 4.1 Finite Element Model of Wall...............................69 4.2 Structure and Element Degrees of Freedom...................70 4.3 Finite Element For Brick...................................72 4.4 Finite Element For Collar Joint............................ 73 4.5 Finite Element For Block..................................75 4.6 Experimental Determination of Brick Element Axial Stiffness Factor...........................................77 4.7 Experimental Determination of Brick Element Rotational Stiffness Factor.......................................79 4.8 Brick Element Stiffness Matrix Formulation Using Experimentally Determined Axial and Rotational Stiffness Factors................................ ...... 81 4.9 Typical P Versus M Interaction Diagram For Brick Element ....................................................82 4.10 Experimental Determination of Collar Joint Shear Spring Stiffness Factor.....................................84 4.11 Determination of Collar Joint Moment Spring Stiffness Factor ................... .............................85 4.12 Structure Force Application Through Test Plate.............89 4.13 Program Algorithm....................................... 99 5.1 Experimental Source of Brick Element P-A Curve (4)........105 5.2 Brick Element P-A Curve................................... 107 5.3 Test Assembly Used By Williams and Geschwinder For Collar Joint Tests (18)....................................108 5.4 Experimental Source of Collar Joint Element P-A Curve (18) ..............................................109 5.5 Collar Joint Element P-A Curve............................111 5.6 Experimental Source of Concrete Block Element P-A Curve (4)......................................... 112 5.7 Block Element P-A Curve..................................114 5.8 Experimental Source of Brick Element M-9 Curves (4).......115 5.9 Relationships Used to Obtain Desired Data From Experimental Results For M-9 Curves.......................117 5.10 Brick Element M-9 Curves..................................125 5.11 Collar Joint Element M-9 Curve...........................126 5.12 Experimental Source of Concrete Block Element M-9 Curves (4)..................................... ......127 5.13 Block Element M-9 Curves................................... 31 5.14 Experimental Source of Brick Element P-M Interaction Diagram (4)...............................................132 5.15 Brick Element P-M Interaction Diagram......................134 5.16 Experimental Source of Concrete Block Element P-M Interaction Diagram (4)....................................135 5.17 Block Element P-M Interaction Diagram......................136 5.18 Example Number 1..........................................138 5.19 Effective Column Length Factors Based On End Conditions (7)............................................140 5.20 Structure Degrees of Freedom For Example Number 1.........142 5.21 Example Number 2.........................................14 5.22 Structure Degrees of Freedom For Example Numbers 2 and 3.......................................... ........ 144 5.23 Example Number 3...... ................................. 146 5.24 Example Number 4......................................... ....147 5.25 Structure Degrees of Freedom For Example Numbers 4 and 5.................... .... ..................... 148 5.26 Example Number 5......................................... 149 6.1 Lateral Wall Deflection Versus Height For Example Number 2............... ............................ 152 6.2 Lateral Wall Deflection Versus Height For Example Number 3........................................ ..... 153 6.3 Lateral Wall Deflection Versus Height For Example Number 4.............................................. 154 6.4 Lateral Wall Deflection Versus Height For Example Number 5.......................................... ......155 6.5 Brick Wythe Vertical Deflection Versus Height For Example Number 2....................................... 156 6.6 Brick Wythe Vertical Deflection Versus Height For Example Number 3.......................................... 157 6.7 Brick Wythe Vertical Deflection Versus Height For Example Number 4.......................... .............. 158 6.8 Brick Wythe Vertical Deflection Versus Height For Example Number 5..........................................159 6.9 Block Wythe Vertical Deflection Versus Height For Example Number 2......................... ................ 161 6.10 Block Wythe Vertical Deflection Versus Height For Example Number 3.............................. .. ....... 162 6.11 Block Wythe Vertical Deflection Versus Height For Example Number 4..........................................163 6.12 Block Wythe Vertical Deflection Versus Height For Example Number 5....................................... 164 6.13 Plate Load Versus End Rotation For Example Number 2.......165 6.14 Plate Load Versus End Rotation For Example Number 3.......166 6.15 Plate Load Versus End Rotation For Example Number 4.......167 6.16 Plate Load Versus End Rotation For Example Number 5.......168 6.17 Wall Wythe Vertical Load Versus Height At 0.30 Pmax For Example Number 2................................169 6.18 Wall Wythe Vertical Load Versus Height At 0.60 Pmax For Example Number 2...............................170 6.19 Wall max 6.20 Wall Pmax 6.21 Wall Pmax 6.22 Wall Pmax 6.23 Wall max 6.24 Wall Pmax 6.25 Wall Pmax 6.26 Wall Pmax 6.27 Wall Pmax 6.28 Wall Pmax 6.29 Wall Pmax 6.30 Wall Pmax 6.31 Wall Pmax 6.32 Wall Pmax Wythe Vertical Load Versus Height At 0.90 For Example Number 2................................ 171 Wythe Vertical Load Versus Height At For Example Number 2................................ 172 Wythe Vertical Load Versus Height At 0.30 For Example Number 3................................. 173 Wythe Vertical Load Versus Height At 0.60 For Example Number 3.................................174 Wythe Vertical Load Versus Height At 0.90 For Example Number 3........ ................... ..175 Wythe Vertical Load Versus Height At For Example Number 3................................. 176 Wythe Vertical Load Versus Height At 0.30 For Example Number 4..................................177 Wythe Vertical Load Versus Height At 0.60 For Example Number 4...............................178 Wythe Vertical Load Versus Height At 0.90 For Example Number 4................................. 179 Wythe Vertical Load Versus Height At For Example Number 4................................ 180 Wythe Vertical Load Versus Height At 0.30 For Example Number 5.................................181 Wythe Vertical Load Versus Height At 0.60 For Example Number 5................................ 182 Wythe Vertical Load Versus Height At 0.90 For Example Number 5................................183 Wythe Vertical Load Versus Height At For Example Number 5.................................184 6.33 Brick Wythe Moment Versus Height For Example Number 2.................................................. 186 6.34 Brick Wythe Moment Versus Height For Example Number 3................................................ 187 6.35 Brick Wythe Moment Versus Height For Example Number 4............................................... 188 6.36 Brick Wythe Moment Versus Height For Example Number 5............. ............................ ......189 6.37 Block Wythe Moment Versus Height For Example Number 2.......... .... ................................ 190 6.38 Block Wythe Moment Versus Height For Example Number 3......................................... ....... 191 6.39 Block Wythe Moment Versus Height For Example Number 4.................... ......o ............ .... .192 6.40 Block Wythe Moment Versus Height For Example Number 5.................................................193 6.41 Collar Joint Shear Stress Versus Height For Example Number 2................ .................. ..... .... .. 194 6.42 Collar Joint Shear Stress Versus Height For Example Number 3................................. .. .............. 195 6.43 Collar Joint Shear Stress Versus Height For Example Number 4............................................ ...... 196 6.44 Collar Joint Shear Stress Versus Height For Example Number 5........... ... ... ............... ..... ....... 197 A.1 Detailed Program Flowchart........................... 207 B.1 Algorithm For Subroutine NULL............................284 Algorithm Algorithm Algorithm Algorithm Algorithm Algorithm Algorithm Algorithm Algorithm Algorithm Algorithm Algorithm For Subroutine For Subroutine For Subroutine For Subroutine For Subroutine For Subroutine For Subroutine For Subroutine For Subroutine For Subroutine For Subroutine For Subroutine EQUAL............................286 ADD.............................. 288 MULT.............................290 SMULT........................... 293 BMULT............................ 295 INSERT ........................... 297 BNSERT .......................... 00 EXTRAK.......................... 302 PULROW.......................... 305 PULMAT............... ............ 307 GAUSS1...........................3.09 STACON........................... 313 B.14 Algorithm For Subroutine TITLE............................ 315 B.2 B.3 B.4 B.5 B.6 B.7 B.8 B.9 B.10 B.11 B.12 B.13 B.15 B.16 B.17 B.18 B.19 B. 20 B.21 B.22 B.23 B.24 B.25 B.26 B.27 B.28 B.29 B. 30 B.31 B.32 B.33 B.34 B.35 B. 36 Algorithm Algorithm Algorithm Algorithm Algorithm Algorithm Algorithm Algorithm Algorithm Algorithm Algorithm Algorithm Algorithm Algorithm Algorithm Algorithm Algorithm Algorithm Algorithm Algorithm Algorithm Algorithm For For For For For For For For For For For For For For For For For For For For For For Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine Subroutine READ............................. 317 COORD............................ 320 CURVES........................... 322 PRINT........................... 324 WRITE............... ............. 326 BRPDMT........................... 330 CJPDMT.......................... 332 BLPDT .......................... 335 STIFAC........................... 338 BRESM.........................3.. 40 CJESM........................ .... 342 BLESM.......................... 344 INDXBR................... ......346 INDXCJ..........................349 INDXBL.............. ............ 351 FORCES........................... 353 APPLYF ...........................355 PLATEK.............................358 DISPLA .......... ............... 360 CHKTOL.......................... 62 CHKFAI.......................... 364 WYTHE ............................367 Abstract of Dissertation Presented to the Graduate School of the University of Florida in Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy FINITE ELEMENT MODEL FOR COMPOSITE MASONRY WALLS By GEORGE X. BOULTON December 1984 Chairman: Dr. Morris W. Self Major Department: Civil Engineering Composite masonry design standards are at an early stage of development. To improve the understanding of composite masonry wall behavior in response to load application, a two-dimensional finite element model has been developed. The model considers a wall subjected to vertical compression and out of plane bending. It takes into account the different strength- deformation properties of the concrete block, collar joint, and clay brick, as well as the nonlinear nature of these properties for each material. The effects of moment magnification and shear deformation in both the brick wythe and the block wythe of a composite wall are also considered. The primary purpose of this model is to analytically represent typical composite masonry walls that might be tested in a laboratory. Wall tests attempt to duplicate conditions found in prototype walls. By comparing the results of the analytical and experimental tests, needed insight into composite wall behavior can be obtained, and design standards can be modified to reflect this increased understanding. Every effort has been made to consider the factors that will most strongly influence composite wall behavior in the development of the model, so that its usefulness as an analytical research tool will not be compromised. This study describes the model in detail, provides information on how it can be used, and contains several numerical examples that illustrate the information its use makes available. CHAPTER ONE INTRODUCTION 1.1 Background A composite masonry wall is a wall which consists of a clay brick wythe, a concrete block wythe, and a collar joint which forms a bond between the two wythes. A section of a typical composite masonry wall is shown in Figure 1.1. The collar joint between the two wythes consists of either masonry mortar or concrete grout. It forces the wall to behave as one structural unit, even though the wall consists of different materials. Composite masonry walls are frequently used as exterior bearing walls with the brick exposed as an architectural surface and the concrete block used as the load-bearing material. Thus, when a floor slab or roof truss bears on a composite masonry wall, it transfers vertical load directly to the block wythe. A typical composite masonry wall load-bearing detail is shown in Figure 1.2. Two design standards have been widely used in the United States as references for the design of engineered masonry construction. These are the Brick Institute of America (BIA) "Building Code Requirements for Engineered Brick Masonry" (2) and the National Concrete Masonry Association (NCMA) "Specifications for the Design and Construction of Load-Bearing Concrete Masonry" (14). A third standard on concrete masonry was recently published by the American Concrete Institute (ACI) entitled ACI 531-79R, "Building Code Requirements for Concrete Masonry CLAY BRICK WYTHE -CONCRETE BLOCK WYTHE COLLAR JOINT Figure 1.1 CLAY BRICK WYTHE COLLAR JOINT Typical Composite Masonry Wall Section REINFORCED CONCRETE FLOOR SLAB Typical Composite Masonry Wall Loadbearing Detail Figure 1.2 Structures" (1). It contains a brief chapter on composite masonry. Finally, a joint American Society of Civil Engineers and American Concrete Institute committee, Committee 530, is currently developing a comprehensive standard to include provisions for both clay and concrete products. Unfortunately, these design standards are limited by a lack of laboratory test data as well as a lack of understanding of the behavior and response of composite masonry. Rational analyses to predict the failure loads and load-deformation properties of composite masonry walls do not presently exist (6). To improve the reliability of the design procedure for composite masonry, several factors not currently considered must be taken into account. The first is that masonry is not linearly elastic, it is nonlinear. In other words, its modulus of elasticity changes depending on the stress level to which it is subjected. Figure 1.3 illustrates some of the complications that result due to the different and nonlinear material properties of brick and block which are present in composite masonry. As evident from the stress-strain curves of the two materials, the block is weaker. Stage 1 depicts the stresses and strains that are generated as a vertical load is applied through the centroid of a composite prism above a 50 percent stress level. As the load is increased further to stage 2, the stiffness of the concrete deteriorates more rapidly than that of the brick. This causes the center of resistance of the section to shift toward the brick. The eccentricity between the point of load application and the effective resistance of the section results in an effective bending of the prism, causing the strains in the block to increase faster than those in the brick. As a CLAY CONCRETE CLAY CONCRETE CLAY CONCRETE 2 A STRAIN POSITION OF APPLIED LOAD POSITION OF ELASTIC CENTER Figure 1.3 Stress and Strain Distribution For a Composite Prism Under Vertical Load CLAY CONCRETE STRAIN STRESS result, the stiffness of the block deteriorates even faster than that of the brick, the center of resistance shifts even further, and the eccentricity and bending it produces are further aggravated. Stage 3 denotes failure of the prism which is characterized by vertical splitting of the concrete. This phenomenon was verified by experimental tests at the University of Florida (11,13,15). The failure mechanism for composite walls under vertical loads should reflect this behavior. A second factor meriting consideration is that roof trusses or floor slabs, as mentioned previously, generally bear on the block wythe causing the wall to be loaded eccentrically toward the block. This will aggravate the failure mechanism just discussed. A third consideration is that with increasing slenderness and height, lateral wall deflections due to bending will increase. As these deflections increase, the additional moment caused by the vertical load acting through these deflections takes on increasing importance and can lead to a stability problem. All three of these aspects of behavior should be affected by the wall's height to thickness ratio as well as the thickness of the block wythe. In response to the need for more experimental work on composite masonry walls and an improved understanding of wall behavior, laboratory tests of composite walls have been performed by Redmond and Allen (10), Yokel, Mathey, and Dikkers (19), and Fattal and Cattaneo (4). Nonetheless, as a whole, the three studies consider limited combinations of wall geometry, masonry unit properties, and height to thickness ratios. Additional tests are needed so that design standards can be supported by a large data base. Wall test results also need to be related to analytical models. 1.2 Objective Lybas and Self (6) have submitted a research proposal to the National Science Foundation aimed at addressing some of these needs. Specifically, they seek to explore both experimentally and analytically 1) The nonlinear load deformation properties of composite masonry walls under compression and out of plane bending. The effect of transverse loading, eccentric compressive loading, slenderness ratio and different masonry unit properties will be considered. 2) The failure mechanisms of composite masonry walls under these types of loading conditions. 3) The transfer of vertical force across the collar joint from block wythe to brick wythe. 4) The suitability of current standards for composite masonry wall design. 5) The development of improved design equations and procedures for composite walls, based on the results of the research. This study essentially consists of the development of an analytical model which will be used, once the experimental phase has been completed, to examine the factors cited above. The model will consider a composite masonry wall subject to compression and out of plane bending, due to either eccentric load application or transverse loading. The two-dimensional finite element model will take into account the different nonlinear load-deformation properties of the brick and block wythes, the load transfer properties of the collar joint, the effect of moment magnification that, as previously mentioned, will result as the lateral wall deflections increase, and the effect of shear deformations. CHAPTER TWO DEVELOPMENT OF THE FINITE ELEMENT MODEL 2.1 Matrix Analysis of Structures by the Direct Stiffness Method The Direct Stiffness Method, like most matrix methods of structural analysis, is a method of combining elements of known behavior to describe the behavior of a structure that is a system of such ele- ments. The following is a summary of the basic relationships used in this technique. It is presented only as a quick review of the Stiffness Method and not as an exhaustive presentation which develops the rela- tionships stated. For that purpose, one of the fine textbooks on matrix analysis, such as Rubinstein's (12), is recommended. A degree of freedom is an independent displacement. Recall that the force displacement equations for an element i, which has n element degrees of freedom, can be written as [f]i [k]i [w]i + [f]i (2.1) where [w]i n x 1 matrix of independent element displacements measured in element coordinates [f]i = n x 1 matrix of corresponding element forces measured in element coordinates [fo]i = n x 1 matrix of corresponding element fixed-end forces measured in element coordinates [k]i = n x n element stiffness matrix measured in element coordinates. In general, [k]i and [fo]i can be found from standard cases. Since this model only considers nodal loads, [f]i matrices will not exist. The force displacement equations for an element i, therefore, reduce to [f]i = [k]i [w]i (2.2) Suppose an element i is connected to other elements to form a structure with N structure degrees of freedom. The structure force displacement (or equilibrium) equations can be expressed as [F] = [K] [W] + [F] (2.3) where [W] = N x 1 matrix of independent structure displacements measured in structure coordinates [F] = N x 1 matrix of corresponding structure forces measured in structure coordinates [Fo] = N x 1 matrix of corresponding structure fixed-end forces measured in structure coordinates [K] = N x N structure stiffness matrix measured in structure coordinates. Again, fixed-end forces are not in the model so these equations reduce to [F] = [K] [W] (2.4) If m equals the total number of structure degrees of freedom that are related to the element degrees of freedom for element i, an index matrix [I]i can be defined as [I] = m x 1 matrix whose elements are the numbers of the structure degrees of freedom that are related to the element degrees of freedom for element i. Usually an n x m transformation matrix [T] is also required to transform structure displacements into element displacements. However, if the element and structure coordinate systems are coincident, a transformation matrix is not required. Such is the case in this model. The solution procedure is generally as follows. For each element, 1) Construct the index matrix [I]. 2) Construct the element stiffness matrix [k]. 3) Insert the element stiffness matrix [k] into the structure stiffness matrix [K] using [I]. Once the structure stiffness matrix has been formed, 4) Construct the structure force matrix [F]. 5) Solve for the structure displacements by solving [F] = [K] [W] for [W]. Finally, for each element 6) Extract the element displacement matrix [w] from the structure displacement matrix [W]. 7) Multiply the element stiffness matrix by the element displacement matrix to yield the matrix of element forces, i.e., [f] = [k] [w]. Sections 2.2 through 2.8 discuss these matrices in more detail. 2.2 Construction of the Element Stiffness Matrix Consider the element shown in Figure 2.1a. This is the basic flexural element found in any text on matrix structural analysis. It contains the six degrees of freedom shown which include a rotation, axial displacement, and lateral displacement at each end. Recall that the stiffness coefficient kij is defined as the force developed at the ith degree of freedom (DOF) due to a unit displacement at the jth DOF of the element while all other nodal displacements are maintained at zero. For example, the stiffness coefficient k11 is the force developed at the first DOF due to a unit displacement at the first 10 W4 I "6 W3 W2 wl (a) Basic Flexural Element (a) Basic Flexural Element w3=1 --6EI 2EI L2 L 4EI 6EI L L2 t + ^l_ 12EI 6EI L/ 6E .- 12EI L -7L3 w6=1 6EI 4EI 2EI LL2 2ElE L k- i L (b) Application of Unit Displacements to Establish Stiffness Coefficients Figure 2.1 Development of Element Stiffness Matrix wI =1 w2=1 6EI L2 SAE L tAE L - 12EI L3 - 12EI L3 w41 1 w5=1 DOF. Similarly, the stiffness coefficient k12 is the force developed at the first DOF due to a unit displacement at the second DOF. The total force F at the first DOF can, therefore, be represented as fl = klw1 + k12w2 + k13w3 + k14w4 + k15w5 + k16w6 where wi equals the element displacement at the ith DOF. Analogously, the forces at the other degrees of freedom are: f2 = k21wl + k22w2 + k23w3 + k24w4 + k25w5 + k26w6 f6 = k61w1 + k62w2 + k63w3 + k64w4 + k65w5 + k66w6 * These equations can be written conveniently in matrix form as k12 k22 k32 k42 k52 k62 k 1 k23 k 33 k43 k53 k63 k 14 k24 k34 k44 k54 k64 k15 k25 k35 k45 k55 k65 k16 k26 k36 k46 k56 k66 or simply as [f] = [k] [w] where [w] = element displacement matrix [f] = element force matrix [k] = element stiffness matrix. Figure 2.1b shows the application of unit displacements to the basic flexural element in order to establish the stiffness coefficients. Figure 2.2 shows the resulting element stiffness matrix. It is a 6 x 6 matrix since the element contains 6 DOF. 2.3 Construction of the Element Index Matrix The index matrix [I] was previously defined in section 2.1 as the matrix which consists of the numbers of the structure degrees of freedom that are related to the element degrees of freedom for a particular element. To illustrate what this means, consider the frame shown in Figure 2.3a. If the structure and element degrees of freedom for this frame are as shown in Figure 2.3b and 2.3c, respectively, it is possible to construct the index matrix for each element simply by noting which structure degrees of freedom correspond to which element degrees of freedom. For example, to construct the index matrix for element number 1, observe that wl, w2, and w3 have no corresponding structure degrees of freedom but W1 corresponds to w4, W3 corresponds to w5, and W4 corresponds to w6. Thus, the index matrix for element number 1 is 0 0 [I] = 0 1 1 3 4 Similarly, the index matrices for elements 2 and 3 are -AE L 0 12EI -6EI 0 -12EI -6EI L3 L2 L3 L2 O -6EI 4EI O 6EI 2EI L2 L L2 L -AE O O AE O 0 L L 0 -12EI 6EI 0 12EI 6EI L3 L2 L L2 -6EI L2 2EI L 6EI L2 4EI L WHERE: A = AREA OF ELEMENT CROSS-SECTION E = MODULUS OF ELASTICITY L = ELEMENT LENGTH I = MOMENT OF INERTIA OF ELEMENT CROSS-SECTION Figure 2.2 Basic Element Stiffness Matrix [k] = 0 /r777 (a) Frame W w4 . t 2 0 -3 '5 "5 (b) Structure Degrees of Freedom w 4 t t w4 W t4 w6QV 5 w 6 W2 3 W2 w w W1 W1 (c) Element Degrees of Freedom Figure 2.3 Structure and Element Degrees of Freedom For a Frame /777 0 1 0 4I = 0 2 21 3 2 5 3 5 The index matrix plays a vital role in the proper assemblage of the structure stiffness matrix. This is discussed in the next section. 2.4 Construction of the Structure Stiffness Matrix The structure stiffness matrix is constructed by using the element stiffness matrices and the element index matrices. To obtain a term in the structure stiffness matrix, it is necessary to add up the appropriate terms from the element stiffness matrices. The procedure for doing this is best illustrated by an example. Consider the frame of Figure 2.3. To identify where each coefficient in each element stiffness matrix belongs in the structure stiffness matrix, place the numbers in the index matrix for an element along the sides of the element stiffness matrix as shown below. 0 0 0 1 3 4 k1 1 k12 k13 k14 k15 k16 0 [k]i = k21 k22 k23 k24 k25 k26 k3 k32 k33 k34 k35 k36 k41 k42 k43 k44 k45 k46 k51 k52 k53 k54 k55 k56 k63 k66 1 4 2 5 1il k13 k21 k22 k23 k24 k31 k32 k33 k34 k42 k43 0 0 0 2 3 5 k13 k15 k16 k21 k22 k23 k24 k25 k26 k31 k32 k33 k34 k35 k36 k41 k42 k43 k44 k45 k46 k51 k52 k53 k54 k55 k56 k63 k64 k65 k66 Define [kij]m as stiffness matrix the stiffness coefficient in for element m. The numbers row i, column j of the in the index matrix along the sides of the element stiffness matrix for each element identify the rows and columns in the structure stiffness matrix in which each coefficient belongs. Since the frame has 5 DOF, the structure stiffness matrix will be a 5 x 5 matrix. The term in row 1, column 1 of the structure stiffness matrix is, therefore, [k]2 = [k] = k12 kil k12 K11 = [k44] + [k11] Similarly, the other terms are as shown below: Similarly, the other terms are as shown below: 1 [k 44] + [k11] 2 [kj 1 2 3 [k45] 4 [k46 ] + [k12] [k14] [ ] [k 44] [k34] [k3,] 4 [k45] [k32] 2 2 + [k332 3 2 + [k46 3 1 3 + [k55] 1 3 [641 [k2 [k65 66 1 [k 4 + [k21J 2 1 + [k 221 2 2 [k 2 [k41] 2 [k65] [k42] [k44] + [k66] Thus, once the stiffness and index matrices are known for each element, the structure stiffness matrix can easily be arrived at. 2.5 Construction of the Structure Force Matrix To construct the structure force matrix, it is only necessary to assign the value of the force to the term in the matrix which corresponds to the degree of freedom at which the force is applied. If the force has the same direction as the degree of freedom, it is [K] = considered positive. If it acts in the opposite direction, it is considered negative. Figure 2.4 shows two examples of structure force matrices for the frame of Figure 2.3. 2.6 Solving For the Structure Displacement Matrix Like the structure force matrix, the structure displacement matrix will also be an N x 1 matrix where N is the number of structure degrees of freedom. Thus, the structure displacement matrix for the frame of Figure 2.3 will be W1 w2 [W] = W W4 W 5 Solving for the structure displacement matrix will entail solving the matrix equation [F] = [K] [W] for [W]. This will consist of solving N simultaneous equations if N is the number of structure DOF. One way to solve this matrix equation is by matrix inversion, or [W] = [K]-1 [F] (2.5) For large values of N, however, the structure stiffness matrix of dimensions N x N will become large and calculating the inverse of a large matrix is very cumbersome and inefficient. Two much more efficient techniques for solving for the structure displacement matrix are Gauss Elimination and Static Condensation. These methods are discussed in detail in sections 3.6.1 and 3.6.2. F3 SF5 - > [F] = 200 [F] = Figure 2.4 Construction of the Structure Force Matrix F4q 50 20 - 0 -50 20 0 0 'I /777 0 0 -100 200 /7777 /7Z7 1) 20 2.7 Solving For the Element Displacement Matrix Once the structure displacement matrix has been solved for, it is very easy to obtain the displacement matrix for each element. Since the index matrix for an element identifies which structure DOF correspond to which element DOF, it also identifies which structure displacements correspond to which element displacements. For example, consider the frame of Figure 2.3 again. It was previously observed that for element 1, Wl, w2, and w3 have no corresponding structure DOF but W1 corresponds to w4, W3 corresponds to w5, and W4 corresponds to w6. This means that the element displacement matrix for element 1 equals w2 0 0 w2 0 0 w 0 0 [w]1 = = where [Il] = w4 Wi 1 W5 W3 3 w6 4 4 Using the index matrices for elements 2 and 3 which were previously developed in section 2.3, the element displacement matrices for elements 2 and 3 are, therefore, wi 0 wI W1 w2 0 w2 W4 w 3 0 [w]2 = = and [w]3 = w3 W2 w4 W w4 W5 W5 W3 w6 W5 Thus, once the structure displacement matrix values are available, the displacement matrix for each element is easily obtained with the help of the index matrix for each element. 2.8 Solving For the Element Force Matrix In section 2.1, it was mentioned that the force displacement equations for an element i can be expressed as [f]i = [k]i [w]i Thus, the force matrix for each element is calculated simply by multiplying the stiffness matrix for that element by the displacement matrix for that element. The values in the element force matrix for an element will correspond to the element displacements for the element. For example, considering element 1 in the frame of Figure 2.3, the force matrix for element 1 will take the form fl f2 [f]1 4 f5 f6 A positive value in the element force matrix implies that the resulting force for that DOF acts in the direction shown for that DOF. Conversely, a negative value represents a force which acts opposite to the direction shown. CHAPTER THREE SPECIAL CONSIDERATIONS 3.1 Different Materials The stiffness matrix for an element is dependent on the geometric, cross-sectional, and material properties of that element. This is evidenced by the nature of the variables in the element stiffness matrix shown in Figure 2.2. A structure which is comprised of different materials can be analyzed by the Direct Stiffness Method. The presence of different materials is accounted for by using the appropriate material property values when constructing the stiffness matrix for each element. Since the structure stiffness matrix is assembled using the stiffness matrix for each element, the solution for the structure will then reflect the presence of material differences among the different elements into which the structure is divided. In short, the presence of different materials in a structure is taken into account during the construction of the stiffness matrix for each element in the structure. 3.2 Different Types of Elements Occasionally, the accurate matrix analysis of a structure involves the use of more than one type of element in the analytical model. This presents no particular difficulty and, in fact, is handled in much the same way as the presence of different materials in the structure. In other words, it too is accounted for during the construction of the stiffness matrix for each element. As mentioned in section 2.2, the stiffness matrix for an element is constructed by applying unit displacements, one at a time, at each DOF. If a structure is modeled by more than one type of element, this only means that the coefficients and variables in the element stiffness matrix of each element type will be different. Once all of the values in an element stiffness matrix are calculated, the element stiffness matrix is inserted into the structure stiffness matrix in the same fashion as discussed previously in section 2.4. Because the structure stiffness matrix is assembled using the element stiffness matrices, the solution of structure displacements and element displacements and forces will reflect the presence of different types of elements in the model. Consider the frame with the structure degrees of freedom, element degrees of freedom, and property values shown in Figure 3.1. Notice that this frame is similar to the one in Figure 2.3, which was previously discussed, except that: 1) Elements 1 and 3 have different property values 2) Element 2 is of a different type than the previous element 2 and is different from the present elements 1 and 3. Assume it is desired to analyze this structure by the Direct Stiffness Method. From the preceding discussion, it was learned that the procedure is identical to the one outlined at the end of section 2.1, but that the stiffness matrix for each element will be different due to the presence of different materials and different element types. To illustrate, the stiffness matrix for each element will be constructed. k s 0 -3 /-- 5W 3 W5 A=A2 E=E2 L=L 1=12 /777 where: k = Shear Spring Stiffness km = Moment Spring Stiffness Other variables defined in Figure 2.2 (a) Structure Degrees of Freedom w w I t SW2 W4 w w w6 (w3 ^w2 w4 w6r-!w5 Wl (b) Element Degrees of Freedom Figure 3.1 Structure and Element Degrees of Freedom For a Frame With Different Materials and Element Types W- 4Q A=A1 E=E1 L=L ='I1 10 1 i 2 First, for elements 1 and 3, the element DOF are identical to what they were in Figure 2.3. Therefore, the derivation of stiffness coefficients for this basic flexural element, shown in Figure 2.1, is valid. The stiffness matrix for elements 1 and 3 will thus take the form shown in Figure 2.2, but the coefficients will recognize the differences in the values of the variables. Figure 3.2 gives the element stiffness matrix for element 1 and the element stiffness matrix for element 3. To construct the element stiffness matrix for element 2, the stiffness coefficients must be derived by the application of unit displacements at each DOF. This is shown in Figure 3.3. The resulting stiffness matrix for element number 2 is illustrated in Figure 3.4. 3.3 Shear Deformation As the depth to span ratio for a member increases, the effect of shear deformation becomes more pronounced and important to consider in the analysis. Figure 3.5a illustrates the shear deformation and bending components of the lateral deflection at the free end of a column in response to a lateral concentrated load. Figure 3.5b, taken from Wang (16), shows the ratio of shear deflection As to bending deflection Ab at the midspan of a simple beam with a rectangular cross-section. Notice that, for a depth to span ratio of 0.25, the shear deflection can be up to 18.75% of the bending deflection. If it is desired to take shear deformation into consideration in the analysis of a structure by the Direct Stiffness Method, this is accomplished by altering the standard terms in the stiffness matrix of the elements for which this effect may be significant. The terms in the stiffness matrix for the standard 6 DOF element, which were derived in AIEI L -AIEI L 0 12E1I1 -6E1I1 0 -12E1I1 -6E1I1 L L2 L L2 0 -6E111 4EI11 0 6E1I1 2E11 L2 L L2 L -AIE1 0 0 A IE 0 0 L L 0 -12E1 6E1I1 0 12E1I1 6E1I1 L3 L2 L3 L2 0 -6E1I1 2E1I1 0 6E1I1 4EI11 L2 L L2 L (a) Element Stiffness Matrix For Element 1 A2E2 0 0 -A2E2 O 0 L L 0 12E212 -6E212 O -12E212 -6E212 L L2 L3 L2 O -6E212 4E212 0 6E212 2E212 L2 L L2 L -A2E2 0 0 A2E2 O 0 L L O -12E212 6E212 0 12E212 6E212 3 L2 L L2 0 -6E2I2 L2 2E212 L 6E2I2 L2 4E2I2 L (b) Element Stiffness Matrix For Element 3 Figure 3.2 Element Stiffness Matrices Figure 3.1 For Elements 1 and 3 in [k] [k]3 = wI w3 W1 W3 f k w2 m W4 (a) Element w1 = 1 ksk ki ks s)ks2 oc-" w3 = 1 S0 ks"2 ks ks k_ fk k kS 0 s s1 w =1 kt km kl k k k + k Zm ks k S kzs 12 km sl w4 = ks km ks t s s2 k m (b) Application of Unit Displacements to Establish Stiffness Coefficients Figure 3.3 Development of Stiffness Matrix For Element 2 of Figure 3.1 ks1 ks + km -k 1 ks21 km -ks -ks1 ks -ks2 ks1 ks 12 km -ks2 s2+ k WHERE: ks km 2 X2 = SHEAR SPRING STIFFNESS = MOMENT SPRING STIFFNESS = LENGTH OF LEFT PART OF ELEMENT = LENGTH OF RIGHT PART OF ELEMENT Figure 3.4 Element Stiffness Matrix For Element 2 in Figure 3.1 [k]2 = ksX2 A-Hs P II /i} II I I (a) Ab (bending) And As (shear) Components of Lateral Deflection DEPTH d CONCENTRATED UNIFORM LOAD SPAN L LOAD AT MIDSPAN 1/12 0.0167 0.0208 1/10 0.0240 0.0300 1/8 0.0375 0.0469 1/6 0.0667 0.0833 1/4 0.1500 0.1875 (b) Ratio of Shear Deflection As to Bending Deflection Ab Figure 3.5 Shear Deformation and Its Importance Figure 2.1, neglect shear deformation. The following derivation, taken from Przemieniecki (9), shows how the element stiffness matrix is obtained for the standard 6 DOF element considering shear deformation. Note that displacements for element degrees of freedom 1 and 4 are not considered below because they are not affected by the consideration of shear deformation. Thus, columns 1 and 4 in the element stiffness matrix shown in Figure 2.2 will remain unchanged. Consider Figure 3.6a. The lateral deflection v on the element subjected to the shearing forces and associated moments shown, is given by v = vb + vs (3.1) where vb is the lateral deflection due to bending strains and vs is the additional deflection due to shearing strains, such that s f5 (3.2) dx GAS with As representing the element cross-sectional area effective in shear, and G representing the shear modulus, where GE (3.3) 2(1 + v) and E equals the modulus of elasticity and v the Poisson's ratio of the material. The bending deflection for the element shown in Figure 3.6a is governed by the differential equation 2 Eld vb f5x f6 MT (3.4) dx2 W4 . W5 W6 Sw4 2 W Figure 3.6 Application of Displacements to Establish Stiffness Coefficients Considering Shear Deformation f5 f5 f5 f6 f 2 f (b) where MT = faETydA (3.5) A From integration of Equations (3.2) and (3.4), it follows that 3 2 2 f5x f6x Mx2 f5 EI EIV = ---- + C + 6 2 2 1 jX + C2 (3.6) s/ where CI and C2 are the constants of integration. Using the boundary conditions in Figure 3.6a, v dv -f v s= 5 at x = o, x = 1 (3.7) dx dx GAs and v = o at x = 1 (3.8) Equation (3.6) becomes 3 2 2 2 3 fx3 f6x Mx2 f5cxl 1 f5 Elyv 6 + (1 + P) (3.9) 6 2 2 12 12 f51l where f6 - MT (3.10) and = 12EI (3.11) GA 1 s It should be noted here that the boundary conditions for the fixed end in the engineering theory of bending when shear deformations vs are included is taken as dvb/dx = 0; that is, slope due to bending deformation is equal to zero. The remaining forces acting on the beam can be determined from the equations of equilibrium; thus, we have f2 -f5 (3.12 and f3 = -f + f51 (3.13: Now at x = O, v = w5, and hence from Equation (3.9) w = ( + ) f5 (3.14: 2 12EI Using Equations (3.10) and (3.12) to (3.14), we have k -5 5 k6' 2,5 T=O 3 k T 3,5 w5 = 0 S 12EI (1 + )13 5- 2w T=0 .M -12EI (1 + )13 -f + fl 1 w 5 T T=0 with the remaining coefficients in column 5 equal to zero. The variable T stands for temperature change. S 6EI (1 + O)12 (3.15) (3.16) (3.17) (3.18) 6EI (1 + )12 ) ) ) Similarly, if the bottom end of the element is fixed, as shown in Figure 3.6b, then by use of the differential equations for the beam deflections or the condition of symmetry it can be demonstrated that k = k 12EI (3.19) 2,2 5,5 (1 + 4)13 k = -k = -6EI (3.20) 3,2 6,5 (1 + t)12 k = k = -12EI (3.21) 5,2 2,5 (1 + t)13 k = -k -6EI (3.22) 6,2 3,5 (1 + ')12 with the remaining coefficients in column 2 equal to zero. In order to determine the stiffness coefficients associated with the rotations wg and w3, the element is subjected to bending moments and the associated shears, as shown in Figure 3.6c and d. The deflections can be determined from Equation (3.6), but the constants C1 and C2 in these equations must now be evaluated from a different set of boundary conditions. With the boundary conditions (Figure 3.6c) v = O at x = O, x = 1 (3.23) and dv dvs 5 at x = 1, (3.24) dx dx GA5 Equation (3.6) becomes El = 12x) + (x x2) + (lx x2) (3.25) 6 2 2 6f6 6MT 5 (4 + )1+ (4 + t)1 (3.26) As before, the remaining forces acting on the element can be determined from the equations of equilibrium, i.e., Equations (3.12) and (3.13). Now at x = 0 db = dv dx dx dv dx 6 (3.27) so that = f6 (1 + -)1 w (4 6 EI (4 + 9) (3.28) MT (1 + P)1 El (4 + ) Hence, from Equations (3.12), (3.13), (3.26), and (3.28) k 6,6 w )6 T=O S(4 + D)EI (1 + 4)1 k 2 = 5= T2=0 T=O k 3 3,6 w= ; T=0 S-f6 1+ f w6 ) T=0 S(2 ) E . (1 + 4)1 If the deflection of the left-hand end of the beam is equal to zero, as shown in Figure 3.6d, it is evident from symmetry that (3.32) k = k = (4 + )EI 3,3 6,6 (1 + T)1 and -6EI (1 + 4)12 (3.29) (3.30) (3.31) k = k 6EI (3.33) 5,6 =6,5 (1 + )I2 k = -6EI (3.34) 2,3 3,2 (1 + $)12 k = k 6EI (3.35) 5,3 3,5 (1 + )12 k = k = (2 )EI (3.36) 6,3 3,6 (1 + 0)1 with the remaining coefficients in columns 3 and 6 equal to zero. Thus, the stiffness matrix for the basic 6 DOF element, shown in Figure 2.1a, takes the form shown in Figure 3.7 when shear deformation is considered. 3.4 Moment Magnification With increasing slenderness and height, the lateral deflections of a vertical member due to bending will increase. As these deflections increase, an additional moment is caused by the vertical load acting through these deflections. This additional moment is often referred to as a secondary bending moment. Moment magnification is one term used to describe this effect. Figure 3.8 shows how moment magnification occurs. The technique for including the effect of moment magnification in the Direct Stiffness Method analysis of a structure also involves altering the standard terms in the stiffness matrix of the elements for which this effect is to be considered. The terms in the stiffness matrix for the standard 6 DOF element, shown in Figure 2.2, neglect moment magnification as well as shear deformation, which was discussed in the previous section. To illustrate how the new terms in the element -AE L 0 12EI -6EI 0 -12EI -6EI L3 (1 + 0) L2 (1 + P) L3 (1 + ) L2 (1 + $) 0 6EI (4 + D) EI 0 6EI (2 4) EI L2 (1 +) *) L2 (1 + L) L (1 + ) -AE 0 0 AE 0 0 L L 0 -12EI 6EI 0 12EI 6EI L3 (1 + ) L2 (1 + ) L3 (1 + L) L2 (1 + ) -6EI L2 (1 + t) (2 4) EI L (1 + ) 6EI L2 (1 + ) (4 + t) EI L (1 + ) WHERE: A E L I AREA OF ELEMENT CROSS-SECTION MODULUS OF ELASTICITY ELEMENT LENGTH MOMENT OF INERTIA OF ELEMENT CROSS-SECTION 4 = FACTOR USED IN ACCOUNTING FOR SHEAR DEFORMATION = 12EI E GAsL2 G = SHEAR MODULUS = E GAL 2(1+v) v = POISSON'S RATIO As = AREA IN SHEAR = .84 x (NET AREA) Figure 3.7 Element Stiffness Matrix Considering Shear Deformation [k] = P A 7 M = PA Moment Magnification Figure 3.8 stiffness matrix which considers moment magnification are developed, the following derivation is presented. It was taken from Chajes (3). Once again, element degrees of freedom 1 and 4 are not considered below because they are not affected by the consideration of moment magnification. Columns 1 and 4 in the element stiffness matrix shown in Figure 2.2 will, therefore, remain intact. Consider an element of a beam column subject to an axial load P and a set of loads [f], as shown in Figure 3.9d. The corresponding element displacements [w] are depicted in Figure 3.9e. It is our purpose to find a matrix relationship between the loads [f] and the deformations [w] in the presence of the axial load P. As long as the deformations are small and the material obeys Hooke's law, the deformations corre- sponding to a given set of loads [f] and P are uniquely determined, regardless of the order of application of the loads. The deformations [wj can, therefore, be determined by applying first the entire axial load P and then the loads [f]. Under these circumstances, the relation of [f] to [w] is linear, and the stiffness matrix can be evaluated using the principle of conservation of energy. The element is assumed to be loaded in two stages. During the first stage, only the axial load P is applied and, during the second stage, the element is bent by the [f] forces while P remains constant. Since the element is in equilibrium at the end of stage one as well as at the end of stage two, the external work must be equal to the strain energy not only for the entire loading process but also for stage two by itself. The external work corresponding to the second loading stage is w4 w6 - w2 w3 w5 w6 3- W2 W3 w3 w4 ,2- w1 w2 4 --- f3 f4 f f I x :A w w4 I w2 Y- I f-wH (a) Original DOF (b) DOF Which Influence Moment Magnification (c) DOF Renumbered for Derivation Only (d) Element Forces (e) Element Deformations Figure 3.9 Element Used in Derivation of Moment Magnification Terms S= [w]T [f] +P I 1 (y,)2 dx (3.7) e 2 2 0 in which the first term represents the work of the [f] forces and the second term the work due to P. Since the ends of the member approach each other during bending, the axial force does positive work when it is a compression force and negative work if it is a tension force. The strain energy stored in the member during stage two is due only to bending. Thus U = EI fl (y)2 dx (338) 2 0 Equating the strain energy to the external work gives 1 [w]T [f] + fI (y')2 dx = E_ i (y,)2 dx (3.39) 2 2 20 Making use of the relationship [f] = [k] [w], in which [k] is the element stiffness matrix, Equation (3.39) becomes [w]T [k] [w] = E fl (y,,)2 dx P fl (y')2 dx (3.40) 0 0 To evaluate [k], it is necessary to put the right-hand side of Equation (3.40) into matrix form. This can be accomplished if the deflection y is assumed to be given by y = A + Bx + Cx2 + Dx3 (3.41) The choice of a deflection function is an extremely important step. A cubic is chosen in this instance because such a function satisfies the conditions of constant shear and linearly varying bending moment that 42 exist in the beam element. Taking the coordinate axes in the direction shown in Figure 3.9e, the boundary conditions for the element are y = y' = w2 at x = 0 Y -wI W and y = -w3 y' = w4 at x = 1 . Substitution of these conditions into Equation (3.41) makes it possible to evaluate the four arbitrary constants and to obtain the following expression for y: -w w 3(w w3) 2 2w2 + 4 x2 y = -w + w x + x x 1 2 12 1 (3.42) + 2 + w4 x + 2(w3 w1 x3 12 1 Equation (3.42) can be rewritten in matrix form as (3.43) y = [A] [w] . Differentiating the expression in (3.43) gives y' = [C] [w] " = [D] [w] and (3.44) (3.45) in which i3- 25x3 2x 252 /5 x2 3 2) 2 13 1 12 1 I 2) (12 2 [ci- 12 13 (1 4x x (6x2 6\ (3x2 x 12 3 12) 2 1 and [D] = 6 12x 6x 12x L12 1 3 +12) 1 12 (6x 1)j. In view of (3.44) and (3.45), one can write (y,)2 = [w]T [c]T [C] [w] and (y")2 = [wT [D]T [D] [w] Substitution of these relations into (3.40) gives [w]T[k][w] = [w]T El 1f [D]T[D]dx P 1f [C]T[C]d w] 0 from which [k] = El f l T D][]dx P fl [C]T[C]dx (3.50) Using the expressions given in (3.46) and (3.47) for [C] and [D] and carrying out the operations indicated in (3.50), one obtains 12 1 6 12 6 4 6 2 12 T 12 1 12 6 12 6 13 2 1 12 6 2 6 4 12 1 12 1 1 6 1 6 1 51 10 51 10 1 21 1 1 10 15 10 30 6 +1 6 1 51 10 51 1( 1 30 (3.46) (3.47) (3.48) (3.49) [k] = El . (3.51) S1 10 Equation (3.51) gives the stiffness matrix of a beam column element with the 4 DOF shown in Figure 3.9c. The matrix consists of two parts: the first is the conventional stiffness matrix of a flexural element and the second is a matrix representing the effect of axial loading on the bending stiffness. Figure 3.10 shows the stiffness matrix for the 6 DOF element shown in Figure 2.1a and Figure 3.9a considering moment magnification. Figure 3.11 shows the stiffness matrix for the same element considering both shear deformation and moment magnification. 3.5 Material Nonlinearity Sometimes it is necessary to analyze a structure which is made up of a material whose load-deformation response is nonlinear. In such a material, the modulus of elasticity will vary and will be a function of the level of loading which the material is subjected to. A stiffness analysis of a structure composed of a nonlinear material can be performed if provision is made for the variation in elastic modulus. This can be done as follows. Recall that the modulus of elasticity is one of the variables required to construct the stiffness matrix for each element in the structure. As mentioned above, a nonlinear material's modulus of elasticity depends on the level the material is loaded to. To properly account for nonlinearity, three things are done. 1) Modulus of elasticity values are made dependent on load level. 2) The load is applied to the structure in increments up to the load for which a solution is desired. 3) The modulus of elasticity value used in constructing the stiffness matrix of an element is based on the element forces resulting from the application of the previous load increment. -AE L 0 12EI 6P -6EI + P O -12EI + 6P -6EI + P L3 5L 2 10 L3 5L L2 10 0 -6EI + P 4EI 2PL 0 6EI P 2EI + PL L2 10 L 15 L2 10 L 30 -AE O 0 AE O 0 L L O -12EI + 6P 6EI P 0 12EI 6P 6EI P L3 5L L2 10 L3 5L L2 10 -6EI L2 2EI PL L 30 6EI L2 P 10 4EI 2PL L 15 WHERE: A E L I P AREA OF ELEMENT CROSS-SECTION MODULUS OF ELASTICITY ELEMENT LENGTH MOMENT OF INERTIA OF ELEMENT CROSS-SECTION ELEMENT AXIAL FORCE Figure 3.10 Element Stiffness Matrix Considering Moment Magnification [k] - -AE L 0 12EI 6P -6EI + P 0 -12EI + 6P -6EI P L3 (1 + ) 5L L2 (1 + ) 10 L3 (1 + ) L L2 ( + 4) 10 0 -6EI + P (4 + ) EI 2PL 0 6EI P (2 ) EI + PL L2 (1 + ) 10 L (1 + ) 15 L2 (1 + $) 10 L (1 + ) 30 -AE O 0 AE 0 0 L _L 0 -12EI + 6P 6EI P O 12EI 6P 6EI P L3 (1 + ) 5L 2 (1 ) 10 L3 (1 + ) 5L L2 (1 + t) 10 -6EI L2 (1 + ) (2 $) El L (1 +) +PL 30 6EI L2 (1 + ) P 10 (4 + I) EI 2PL L (1 + P) 15 WHERE: A = AREA OF ELEMENT CROSS-SECTION E = MODULUS OF ELASTICITY L = ELEMENT LENGTH I = MOMENT OF INERTIA OF ELEMENT CROSS-SECTION = FACTOR USED IN ACCOUNTING FOR SHEAR DEFORMATION = 12EI E GAsL2 G = SHEAR MODULUS = E 2(1+V) v = POISSON'S RATIO As = AREA IN SHEAR = .84 x (NET AREA) P = ELEMENT AXIAL FORCE Element Stiffness Matrix Considering Shear Deformation and Moment Magnification [k] - + P 10 Figure 3.11 Consider the nonlinear load-deformation curve in Figure 3.12 which is approximated by three straight line segments. Notice that three modulus of elasticity (E) values exist, and each is valid only over a certain region of load. Assume it is desired to load a structure to a value in load region 3. First of all, the load would be divided into increments. This is necessary since the solution to the application of one load increment will affect the response to the next load increment, and so forth. Now, apply the load to the structure in increments. After the application of each load step, go through the entire process of constructing the stiffness matrix for each element, constructing the structure stiffness matrix, solving for structure displacements, and obtaining element forces and displacements. To decide what value of E to use in constructing the stiffness matrix of an element, determine which load region the element forces fall in, based on the solution to the previous load increment. Since, in this fashion, the modulus of elasticity value is indeed related to the load each element experiences, the solution for the analysis of the structure will reflect the true load-deformation properties of the material from which it is made. Two additional points should be considered. First of all, when an element changes from one load region to the other, say from region 1 to region 2 in Figure 3.12, its modulus of elasticity will decrease from a value of E, to a value of E2. This means the stiffness of this element has decreased. Loads in the structure are resisted by the elements in accordance with their stiffnesses such that stiffer elements resist a larger part of the load and, therefore, develop larger element forces. After the application of the next load increment, since the modulus of elasticity for this element has gone down from El to E2, this element 48 E2 DEFORMATION Nonlinear Load Deformation Curve I-z cZ 0 0-J '-1 Figure 3.12 will now resist a smaller part of the total load, and a "load redistribution" occurs. If the load increment just applied is small relative to the decrease in element stiffness, this element will develop forces smaller than its previous element forces and go back to region 1. The problem is that now, the stiffness increases from E2 to E1 and after the next iteration will go back to E2 as before and cycle back and forth. To prevent this, care must be taken to insure that once the stiffness of an element decreases, i.e., the next smaller value of E is used to construct its stiffness matrix, the modulus of elasticity is not permitted to increase due to a drop in the element's load level. This is best accomplished by never selecting values of E for an element which are based on a load level lower than the highest experienced up to that point. The second point also is related to the load redistribution which occurs as stiffnesses of individual elements change. During the incremental loading of a structure, it is often desired to examine the structural response after each load step is applied. This permits an examination of how the structural behavior varies as the external loads on the structure are increased. To insure each intermediate solution is accurate, a provision can be made that if any element experienced a change in stiffness during the application of the previous load increment, a new solution with the current element stiffnesses should be calculated before applying the next load increment. This will provide a steady-state solution for each load level; that is, one in which no load redistribution occurred. Thus, accounting for material nonlinearity in the analysis of a structure by the Direct Stiffness Method involves applying the load in increments and carefully monitoring the selection of modulus of elasticity values used in constructing the stiffness matrix of each element. The remainder of the general procedure outlined in section 2.1 remains the same. 3.6 Equation Solving Techniques Recall from Chapter Two that once the structure stiffness matrix [K] and the structure force matrix [F] have been constructed, solving for the structure displacement matrix [W] entails solving the matrix equation [F] = [K] [W] for [W]. This requires solving N simultaneous equations, where N is the number of structure degrees of freedom. Two efficient techniques for solving simultaneous equations are Gauss Elimination and Static Condensation. Meyer (8) discusses both of these methods and was used as a reference for the following two sections. 3.6.1 Gauss Elimination As previously discussed, the equation [F] = [K] [W] takes the form shown below. K11 K12 K1 F1 K21 K22 K2N W2 2 KN1 KN2 KNN N FN Gauss Elimination essentially consists of two steps: 1) Forward elimination to force zeros in all positions below the diagonal of [K] by performing legal row operations on [K] and [F] 2) Backward substitution to solve for [W]. The following simple example helps illustrate how this is done. Assume it is desired to solve the four simultaneous equations: 2W1 2W1 + W2 + 6W2 6W3 - 6W2 + 93 7W4 = 7W3 + 5 W4 = 10 0 -28 0 They can be rewritten in matrix form as 10 0 O -28 0 Forward elimination is performed by (-1 x row 1) + row 2 10 0 -28 0 which yields then (6/5 x row 2') + row 3 yields and finally (35/9 x row 3') + row 4 produces 10 -10 -28 0 10 -10 -28 0 10 -10 -40 0 10 -10 -40 0 0 -6 9/5 -7 0 0 -7 5 0 -6 9/5 -7 2 1 0 0 10 0 5 -6 0 -10 0 0 9/5 -7 -40 0 0 0 -200/9 -1400/9 Now from back substitution, -200/9 W4 = -1400/9 therefore W4 = 7 , 9/5 W3 7W4 = -40 since W4 is known, W3 can be found directly W3 = 5 . Similarly, 5W2 6W3 = -10 yields W2 = 4 and 2W1 + W2 = 10 produces W1 = 3 . Therefore, matrix [W] has been solved for and found to be Wi 3 W2 4 [W] W3 5 W-4 7 Two properties of stiffness matrices can be used to further reduce the computational effort required to solve for the structure displacements. Stiffness matrices are symmetric and frequently will be tightly banded around the diagonal. Due to symmetry, the upper right triangular part of the matrix will be identical to the lower left triangular portion. Being banded around the diagonal means all nonzero coefficients will be concentrated near the diagonal. Since only the nonzero terms need to be considered for Gauss Elimination, and since only half of the nonzero terms need to be stored due to symmetry, tremendous savings in storage and computational efficiency are possible. Figure 3.13 qualitatively shows what an actual stiffness matrix might look like versus the stiffness matrix which is stored and used by a modified Gauss Elimination procedure that requires only half of the symmetric nonzero coefficients. Half the bandwidth (including the diagonal term) is shown as HBW. An idea of the storage savings that result from only storing half of the symmetric nonzero values in the stiffness matrix can be obtained by considering a simple example. For a structure with 194 degrees of freedom, there will be 194 simultaneous equations to solve, and the structure stiffness matrix will be a 194 x 194 matrix consisting of HBW HBW I---- I I ----- ********** -0- ****** WWWWWWWWWW# W 44444 ** **** x*** *K ** [K] = *S ^ [K] = * -0- " -o- ^tsm'li-Sti-o-* *****SfnsMt* -0-S (a) Actual Stiffness Matrix (b) Stiffness Matrix Stored Figure 3.13 Advantage of Symmetry and Bandedness of Structure Stiffness Matrix 37,636 numbers. If half the bandwidth of this matrix is 9, then only 194 x 9, or 1746 numbers need to be stored. This is a savings of 95.4%! Figure 3.14 compares the number of operations required by the regular Gauss-Elimination procedure to the number required by the modified Gauss-Elimination technique for a structure stiffness matrix with a value of 9 for half the bandwidth. Figure 3.15 shows the percent savings in computational effort that result by using the modified Gauss- Elimination method on half of the symmetric nonzero values in the stiffness matrix where half the bandwidth is equal to 9. As shown, up to 95.5% savings are possible! 3.6.2 Static Condensation Static condensation is another equation solving technique which can be used to solve the matrix equation [F] = [K] [W]. It involves partitioning each of the three matrices and is performed as follows: 1) Partition K W s= F into Kaa Kab a Fa Kba Kbbj Wbj 2) Perform raa forward eliminations on [K] where raa = number of rows in [Kaa]. This will yield 2,500,000 2,000,000 1,500,000 1,000,000 500,000 Standard Gauss Elimination Modified Gauss Elimination --- 0 50 100 150 NUMBER OF EQUATIONS Figure 3.14 Comparison of the Equation Solving Operations of Standard Gauss Elimination Versus Modified Gauss Elimination NUMBER OF EQUATIONS Figure 3.15 Computational Savings of Modified Gauss Elimination Over Standard Gauss Elimination K K' W F' aa ab a a LKa Kb Wb F Note that [Ka] [Wa] + [Kb] [Wb] = [Fe] but because of the forward eliminations [Ka] = 0 so this equation reduces to [K'b] [Wb] = [F]] (3.52) 3) Solve [K b] [Wb] [F ] for [Wb] 4) Note that [Ka [Wa] + [Kb] [Wb] = [Fa] and only [Wa] is unknown, so solve [K] [Wa] = ([F;] [Klb] Wb]) (3.53) for [Wa] by backward substitution. 5) Construct [w] =- ] In performing Static Condensation, best efficiency is obtained if: 1) [K] is partitioned into four equal quarters and [W] and [F], therefore, are each divided in half, and 2) Gauss Elimination is used to solve Equation (3.52). The numerical example of section 3.6 is solved below using Static Condensation. Recall that [K] [W] = [F] was written in matrix form as After partitioning, Performing 2 forward eliminations on [K] entails (-1 x row 1) + row 2 (6/5 x row 2') row 3 and yields 0 0 0 -6 0 0 9/5 -7 0 0 -7 5 W1I W2 w3 w4 10 0 -28 0 W1 W2 W3 W 10 0 -28 0 W1 W2 W3 W4 10 0 -28 0 9 -7 WI W2 W3 W4 10 -10 -40 0 -- ---i-r-r-i iiL^Ll .- --- Note that [K b] = O Solve [Kb] [Wb] = [F] for [Wb] using Gauss Elimination. 9 x rw 3 r/5 4 (35/9 x row 3') + row 4 -7 -40 0 gives -7 W3 -20019 V4 1 40 -1400/9 From backward substitution W = 7 W3 = 5 and therefore [WbI = ] Multiplication of [Kab] [Wb] gives 5 7 0 0 0 30 0 -6 Then [F;] [K b] [Wb] produces 10 0] 10 -10 30 20 Finally, [Kaa] [Wa = ([Fa] - [Kab] [b]) can be solved for [Wa] by back substitution. 1 W 10 5 W2 20 W2 = 4 W = 3 Therefore, [Wal = ] 4 [w] = Thus, Static Condensation uses Gauss Elimination as part of its procedure for solving simultaneous equations. Section 3.6.1 addressed the large storage and computational savings that result from using a modified Gauss Elimination technique which only uses half of the symmetric nonzero terms in the stiffness matrix. The use of Static Condensation in conjunction with the modified Gauss Elimination procedure was explored. This technique was found to be less efficient than just modified Gauss Elimination for small matrices, but for large matrices it was up to 11.5% more efficient than even modified Gauss Elimination. Figure 3.16 compares the number of operations required by modified Static Condensation to the number required by modified Gauss Elimination for a structure stiffness matrix with a value of 9 for half the bandwidth. The percent savings (or loss) in computational efficiency that results from the use of modified Static Condensation is shown in Figure 3.17. Since both of these methods each require storage of half of the symmetric nonzero values in the stiffness matrix, the storage requirements of each technique are the same. 3.7 Solution Convergence In the standard use of the Direct Stiffness Method, convergence of the solution rarely presents a problem. However, with consideration of the special items discussed in sections 3.1 through 3.5 immediate convergence of the solution is not guaranteed. One method of monitoring the accuracy of the solution is to add a step to the procedure detailed at the end of section 2.1. After the force matrix for each element is calculated, an equilibrium check can be made by multiplying every element force matrix by -1.0 and inserting each one into the structure force matrix. The resulting values should Modified Gauss Elimination Modified Static Condensation ----- Z 0 50 100 150 NUMBER OF EQUATIONS Figure 3.16 Comparison of Equation Solving Operations of Modified Gauss Elimination Versus Modified Static Condensation 10,000 7,500 5,000 2,500 200 100 - 75- 50- 25 Ij-j az w-j -25 0 50 100 150 200 NUMBER OF EQUATIONS Figure 3.17 Computational Savings of Modified Static Condensation Over Modified Gauss Elimination be very small, and the closer they are to zero, the better the solution. Thus, one can speak of the degree to which the solution converged. For example, if m is the number of elements and the previous operation, shown below, is carried out m [ERROR] = [F] + E (-1.0 x [f]i) i=1 and the largest value in the matrix [ERROR] is 1 x 10-14, this solution converged better than one which produced a value in [ERROR] equal to 9 x 10-1 To insure convergence of the solution to the matrix equation [F] = [K] [W] within a specified tolerance, a reasonable tolerance value of, say 0.001, should be selected and all values in matrix [ERROR] compared to it. If no value exceeds the tolerance, the solution is acceptable. If any value exceeds the tolerance, then all the values can be treated as an incremental force matrix [AF] and then used to solve for the incremental structure displacement matrix [AW] for the previous structure stiffness matrix [K]. In other words, set [AF] = [ERROR] and solve [AF] = [K] [AV] (3.54) for [AW] . The total structure displacements then become equal to [W] + [AW], the total displacements for each element become [w] + [Aw], and the total forces for each element [f] + [Af]. Once again, each element force matrix should be multiplied by -1.0, inserted into the original structure force matrix, and the values compared with the allowable tolerance once more. This procedure, therefore, monitors convergence and also promotes it. Naturally, cases may arise where the tolerance is set below the value that a satisfactory solution can be arrived at, even with the use of the incremental structure force matrix concept, so the number of cycles in which an attempt is made to converge on a solution should be limited to some fairly small value, such as 9. CHAPTER FOUR USING THE MODEL TO ANALYZE COMPOSITE MASONRY WALLS 4.1 Structural Idealization of Wall The analytical model considers the lowest story of a composite wall, the same portion that will be represented by the laboratory test walls. The wall is divided into a series of three types of elements, one for the brick, one for the block, and one for the collar joint. Each element extends the entire depth of the wall, which for the test walls will be 24 inches. Figure 4.1 shows a wall divided into these elements. As shown, the lowest nodes of both wythes are fixed to the foundation and the top nodes are restrained from lateral displacement as they will be in the laboratory test fixture. Thus, the wall is modeled as a frame with two lines of columns connected at 8 inch intervals by shear beams with rigid ends. Figure 4.2 shows the structure and element degrees of freedom used in the model. The manner of numbering the structure degrees of freedom follows the pattern shown regardless of wall height. Of course, the element degrees of freedom for each element of a given type are as shown. For the highest test wall, which will be 26 feet in height, there are 194 structure degrees of freedom and 117 elements, 39 of each type. 8" TYPICAL BRICK ELEMENT BRICK WYTHE RIGID COLLAR JOINT ELEMENT CONCRETE BLOCK ELEMENT CONCRETE BLOCK WYTHE Figure 4.1 Finite Element Kodel of Wall W16 W17 I I fi < 13 W14 W15 W4 W5 Q 2 11 W3 W4 t f w w U 5 W6 W6 BRICK w6 COLLAR JOINT BLOCK ELEMENT ELEMENT ELEMENT ,, 2 " fW22 w3 wT3 W1 W1 Figure 4.2 Structure and Element Degrees of Freedom 4.2 Types of Elements As previously noted, the model considers the wall divided into a series of three types of elements. These include a brick element, a collar joint element, and a block element. 4.2.1 Brick Element The basic finite element for the brick wythe is shown in Figure 4.3. It models a brick prism either two or three bricks high, depending on brick type, and the mortar joints adjacent to those bricks. The prism and element are each 8 inches tall. The element extends the entire 24 inch depth of the test wall, with uniformly distributed forces over the wall depth. Experimental tests will be done on brick prisms, 24 inches deep, in order to determine the load-deformation properties of the brick element to be used by the model. Like the brick in the wall, the brick in the prisms will contain running bond. The brick element will have the six degrees of freedom shown, which include a rotation, axial displacement, and lateral displacement at each end. This is the basic flexural element previously illustrated and discussed in section 2.2. The effects of shear deformation and moment magnification will be considered in the brick wythe of the model in accordance with the pro- cedures outlined in sections 3.3 and 3.4. In other words, the stiffness matrix for a brick element will take the form shown in Figure 3.11. 4.2.2 Collar Joint Element The basic finite element for the collar joint is shown in Figure 4.4. The element will have the four degrees of freedom shown which include a rotation and vertical displacement at each end. The element is rigid axially which forces the brick and block wythes to have the MORTAR JOINT 4" BRICK PRISM WITH 3 5/8" x 7 5/8" x 2 1/4" BRICK BRICK PRISM WITH 3 5/8" x 7 5/8" x 3 1/2" BRICK Figure 4.3 Finite Element For Brick 8" 8"[> w5 W6 (--W2 3 1 2 ?- Ji f3 -I- -2 ks w2 m w 4 RIGID 1 = ONE-HALF THE THICKNESS OF THE BRICK WYTHE (= 2") 2 = ONE-HALF THE THICKNESS OF S THE BLOCK WYTHE (= 2", 3" or 4") k = SHEAR SPRING STIFFNESS FACTOR k = MOMENT SPRING STIFFNESS FACTOR m Finite Element For Collar Joint Figure 4.4 same horizontal displacement at each node. Rigid links at each end represent one half of the width of each wythe. The two springs are situated at the wythe to wythe interface. The primary function of the collar joint in a composite wall is to serve as a continuous connection between the two wythes, transferring the shear stress between them. This transfer of shear is modeled by the vertical spring, called the shear spring. At this stage, it is uncertain how much moment the collar joint transfers between the wythes. The rotational spring, called the moment spring, has been included in the collar joint element so the effect of moment transfer can be included in the model if, at a later date, this proves necessary. The stiffness matrix for the collar joint element was developed previously in section 3.2 (see Figure 3.3) and takes the form shown in Figure 3.4. The values of the shear spring stiffness factor (ks) and the moment spring stiffness factor (km) will be obtained from experimental tests. 4.2.3 Concrete Block Element The basic finite element for the concrete block is shown in Figure 4.5. It models a block prism which consists of one block and a mortar joint. The prism and element are each 8 inches tall, the same height as the brick element. The block element also extends the entire 24 inch depth of the test wall, with uniformly distributed forces over the wall depth. Experimental tests will be done on block prisms, 24 inches deep, to determine the load-deformation properties of the block element to be used by the model. The block element will have the six degrees of freedom shown, which include a rotation, axial displacement, and lateral displacement at each MORTAR JOINTS .. . . w4 w6 LI$-: L I I 4 6" 8" BLOCK PRISM BLOCK PRISM WITH BLOCK PRISM WITH WITH 5 5/8" x 15 5/8" x 7 5/8" 7 5/8" x 15 5/8" x 7 5/8" 3 5/8" x 15 5/8" BLOCK BLOCK x 7 5/8" BLOCK W 2 W3 Wi Figure 4.5 Finite Element For Block end. It is identical to the brick element which is the same as the basic flexural element previously illustrated and discussed in section 2.2. The effects of shear deformation and moment magnification will also be considered in the block wythe of the model according to the procedures outlined in sections 3.3 and 3.4. This means that the stiffness matrix for a block element, like the stiffness matrix for a brick element, will also take the form shown in Figure 3.11. 4.3 Experimental Determination of Material Properties As mentioned earlier, experimental tests will be done to establish the material properties of each type of element. 4.3.1 Brick Brick prisms, like those shown previously in Figure 4.3, will be tested experimentally to obtain the values of some of the variables that appear in the terms of the brick element stiffness matrix. Two types of tests will be performed to determine strength and deformation properties for the brick element. In the first type of test, prisms will be loaded axially to failure and the axial deformation noted for each level of load. This will establish the relationship between axial load and axial deformation. The axial load will then be plotted against the axial deformation to obtain a curve. This curve will be approximated in a piecewise-linear fashion, i.e., divided into a series of straight line segments. The slope of each straight line segment will be equivalent to the axial stiffness factor AE/L for the region of load values established by the load coordinates of the end points of each line segment. This is shown in Figure 4.6. P= 0O A-t IA =I AE L BASIC RELATIONSHIP k, = AXIAL STIFFNESS FACTOR = AE a L Figure 4.6 Experimental Determination of Brick Element Axial Stiffness Factor P>0 The second type of test will involve loading the prisms eccentrically and measuring the end rotation due to the applied end moment for various eccentricities and levels of loading. As before, the moment will be plotted against the rotation and the resulting curves approximated by straight line segments. The slope of each segment, this time, will equal the rotational stiffness factor 3EI/L for the region of moment values established by the moment coordinates of each line segment for the corresponding axial load. This is shown in Figure 4.7. The prism sketches in Figures 4.6 and 4.7 are intended only to give a general idea of how these tests will be done, and are not meant to be detailed representations of the test set-up and instrumentation required to measure deflections and rotations. The remainder of the variables needed to construct the element stiffness matrix for the brick element can be obtained without further tests. Table 4.1 lists all the variables needed and their sources. As indicated, the modulus of elasticity value for the brick was taken from Tabatabai (15) and the brick Poisson's ratio from Grimm (5). Figure 4.8 shows how the element stiffness matrix for a brick element is calculated using the stiffness factors from the prism tests. Since the prisms will be loaded to failure in each type of test, the maximum axial compressive load capacity as well as the relationship between axial load and moment carrying capacity will be established. This will enable an axial load versus moment interaction diagram to be drawn for the brick element. Figure 4.9 shows the general form this diagram will take. The curve in this diagram will also be approximated by straight line segments. It will be used by the model to determine when a brick element has failed. Brick prism failure will be defined as P>O I 3EI L il e 6=1 BASIC RELATIONSHIP 3El k = ROTATIONAL STIFFNESS FACTOR - r L Figure 4.7 Experimental Determination of Brick Element Rotational Stiffness Factor Variables Used in Constructing Brick Element Stiffness Matrix VARIABLE DEFINITION SOURCE 4 FACTOR USED IN CALCULATED, = 12E ACCOUNTING FOR GA L2 SHEAR DEFORMATION s E BRICK MODULUS OF TABATABAI (15), E = 2,918x106 PSI ELASTICITY (ONLY USED FOR SHEAR DEFORMATION) I MOMENT OF INERTIA NOT USED DIRECTLY; 3EI/L FACTOR OF BRICK ELEMENT FROM TESTS USED CROSS-SECTION G BRICK SHEAR MODULUS CALCULATED, G = E 2(1+v) v BRICK POISSON'S RATIO GRIMM (5), v = 0.15 As BRICK AREA IN SHEAR CALCULATED, As = .84 Anet As = 17.19 IN2 Anet BRICK NET AREA MEASURED, Anet = 20.47 IN2 L BRICK ELEMENT LENGTH LENGTH DEFINED BY MODEL = 8 IN A AREA OF BRICK ELEMENT NOT USED DIRECTLY; AE/L FACTOR FROM CROSS-SECTION TESTS USED P BRICK ELEMENT AXIAL BRICK ELEMENT FORCE MATRIX DEGREE FORCE OF FREEDOM NUMBER 1; CALCULATED BY PROGRAM FOR EACH LOAD LEVEL Table 4.1 VALUES KNOWN: k a 0 0 [k] = ,, E, L, P, AE, 3EI L L LET ka = AE, L Figure 4.8 Brick Element Stiffness Matrix Formulation Using Experimentally Determined Axial and Rotational Stiffness Factors kr = 3E L Figure 4.9 Typical P Versus K Interaction Diagram For Brick Element |

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AE L 0 0 -AE L 0 0 0 12EI 6P -6EI P 0 -12EI 6P -6EI P 5L L2 10 l3 5L L2 10 0 -6EI * P 4EI 2PL 0 6EI P 2EI + PL L2 10 L 15 L2 10 L 30 -AE 0 0 AE 0 0 L L 0 -12EI A 6P 6EI P 0 12EI 6P 6EI P 1? 5L L2 10 l3 5L L2 10 0 -6EI * P 2EI + PL 0 6 El P 4EI 2PL L2 10 L 30 L2 10 L 15 WHERE: A = AREA OF ELEMENT CROSS-SECTION E = MODULUS OF ELASTICITY L = ELEMENT LENGTH I = MOMENT OF INERTIA OF ELEMENT CROSS-SECTION P = ELEMENT AXIAL FORCE Figure 3*10 Element Stiffness Matrix Considering Moment Magnification STOMST DOUBLE PRECISION STOVF DOUBLE PRECISION STOVST DOUBLE PRECISION STRF DOUBLE PRECISION STRK DOUBLE PRECISION STRW DOUBLE PRECISION TOTEIN INTEGER TOTEK DOUBLE PRECISION W1 DOUBLE PRECISION W2 DOUBLE PRECISION STORES THE ROTATIONAL STIFFNESS FACTOR FOR EACH ELEMENT STORES THE ABSOLUTE VALUE OF THE VERTICAL FORCE FOR EACH ELEMENT STORES THE VERTICAL STIFFNESS FACTOR FOR EACH ELEMENT STRUCTURE FORCE MATRIX STRUCTURE STIFFNESS MATRIX STRUCTURE DISPLACEMENT MATRIX STORES ALL OF THE ELEMENT INDEX MATRICES STORES ALL OF THE ELEMENT STIFFNESS MATRICES TOP HALF OF STRUCTURE DISPLACEMENT MATRIX BOTTOM HALF OF STRUCTURE DISPLACEMENT MATRIX SUBROUTINE DESCRIPTION ADD APPLYF BLESM BLPDMT BMULT ADDS MATRICES APPLIES FORCES ON STRUCTURE CONSTRUCTS BLOCK ELEMENT STIFFNESS MATRIX CALCULATES BLOCK AXIAL AND ROTATIONAL STIFFNESS FACTORS MULTIPLIES A REGULAR MATRIX BY A SYMMETRIC BANDED MATRIX BNSERT INSERTS A MATRIX INTO A SYMMETRIC BANDED MATRIX 224 B.6 Subroutine BMULT 292 B.7 Subroutine INSERT 292 B.8 Subroutine BNSERT 292 B.9 Subroutine EXTRAK 298 B.10 Subroutine PULROW 303 B.11 Subroutine PULMAT 303 B.12 Subroutine GAUSS1 303 B. 13 Subroutine STACON 310 B.14 Subroutine TITLE 310 B.15 Subroutine READ 310 B.16 Subroutine COORD 310 B-17 Subroutine CURVES 318 B.18 Subroutine PRINT 318 B. 19 Subroutine WRITE 318 B.20 Subroutine BRPDMT 318 B. 21 Subroutine CJPDMT 327 B.22 Subroutine BLPDMT 327 B. 23 Subroutine STIFAC 336 B. 24 Subroutine BRESM 336 B. 25 Subroutine CJESM 336 B.26 Subroutine BLESM 336 B. 27 Subroutine INDXBR 336 B.28 Subroutine INDXCJ 347 B.29 Subroutine INDXBL 347 B. 30 Subroutine FORCES 347 B.31 Subroutine APPLYF 347 B.32 Subroutine PLATEK 347 B.33 Subroutine DISPLA 356 B.34 Subroutine CHKTOL 356 B.35 Subroutine CHKFAI 356 B.36 Subroutine WYTHE 356 C USERS MANUAL 368 C.1 General Information 368 C.2 Data Input Guide 370 D DATA FILES FOR NUMERICAL EXAMPLES 386 D.1 Example Number 1 386 D.2 Example Number 2 391 D.3 Example Number 3 396 D.4 Example Number 4 401 D.5 Example Number 5 406 REFERENCES ...410 BIOGRAPHICAL SKETCH 412 C H0I1IJ5II [ K2* ][W2 j=Â£K2H2]. 165 CALI NULL (K2N2 ,NETOP, 1) NSTAB^NMOP-MHH IF (NSIAfil LI 1) MSTABT= 1 NCCL=NBBG1 IF(NEBOT.GT.Hi)NCCL=M1 DG 1Â£0 I=iiÂ£'IAEI#NBIOP K2W2 (I) =0 0 EC 170 K= 1,BCGI K22 (I) = K2N2 (I) +K2 (I, K) *H2 (K) 170 CC MINUE 180 CONTINUE C CCNSTBUC1 [F1] FBCK [FJ. DO 190 1= 1 NBTCF f 1(I)=B(I) 190 CONTINUE C PEBFOBH [FT J-[ K2 ]Â£ W2 ]=[ F1 HKW2 ] . CALI NUIL (F 1 HKW2, NBTOP, 1) NSIABT-NBIGP-HUI IF (NSTABT.LT. 1) NSTABT= 1 DO 2Cti I=NSTAE1,NB1GP f 1MKW2 (I) = F 1 (I) -K2M2 (I) 200 COMINUE IF (NSTAfiT.fig, 1) GO 10 2 15 NS1M 1=11 S1AB1- 1 DC 210 J=l,K5Tfl1 F 1HKH2(J)=F1 (J) 210 CCNIINUE C CCNSTBUCI [KT] FBGH [C]. 215 NCI N1=NBTCE IF(N1.L1.NB10P)NCCLK1=K1 NSTAFl-NBTCE-Kl+1 IF(NSTAHT.LI.1)NSTABT=1 DO 230 1=1,ESIAEl EC 220 J=1,NCOIK1 K1 (I,J)=C(I,J) 254 Table A.2-continued MATRIX TYPE BRIDP DOUBLE PRECISION BRMCOR DOUBLE PRECISION BRPCOR DOUBLE PRECISION BRPCV DOUBLE PRECISION BRTCOR DOUBLE PRECISION BRWYLD DOUBLE PRECISION CJDCOR DOUBLE PRECISION CJEF DOUBLE PRECISION CJEK DOUBLE PRECISION CJEW DOUBLE PRECISION CJMCOR DOUBLE PRECISION CJPCOR DOUBLE PRECISION CJTCOR DOUBLE PRECISION DELTAF DOUBLE PRECISION DELTAW DOUBLE PRECISION ERROR DOUBLE PRECISION DESCRIPTION STORES BRICK INTERACTION DIAGRAM AXIAL LOAD COORDINATES STORES BRICK MOMENT COORDINATES FOR EACH M-THETA CURVE STORES BRICK AXIAL LOAD COORDINATES STORES THE VALUE OF P BY WHICH EACH BRICK M-THETA CURVE IS IDENTIFIED STORES BRICK THETA COORDINATES FOR EACH M-THETA CURVE MATRIX THAT STORES THE VERTICAL LOAD CARRIED BY THE BRICK WYTHE AT EACH WALL LEVEL STORES COLLAR JOINT DELTA COORDINATES COLLAR JOINT ELEMENT FORCE MATRIX COLLAR JOINT ELEMENT STIFFNESS MATRIX COLLAR JOINT ELEMENT DISPLACEMENT MATRIX STORES COLLAR JOINT MOMENT COORDINATES STORES COLLAR JOINT VERTICAL LOAD COORDINATES STORES COLLAR JOINT THETA COORDINATES INCREMENTAL STRUCTURE FORCE MATRIX INCREMENTAL STRUCTURE DISPLACEMENT MATRIX MATRIX THAT MONITORS SOLUTION CONVERGENCE tu rv> WALL HEIGHT, INCHES 197 Figure 6.44 Collar Joint Shear Stress Versus Height For Example Number 5 WALL HEIGHT, INCHES 172 Figure 6.20 Wall Wythe Vertical Load Versus Height At Pmax For Example Humber 2 6.19 6.20 6.21 6.22 6.23 6.24 6.25 6.26 6.27 6.28 6.29 6.30 6.31 6.32 6.33 Wall Wythe Vertical Load Versus Height At 0.90 P For Example Number 2 max r Wall Wythe Vertical Load Versus Height At Pmax For Example Number 2 Wall Wythe Vertical Load Versus Height At 0.30 Pmax For Example Number 3 Wall Wythe Vertical Load Versus Height At 0.60 P_nv For Example Number 3 Wall Wythe Vertical Load Versus Height At 0.90 P For Example Number 3 max * Wall Wythe Vertical Load Versus Height At Pmax For Example Number 3 Wall Wythe Vertical Load Versus Height At 0.30 Pmax For Example Number 4 Wall Wythe Vertical Load Versus Height At 0.60 P For Example Number 4 max r Wall Wythe Vertical Load Versus Height At 0.90 P_ For Example Number 4 max Wall Wythe Vertical Load Versus Height At Pmax For Example Number 4 Wall Wythe Vertical Load Versus Height At 0.30 Pmax For Example Number 5 Wall Wythe Vertical Load Versus Height At 0.60 P x For Example Number 5 Wall Wythe Vertical Load Versus Height At 0.90 Pmax For Example Number 5 Wall Wythe Vertical Load Versus Height At Pmax For Example Number 5 Brick Wythe Moment Versus Height For Example Number 2 171 172 173 174 175 176 177 178 179 180 181 182 183 184 186 6.34 Brick Wythe Moment Versus Height For Example Number 3 137 6.35 Brick Wythe Moment Versus Height For Example Number 4 188 6.36 Brick Wythe Moment Versus Height For Example Number 5 139 67 forces for each element [f] + [Af]. Once again, each element force matrix should be multiplied by -1.0, inserted into the original structure force matrix, and the values compared with the allowable tolerance once more. This procedure, therefore, monitors convergence and also promotes it. Naturally, cases may arise where the tolerance is set below the value that a satisfactory solution can be arrived at, even with the use of the incremental structure force matrix concept, so the number of cycles in which an attempt is made to converge on a solution should be limited to some fairly small value, such as 9. WALL HEIGHT, INCHES 155 Figure 6.4 Lateral Wall Deflection Versus Height For Example Number 5 FINITE ELEMENT MODEL FOR COMPOSITE MASONRY WALLS By GEORGE X. BOULTON A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY UNIVERSITY OF FLORIDA 1984 This dissertation is dedicated to Almighty God in thanksgiving for all of His blessings. ACKNOWLEDGMENTS I would like to thank Dr. James H. Schaub for his active personal interest and support which encouraged me to attend the University of Florida, and for his willingness to serve on my supervisory committee. Dr. Morris W. Self deserves special thanks for serving as chairman of my supervisory committee, being my graduate advisor, and affording me the opportunity to work on this project. Special thanks also go to Dr. John M. Lybas for providing much invaluable aid and guidance during the development of this model. Further gratitude is extended to Professor William W. Coons, Dr. Fernando E. Fagundo, and Dr. Harry J. Reitz, Jr. for also serving on my supervisory committee. Each individual has greatly contributed to the value of my graduate studies and is an asset to their profession and to the University of Florida. Appreciation is expressed to Dr. Clifford 0. Hays. From his classes and notes, the author acquired the technical background necessary to undertake this effort. Thanks also go to Dr. Mang Tia for his technical assistance and recommendations. Randall Brown and Kevin Toye were valuable sources of advice, wisdom, and friendship throughout the author's graduate studies and deserve special mention. The assistance received from Krai Soongswang is also gratefully acknowledged. The unselfish sacrifice, constant support, and endless love provided by the author's parents have been a tremendous source of inspiration and must not go unnoticed. The encouragement of family, friends, and former teachers is also appreciated. Finally, the author extends heartfelt thanks to Ms. Cindy Zimmerman and Ms. Joanne Stevens for typing the manuscript and helping with its preparation. TABLE OF CONTENTS Page ACKNOWLEDGMENTS iii LIST OF TABLES viii LIST OF FIGURES x ABSTRACT xvii CHAPTER ONE INTRODUCTION 1 1.1 Background 1 1.2 Objective 6 TWO DEVELOPMENT OF THE FINITE ELEMENT MODEL 7 2.1 Matrix Analysis of Structures by the Direct Stiffness Method .....7 2.2 Construction of the Element Stiffness Matrix 9 2.3 Construction of the Element Index Matrix 12 2.4 Construction of the Structure Stiffness Matrix 15 2.5 Construction of the Structure Force Matrix 17 2.6 Solving For the Structure Displacement Matrix 18 2.7 Solving For the Element Displacement Matrix 20 2.8 Solving For the Element Force Matrix 21 THREE SPECIAL CONSIDERATIONS 22 3.1 Different Materials .....22 3.2 Different Types of Elements 22 3.3 Shear Deformation 25 3.4 Moment Magnification 36 3.5 Material Nonlinearity 44 3.6 Equation Solving Techniques 50 3.6.1 Gauss Elimination 50 3.6.2 Static Condensation 56 3*7 Solution Convergence .....63 FOUR USING THE MODEL TO ANALYZE COMPOSITE MASONRY WALLS 68 4.1 Structural Idealization of Wall 68 4.2 Types of Elements ..........71 4.2.1 Brick Element 71 4.2.2 Collar Joint Element 71 4.2.3 Concrete Block Element 74 4.3 Experimental Determination of Material Properties 76 4.3.1 Brick 76 4.3.2 Collar Joint 83 4.3.3 Concrete Block 83 4*4 Load Application 86 4.5 Solution Procedure 97 FIVE NUMERICAL EXAMPLES 102 5.1 General Comments 102 5.2 Material Property Data..... 103 5.2.1 P-A Curves 103 5.2.2 M- Curves 113 5.2.3 P-M Interaction Diagrams 130 5*3 Example Number 1 Finite Element Analysis of a Test Wall 137 5.4 Illustrative Examples 141 5.4.1 Example Number 2 141 5.4.2 Example Number 3 145 5.4.3 Example Number 4 145 5.4.4 Example Number 5.... 145 SIX RESULTS OF ANALYSIS 150 6.1 Wall Failure 150 6.2 Lateral Wall Deflection Versus Height 150 6.3 Brick Wythe Vertical Deflection Versus Height 150 6.4 Block Wythe Vertical Deflection Versus Height 160 6.5 Plate Load Versus End Rotation 160 6.6 Wall Wythe Vertical Load Versus Height 160 6.7 Brick Wythe Moment Versus Height 185 6.8 Block Wythe Moment Versus Height 185 6.9 Collar Joint Shear Stress Versus Height 185 SEVEN CONCLUSIONS AND RECOMMENDATIONS ..200 APPENDIX A COMPUTER PROGRAM 202 A.1 Introduction 202 A. 2 Detailed Program Flowchart 204 A. 3 Program Nomenclature 204 A.4 Listing of Program and Subroutines..... ...227 B SUBROUTINES 232 B. 1 Subroutine NULL 282 B.2 Subroutine EQUAL ..282 B. 3 Subroutine ADD 282 B.4 Subroutine MULT 282 B.5 Subroutine SMULT 282 B.6 Subroutine BMULT 292 B.7 Subroutine INSERT 292 B.8 Subroutine BNSERT 292 B.9 Subroutine EXTRAK 298 B.10 Subroutine PULROW 303 B.11 Subroutine PULMAT 303 B.12 Subroutine GAUSS1 303 B. 13 Subroutine STACON 310 B.14 Subroutine TITLE 310 B.15 Subroutine READ 310 B.16 Subroutine COORD 310 B-17 Subroutine CURVES 318 B.18 Subroutine PRINT 318 B. 19 Subroutine WRITE 318 B.20 Subroutine BRPDMT 318 B. 21 Subroutine CJPDMT 327 B.22 Subroutine BLPDMT 327 B. 23 Subroutine STIFAC 336 B. 24 Subroutine BRESM 336 B. 25 Subroutine CJESM 336 B.26 Subroutine BLESM 336 B. 27 Subroutine INDXBR 336 B.28 Subroutine INDXCJ 347 B.29 Subroutine INDXBL 347 B. 30 Subroutine FORCES 347 B.31 Subroutine APPLYF 347 B.32 Subroutine PLATEK 347 B.33 Subroutine DISPLA 356 B.34 Subroutine CHKTOL 356 B.35 Subroutine CHKFAI 356 B.36 Subroutine WYTHE 356 C USERS MANUAL 368 C.1 General Information 368 C.2 Data Input Guide 370 D DATA FILES FOR NUMERICAL EXAMPLES 386 D.1 Example Number 1 386 D.2 Example Number 2 391 D.3 Example Number 3 396 D.4 Example Number 4 401 D.5 Example Number 5 406 REFERENCES ...410 BIOGRAPHICAL SKETCH 412 LIST OF TABLES Table Page 4.1 Variables Used in Constructing Brick Element Stiffness Matrix 30 4.2 Variables Used in Constructing Block Element Stiffness Matrix 87 6.1 Summary of Wall Failure For Examples Number 1 Through 5 151 A. 1 Variables Used in Detailed Program Flowchart 205 A. 2 Program Nomenclature 215 B. 1 Nomenclature For Subroutine NULL 283 B.2 Nomenclature For Subroutine EQUAL 285 B.3 Nomenclature For Subroutine ADD 287 B.4 Nomenclature For Subroutine MULT .289 B.5 Nomenclature For Subroutine SMULT 291 B.6 Nomenclature For Subroutine BMULT 294 B.7 Nomenclature For Subroutine INSERT 296 B.8 Nomenclature For Subroutine BNSERT 299 B.9 Nomenclature For Subroutine EXTRAK 301 B.10 Nomenclature For Subroutine PULROW 304 B. 11 Nomenclature For Subroutine PULMAT 306 B.12 Nomenclature For Subroutine GAUSS1 3CQ B.13 Nomenclature For Subroutine STACON 311 B.14 Nomenclature For Subroutine TITLE 314 B.15 Nomenclature For Subroutine READ 316 B.16 Nomenclature For Subroutine COORD 319 B.17 Nomenclature For Subroutine CURVES 321 B. 18 Nomenclature For Subroutine PRINT 323 B.19 Nomenclature For Subroutine WRITE 325 B.20 Nomenclature For Subroutine BRPDMT 328 B.21 Nomenclature For Subroutine CJPDMT 331 B.22 Nomenclature For Subroutine BLPDMT 333 B.23 Nomenclature For Subroutine STIFAC 337 B.24 Nomenclature For Subroutine BRESM 339 B.25 Nomenclature For Subroutine CJESM 341 B.26 Nomenclature For Subroutine BLESM 343 B.27 Nomenclature For Subroutine IHDXBR 345 B.28 Nomenclature For Subroutine INDXCJ 348 B.29 Nomenclature For Subroutine INBXBL 350 B. 30 Nomenclature For Subroutine FORCES 352 B. 31 Nomenclature For Subroutine APPLYF 354 B.32 Nomenclature For Subroutine PLATEK 357 B.33 Nomenclature For Subroutine DISPLA 359 B.34 Nomenclature For Subroutine CHKTOL 361 B.35 Nomenclature For Subroutine CHKFAI 363 B.36 Nomenclature For Subroutine WYTHE 365 C.1 Data Input Guide 375 LIST OF FIGURES Figure Page 1.1 Typical Composite Masonry Wall Section 2 1.2 Typical Composite Masonry Wall Loadbearing Detail 2 1.3 Stress and Strain Distribution For a Composite Prism Under Vertical Load 4 2.1 Development of Element Stiffness Matrix 10 2.2 Basic Element Stiffness Matrix 13 2.3 Structure and Element Degrees of Freedom For a Frame 14 2.4 Construction of the Structure Force Matrix 19 3.1 Structure and Element Degrees of Freedom For a Frame With Different Materials and Element Types.. 24 3.2 Element Stiffness Matrices For Elements 1 and 3 in Figure 3*1 26 3.3 Development of Stiffness Matrix For Element 2 of Figure 3*1 27 3.4 Element Stiffness Matrix For Element 2 in Figure 3.1 28 3.5 Shear Deformation and Its Importance 29 3.6 Application of Displacements to Establish Stiffness Coefficients Considering Shear Deformation 31 3.7 Element Stiffness Matrix Considering Shear Deformation 37 3.8 Moment Magnification 33 3*9 Element Used in Derivation of Moment Magnification Terms 40 3.10 Element Stiffness Matrix Considering Moment Magnification 45 3*11 Element Stiffness Matrix Considering Shear Deformation and Moment Magnification 46 3.12 Nonlinear Load Deformation Curve 48 3.13 Advantage of Symmetry and Bandedness of Structure Stiffness Matrix 55 3.14 Comparison of the Equation Solving Operations of Standard Gauss Elimination Versus Modified Gauss Elimination 37 3.15 Computational Savings of Modified Gauss Elimination Over Standard Gauss Elimination 58 3.16 Comparison of Equation Solving Operations of Modified Gauss Elimination Versus Modified Static Condensation 64 3.17 Computational Savings of Modified Static Condensation Over Modified Gauss Elimination 65 4.1 Finite Element Model of Wall 69 4.2 Structure and Element Degrees of Freedom 70 4.3 Finite Element For Brick 72 4.4 Finite Element For Collar Joint 73 4.5 Finite Element For Block... 75 4.6 Experimental Determination of Brick Element Axial Stiffness Factor 77 4.7 Experimental Determination of Brick Element Rotational Stiffness Factor 79 4.8 Brick Element Stiffness Matrix Formulation Using Experimentally Determined Axial and Rotational Stiffness Factors 81 4.9 Typical P Versus M Interaction Diagram For Brick Element .82 4.10 Experimental Determination of Collar Joint Shear Spring Stiffness Factor 84 4.11 Determination of Collar Joint Moment Spring Stiffness Factor 85 4.12 Structure Force Application Through Test Plate 89 4.13 Program Algorithm 99 5.1 Experimental Source of Brick Element P-A Curve (4) 105 5.2 Brick Element P-A Curve 107 5*3 Test Assembly Used By Williams and Geschwinder For Collar Joint Tests (18) 108 5.4 Experimental Source of Collar Joint Element P-A Curve (18) 109 5*5 Collar Joint Element P-A Curve 111 5.6 Experimental Source of Concrete Block Element P-A Curve (4) 112 5*7 Block Element P-A Curve 114 5.8 Experimental Source of Brick Element M-9 Curves (4) 115 5.9 Relationships Used to Obtain Desired Data From Experimental Results For M-9 Curves 117 5.10 Brick Element M-9 Curves 125 5-11 Collar Joint Element M-9 Curve 126 5.12 Experimental Source of Concrete Block Element M-9 Curves (4) 127 5-13 Block Element M-9 Curves 131 5.14 Experimental Source of Brick Element P-M Interaction Diagram (4) 132 5.15 Brick Element P-M Interaction Diagram 134 5.16 Experimental Source of Concrete Block Element P-M Interaction Diagram (4) 135 5.17 Block Element P-M Interaction Diagram 136 5*18 Example Number 1 138 5.19 Effective Column Length Factors Based On End Conditions (7) 140 5.20 Structure Degrees of Freedom For Example Number 1 ...142 5.21 Example Number 2 143 5*22 Structure Degrees of Freedom For Example Numbers 2 and 3 144 5*23 Example Number 3 146 5> 24 Example Number 4 147 5.25 Structure Degrees of Freedom For Example Numbers 4 and 5 148 5*26 Example Number 5 149 6.1 Lateral Wall Deflection Versus Height For Example Number 2 152 6.2 Lateral Wall Deflection Versus Height For Example Number 5 193 6.3 Lateral Wall Deflection Versus Height For Example Number 4 154 6.4 Lateral Wall Deflection Versus Height For Example Number 5 153 6.5 Brick Wythe Vertical Deflection Versus Height For Example Number 2 156 6.6 Brick Wythe Vertical Deflection Versus Height For Example Number 3 157 6.7 Brick Wythe Vertical Deflection Versus Height For Example Number 4 158 6.8 Brick Wythe Vertical Deflection Versus Height For Example Number 5 159 6.9 Block Wythe Vertical Deflection Versus Height For Example Number 2 161 6.10 Block Wythe Vertical Deflection Versus Height For Example Number 3 162 6.11 Block Wythe Vertical Deflection Versus Height For Example Number 4 163 6.12 Block Wythe Vertical Deflection Versus Height For Example Number 5 164 6.15 Plate Load Versus End Rotation For Example Number 2 165 6.14 Plate Load Versus End Rotation For Example Number 3 166 6.15 Plate Load Versus End Rotation For Example Number 4 167 6.16 Plate Load Versus End Rotation For Example Number 5 168 6.17 Wall Wythe Vertical Load Versus Height At 0.30 Pmax Por Example Number 2 169 6.18 Wall Wythe Vertical Load Versus Height At 0.60 PmaX Por Example Number 2 170 6.19 6.20 6.21 6.22 6.23 6.24 6.25 6.26 6.27 6.28 6.29 6.30 6.31 6.32 6.33 Wall Wythe Vertical Load Versus Height At 0.90 P For Example Number 2 max r Wall Wythe Vertical Load Versus Height At Pmax For Example Number 2 Wall Wythe Vertical Load Versus Height At 0.30 Pmax For Example Number 3 Wall Wythe Vertical Load Versus Height At 0.60 P_nv For Example Number 3 Wall Wythe Vertical Load Versus Height At 0.90 P For Example Number 3 max * Wall Wythe Vertical Load Versus Height At Pmax For Example Number 3 Wall Wythe Vertical Load Versus Height At 0.30 Pmax For Example Number 4 Wall Wythe Vertical Load Versus Height At 0.60 P For Example Number 4 max r Wall Wythe Vertical Load Versus Height At 0.90 P_ For Example Number 4 max Wall Wythe Vertical Load Versus Height At Pmax For Example Number 4 Wall Wythe Vertical Load Versus Height At 0.30 Pmax For Example Number 5 Wall Wythe Vertical Load Versus Height At 0.60 P x For Example Number 5 Wall Wythe Vertical Load Versus Height At 0.90 Pmax For Example Number 5 Wall Wythe Vertical Load Versus Height At Pmax For Example Number 5 Brick Wythe Moment Versus Height For Example Number 2 171 172 173 174 175 176 177 178 179 180 181 182 183 184 186 6.34 Brick Wythe Moment Versus Height For Example Number 3 137 6.35 Brick Wythe Moment Versus Height For Example Number 4 188 6.36 Brick Wythe Moment Versus Height For Example Number 5 139 6.37 Block Wythe Moment Versus Height For Example Number 2 190 6.38 Block Wythe Moment Versus Height For Example Number 3 191 6.39 Block Wythe Moment Versus Height For Example Number 4 192 6.40 Block Wythe Moment Versus Height For Example Number 5 .....193 6.41 Collar Joint Shear Stress Versus Height For Example Number 2.. 194 6.42 Collar Joint Shear Stress Versus Height For Example Number 3 195 6.43 Collar Joint Shear Stress Versus Height For Example Number 4 196 6.44 Collar Joint Shear Stress Versus Height For Example Number 5 ..197 A.1 Detailed Program Flowchart 207 B.1 Algorithm For Subroutine NULL 284 B. 2 Algorithm For Subroutine EQUAL 286 B.3 Algorithm For Subroutine ADD 288 B.4 Algorithm For Subroutine MULT 290 B.5 Algorithm For Subroutine SMULT 293 B.6 Algorithm For Subroutine BMULT 295 B.7 Algorithm For Subroutine INSERT 297 B.8 Algorithm For Subroutine BNSERT 300 B.9 Algorithm For Subroutine EXTRAK 302 B.10 Algorithm For Subroutine PULROW 305 B.11 Algorithm For Subroutine PULMAT 307 B.12 Algorithm For Subroutine GAUSS 1 309 B.13 Algorithm For Subroutine STACON 313 B.14 Algorithm For Subroutine TITLE 315 B. 15 Algorithm For Subroutine READ 517 B.16 Algorithm For Subroutine COORD 520 B. 17 Algorithm For Subroutine CURVES 522 B.18 Algorithm For Subroutine PRINT 524 B. 19 Algorithm For Subroutine WRITE 526 B.20 Algorithm For Subroutine BRPDMT 530 B. 21 Algorithm For Subroutine CJPDMT 532 B.22 Algorithm For Subroutine BLPDMT 335 B.25 Algorithm For Subroutine STIFAC 338 B.24 Algorithm For Subroutine BRESM 340 B.25 Algorithm For Subroutine CJESM 542 B.26 Algorithm For Subroutine BLESM 544 B.27 Algorithm For Subroutine IHDXBR 546 B.28 Algorithm For Subroutine INDXCJ 549 B.29 Algorithm For Subroutine INDXBL 551 B.50 Algorithm For Subroutine FORCES 555 B.51 Algorithm For Subroutine APPLYF 555 B.52 Algorithm For Subroutine PLATEK 558 B.53 Algorithm For Subroutine DISPLA 560 B.54 Algorithm For Subroutine CHKTOL 562 B.55 Algorithm For Subroutine CHKFAI 564 B.56 Algorithm For Subroutine WYTHE 567 Abstract of Dissertation Presented to the Graduate School of the University of Florida in Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy FINITE ELEMENT MODEL FOR COMPOSITE MASONRY WALLS By GEORGE X. BOULTON December 1984 Chairman: Dr. Morris W. Self Major Department: Civil Engineering Composite masonry design standards are at an early stage of development. To improve the understanding of composite masonry wall behavior in response to load application, a two-dimensional finite element model has been developed. The model considers a wall subjected to vertical compression and out of plane bending. It takes into account the different strength- deformation properties of the concrete block, collar joint, and clay brick, as well as the nonlinear nature of these properties for each material. The effects of moment magnification and shear deformation in both the brick wythe and the block wythe of a composite wall are also considered. The primary purpose of this model is to analytically represent typical composite masonry walls that might be tested in a laboratory. Wall tests attempt to duplicate conditions found in prototype walls. By comparing the results of the analytical and experimental tests, needed insight into composite wall behavior can be obtained, and design standards can be modified to reflect this increased understanding. Every effort has been made to consider the factors that will most strongly influence composite wall behavior in the development of the model, so that its usefulness as an analytical research tool will not be compromised. This study describes the model in detail, provides information on how it can be used, and contains several numerical examples that illustrate the information its use makes available. CHAPTER ONE INTRODUCTION 1.1 Background A composite masonry wall is a wall which consists of a clay brick wythe, a concrete block wythe, and a collar joint which forms a bond between the two wythes. A section of a typical composite masonry wall is shown in Figure 1.1. The collar joint between the two wythes consists of either masonry mortar or concrete grout. It forces the wall to behave as one structural unit, even though the wall consists of different materials. Composite masonry walls are frequently used as exterior bearing walls with the brick exposed as an architectural surface and the concrete block used as the load-bearing material. Thus, when a floor slab or roof truss bears on a composite masonry wall, it transfers vertical load directly to the block wythe. A typical composite masonry wall load-bearing detail is shown in Figure 1.2. Two design standards have been widely used in the United States as references for the design of engineered masonry construction. These are the Brick Institute of America (BIA) "Building Code Requirements for Engineered Brick Masonry" (2) and the National Concrete Masonry Association (NCMA) "Specifications for the Design and Construction of Load-Bearing Concrete Masonry" (14). A third standard on concrete masonry was recently published by the American Concrete Institute (ACl) entitled ACI 531-79R, "Building Code Requirements for Concrete Masonry 1 2 CLAY BRICK WYTHE Figure 1.1 Typical Composite Masonry Vail Section Figure 1.2 Typical Composite Masonry Vail Loadbearing Detail 3 Structures" (1). It contains a brief chapter on composite masonry. Finally, a joint American Society of Civil Engineers and American Concrete Institute committee, Committee 530, is currently developing a comprehensive standard to include provisions for both clay and concrete products. Unfortunately, these design standards are limited by a lack of laboratory test data as well as a lack of understanding of the behavior and response of composite masonry. Rational analyses to predict the failure loads and load-deformation properties of composite masonry walls do not presently exist (6). To improve the reliability of the design procedure for composite masonry, several factors not currently considered must be taken into account. The first is that masonry is not linearly elastic, it is nonlinear. In other words, its modulus of elasticity changes depending on the stress level to which it is subjected. Figure 1.3 illustrates some of the complications that result due to the different and nonlinear material properties of brick and block which are present in composite masonry. As evident from the stress-strain curves of the two materials, the block is weaker. Stage 1 depicts the stresses and strains that are generated as a vertical load is applied through the centroid of a composite prism above a 50 percent stress level. As the load is increased further to stage 2, the stiffness of the concrete deteriorates more rapidly than that of the brick. This causes the center of resistance of the section to shift toward the brick. The eccentricity between the point of load application and the effective resistance of the section results in an effective bending of the prism, causing the strains in the block to increase faster than those in the brick. As a POSITION OF \/ APPLIED LOAD A POSITION OF ^ ELASTIC CENTER Figure 1.3 Stress and Strain Distribution For a Composite Prism Under Vertical Load 5 result, the stiffness of the block deteriorates even faster than that of the brick, the center of resistance shifts even further, and the eccentricity and bending it produces are further aggravated. Stage 3 denotes failure of the prism which is characterized by vertical splitting of the concrete. This phenomenon was verified by experimental tests at the University of Florida (11,13,15). The failure mechanism for composite walls under vertical loads should reflect this behavior. A second factor meriting consideration is that roof trusses or floor slabs, as mentioned previously, generally bear on the block wythe causing the wall to be loaded eccentrically toward the block. This will aggravate the failure mechanism just discussed. A third consideration is that with increasing slenderness and height, lateral wall deflections due to bending will increase. As these deflections increase, the additional moment caused by the vertical load acting through these deflections takes on increasing importance and can lead to a stability problem. All three of these aspects of behavior should be affected by the wall's height to thickness ratio as well as the thickness of the block wythe. In response to the need for more experimental work on composite masonry walls and an improved understanding of wall behavior, laboratory tests of composite walls have been performed by Redmond and Allen (10), Yokel, Mathey, and Dikkers (19), and Fattal and Gattaneo (4). Nonetheless, as a whole, the three studies consider limited combinations of wall geometry, masonry unit properties, and height to thickness ratios. Additional tests are needed so that design standards can be supported by a large data base. Wall test results also need to be related to analytical models. 6 1.2 Objective Lybas and Self (6) have submitted a research proposal to the National Science Foundation aimed at addressing some of these needs. Specifically, they seek to explore both experimentally and analytically 1) The nonlinear load deformation properties of composite masonry walls under compression and out of plane bending. The effect of transverse loading, eccentric compressive loading, slenderness ratio and different masonry unit properties will be considered. 2) The failure mechanisms of composite masonry walls under these types of loading conditions. 3) The transfer of vertical force across the collar joint from block wythe to brick wythe. 4) The suitability of current standards for composite masonry wall design. 5) The development of improved design equations and procedures for composite walls, based on the results of the research. This study essentially consists of the development of an analytical model which will be used, once the experimental phase has been completed, to examine the factors cited above. The model will consider a composite masonry wall subject to compression and out of plane bending, due to either eccentric load application or transverse loading. The two-dimensional finite element model will take into account the different nonlinear load-deformation properties of the brick and block wythes, the load transfer properties of the collar joint, the effect of moment magnification that, as previously mentioned, will result as the lateral wall deflections increase, and the effect of shear deformations. CHAPTER TWO DEVELOPMENT OF THE FINITE ELEMENT MODEL 2.1 Matrix Analysis of Structures by the Direct Stiffness Method The Direct Stiffness Method, like most matrix methods of structural analysis, is a method of combining elements of known behavior to describe the behavior of a structure that is a system of such ele ments. The following is a summary of the basic relationships U3ed in this technique. It is presented only as a quick review of the Stiffness Method and not as an exhaustive presentation which develops the rela tionships stated. For that purpose, one of the fine textbooks on matrix analysis, such as Rubinstein's (12), is recommended. A degree of freedom is an independent displacement. Recall that the force displacement equations for an element i, which has n element degrees of freedom, can be written as Mi Mi [v]. [f0^ (2.1) where [w]i n x 1 matrix of independent element displacements measured in element coordinates Mi = n x 1 matrix of corresponding element forces measured in element coordinates [f0]^ = n x 1 matrix of corresponding element fixed-end forces measured in element coordinates [k]^ = n x n element stiffness matrix measured in element coordinates. 7 8 In general, [k]^ and [f]j_ can be found from standard cases. Since this model only considers nodal loads, [f0]^ matrices will not exist. The force displacement equations for an element i, therefore, reduce to [f]i = Mi Mi (2.2) Suppose an element i is connected to other elements to form a structure with N structure degrees of freedom. The structure force displacement (or equilibrium) equations can be expressed as !>] = [K] [w] + [F] (2.3) where [w] = N x 1 matrix of independent structure displacements measured in structure coordinates [f] = N x 1 matrix of corresponding structure forces measured in structure coordinates [f] = II x 1 matrix of corresponding structure fixed-end forces measured in structure coordinates [k] = N x N structure stiffness matrix measured in structure coordinates. Again, fixed-end forces are not in the model so these equations reduce to [f] = M M (2.4) If m equals the total number of structure degrees of freedom that are related to the element degrees of freedom for element i, an index matrix [l]j_ can be defined as [l]j_ = m x 1 matrix whose elements are the numbers of the structure degrees of freedom that are related to the element degrees of freedom for element i. 9 Usually an n x m transformation matrix [t] is also required to transform structure displacements into element displacements. However, if the element and structure coordinate systems are coincident, a transformation matrix is not required. Such is the case in this model. The solution procedure is generally as follows. For each element, 1) Construct the index matrix [i]. 2) Construct the element stiffness matrix [k]. 3) Insert the element stiffness matrix [k] into the structure stiffness matrix [k] using [i]. Once the structure stiffness matrix has been formed, 4) Construct the structure force matrix [f]. 5) Solve for the structure displacements by solving [f] = [k] [w] for [w]. Finally, for each element 6) Extract the element displacement matrix [w] from the structure displacement matrix [w]. 7) Multiply the element stiffness matrix by the element displacement matrix to yield the matrix of element forces, i.e., [f] = M M. Sections 2.2 through 2.8 discuss these matrices in more detail. 2.2 Construction of the Element Stiffness Matrix Consider the element shown in Figure 2.1a. This is the basic flexural element found in any text on matrix structural analysis. It contains the six degrees of freedom shown which include a rotation, axial displacement, and lateral displacement at each end. Recall that the stiffness coefficient k^j is defined as the force developed at the ith degree of freedom (DOF) due to a unit displacement at the jth DOF of the element while all other nodal displacements are maintained at zero. For example, the stiffness coefficient k^ is the force developed at the first DOF due to a unit displacement at the first 10 w 4 w 3 w. (a) Basic Flexural Element (b) Application of Unit Displacements to Establish Stiffness Coefficients Figure 2.1 Development of Element Stiffness Matrix DOF. Similarly, the stiffness coefficient is the force developed at the first DOF due to a unit displacement at the second DOF. The total force F at the first DOF can, therefore, be represented as f1 = k11w1 + k12w2 + k13w3 + k14w4 + k15w5 + kl6w6 where w^ equals the element displacement at the ith DOF. Analogously, the forces at the other degrees of freedom are: f2 = k21w1 + k22w2 + k23w3 + k24w4 + k25w5 + k26w6 f6 = k6lw1 + k62w2 + k63w3 + k64w4 + k65w5 + k66w6 These equations can be written conveniently in matrix form as f1 k11 k1 2 k13 k14 k15 kl6 f2 k21 k22 k23 k24 k25 k26 f3 k31 k32 k33 k34 k35 k36 f4 k41 k42 k43 k44 k45 k46 f5 k51 k52 k53 k54 k55 k56 _f6_ k6l k62 k63 k64 k65 k66 or simply as [f] [k] [w] [w] = element [f] = element [k] = element displacement matrix force matrix stiffness matrix. where 12 Figure 2.1b shows the application of unit displacements to the basic flexural element in order to establish the stiffness coefficients. Figure 2.2 shows the resulting element stiffness matrix. It is a 6 x 6 matrix since the element contains 6 DOF. 2.3 Construction of the Element Index Matrix The index matrix [i] was previously defined in section 2.1 as the matrix which consists of the numbers of the structure degrees of freedom that are related to the element degrees of freedom for a particular element. To illustrate what this means, consider the frame shown in Figure 2.3a. If the structure and element degrees of freedom for this frame are as shown in Figure 2.3b and 2.3c, respectively, it is possible to construct the index matrix for each element simply by noting which structure degrees of freedom correspond to which element degrees of freedom. For example, to construct the index matrix for element number 1, observe that w^, W2> and w^ have no corresponding structure degrees of freedom but corresponds to w^, corresponds to w^, and corresponds to Wg. Thus, the index matrix for element number 1 is 0 0 3 4 Similarly, the index matrices for elements 2 and 3 are 13 M - AE L 0 0 -AE L 0 0 0 1 2EI -6EI 0 -12EI -6EI L2 l? L2 . 0 -6EI 4EI 0 6EI 2EI L2 L L2 L -AE 0 0 AE 0 0 L L 0 -12EI 6EI 0 1 2EI 6EI l3 L2 L^ L2 0 -6EI 2EI 0 6EI 4EI L2 L L2 L WHERE: A = AREA OF ELEMENT CROSS-SECTION E = MODULUS OF ELASTICITY L = ELEMENT LENGTH I = MOMENT OF INERTIA OF ELEMENT CROSS-SECTION Figure 2.2 Basic Element Stiffness Matrix 14 (a) Frame (b) Structure Degrees of Freedom w (c) Element Degrees of Freedom w. w 2 Figure 2.3 Structure and Element Degrees of Freedom For a Frame 15 [I] 2 1 4 2 5 O O O 2 3 5 The index matrix plays a vital role in the proper assemblage of the structure stiffness matrix. This is discussed in the next section. 2.4 Construction of the Structure Stiffness Matrix The structure stiffness matrix is constructed by using the element stiffness matrices and the element index matrices. To obtain a term in the structure stiffness matrix, it is necessary to add up the appropriate terms from the element stiffness matrices. The procedure for doing this is best illustrated by an example. Consider the frame of Figure 2.3. To identify where each coefficient in each element stiffness matrix belongs in the structure stiffness matrix, plpce the numbers in the index matrix for an element along the sides of the element stiffness matrix as shown below. 0 0 0 1 34 W1 - k11 k12 k13 k14 k15 kl6 k21 k22 k23 k24 k25 k26 k31 k32 k33 k34 k35 k36 k41 k42 k43 k44 k45 k46 k51 k52 k53 k54 k55 k56 VO k62 k63 k64 k65 k66 0 0 0 1 3 4 16 [k]2 - 1 4 2 5 k11 k1 2 k13 k14 k21 k22 K\ C\J k24 k31 k32 k33 k34 k41 k42 k43 k44 1 4 2 5 6 [k] 3 " 0 0 0 2 5 5 k11 k12 k13 k14 k15 kl6 k21 k22 k23 k24 k25 k26 k31 k32 k33 k34 k35 k36 k41 k42 k43 k44 k45 k46 k51 k52 k53 k54 k55 k56 k6l k62 k63 k64 k65 k66 0 0 0 2 3 5 Define LKijJm as the stiffness coefficient in row i, column j of the stiffness matrix for element m. The numbers in the index matrix along the sides of the element stiffness matrix for each element identify the rows and columns in the structure stiffness matrix in which each coefficient belongs. Since the frame has 5 DOF, the structure stiffness matrix will be a 5 x 5 matrix. The term in row 1, column 1 of the structure stiffness matrix is, therefore, 17 K11 fk44^ + ^k11^2 * Similarly, the other terms are as shown below: [K] - 1 2 3 4 5 * &]2 ^13-^ Ck453, ^3, + ^-k12^ 2 [*u]2 Ck31 J2 3 + [k ,,] " 2 [k,2] ^ 2 ^34^2 M3 [kcc3 55 1 + c*55], ^3, + ^21^ ^3, + [kp?] 2 ^2 2 ^43^2 * M3 5 fr3 4 2 * [k66^ 1 2 3 4 5 Thus, once the stiffness and index matrices are known for each element, the structure stiffness matrix can easily be arrived at. 2.5 Construction of the Structure Force Matrix To construct the structure force matrix, it is only necessary to assign the value of the force to the term in the matrix which corresponds to the degree of freedom at which the force is applied. If the force has the same direction as the degree of freedom, it is 18 considered positive. If it acts in the opposite direction, it is considered negative. Figure 2.4 shows two examples of structure force matrices for the frame of Figure 2.3* 2.6 Solving For the Structure Displacement Matrix Like the structure force matrix, the structure displacement matrix will also be an N x 1 matrix where N is the number of structure degrees of freedom. Thus, the structure displacement matrix for the frame of Figure 2.3 will be [W] = Solving for the structure displacement matrix will entail solving the matrix equation [f] = [k] [w] for [w]. This will consist of solving N simultaneous equations if N is the number of structure DOF. One way to solve this matrix equation is by matrix inversion, or [W] = [K]-1 [F] (2.5) For large values of N, however, the structure stiffness matrix of dimensions N x N will become large and calculating the inverse of a large matrix is very cumbersome and inefficient. Two much more efficient techniques for solving for the structure displacement matrix are Gauss Elimination and Static Condensation. These methods are W, V/ Wc discussed in detail in sections 3.6.1 and 3.6.2. 19 > " F1 0 F2 -50 F3 = 20 F4 0 F5 0 200 \ [F] 0 0 0 -100 200 Figure 2.4 Construction of the Structure Force Matrix 20 2.7 Solving For the Element Displacement Matrix Once the structure displacement matrix has been solved for, it is very easy to obtain the displacement matrix for each element. Since the index matrix for an element identifies which structure DOF correspond to which element DOF, it also identifies which structure displacements correspond to which element displacements. For example, consider the frame of Figure 2.3 again. It was previously observed that for element 1, w.|, w2, and w^ have no corresponding structure DOF but corresponds to w^, corresponds to w^, and corresponds to wg. This means that the element displacement matrix for element 1 equals W1 0 0 W2 0 0 w3 0 where [i]^ = 0 = w4 W1 1 w5 W3 3 w6 _W4 4 Using the index matrices for elements 2 and 3 which were previously developed in section 2.3 the element displacement matrices for elements 2 and 3 are, therefore, W1 0 W1 ~w1~ w2 0 w2 ss W4 and [w]j = w3 = 0 w3 w2 w4 W2 w4 s_ w5 w3 w6 _W5_ 21 Thus, once the structure displacement matrix values are available, the displacement matrix for each element is easily obtained with the help of the index matrix for each element. 2.8 Solving For the Element Force Matrix In section 2.1, it was mentioned that the force displacement equations for an element i can be expressed as Mi = Mi Mi Thus, the force matrix for each element is calculated simply by multiplying the stiffness matrix for that element by the displacement matrix for that element. The values in the element force matrix for an element will correspond to the element displacements for the element. For example, considering element 1 in the frame of Figure 2.3, the force matrix for element 1 will take the form [f] 1 f f f f f f 1 2 3 4 5 6 A positive value in the element force matrix implies that the resulting force for that DOF acts in the direction shown for that DOF. Conversely, a negative value represents a force which acts opposite to the direction shown. CHAPTER THREE SPECIAL CONSIDERATIONS 31 Different Materials The stiffness matrix for an element is dependent on the geometric, cross-sectional, and material properties of that element. This is evidenced by the nature of the variables in the element stiffness matrix shown in Figure 2.2. A structure which is comprised of different materials can be analyzed by the Direct Stiffness Method. The presence of different materials is accounted for by using the appropriate material property values when constructing the stiffness matrix for each element. Since the structure stiffness matrix is assembled using the stiffness matrix for each element, the solution for the structure will then reflect the presence of material differences among the different elements into which the structure is divided. In short, the presence of different materials in a structure is taken into account during the construction of the stiffness matrix for each element in the structure. 3.2 Different Types of Elements Occasionally, the accurate matrix analysis of a structure involves the use of more than one type of element in the analytical model. This presents no particular difficulty and, in fact, is handled in much the same way as the presence of different materials in the structure. In 23 other words, it too is accounted for during the construction of the stiffness matrix for each element. As mentioned in section 2.2, the stiffness matrix for an element is constructed by applying unit displacements, one at a time, at each DOF. If a structure is modeled by more than one type of element, this only means that the coefficients and variables in the element stiffness matrix of each element type will be different. Once all of the values in an element stiffness matrix are calculated, the element stiffness matrix is inserted into the structure stiffness matrix in the same fashion as discussed previously in section 2.4. Because the structure stiffness matrix is assembled using the element stiffness matrices, the solution of structure displacements and element displacements and forces will reflect the presence of different types of elements in the model. Consider the frame with the structure degrees of freedom, element degrees of freedom, and property values shown in Figure 3*1 Notice that this frame is similar to the one in Figure 2.3, which was previously discussed, except that: 1) Elements 1 and 3 have different property values 2) Element 2 is of a different type than the previous element 2 and is different from the present elements 1 and 3 Assume it is desired to analyze this structure by the Direct Stiffness Method. From the preceding discussion, it was learned that the procedure is identical to the one outlined at the end of section 2.1, but that the stiffness matrix for each element will be different due to the presence of different materials and different element types. To illustrate, the stiffness matrix for each element will be constructed. 24 L Figure 3. where: k = Shear Spring s Stiffness k = Moment Spring Stiffness Other variables defined in Figure 2.2 (a) Structure Degrees of Freedom W] w1 (b) Element Degrees of Freedom 1 Structure and Element Degrees of Freedom For a Frame With Different Materials and Element Types 25 First, for elements 1 and 3, the element DOF are identical to what they were in Figure 2.3- Therefore, the derivation of stiffness coefficients for this basic flexural element, shown in Figure 2.1, is valid. The stiffness matrix for elements 1 and 3 will thus take the form shown in Figure 2.2, but the coefficients will recognize the differences in the values of the variables. Figure 3-2 gives the element stiffness matrix for element 1 and the element stiffness matrix for element 3 To construct the element stiffness matrix for element 2, the stiffness coefficients must be derived by the application of unit displacements at each DOF. This is shown in Figure 3.3- The resulting stiffness matrix for element number 2 is illustrated in Figure 3.4. 3 3 Shear Deformation As the depth to span ratio for a member increases, the effect of shear deformation becomes more pronounced and important to consider in the analysis. Figure 3*5a illustrates the shear deformation and bending components of the lateral deflection at the free end of a column in response to a lateral concentrated load. Figure 3-5b, taken from Wang (16), shows the ratio of shear deflection as to bending deflection a^, at the midspan of a simple beam with a rectangular cross-section. Notice that, for a depth to span ratio of 0.25, the shear deflection can be up to 18.75$ of the bending deflection. If it is desired to take shear deformation into consideration in the analysis of a structure by the Direct Stiffness Method, this is accomplished by altering the standard terms in the stiffness matrix of the elements for which this effect may be significant. The terms in the stiffness matrix for the standard 6 DOF element, which were derived in 26 [k], - [k] 3 ~ A1E1 0 0 'A1E1 0 0 L L 0 12E11! -6E1I1 0 -12E1I1 -6E1I1 L2 l? L2 0 -6E,I, 4E111 0 6E1I1 2ElIl L2 L L2 L -A1E1 0 0 A1E1 0 0 L L 0 -1 2E111 6E,I, 0 12E1I1 6E,I, L3 L2 L2 0 -6E,I, 2E111 0 6E1I1 4EiIi L2 L L2 L (a) Element Stiffness Matrix For Element 1 A 2^*2 0 0 "A 2^2 0 0 L L 0 12E2I2 -6E2I2 0 -12E2I2 -6E2I2 l? L2 L2 0 6E2I2 4E2I2 0 6E2I2 2E2J2 L2 L L2 L A 2 0 0 2^2 0 0 L L 0 -12E2I2 6E2I2 0 12E2I2 6E2I2 1? L2 L2 0 -6E2I2 2E2^2 0 6E2I2 4E2I2 L2 L L2 L (b) Element Stiffness Matrix For Element 3 Figure 3*2 Element Stiffness Matrices For Elements 1 and 3 in Figure 3.1 27 (a) Element Figure 3.3 Development of Stiffness Matrix For Element 2 of Figure 3.1 28 ks ksÂ£1 _ks ksÂ£2 ksÂ£1 ksÂ£f + km *ksÂ£1 ksÂ£2Â£1 km "ks ~ksÂ£1 ks ksÂ£2 ksÂ£2 ksÂ£1Â£2 ~ km "ksÂ£2 ksÂ£2 + km WHERE: k = SHEAR SPRING STIFFNESS b km = MOMENT SPRING STIFFNESS Â£1 = LENGTH OF LEFT PART OF ELEMENT Â£2 = LENGTH OF RIGHT PART OF ELEMENT Figure 3*4 Element Stiffness Matrix For Element 2 in Figure 31 29 (a) (bending) And (shear) Components of Lateral Deflection DEPTH d SPAN L UNIFORM LOAD CONCENTRATED LOAD AT MIDSPAN 1/12 0.0167 0.0208 1/10 0.0240 0.0300 1/8 0.0375 0.0469 1/6 0.0667 0.0833 1/4 0.1500 0.1875 (b) Ratio of Shear Deflection Ag to Bending Deflection Ab Figure 3.5 Shear Deformation and Its Importance 30 Figure 2.1, neglect shear deformation. The following derivation, taken from Przemieniecki (9), shows how the element stiffness matrix is obtained for the standard 6 DOF element considering shear deformation. Note that displacements for element degrees of freedom 1 and 4 are not considered below because they are not affected by the consideration of shear deformation. Thus, columns 1 and 4 in the element stiffness matrix shown in Figure 2.2 will remain unchanged. Consider Figure 3.6a. The lateral deflection v on the element subjected to the shearing forces and associated moments shown, is given by v vb + vs (3.1) where v^ is the lateral deflection due to bending strains and vg is the additional deflection due to shearing strains, such that dvc -fe s p dx GA b (3.2) with A_ representing the element cross-sectional area effective in s shear, and G representing the shear modulus, where G E 2(1 + v) (3-3) and E equals the modulus of elasticity and v the Poisson's ratio of the material. The bending deflection for the element shown in Figure 5*6a is governed by the differential equation 2 El fJ. = f5x f6 MT dx2 (3.4) Figure 35.6 Application of Displacements to Establish Stiffness Coefficients Considering Shear Deformat 32 where Mt = /aETydA A (3.5) From integration of Equations (3.2) and (3.4), it follows that EIv = 3 2 2 f5x f6x MTX + C, - fr-EI^ 5 1 GA x + C, (3.6) where and C2 are the constants of integration. Using the boundary- conditions in Figure 3.6a, dv _ 5. IL at x dx dx GAS = o, x = 1 and v = 0 at x = 1 (3.7) (3.8) Equation (3*6) becomes 3 2 2 2 3 f r-X^ tcx M_x f c$xl l^fc EIv = -1 Â§ I l + (1 + *) 1 6 2 2 12 v 12 (3.9) where f51 f 6 Hr ht (3.10) j 12EI / .. \ and $ = K- (3.11) GA 1 s It should be noted here that the boundary conditions for the fixed end in the engineering theory of bending when shear deformations vg are included is taken as dv^/dx = 0; that is, slope due to bending deformation is equal to zero. 33 The remaining forces acting on the beam can be determined from the equations of equilibrium; thus, we have f2 = -f5 (5.12) and f3 = -f6 + ^ * Now at x = 0, v = Wcj, and hence from Equation (3*9) i5f5 *2 0 W Using Equations (3*10) and (3-12) to (3*14), we have 1 2EI k55' [%.o k _ f5lN 6,51 H-O 6EI T=0 (1 + *)1 = I _1\ = -12EI 2,5 W5/t=0 (1 + $)13 '~^6 + f5lN k3,5 w 5/t=0 wr 6EI T=0 (1 + *)1 (3.13) (3.14) (3.15) (3.16) (3.17) (3.18) with the remaining coefficients in column 5 equal to zero. The variable T stands for temperature change. 34 Similarly, if the bottom end of the element is fixed, as shown in Figure 36b, then by use of the differential equations for the beam deflections or the condition of symmetry it can be demonstrated that k = k 12EI 2,2 5,5 (1 + *)!3 (3.19) -k -6EI 3,2 6,5 (1 + #)li (3.20) -12EI 5,2 2,5 (1 + #)13 (3.21) k = -k -6EI 6,2 3,5 (1 + #)li (3.22) with the remaining coefficients in column 2 equal to zero. In order to determine the stiffness coefficients associated with the rotations wg and w^, the element is subjected to bending moments and the associated shears, as shown in Figure 3.6c and d. The deflections can be determined from Equation (3*6), but the constants C1 and C2 in these equations must now be evaluated from a different set of boundary conditions. With the boundary conditions (Figure 3.6c) v = 0 at x = 0, x = 1 (3.23) and dv dvs dx dx GAc at x = 1 , (3.24) Equation (3*6) becomes Elv _5(X3 A) lT(lx x2) + Â£6{1i . 6 2 2 (3.25) 35 and f 6y 61% . 5 (4 + $)1 (4 + $)1 (3.26) As before, the remaining forces acting on the element can be determined from the equations of equilibrium, i.e., Equations (3.12) and (3.13)* How at x = 0 dv-L dv b _dv_ s_ _ dx dx dx 6 (3.27) so that w = ffi (1 + *)1 Mrn (1 + *)1 6 El (4 + *) El (4 + i>) (3.28) Hence, from Equations (3.12), (3-13), (3*26), and (3-28) 6,6 \ vf r6 \ = (4 + $ )EI ) (1 + *)1 'T0 (3.29) f-U -6EI 26 \w6/ lw6 ) (1 + $)l2 ' 't=o 't=o (3.30) 3,6 lw T=0 + ^ w6 ) (2 0) El . (1 + *)l [=0 (3.31) If the deflection of the left-hand end of the beam is equal to zero, as shown in Figure 3.6d, it is evident from symmetry that k 3,3 k 6,6 (4 + $)EI (1 + *)1 (3-32) 36 k 5,6 k 2,3 k 5,3 k 6,3 6EI (3.33) 6,5 0 + $)12 -6EI (3.34) 3,2 (1 + $)12 6EI (3.35) 3,5 (1 + $)l2 (2 - *)EI (3.36) 3,6 (1 + $)1 with the remaining coefficients in columns 3 and 6 equal to zero. Thus, the stiffness matrix for the basic 6 DOF element, shown in Figure 2.1a, takes the form shown in Figure 3*7 when shear deformation is considered. 3.4 Moment Magnification With increasing slenderness and height, the lateral deflections of a vertical member due to bending will increase. As these deflections increase, an additional moment is caused by the vertical load acting through these deflections. This additional moment is often referred to as a secondary bending moment. Moment magnification is one term used to describe this effect. Figure 3.8 shows how moment magnification occurs. The technique for including the effect of moment magnification in the Direct Stiffness Method analysis of a structure also involves altering the standard terms in the stiffness matrix of the elements for which this effect is to be considered. The terms in the stiffness matrix for the standard 6 DOF element, shown in Figure 2.2, neglect moment magnification as well as shear deformation, which was discussed in the previous section. To illustrate how the new terms in the element [k] AE L 0 0 -AE L 0 0 0 12EI -6EI 0 -12EI -6EI L3 (1 + $) L2 (1 + $) (1 + $) L2 (1 + *) 0 6EI (4 + *) El 0 6EI (2 $) El L2 (1 + *) L (1 + 4) L2 (1 + $) L (1 + *) -AE 0 0 AE 0 0 L L 0 -12EI 6EI 0 12EI 6EI L3 (1 + $) L2 (1 + #) L3 (1 + *) L2 (1 + *) 0 -6EI (2 *) El 0 6EI (4 + *) El L2 (1 + *) L (1 + 4>) L2 (1 + *) L (1 + *) WHERE: A E L I 4> G v AREA OF ELEMENT CROSS-SECTION MODULUS OF ELASTICITY ELEMENT LENGTH MOMENT OF INERTIA OF ELEMENT CROSS-SECTION FACTOR USED IN ACCOUNTING FOR SHEAR DEFORMATION = 12EI GASL2 SHEAR MODULUS = y ..E. . 2(1+v) POISSON'S RATIO AREA IN SHEAR = .84 x (NET AREA) Figure 3.7 Element Stiffness Matrix Considering Shear Deformation 3S P Figure 3.8 Moment Magnification stiffness matrix which considers moment magnification are developed, the following derivation is presented. It was taken from Chajes (3). Once again, element degrees of freedom 1 and 4 are not considered below because they are not affected by the consideration of moment magnification. Columns 1 and 4 in the element stiffness matrix shown in Figure 2.2 will, therefore, remain intact. Consider an element of a beam column subject to an axial load P and a set of loads [f], as shown in Figure 3*9d. The corresponding element displacements [w] are depicted in Figure 3*9e. It is our purpose to find a matrix relationship between the loads [f] and the deformations [w] in the presence of the axial load P. As long as the deformations are small and the material obeys Hooke's law, the deformations corre sponding to a given set of loads [f] and P are uniquely determined, regardless of the order of application of the loads. The deformations [w] can, therefore, be determined by applying first the entire axial load P and then the loads [f]. Under these circumstances, the relation of [f] to [w] is linear, and the stiffness matrix can be evaluated using the principle of conservation of energy. The element is assumed to be loaded in two stages. During the first stage, only the axial load P is applied and, during the second stage, the element is bent by the [f] forces while P remains constant. Since the element is in equilibrium at the end of stage one as well as at the end of stage two, the external work must be equal to the strain energy not only for the entire loading process but also for stage two by itself. The external work corresponding to the second loading stage is w 4 Wr (a) Original DOF (b) DOF Which Influence Moment Magnification P P (c) DOF Renumbered (d) Element Forces (e) Element for Derivation Deformations Only o Figure 3.9 Element Used in Derivation of Moment Magnification Terms 41 V 1 [w]T [f] I1 (y)2 dx e 2 2 0 (3.37) in which the first term represents the work of the [f] forces and the second term the work due to P. Since the ends of the member approach each other during bending, the axial force does positive work when it is a compression force and negative work if it is a tension force. The strain energy stored in the member during stage two is due only to bending. Thus U-S. I1 (y")2dx. 5-38) 2 0 Equating the strain energy to the external work gives 4 MT [f] +4 i1 (y')2 dx = .M. J1 (y)2 dx (3.39) Making use of the relationship [f] = [k] [w], in which [k] is the element stiffness matrix, Equation (3*39) becomes [w]T [k] [w] = El f1 (y")2 dx P f1 (y*)2 dx (3-40) 0 0 To evaluate [k], it is necessary to put the right-hand side of Equation (340) into matrix form. This can be accomplished if the deflection y is assumed to be given by y = A + Bx + Cx^ + Dx-5 (3-41) The choice of a deflection function is an extremely important step. A cubic is chosen in this instance because such a function satisfies the conditions of constant shear and linearly varying bending moment that 42 exist in the beam element. Taking the coordinate axes in the direction shown in Figure 3-9e, the boundary conditions for the element are and y = -w1 y' = w2 at x = 0 y = -w-j y' = w4 at x = 1 . Substitution of these conditions into Equation (5-41) makes it possible to evaluate the four arbitrary constants and to obtain the following expression for y: 3(w< w,) p 2wp + w4 p y = -w + w x + x - 2- x 1 2 ,2 1 + W4 x3 + 2(w3 ~ w1^ x3 (3.42) Equation (3*42) can be rewritten in matrix form as ',y2 or - 1) (x *Â£ + \(2^ . x? y = [a] [w] . W1 w2 w3 W (3.43) Differentiating the expression in (3.43) gives and y [c] W y" [d] [w] (3.44) (3.45) in which 43 In view of (3*44) and (3*45), one can write (y'>2 MT [c]T [c] [v] and (y")2 []T MT [B] [] . Substitution of these relations into (3*40) gives (3.46) (3.47) (3.48) (3.49) H'MM MT (El I1 [D]T[D]dX P /! [c]T[c]dx [] 0 0 / from which [k] = El I1 [D]T[D]dx P f1 [c]T[c]dx . 0 0 (3.50) Using the expressions given in (3-46) and (3-47) for [c] and [d] and carrying out the operations indicated in (3*50), one obtains [k] = El 12 6 12 6 1? l2 l5 l2 6 4 6 2 l2 1 l2 1 _ 12 6 12 6 l3 l2 l2 6 2 6 4 l2 1 l2 1 6 1 6 1 51 10 51 10 1 21 + 1 1 10 15 10 30 _ 6 + 1 6 1 51 10 51 10 1 1 1 21 10 30 10 15 (3.51) 44 Equation (3*51) gives the stiffness matrix of a beam column element with the 4 DOF shown in Figure 3*9c. The matrix consists of two parts: the first is the conventional stiffness matrix of a flexural element and the second is a matrix representing the effect of axial loading on the bending stiffness. Figure 3*10 shows the stiffness matrix for the 6 DOF element shown in Figure 2.1a and Figure 3*9a considering moment magnification. Figure 3.11 shows the stiffness matrix for the same element' considering both shear deformation and moment magnification. 3.3 Material Nonlinearity Sometimes it is necessary to analyze a structure which is made up of a material whose load-deformation response is nonlinear. In such a material, the modulus of elasticity will vary and will be a function of the level of loading which the material is subjected to. A stiffness analysis of a structure composed of a nonlinear material can be performed if provision is made for the variation in elastic modulus. This can be done as follows. Recall that the modulus of elasticity is one of the variables required to construct the stiffness matrix for each element in the structure. As mentioned above, a nonlinear material's modulus of elasticity depends on the level the material is loaded to. To properly account for nonlinearity, three things are done. 1) Modulus of elasticity values are made dependent on load level. 2) The load is applied to the structure in increments up to the load for which a solution is desired. 3) The modulus of elasticity value used in constructing the stiffness matrix of an element is based on the element forces resulting from the application of the previous load increment. [k]- AE L 0 0 -AE L 0 0 0 12EI 6P -6EI P 0 -12EI 6P -6EI P 5L L2 10 l3 5L L2 10 0 -6EI * P 4EI 2PL 0 6EI P 2EI + PL L2 10 L 15 L2 10 L 30 -AE 0 0 AE 0 0 L L 0 -12EI A 6P 6EI P 0 12EI 6P 6EI P 1? 5L L2 10 l3 5L L2 10 0 -6EI * P 2EI + PL 0 6 El P 4EI 2PL L2 10 L 30 L2 10 L 15 WHERE: A = AREA OF ELEMENT CROSS-SECTION E = MODULUS OF ELASTICITY L = ELEMENT LENGTH I = MOMENT OF INERTIA OF ELEMENT CROSS-SECTION P = ELEMENT AXIAL FORCE Figure 3*10 Element Stiffness Matrix Considering Moment Magnification [k] AE L 0 0 -AE L 0 0 0 1 2EI 6P -6EI + P 0 -12EI + 6P -6EI A P L3 (1 + $) 5L L2 (1 + *) 10 L3 (1 + *) 5L L2 (1 + $) 10 0 -6EI + P (4 + $) El 2PL 0 6EI P (2 $) El + PL L2 (1 + *) 10 L (1 + *) 15 L2 (1 + <*>) 10 L (1 + 30 -AE 0 0 AE 0 0 L L 0 -12EI + 6P 6EI P 0 12EI 6P 6EI P L3 (1 + $) 5L L2 (1 + $) 10 L3 (1 + $) 5L L2 (1 + *) 10 0 -6EI + P (2 *) El + PL 0 6EI P (4 + *) El 2PL L2 (1 + $) 10 L (1 + *) 30 L2 (1 + *) 10 lT+T)~ 15 WHERE: A E L I $ G v As p = AREA OF ELEMENT CROSS-SECTION = MODULUS OF ELASTICITY = ELEMENT LENGTH = MOMENT OF INERTIA OF ELEMENT CROSS-SECTION = FACTOR USED IN ACCOUNTING FOR SHEAR DEFORMATION = -l2!! GAL2 = SHEAR MODULUS = ... E . 2(1+v) = POISSONS RATIO = AREA IN SHEAR = .84 x (NET AREA) = ELEMENT AXIAL FORCE Figure 3.11 Element Stiffness Matrix Considering Shear Deformation and Moment Magnification 47 Consider the nonlinear load-deformation curve in Figure 3*12 which is approximated by three straight line segments. Notice that three modulus of elasticity (E) values exist, and each is valid only over a certain region of load. Assume it is desired to load a structure to a value in load region 3 First of all, the load would be divided into increments. This is necessary since the solution to the application of one load increment will affect the response to the next load increment, and so forth. How, apply the load to the structure in increments. After the application of each load step, go through the entire process of constructing the stiffness matrix for each element, constructing the structure stiffness matrix, solving for structure displacements, and obtaining element forces and displacements. To decide what value of E to use in constructing the stiffness matrix of an element, determine which load region the element forces fall in, based on the solution to the previous load increment. Since, in this fashion, the modulus of elasticity value is indeed related to the load each element experiences, the solution for the analysis of the structure will reflect the true load-deformation properties of the material from which it is made. Two additional points should be considered. First of all, when an element changes from one load region to the other, say from region 1 to region 2 in Figure 3.12, its modulus of elasticity will decrease from a value of E^ to a value of EÂ£. This means the stiffness of this element has decreased. Loads in the structure are resisted by the elements in accordance with their stiffnesses such that stiffer elements resist a larger part of the load and, therefore, develop larger element forces. After the application of the next load increment, since the modulus of elasticity for this element has gone down from E^ to E2 this element Figure 3-12 Nonlinear Load Deformation Curve LOAD REGION 1 o LOAD REGION 2 4- LOAD REGION 3 LOAD 49 will now resist a smaller part of the total load, and a "load redistribution" occurs. If the load increment just applied is small relative to the decrease in element stiffness, this element will develop forces smaller than its previous element forces and go back to region 1. The problem is that now, the stiffness increases from EÂ£ to E-Â¡ and after the next iteration will go back to S2 as before and cycle back and forth. To prevent this, care must be taken to insure that once the stiffness of an element decreases, i.e., the next smaller value of E is used to construct its stiffness matrix, the modulus of elasticity is not permitted to increase due to a drop in the element's load level. This is best accomplished by never selecting values of E for an element which are based on a load level lower than the highest experienced up to that point. The second point also is related to the load redistribution which occurs as stiffnesses of individual elements change. During the incremental loading of a structure, it is often desired to examine the structural response after each load step is applied. This permits an examination of how the structural behavior varies as the external loads on the structure are increased. To insure each intermediate solution is accurate, a provision can be made that if any element experienced a change in stiffness during the application of the previous load increment, a new solution with the current element stiffnesses should be calculated before applying the next load increment. This will provide a steady-state solution for each load level; that is, one in which no load redistribution occurred. 50 Thus, accounting for material nonlinearity in the analysis of a structure by the Direct Stiffness Method involves applying the load in increments and carefully monitoring the selection of modulus of elasticity values used in constructing the stiffness matrix of each element. The remainder of the general procedure outlined in section 2.1 remains the same. 3.6 Equation Solving Techniques Recall from Chapter Two that once the structure stiffness matrix [k] and the structure force matrix [f] have been constructed, solving for the structure displacement matrix [w] entails solving the matrix equation [f] = [k] [w] for [w]. This requires solving N simultaneous equations, where N is the number of structure degrees of freedom. Two efficient techniques for solving simultaneous equations are Gauss Elimination and Static Condensation. Meyer (8) discusses both of these methods and was used as a reference for the following two sections. 5.6.1 Gauss Elimination As previously discussed, the equation [f] = [k] [w] takes the form shown below. K11 K21 KN1 K12 * k1n *1 *r K22 K2N w2 F2 %2 %N % % Gauss Elimination essentially consists of two steps: 51 1) Forward elimination to force zeros in all positions below the diagonal of [k] by performing legal row operations on [k] and [F] 2) Backward substitution to solve for [w]. The following simple example helps illustrate how this is done. Assume it is desired to solve the four simultaneous equations: 2W1 + W2 =10 2Â¥1 + 6W2 6W5 =0 - 6W2 + 9W5 7W4 = -28 - 7W3 + 5 W4 = 0 They can be rewritten in matrix form as 2 10 0 *1 10 2 6-60 w2 0 0-6 9-7 Â¥3 -28 0 0-75 *4 0 __ Forward elimination is performed by 2 10 0 10 (-1 x row 1) + row 2 2 6-60 0 0-6 9-7 -28 0 0-75 0 which yields 52 2 10 0 10 0 5-60 -10 0-6 9-7 -28 0 0-75 0 then 2 1 0 0~ ~10~ 0 5-60 -10 (6/5 x row 2') + row 3 0-6 9-7 -28 0 0-75 0 yields 2 10 0 10 0 5-60 -10 0 0 9/5 _7 -40 I ITv t- 1 o 0 and finally 2 10 0 10 0 5-60 -10 0 0 9/5 -7 -40 x row 3') + row 4 0 0-75 0 produces 53 2 10 0 10 0 5-60 -10 0 0 9/5 -7 -40 000 200/9 J400/g Now from back substitution, ~200/9 W4 = "14/g therefore W4 = 7 , 9/5 W3 7V4 = -40 since W4 is known, W3 can be found directly W3 = 5 Similarly, 5W2 6W3 = -10 yields W2 = 4 and 2W1 + \Â¡2 10 produces = 3 Therefore, matrix [w] has been solved for and found to be 54 [w] v1" 3 w2 4 W3 5 _V4_ 7 Two properties of stiffness matrices can be used to further reduce the computational effort required to solve for the structure displacements. Stiffness matrices are symmetric and frequently will be tightly banded around the diagonal. Due to symmetry, the upper right triangular part of the matrix will be identical to the lower left triangular portion. Being banded around the diagonal means all nonzero coefficients will be concentrated near the diagonal. Since only the nonzero terms need to be considered for Gauss Elimination, and since only half of the nonzero terms need to be stored due to symmetry, tremendous savings in storage and computational efficiency are possible. Figure 3.15 qualitatively shows what an actual stiffness matrix might look like versus the stiffness matrix which is stored and used by a modified Gauss Elimination procedure that requires only half of the symmetric nonzero coefficients. Half the bandwidth (including the diagonal term) is shown as HBW. An idea of the storage savings that result from only storing half of the symmetric nonzero values in the stiffness matrix can be obtained by considering a simple example. For a structure with 194 degrees of freedom, there will be 194 simultaneous equations to solve, and the structure stiffness matrix will be a 194 x 194 matrix consisting of ****** ****** ******* ******** ********* ********** _n_ *********** ^ ****** ****** ****** ****** ****** *********** ****** *********** ****** *********** [K] = ****** ** ********* ****** *********** ****** *********** ****** *********** ****** *********** ****** *********** ****** ********** ********* ***** **** ******** *** ******* ****** l* -0- (a) Actual Stiffness Matrix (b) Stiffness Matrix Stored Figure 3.13 Advantage of Symmetry and Bandedness of Structure Stiffness Matrix 56 57,636 numbers. If half the bandwidth of this matrix is 9, then only 194 x 9, or 1746 numbers need to be stored. This is a savings of 95.4$! Figure 3*14 compares the number of operations required by the regular Gauss-Elimination procedure to the number required by the modified Gauss-Elimination technique for a structure stiffness matrix with a value of 9 for half the bandwidth. Figure 3*15 shows the percent savings in computational effort that result by using the modified Gauss- Elimination method on half of the symmetric nonzero values in the stiffness matrix where half the bandwidth is equal to 9* As shown, up to 95.5$ savings are possible! 5.6.2 Static Condensation Static condensation is another equation solving technique which can be used to solve the matrix equation [f] = [k] [w]. It involves partitioning each of the three matrices and is performed as follows: 1) Partition K W F into Kaa Kab Wa Fa _Kba Kbb Wb Fb 2) Perform r__ forward eliminations on [k] where rn act act rows in [K a3. This will yield = number of NUMBER OF OPERATIONS 57 Figure 3*14 Comparison of the Equation Solving Operations of Standard Gauss Elimination Versus Modified Gauss Elimination PERCENT SAVINGS 58 Figure 3.15 Computational Savings of Modified Gauss Elimination Over Standard Gauss Elimination 59 KÂ¡a KÂ¡b Ka Kb W f: a = a Wb n Note that [Ka] [WQ] + [Kb] [wb] = [F] but because of the forward eliminations [Kb ] = 0 so this equation reduces to [Rbb] W = ^ (3-52) 3) Solve [K{J [Wb] = [F] for [wfc] . 4) Note that [Ka] [wj + [Kb] [wb] [F] and only [Wfl] is unknown, so solve ka) W G>] Kl>] KB <5-53) for [W ] by backward substitution. 5) Construct j^j _ In performing Static Condensation, best efficiency is obtained if: D [K] is partitioned into four equal quarters and [w] and [f], therefore, are each divided in half, and 2) Gauss Elimination is used to solve Equation (3.52). The numerical example of section 3*6 is solved below using Static Condensation. Recall that [k] [w] = [f] was written in matrix form as 60 2 10 0 Â¥1~ 10 2 6-60 W2 0 0-6 9-7 V3 -28 0 0-75 A 0 After partitioning, 2 1 0 0 Â¥1 10 2 6 -6 0 w2 0 C5 0 -6 9 -7 w3 -28 0 0 -7 5 Â¥4 0 _ Performing 2 forward eliminations on [k] entails 2 1 0 0 Â¥1 10 (-1 x row 1) + row 2 2 6 -6 0 w2 0 (6/5 x row 2') + row > 0 -6 9 -7 W3 -28 0 0 -7 5 w4 0 and yields 2 1 0 0 2 5 -6 0 0 0 9/5 -7 0 0 -7 5 Â¥1 10 Â¥2 -10 Â¥3 -40 Â¥4 0 61 Note that [Ka] = O Solve [Kb] [wb] = [F] for [wb] using Gauss Elimination. 9/5 -7 -40 (^/9 x row 3') + row 4 -7 5 0 gives 9/5 -7 W3 40 0 -200/9 _W4_ -1400/9 From backward substitution W4 = 7 and = 5 therefore Ob] 5 7 Multiplication of K] gives 5 7 0 -6 0 -30 62 Then [P] [Kb] [wb] produces 10~ 0~ ~io" - c: -10 30 20 Finally, [KJ [vj ([f] [KÂ¡b] [wj) can be solved for [WQ1 by back substitution. 2 1 Â¥1 10 0 5 _V2_ 20 W2 4 W1 3 . Therefore, [wal 3 4 and 3 [W] = 4 5 7 63 Thus, Static Condensation uses Gauss Elimination as part of its procedure for solving simultaneous equations. Section 3*6.1 addressed the large storage and computational savings that result from using a modified Gauss Elimination technique which only uses half of the symmetric nonzero terms in the stiffness matrix. The use of Static Condensation in conjunction with the modified Gauss Elimination procedure was explored. This technique was found to be less efficient than just modified Gauss Elimination for small matrices, but for large matrices it was up to 11.5$ more efficient than even modified Gauss Elimination. Figure 3*16 compares the number of operations required by modified Static Condensation to the number required by modified Gauss Elimination for a structure stiffness matrix with a value of 9 for half the bandwidth. The percent savings (or loss) in computational efficiency that results from the use of modified Static Condensation is shown in Figure 3.17. Since both of these methods each require storage of half of the symmetric nonzero values in the stiffness matrix, the storage requirements of each technique are the same. 3.7 Solution Convergence In the standard use of the Direct Stiffness Method, convergence of the solution rarely presents a problem. However, with consideration of the special items discussed in sections 3*1 through 3.5 immediate convergence of the solution is not guaranteed. One method of monitoring the accuracy of the solution is to add a step to the procedure detailed at the end of section 2.1. After the force matrix for each element is calculated, an equilibrium check can be made by multiplying every element force matrix by -1.0 and inserting each one into the structure force matrix. The resulting values should NUMBER OF OPERATIONS 64 Figure 3.16 Comparison of Equation Solving Operations of Modified Gauss Elimination Versus Modified Static Condensation PERCENT PERCENT SAVINGS LOSS 65 Figure 5.17 Computational Savings of Modified Static Condensation Over Modified Gauss Elimination 66 be very small, and the closer they are to zero, the better the solution. Thus, one can speak of the degree to which the solution converged. For example, if m is the number of elements and the previous operation, shown below, is carried out m [error] = [f] + Z (-1.0 x [f]i) i=1 and the largest value in the matrix [ERROR] is 1 x 10^, this solution converged better than one which produced a value in [ERROR] equal to 9 x 10~1. To insure convergence of the solution to the matrix equation [f] = [k] [v] within a specified tolerance, a reasonable tolerance value of, say 0.001, should be selected and all values in matrix [ERROR] compared to it. If no value exceeds the tolerance, the solution is acceptable. If any value exceeds the tolerance, then all the values can be treated as an incremental force matrix [aF] and then used to solve for the incremental structure displacement matrix [aw] for the previous structure stiffness matrix [k]. In other words, set [af] = [error] and solve [af] = [k] [aw] (3.54) for [aw] . The total structure displacements then become equal to [w] + [aW], the total displacements for each element become [w] + [aw], and the total 67 forces for each element [f] + [Af]. Once again, each element force matrix should be multiplied by -1.0, inserted into the original structure force matrix, and the values compared with the allowable tolerance once more. This procedure, therefore, monitors convergence and also promotes it. Naturally, cases may arise where the tolerance is set below the value that a satisfactory solution can be arrived at, even with the use of the incremental structure force matrix concept, so the number of cycles in which an attempt is made to converge on a solution should be limited to some fairly small value, such as 9. CHAPTER FOUR USING THE MODEL TO ANALYZE COMPOSITE MASONRY WALLS 4.1 Structural Idealization of Wall The analytical model considers the lowest story of a composite wall, the same portion that will be represented by the laboratory test walls. The wall is divided into a series of three types of elements, one for the brick, one for the block, and one for the collar joint. Each element extends the entire depth of the wall, which for the test walls will be 24 inches. Figure 4.1 shows a wall divided into these elements. As shown, the lowest nodes of both wythes are fixed to the foundation and the top nodes are restrained from lateral displacement as they will be in the laboratory test fixture. Thus, the wall is modeled as a frame with two lines of columns connected at 8 inch intervals by shear beams with rigid ends. Figure 4.2 shows the structure and element degrees of freedom used in the model. The manner of numbering the structure degrees of freedom follows the pattern shown regardless of wall height. Of course, the element degrees of freedom for each element of a given type are as shown. For the highest test wall, which will be 26 feet in height, there are 194 structure degrees of freedom and 117 elements, 59 of each type. 68 RIGID WYTHE COLLAR JOINT ELEMENT CONCRETE BLOCK ELEMENT Figure 4.1 Finite Element Model of Wall 7C BRICK ELEMENT W/ W 16 W 17 K w w, 18 ^:%rtr-Wl3 W14 W15 wfi i W, oj a M /777 /V77 \ 19 12 W- WÂ£ 10 r w Wo w, V-r Wo * d w6 COLLAR JOINT ELEMENT Cf Wr U- W, W- w. w. w 5 BLOCK ELEMENT Figure 4.2 Structure and Element Degrees of Freedom 71 4.2 Types of Elements As previously noted, the model considers the wall divided into a series of three types of elements. These include a brick element, a collar joint element, and a block element. 4.2.1 Brick Element The basic finite element for the brick wythe is shown in Figure 4. 3 It models a brick prism either two or three bricks high, depending on brick type, and the mortar joints adjacent to those bricks. The prism and element are each 8 inches tall. The element extends the entire 24 inch depth of the test wall, with uniformly distributed forces over the wall depth. Experimental tests will be done on brick prisms, 24 inches deep, in order to determine the load-deformation properties of the brick element to be used by the model. Like the brick in the wall, the brick in the prisms will contain running bond. The brick element will have the six degrees of freedom shown, which include a rotation, axial displacement, and lateral displacement at each end. This is the basic flexural element previously illustrated and discussed in section 2.2. The effects of shear deformation and moment magnification will be considered in the brick wythe of the model in accordance with the pro cedures outlined in sections 3.3 and 3.4. In other words, the stiffness matrix for a brick element will take the form shown in Figure 311 4.2.2 Collar Joint Element The basic finite element for the collar joint is shown in Figure 4.4. The element will have the four degrees of freedom shown which include a rotation and vertical displacement at each end. The element is rigid axially which forces the brick and block wythes to have the 72 MORTAR JOINT w, 4" BRICK PRISM WITH 8" 4" BRICK PRISM WITH 3 5/8"x 7 5/8" x 2 1/4" 3 5/8" x 7 5/8"x 3 1/2* BRICK BRICK (1- Wr Wc 3 w Figure 4.3 Finite Element For Brick 73 *, = ONE-HALF THE THICKNESS OF 1 THE BRICK WYTHE (= 2") = ONE-HALF THE THICKNESS OF THE BLOCK WYTHE (= 2", 3" or 4") ks = SHEAR SPRING STIFFNESS FACTOR k = MOMENT SPRING STIFFNESS FACTOR m Figure 4.4 Finite Element For Collar Joint 74 same horizontal displacement at each node. Rigid links at each end represent one half of the width of each wythe. The two springs are situated at the wythe to wythe interface. The primary function of the collar joint in a composite wall is to serve as a continuous connection between the two wythes, transferring the shear stress between them. This transfer of shear is modeled by the vertical spring, called the shear spring. At this stage, it is uncertain how much moment the collar joint transfers between the wythes. The rotational spring, called the moment spring, has been included in the collar joint element so the effect of moment transfer can be included in the model if, at a later date, this proves necessary. The stiffness matrix for the collar joint element was developed previously in section 3*2 (see Figure 3*3) and takes the form shown in Figure 3-4. The values of the shear spring stiffness factor (kg) and the moment spring stiffness factor (km) will be obtained from experimental tests. 4.2.3 Concrete Block Element The basic finite element for the concrete block is shown in Figure 4.5. It models a block prism which consists of one block and a mortar joint. The prism and element are each 8 inches tall, the same height as the brick element. The block element also extends the entire 24 inch depth of the test wall, with uniformly distributed forces over the wall depth. Experimental tests will be done on block prisms, 24 inches deep, to determine the load-deformation properties of the block element to be used by the model. The block element will have the six degrees of freedom shown, which include a rotation, axial displacement, and lateral displacement at each MORTAR JOINTS BLOCK PRISM BLOCK PRISM WITH BLOCK PRISM WITH WITH 5 5/8" x 15 5/8" x 7 5/8" 7 5/8" x 15 5/8" X 7 3 5/8" x 155/8" BLOCK BLOCK x 7 5/8" BLOCK 4" 6" H r Figure 4.5 Finite Element For Block 76 end. It is identical to the brick element which is the same as the basic flexural element previously illustrated and discussed in section 2.2. The effects of shear deformation and moment magnification will also be considered in the block wythe of the model according to the procedures outlined in sections 3.3 and 3*4. This means that the stiffness matrix for a block element, like the stiffness matrix for a brick element, will also take the form shown in Figure 3.11* 4.3 Experimental Determination of Material Properties As mentioned earlier, experimental tests will be done to establish the material properties of each type of element. 4.3.1 Brick Brick prisms, like those shown previously in Figure 4.3, will be tested experimentally to obtain the values of some of the variables that appear in the terms of the brick element stiffness matrix. Two types of tests will be performed to determine strength and deformation properties for the brick element. In the first type of test, prisms will be loaded axially to failure and the axial deformation noted for each level of load. This will establish the relationship between axial load and axial deformation. The axial load will then be plotted against the axial deformation to obtain a curve. This curve will be approximated in a piecewise-linear fashion, i.e., divided into a series of straight line segments. The slope of each straight line segment will be equivalent to the axial stiffness factor AE/L for the region of load values established by the load coordinates of the end points of each line segment. This is shown in Figure 4.6. I ^77777777777 3> = i AE_ L BASIC RELATIONSHIP k. = AXIAL STIFFNESS FACTOR = ^ a L Figure 4.6 Experimental Determination of Brick Element Axial Stiffness Factor 78 The second type of test will involve loading the prisms eccentrically and measuring the end rotation due to the applied end moment for various eccentricities and levels of loading. As before, the moment will be plotted against the rotation and the resulting curves approximated by straight line segments. The slope of each segment, this time, will equal the rotational stiffness factor 3EI/L for the region of moment values established by the moment coordinates of each line segment for the corresponding axial load. This is shown in Figure 4.7. The prism sketches in Figures 4.6 and 4.7 are intended only to give a general idea of how these tests will be done, and are not meant to be detailed representations of the test set-up and instrumentation required to measure deflections and rotations. The remainder of the variables needed to construct the element stiffness matrix for the brick element can be obtained without further tests. Table 4.1 lists all the variables needed and their sources. As indicated, the modulus of elasticity value for the brick was taken from Tabatabai (15) and the brick Poissons ratio from Grimm (5). Figure 4.8 shows how the element stiffness matrix for a brick element is calculated using the stiffness factors from the prism tests. Since the prisms will be loaded to failure in each type of test, the maximum axial compressive load capacity as well as the relationship between axial load and moment carrying capacity will be established. This will enable an axial load versus moment interaction diagram to be drawn for the brick element. Figure 4.9 shows the general form this diagram will take. The curve in this diagram will also be approximated by straight line segments. It will be used by the model to determine when a brick element has failed. Brick prism failure will be defined as 79 P = O rTTTm^mhm kr = ROTATIONAL STIFFNESS FACTOR 3EI L Figure 4.7 Experimental Determination of Brick Element Rotational Stiffness Factor Table 4.1 Variables Used in Constructing Brick Element Stiffness Matrix VARIABLE DEFINITION SOURCE 4 FACTOR USED IN ACCOUNTING FOR SHEAR DEFORMATION CALCULATED, $ = 1 2EI GA L2 s E BRICK MODULUS OF ELASTICITY TABATABAI (l5), E = 2,918x106 PSI (ONLY USED FOR SHEAR DEFORMATION) I MOMENT OF INERTIA OF BRICK ELEMENT CROSS-SECTION NOT USED DIRECTLY; 3EI/L FACTOR FROM TESTS USED G BRICK SHEAR MODULUS CALCULATED, G = .E, 2(1+v) V BRICK POISSON'S RATIO GRIM (5), v = 0.15 As BRICK AREA IN SHEAR CALCULATED, As = .84 Anet A = 17-19 IN2 b Anet BRICK NET AREA MEASURED, Anet = 20.47 IN2 L BRICK ELEMENT LENGTH LENGTH DEFINED BY MODEL = 8 IN A AREA OF BRICK ELEMENT CROSS-SECTION NOT USED DIRECTLY; AE/L FACTOR FROM TESTS USED P BRICK ELEMENT AXIAL FORCE ERICK ELEMENT FORCE MATRIX DEGREE OF FREEDOM NUMBER 1; CALCULATED BY PROGRAM FOR EACH LOAD LEVEL VALUES [k] = Figure KNOWN: #t E, L, P, AE, 3EI L L LET ka = AE, kr = 3EI L L ka 0 0 -ka 0 0 0 4 x kr 6P -2 x kr A P 0 -4 x kr + 6P -2 x kr + P L2 (1 + $) 5L L( 1 + #) 10 L2 (1 + *) 5L L(1 + ) 10 0 -2 x kr P (4 + *) x kr 2PL 0 -2 x kr P (2 *) x kr + PL L(1 + 10 3(1 + *) 15 l(i + *) 10 3(1 + *) 30 -ka 0 0 ka 0 0 0 -4 x kr A 6P 2 x kr P 0 4 x kr 6P 2 x kr A P L2 (1 + *) 5L L(1 + $) 10 L2 (1 + *) 5L L(1 + *) 10 0 -2 x kr + P (2 $) x kr A PL 0 2 x kr P (4 + ) x kr 2PL L(1 + $) 10 3(1 + $) 30 L(1 + $) 10 3(1 + '*) 15 4.8 Brick Element Stiffness Matrix Formulation Using Experimentally Determined Axial and Rotational Stiffness Factors 82 M Figure 4.9 Typical P Versus K Interaction Diagram For Brick Element 83 the point in the loading process at which the brick prism is unable to support additional load. 4.3.2 Collar Joint To determine the shear spring stiffness factor (kg) necessary for constructing the collar joint element stiffness matrix, tests will be performed to establish the relationship between vertical load and vertical deformation for the collar joint when loaded in shear. The vertical load will be plotted against the vertical deformation to obtain a curve. This curve will be approximated in a piecewise-linear fashion (divided into a series of straight line segments). The slope of each straight line segment will be equivalent to the shear spring stiffness factor k for the region of load values established by the load coordinates of the end points of each line segment. This is shown in Figure 4.10. In these tests, the collar joint will be loaded to failure (inability to carry additional load) so the maximum shear load capacity of the collar joint will be determined. If the effect of moment transfer by the collar joint is found to merit consideration, a test will be devised to determine the relationship between moment and rotation for the collar joint. The moment will be plotted against the rotation and the resulting curve approximated by straight line segments. The slope of each segment will equal the moment spring stiffness factor (km) for the region of moment values established by the moment coordinates of each line segment. Figure 4.11 shows what this curve might look like. 4.33 Concrete Block Experimental tests will be performed on the block prisms identical to those done on the brick prisms, to obtain the axial and rotational 84 P = O Al P > O k = SHEAR SPRING STIFFNESS FACTOR s Figure 4.10 Experimental Determination of Collar Joint Shear Spring Stiffness Factor 85 k = MOMENT SPRING STIFFNESS FACTOR m Figure 4.11 Determination of Collar Joint Moment Spring Stiffness Factor 86 stiffness factors, as well as the maximum axial compressive load capacity and the axial load versus moment interaction diagram for the block element. Table 4.2 shows all the variables needed to construct the block element stiffness matrix which will be identical to the brick element stiffness matrix shown in Figure 4.8. The block element axial load versus moment interaction diagram should have the same general shape shown in Figure 4.9. It will be approximated by straight line segments and used by the model to determine when a block element has failed. Block prism failure will be defined as the point in the loading process at which the block prism is unable to support additional load. 4.4 Load Application This model is an analytical representation of the test walls that will be built and tested in the experimental phase of the project. In the tests, vertical compressive loads will be transmitted to the top of the wall through a rigid steel plate. When the load is applied eccentrically, it will also cause a moment to be applied to the top of the wall. The external axial load and moment will be applied through the vertical force and moment degrees of freedom at the top of each wythe. The magnitude of these, however, will depend on the relative stiffness of each wythe as well as on the relative magnitude between the axial and rotational stiffness for a wythe. The stiffnesses though, are a function of the level of loading. In short, the problem is that two loads (a force and a moment) are applied to the test plate, but there are four degrees of freedom (a force and a moment for each wythe) at the top of the wall through which these loads can be applied. Since properly converting the two loads into four loads requires consideration Table 4.2 Variables Used in Constructing Block Element Stiffness Matrix VARIABLE DEFINITION SOURCE $ FACTOR USED IN ACCOUNTING FOR SHEAR DEFORMATION CALCULATED, $ = 12EI GA L2 s E BLOCK MODULUS OF ELASTICITY TABATABAI (15), E = 2.023x106 PSI FOR 4" BLOCK, E 1.807x10 PSI FOR 6" BLOCK, E = 1.622x10^ PSI FOR 8" BLOCK (ONLY USED FOR SHEAR DEFORMATION) I MOMENT OF INERTIA OF BLOCK ELEMENT CROSS-SECTION NOT USED DIRECTLY; 3EI/L FACTOR FROM TESTS USED G BLOCK SHEAR MODULUS CALCULATED, G = 2(l+v) V BLOCK POISSON'S RATIO VALUE FOR CONCRETE, v = 0.15 As BLOCK AREA IN SHEAR CALCULATED, As = .84 Anet A = 14.85 IN2 FOR 4 BLOCK, A = 27.87 IN2 FOR 6" BLOCK, AND A_ = 39-74 IN2 FOR 8" BLOCK 9 Anet BLOCK NET AREA MEASURED CONSIDERING ONLY WEBS OF BLOCK, Ane. = 17.69 IN2 FOR 4" BLOCK, Anet = 33.18 IN2 FOR 6" BLOCK AND Anet = 47,30 IIj2 F0R 8" BL0CK L BLOCK ELEMENT LENGTH LENGTH DEFINED BY MODEL = 8 IN A AREA OF BLOCK ELEMENT CROSS-SECTION NOT USED DIRECTLY; AE/L FACTOR FROM TESTS USED P BLOCK ELEMENT AXIAL FORCE BLOCK ELEMENT FORCE MATRIX DEGREE OF FREEDOM NUMBER 1; CALCULATED BY PROGRAM FOR EACH LOAD LEVEL 88 of the stiffnesses, which depend on the load values, an unusual challenge exists. This problem is solved by temporarily reducing the four degrees of freedom at the top of the wall to the two which correspond to the test plate for load application purposes. By taking advantage of the force and displacement relationships between the two plate degrees of freedom and the four nodal degrees of freedom at the wall top, the model will solve for the equivalent four nodal forces that the plate degrees of freedom produce, as well as the structure displacements and element displacements and forces considering all four degrees of freedom at the top of the wall. The technique by which this is done is shown below. Figure 4.12a shows the four nodal degrees of freedom at the top of a wall. The wall wythe axial loads and moments which transmit the vertical load and moment from the test plate to the wall wythes are shown in Figure 4.12b. A freebody diagram of the test plate is illustrated in Figure 4.12c. From equilibrium (ZF=0 and ZM=0), PV = P16 + P17 Â¥ = M18 + Miq + P17 ~ P16 (1) . 2 These force relationships between the two plate degrees of freedom and the four nodal degrees of freedom at the wall top can be expressed in matrix form as 89 W16 W17 W W 19 H -Nr-1 (a) Nodal Degrees of Freedom at Wall Top 16 17 "-0 H" V- (b) Wall Wythe Axial Loads and Moments at Wall Top Figure 4.12 Structure Force Application Through Test Plate 90 P16+P17 P17"P16 H- z + P16+P17 P17-P16 M18 M 19 (c) Freebody Diagram of Test Plate A 16 A 17 018 19 (d) Wall Wythe Axial Displacements and Rotations at Wall Top Figure 4.12-continued. 91 r 7^ CD -o a J U a 1 % a16 a17 (e) Test Plate Displacements (f) Test Plate Displacement Geometry Figure 412-continued. 92 110 0 ~P16 pv 1 + 1 1 1 P17 2 2 V 7 M18 M [a] 19 (4.1) From this, a matrix [x] can be constructed such that [Fne] [X] [Fold] (4.2) where [F ] = new force matrix which only considers 2 plate DOF at wall top [x] = transformation matrix used to transform the old force matrix with 4 DOF at the wall top, to the new force matrix with 2 DOF at the wall top [FQid] = original force matrix which considers all 4 DOF at the wall top. The matrix [x] is shown below where this matrix equation is illustrated. F I -0- F (z x 1 ) (z X z) (z x 4) (z x 1) pv 0 a P16 M (2 x z) (2 x 4) P17 V. J 18 IM 19 (4.3) where 93 z = total number of structure DOF 4 [f] = force matrix excluding the 4 DOF at the top of the wall [i] = identity matrix (1's along the diagonal, 0's elsewhere) [a] = transformation part of [x] matrix, shown in Equation (4.1) Figure 4.12d shows the wall wythe axial displacements and rotations corresponding to the wall wythe axial loads and moments at the wall top shown in Figure 4.12b. The test plate displacements are shown in Figure 4.12e. From compatibility, the test plate displacement and rotation can be expressed in terms of the wall wythe displacements and rotations as 9 p1 18 ei9 A P1 A16 + A17 2 The test plate displacement geometry is illustrated in Figure 4.12f. Consider the centerline of the plate moving from position a-a to position a'-a' as shown. The wall wythe axial displacements and rotations can be expressed in terms of the plate displacement and rotation as A 16 A 17 + 0 2 pi 918 = 9p1 919 = 9p1 * These displacement relationships between the four nodal degrees of freedom at the wall top and the two plate degrees of freedom can be 94 expressed in matrix form as A16 1 1 1 2 Ap1 a17 1 - 2 0 1 9P1 919 0 1 ^ [<.T] (4.4) From this, a matrix [x^] can be constructed such that [old] [*T] [neJ (4.5) where [W0iJ = original displacement matrix which considers all 4 DOF at the wall top [xT] = transformation matrix used to transform the new displacement matrix with 2 DOF at the wall top, to the old displacement matrix with 4 DOF at the wall top [Wnew] = new displacement matrix which only considers 2 plate DOF at wall top. The matrix [x^] is shown below where this matrix equation is illustrated. 95 Â¥ (z x 1) I (z X z) -0- (z x 2) W (z x 1) a16 *P1 a17 -0- aT - Sp' 918 ei9 (4xz) (4x2) v J 0T] where (4.6) z = total number of structure DOF 4 [w] = displacement matrix excluding the 4 DOF at the top of the wall [i] = identity matrix (1's along the diagonal, 0's elsewhere) [a^] = transformation part of [x^] matrix; shown in Equation (4.4). Note that [o^] is the transpose of [a] and [x^] is the transpose of [x]. From Equation (2.4) recall that [>] [K] [V] where [f] = structure force matrix [k] = structure stiffness matrix [W] = structure displacement matrix. By considering all original DOF (all 4 DOF at the wall top), this equation can be expressed as [W Kid] Kid] (4.7) 96 By substituting Equation (4.7) into Equation (4.2), one obtains ^Fnew^ ~ W fKold^ fWold^ * (4.8) Then by substituting Equation (4.5) into Equation (4.8), (4.9) the "new" structure stiffness matrix considering only the two plate DOF at the wall top can be constructed from the original or "old" structure stiffness matrix which considers all 4 DOF at the top of the wall. The procedure then, to solve for the original forces and displacements in the model while only considering the plate degrees of freedom at the wall top for load application purposes, is as follows: 1) Construct the original structure stiffness matrix [Kq^] as described in section 2.4. 2) Obtain the new structure stiffness matrix by [^new] = [l] [*T] 3) Construct the new structure force matrix [Fnew] considering only the axial force and moment delivered to the test plate at the top of the wall (plus other forces elsewhere, if any). 4) Solve Equation (4.9) for the new structure displacement matrix [w,! L newJ 5) Calculate the original structure displacement matrix from Equation (4.5). 6) Calculate the original structure force matrix [F0i[] from Equation (4-7). 7) Solve for element displacements and forces as usual (described in sections 2.7 and 2.8). Efficient handling of the structure stiffness matrix by taking advantage of its symmetry and banding as discussed previously in section 3.6 97 presents no difficulty in executing the previous procedure. In fact, the nature of the transformation matrices [xj and [x^] are such that all multiplications in which they are involved can be performed very efficiently. The model takes advantage of this additional opportunity to keep numerical computations to a minimum. 4.9 Solution Procedure The solution procedure used by this finite element model is basically the general procedure for the Direct Stiffness Method outlined at the end of section 2.1, modified to include all of the special considerations discussed in chapters 3 and 4 in the fashion indicated there. The presence of different materials is handled by having three types of elements as discussed in section 4.2 and using the material properties of each element type in constructing the respective element stiffness matrices. Shear deformation in the brick and block wythes is accounted for by special terms in the stiffness matrices of the brick and block elements. Similarly, moment magnification is considered in the brick and block wythes by special terms in the stiffness matrices of the brick and block elements. Material nonlinearity is accounted for by applying the loads on the structure in increments, obtaining a solution after each load step is applied, and using the information obtained to determine the material properties from the nonlinear curves for the next load step. Static Condensation with Gauss Elimination is used for solving the structure matrix equation while taking advantage of the symmetry and bandedness of the structure stiffness matrix. Solution convergence is monitored and promoted by the method indicated in section 3.6. The application of loads on the wall through the test plate at the 98 wall top is analytically handled in the fashion described in section 4.4. Agreement between experimental wall tests and the results obtained from the model is aided by the experimental determination of load- deformation properties for each element type. Element failure for the brick and block elements is considered to occur when an element exceeds the allowable combination of axial load and moment defined by the axial load versus moment interaction diagram for that type of element. The axial forces are always equal and opposite for the brick and block elements, but the end moments may be different. In checking for failure, the maximum end moment is used. Collar joint element failure occurs when the vertical collar joint element force exceeds the maximum collar joint shear load capacity. Wall failure in the model takes place when any element fails according to the aforementioned criteria. In physical prism and wall tests, failure is defined as the inability of the specimen to resist further load. An algorithm for the program is shown in Figure 4.13. All major operations, as well as special features, are discussed in detail in chapters 2, 3, and 4. The program uses 36 modules or subroutines, where each subroutine performs a unique operation. This has the advantages of: * Improving the understanding of the logic " Making program modifications easier * Preventing the program from becoming overwhelming by dividing it into manageable sections. The program is discussed in more detail in Appendix A and each subroutine is discussed in Appendix B. The manner in which these 36 subroutines are used is dealt with in these two locations. Appendix C 99 (start) READ WALL GEOMETRY, STRENGTH AND DEFORMATION PROPERTIES OF EACH ELEMENT, AND STRUCTURE LOAD APPLICATION INFORMATION CALCULATE ELEMENT INDEX MATRIX [i] CONSTRUCT INITIAL ELEMENT STIFFNESS MATRIX [k] RECALL ELEMENT INDEX MATRIX [i] EXTRACT ELEMENT DISPLACEMENT MATRIX [w] FROM STRUCTURE DISPLACEMENT MATRIX [w] RECALL ELEMENT STIFFNESS MATRIX [k] ' 1 ' MULTIPLY ELEMENT STIFFNESS MATRIX [k] BY ELEMENT DISPLACEMENT MATRIX [w] TO OBTAIN ELEMENT FORCE MATRIX [f] OR [k][w] = [f] I CHECK FOR ELEMENT FAILURE (IF ELEMENT FAILS STOP AFTER CHECKING ALL ELEMENTS) i ~ CONSTRUCT NEW ELEMENT STIFFNESS MATRIX [k] BASED ON ELEMENT LOAD LEVEL AND LOAD DEFORMATION PROPERTIES 1 ' ' INSERT -1.0 TIMES ELEMENT FORCE MATRIX [f] INTO MATRIX [ERROR] THAT MONITORS CONVERGENCE Figure 4.13 Program Algorithm 100 Figure 4.15-continued 101 provides detailed information on the use of the program, such as how the data input is prepared. The data input files for the numerical examples presented in Chapter Five are listed in Appendix D. All program input and output is in basic units of inches, pounds, or radians. CHAPTER FIVE NUMERICAL EXAMPLES 5.1 General Comments In this chapter, five numerical examples are presented to illustrate the use of the analytical model and show how the information it generates may be used. Since the results of the experimental phase of this project are not yet available, strength and deformation properties very similar to what might be expected from actual tests of each element type were extracted from similar tests done by Fattal and Cattaneo (4) and Williams and Geschwinder (18). This is dealt with in the next section. Once the experimental phase is complete, and the results of the actual wall tests are compared with those from the analytical model, it may be desirable to perform an extensive parametric analysis by systematically ranging the values of some variables, while keeping the others constant. This should give a good indication of the relative sensitivity of the variables that affect composite masonry wall behavior and isolate which factors are the most important. It would also provide much needed insight into composite masonry wall behavior, and would likely be a source for further experimental research. The current model is based on sound principles and has been developed to consider what are believed to be the most important effects that take place in composite walls. Nonetheless, until actual data are available, it would not be prudent to perform a parameter analysis, since the results might be 102 1C3 tainted due to the use of only approximate test data, and at best would always be questionable. Example number 1 is an analysis of a wall that was actually tested by Fattal and Cattaneo (4). Section 5.3 compares the failure load of the actual wall with the failure load predicted from this finite element model. The remaining four examples consider walls which vary in slenderness, or H/T, ratio and eccentricity of load application. Thus, two sources of evidence are presented to confirm the capability of the analytical model to generate reasonable results. The first is the comparison of analytical results with actual test results. The second is the presentation of analytical results for walls with different slenderness ratios and eccentricity of load application. General trends in the behavior of walls due to these two effects are known and will be used to gauge the plausibility of the results generated. 5.2 Material Property Data As previously mentioned in section 4.3, prism tests will be performed to estabish the load-deformation properties of the brick, collar joint, and concrete block elements. The information obtained from these tests will be expressed in the form of P-A curves, M-9 curves, and P-M interaction diagrams. 5.2.1 P-A Curves The P-A curve for the brick element is generated in the manner described in section 4.3*1 The P-A curve for the brick element used in the numerical examples was obtained from tests very similar to those described in section 4.3*1 which were performed by Fattal and Cattaneo (4) on brick prisms. They tested three 4x32x16 in brick prisms by 104 loading them axially and noting the vertical strain variation with load. The results of their tests are shown in Figure 5.1* In order to use their results, it is first necessary to convert their data, which is for 4x32x16 in prisms, to equivalent data for 4x24x8 in prisms. This is done as follows. The actual prism height is 15.7 in. The desired prism height is 8.0 in. Let = P on actual prism Pp s P on desired prism p Aqa gross area of actual prism = 4x32 = 128 in O Aqp = gross area of desired prism = 4x24 = 96 in . The vertical stress c is given by a AGA agd therefore (5.1) The vertical displacement A for 15.7 in high prisms is not given directly, instead the vertical strain e is given. Recall that e is given by therefore A = e L (5.2) 105 Relationship between vertical compressive load and vertical strain for 4 X 32 x 16-in brick prisms at e= 0. Figure 5*1 Experimental Source of Brick Element P-A Curve (4) 106 where L is the prism height, which for the desired prism is 8 in. By- applying Equations (5.1) and (5.2) on several points in Figure 5.1, a new P-A curve can be generated which will be representative of the results expected for a 4x24x8 in prism. Figure 5.2 illustrates the brick element P-A curve which was obtained in this fashion. The P-A curve for the collar joint element is generated in the manner discussed in section 4*3.2. Tests performed by Williams and Geschwinder (18) were used as the basis for the collar joint P-A curve used in the examples. Figure 5.3a shows an isometric of their test assembly while the geometry of the test assembly is depicted in Figure 5.3b. They tested two prisms of this size by vertically loading the concrete masonry wythe while monitoring the variation of vertical displacement with load. The results of their tests are shown in Figure 5.4. To use their results, it is necessary to convert their data, based on an area in shear As of 15 ^/8 x 15 ^/8 or 488 in^ to equivalent data p for an area in shear of 24x8 or 192 in This is simplified by Williams and Geschwinders observation that the first few points in Figure 5.4 should not be used, since they reflect the seating effect of the assembly on the test plate. This results in a linear P versus A relation. Since the collar joint shear stiffness is equal to the slope of the P-A curve, provided the slope of the line in Figure 5.4 is used, the stiffness will be indicative of that obtained from their tests. Their data is converted as follows. Recall that shear stress t is given by T 107 400,000, 300,000 CQ . 200,000- Q. 100,000- JA t n mum T"'1i1T .005 Data Converted to Tests of fmbr = 5039 si 3 Prisms4x24x8 in Brick 4x8x2% in Mortar Type S Mil .010 .015 A, IN .020 .025 Figure 5.2 Erick Element P-4 Curve 108 (a) Isometric of Typical Assembly. 17 5/8" 15 5/8" L I a. Side View 5/8" 5 5/8" ? 5/S / ' 3/8^ (oj 00 CO m ro b. Front View (b) Geometry of Test Assembly. Figure 5*3 Test Assembly Used By Williams and Geschwinder For Collar Joint Tests (18) 109 Vertical Displacement (x 0.001 in.) 60 50 60 30 20 10 0 Figure 5-4 Experimental Source of Collar Joint Element P-A Curve (18) Shear Bond Stress, psi 110 where P = vertical load Ag = area in shear therefore, P t Aa (5.3) By taking the average shear stress from their tests and multiplying it O by the desired area in shear of 192 in a new maximum value of P is obtained. Then by dividing the new Pmax by the average slope, a new Amax is obtained. Figure 5*5 shows the collar joint element P-A curve obtained in this fashion. The P-A curve for the block element is generated in the fashion described in section 4.3.3. Tests similar, to those, which were performed by Fattal and Cattaneo (4) on block prisms, were used to obtain the P-A curve for the block element which is used in the numerical examples. They tested three 6x32x24 in block prisms by loading them axially and noting the vertical strain variation with load. The results of their tests are shown in Figure 5.6. To use their results, it is necessary to convert their data, for 6x32x24 in prisms, to equivalent data for 6x24x8 in prisms. This is done in the same way that was discussed earlier for the brick prism results. In other words, by applying Equations (5.1) and (5.2) on several points in Figure 5.6, a new P-A curve can be generated which is representative of the results that would be obtained for a 6x24x8 in 111 Figure 5*5 Collar Joint Element P-A Curve 112 Figure 5*6 Experimental Source of Concrete Block Element P-A Curve (4) 113 prism. Figure 5-7 depicts the concrete block element P-A curve which was obtained in this fashion. 5.2.2 M-9 Curves The M-9 curves for the brick element are generated in the manner explained in section 4.3*1* The M-9 curves for the brick element used in the numerical examples were obtained from tests very similar to those described in section 4.3-1 which were performed by Fattal and Cattaneo (4) on brick prisms. They tested twelve 4x32x16 in prisms by loading them eccentrically and recording the vertical strain on each side of the prism in the plane of the applied end moment. The results of their tests are shown in Figure 5*8. Before using their results, it is first necessary to convert their data, which is for 4x32x16 in prisms, to equivalent data for 4x24x8 in prisms. This is done as follows. Consider Figure 5.9a. Line b-b represents a horizontal plane through a 4x32x16 in prism. Line b'-b shows the position of that plane after the prism bends in response to an end moment applied during a test. From small angle theory, = tan 9 9 t (5.4) Define the difference in vertical strain Ae as Ag Â£ ^ *" Â£ -j (5.5) where e = vertical strain in the prism measured on side 1 = vertical strain in the prism measured on side 2. 114 Figure 5.7 Block Element P-A Curve VERTICAL COMPRESStVE LOAD, kip ( 3 ) Relationship between vertical compressive load and vertical strain for 4 X 32 X 16-in brick prisms at e = tl 12. vertical strain (b), Relationship between vertical compressive toad and vertical strain for 4 X 32 X 16-in brick prisms ate = t/6. Figure 5*8 Experimental Source of Brick Element M-9 Curves (4) 116 (c) Relationship between vertical compressive load and vertical strain for 4 X 42 X 16-in brick prisms at e t/4. (d) Relationship between vertical compressive load and vertical strain for 4 x 32 X t6-in brick prisms at e t/3. Figure 5.8-continued 117 (a) I 1/2 1/2 El 1 A-j i A2 (b) Figure 5.9 Relationships Used to Obtain Desired Data From Experimental Results For M-0 Curves 118 (c) Figure 59-continued. 119 Figure 59-continued 120 Side 1 represents the left side of the prism and side 2 the right side. Note from Figure 5.8 that vertical strains were measured on each side of the prism during testing. By definition, (5.6) (5.7) where <$1 = vertical displacement on side 1 of the prism 2 = vertical displacement on side 2 of the prism L = prism height. From Figure 5*9a, 6 Plugging Equations (5.6) and (5.7) into Equation (5-5), Ac = 2~1 L L L From Equation (5.8), this reduces to Ac A L (5.8) (5-9) (5.10) Therefore, 5 = (Ae)L . (5.11) 121 Substituting Equation (5-11) into (5.4), 0 = eL * t From Equation (5*5), 9 can also be expressed as (5.12) (5.13) Thus, from the vertical strain measurements made on each side of the prism during the tests by Fattal and Cattaneo, it is possible to calculate the end rotation 9 of a prism in response to applied end moment. For the prisms they tested, L has a value of 15.7 in and t = 3-56 in. The only thing that is required now is a means of relating the end rotation 9 for the 15.7 in high prisms they used to the end rotation that would be produced if 8 in high prisms were subjected to the same end moment. Recall the first Moment Area Theorem, presented in West (17). In reference to Figure 5.9b, it would state that the angle change between points a and b on the deflected structure, or the slope at point b relative to the slope at point a, is given by the area under the K/EI diagram between these two points. In other words, 9 a/b (5.H) Figure 5.9c is a schematic representation of the prism size tested (4x32x16 in) for which results are available and the prism size for which results are desired (4x24x8 in). From Equation (5.14) note that end rotation 9 will be affected by differences in 1, or prism height, as 122 well as differences in moment of inertia I. Since "both prisms consist of the same materials, there will be no difference in the modulus of elasticity E for each. The difference in moment of inertia results from the difference in cross-section of the two prism sizes, which is due to the different depths of each prism. The 4x52x16 in prism, for moment of inertia calculations, has a 4x52 in cross-section and the 4x24x8 in prisms, a 4x24 in cross-section. Consider Figure 5*9d. From Equation (5.14), e ,f?16 (5.15) 16 EI16 and q (5.16) 8 EIq Expressing 9g in terms of 9-|g then, 9q = (x) 916 or J-(x)fll- (5-17) EIq EI16 Recall that the equation for the moment of inertia of a rectangular cross-section is I = bh5 (5.18) 12 For the 16 in high prism (actually 15*7 in but height does not enter into the calculation of I), 123 I = 32(4)3 = 16 12 170.667 in4 and for the 8 in high prisms, z = 24(4)3 = 8 12 128 in4 . For equal end moments (M = Mg = M.Â¡g), if these values are inserted into Equation (5*17), it reduces to 9g = 0.6667 16 # (5.19) This means that if an end moment M is applied to a 16 in high prism and to an 8 in high prism with the cross-sections shown in Figure 5.9c, the end rotation produced in the 8 in high prism will be 0.6667 times the end rotation in the 16 in high prism. The M-9 curves for the brick element were obtained from the test data shown in Figure 5*8 as follows. For a given value of P, the end moment M, which equals P x e, was calculated for the different eccentricities in parts (a), (b), (c) and (d) of Figure 5-8. Using Equation (5-13)* where L = 15.7 in t = 3*56 in Â£2 Â£] read from the curves, the end rotation 9 for the prism was calculated for each value of applied end moment M. This end rotation is for the 4x32x16 in prisms used in the tests, so using Equation (5*19) an equivalent end rotation for 4x24x8 in prisms was obtained. This procedure was carried out for 124 several values of P, and the results shown in Figure 5.10 were obtained. For values of P higher than 200,000 lb and lower than 100,000 lb, the M-Q relation did not differ from that shown for those values. Back in section 4.3*2, it was mentioned that, if the effect of moment transfer in the collar joint is found to merit consideration, a test will be devised to establish the relationship between moment and rotation. Figure 4.11 showed what this relationship might look like, and how the moment spring stiffness factor would be calculated. At this point, it is not known how important the collar joint moment transfer effect is and, as a result, it is desired to not consider it in the analysis. This is done by insuring that the moment spring stiffness factor kffl has a value of zero. The best way to accomplish this is by using the M-9 curve for the collar joint shown in Figure 5.11 which was used in the numerical examples. The M-9 curves for the concrete block element are developed in the fashion discussed in section 4*3.3. Tests similar to those were performed by Fattal and Cattaneo (4) on block prisms. Their results were used to obtain the M-9 curves used in the numerical examples. They tested twelve 6x32x24 in prisms by loading them eccentrically and recording the vertical strain on each side of the prism in the plane of the applied end moment. The results of their tests are shown in Figure 5.12. To use their results, it is necessary to convert their data, for 6x32x24 in prisms to equivalent data for 6x24x8 in prisms. This is done in a fashion similar to the technique used for converting brick prism results discussed previously. The M-9 curves for the concrete block element were obtained from the test data shown in Figure 5*12 as IN-LB 125 Figure 5*10 Brick Element M-9 Curves IN-LB 126 150,000 i 120,000 No Test; Collar Joint Moment Stiffness = 0 90,000 3E 60,000 30,000 0 | I I I I | 0 .20 .40 .60 0, RADIANS !' I I I I | .80 1.0 Figure 5-11 Collar Joint Element M-9 Curve 127 ( 3 ) Relationship between vertical compressive load and vertical strain for 6 X 32 X 24-in hollow block prisms at e= t/12. (b) Relationship between vertical compressive load and vertical strain for 6 X .32 X 24-in hollow block prisms at e 1/6. Figure 5.12 Experimental Source of Concrete Block Element M-Q Curves (4) 128 (c) Relationship between vertical compressive load and vertical strain for 6 X 32 X 24-in hollow block prisms at e= t/4. .(d) Relationship between vertical compressive load and vertical strain for 6 X 32 X 24-in hollow block prisms at e= t/3. Figure 512-continued 129 follows. For a given value of P, the end moment M, which equals P x e, was calculated for the different eccentricities in parts (a), (b), (c) and (d) of Figure 5.12. Using Equation (5*13), where L = 23.7 in t = 5.6 in e2, ei read from the curves, the end rotation 9 for the prism was calculated for each value of applied end moment M. This end rotation is for the 6x32x24 in prisms used in the tests. An equation similar to Equation (5.19) which is valid for calculating an equivalent end rotation for 6x24x8 in block prisms is necessary. This is developed in the same way as before. From Figure 5-9e and the first Moment Area Theorem, Equation (5.14) becomes 0 = Z 24 EI24 for the 24 in high prisms, and 4Ma 9 = 8 EIq for the 8 in high prisms. Expressing 9g in terms of 24 98 = x 924 or 8 = (x) _12M24 . ETg EI24 (5.20) (5.21) (5.22) 130 The moment of inertia for each prism is and ! = 24(6)3 _ 8 12 432 in4 . For equal end moments (M = Mg = if these values are inserted into Equation (5.22), it reduces to (5.23) 8 = 0.444424 Thus, by using Equation (5.23), an equivalent end rotation for 6x24x8 in prisms was obtained. This procedure was carried out for several values of P, and the results illustrated in Figure 5.13 were obtained. For values of P higher than 75,000 lb and lower than 25,000 lb, the M-9 relation did not differ from that shown for these values. 5.2.3 P-M Interaction Diagrams The P-M interaction diagram for the brick element is generated in the manner discussed in section 4.3.1. The P-M interaction diagram for the brick element used in the numerical examples was obtained from tests performed by Fattal and Cattaneo (4) on brick prisms. They tested fifteen 4x32x16 in brick prisms and obtained the results shown in Figure 5.14. By converting their results to equivalent data for 4x24x8 in prisms, their P-M interaction diagram can be used to generate a valid one for the 4x24x8 in prisms. This is done by applying Equation (5.1) to obtain a value of P for the desired prism size. The eccentricity e IN-LB P = 25,000 LB 6, RADIANS Figure 5.13 Block Element M-Q Curves 132 Cross-sectional capacity of brick prisms. Figure 5*14 Experimental Source of Brick Element P-M Interaction Diagram (4) 135 can be calculated from (5.24) Then, the moment for the desired prism size is obtained from (5.25) By using this technique for several values of P, Figure 5.15, which was used for the examples, was generated. In the above equations, Pj) = P on desired prism P^ = P on actual prism Mjj = M on desired prism M^ = M on actual prism. The P-M interaction diagram for the block element is generated in the manner discussed in section 4.3*3. Once again, block prism tests performed by Fattal and Cattaneo (4) were used as a basis for the block element P-M interaction diagram used in the numerical examples. They tested fifteen 6x32x24 in block prisms and obtained the results shown in Figure 5.16. As for the brick element, an equivalent P-M interaction diagram for 6x24x8 in prisms was generated by applying Equations (5.1), (5.24) and (5.25) on several values of P. This new P-M interaction diagram is shown in Figure 5.17. Because, for both types of elements, the P-M interaction diagram is a measure of cross-sectional capacity, only the differences in prism cross-section need to be considered when converting results from 6x32x24 in prisms to results for 6x24x8 in prisms. p Figure 5*15 Brick Element P-M Interaction Diagram 135 Cross-sectional capacity of concrete block prisms. Figure 5-16 Experimental Source of Concrete Block Element P-M Interaction Diagram (4) 136 P ^ M W M Figure 5.17 Block Element P-K Interaction Diagram 137 5.3 Example Number 1 Finite Element Analysis of a Test Wall Figure 5*18 shows a composite masonry wall which was tested by Fattal and Cattaneo (4) and also analyzed using the finite element model. As shown in the figure, the wall is 8 feet high and 10 inches thick. Both were loaded at the same eccentricity. Two differences between the test wall and the model wall exist. The cross-section of the test wall is 10x31.6 in and the cross-section of the model wall is 10x24 in. The second difference is that the test wall was free to rotate at the top and the bottom, but the model wall is free to rotate at the top but fixed at the bottom. Both of these differences can be taken into account so the failure load of the actual wall and the failure load predicted by the model can be compared. Fattal and Cattaneo (4) tested two wall specimens with the characteristics just described and one failed at 170,000 lb and the other at 190,000 lb. Failure was characterized by vertical splitting in the webs of the block, due to compression, at the top or bottom three courses of the specimen. The analytical model yielded a wall failure load of 132,000 lb characterized by a compression failure in the block 4 in from the top of the wall. In order to make a valid comparison of the failure loads, the differences in cross-section and end conditions must be considered. Recall Equation (5.1), which was developed assuming equal vertical compressive stress and stated P X P. . D Aga A Redefine the variables as 8.00 FT 6.25 IN II 1-3.75 IN P 4 IN H I16 IN TEST WALL Sign Convention H/T = 9.6 P P + AP AP 4000 LB e 1.25 IN M Pe 1.25P ^ m is 1 8 IN TYPICAL 2 IN /7T7fi77 -I M3 IN MODEL Figure 5*18 Example Number 1 139 Pp = P on desired wall cross-section P^ = P on actual wall cross-section Aqj, = gross area of desired wall cross-section Aqa = gross area of actual wall cross-section. p From the cross-section dimensions given earlier, Aqjj=240 in and Aq^=316 p in Neglecting the difference in end conditions, this means the wall test failure load value to be compared with the model failure load is P = 240 x 18o,000 136,700 lb if the average failure load of the two tests is used. To account for the difference in end conditions, an approximate wall failure load which might have been obtained from the tests if the end conditions in the model had been used, can be calculated using Figure 5*19 taken from the Manual of Steel Construction (7). The end conditions shown in (d) were those used in the test walls. The model considers the end conditions given in (b). Note that K for the test walls equals 1.0 and K for the model wall equals 0.80. The wall test failure load value expected for the end conditions in the model would be P (approx) 136,700 x -1aÂ£_ = 170,900 lb . 0.80 Since the model predicted failure at 132,000 lb, it underestimated the failure load by about 23$. Nonetheless, this is not excessive since a 12% difference in wall failure loads was obtained in the two wall tests. Furthermore, on the basis of the experimental testing program to be conducted in the future, further refinements to the analytical model 140 Buckled shape of column is shown by dashed line (a) 1 / / / 1 1 l 1 \ \ \ 1 JTnW t (b) Y / / t i i \ \ \ \ \ VTrr t (c) 2 s :ji 1 1 1 / / / / t / t rmr t (d) t Y \ \ \ \ \ i ( i i / / nmt \ (e) 1 1 ? P / / / / / i f rmr t (f) ea 1 1 > 1 1 / / 1 l f / I Theoretical K value 0.5 0.7 1.0 1.0 2.0 2.0 Recommended design value when ideal condi tions are approximated 0.65 0.80 1.2 1.0 2.10 2.0 End condition code Y ? Rotation fixed and translation fixed Rotation free and translation fixed Rotation fixed and translation free Rotation free and translation free Figure 5-19 Effective Column Length Factors Based On End Conditions (7) 141 may be made. Figure 5.20 shows the structure degrees of freedom used in the model wall. 54 Illustrative Examples The following four examples consider walls which vary in slenderness, or H/T, ratio and eccentricity of load application. Example number 2 considers a wall with a slenderness ratio of 12 and a plate load eccentricity of 0 in. In example number 3, the wall contains a slenderness ratio of 12 but a plate load eccentricity of 2 in. Example number 4 models a wall with a slenderness ratio of 20 and a plate load eccentricity of 0 in. Finally, example 5 considers a wall with a slenderness ratio of 20 and a plate load eccentricity of 2 in. In all four examples, 6 inch block is used, resulting in a wall thickness of 10 inches since the brick wythe is always 4 inches thick. The material property curves from section 5.2 were used for all of the examples. Intermediate printouts for each example were generated at load levels of 0.30 Pmax, 0.60 Pmax, 0.90 F^ and Pmax. The results of the finite element analysis for each wall are presented in chapter 6. 5.4.1 Example Humber 2 Example number 2 considers a wall 10 feet or 120 inches high, 10 inches thick, and 24 inches deep. The plate axial load is applied in increments of 4000 lb at 0 eccentricity up to failure. Figure 5.21 shows the wall of example 2. The degrees of freedom for external wall loads, equivalent wall loads, and wall displacements for examples 2 and 3 are shown in Figure 5.22. The data deck for this example can be found in Appendix D.2. 142 Ws 6 W57 Wx W4 W56 W5: w56 w 57 w. w 58 v> Wx f Wt " mVs V.> t w2 w a Wb /777 rrn EXTERNAL WALL LOADS EQUIVALENT WALL LOADS WALL DISPLACEMENTS Figure 5*20 Structure Degrees of Freedom For Example Number 1 143 5 IN l-M 5 IN T inT/UW 4 IN H H 6 IN TEST WALL Sign Convention H/T = 12 P = P + AP AP = 4000 LB e 0 IN M Pe 0 P '"'H ^ m k /nrrfn I 8 IN TYPICAL 2 IN111 3 IN MODEL Figure 5*21 Example Number 2 144 W7i t ^72 rft t t v*1 s -i-!^ t. V* 41 V. 4 t W1 t w u t U t vp-or H_U virl v # t H- V jT v V.#' w V#Tv 7777/777 t ,t_ t2 w, WR 3 w 71 w 72 W 73 vj t V t VI' t Vp t V' V t V V t M N* W. V^s. rl V' vi v tfi >. V1 v W v. /777/m W 74 u * u 1 ,L U jt t_ 4 U 4 ;t_ 4 4 t tw2 * W3 w, 3 w 73 W71 w72 'iff t_ M * ,t_ 4 4 t_ 4 t_ * ,t_ 4 U 4 > u Â¡r * U jWj t V t W11 w V?77 74 2 W. w 3 /7?7 EXTERNAL WALL LOADS EQUIVALENT WALL LOADS WALL DISPLACEMENTS Figure 5.22 Structure Degrees of Freedom For Example Numbers 2 and 5 145 5.4.2 Example Number 5 Example number 5 considers a wall 10 feet or 120 inches high, 10 inches thick, and 24 inches deep. The plate axial load is applied in increments of 4000 lb at 2 inches of eccentricity toward the block wythe, up to failure. The wall of example 5 is shown in Figure 5.23. The data file for this example is given in Appendix D.3. 5.4.3 Example Number 4 Example number 4 considers a wall 16.67 feet or 200 inches high, 10 inches thick, and 24 inches deep. The plate axial load is applied in increments of 4000 lb at 0 eccentricity up to failure. Figure 5.24 shows the wall of example 4. The degrees of freedom for external wall loads, equivalent-wall loads, and wall displacements for examples 4 and 5 are shown in Figure 5.25. The data deck for this example is given in Appendix D.4. 5.4.4 Example Humber 5 Example number 5 considers a wall with a height of 16.67 feet or 200 inches, a thickness of 10 inches, and a depth of 24 inches. The plate axial load is applied in increments of 4000 lb at 2 inches of eccentricity toward the block wythe up to failure. Figure 5.26 illustrates the wall of example 5. The data file for this example is provided in Appendix D.5. 10.00 FT 146 7 IM Wi I3 IN P rN ni'nin// 4 IN H 6 IN TEST WALL Sign Convention H/T = 12 P = P + P P = 4000 LB e = 2 IN M = Pe = 2P mi*. I 8 IN TYPICAL /mnv 2 IN nI 3 IN MODEL Figure 5*23 Example Number 3 147 5 IN HH5 IN P t fN rnfTnw 4 IN III 6 IN Sign Convention H/T = 20 P P + AP P = 4000 LB e = 0 IN M = Pe = 0 mis /777/777 2 INHII 3 IN J8 IN TYPICAL TEST WALL MODEL Figure 5*24 Example Number 4 148 Wi21 vt *, r* 7 V Tv > v+'r'+'' 4 4 VET^ U V47~ 4 V7~/ I f W1 t. 4^ /& 4 4 4 virra vF'Tji 4 '4Tr3# 4 v^rq* 4 iw uThn* +3 A? - 4 4 4_ / t i#-** I rf#P 4 rf** * 4 4 4_ t 4__ * w 2 ~ w, w 3 EXTERNAL WALL LOADS ^121 W122 w 1 * w123 W124 V.#1 V?vf t i V V 4 v. 4 v 4 vi# 4 vp 4 V|# 4 4 VJ# 4 vj* 4 Wi4 w,v 4_ 4_^ t 4 if .4_ if 4 V 4 ir 4__ r 4_^ if 4 if"* 4_^ if 4 If 4_<_ f 4__ if i- t t f W3 w5 3 /77 /77 EQUIVALENT WALL LOADS 2l 2 2 4 4 ^1 2 3 v. 7"v ^124 V >. 9 4 V tLJU 4 v. <# . r 'i '< 4 4 t w Jt 1 4 *4- 4 ^ VUf N. # H V4f s+f 4. 4- sir <4# 4 +- 4- v+# 4 'JT~3i 4 v 4 V 4 4 . f 4 f 4_ < 4_ f 4_ # 4^ 4_ * 4_ 4_ f 4 f T 4 4_ 4 4_ 4_ }"w2 w, w. /7r/rr WALL DISPLACEMENTS Figure 5.25 Structure Degrees of Freedom For Example Numbers 4 and 5 149 7 IN 3 IN f/m////! 4 IN -ni 6 IN TEST WALL Sign Convention H/T = 20 P = P + AP AP = 4000 LB e = 2 IN M = Pe = 2P V' P L K I 8 IN TYPICAL /7T7/7T7 2 INHW3 IN MODEL Figure 5.26 Example Number 5 CHAPTER SIX RESULTS OF ANALYSIS 6.1Wall Failure Wall failure load values and locations of failure are shown in Table 6.1 for all five examples. As expected, walls with higher slenderness ratios fail at lower loads than walls with smaller slenderness ratios. Also, walls loaded eccentrically fail before walls loaded axially. These results are, therefore, in agreement with generally expected patterns of structural behavior. 6.2Lateral Wall Deflection Versus Height The lateral wall deflection versus height for examples 2 through 5 is shown in Figures 6.1 through 6.4* Note that the deflected shape of the wall agrees with what would be expected for the end conditions considered by the model. Not surprisingly, for walls of equal height, the wall with a higher failure load experiences larger lateral deflections. 6.3Brick Wythe Vertical Deflection Versus Height Figures 6.5 through 6.8 show the brick wythe vertical deflection as a function of height for examples 2 through 5. As expected, the brick wythe vertical deflection increases as the load increases and the maximum brick wythe vertical deflection is greater for walls which fail at higher loads. 150 151 Table 6.1 Summary of Wall Failure for Examples Number 1 Through Number 5 WALL FAILURE LOADS FAILURE MODE EXAMPLE NUMBER P, LB M, IN-LB ELEMENT TYPE ELEMENT NO. LOCATION 1 132,000 165,000 BLOCK 56 4" FROM TOP OF WALL 2 204,000 0 BLOCK 59 20" FROM 42 12" TOP OF 45 4" WALL 5 106,000 216,000 BLOCK 5 4" FROM BOTTOM OF WALL 45 4" FROM TOP OF WALL 4 168,000 0 BLOCK 3 4" FROM BOTTOM OF WALL 5 36,000 72,000 BLOCK 3 4" FROM BOTTOM OF WALL WALL HEIGHT, INCHES 152 Figure 6.1 Lateral V/all Deflection Versus Height For Example Number 2 WALL HEIGHT, INCHES Figure 6.2 Lateral Wall reflection Versus Height For Example Number 3 WALL HEIGHT, INCHES 154 Figure 6.3 Lateral Wall Deflection Versus Height For Example Number 4 WALL HEIGHT, INCHES 155 Figure 6.4 Lateral Wall Deflection Versus Height For Example Number 5 WALL HEIGHT, INCHES 156 Figure 1 I I I j I I I | I I 1 1 | 0 .075 .150 .225 .300 BRICK WYTHE VERTICAL DEFLECTION, INCHES 6.5 Brick Wythe Vertical Deflection Versus Height For Example Number 2 WALL HEIGHT, INCHES 157 120.0 112.0 P v- I T I'" I 0 .075 .150 .225 BRICK WYTHE VERTICAL DEFLECTION, INCHES .300 Figure 6.6 Brick 'Wythe Vertical Deflection Versus Height For Example Number 3 WALL HEIGHT, INCHES 158 200.0 192.0 184.0 176.0 168.0 160.0 152.0 144.0 136.0 128.0 120.0 112.0 104.0 96.0 88.0 80.0 72.0 64.0 56.0 48.0 40.0 32.0 24.0 16.0 8.0 0 TTTT .075 .150 .225 BRICK WYTHE VERTICAL DEFLECTION, INCHES T~| .300 Figure 6.7 Brick Wythe Vertical Deflection Versus Height For Example Number 4 WALL HEIGHT, INCHES 159 Figure 6.8 Brick Wythe Vertical Deflection Versus Height For Example Number 5 160 6.4Block Wythe Vertical Deflection Versus Height The block wythe vertical deflection versus height for examples 2 through 5 is illustrated in Figures 6.9 through 6.12. Once again, the block wythe vertical deflection increases as the load increases and the maximum block wythe vertical deflection is greater for walls which fail at higher loads. This agrees with what would be expected. Also notice that for each example, at equal load levels the block wythe vertical deflection is greater. This happens because the brick wythe is stiffer than the block wythe and, in some cases, the wall is loaded eccentrically towards the block. 6.5Plate Load Versus End Rotation The variation of end rotation at the top of the wall as a function of vertical plate load is depicted in Figures 6.15 through 6.16 for examples 2 through 5* The end rotation, of course, increases with increase in load and, not surprisingly, at equal load values for walls of the same height, the end rotation is greater for the wall in which the load is applied eccentrically. At high load levels, the block stiffness is considerably less than the brick stiffness so the increase in end rotation per unit increase in load becomes greater. This is depicted by the flattening out of the curve at high load values in Figures 6.15 and 6.15. 6.6Wall Wythe Vertical Load Versus Height Figures 6.17 through 6.52 show the wall wythe vertical load versus height for four load levels for each of examples 2 through 5* Notice that with decrease in height from the top of the wall, more of the load is transferred to the brick wythe. For the walls loaded eccentrically, WALL HEIGHT, INCHES Figure 6.9 Block Wythe Vertical Deflection Versus Height For Example Number 2 WALL HEIGHT, INCHES 162 Figure 6.10 Block tfythe Vertical Deflection Versus Height For Example Humber 3 WALL HEIGHT, INCHES 163 0 .075 .150 .225 .300 BLOCK WYTHE VERTICAL DEFLECTION, INCHES Figure 6.11 Block Wythe Vertical Deflection Versus Height For Example Number 4 WALL HEIGHT, INCHES 164 Figure 6.12 Block Wythe Vertical Deflection Versus Height For Example Number 5 165 Figure 6.13 Plate Load Versus End Rotation For Example Number 2 Figure 6.14 Plate Load Versus End Rotation For Example Humber 167 Figure 6.15 Plate Load Versus End Rotation For Example Number 4 168 Figure 6.16 Plate Load Versus End Rotation For Example Number 5 WALL HEIGHT, INCHES Figure 6.17 Wall Wythe Vertical Load Versus Height At 0.JO ?max Example Number 2 For WALL HEIGHT, INCHES 170 Figure 6.18 Wall Wythe Vertical Load Versus Height At 0.60 Pmax Example Number 2 For WALL HEIGHT, INCHES 171 Figure 6.19 Wall Wythe Vertical Load Versus Height At 0.9C P__, For Example Number 2 WALL HEIGHT, INCHES 172 Figure 6.20 Wall Wythe Vertical Load Versus Height At Pmax For Example Humber 2 WALL HEIGHT, INCHES 173 Figure 6.21 Wall Wythe Vertical Load Versus Height At 0.30 Pmax For Example Humber 3 WALL HEIGHT, INCHES 174 Figure 6.22 Wall Wythe Vertical Load Versus Height At G.6G Pnax For Example Number 3 WALL HEIGHT, INCHES 175 Figure 6.25 Wall Wythe Vertical Load Versus Height At 0.90 Pmax For Example Number 3 WALL HEIGHT, INCHES 176 120.0 112.0 104.0 96.0 88.0 80.0 72.0 64.0 56.0 48.0 40.0 32.0 24.0 16.0 8.0 0 1 Figure 6.24 V/all Wythe Vertical Load Versus Height At P Example Number 3 100 For WALL HEIGHT, INCHES 177 Figure 6.25 Wall Wythe Vertical Load Versus Height At 0.30 Pnax Example Number 4 For WALL HEIGHT, INCHES 178 Figure 6.26 Wall Wythe Vertical Load Versus Height At C.60 Praax For Example Humber 4 WALL HEIGHT, INCHES 179 Figure 6.27 Wall Wythe Vertical Load Versus Height At 0.00 P For max Example Number 4 WALL HEIGHT, INCHES 180 Figure 6.28 Vail Wythe Vertical Load Versus Height At Pmax Example Number 4 For WALL HEIGHT, INCHES 181 Figure 6.29 Wall Wythe Vertical Load Versus Height At 0.30 P For ITiclX Example Humber 5 WALL HEIGHT, INCHES 182 Figure 6.30 Wall Wythe Vertical Load Versus Height At 0.60 Pmax Example Number 5 For WALL HEIGHT, INCHES 183 Figure 6.31 Wall Wythe Vertical Load Versus Height At 0.90 Pmax For Example Number 5 WALL HEIGHT, INCHES 184 Figure 6.22 Wall Wythe Vertical Load Versus Height At Pmax Example Number 5 For 185 the block wythe carries a higher percentage of the load near the top as might be expected, due to the location of load application. 6.7 Brick Wythe Koment Versus Height The brick wythe moment as a function of wall height is plotted for examples 2 through 5 in Figures 6.33 through 6.36. Notice that the moment diagrams and the deflected shapes shown in Figures 6.1 through 6.4 agree. 6.8 Block Wythe Koment Versus Height Figures 6.37 through 6.40 show the block wythe moment versus height for examples 2 through 5. Once again, the moment diagrams and the deflected shapes shown previously in Figures 6.1 through 6.4 agree. For each example, the moment in the block wythe is greater than the moment in the brick wythe for equal wall load levels. This is due to the higher rotational stiffness of the block prisms compared with the brick prisms. 6.9Collar Joint Shear Stress Versus Height The collar joint shear stress as a function of wall height for several load levels in each of examples 2 through 5 is illustrated in Figures 6.41 through 6.44. Like the rest of the data plotted in chapter 6, it too was obtained directly from the output generated by the finite element analysis program. Note that in example 2 at a height of 72" and in example 4 at a height of 136" there are discontinuities in the collar joint shear stress in each wall at maximum load levels. These are shown in Figures 6.41 and 6.43, respectively. This coincides with two other observations. At the same heights and load levels for these examples, the moment diagram changes in curvature as seen in Figures 6.33 and 6.37 WALL HEIGHT, INCHES 186 MOMENT, INCH-POUNDS Figure 6.33 Brick Wythe Moment Versus Height For Example Number 2 WALL HEIGHT, INCHES 187 Figure 6.34 Brick Wythe Moment Versus Height For Example Number 3 WALL HEIGHT, INCHES 188 TT 75,000 Figure 6.35 Brick Wythe Moment Versus Height For Example Number 4 WALL HEIGHT, INCHES 189 MOMENT, INCH-POUNDS Figure 6.36 Brick Wythe Moment Versus Height For Example Number 5 WALL HEIGHT, INCHES 190 Figure 6.37 Block Wythe Moment Versus Height For Example Number 2 WALL HEIGHT, INCHES 191 Figure 6.38 Block Wythe Moment Versus Height For Example Number 3 WALL HEIGHT, INCHES 192 Figure 6.39 Block Wythe Moment Versus Height For Example Number 4 WALL HEIGHT, INCHES 193 Figure 6.40 Block Wythe Moment Versus Height For Example Number 5 WALL HEIGHT. INCHES 194 Figure 6.41 Collar Joint Shear Stress Versus Height For Example Number 2 195 Figure 6.42 Collar Joint Shear Stress Versus Height For Example Number 3 WALL HEIGHT, INCHES 196 Figure 6.43 Collar Joint Shear Stress Versus Height For Example Number 4 WALL HEIGHT, INCHES 197 Figure 6.44 Collar Joint Shear Stress Versus Height For Example Number 5 198 respectively, for the brick wythe, and Figures 6.35 and 6.39 respectively, for the block wythe. Also, at Pmax> the block wythe vertical deflection increases nonlinearly above these heights for each example as shown in Figures 6.9 and 6.11, respectively. Examination of the program output revealed that at these heights and load levels in each example the block element axial force is very close to 75,375 lb and, in fact, above this height it exceeds that amount. At this value, a sharp decrease occurs in the block element axial stiffness as shown in Figure 5*7. A possible explanation of these occurrences is the following. In response to the compressive load applied at the top of each wall, the walls assume the deflected shapes shown in Figure 6.1 for example 2 and Figure 6.3 for example 4. The bending action which produces these shapes tends to put the outer, or brick, wythe of the wall in tension and the inner, or block, wythe of the wall in compression. Similarly, considering each wythe separately, the inner or right side of the brick wythe is subject to compression and the inner or left side of the block wythe is subject to tension. This creates a positive collar joint shear stress. When the block elements near the top of each wall suddenly decrease in axial stiffness, this causes the vertical deflection in the block wythe to increase rapidly at this location. This leads to a sudden reduction in shear stress in the collar joint, producing the discontinuities previously observed. Since all elements must remain in equilibrium, the shear or vertical forces in the collar joint element directly affect the end moments on the left and right ends of the collar joint element. These end moments are also the value by which the end moments due to the other two contributing 199 elements at a node are not in equilibrium. When the collar joint element end moments decrease at a node due to a decrease in shear stress, this causes the moments in the brick (or block) wythe at that node to be lower than they would have been otherwise. This produces the change in curvature in the moment diagrams observed previously. Eventually, if the collar joint shear stress becomes negative, as it does in Figure 6.43, this causes the moment in the brick and block wythes to actually decrease as shown in Figures 6.35 and 6.39, respectively. Not surprisingly, the maximum moment in each wythe for example 4 occurs at a height of 168" which is the height at which the collar joint shear stress is zero. CHAPTER SEVEN CONCLUSIONS AND RECOMMENDATIONS A two-dimensional finite element model was developed to analyze composite masonry walls subject to compression and out of plane bending. The model considers the factors that most strongly influence composite wall behavior. These factors include the different strength- deformation properties of the concrete block, collar joint, and clay brick, the nonlinear nature of these properties, the load transfer properties of the collar joint, and the effects of shear deformation and moment magnification in the brick and block wythes. The accuracy of the model was verified by comparing the results of a wall analyzed by the model with test results, taken from the literature, of a similar wall. Furthermore, four examples were done in which the slenderness ratio and eccentricity of load application were varied for four walls. The analytical results of these examples were found to conform with accepted patterns of structural behavior. Originally, it was intended to conduct extensive experimental testing and compare the results obtained with those predicted by the model. Due to problems associated with funding, this was not possible. Therefore, it is recommended that physical testing be performed next and that the results generated be correlated with those obtained from the model. Based on the findings acquired from this comparison, it will be possible to make further refinements to the model and consider such effects as temperature changes and shrinkage. Once 200 201 very close agreement is reached between the experimental and analytical results, the model may be used to explore variables not considered in the testing program. Ultimately, these efforts will yield valuable information, not currently available, that will enable modifications to be made to the current design standards. More accurate and reliable design procedures will make it possible for structural engineers to design safer, more efficient, and more economical composite masonry structures. It is hoped that this study has brought the realization of that goal one step closer. APPENDIX A COMPUTER PROGRAM A.1 Introduction The finite element program for composite masonry walls was written in the Fortran-77 language and run on a VAX 11/780 computer. A deliberate effort was made to avoid the sophisticated features available in Fortran-77 but not permitted in the WATFIV version of Fortran, in order to increase the program's usability. Customary composite wall failure is characterized by a failure of the block in compression. If loads are placed on the wall at too great an eccentricity towards the block face (off of the cross-section), this may cause the wall to fail by tension induced in the brick wythe. If in the analysis the wall fails because either a brick or block element goes into axial tension, this will be identified by the program and a recommendation to check for unusual loading made, since this should only occur if the vertical load is placed at an eccentricity greater than that at which it could be physically placed in the prototype wall or test wall. Analytically, wall failure is defined by the model as: 1) A brick or block element going into tension as mentioned above 2) A brick or block element exceeding the allowable load and moment combination permitted by the interaction diagram for that element type 3) A collar joint element exceeding its allowable shear load capacity. 201 When a failure of one of the latter two types occurs, the program identifies the wall loads at which this takes place, as well as which element(s) has (have) failed. Once the wall has failed, or all the loads reached their maximum value, the program will output: 1) The structure forces and displacements 2) The lateral wall deflection versus height 3) The vertical wall deflection versus height 4) The element forces and displacements 5) The wall wythe vertical load versus height. This information may also be displayed for intermediate load values as described in the data input section. In listing the structure forces, a subheading of external wall loads will identify the degrees of freedom loaded when the load plate degrees of freedom are used. A subheading of equivalent wall loads will represent the resulting degrees of freedom loaded when the original structure degrees of freedom are considered. The program uses 36 subroutines each of which performs a unique operation. Throughout the program and subroutines, comments have been strategically interspersed to explain how the analytical process is undertaken. Furthermore, extensive documentation in the four appendices provides detailed information on the finite element program and its usage. All program input and output is in basic units of inches, pounds, or radians 204 A.2 Detailed Program Flowchart Table A.1 defines the variables used in the detailed program flowchart. The flowchart is presented in Figure A.1. As evident, it goes into considerable more detail than the program algorithm shown in Figure 4.13- The call statements in parentheses identify the subroutines which are called at that point in the program to perform the operation described in the box. Appendix B contains individual algorithms for each subroutine. These are analagous to the program algorithm and, therefore, are not intended to show the detail which is provided in the detailed program flowchart. Detailed flowcharts for each subroutine are not furnished because, since each subroutine is short and only performs one major operation, this type of flowchart would be little more than a repetition of the subroutine listing. It would add little additional understanding over that available from examining the listing of the subroutines furnished in appendix A.4 and would contribute more to confusion than to clarity. A.3 Program Nomenclature A complete alphabetical listing of the program nomenclature is given in Table A.2. It identifies all of the variables, matrices, and subroutines used in the program. Except where standard variable names are used, most variable names are acronyms. For example, BRESM stands for 'Brick Element .Stiffness Matrix' and NBRIDP stands for 'Number of Brick .Interaction .Diagram .Points'. Appendix B contains a table for each subroutine which identifies the main variables the subroutine uses. All of the variables in the argument list of each subroutine are defined. 205 Table A.1 Variables Used in Detailed Program Flowchart VARIABLE DEFINITION [K] STRUCTURE STIFFNESS MATRIX [w] STRUCTURE DISPLACEMENT MATRIX [F] STRUCTURE FORCE MATRIX [aw] INCREMENTAL STRUCTURE DISPLACEMENT MATRIX [ af] INCREMENTAL STRUCTURE FORCE MATRIX [K*] MODIFIED STRUCTURE STIFFNESS MATRIX WHICH AT THE TOP OF THE WALL ONLY CONSIDERS THE TWO PLATE DEGREES OF FREEDOM [V] MODIFIED STRUCTURE DISPLACEMENT MATRIX WHICH AT THE TOP OF THE WALL ONLY CONSIDERS THE TWO PLATE DEGREES OF FREEDOM [F] MODIFIED STRUCTURE FORCE MATRIX WHICH AT THE TOP OF THE WALL ONLY CONSIDERS THE TWO PLATE DEGREES OF FREEDOM [I] ELEMENT INDEX MATRIX 1 1 ELEMENT STIFFNESS MATRIX [w] ELEMENT DISPLACEMENT MATRIX [f] ELEMENT FORCE MATRIX [error] MATRIX THAT MONITORS SOLUTION CONVERGENCE TOLER TOLERANCE OF VALUE AGAINST WHICH THE NUMBERS IN [ERROR] ARE COMPARED NCONV NUMBER OF CURRENT CONVERGENCE ATTEMPT; A MAXIMUM OF 10 ATTEMPTS ARE MADE NSTAT STATUS OF AN ELEMENT AFTER IT IS CHECKED FOR FAILURE; 0 = O.K.; 1 = FAILED; 9 = IN TENSION TRSTIF VARIABLE THAT IDENTIFIES WHETHER OR NOT ANY ELEMENT STIFF NESS FACTORS HAVE CHANGED; 0 = NO; 1 = YES, THEREFORE RUN IS REPEATED WITHOUT INCREMENTING THE STRUCTURE FORCES 2C6 Table A.1- -continued. VARIABLE DEFINITION TRCONV VARIABLE THAT IDENTIFIES IF A CONVERGENCE ITERATION IS NECESSARY; 0 = NO, MEANING [ERROR] < TOLER AND SOLUTION IS ACCEPTABLE; 1 = YES, MEANING [ERROR] > TOLER AND AN ATTEMPT WILL BE MADE TO CONVERGE ON THE SOLUTION BY USING THE INCREMENTAL FORCES TO SOLVE FOR THE INCREMENTAL DISPLACEMENTS TRELPR VARIABLE THAT IDENTIFIES WHETHER OR NOT ELEMENT FORCES AND DISPLACEMENTS ARE TO BE PRINTED; 0 = NO; 1 YES TPRINT VARIABLE THAT CONTROLS WHETHER OR NOT A STATEMENT IDENTIFYING EACH ELEMENT THAT HAS FAILED IS TO BE PRINTED; 0 = NO; 2 = YES TRFAIL VARIABLE THAT IDENTIFIES THE FAILURE STATUS OF THE WALL; 0 = HAS NOT FAILED; 1 = HAS FAILED, THEREFORE PRINT STRUCTURE FORCES AND DISPLACEMENTS; 4 = STOP AFTER PRINTING ELEMENT FORCES AND DISPLACEMENTS 207 (start) DECLARE MATRICES AND VARIABLES REAL OR INTEGER AS NEEDED DOUBLE PRECISION MATRICES AND VARIABLES DIMENSION MATRICES /READ AND PRINT PROBLEM DESCRIPTION (CALL READ)/ NULL MATRICES AS NEEDED (CALL NULL) /read AND PRINT P-A CURVES FOR brick, collar/ / JOINT, AND BLOCK ELEMENTS (CALL COORD) / /READ AND PRINT M-0 CURVES FOR BRICK, collar/ / JOINT, AND BLOCK ELEMENTS (CALL CURVES) / / READ AND PRINT P-M INTERACTION DIAGRAMS FOR BRICK / AND BLOCK ELEMENTS (CALL COORD) Â¡ ' READ AND PRINT MAXIMUM VERTICAL LOAD CAPACITY FOR BRICK, COLLAR JOINT, AND BLOCK ELEMENTS (CALL TITLE TO READ HEADINGS) /READ AND PRINT FORCE APPLICATION INFORMATION (CALL FORCES)/ r T / /READ, PROCESS, AND PRINT INSTRUCTIONS FOR INTERMEDIATE/ I PRINTOUTS (CALL TITLE TO READ HEADINGS) / DO FOR ITER-1 to 10000) I Figure A.1 Detailed Program Flowchart CALCULATE [i] (CALL INDXBR) T SELECT INITIAL STIFFNESS FACTORS (CALL BRPDMT) T" 'PRINT 'ELEMENT IS IN TENSION' YES ''NO TRFAIL=1 (stop) YES STORE EACH STIFFNESS FACTOR (CALL STIFAC) ~(V) CALCULATE STIFFNESS FACTORS (CALL BRPDMT) I PRINT 'ELEMENT/ HAS FAILED / SELECT AND STORE APPROPRIATE STIFFNESS FACTORS (CALL STIFAC) Figure A.1-continued 209 Figure A.1-continued. 210 Figure A.1-continued 211 Figure A.1-continued 212 Figure A.1-continued x 213 Figure A.1-continued 214 Figure A.1-continued Table A. 2 Program Nomenclature VARIABLE TYPE ARSHBL DOUBLE PRECISION ARSHBR DOUBLE PRECISION BL INTEGER BLAEL DOUBLE PRECISION BLMAXP DOUBLE PRECISION BLMOM DOUBLE PRECISION BL3EIL DOUBLE PRECISION BLVERF DOUBLE PRECISION BR INTEGER BRAEL DOUBLE PRECISION BRMAXP DOUBLE PRECISION BRMOM DOUBLE PRECISION BR3EIL DOUBLE PRECISION BRVERF DOUBLE PRECISION CJ INTEGER CJMAXP DOUBLE PRECISION DEFINITION AREA IN SHEAR FOR THE BLOCK AREA IN SHEAR FOR THE BRICK NUMBER OF BLOCK ELEMENT DEGREES OF FREEDOM BLOCK AE/L (AXIAL STIFFNESS FACTOR) BLOCK MAXIMUM P (COMPRESSIVE LOAD CARRYING CAPACITY) ABSOLUTE VALUE OF AVERAGE OF END MOMENTS FOR A BLOCK ELEMENT BLOCK 3EI/L (ROTATIONAL STIFFNESS FACTOR) VERTICAL (AXIAL) FORCE AT THE BOTTOM OF A BLOCK ELEMENT NUMBER OF BRICK ELEMENT DEGREES OF FREEDOM BRICK AE/L (AXIAL STIFFNESS FACTOR) BRICK MAXIMUM P (COMPRESSIVE LOAD CARRYING CAPACITY) ABSOLUTE VALUE OF AVERAGE OF END MOMENTS FOR A BRICK ELEMENT BRICK 3EI/L (ROTATIONAL STIFFNESS FACTOR) VERTICAL (AXIAL) FORCE AT THE BOTTOM OF A BRICK ELEMENT NUMBER OF COLLAR JOINT ELEMENT DEGREES OF FREEDOM COLLAR JOINT MAXIMUM P (VERTICAL LOAD CARRYING CAPACITY) 215 CJMOM CJMOMS CJSHRS CJVERF COUNT ELASBL ELASBR FMULT HEIGHT I ITER J K LNHBL LNHBR LNVER HI M1P1 DOUBLE PRECISION DOUBLE PRECISION DOUBLE PRECISION DOUBLE PRECISION INTEGER DOUBLE PRECISION DOUBLE PRECISION DOUBLE PRECISION REAL INTEGER INTEGER INTEGER INTEGER DOUBLE PRECISION DOUBLE PRECISION DOUBLE PRECISION INTEGER INTEGER ABSOLUTE VALUE OF AVERAGE OF END MOMENTS FOR A COLLAR JOINT ELEMENT COLLAR JOINT MOMENT SPRING STIFFNESS COLLAR JOINT SHEAR SPRING STIFFNESS ABSOLUTE VALUE OF COLLAR JOINT SHEAR FORCE COUNTING VARIABLE USED TO HELP STORE THE BRICK AND BLOCK WYTHE VERTICAL LOAD PER HEIGHT MODULUS OF ELASTICITY OF THE BLOCK MODULUS OF ELASTICITY OF THE BRICK VARIABLE USED TO STORE THE MULTIPLES OF A STRUCTURE FORCE FOR WHICH AN INTERMEDIATE PRINTOUT IS DESIRED VARIABLE USED TO HELP PRINT THE LATERAL WALL DEFLECTION VERSUS HEIGHT DO-LOOP PARAMETER DO-LOOP PARAMETER DO-LOOP PARAMETER DO-LOOP PARAMETER HORIZONTAL LENGTH OF THE BLOCK HALF OF THE COLLAR JOINT ELEMENT HORIZONTAL LENGTH OF THE BRICK HALF OF THE COLLAR JOINT ELEMENT VERTICAL LENGTH OF THE BRICK AND BLOCK ELEMENTS HALF THE BANDWIDTH (INCLUDING THE DIAGONAL) OF THE STRUCTURE STIFFNESS MATRIX HALF THE BANDWIDTH PLUS ONE Table A.2-continued VARIABLE TYPE NBRIDP INTEGER NBRMTP INTEGER NBRPDP INTEGER NBLIDP INTEGER NBLMTP INTEGER NBLPDP INTEGER NCJMTP INTEGER NCJPDP INTEGER NCONV INTEGER NDOF INTEGER NELEM INTEGER NEMO INTEGER NEMT INTEGER NFORCE INTEGER NOBLPC INTEGER DEFINITION NUMBER OF POINTS IN THE BRICK INTERACTION DIAGRAM NUMBER OF POINTS IN EACH BRICK M-THETA CURVE NUMBER OF POINTS IN THE BRICK P-DELTA CURVE NUMBER OF POINTS IN THE BLOCK INTERACTION DIAGRAM NUMBER OF POINTS IN EACH BLOCK M-THETA CURVE NUMBER OF POINTS IN THE BLOCK P-DELTA CURVE NUMBER OF POINTS IN THE COLLAR JOINT M-THETA CURVE NUMBER OF POINTS IN THE COLLAR JOINT P-DELTA CURVE NUMBER OF CURRENT CONVERGENCE ATTEMPT NUMBER OF THE DEGREE OF FREEDOM OF A PARTICULAR FORCE NUMBER OF ELEMENTS NUMBER OF ELEMENTS MINUS ONE NUMBER OF ELEMENTS MINUS TWO NUMBER OF FORCE DEGREE OF FREEDOM, MULTIPLES OF WHICH WILL CAUSE AN INTERMEDIATE PRINTOUT HUMBER OF BLOCK M-THETA CURVES (EACH IS IDENTIFIED BY THE P OR AXIAL LOAD LEVEL FOR WHICH THE MOMENT CURVATURE RELATIONSHIP IS VALID) 217 NOBRPC INTEGER HOF INTEGER NOFN INTEGER NRBM1 INTEGER NRBOT INTEGER NRTM1 INTEGER NRTMP1 INTEGER NRTOP INTEGER NRTP1 INTEGER NSDFMS INTEGER NSDFMT INTEGER NSDM2 INTEGER NS DOF INTEGER NSTAT INTEGER NSTORY INTEGER PROUT INTEGER SHEARL DOUBLE PRECISION NUMBER OF BRICK M-THETA CURVES (EACH IS IDENTIFIED BY THE P OR AXIAL LOAD LEVEL FOR WHICH THE MOMENT CURVATURE RELATIONSHIP IS VALID) NUMBER OF FORCES FORCE NUMBER NUMBER OF ROWS IN THE BOTTOM HALF OF THE STRUCTURE STIFFNESS MATRIX MINUS ONE NUMBER OF ROWS IN THE BOTTOM HALF OF THE STRUCTURE STIFFNESS MATRIX NUMBER OF ROWS IN THE TOP HALF OF THE STRUCTURE STIFFNESS MATRIX MINUS ONE NUMBER OF ROWS IN THE TOP HALF OF THE STRUCTURE STIFFNESS MATRIX MINUS ONE PLUS ONE NUMBER OF ROWS IN THE TOP HALF OF THE STRUCTURE STIFFNESS MATRIX NUMBER OF ROWS IN THE TOP HALF OF THE STRUCTURE STIFFNESS MATRIX PLUS ONE NUMBER OF STRUCTURE DEGREES OF FREEDOM MINUS SIX NUMBER OF STRUCTURE DEGREES OF FREEDOM MINUS THREE NUMBER OF STRUCTURE DEGREES OF FREEDOM MINUS TWO NUMBER OF STRUCTURE DEGREES OF FREEDOM STATUS OF AN ELEMENT AFTER IT IS CHECKED FOR FAILURE; O = O.K., 1 = FAILED; 9 = IN TENSION NUMBER OF STORIES OF WALL WHERE EACH STORY IS 8 INCHES TALL VARIABLE THAT IDENTIFIES WHETHER OR NOT AN INTERMEDIATE PRINTOUT IS DESIRED; 0 = NO; 1 = YES COLLAR JOINT SHEAR STRESS ON THE LEFT OR BRICK FACE OF THE COLLAR JOINT 218 Table A.2-continued VARIABLE SHEARR SHOWF TOLER TRCONV TRELPR TRFAIL TPRINT TRSHOW TRSTIF WDEPTH WHI TYPE DEFINITION DOUBLE PRECISION DOUBLE PRECISION DOUBLE PRECISION INTEGER COLLAR JOINT SHEAR STRESS ON THE RIGHT OR BLOCK FACE OF THE COLLAR JOINT FORCE VALUE FOR WHICH AN INTERMEDIATE PRINTOUT IS TO BE MADE TOLERANCE SET ON EQUILIBRIUM ERROR VARIABLE THAT IDENTIFIES IF A CONVERGENCE ITERATION IS NECESSARY; 0 = NO; 1 = YES INTEGER VARIABLE THAT IDENTIFIES WHETHER OR NOT ELEMENT FORCES AND DISPLACEMENTS ARE TO BE PRINTED; 0 = NO; 1 = YES INTEGER VARIABLE THAT IDENTIFIES THE FAILURE STATUS OF THE WALL; 0 = HAS NOT FAILED; 1 = HAS FAILED; 4 = STOP AFTER PRINTING ELEMENT FORCES AND DISPLACEMENTS - INTEGER VARIABLE THAT IDENTIFIES WHETHER OR NOT AN ELEMENT FAILURE IDENTIFICATION STATEMENT IS TO BE PRINTED; 0 = NO; 2 = YES INTEGER VARIABLE THAT IDENTIFIES WHETHER A CHECK FOR AN INTERMEDIATE PRINTOUT IS TO BE MADE; 0 = YES; 1 NO INTEGER VARIABLE THAT IDENTIFIES WHETHER OR HOT ANY ELEMENT STIFFNESS FACTORS HAVE CHANGED; 0 = NO; 1 = YES INTEGER WALL DEPTH IN INCHES INTEGER WALL HEIGHT IN INCHES MATRIX TYPE DESCRIPTION BLDCOR DOUBLE PRECISION BLEF DOUBLE PRECISION BLEK DOUBLE PRECISION BLEW DOUBLE PRECISION BLIDM DOUBLE PRECISION BLIDP DOUBLE PRECISION BLMCOR DOUBLE PRECISION BLPCOR DOUBLE PRECISION BLPCV DOUBLE PRECISION BLTCOR DOUBLE PRECISION BLWYLD DOUBLE PRECISION BRDCOR DOUBLE PRECISION BREF DOUBLE PRECISION BREK DOUBLE PRECISION BREW DOUBLE PRECISION BRIDM DOUBLE PRECISION STORES BLOCK DELTA COORDINATES BLOCK ELEMENT FORCE MATRIX BLOCK ELEMENT STIFFNESS MATRIX BLOCK ELEMENT DISPLACEMENT MATRIX STORES BLOCK INTERACTION DIAGRAM MOMENT COORDINATES STORES BLOCK INTERACTION DIAGRAM AXIAL LOAD COORDINATES STORES BLOCK MOMENT COORDINATES FOR EACH M-THETA CURVE STORES BLOCK AXIAL LOAD COORDINATES STORES THE VALUE OF P BY WHICH EACH BLOCK M-THETA CURVE IS IDENTIFIED STORES BLOCK THETA COORDINATES FOR EACH M-THETA CURVE MATRIX THAT STORES THE VERTICAL LOAD CARRIED BY THE BLOCK WYTHE AT EACH WALL LEVEL STORES BRICK DELTA COORDINATES BRICK ELEMENT FORCE MATRIX BRICK ELEMENT STIFFNESS MATRIX BRICK ELEMENT DISPLACEMENT MATRIX STORES BRICK INTERACTION DIAGRAM MOMENT COORDINATES 220 Table A.2-continued MATRIX TYPE BRIDP DOUBLE PRECISION BRMCOR DOUBLE PRECISION BRPCOR DOUBLE PRECISION BRPCV DOUBLE PRECISION BRTCOR DOUBLE PRECISION BRWYLD DOUBLE PRECISION CJDCOR DOUBLE PRECISION CJEF DOUBLE PRECISION CJEK DOUBLE PRECISION CJEW DOUBLE PRECISION CJMCOR DOUBLE PRECISION CJPCOR DOUBLE PRECISION CJTCOR DOUBLE PRECISION DELTAF DOUBLE PRECISION DELTAW DOUBLE PRECISION ERROR DOUBLE PRECISION DESCRIPTION STORES BRICK INTERACTION DIAGRAM AXIAL LOAD COORDINATES STORES BRICK MOMENT COORDINATES FOR EACH M-THETA CURVE STORES BRICK AXIAL LOAD COORDINATES STORES THE VALUE OF P BY WHICH EACH BRICK M-THETA CURVE IS IDENTIFIED STORES BRICK THETA COORDINATES FOR EACH M-THETA CURVE MATRIX THAT STORES THE VERTICAL LOAD CARRIED BY THE BRICK WYTHE AT EACH WALL LEVEL STORES COLLAR JOINT DELTA COORDINATES COLLAR JOINT ELEMENT FORCE MATRIX COLLAR JOINT ELEMENT STIFFNESS MATRIX COLLAR JOINT ELEMENT DISPLACEMENT MATRIX STORES COLLAR JOINT MOMENT COORDINATES STORES COLLAR JOINT VERTICAL LOAD COORDINATES STORES COLLAR JOINT THETA COORDINATES INCREMENTAL STRUCTURE FORCE MATRIX INCREMENTAL STRUCTURE DISPLACEMENT MATRIX MATRIX THAT MONITORS SOLUTION CONVERGENCE tu rv> F1 DOUBLE PRECISION F1MKW2 DOUBLE PRECISION F2 DOUBLE PRECISION GDELTA DOUBLE PRECISION GSTRK DOUBLE PRECISION GSTRW DOUBLE PRECISION IBL INTEGER IBR INTEGER ICJ INTEGER K1 DOUBLE PRECISION K2 DOUBLE PRECISION K2W2 DOUBLE PRECISION K4 DOUBLE PRECISION MBLEF DOUBLE PRECISION MBREF DOUBLE PRECISION MCJEF DOUBLE PRECISION NDOF INTEGER NGSTRF DOUBLE PRECISION TOP PORTION OF STRUCTURE FORCE MATRIX STORES VALUES OF [Fl] [k] [W2] BOTTOM PORTION OF STRUCTURE FORCE MATRIX INCREMENTAL STRUCTURE FORCE MATRIX USED IN SOLVING FOR STRUCTURE DISPLACEMENTS STRUCTURE STIFFNESS MATRIX USED IN SOLVING FOR STRUCTURE DISPLACEMENTS STRUCTURE DISPLACEMENT MATRIX INDEX MATRIX FOR A BLOCK ELEMENT INDEX MATRIX FOR A BRICK ELEMENT INDEX MATRIX FOR A COLLAR JOINT ELEMENT TOP LEFT HAND PORTION OF STRUCTURE STIFFNESS MATRIX TOP RIGHT HAND PORTION OF STRUCTURE STIFFNESS MATRIX STORES VALUES OF [K2] [W2] BOTTOM RIGHT HAND PORTION OF STRUCTURE STIFFNESS MATRIX MINUS 1.0 TIMES BLOCK ELEMENT FORCE MATRIX MINUS 1.0 TIMES BRICK ELEMENT FORCE MATRIX MINUS 1.0 TIMES COLLAR JOINT ELEMENT FORCE MATRIX NUMBER OF DEGREE OF FREEDOM OF FORCE STRUCTURE FORCE MATRIX USED IN SOLVING FOR STRUCTURE DISPLACEMENTS 222 Table A.2-continued MATRIX TYPE NK1 DOUBLE PRECISION NK2 DOUBLE PRECISION NK4 DOUBLE PRECISION NSTRF DOUBLE PRECISION NSTRK DOUBLE PRECISION NS TRW DOUBLE PRECISION SBLMT DOUBLE PRECISION SBLPD DOUBLE PRECISION SBRMT DOUBLE PRECISION SBRPD DOUBLE PRECISION SCJMT DOUBLE PRECISION SCJPD DOUBLE PRECISION SFINCR DOUBLE PRECISION SFINIT DOUBLE PRECISION SFMAX DOUBLE PRECISION STOMF DOUBLE PRECISION DESCRIPTION TOP LEFT HAND PORTION OF STRUCTURE STIFFNESS MATRIX TOP RIGHT HAND PORTION OF STRUCTURE STIFFNESS MATRIX BOTTOM RIGHT PORTION OF STRUCTURE STIFFNESS MATRIX STRUCTURE FORCE MATRIX USED IN SOLVING FOR STRUCTURE DISPLACEMENTS STRUCTURE STIFFNESS MATRIX USED IN SOLVING FOR STRUCTURE DISPLACEMENTS STRUCTURE DISPLACEMENT MATRIX STORES THE SLOPES OF THE BLOCK MOMENT THETA CURVES STORES THE SLOPES OF THE BLOCK AXIAL LOAD DELTA CURVE STORES THE SLOPES OF THE BRICK MOMENT THETA CURVES STORES THE SLOPES OF THE BRICK AXIAL LOAD DELTA CURVE STORES THE SLOPES OF THE COLLAR JOINT MOMENT THETA CURVE STORES THE SLOPES OF THE COLLAR JOINT AXIAL LOAD DELTA CURVE STORES THE STRUCTURE FORCE INCREMENT FOR EACH LOADED DEGREE OF FREEDOM STORES THE INITIAL STRUCTURE FORCE FOR EACH LOADED DEGREE OF FREEDOM STORES THE MAXIMUM STRUCTURE FORCE FOR EACH LOADED DEGREE OF FREEDOM STORES THE ABSOLUTE VALUE OF AVERAGE OF END MOMENTS FOR EACH ELEMENT ro r\j STOMST DOUBLE PRECISION STOVF DOUBLE PRECISION STOVST DOUBLE PRECISION STRF DOUBLE PRECISION STRK DOUBLE PRECISION STRW DOUBLE PRECISION TOTEIN INTEGER TOTEK DOUBLE PRECISION W1 DOUBLE PRECISION W2 DOUBLE PRECISION STORES THE ROTATIONAL STIFFNESS FACTOR FOR EACH ELEMENT STORES THE ABSOLUTE VALUE OF THE VERTICAL FORCE FOR EACH ELEMENT STORES THE VERTICAL STIFFNESS FACTOR FOR EACH ELEMENT STRUCTURE FORCE MATRIX STRUCTURE STIFFNESS MATRIX STRUCTURE DISPLACEMENT MATRIX STORES ALL OF THE ELEMENT INDEX MATRICES STORES ALL OF THE ELEMENT STIFFNESS MATRICES TOP HALF OF STRUCTURE DISPLACEMENT MATRIX BOTTOM HALF OF STRUCTURE DISPLACEMENT MATRIX SUBROUTINE DESCRIPTION ADD APPLYF BLESM BLPDMT BMULT ADDS MATRICES APPLIES FORCES ON STRUCTURE CONSTRUCTS BLOCK ELEMENT STIFFNESS MATRIX CALCULATES BLOCK AXIAL AND ROTATIONAL STIFFNESS FACTORS MULTIPLIES A REGULAR MATRIX BY A SYMMETRIC BANDED MATRIX BNSERT INSERTS A MATRIX INTO A SYMMETRIC BANDED MATRIX 224 Table A.2-continued SUBROUTINE DESCRIPTION BRESM CONSTRUCTS BRICK ELEMENT STIFFNESS MATRIX BRPDMT CALCULATES BRICK AXIAL AND ROTATIONAL STIFFNESS FACTORS CHKFAI CHECKS AN ELEMENT FOR FAILURE CHKTOL CHECKS TO SEE IF ALL THE VALUES IN [ERROR] ARE LESS THAN THE VALUE OF TOLER WHICH IS THE TOLERANCE CJESM CONSTRUCTS COLLAR JOINT ELEMENT STIFFNESS MATRIX CJPDMT CALCULATES COLLAR JOINT AXIAL AND ROTATIONAL STIFFNESS FACTORS COORD READS THE COORDINATES OF A CURVE CURVES READS THE COORDINATES OF UP TO TWENTY CURVES DISPLA CONVERTS THE STRUCTURE DISPLACEMENT MATRIX WITH PLATE DEGREES OF FREEDOM TO THE REGULAR DISPLACEMENT MATRIX EQUAL MAKES TWO MATRICES EQUAL BY COPYING THE VALUES OF THE FIRST MATRIX INTO THE SECOND MATRIX EXTRAK EXTRACTS A MATRIX OUT OF ANOTHER MATRIX FORCES READS AND STORES THE STRUCTURE FORCE APPLICATION INFORMATION GAUSS1 SOLVES A MATRIX EQUATION OF THE FORM [a] [b] = [c] USING GAUSS ELIMINATION WHERE [a] IS A SYMMETRIC BANDED MATRIX INDXBL CONSTRUCTS THE INDEX MATRIX FOR A BLOCK ELEMENT INDXBR CONSTRUCTS THE INDEX MATRIX FOR A BRICK ELEMENT 225 INDXCJ CONSTRUCTS THE INDEX MATRIX FOR A COLLAR JOINT ELEMENT INSERT INSERTS A MATRIX INTO A SECOND ONE MULT MULTIPLIES TWO MATRICES NULL SETS ALL VALUES OF A MATRIX EQUAL TO ZERO PLATEK TRANSFORMS A REGULAR STRUCTURE STIFFNESS MATRIX INTO A STRUCTURE STIFFNESS MATRIX THAT CONSIDERS PLATE DEGREES OF FREEDOM PRINTS A MATRIX AND IDENTIFIES THE ROWS AS DOF PULMAT PULLS A MATRIX OUT OF ANOTHER MATRIX PULROW PULLS A ROW OUT OF A MATRIX READ READS PROBLEM DESCRIPTION DATA AND PRINTS IT SMULT MULTIPLES A MATRIX BY A SCALAR STACON SOLVES A MATRIX EQUATION OF THE FORM [a] [b] = [c] USING GAUSS ELIMINATION AND STATIC CONDENSATION WHERE [a] IS A SYMMETRIC BANDED MATRIX STIFAC SELECTS AND STORES THE PROPER STIFFNESS FACTORS AND DETECTS ANY CHANGES IN THE STIFFNESS FACTORS FOR AN ELEMENT TITLE READS AND PRINTS A COMMENT CARD WRITE PRINTS A MATRIX AND IDENTIFIES THE DIMENSIONS AND ROWS WYTHE CALCULATES THE VERTICAL LOAD FOR EACH WYTHE AND PRINTS IT 226 227 A.4 Listing of Program and Subroutines The following is a listing of the program and subroutines. As previously noted, comments will be found throughout each to identify what is taking place. c c c c c c c c c c c c c c c c c c c c c c c c c c c *************************************************************** * * * FINITE ELEMENT MODEL * * * * FOR * * * * COMPOSITE MASONRY RAILS * * * * * DEVELOPED BY * * * * * * GEORGE X. BOULTON * * CIVIL ENGINEERING DEPARTMENT * * UNIVERSITY Of fLOSIDA * * SPRING 1984 * * * *************************************************************** PROGRAM AS PRESENTLY DIMENSIONED ILL HANDLE UP TO: 194 STRUCTURE DEGREES CE FREEDOM 117 ELEMENTS (MAXIMUM WALL HEIGHT OF 3 12 INCHES). TO CHANGE DIMENSIONS CHANGE DIMENSION STATEMENTS. DECLARE MAT REAI EE INTEGER INTEGER DOUBLE PEEC DOUBLE DOUBLE DCUBIE DOUBLE DCUBIE RICES AND VARIABLES REAL CE INTEGER AS NEEDED. IGHT TOTEIN,WHI,BR,CJ,BL,PRCUT,TESTIF,T ECONV,T RELEE THFAIL,TRSHOW,COUNT, IPRINI ISIGN MATRICES AND VARIABLES AS NEEDED. PRECISICN TOTEK,STRK,GSTEK,NSTRK,NGSTHK,BREK,CJFK PRECISION BLEK,SiaW,GSTBS,DELTA,NS1BW,ERES,CJEW PRECISION BLEW,STEF,DELTAE,GDEITF,ERROR,NSTRF,NGSTRE PRECISION SFINIT,SFINCR,SFMAX,FUIT,SHONE,EEEE,MUBEE PRECISION CJEF,MCJEF,BLEE,MBLEF,K1,K2,K4,NK1,NK2,NK4 228 DOUELE DOUBLE DOUELE DOUELE DOUBLE DOUBLE DOUELE DOUBLE DOUBLE DOUBLE C DIMENSION PRECISION PRECISION PRECISION PRECISION PRECISION PRECISION PRECISION PRECISION PRECISION PRECISION MATRICES. W1,W2,K2W2,F1,F2,F1MKW2,BEPCOR,BRDCGR S BRPD/CJPCOfi,CJDCOR ,SCJPD,BLPCCR,BLDCCR SBLPO,BRMCCB,BÂ£TCCR,SBREl,CJMCCR,CJTCOR SCJHT,BLHCOR, ELTCGR,SBLMT,BRMAXP,CJHAXP BLMAXP ,BRIDP,BEIDH,BLIDF,ELIDM,EfiPCV,ELPCV EBAEL,BR3EIL,COSHES, CJHOMS,BLAEL,BL3EIL BRHYLD,BL8YLD,ELASBR,KLASEL,AESHBE, ARSHEL LNHBR,LNHBL,LNVER,TOLÂ£R,SHEARL,SHEARE BRHOM,BRVERF,CJMCM,CJVEBF,BLMCK,BLVÂ£RF STOVF, STOMF,SIQVS1,STOMSI TCT EK (6,6 117) ,STRK( 194,9) GSTRK (194,9) NSTRK (192,10) NG STM (192,10) BK EK (6,6) ,CJEK (4,4) SEEK (6,6) ,STRW (19 4) GSTR8 (1 94) DELI AW (194) NSTBi (192) ,BREW(6) ,CJEW(4) ,BÂ£EW (6) ,STBF(194) DELTAE(194), GDELTF (194) ,ERRCE(194) NS1BF (192) NGSTRF (192) ,NDCF( 192) ,SFINIT(192) ,SFINCR (192) SFMAX (192) ,BRÂ£F (6) ,MBEEF (6) ,CJEF (4) ,MCJEF(4) BLEF (6) MBLEF (6) K 1 ( 100,9) K2 ( 1CC, 9) K4 ( 10 0,9) NK1 (100,10) ,NK2( 100,10) ,NK4 (100,10) ,81(100) 2 (100) ,K2W2 (100) ,F1 (100),F2 (10C),F1HKS2 (10 0) TOTEIN (117,6) IBR (6) ,ICJ(4) ,IBL (6) ,BEPCOR(2 0) BEDCCB (20) ,SBRPD (20) ,CJPCOR (20) ,CJDCCE (20) SCJPD (20) ,BLPCOR(20) ,BLICOR (20) ,SBLPD (20) BEMCOR (20,20) ,BRTCCR (20 ,20) ,CJMCCE(20) CJTCCB (20) SCJMT (20) BLMCOR (20,20) ,BI1C0R ( 2C,2 0) BRIDP (20) ,BRIDM (2 0) ,BBPCy(2 0) ,B1PCV(20) BLIDP(20),BLICH(20),STOVF(3,117),STOMF (3,117) STOVST (3, 117) ,STOMST (3, 117) ,ERWYLD(39) BLtYLD(39),SBRMT(20,20),SBLMT(20,20) C OPEN 6 IDENTIFY WHICH FILE CONTAINS THE DATA 6 WHICH FILE IS TO C STORE TEE OUTPUT. OPEN (U NIT= 5,FIIE= *SMURF.DAT,,STATUS=*OLD*) OPEN (UNIT=6,FILE= OUT.DAT',STATUS=* NEW*) C GO TO SUBROUTINE, BEAD THE WALL DESCRIPTION DATA Â£ PRINT IT. CALL READ (HHI LNVER,LNHBfi,LNHBL,NSTORY,NS DCF,NEIEH, DIMENSION DIMENSION DIMENSION DIMENSION DIMENSION DIMENSION DIMENSION DIMENSION DIMENSION DIMENSION DIMENSION DIMENSION DIME N5I0N DIMENSION DIMENSION DIMENSION DIMENSION DIMENSION DIMENSION 229 n tr. o u> n n n m $ ELASBR ,ARSHBR,ELASBL,ARSHBL) C NULL CUT MATRICES AS NEEDED. CAII NUIL (STRK,NSDOF, 9) CALL NULL (STR 4i ,N SDOF, 1) CAII NULL (SIRE,NS EOF,1) CALL NULL (BREK,6,6) CAII NULL (EREW,6,1) CALL NULL (BEEF,6,1) CAII NULL (ElÂ£K,b,6) CALL NULL (BLE h,6,1) CAII NII (ELE E,6, 1) CALL NULL (CJEW,4,1) CAII NUIL (C JEE 4 1) C BEAD Â£ PRINT BRICK ELEMENT P-DEL1A CURVE DESC E.IFTIC K. C ALI COCED (NEBFDE,BRDCOR,BE ECOS) WRITE (6, 1) 1 FCRM AI (1 11) C READ & PRINT BRICK ELEMENT M-THE1A CURVES DESCEIPTICN. CALI CURVES (NCEBFC,NBRMTP,B'PCV,BfiTCOH,BBMCOR) WRITE (6,2) FORKAT (* 11) SEAL S PRINT COLLAR JOINT ELEMENT P-DELTA CURVE DESCRIPTION. CALI CCCRD (NCJEDP,CJDCOR,CJECOR) READ & PRINT COLLAR JOINT ELEMENT M-THETA CUBVE DESCEIPTICN. CAII CCCRD (NCJ Ml Â£ CJT COR, C J ECCB) BEAD & PRINT BLOCK ELEMENT P-DEITA CUBVE DESCEIPTICN. CAII CCCRD (NELPBE,BLDCCR,BLPCCH) WHITE (6,3) FOB MAT (* 1 *) HEAD 6 PRINT BLCCK ELEMENT M-THEIA CURVES DESCEIPTICN. CAII CURVES (NGELEC,NBLMIP,BLPCV,BLTCOB,BLMCQR) WRITE (6,5) FCBMA1 ( 1 *) READ & PRINT BBICK ELEMENT p-M INTERACTION DIAGRAM DESCRIPTION. CAII CCCRD (N EBIDE BRIDM, BR IEF) C READ Â£ PRINT BLOCK ELEMENT P-M INTERACTION DIAGRAM DESCRIPTION. OZ CALL COORD (NBLIDP,BLIDM ,BLIDP) C BEAD Â£ FEINT MAXIMUM BEICK ELEMENT COMPRESSIVE LOAD CAPACITY. CALL TITLE HEAD (5,10) BEHAXP 10 FORMAT (D13. 6) BITE (6,20)ERMAXP 20 FOB BAT ( ',17X,D13.6) C BEAD 6 PRINT BAXIBUM COLLAR JOINT ELEMENT SHEAR LOAD CAPACITY. CALL TITLE REAL (5,30)CJMAXE 30 FORMAT(D13.6) WRITE (6 ,40) CJKAXP 40 EOBMAT(1 ,17X,D13.6) C READ 6 PRINT BAXIBUM ALLOWABLE BLOCK ELEMENT COMPRESSIVE LOAD. CALI TITLE REAE (5,50) BIMAXP 50 FCRMAT(D13.6) WRITE (6,60)ELBAXP 60 FOHMAT(* 17X ,D 13.6/) ^ C READ 6 PRINT FORCE APPLICATION INFORMATION. WRITE(6,65) 65 FCRKAT('I') CALL FORCES (EOF, NDQF,SPIRIT,5FINCÂ£,SFMAX,KSBCF) C READ 6 PRINT INSTRUCTIONS ON INTERMEDIATE PRINTOUTS. CALL TITLE READ (5,70)FfiGUI 70 FORMAT (11) IF (FECUT. EQ. 0) GO TO 110 WRITE (6,80) 80 FORMAT(0*,INTERMEDIAIS RESULTS WILL BE PRINTED FOR , 'STRUCTURE FORCE*) READ (5, SO) NGFN NFORCE, FMU1T 90 F0RMAT(2I3,D13.6) WRITE (6,100) FKULT, NFORCE 100 FORMAT(' ','VALUES WHICH ARE MULTIPLES CF ,1X,D 13.6, 1X , $ 'FCfi DCF NO. ,14) SHCfcF=SFINIT (NCFN) GC 1C 1JO 110 R8I1E(6,120) 120 FORMAT ('O','INTERMEDIATE RESULTS MILL KOI EE PRINTED') SHCHF=1.0D*2G C DEFINE EANGE OF BEICK ELEMENTS AND COLLAR JOINT ELEMENTS. 130 NEM1-NELEM-2 NEMC=NEIEM-1 KSEFÂ£lS=NSCOF-6 NSLFM1= NSDGE-3 NSDM2=NSDOF2 C DEFINE NUMBER OF DOF FOR EACH ELEMENT TYPE. Bfi=fc CJ=4 BL = E C DEFINE 1HE SIZE OF HALF THE BANDWIDTH (INCLUDING THE DIAGONAL) C FOR THE STRUCTURE STIFFNESS MATRIX. M1 = S M 1P 1=M 1 + 1 C DEFINE THE VARIABLES, NEEDED BY TEE SUBROUTINE THAT PERFORMS C STATIC CONDENSATION, WHICH DIVIDE AND Â£F] IKTC HALVES. NRICE=NSDCE/2 NRBGT=N SDCF-NBIOP NRII 1 = NETCE + 1 N BTM 1=N SDM2/2 NEBM1=NSDM2~NETM1 NR1ME 1=NHIM 1+1 C ENTER TEE MAIN EBCGEAM LOOP. 140 DO 7C0 ITER = 1, 1 COC0 CALL NULL (STRK,NSDOF, M1) C FOB EACH BRICK ELEMENT: C (THE VEBY EIRS1 TIME) C 1. CONSTRUCT THE INDEX MATRIX 6 STCEE IT. C 2. GET INITIAL STIFFNESS FACTORS 5 STORE THEM. C 3. CONSTRUCT THE ELEMENT STIFFNESS MATRIX. C 4. INSEET THE ELEMENT STIFFNESS MATRIX INTO THE STRUCTURE STIFFNESS MATRIX C C C (EVEEY CIHER TIBE) C 1. RECALL THE INDEX MATRIX. C 2. EXTBACT T BE ELEMENT DISPLACEMENT MATKIX FROM THE C SIBOCTUHE DISPLACEMENT MATEIX. C 3. BECALL THE ELEMENT STIFFNESS MATEIX. C 4. MULTIPLY THE ELEMENT STIFFNESS MATBIX EY THE ELEMENT C DISPLACEMENT MATBIX TO YIELD THE ELEMENT FOBCE MATBIX. C 5. PBINT THE ELEMENT FOBCES WHEN AEEROEEIAIE. C 6. POINT THE ELEMENT DISPLACEMENTS WHEN APPROPRIATE. C 7. CHECK FOB ELEMENT (AND THEREFORE, WALL) FAILURE USING C IKE P-M INTEBACT ION DIAGRAM. C 6. GET STIFFNESS FACTORS BASED CN THE ELEMENT FCBCES FBCM C THE P-DELTA S M-THETA CURVES S COMPARE THEM WITH THCSE C OF THE PREVIOUS CYCLE TO DETECT ANY CHANGES. C 9. CCNSTEUCT TEE ELEMENT STIFFNESS MATRIX. C 10. INSERI THE ELEMENT STIFFNESS MATRIX INTC THE C STRUCTURE SIIEENESS MATRIX. C 11. MULTIPLY THE ELEMENT FORCE MATRIX BY -1.0 & INSERT IT C INTO THE STRUCTURE FORCE MATRIX FOR CONVERGENCE CHECK. 142 CON1=0 DC 2 80 1=1,NEMI,3 IF (IIER.NE.1) GO TO 145 CALI INEXER (IBR,TOTEIN,I,NEMT,BE,NELEM) GO TO 2 5 C 145 CALL PULEOW (IBB TOT EIN, I, UB N ELEM) CALL NULL (BREW,BE,1) IF (I,EQ.N EMT)CALL EXTRAK (STEW,BEEN,2,3,NSDCF,BE,1BB) IF (I.EQ.NEMT) GO TO 170 ' IF(I. EQ. 1) CALL EXTRAK (SIRW,BE EH,2, 1,NSDGF,BR,1BO) IF (I N Â£. 1) CALL EXTRAK (SIRN,BREW ,2,2, NSDCF, BR,.IÂ£fi) 17 CALL NULL (EBEK,BR,BB) CALL EULHAT (BEEK,TOTEK,I,BB,NELEM) CALL MULT (EflEK,BREM,EEEF,BS,BR, 1) CGUNT=CCUNT* 1 i\) V>) 160 185 190 200 210 220 230 240 250 270 BRWXLD (COUNT) =BREF ( 1) II(TRCONV.EQ.1) 00 TO 275 IF (TRELPB. HE. 1) GOTO 190 WHITE (6,160)1 F CRM AT(10' ','FORCES FCH BDICK ELEMENT NUMBER,14) CALL PHINT (BEEF,BR) WBIIE (6,185) I FOBMAT('0','DISPLACEMENTS FCR BRICK ELEMENT NMIER, $ 14) CALL PRINT (BREW,BE) CALL CHKFAI (BREF,NBBIEP,BRIDM, BHIDE,BRMAXP,1,NSIAI) IF(NSTA1.EQ.0) GO TO 250 IF (HS1A1.EQ. 1) GO TO 220 WHITE (6,2C0)I FORMAT('0,'***WARNING*** BBICK ELEMENT N0.',I4,1X, $ 'IS IN AXIAL) WRITE (6,2 10) FORMAT ( ,14X,*TENSION, CHECK FCR GNUSUAL LCAE1NG *) STOP IF(THFAIL.Sg.4) GO TO 230 IEFAII= 1 IF (TPRINT.NE.2) GO TO 250 WRITE (6,240) I FORMAT( 0 , *** ELEMENT NO. ,14,1 X HAS FAILED ***) CALL BRPBMT (NBRPDP,NOBRPC,NBRMTP,BRECOR, BRPCOH,BRPCV , BRICOR,BEMCOR,BREF,SBRPD,SBRiiT,BEVERF,BRKCM,BIAEL,BR3EIL,ITEE) CALL STIF AC (BRVERF,BRMOM,STGVF,STOME,BRA El,ER3EIL, 3 STO VST,STCMST,TESTIF,ITER,1,1) CALL BSESH (BREK,EREF,TOTEK,ERAEL,BR3EIL,LNVER , $ ELASBB,ARSHDB,I,EB,KEIEM) IF (I Â£Q. NEMT) CALL BNSERT (STRK,BREK,J,NSDCF,M1,ER, $ IBE) IF (I, EQ. NEMT) GO TO 270 IF(I.EQ.1) CALL BNSERT (STBK,BREK,1,FSBCF,M1,BR,IDR) IF (I. HE. 1) CALL BNSERT (STRK, BR EK 2 NSDOF M 1 ,B H IB 8) IF(ITER.Ey.1) GO TO 280 275 CALL SMULT (- 1.0 D + 00 BEEF MBEEF BE, 1) IF(I.EQ.NEM1)CALL INSERT (ERROR,MBBEE,2,3,NSBGF,EH, $ IBB) IF(I.Â£Q.NÂ£BT) GO TO 280 IF (I.EC. 1)CALL INSERT (ERHOR,MBREF,2,1,NSDCF,BB,IBn) IF(INE. 1)CALL INSERT (ERBCR,BBREF,2,2,NSDCF,BB,IBR) 280 CONTINUE C FOR FACE COLLAR JOINT ELEMENT: C IT HE VERY FIES1 TIKE) C 1. CONSTRUCT THE INDEX MATRIX 8 STORE IT. C 2. GET INITIAL STIEFNESS FACTORS 8 STORE TEEM. C 3. CONSTRUCT THE ELEMENT STIFFNESS MATRIX. C 4. INSERT THE ELEMENT STIFFNESS MATRIX INTO THE C STRUCTURE STIFFNESS MATRIX. C C (EVERY OTHER TIME) C 1. RECALL THE INDEX MATRIX. C 2. EX1EACT TEE ELEMENT DISPLACEMENT MATRIX FROM THE C STRUCTURE DISPLACEMENT MATRIX. C 3. RECALL THE ELEMENT STIFFNESS MATRIX. C 4. MUITI FLY THE ELEMENT ST IFENESS MATRIX BY THE ELEMENT C DISPLACEMENT MATRIX TO YIELD THE ELEMENT FORCE MATRIX. C 5. PRINT THE ELEMENT FORCES WEEN APPROPRIATE. C 6. PRINT THE ELEMENT DISPLACEMENTS BHEN AEPIC PEI ATE. C 7. CEECK FCB ELEMENT (AND THEREFORE, KAIL) FAILURE. C fi. GET STIFFNESS FACTORS BASED CN THE Â£ IEME NT FORCES EBCM C TEE P-DELTA 6 B-THETA CURVES 8 COMPARE THEM KITH THOSE C 01 THE PREVIOUS CYCLE TO DETECT ANY CHANGES. C 9. CONSTRUCT TEE ELEMENT STIEENESS MATRIX. C 10. INSERT THE ELEMENT STIFFNESS MATRIX INTC THE C STRUCTURE STIEFNESS MATRIX. C 11. MULTIPLY THE ELEMENT FORCE MATRIX BY -1.0 6 INSERT IT C INTO THE STRUCTURE FORCE MATRIX FOR CONVERGENCE CHECK. DO 360 J=2,NEMO,3 IF (ITER. NE. 1) GO TO 290 CALL INDXCJ (ICJ,IOTEIN,J,NEMO,CJ,KEIEM) 290 320 321 322 323 325 330 340 350 360 GO TO 36C CAII FULSCW (ICJ,TOTEIE,J,CJ,NElEK) CALI NULL (CJÂ£W,CJ, 1) CALL EXTBAK (STBW ,CJEW ,2 ,2 NSDOF ,CJ ,ICJ) CAII NULL (CJEK, CJ CJ) CALL PULMAT (CJEK ,TOTEK, J ,C J ,NEIEM) CALI MULT (C J EK, CJ EW C JEF, C J,CJ 1) IF (TfiCGNV.EQ. 1) GO TO 375 IF (TBELEB. NE. 1) GO TO 330 BITE (6,320)J FGBMAT(*0/' ',* F0BCE5 FGB CCLLAB JOINT ELEMENT *, ' HUMI3EB ,14) CAII PHINT (CJEF,CJ) HITE (6,321)J FCBMAT ('0, 'SHEAR STHESSES FGB CCLLAB JOINT ', ELEMENT NUMBER',14) S HE AJB1= CJ EF (1) /19 2. 0 SHEAHB=CJEF (3)/192.0 WHITE (6,322) SHEABL vS FHMA1 ('O' 'BRICK FACE 1 ,2X, D 13. 6) WBITE (6,323)SHEABR FORMAT(' *,'BLOCK FACE ,2X,D13.6) WHITE(6,325)J FORMAT(0*,'DISPLACEME STS FOE CCLLAB JCIKT ELEMENT 'NUMBER', 14) CALL PRINT (CJEW,CJ) CALL CHKFAI (CJEF,NBHIÂ£P,BBIDM,BÂ£IDE,CJMAXP,2,NSTAT) IF(NSTAT.EQ.G) GO TO 36 IF (TBFAIl. Et4) GO TO 340 1RFAIL=1 If (Tf BINT. BE. 2) GO TO 360 WRITE(6,350)J FOBMAT ('0',* *** ELEMENT NO. ',14, 1X,* HAS FAILED ***') CALL CJPDMT (NCJPDP,NCJETE,CJDCCB,CJECCB,CJTCOE, CJMCCB,CJBF,SCJPD,SCJMT,CJVEBF,CJMOM,CJSHRS,CJMOMS ,ITEB) CALL ST IF AC (CJ 7EBF ,C J MOM ,STCVF ,STC IS C JSfi ES C J KCMS , $ STOVST,STCMST#TESTIF,I1EB,2,J) CALI CJESM (CJEK ,TOTEK,CJSHRS,CJMOMS,LNHÂ£E,LNHEL, $ J,CJ, NELEM) CALL BNSERT (STRK ,C JE K ,2 NSDCF M1 ,C 0 IC J) I F (IT Â£R. EQ. 1) GO TO 380 375 CALL SMULT (-1.0D+00 ,C JEF ,MC JEF ,C J, 1) CALL INSIST {EfROR,MCJEF,2,2,NSDCF,CJ,ICJ) 380 CONTINUE C FOE EACH ELCCK ELEMENT: C (THE VERY FIRST TIME) C 1, CONSTRUCT THE INDEX MATRIX S STORE IT. C 2. GE1 INITIAL STIFFNESS FACTORS 6 STORE THEM. C 3. CONSTRUCT THE ELEMENT STIFFNESS MATRIX. C U. INSERT THE ELEMENT STIFFNESS MATRIX INTO THE C STRUCTURE STIFFNESS MATRIX. C C (EVEEX ClfiEfi TIME) C 1. RECALL THE INDEX MATRIX. C 2. EXTRACT TEE ELEMENT DISPLACEMENT MATRIX FROM THE C STRUCTDR E DISPLACEMENT MATRIX. C 3. RECALL THE ELEMENT STIFFNESS MATRIX. C 4. MULTIPLY THE ELEMENT STIFFNESS MATRIX EY THE ELEMENT C DISPLACEMENT MATRIX TO YIEID THE ELEMENT FORCE MATRIX. C 5. PRINT THE ELEMENT FORCE MATRIX WHEN APPROPRIATE. C 6. PRINT THE ELEMENT DISPLACEMENTS WHEN APPROPRIATE. C 7. CHECK FCR ELEMENT (AND THEREFORE, WAIl) FAILURE USING C THE P-M INTERACTION DIAGRAM. C Â£. GET STIFFNESS FACTORS BASED CN THE ELEMENT FORCES FROM C TEE P-DELTA & H-1HETA CURVES 8 COMPARE 1HEM WITH THOSE C OF THE PREVIOUS CYCLE TO DETECT ANY CHANGES. C 9. CONSTRUCT THE ELEMENT STIFFNESS MATRIX. C 10. INSERT THE ELEMENT STIFFNESS MATRIX INTO THE C STRUCTURE STIFFNESS MATRIX. C 11. MULTIPLY THE ELEMENT FORCE MATRIX BY -1.0 8 INSERT IT C INTO THE STRUCTURE FORCE MATRIX FOR CONVERGENCE CHECK. CO UN I-0 390 $ 420 430 435 440 450 460 470 DC 530 K=3,NELEM, 3 IF (IIEB.NE.1) GO 10 390 CALL INEXBL (IBL,TOTEIN,K,EL,NELEM) GO TO 5CC CALL EULfiCW (IBL,TOTElN,K,BI,NELEM) CALL DULL (BLEW,BL,1) IF (K, EQ NELEM) CALL EXTRAE (STRW,BLÂ£W,2,3,NSDOF,BL , IBL) IF (K. EQ. NELEM) GOTO 420 IF (K.EQ.3) CALL EX1RAK (SIRW,BLEW,2,1,NSDCF,EL,I EL) IF(K. BÂ£.3) CALL EXTRAK (STBW,BLEW,2,2,NSDCF,BL,IBL) CALI NULL (BLEK,BL,BL) CAII 10IB AT (BLEK,TOTEK,K,BL,NELEM) CALL BOLT (BLEK,BLEW,BLEE,BL,BL,1) CCUNT-CCUNT+1 BLWLD(COUNT)=BLEF(1) IF (TBCONV. Â£C> 1) GO TO 525 IF (IRELPR.NE.1) GO TO 440 WRITE (6,430) K FORMAI(* 01/' 'FORCES FCR BLOCK ELEMENT NUMEEE',14) CALI PRINT (BLEF,EL) WRITE (6,4 35)K FORMAT (*0', DISPLACEMENTS FOR BLOCK ELEMENT NUMBER*, 15) CALL ERINT (BLEW,EL) CALL CUKFAI (BLEF NBLID E BLIDM BIIE Â£ ELM A XP 3, NST AT) IF (NSTAT.EQ.O) GO TO SCO IF (NSTAI,EQ.1) GO TO 470 WRITE (6,450) K FORMAT ( 0 1 WARNING*4'* BICCK ELEMENT SC. ',14, IX, 'IS IN AXIAL*) WRITE(6,460) FORMAT (' 14X, 'TENSION, CHECK FOR UNUSUAL LOADING') STOP IF (TRFAIL. IQ, 4) GO TO 480 TRFAIL=1 480 IF (TPfilNT. NE. 2) GO TO 500 WRITE(6,490)K 490 FORMAT ('O',' *** ELEMENT NO. ,14, 1X, MS FAILED ***) 500 CALL BLPDMT (NBLPDP, NCELEC, RELMTE HLDCC.B, ELPCOJ8 ELPCV, $ BI1CCR,BLMCCÂ£, ELEF, SBLPC, SELMT,BLVEBF ,ELMCM, BLAEL ,BL3EIL ,ITER) CALL STIFAC (BLVEBF,BIMGM,STCVF,STCEE,ElAEI,EL3EIL, $ STOVST,STGMSI,TEST IF,ITER,3,K) CALL BLESM (BLEK,BLEF,10TEK,BLAEL,B13EIL,LNVEB, ELASBL,AES EEL,K,EL,NELEM) IF(K.EQ.NELEM) CALL BNS2RT (STBK,BIEK,3,NSDCF,M1,EL, 1BL) IF (K.EQ.NELEM) GO TO 520 IF (K.EQ.3) CALL ENSEBT (STRK,BLÂ£K,1,NSDOF,M1,BL,IBL) IF (K. NE 3) CALL BNSEBT (STBK ,BLEK 2 BSD CF M1 BL IBL) IF (ITEB.EC. 1) GO TO 530 CALL SMULT {-1.0D + 00,BIEF,MBIEF ,EL,1) IF (K. EQ. NELEM) CALL INSERT ( ERROR, M ELEF 2,3, NSDCF ,BL , IBL) IF (K. EQ, NELEM) GO TO 530 IF (K. EQ. 3) CALL INSEBT (EBRCB,MRIEF ,2, 1 NSDCF, El, IBL) IF(K. NE.3) CALL INSEBT IF (lEELPfi. NÂ£. 1) GO TO 539 CALL WYTHE (BERYLD,BLYLD,NSTC&Y) $ $ 520 525 $ C c c c c c c c c c c c FINALLY : (TEE VEBY FIBSI TIME) 1, CCNSIBUCT TEE STRUCTURE FOFCE MATRIX. 2. SOLVE THE EQUATION Â£K]*Â£W]=Â£F] FCB Â£N], WHERE Â£K] IS THE STBCTUBE STIFFNESS MATRIX, Â£H] IS TEE STBUCTUBE DISPLACEMENT MATRIX, AND Â£F] IS THE STBUCTUBE FCECE MATRIX. USE A GAUSS ELIMINATION IYPE OF SOLUTION TECHNIQUE WHICH TAKES ADVANTAGE CP THE E ANDEENESS AND SYMMETRY CF THE STRUCTURE STIFFNESS MATRIX 'AND ALSO USES STATIC CONDENSATION. . SET Â£ F]=Â£ EBBOfi ] SO IT CAN BE USED TO INSURE CONVERGENCE. 3 c c c c c c c c c c c c c c c c c c c c c 539 540 545 550 (EVERY OTHER TIME) 1. CHECK THE SOLUTION FOB CONVERGENCE 5 ITiBATE IF HEQUIRED. 2. INCREMENT THE STRUCTURE LOADS. 3. SOLVE [K3*Â£WJ=Â£F] FOR Â£W] THE STRUCTURE EISELACEMENTS, AS DESCRIBED AECVI. 4. IE THE WALL IS AT THE HIGHEST LCADING LEVEL DESIRED, IDENTIFY THIS AND PRINT THE WALL LOADS & DISPLACEMENTS, THE LATERAL HALL DEFLECTION VERSUS HEIGHT, THE VERTICAL WAIL DEFLECTION VERSUS HEIGHT, TEE ELEMENT FORCES AND DISPLACEMENTS, AND THE WALI HYTI1E VERTICAL LOAD VERSUS HEIGHT. 5. IE AN INTERMEDIATE PRINTOUT WAS REQUESTED FOE THF CURRENT LEVEL OF WAIL LOADING, IDENTIFY THIS AND PRINT THE INFORMATION DESCRIBED ABOVE. 6. IF THE WALL HAS FAILED, IDENTIFY THIS AND PRINT THE WALL FAILURE LOADS Â£ DISPLACEMENTS, AS WEIL AS THF OTHER INFORMATION DESCRIBED ABOVE. 1. UNLESS THE WAIL HAS FAILED CE THE LOADS REACHED THE HIGHEST IEVEL DESIRED, CONTINUE INCREASING THE LOADS ONE INCREMENT AT A TIME AND SOLVING FOR EIE E E NT ECfiCES AND DISPLACEMENTS UNTIL EITHER OF THESE HAPPEN. IF (ITER.NE. 1) GO TO 540 CALL APPLY F (NSIRF, NS DM2 NC F NDCF, SE'IN 11, SPI NCR ,S E KAX , I IT ER,KE Y1) GO TO 690 TRELPR=C IF (TEFAIL, NE.4) GO TO 550 WRITE(6,545) FORMAT i'OV'O1 '*** END OF FINITE ELEMENT ANALYSIS ***) STOP TO IE E= 1. 0D-03 TBCONV=C CALL CBKTCI (ERROR,N3D0F,TOLER,KEY) IF (KEY.EQ.C) GO TO 560 TRCC NV=1 CALL EQUAL (DEITAF,ERROR,NSDGF,1) 240 CALL EQUAL (GDELLF, DELTAF NSDGF 1) CALL EQUAL (GELRK,STRK,NSIGÂ£,M1) CALL NULL (DELL AW, NS DCF, 1) CALL SIACON (DELTA W,GSTRK ,GDÂ£LTF,K4,F2,W2,K2,K2W2,F1, $ F 1 MKW 2,K1,W1,H1,NSD0F,NGIP1,NRBCI,NR'ICE) CALL ADD (DELLA W,STE K,GSIEN,NSDCF,1) CALL EQUAL (STEW, GSTR'W, NS EOF 1) CALL ADD (DELTAF,STRF,ERRC5 NSDCF 1) NCC NV= NCC N V + 1 IF(NCONV.L1.1) GO TO 142 SUIT 1 (6,555) 555 FORMAL ('0','*-** WARNING*** SOLUTION WLLL NCT CCNVEEGE, ', $ 'CHANGE',/* ',14X, EITHER TOLERANCE OR SLR UCTUfi E LOADS') SLOP 560 KCC N V=0 IF (1ESTIF.EQ.1) GO TC 69 IF (I EF AIL. EQ. 0 ) GO TO 595 BfiITE(6,59C) 590 FORMAT ( 1 ,66 ( = ')/ ',14X,'WALL FAILURE LOADS ', $ 'AND DISL PLACE KENTS / ',66 ( = *)) LERI NT-2 1EFAIL= 4 GC TC 640 595 IF (LRSHOW.EQ.O) GO TC 600 GC TO 620 600 IF(NSTRF(UFOBCE)* NE.SHOWF) GO TC 620 IRSHCW=1 HITE (6,61C) 610 FCEMAT ( 1',66 { = )/ ',12X, 1 'INTERMEDIALE ALL LOADS AND DISPLACEMENTS'/* ',66 (' = *)) SHCWF=SHCWF+FMULT GC LO 640 620 CALL AEPLYF (NSTRF,NS DM2,NOF,NDCF,SFINI1,SFINCR,SFMAX, $ TIER ,KE 1) I RSHCj=0 GC LO 6SC 625 630 631 6 40 645 649 650 651 652 653 654 655 656 660 WRITE (6,63 C) FCBI$A1(*1*,66 (' = ')/' ',19X,'WAIL LOADS AND , 'DISPLACEMENTS ) WRITE (6,631) ECRMAT {* *,18X,'(LOADS AT TEAK VALUES DESIRED]'/* '# 66('=)) IREA1L=4 WRITE(6,645) FGBMAE (*0 /' 1X,EXTERNAL WALL LOADS') CALL PRINT (NSliiF ,NSDM2) W BITE (6,649) FORMAT(* 0'/' ','EQUIVALENT WALL LOADS') CAIL PRINT (STB F, NS DOF) WRITE (6,65C) FORMAT (0/' *, 2X,'W ALL DISPLACEMENTS') CALL PRINT (STS W,NS DOF) BEITE (6,651) ECRMAT ('1* ,66 ( = )/* *, 14X,' LATERAL WAII ', DEFLECTION VERSUS EEIGHT */ ',66(' = *)) kfiITE (6,652) FORMAT (' 0'/' ',3X,HEIGHT*,6X,'IATERAI DEFLECTION') WRITE (6,653) FORMAT(* C,5X,'0.0',10X,* C.GOOOOOD+OO ) EFIGHT=0,0 EC 655 L=3,NSDFMS,5 HEIGHT= BEIGH1+8.0 WRITE(6,654)HEIGHT,STBS (I) FORMAT (* ',3X,F5. 1,9X,D 13.6) CONTINUE EEIG HT= HEIGH1 + 8.0 WHITE (6,656)HEIGHT ECRMAT (' ,3X,E5. 1, 10X,* 0.OCOCCCD+00) KB ITE (6,66 C) FORMAT (1',66 ( = )/ ,14X,'VERTICAI KAIL ', DEFLECTION VERSUS HEIGHT'/* ',66 (' = )) WRITE (6,661) 242 661 662 664 665 66B 690 695 700 FORMAT ( 0 / ,2X, HEIGHT ,6X, 'BRICK WYTHE DEFLECT! CN' , $ 6X,'BLOCK WYTHE DEFLECTION) KBITE(6,662) ECOHAT (*G* ,4X,*0,0*,12X, *0.CCOOOOD + OO, 16X, $ 0.CCOOOOD+OO) EEIGIiT=0, 0 CO 665 L=1,NSDFMT,5 LEL=L+1 HEIGHT=flSIGHT+8. 0 BITE(6,664) BEIGHT,STB(L) ,SlfiW (LEL) FORMAT(* ,2X,F5.1, 11 X ,D 13.6,15 X D 13.6) CC MINUE BITE (6, 668) FORMAT ( 1 ,66 (' = )/ 17X, $ 'ELEMENT FORCES AND DISPLACEMENTS'/* ,66( = )) T EI F E= 1 GC TO 695 CALI ECUAL (NGSXRF, NSTRF, NSDM2, 1) CALL PLATER (STEK,NSTRK,M1,M1P1,NSDCF,NSDM2,1NHER,INHBL) CA LL Â£UAL (NGSTRK, NSTRK NSEM2 N IP 1) CALL NOLL (NSTR ts,NSDM2,1) TÂ£STIF=0 CALL STACON (NSTEW, NGSTRK,NGSTEF NK4 ,F2, H 2 NK2, K2W 2, F 1, $ F1MKW2,NK1,W 1,MIP 1,NSDM2,NRTMP1,NRBM1,NRTM1) CALL DISPLA (STEW,NSIRH,NSDCF,NSDH2 ,iNHER,INfiEL) CALL BML1 (ST BK, STEW STfif, NSDCF, M1) IE (KEX1.EU.1) GC TO 625 CALL ECOAL ( ERROR,STEF,NS EOF,1) CONTINUE ENE ro -P* o n o -* non on S UBECUTINE MULL (A,N,M) C SUBROUTINE NULL HILL CREATE A MULL OB ZERO MAT RIX Â£ A ]. DISENSION A(N,M) DOUELE PRECISION A DC 10 1=1,N DC 1C J=1,M 10 A(I,J)=0,0 RETURN EKE C SUBROUTINE EQUAL (A,B,N,M) SUBROUTINE EQUAL HILL CREATE AN KXS MATRIX [A] THAT IS EQUAL TC MATRIX [BJ. DIMENSION A (N,H) ,B (N,M) DOUELE PRECISION A, B DO 1C 1=1,N EC 10 J=1,K 0 A (I J) =B(I,J) RETURN ENE SUBROUTINE ADD (A,B,C,N,M) SUBROUTINE ADD WILL ADD AN NXM MATRIX [B ] TC AN NXM MATRIX [A] TO YIELD AN NXM MATRIX Â£C]. DCUEIE PRECISION A,B,C DIMENSION A(U,M) ,B(N,M) ,C(N,M) DC 10 1=1, N EC 10 J=1,H 0 C (I,J) = A (I, J)+Â£ (I, J) RETURN EKE SUE ROUTINE MUIT (A, B,C,N 1, N2, N3) SUBROUTINE MULT WILL MULTIPLY AN N1XK2 MATEIX [A] TIMES AN K2XN3 MATRIX Â£BJ TO 0ETA IN AN N1XN3 MATRIX Â£C]. 244 DCUEIE PRECISION A,B,C DIMENSION A (N1,N2) ,B (N2 ,N3) #C (N1 ,K3) DC 10 1=1,N1 EC 10 J= 1, N 3 C <1, J)=0,0 DC 10 K=1,N2 10 C(I,J)=C(1,J) + A(I,KJ*B(K,J) RETURN END C SUBROUTINE SMUI1 (A,B,C,N,M) C SUBROUTINE SMDLT WILL MU1TIPIY A SCAIAE A TIBES AN tiXM C MATRIX Â£Ej TO OBTAIN AN NX H MATRIX C. DOUBLE PRECISION A,B,C DIMENSION E (N,M) ,C (N,H) DO 1CC 1=1,N EC 100 J=1,H 100 C(1,J)=A*B(I,J) RETURN ENE C SUBROUTINE BMUIT (A,B,C,N,M1) C SUBROUTINE EMULT RILL MULTIPLY A SYMMETRIC BANDED C MATRIX EY A VECTOR TO OBTAIN ANCTBER VECTCE. IT WILL C I ERF CRM Â£ A ]*Â£ E ]=Â£C ], Â£A] IS AN NXM 1 MATRIX WHICH C CCNTAINS THE UPPER TRIANGULAR PORTION CF A SYMMETRIC C NX N MATRIX THAT HAS A BANDNIDIfc OF (2*M1)-1. Â£B] IS AN C NX 1 MATRIX. Â£ C] IS AN NX1 MATRIX KUCSE VALUES ARE THE C PRODUCT Cf Â£ A ]*Â£ B J. DOUBLE PRECISION A,B,C DIMENSION A (N,B1) ,U(N) C(N) N E Â£ G = 1 ICC 1= M1 N5T0P = N-MH-1 DC 10 1=1, N C (I) =0.CDfCC ro VJI nnn 1M 1 = 1-1 IF (1 .El* 1) GC 1C 35 IF (I.GT. HI) GO TO 7 0 NC = I DC 30 L= 1, IM 1 C{I) =C (I) +A |L,NC)*B (I) NC=NC-1 30 CCN1INUE IF {I. GI NS1CE) GO TO 50 35 EO 4 C J=1,H1 L=J+IH1 C (I) =C (I) +A (I,J) *B (L) 40 CCNTINUE GC TO 1C C 50 ICCI=ICCI-1 55 LO 60 J=1,LC01 1=J+IE1 C (I) =C(I) +A (I,J) *B (L) 60 CCNTINUE GC TG 10 C 70 Nt EG= NE EG+ 1 NC=H 1 DC 80 L=NBEG/IN1 C(I) =C (I) +A (.L,NC) *B (L) NC= NC -1 80 CONTINUE IF (I. GT. NSTOE) GO TO 50 GC TO Â£5 100 CONTINUE BE1UEN EKE C SBFCUTINE INSEBT (A,B,KEY,TYPE,N,M, INDEX) SUBBOU1INE INSEBT BILL INSEBT A MA1BIX [Â£] INTC A LAfiGEfi HAIBIX [A]. TEE VALUES CF Â£B] ABE ADDED 10 1UE VALUES OF [Aj. IF KÂ£Y=1, THEN [A] IS A K NXN HATBIX AND non u> to cd a. C Â£B] IS AH axil MATRIX, IF KEY=2, THEM Â£ A ] IS AN H X1 C MATRIX AND [B] IS AN MX1 NATH IX. THE VECTOR [INDEX] C HAS a EIEKEKT3 WHICH GIVI THE POSITIONS OF Â£B] IN [A]. C IF TYPE=1, THEN THE FIES1 3 NUMBERS IK THE [INDEX] C MATRIX ABE MOT USED. IF TYPE=2, THEN AIL THE NUMBERS C IN THE [INDEX] MATRIX ARE USED. IE TYEE=3, THEN THE C FIFTH NDHEEE IN THE [INDEX] MATRIX IS NOT USED. INTEGER TYPE DCUEIE PRECISION A,Q DIMENSION A (N,N) ,B (M, H) #INDEX (M) 1=1 IF (TYPE.EQ. 1) 1 = 4 DC 3 I = L,H II=INDEX (I) IE (I. EC. 5) GC TO 1 GC TO 4 1 IF (TYPE.EQ.3) GC TO 30 4 IE (KEY.EC.2) GO TO 20 DC 10 0=1, K JJ=INDSX(J) IF (J.EC.5) GC TO 6 GC TO 8 IE (TYPE. EC. 3) GO TO 10 A(II,JJ)=A(II,JJ)+B (I J) 0 CONTINUE GC TO 3 G 0 JJ= 1 J=1 A (1I,JJ)=A (II,00)+ E (I,J) 0 CONTINUE RETURN END SUBROUTINE BNSEflT (C,B,TYPE,ft,Ml,fi,INDEX) SUBROUTINE BMSEBT WILL INSERT THE UPPER TRIANGULAR PORTION OF AN tiXH SYMMETRIC MATRIX [Â£] INTO A MATRIX 247 C Â£C]. [C] CONTAINS THE UPEEB TRIANGULAI PORTION OF AN C NXN NATHI X TH 21 HAS A BANDWIDTH OF (2*H1)-1. THE ACTUAL C NXN MATRIX IS NOT STORED. If IXPE=1, THE FIRST 3 C NUMBERS IN THE [INDEXj MATBIX ABE NOT USED. IF TYFE=2, C ALL THE NUMBEBS IN THE Â£ INDEX ] MATBIX ABE USED. IF C IÂ¥PE=3, THEN THE FIFTH NUMBER IN THE [INDEX] MATBIX IS C SCI USEE. INTEGER TYPE DIMENSION C (N,M1) B (M, M) INDEX (M) DOUELE PRECISION C,B 1=1 IF (TYPE.EQ. 1) L = 4 DO 20 1=1,M II = INDEX (I) IF (I. EC. 5) GC TO 1 GO TO 4 1 IE (TYPE.EQ.3) GO TO 2 4 DO 10 J=L#M *3 J=INDEX (J) IF (J.EQ.5) GO TO 6 GC TO 8 6 IF(1YPE.EQ.3) GO TO 10 8 IF(JJ.LI.II) GO TO 10 JJJ=JJ-II+1 C (II,JJJ)=C (II,JJJ) +B(I,J) 10 CONTINUE 20 CONTINUE fiElUEN END C SUBECUTINE EXTBAK (A,B,KEY,TYPE,N,H,INDEX) C SUBBOUTINE EXTBAK HILL PICK UP A MATRIX [B] CUT OF A C IAEGEE MATBIX Â£ A ]. THE VALUES OF [ A ] ABE NOT AFFECTED. C THE OLD VALUES OF [B], If ANY, ABE BEELACEE EY THE C VALUES FOUND IN Â£ AJ. IF KEÂ¥=1, THEN Â£A] IS AN NXN C MATBIX AND Â£B] IS AN MXM MATBIX. IF KEY=2, THEN [A ] IS n n rÂ¡ n uÂ¡ c o1' C AH NX1 MATRIX AND [] IS AN KX1 MATRIX. IF 1YPE=1, THEN C THE FIESI 3 NUMBERS IN I EE [INDEX] MATRIX A EE NOT USED. C IF TÂ¥PE=2, THEN ALL THE NUMBERS IN TEE INDEX MATRIX ARE C USED. If TÂ¥PÂ£=3, THEN THE FIFTH NUMEEB IN THE INDEX C MATRIX IS NCT USED. INTEGER TYPE DOUBIE PRECISIC N A B DIMENSION A (N ,N) ,B (M,M) ,INDEX (M) L= 1 IF (TYPE.EQ. 1) L = 4 DC 30 I=L,M II=INDEX(I) IF (I. EC. 5) GC TO 1 GO TO 4 1 IF (T YPE. EC. 3) GO TO 30 4 IF (KEY.EQ. 2) GO TO 20 EC 10 0=1,K 00=INDEX (0) IF (J. EQ. 5) GO TO 6 GO TO 8 IF (TYEE. EQ-3) GO TO 10 B (I, J) =A (11,00) CONTINUE GC TO 30 JJ=1 0= 1 B (I,0) = A (11,00) CONTINUE RETUBN END ru *Â£> SUBROUTINE PULRGW (A,B,I,T,N) SUBROUTINE EULROW WILL PULL THE FIRST T NUMBERS IN ROW I OUT OF AN NX6 MATRIX [Bj AND STORE IDEM IK A TX1 MATRIX [Aj. INTEGER A,B,T n n n I)If]ENSIGN B (N f 6) A (I) DC 1C J = 1,T A |J)=B(I,J) 10 CONTINUE RETURN ENE C SUBROUTINE PUL-MAT (A,B,I,T,N) SUBROUT IN E PUL WAT WILL PULL A TXT MATRIX OUT OF A 6XXN MATRIX Â£ B] AND STORE THE VALUES IN A TXT MATRIX Â£ A]. *** WARNING *** T MUST EE LESS THAN GE EQUAL TO 6. INTEGER I BCUEIE PRECISION A B DIMENSION 8(6,6,]]) ,A (T, I) DC 20 K=1,I EC 1C L= 1, T A (K, L) = B (K,L,I) 10 CONTINUE 20 CONTINUE RETURN EKE C SUBROUTINE GAUSS 1 (X,C,B,M 1,N,KEY) C SUBROUTINE GAUSS 1 SOLVES A MATRIX EQUATION OF THE EORE C Â£A ]*Â£X ]=Â£B ], WHERE Â£Aj = AN NXN MATRIX WHICH HAS A BAND C WIDTH OF (2*M1)-1. THE SOLUTION IS A STANDARD GAUSS C ELIMINATION FOR A SYMMETRIC BANDED MATRIX. Â£X] IS THE C MATRIX THAT STORES THE SOLUTION, Â£C] IS THE MATRIX USEE C TO STORE THE SYMMETRIC NON-ZERO VALUES IN THE Â£A] C MATRIX, Â£ B] IS THE MATRIX THAT STORES THE NUMBERS C THAT = Â£A]*Â£X], Ml EQUALS HALF THE BANDWIDTH C (INCLUDING THE DIAGONAL)-} AND N IS THE NUKE EE OF C EQUATIONS. KEY = 1 IS EOR A REGULAR P20ELEM AND KE Y = 2 IS C FOR MULTIPLE LOAD GAUSS USE WHEN Â£ A ] MATRIX SAME AS C IN IAST CALL TO GAUSSl. DOUBLE PRECISION X,C,B,COEFF 250 DIMENSION X (N) ,C(N,H1) ,B (N) C FCfifcAED ELICIKATICN. fl = H 1-1 Sfl 1 = N-1 DC 98 J= 1,11 HI JK=J+M IF (Jti.GT. N) JH=N df 1=J + 1 K=M + 1 IF (tl+J.GT.N) tM = N-J+ 1 DC 97 K=JP 1 f JM KK=K-JP1+2 CCEEI=-C (J,KK)/C (J, 1) fl (K) = B (K) +CGEFF+B (J) ME=Kfi-1 IF(KEi.EQ.2)GO TO 97 DC 96 1=1, MM II=I+K-J C (K, I)=C (K, I) +COEFF *C ( J, 11) 96 CONTINUE 97 CCNTINUE 98 CONTINUE C BACK SUBSTITUTION* X(N)=B(N)/C(N,1) N K= 0 DO 190 KK= 2#N K= N-KK +1 IF (KK.GT.M)GC TO 150 KK=NM+ 1 GC TO 16C 150 KF=K 160 X(K)=B(K) h 1=1 N 2=K DC 170 11=1,KM N 1=N 1 + 1 r\> 170 N 2=N2 + 1 X( JR) =X(K)-C(K,N1)*X(N2) X(K)=X(K)/C(K,1) 190 CON1INUE RETURN ENI C S UBRUT INE STACCN (X ,C B, K4 ,F2 W2 #K2 K2H2 ,11 F 1 MKW 2, K 1, H 1, $ M 1, N NRTP 1, N RBC1, N filOP) C SUBROUTINE STACON Bill SOLVE A MAT BIX EQUATION OF THE C FORM Â£ A ]*Â£X]=Â£ B], WHERE Â£A ] = AN NXN MATRIX WHICH HAS A C BANDWIDTH OF (2*M1)-1. THE SOLUTION USES STATIC C CONDENSATION AND STANDAR! GAUSS ELIMINATION FO A C SYMMETRIC EANDED MATRIX. Â£X] IS THE MATRIX THAT STORES C THE SOLUTION, Â£C] IS THE MATRIX USED TO STORE THE C SYMMETRIC NGN-ZERO VALUES IN THE Â£A] MATRIX, Â£E] IS THE C MATRIX THAT STORES THE NUMBERS THAT = [A]*[X], Ml C EQUALS HALF THE BANDWIDTH (INCLUDING THE DIAGONAL), AN! C N IS THE NUMBER OF EQUATIONS. DOUBLE PRECISION X,C,B,COEE\F ,K4,F2,W2 ,K2, R2W2 f 1, F 1MKW 2, K 1 DCUEIE EBECISIC N W 1 DIMENSION X (N) ,C (N,M1) ,B (N) K4 ( N R B OT Mi 1) f 2 ( NRECT) DIMENSION W2 (NREOI) FI (NRTOE) 1MKW2 ( NRTOE) K 1 (N STOP, Ml) DIMENSION W 1 ( NRTO!) K2 ( NRIGE ,M 1) ,K2W2 (NRTCE) C PERFORM NRTCE FORWARD ELIMINATIONS ON Â£C], M = M 1- 1 DC S8 J=1,NRTCE JM=J+M IF (JM.GT. N) JM=N J E1=J + 1 t M=M+1 IF (MS+J.GT.N) HM=N-J+ 1 EC 97 K=JE1,JM KK=K-JP 1 + 2 COEFF=-C (J,KK)/C (J, 1) B (K) =B (K) +CCÂ£FF*B (J) 252 DC 96 1=1, II=I+K-J C (K,I)=C (K,I) + COEE E*C(J, II) 96 CONTINUE 97 CONTINUE 98 CONTINUE C CCNSTEUCT [K4' ] EEC Â£C]. K=C DC 120 1= NUT P1, N K = K+ 1 DC 110 J=1,M1 K4 (K, J) =C (I, J) 110 CONTINUE 120 CO N1INUE C CCNSIEUCl Â£ JE2 j FECH Â£ E ]. D= 0 DO 140 1= NETP 1, N L=L + 1 2(I) = E(I) 140 CONTINUE C SOLVE Â£ K4 ]Â£ W2 ]=Â£ F2 ] FOE Â£W2J USING GAUSS1 CALL GAUSS1 (W2 ,K4 ,F'2, M 1 NEBC1 1) C CCNSIEUCl Â£ K2' ] Â£C Â£C]. CALL NULL (K2,NRTCP,H1) DC 160 NE=1,N ETCP I=NK1P1-Nfi J=0 N EP1=NR+ 1 IF (NEP1.GI. Ml) GO TO 165 DC 150 L=NBP 1,M1 J=J+1 IF (J .Gl. NBBC1) GO TO 160 K2 (I,)=C (I, L) 150 CONTINUE 160 CONTINUE f\3 OI OI C H0I1IJ5II [ K2* ][W2 j=Â£K2H2]. 165 CALI NULL (K2N2 ,NETOP, 1) NSTAB^NMOP-MHH IF (NSIAfil LI 1) MSTABT= 1 NCCL=NBBG1 IF(NEBOT.GT.Hi)NCCL=M1 DG 1Â£0 I=iiÂ£'IAEI#NBIOP K2W2 (I) =0 0 EC 170 K= 1,BCGI K22 (I) = K2N2 (I) +K2 (I, K) *H2 (K) 170 CC MINUE 180 CONTINUE C CCNSTBUC1 [F1] FBCK [FJ. DO 190 1= 1 NBTCF f 1(I)=B(I) 190 CONTINUE C PEBFOBH [FT J-[ K2 ]Â£ W2 ]=[ F1 HKW2 ] . CALI NUIL (F 1 HKW2, NBTOP, 1) NSIABT-NBIGP-HUI IF (NSTABT.LT. 1) NSTABT= 1 DO 2Cti I=NSTAE1,NB1GP f 1MKW2 (I) = F 1 (I) -K2M2 (I) 200 COMINUE IF (NSTAfiT.fig, 1) GO 10 2 15 NS1M 1=11 S1AB1- 1 DC 210 J=l,K5Tfl1 F 1HKH2(J)=F1 (J) 210 CCNIINUE C CCNSTBUCI [KT] FBGH [C]. 215 NCI N1=NBTCE IF(N1.L1.NB10P)NCCLK1=K1 NSTAFl-NBTCE-Kl+1 IF(NSTAHT.LI.1)NSTABT=1 DO 230 1=1,ESIAEl EC 220 J=1,NCOIK1 K1 (I,J)=C(I,J) 254 220 CONTINUE 230 CONTINUE N S T11 = N ST A BT + 1 NCCLH1 = NCOLK1 DC 250 1= NS1F 1,N B1GP NCCL M 1 = NCQLM1 1 EC 240 J=1,NCOIfl1 K1(I,J)=C(I,J) 240 CONTINUE 250 CONTINUE C SOLVE Â£ K 1 ]Â£ M1 ]=Â£ E 1 1 J-Â£ K2* ]Â£ W2 ] FOB Â£ 'ill ] BY BACK SUBSTITUTION C (SINCE FOHWABD ELIMINATION PORTION OF GAUSS 1 A IDEADY DCNE ON C Â£ K 1 ] IAET CF Â£ C j) , 9 1 (NBTOP) = F1MKJi2 (NBTOP)/K1 (NBTCE, 1 ) EB=C DO 290 KK=2,NBTOP B=NB1CP-KK+1 IF (KK.GT.M)GO TO 260 NM=NK+1 GO TO 27C 260 KE=M 270 1 (K) =F1M02 (K) 1=1 N 2=K DC 280 11=1,KM N1=N1+1 N2=N2+1 280 W1 (K}=N1 (K)-K1 (K,N1) *N1 (N2) 1*1 (K) = K1 (K) / K 1 (K, 1) 290 CONTINUE C PUT Â£ HI 3 Â£ Â£W2 j IKTC Â£X ]. DO 3C0 1=1,NBTOP X (I)=W1 (I) 300 CONTINUE J=0 DO 310 I=NBTP1,N non o to nron 310 J=J + 1 X (I) =H2 (J) CGNTINUE REIDBN END SUEECUTINE TITLE SUBROUTINE TITLE HILL BEAD AND PRINT CNE DATA CABE OF ALPHA NUMERIC COMMENTS, IT WILL SKIP 2 LINES BEFORE PRINTING THE COMMENTS. THE COMMENTS CANNOT FXCEE COLUMN 80. DIMENSION ALPHA (20) BE AD (5,10) (ALPHA(I) ,1=1,20) FOBMAI(20A4) WRITE (6,20) (AIEEA (I) 1= 1, 20) FOBMAT(* 0*,20A4) BET UR K END IV) vn SBEOUTINE BEAD (HHI,LNVEB,LKHBE,IKHBI,NSTORY,NSDCF,NFLEM, ^ $ elasbh,arsf:eh,elasbl,arshel) SUBBOUTIN E BEAD HILL PRINT OUT A HEADING, BEAD ANE FEINT TWO COMMENT CARDS, AND BEAD AND PRINT THE WAIL DESCRIPTION DATA. IN1EGEE HHI,DEPTH DOUBLE PRECISION LN VER LNHB R IN HBL EL ASBJ5, ABS H E F E IAS El DCUBIE PRECISION ABSHUL WRITE{6,1C) 10 FORMAT (* ,30 (**)/ *,** RESULTS OF LATEST ANALYSIS **/ $ *,30(***)) WEI1E(6,15) 15 FOBMAT(0'/) CAII TITLE HBITE(6,17) 17 FORMAT (*0 / ', UNITS: CALL TITLE INCH/LE,RADIAN*/' ') 20 30 40 45 50 60 70 75 80 90 100 105 READ (5,20) SOI FCHEAT (14) READ (5, 30) LEVER FCEEAT (Â£10,4) READ (5,40) LNHER FCEEAT (D1G. 4) REAL (5,45)IKHEL FORMAT (D1 0.4) BEAE (5,50) WLEETH FORMAT (14) WHF=WHI/12. 0 THICK=2.0*LE1IBR + 2.G*LEHBL BRW YTH=2, Q*INEBB BLWYTH=20*LNHBL NST CÂ£ Y= Will/ 8 NSBGF=(ESTORY* 5)- 1 NEIEE=NSTCBY*3 Â£LEE=LE VER CJIEK=LKHBfi+LNHEL WRITE (6,60) Will, WHE FORMAT ('0*,'WAIL BRIGHT =,14,1X,INC BES* ,3X [ ,F,2,1X, l F E EI ]) WRITE (6,70)THICK FORMAT(O',* HALL THICKNESS =,F6.2,1X MECHES ) WRITE (6,75)WDEETH FORMAT(*0*,'WALL DEPTH =*,I4,1X,*IECHES') WRITE (6,80) ERWYTH FO R M AT ( 'O','BRICK WYTHE THICKNESS =',F5.2,1X,I ECHES ) WRITE (6,90)ELWYTH FORMAT(*0','BLOCK WYTHE THICKNESS =,F5.2,1X, I ECHES') WRITE (6,100)NELFM FOBHATpO* ,'HOUBEH OF ELEMENTS IE FINITE EIEKEKT MODEL =', $ 15) WRITE(6,105)ESDCF FORMAT(* 0 *,NUMBER OF STRUCTURE DEGREES CF FREEDOM =',I4) WRITE(6,110)BLEN 110 FBJ3AT ( 0 'BRICK ELEMENT LENGTH =*,*5.2 SEITE (6, 120) CJLEN 120 FOBSAT {'O','CCILAE JOINT ELEMENT LENGTH $ INCHES *) WHITE (6,130)ELEN 130 FOBMAT('0*,'BLGCK ELEMENT LENGTH =',F5.2 CAII TITLE WHITE (6, 135) 135 FCBKAT (' ) ELASBfi=2.S16D+C6 AESDEB=5 1, 57D+00 IF (ELWYTH.EQ. 4) GC TO 140 IF (BIWYIfl.EÂ£. 6) GC TO 150 IF (BLHY1H.EQ.8) GC TO 160 140 EIASEI=2.023D+06 AHSHEL=14.85D+CC GC TC 170 150 ELASBL=1.6C7D+06 ABSHEI=27, B7D+G0 GO TO 170 160 EIASÂ£I=1.622D+G6 ARSHBL=35.74D+CC 170 BETIN ENC C SUBROUTINE COOED ( NPNTS,XCOGR,YCGCE) C SUBROUTINE COOED BILL BEAD AND PRINT C COORDINATES OE UP TO 20 ECINTS. DOUBLE EBECISICN XCCOR,YCOB DIMENSION XCCOR (2C) ,YCOCR (20) CAII TITLE BEAD (5, 10) NPNTS 10 FCBMAT (13) CALL TITLE DC 30 1=1,KENTS BEAD (5, 2C) J ,XCCGR (I) YCQCB (I) 1X,'INCHES*) *,F5.2,1X, 1X, 'INCHES V) TEE X AND Y 20 FORMAT (13,2D 13.6} WR1TE(6,25)J,XCCOR(I),YCCCR (I) 25 F CBM AT ( ,13,3X, D13. 6, 3X D 13. c) 30 CONTINUE WRITE (6 ,40) 40 FORMAT(* ) EETUEN END C SUBBOUTINE CURVES (NOCUBV,NOPPC,CURVAl,XCCCB,YCCCR) C SUBBOUTINE CUEVES WILL BEAD AND PEINT THE X AND Y C COOBDINATES OF UP TO 20 POINTS FOE UP TG 20 DIFFEBENT C CUBVES. NOCUBtf IS THE NUMBER OF CURVES, NOPPC IS THE C NUMBER CF POINTS PEB CUBVE, CUBVAI IS THE VALUE EACH C CUBVE IS IDENTIFIED BY, XCCOR STORES TEE X COORDINATE C OF EACH CURVE WHERE THE FIRST SUBSCRIPT IS THE CUBVE C NUHBEB AND THE SECOND SUESCBIPT IS THE POINT NUMBER, C AND YCOOR STORES THE Y CCOBDINATE CF EACH CUBVE WHERE C THE FIBST SUBSCRIPT IS TEE CURVE NUHBEB AND THE SECOND C SUBSCRIPT IS THE POINT NUMEER. DOUBLE PRECISION CURVAL,XCOOE,YCOOH DIMENSION CUR VAL (2 0) XCCOR (2 0,2 0) ,YCOCfi (20,20) CAII TITLE READ (5, 10) NOCURV 10 FCB CAT (13) READ (Â£, 20) NOPPC 20 F CEE AT (13) DO Â£G 1=1,NOCURV BEAD (5,30) CUBVAI (I) 30 FORMAT(D136) WEITE (6,35) CURV AL (I) 35 FCflM AT('C,9 X,'P = ',013.6) CALL TITLE DC 50 J=1,NOPPC BEAD(5,40)K,XCOOB (I,J) ,YCOOR(I,J) FORMAT(13,2D 13.6) 40 non lj k: - n n HITE ( 6,45) K,XCOOR (I, 0) ,50000(1,0) 45 FORMAT(1 *,13,3X,Â£13.6,35,013.6) 50 CONTINUE 60 CONTINUE WHITE (6,70) 70 FORMAT (* ) BET DEN ENE C SUBROUTINE FBI NT (A H) SDBfiOUTINE PRINT WILL EfilNI AN NX! MATRIX [A] AND IDENTIFY EACH BOW AS A DEGREE CF EBEEECM. DCOEIE EBECISICK A DIMENSION A(N) WHITE (6,10) 0 FORMAT ( ) DC 30 1=1,N WHITE (6,20) I,A (I) ECEM AT (' ,COE *,I4,2X,D13.6) 0 CONTINUE BETUHN ENE SUEBOUTINE WHITE (A N,M) SUBROUTINE WRITE WILL PRINT OUT THE DIMENSIONS N AND M OF A MATRIX Â£ A J AND THE MATRIX OF CCEEEICIE NTS IN EOWS. DOUEIE E BECISIC N A DIMENSION A (N ,M) WBITE (6,20) K,M DO 1C 1=1,N IE (M. LE. b) THEN WRITE (6,30)1, (A (I,J) ,0=1 M) ELSE WRITE (6,40)I,(A (1,0),0=1,6) WfilTE (6,50) (A ( I, 0) J = 7,M) END IF 260 10 CONTINUE WHITE (6,0) 20 FO EM AT (// MATRIX SIZE: ',13,' X',I3/) 30 FCEMAT( HOW* ,I3,6C12, 4) 40 FORM AT(/* HOW ,13,6D12.4) 50 FGEiiAI(7X,6D 12, 4) 60 FORMAT(//) HETUEN END C SBBCUTIUE BRPDMT (NBRPDP,NOERPC,NERMTP,BEDCCB,EEPCOR, ERPCV, $ Â£B1CCB,BEMCCE, EREF, SBRPD, S EBMT BRVER F B.R MCM B RAEL ,BB3 El L IT Eli) C S UBHOUTIN E BRPDMT WILL CALCULATE THE SLOPES CF THE C IINEABLY APPROXIMATED P-UELTA AND M-1HETA CURVES FOB C THE BRICK ELEMENT THE FIBS1 TIKE IT IS CALLED. IT C ALWAYS FINDS THE SLOPE FOB THE REGION IN WHICH THE C ELEMENT P FALLS AND THEREFORE GENE BATES THE AEPECERIAT E C AXIAL STIFFNESS FACTOR OF A*E/L. SIMILARLY, IT FINDS C THE SLOPE FOB THE REGION IN WHICH THE AVERAGE MOMENT C FAILS GN THE CURVE AND GENERATES TEE RC1A1ICNAL C STIFFNESS FACTOR 3*E*I/L FCR THE ELEMENT E, INTEGER UIE DGUELE PRECISION B.SDCOE, 8HPCCE BfilCOR DEM OCR Efi F F EE A FI DCUEIE PRECISION ER3EIL,SBRPD,SERMl, A VGMOM,BRPC V,LLSMI,ULSMl DOUBLE PRECISION BRMOM,SEVERE DIMENSION BEECOE (20) BBPOOR (20) BfilCOR (20,20) ,BEMCGR (20,20) DIMENSION BEEF (6) ,SBRPD (20) ,SBEMT (20,20) ERPCV (20) IF (ITER, HE, 1) GO TO 60 N PL S 1=N ERPDP1 DO 20 1= 1,NPLS 1 J = I + 1 SBRPD (I)= (EEPCGR(J) -ESPCOE (I) ) / (BRDCOR (J) -BEDCOR(I) ) 20 CONTINUE NEIS 1 = NEEMTP-1 DO 45 J=1,NGBRPC EC 40 1= 1, NfiLS 1 CTv K=L+1 SBlflT (J,L) = (OfiHCOH (J, K)-BRMCCR ( J, L) )/ (BBICOB (J K) - I BBTCOR (J ,L) ) 40 CCN1INDE 45 CONTINUE B EAEI=S EEPD(1) BB3EIL=SBB21T ( 1, 1) GC 1C 150 60 DO EC I=1,HPLS1 If (BEEF (1) .GE. EEPCOH (I) ) BRA FL=SBRPD (I) If (BEEF (1) .LT.BBPCOR (I) ) GO 10 90 80 CCEliNUE 90 A VGECÂ£1 = DABS ( (BREF (3)+B BSF (6) )/2) IF {EEEF (1) GT. ERPCV (1) ) GO TC 105 DO 1C0 L=1,NMLS1 IF (AVGMCE, GE. BB MCOR ( 1,L) ) EB3EIL=SBHM1( 1 ,L) IF(AVGOl.LT. BBHCOH (1 L) ) GC TC 150 100 CCEIIDIUE GO 1C 150 105 IF (BEEF (1) GE. BBPCV (NOBBPC) ) GO TO 125 NOPB 1=NOBBPC1 DC 110 K=1,K0Â£W1 L=K+ 1 IF (BBEF (1) GE, EEPCV (K) ) LLÂ£=K IF (BEEF (1) .LI. BBPCV (L) ) OIf = I IF (BEEF (1) .LT. EEPCV (L) ) GOTO 115 110 CONTINUE 115 DC 120 I=1,NHLS1 IF (AVGMOFl. GE. BBMCOR (LLP,I) ) ILSM1=SBB3T (1LE/I) IF (AVGMOM.GE.BBNCOE(ULP,1})ULSMT=SBENT(ULP,I) 120 CONTINUE Bfi3EIL= DLSMT- ( ( (BBPCV (ULP) -BEEF ( 1) )/(BBPCV (ULP) -BBPCV (ILP) ) ) $ (ULSfll-LLBHT)) GC 1C 150 125 DO 140 1= 1, NMLS1 IF (AVGMCH.GE.BF ACOR (NOBBPC,!) ) ER 3E IL = SBfiMT(NOBBPC,I) 262 IF (AVGHCM. II. BEMCOR (NOBRPC, 3) ) GO TO 150 140 CGN1INUE 150 BfiMCM=AVGKOM BRVEEE=BREF(1) BE1DRN END C SUBEGUTINE CJPEMT (NCJPDP,NCJMTP,CJDCOB,CJPCOR,CJTCOfi, i CJMCOR,CJEF,SCJPD,SCJMT,CJVERF,CJMOM,CJSfiES,CJKCMS, ITER) C SUBROUTINE CJEEMT WILL CALCULATE I BE SLOPES OF I HE C LINEALLY APPROXIMATED P-DEITA AND E-THETA CURVES FOB C THE COLLAB JOINT ELEMENT THE FIRST TIME IT IS CALLED. C IT ALWAYS FINDS THE SLOPE FOB THE REGION IN WHICH THE C ELEMENT P FALLS AND THEBEFORE GENERATES THE APPROPRIATE C SHEAR SPBING STIFFNESS FACTOR. SIKILAIIY, IT FINES THE C SICEE FCB THE REGION IN WHICH THE AVERAGE MOMENT FALLS C CN THE CURVE, AND GENEEATES THE MOMENT SEEING C STIEFNES3 EACTOR. DOUBLE PRECISION CJDCOR,CJPCCS,CJTCOR,CJKCCR,CJEF,CJSHRS DOUEIE PRECISION 0JM0M3,SCJPE,SCJMI,CJAXF,AVGMOM,CJMCM DOUBLE PRECISION CJVERF DIMENSION CJDCCE(20) CJPCOR (20) ,CJ1C0R (20) ,CJMCOR(20) DIMENSION CJEF (4) ,SCJPD (20) ,SCJMT (20) IF (ITER. NE. 1) GO TG 60 NPLS 1 = NCJPDP-1 DC 20 1= 1 ,N ELS 1 J=I* 1 SCJPD (I) = (CJECCH(J) -CJPCOR (I) )/(CJBCOR (J) -CJDCOR(I) ) 20 CONTINUE NMIS 1= NC JMT P- 1 DC 4C L = 1 MLS 1 R=I+ 1 SCJMT (L) = (CJMCCR (K) -CJMCCR (L) )/ (CJTCCR (K) -CJTCCB(L) ) CONTINUE CJSHBS=SCJPD ( 1) CJ MC ES=SCJ Ml (1) 40 60 GC 1C 120 CJAXF=DABS(CJEF (1) ) DC 80 I=1,NILS1 IF (CJAXF.GE.CJPCGR (I) ) CJSHRS=SCJPD (I) IF (CJAXF.L1.CJECOR(I) ) GO 1C 90 80 CONTINUE 90 AVGMCM=DABS ( (CJEF (2) +CJEF(4) )/2.0) DO 1GC L-1,NMLS1 IF (AVGMGM. GF. CJMCOfi (L) ) CJMGMS=SCJMT (L) IF (AVGHM.LI.CJMCOR(L) ) GC 1C 120 100 CONTINUE 120 CJMCM=AVGMGM C]V EEF=CJAXF EE1UBN END C SOEEGUTINE ELIDM1' (NBLPDP, NOBLPC, NBLMTP,BLDCOE ,BLPCOB ,BLPC V , $ BLTCCfi,BLMCOB,BLEF,SBLPD,SBLMI,BIVERF,BLi!CM,BIAEL,BI3EIL,ITEE) C SUBBCUTINE ELPDMT HILL CALCULATE 1 EE ELOPES CF THE C LINEARLY APPROXIMATED P-DEI1A AND K-THETA CURVES FCB C THE BLCCK ELEMENT THE FIBBT TIME 31 IS CALLED. IT C ALWAYS FINDS THE SLCPE FCB THE REGION IN WHICH THE C ELEMENT P FALLS AND THEREFORE GENERATES THE APPROPRIATE C AXIAL STIFFNESS FACTOR OF A*E/I. SIMILARLY, IT FINDS C THE SLOPE FCB THE REGION IN WHICH THE AVERAGE MOMENT C FALLS CN THE CURVE AND GENERATES TEE FCTATICNAL C STIFFNESS F ACTOB 3*E*I/I FCB THE ELEMENT P. INTEGER ULP DOUEIE PRECISIC N BLDCOR,BLPCCR,BLTCOR,BLMCCR,BIFF,BLAEL DOUBLE PRECISION BL3EIL,SBLED,SBLEl,AVGMCt,BLECV,LLSMl,ULSMT DOUBLE EBECISIC N ELMOM,ELV ER F DIMENSION BLDCCB (2 C) ,BLPCOR (20) ,BLTCOR (20,20) ,BIMCCR (20, 20) DIMENSION ELEF (6) ,SBLPD (20) ,SBLMT (20,20) ,BIPCV (20) IF (ITER.NE. 1) GC TO 60 NPLS1=NELPDÂ£-1 DC 20 1 = 1,NPL S1 264 20 J=I+1 SBIJED (I) = (ELECCS (J) -ELPCGL (I) )/ (BLDCCÂ£ (J) -BICCCS (I) ) CONTINUE Hf5LÂ£ 1 = NBLM1P-1 DC 45 J=1,NCBIEC DC 40 L=1,NKLS1 K=L+1 SBLMT (J,L) = (BLMCOB (J,K)-BIMCCR (J,L) )/ (EITCCB (J,K) - $ ELTCOR(J,l)) 40 CONTINUE 45 CONTINUE ELAEL = SBLPD (1) BI3EIl=SBIBT (1,1) GO 10 150 60 DC 80 1= 1,N ELS 1 IF (BLEF (1) GE. BLPCOB (I) )BIAEI=SBLPD (I) If (BLEF (1) .LI. ELPCOR (I) ) GO 10 90 80 COM1IN 0 E 90 AVGKCM=DAES ( (EIEF (3) + BLEF (b) )/2) IF (BLEF (1) .GT.BLPCV (1) ) GO 1C 105 DC 100 L= 1, N iS L 3 1 IF(AVGMCH.GE.BLUCOR (1,L) ) EL3EIL=SBLET (1,L) IF (AVGMOM. IT. BIKCOR (1,L) ) GO 10 150 100 CONTINUE GC 1C 150 105 IF (BLEF (1) .GE.BLPCV (NOBLPC) ) GC 1C 125 NGEE1=NCBLEC-1 DO 110 K= 1,NOP 1 I = K+ 1 IF (BLEF ( 1) .GE.BLPCV (K) ) LI E=K IF (BLEF (1) LI. ELPCV (L) ) ULE=L IF (BLEF (1) .Ll.BLPCV(L) ) GC 1C 115 110 CONTINUE 115 DO 120 I=1,NiviLS1 IF(AVGMOft.GE.BLECCR(LLP,I))LLSM1=SBLMT(LLP,I) IF(AVGMON.GE.BLMCOB(0LP,Ij)UISE1=SBLET (ULP,I) 120 CONTINUE BI3 EIL= LLSM1- l ( (BIPCV (LLP) -BIEF 1) ) / ( BLPC V (ULP) -BLPCV (LLP) ) ) $ *(OLSMl-ILSMT)) GC 1C 150 125 DO 1C0 1 = 1,NMIÂ£ 1 IF (AVGMCM. GE.BIMCOR(NOBLPC, I) ) BL 3KIL=SBLMT (NCBLFC ,1) IF (AVGMOH.II.BIMCOR(NOBIPC,I)) GG 1C 150 140 CC MINUE 150 BLMGM=AVGMGM BIVEbF=ELEF (1) RETURN ENE C SUBROUTINE STIFAC (VERTF,MOME,STOVE,STOMF,VSTIF,MSTIF, $ ST0VS1,S1CMS1,1RSTIF ,11 EÂ£ KE Y E IE H N C) C SUBROUTINE STIFAC ilLL SELECT THE APPROPRIATE STIFFNESS C FACTORS FOR AN ELEMENT BASED CN THE VAIUFS CE THESE C FACTORS AS DETERMINED FROM THE P-CELIA AND B-1BE1A C CURVES, THE CURRENT LEVEL CF ELEMENT LCADING, AND THE C EREVIOUS LEVEL OF ELEMENT LCADING. USE KEY = 1 FOE C BRICK ELEMENTS, KEY = 2 FOR CCILAE JOINT ELEMENTS, AND C KEY = 3 FCfi BICCK ELEMENTS. INT EGER IR STIE,ELEMNO DOUBLE PRECLSLC N VERTF,MOMF,STOVF,STOMF,VSTIF,MSILF,S10VST DUELE PRECISION STOMST DIMENSION STOVE (3,117) ,STOMF (3 1 1 7) S TG VS1 (3 1 1 7) DIMENSION STOMST (3,117) 1 = KEY J =EIEMNG IF (ITFB. NE. 1) GC TC 10 STGVF (I,J) = VSRTF S1CMF (I J) = MCKÂ£ STQ V ST (I J) = V S11F STOMST (I J) = MST IF GO TO 50 10 IF(VERIF.GT.STGVF (I,J)) GO TC 20 o n 20 VSlIf=SlGVST (I/O) GO 10 30 S1GVF (I,J) = VEETf IF (VSTIF.EQ.SlOSl (I,J) ) GO 10 30 1BS1IF= 1 S10VST(I,J)=VS1IF 30 IF (E C M F. G1. SIC M F (I J) } GO TO 40 HSlIE=S10HSl{I,d) GC 1C 50 40 SICME (I,J)= MOHF IF(MSTIE.EQ.S1CMST (I,J)) GO 10 50 IBS1IF=1 SICtSI (I,J)=HSTIF 50 RETURN ESC SUBfiCUTINE ERESM (EBEK,EREF,101EK,ERAEL,BE3EIL,LN,ELASBR, ABSfiBB,I,EE,KEIE) S UBBOU'IIN E ERESM HILL CONSTRUCT TEE STIFFNESS MA1EIX EOS EACH BRICK ELEMENT. IN1EGEE BB BOU ELE PRECISION BB EK BEEF,1CTEK,LN,BÂ£AEI,ES3EIL,ELASÂ£E DOUBLE PRECISION A ES HR GBR 1? HIR LO ISBR DIMENSION 10TEK(6,6,NELEK) BEEK (Efi.DR) ,BE E E(EB) E CIS EB=0, 1 5D + 00 GBR=ELASBR/(2.0*(1,0+PQISBR)) ?HIÂ£E=Â£B3Â£I 1*4.0/ (GBR* ABSHBR+LN) BREK (1, 1) =BEAEL BEEK (2,1) = 0. 0 BHEK (3, 1)=0.0 BHEK (4, 1) =-1, 0*EBÂ£K (1, 1) BREK ( S, 1) =0. 0 BREE (, 1)=0.0 BREK ( 2, 2) = (BR3EIL* 4.0/ ( (1.0 + PUIBR) *(LN**2)))- $ (EBEE (1) *6. 0/(5. 0*LN) ) BHEK (3, 2) = {-DH3EIL*2,/ ( (1.0 + PHIBR) *LN) ) + (EREF (1) / 10.0) 20 30 C c c BBEK (4,2) =0, 0 BREK (5, 2) =-1.C*BHEK (2,2) BREK (6,2) = EBEK (3,2) BREK (3, 3) = (BR3EIL* (4.0+PRIBR)/ (3, 0* (1.0 + EiiIBS) ) ) - $ (2. 0 ER E E (1) IB/15.0) BREK (4, 3) =0.0 BREK (5,3)=-lQ*EREK(3,2) BREK (6, 3) = (BR3EIL* (2.0-PHIBR)/ (3. 0* (1.0+IHIBB) ) ) + $ (EBEE (1) +LN/30.0) BREK (4, 4) = BREK(1, 1) BREK (5,4) =0 .0 BREK (6, 4) =0. C BREK (5,5) = E BEK (2,2) BHEK(5,6)=-1.G*BEEK(3,2) BREK (6,5) =EEÂ£K (5,6) BREK (6, 6) =BHEK (3,3) DC 30 J= 1 ,EE EC 20 K=1,B2 BREK (J,K)=BEER (K, J) 10TEK (J, K,1) =BKKK (J,K) CONTINUE CONTINUE RETUEN END SUBROUTINE CJESM (CJEK ,TCTE K ,C JSH ES, C J f C MS ,L NH Â£E L Nil EX , $ I,CJ,NELEM) SUBROUTINE CJESM WILL CONSTRUCT TflE STIFFNESS MATRIX ECB EACH COLLAR JOINT ELEMENT. INTEGER CJ DGUEIE PRECISION CJEK,TGTEK,IN HER,IN UBI,CJSHRS,CJMOMS DIMENSION TOTEK(6,6, NELEM) ,CJEK (CJ,CJ) CJEK (1, 1)=CJSHES CJEK (2, 1) =CJSIifiS*INHBR CJEK (3,1)=-1.Q*CJÂ£K(1, 1) CJEK (J, 1)=CJSHBS*LNHBL 268 n r. CJEK (2,2) = CJSHRS* (LNHBR**2) +CJMOMS CJEK (3, 2) =-1.G*CJEK (2, 1) CJEK (4,2)=CJSBBS*LNHBR*LNHEL-CJM0MS CJEK (3, 3) =CJEK (1 1) CJEK (4,3) =-1.Q*CJEK (4, 1) CJEK (3, 4) =CJEK (4,3) CJEK (4,4) = CJS HRS* (1NHBL**2) +CJMOMS DO 3C J=1,CJ EC 20 K= 1,CJ CJEK (J K) =C J EK (K, J) TCTEK (J,K,I) =CJEK(J, K) 20 CONTINUE 30 CONTINUE RETURN EKE SUBROUTINE ELESM (BLEK,ELÂ£F,TOTIK,ELAEL,BL3EIL,LN,ELASBL, $ AKSHBL,I,BI,KEIEE) SUBROUTINE ELESM NHL CCNSTBUCT TEE STIFFNESS MATRIX FOR EACH CGNCE ETE BLOCK BLE EE NT. INTEGER BL ECU ELE PRECISION BLEK,BLEF,TCTEK,LK,BLAEI,E13 EIL,ElAS EL DCUELE PRECISICN A ES HBL,GBL,PHIBL,POISBL DIMENSION TOTEK(6,6,NÂ£LEM),BLEK (BL ,BI) ,EIEF(EL) fCISEL=0,15 E 0 0 GBL=ELASBL/(2.C*(1.+POISBL)) PIiIEI=EL3EII*4. 0/ (GBL*ARSHEL*LN) BLEK (1, 1) =ELAEL BIEK (2, 1) =0. 0 ELEK (3, 1) = C. 0 BLEK (4, 1) = -10*ELEK(1, 1) ELEK (5, 1) =0.0 BIEK (6, 1) =0.0 BLEK (2, 2) = (BL3EIL* 4.0/ { (1.0+PHIBL) (LN**2.0) ) ) - $ (ELE.E(1) *6. 0/(5.O+LN)) BLEK (3, 2) = (-BL3EIL*:2.0/ ( (1. O+PHIBl) *LN) ) + (ELEF (1) / 10.0) o o BIEK (4,2) = 0.0 BLEK (5,2) =- 1.0*BLEK (2,2) BIEK (6,2) = Â£IEK (3,2) BLEK (3, 3) = (BL3EIL* (4. O+PHI BL)/ (3. O* (1. 0 + FHIBL) ) ) - $ (2.0*ELEF(1) *LN/15.0) BLEK (4, 3) = C. G BIEK (5,3)=-1.0*ELEK(3,2) BIEK (6, 3) =(BL3EIL* (2-O-PHIBL)/ (3. O* (1 O+FBIBL) ) ) + $ (ELBE (1) *LN/30, 0) ELEK(4,4) =BLEK (1, 1) BIEK (5,4)=0.0 BIEK (6, 4) =C. O BIEK (5,5) = EIÂ£K (2,2) BLEK(5, 6)=-1.C*BLEK (3,2) BIEK (6,5) =EIEK (5,6) BIEK (6, 6) =BLEK (3,3) DC 30 d= 1,El EO 20 K=1,BI BIEK (J,K) = BIEK(K, J) 1TEK (J,K,I) =BLEK (J,K) 20 C0N1INU E 30 CCN1INE E1UEN EME SUBRCOTINE IM DXBt (IBD TOTEI K,I MEET, BE, 8 EIEH) SUBROUTINE INIXBR MILI CCHSIfiCT I BE INDEX MATRIX FOR EACH BRICK ELEMENT. I MI EGER TCTÂ£IN,ER DIM EM SICi 10TEIH (KKLEH,6) ,IBE(6) IF (I.Eg. 1) GC 1C 5 IE(I.EQ.4) GO 10 15 IF(I,GT.4) GC TC 30 IBB (1) =0 IBE (2) = 0 IBB (3) =0 270 15 20 25 30 40 45 50 70 C C C IBB (4) = 1 IBE (Â£) =3 IBB J6) = 4 GC 1C 50 DC 20 J= 1,3 Â£=J + 3 IBB {J ) = IE F< (!) CONTINUE DC 25 K=4,6 IBB (K) =IBB (K) +5 CGK1INUE GO 10 50 IBB (1) = IBB (4) DO 4C M=2, IEB(K) = IÂ£ B(H)+5 CONTINUE IF (I,EC,NEBI) GC TO 45 GO 10 50 I EE (6) = IBB (5) IBB (5) = 0 DC 70 L=1,BE 101EIN (1, L) =IBB (L) CONTINUE BE10BN END 3UBE0T1NE INLXCJ {ICJ,TOTEIN,I,NEDO,CJ,NELEM) SUBROUTINE INDXCJ WILL CCNS1NUCT IBE INDEX EATBIX FOE EACH CCILAE JCINX ELEMENT, INTEGEB TG1EIN,CJ DICE NSION TCTEIN (N ELEM, 6) ICJ (4) IF (I. EQ.NEMO) GO TO 35 IF (I,NE,2) GO TC 5 ICJ {1) =1 ICJ (2) =4 ICJ (3) =2 5 ICJ (4) =5 GC 1C 5 DC 30 N = 1,4 ICJ(N) =ICJ (N) +5 30 CONTINUE GG 10 50 .35 ICJ (1) = ICJ (1) +5 ICJ (2) =ICJ (2) *4 ICJ {3) = ICJ (3) +5 ICJ (4) =ICJ (4) +4 50 DC 70 1=1,CJ 10IE.IN (I,L) =ICJ (L) 70 CONTINUE EE1UBN END C SUBECUIINE INLXEL (IBL, TOTEIN, I, BL/NELEM) C SUBBOUTIN E INDXBL HILL CCNS1B0CT 1LE INDEX EAT BIX FOB C EACH CCN'CBEIE ELOCK ELEMENT. IN1EGEB TO1EIN ,EL DIKE NSIC N TCIEIN (NELEM,6),IBL(fa) IF (I EQ3) GO 1G 5 IF (I,EC. 6) GO i'C 15 IF (I.G1.5) GO 10 3 0 5 IEI (1) =0 IBL (2) =0 IEI (3) =0 IEL (4) =2 Ifil (5) =3 IEL (6) =5 GC 1C 50 15 DO 2C J = 1,3 E=J+3 IBL(J)=IBL (P) CONTINUE DO 25 K=4,6 20 272 25 JO 40 45 50 70 C c c c c c c c 10 15 IEL(K) = IBL(K)+5 CONTINUE GC 1C 50 IBL (1) =IBL (4) DC 40 M=2,6 IBL(M)=IBL (M)+5 CCK1INUE IF(I.EQ.NELEM) GO 10 45 GC 1C 50 IBL (E) =0 IEI (6) = IBL (6) -1 DO 70 L=1#BL 1G1EIN (I,L)=IBL (L) CONTINUE RE1UEN ENE SUEBGOTINE FORCES (NGF,NDOE,SFINI1,5FINCR,SFMAX,NSECF) SUBROUTINE FOECES READS AND PHIH1S 1BF NUMBER OF 3 DEGREES OF FREEDOM 1HAT ARE LCADED, 1 HE INITIAL LOAD AT ^ EACH DCF, THE INCREMENTAL LOAD AT EACH DOF (AMOUN3 BY WHICH LOAD Al 1HAT DOF IS TO BE INCRE EE NT EE) AND THE MAXIMUM LGA E AT EACH DOF. IF NO MAXIMUM LOAD IS DESIRED AT A DOF, PUT A LARGE NUMBER IN THE EA1A SET, SAY 1. 0E + 2, DOUBLE PRECISION SFINIT,SFINCR,SFMAX DIMENSION SFINIT (N5DOF) SFINCE { NS DCF) SFM A X (N SDOF) DIMENSION NDOF(NSDCF) CAII TITLE READ (5, 10) NCF FORMAT (13) WRUE (6,15)NOP FORMAT ('O' NC, GF FORCES ON STRUCTURE =',I4) CALL TITLE DC 30 I=1,NCF READ (5, 20) NDOF (1) ,SFINI1 (I) ,SFINCR (I) ,SFHAX (I) 20 FCRMAT (13,3D 13.6) WRITE (6,25) NUCE (I) ,SFIUII (I) ,SFINCfi (I) ,SFMAX (I) 25 FCBMAT ('0* ,13,3X,Di 3.6,4X,D13,6,4X,D13,6) 30 CONTINUE WBITE (6,35) 35 FORMA1C ') BE1UEN ENE C S BBCU1IN E APP1YF (STRF NSDC NCF NDOF S.F I KIT SFIN CB, S EM AX , $ ITEB KEY 1) C SUBROUTINE APPLY? WILL EIACE THE INITIAL LCADS ON THE C SIRUCTUSE IN THE STRUCTURE FORCE MATRIX THE FIRST TIME C IT IS CALLED. THEREAFTER IT KILL INCREMENT EACH ICAD C ACCCEDING 1C THE INFORMATION BEAD BY SUBROUTINE FORCES C FROM THE DATA SET. IT WILL HOT INC BE EE FT AM LOAD C EEYCND ITS SPECIFIED MAXIMUM VALUE. IE KEY 1 = 0, THE C LOADS HAVE NOT REACHED THEIR MAXIMUM VALUE. IF KEY 1=1, C THE LOADS ARE ALL AT THEIR RESPECTIVE MAXIMUMS AND NO C E URTHER INCREASES ARE REQUIRED. DCUEIE ERECISIC N STEF,SiIN IT,SFINCR,SEHAX,DSTKE,DSFMA X DIMENSION SIRE (NSDOF),SFINIT (NSDOF),SEINCE(NSDCF) DIMENSION SFMAX (NSDOF) NDOF (NOF) KEY 1= 1 IE (ITEE. NE. 1) GC TC 20 DO 1C 1=1,NCF F= NDCF (I) STRF (K) = Sf 1NI.1 (I) 10 CONTINUE KEY 1 = 0 GC 1C 8U 20 IF (KEY2.NE. C) GO TC 40 DC 3U 1=1,NCF =NDGF(I) STEF (M) = SIR? (M) +SFINCR(I) ESTBF=DABS(STRF (M)) 274 30 40 50 60 70 80 C C c c c c 10 20 ESFHAX = DABÂ¡~ (SFMAX (I) ) IF (DSTRF.GT. E5FMAX)STBF (K)=SFMAX (I) IF (DSTRF.LT.DSFMAX) KÂ£X1=0 CONTINUE GC TC 80 IF (KEX2.NE.3) GO 30 60 CO 50 I = 1, N C F J=NDOF (I) STBF (J) = STÂ£F (J) -SFINCR (I) / 1 0 CONTINUE HEX 1 = 0 GO 30 80 DC 70 1=1,NCF I=NDOF(I) STBF (LJ=STBÂ£ (!) +SFI NCR (I) / 1 0 CONTINUE KE X 1 = 0 RETURN END 8 UB ECU TINE FLATEK (STRK NSTRK, M 1, M 1P 1, NSDGE, NSDM2,LNHBN, $ LNHBL) SUBROUTINE PLATER WILL CONSTRUCT A STRUCTURE STIFFNESS MATRIX WHICH ONLY CONSIDERS THE TBC STEUCTUFE DEGREES CF FREEDOM AT THE TOP OF TEE WALL THAI CORRESPOND TO THE TEST PLATE LUTHER THAN THE ORIGINAL FCUF DEGREES CF FREEDOM AT THAT LOCATION. DCUELE PRECISION SISK NSIE K, LNHiJB INH131,LC2, HLC2, P REI DIMENSION STRK (NSÂ£OF,M 1) NST EK (NS DM2, M 1P 1) ,PR EL (2,4) CALL NULL (NSTRK,NSDM2,M1P1) N E M1M3= NSDCE-Kl-3 DO 2C I = 1, N M M 1M 3 EC 10 J=1,M 1 NSTRK (I,J)=STRK(1,J) COM I NOE CONTINUE iv> vn H SI Afil=lH1M 1M3 + 1 NSDIi4=KSDGT-4 NS1CE=8 DO 40 K=NSTABl#NSDfl4 DC 30 L=1,flSTOÂ£ NSTBK (K,L) =SIBK (K ,L) 30 CCKTINE 11 Â£10 P=N STOP- 1 40 CGSIIKUE NSTBP1=BSIAHT+1 NSI6 E2= N3T ABT + 2 NSlBE3=NSlABT+3 NHIE= 10 WHICH 1=9 IC2= (OHBE + iWHI:I)/2.0 MLC2=-1.0*L02 NSTBK ( NSTAfiT, rfIDfi 1) =SXBK (NSIABT, 9) 11S1BK (N STAB! N KID) =JSL02* SXRK (NSlABIf9) NHIE=NHID-1 NWICM 1=NHIDH1-1 NSIEK (NSTBP1,NWIE1 1) =ST BK ( N SIS P 1, 8) + SIKK (NSTB.P 1,9) NSIBK (WSTBP1#NHID) =HL02 *STfi K (ESTE E 1,8) +LC2*STfiK (NSTRPl,y) Â£UIE=KUID-1 WHICH 1=NHIDH1-1 BS1EK (NSIRE2, MIDE 1) =STRK (NSIRP2,7) +SIBK (NSTBP2,8) NS1RK (USURP 2, MUID) =HL02*STRK (BSTBÂ£2,7 ) +102*STBK 11 UIC = NHID 1 NICn=NWIDHl-1 H C C L = 6 DC 60 I=NSTBE3,RSÂ£M4 RCQLE 1 = 11C0L+ 1 KCCLÂ£2= KCCI + 2 NCCLP3=RCOI+3 KSIBK (I, MICM 1) =STRK (I,NCC1) +51RK (I, NCCIP 1) NS1RK (I,MID) =KL02*STRK (I ,NCOL) +L02+STBK (I, KCCLE1) +STRK (I NCCIP2) + SIRK (I,NC0LP3) $ KKID=NHID- 1 NHIDM^NHIDE 1-1 NCOL=NCCL- 1 60 CONTINUE NSDM3=NSDOF3 KSDE2=NSDCF-2 N SUM 1 =N SDOF- 1 PEEL (1# 1)=STRK (NSEM3, 1) +STHK (NSDM3,2) PREL (1,2) = STRK (NSEM3,2) + STRK (NSDM2,1) PBEI (1,3)=STEK (NS DM3,3) +STEK (NSDM2,2) PREL (1, 4) = STRK (NSDM3,4) +STRK (NSDB2,3) PBEI (2, 1) = ai02*STÂ£K(N3EM3, 1) L02*STRK{NSDM3,2) 1 + SIHK (NÂ£DH3,3)+ STEK (NSDK3,4) PBEI (2,2)=MI02*STEK (NSDM3,2) L02*S'IBK (NSDM2, 1) $ + STRK (NSDH2,2)+SIEK(NSDK2,3) PBEI (2, 3) =FIC2*STRK (NSDM3,3) +LC2*STBK (BSD Â£2,2) $ +STRK (NSDMl, 1) + STRK ( N5DM 1, 2) PEEL (2, 4) =ML02*STBK (NSDM3,4) +IC2+SIHK (NSDK2,3) $ + STBK (NS EM 1,2) +STHK (NSBCE> 1) NSTBK (NSDM3, 1) =PREL (1, 1) + PEEL (1,2) NSTRK (NSDM3,2)=MLC2*PREL (1,1) +I02*PREL (1,2) +PEEL (1,3) $ + PBEL (1,4) NS1RK (NSDM2,1) =MLC2*PEEL(2, 1) +I02*PREL(2, 2) PBEI (2,3) +PBEL (2,4) RETURN ENE C SUBROUTINE DISPIA (STRK,NSTHÂ£ NSDCF,NSDM2 INHBR,INUBI) C SUBEOUTINE DISPLA ILL CALCULATE THE ACTUAL STRUCTURE C LISPLACEMENT MATRIX FROM TEE STRUCTURE DISPLACEMEET C FAT BIX THAT ONLY CONSIDERS TWO DEGREES OF FREEDOM AT C THE IOP OF THE HALL. DCUEIE EBECISIC N STEW,NSTRW,LNRBB,LNHBL,LG2,MLC2 DIMENSION STEW (NSBOP),NSTEW (NSDM2) NSDE4=NSDCF-4 277 o rÂ¡ n NS D E 3= NS DCF -3 HSEM1=NSD0F-1 LO 2= (LNHB.E + LNBBL) /2.0 NIC2=-1,0*LC2 EO 2C I=1,NSDM4 STRW (I) = NS RN (I) 20 COSUSE STRR (NSDM3)=NSTSW ( NSDM3) + (MLC2 + NS1EW (NSDM2)) STB K (NSDM2) =N STRW { NSDM3) + (LC2* NSTEH ( N SDM2) ) STEii {NSEM1 )=NSI fiW (NSDM2) SlEfi (NSDGF)=NSTBH{NSDM2) RETURN ENE C SUBROUTINE CHK2CL (A,N,TOLES,KEY) SUBROUTINE CHKTOL CHECKS TO SEE IE EACH NUMBER IN MATRIX [A] IS LESS IHAN ICLEB. IF IT IS, KEÂ¥=0. If AT LEAST ONE NUMEER IS NOT, KEÂ¥= 1. Â£ A] IS AN NX1 MATRIX. DOUBLE PRECISION A,TOLER,ABSCL J DIMENSION A ( N) 03 KÂ£Â¥= C DC 10 1=1,N AESOL=DABS (A (I)) IE (ABSCI.Gi. ICIER) KEÂ¥=1 IF (ABSOL.I.TOLER) GO IC 20 10 CONTINUE 20 RETURN EKE C SUBROUTINE CHKFAI (F,NPMPTS,COOEM,COORP,PMAX,KEÂ¥,NSTAT) C SUBROUTINE CHKFAI HILL CHECK TC SEE IE AN EIEMENI HAS C FAILED. FOR THE BRICK OR BICCK ELEMENT, THE ACTUAL C AXIAL LOAD AND MOMENT IS COMPARED fcllH THE ALLCHAEIE C AXIAL 1CAE AND MOMENT FOB TEAT TXPE OF ELEMENT. FOR C THE COLLAR JOINT ELEMENT, THE ACTUAL SEEAR FCECE IS C CCMPABED WITH THE ALEONARLE SHEAR FORCE. FOR THE BRICK ELEMENT OSE KFY=1, FOB TEE COLLAR JOINT ELEMENT OSE KEY=2, AND FOE THE BLOCK ELEMENT USE KFY=3. IF MSIAT=0, THE ELEMENT HAS NCT FAILED. IF NSTAT= 1, THE ELEMENT HAS FAILED. IF NSTAT=9, THE ELEMENT IS IN AXIAL I ENSIGN. CCUEIE PRECISION F, COOBP, COGHM, PMAX, ALLOW B ,HIM 1 MIP1 FI DOUBLE PBECISION PIP 1,PIM1,MAXKCM ,FGNE DIMENSION F (6) ,CCGfiP (NPMPTS) CCCB M (N P MPT S) NSTAT=0 IF (KEY. KE.2) GC TC 10 IF (DABS (F ( 1) ) .GT. PMAX) GO TC 25 GC 1C 80 FGN E=F (1) IF (F (1) LT. 0) GC TC 15 GO TO 2 G NST AT=9 GC TC 80 IF (F (1) .GT. PMAX) GG TO 25 GC TC 27 NSTAI= 1 GO TO 8 C DC 50 I=1,KEHE1S IF (FCNE. EQ.CCOfiP (I) ) GO TC 30 IF (FONE. GT. CCCEP (I) ) GO TC 10 GC TO 6G AILCNH=CCORK (I) GC TO 7G EIH1 = CCCEiF (I) J = I IE 1=1+ 1 CONTINUE PIE 1 = COCBP(IP 1) PI = FC NE Hit 1 = C C C R K (J) HIE 1=CCOBM (IP1) IF (PIPI.EC.E1ft 1)ALLOWM=MIM1 70 IF(E2Â£1.Â£Â£.Â£I!S1) GC TO 70 AIÂ£CBH=Â£lIfl1 + (P1-PIH1)* (MIP1-KIM 1) / (PI P1-PIM 1) IF(KEY.NE.2) GG TC 75 KAXÂ£CE=DAES (F (2) ) If (DABS (F(4)) .GT.MAXMOM)MAXMCM=DABS(F (4)) GC TC 78 75 MAXMCM=DABS (F (3)) IF (LABS (F (6) ) GT K AX MOB) MAXMCM= CABS ( F ( 6) ) 78 IF (KAXMCM.G1. ALLQSH) NSTAT=1 80 RETURN ENE C S DEBOOTINE WYTHE (BRWYLD,BLWYLD,NSTOHY) C SEEOTIHE WYTHE WILL POINT, FOR EACH WALL LEVEL (EVERY C Â£ INCHES), THE VERTICAL LOAD CARRIED EY EACH WALL WYTHE C AND THE PERCENT OF THE TOTAL LOAD WHICH THAT C REPRESENTS. RE AI HEIGHT DCELE PRECISION BRKYLD,BLWYID,TCTLD DISENSION BEW YLE (NSTORY) ,BLW Y L D (NSIOR Y) WRITE(6,10) 10 FCEAT ('1',66 { )/ ',14X,WALL WYTHE VERTICAL LOAD $ VERSUS HEIGHT/* ,66 (=)) WRITE (6,20) 20 FOBHAT (10/ ,HEIGHT*,2X,*BRICK WYTHE 1CAD',2X * CE , $ TCTAI' ,2X,'BLOCK WYTHE ICAD,2X,*54 C TOTAL '/' ') HEIGHT=0,0 DC EC L = 1,N STOR Y ICTLB=BRW YLE (L) +BLW YLD (L) BfiPEH=10C.0*BRWYLD(L)/TOTID ELPEB=10 0.0*Â£LWYLD(L)/TCTLD WRITE(6,30)HEIGHT,BR fcYLD (L) ,BRPER,BIWYIE (I) ,ELPER 30 F CBM AT (* F 5. 1,4X, D13. b 6X, F5.2, 6X D 13.6,6X F 5.2) EEIGHT=HEIGHT+Â£.0 CCNT1NUE RETURN 50 280 END 281 APPENDIX B SUBROUTINES B.1 Subroutine NULL Subroutine NULL will set all values in an n x m matrix [a] equal to zero. Table B.1 defines the nomenclature used in this subroutine. Figure B.1 is an algorithm for this subroutine. B.2 Subroutine EQUAL Subroutine EQUAL will create an n x m matrix [a] that is equal to an n x m matrix [b]. In other words, it will create a matrix [a] which is identical to a second matrix [b]. Table B.2 defines the nomenclature used in this subroutine. Figure B.2 is an algorithm for this subroutine. B.5 Subroutine ADD Subroutine ADD will add an n x m matrix [b] to an n x m matrix [a] to yield an n x m matrix [c]. Table B.3 defines the nomenclature used in this subroutine. Figure B.3 is an algorithm for this subroutine. B.4 Subroutine MULT Subroutine MULT will multiply an n1 x n2 matrix [a] times an n2 x n3 matrix [b] to obtain an n1 x n3 matrix [c]. Table B.4 defines the nomenclature used in this subroutine. Figure B.4 is an algorithm for this subroutine. B.5 Subroutine SMULT Subroutine SMULT will multiply all values in an n x m matrix [b] by a number A to obtain an n x m matrix [C]. Table B.5 defines the 282 283 Table B.1 Nomenclature For Subroutine NULL VARIABLE TYPE DEFINITION A DOUBLE PRECISION H x M MATRIX WHOSE VALUES ARE SET EQUAL TO ZERO N INTEGER NUMBER OF ROWS IN MATRIX [a] M INTEGER NUMBER OF COLUMNS IN MATRIX [a] 284 (start) r { FOR EACH ~ROw) -|for each column) I L_- SET [A] = 0 (return) Figure B.1 Algorithm For Subroutine NULL 285 Table B.2 Nomenclature For Subroutine EQUAL VARIABLE TYPE DEFINITION A DOUBLE PRECISION N x M MATRIX WHOSE VALUES ARE SET EQUAL TO THE CORRESPONDING VALUES IN MATRIX [B] B DOUBLE PRECISION N x M MATRIX WHOSE VALUES ARE COPIED INTO THE CORRESPONDING VALUES IN MATRIX [a] N INTEGER NUMBER OF ROWS IN MATRICES [a], [b] M INTEGER NUMBER OF COLUMNS IN. MATRICES [a], [b] 286 (start) r | FOR EACH ROT) r' -*j FOR EACH COLUMN) SET [A - [B] (return) Figure B.2 Algorithm For Subroutine EQUAL 287 Table B.3 Nomenclature For Subroutine ADD VARIABLE TYPE DEFINITION A DOUBLE PRECISION N x M MATRIX WHOSE VALUES ARE ADDED TO THOSE OF MATRIX [b] TO OBTAIN MATRIX [c] B C N M DOUBLE PRECISION DOUBLE PRECISION INTEGER INTEGER N x M MATRIX WHOSE VALUES ARE ADDED TO THOSE OF MATRIX [a] TO OBTAIN MATRIX [c] N x M MATRIX WHOSE VALUES ARE THE SUM OF THE CORRESPONDING VALUES IN [a], [b] NUMBER OF ROWS IN MATRICES [a], [b], [c] NUMBER OF COLUMNS IN MATRICES [A], [B], [C] 238 (start) r~ I I l_L- * FOR EACH ROW) FOR EACH COLUMN) : SET [c] = [a] + [b] (return) Figure B.3 Algorithm For Subroutine ADD 289 Table B.4 Nomenclature For Subroutine MULT VARIABLE TYPE DEFINITION A DOUBLE PRECISION N1 x N2 MATRIX WHOSE VALUES ARE MULTIPLED BY MATRIX [b] TO OBTAIN MATRIX [c] B DOUBLE PRECISION N2 x N3 MATRIX WHOSE VALUES ARE MULTIPLED BY MATRIX [a] TO OBTAIN MATRIX [C] C DOUBLE PRECISION N1 x N3 MATRIX WHOSE VALUES ARE THE PRODUCT OF THE CORRESPONDING VALUES IN [A], [B] N1 INTEGER NUMBER OF ROWS IN MATRICES [a], [c] N2 INTEGER NUMBER OF COLUMNS IN MATRIX [a], ALSO NUMBER OF ROWS IN MATRIX [b] N3 INTEGER NUMBER OF COLUMNS IN MATRICES [b], [c] 290 (start) FOR EACH ROW IN MATRIX [cj) | | -FOR EACH COLUMN IN MATRIX [c]) 1 !' 1 1 *-Z_T'Z_rZ: SET [C] = [A] x [B] (return) Figure B.4 Algorithm For Subroutine MULT 291 Table B.5 Nomenclature For Subroutine SMULT VARIABLE TYPE DEFINITION A DOUBLE PRECISION NUMBER BY WHICH ALL VALUES IN MATRIX [b] ARE MULTIPLIED TO OBTAIN MATRIX [c] B C N M DOUBLE PRECISION DOUBLE PRECISION INTEGER INTEGER N x M MATRIX WHOSE VALUES ARE MULTIPLIED BY THE NUMBER A TO OBTAIN MATRIX [c] N x M MATRIX WHOSE VALUES ARE THE PRODUCT OF A TIMES [c] NUMBER OF ROWS IN MATRICES [b], [c] NUMBER OF COLUMNS IN MATRICES [b], [c] 292 nomenclature used in this subroutine. Figure B.5 is an algorithm for this subroutine. B.6 Subroutine BMULT Subroutine BMULT will multiply a symmetric banded matrix by a vector to obtain another vector. It will perform the operation [a] x [b] = [c], where [a] is an n x ml matrix which contains the upper triangular portion of a symmetric n x n matrix that has a bandwidth of (2 x ml) 1 and ml equals half the bandwidth (including the diagonal) of matrix [a]. Matrix [b] is an n x 1 matrix whose values are multiplied by those in matrix [a] to obtain the n x 1 matrix [c]. Table B.6 defines the nomenclature used in this subroutine. Figure B.6 is an algorithm for this subroutine. B.7 Subroutine INSERT Subroutine INSERT will insert a matrix [b] into a larger matrix [a]. In other words, it will add the values of [b] to certain values in [a]. If KEY equals 1, then [a] is an n x n matrix and [b] is an m x m matrix. If KEY equals 2, then [a] is an n x 1 matrix and [b] is an m x 1 matrix. The m x 1 matrix [INDEX] contains the numbers which identify the positions in [a] to which the appropriate values in [b] should be added. If TYPE equals 2, then all the numbers in the [INDEX] matrix are used. If TYPE equals 3, then the fifth number in the [INDEX] matrix is not used. Table B.7 defines the nomenclature used in this subroutine. Figure B.7 is an algorithm for this subroutine. B.8 Subroutine BNSERT Subroutine BNSERT will insert the upper triangular portion of an m x m symmetric matrix [b] into a matrix [c]. Matrix [c] contains the upper triangular portion of an n x n matrix that has a bandwidth of 293 i I r~ I I LJr-^ (start) - FOR EACH ROW IN MATRIX [c]) FOR EACH COLUMN IN MATRIX [c] SET [c] A x [b] (return) Figure B.5 Algorithm For Subroutine SMULT 294 Table B.6 Nomenclature For Subroutine BMULT VARIABLE TYPE DEFINITION A DOUBLE PRECISION NxHI MATRIX WHICH CONTAINS THE UPPER TRIANGULAR PORTION OF A SYMMETRIC N x N MATRIX THAT HAS A BANDWIDTH OF (2 x M1) 1 B DOUBLE PRECISION N x 1 MATRIX WHOSE VALUES ARE MULTIPLIED BY THOSE IN MATRIX [a] TO OBTAIN MATRIX [c] C DOUBLE PRECISION N x 1 MATRIX WHOSE VALUES ARE THE PRODUCT OF [A] x [b] N INTEGER NUMBER OF ROWS IN MATRIX [a], ALSO NUMBER OF ROWS IN COLUMN VECTORS [b], [c] M1 INTEGER HALF THE BANDWIDTH (INCLUDING THE DIAGONAL) OF MATRIX [a] 295 (start) r jFOR EACH ROW IN COLUMN VECTOR [c]) I SET [c] = [A] x [B] BUT PERFORM MULTIPLICATION AS IF [a] WERE AN N x N MATRIX (return) Figure B.6 Algorithm For Subroutine BMULT 296 Table B.7 Nomenclature For Subroutine INSERT VARIABLE TYPE DEFINITION A DOUBLE PRECISION MATRIX INTO WHICH MATRIX [b] IS INSERTED; IF KEY = 1, THEN [a] IS AN N x N MATRIX AND IF KEY = 2, THEN [a] IS AN N x 1 MATRIX B DOUBLE PRECISION MATRIX WHICH IS INSERTED INTO MATRIX [A]; IF KEY = 1, THEN [b] IS AN M x M MATRIX AND IF KEY = 2, THEN [b] IS AN M x 1 MATRIX KEY INTEGER VARIABLE THAT KEEPS TRACK OF DIMENSIONS OF MATRICES [a], [b] TYPE INTEGER VARIABLE THAT KEEPS TRACK OF WHICH NUMBERS IN THE INDEX MATRIX ARE USED; IF TYPE = 1, THEN THE FIRST THREE NUMBERS IN THE INDEX MATRIX ARE NOT USED; IF TYPE = 2, THEN ALL THE NUMBERS IN THE INDEX MATRIX ARE USED; IF TYPE = 3, THEN THE FIFTH NUMBER IN THE INDEX MATRIX IS NOT USED N INTEGER NUMBER OF ROWS (AND POSSIBLY NUMBER OF COLUMNS) IN MATRIX [a] M INTEGER NUMBER OF ROWS (AND POSSIBLY NUMBER OF COLUMNS) IN MATRIX [b] INDEX INTEGER M x 1 INDEX MATRIX WHOSE VALUES GIVE THE POSITIONS IN MATRIX [a] INTO WHICH THE VALUES IN MATRIX [b] ARE INSERTED 297 (start) L FOR EACH NUMBER IN THE INDEX MATRIX) INSERT THE VALUE IN MATRIX [b] INTO THE PROPER LOCATION IN MATRIX [a] (return) Figure B.7 Algorithm For Subroutine INSERT 298 (2 x ml) 1, where ral equals half the bandwidth (including the diagonal) of the n x n matrix. The actual n x n matrix is not stored but after matrix [b] is inserted into the n x ml matrix [c], the result is equivalent to what would be obtained if the m x m matrix [b] were inserted into the actual n x n matrix. The m x 1 matrix [INDEX] contains the numbers which identify the positions in [c] to which the appropriate values in [b] should be added. If TYPE equals 1, then the first three numbers in the [INDEX] matrix are not used. If TYPE equals 2, then all the numbers in the [INDEX] matrix are used. If TYPE equals 3, then the fifth number in the [INDEX] matrix is not used. Table B.8 defines the nomenclature used in this subroutine. Figure B.8 is an algorithm for this subroutine. B.9 Subroutine EXTRAK Subroutine EXTRAK will pick up a matrix [b] out of a larger matrix [a]. The values of [a] are not affected. The old values of [b], if any, are replaced by the values found in [a]. If KEY equals 1, then [a] is an n x n matrix and [b] is an m x m matrix. If KEY equals 2, then [a] is an n x 1 matrix and [b] is an m x 1 matrix. The mx1 matrix [INDEX] contains the numbers which identify the positions in [a] from which matrix [b] is constructed. If TYPE equals 1, then the first three numbers in the [INDEX] matrix are not used. If TYPE equals 2, then all the numbers in the [INDEX] matrix are used. If TYPE equals 3, then the fifth number in the [INDEX] matrix is not used. Table B.9 defines the nomenclature used in this subroutine. Figure B.9 is an algorithm for this subroutine. 299 Table B.8 Nomenclature For Subroutine BNSERT VARIABLE TYPE DEFINITION C DOUBLE PRECISION N x M1 MATRIX WHICH CONTAINS THE UPPER TRIANGULAR PORTION OF A SYMMETRIC N x N MATRIX THAT HAS A BANDWIDTH OF (2 x Ml) 1 B DOUBLE PRECISION M x M SYMMETRIC MATRIX, THE UPPER TRIANGULAR PORTION OF WHICH IS INSERTED INTO MATRIX [c] TYPE INTEGER VARIABLE THAT KEEPS TRACK OF WHICH NUMBERS IN THE INDEX MATRIX ARE USED; IF TYPE 1, THEN THE FIRST THREE NUMBERS IN THE INDEX MATRIX ARE NOT USED; IF TYPE 2, THEN ALL THE NUMBERS IN THE INDEX MATRIX ARE USED; IF TYPE 3, THEN THE FIFTH NUMBER IN THE INDEX MATRIX IS NOT USED N INTEGER HUMBER OF ROWS IN MATRIX [c] M1 INTEGER HALF THE BANDWIDTH (INCLUDING THE DIAGONAL) OF MATRIX [c] M INTEGER NUMBER OF ROWS AND NUMBER OF COLUMNS IN MATRIX [B] INDEX INTEGER M x 1 INDEX MATRIX WHOSE VALUES GIVE THE POSITIONS IN MATRIX [c] INTO WHICH THE VALUES IN THE UPPER TRIANGULAR PORTION OF MATRIX [B] ARE INSERTED 300 (start) r I FOR EACH NUMBER IN THE INDEX MATRIX) INSERT THE VALUE IN THE UPPER TRIANGULAR PORTION OF MATRIX [b] INTO THE PROPER LOCATION IN MATRIX [c] (return) Figure B.8 Algorithm For Subroutine BNSERT 301 Table B.9 Nomenclature For Subroutine EXTRAK VARIABLE TYPE DEFINITION A DOUBLE PRECISION MATRIX FROM WHICH MATRIX [b] IS EXTRACTED; IF KEY = 1, THEN [a] IS AN N x N MATRIX AND IF KEY = 2, THEN [a] IS AN N x 1 MATRIX B DOUBLE PRECISION MATRIX WHICH IS EXTRACTED FROM MATRIX [A]; IF KEY = 1, THEN [b] IS AN M x M MATRIX AND IF KEY = 2, THEN [b] IS AN M x 1 MATRIX KEY INTEGER VARIABLE THAT KEEPS TRACK OF DIMENSIONS OF MATRICES [a], [b] TYPE INTEGER VARIABLE THAT KEEPS TRACK OF WHICH NUMBERS IN THE INDEX MATRIX ARE USED; IF TYPE = 1, THEN THE FIRST THREE NUMBERS IN THE INDEX MATRIX ARE NOT USED; IF TYPE 2, THEN ALL THE NUMBERS IN THE INDEX MATRIX ARE USED; IF TYPE = 3, THEN THE FIFTH NUMBER IN THE INDEX MATRIX IS NOT USED N INTEGER NUMBER OF ROWS (AND POSSIBLY NUMBER OF COLUMNS) IN MATRIX [a] M INTEGER NUMBER OF ROWS (AND POSSIBLY NUMBER OF COLUMNS) IN MATRIX [b] INDEX INTEGER M x 1 INDEX MATRIX WHOSE VALUES GIVE THE POSITIONS IN MATRIX [a] FROM WHICH THE VALUES FOR MATRIX [b] ARE EXTRACTED 302 r~ i l l Figure (start) FOR EACH NUMBER IN THE INDEX MATRIX) EXTRACT FROM THE PROPER LOCATION IN MATRIX [A] THE VALUE FOR MATRIX [b] (return) .9 Algorithm For Subroutine EXTRAK 303 B.10 Subroutine PULROW Subroutine PULROW will pull the first T numbers in row I out of an n x 6 matrix [b] and store them in a T x 1 matrix [a]. Table B.10 defines the nomenclature used in this subroutine. Figure B.10 is an algorithm for this subroutine. B.11 Subroutine PULMAT Subroutine PULMAT will pull a T x T matrix out of a 6 x 6 x n matrix [b] and store the values in a T x T matrix [a]. The value of T must be less than or equal to 6. The value of I identifies which 6x6 matrix in [b] matrix [a] is to be obtained from. Table B.11 defines the nomenclature used in this subroutine. Figure B.11 is an algorithm for this subroutine. B.12 Subroutine GAUSS 1 Subroutine GAUSS1 solves a matrix equation of the form [a] x [x] = [b], where [a] is an n x n matrix which has a bandwidth of (2 x ml) 1. The solution is a standard Gauss Elimination for a symmetric banded matrix. Matrix [x] is the n x 1 matrix that stores the solution, [c] is the n x ml matrix used to store the upper triangular portion of the symmetric nxn matrix [A] with a bandwidth of (2 x ml) 1, and [b] is the n x 1 matrix that stores the product of [a] x [x]. The variable ml equals half the bandwidth (including the diagonal) of matrix [a]. For a regular problem, KEY should equal 1. When subroutine GAUSS1 is called and matrix [a] is the same as in the last call to GAUSS1, KEY should equal 2. Table B.12 defines the nomenclature used in this subroutine. Figure B.12 is an algorithm for this subroutine. 304 Table B.10 Nomenclature For Subroutine PULROW VARIABLE TYPE DEFINITION A INTEGER T x 1 MATRIX WHOSE VALUES CORRESPOND TO THOSE IN ROW I OF MATRIX [b] B INTEGER N x 6 MATRIX FROM WHICH A ROW IS COPIED INTO MATRIX [a] I INTEGER ROW IN MATRIX [b] WHICH IS COPIED INTO MATRIX [A] T INTEGER NUMBER OF NUMBERS IN ROW I OF MATRIX [b] WHICH ARE COPIED INTO MATRIX [a] N INTEGER NUMBER OF ROWS IN MATRIX [b] 305 (start) r FOR FIRST T NUMBERS IN ROW I OF MATRIX [b]) COPY VALUE INTO MATRIX [a] (return) Figure B.10 Algorithm For Subroutine PULROW 306 Table B.11 Nomenclature For Subroutine PULMAT VARIABLE TYPE DEFINITION A DOUBLE PRECISION T x T MATRIX WHOSE VALUES ARE EXTRACTED FROM MATRIX [b] B DOUBLE PRECISION 6 x 6 x N MATRIX FROM WHICH MATRIX [a] IS EXTRACTED I INTEGER NUMBER OF 6 x 6 MATRIX IN MATRIX [b] FROM WHICH T x T MATRIX [a] IS EXTRACTED (MATRIX [A] IS EXTRACTED FROM MATRIX 6 x 6 x I IN MATRIX [b]) T INTEGER NUMBER OF ROWS AND NUMBER OF COLUMNS IN MATRIX [A] N INTEGER NUMBER OF 6 x 6 MATRICES IN MATRIX [b] 307 I- I L Figure (start) FOR FIRST T x T VALUES IN 6 x 6 x MATRIX IN MATRIX [b] COPY VALUE INTO MATRIX [a] (return) B.11 Algorithm For Subroutine PULMAT 3C8 Table B.12 Nomenclature For Subroutine GAUSS 1 VARIABLE TYPE DEFINITION X DOUBLE PRECISION N x 1 MATRIX WHICH STORES THE SOLUTION TO THE MATRIX EQUATION OF THE FORM [a] [x] [b] WHERE THE SOLUTION IS CALCULATED BY THE SUBROUTINE C DOUBLE PRECISION N x Ml MATRIX WHICH STORES THE UPPER TRIANGULAR PORTION OF A SYMMETRIC N x N MATRIX [A] THAT HAS A BANDWIDTH OF (2 x Ml) 1 B DOUBLE PRECISION N x 1 MATRIX THAT STORES THE NUMBERS THAT EQUAL [a] [x] M1 INTEGER HALF THE BANDWIDTH (INCLUDING THE DIAGONAL) OF MATRIX [a] N INTEGER NUMBER OF ROWS IN MATRICES [x], [cj, [b] KEY INTEGER VARIABLE THAT INDICATES TYPE OF PROBLEM; KEY = 1 IS FOR A REGULAR PROBLEM; KEY =* 2 IS FOR THE CASE IN WHICH MATRIX [a] (AND THEREFORE MATRIX [c]) IS THE SAME AS IN THE LAST CALL TO GAUSS 1 BUT MATRIX [B] IS DIFFERENT 309 Figure B.12 Algorithm For Subroutine GAUSS1 310 B.13 Subroutine STACON Subroutine STACON solves a matrix equation of the form [a] x [x] = [b], where [a] is an n x n matrix which has a bandwidth of (2 x ml) 1. The solution uses Static Condensation and standard Gauss Elimination for a symmetric banded matrix. Matrix [x] is the n x 1 matrix that stores the solution, [c] is the n x ml matrix used to store the upper triangular portion of the symmetric n x n matrix [a] with a bandwidth of (2 x ml) 1, and [B] is the n x 1 matrix that stores the product of [a] x [x]. The variable ml equals half the bandwidth (including the diagonal) of matrix [a]. Table B.13 defines the nomenclature used in this subroutine. Figure B.13 is an algorithm for this subroutine. B.14 Subroutine TITLE Subroutine TITLE will read and print one comment card of alphanumeric or character data. It will skip 2 lines before printing the comments. The comments cannot exceed column 80. Table B.14 defines the nomenclature used in this subroutine. Figure B.14 is an algorithm for this subroutine. B.15 Subroutine READ Subroutine READ will print a heading, read and print two comment cards, read and print the wall description data and calculate the modulus of elasticity and area in shear for the brick and for the block. Table B.15 defines the nomenclature used in this subroutine. Figure B.15 is an algorithm for this subroutine. B.16 Subroutine COORD Subroutine COORD will read and print the x and y coordinates of up to 20 points for a curve. The variable NPHTS is the number of points in 311 Table B.13 Nomenclature For Subroutine STACON VARIABLE TYPE DEFINITION X C B K4 F2 W2 K2 K2W2 DOUBLE PRECISION N x 1 MATRIX WHICH STORES THE SOLUTION TO THE MATRIX EQUATION OF THE FORM [A] [X] [B] WHERE THE SOLUTION IS CALCULATED BY THE SUBROUTINE DOUBLE PRECISION N x HI MATRIX WHICH STORES THE UPPER TRIANGULAR PORTION OF A SYMMETRIC N x N MATRIX [a] THAT HAS A BANDWIDTH OF (2 x M1) 1 DOUBLE PRECISION N x 1 MATRIX THAT STORES THE NUMBERS THAT EQUAL [a] [x] DOUBLE PRECISION MATRIX WHICH EQUALS THE PORTION OF THE [c] MATRIX THAT IS EQUIVALENT TO THE BOTTOM LEFT FOURTH OF THE N x N MATRIX [A] DOUBLE PRECISION MATRIX WHICH EQUALS THE BOTTOM HALF OF THE N x 1 MATRIX [b] DOUBLE PRECISION MATRIX WHICH EQUALS THE BOTTOM HALF OF THE N x 1 MATRIX [x] DOUBLE PRECISION MATRIX WHICH EQUALS THE PORTION OF THE [C] MATRIX THAT IS EQUIVALENT TO THE TOP RIGHT FOURTH OF THE N x N MATRIX [a] DOUBLE PRECISION MATRIX WHICH EQUALS [K2] [W2] F1 F1MKW2 K1 W1 M1 N DOUBLE PRECISION MATRIX WHICH EQUALS THE TOP HALF OF THE N x 1 MATRIX [B] DOUBLE PRECISION MATRIX WHICH EQUALS [Fl] ([K2] [W2]) DOUBLE PRECISION MATRIX WHICH EQUALS THE PORTION OF THE [c] MATRIX THAT IS EQUIVALENT TO THE TOP LEFT FOURTH OF THE N x N MATRIX [a] DOUBLE PRECISION MATRIX WHICH EQUALS THE TOP HALF OF THE H x 1 MATRIX [x] INTEGER HALF THE BANDWIDTH (INCLUDING THE DIAGONAL) OF MATRIX [a] INTEGER NUMBER OF ROWS IN MATRICES [x], [c], [b] 312 Table B.13-continued. VARIABLE TYPE DEFINITION NRTP1 INTEGER NUMBER OF ROWS IN THE TOP HALF OF MATRICES [X], [c], [b] PLUS ONE; I.E. NRTP1 = (0.5N) + 1 NRBOT INTEGER NUMBER OF ROWS IN THE BOTTOM HALF OF MATRICES [X], [c], [b]; I.E. NRBOT - N NRTOP NRTOP INTEGER NUMBER OF ROWS IN THE TOP HALF OF MATRICES [X], [c], [b]; I.E. NRTOP = 0.5N 313 Figure B.13 Algorithm For Subroutine STACON 314 Table B.14 Nomenclature For Subroutine TITLE VARIABLE TYPE DEFINITION ALPHA REAL VARIABLE CHARACTER HEADINGS WHICH READS ALPHANUMERIC OR DATA USED TO READ AND PRINT 315 . I'll vrt) /read one line of alphanumeric comments/ SKIP TWO LINES /print the line of alphanumeric comments/ (ret urn) Figure B.14 Algorithm For Subroutine TITLE 316 Table B.15 Nomenclature For Subroutine READ VARIABLE TYPE DEFINITION WHI INTEGER WALL HEIGHT IN INCHES LNVER DOUBLE PRECISION VERTICAL LENGTH OF THE BRICK ELEMENTS AND BLOCK LNHBR DOUBLE PRECISION HORIZONTAL LENGTH OF THE BRICK HALF OF THE COLLAR JOINT ELEMENT LNHBL DOUBLE PRECISION HORIZONTAL LENGTH OF THE BLOCK HALF OF THE COLLAR JOINT ELEMENT NSTORY INTEGER NUMBER OF STORIES OF WALL WHERE EACH STORY IS 8 INCHES TALL NSDOP INTEGER NUMBER OF STRUCTURE DEGREES OF FREEDOM NELEM INTEGER NUMBER OF ELEMENTS ELASBR DOUBLE PRECISION MODULUS OF ELASTICITY OF THE BRICK ARSHBR DOUBLE PRECISION AREA IN SHEAR FOR THE BRICK ELASBL DOUBLE PRECISION MODULUS OF ELASTICITY OF THE BLOCK ARSHBL DOUBLE PRECISION AREA IN SHEAR FOR THE BLOCK 317 iil \rt) /read the wall description data/ /print the wall description data/ calculate the modulus of elasticity AND AREA IN SHEAR FOR THE BRICK CALCULATE THE MODULUS OF ELASTICITY AND AREA IN SHEAR FOR THE BLOCK urn) Figure B.15 Algorithm For Subroutine READ 318 the curve. Matrix [XCOOR] stores the x coordinate of each point and matrix [YCOOR] stores the y coordinate of each point. Table B.16 defines the nomenclature used in this subroutine. Figure B.16 is an algorithm for this subroutine. B.17 Subroutine CURVES Subroutine CURVES will read and print the x and y coordinates of up to 20 points for up to 20 different curves. The variable NOCURV is the number of curves, while the number of points per curve is denoted by NOPPC and CURVAL is the value each curve is identified by. The matrix [XCOOR] stores the x coordinate of each curve where the first subscript is the curve number and the second subscript is the point number. Similarly, matrix [YCOOR] stores the y coordinate of each curve where the first subscript is the curve number and the second subscript is the point number. Table B.17 defines the nomenclature used in this subroutine. Figure B.17 is an algorithm for this subroutine. B.18 Subroutine PRINT Subroutine PRINT will print an n x 1 matrix [a] and identify each row as a degree of freedom. Table B.18 defines the nomenclature used in this subroutine. Figure B.18 is an algorithm for this subroutine. B.19 Subroutine WRITE Subroutine WRITE will print the dimensions n and a of a matrix [a] then print the n x m matrix. Table B.19 defines the nomenclature used in this subroutine. Figure B.19 is an algorithm for this subroutine. B.20 Subroutine BRPDMT Subroutine BRPDMT will calculate the slopes of the linearly approximated P-A and M-Q curves for the brick element the first time it is called. It always finds the slope for the region in which the 319 Table B.16 Nomenclature For Subroutine COORD VARIABLE TYPE DEFINITION NPNTS INTEGER NUMBER OF POINTS IN CURVE XCOOR DOUBLE PRECISION MATRIX WHICH STORES THE X COORDINATE OF EACH POINT YCOOR DOUBLE PRECISION MATRIX WHICH STORES THE Y COORDINATE OF EACH POINT 320 (start) /read the number of points in the curve/ /read the X and y coordinate of each point/ (return) Figure B.16 Algorithm For Subroutine COORD 321 Table B.17 Nomenclature For Subroutine CURVES VARIABLE TYPE DEFINITION NOCURV INTEGER NUMBER OF CURVES NOPPC INTEGER NUMBER OF POINTS PER CURVE CURVAL DOUBLE PRECISION MATRIX WHICH STORES THE VALUE EACH CURVE IS IDENTIFIED BY XCOOR DOUBLE PRECISION MATRIX WHICH STORES THE X COORDINATE OF EACH CURVE; THE FIRST SUBSCRIPT IS THE CURVE NUMBER AND THE SECOND SUBSCRIPT THE POINT NUMBER YCOOR DOUBLE PRECISION MATRIX WHICH STORES THE Y COORDINATE OF EACH CURVE; THE FIRST SUBSCRIPT IS THE CURVE NUMBER AND THE SECOND SUBSCRIPT THE POINT NUMBER 322 (start) /read the number of curves/ / READ THE NUMBER OE POINTS PER CURVE / r FOR EACH CURVE) r /read the curve value/ FOR EACH POINT) /read the x coordinate/ zj READ THE Y COORDINATE / (return) Figure B.17 Algorithm For Subroutine CURVES 323 Table B.18 Nomenclature For Subroutine PRINT VARIABLE TYPE DEFINITION A DOUBLE PRECISION N x 1 MATRIX WHICH THE SUBROUTINE PRINTS N INTEGER NUMBER OF ROWS IN MATRIX [a] 524 (start) PRINT MATRIX [a] IDENTIFYING EACH ROW AS A DEGREE OF FREEDOM (return) Figure B.18 Algorithm For Subroutine PRINT Table B.19 Nomenclature For Subroutine WRITE VARIABLE TYPE DEFINITION A DOUBLE PRECISION N x M MATRIX WHICH THE SUBROUTINE PRINTS N INTEGER NUMBER OF ROWS IN MATRIX [a] M INTEGER NUMBER OF COLUMNS IN MATRIX [a] 326 (start) /PRINT THE DIMENSIONS N AND M OF MATRIX [a] / /print matrix [a]/ (return) Figure B.19 Algorithm For Subroutine WRITE 327 element P value falls on the P-A curve and, therefore, generates the appropriate axial stiffness factor of AE/L. Similarly, it finds the slope for the region in which the average moment falls on the M-9 curve which corresponds to the element P value and generates the rotational stiffness factor of 3EI/L. Table B.20 defines the nomenclature used in this subroutine. Figure B.20 is an algorithm for this subroutine. B.21 Subroutine CJPDMT Subroutine CJPDMT will calculate the slopes of the linearly approximated P-A and M-9 curves for the collar joint element the first time it is called. It always finds the slope for the region in which the element P falls on the P-A curve and, therefore, generates the appropriate shear spring stiffness factor. Similarly, it finds the slope for the region in which the average moment falls on the M-9 curve and generates the moment spring stiffness factor. Table B.21 defines the nomenclature used in this subroutine. Figure B.21 is an algorithm for this subroutine. B.22 Subroutine BLPDMT Subroutine BLPDMT will calculate the slopes of the linearly approximated P-A and M-9 curves for the block element the first time it is called. It always finds the slope for the region in which the element P values falls on the P-A curve and, therefore, generates the appropriate axial stiffness factor of AE/L. Similarly, it finds the slope for the region in which the average moment falls on the M-9 curve which corresponds to the element P value and generates the rotational stiffness factor of 3EI/L. Table B.22 defines the nomenclature used in this subroutine. Figure B.22 is an algorithm for this subroutine. Table B.20 Nomenclature For Subroutine BRPDMT VARIABLE TYPE DEFINITION NBRPDP INTEGER NUMBER OF POINTS IN THE BRICK P-DELTA CURVE NOBRPC INTEGER NUMBER OF BRICK M-THETA CURVES (EACH IS IDENTIFIED BY THE P OR AXIAL LOAD LEVEL FOR WHICH THE MOMENT CURVATURE RELATION SHIP IS VALID) NBRMTP INTEGER NUMBER OF POINTS IN EACH BRICK M-THETA CURVE BRDCOR DOUBLE PRECISION MATRIX THAT STORES BRICK DELTA COORDINATES BRPCOR DOUBLE PRECISION MATRIX THAT STORES BRICK AXIAL LOAD COORDINATES BRPCV DOUBLE PRECISION MATRIX THAT STORES THE VALUE OF P BY WHICH EACH BRICK M-THETA CURVE IS IDENTIFIED BRTCOR DOUBLE PRECISION MATRIX THAT STORES BRICK THETA COORDINATES FOR EACH M-THETA CURVE BRMCOR DOUBLE PRECISION MATRIX THAT STORES BRICK MOMENT COORDINATES FOR EACH M-THETA CURVE BREF DOUBLE PRECISION BRICK ELEMENT FORCE MATRIX SBRPD DOUBLE PRECISION MATRIX THAT STORES THE SLOPES OF THE BRICK AXIAL LOAD DELTA CURVE SBRMT DOUBLE PRECISION MATRIX THAT STORES THE SLOPES OF THE BRICK MOMENT THETA CURVE BRVERF DOUBLE PRECISION VERTICAL (AXIAL) FORCE AT THE BOTTOM OF A BRICK ELEMENT BRMOM DOUBLE PRECISION ABSOLUTE VALUE OF AVERAGE OF END MOMENTS FOR A BRICK ELEMENT BRAEL DOUBLE PRECISION BRICK AE/L (AXIAL STIFFNESS FACTOR) 329 Table B.20-continued. VARIABLE TYPE DEFINITION BR3EIL DOUBLE PRECISION BRICK 3EI/L (ROTATIONAL STIFFNESS FACTOR) ITER INTEGER DO-LOOP PARAMETER 330 Figure B.20 Algorithm For Subroutine BRPDMT 331 Table B.21 Nomenclature For Subroutine CJPDMT VARIABLE TYPE DEFINITION NCJPDP INTEGER NUMBER OF POINTS IN THE COLLAR JOINT P-DELTA CURVE NCJMTP INTEGER NUMBER OF POINTS IN THE COLLAR JOINT M-THETA CURVE CJDCOR DOUBLE PRECISION MATRIX THAT STORES COLLAR JOINT DELTA COORDINATES CJPCOR DOUBLE PRECISION MATRIX THAT STORES COLLAR JOINT VERTICAL LOAD COORDINATES CJTCOR DOUBLE PRECISION MATRIX THAT STORES COLLAR JOINT THETA COORDINATES CJMCOR DOUBLE PRECISION MATRIX THAT STORES COLLAR JOINT MOMENT COORDINATES CJEF DOUBLE PRECISION COLLAR JOINT ELEMENT FORCE MATRIX SCJPD DOUBLE PRECISION MATRIX THAT STORES THE SLOPES OF THE COLLAR JOINT AXIAL LOAD DELTA CURVE SCJMT DOUBLE PRECISION MATRIX THAT STORES THE SLOPES OF THE COLLAR JOINT MOMENT THETA CURVE CJVERF DOUBLE PRECISION ASOLUTE VALUE OF COLLAR JOINT SHEAR FORCE CJMOM DOUBLE PRECISION ABSOLUTE VALUE OF AVERAGE OF END MOMENTS FOR A COLLAR JOINT ELEMENT CJSHRS DOUBLE PRECISION COLLAR JOINT SHEAR SPRING STIFFNESS CJMOMS DOUBLE PRECISION COLLAR JOINT MOMENT SPRING STIFFNESS ITER INTEGER DO-LOOP PARAMETER 332 Figure B.21 Algorithm For Subroutine CJPDMT 333 Table B.22 Nomenclature For Subroutine BLPDMT VARIABLE TYPE DEFINITION NBLPDP INTEGER NUMBER OF POINTS IN THE BLOCK P-DELTA CURVE NOBLPC INTEGER NUMBER OF BLOCK M-THETA CURVES (EACH IS IDENTIFIED BY THE P OR AXIAL LOAD LEVEL FOR WHICH THE MOMENT CURVATURE RELATIONSHIP IS VALID) NBLMTP INTEGER NUMBER OF POINTS IN EACH BLOCK M-THETA CURVE BLDCOR DOUBLE PRECISION MATRIX THAT STORES BLOCK DELTA COORDINATES BLPCOR DOUBLE PRECISION MATRIX THAT STORES BLOCK AXIAL LOAD COORDINATES BLPCV DOUBLE PRECISION MATRIX THAT STORES THE VALUE OF P BY WHICH EACH BLOCK M-THETA CURVE IS IDENTIFIED BLTCOR DOUBLE PRECISION MATRIX THAT STORES BLOCK THETA COORDINATES FOR EACH M-THETA CURVE BLMCOR DOUBLE PRECISION MATRIX THAT STORES BLOCK MOMENT COORDINATES FOR EACH M-THETA CURVE BLEF DOUBLE PRECISION BLOCK ELEMENT FORCE MATRIX SBLPD DOUBLE PRECISION MATRIX THAT STORES THE SLOPES OF THE BLOCK AXIAL LOAD DELTA CURVE SBLMT DOUBLE PRECISION MATRIX THAT STORES THE SLOPES OF THE BLOCK MOMENT THETA CURVES BLVERF DOUBLE PRECISION VERTICAL (AXIAL) FORCE AT THE BOTTOM OF A BLOCK ELEMENT BLMOM DOUBLE PRECISION ABSOLUTE VALUE OF AVERAGE OF END MOMENTS FOR A BLOCK ELEMENT BLAEL DOUBLE PRECISION BLOCK AE/L (AXIAL STIFFNESS FACTOR) Table B.22-continued VARIABLE TYPE DEFINITION BL3EIL DOUBLE PRECISION ITER INTEGER BLOCK 3EI/L (ROTATIONAL STIFFNESS FACTOR) DO-LOOP PARAMETER Figure B.22 Algorithm For Subroutine BLPDMT 336 B.23 Subroutine STIFAC Subroutine STIFAC will select the appropriate stiffness factors for an element based on the values of these factors as determined from the P-A and M-9 curves, the current level of element loading, and the previous level of element loading. Table B.23 defines the nomenclature used in this subroutine. Figure B.23 is an algorithm for this subroutine. B.24 Subroutine BRESM Subroutine BRESM will construct the stiffness matrix for each brick element considering shear deformation and moment magnification. Table B.24 defines the nomeclature used in this subroutine. Figure B.24 is an algorithm for this subroutine. B.25 Subroutine CJESM Subroutine CJESM will construct the stiffness matrix for each collar joint element. Table B.25 defines the nomenclature used in this subroutine. Figure B.25 is an algorithm for this subroutine. B.26 Subroutine BLESM Subroutine BLESM will construct the stiffness matrix for each block element considering shear deformation and moment magnification. Table B.26 defines the nomenclature used in this subroutine. Figure B.26 is an algorithm for this subroutine. B.27 Subroutine INDXBR Subroutine INDXBR will construct the index matrix for each brick element. Table B.27 defines the nomenclature used in this subroutine. Figure B.27 is an algorithm for this subroutine. 371 Table B.23 Nomenclature For Subroutine STIFAC VARIABLE TYPE DEFINITION VERTF DOUBLE PRECISION ELEMENT VERTICAL FORCE MOMF DOUBLE PRECISION ABSOLUTE VALUE OF AVERAGE OF END MOMENTS FOR AN ELEMENT STOVF DOUBLE PRECISION MATRIX THAT STORES THE ABSOLUTE VALUE OF THE VERTICAL FORCE FOR EACH ELEMENT STOMF DOUBLE PRECISION MATRIX THAT STORES THE ABSOLUTE VALUE OF THE AVERAGE OF END MOMENTS FOR EACH ELEMENT VSTIF DOUBLE PRECISION ELEMENT VERTICAL STIFFNESS FACTOR MSTIF DOUBLE PRECISION ELEMENT ROTATIONAL STIFFNESS FACTOR STOVST DOUBLE PRECISION MATRIX THAT STORES THE VERTICAL STIFFNESS FACTOR FOR EACH ELEMENT STOMST DOUBLE PRECISION MATRIX THAT STORES' THE ROTATIONAL STIFFNESS FACTOR FOR EACH ELEMENT TRSTIF INTEGER VARIABLE THAT IDENTIFIES WHETHER OR NOT ANY ELEMENT STIFFNESS FACTORS HAVE CHANGED; 0 = NO; 1 = YES ITER INTEGER DO-LOOP PARAMETER KEY INTEGER VARIABLE THAT IDENTIFIES THE TYPE OF ELEMENT; 1 = BRICK; 2 = COLLAR JOINT; 3 = BLOCK ELEMNO INTEGER ELEMENT NUMBER 338 SELECT VERTICAL STIFFNESS FACTOR BASED ON CURRENT LEVEL OF ELEMENT LOADING SELECT PREVIOUS VERTICAL STIFFNESS FACTOR Figure B.23 Algorithm For Subroutine STIFAC 339 Table B.24 Nomenclature For Subroutine BRESM VARIABLE TYPE DEFINITION BREK DOUBLE PRECISION BREF DOUBLE PRECISION TOTEK DOUBLE PRECISION BRAEL DOUBLE PRECISION BR3EIL DOUBLE PRECISION LN DOUBLE PRECISION ELASBR DOUBLE PRECISION ARSHBR DOUBLE PRECISION I INTEGER BR INTEGER NELEM INTEGER BRICK ELEMENT STIFFNESS MATRIX BRICK ELEMENT FORCE MATRIX MATRIX THAT STORES ALL OF THE ELEMENT STIFFNESS MATRICES BRICK AE/L (AXIAL STIFFNESS FACTOR) BRICK 3EI/L (ROTATIONAL STIFFNESS FACTOR) VERTICAL LENGTH OF THE BRICK ELEMENT MODULUS OF ELASTICITY OF THE BRICK AREA IN SHEAR FOR THE BRICK ELEMENT NUMBER NUMBER OF BRICK ELEMENT DEGREES OF FREEDOM NUMBER OF ELEMENTS 340 (start) CONSTRUCT THE STIFFNESS MATRIX FOR THE BRICK ELEMENT (return) Figure B.24 Algorithm For Subroutine BRESM 341 Table B.25 Nomenclature For Subroutine CJESK VARIABLE TYPE DEFINITION CJEK DOUBLE PRECISION COLLAR JOINT ELEMENT STIFFNESS MATRIX TOTEK DOUBLE PRECISION MATRIX THAT STORES ALL OF THE ELEMENT STIFFNESS MATRICES CJSHRS DOUBLE PRECISION COLLAR JOINT SHEAR SPRING STIFFNESS CJMOMS DOUBLE PRECISION COLLAR JOINT MOMENT SPRING STIFFNESS LNHBR DOUBLE PRECISION HORIZONTAL LENGTH OF THE BRICK HALF OF THE COLLAR JOINT ELEMENT LNHBL DOUBLE PRECISION HORIZONTAL LENGTH OF THE BLOCK HALF OF THE COLLAR JOINT ELEMENT I INTEGER ELEMENT NUMBER CJ INTEGER NUMBER OF COLLAR JOINT ELEMENT DEGREES OF FREEDOM NELEM INTEGER NUMBER OF ELEMENTS 342 (start) CONSTRUCT THE STIFFNESS MATRIX FOR THE COLLAR JOINT ELEMENT (return) Figure B.25 Algorithm For Subroutine CJESM 343 Table B.26 Nomenclature For Subroutine BLESM VARIABLE TYPE DEFINITION BLEK DOUBLE PRECISION BLEF DOUBLE PRECISION TOTEK DOUBLE PRECISION BLAEL DOUBLE PRECISION BL3EIL DOUBLE PRECISION LN DOUBLE PRECISION ELASBL DOUBLE PRECISION ARSHBL DOUBLE PRECISION I INTEGER BL INTEGER NELEM INTEGER BLOCK ELEMENT STIFFNESS MATRIX BLOCK ELEMENT FORCE MATRIX MATRIX THAT STORES ALL OF THE ELEMENT STIFFNESS MATRICES BLOCK AE/L (AXIAL STIFFNESS FACTOR) BLOCK 3EI/L (ROTATIONAL STIFFNESS FACTOR) VERTICAL LENGTH OF THE BLOCK ELEMENT MODULUS OF ELASTICITY OF THE BLOCK AREA IN SHEAR FOR THE BLOCK ELEMENT NUMBER NUMBER OF BLOCK ELEMENT DEGREES OF FREEDOM NUMBER OF ELEMENTS 344 (start) CONSTRUCT THE STIFFNESS MATRIX FOR THE BLOCK ELEMENT (return) Figure B.26 Algorithm For Subroutine BLESM 345 Table B.27 Nomenclature For Subroutine INDXBR VARIABLE TYPE DEFINITION IBR INTEGER INDEX MATRIX FOR A BRICK ELEMENT TOTEM INTEGER MATRIX THAT STORES INDEX MATRICES ALL OF THE ELEMENT I INTEGER ELEMENT NUMBER NEMT INTEGER NUMBER OF ELEMENTS MINUS TWO BR INTEGER NUMBER OF BRICK ELEMENT DEGREES OF FREEDOM NELEM INTEGER NUMBER OF ELEMENTS 34b (start) CONSTRUCT THE INDEX MATRIX FOR THE BRICK ELEMENT (return) Figure B.27 Algorithm For Subroutine INDXBR 347 B.28 Subroutine INDXCJ Subroutine INDXCJ will construct the index matrix for each collar joint element. Table B.28 defines the nomenclature used in this subroutine. Figure B.28 is an algorithm for this subroutine. B.29 Subroutine IHDXBL Subroutine IHDXBL will construct the index matrix for each block element. Table B.29 defines the nomenclature used in this subroutine. Figure B.29 is an algorithm for this subroutine. B.30 Subroutine FORCES Subroutine FORCES reads and prints the number of degrees of freedom that are loaded, the initial load at each degree of freedom, the load increment at each degree of freedom, and the maximum load at each degree of freedom. Table B.30 defines the nomenclature used in this subroutine. Figure B.30 is an algorithm for this subroutine. B.31 Subroutine APPLYF Subroutine APPLYF will place the initial loads in the structure force matrix the first time it is called. Thereafter, it will increment each load according to the information read by subroutine FORCES from the data set. It will not increment any load beyond its specified maximum value. Table B.31 defines the nomenclature used in this subroutine. Figure B.31 is an algorithm for this subroutine. B.32 Subroutine PLATEK Subroutine PLATEK will construct a structure stiffness matrix which only considers the two structure degrees of freedom at the top of the wall that correspond to the test plate. It will do this by modifying the original structure stiffness matrix which considers four degrees of 348 Table B.28 Nomenclature For Subroutine INDXCJ VARIABLE TYPE DEFINITION ICJ INTEGER INDEX MATRIX FOR A COLLAR JOINT ELEMENT TOTEIN INTEGER MATRIX THAT STORES ALL OF THE ELEMENT INDEX MATRICES I INTEGER ELEMENT NUMBER NEMO INTEGER NUMBER OF ELEMENTS MINUS ONE CJ INTEGER NUMBER OF COLLAR JOINT ELEMENT DEGREES OF FREEDOM NELEM INTEGER NUMBER OF ELEMENTS 349 (start) CONSTRUCT THE INDEX MATRIX FOR THE COLLAR JOINT ELEMENT (return) Figure B.28 Algorithm For Subroutine INDXCJ 350 Table B.29 Nomenclature For Subroutine INDXBL VARIABLE TYPE DEFINITION IBL INTEGER INDEX MATRIX FOR A BLOCK ELEMENT TOTEIN INTEGER MATRIX THAT STORES ALL OF THE ELEMENT INDEX MATRICES I INTEGER ELEMENT NUMBER BL INTEGER NUMBER OF BLOCK ELEMENT DEGREES OF FREEDOM NELEM INTEGER NUMBER OF ELEMENTS 351 (start) CONSTRUCT THE INDEX MATRIX FOR THE BLOCK ELEMENT (return) Figure B.29 Algorithm For Subroutine INDXBL 352 Table B.30 Nomenclature For Subroutine FORCES VARIABLE TYPE DEFINITION NOF INTEGER NUMBER OF FORCES NDOF INTEGER NUMBER OF THE DEGREE OF FREEDOM OF A PARTICULAR FORCE SFINIT DOUBLE PRECISION STORES THE INITIAL STRUCTURE FORCE FOR EACH LOADED DEGREE OF FREEDOM SFINCR DOUBLE PRECISION STORES THE STRUCTURE FORCE INCREMENT FOR EACH LOADED DEGREE OF FREEDOM SFMAX DOUBLE PRECISION STORES THE MAXIMUM STRUCTURE FORCE FOR EACH LOADED DEGREE OF FREEDOM NSDOF INTEGER NUMBER OF STRUCTURE DEGREES OF FREEDOM 'Si 353 (start) READ THE STRUCTURE LOAD APPLICATION INFORMATION AND PRINT IT / (return) Figure B.30 Algorithm For Subroutine FORCES 354 Table B-31 Nomenclature For Subroutine APPLYF VARIABLE TYPE DEFINITION STRF DOUBLE PRECISION STRUCTURE FORCE MATRIX NSDOF INTEGER NUMBER OF STRUCTURE DEGREES OF FREEDOM NOF INTEGER NUMBER OF FORCES NDOF INTEGER NUMBER OF THE DEGREE OF FREEDOM OF A PARTICULAR FORCE SFINIT DOUBLE PRECISION STORES THE INITIAL STRUCTURE FORCE FOR EACH LOADED DEGREE OF FREEDOM SFINCR DOUBLE PRECISION STORES THE STRUCTURE FORCE INCREMENT FOR EACH LOADED DEGREE OF FREEDOM SFMAX DOUBLE PRECISION STORES THE MAXIMUM STRUCTURE FORCE FOR EACH LOADED DEGREE OF FREEDOM ITER INTEGER DO-LOOP PARAMETER KEY1 INTEGER VARIABLE THAT KEEPS TRACK OF THE APPLI CATION OF STRUCTURE LOADS; 0 = THE LOADS HAVE NOT REACHED THEIR MAXIMUM VALUE; 1 THE LOADS ARE ALL AT THEIR RESPECTIVE MAXIMUMS AND NO FURTHER INCREASES ARE REQUIRED 355 Figure B.31 Algorithm For Subroutine APPLYF 356 freedom at the wall top. Table B.32 defines the nomenclature used in this subroutine. Figure B.32 is an algorithm for this subroutine. B.33 Subroutine DISPLA Subroutine DISPLA will calculate the actual structure displacement matrix from the structure displacement matrix that only considers two degrees of freedom at the top of the wall. Table B.33 defines the nomenclature used in this subroutine. Figure B.33 is an algorithm for this subroutine. B.34 Subroutine CHKTOL Subroutine CHKTOL checks to see if each number in an n x 1 matrix [A] is less than the value of TOLER. If each number is, the variable KEY is assigned a value of 0. If at least one number is not, KEY is set equal to 1. Table B.34 defines the nomenclature used in this subrou tine. Figure B.34 is an algorithm for this subroutine. B.35 Subroutine CHKFAI Subroutine CHKFAI will check to see if an element has failed. For a brick and block element, the actual axial load and moment is compared with the allowable axial load and moment as provided by the P-M interaction diagram for that type of element. For the collar joint element, the actual shear force is compared with the allowable shear force. Table B.35 defines the nomenclature used in this subroutine. Figure B.35 is an algorithm for this subroutine. B.36 Subroutine WYTHE Subroutine WYTHE will print, for each wall level (every 8 inches), the vertical load carried by each wall wythe and the percent of the total vertical load which that represents. Table B.36 defines the 357 Table B.32 Nomenclature For Subroutine PLATEK VARIABLE TYPE DEFINITION STRK DOUBLE PRECISION STRUCTURE STIFFNESS MATRIX NSTRK DOUBLE PRECISION STRUCTURE STIFFNESS MATRIX USED IN SOLVING FOR STRUCTURE DISPLACEMENTS M1 INTEGER HALF THE BANDWIDTH (INCLUDING THE DIAGONAL) OF THE STRUCTURE STIFFNESS MATRIX M1P1 INTEGER HALF THE BANDWIDTH PLUS ONE NSDOF INTEGER NUMBER OF STRUCTURE DEGREES OF FREEDOM NSDH2 INTEGER NUMBER OF STRUCTURE DEGREES OF FREEDOM MINUS TWO LNHBR DOUBLE PRECISION HORIZONTAL LENGTH OF THE BRICK HALF OF THE COLLAR JOINT ELEMENT LNHBL DOUBLE PRECISION HORIZONTAL LENGTH OF THE BLOCK HALF OF THE COLLAR JOINT ELEMENT 358 (start) CONSTRUCT STRUCTURE STIFFNESS MATRIX THAT ONLY CONSIDERS TWO DEGREES OF FREEDOM AT THE TOP OF THE WALL (CORRESPONDING TO TEST PLATE) FROM THE ORIGINAL STRUCTURE STIFFNESS MATRIX THAT CONSIDERS FOUR DEGREES OF FREEDOM AT THE WALL TOP (return) Figure B.32 Algorithm For Subroutine PLATEK 359 Table B.33 Nomenclature For Subroutine DISPLA VARIABLE TYPE DEFINITION STRW DOUBLE PRECISION STRUCTURE DISPLACEMENT MATRIX NSTRW DOUBLE PRECISION STRUCTURE DISPLACEMENT MATRIX NSDOF INTEGER NUMBER OF STRUCTURE DEGREES OF FREEDOM NSDM2 INTEGER NUMBER OF STRUCTURE DEGREES OF FREEDOM MINUS TWO LNHBR DOUBLE PRECISION HORIZONTAL LENGTH OF THE BRICK HALF OF THE COLLAR JOINT ELEMENT LNHBL DOUBLE PRECISION HORIZONTAL LENGTH OF THE BLOCK HALF OF THE COLLAR JOINT ELEMENT 360 (start) CONSTRUCT THE ACTUAL STRUCTURE DISPLACEMENT MATRIX FROM THE STRUCTURE DISPLACEMENT MATRIX THAT ONLY CONSIDERS TWO DEGREES OF FREEDOM AT THE WALL TOP (return) Figure B.33 Algorithm For Subroutine DISPLA 361 Table B.34 Nomenclature For Subroutine CHKTOL VARIABLE TYPE DEFINITION A DOUBLE PRECISION N x 1 MATRIX N INTEGER NUMBER OF ROWS IN MATRIX [a] TOLER DOUBLE PRECISION NUMBER WHICH EACH VALUE IN MATRIX [a] IS COMPARED WITH KEY INTEGER VARIABLE WHICH IDENTIFIES WHETHER OR NOT EACH VALUE IN MATRIX [a] IS LESS THAN TOLER; 0 = EACH VALUE IS LESS THAN TOLER; 1 = AT LEAST ONE VALUE IS NOT LESS THAN TOLER 362 (start) DETERMINE IF EACH VALUE IN MATRIX [A] IS LESS THAN TOLER (return) Figure B.34 Algorithm For Subroutine CHKTOL 363 Table B.35 Nomenclature For Subroutine CHKFAI VARIABLE TYPE DEFINITION F DOUBLE PRECISION ELEMENT FORCE MATRIX NPMPTS INTEGER NUMBER OF POINTS IN THE ELEMENT INTERACTION DIAGRAM COORM DOUBLE PRECISION MATRIX THAT CONTAINS ELEMENT MOMENT COORDINATES COORP DOUBLE PRECISION MATRIX THAT CONTAINS ELEMENT VERTICAL LOAD COORDINATES PMAX DOUBLE PRECISION MAXIMUM ELEMENT P (VERTICAL LOAD CARRYING CAPACITY) KEY INTEGER VARIABLE THAT IDENTIFIES THE TYPE OF ELEMENT; 1 BRICK; 2 = COLLAR JOINT; 3 = BLOCK NSTAT INTEGER VARIABLE THAT IDENTIFIES THE FAILURE STATUS OF THE ELEMENT; 0 = ELEMENT HAS NOT FAILED; 1 = ELEMENT HAS FAILED, 9 = ELEMENT IS IN AXIAL TENSION 364 (start) CHECK ELEMENT FOR FAILURE (return) Figure B.35 Algorithm For Subroutine CHKFA 365 Table B.36 Nomenclature For Subroutine WYTHE VARIABLE TYPE DEFINITION BRWYLD DOUBLE PRECISION MATRIX THAT STORES THE VERTICAL LOAD CARRIED BY THE BRICK WYTHE AT EACH WALL LEVEL BLWYLD DOUBLE PRECISION MATRIX THAT STORES THE VERTICAL LOAD CARRIED BY THE BLOCK WYTHE AT EACH WALL LEVEL HSTORY INTEGER NUMBER OF STORIES OF WALL WHERE EACH STORY IS 8 INCHES TALL 366 nomenclature used in this subroutine. Figure B.36 is an algorithm for this subroutine. 367 (start) PRINT THE VERTICAL LOAD CARRIED BY EACH WALL WYTHE AND THE PERCENT 'OF THE TOTAL LOAD WHICH THAT REPRESENTS, (return) Figure B.36 Algorithm For Subroutine WYTHE APPENDIX C USER'S MANUAL C.1 General Information To use the program, the user must create a data file in the manner outlined in the next section. As previously mentioned, all input (and output) is in basic units of inches, pounds, or radians. The data describing the application of forces consider the structure degrees of freedom with the test plate, i.e., only two degrees of freedom at the top of the wall. The general structure degrees of freedom were shown in Figure 4.2. As an example, suppose it is desired to apply a compressive force to that wall at an eccentricity of 2 inches towards the block wythe (the right hand side). The test plate degrees of freedom simply replace and with W^, and W1Q and with W-jy. The new is present midway between the old and (at "the geometric center of the wall) and the new is present midway between the old W^0 and The original direction of each is maintained. If the compressive force has a value of 4000 lb, then one would place in the data set the information that degree of freedom 16 is loaded with a value of -4000 (since a compressive force acts down on the wall) and degree of freedom 17 is loaded with a value of -1000 x 2 = -2000 (since the moment is equal to P x e and would be clockwise). Earlier, it was mentioned that the pattern for numbering structure degrees of freedom is valid regardless of wall height. This makes determining what degrees of freedom to apply a load at almost trivial. 568 369 Loads will normally be applied at the top of a wall, and when it is desired to apply transverse loads, also at the lateral structure degrees of freedom. The pertinent equations are as follows: No. of 8 inch levels = (Wall hgt. in inches) x (1/8) (C.1) No. of stories = (No. of 8 inch levels) (C.2) No. of structure DOF without test plate = [(No. of stories) x 5] 1 (C.3) No. of structure DOF with test plate = (No. of structure DOF without test plate) 2 (C.4) Axial load DOF on test plate = (No. of structure DOF with test plate) 1 (C.5) Moment DOF on test plate = (No. of structure DOF with test plate) (C.6) Lateral DOF = 3f 8, 13 n+5 up to [(No. of structure DOF with test plate) 4] (C.7) Thus, if a wall is 10 feet high, it is 120 in high, has 15 stories, 74 structure DOF without test plate, 72 structure DOF with test plate, the axial load DOF on the test plate occurs at DOF number 71, the moment DOF on the test plates occurs at DOF number 72, and the lateral DOF = 3, 8, 13, 18, 23, 28, 33, 38, 43, 48, 53, 58, 63 and 68. Because the stiffness for each element is a function of the load level the element experiences, the load should be applied in relatively small increments, say 4000 pounds. The data input guide will indicate that a load at a degree of freedom is read as the initial force, the force increment, and the maximum force. To load the wall to failure, a very large maximum force should be specified, like 1x10^. J70 C.2 Data Input Guide Table C.1 shows how the data are to be placed in the data file. It is preceded by a copy of the data file used for example number 2 which contains line numbers for illustrative purposes. Care must be taken to insert the data in the proper columns and lines, but once a file is created, the data can be altered for a new run by simply editing the file. In short, the data file demands care to construct, but is very easily changed. The user should verify from the output that the data read and used are the same as originally intended. LINE EXAMPLE #2 H/T =12 PLATE LOAD ECCENTRICITY=0 IN. 1. DESCRIPTION OF HALL 2, 120 3. 8. 0D + 00 4. 2, OD + OO 5. 3. OD + OO 6, 24 7. MATERIAL PROPERTIES 8, BRICK ELEMENT P-DELTA CORVE 9. 6 10. PT DELTA COORDINATE P COORDINATE 11. 1 0. OD + OO 0.OD+OO 12, 2 4. OD -0 3 9.2175D+04 13. 3 8;, 0D-03 1.67175D+05 14, 4 1.2D-02 2.32725D+05 15. 5 1.6D-02 2,91450D+05 16, 6 2.COD-02 3.38850D+05 17. B RICK ELEMENT M-TilEIA CURVES 18. 3 19. 4 20, 1.0D + 05 21. PT THETA COORDINATE M COORDINATE 22. 1 0.OD+OO 0.OD+OO 23. 2 2. 117D-03 5.93D+04 24. 3 9,7 02D-03 1.187D+05 25. 4 1 .D-02 1.53591D+05 26. 1.5D+05 27. PT THETA COORDINATE H COORDINATE 28, 1 0.OD+OO 0.OD+OO 29. 2 3,13 ID-03 8 895D+04 30. 3 7.865D-03 1.3 35D+05 31. 4 1.0D-02 1,53591D+05 32, 2.D+05 33. PT THETA COORDINATE ft COORDINATE 1 O.OD+OO 0.0D+00 2 4 1G9D-03 1 186D+05 3 6.616D-03 1.78D+05 4 1.QD-02 2,58179D+ 05 COLLAR JOINT ELEMENT P-DELTA CURVE 2 PT DELTA COORDINATE P COORDINATE 1 0.0D + 00 0,D+00 2 4.613D-02 9.777D+03 COLLAR JOINT ELEMENT M-THETA CURVE 2 PT THETA COORDINATE H COORDINATE 1 0.0D+00 0.0D+00 2 1.0D + 0 0, 0D+00 BLOCK ELEMENT P-DELTA CURVE 5 PT DELTA COORDINATE P COORDINATE 1 0, 0D + 00 2 4.0D-03 3 8,0D-03 4 1.2D-02 5 1 68D-02 BLOCK ELEMENT 3 5 2.5D+4 O.OD+OO 2.5125D+04 5,025D+04 7.5375D+04 9,315D+04 fi-THETA CURVES PT 1 2 3 4 5 3,50-03 5. 0D+04 H COORDINATE 0.0D+00 2.3325D+04 3,5D+04 4.6675D+04 8.3114D+04 THETA COORDINATE 0. 0D + 00 5.63D-04 8,46D-04 1.49D-03 PT THETA COORDINATE M COORDINATE 1 0.0D+00 0.0D+00 34. 35. 36. 3 7. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 372 2 1.035D-03 4.665D+04 3 1.674D-03 7.0D+04 4 2.962D-03 9,335D+04 5 3.5D-03 1.03103D+05 7.5D+04 PT THETA COORDINATE H COORDINATE 1 0, GD + 00 0,OD+OO 2 7. 52D-04 3.5025D+04 3 1,505D-03 6.9975D+04 4 3.245D-03 1.05D+05 5 3.5D-03 1.10132D+05 BRICK ELEMENT P-H INTERACTION DIAGRAM 10 PT M COORDINATE P COORDINATE 1 0, OD+OO 0,OD+OO 2 3.75D+04 1.875D+04 3 1,0725D+05 7.5D+04 4 1.5D+05 1.2D+05 5 1.65D + 05 1.5D+05 6 1.73475D+05 1.9275D+05 7 1.6875D+05 2,25D+05 8 1.26D+05 3.0D+05 9 1, 08 *1320 + 05 3.3885D+05 10 0.0D+0 3.3885D+05 BLOCK 8 PT ELEMENT P-M INTERACTION DIAGRAM M COORDINATE P COORDINATE 1 7.5D+03 0.OD+OO 2 4.875D+04 1,875D+04 3 8.625D+04 3.75D+04 4 9,996D + 04 5.1D+04 5 9.675D+04 5.625D+04 6 6,75D+04 7.5D+04 7 3.9123D+04 9.315D+04 8 0.OD + 00 9,3 15D+04 MAXIMUM DRICE ELE BENT COMPRESSIVE LOAD CAPACITY 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. 80. 81 . 82. 8 3. 84. 85. 86. 89. 90. 91. 92. 93. 94. 95. 96. 97. 98. 99. 100, 101, 102. 103. 104. 105. 3. 3885D + 05 MAXIMUM CO LIAR JOINT ELEMENT SHEAS LOAD CAPACITY 9,777D+03 MAXIMUM BLOCK ELEMENT COMPRESSIVE LOAD CAPACITY 9 3 15D + 04 STRUCTUSE FORCE APPLICATION INFORMATION 1 DOF INITIAL FORCE FORCE INCREMENT MAXIMUM FORCE 71 -4.QD+03 -4.0D+03 -4.0D+05 INSTRUCTIONS ON PRINTING INTERMEDIATE RESULTS: 1 171 -5.6D+04 106. 107. 108. 109. 110. 111. 112. 113. 114. 115. 116. 117. Table C.1 Data Input Guide LINE VARIABLE DESCRIPTION FIELD START IN TYPE DESCRIPTOR COLUMN 1 2 3 WHI 4 LNVER COMMENT CARD WITH DESCRIPTION OF PROBLEM COMMENT OF DESCRIPTION OF WALL WALL HEIGHT IN INCHES; MUST BE A MULTIPLE OF 8 VERTICAL LENGTH OF THE BRICK AND BLOCK ELEMENTS 5 LNHBR 6 LNHBL 7 WDEPTH 8 9 10 NBRPDP 11 12-17 J BRDCOR HORIZONTAL LENGTH OF THE BRICK HALF OF THE COLLAR JOINT ELEMENT HORIZONTAL LENGTH OF THE BLOCK HALF OF THE COLLAR JOINT ELEMENT WALL DEPTH IN INCHES COMMENT OF 'MATERIAL PROPERTIES' COMMENT OF 'BRICK ELEMENT P-DELTA CURVE' NUMBER OF POINTS IN THE BRICK P-DELTA CURVE COMMENT OF 'PT' COMMENT OF 'DELTA COORDINATE' COMMENT OF 'P COORDINATE' POINT NUMBER BRICK DELTA COORDINATE 9 INTEGER 14 1 DOUBLE D10.4 1 PRECISION DOUBLE DIO.4 1 PRECISION DOUBLE DIO.4 1 PRECISION INTEGER 14 1 9 6 INTEGER 13 1 2 6 24 INTEGER 13 1 DOUBLE D13.6 4 PRECISION BRPCOR BRICK P COORDINATE DOUBLE PRECISION D13.6 18 - COMMENT OF 'BRICK ELEMENT M-THETA CURVES - - 19 NOBRPC NUMBER OF BRICK M-THETA CURVES INTEGER 13 20 NBRMTP NUMBER OF POINTS IN EACH BRICK M-THETA CURVE INTEGER 13 21 BRPCV VALUE OF P WHICH IDENTIFIES THE BRICK M-THETA THAT FOLLOWS CURVE DOUBLE PRECISION D13.6 22 - COMMENT OF 'PT' - - - COMMENT OF THETA COORDINATE' - - - COMMENT OF 'M COORDINATE - - 23-26 K POINT NUMBER INTEGER 13 BRTCOR BRICK THETA COORDINATE DOUBLE PRECISION D13.6 BRMCOR BRICK MOMENT COORDINATE DOUBLE PRECISION D13.6 27 BRPCV VALUE OF P WHICH IDENTIFIES THE BRICK M-THETA THAT FOLLOWS CURVE DOUBLE PRECISION D13.6 28 - COMMENT OF PT - - COMMENT OF 'THETA COORDINATE __ COMMENT OP 'M COORDINATE Table C.1-continued LINE VARIABLE DESCRIPTION TYPE FIELD DESCRIPTOR START IN COLUMN 29-32 K POINT NUMBER INTEGER 13 1 BRTCOR BRICK THETA COORDINATE DOUBLE PRECISION D13-6 4 BRMCOR BRICK MOMENT COORDINATE DOUBLE PRECISION D13.6 17 33 BRPCV VALUE OF P WHICH IDENTIFIES THE BRICK M-THETA CURVE THAT FOLLOWS DOUBLE PRECISION D13.6 1 34 - COMMENT OF 'PT' - - 2 - COMMENT OF 'THETA COORDINATE' - - 6 - COMMENT OF 'M COORDINATE' - - 24 35-38 K POINT NUMBER INTEGER 13 1 BRTCOR BRICK THETA COORDINATE DOUBLE PRECISION D13.6 4 BRMCOR BRICK MOMENT COORDINATE DOUBLE PRECISION D13.6 17 39 - COMMENT OF 'COLLAR JOINT ELEMENT P-DELTA CURVE' - - 2 40 NCJPDP NUMBER OF POINTS IN THE COLLAR JOINT P-DELTA CURVE INTEGER 13 1 41 COMMENT OF 'PT' 2 COMMENT OF 'DELTA COORDINATE COMMENT OF *P COORDINATE' 42-43 J POINT NUMBER CJDCOR COLLAR JOINT DELTA COORDINATE CJPCOR COLLAR JOINT P COORDINATE 44 45 NCJMTP 46 47-48 J CJTCOR COMMENT OF 'COLLAR JOINT ELEMENT M-THETA CURVE' HUMBER OF POINTS IN THE COLLAR JOINT M-THETA CURVE COMMENT OF 'PT' COMMENT OF 'THETA COORDINATE' COMMENT OF 'M COORDINATE' POINT NUMBER COLLAR JOINT THETA COORDINATE CJMCOR COLLAR JOINT MOMENT COORDINATE 49 COMMENT OF 'BLOCK ELEMENT P-DELTA CURVE' 50 NBLPDP NUMBER OF POINTS IN THE BLOCK P-DELTA CURVE 51 COMMENT OF 'PT' INTEGER 13 DOUBLE PRECISION D13*6 DOUBLE PRECISION D13.6 - - INTEGER 13 - - - - INTEGER 13 DOUBLE PRECISION D13.6 DOUBLE PRECISION D13.6 - - INTEGER 13 COMMENT OF 'DELTA COORDINATE' Table C.1-continued LINE VARIABLE DESCRIPTION FIELD START IN TYPE DESCRIPTOR COLUMN 52-56 J BLDCOR COMMENT OF 'P COORDINATE POINT NUMBER BLOCK DELTA COORDINATE BLPCOR BLOCK P COORDINATE 57 58 NOBLPC 59 NBLMTP 60 BLPCV 61 62-66 K BLTCOR COMMENT OF BLOCK ELEMENT M-THETA CURVES' NUMBER OF BLOCK M-THETA CURVES NUMBER OF POINTS IN EACH BLOCK M-THETA CURVE VALUE OF P WHICH IDENTIFIES THE BLOCK M-THETA CURVE THAT FOLLOWS COMMENT OF PT COMMENT OF 'THETA COORDINATE' COMMENT OF 'M COORDINATE POINT NUMBER BLOCK THETA COORDINATE 24 INTEGER 13 1 DOUBLE D13.6 4 PRECISION DOUBLE D13.6 17 PRECISION 5 INTEGER 13 1 INTEGER 13 1 DOUBLE D13.6 1 PRECISION 2 6 24 INTEGER 13 1 DOUBLE D13.6 4 PRECISION BLMCOR BLOCK M COORDINATE DOUBLE PRECISION D13.6 67 BLPCV VALUE OF P WHICH IDENTIFIES THE THAT FOLLOWS BLOCK M-THETA CURVE DOUBLE PRECISION D13-6 68 - COMMENT OF 'PT' - - - COMMENT OF 'THETA COORDINATE' - - - COMMENT OF 'M COORDINATE' - - 69-73 K POINT NUMBER INTEGER 13 BLTCOR BLOCK THETA COORDINATE DOUBLE PRECISION D1 3.6 BLMCOR BLOCK M COORDINATE DOUBLE PRECISION D13-6 74 BLPCV VALUE OF P WHICH IDENTIFIES THE THAT FOLLOWS BLOCK M-THETA CURVE DOUBLE PRECISION D13.6 75 - COMMENT OF 'PT' - - - COMMENT OF 'THETA COORDINATE' - - - COMMENT OF 'M COORDINATE - - 76-80 K POINT NUMBER INTEGER 13 BLTCOR BLOCK THETA COORDINATE DOUBLE PRECISION D13*6 BLMCOR BLOCK M COORDINATE DOUBLE PRECISION D13.6 81 COMMENT OF 'BRICK ELEMENT P-M INTERACTION DIAGRAM' 17 1 2 6 24 1 4 17 1 2 6 24 1 4 17 1 oeÂ£ Table C.1-continued LINE VARIABLE DESCRIPTION TYPE FIELD DESCRIPTOR START IN COLUMN 82 NBRIDP NUMBER OF POINTS IN THE BRICK INTERACTION DIAGRAM INTEGER 13 1 83 - COMMENT OF 'PT' - - 2 - COMMENT OF 'M COORDINATE - - 8 - COMMENT OF P COORDINATE' - - 24 84-93 J POINT NUMBER INTEGER 13 1 BRIDH BRICK INTERACTION DIAGRAM M COORDINATE DOUBLE PRECISION D13.6 4 BRIDP BRICK INTERACTION DIAGRAM P COORDINATE DOUBLE PRECISION D13.6 17 94 - COMMENT OF 'BLOCK ELEMENT P-M INTERACTION DIAGRAM' - - 1 95 NBLIDP NUMBER OF POINTS IN THE BLOCK INTERACTION DIAGRAM INTEGER 13 1 96 - COMMENT OF 'PT' - - 2 - COMMENT OF 'M COORDINATE' - - 8 - COMMENT OF *P COORDINATE - - 24 97-104 J POINT NUMBER INTEGER 13 1 BLIDM BLOCK INTERACTION DIAGRAM M COORDINATE DOUBLE PRECISION D13-6 4 BLIDP BLOCK INTERACTION DIAGRAM P COORDINATE DOUBLE PRECISION D13.6 17 105 - COMMENT OF 'MAXIMUM BRICK ELEMENT COMPRESSIVE LOAD CAPACITY' - - 1 106 BRMAXP BRICK MAXIMUM P DOUBLE PRECISION D13-6 1 107 - COMMENT OF 'MAXIMUM COLLAR JOINT ELEMENT SHEAR LOAD CAPACITY - - 1 108 CJMAXP COLLAR JOINT MAXIMUM P DOUBLE PRECISION D13-6 1 109 - COMMENT OF 'MAXIMUM BLOCK ELEMENT COMPRESSIVE LOAD CAPACITY - - 1 110 BLMAXP BLOCK MAXIMUM P DOUBLE PRECISION D13*6 1 111 - COMMENT OF 'STRUCTURE FORCE APPLICATION INFORMATION' - - 7 112 NOF NUMBER OF FORCES INTEGER 13 1 113 - COMMENT OF 'DOF' - - 1 - COMMENT OF 'INITIAL FORCE' - - 7 - COMMENT OF 'FORCE INCREMENT' - - 23 - COMMENT OF 'MAXIMUM FORCE - - 41 114 NDOF NUMBER OF THE DEGREE OF FREEDOM OF THE FORCE INTEGER 13 1 SFINIT INITIAL FORCE VALUE FOR THAT DEGREE OF FREEDOM DOUBLE PRECISION D13.6 4 Table C.1-continued LINE VARIABLE DESCRIPTION FIELD START IN TYPE DESCRIPTOR COLUMN SFINCR FORCE INCREMENT FOR THAT DEGREE OF FREEDOM DOUBLE PRECISION D13.6 17 SFMAX MAXIMUM FORCE VALUE FOR THAT DEGREE OF FREEDOM DOUBLE PRECISION D13.6 30 115 - COMMENT OF 'INSTRUCTIONS ON PRINTING INTERMEDIATE RESULTS' - - 1 116 PROUT VARIABLE THAT IDENTIFIES WHETHER OR NOT AN INTERMEDIATE PRINTOUT IS DESIRED; 0 = NO; 1 = YES INTEGER 11 1 117 NOFN FORCE NUMBER (1 FOR FIRST FORCE LISTED, 2 FOR SECOND, ETC.) INTEGER 13 1 NFORCE NUMBER OF FORCE DEGREE OF FREEDOM, MULTIPLES OF WHICH WILL CAUSE AN INTERMEDIATE PRINTOUT INTEGER 13 4 FMULT MULTIPLIES OF NFORCE FOR WHICH AN INTERMEDIATE PRINTOUT IS DESIRED DOUBLE PRECISION D13.6 7 APPENDIX D DATA FILES FOR NUMERICAL EXAMPLES D.1 Example Number 1 GAD ECCENTRIC!T Y EXAMP LE #1 11/1=3, 6 PLATE L DESCRIPTION OF WALL 96 8,OD+OU 2. ODfOO 3.0D+00 24 MATERIAL PROPERTIES Eli ICE ELEMENT P-DELTA CURVE 6 PT 1 2 3 4 5 6 3 4 PT 1 2 3 4 PT 1 2 3 4 PT 1 DELTA COORDINATE P COORDINATE 0, 0D+00 4.0D-03 8,00-03 1.2D-02 1.6D-02 2.08D-02 O,D+00 9.2175D+04 1.671750+05 2.32725D+05 2.91450D+05 3.3885GD+5 BRICK ELEMENT H-IHETA CURVES 1.0D+05 THETA COORDINATE 0.0B+00 2.117D-03 5 9.702D-03 1. 1,00-02 1,21 1.5D+05 THETA COORDINATE 0.0D+00 3, 131D-03 8. 7,865D-03 1. 1.0D-02 1,53 2, 0D+05 THETA COORDINATE 0.0D+00 M COORDINATE 0.0D+00 ,93D+0 4 187D+05 033D+05 H COORDINATE 0.OD+0 895D+04 3 35D+0 5 591D+05 M COORDINATE 0,0D+00 1.25 IN 2 4,109D-Q3 1186D+05 3 6.616D-3 1.70D + 0 5 4 1,OD-02 2.58179D+05 COLLAR JCINT ELEMENT P-DELTA CURVE 2 PT DELTA COORDINATE P COORDINATE 1 0.0D+00 O.OD+OO 2 4,613D-02 9,777D+03 COLLAR JOINT ELE HE NT H-THETA CURVE 2 PT THETA COORDINATE M COORDINATE 1 0.0D+00 .0D+0 2 1.0D+00 0.D+00 BLOCK ELEMENT P-DELTA CURVE 5 PT DELTA COORDINATE P COORDINATE 1 0.D+00 2 4.0D-03 3 8,OD-03 4 1,20-02 5 1.68D-02 BLOCK ELEMENT 3 0.QD + OO 2.5125D+04 5025D+04 7.5375D+04 9,3 15D+0 4 K-THETA CURVES d PT 1 2 3 4 5 PT 1 2 3 2,5D + 04 THETA COORDINATE II COORDINATE G. CD+QO 0.0D+00 5,63D-04 2,3325D+04 8.46D-04 3.5D+04 1.49D-03 4.6675D+04 3.5D-03 8,31140+04 5. OD +04 THETA COORDINATE E COORDINATE 0, 00 + 00 0. 01)+0 0 1, 035D-03 4.665D+04 1.674D-03 7.0D+04 4 PT 1 2 3 4 5 2.9620-03 9.3350+04 3.5D-03 1.Q3103D+05 7.5D+04 THETA COORDINATE H COORDINATE 0.OD + 00 7.52D-04 1,505D-03 3.245D-03 3.5D-3 0,OD+OO 3.5025D+04 6.9975D+04 1. G5D+0 5 1.10132D+05 BRICK ELEMENT P-M 10 PT H COORDINATE 1 0. OD+OO 2 3.75D+4 3 1.0 725D + 05 4 1.5D+05 5 1.65D+05 6 1. 73475D + 05 7 1 ,68750 + 05 8 1.26D+05 9 1,00 4 3 2D+0 5 10 0.OD+OO BLOCK ELEMENT P-M I INTERACTION DIAGRAM P COORDINATE 0.OD+OO 1.075D+04 7,5Df04 1. 2D+05 1.50+05 1.9275D+05 2,25D+05 3.0D+05 3,38850+05 3.38S5D+05 NTERACTION DIAGRAM 8 PT 1 2 3 4 5 6 7 8 H CCCRDIMA1E 7.5D+03 4.875D+04 8.625D+04 9,996D+04 9,675D+04 6.75D+04 3.912 3 D + 0 4 0. OD+OO P COORDINATE 0.OD+OO 1.875D+04 3.75D+04 5. 1D + 04 5.6 25D + 04 7,5D+04 9.315D+04 9.315D+04 MAXIMUM ERICK ELEMENT COMPRESSIVE LOAD CAPACITY 3. 385D + G5 MAXIMUM COLLAR JOINT ELEMENT SHEAR LOAD CAPACITY oo 9.777D+03 MAXIMUM BLOCK ELEMENT COMPRESSIVE LOAD CAPACITY 9.315D+04 STRUCTURE FORCE APPLICATION INFORMATION 2 DOF INITIAL FORCE FORCE INCREMENT MAXIMUM 56 -4,00+03 -4.D+Q3 -4.0D+05 57 -5,00+03 -5,0D+03 -5.0D+05 INSTRUCTIONS ON PRINTING INTERMEDIATE RESULTS: 0 FORCE D.2 Example Number 2 EXAMPLE #2 H/T=12 PLATE LOAD ECCENTBICITY=G IK. DESCElPITON Of WALL 120 8.CD+GQ 2. OD + OO 3, D + QC 24 6 PT 1 2 3 4 5 6 3 4 PT 1 2 3 4 PT 1 2 3 4 PT 1 MATERIAL PROPERTIES BRICK ELEMENT P-JCELTA CURVE DELTA COORDINATE P COORDINATE 0, OD + OO 4.0D-03 6.0D-03 1. 2D-02 1.6D-02 2.C8D-2 BRICK ELEMENT 0,00+00 9.2175D+04 1,o7175D+U3 2,32725D+05 2.9145OD+05 3.38850D+05 M-THETA CURVES 1,0D+05 THETA COORDINATE M COORDINATE 0.GD+00 5,93D+04 1.187D+05 1.21033D+05 0.OD+OO 2. 117D-3 9.702D-03 1,00-02 1.5D+05 THETA COORDINATE B COORDINATE C,OD+OO 0,OD+OO 3.131D-03 8.895D+04 7.865D-3 1.335D+05 1.OD-02 1.53591D+05 2. 0E+05 THETA COORDINATE fl COORDINATE 0-OD+OO 0.OD+OO VjJ VO 2 4.1C9D-Q3 1186D+05 3 6.616D-03 1.780+05 4 1.0D-02 2,58179D+05 COLLAR JOINT ELEMENT P-DELTA CORVE 2 PT DELTA COORDINATE P COORDINATE 1 O.OD + OO O.OD+OO 2 4,6 13D-02 9 777D+03 COLLAR JOINT ELEMENT B-THETA CURVE 2 PT THETA COORDINATE F COORDINATE 1 0.004-00 U.D+Q0 2 1. OD4-00 0.0D+00 BLOCK ELEMENT P-DELTA CURVE PT 1 2 3 4 5 3 5 PT 1 2 3 4 5 PT 1 2 3 DELTA COORDINATE P COORDINATE 0,QD+00 4.0D-03 .0D-03 1.20-02 1,60D-02 0.0D+00 2.5125D+04 5.0250+04 7.5375D+04 9,315D+04 BLOCK ELEMENT M-THETA CURVES 2.5E+04 THETA COORDINATE M COORDINATE 0.0D+00 0.0D+00 5.63D-04 2.3325D+04 8,460-04 3.5D+04 1.49D-03 4.6675D+04 3.5D-03 8.3114D+4 5.0D+04 THETA COORDINATE H COORDINATE O.OD+OO O.OD+OO 1.0350-03 4,665D+04 1.674P-03 7.0D+04 vi KO r\> 03 -03 1 PI 1 2 3 4 5 BRICK 10 PT 1 2 3 4 5 6 7 8 9 10 BLOCK 8 PT 1 2 3 4 5 6 1 8 2.S62D 3.5D 7.5E + 04 THETA COORDINATE 0,0D+00 7.52D-04 1.5G5D-03 3.245D-03 3.50-03 ELEMENT P-H 9.335D+04 031C3D+05 H COORDINATE 0,D+00 5025D+04 9975D+04 1.05D+05 1 10132D+05 INTERACTION DIAGRAM 3, 6 M 1 COORDINATE 0. CD+00 3.75D+04 0725D+05 1.5D+05 1.65D+05 1.73475D+05 U6875D + 05 1.26D+05 1. 08432D+ 05 0.D+00 ELEMENT P M P COORDINATE 0,OD+OO 1.B75D+04 7,50+04 1. 2D+0 5 1,5D+05 1.9275D+05 2,25D+05 3.0D+05 3.3885D+05 3,3885D+05 INTERACTION M COORDINATE 7.5D+03 4,875D+04 8.625D+4 9, S96D+04 9,675D+04 6,75D + 04 .9123D+04 0,OD+OO DIAGRAM E COORDINATE 0.OD+OO 1,875D+04 3.75D+04 5,1D+04 5,625D+04 7.5D+04 9.315D+04 9.315D+04 COMPRESSIVE LOAD CAPACITY MAXIMUM BRICK ELEMENT 3,38850+05 MAXIMUM CC LLAR JOINT ELEMENT SHEAR LOAD CAPACITY UD 9.777C+03 MAXIMUM BICCK ELEMENT COMPRESSIVE LOAD CAPACITY 9,31513 + 04 STRUCTURE FORCE APPLICATION INFORMATICS 1 DOF INITIAL FORCE FORCE INCREMENT MAXIMUM 71 -4.0D+03 -4,0D+U3 -4.0D+05 INSTRUCTIONS ON PRINTING INTERMEDIATE RESULTS: 1 1 71 -5.6D+04 FORCE VjJ -P D.5 Example Number 3 EXAMPLE #3 H/T=12 PLATE LOAD ECCENTBICITY=2 IN. DESCRIPTION OF WALL 120 8. OD + CO 2. OD + CO 3. 0D + 00 2a RATERIAL PROPERTIES BRICK ELEMENT P-DELTA CURVE 6 FT 1 2 3 4 5 6 3 4 PT 1 2 3 4 PT 1 2 3 4 PT 1 EELIA COORD I 0.0D+00 4.OD-03 8.0D-03 NATE t COORDINATE 0.0D+00 9.2175D+04 1.6717 5D+0 5 1.2D-02 1.6D-02 2. C8D-02 BRICK ELEMENT 2.32725D+05 2.9 145QD+Q5 3 .38850D+05 H-1HETA CURVES 1.GE + 05 THETA COORDINATE COORDINATE O.OD + OO O.OD+OO 2.117D-03 5.93D+04 9 7G2D-03 1.187D+05 1.0D-02 1.2 1033D+05 1.5D + 05 THETA COORDINATE H COORDINA! 0.0D+00 3. 131D-03 7.865D-03 1.0D-02 2.GD + 05 0.0D+00 8.895D+04 1.335D+05 1 5359 1D+05 E THETA COORDINATE fl COORDINATE 0.0D+00 0.0D+00 v>) a> 2 4, 1C9D-03 1.186D+05 3 6.616D-03 1. 78D+05 4 1,0D-02 2,5 81790+05 COLLAR JOINT ELEMENT P-DELTA CORVE 2 PT DELTA COORDINATE P COORDINATE 1 O.OD+OO O.OD+OO 2 4,6130-02 9.777D+03 COLLAR JOINT ELEMENT M-THETA CURVE 2 PT THETA COORDINATE H COORDINATE 1 O.OD+OO O.OD+OO 2 1.0D+00 O.OD+OO BLOCK ELEMENT P-DELTA CURVE 5 PT 1 2 3 4 5 3 5 PT 1 2 3 4 5 PT 1 2 3 DELTA COORDINATE I COORDINATE 0.0D+00 4.0D-03 8,0D-03 1.2D-02 1.68D-02 BLOCK ELEMENT O.OD+OO 2.5125D+04 5.025D+04 7.5375D+0 4 9,315D+04 M-THETA CURVES 2.5L+04 THETA COORDINATE H COORDINATE O.OD+OO O.OD+OO 5.63D-04 2.3325D+04 8.46D-04 3.5D+04 1.49D-03 4.6675D+0 4 3.5D-03 8,3114D+04 5.0D+04 THETA COORDINATE fl COORDINATE O.OD+OO O.OD+OO 1.35D-03 4.665D+04 1,674D-03 7. 0D + 04 VjJ -a 4 5 PT 1 2 3 4 5 BRICK 10 PT 1 2 3 4 5 6 7 8 9 10 BLOCK 8 PT 1 2 3 4 5 6 7 8 2.962D-03 3.5D-3 7.50*04 THETA COORD1 0.0D+00 7.52D-04 1.505D-03 3*245D-03 3.50-03 ELEMENT P-M M 1 9. 335D+04 031G3D+Q5 NATE M COORDINATE 0.OD+OO 50250+04 9975D+04 1,C5D+05 1.10132D+05 INTERACTION 3 6 DIAGRAM COORDINATE 0.OD *00 3,750+04 1. 07250 + 05 1,5D+05 1.65D+05 1,734750+05 1.68750+05 1.26D+05 1. G8432D+05 0,OD+OO ELEMENT P-H M P COORDINATE 0,OD+OO 1,8750+04 7.5D+04 1,20+05 1.5D+05 1,92 7 50 + 05 2. 2 5D+05 3.00+05 3.38850+05 3.3885D+05 INTERACTION CCORDIN ATE 7.50+03 4,875D+04 8,6250+04 9.9960+04 9.675D+04 6.75D+04 3.91230+04 0.OD+OO DIAGRAM P COORDINATE 0,OD+OO 1.875D+04 3.75D+04 5, 1D+04 5.625D+04 7,5D+04 9.315D+04 9.315D+04 MAXIMUM BRICK ELEMENT COMPRESSIVE LOAD CAPACITY 3. 38 85C + 05 MAXIMUM CC1LAR JOINT ELEMENT SHEAR LOAD CAPACITY V>1 05 9777B + Q3 MAXIMUM BLOCK ELEMENT COMPRESSIVE LOAD CAPACITY 9.315D+04 STRUCTURE FORCE APPLICATION INFORMATION 2 DOF INITIAL FORCE FORCE INCREMENT MAXIMUM 71 -4.0D+03 -4.0D+03 -4.0D+C5 72 -8,0D+03 -8,0D+03 -8.0D+05 INSTRUCTIONS CN PRINTING INTERMEDIATE RESULTS: 1 1 71 -3.2D+4 FORCE Vjl VO VO D.4 Example Number 4 " > EXAMPLE #4 H/T=20 PLATE LOAD ECCÂ£NTBICITÂ¥=0 DESCRLPTIGN OF WALL 200 8,OD +00 2. CD + OC 3. OD + OO 24 6 PT 1 2 3 4 5 6 3 4 PT 1 2 3 4 PT 1 2 3 4 PT 1 MATESIAL PHCEERTIES BRICK ELEMENT P-DELTA CURVE CELTA COORDINATE P COORDINATE 0,OD+OO 0.OD+OO 4.OD-03 8.0D-03 1. 2D-02 1.6D-02 2.8D-02 BRICK ELEMENT S.2175D+04 1*67175D+05 2* 3272 5D+ 05 2.9 1450D + 05 3.3685OD+ 05 H-THETA CURVES 1.GC+05 THETA COORDINATE 0. OD + CO 2,117D-03 9.7C2D-3 1.0D-02 1.ED+05 THETA COORDINATE 0.OD+OO 3. 131D-03 7.85D-03 1,OD-02 2.0E+G5 M COORDINATE 0.OD+OO 5 93D+04 1.187D+05 1 2 1033D+0 5 M COORDINATE 0.OD+OO 8.895D+04 1.3350+05 1 > 5359 1D+05 THETA COORDINATE R COORDINATE 0,OD+OO 0.OD+OO IN -p* o 2 4.1C9D-Q3 1.186D+05 3 6.616D-03 1.78D+05 4 1,0D-02 2,58179D+05 COLLAR JOINT ELEMENT P-DELTA CURVE 2 PT DELTA COORDINATE P COORDINATE 1 0.D+00 0.0D+00 2 4.613D-02 Â§.7 77D + 03 COLLAS JCLNT ELEMENT M-THETA CURVE PT 1 2 5 PT 1 2 3 4 5 3 5 PT 1 2 3 4 5 PT 1 2 3 THETA COORDINATE K COORDINATE 0.D+00 .0D + 00 1.D+00 .D+0 BLOCK ELEMENT P-DELTA CURVE DELTA COORDINATE P COORDINATE 0.0D+00 0.D+0 4.0D-03 8,OD-03 1.2D-02 1.68D-02 BLOCK ELEMENT 2.5125D+4 5.025D+04 7.5375D+04 9.3 15D+04 M-THETA CORVES 2.5D+04 THETA COORDINATE M COORDINATE 0.OD + OO 5.63D-04 8.46D-04 1.49D-03 3.5D-03 0,OD+OO 2.3325D+04 3.5D+04 4.6675D+04 8.31 14D+04 5.0D+Q4 THETA COORDINATE K COORDINATE C.OD+OO 0.OD + OO 1.035D-03 4.665D+04 1.674D-03 7.0D+04 402 4 5 PI 1 2 3 4 5 BH ICK 10 P'E 1 2 3 4 5 6 7 8 9 10 BLOCK 8 PT 1 2 3 4 c 7 8 9 1 2.S62D-03 3.5D-03 7.5D+04 THETA COORDINATE 0 D + QQ 7.52D-04 1.505D-03 3.245D-03 3.5D-03 ELEMENI P-M 9.335D+04 C31C3D+05 H COORDINATE 0 D+00 5025D+04 9975D+04 1.C5D+05 1.1G132D+05 INTERACTION DIAGRAM 3, 6, M 1 COORDINATE 0.D + 0 3.75D+Q4 0725D+05 1. ED+05 1.65D+05 1. 73 47 D + 05 1.6875D + 05 1.26D+Q5 1. 08 432D +05 0. CD+GO ELEMENT P-M P COORDINATE 0.0D+00 1.875D+4 7.5D+04 1.2D+05 1.5D+05 1.92750+05 2. 25D + 05 3.0D+05 3.3865D+05 3.38 Â£5D* 05 INTERACTION Â¡ COORDINATE 7.5D + G3 4.875D+04 6.62 ED+ 04 9.996D+04 9.675D+04 fa.75D+04 I.9123D+04 0.0D+00 DIAGRAM D COORDINATE 0.OD+OO 1.875D+04 3.75D+04 5. 1D + 04 5.25D+04 7.5D+04 9.315D+04 9.3 15D+04 ELEMENT COMPRESSIVE LOAD CAPACITY MAXIMUM ERICK 3. 33E5C + C5 MAXIMUM COLLAR JOINT ELEMENT SHEAR LOAD CAPACITY o 9.77D+C3 fiAXIflOH EICCK ELEEEKT COMPRESSIVE LOAD CAPACITY 9.315E+4 STRUCTURE FORCE APPLICATION INFCBKATICN 1 DOF INITIAL FORCE FORCE INCREMENT HAXIKUH 121 -4,0D+U3 -4.0D+03 -4.0D+05 INSTRUCTIONS ON PRINTING INTERMEDIATE' RESULTS: 1 1121 -4.8D+04 FORCE 404 D.5 Example Number 5 EXAMPLE #5 H/T=2Q PLATE LOAD ECCE NT R.ICI T Â¥= 2 DESCRIPTION CE WALL 200 8.D + 00 2. GD+CO 3.D + 00 24 6 PT 1 2 3 4 5 6 3 4 PT 1 2 3 4 PT 1 2 3 4 PI 1 MATERIAL PROPERTIES BRICK ELEMENT P-DELTA CURVE DELTA COORDINATE P COORDINATE 0,GD+00 4.GD-03 8,D-3 1. 2D-02 1.6D-02 2- C8D-02 0.0D+00 9.2175D+04 1,671750+05 2.32725D+05 2,9 1450D+05 3.3885OD+ 05 BRICK ELEMENT M-IHETA CURVES 1.CC IN El A G 2. 1 9.7 1 1.ED THETA C J. 1 7.8 1 2. CE THETA C + 05 CCCRDIN AT , CD + CO 17D-03 C2D-G3 ,0D-02 1. + 05 COORDINA! .CD+OO 31D -03 t SD-3 .GD-02 1, + 05 COORDINA! .CD+CO M COORDINATE O.GD+OO 5.93D+04 1.1870 + 05 2 1033D+05 E M COORDINATE 0. CD + OO 8.895D+04 1 335D + Q5 5359 1D+05 E M COORDINATE 0. CD + OO IN. 406 2 4.1C9D-3 1.186D+05 3 6.Â£ 16D-03 1,730+05 4 1.CD-02 2.58179D+05 COLLAR JCIKT ELEMENT P- DELTA CURVE 2 PT DELTA COCEDI NATE Â£ COORDINATE 1 0.0D+00 0.0D+00 2 4.C13D-2 S .777D+03 COLLAR JCINT ELEMENT H-THETA CURVE 2 PT THETA COORDINATE K COORDINATE 1 C. CD + 00 0.GD+00 2 1.0D+00 O.OD+OO BLOCK ELEMENT P-DELTA CURVE PT 1 2 3 4 5 3 5 PT 1 2 3 4 5 PI 1 2 3 DELTA COORDINATE P COORDINATE C.0D+00 4. CD-03 8.0D-3 1.2D-02 1.68D-02 0.0D+00 2.5125D+04 5,025D+04 7.5375D+04 9.315D+04 BLOCK ELEMENT M-THETA CURVES 2,5E+04 THETA COORDINATE M COORDINATE .0D+0 5.63D-04 8.46D-04 1.49D-03 3.5D-03 0.0D+00 2.3325D+04 3.5D+04 4.6675D+04 8.31 14D+04 5.OD + 04 THETA COORDINATE K COORDINATE C.CD+CO C.CD+00 1.035D-03 4.665D+04 1.674D-C3 7.D+04 407 4 2.962D-G3 9.335D+04 5 3.5D-03 1.C31C3D+05 7.5E+Q4 PT THETA COORDINATE E COORDINATE 1 0.0D+0 0.0D+00 2 7.52D-04 3.5025D+4 3 1,505D-03 6.9975D+04 4 3.245D-03 1.C5D+05 5 3.5D-03 1, 10132D+05 BRICK ELEMENT P-M INTERACTION DIAGRAM 10 PT M COORDINATE P COORDINATE 1 0.QD+00 0.0D+00 2 3.75D+Q4 1.875D+04 3 1,07250+05 7,5D+04 4 1-5D +05 1.2D+05 5 1 65D+05 1,50+05 6 1.7347ED+05 1.S275D+05 7 1,6Â£75D+05 2 25D + 05 8 1.26D+05 3, CD+05 9 1. 8432D+05 3.3865D+05 10 C.CD + 00 3.38 Â£50 + 05 BLOCK 8 PT ELEMENT P-fl INTERACTION DIAGRAM M COORDINATE E COORDINATE 1 7.5D+03 0.OD+OO 2 4 875D + 04 1.8 75D+04 3 Â£.E25D +04 3.75D+04 4 9* 996D+04 5. 1D+04 c 9.675D+04 5.625D+04 6 o,75D+04 7.5D+04 7 3.9123D+04 S.315D+04 8 0,0D+00 9,3 15D+04 MAXIMUM BRICK ELEMENT COMPRESSIVE ICAD CAPACITY 3. 38E5E + C5 MAXIEUH COLLAR JOINT ELEMENT SHEAR LCAE CAPACITY COMPRESSIVE IGAD CAPACITY 9.7771*03 MAXIMUM EICCK ELEMENT 9.3 15D + 04 STRUCTURE FORCE APPLICATION INFORMATION 2 DO? INITIAL FORCE FORCE INCREMENT MAXIMUM 121 -4.0D+03 -4. QD*-03 -4.QD+05 122 -fi.OD+03 -8.0D+03 -8.0D+05 INSTEOCTICNS ON PRINTING INTERMEDIATE RESULTS: 1 1121 -8.0D*03 FORCE s REFERENCES 1. Building Code Requirements for Concrete Masonry Structures, ACI 531- 79R, American Concrete Institute, Detroit, Michigan, 1979 (revised 1983). 2. Building Code Requirements for Engineered Brick Masonry, Structural Clay Products Institute, McLean, Virginia, 1969. 3* Chajes, Alexander, Principles of Structural Stability Theory, Prentice-Hall, Inc., Englewood Cliffs, New Jersey, 1974. 4. Fattal, S.G., Cattaneo, L.E., "Structural Performance of Masonry Walls Under Compression and Flexure," Building Science Series 73, National Bureau of Standards, Washington, D.C., June 1976. 5- Grimm, Clayford T., "Strength and Related Properties of Brick Masonry," Journal of the Structural Division, Proceedings of the American Society of Civil Engineers, No. 11066 ST1, New York, New York, January 1975 6. Lybas, John M., Self, M.W., "Design Guidelines for Composite Clay Brick and Concrete Block Masonry," National Science Foundation Research Proposal, Department of Civil Engineering, University of Florida, Gainesville, Florida, January 1984. 7. Manual of Steel Construction, 8th edition, American Institute of Steel Construction, Chicago, Illinois, 1980. 8. Meyer, V.J., Matrix Analysis of Structures, Harper & Row Publishers, New York, New York, 1983* 9* Przemieniecki, J.S., Theory of Matrix Structural Analysis, McGraw- Hill, New York, New York, 1968. 10. Redmond, T.B., Allen, M.H., "Compressive Strength of Composite Brick and Concrete Masonry Walls," Masonry; Past and PresentASTM Technical Publication 589, American Society for Testing and Materials, Philadelphia, Pennsylvania, June 1974. 11. Roman, Oswaldo, "Compressive Strength of Composite Concrete Block and Clay Brick Prisms," Master of Engineering Thesis, University of Florida, Gainesville, Florida, 1982. 12. Rubinstein, Moshe F., Matrix Computer Analysis of Structures, Prentice-Hall Inc., Englewood Cliffs, New Jersey, 1966. 411 13. Self, M.W., "Design Guidelines for Composite Clay Brick and Concrete Block Masonry: Part 1Composite Masonry Prisms," Masonry Research Foundation Research Report, Department of Civil Engineering, University of Florida, Gainesville, Florida, April 1983- 14. Specifications For Design and Construction of Loadbearing Concrete Masonry, National Concrete Masonry Association, Herndon, Virginia, 1976. 15. Tabatabai, Habibollah, "Modulus of Elasticity of Clay Brick and Ungrouted Concrete Block Prisms," Master of Engineering Report, University of Florida, Gainesville, Florida, 1982. 16. Wang, C.K., Intermediate Structural Analysis, McGraw-Hill, New York, New York, 1983* 17. West, Harry H., Analysis of Structures, John Wiley and Sons, Inc., New York, New York, 1980. 18. Williams, Robert T., Geschwinder, Louis F., "Shear Stress Across Collar Joints in Composite Masonry Walls," Proceedings of the Second North American Masonry Conference, College Park, Maryland, August 1982. 19- Yokel, F.Y., Mathey, R.G., Dikkers, R.D., "Strength of Masonry Walls Under Compressive and Transverse Loads," Building Science Series 34, National Bureau of Standards, Washington, D.C., March 1971. ML BIOGRAPHICAL SKETCH George Xavier Boulton was born April 13, 1959, in Havana, Cuba. To escape communist rule, he moved with his parents to the United States in 1961. They lived in New York City until he was five, then moved to Mobile, Alabama. There he attended parochial schools and graduated from McGill-Toolen High School in 1977. He then enrolled at the University of South Alabama, where he received his Bachelor of Science in Civil Engineering degree in 1991, graduating with high honors. In August of 1981, he moved to Gainesville, Florida, to pursue graduate studies at the University of Florida. In August of 1982, he received his Master of Engineering Degree in civil engineering from the University of Florida. He then enrolled in the doctoral program of the structures area of the Civil Engineering Department. George is an American citizen. He is also a member of the American Society of Civil Engineers, the National Society of Professional Engineers, and is registered as an Engineer in Training in the State of Alabama. He speaks fluent Spanish and enjoys most outdoor sports. Since his sophomore year in high school, George has been working part-time while also going to school full-time. He is anxiously anticipating entering professional practice upon graduation, and plans to become registered as a professional engineer as soon as possible thereafter. I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. 5 Morris V. Self, Chairmari' Professor of Civil Engineering I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. y // Jr James H. Schaub rofessor of Civil Engineering I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. /o * 7 Fernando E. Fagundi)/ Assistant Professor of Civil Engineering I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. Professor of Management This dissertation was submitted to the Graduate Faculty of the College of Engineering and to the Graduate School, and was accepted as partial fulfillment of the requirements for the degree of Doctor of Philosophy. December 1984 Cl' Dean, College of Engineering Dean for Graduate Studies and Research Page 2 of 2 Internet Distribution Consent Agreement In reference to the following dissertation: AUTHOR: Boulton, George TITLE: Finite element model for composite masonry walls / (record number: 487104) PUBLICATION DATE: 1984 I, %, as copyright holder for the aforementioned dissertation, hereby grant specific and limited archive and distribution rights to the Board of Trustees of the University of Florida and its agents. I authorize the University of Florida to digitize and distribute the dissertation described above for nonprofit, educational purposes via the Internet or successive technologies. This is a non-exclusive grant of permissions for specific off-line and on-line uses for an indefinite term. Off-line uses shall be limited to those specifically allowed by "Fair Use" as prescribed by the terms of United States copyright legislation (cf, Title 17, U.S. Code) as well as to the maintenance and preservation of a digital archive copy. Digitization allows the University of Elorida to generate image- and text-based versions as appropriate and to provide and enhance access using search software. This grant of permissions prohibits use of the digitized versions for commercial use or profit. Signature of Copyright Holder Georgiy, feoU H~gw Printed or Typed Name of Copyright Holder/Licensee Personal information blurred e-2(-zoe> Date of Signature Please print, sign and return to: Cathleen Martyniak UF Dissertation Project Preservation Department University of Florida Libraries P.O.Box 117007 Gainesville, FL 32611-7007 6/21/2008 POSITION OF \/ APPLIED LOAD A POSITION OF ^ ELASTIC CENTER Figure 1.3 Stress and Strain Distribution For a Composite Prism Under Vertical Load WALL HEIGHT, INCHES 189 MOMENT, INCH-POUNDS Figure 6.36 Brick Wythe Moment Versus Height For Example Number 5 110 FBJ3AT ( 0 'BRICK ELEMENT LENGTH =*,*5.2 SEITE (6, 120) CJLEN 120 FOBSAT {'O','CCILAE JOINT ELEMENT LENGTH $ INCHES *) WHITE (6,130)ELEN 130 FOBMAT('0*,'BLGCK ELEMENT LENGTH =',F5.2 CAII TITLE WHITE (6, 135) 135 FCBKAT (' ) ELASBfi=2.S16D+C6 AESDEB=5 1, 57D+00 IF (ELWYTH.EQ. 4) GC TO 140 IF (BIWYIfl.EÂ£. 6) GC TO 150 IF (BLHY1H.EQ.8) GC TO 160 140 EIASEI=2.023D+06 AHSHEL=14.85D+CC GC TC 170 150 ELASBL=1.6C7D+06 ABSHEI=27, B7D+G0 GO TO 170 160 EIASÂ£I=1.622D+G6 ARSHBL=35.74D+CC 170 BETIN ENC C SUBROUTINE COOED ( NPNTS,XCOGR,YCGCE) C SUBROUTINE COOED BILL BEAD AND PRINT C COORDINATES OE UP TO 20 ECINTS. DOUBLE EBECISICN XCCOR,YCOB DIMENSION XCCOR (2C) ,YCOCR (20) CAII TITLE BEAD (5, 10) NPNTS 10 FCBMAT (13) CALL TITLE DC 30 1=1,KENTS BEAD (5, 2C) J ,XCCGR (I) YCQCB (I) 1X,'INCHES*) *,F5.2,1X, 1X, 'INCHES V) TEE X AND Y 4 5 PI 1 2 3 4 5 BH ICK 10 P'E 1 2 3 4 5 6 7 8 9 10 BLOCK 8 PT 1 2 3 4 c 7 8 9 1 2.S62D-03 3.5D-03 7.5D+04 THETA COORDINATE 0 D + QQ 7.52D-04 1.505D-03 3.245D-03 3.5D-03 ELEMENI P-M 9.335D+04 C31C3D+05 H COORDINATE 0 D+00 5025D+04 9975D+04 1.C5D+05 1.1G132D+05 INTERACTION DIAGRAM 3, 6, M 1 COORDINATE 0.D + 0 3.75D+Q4 0725D+05 1. ED+05 1.65D+05 1. 73 47 D + 05 1.6875D + 05 1.26D+Q5 1. 08 432D +05 0. CD+GO ELEMENT P-M P COORDINATE 0.0D+00 1.875D+4 7.5D+04 1.2D+05 1.5D+05 1.92750+05 2. 25D + 05 3.0D+05 3.3865D+05 3.38 Â£5D* 05 INTERACTION Â¡ COORDINATE 7.5D + G3 4.875D+04 6.62 ED+ 04 9.996D+04 9.675D+04 fa.75D+04 I.9123D+04 0.0D+00 DIAGRAM D COORDINATE 0.OD+OO 1.875D+04 3.75D+04 5. 1D + 04 5.25D+04 7.5D+04 9.315D+04 9.3 15D+04 ELEMENT COMPRESSIVE LOAD CAPACITY MAXIMUM ERICK 3. 33E5C + C5 MAXIMUM COLLAR JOINT ELEMENT SHEAR LOAD CAPACITY o 168 Figure 6.16 Plate Load Versus End Rotation For Example Number 5 332 Figure B.21 Algorithm For Subroutine CJPDMT nnn 1M 1 = 1-1 IF (1 .El* 1) GC 1C 35 IF (I.GT. HI) GO TO 7 0 NC = I DC 30 L= 1, IM 1 C{I) =C (I) +A |L,NC)*B (I) NC=NC-1 30 CCN1INUE IF {I. GI NS1CE) GO TO 50 35 EO 4 C J=1,H1 L=J+IH1 C (I) =C (I) +A (I,J) *B (L) 40 CCNTINUE GC TO 1C C 50 ICCI=ICCI-1 55 LO 60 J=1,LC01 1=J+IE1 C (I) =C(I) +A (I,J) *B (L) 60 CCNTINUE GC TG 10 C 70 Nt EG= NE EG+ 1 NC=H 1 DC 80 L=NBEG/IN1 C(I) =C (I) +A (.L,NC) *B (L) NC= NC -1 80 CONTINUE IF (I. GT. NSTOE) GO TO 50 GC TO Â£5 100 CONTINUE BE1UEN EKE C SBFCUTINE INSEBT (A,B,KEY,TYPE,N,M, INDEX) SUBBOU1INE INSEBT BILL INSEBT A MA1BIX [Â£] INTC A LAfiGEfi HAIBIX [A]. TEE VALUES CF Â£B] ABE ADDED 10 1UE VALUES OF [Aj. IF KÂ£Y=1, THEN [A] IS A K NXN HATBIX AND 352 Table B.30 Nomenclature For Subroutine FORCES VARIABLE TYPE DEFINITION NOF INTEGER NUMBER OF FORCES NDOF INTEGER NUMBER OF THE DEGREE OF FREEDOM OF A PARTICULAR FORCE SFINIT DOUBLE PRECISION STORES THE INITIAL STRUCTURE FORCE FOR EACH LOADED DEGREE OF FREEDOM SFINCR DOUBLE PRECISION STORES THE STRUCTURE FORCE INCREMENT FOR EACH LOADED DEGREE OF FREEDOM SFMAX DOUBLE PRECISION STORES THE MAXIMUM STRUCTURE FORCE FOR EACH LOADED DEGREE OF FREEDOM NSDOF INTEGER NUMBER OF STRUCTURE DEGREES OF FREEDOM 'Si 160 185 190 200 210 220 230 240 250 270 BRWXLD (COUNT) =BREF ( 1) II(TRCONV.EQ.1) 00 TO 275 IF (TRELPB. HE. 1) GOTO 190 WHITE (6,160)1 F CRM AT(10' ','FORCES FCH BDICK ELEMENT NUMBER,14) CALL PHINT (BEEF,BR) WBIIE (6,185) I FOBMAT('0','DISPLACEMENTS FCR BRICK ELEMENT NMIER, $ 14) CALL PRINT (BREW,BE) CALL CHKFAI (BREF,NBBIEP,BRIDM, BHIDE,BRMAXP,1,NSIAI) IF(NSTA1.EQ.0) GO TO 250 IF (HS1A1.EQ. 1) GO TO 220 WHITE (6,2C0)I FORMAT('0,'***WARNING*** BBICK ELEMENT N0.',I4,1X, $ 'IS IN AXIAL) WRITE (6,2 10) FORMAT ( ,14X,*TENSION, CHECK FCR GNUSUAL LCAE1NG *) STOP IF(THFAIL.Sg.4) GO TO 230 IEFAII= 1 IF (TPRINT.NE.2) GO TO 250 WRITE (6,240) I FORMAT( 0 , *** ELEMENT NO. ,14,1 X HAS FAILED ***) CALL BRPBMT (NBRPDP,NOBRPC,NBRMTP,BRECOR, BRPCOH,BRPCV , BRICOR,BEMCOR,BREF,SBRPD,SBRiiT,BEVERF,BRKCM,BIAEL,BR3EIL,ITEE) CALL STIF AC (BRVERF,BRMOM,STGVF,STOME,BRA El,ER3EIL, 3 STO VST,STCMST,TESTIF,ITER,1,1) CALL BSESH (BREK,EREF,TOTEK,ERAEL,BR3EIL,LNVER , $ ELASBB,ARSHDB,I,EB,KEIEM) IF (I Â£Q. NEMT) CALL BNSERT (STRK,BREK,J,NSDCF,M1,ER, $ IBE) IF (I, EQ. NEMT) GO TO 270 IF(I.EQ.1) CALL BNSERT (STBK,BREK,1,FSBCF,M1,BR,IDR) IF (I. HE. 1) CALL BNSERT (STRK, BR EK 2 NSDOF M 1 ,B H IB 8) IF(ITER.Ey.1) GO TO 280 32 where Mt = /aETydA A (3.5) From integration of Equations (3.2) and (3.4), it follows that EIv = 3 2 2 f5x f6x MTX + C, - fr-EI^ 5 1 GA x + C, (3.6) where and C2 are the constants of integration. Using the boundary- conditions in Figure 3.6a, dv _ 5. IL at x dx dx GAS = o, x = 1 and v = 0 at x = 1 (3.7) (3.8) Equation (3*6) becomes 3 2 2 2 3 f r-X^ tcx M_x f c$xl l^fc EIv = -1 Â§ I l + (1 + *) 1 6 2 2 12 v 12 (3.9) where f51 f 6 Hr ht (3.10) j 12EI / .. \ and $ = K- (3.11) GA 1 s It should be noted here that the boundary conditions for the fixed end in the engineering theory of bending when shear deformations vg are included is taken as dv^/dx = 0; that is, slope due to bending deformation is equal to zero. 5.25 Structure Degrees of Freedom For Example Numbers 4 and 5 148 5*26 Example Number 5 149 6.1 Lateral Wall Deflection Versus Height For Example Number 2 152 6.2 Lateral Wall Deflection Versus Height For Example Number 5 193 6.3 Lateral Wall Deflection Versus Height For Example Number 4 154 6.4 Lateral Wall Deflection Versus Height For Example Number 5 153 6.5 Brick Wythe Vertical Deflection Versus Height For Example Number 2 156 6.6 Brick Wythe Vertical Deflection Versus Height For Example Number 3 157 6.7 Brick Wythe Vertical Deflection Versus Height For Example Number 4 158 6.8 Brick Wythe Vertical Deflection Versus Height For Example Number 5 159 6.9 Block Wythe Vertical Deflection Versus Height For Example Number 2 161 6.10 Block Wythe Vertical Deflection Versus Height For Example Number 3 162 6.11 Block Wythe Vertical Deflection Versus Height For Example Number 4 163 6.12 Block Wythe Vertical Deflection Versus Height For Example Number 5 164 6.15 Plate Load Versus End Rotation For Example Number 2 165 6.14 Plate Load Versus End Rotation For Example Number 3 166 6.15 Plate Load Versus End Rotation For Example Number 4 167 6.16 Plate Load Versus End Rotation For Example Number 5 168 6.17 Wall Wythe Vertical Load Versus Height At 0.30 Pmax Por Example Number 2 169 6.18 Wall Wythe Vertical Load Versus Height At 0.60 PmaX Por Example Number 2 170 315 . I'll vrt) /read one line of alphanumeric comments/ SKIP TWO LINES /print the line of alphanumeric comments/ (ret urn) Figure B.14 Algorithm For Subroutine TITLE 54 [w] v1" 3 w2 4 W3 5 _V4_ 7 Two properties of stiffness matrices can be used to further reduce the computational effort required to solve for the structure displacements. Stiffness matrices are symmetric and frequently will be tightly banded around the diagonal. Due to symmetry, the upper right triangular part of the matrix will be identical to the lower left triangular portion. Being banded around the diagonal means all nonzero coefficients will be concentrated near the diagonal. Since only the nonzero terms need to be considered for Gauss Elimination, and since only half of the nonzero terms need to be stored due to symmetry, tremendous savings in storage and computational efficiency are possible. Figure 3.15 qualitatively shows what an actual stiffness matrix might look like versus the stiffness matrix which is stored and used by a modified Gauss Elimination procedure that requires only half of the symmetric nonzero coefficients. Half the bandwidth (including the diagonal term) is shown as HBW. An idea of the storage savings that result from only storing half of the symmetric nonzero values in the stiffness matrix can be obtained by considering a simple example. For a structure with 194 degrees of freedom, there will be 194 simultaneous equations to solve, and the structure stiffness matrix will be a 194 x 194 matrix consisting of J70 C.2 Data Input Guide Table C.1 shows how the data are to be placed in the data file. It is preceded by a copy of the data file used for example number 2 which contains line numbers for illustrative purposes. Care must be taken to insert the data in the proper columns and lines, but once a file is created, the data can be altered for a new run by simply editing the file. In short, the data file demands care to construct, but is very easily changed. The user should verify from the output that the data read and used are the same as originally intended. 82 M Figure 4.9 Typical P Versus K Interaction Diagram For Brick Element Table C.1-continued LINE VARIABLE DESCRIPTION TYPE FIELD DESCRIPTOR START IN COLUMN 29-32 K POINT NUMBER INTEGER 13 1 BRTCOR BRICK THETA COORDINATE DOUBLE PRECISION D13-6 4 BRMCOR BRICK MOMENT COORDINATE DOUBLE PRECISION D13.6 17 33 BRPCV VALUE OF P WHICH IDENTIFIES THE BRICK M-THETA CURVE THAT FOLLOWS DOUBLE PRECISION D13.6 1 34 - COMMENT OF 'PT' - - 2 - COMMENT OF 'THETA COORDINATE' - - 6 - COMMENT OF 'M COORDINATE' - - 24 35-38 K POINT NUMBER INTEGER 13 1 BRTCOR BRICK THETA COORDINATE DOUBLE PRECISION D13.6 4 BRMCOR BRICK MOMENT COORDINATE DOUBLE PRECISION D13.6 17 39 - COMMENT OF 'COLLAR JOINT ELEMENT P-DELTA CURVE' - - 2 40 NCJPDP NUMBER OF POINTS IN THE COLLAR JOINT P-DELTA CURVE INTEGER 13 1 41 COMMENT OF 'PT' 2 WALL HEIGHT, INCHES 163 0 .075 .150 .225 .300 BLOCK WYTHE VERTICAL DEFLECTION, INCHES Figure 6.11 Block Wythe Vertical Deflection Versus Height For Example Number 4 WALL HEIGHT, INCHES 164 Figure 6.12 Block Wythe Vertical Deflection Versus Height For Example Number 5 20 FCRMAT (13,3D 13.6) WRITE (6,25) NUCE (I) ,SFIUII (I) ,SFINCfi (I) ,SFMAX (I) 25 FCBMAT ('0* ,13,3X,Di 3.6,4X,D13,6,4X,D13,6) 30 CONTINUE WBITE (6,35) 35 FORMA1C ') BE1UEN ENE C S BBCU1IN E APP1YF (STRF NSDC NCF NDOF S.F I KIT SFIN CB, S EM AX , $ ITEB KEY 1) C SUBROUTINE APPLY? WILL EIACE THE INITIAL LCADS ON THE C SIRUCTUSE IN THE STRUCTURE FORCE MATRIX THE FIRST TIME C IT IS CALLED. THEREAFTER IT KILL INCREMENT EACH ICAD C ACCCEDING 1C THE INFORMATION BEAD BY SUBROUTINE FORCES C FROM THE DATA SET. IT WILL HOT INC BE EE FT AM LOAD C EEYCND ITS SPECIFIED MAXIMUM VALUE. IE KEY 1 = 0, THE C LOADS HAVE NOT REACHED THEIR MAXIMUM VALUE. IF KEY 1=1, C THE LOADS ARE ALL AT THEIR RESPECTIVE MAXIMUMS AND NO C E URTHER INCREASES ARE REQUIRED. DCUEIE ERECISIC N STEF,SiIN IT,SFINCR,SEHAX,DSTKE,DSFMA X DIMENSION SIRE (NSDOF),SFINIT (NSDOF),SEINCE(NSDCF) DIMENSION SFMAX (NSDOF) NDOF (NOF) KEY 1= 1 IE (ITEE. NE. 1) GC TC 20 DO 1C 1=1,NCF F= NDCF (I) STRF (K) = Sf 1NI.1 (I) 10 CONTINUE KEY 1 = 0 GC 1C 8U 20 IF (KEY2.NE. C) GO TC 40 DC 3U 1=1,NCF =NDGF(I) STEF (M) = SIR? (M) +SFINCR(I) ESTBF=DABS(STRF (M)) 274 119 Figure 59-continued 90 P16+P17 P17"P16 H- z + P16+P17 P17-P16 M18 M 19 (c) Freebody Diagram of Test Plate A 16 A 17 018 19 (d) Wall Wythe Axial Displacements and Rotations at Wall Top Figure 4.12-continued. 03 -03 1 PI 1 2 3 4 5 BRICK 10 PT 1 2 3 4 5 6 7 8 9 10 BLOCK 8 PT 1 2 3 4 5 6 1 8 2.S62D 3.5D 7.5E + 04 THETA COORDINATE 0,0D+00 7.52D-04 1.5G5D-03 3.245D-03 3.50-03 ELEMENT P-H 9.335D+04 031C3D+05 H COORDINATE 0,D+00 5025D+04 9975D+04 1.05D+05 1 10132D+05 INTERACTION DIAGRAM 3, 6 M 1 COORDINATE 0. CD+00 3.75D+04 0725D+05 1.5D+05 1.65D+05 1.73475D+05 U6875D + 05 1.26D+05 1. 08432D+ 05 0.D+00 ELEMENT P M P COORDINATE 0,OD+OO 1.B75D+04 7,50+04 1. 2D+0 5 1,5D+05 1.9275D+05 2,25D+05 3.0D+05 3.3885D+05 3,3885D+05 INTERACTION M COORDINATE 7.5D+03 4,875D+04 8.625D+4 9, S96D+04 9,675D+04 6,75D + 04 .9123D+04 0,OD+OO DIAGRAM E COORDINATE 0.OD+OO 1,875D+04 3.75D+04 5,1D+04 5,625D+04 7.5D+04 9.315D+04 9.315D+04 COMPRESSIVE LOAD CAPACITY MAXIMUM BRICK ELEMENT 3,38850+05 MAXIMUM CC LLAR JOINT ELEMENT SHEAR LOAD CAPACITY UD 363 Table B.35 Nomenclature For Subroutine CHKFAI VARIABLE TYPE DEFINITION F DOUBLE PRECISION ELEMENT FORCE MATRIX NPMPTS INTEGER NUMBER OF POINTS IN THE ELEMENT INTERACTION DIAGRAM COORM DOUBLE PRECISION MATRIX THAT CONTAINS ELEMENT MOMENT COORDINATES COORP DOUBLE PRECISION MATRIX THAT CONTAINS ELEMENT VERTICAL LOAD COORDINATES PMAX DOUBLE PRECISION MAXIMUM ELEMENT P (VERTICAL LOAD CARRYING CAPACITY) KEY INTEGER VARIABLE THAT IDENTIFIES THE TYPE OF ELEMENT; 1 BRICK; 2 = COLLAR JOINT; 3 = BLOCK NSTAT INTEGER VARIABLE THAT IDENTIFIES THE FAILURE STATUS OF THE ELEMENT; 0 = ELEMENT HAS NOT FAILED; 1 = ELEMENT HAS FAILED, 9 = ELEMENT IS IN AXIAL TENSION 3C8 Table B.12 Nomenclature For Subroutine GAUSS 1 VARIABLE TYPE DEFINITION X DOUBLE PRECISION N x 1 MATRIX WHICH STORES THE SOLUTION TO THE MATRIX EQUATION OF THE FORM [a] [x] [b] WHERE THE SOLUTION IS CALCULATED BY THE SUBROUTINE C DOUBLE PRECISION N x Ml MATRIX WHICH STORES THE UPPER TRIANGULAR PORTION OF A SYMMETRIC N x N MATRIX [A] THAT HAS A BANDWIDTH OF (2 x Ml) 1 B DOUBLE PRECISION N x 1 MATRIX THAT STORES THE NUMBERS THAT EQUAL [a] [x] M1 INTEGER HALF THE BANDWIDTH (INCLUDING THE DIAGONAL) OF MATRIX [a] N INTEGER NUMBER OF ROWS IN MATRICES [x], [cj, [b] KEY INTEGER VARIABLE THAT INDICATES TYPE OF PROBLEM; KEY = 1 IS FOR A REGULAR PROBLEM; KEY =* 2 IS FOR THE CASE IN WHICH MATRIX [a] (AND THEREFORE MATRIX [c]) IS THE SAME AS IN THE LAST CALL TO GAUSS 1 BUT MATRIX [B] IS DIFFERENT 100 Figure 4.15-continued 136 P ^ M W M Figure 5.17 Block Element P-K Interaction Diagram WALL HEIGHT, INCHES 158 200.0 192.0 184.0 176.0 168.0 160.0 152.0 144.0 136.0 128.0 120.0 112.0 104.0 96.0 88.0 80.0 72.0 64.0 56.0 48.0 40.0 32.0 24.0 16.0 8.0 0 TTTT .075 .150 .225 BRICK WYTHE VERTICAL DEFLECTION, INCHES T~| .300 Figure 6.7 Brick Wythe Vertical Deflection Versus Height For Example Number 4 95 Â¥ (z x 1) I (z X z) -0- (z x 2) W (z x 1) a16 *P1 a17 -0- aT - Sp' 918 ei9 (4xz) (4x2) v J 0T] where (4.6) z = total number of structure DOF 4 [w] = displacement matrix excluding the 4 DOF at the top of the wall [i] = identity matrix (1's along the diagonal, 0's elsewhere) [a^] = transformation part of [x^] matrix; shown in Equation (4.4). Note that [o^] is the transpose of [a] and [x^] is the transpose of [x]. From Equation (2.4) recall that [>] [K] [V] where [f] = structure force matrix [k] = structure stiffness matrix [W] = structure displacement matrix. By considering all original DOF (all 4 DOF at the wall top), this equation can be expressed as [W Kid] Kid] (4.7) +STRK (I NCCIP2) + SIRK (I,NC0LP3) $ KKID=NHID- 1 NHIDM^NHIDE 1-1 NCOL=NCCL- 1 60 CONTINUE NSDM3=NSDOF3 KSDE2=NSDCF-2 N SUM 1 =N SDOF- 1 PEEL (1# 1)=STRK (NSEM3, 1) +STHK (NSDM3,2) PREL (1,2) = STRK (NSEM3,2) + STRK (NSDM2,1) PBEI (1,3)=STEK (NS DM3,3) +STEK (NSDM2,2) PREL (1, 4) = STRK (NSDM3,4) +STRK (NSDB2,3) PBEI (2, 1) = ai02*STÂ£K(N3EM3, 1) L02*STRK{NSDM3,2) 1 + SIHK (NÂ£DH3,3)+ STEK (NSDK3,4) PBEI (2,2)=MI02*STEK (NSDM3,2) L02*S'IBK (NSDM2, 1) $ + STRK (NSDH2,2)+SIEK(NSDK2,3) PBEI (2, 3) =FIC2*STRK (NSDM3,3) +LC2*STBK (BSD Â£2,2) $ +STRK (NSDMl, 1) + STRK ( N5DM 1, 2) PEEL (2, 4) =ML02*STBK (NSDM3,4) +IC2+SIHK (NSDK2,3) $ + STBK (NS EM 1,2) +STHK (NSBCE> 1) NSTBK (NSDM3, 1) =PREL (1, 1) + PEEL (1,2) NSTRK (NSDM3,2)=MLC2*PREL (1,1) +I02*PREL (1,2) +PEEL (1,3) $ + PBEL (1,4) NS1RK (NSDM2,1) =MLC2*PEEL(2, 1) +I02*PREL(2, 2) PBEI (2,3) +PBEL (2,4) RETURN ENE C SUBROUTINE DISPIA (STRK,NSTHÂ£ NSDCF,NSDM2 INHBR,INUBI) C SUBEOUTINE DISPLA ILL CALCULATE THE ACTUAL STRUCTURE C LISPLACEMENT MATRIX FROM TEE STRUCTURE DISPLACEMEET C FAT BIX THAT ONLY CONSIDERS TWO DEGREES OF FREEDOM AT C THE IOP OF THE HALL. DCUEIE EBECISIC N STEW,NSTRW,LNRBB,LNHBL,LG2,MLC2 DIMENSION STEW (NSBOP),NSTEW (NSDM2) NSDE4=NSDCF-4 277 NOBRPC INTEGER HOF INTEGER NOFN INTEGER NRBM1 INTEGER NRBOT INTEGER NRTM1 INTEGER NRTMP1 INTEGER NRTOP INTEGER NRTP1 INTEGER NSDFMS INTEGER NSDFMT INTEGER NSDM2 INTEGER NS DOF INTEGER NSTAT INTEGER NSTORY INTEGER PROUT INTEGER SHEARL DOUBLE PRECISION NUMBER OF BRICK M-THETA CURVES (EACH IS IDENTIFIED BY THE P OR AXIAL LOAD LEVEL FOR WHICH THE MOMENT CURVATURE RELATIONSHIP IS VALID) NUMBER OF FORCES FORCE NUMBER NUMBER OF ROWS IN THE BOTTOM HALF OF THE STRUCTURE STIFFNESS MATRIX MINUS ONE NUMBER OF ROWS IN THE BOTTOM HALF OF THE STRUCTURE STIFFNESS MATRIX NUMBER OF ROWS IN THE TOP HALF OF THE STRUCTURE STIFFNESS MATRIX MINUS ONE NUMBER OF ROWS IN THE TOP HALF OF THE STRUCTURE STIFFNESS MATRIX MINUS ONE PLUS ONE NUMBER OF ROWS IN THE TOP HALF OF THE STRUCTURE STIFFNESS MATRIX NUMBER OF ROWS IN THE TOP HALF OF THE STRUCTURE STIFFNESS MATRIX PLUS ONE NUMBER OF STRUCTURE DEGREES OF FREEDOM MINUS SIX NUMBER OF STRUCTURE DEGREES OF FREEDOM MINUS THREE NUMBER OF STRUCTURE DEGREES OF FREEDOM MINUS TWO NUMBER OF STRUCTURE DEGREES OF FREEDOM STATUS OF AN ELEMENT AFTER IT IS CHECKED FOR FAILURE; O = O.K., 1 = FAILED; 9 = IN TENSION NUMBER OF STORIES OF WALL WHERE EACH STORY IS 8 INCHES TALL VARIABLE THAT IDENTIFIES WHETHER OR NOT AN INTERMEDIATE PRINTOUT IS DESIRED; 0 = NO; 1 = YES COLLAR JOINT SHEAR STRESS ON THE LEFT OR BRICK FACE OF THE COLLAR JOINT 218 9.777D+03 MAXIMUM BLOCK ELEMENT COMPRESSIVE LOAD CAPACITY 9.315D+04 STRUCTURE FORCE APPLICATION INFORMATION 2 DOF INITIAL FORCE FORCE INCREMENT MAXIMUM 56 -4,00+03 -4.D+Q3 -4.0D+05 57 -5,00+03 -5,0D+03 -5.0D+05 INSTRUCTIONS ON PRINTING INTERMEDIATE RESULTS: 0 FORCE END 281 ELEMENT OSE KFY=1, FOB TEE COLLAR JOINT ELEMENT OSE KEY=2, AND FOE THE BLOCK ELEMENT USE KFY=3. IF MSIAT=0, THE ELEMENT HAS NCT FAILED. IF NSTAT= 1, THE ELEMENT HAS FAILED. IF NSTAT=9, THE ELEMENT IS IN AXIAL I ENSIGN. CCUEIE PRECISION F, COOBP, COGHM, PMAX, ALLOW B ,HIM 1 MIP1 FI DOUBLE PBECISION PIP 1,PIM1,MAXKCM ,FGNE DIMENSION F (6) ,CCGfiP (NPMPTS) CCCB M (N P MPT S) NSTAT=0 IF (KEY. KE.2) GC TC 10 IF (DABS (F ( 1) ) .GT. PMAX) GO TC 25 GC 1C 80 FGN E=F (1) IF (F (1) LT. 0) GC TC 15 GO TO 2 G NST AT=9 GC TC 80 IF (F (1) .GT. PMAX) GG TO 25 GC TC 27 NSTAI= 1 GO TO 8 C DC 50 I=1,KEHE1S IF (FCNE. EQ.CCOfiP (I) ) GO TC 30 IF (FONE. GT. CCCEP (I) ) GO TC 10 GC TO 6G AILCNH=CCORK (I) GC TO 7G EIH1 = CCCEiF (I) J = I IE 1=1+ 1 CONTINUE PIE 1 = COCBP(IP 1) PI = FC NE Hit 1 = C C C R K (J) HIE 1=CCOBM (IP1) IF (PIPI.EC.E1ft 1)ALLOWM=MIM1 WALL HEIGHT. INCHES 194 Figure 6.41 Collar Joint Shear Stress Versus Height For Example Number 2 360 (start) CONSTRUCT THE ACTUAL STRUCTURE DISPLACEMENT MATRIX FROM THE STRUCTURE DISPLACEMENT MATRIX THAT ONLY CONSIDERS TWO DEGREES OF FREEDOM AT THE WALL TOP (return) Figure B.33 Algorithm For Subroutine DISPLA 36 k 5,6 k 2,3 k 5,3 k 6,3 6EI (3.33) 6,5 0 + $)12 -6EI (3.34) 3,2 (1 + $)12 6EI (3.35) 3,5 (1 + $)l2 (2 - *)EI (3.36) 3,6 (1 + $)1 with the remaining coefficients in columns 3 and 6 equal to zero. Thus, the stiffness matrix for the basic 6 DOF element, shown in Figure 2.1a, takes the form shown in Figure 3*7 when shear deformation is considered. 3.4 Moment Magnification With increasing slenderness and height, the lateral deflections of a vertical member due to bending will increase. As these deflections increase, an additional moment is caused by the vertical load acting through these deflections. This additional moment is often referred to as a secondary bending moment. Moment magnification is one term used to describe this effect. Figure 3.8 shows how moment magnification occurs. The technique for including the effect of moment magnification in the Direct Stiffness Method analysis of a structure also involves altering the standard terms in the stiffness matrix of the elements for which this effect is to be considered. The terms in the stiffness matrix for the standard 6 DOF element, shown in Figure 2.2, neglect moment magnification as well as shear deformation, which was discussed in the previous section. To illustrate how the new terms in the element 285 Table B.2 Nomenclature For Subroutine EQUAL VARIABLE TYPE DEFINITION A DOUBLE PRECISION N x M MATRIX WHOSE VALUES ARE SET EQUAL TO THE CORRESPONDING VALUES IN MATRIX [B] B DOUBLE PRECISION N x M MATRIX WHOSE VALUES ARE COPIED INTO THE CORRESPONDING VALUES IN MATRIX [a] N INTEGER NUMBER OF ROWS IN MATRICES [a], [b] M INTEGER NUMBER OF COLUMNS IN. MATRICES [a], [b] c c c c c c c c c c c c c c c c c c c c c c c c c c c *************************************************************** * * * FINITE ELEMENT MODEL * * * * FOR * * * * COMPOSITE MASONRY RAILS * * * * * DEVELOPED BY * * * * * * GEORGE X. BOULTON * * CIVIL ENGINEERING DEPARTMENT * * UNIVERSITY Of fLOSIDA * * SPRING 1984 * * * *************************************************************** PROGRAM AS PRESENTLY DIMENSIONED ILL HANDLE UP TO: 194 STRUCTURE DEGREES CE FREEDOM 117 ELEMENTS (MAXIMUM WALL HEIGHT OF 3 12 INCHES). TO CHANGE DIMENSIONS CHANGE DIMENSION STATEMENTS. DECLARE MAT REAI EE INTEGER INTEGER DOUBLE PEEC DOUBLE DOUBLE DCUBIE DOUBLE DCUBIE RICES AND VARIABLES REAL CE INTEGER AS NEEDED. IGHT TOTEIN,WHI,BR,CJ,BL,PRCUT,TESTIF,T ECONV,T RELEE THFAIL,TRSHOW,COUNT, IPRINI ISIGN MATRICES AND VARIABLES AS NEEDED. PRECISICN TOTEK,STRK,GSTEK,NSTRK,NGSTHK,BREK,CJFK PRECISION BLEK,SiaW,GSTBS,DELTA,NS1BW,ERES,CJEW PRECISION BLEW,STEF,DELTAE,GDEITF,ERROR,NSTRF,NGSTRE PRECISION SFINIT,SFINCR,SFMAX,FUIT,SHONE,EEEE,MUBEE PRECISION CJEF,MCJEF,BLEE,MBLEF,K1,K2,K4,NK1,NK2,NK4 228 D.1 Example Number 1 Table B.19 Nomenclature For Subroutine WRITE VARIABLE TYPE DEFINITION A DOUBLE PRECISION N x M MATRIX WHICH THE SUBROUTINE PRINTS N INTEGER NUMBER OF ROWS IN MATRIX [a] M INTEGER NUMBER OF COLUMNS IN MATRIX [a] 359 Table B.33 Nomenclature For Subroutine DISPLA VARIABLE TYPE DEFINITION STRW DOUBLE PRECISION STRUCTURE DISPLACEMENT MATRIX NSTRW DOUBLE PRECISION STRUCTURE DISPLACEMENT MATRIX NSDOF INTEGER NUMBER OF STRUCTURE DEGREES OF FREEDOM NSDM2 INTEGER NUMBER OF STRUCTURE DEGREES OF FREEDOM MINUS TWO LNHBR DOUBLE PRECISION HORIZONTAL LENGTH OF THE BRICK HALF OF THE COLLAR JOINT ELEMENT LNHBL DOUBLE PRECISION HORIZONTAL LENGTH OF THE BLOCK HALF OF THE COLLAR JOINT ELEMENT 66 be very small, and the closer they are to zero, the better the solution. Thus, one can speak of the degree to which the solution converged. For example, if m is the number of elements and the previous operation, shown below, is carried out m [error] = [f] + Z (-1.0 x [f]i) i=1 and the largest value in the matrix [ERROR] is 1 x 10^, this solution converged better than one which produced a value in [ERROR] equal to 9 x 10~1. To insure convergence of the solution to the matrix equation [f] = [k] [v] within a specified tolerance, a reasonable tolerance value of, say 0.001, should be selected and all values in matrix [ERROR] compared to it. If no value exceeds the tolerance, the solution is acceptable. If any value exceeds the tolerance, then all the values can be treated as an incremental force matrix [aF] and then used to solve for the incremental structure displacement matrix [aw] for the previous structure stiffness matrix [k]. In other words, set [af] = [error] and solve [af] = [k] [aw] (3.54) for [aw] . The total structure displacements then become equal to [w] + [aW], the total displacements for each element become [w] + [aw], and the total 18 considered positive. If it acts in the opposite direction, it is considered negative. Figure 2.4 shows two examples of structure force matrices for the frame of Figure 2.3* 2.6 Solving For the Structure Displacement Matrix Like the structure force matrix, the structure displacement matrix will also be an N x 1 matrix where N is the number of structure degrees of freedom. Thus, the structure displacement matrix for the frame of Figure 2.3 will be [W] = Solving for the structure displacement matrix will entail solving the matrix equation [f] = [k] [w] for [w]. This will consist of solving N simultaneous equations if N is the number of structure DOF. One way to solve this matrix equation is by matrix inversion, or [W] = [K]-1 [F] (2.5) For large values of N, however, the structure stiffness matrix of dimensions N x N will become large and calculating the inverse of a large matrix is very cumbersome and inefficient. Two much more efficient techniques for solving for the structure displacement matrix are Gauss Elimination and Static Condensation. These methods are W, V/ Wc discussed in detail in sections 3.6.1 and 3.6.2. o rÂ¡ n NS D E 3= NS DCF -3 HSEM1=NSD0F-1 LO 2= (LNHB.E + LNBBL) /2.0 NIC2=-1,0*LC2 EO 2C I=1,NSDM4 STRW (I) = NS RN (I) 20 COSUSE STRR (NSDM3)=NSTSW ( NSDM3) + (MLC2 + NS1EW (NSDM2)) STB K (NSDM2) =N STRW { NSDM3) + (LC2* NSTEH ( N SDM2) ) STEii {NSEM1 )=NSI fiW (NSDM2) SlEfi (NSDGF)=NSTBH{NSDM2) RETURN ENE C SUBROUTINE CHK2CL (A,N,TOLES,KEY) SUBROUTINE CHKTOL CHECKS TO SEE IE EACH NUMBER IN MATRIX [A] IS LESS IHAN ICLEB. IF IT IS, KEÂ¥=0. If AT LEAST ONE NUMEER IS NOT, KEÂ¥= 1. Â£ A] IS AN NX1 MATRIX. DOUBLE PRECISION A,TOLER,ABSCL J DIMENSION A ( N) 03 KÂ£Â¥= C DC 10 1=1,N AESOL=DABS (A (I)) IE (ABSCI.Gi. ICIER) KEÂ¥=1 IF (ABSOL.I.TOLER) GO IC 20 10 CONTINUE 20 RETURN EKE C SUBROUTINE CHKFAI (F,NPMPTS,COOEM,COORP,PMAX,KEÂ¥,NSTAT) C SUBROUTINE CHKFAI HILL CHECK TC SEE IE AN EIEMENI HAS C FAILED. FOR THE BRICK OR BICCK ELEMENT, THE ACTUAL C AXIAL LOAD AND MOMENT IS COMPARED fcllH THE ALLCHAEIE C AXIAL 1CAE AND MOMENT FOB TEAT TXPE OF ELEMENT. FOR C THE COLLAR JOINT ELEMENT, THE ACTUAL SEEAR FCECE IS C CCMPABED WITH THE ALEONARLE SHEAR FORCE. FOR THE BRICK APPENDIX C USER'S MANUAL C.1 General Information To use the program, the user must create a data file in the manner outlined in the next section. As previously mentioned, all input (and output) is in basic units of inches, pounds, or radians. The data describing the application of forces consider the structure degrees of freedom with the test plate, i.e., only two degrees of freedom at the top of the wall. The general structure degrees of freedom were shown in Figure 4.2. As an example, suppose it is desired to apply a compressive force to that wall at an eccentricity of 2 inches towards the block wythe (the right hand side). The test plate degrees of freedom simply replace and with W^, and W1Q and with W-jy. The new is present midway between the old and (at "the geometric center of the wall) and the new is present midway between the old W^0 and The original direction of each is maintained. If the compressive force has a value of 4000 lb, then one would place in the data set the information that degree of freedom 16 is loaded with a value of -4000 (since a compressive force acts down on the wall) and degree of freedom 17 is loaded with a value of -1000 x 2 = -2000 (since the moment is equal to P x e and would be clockwise). Earlier, it was mentioned that the pattern for numbering structure degrees of freedom is valid regardless of wall height. This makes determining what degrees of freedom to apply a load at almost trivial. 568 Table C.1-continued LINE VARIABLE DESCRIPTION FIELD START IN TYPE DESCRIPTOR COLUMN 52-56 J BLDCOR COMMENT OF 'P COORDINATE POINT NUMBER BLOCK DELTA COORDINATE BLPCOR BLOCK P COORDINATE 57 58 NOBLPC 59 NBLMTP 60 BLPCV 61 62-66 K BLTCOR COMMENT OF BLOCK ELEMENT M-THETA CURVES' NUMBER OF BLOCK M-THETA CURVES NUMBER OF POINTS IN EACH BLOCK M-THETA CURVE VALUE OF P WHICH IDENTIFIES THE BLOCK M-THETA CURVE THAT FOLLOWS COMMENT OF PT COMMENT OF 'THETA COORDINATE' COMMENT OF 'M COORDINATE POINT NUMBER BLOCK THETA COORDINATE 24 INTEGER 13 1 DOUBLE D13.6 4 PRECISION DOUBLE D13.6 17 PRECISION 5 INTEGER 13 1 INTEGER 13 1 DOUBLE D13.6 1 PRECISION 2 6 24 INTEGER 13 1 DOUBLE D13.6 4 PRECISION 4 2.962D-G3 9.335D+04 5 3.5D-03 1.C31C3D+05 7.5E+Q4 PT THETA COORDINATE E COORDINATE 1 0.0D+0 0.0D+00 2 7.52D-04 3.5025D+4 3 1,505D-03 6.9975D+04 4 3.245D-03 1.C5D+05 5 3.5D-03 1, 10132D+05 BRICK ELEMENT P-M INTERACTION DIAGRAM 10 PT M COORDINATE P COORDINATE 1 0.QD+00 0.0D+00 2 3.75D+Q4 1.875D+04 3 1,07250+05 7,5D+04 4 1-5D +05 1.2D+05 5 1 65D+05 1,50+05 6 1.7347ED+05 1.S275D+05 7 1,6Â£75D+05 2 25D + 05 8 1.26D+05 3, CD+05 9 1. 8432D+05 3.3865D+05 10 C.CD + 00 3.38 Â£50 + 05 BLOCK 8 PT ELEMENT P-fl INTERACTION DIAGRAM M COORDINATE E COORDINATE 1 7.5D+03 0.OD+OO 2 4 875D + 04 1.8 75D+04 3 Â£.E25D +04 3.75D+04 4 9* 996D+04 5. 1D+04 c 9.675D+04 5.625D+04 6 o,75D+04 7.5D+04 7 3.9123D+04 S.315D+04 8 0,0D+00 9,3 15D+04 MAXIMUM BRICK ELEMENT COMPRESSIVE ICAD CAPACITY 3. 38E5E + C5 MAXIEUH COLLAR JOINT ELEMENT SHEAR LCAE CAPACITY c c c c c c c c c c c c c c c c c c c c c 539 540 545 550 (EVERY OTHER TIME) 1. CHECK THE SOLUTION FOB CONVERGENCE 5 ITiBATE IF HEQUIRED. 2. INCREMENT THE STRUCTURE LOADS. 3. SOLVE [K3*Â£WJ=Â£F] FOR Â£W] THE STRUCTURE EISELACEMENTS, AS DESCRIBED AECVI. 4. IE THE WALL IS AT THE HIGHEST LCADING LEVEL DESIRED, IDENTIFY THIS AND PRINT THE WALL LOADS & DISPLACEMENTS, THE LATERAL HALL DEFLECTION VERSUS HEIGHT, THE VERTICAL WAIL DEFLECTION VERSUS HEIGHT, TEE ELEMENT FORCES AND DISPLACEMENTS, AND THE WALI HYTI1E VERTICAL LOAD VERSUS HEIGHT. 5. IE AN INTERMEDIATE PRINTOUT WAS REQUESTED FOE THF CURRENT LEVEL OF WAIL LOADING, IDENTIFY THIS AND PRINT THE INFORMATION DESCRIBED ABOVE. 6. IF THE WALL HAS FAILED, IDENTIFY THIS AND PRINT THE WALL FAILURE LOADS Â£ DISPLACEMENTS, AS WEIL AS THF OTHER INFORMATION DESCRIBED ABOVE. 1. UNLESS THE WAIL HAS FAILED CE THE LOADS REACHED THE HIGHEST IEVEL DESIRED, CONTINUE INCREASING THE LOADS ONE INCREMENT AT A TIME AND SOLVING FOR EIE E E NT ECfiCES AND DISPLACEMENTS UNTIL EITHER OF THESE HAPPEN. IF (ITER.NE. 1) GO TO 540 CALL APPLY F (NSIRF, NS DM2 NC F NDCF, SE'IN 11, SPI NCR ,S E KAX , I IT ER,KE Y1) GO TO 690 TRELPR=C IF (TEFAIL, NE.4) GO TO 550 WRITE(6,545) FORMAT i'OV'O1 '*** END OF FINITE ELEMENT ANALYSIS ***) STOP TO IE E= 1. 0D-03 TBCONV=C CALL CBKTCI (ERROR,N3D0F,TOLER,KEY) IF (KEY.EQ.C) GO TO 560 TRCC NV=1 CALL EQUAL (DEITAF,ERROR,NSDGF,1) 240 DOUELE DOUBLE DOUELE DOUELE DOUBLE DOUBLE DOUELE DOUBLE DOUBLE DOUBLE C DIMENSION PRECISION PRECISION PRECISION PRECISION PRECISION PRECISION PRECISION PRECISION PRECISION PRECISION MATRICES. W1,W2,K2W2,F1,F2,F1MKW2,BEPCOR,BRDCGR S BRPD/CJPCOfi,CJDCOR ,SCJPD,BLPCCR,BLDCCR SBLPO,BRMCCB,BÂ£TCCR,SBREl,CJMCCR,CJTCOR SCJHT,BLHCOR, ELTCGR,SBLMT,BRMAXP,CJHAXP BLMAXP ,BRIDP,BEIDH,BLIDF,ELIDM,EfiPCV,ELPCV EBAEL,BR3EIL,COSHES, CJHOMS,BLAEL,BL3EIL BRHYLD,BL8YLD,ELASBR,KLASEL,AESHBE, ARSHEL LNHBR,LNHBL,LNVER,TOLÂ£R,SHEARL,SHEARE BRHOM,BRVERF,CJMCM,CJVEBF,BLMCK,BLVÂ£RF STOVF, STOMF,SIQVS1,STOMSI TCT EK (6,6 117) ,STRK( 194,9) GSTRK (194,9) NSTRK (192,10) NG STM (192,10) BK EK (6,6) ,CJEK (4,4) SEEK (6,6) ,STRW (19 4) GSTR8 (1 94) DELI AW (194) NSTBi (192) ,BREW(6) ,CJEW(4) ,BÂ£EW (6) ,STBF(194) DELTAE(194), GDELTF (194) ,ERRCE(194) NS1BF (192) NGSTRF (192) ,NDCF( 192) ,SFINIT(192) ,SFINCR (192) SFMAX (192) ,BRÂ£F (6) ,MBEEF (6) ,CJEF (4) ,MCJEF(4) BLEF (6) MBLEF (6) K 1 ( 100,9) K2 ( 1CC, 9) K4 ( 10 0,9) NK1 (100,10) ,NK2( 100,10) ,NK4 (100,10) ,81(100) 2 (100) ,K2W2 (100) ,F1 (100),F2 (10C),F1HKS2 (10 0) TOTEIN (117,6) IBR (6) ,ICJ(4) ,IBL (6) ,BEPCOR(2 0) BEDCCB (20) ,SBRPD (20) ,CJPCOR (20) ,CJDCCE (20) SCJPD (20) ,BLPCOR(20) ,BLICOR (20) ,SBLPD (20) BEMCOR (20,20) ,BRTCCR (20 ,20) ,CJMCCE(20) CJTCCB (20) SCJMT (20) BLMCOR (20,20) ,BI1C0R ( 2C,2 0) BRIDP (20) ,BRIDM (2 0) ,BBPCy(2 0) ,B1PCV(20) BLIDP(20),BLICH(20),STOVF(3,117),STOMF (3,117) STOVST (3, 117) ,STOMST (3, 117) ,ERWYLD(39) BLtYLD(39),SBRMT(20,20),SBLMT(20,20) C OPEN 6 IDENTIFY WHICH FILE CONTAINS THE DATA 6 WHICH FILE IS TO C STORE TEE OUTPUT. OPEN (U NIT= 5,FIIE= *SMURF.DAT,,STATUS=*OLD*) OPEN (UNIT=6,FILE= OUT.DAT',STATUS=* NEW*) C GO TO SUBROUTINE, BEAD THE WALL DESCRIPTION DATA Â£ PRINT IT. CALL READ (HHI LNVER,LNHBfi,LNHBL,NSTORY,NS DCF,NEIEH, DIMENSION DIMENSION DIMENSION DIMENSION DIMENSION DIMENSION DIMENSION DIMENSION DIMENSION DIMENSION DIMENSION DIMENSION DIME N5I0N DIMENSION DIMENSION DIMENSION DIMENSION DIMENSION DIMENSION 229 WALL HEIGHT, INCHES 175 Figure 6.25 Wall Wythe Vertical Load Versus Height At 0.90 Pmax For Example Number 3 B.17 Nomenclature For Subroutine CURVES 321 B. 18 Nomenclature For Subroutine PRINT 323 B.19 Nomenclature For Subroutine WRITE 325 B.20 Nomenclature For Subroutine BRPDMT 328 B.21 Nomenclature For Subroutine CJPDMT 331 B.22 Nomenclature For Subroutine BLPDMT 333 B.23 Nomenclature For Subroutine STIFAC 337 B.24 Nomenclature For Subroutine BRESM 339 B.25 Nomenclature For Subroutine CJESM 341 B.26 Nomenclature For Subroutine BLESM 343 B.27 Nomenclature For Subroutine IHDXBR 345 B.28 Nomenclature For Subroutine INDXCJ 348 B.29 Nomenclature For Subroutine INBXBL 350 B. 30 Nomenclature For Subroutine FORCES 352 B. 31 Nomenclature For Subroutine APPLYF 354 B.32 Nomenclature For Subroutine PLATEK 357 B.33 Nomenclature For Subroutine DISPLA 359 B.34 Nomenclature For Subroutine CHKTOL 361 B.35 Nomenclature For Subroutine CHKFAI 363 B.36 Nomenclature For Subroutine WYTHE 365 C.1 Data Input Guide 375 310 B.13 Subroutine STACON Subroutine STACON solves a matrix equation of the form [a] x [x] = [b], where [a] is an n x n matrix which has a bandwidth of (2 x ml) 1. The solution uses Static Condensation and standard Gauss Elimination for a symmetric banded matrix. Matrix [x] is the n x 1 matrix that stores the solution, [c] is the n x ml matrix used to store the upper triangular portion of the symmetric n x n matrix [a] with a bandwidth of (2 x ml) 1, and [B] is the n x 1 matrix that stores the product of [a] x [x]. The variable ml equals half the bandwidth (including the diagonal) of matrix [a]. Table B.13 defines the nomenclature used in this subroutine. Figure B.13 is an algorithm for this subroutine. B.14 Subroutine TITLE Subroutine TITLE will read and print one comment card of alphanumeric or character data. It will skip 2 lines before printing the comments. The comments cannot exceed column 80. Table B.14 defines the nomenclature used in this subroutine. Figure B.14 is an algorithm for this subroutine. B.15 Subroutine READ Subroutine READ will print a heading, read and print two comment cards, read and print the wall description data and calculate the modulus of elasticity and area in shear for the brick and for the block. Table B.15 defines the nomenclature used in this subroutine. Figure B.15 is an algorithm for this subroutine. B.16 Subroutine COORD Subroutine COORD will read and print the x and y coordinates of up to 20 points for a curve. The variable NPHTS is the number of points in LIST OF FIGURES Figure Page 1.1 Typical Composite Masonry Wall Section 2 1.2 Typical Composite Masonry Wall Loadbearing Detail 2 1.3 Stress and Strain Distribution For a Composite Prism Under Vertical Load 4 2.1 Development of Element Stiffness Matrix 10 2.2 Basic Element Stiffness Matrix 13 2.3 Structure and Element Degrees of Freedom For a Frame 14 2.4 Construction of the Structure Force Matrix 19 3.1 Structure and Element Degrees of Freedom For a Frame With Different Materials and Element Types.. 24 3.2 Element Stiffness Matrices For Elements 1 and 3 in Figure 3*1 26 3.3 Development of Stiffness Matrix For Element 2 of Figure 3*1 27 3.4 Element Stiffness Matrix For Element 2 in Figure 3.1 28 3.5 Shear Deformation and Its Importance 29 3.6 Application of Displacements to Establish Stiffness Coefficients Considering Shear Deformation 31 3.7 Element Stiffness Matrix Considering Shear Deformation 37 3.8 Moment Magnification 33 3*9 Element Used in Derivation of Moment Magnification Terms 40 3.10 Element Stiffness Matrix Considering Moment Magnification 45 3*11 Element Stiffness Matrix Considering Shear Deformation and Moment Magnification 46 20 J=I+1 SBIJED (I) = (ELECCS (J) -ELPCGL (I) )/ (BLDCCÂ£ (J) -BICCCS (I) ) CONTINUE Hf5LÂ£ 1 = NBLM1P-1 DC 45 J=1,NCBIEC DC 40 L=1,NKLS1 K=L+1 SBLMT (J,L) = (BLMCOB (J,K)-BIMCCR (J,L) )/ (EITCCB (J,K) - $ ELTCOR(J,l)) 40 CONTINUE 45 CONTINUE ELAEL = SBLPD (1) BI3EIl=SBIBT (1,1) GO 10 150 60 DC 80 1= 1,N ELS 1 IF (BLEF (1) GE. BLPCOB (I) )BIAEI=SBLPD (I) If (BLEF (1) .LI. ELPCOR (I) ) GO 10 90 80 COM1IN 0 E 90 AVGKCM=DAES ( (EIEF (3) + BLEF (b) )/2) IF (BLEF (1) .GT.BLPCV (1) ) GO 1C 105 DC 100 L= 1, N iS L 3 1 IF(AVGMCH.GE.BLUCOR (1,L) ) EL3EIL=SBLET (1,L) IF (AVGMOM. IT. BIKCOR (1,L) ) GO 10 150 100 CONTINUE GC 1C 150 105 IF (BLEF (1) .GE.BLPCV (NOBLPC) ) GC 1C 125 NGEE1=NCBLEC-1 DO 110 K= 1,NOP 1 I = K+ 1 IF (BLEF ( 1) .GE.BLPCV (K) ) LI E=K IF (BLEF (1) LI. ELPCV (L) ) ULE=L IF (BLEF (1) .Ll.BLPCV(L) ) GC 1C 115 110 CONTINUE 115 DO 120 I=1,NiviLS1 IF(AVGMOft.GE.BLECCR(LLP,I))LLSM1=SBLMT(LLP,I) IF(AVGMON.GE.BLMCOB(0LP,Ij)UISE1=SBLET (ULP,I) 121 Substituting Equation (5-11) into (5.4), 0 = eL * t From Equation (5*5), 9 can also be expressed as (5.12) (5.13) Thus, from the vertical strain measurements made on each side of the prism during the tests by Fattal and Cattaneo, it is possible to calculate the end rotation 9 of a prism in response to applied end moment. For the prisms they tested, L has a value of 15.7 in and t = 3-56 in. The only thing that is required now is a means of relating the end rotation 9 for the 15.7 in high prisms they used to the end rotation that would be produced if 8 in high prisms were subjected to the same end moment. Recall the first Moment Area Theorem, presented in West (17). In reference to Figure 5.9b, it would state that the angle change between points a and b on the deflected structure, or the slope at point b relative to the slope at point a, is given by the area under the K/EI diagram between these two points. In other words, 9 a/b (5.H) Figure 5.9c is a schematic representation of the prism size tested (4x32x16 in) for which results are available and the prism size for which results are desired (4x24x8 in). From Equation (5.14) note that end rotation 9 will be affected by differences in 1, or prism height, as 6.37 Block Wythe Moment Versus Height For Example Number 2 190 6.38 Block Wythe Moment Versus Height For Example Number 3 191 6.39 Block Wythe Moment Versus Height For Example Number 4 192 6.40 Block Wythe Moment Versus Height For Example Number 5 .....193 6.41 Collar Joint Shear Stress Versus Height For Example Number 2.. 194 6.42 Collar Joint Shear Stress Versus Height For Example Number 3 195 6.43 Collar Joint Shear Stress Versus Height For Example Number 4 196 6.44 Collar Joint Shear Stress Versus Height For Example Number 5 ..197 A.1 Detailed Program Flowchart 207 B.1 Algorithm For Subroutine NULL 284 B. 2 Algorithm For Subroutine EQUAL 286 B.3 Algorithm For Subroutine ADD 288 B.4 Algorithm For Subroutine MULT 290 B.5 Algorithm For Subroutine SMULT 293 B.6 Algorithm For Subroutine BMULT 295 B.7 Algorithm For Subroutine INSERT 297 B.8 Algorithm For Subroutine BNSERT 300 B.9 Algorithm For Subroutine EXTRAK 302 B.10 Algorithm For Subroutine PULROW 305 B.11 Algorithm For Subroutine PULMAT 307 B.12 Algorithm For Subroutine GAUSS 1 309 B.13 Algorithm For Subroutine STACON 313 B.14 Algorithm For Subroutine TITLE 315 151 Table 6.1 Summary of Wall Failure for Examples Number 1 Through Number 5 WALL FAILURE LOADS FAILURE MODE EXAMPLE NUMBER P, LB M, IN-LB ELEMENT TYPE ELEMENT NO. LOCATION 1 132,000 165,000 BLOCK 56 4" FROM TOP OF WALL 2 204,000 0 BLOCK 59 20" FROM 42 12" TOP OF 45 4" WALL 5 106,000 216,000 BLOCK 5 4" FROM BOTTOM OF WALL 45 4" FROM TOP OF WALL 4 168,000 0 BLOCK 3 4" FROM BOTTOM OF WALL 5 36,000 72,000 BLOCK 3 4" FROM BOTTOM OF WALL SHCfcF=SFINIT (NCFN) GC 1C 1JO 110 R8I1E(6,120) 120 FORMAT ('O','INTERMEDIATE RESULTS MILL KOI EE PRINTED') SHCHF=1.0D*2G C DEFINE EANGE OF BEICK ELEMENTS AND COLLAR JOINT ELEMENTS. 130 NEM1-NELEM-2 NEMC=NEIEM-1 KSEFÂ£lS=NSCOF-6 NSLFM1= NSDGE-3 NSDM2=NSDOF2 C DEFINE NUMBER OF DOF FOR EACH ELEMENT TYPE. Bfi=fc CJ=4 BL = E C DEFINE 1HE SIZE OF HALF THE BANDWIDTH (INCLUDING THE DIAGONAL) C FOR THE STRUCTURE STIFFNESS MATRIX. M1 = S M 1P 1=M 1 + 1 C DEFINE THE VARIABLES, NEEDED BY TEE SUBROUTINE THAT PERFORMS C STATIC CONDENSATION, WHICH DIVIDE AND Â£F] IKTC HALVES. NRICE=NSDCE/2 NRBGT=N SDCF-NBIOP NRII 1 = NETCE + 1 N BTM 1=N SDM2/2 NEBM1=NSDM2~NETM1 NR1ME 1=NHIM 1+1 C ENTER TEE MAIN EBCGEAM LOOP. 140 DO 7C0 ITER = 1, 1 COC0 CALL NULL (STRK,NSDOF, M1) C FOB EACH BRICK ELEMENT: C (THE VEBY EIRS1 TIME) C 1. CONSTRUCT THE INDEX MATRIX 6 STCEE IT. C 2. GET INITIAL STIFFNESS FACTORS 5 STORE THEM. C 3. CONSTRUCT THE ELEMENT STIFFNESS MATRIX. C 4. INSEET THE ELEMENT STIFFNESS MATRIX INTO THE 369 Loads will normally be applied at the top of a wall, and when it is desired to apply transverse loads, also at the lateral structure degrees of freedom. The pertinent equations are as follows: No. of 8 inch levels = (Wall hgt. in inches) x (1/8) (C.1) No. of stories = (No. of 8 inch levels) (C.2) No. of structure DOF without test plate = [(No. of stories) x 5] 1 (C.3) No. of structure DOF with test plate = (No. of structure DOF without test plate) 2 (C.4) Axial load DOF on test plate = (No. of structure DOF with test plate) 1 (C.5) Moment DOF on test plate = (No. of structure DOF with test plate) (C.6) Lateral DOF = 3f 8, 13 n+5 up to [(No. of structure DOF with test plate) 4] (C.7) Thus, if a wall is 10 feet high, it is 120 in high, has 15 stories, 74 structure DOF without test plate, 72 structure DOF with test plate, the axial load DOF on the test plate occurs at DOF number 71, the moment DOF on the test plates occurs at DOF number 72, and the lateral DOF = 3, 8, 13, 18, 23, 28, 33, 38, 43, 48, 53, 58, 63 and 68. Because the stiffness for each element is a function of the load level the element experiences, the load should be applied in relatively small increments, say 4000 pounds. The data input guide will indicate that a load at a degree of freedom is read as the initial force, the force increment, and the maximum force. To load the wall to failure, a very large maximum force should be specified, like 1x10^. 132 Cross-sectional capacity of brick prisms. Figure 5*14 Experimental Source of Brick Element P-M Interaction Diagram (4) non u> to cd a. C Â£B] IS AH axil MATRIX, IF KEY=2, THEM Â£ A ] IS AN H X1 C MATRIX AND [B] IS AN MX1 NATH IX. THE VECTOR [INDEX] C HAS a EIEKEKT3 WHICH GIVI THE POSITIONS OF Â£B] IN [A]. C IF TYPE=1, THEN THE FIES1 3 NUMBERS IK THE [INDEX] C MATRIX ABE MOT USED. IF TYPE=2, THEN AIL THE NUMBERS C IN THE [INDEX] MATRIX ARE USED. IE TYEE=3, THEN THE C FIFTH NDHEEE IN THE [INDEX] MATRIX IS NOT USED. INTEGER TYPE DCUEIE PRECISION A,Q DIMENSION A (N,N) ,B (M, H) #INDEX (M) 1=1 IF (TYPE.EQ. 1) 1 = 4 DC 3 I = L,H II=INDEX (I) IE (I. EC. 5) GC TO 1 GC TO 4 1 IF (TYPE.EQ.3) GC TO 30 4 IE (KEY.EC.2) GO TO 20 DC 10 0=1, K JJ=INDSX(J) IF (J.EC.5) GC TO 6 GC TO 8 IE (TYPE. EC. 3) GO TO 10 A(II,JJ)=A(II,JJ)+B (I J) 0 CONTINUE GC TO 3 G 0 JJ= 1 J=1 A (1I,JJ)=A (II,00)+ E (I,J) 0 CONTINUE RETURN END SUBROUTINE BNSEflT (C,B,TYPE,ft,Ml,fi,INDEX) SUBROUTINE BMSEBT WILL INSERT THE UPPER TRIANGULAR PORTION OF AN tiXH SYMMETRIC MATRIX [Â£] INTO A MATRIX 247 Table 4.2 Variables Used in Constructing Block Element Stiffness Matrix VARIABLE DEFINITION SOURCE $ FACTOR USED IN ACCOUNTING FOR SHEAR DEFORMATION CALCULATED, $ = 12EI GA L2 s E BLOCK MODULUS OF ELASTICITY TABATABAI (15), E = 2.023x106 PSI FOR 4" BLOCK, E 1.807x10 PSI FOR 6" BLOCK, E = 1.622x10^ PSI FOR 8" BLOCK (ONLY USED FOR SHEAR DEFORMATION) I MOMENT OF INERTIA OF BLOCK ELEMENT CROSS-SECTION NOT USED DIRECTLY; 3EI/L FACTOR FROM TESTS USED G BLOCK SHEAR MODULUS CALCULATED, G = 2(l+v) V BLOCK POISSON'S RATIO VALUE FOR CONCRETE, v = 0.15 As BLOCK AREA IN SHEAR CALCULATED, As = .84 Anet A = 14.85 IN2 FOR 4 BLOCK, A = 27.87 IN2 FOR 6" BLOCK, AND A_ = 39-74 IN2 FOR 8" BLOCK 9 Anet BLOCK NET AREA MEASURED CONSIDERING ONLY WEBS OF BLOCK, Ane. = 17.69 IN2 FOR 4" BLOCK, Anet = 33.18 IN2 FOR 6" BLOCK AND Anet = 47,30 IIj2 F0R 8" BL0CK L BLOCK ELEMENT LENGTH LENGTH DEFINED BY MODEL = 8 IN A AREA OF BLOCK ELEMENT CROSS-SECTION NOT USED DIRECTLY; AE/L FACTOR FROM TESTS USED P BLOCK ELEMENT AXIAL FORCE BLOCK ELEMENT FORCE MATRIX DEGREE OF FREEDOM NUMBER 1; CALCULATED BY PROGRAM FOR EACH LOAD LEVEL n n n I)If]ENSIGN B (N f 6) A (I) DC 1C J = 1,T A |J)=B(I,J) 10 CONTINUE RETURN ENE C SUBROUTINE PUL-MAT (A,B,I,T,N) SUBROUT IN E PUL WAT WILL PULL A TXT MATRIX OUT OF A 6XXN MATRIX Â£ B] AND STORE THE VALUES IN A TXT MATRIX Â£ A]. *** WARNING *** T MUST EE LESS THAN GE EQUAL TO 6. INTEGER I BCUEIE PRECISION A B DIMENSION 8(6,6,]]) ,A (T, I) DC 20 K=1,I EC 1C L= 1, T A (K, L) = B (K,L,I) 10 CONTINUE 20 CONTINUE RETURN EKE C SUBROUTINE GAUSS 1 (X,C,B,M 1,N,KEY) C SUBROUTINE GAUSS 1 SOLVES A MATRIX EQUATION OF THE EORE C Â£A ]*Â£X ]=Â£B ], WHERE Â£Aj = AN NXN MATRIX WHICH HAS A BAND C WIDTH OF (2*M1)-1. THE SOLUTION IS A STANDARD GAUSS C ELIMINATION FOR A SYMMETRIC BANDED MATRIX. Â£X] IS THE C MATRIX THAT STORES THE SOLUTION, Â£C] IS THE MATRIX USEE C TO STORE THE SYMMETRIC NON-ZERO VALUES IN THE Â£A] C MATRIX, Â£ B] IS THE MATRIX THAT STORES THE NUMBERS C THAT = Â£A]*Â£X], Ml EQUALS HALF THE BANDWIDTH C (INCLUDING THE DIAGONAL)-} AND N IS THE NUKE EE OF C EQUATIONS. KEY = 1 IS EOR A REGULAR P20ELEM AND KE Y = 2 IS C FOR MULTIPLE LOAD GAUSS USE WHEN Â£ A ] MATRIX SAME AS C IN IAST CALL TO GAUSSl. DOUBLE PRECISION X,C,B,COEFF 250 299 Table B.8 Nomenclature For Subroutine BNSERT VARIABLE TYPE DEFINITION C DOUBLE PRECISION N x M1 MATRIX WHICH CONTAINS THE UPPER TRIANGULAR PORTION OF A SYMMETRIC N x N MATRIX THAT HAS A BANDWIDTH OF (2 x Ml) 1 B DOUBLE PRECISION M x M SYMMETRIC MATRIX, THE UPPER TRIANGULAR PORTION OF WHICH IS INSERTED INTO MATRIX [c] TYPE INTEGER VARIABLE THAT KEEPS TRACK OF WHICH NUMBERS IN THE INDEX MATRIX ARE USED; IF TYPE 1, THEN THE FIRST THREE NUMBERS IN THE INDEX MATRIX ARE NOT USED; IF TYPE 2, THEN ALL THE NUMBERS IN THE INDEX MATRIX ARE USED; IF TYPE 3, THEN THE FIFTH NUMBER IN THE INDEX MATRIX IS NOT USED N INTEGER HUMBER OF ROWS IN MATRIX [c] M1 INTEGER HALF THE BANDWIDTH (INCLUDING THE DIAGONAL) OF MATRIX [c] M INTEGER NUMBER OF ROWS AND NUMBER OF COLUMNS IN MATRIX [B] INDEX INTEGER M x 1 INDEX MATRIX WHOSE VALUES GIVE THE POSITIONS IN MATRIX [c] INTO WHICH THE VALUES IN THE UPPER TRIANGULAR PORTION OF MATRIX [B] ARE INSERTED 4.2.2 Collar Joint Element 71 4.2.3 Concrete Block Element 74 4.3 Experimental Determination of Material Properties 76 4.3.1 Brick 76 4.3.2 Collar Joint 83 4.3.3 Concrete Block 83 4*4 Load Application 86 4.5 Solution Procedure 97 FIVE NUMERICAL EXAMPLES 102 5.1 General Comments 102 5.2 Material Property Data..... 103 5.2.1 P-A Curves 103 5.2.2 M- Curves 113 5.2.3 P-M Interaction Diagrams 130 5*3 Example Number 1 Finite Element Analysis of a Test Wall 137 5.4 Illustrative Examples 141 5.4.1 Example Number 2 141 5.4.2 Example Number 3 145 5.4.3 Example Number 4 145 5.4.4 Example Number 5.... 145 SIX RESULTS OF ANALYSIS 150 6.1 Wall Failure 150 6.2 Lateral Wall Deflection Versus Height 150 6.3 Brick Wythe Vertical Deflection Versus Height 150 6.4 Block Wythe Vertical Deflection Versus Height 160 6.5 Plate Load Versus End Rotation 160 6.6 Wall Wythe Vertical Load Versus Height 160 6.7 Brick Wythe Moment Versus Height 185 6.8 Block Wythe Moment Versus Height 185 6.9 Collar Joint Shear Stress Versus Height 185 SEVEN CONCLUSIONS AND RECOMMENDATIONS ..200 APPENDIX A COMPUTER PROGRAM 202 A.1 Introduction 202 A. 2 Detailed Program Flowchart 204 A. 3 Program Nomenclature 204 A.4 Listing of Program and Subroutines..... ...227 B SUBROUTINES 232 B. 1 Subroutine NULL 282 B.2 Subroutine EQUAL ..282 B. 3 Subroutine ADD 282 B.4 Subroutine MULT 282 B.5 Subroutine SMULT 282 283 Table B.1 Nomenclature For Subroutine NULL VARIABLE TYPE DEFINITION A DOUBLE PRECISION H x M MATRIX WHOSE VALUES ARE SET EQUAL TO ZERO N INTEGER NUMBER OF ROWS IN MATRIX [a] M INTEGER NUMBER OF COLUMNS IN MATRIX [a] 291 Table B.5 Nomenclature For Subroutine SMULT VARIABLE TYPE DEFINITION A DOUBLE PRECISION NUMBER BY WHICH ALL VALUES IN MATRIX [b] ARE MULTIPLIED TO OBTAIN MATRIX [c] B C N M DOUBLE PRECISION DOUBLE PRECISION INTEGER INTEGER N x M MATRIX WHOSE VALUES ARE MULTIPLIED BY THE NUMBER A TO OBTAIN MATRIX [c] N x M MATRIX WHOSE VALUES ARE THE PRODUCT OF A TIMES [c] NUMBER OF ROWS IN MATRICES [b], [c] NUMBER OF COLUMNS IN MATRICES [b], [c] 99 (start) READ WALL GEOMETRY, STRENGTH AND DEFORMATION PROPERTIES OF EACH ELEMENT, AND STRUCTURE LOAD APPLICATION INFORMATION CALCULATE ELEMENT INDEX MATRIX [i] CONSTRUCT INITIAL ELEMENT STIFFNESS MATRIX [k] RECALL ELEMENT INDEX MATRIX [i] EXTRACT ELEMENT DISPLACEMENT MATRIX [w] FROM STRUCTURE DISPLACEMENT MATRIX [w] RECALL ELEMENT STIFFNESS MATRIX [k] ' 1 ' MULTIPLY ELEMENT STIFFNESS MATRIX [k] BY ELEMENT DISPLACEMENT MATRIX [w] TO OBTAIN ELEMENT FORCE MATRIX [f] OR [k][w] = [f] I CHECK FOR ELEMENT FAILURE (IF ELEMENT FAILS STOP AFTER CHECKING ALL ELEMENTS) i ~ CONSTRUCT NEW ELEMENT STIFFNESS MATRIX [k] BASED ON ELEMENT LOAD LEVEL AND LOAD DEFORMATION PROPERTIES 1 ' ' INSERT -1.0 TIMES ELEMENT FORCE MATRIX [f] INTO MATRIX [ERROR] THAT MONITORS CONVERGENCE Figure 4.13 Program Algorithm WALL HEIGHT, INCHES 192 Figure 6.39 Block Wythe Moment Versus Height For Example Number 4 366 nomenclature used in this subroutine. Figure B.36 is an algorithm for this subroutine. B. 15 Algorithm For Subroutine READ 517 B.16 Algorithm For Subroutine COORD 520 B. 17 Algorithm For Subroutine CURVES 522 B.18 Algorithm For Subroutine PRINT 524 B. 19 Algorithm For Subroutine WRITE 526 B.20 Algorithm For Subroutine BRPDMT 530 B. 21 Algorithm For Subroutine CJPDMT 532 B.22 Algorithm For Subroutine BLPDMT 335 B.25 Algorithm For Subroutine STIFAC 338 B.24 Algorithm For Subroutine BRESM 340 B.25 Algorithm For Subroutine CJESM 542 B.26 Algorithm For Subroutine BLESM 544 B.27 Algorithm For Subroutine IHDXBR 546 B.28 Algorithm For Subroutine INDXCJ 549 B.29 Algorithm For Subroutine INDXBL 551 B.50 Algorithm For Subroutine FORCES 555 B.51 Algorithm For Subroutine APPLYF 555 B.52 Algorithm For Subroutine PLATEK 558 B.53 Algorithm For Subroutine DISPLA 560 B.54 Algorithm For Subroutine CHKTOL 562 B.55 Algorithm For Subroutine CHKFAI 564 B.56 Algorithm For Subroutine WYTHE 567 338 SELECT VERTICAL STIFFNESS FACTOR BASED ON CURRENT LEVEL OF ELEMENT LOADING SELECT PREVIOUS VERTICAL STIFFNESS FACTOR Figure B.23 Algorithm For Subroutine STIFAC 120 Side 1 represents the left side of the prism and side 2 the right side. Note from Figure 5.8 that vertical strains were measured on each side of the prism during testing. By definition, (5.6) (5.7) where <$1 = vertical displacement on side 1 of the prism 2 = vertical displacement on side 2 of the prism L = prism height. From Figure 5*9a, 6 Plugging Equations (5.6) and (5.7) into Equation (5-5), Ac = 2~1 L L L From Equation (5.8), this reduces to Ac A L (5.8) (5-9) (5.10) Therefore, 5 = (Ae)L . (5.11) WALL HEIGHT, INCHES 170 Figure 6.18 Wall Wythe Vertical Load Versus Height At 0.60 Pmax Example Number 2 For 13 M - AE L 0 0 -AE L 0 0 0 1 2EI -6EI 0 -12EI -6EI L2 l? L2 . 0 -6EI 4EI 0 6EI 2EI L2 L L2 L -AE 0 0 AE 0 0 L L 0 -12EI 6EI 0 1 2EI 6EI l3 L2 L^ L2 0 -6EI 2EI 0 6EI 4EI L2 L L2 L WHERE: A = AREA OF ELEMENT CROSS-SECTION E = MODULUS OF ELASTICITY L = ELEMENT LENGTH I = MOMENT OF INERTIA OF ELEMENT CROSS-SECTION Figure 2.2 Basic Element Stiffness Matrix Table A.2-continued MATRIX TYPE NK1 DOUBLE PRECISION NK2 DOUBLE PRECISION NK4 DOUBLE PRECISION NSTRF DOUBLE PRECISION NSTRK DOUBLE PRECISION NS TRW DOUBLE PRECISION SBLMT DOUBLE PRECISION SBLPD DOUBLE PRECISION SBRMT DOUBLE PRECISION SBRPD DOUBLE PRECISION SCJMT DOUBLE PRECISION SCJPD DOUBLE PRECISION SFINCR DOUBLE PRECISION SFINIT DOUBLE PRECISION SFMAX DOUBLE PRECISION STOMF DOUBLE PRECISION DESCRIPTION TOP LEFT HAND PORTION OF STRUCTURE STIFFNESS MATRIX TOP RIGHT HAND PORTION OF STRUCTURE STIFFNESS MATRIX BOTTOM RIGHT PORTION OF STRUCTURE STIFFNESS MATRIX STRUCTURE FORCE MATRIX USED IN SOLVING FOR STRUCTURE DISPLACEMENTS STRUCTURE STIFFNESS MATRIX USED IN SOLVING FOR STRUCTURE DISPLACEMENTS STRUCTURE DISPLACEMENT MATRIX STORES THE SLOPES OF THE BLOCK MOMENT THETA CURVES STORES THE SLOPES OF THE BLOCK AXIAL LOAD DELTA CURVE STORES THE SLOPES OF THE BRICK MOMENT THETA CURVES STORES THE SLOPES OF THE BRICK AXIAL LOAD DELTA CURVE STORES THE SLOPES OF THE COLLAR JOINT MOMENT THETA CURVE STORES THE SLOPES OF THE COLLAR JOINT AXIAL LOAD DELTA CURVE STORES THE STRUCTURE FORCE INCREMENT FOR EACH LOADED DEGREE OF FREEDOM STORES THE INITIAL STRUCTURE FORCE FOR EACH LOADED DEGREE OF FREEDOM STORES THE MAXIMUM STRUCTURE FORCE FOR EACH LOADED DEGREE OF FREEDOM STORES THE ABSOLUTE VALUE OF AVERAGE OF END MOMENTS FOR EACH ELEMENT ro r\j 313 Figure B.13 Algorithm For Subroutine STACON Figure 3-12 Nonlinear Load Deformation Curve LOAD REGION 1 o LOAD REGION 2 4- LOAD REGION 3 LOAD o o BIEK (4,2) = 0.0 BLEK (5,2) =- 1.0*BLEK (2,2) BIEK (6,2) = Â£IEK (3,2) BLEK (3, 3) = (BL3EIL* (4. O+PHI BL)/ (3. O* (1. 0 + FHIBL) ) ) - $ (2.0*ELEF(1) *LN/15.0) BLEK (4, 3) = C. G BIEK (5,3)=-1.0*ELEK(3,2) BIEK (6, 3) =(BL3EIL* (2-O-PHIBL)/ (3. O* (1 O+FBIBL) ) ) + $ (ELBE (1) *LN/30, 0) ELEK(4,4) =BLEK (1, 1) BIEK (5,4)=0.0 BIEK (6, 4) =C. O BIEK (5,5) = EIÂ£K (2,2) BLEK(5, 6)=-1.C*BLEK (3,2) BIEK (6,5) =EIEK (5,6) BIEK (6, 6) =BLEK (3,3) DC 30 d= 1,El EO 20 K=1,BI BIEK (J,K) = BIEK(K, J) 1TEK (J,K,I) =BLEK (J,K) 20 C0N1INU E 30 CCN1INE E1UEN EME SUBRCOTINE IM DXBt (IBD TOTEI K,I MEET, BE, 8 EIEH) SUBROUTINE INIXBR MILI CCHSIfiCT I BE INDEX MATRIX FOR EACH BRICK ELEMENT. I MI EGER TCTÂ£IN,ER DIM EM SICi 10TEIH (KKLEH,6) ,IBE(6) IF (I.Eg. 1) GC 1C 5 IE(I.EQ.4) GO 10 15 IF(I,GT.4) GC TC 30 IBB (1) =0 IBE (2) = 0 IBB (3) =0 270 117 (a) I 1/2 1/2 El 1 A-j i A2 (b) Figure 5.9 Relationships Used to Obtain Desired Data From Experimental Results For M-0 Curves 227 A.4 Listing of Program and Subroutines The following is a listing of the program and subroutines. As previously noted, comments will be found throughout each to identify what is taking place. 9777B + Q3 MAXIMUM BLOCK ELEMENT COMPRESSIVE LOAD CAPACITY 9.315D+04 STRUCTURE FORCE APPLICATION INFORMATION 2 DOF INITIAL FORCE FORCE INCREMENT MAXIMUM 71 -4.0D+03 -4.0D+03 -4.0D+C5 72 -8,0D+03 -8,0D+03 -8.0D+05 INSTRUCTIONS CN PRINTING INTERMEDIATE RESULTS: 1 1 71 -3.2D+4 FORCE Vjl VO VO 2 CLAY BRICK WYTHE Figure 1.1 Typical Composite Masonry Vail Section Figure 1.2 Typical Composite Masonry Vail Loadbearing Detail 365 Table B.36 Nomenclature For Subroutine WYTHE VARIABLE TYPE DEFINITION BRWYLD DOUBLE PRECISION MATRIX THAT STORES THE VERTICAL LOAD CARRIED BY THE BRICK WYTHE AT EACH WALL LEVEL BLWYLD DOUBLE PRECISION MATRIX THAT STORES THE VERTICAL LOAD CARRIED BY THE BLOCK WYTHE AT EACH WALL LEVEL HSTORY INTEGER NUMBER OF STORIES OF WALL WHERE EACH STORY IS 8 INCHES TALL 170 N 2=N2 + 1 X( JR) =X(K)-C(K,N1)*X(N2) X(K)=X(K)/C(K,1) 190 CON1INUE RETURN ENI C S UBRUT INE STACCN (X ,C B, K4 ,F2 W2 #K2 K2H2 ,11 F 1 MKW 2, K 1, H 1, $ M 1, N NRTP 1, N RBC1, N filOP) C SUBROUTINE STACON Bill SOLVE A MAT BIX EQUATION OF THE C FORM Â£ A ]*Â£X]=Â£ B], WHERE Â£A ] = AN NXN MATRIX WHICH HAS A C BANDWIDTH OF (2*M1)-1. THE SOLUTION USES STATIC C CONDENSATION AND STANDAR! GAUSS ELIMINATION FO A C SYMMETRIC EANDED MATRIX. Â£X] IS THE MATRIX THAT STORES C THE SOLUTION, Â£C] IS THE MATRIX USED TO STORE THE C SYMMETRIC NGN-ZERO VALUES IN THE Â£A] MATRIX, Â£E] IS THE C MATRIX THAT STORES THE NUMBERS THAT = [A]*[X], Ml C EQUALS HALF THE BANDWIDTH (INCLUDING THE DIAGONAL), AN! C N IS THE NUMBER OF EQUATIONS. DOUBLE PRECISION X,C,B,COEE\F ,K4,F2,W2 ,K2, R2W2 f 1, F 1MKW 2, K 1 DCUEIE EBECISIC N W 1 DIMENSION X (N) ,C (N,M1) ,B (N) K4 ( N R B OT Mi 1) f 2 ( NRECT) DIMENSION W2 (NREOI) FI (NRTOE) 1MKW2 ( NRTOE) K 1 (N STOP, Ml) DIMENSION W 1 ( NRTO!) K2 ( NRIGE ,M 1) ,K2W2 (NRTCE) C PERFORM NRTCE FORWARD ELIMINATIONS ON Â£C], M = M 1- 1 DC S8 J=1,NRTCE JM=J+M IF (JM.GT. N) JM=N J E1=J + 1 t M=M+1 IF (MS+J.GT.N) HM=N-J+ 1 EC 97 K=JE1,JM KK=K-JP 1 + 2 COEFF=-C (J,KK)/C (J, 1) B (K) =B (K) +CCÂ£FF*B (J) 252 I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. 5 Morris V. Self, Chairmari' Professor of Civil Engineering I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. y // Jr James H. Schaub rofessor of Civil Engineering I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. /o * 7 Fernando E. Fagundi)/ Assistant Professor of Civil Engineering I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. Professor of Management This dissertation was submitted to the Graduate Faculty of the College of Engineering and to the Graduate School, and was accepted as partial fulfillment of the requirements for the degree of Doctor of Philosophy. December 1984 Cl' Dean, College of Engineering Dean for Graduate Studies and Research 185 the block wythe carries a higher percentage of the load near the top as might be expected, due to the location of load application. 6.7 Brick Wythe Koment Versus Height The brick wythe moment as a function of wall height is plotted for examples 2 through 5 in Figures 6.33 through 6.36. Notice that the moment diagrams and the deflected shapes shown in Figures 6.1 through 6.4 agree. 6.8 Block Wythe Koment Versus Height Figures 6.37 through 6.40 show the block wythe moment versus height for examples 2 through 5. Once again, the moment diagrams and the deflected shapes shown previously in Figures 6.1 through 6.4 agree. For each example, the moment in the block wythe is greater than the moment in the brick wythe for equal wall load levels. This is due to the higher rotational stiffness of the block prisms compared with the brick prisms. 6.9Collar Joint Shear Stress Versus Height The collar joint shear stress as a function of wall height for several load levels in each of examples 2 through 5 is illustrated in Figures 6.41 through 6.44. Like the rest of the data plotted in chapter 6, it too was obtained directly from the output generated by the finite element analysis program. Note that in example 2 at a height of 72" and in example 4 at a height of 136" there are discontinuities in the collar joint shear stress in each wall at maximum load levels. These are shown in Figures 6.41 and 6.43, respectively. This coincides with two other observations. At the same heights and load levels for these examples, the moment diagram changes in curvature as seen in Figures 6.33 and 6.37 122 well as differences in moment of inertia I. Since "both prisms consist of the same materials, there will be no difference in the modulus of elasticity E for each. The difference in moment of inertia results from the difference in cross-section of the two prism sizes, which is due to the different depths of each prism. The 4x52x16 in prism, for moment of inertia calculations, has a 4x52 in cross-section and the 4x24x8 in prisms, a 4x24 in cross-section. Consider Figure 5*9d. From Equation (5.14), e ,f?16 (5.15) 16 EI16 and q (5.16) 8 EIq Expressing 9g in terms of 9-|g then, 9q = (x) 916 or J-(x)fll- (5-17) EIq EI16 Recall that the equation for the moment of inertia of a rectangular cross-section is I = bh5 (5.18) 12 For the 16 in high prism (actually 15*7 in but height does not enter into the calculation of I), Table C.1-continued LINE VARIABLE DESCRIPTION TYPE FIELD DESCRIPTOR START IN COLUMN 82 NBRIDP NUMBER OF POINTS IN THE BRICK INTERACTION DIAGRAM INTEGER 13 1 83 - COMMENT OF 'PT' - - 2 - COMMENT OF 'M COORDINATE - - 8 - COMMENT OF P COORDINATE' - - 24 84-93 J POINT NUMBER INTEGER 13 1 BRIDH BRICK INTERACTION DIAGRAM M COORDINATE DOUBLE PRECISION D13.6 4 BRIDP BRICK INTERACTION DIAGRAM P COORDINATE DOUBLE PRECISION D13.6 17 94 - COMMENT OF 'BLOCK ELEMENT P-M INTERACTION DIAGRAM' - - 1 95 NBLIDP NUMBER OF POINTS IN THE BLOCK INTERACTION DIAGRAM INTEGER 13 1 96 - COMMENT OF 'PT' - - 2 - COMMENT OF 'M COORDINATE' - - 8 - COMMENT OF *P COORDINATE - - 24 97-104 J POINT NUMBER INTEGER 13 1 BLIDM BLOCK INTERACTION DIAGRAM M COORDINATE DOUBLE PRECISION D13-6 4 345 Table B.27 Nomenclature For Subroutine INDXBR VARIABLE TYPE DEFINITION IBR INTEGER INDEX MATRIX FOR A BRICK ELEMENT TOTEM INTEGER MATRIX THAT STORES INDEX MATRICES ALL OF THE ELEMENT I INTEGER ELEMENT NUMBER NEMT INTEGER NUMBER OF ELEMENTS MINUS TWO BR INTEGER NUMBER OF BRICK ELEMENT DEGREES OF FREEDOM NELEM INTEGER NUMBER OF ELEMENTS EXAMPLE #3 H/T=12 PLATE LOAD ECCENTBICITY=2 IN. DESCRIPTION OF WALL 120 8. OD + CO 2. OD + CO 3. 0D + 00 2a RATERIAL PROPERTIES BRICK ELEMENT P-DELTA CURVE 6 FT 1 2 3 4 5 6 3 4 PT 1 2 3 4 PT 1 2 3 4 PT 1 EELIA COORD I 0.0D+00 4.OD-03 8.0D-03 NATE t COORDINATE 0.0D+00 9.2175D+04 1.6717 5D+0 5 1.2D-02 1.6D-02 2. C8D-02 BRICK ELEMENT 2.32725D+05 2.9 145QD+Q5 3 .38850D+05 H-1HETA CURVES 1.GE + 05 THETA COORDINATE COORDINATE O.OD + OO O.OD+OO 2.117D-03 5.93D+04 9 7G2D-03 1.187D+05 1.0D-02 1.2 1033D+05 1.5D + 05 THETA COORDINATE H COORDINA! 0.0D+00 3. 131D-03 7.865D-03 1.0D-02 2.GD + 05 0.0D+00 8.895D+04 1.335D+05 1 5359 1D+05 E THETA COORDINATE fl COORDINATE 0.0D+00 0.0D+00 v>) a> 101 provides detailed information on the use of the program, such as how the data input is prepared. The data input files for the numerical examples presented in Chapter Five are listed in Appendix D. All program input and output is in basic units of inches, pounds, or radians. 661 662 664 665 66B 690 695 700 FORMAT ( 0 / ,2X, HEIGHT ,6X, 'BRICK WYTHE DEFLECT! CN' , $ 6X,'BLOCK WYTHE DEFLECTION) KBITE(6,662) ECOHAT (*G* ,4X,*0,0*,12X, *0.CCOOOOD + OO, 16X, $ 0.CCOOOOD+OO) EEIGIiT=0, 0 CO 665 L=1,NSDFMT,5 LEL=L+1 HEIGHT=flSIGHT+8. 0 BITE(6,664) BEIGHT,STB(L) ,SlfiW (LEL) FORMAT(* ,2X,F5.1, 11 X ,D 13.6,15 X D 13.6) CC MINUE BITE (6, 668) FORMAT ( 1 ,66 (' = )/ 17X, $ 'ELEMENT FORCES AND DISPLACEMENTS'/* ,66( = )) T EI F E= 1 GC TO 695 CALI ECUAL (NGSXRF, NSTRF, NSDM2, 1) CALL PLATER (STEK,NSTRK,M1,M1P1,NSDCF,NSDM2,1NHER,INHBL) CA LL Â£UAL (NGSTRK, NSTRK NSEM2 N IP 1) CALL NOLL (NSTR ts,NSDM2,1) TÂ£STIF=0 CALL STACON (NSTEW, NGSTRK,NGSTEF NK4 ,F2, H 2 NK2, K2W 2, F 1, $ F1MKW2,NK1,W 1,MIP 1,NSDM2,NRTMP1,NRBM1,NRTM1) CALL DISPLA (STEW,NSIRH,NSDCF,NSDH2 ,iNHER,INfiEL) CALL BML1 (ST BK, STEW STfif, NSDCF, M1) IE (KEX1.EU.1) GC TO 625 CALL ECOAL ( ERROR,STEF,NS EOF,1) CONTINUE ENE ro -P* 343 Table B.26 Nomenclature For Subroutine BLESM VARIABLE TYPE DEFINITION BLEK DOUBLE PRECISION BLEF DOUBLE PRECISION TOTEK DOUBLE PRECISION BLAEL DOUBLE PRECISION BL3EIL DOUBLE PRECISION LN DOUBLE PRECISION ELASBL DOUBLE PRECISION ARSHBL DOUBLE PRECISION I INTEGER BL INTEGER NELEM INTEGER BLOCK ELEMENT STIFFNESS MATRIX BLOCK ELEMENT FORCE MATRIX MATRIX THAT STORES ALL OF THE ELEMENT STIFFNESS MATRICES BLOCK AE/L (AXIAL STIFFNESS FACTOR) BLOCK 3EI/L (ROTATIONAL STIFFNESS FACTOR) VERTICAL LENGTH OF THE BLOCK ELEMENT MODULUS OF ELASTICITY OF THE BLOCK AREA IN SHEAR FOR THE BLOCK ELEMENT NUMBER NUMBER OF BLOCK ELEMENT DEGREES OF FREEDOM NUMBER OF ELEMENTS WALL HEIGHT, INCHES 180 Figure 6.28 Vail Wythe Vertical Load Versus Height At Pmax Example Number 4 For 83 the point in the loading process at which the brick prism is unable to support additional load. 4.3.2 Collar Joint To determine the shear spring stiffness factor (kg) necessary for constructing the collar joint element stiffness matrix, tests will be performed to establish the relationship between vertical load and vertical deformation for the collar joint when loaded in shear. The vertical load will be plotted against the vertical deformation to obtain a curve. This curve will be approximated in a piecewise-linear fashion (divided into a series of straight line segments). The slope of each straight line segment will be equivalent to the shear spring stiffness factor k for the region of load values established by the load coordinates of the end points of each line segment. This is shown in Figure 4.10. In these tests, the collar joint will be loaded to failure (inability to carry additional load) so the maximum shear load capacity of the collar joint will be determined. If the effect of moment transfer by the collar joint is found to merit consideration, a test will be devised to determine the relationship between moment and rotation for the collar joint. The moment will be plotted against the rotation and the resulting curve approximated by straight line segments. The slope of each segment will equal the moment spring stiffness factor (km) for the region of moment values established by the moment coordinates of each line segment. Figure 4.11 shows what this curve might look like. 4.33 Concrete Block Experimental tests will be performed on the block prisms identical to those done on the brick prisms, to obtain the axial and rotational 4 5 PT 1 2 3 4 5 BRICK 10 PT 1 2 3 4 5 6 7 8 9 10 BLOCK 8 PT 1 2 3 4 5 6 7 8 2.962D-03 3.5D-3 7.50*04 THETA COORD1 0.0D+00 7.52D-04 1.505D-03 3*245D-03 3.50-03 ELEMENT P-M M 1 9. 335D+04 031G3D+Q5 NATE M COORDINATE 0.OD+OO 50250+04 9975D+04 1,C5D+05 1.10132D+05 INTERACTION 3 6 DIAGRAM COORDINATE 0.OD *00 3,750+04 1. 07250 + 05 1,5D+05 1.65D+05 1,734750+05 1.68750+05 1.26D+05 1. G8432D+05 0,OD+OO ELEMENT P-H M P COORDINATE 0,OD+OO 1,8750+04 7.5D+04 1,20+05 1.5D+05 1,92 7 50 + 05 2. 2 5D+05 3.00+05 3.38850+05 3.3885D+05 INTERACTION CCORDIN ATE 7.50+03 4,875D+04 8,6250+04 9.9960+04 9.675D+04 6.75D+04 3.91230+04 0.OD+OO DIAGRAM P COORDINATE 0,OD+OO 1.875D+04 3.75D+04 5, 1D+04 5.625D+04 7,5D+04 9.315D+04 9.315D+04 MAXIMUM BRICK ELEMENT COMPRESSIVE LOAD CAPACITY 3. 38 85C + 05 MAXIMUM CC1LAR JOINT ELEMENT SHEAR LOAD CAPACITY V>1 05 WALL HEIGHT, INCHES 190 Figure 6.37 Block Wythe Moment Versus Height For Example Number 2 33 The remaining forces acting on the beam can be determined from the equations of equilibrium; thus, we have f2 = -f5 (5.12) and f3 = -f6 + ^ * Now at x = 0, v = Wcj, and hence from Equation (3*9) i5f5 *2 0 W Using Equations (3*10) and (3-12) to (3*14), we have 1 2EI k55' [%.o k _ f5lN 6,51 H-O 6EI T=0 (1 + *)1 = I _1\ = -12EI 2,5 W5/t=0 (1 + $)13 '~^6 + f5lN k3,5 w 5/t=0 wr 6EI T=0 (1 + *)1 (3.13) (3.14) (3.15) (3.16) (3.17) (3.18) with the remaining coefficients in column 5 equal to zero. The variable T stands for temperature change. NUMBER OF OPERATIONS 64 Figure 3.16 Comparison of Equation Solving Operations of Modified Gauss Elimination Versus Modified Static Condensation WALL HEIGHT, INCHES Figure 6.2 Lateral Wall reflection Versus Height For Example Number 3 78 The second type of test will involve loading the prisms eccentrically and measuring the end rotation due to the applied end moment for various eccentricities and levels of loading. As before, the moment will be plotted against the rotation and the resulting curves approximated by straight line segments. The slope of each segment, this time, will equal the rotational stiffness factor 3EI/L for the region of moment values established by the moment coordinates of each line segment for the corresponding axial load. This is shown in Figure 4.7. The prism sketches in Figures 4.6 and 4.7 are intended only to give a general idea of how these tests will be done, and are not meant to be detailed representations of the test set-up and instrumentation required to measure deflections and rotations. The remainder of the variables needed to construct the element stiffness matrix for the brick element can be obtained without further tests. Table 4.1 lists all the variables needed and their sources. As indicated, the modulus of elasticity value for the brick was taken from Tabatabai (15) and the brick Poissons ratio from Grimm (5). Figure 4.8 shows how the element stiffness matrix for a brick element is calculated using the stiffness factors from the prism tests. Since the prisms will be loaded to failure in each type of test, the maximum axial compressive load capacity as well as the relationship between axial load and moment carrying capacity will be established. This will enable an axial load versus moment interaction diagram to be drawn for the brick element. Figure 4.9 shows the general form this diagram will take. The curve in this diagram will also be approximated by straight line segments. It will be used by the model to determine when a brick element has failed. Brick prism failure will be defined as Table A. 2 Program Nomenclature VARIABLE TYPE ARSHBL DOUBLE PRECISION ARSHBR DOUBLE PRECISION BL INTEGER BLAEL DOUBLE PRECISION BLMAXP DOUBLE PRECISION BLMOM DOUBLE PRECISION BL3EIL DOUBLE PRECISION BLVERF DOUBLE PRECISION BR INTEGER BRAEL DOUBLE PRECISION BRMAXP DOUBLE PRECISION BRMOM DOUBLE PRECISION BR3EIL DOUBLE PRECISION BRVERF DOUBLE PRECISION CJ INTEGER CJMAXP DOUBLE PRECISION DEFINITION AREA IN SHEAR FOR THE BLOCK AREA IN SHEAR FOR THE BRICK NUMBER OF BLOCK ELEMENT DEGREES OF FREEDOM BLOCK AE/L (AXIAL STIFFNESS FACTOR) BLOCK MAXIMUM P (COMPRESSIVE LOAD CARRYING CAPACITY) ABSOLUTE VALUE OF AVERAGE OF END MOMENTS FOR A BLOCK ELEMENT BLOCK 3EI/L (ROTATIONAL STIFFNESS FACTOR) VERTICAL (AXIAL) FORCE AT THE BOTTOM OF A BLOCK ELEMENT NUMBER OF BRICK ELEMENT DEGREES OF FREEDOM BRICK AE/L (AXIAL STIFFNESS FACTOR) BRICK MAXIMUM P (COMPRESSIVE LOAD CARRYING CAPACITY) ABSOLUTE VALUE OF AVERAGE OF END MOMENTS FOR A BRICK ELEMENT BRICK 3EI/L (ROTATIONAL STIFFNESS FACTOR) VERTICAL (AXIAL) FORCE AT THE BOTTOM OF A BRICK ELEMENT NUMBER OF COLLAR JOINT ELEMENT DEGREES OF FREEDOM COLLAR JOINT MAXIMUM P (VERTICAL LOAD CARRYING CAPACITY) 215 stiffness matrix which considers moment magnification are developed, the following derivation is presented. It was taken from Chajes (3). Once again, element degrees of freedom 1 and 4 are not considered below because they are not affected by the consideration of moment magnification. Columns 1 and 4 in the element stiffness matrix shown in Figure 2.2 will, therefore, remain intact. Consider an element of a beam column subject to an axial load P and a set of loads [f], as shown in Figure 3*9d. The corresponding element displacements [w] are depicted in Figure 3*9e. It is our purpose to find a matrix relationship between the loads [f] and the deformations [w] in the presence of the axial load P. As long as the deformations are small and the material obeys Hooke's law, the deformations corre sponding to a given set of loads [f] and P are uniquely determined, regardless of the order of application of the loads. The deformations [w] can, therefore, be determined by applying first the entire axial load P and then the loads [f]. Under these circumstances, the relation of [f] to [w] is linear, and the stiffness matrix can be evaluated using the principle of conservation of energy. The element is assumed to be loaded in two stages. During the first stage, only the axial load P is applied and, during the second stage, the element is bent by the [f] forces while P remains constant. Since the element is in equilibrium at the end of stage one as well as at the end of stage two, the external work must be equal to the strain energy not only for the entire loading process but also for stage two by itself. The external work corresponding to the second loading stage is 303 B.10 Subroutine PULROW Subroutine PULROW will pull the first T numbers in row I out of an n x 6 matrix [b] and store them in a T x 1 matrix [a]. Table B.10 defines the nomenclature used in this subroutine. Figure B.10 is an algorithm for this subroutine. B.11 Subroutine PULMAT Subroutine PULMAT will pull a T x T matrix out of a 6 x 6 x n matrix [b] and store the values in a T x T matrix [a]. The value of T must be less than or equal to 6. The value of I identifies which 6x6 matrix in [b] matrix [a] is to be obtained from. Table B.11 defines the nomenclature used in this subroutine. Figure B.11 is an algorithm for this subroutine. B.12 Subroutine GAUSS 1 Subroutine GAUSS1 solves a matrix equation of the form [a] x [x] = [b], where [a] is an n x n matrix which has a bandwidth of (2 x ml) 1. The solution is a standard Gauss Elimination for a symmetric banded matrix. Matrix [x] is the n x 1 matrix that stores the solution, [c] is the n x ml matrix used to store the upper triangular portion of the symmetric nxn matrix [A] with a bandwidth of (2 x ml) 1, and [b] is the n x 1 matrix that stores the product of [a] x [x]. The variable ml equals half the bandwidth (including the diagonal) of matrix [a]. For a regular problem, KEY should equal 1. When subroutine GAUSS1 is called and matrix [a] is the same as in the last call to GAUSS1, KEY should equal 2. Table B.12 defines the nomenclature used in this subroutine. Figure B.12 is an algorithm for this subroutine. WALL HEIGHT, INCHES 188 TT 75,000 Figure 6.35 Brick Wythe Moment Versus Height For Example Number 4 2 4,109D-Q3 1186D+05 3 6.616D-3 1.70D + 0 5 4 1,OD-02 2.58179D+05 COLLAR JCINT ELEMENT P-DELTA CURVE 2 PT DELTA COORDINATE P COORDINATE 1 0.0D+00 O.OD+OO 2 4,613D-02 9,777D+03 COLLAR JOINT ELE HE NT H-THETA CURVE 2 PT THETA COORDINATE M COORDINATE 1 0.0D+00 .0D+0 2 1.0D+00 0.D+00 BLOCK ELEMENT P-DELTA CURVE 5 PT DELTA COORDINATE P COORDINATE 1 0.D+00 2 4.0D-03 3 8,OD-03 4 1,20-02 5 1.68D-02 BLOCK ELEMENT 3 0.QD + OO 2.5125D+04 5025D+04 7.5375D+04 9,3 15D+0 4 K-THETA CURVES d PT 1 2 3 4 5 PT 1 2 3 2,5D + 04 THETA COORDINATE II COORDINATE G. CD+QO 0.0D+00 5,63D-04 2,3325D+04 8.46D-04 3.5D+04 1.49D-03 4.6675D+04 3.5D-03 8,31140+04 5. OD +04 THETA COORDINATE E COORDINATE 0, 00 + 00 0. 01)+0 0 1, 035D-03 4.665D+04 1.674D-03 7.0D+04 PERCENT PERCENT SAVINGS LOSS 65 Figure 5.17 Computational Savings of Modified Static Condensation Over Modified Gauss Elimination 201 very close agreement is reached between the experimental and analytical results, the model may be used to explore variables not considered in the testing program. Ultimately, these efforts will yield valuable information, not currently available, that will enable modifications to be made to the current design standards. More accurate and reliable design procedures will make it possible for structural engineers to design safer, more efficient, and more economical composite masonry structures. It is hoped that this study has brought the realization of that goal one step closer. CJMOM CJMOMS CJSHRS CJVERF COUNT ELASBL ELASBR FMULT HEIGHT I ITER J K LNHBL LNHBR LNVER HI M1P1 DOUBLE PRECISION DOUBLE PRECISION DOUBLE PRECISION DOUBLE PRECISION INTEGER DOUBLE PRECISION DOUBLE PRECISION DOUBLE PRECISION REAL INTEGER INTEGER INTEGER INTEGER DOUBLE PRECISION DOUBLE PRECISION DOUBLE PRECISION INTEGER INTEGER ABSOLUTE VALUE OF AVERAGE OF END MOMENTS FOR A COLLAR JOINT ELEMENT COLLAR JOINT MOMENT SPRING STIFFNESS COLLAR JOINT SHEAR SPRING STIFFNESS ABSOLUTE VALUE OF COLLAR JOINT SHEAR FORCE COUNTING VARIABLE USED TO HELP STORE THE BRICK AND BLOCK WYTHE VERTICAL LOAD PER HEIGHT MODULUS OF ELASTICITY OF THE BLOCK MODULUS OF ELASTICITY OF THE BRICK VARIABLE USED TO STORE THE MULTIPLES OF A STRUCTURE FORCE FOR WHICH AN INTERMEDIATE PRINTOUT IS DESIRED VARIABLE USED TO HELP PRINT THE LATERAL WALL DEFLECTION VERSUS HEIGHT DO-LOOP PARAMETER DO-LOOP PARAMETER DO-LOOP PARAMETER DO-LOOP PARAMETER HORIZONTAL LENGTH OF THE BLOCK HALF OF THE COLLAR JOINT ELEMENT HORIZONTAL LENGTH OF THE BRICK HALF OF THE COLLAR JOINT ELEMENT VERTICAL LENGTH OF THE BRICK AND BLOCK ELEMENTS HALF THE BANDWIDTH (INCLUDING THE DIAGONAL) OF THE STRUCTURE STIFFNESS MATRIX HALF THE BANDWIDTH PLUS ONE 110 where P = vertical load Ag = area in shear therefore, P t Aa (5.3) By taking the average shear stress from their tests and multiplying it O by the desired area in shear of 192 in a new maximum value of P is obtained. Then by dividing the new Pmax by the average slope, a new Amax is obtained. Figure 5*5 shows the collar joint element P-A curve obtained in this fashion. The P-A curve for the block element is generated in the fashion described in section 4.3.3. Tests similar, to those, which were performed by Fattal and Cattaneo (4) on block prisms, were used to obtain the P-A curve for the block element which is used in the numerical examples. They tested three 6x32x24 in block prisms by loading them axially and noting the vertical strain variation with load. The results of their tests are shown in Figure 5.6. To use their results, it is necessary to convert their data, for 6x32x24 in prisms, to equivalent data for 6x24x8 in prisms. This is done in the same way that was discussed earlier for the brick prism results. In other words, by applying Equations (5.1) and (5.2) on several points in Figure 5.6, a new P-A curve can be generated which is representative of the results that would be obtained for a 6x24x8 in 53 2 10 0 10 0 5-60 -10 0 0 9/5 -7 -40 000 200/9 J400/g Now from back substitution, ~200/9 W4 = "14/g therefore W4 = 7 , 9/5 W3 7V4 = -40 since W4 is known, W3 can be found directly W3 = 5 Similarly, 5W2 6W3 = -10 yields W2 = 4 and 2W1 + \Â¡2 10 produces = 3 Therefore, matrix [w] has been solved for and found to be 143 5 IN l-M 5 IN T inT/UW 4 IN H H 6 IN TEST WALL Sign Convention H/T = 12 P = P + AP AP = 4000 LB e 0 IN M Pe 0 P '"'H ^ m k /nrrfn I 8 IN TYPICAL 2 IN111 3 IN MODEL Figure 5*21 Example Number 2 480 IF (TPfilNT. NE. 2) GO TO 500 WRITE(6,490)K 490 FORMAT ('O',' *** ELEMENT NO. ,14, 1X, MS FAILED ***) 500 CALL BLPDMT (NBLPDP, NCELEC, RELMTE HLDCC.B, ELPCOJ8 ELPCV, $ BI1CCR,BLMCCÂ£, ELEF, SBLPC, SELMT,BLVEBF ,ELMCM, BLAEL ,BL3EIL ,ITER) CALL STIFAC (BLVEBF,BIMGM,STCVF,STCEE,ElAEI,EL3EIL, $ STOVST,STGMSI,TEST IF,ITER,3,K) CALL BLESM (BLEK,BLEF,10TEK,BLAEL,B13EIL,LNVEB, ELASBL,AES EEL,K,EL,NELEM) IF(K.EQ.NELEM) CALL BNS2RT (STBK,BIEK,3,NSDCF,M1,EL, 1BL) IF (K.EQ.NELEM) GO TO 520 IF (K.EQ.3) CALL ENSEBT (STRK,BLÂ£K,1,NSDOF,M1,BL,IBL) IF (K. NE 3) CALL BNSEBT (STBK ,BLEK 2 BSD CF M1 BL IBL) IF (ITEB.EC. 1) GO TO 530 CALL SMULT {-1.0D + 00,BIEF,MBIEF ,EL,1) IF (K. EQ. NELEM) CALL INSERT ( ERROR, M ELEF 2,3, NSDCF ,BL , IBL) IF (K. EQ, NELEM) GO TO 530 IF (K. EQ. 3) CALL INSEBT (EBRCB,MRIEF ,2, 1 NSDCF, El, IBL) IF(K. NE.3) CALL INSEBT IF (lEELPfi. NÂ£. 1) GO TO 539 CALL WYTHE (BERYLD,BLYLD,NSTC&Y) $ $ 520 525 $ C c c c c c c c c c c c FINALLY : (TEE VEBY FIBSI TIME) 1, CCNSIBUCT TEE STRUCTURE FOFCE MATRIX. 2. SOLVE THE EQUATION Â£K]*Â£W]=Â£F] FCB Â£N], WHERE Â£K] IS THE STBCTUBE STIFFNESS MATRIX, Â£H] IS TEE STBUCTUBE DISPLACEMENT MATRIX, AND Â£F] IS THE STBUCTUBE FCECE MATRIX. USE A GAUSS ELIMINATION IYPE OF SOLUTION TECHNIQUE WHICH TAKES ADVANTAGE CP THE E ANDEENESS AND SYMMETRY CF THE STRUCTURE STIFFNESS MATRIX 'AND ALSO USES STATIC CONDENSATION. . SET Â£ F]=Â£ EBBOfi ] SO IT CAN BE USED TO INSURE CONVERGENCE. 3 28 ks ksÂ£1 _ks ksÂ£2 ksÂ£1 ksÂ£f + km *ksÂ£1 ksÂ£2Â£1 km "ks ~ksÂ£1 ks ksÂ£2 ksÂ£2 ksÂ£1Â£2 ~ km "ksÂ£2 ksÂ£2 + km WHERE: k = SHEAR SPRING STIFFNESS b km = MOMENT SPRING STIFFNESS Â£1 = LENGTH OF LEFT PART OF ELEMENT Â£2 = LENGTH OF RIGHT PART OF ELEMENT Figure 3*4 Element Stiffness Matrix For Element 2 in Figure 31 144 W7i t ^72 rft t t v*1 s -i-!^ t. V* 41 V. 4 t W1 t w u t U t vp-or H_U virl v # t H- V jT v V.#' w V#Tv 7777/777 t ,t_ t2 w, WR 3 w 71 w 72 W 73 vj t V t VI' t Vp t V' V t V V t M N* W. V^s. rl V' vi v tfi >. V1 v W v. /777/m W 74 u * u 1 ,L U jt t_ 4 U 4 ;t_ 4 4 t tw2 * W3 w, 3 w 73 W71 w72 'iff t_ M * ,t_ 4 4 t_ 4 t_ * ,t_ 4 U 4 > u Â¡r * U jWj t V t W11 w V?77 74 2 W. w 3 /7?7 EXTERNAL WALL LOADS EQUIVALENT WALL LOADS WALL DISPLACEMENTS Figure 5.22 Structure Degrees of Freedom For Example Numbers 2 and 5 72 MORTAR JOINT w, 4" BRICK PRISM WITH 8" 4" BRICK PRISM WITH 3 5/8"x 7 5/8" x 2 1/4" 3 5/8" x 7 5/8"x 3 1/2* BRICK BRICK (1- Wr Wc 3 w Figure 4.3 Finite Element For Brick 15 [I] 2 1 4 2 5 O O O 2 3 5 The index matrix plays a vital role in the proper assemblage of the structure stiffness matrix. This is discussed in the next section. 2.4 Construction of the Structure Stiffness Matrix The structure stiffness matrix is constructed by using the element stiffness matrices and the element index matrices. To obtain a term in the structure stiffness matrix, it is necessary to add up the appropriate terms from the element stiffness matrices. The procedure for doing this is best illustrated by an example. Consider the frame of Figure 2.3. To identify where each coefficient in each element stiffness matrix belongs in the structure stiffness matrix, plpce the numbers in the index matrix for an element along the sides of the element stiffness matrix as shown below. 0 0 0 1 34 W1 - k11 k12 k13 k14 k15 kl6 k21 k22 k23 k24 k25 k26 k31 k32 k33 k34 k35 k36 k41 k42 k43 k44 k45 k46 k51 k52 k53 k54 k55 k56 VO k62 k63 k64 k65 k66 0 0 0 1 3 4 6 1.2 Objective Lybas and Self (6) have submitted a research proposal to the National Science Foundation aimed at addressing some of these needs. Specifically, they seek to explore both experimentally and analytically 1) The nonlinear load deformation properties of composite masonry walls under compression and out of plane bending. The effect of transverse loading, eccentric compressive loading, slenderness ratio and different masonry unit properties will be considered. 2) The failure mechanisms of composite masonry walls under these types of loading conditions. 3) The transfer of vertical force across the collar joint from block wythe to brick wythe. 4) The suitability of current standards for composite masonry wall design. 5) The development of improved design equations and procedures for composite walls, based on the results of the research. This study essentially consists of the development of an analytical model which will be used, once the experimental phase has been completed, to examine the factors cited above. The model will consider a composite masonry wall subject to compression and out of plane bending, due to either eccentric load application or transverse loading. The two-dimensional finite element model will take into account the different nonlinear load-deformation properties of the brick and block wythes, the load transfer properties of the collar joint, the effect of moment magnification that, as previously mentioned, will result as the lateral wall deflections increase, and the effect of shear deformations. 327 element P value falls on the P-A curve and, therefore, generates the appropriate axial stiffness factor of AE/L. Similarly, it finds the slope for the region in which the average moment falls on the M-9 curve which corresponds to the element P value and generates the rotational stiffness factor of 3EI/L. Table B.20 defines the nomenclature used in this subroutine. Figure B.20 is an algorithm for this subroutine. B.21 Subroutine CJPDMT Subroutine CJPDMT will calculate the slopes of the linearly approximated P-A and M-9 curves for the collar joint element the first time it is called. It always finds the slope for the region in which the element P falls on the P-A curve and, therefore, generates the appropriate shear spring stiffness factor. Similarly, it finds the slope for the region in which the average moment falls on the M-9 curve and generates the moment spring stiffness factor. Table B.21 defines the nomenclature used in this subroutine. Figure B.21 is an algorithm for this subroutine. B.22 Subroutine BLPDMT Subroutine BLPDMT will calculate the slopes of the linearly approximated P-A and M-9 curves for the block element the first time it is called. It always finds the slope for the region in which the element P values falls on the P-A curve and, therefore, generates the appropriate axial stiffness factor of AE/L. Similarly, it finds the slope for the region in which the average moment falls on the M-9 curve which corresponds to the element P value and generates the rotational stiffness factor of 3EI/L. Table B.22 defines the nomenclature used in this subroutine. Figure B.22 is an algorithm for this subroutine. 4 PT 1 2 3 4 5 2.9620-03 9.3350+04 3.5D-03 1.Q3103D+05 7.5D+04 THETA COORDINATE H COORDINATE 0.OD + 00 7.52D-04 1,505D-03 3.245D-03 3.5D-3 0,OD+OO 3.5025D+04 6.9975D+04 1. G5D+0 5 1.10132D+05 BRICK ELEMENT P-M 10 PT H COORDINATE 1 0. OD+OO 2 3.75D+4 3 1.0 725D + 05 4 1.5D+05 5 1.65D+05 6 1. 73475D + 05 7 1 ,68750 + 05 8 1.26D+05 9 1,00 4 3 2D+0 5 10 0.OD+OO BLOCK ELEMENT P-M I INTERACTION DIAGRAM P COORDINATE 0.OD+OO 1.075D+04 7,5Df04 1. 2D+05 1.50+05 1.9275D+05 2,25D+05 3.0D+05 3,38850+05 3.38S5D+05 NTERACTION DIAGRAM 8 PT 1 2 3 4 5 6 7 8 H CCCRDIMA1E 7.5D+03 4.875D+04 8.625D+04 9,996D+04 9,675D+04 6.75D+04 3.912 3 D + 0 4 0. OD+OO P COORDINATE 0.OD+OO 1.875D+04 3.75D+04 5. 1D + 04 5.6 25D + 04 7,5D+04 9.315D+04 9.315D+04 MAXIMUM ERICK ELEMENT COMPRESSIVE LOAD CAPACITY 3. 385D + G5 MAXIMUM COLLAR JOINT ELEMENT SHEAR LOAD CAPACITY oo DIMENSION X (N) ,C(N,H1) ,B (N) C FCfifcAED ELICIKATICN. fl = H 1-1 Sfl 1 = N-1 DC 98 J= 1,11 HI JK=J+M IF (Jti.GT. N) JH=N df 1=J + 1 K=M + 1 IF (tl+J.GT.N) tM = N-J+ 1 DC 97 K=JP 1 f JM KK=K-JP1+2 CCEEI=-C (J,KK)/C (J, 1) fl (K) = B (K) +CGEFF+B (J) ME=Kfi-1 IF(KEi.EQ.2)GO TO 97 DC 96 1=1, MM II=I+K-J C (K, I)=C (K, I) +COEFF *C ( J, 11) 96 CONTINUE 97 CCNTINUE 98 CONTINUE C BACK SUBSTITUTION* X(N)=B(N)/C(N,1) N K= 0 DO 190 KK= 2#N K= N-KK +1 IF (KK.GT.M)GC TO 150 KK=NM+ 1 GC TO 16C 150 KF=K 160 X(K)=B(K) h 1=1 N 2=K DC 170 11=1,KM N 1=N 1 + 1 r\> K=L+1 SBlflT (J,L) = (OfiHCOH (J, K)-BRMCCR ( J, L) )/ (BBICOB (J K) - I BBTCOR (J ,L) ) 40 CCN1INDE 45 CONTINUE B EAEI=S EEPD(1) BB3EIL=SBB21T ( 1, 1) GC 1C 150 60 DO EC I=1,HPLS1 If (BEEF (1) .GE. EEPCOH (I) ) BRA FL=SBRPD (I) If (BEEF (1) .LT.BBPCOR (I) ) GO 10 90 80 CCEliNUE 90 A VGECÂ£1 = DABS ( (BREF (3)+B BSF (6) )/2) IF {EEEF (1) GT. ERPCV (1) ) GO TC 105 DO 1C0 L=1,NMLS1 IF (AVGMCE, GE. BB MCOR ( 1,L) ) EB3EIL=SBHM1( 1 ,L) IF(AVGOl.LT. BBHCOH (1 L) ) GC TC 150 100 CCEIIDIUE GO 1C 150 105 IF (BEEF (1) GE. BBPCV (NOBBPC) ) GO TO 125 NOPB 1=NOBBPC1 DC 110 K=1,K0Â£W1 L=K+ 1 IF (BBEF (1) GE, EEPCV (K) ) LLÂ£=K IF (BEEF (1) .LI. BBPCV (L) ) OIf = I IF (BEEF (1) .LT. EEPCV (L) ) GOTO 115 110 CONTINUE 115 DC 120 I=1,NHLS1 IF (AVGMOFl. GE. BBMCOR (LLP,I) ) ILSM1=SBB3T (1LE/I) IF (AVGMOM.GE.BBNCOE(ULP,1})ULSMT=SBENT(ULP,I) 120 CONTINUE Bfi3EIL= DLSMT- ( ( (BBPCV (ULP) -BEEF ( 1) )/(BBPCV (ULP) -BBPCV (ILP) ) ) $ (ULSfll-LLBHT)) GC 1C 150 125 DO 140 1= 1, NMLS1 IF (AVGMCH.GE.BF ACOR (NOBBPC,!) ) ER 3E IL = SBfiMT(NOBBPC,I) 262 301 Table B.9 Nomenclature For Subroutine EXTRAK VARIABLE TYPE DEFINITION A DOUBLE PRECISION MATRIX FROM WHICH MATRIX [b] IS EXTRACTED; IF KEY = 1, THEN [a] IS AN N x N MATRIX AND IF KEY = 2, THEN [a] IS AN N x 1 MATRIX B DOUBLE PRECISION MATRIX WHICH IS EXTRACTED FROM MATRIX [A]; IF KEY = 1, THEN [b] IS AN M x M MATRIX AND IF KEY = 2, THEN [b] IS AN M x 1 MATRIX KEY INTEGER VARIABLE THAT KEEPS TRACK OF DIMENSIONS OF MATRICES [a], [b] TYPE INTEGER VARIABLE THAT KEEPS TRACK OF WHICH NUMBERS IN THE INDEX MATRIX ARE USED; IF TYPE = 1, THEN THE FIRST THREE NUMBERS IN THE INDEX MATRIX ARE NOT USED; IF TYPE 2, THEN ALL THE NUMBERS IN THE INDEX MATRIX ARE USED; IF TYPE = 3, THEN THE FIFTH NUMBER IN THE INDEX MATRIX IS NOT USED N INTEGER NUMBER OF ROWS (AND POSSIBLY NUMBER OF COLUMNS) IN MATRIX [a] M INTEGER NUMBER OF ROWS (AND POSSIBLY NUMBER OF COLUMNS) IN MATRIX [b] INDEX INTEGER M x 1 INDEX MATRIX WHOSE VALUES GIVE THE POSITIONS IN MATRIX [a] FROM WHICH THE VALUES FOR MATRIX [b] ARE EXTRACTED 318 the curve. Matrix [XCOOR] stores the x coordinate of each point and matrix [YCOOR] stores the y coordinate of each point. Table B.16 defines the nomenclature used in this subroutine. Figure B.16 is an algorithm for this subroutine. B.17 Subroutine CURVES Subroutine CURVES will read and print the x and y coordinates of up to 20 points for up to 20 different curves. The variable NOCURV is the number of curves, while the number of points per curve is denoted by NOPPC and CURVAL is the value each curve is identified by. The matrix [XCOOR] stores the x coordinate of each curve where the first subscript is the curve number and the second subscript is the point number. Similarly, matrix [YCOOR] stores the y coordinate of each curve where the first subscript is the curve number and the second subscript is the point number. Table B.17 defines the nomenclature used in this subroutine. Figure B.17 is an algorithm for this subroutine. B.18 Subroutine PRINT Subroutine PRINT will print an n x 1 matrix [a] and identify each row as a degree of freedom. Table B.18 defines the nomenclature used in this subroutine. Figure B.18 is an algorithm for this subroutine. B.19 Subroutine WRITE Subroutine WRITE will print the dimensions n and a of a matrix [a] then print the n x m matrix. Table B.19 defines the nomenclature used in this subroutine. Figure B.19 is an algorithm for this subroutine. B.20 Subroutine BRPDMT Subroutine BRPDMT will calculate the slopes of the linearly approximated P-A and M-Q curves for the brick element the first time it is called. It always finds the slope for the region in which the 124 several values of P, and the results shown in Figure 5.10 were obtained. For values of P higher than 200,000 lb and lower than 100,000 lb, the M-Q relation did not differ from that shown for those values. Back in section 4.3*2, it was mentioned that, if the effect of moment transfer in the collar joint is found to merit consideration, a test will be devised to establish the relationship between moment and rotation. Figure 4.11 showed what this relationship might look like, and how the moment spring stiffness factor would be calculated. At this point, it is not known how important the collar joint moment transfer effect is and, as a result, it is desired to not consider it in the analysis. This is done by insuring that the moment spring stiffness factor kffl has a value of zero. The best way to accomplish this is by using the M-9 curve for the collar joint shown in Figure 5.11 which was used in the numerical examples. The M-9 curves for the concrete block element are developed in the fashion discussed in section 4*3.3. Tests similar to those were performed by Fattal and Cattaneo (4) on block prisms. Their results were used to obtain the M-9 curves used in the numerical examples. They tested twelve 6x32x24 in prisms by loading them eccentrically and recording the vertical strain on each side of the prism in the plane of the applied end moment. The results of their tests are shown in Figure 5.12. To use their results, it is necessary to convert their data, for 6x32x24 in prisms to equivalent data for 6x24x8 in prisms. This is done in a fashion similar to the technique used for converting brick prism results discussed previously. The M-9 curves for the concrete block element were obtained from the test data shown in Figure 5*12 as I ^77777777777 3> = i AE_ L BASIC RELATIONSHIP k. = AXIAL STIFFNESS FACTOR = ^ a L Figure 4.6 Experimental Determination of Brick Element Axial Stiffness Factor 625 630 631 6 40 645 649 650 651 652 653 654 655 656 660 WRITE (6,63 C) FCBI$A1(*1*,66 (' = ')/' ',19X,'WAIL LOADS AND , 'DISPLACEMENTS ) WRITE (6,631) ECRMAT {* *,18X,'(LOADS AT TEAK VALUES DESIRED]'/* '# 66('=)) IREA1L=4 WRITE(6,645) FGBMAE (*0 /' 1X,EXTERNAL WALL LOADS') CALL PRINT (NSliiF ,NSDM2) W BITE (6,649) FORMAT(* 0'/' ','EQUIVALENT WALL LOADS') CAIL PRINT (STB F, NS DOF) WRITE (6,65C) FORMAT (0/' *, 2X,'W ALL DISPLACEMENTS') CALL PRINT (STS W,NS DOF) BEITE (6,651) ECRMAT ('1* ,66 ( = )/* *, 14X,' LATERAL WAII ', DEFLECTION VERSUS EEIGHT */ ',66(' = *)) kfiITE (6,652) FORMAT (' 0'/' ',3X,HEIGHT*,6X,'IATERAI DEFLECTION') WRITE (6,653) FORMAT(* C,5X,'0.0',10X,* C.GOOOOOD+OO ) EFIGHT=0,0 EC 655 L=3,NSDFMS,5 HEIGHT= BEIGH1+8.0 WRITE(6,654)HEIGHT,STBS (I) FORMAT (* ',3X,F5. 1,9X,D 13.6) CONTINUE EEIG HT= HEIGH1 + 8.0 WHITE (6,656)HEIGHT ECRMAT (' ,3X,E5. 1, 10X,* 0.OCOCCCD+00) KB ITE (6,66 C) FORMAT (1',66 ( = )/ ,14X,'VERTICAI KAIL ', DEFLECTION VERSUS HEIGHT'/* ',66 (' = )) WRITE (6,661) 242 62 Then [P] [Kb] [wb] produces 10~ 0~ ~io" - c: -10 30 20 Finally, [KJ [vj ([f] [KÂ¡b] [wj) can be solved for [WQ1 by back substitution. 2 1 Â¥1 10 0 5 _V2_ 20 W2 4 W1 3 . Therefore, [wal 3 4 and 3 [W] = 4 5 7 286 (start) r | FOR EACH ROT) r' -*j FOR EACH COLUMN) SET [A - [B] (return) Figure B.2 Algorithm For Subroutine EQUAL o n 20 VSlIf=SlGVST (I/O) GO 10 30 S1GVF (I,J) = VEETf IF (VSTIF.EQ.SlOSl (I,J) ) GO 10 30 1BS1IF= 1 S10VST(I,J)=VS1IF 30 IF (E C M F. G1. SIC M F (I J) } GO TO 40 HSlIE=S10HSl{I,d) GC 1C 50 40 SICME (I,J)= MOHF IF(MSTIE.EQ.S1CMST (I,J)) GO 10 50 IBS1IF=1 SICtSI (I,J)=HSTIF 50 RETURN ESC SUBfiCUTINE ERESM (EBEK,EREF,101EK,ERAEL,BE3EIL,LN,ELASBR, ABSfiBB,I,EE,KEIE) S UBBOU'IIN E ERESM HILL CONSTRUCT TEE STIFFNESS MA1EIX EOS EACH BRICK ELEMENT. IN1EGEE BB BOU ELE PRECISION BB EK BEEF,1CTEK,LN,BÂ£AEI,ES3EIL,ELASÂ£E DOUBLE PRECISION A ES HR GBR 1? HIR LO ISBR DIMENSION 10TEK(6,6,NELEK) BEEK (Efi.DR) ,BE E E(EB) E CIS EB=0, 1 5D + 00 GBR=ELASBR/(2.0*(1,0+PQISBR)) ?HIÂ£E=Â£B3Â£I 1*4.0/ (GBR* ABSHBR+LN) BREK (1, 1) =BEAEL BEEK (2,1) = 0. 0 BHEK (3, 1)=0.0 BHEK (4, 1) =-1, 0*EBÂ£K (1, 1) BREK ( S, 1) =0. 0 BREE (, 1)=0.0 BREK ( 2, 2) = (BR3EIL* 4.0/ ( (1.0 + PUIBR) *(LN**2)))- $ (EBEE (1) *6. 0/(5. 0*LN) ) BHEK (3, 2) = {-DH3EIL*2,/ ( (1.0 + PHIBR) *LN) ) + (EREF (1) / 10.0) 356 freedom at the wall top. Table B.32 defines the nomenclature used in this subroutine. Figure B.32 is an algorithm for this subroutine. B.33 Subroutine DISPLA Subroutine DISPLA will calculate the actual structure displacement matrix from the structure displacement matrix that only considers two degrees of freedom at the top of the wall. Table B.33 defines the nomenclature used in this subroutine. Figure B.33 is an algorithm for this subroutine. B.34 Subroutine CHKTOL Subroutine CHKTOL checks to see if each number in an n x 1 matrix [A] is less than the value of TOLER. If each number is, the variable KEY is assigned a value of 0. If at least one number is not, KEY is set equal to 1. Table B.34 defines the nomenclature used in this subrou tine. Figure B.34 is an algorithm for this subroutine. B.35 Subroutine CHKFAI Subroutine CHKFAI will check to see if an element has failed. For a brick and block element, the actual axial load and moment is compared with the allowable axial load and moment as provided by the P-M interaction diagram for that type of element. For the collar joint element, the actual shear force is compared with the allowable shear force. Table B.35 defines the nomenclature used in this subroutine. Figure B.35 is an algorithm for this subroutine. B.36 Subroutine WYTHE Subroutine WYTHE will print, for each wall level (every 8 inches), the vertical load carried by each wall wythe and the percent of the total vertical load which that represents. Table B.36 defines the 3.12 Nonlinear Load Deformation Curve 48 3.13 Advantage of Symmetry and Bandedness of Structure Stiffness Matrix 55 3.14 Comparison of the Equation Solving Operations of Standard Gauss Elimination Versus Modified Gauss Elimination 37 3.15 Computational Savings of Modified Gauss Elimination Over Standard Gauss Elimination 58 3.16 Comparison of Equation Solving Operations of Modified Gauss Elimination Versus Modified Static Condensation 64 3.17 Computational Savings of Modified Static Condensation Over Modified Gauss Elimination 65 4.1 Finite Element Model of Wall 69 4.2 Structure and Element Degrees of Freedom 70 4.3 Finite Element For Brick 72 4.4 Finite Element For Collar Joint 73 4.5 Finite Element For Block... 75 4.6 Experimental Determination of Brick Element Axial Stiffness Factor 77 4.7 Experimental Determination of Brick Element Rotational Stiffness Factor 79 4.8 Brick Element Stiffness Matrix Formulation Using Experimentally Determined Axial and Rotational Stiffness Factors 81 4.9 Typical P Versus M Interaction Diagram For Brick Element .82 4.10 Experimental Determination of Collar Joint Shear Spring Stiffness Factor 84 4.11 Determination of Collar Joint Moment Spring Stiffness Factor 85 4.12 Structure Force Application Through Test Plate 89 4.13 Program Algorithm 99 5.1 Experimental Source of Brick Element P-A Curve (4) 105 5.2 Brick Element P-A Curve 107 WALL HEIGHT, INCHES 193 Figure 6.40 Block Wythe Moment Versus Height For Example Number 5 $ STOVST,STCMST#TESTIF,I1EB,2,J) CALI CJESM (CJEK ,TOTEK,CJSHRS,CJMOMS,LNHÂ£E,LNHEL, $ J,CJ, NELEM) CALL BNSERT (STRK ,C JE K ,2 NSDCF M1 ,C 0 IC J) I F (IT Â£R. EQ. 1) GO TO 380 375 CALL SMULT (-1.0D+00 ,C JEF ,MC JEF ,C J, 1) CALL INSIST {EfROR,MCJEF,2,2,NSDCF,CJ,ICJ) 380 CONTINUE C FOE EACH ELCCK ELEMENT: C (THE VERY FIRST TIME) C 1, CONSTRUCT THE INDEX MATRIX S STORE IT. C 2. GE1 INITIAL STIFFNESS FACTORS 6 STORE THEM. C 3. CONSTRUCT THE ELEMENT STIFFNESS MATRIX. C U. INSERT THE ELEMENT STIFFNESS MATRIX INTO THE C STRUCTURE STIFFNESS MATRIX. C C (EVEEX ClfiEfi TIME) C 1. RECALL THE INDEX MATRIX. C 2. EXTRACT TEE ELEMENT DISPLACEMENT MATRIX FROM THE C STRUCTDR E DISPLACEMENT MATRIX. C 3. RECALL THE ELEMENT STIFFNESS MATRIX. C 4. MULTIPLY THE ELEMENT STIFFNESS MATRIX EY THE ELEMENT C DISPLACEMENT MATRIX TO YIEID THE ELEMENT FORCE MATRIX. C 5. PRINT THE ELEMENT FORCE MATRIX WHEN APPROPRIATE. C 6. PRINT THE ELEMENT DISPLACEMENTS WHEN APPROPRIATE. C 7. CHECK FCR ELEMENT (AND THEREFORE, WAIl) FAILURE USING C THE P-M INTERACTION DIAGRAM. C Â£. GET STIFFNESS FACTORS BASED CN THE ELEMENT FORCES FROM C TEE P-DELTA & H-1HETA CURVES 8 COMPARE 1HEM WITH THOSE C OF THE PREVIOUS CYCLE TO DETECT ANY CHANGES. C 9. CONSTRUCT THE ELEMENT STIFFNESS MATRIX. C 10. INSERT THE ELEMENT STIFFNESS MATRIX INTO THE C STRUCTURE STIFFNESS MATRIX. C 11. MULTIPLY THE ELEMENT FORCE MATRIX BY -1.0 8 INSERT IT C INTO THE STRUCTURE FORCE MATRIX FOR CONVERGENCE CHECK. CO UN I-0 GAD ECCENTRIC!T Y EXAMP LE #1 11/1=3, 6 PLATE L DESCRIPTION OF WALL 96 8,OD+OU 2. ODfOO 3.0D+00 24 MATERIAL PROPERTIES Eli ICE ELEMENT P-DELTA CURVE 6 PT 1 2 3 4 5 6 3 4 PT 1 2 3 4 PT 1 2 3 4 PT 1 DELTA COORDINATE P COORDINATE 0, 0D+00 4.0D-03 8,00-03 1.2D-02 1.6D-02 2.08D-02 O,D+00 9.2175D+04 1.671750+05 2.32725D+05 2.91450D+05 3.3885GD+5 BRICK ELEMENT H-IHETA CURVES 1.0D+05 THETA COORDINATE 0.0B+00 2.117D-03 5 9.702D-03 1. 1,00-02 1,21 1.5D+05 THETA COORDINATE 0.0D+00 3, 131D-03 8. 7,865D-03 1. 1.0D-02 1,53 2, 0D+05 THETA COORDINATE 0.0D+00 M COORDINATE 0.0D+00 ,93D+0 4 187D+05 033D+05 H COORDINATE 0.OD+0 895D+04 3 35D+0 5 591D+05 M COORDINATE 0,0D+00 1.25 IN PT THETA COORDINATE ft COORDINATE 1 O.OD+OO 0.0D+00 2 4 1G9D-03 1 186D+05 3 6.616D-03 1.78D+05 4 1.QD-02 2,58179D+ 05 COLLAR JOINT ELEMENT P-DELTA CURVE 2 PT DELTA COORDINATE P COORDINATE 1 0.0D + 00 0,D+00 2 4.613D-02 9.777D+03 COLLAR JOINT ELEMENT M-THETA CURVE 2 PT THETA COORDINATE H COORDINATE 1 0.0D+00 0.0D+00 2 1.0D + 0 0, 0D+00 BLOCK ELEMENT P-DELTA CURVE 5 PT DELTA COORDINATE P COORDINATE 1 0, 0D + 00 2 4.0D-03 3 8,0D-03 4 1.2D-02 5 1 68D-02 BLOCK ELEMENT 3 5 2.5D+4 O.OD+OO 2.5125D+04 5,025D+04 7.5375D+04 9,315D+04 fi-THETA CURVES PT 1 2 3 4 5 3,50-03 5. 0D+04 H COORDINATE 0.0D+00 2.3325D+04 3,5D+04 4.6675D+04 8.3114D+04 THETA COORDINATE 0. 0D + 00 5.63D-04 8,46D-04 1.49D-03 PT THETA COORDINATE M COORDINATE 1 0.0D+00 0.0D+00 34. 35. 36. 3 7. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 372 195 Figure 6.42 Collar Joint Shear Stress Versus Height For Example Number 3 WALL HEIGHT, INCHES 177 Figure 6.25 Wall Wythe Vertical Load Versus Height At 0.30 Pnax Example Number 4 For ACKNOWLEDGMENTS I would like to thank Dr. James H. Schaub for his active personal interest and support which encouraged me to attend the University of Florida, and for his willingness to serve on my supervisory committee. Dr. Morris W. Self deserves special thanks for serving as chairman of my supervisory committee, being my graduate advisor, and affording me the opportunity to work on this project. Special thanks also go to Dr. John M. Lybas for providing much invaluable aid and guidance during the development of this model. Further gratitude is extended to Professor William W. Coons, Dr. Fernando E. Fagundo, and Dr. Harry J. Reitz, Jr. for also serving on my supervisory committee. Each individual has greatly contributed to the value of my graduate studies and is an asset to their profession and to the University of Florida. Appreciation is expressed to Dr. Clifford 0. Hays. From his classes and notes, the author acquired the technical background necessary to undertake this effort. Thanks also go to Dr. Mang Tia for his technical assistance and recommendations. Randall Brown and Kevin Toye were valuable sources of advice, wisdom, and friendship throughout the author's graduate studies and deserve special mention. The assistance received from Krai Soongswang is also gratefully acknowledged. The unselfish sacrifice, constant support, and endless love provided by the author's parents have been a tremendous source of MORTAR JOINTS BLOCK PRISM BLOCK PRISM WITH BLOCK PRISM WITH WITH 5 5/8" x 15 5/8" x 7 5/8" 7 5/8" x 15 5/8" X 7 3 5/8" x 155/8" BLOCK BLOCK x 7 5/8" BLOCK 4" 6" H r Figure 4.5 Finite Element For Block 17 K11 fk44^ + ^k11^2 * Similarly, the other terms are as shown below: [K] - 1 2 3 4 5 * &]2 ^13-^ Ck453, ^3, + ^-k12^ 2 [*u]2 Ck31 J2 3 + [k ,,] " 2 [k,2] ^ 2 ^34^2 M3 [kcc3 55 1 + c*55], ^3, + ^21^ ^3, + [kp?] 2 ^2 2 ^43^2 * M3 5 fr3 4 2 * [k66^ 1 2 3 4 5 Thus, once the stiffness and index matrices are known for each element, the structure stiffness matrix can easily be arrived at. 2.5 Construction of the Structure Force Matrix To construct the structure force matrix, it is only necessary to assign the value of the force to the term in the matrix which corresponds to the degree of freedom at which the force is applied. If the force has the same direction as the degree of freedom, it is CHAPTER SIX RESULTS OF ANALYSIS 6.1Wall Failure Wall failure load values and locations of failure are shown in Table 6.1 for all five examples. As expected, walls with higher slenderness ratios fail at lower loads than walls with smaller slenderness ratios. Also, walls loaded eccentrically fail before walls loaded axially. These results are, therefore, in agreement with generally expected patterns of structural behavior. 6.2Lateral Wall Deflection Versus Height The lateral wall deflection versus height for examples 2 through 5 is shown in Figures 6.1 through 6.4* Note that the deflected shape of the wall agrees with what would be expected for the end conditions considered by the model. Not surprisingly, for walls of equal height, the wall with a higher failure load experiences larger lateral deflections. 6.3Brick Wythe Vertical Deflection Versus Height Figures 6.5 through 6.8 show the brick wythe vertical deflection as a function of height for examples 2 through 5. As expected, the brick wythe vertical deflection increases as the load increases and the maximum brick wythe vertical deflection is greater for walls which fail at higher loads. 150 98 wall top is analytically handled in the fashion described in section 4.4. Agreement between experimental wall tests and the results obtained from the model is aided by the experimental determination of load- deformation properties for each element type. Element failure for the brick and block elements is considered to occur when an element exceeds the allowable combination of axial load and moment defined by the axial load versus moment interaction diagram for that type of element. The axial forces are always equal and opposite for the brick and block elements, but the end moments may be different. In checking for failure, the maximum end moment is used. Collar joint element failure occurs when the vertical collar joint element force exceeds the maximum collar joint shear load capacity. Wall failure in the model takes place when any element fails according to the aforementioned criteria. In physical prism and wall tests, failure is defined as the inability of the specimen to resist further load. An algorithm for the program is shown in Figure 4.13. All major operations, as well as special features, are discussed in detail in chapters 2, 3, and 4. The program uses 36 modules or subroutines, where each subroutine performs a unique operation. This has the advantages of: * Improving the understanding of the logic " Making program modifications easier * Preventing the program from becoming overwhelming by dividing it into manageable sections. The program is discussed in more detail in Appendix A and each subroutine is discussed in Appendix B. The manner in which these 36 subroutines are used is dealt with in these two locations. Appendix C 140 Buckled shape of column is shown by dashed line (a) 1 / / / 1 1 l 1 \ \ \ 1 JTnW t (b) Y / / t i i \ \ \ \ \ VTrr t (c) 2 s :ji 1 1 1 / / / / t / t rmr t (d) t Y \ \ \ \ \ i ( i i / / nmt \ (e) 1 1 ? P / / / / / i f rmr t (f) ea 1 1 > 1 1 / / 1 l f / I Theoretical K value 0.5 0.7 1.0 1.0 2.0 2.0 Recommended design value when ideal condi tions are approximated 0.65 0.80 1.2 1.0 2.10 2.0 End condition code Y ? Rotation fixed and translation fixed Rotation free and translation fixed Rotation fixed and translation free Rotation free and translation free Figure 5-19 Effective Column Length Factors Based On End Conditions (7) REFERENCES 1. Building Code Requirements for Concrete Masonry Structures, ACI 531- 79R, American Concrete Institute, Detroit, Michigan, 1979 (revised 1983). 2. Building Code Requirements for Engineered Brick Masonry, Structural Clay Products Institute, McLean, Virginia, 1969. 3* Chajes, Alexander, Principles of Structural Stability Theory, Prentice-Hall, Inc., Englewood Cliffs, New Jersey, 1974. 4. Fattal, S.G., Cattaneo, L.E., "Structural Performance of Masonry Walls Under Compression and Flexure," Building Science Series 73, National Bureau of Standards, Washington, D.C., June 1976. 5- Grimm, Clayford T., "Strength and Related Properties of Brick Masonry," Journal of the Structural Division, Proceedings of the American Society of Civil Engineers, No. 11066 ST1, New York, New York, January 1975 6. Lybas, John M., Self, M.W., "Design Guidelines for Composite Clay Brick and Concrete Block Masonry," National Science Foundation Research Proposal, Department of Civil Engineering, University of Florida, Gainesville, Florida, January 1984. 7. Manual of Steel Construction, 8th edition, American Institute of Steel Construction, Chicago, Illinois, 1980. 8. Meyer, V.J., Matrix Analysis of Structures, Harper & Row Publishers, New York, New York, 1983* 9* Przemieniecki, J.S., Theory of Matrix Structural Analysis, McGraw- Hill, New York, New York, 1968. 10. Redmond, T.B., Allen, M.H., "Compressive Strength of Composite Brick and Concrete Masonry Walls," Masonry; Past and PresentASTM Technical Publication 589, American Society for Testing and Materials, Philadelphia, Pennsylvania, June 1974. 11. Roman, Oswaldo, "Compressive Strength of Composite Concrete Block and Clay Brick Prisms," Master of Engineering Thesis, University of Florida, Gainesville, Florida, 1982. 12. Rubinstein, Moshe F., Matrix Computer Analysis of Structures, Prentice-Hall Inc., Englewood Cliffs, New Jersey, 1966. Table A.2-continued VARIABLE SHEARR SHOWF TOLER TRCONV TRELPR TRFAIL TPRINT TRSHOW TRSTIF WDEPTH WHI TYPE DEFINITION DOUBLE PRECISION DOUBLE PRECISION DOUBLE PRECISION INTEGER COLLAR JOINT SHEAR STRESS ON THE RIGHT OR BLOCK FACE OF THE COLLAR JOINT FORCE VALUE FOR WHICH AN INTERMEDIATE PRINTOUT IS TO BE MADE TOLERANCE SET ON EQUILIBRIUM ERROR VARIABLE THAT IDENTIFIES IF A CONVERGENCE ITERATION IS NECESSARY; 0 = NO; 1 = YES INTEGER VARIABLE THAT IDENTIFIES WHETHER OR NOT ELEMENT FORCES AND DISPLACEMENTS ARE TO BE PRINTED; 0 = NO; 1 = YES INTEGER VARIABLE THAT IDENTIFIES THE FAILURE STATUS OF THE WALL; 0 = HAS NOT FAILED; 1 = HAS FAILED; 4 = STOP AFTER PRINTING ELEMENT FORCES AND DISPLACEMENTS - INTEGER VARIABLE THAT IDENTIFIES WHETHER OR NOT AN ELEMENT FAILURE IDENTIFICATION STATEMENT IS TO BE PRINTED; 0 = NO; 2 = YES INTEGER VARIABLE THAT IDENTIFIES WHETHER A CHECK FOR AN INTERMEDIATE PRINTOUT IS TO BE MADE; 0 = YES; 1 NO INTEGER VARIABLE THAT IDENTIFIES WHETHER OR HOT ANY ELEMENT STIFFNESS FACTORS HAVE CHANGED; 0 = NO; 1 = YES INTEGER WALL DEPTH IN INCHES INTEGER WALL HEIGHT IN INCHES CALL EQUAL (GDELLF, DELTAF NSDGF 1) CALL EQUAL (GELRK,STRK,NSIGÂ£,M1) CALL NULL (DELL AW, NS DCF, 1) CALL SIACON (DELTA W,GSTRK ,GDÂ£LTF,K4,F2,W2,K2,K2W2,F1, $ F 1 MKW 2,K1,W1,H1,NSD0F,NGIP1,NRBCI,NR'ICE) CALL ADD (DELLA W,STE K,GSIEN,NSDCF,1) CALL EQUAL (STEW, GSTR'W, NS EOF 1) CALL ADD (DELTAF,STRF,ERRC5 NSDCF 1) NCC NV= NCC N V + 1 IF(NCONV.L1.1) GO TO 142 SUIT 1 (6,555) 555 FORMAL ('0','*-** WARNING*** SOLUTION WLLL NCT CCNVEEGE, ', $ 'CHANGE',/* ',14X, EITHER TOLERANCE OR SLR UCTUfi E LOADS') SLOP 560 KCC N V=0 IF (1ESTIF.EQ.1) GO TC 69 IF (I EF AIL. EQ. 0 ) GO TO 595 BfiITE(6,59C) 590 FORMAT ( 1 ,66 ( = ')/ ',14X,'WALL FAILURE LOADS ', $ 'AND DISL PLACE KENTS / ',66 ( = *)) LERI NT-2 1EFAIL= 4 GC TC 640 595 IF (LRSHOW.EQ.O) GO TC 600 GC TO 620 600 IF(NSTRF(UFOBCE)* NE.SHOWF) GO TC 620 IRSHCW=1 HITE (6,61C) 610 FCEMAT ( 1',66 { = )/ ',12X, 1 'INTERMEDIALE ALL LOADS AND DISPLACEMENTS'/* ',66 (' = *)) SHCWF=SHCWF+FMULT GC LO 640 620 CALL AEPLYF (NSTRF,NS DM2,NOF,NDCF,SFINI1,SFINCR,SFMAX, $ TIER ,KE 1) I RSHCj=0 GC LO 6SC 2 4, 1C9D-03 1.186D+05 3 6.616D-03 1. 78D+05 4 1,0D-02 2,5 81790+05 COLLAR JOINT ELEMENT P-DELTA CORVE 2 PT DELTA COORDINATE P COORDINATE 1 O.OD+OO O.OD+OO 2 4,6130-02 9.777D+03 COLLAR JOINT ELEMENT M-THETA CURVE 2 PT THETA COORDINATE H COORDINATE 1 O.OD+OO O.OD+OO 2 1.0D+00 O.OD+OO BLOCK ELEMENT P-DELTA CURVE 5 PT 1 2 3 4 5 3 5 PT 1 2 3 4 5 PT 1 2 3 DELTA COORDINATE I COORDINATE 0.0D+00 4.0D-03 8,0D-03 1.2D-02 1.68D-02 BLOCK ELEMENT O.OD+OO 2.5125D+04 5.025D+04 7.5375D+0 4 9,315D+04 M-THETA CURVES 2.5L+04 THETA COORDINATE H COORDINATE O.OD+OO O.OD+OO 5.63D-04 2.3325D+04 8.46D-04 3.5D+04 1.49D-03 4.6675D+0 4 3.5D-03 8,3114D+04 5.0D+04 THETA COORDINATE fl COORDINATE O.OD+OO O.OD+OO 1.35D-03 4.665D+04 1,674D-03 7. 0D + 04 VjJ -a 89 W16 W17 W W 19 H -Nr-1 (a) Nodal Degrees of Freedom at Wall Top 16 17 "-0 H" V- (b) Wall Wythe Axial Loads and Moments at Wall Top Figure 4.12 Structure Force Application Through Test Plate 56 57,636 numbers. If half the bandwidth of this matrix is 9, then only 194 x 9, or 1746 numbers need to be stored. This is a savings of 95.4$! Figure 3*14 compares the number of operations required by the regular Gauss-Elimination procedure to the number required by the modified Gauss-Elimination technique for a structure stiffness matrix with a value of 9 for half the bandwidth. Figure 3*15 shows the percent savings in computational effort that result by using the modified Gauss- Elimination method on half of the symmetric nonzero values in the stiffness matrix where half the bandwidth is equal to 9* As shown, up to 95.5$ savings are possible! 5.6.2 Static Condensation Static condensation is another equation solving technique which can be used to solve the matrix equation [f] = [k] [w]. It involves partitioning each of the three matrices and is performed as follows: 1) Partition K W F into Kaa Kab Wa Fa _Kba Kbb Wb Fb 2) Perform r__ forward eliminations on [k] where rn act act rows in [K a3. This will yield = number of CHAPTER TWO DEVELOPMENT OF THE FINITE ELEMENT MODEL 2.1 Matrix Analysis of Structures by the Direct Stiffness Method The Direct Stiffness Method, like most matrix methods of structural analysis, is a method of combining elements of known behavior to describe the behavior of a structure that is a system of such ele ments. The following is a summary of the basic relationships U3ed in this technique. It is presented only as a quick review of the Stiffness Method and not as an exhaustive presentation which develops the rela tionships stated. For that purpose, one of the fine textbooks on matrix analysis, such as Rubinstein's (12), is recommended. A degree of freedom is an independent displacement. Recall that the force displacement equations for an element i, which has n element degrees of freedom, can be written as Mi Mi [v]. [f0^ (2.1) where [w]i n x 1 matrix of independent element displacements measured in element coordinates Mi = n x 1 matrix of corresponding element forces measured in element coordinates [f0]^ = n x 1 matrix of corresponding element fixed-end forces measured in element coordinates [k]^ = n x n element stiffness matrix measured in element coordinates. 7 111 Figure 5*5 Collar Joint Element P-A Curve 49 will now resist a smaller part of the total load, and a "load redistribution" occurs. If the load increment just applied is small relative to the decrease in element stiffness, this element will develop forces smaller than its previous element forces and go back to region 1. The problem is that now, the stiffness increases from EÂ£ to E-Â¡ and after the next iteration will go back to S2 as before and cycle back and forth. To prevent this, care must be taken to insure that once the stiffness of an element decreases, i.e., the next smaller value of E is used to construct its stiffness matrix, the modulus of elasticity is not permitted to increase due to a drop in the element's load level. This is best accomplished by never selecting values of E for an element which are based on a load level lower than the highest experienced up to that point. The second point also is related to the load redistribution which occurs as stiffnesses of individual elements change. During the incremental loading of a structure, it is often desired to examine the structural response after each load step is applied. This permits an examination of how the structural behavior varies as the external loads on the structure are increased. To insure each intermediate solution is accurate, a provision can be made that if any element experienced a change in stiffness during the application of the previous load increment, a new solution with the current element stiffnesses should be calculated before applying the next load increment. This will provide a steady-state solution for each load level; that is, one in which no load redistribution occurred. 300 (start) r I FOR EACH NUMBER IN THE INDEX MATRIX) INSERT THE VALUE IN THE UPPER TRIANGULAR PORTION OF MATRIX [b] INTO THE PROPER LOCATION IN MATRIX [c] (return) Figure B.8 Algorithm For Subroutine BNSERT 214 Figure A.1-continued 213 Figure A.1-continued 1C3 tainted due to the use of only approximate test data, and at best would always be questionable. Example number 1 is an analysis of a wall that was actually tested by Fattal and Cattaneo (4). Section 5.3 compares the failure load of the actual wall with the failure load predicted from this finite element model. The remaining four examples consider walls which vary in slenderness, or H/T, ratio and eccentricity of load application. Thus, two sources of evidence are presented to confirm the capability of the analytical model to generate reasonable results. The first is the comparison of analytical results with actual test results. The second is the presentation of analytical results for walls with different slenderness ratios and eccentricity of load application. General trends in the behavior of walls due to these two effects are known and will be used to gauge the plausibility of the results generated. 5.2 Material Property Data As previously mentioned in section 4.3, prism tests will be performed to estabish the load-deformation properties of the brick, collar joint, and concrete block elements. The information obtained from these tests will be expressed in the form of P-A curves, M-9 curves, and P-M interaction diagrams. 5.2.1 P-A Curves The P-A curve for the brick element is generated in the manner described in section 4.3*1 The P-A curve for the brick element used in the numerical examples was obtained from tests very similar to those described in section 4.3*1 which were performed by Fattal and Cattaneo (4) on brick prisms. They tested three 4x32x16 in brick prisms by ****** ****** ******* ******** ********* ********** _n_ *********** ^ ****** ****** ****** ****** ****** *********** ****** *********** ****** *********** [K] = ****** ** ********* ****** *********** ****** *********** ****** *********** ****** *********** ****** *********** ****** ********** ********* ***** **** ******** *** ******* ****** l* -0- (a) Actual Stiffness Matrix (b) Stiffness Matrix Stored Figure 3.13 Advantage of Symmetry and Bandedness of Structure Stiffness Matrix 319 Table B.16 Nomenclature For Subroutine COORD VARIABLE TYPE DEFINITION NPNTS INTEGER NUMBER OF POINTS IN CURVE XCOOR DOUBLE PRECISION MATRIX WHICH STORES THE X COORDINATE OF EACH POINT YCOOR DOUBLE PRECISION MATRIX WHICH STORES THE Y COORDINATE OF EACH POINT WALL HEIGHT, INCHES 181 Figure 6.29 Wall Wythe Vertical Load Versus Height At 0.30 P For ITiclX Example Humber 5 93 z = total number of structure DOF 4 [f] = force matrix excluding the 4 DOF at the top of the wall [i] = identity matrix (1's along the diagonal, 0's elsewhere) [a] = transformation part of [x] matrix, shown in Equation (4.1) Figure 4.12d shows the wall wythe axial displacements and rotations corresponding to the wall wythe axial loads and moments at the wall top shown in Figure 4.12b. The test plate displacements are shown in Figure 4.12e. From compatibility, the test plate displacement and rotation can be expressed in terms of the wall wythe displacements and rotations as 9 p1 18 ei9 A P1 A16 + A17 2 The test plate displacement geometry is illustrated in Figure 4.12f. Consider the centerline of the plate moving from position a-a to position a'-a' as shown. The wall wythe axial displacements and rotations can be expressed in terms of the plate displacement and rotation as A 16 A 17 + 0 2 pi 918 = 9p1 919 = 9p1 * These displacement relationships between the four nodal degrees of freedom at the wall top and the two plate degrees of freedom can be 92 110 0 ~P16 pv 1 + 1 1 1 P17 2 2 V 7 M18 M [a] 19 (4.1) From this, a matrix [x] can be constructed such that [Fne] [X] [Fold] (4.2) where [F ] = new force matrix which only considers 2 plate DOF at wall top [x] = transformation matrix used to transform the old force matrix with 4 DOF at the wall top, to the new force matrix with 2 DOF at the wall top [FQid] = original force matrix which considers all 4 DOF at the wall top. The matrix [x] is shown below where this matrix equation is illustrated. F I -0- F (z x 1 ) (z X z) (z x 4) (z x 1) pv 0 a P16 M (2 x z) (2 x 4) P17 V. J 18 IM 19 (4.3) where 160 6.4Block Wythe Vertical Deflection Versus Height The block wythe vertical deflection versus height for examples 2 through 5 is illustrated in Figures 6.9 through 6.12. Once again, the block wythe vertical deflection increases as the load increases and the maximum block wythe vertical deflection is greater for walls which fail at higher loads. This agrees with what would be expected. Also notice that for each example, at equal load levels the block wythe vertical deflection is greater. This happens because the brick wythe is stiffer than the block wythe and, in some cases, the wall is loaded eccentrically towards the block. 6.5Plate Load Versus End Rotation The variation of end rotation at the top of the wall as a function of vertical plate load is depicted in Figures 6.15 through 6.16 for examples 2 through 5* The end rotation, of course, increases with increase in load and, not surprisingly, at equal load values for walls of the same height, the end rotation is greater for the wall in which the load is applied eccentrically. At high load levels, the block stiffness is considerably less than the brick stiffness so the increase in end rotation per unit increase in load becomes greater. This is depicted by the flattening out of the curve at high load values in Figures 6.15 and 6.15. 6.6Wall Wythe Vertical Load Versus Height Figures 6.17 through 6.52 show the wall wythe vertical load versus height for four load levels for each of examples 2 through 5* Notice that with decrease in height from the top of the wall, more of the load is transferred to the brick wythe. For the walls loaded eccentrically, 3 Structures" (1). It contains a brief chapter on composite masonry. Finally, a joint American Society of Civil Engineers and American Concrete Institute committee, Committee 530, is currently developing a comprehensive standard to include provisions for both clay and concrete products. Unfortunately, these design standards are limited by a lack of laboratory test data as well as a lack of understanding of the behavior and response of composite masonry. Rational analyses to predict the failure loads and load-deformation properties of composite masonry walls do not presently exist (6). To improve the reliability of the design procedure for composite masonry, several factors not currently considered must be taken into account. The first is that masonry is not linearly elastic, it is nonlinear. In other words, its modulus of elasticity changes depending on the stress level to which it is subjected. Figure 1.3 illustrates some of the complications that result due to the different and nonlinear material properties of brick and block which are present in composite masonry. As evident from the stress-strain curves of the two materials, the block is weaker. Stage 1 depicts the stresses and strains that are generated as a vertical load is applied through the centroid of a composite prism above a 50 percent stress level. As the load is increased further to stage 2, the stiffness of the concrete deteriorates more rapidly than that of the brick. This causes the center of resistance of the section to shift toward the brick. The eccentricity between the point of load application and the effective resistance of the section results in an effective bending of the prism, causing the strains in the block to increase faster than those in the brick. As a WALL HEIGHT, INCHES 191 Figure 6.38 Block Wythe Moment Versus Height For Example Number 3 19 > " F1 0 F2 -50 F3 = 20 F4 0 F5 0 200 \ [F] 0 0 0 -100 200 Figure 2.4 Construction of the Structure Force Matrix APPENDIX B SUBROUTINES B.1 Subroutine NULL Subroutine NULL will set all values in an n x m matrix [a] equal to zero. Table B.1 defines the nomenclature used in this subroutine. Figure B.1 is an algorithm for this subroutine. B.2 Subroutine EQUAL Subroutine EQUAL will create an n x m matrix [a] that is equal to an n x m matrix [b]. In other words, it will create a matrix [a] which is identical to a second matrix [b]. Table B.2 defines the nomenclature used in this subroutine. Figure B.2 is an algorithm for this subroutine. B.5 Subroutine ADD Subroutine ADD will add an n x m matrix [b] to an n x m matrix [a] to yield an n x m matrix [c]. Table B.3 defines the nomenclature used in this subroutine. Figure B.3 is an algorithm for this subroutine. B.4 Subroutine MULT Subroutine MULT will multiply an n1 x n2 matrix [a] times an n2 x n3 matrix [b] to obtain an n1 x n3 matrix [c]. Table B.4 defines the nomenclature used in this subroutine. Figure B.4 is an algorithm for this subroutine. B.5 Subroutine SMULT Subroutine SMULT will multiply all values in an n x m matrix [b] by a number A to obtain an n x m matrix [C]. Table B.5 defines the 282 123 I = 32(4)3 = 16 12 170.667 in4 and for the 8 in high prisms, z = 24(4)3 = 8 12 128 in4 . For equal end moments (M = Mg = M.Â¡g), if these values are inserted into Equation (5*17), it reduces to 9g = 0.6667 16 # (5.19) This means that if an end moment M is applied to a 16 in high prism and to an 8 in high prism with the cross-sections shown in Figure 5.9c, the end rotation produced in the 8 in high prism will be 0.6667 times the end rotation in the 16 in high prism. The M-9 curves for the brick element were obtained from the test data shown in Figure 5*8 as follows. For a given value of P, the end moment M, which equals P x e, was calculated for the different eccentricities in parts (a), (b), (c) and (d) of Figure 5-8. Using Equation (5-13)* where L = 15.7 in t = 3*56 in Â£2 Â£] read from the curves, the end rotation 9 for the prism was calculated for each value of applied end moment M. This end rotation is for the 4x32x16 in prisms used in the tests, so using Equation (5*19) an equivalent end rotation for 4x24x8 in prisms was obtained. This procedure was carried out for WALL HEIGHT, INCHES 152 Figure 6.1 Lateral V/all Deflection Versus Height For Example Number 2 D.4 Example Number 4 70 IF(E2Â£1.Â£Â£.Â£I!S1) GC TO 70 AIÂ£CBH=Â£lIfl1 + (P1-PIH1)* (MIP1-KIM 1) / (PI P1-PIM 1) IF(KEY.NE.2) GG TC 75 KAXÂ£CE=DAES (F (2) ) If (DABS (F(4)) .GT.MAXMOM)MAXMCM=DABS(F (4)) GC TC 78 75 MAXMCM=DABS (F (3)) IF (LABS (F (6) ) GT K AX MOB) MAXMCM= CABS ( F ( 6) ) 78 IF (KAXMCM.G1. ALLQSH) NSTAT=1 80 RETURN ENE C S DEBOOTINE WYTHE (BRWYLD,BLWYLD,NSTOHY) C SEEOTIHE WYTHE WILL POINT, FOR EACH WALL LEVEL (EVERY C Â£ INCHES), THE VERTICAL LOAD CARRIED EY EACH WALL WYTHE C AND THE PERCENT OF THE TOTAL LOAD WHICH THAT C REPRESENTS. RE AI HEIGHT DCELE PRECISION BRKYLD,BLWYID,TCTLD DISENSION BEW YLE (NSTORY) ,BLW Y L D (NSIOR Y) WRITE(6,10) 10 FCEAT ('1',66 { )/ ',14X,WALL WYTHE VERTICAL LOAD $ VERSUS HEIGHT/* ,66 (=)) WRITE (6,20) 20 FOBHAT (10/ ,HEIGHT*,2X,*BRICK WYTHE 1CAD',2X * CE , $ TCTAI' ,2X,'BLOCK WYTHE ICAD,2X,*54 C TOTAL '/' ') HEIGHT=0,0 DC EC L = 1,N STOR Y ICTLB=BRW YLE (L) +BLW YLD (L) BfiPEH=10C.0*BRWYLD(L)/TOTID ELPEB=10 0.0*Â£LWYLD(L)/TCTLD WRITE(6,30)HEIGHT,BR fcYLD (L) ,BRPER,BIWYIE (I) ,ELPER 30 F CBM AT (* F 5. 1,4X, D13. b 6X, F5.2, 6X D 13.6,6X F 5.2) EEIGHT=HEIGHT+Â£.0 CCNT1NUE RETURN 50 280 298 (2 x ml) 1, where ral equals half the bandwidth (including the diagonal) of the n x n matrix. The actual n x n matrix is not stored but after matrix [b] is inserted into the n x ml matrix [c], the result is equivalent to what would be obtained if the m x m matrix [b] were inserted into the actual n x n matrix. The m x 1 matrix [INDEX] contains the numbers which identify the positions in [c] to which the appropriate values in [b] should be added. If TYPE equals 1, then the first three numbers in the [INDEX] matrix are not used. If TYPE equals 2, then all the numbers in the [INDEX] matrix are used. If TYPE equals 3, then the fifth number in the [INDEX] matrix is not used. Table B.8 defines the nomenclature used in this subroutine. Figure B.8 is an algorithm for this subroutine. B.9 Subroutine EXTRAK Subroutine EXTRAK will pick up a matrix [b] out of a larger matrix [a]. The values of [a] are not affected. The old values of [b], if any, are replaced by the values found in [a]. If KEY equals 1, then [a] is an n x n matrix and [b] is an m x m matrix. If KEY equals 2, then [a] is an n x 1 matrix and [b] is an m x 1 matrix. The mx1 matrix [INDEX] contains the numbers which identify the positions in [a] from which matrix [b] is constructed. If TYPE equals 1, then the first three numbers in the [INDEX] matrix are not used. If TYPE equals 2, then all the numbers in the [INDEX] matrix are used. If TYPE equals 3, then the fifth number in the [INDEX] matrix is not used. Table B.9 defines the nomenclature used in this subroutine. Figure B.9 is an algorithm for this subroutine. 107 400,000, 300,000 CQ . 200,000- Q. 100,000- JA t n mum T"'1i1T .005 Data Converted to Tests of fmbr = 5039 si 3 Prisms4x24x8 in Brick 4x8x2% in Mortar Type S Mil .010 .015 A, IN .020 .025 Figure 5.2 Erick Element P-4 Curve 109 Vertical Displacement (x 0.001 in.) 60 50 60 30 20 10 0 Figure 5-4 Experimental Source of Collar Joint Element P-A Curve (18) Shear Bond Stress, psi 275 CALL SMULT (- 1.0 D + 00 BEEF MBEEF BE, 1) IF(I.EQ.NEM1)CALL INSERT (ERROR,MBBEE,2,3,NSBGF,EH, $ IBB) IF(I.Â£Q.NÂ£BT) GO TO 280 IF (I.EC. 1)CALL INSERT (ERHOR,MBREF,2,1,NSDCF,BB,IBn) IF(INE. 1)CALL INSERT (ERBCR,BBREF,2,2,NSDCF,BB,IBR) 280 CONTINUE C FOR FACE COLLAR JOINT ELEMENT: C IT HE VERY FIES1 TIKE) C 1. CONSTRUCT THE INDEX MATRIX 8 STORE IT. C 2. GET INITIAL STIEFNESS FACTORS 8 STORE TEEM. C 3. CONSTRUCT THE ELEMENT STIFFNESS MATRIX. C 4. INSERT THE ELEMENT STIFFNESS MATRIX INTO THE C STRUCTURE STIFFNESS MATRIX. C C (EVERY OTHER TIME) C 1. RECALL THE INDEX MATRIX. C 2. EX1EACT TEE ELEMENT DISPLACEMENT MATRIX FROM THE C STRUCTURE DISPLACEMENT MATRIX. C 3. RECALL THE ELEMENT STIFFNESS MATRIX. C 4. MUITI FLY THE ELEMENT ST IFENESS MATRIX BY THE ELEMENT C DISPLACEMENT MATRIX TO YIELD THE ELEMENT FORCE MATRIX. C 5. PRINT THE ELEMENT FORCES WEEN APPROPRIATE. C 6. PRINT THE ELEMENT DISPLACEMENTS BHEN AEPIC PEI ATE. C 7. CEECK FCB ELEMENT (AND THEREFORE, KAIL) FAILURE. C fi. GET STIFFNESS FACTORS BASED CN THE Â£ IEME NT FORCES EBCM C TEE P-DELTA 6 B-THETA CURVES 8 COMPARE THEM KITH THOSE C 01 THE PREVIOUS CYCLE TO DETECT ANY CHANGES. C 9. CONSTRUCT TEE ELEMENT STIEENESS MATRIX. C 10. INSERT THE ELEMENT STIFFNESS MATRIX INTC THE C STRUCTURE STIEFNESS MATRIX. C 11. MULTIPLY THE ELEMENT FORCE MATRIX BY -1.0 6 INSERT IT C INTO THE STRUCTURE FORCE MATRIX FOR CONVERGENCE CHECK. DO 360 J=2,NEMO,3 IF (ITER. NE. 1) GO TO 290 CALL INDXCJ (ICJ,IOTEIN,J,NEMO,CJ,KEIEM) 20 2.7 Solving For the Element Displacement Matrix Once the structure displacement matrix has been solved for, it is very easy to obtain the displacement matrix for each element. Since the index matrix for an element identifies which structure DOF correspond to which element DOF, it also identifies which structure displacements correspond to which element displacements. For example, consider the frame of Figure 2.3 again. It was previously observed that for element 1, w.|, w2, and w^ have no corresponding structure DOF but corresponds to w^, corresponds to w^, and corresponds to wg. This means that the element displacement matrix for element 1 equals W1 0 0 W2 0 0 w3 0 where [i]^ = 0 = w4 W1 1 w5 W3 3 w6 _W4 4 Using the index matrices for elements 2 and 3 which were previously developed in section 2.3 the element displacement matrices for elements 2 and 3 are, therefore, W1 0 W1 ~w1~ w2 0 w2 ss W4 and [w]j = w3 = 0 w3 w2 w4 W2 w4 s_ w5 w3 w6 _W5_ WALL HEIGHT, INCHES 178 Figure 6.26 Wall Wythe Vertical Load Versus Height At C.60 Praax For Example Humber 4 WALL HEIGHT, INCHES 157 120.0 112.0 P v- I T I'" I 0 .075 .150 .225 BRICK WYTHE VERTICAL DEFLECTION, INCHES .300 Figure 6.6 Brick 'Wythe Vertical Deflection Versus Height For Example Number 3 135 Cross-sectional capacity of concrete block prisms. Figure 5-16 Experimental Source of Concrete Block Element P-M Interaction Diagram (4) 5 ICJ (4) =5 GC 1C 5 DC 30 N = 1,4 ICJ(N) =ICJ (N) +5 30 CONTINUE GG 10 50 .35 ICJ (1) = ICJ (1) +5 ICJ (2) =ICJ (2) *4 ICJ {3) = ICJ (3) +5 ICJ (4) =ICJ (4) +4 50 DC 70 1=1,CJ 10IE.IN (I,L) =ICJ (L) 70 CONTINUE EE1UBN END C SUBECUIINE INLXEL (IBL, TOTEIN, I, BL/NELEM) C SUBBOUTIN E INDXBL HILL CCNS1B0CT 1LE INDEX EAT BIX FOB C EACH CCN'CBEIE ELOCK ELEMENT. IN1EGEB TO1EIN ,EL DIKE NSIC N TCIEIN (NELEM,6),IBL(fa) IF (I EQ3) GO 1G 5 IF (I,EC. 6) GO i'C 15 IF (I.G1.5) GO 10 3 0 5 IEI (1) =0 IBL (2) =0 IEI (3) =0 IEL (4) =2 Ifil (5) =3 IEL (6) =5 GC 1C 50 15 DO 2C J = 1,3 E=J+3 IBL(J)=IBL (P) CONTINUE DO 25 K=4,6 20 272 60 2 10 0 Â¥1~ 10 2 6-60 W2 0 0-6 9-7 V3 -28 0 0-75 A 0 After partitioning, 2 1 0 0 Â¥1 10 2 6 -6 0 w2 0 C5 0 -6 9 -7 w3 -28 0 0 -7 5 Â¥4 0 _ Performing 2 forward eliminations on [k] entails 2 1 0 0 Â¥1 10 (-1 x row 1) + row 2 2 6 -6 0 w2 0 (6/5 x row 2') + row > 0 -6 9 -7 W3 -28 0 0 -7 5 w4 0 and yields 2 1 0 0 2 5 -6 0 0 0 9/5 -7 0 0 -7 5 Â¥1 10 Â¥2 -10 Â¥3 -40 Â¥4 0 340 (start) CONSTRUCT THE STIFFNESS MATRIX FOR THE BRICK ELEMENT (return) Figure B.24 Algorithm For Subroutine BRESM 118 (c) Figure 59-continued. 27 (a) Element Figure 3.3 Development of Stiffness Matrix For Element 2 of Figure 3.1 PERCENT SAVINGS 58 Figure 3.15 Computational Savings of Modified Gauss Elimination Over Standard Gauss Elimination 30 40 50 60 70 80 C C c c c c 10 20 ESFHAX = DABÂ¡~ (SFMAX (I) ) IF (DSTRF.GT. E5FMAX)STBF (K)=SFMAX (I) IF (DSTRF.LT.DSFMAX) KÂ£X1=0 CONTINUE GC TC 80 IF (KEX2.NE.3) GO 30 60 CO 50 I = 1, N C F J=NDOF (I) STBF (J) = STÂ£F (J) -SFINCR (I) / 1 0 CONTINUE HEX 1 = 0 GO 30 80 DC 70 1=1,NCF I=NDOF(I) STBF (LJ=STBÂ£ (!) +SFI NCR (I) / 1 0 CONTINUE KE X 1 = 0 RETURN END 8 UB ECU TINE FLATEK (STRK NSTRK, M 1, M 1P 1, NSDGE, NSDM2,LNHBN, $ LNHBL) SUBROUTINE PLATER WILL CONSTRUCT A STRUCTURE STIFFNESS MATRIX WHICH ONLY CONSIDERS THE TBC STEUCTUFE DEGREES CF FREEDOM AT THE TOP OF TEE WALL THAI CORRESPOND TO THE TEST PLATE LUTHER THAN THE ORIGINAL FCUF DEGREES CF FREEDOM AT THAT LOCATION. DCUELE PRECISION SISK NSIE K, LNHiJB INH131,LC2, HLC2, P REI DIMENSION STRK (NSÂ£OF,M 1) NST EK (NS DM2, M 1P 1) ,PR EL (2,4) CALL NULL (NSTRK,NSDM2,M1P1) N E M1M3= NSDCE-Kl-3 DO 2C I = 1, N M M 1M 3 EC 10 J=1,M 1 NSTRK (I,J)=STRK(1,J) COM I NOE CONTINUE iv> vn 333 Table B.22 Nomenclature For Subroutine BLPDMT VARIABLE TYPE DEFINITION NBLPDP INTEGER NUMBER OF POINTS IN THE BLOCK P-DELTA CURVE NOBLPC INTEGER NUMBER OF BLOCK M-THETA CURVES (EACH IS IDENTIFIED BY THE P OR AXIAL LOAD LEVEL FOR WHICH THE MOMENT CURVATURE RELATIONSHIP IS VALID) NBLMTP INTEGER NUMBER OF POINTS IN EACH BLOCK M-THETA CURVE BLDCOR DOUBLE PRECISION MATRIX THAT STORES BLOCK DELTA COORDINATES BLPCOR DOUBLE PRECISION MATRIX THAT STORES BLOCK AXIAL LOAD COORDINATES BLPCV DOUBLE PRECISION MATRIX THAT STORES THE VALUE OF P BY WHICH EACH BLOCK M-THETA CURVE IS IDENTIFIED BLTCOR DOUBLE PRECISION MATRIX THAT STORES BLOCK THETA COORDINATES FOR EACH M-THETA CURVE BLMCOR DOUBLE PRECISION MATRIX THAT STORES BLOCK MOMENT COORDINATES FOR EACH M-THETA CURVE BLEF DOUBLE PRECISION BLOCK ELEMENT FORCE MATRIX SBLPD DOUBLE PRECISION MATRIX THAT STORES THE SLOPES OF THE BLOCK AXIAL LOAD DELTA CURVE SBLMT DOUBLE PRECISION MATRIX THAT STORES THE SLOPES OF THE BLOCK MOMENT THETA CURVES BLVERF DOUBLE PRECISION VERTICAL (AXIAL) FORCE AT THE BOTTOM OF A BLOCK ELEMENT BLMOM DOUBLE PRECISION ABSOLUTE VALUE OF AVERAGE OF END MOMENTS FOR A BLOCK ELEMENT BLAEL DOUBLE PRECISION BLOCK AE/L (AXIAL STIFFNESS FACTOR) 354 Table B-31 Nomenclature For Subroutine APPLYF VARIABLE TYPE DEFINITION STRF DOUBLE PRECISION STRUCTURE FORCE MATRIX NSDOF INTEGER NUMBER OF STRUCTURE DEGREES OF FREEDOM NOF INTEGER NUMBER OF FORCES NDOF INTEGER NUMBER OF THE DEGREE OF FREEDOM OF A PARTICULAR FORCE SFINIT DOUBLE PRECISION STORES THE INITIAL STRUCTURE FORCE FOR EACH LOADED DEGREE OF FREEDOM SFINCR DOUBLE PRECISION STORES THE STRUCTURE FORCE INCREMENT FOR EACH LOADED DEGREE OF FREEDOM SFMAX DOUBLE PRECISION STORES THE MAXIMUM STRUCTURE FORCE FOR EACH LOADED DEGREE OF FREEDOM ITER INTEGER DO-LOOP PARAMETER KEY1 INTEGER VARIABLE THAT KEEPS TRACK OF THE APPLI CATION OF STRUCTURE LOADS; 0 = THE LOADS HAVE NOT REACHED THEIR MAXIMUM VALUE; 1 THE LOADS ARE ALL AT THEIR RESPECTIVE MAXIMUMS AND NO FURTHER INCREASES ARE REQUIRED 304 Table B.10 Nomenclature For Subroutine PULROW VARIABLE TYPE DEFINITION A INTEGER T x 1 MATRIX WHOSE VALUES CORRESPOND TO THOSE IN ROW I OF MATRIX [b] B INTEGER N x 6 MATRIX FROM WHICH A ROW IS COPIED INTO MATRIX [a] I INTEGER ROW IN MATRIX [b] WHICH IS COPIED INTO MATRIX [A] T INTEGER NUMBER OF NUMBERS IN ROW I OF MATRIX [b] WHICH ARE COPIED INTO MATRIX [a] N INTEGER NUMBER OF ROWS IN MATRIX [b] 322 (start) /read the number of curves/ / READ THE NUMBER OE POINTS PER CURVE / r FOR EACH CURVE) r /read the curve value/ FOR EACH POINT) /read the x coordinate/ zj READ THE Y COORDINATE / (return) Figure B.17 Algorithm For Subroutine CURVES 12 Figure 2.1b shows the application of unit displacements to the basic flexural element in order to establish the stiffness coefficients. Figure 2.2 shows the resulting element stiffness matrix. It is a 6 x 6 matrix since the element contains 6 DOF. 2.3 Construction of the Element Index Matrix The index matrix [i] was previously defined in section 2.1 as the matrix which consists of the numbers of the structure degrees of freedom that are related to the element degrees of freedom for a particular element. To illustrate what this means, consider the frame shown in Figure 2.3a. If the structure and element degrees of freedom for this frame are as shown in Figure 2.3b and 2.3c, respectively, it is possible to construct the index matrix for each element simply by noting which structure degrees of freedom correspond to which element degrees of freedom. For example, to construct the index matrix for element number 1, observe that w^, W2> and w^ have no corresponding structure degrees of freedom but corresponds to w^, corresponds to w^, and corresponds to Wg. Thus, the index matrix for element number 1 is 0 0 3 4 Similarly, the index matrices for elements 2 and 3 are 209 Figure A.1-continued. Table C.1 Data Input Guide LINE VARIABLE DESCRIPTION FIELD START IN TYPE DESCRIPTOR COLUMN 1 2 3 WHI 4 LNVER COMMENT CARD WITH DESCRIPTION OF PROBLEM COMMENT OF DESCRIPTION OF WALL WALL HEIGHT IN INCHES; MUST BE A MULTIPLE OF 8 VERTICAL LENGTH OF THE BRICK AND BLOCK ELEMENTS 5 LNHBR 6 LNHBL 7 WDEPTH 8 9 10 NBRPDP 11 12-17 J BRDCOR HORIZONTAL LENGTH OF THE BRICK HALF OF THE COLLAR JOINT ELEMENT HORIZONTAL LENGTH OF THE BLOCK HALF OF THE COLLAR JOINT ELEMENT WALL DEPTH IN INCHES COMMENT OF 'MATERIAL PROPERTIES' COMMENT OF 'BRICK ELEMENT P-DELTA CURVE' NUMBER OF POINTS IN THE BRICK P-DELTA CURVE COMMENT OF 'PT' COMMENT OF 'DELTA COORDINATE' COMMENT OF 'P COORDINATE' POINT NUMBER BRICK DELTA COORDINATE 9 INTEGER 14 1 DOUBLE D10.4 1 PRECISION DOUBLE DIO.4 1 PRECISION DOUBLE DIO.4 1 PRECISION INTEGER 14 1 9 6 INTEGER 13 1 2 6 24 INTEGER 13 1 DOUBLE D13.6 4 PRECISION 220 CONTINUE 230 CONTINUE N S T11 = N ST A BT + 1 NCCLH1 = NCOLK1 DC 250 1= NS1F 1,N B1GP NCCL M 1 = NCQLM1 1 EC 240 J=1,NCOIfl1 K1(I,J)=C(I,J) 240 CONTINUE 250 CONTINUE C SOLVE Â£ K 1 ]Â£ M1 ]=Â£ E 1 1 J-Â£ K2* ]Â£ W2 ] FOB Â£ 'ill ] BY BACK SUBSTITUTION C (SINCE FOHWABD ELIMINATION PORTION OF GAUSS 1 A IDEADY DCNE ON C Â£ K 1 ] IAET CF Â£ C j) , 9 1 (NBTOP) = F1MKJi2 (NBTOP)/K1 (NBTCE, 1 ) EB=C DO 290 KK=2,NBTOP B=NB1CP-KK+1 IF (KK.GT.M)GO TO 260 NM=NK+1 GO TO 27C 260 KE=M 270 1 (K) =F1M02 (K) 1=1 N 2=K DC 280 11=1,KM N1=N1+1 N2=N2+1 280 W1 (K}=N1 (K)-K1 (K,N1) *N1 (N2) 1*1 (K) = K1 (K) / K 1 (K, 1) 290 CONTINUE C PUT Â£ HI 3 Â£ Â£W2 j IKTC Â£X ]. DO 3C0 1=1,NBTOP X (I)=W1 (I) 300 CONTINUE J=0 DO 310 I=NBTP1,N 364 (start) CHECK ELEMENT FOR FAILURE (return) Figure B.35 Algorithm For Subroutine CHKFA 24 L Figure 3. where: k = Shear Spring s Stiffness k = Moment Spring Stiffness Other variables defined in Figure 2.2 (a) Structure Degrees of Freedom W] w1 (b) Element Degrees of Freedom 1 Structure and Element Degrees of Freedom For a Frame With Different Materials and Element Types 112 Figure 5*6 Experimental Source of Concrete Block Element P-A Curve (4) [k] AE L 0 0 -AE L 0 0 0 12EI -6EI 0 -12EI -6EI L3 (1 + $) L2 (1 + $) (1 + $) L2 (1 + *) 0 6EI (4 + *) El 0 6EI (2 $) El L2 (1 + *) L (1 + 4) L2 (1 + $) L (1 + *) -AE 0 0 AE 0 0 L L 0 -12EI 6EI 0 12EI 6EI L3 (1 + $) L2 (1 + #) L3 (1 + *) L2 (1 + *) 0 -6EI (2 *) El 0 6EI (4 + *) El L2 (1 + *) L (1 + 4>) L2 (1 + *) L (1 + *) WHERE: A E L I 4> G v AREA OF ELEMENT CROSS-SECTION MODULUS OF ELASTICITY ELEMENT LENGTH MOMENT OF INERTIA OF ELEMENT CROSS-SECTION FACTOR USED IN ACCOUNTING FOR SHEAR DEFORMATION = 12EI GASL2 SHEAR MODULUS = y ..E. . 2(1+v) POISSON'S RATIO AREA IN SHEAR = .84 x (NET AREA) Figure 3.7 Element Stiffness Matrix Considering Shear Deformation D.5 Example Number 5 113 prism. Figure 5-7 depicts the concrete block element P-A curve which was obtained in this fashion. 5.2.2 M-9 Curves The M-9 curves for the brick element are generated in the manner explained in section 4.3*1* The M-9 curves for the brick element used in the numerical examples were obtained from tests very similar to those described in section 4.3-1 which were performed by Fattal and Cattaneo (4) on brick prisms. They tested twelve 4x32x16 in prisms by loading them eccentrically and recording the vertical strain on each side of the prism in the plane of the applied end moment. The results of their tests are shown in Figure 5*8. Before using their results, it is first necessary to convert their data, which is for 4x32x16 in prisms, to equivalent data for 4x24x8 in prisms. This is done as follows. Consider Figure 5.9a. Line b-b represents a horizontal plane through a 4x32x16 in prism. Line b'-b shows the position of that plane after the prism bends in response to an end moment applied during a test. From small angle theory, = tan 9 9 t (5.4) Define the difference in vertical strain Ae as Ag Â£ ^ *" Â£ -j (5.5) where e = vertical strain in the prism measured on side 1 = vertical strain in the prism measured on side 2. LINE EXAMPLE #2 H/T =12 PLATE LOAD ECCENTRICITY=0 IN. 1. DESCRIPTION OF HALL 2, 120 3. 8. 0D + 00 4. 2, OD + OO 5. 3. OD + OO 6, 24 7. MATERIAL PROPERTIES 8, BRICK ELEMENT P-DELTA CORVE 9. 6 10. PT DELTA COORDINATE P COORDINATE 11. 1 0. OD + OO 0.OD+OO 12, 2 4. OD -0 3 9.2175D+04 13. 3 8;, 0D-03 1.67175D+05 14, 4 1.2D-02 2.32725D+05 15. 5 1.6D-02 2,91450D+05 16, 6 2.COD-02 3.38850D+05 17. B RICK ELEMENT M-TilEIA CURVES 18. 3 19. 4 20, 1.0D + 05 21. PT THETA COORDINATE M COORDINATE 22. 1 0.OD+OO 0.OD+OO 23. 2 2. 117D-03 5.93D+04 24. 3 9,7 02D-03 1.187D+05 25. 4 1 .D-02 1.53591D+05 26. 1.5D+05 27. PT THETA COORDINATE H COORDINATE 28, 1 0.OD+OO 0.OD+OO 29. 2 3,13 ID-03 8 895D+04 30. 3 7.865D-03 1.3 35D+05 31. 4 1.0D-02 1,53591D+05 32, 2.D+05 33. 10 CONTINUE WHITE (6,0) 20 FO EM AT (// MATRIX SIZE: ',13,' X',I3/) 30 FCEMAT( HOW* ,I3,6C12, 4) 40 FORM AT(/* HOW ,13,6D12.4) 50 FGEiiAI(7X,6D 12, 4) 60 FORMAT(//) HETUEN END C SBBCUTIUE BRPDMT (NBRPDP,NOERPC,NERMTP,BEDCCB,EEPCOR, ERPCV, $ Â£B1CCB,BEMCCE, EREF, SBRPD, S EBMT BRVER F B.R MCM B RAEL ,BB3 El L IT Eli) C S UBHOUTIN E BRPDMT WILL CALCULATE THE SLOPES CF THE C IINEABLY APPROXIMATED P-UELTA AND M-1HETA CURVES FOB C THE BRICK ELEMENT THE FIBS1 TIKE IT IS CALLED. IT C ALWAYS FINDS THE SLOPE FOB THE REGION IN WHICH THE C ELEMENT P FALLS AND THEREFORE GENE BATES THE AEPECERIAT E C AXIAL STIFFNESS FACTOR OF A*E/L. SIMILARLY, IT FINDS C THE SLOPE FOB THE REGION IN WHICH THE AVERAGE MOMENT C FAILS GN THE CURVE AND GENERATES TEE RC1A1ICNAL C STIFFNESS FACTOR 3*E*I/L FCR THE ELEMENT E, INTEGER UIE DGUELE PRECISION B.SDCOE, 8HPCCE BfilCOR DEM OCR Efi F F EE A FI DCUEIE PRECISION ER3EIL,SBRPD,SERMl, A VGMOM,BRPC V,LLSMI,ULSMl DOUBLE PRECISION BRMOM,SEVERE DIMENSION BEECOE (20) BBPOOR (20) BfilCOR (20,20) ,BEMCGR (20,20) DIMENSION BEEF (6) ,SBRPD (20) ,SBEMT (20,20) ERPCV (20) IF (ITER, HE, 1) GO TO 60 N PL S 1=N ERPDP1 DO 20 1= 1,NPLS 1 J = I + 1 SBRPD (I)= (EEPCGR(J) -ESPCOE (I) ) / (BRDCOR (J) -BEDCOR(I) ) 20 CONTINUE NEIS 1 = NEEMTP-1 DO 45 J=1,NGBRPC EC 40 1= 1, NfiLS 1 CTv 71 4.2 Types of Elements As previously noted, the model considers the wall divided into a series of three types of elements. These include a brick element, a collar joint element, and a block element. 4.2.1 Brick Element The basic finite element for the brick wythe is shown in Figure 4. 3 It models a brick prism either two or three bricks high, depending on brick type, and the mortar joints adjacent to those bricks. The prism and element are each 8 inches tall. The element extends the entire 24 inch depth of the test wall, with uniformly distributed forces over the wall depth. Experimental tests will be done on brick prisms, 24 inches deep, in order to determine the load-deformation properties of the brick element to be used by the model. Like the brick in the wall, the brick in the prisms will contain running bond. The brick element will have the six degrees of freedom shown, which include a rotation, axial displacement, and lateral displacement at each end. This is the basic flexural element previously illustrated and discussed in section 2.2. The effects of shear deformation and moment magnification will be considered in the brick wythe of the model in accordance with the pro cedures outlined in sections 3.3 and 3.4. In other words, the stiffness matrix for a brick element will take the form shown in Figure 311 4.2.2 Collar Joint Element The basic finite element for the collar joint is shown in Figure 4.4. The element will have the four degrees of freedom shown which include a rotation and vertical displacement at each end. The element is rigid axially which forces the brick and block wythes to have the 141 may be made. Figure 5.20 shows the structure degrees of freedom used in the model wall. 54 Illustrative Examples The following four examples consider walls which vary in slenderness, or H/T, ratio and eccentricity of load application. Example number 2 considers a wall with a slenderness ratio of 12 and a plate load eccentricity of 0 in. In example number 3, the wall contains a slenderness ratio of 12 but a plate load eccentricity of 2 in. Example number 4 models a wall with a slenderness ratio of 20 and a plate load eccentricity of 0 in. Finally, example 5 considers a wall with a slenderness ratio of 20 and a plate load eccentricity of 2 in. In all four examples, 6 inch block is used, resulting in a wall thickness of 10 inches since the brick wythe is always 4 inches thick. The material property curves from section 5.2 were used for all of the examples. Intermediate printouts for each example were generated at load levels of 0.30 Pmax, 0.60 Pmax, 0.90 F^ and Pmax. The results of the finite element analysis for each wall are presented in chapter 6. 5.4.1 Example Humber 2 Example number 2 considers a wall 10 feet or 120 inches high, 10 inches thick, and 24 inches deep. The plate axial load is applied in increments of 4000 lb at 0 eccentricity up to failure. Figure 5.21 shows the wall of example 2. The degrees of freedom for external wall loads, equivalent wall loads, and wall displacements for examples 2 and 3 are shown in Figure 5.22. The data deck for this example can be found in Appendix D.2. WALL HEIGHT, INCHES 174 Figure 6.22 Wall Wythe Vertical Load Versus Height At G.6G Pnax For Example Number 3 411 13. Self, M.W., "Design Guidelines for Composite Clay Brick and Concrete Block Masonry: Part 1Composite Masonry Prisms," Masonry Research Foundation Research Report, Department of Civil Engineering, University of Florida, Gainesville, Florida, April 1983- 14. Specifications For Design and Construction of Loadbearing Concrete Masonry, National Concrete Masonry Association, Herndon, Virginia, 1976. 15. Tabatabai, Habibollah, "Modulus of Elasticity of Clay Brick and Ungrouted Concrete Block Prisms," Master of Engineering Report, University of Florida, Gainesville, Florida, 1982. 16. Wang, C.K., Intermediate Structural Analysis, McGraw-Hill, New York, New York, 1983* 17. West, Harry H., Analysis of Structures, John Wiley and Sons, Inc., New York, New York, 1980. 18. Williams, Robert T., Geschwinder, Louis F., "Shear Stress Across Collar Joints in Composite Masonry Walls," Proceedings of the Second North American Masonry Conference, College Park, Maryland, August 1982. 19- Yokel, F.Y., Mathey, R.G., Dikkers, R.D., "Strength of Masonry Walls Under Compressive and Transverse Loads," Building Science Series 34, National Bureau of Standards, Washington, D.C., March 1971. 29 (a) (bending) And (shear) Components of Lateral Deflection DEPTH d SPAN L UNIFORM LOAD CONCENTRATED LOAD AT MIDSPAN 1/12 0.0167 0.0208 1/10 0.0240 0.0300 1/8 0.0375 0.0469 1/6 0.0667 0.0833 1/4 0.1500 0.1875 (b) Ratio of Shear Deflection Ag to Bending Deflection Ab Figure 3.5 Shear Deformation and Its Importance COMMENT OF 'DELTA COORDINATE COMMENT OF *P COORDINATE' 42-43 J POINT NUMBER CJDCOR COLLAR JOINT DELTA COORDINATE CJPCOR COLLAR JOINT P COORDINATE 44 45 NCJMTP 46 47-48 J CJTCOR COMMENT OF 'COLLAR JOINT ELEMENT M-THETA CURVE' HUMBER OF POINTS IN THE COLLAR JOINT M-THETA CURVE COMMENT OF 'PT' COMMENT OF 'THETA COORDINATE' COMMENT OF 'M COORDINATE' POINT NUMBER COLLAR JOINT THETA COORDINATE CJMCOR COLLAR JOINT MOMENT COORDINATE 49 COMMENT OF 'BLOCK ELEMENT P-DELTA CURVE' 50 NBLPDP NUMBER OF POINTS IN THE BLOCK P-DELTA CURVE 51 COMMENT OF 'PT' INTEGER 13 DOUBLE PRECISION D13*6 DOUBLE PRECISION D13.6 - - INTEGER 13 - - - - INTEGER 13 DOUBLE PRECISION D13.6 DOUBLE PRECISION D13.6 - - INTEGER 13 COMMENT OF 'DELTA COORDINATE' Figure 35.6 Application of Displacements to Establish Stiffness Coefficients Considering Shear Deformat 30 Figure 2.1, neglect shear deformation. The following derivation, taken from Przemieniecki (9), shows how the element stiffness matrix is obtained for the standard 6 DOF element considering shear deformation. Note that displacements for element degrees of freedom 1 and 4 are not considered below because they are not affected by the consideration of shear deformation. Thus, columns 1 and 4 in the element stiffness matrix shown in Figure 2.2 will remain unchanged. Consider Figure 3.6a. The lateral deflection v on the element subjected to the shearing forces and associated moments shown, is given by v vb + vs (3.1) where v^ is the lateral deflection due to bending strains and vg is the additional deflection due to shearing strains, such that dvc -fe s p dx GA b (3.2) with A_ representing the element cross-sectional area effective in s shear, and G representing the shear modulus, where G E 2(1 + v) (3-3) and E equals the modulus of elasticity and v the Poisson's ratio of the material. The bending deflection for the element shown in Figure 5*6a is governed by the differential equation 2 El fJ. = f5x f6 MT dx2 (3.4) IN-LB 125 Figure 5*10 Brick Element M-9 Curves 25 First, for elements 1 and 3, the element DOF are identical to what they were in Figure 2.3- Therefore, the derivation of stiffness coefficients for this basic flexural element, shown in Figure 2.1, is valid. The stiffness matrix for elements 1 and 3 will thus take the form shown in Figure 2.2, but the coefficients will recognize the differences in the values of the variables. Figure 3-2 gives the element stiffness matrix for element 1 and the element stiffness matrix for element 3 To construct the element stiffness matrix for element 2, the stiffness coefficients must be derived by the application of unit displacements at each DOF. This is shown in Figure 3.3- The resulting stiffness matrix for element number 2 is illustrated in Figure 3.4. 3 3 Shear Deformation As the depth to span ratio for a member increases, the effect of shear deformation becomes more pronounced and important to consider in the analysis. Figure 3*5a illustrates the shear deformation and bending components of the lateral deflection at the free end of a column in response to a lateral concentrated load. Figure 3-5b, taken from Wang (16), shows the ratio of shear deflection as to bending deflection a^, at the midspan of a simple beam with a rectangular cross-section. Notice that, for a depth to span ratio of 0.25, the shear deflection can be up to 18.75$ of the bending deflection. If it is desired to take shear deformation into consideration in the analysis of a structure by the Direct Stiffness Method, this is accomplished by altering the standard terms in the stiffness matrix of the elements for which this effect may be significant. The terms in the stiffness matrix for the standard 6 DOF element, which were derived in 353 (start) READ THE STRUCTURE LOAD APPLICATION INFORMATION AND PRINT IT / (return) Figure B.30 Algorithm For Subroutine FORCES 2 1.035D-03 4.665D+04 3 1.674D-03 7.0D+04 4 2.962D-03 9,335D+04 5 3.5D-03 1.03103D+05 7.5D+04 PT THETA COORDINATE H COORDINATE 1 0, GD + 00 0,OD+OO 2 7. 52D-04 3.5025D+04 3 1,505D-03 6.9975D+04 4 3.245D-03 1.05D+05 5 3.5D-03 1.10132D+05 BRICK ELEMENT P-H INTERACTION DIAGRAM 10 PT M COORDINATE P COORDINATE 1 0, OD+OO 0,OD+OO 2 3.75D+04 1.875D+04 3 1,0725D+05 7.5D+04 4 1.5D+05 1.2D+05 5 1.65D + 05 1.5D+05 6 1.73475D+05 1.9275D+05 7 1.6875D+05 2,25D+05 8 1.26D+05 3.0D+05 9 1, 08 *1320 + 05 3.3885D+05 10 0.0D+0 3.3885D+05 BLOCK 8 PT ELEMENT P-M INTERACTION DIAGRAM M COORDINATE P COORDINATE 1 7.5D+03 0.OD+OO 2 4.875D+04 1,875D+04 3 8.625D+04 3.75D+04 4 9,996D + 04 5.1D+04 5 9.675D+04 5.625D+04 6 6,75D+04 7.5D+04 7 3.9123D+04 9.315D+04 8 0.OD + 00 9,3 15D+04 MAXIMUM DRICE ELE BENT COMPRESSIVE LOAD CAPACITY 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. 80. 81 . 82. 8 3. 84. 85. 86. 89. 90. 91. 92. 93. 94. 95. 96. 97. 98. 99. 100, 101, 102. 103. 104. 105. 127 ( 3 ) Relationship between vertical compressive load and vertical strain for 6 X 32 X 24-in hollow block prisms at e= t/12. (b) Relationship between vertical compressive load and vertical strain for 6 X .32 X 24-in hollow block prisms at e 1/6. Figure 5.12 Experimental Source of Concrete Block Element M-Q Curves (4) F1 DOUBLE PRECISION F1MKW2 DOUBLE PRECISION F2 DOUBLE PRECISION GDELTA DOUBLE PRECISION GSTRK DOUBLE PRECISION GSTRW DOUBLE PRECISION IBL INTEGER IBR INTEGER ICJ INTEGER K1 DOUBLE PRECISION K2 DOUBLE PRECISION K2W2 DOUBLE PRECISION K4 DOUBLE PRECISION MBLEF DOUBLE PRECISION MBREF DOUBLE PRECISION MCJEF DOUBLE PRECISION NDOF INTEGER NGSTRF DOUBLE PRECISION TOP PORTION OF STRUCTURE FORCE MATRIX STORES VALUES OF [Fl] [k] [W2] BOTTOM PORTION OF STRUCTURE FORCE MATRIX INCREMENTAL STRUCTURE FORCE MATRIX USED IN SOLVING FOR STRUCTURE DISPLACEMENTS STRUCTURE STIFFNESS MATRIX USED IN SOLVING FOR STRUCTURE DISPLACEMENTS STRUCTURE DISPLACEMENT MATRIX INDEX MATRIX FOR A BLOCK ELEMENT INDEX MATRIX FOR A BRICK ELEMENT INDEX MATRIX FOR A COLLAR JOINT ELEMENT TOP LEFT HAND PORTION OF STRUCTURE STIFFNESS MATRIX TOP RIGHT HAND PORTION OF STRUCTURE STIFFNESS MATRIX STORES VALUES OF [K2] [W2] BOTTOM RIGHT HAND PORTION OF STRUCTURE STIFFNESS MATRIX MINUS 1.0 TIMES BLOCK ELEMENT FORCE MATRIX MINUS 1.0 TIMES BRICK ELEMENT FORCE MATRIX MINUS 1.0 TIMES COLLAR JOINT ELEMENT FORCE MATRIX NUMBER OF DEGREE OF FREEDOM OF FORCE STRUCTURE FORCE MATRIX USED IN SOLVING FOR STRUCTURE DISPLACEMENTS 222 non o to nron 310 J=J + 1 X (I) =H2 (J) CGNTINUE REIDBN END SUEECUTINE TITLE SUBROUTINE TITLE HILL BEAD AND PRINT CNE DATA CABE OF ALPHA NUMERIC COMMENTS, IT WILL SKIP 2 LINES BEFORE PRINTING THE COMMENTS. THE COMMENTS CANNOT FXCEE COLUMN 80. DIMENSION ALPHA (20) BE AD (5,10) (ALPHA(I) ,1=1,20) FOBMAI(20A4) WRITE (6,20) (AIEEA (I) 1= 1, 20) FOBMAT(* 0*,20A4) BET UR K END IV) vn SBEOUTINE BEAD (HHI,LNVEB,LKHBE,IKHBI,NSTORY,NSDCF,NFLEM, ^ $ elasbh,arsf:eh,elasbl,arshel) SUBBOUTIN E BEAD HILL PRINT OUT A HEADING, BEAD ANE FEINT TWO COMMENT CARDS, AND BEAD AND PRINT THE WAIL DESCRIPTION DATA. IN1EGEE HHI,DEPTH DOUBLE PRECISION LN VER LNHB R IN HBL EL ASBJ5, ABS H E F E IAS El DCUBIE PRECISION ABSHUL WRITE{6,1C) 10 FORMAT (* ,30 (**)/ *,** RESULTS OF LATEST ANALYSIS **/ $ *,30(***)) WEI1E(6,15) 15 FOBMAT(0'/) CAII TITLE HBITE(6,17) 17 FORMAT (*0 / ', UNITS: CALL TITLE INCH/LE,RADIAN*/' ') 106 where L is the prism height, which for the desired prism is 8 in. By- applying Equations (5.1) and (5.2) on several points in Figure 5.1, a new P-A curve can be generated which will be representative of the results expected for a 4x24x8 in prism. Figure 5.2 illustrates the brick element P-A curve which was obtained in this fashion. The P-A curve for the collar joint element is generated in the manner discussed in section 4*3.2. Tests performed by Williams and Geschwinder (18) were used as the basis for the collar joint P-A curve used in the examples. Figure 5.3a shows an isometric of their test assembly while the geometry of the test assembly is depicted in Figure 5.3b. They tested two prisms of this size by vertically loading the concrete masonry wythe while monitoring the variation of vertical displacement with load. The results of their tests are shown in Figure 5.4. To use their results, it is necessary to convert their data, based on an area in shear As of 15 ^/8 x 15 ^/8 or 488 in^ to equivalent data p for an area in shear of 24x8 or 192 in This is simplified by Williams and Geschwinders observation that the first few points in Figure 5.4 should not be used, since they reflect the seating effect of the assembly on the test plate. This results in a linear P versus A relation. Since the collar joint shear stiffness is equal to the slope of the P-A curve, provided the slope of the line in Figure 5.4 is used, the stiffness will be indicative of that obtained from their tests. Their data is converted as follows. Recall that shear stress t is given by T VERTICAL COMPRESStVE LOAD, kip ( 3 ) Relationship between vertical compressive load and vertical strain for 4 X 32 X 16-in brick prisms at e = tl 12. vertical strain (b), Relationship between vertical compressive toad and vertical strain for 4 X 32 X 16-in brick prisms ate = t/6. Figure 5*8 Experimental Source of Brick Element M-9 Curves (4) 297 (start) L FOR EACH NUMBER IN THE INDEX MATRIX) INSERT THE VALUE IN MATRIX [b] INTO THE PROPER LOCATION IN MATRIX [a] (return) Figure B.7 Algorithm For Subroutine INSERT 5 result, the stiffness of the block deteriorates even faster than that of the brick, the center of resistance shifts even further, and the eccentricity and bending it produces are further aggravated. Stage 3 denotes failure of the prism which is characterized by vertical splitting of the concrete. This phenomenon was verified by experimental tests at the University of Florida (11,13,15). The failure mechanism for composite walls under vertical loads should reflect this behavior. A second factor meriting consideration is that roof trusses or floor slabs, as mentioned previously, generally bear on the block wythe causing the wall to be loaded eccentrically toward the block. This will aggravate the failure mechanism just discussed. A third consideration is that with increasing slenderness and height, lateral wall deflections due to bending will increase. As these deflections increase, the additional moment caused by the vertical load acting through these deflections takes on increasing importance and can lead to a stability problem. All three of these aspects of behavior should be affected by the wall's height to thickness ratio as well as the thickness of the block wythe. In response to the need for more experimental work on composite masonry walls and an improved understanding of wall behavior, laboratory tests of composite walls have been performed by Redmond and Allen (10), Yokel, Mathey, and Dikkers (19), and Fattal and Gattaneo (4). Nonetheless, as a whole, the three studies consider limited combinations of wall geometry, masonry unit properties, and height to thickness ratios. Additional tests are needed so that design standards can be supported by a large data base. Wall test results also need to be related to analytical models. w 4 Wr (a) Original DOF (b) DOF Which Influence Moment Magnification P P (c) DOF Renumbered (d) Element Forces (e) Element for Derivation Deformations Only o Figure 3.9 Element Used in Derivation of Moment Magnification Terms insight into composite wall behavior can be obtained, and design standards can be modified to reflect this increased understanding. Every effort has been made to consider the factors that will most strongly influence composite wall behavior in the development of the model, so that its usefulness as an analytical research tool will not be compromised. This study describes the model in detail, provides information on how it can be used, and contains several numerical examples that illustrate the information its use makes available. 204 A.2 Detailed Program Flowchart Table A.1 defines the variables used in the detailed program flowchart. The flowchart is presented in Figure A.1. As evident, it goes into considerable more detail than the program algorithm shown in Figure 4.13- The call statements in parentheses identify the subroutines which are called at that point in the program to perform the operation described in the box. Appendix B contains individual algorithms for each subroutine. These are analagous to the program algorithm and, therefore, are not intended to show the detail which is provided in the detailed program flowchart. Detailed flowcharts for each subroutine are not furnished because, since each subroutine is short and only performs one major operation, this type of flowchart would be little more than a repetition of the subroutine listing. It would add little additional understanding over that available from examining the listing of the subroutines furnished in appendix A.4 and would contribute more to confusion than to clarity. A.3 Program Nomenclature A complete alphabetical listing of the program nomenclature is given in Table A.2. It identifies all of the variables, matrices, and subroutines used in the program. Except where standard variable names are used, most variable names are acronyms. For example, BRESM stands for 'Brick Element .Stiffness Matrix' and NBRIDP stands for 'Number of Brick .Interaction .Diagram .Points'. Appendix B contains a table for each subroutine which identifies the main variables the subroutine uses. All of the variables in the argument list of each subroutine are defined. 350 Table B.29 Nomenclature For Subroutine INDXBL VARIABLE TYPE DEFINITION IBL INTEGER INDEX MATRIX FOR A BLOCK ELEMENT TOTEIN INTEGER MATRIX THAT STORES ALL OF THE ELEMENT INDEX MATRICES I INTEGER ELEMENT NUMBER BL INTEGER NUMBER OF BLOCK ELEMENT DEGREES OF FREEDOM NELEM INTEGER NUMBER OF ELEMENTS Table C.1-continued LINE VARIABLE DESCRIPTION FIELD START IN TYPE DESCRIPTOR COLUMN SFINCR FORCE INCREMENT FOR THAT DEGREE OF FREEDOM DOUBLE PRECISION D13.6 17 SFMAX MAXIMUM FORCE VALUE FOR THAT DEGREE OF FREEDOM DOUBLE PRECISION D13.6 30 115 - COMMENT OF 'INSTRUCTIONS ON PRINTING INTERMEDIATE RESULTS' - - 1 116 PROUT VARIABLE THAT IDENTIFIES WHETHER OR NOT AN INTERMEDIATE PRINTOUT IS DESIRED; 0 = NO; 1 = YES INTEGER 11 1 117 NOFN FORCE NUMBER (1 FOR FIRST FORCE LISTED, 2 FOR SECOND, ETC.) INTEGER 13 1 NFORCE NUMBER OF FORCE DEGREE OF FREEDOM, MULTIPLES OF WHICH WILL CAUSE AN INTERMEDIATE PRINTOUT INTEGER 13 4 FMULT MULTIPLIES OF NFORCE FOR WHICH AN INTERMEDIATE PRINTOUT IS DESIRED DOUBLE PRECISION D13.6 7 TABLE OF CONTENTS Page ACKNOWLEDGMENTS iii LIST OF TABLES viii LIST OF FIGURES x ABSTRACT xvii CHAPTER ONE INTRODUCTION 1 1.1 Background 1 1.2 Objective 6 TWO DEVELOPMENT OF THE FINITE ELEMENT MODEL 7 2.1 Matrix Analysis of Structures by the Direct Stiffness Method .....7 2.2 Construction of the Element Stiffness Matrix 9 2.3 Construction of the Element Index Matrix 12 2.4 Construction of the Structure Stiffness Matrix 15 2.5 Construction of the Structure Force Matrix 17 2.6 Solving For the Structure Displacement Matrix 18 2.7 Solving For the Element Displacement Matrix 20 2.8 Solving For the Element Force Matrix 21 THREE SPECIAL CONSIDERATIONS 22 3.1 Different Materials .....22 3.2 Different Types of Elements 22 3.3 Shear Deformation 25 3.4 Moment Magnification 36 3.5 Material Nonlinearity 44 3.6 Equation Solving Techniques 50 3.6.1 Gauss Elimination 50 3.6.2 Static Condensation 56 3*7 Solution Convergence .....63 FOUR USING THE MODEL TO ANALYZE COMPOSITE MASONRY WALLS 68 4.1 Structural Idealization of Wall 68 4.2 Types of Elements ..........71 4.2.1 Brick Element 71 74 same horizontal displacement at each node. Rigid links at each end represent one half of the width of each wythe. The two springs are situated at the wythe to wythe interface. The primary function of the collar joint in a composite wall is to serve as a continuous connection between the two wythes, transferring the shear stress between them. This transfer of shear is modeled by the vertical spring, called the shear spring. At this stage, it is uncertain how much moment the collar joint transfers between the wythes. The rotational spring, called the moment spring, has been included in the collar joint element so the effect of moment transfer can be included in the model if, at a later date, this proves necessary. The stiffness matrix for the collar joint element was developed previously in section 3*2 (see Figure 3*3) and takes the form shown in Figure 3-4. The values of the shear spring stiffness factor (kg) and the moment spring stiffness factor (km) will be obtained from experimental tests. 4.2.3 Concrete Block Element The basic finite element for the concrete block is shown in Figure 4.5. It models a block prism which consists of one block and a mortar joint. The prism and element are each 8 inches tall, the same height as the brick element. The block element also extends the entire 24 inch depth of the test wall, with uniformly distributed forces over the wall depth. Experimental tests will be done on block prisms, 24 inches deep, to determine the load-deformation properties of the block element to be used by the model. The block element will have the six degrees of freedom shown, which include a rotation, axial displacement, and lateral displacement at each WALL HEIGHT, INCHES 156 Figure 1 I I I j I I I | I I 1 1 | 0 .075 .150 .225 .300 BRICK WYTHE VERTICAL DEFLECTION, INCHES 6.5 Brick Wythe Vertical Deflection Versus Height For Example Number 2 o n o -* non on S UBECUTINE MULL (A,N,M) C SUBROUTINE NULL HILL CREATE A MULL OB ZERO MAT RIX Â£ A ]. DISENSION A(N,M) DOUELE PRECISION A DC 10 1=1,N DC 1C J=1,M 10 A(I,J)=0,0 RETURN EKE C SUBROUTINE EQUAL (A,B,N,M) SUBROUTINE EQUAL HILL CREATE AN KXS MATRIX [A] THAT IS EQUAL TC MATRIX [BJ. DIMENSION A (N,H) ,B (N,M) DOUELE PRECISION A, B DO 1C 1=1,N EC 10 J=1,K 0 A (I J) =B(I,J) RETURN ENE SUBROUTINE ADD (A,B,C,N,M) SUBROUTINE ADD WILL ADD AN NXM MATRIX [B ] TC AN NXM MATRIX [A] TO YIELD AN NXM MATRIX Â£C]. DCUEIE PRECISION A,B,C DIMENSION A(U,M) ,B(N,M) ,C(N,M) DC 10 1=1, N EC 10 J=1,H 0 C (I,J) = A (I, J)+Â£ (I, J) RETURN EKE SUE ROUTINE MUIT (A, B,C,N 1, N2, N3) SUBROUTINE MULT WILL MULTIPLY AN N1XK2 MATEIX [A] TIMES AN K2XN3 MATRIX Â£BJ TO 0ETA IN AN N1XN3 MATRIX Â£C]. 244 79 P = O rTTTm^mhm kr = ROTATIONAL STIFFNESS FACTOR 3EI L Figure 4.7 Experimental Determination of Brick Element Rotational Stiffness Factor Figure 6.14 Plate Load Versus End Rotation For Example Humber 330 Figure B.20 Algorithm For Subroutine BRPDMT CALCULATE [i] (CALL INDXBR) T SELECT INITIAL STIFFNESS FACTORS (CALL BRPDMT) T" 'PRINT 'ELEMENT IS IN TENSION' YES ''NO TRFAIL=1 (stop) YES STORE EACH STIFFNESS FACTOR (CALL STIFAC) ~(V) CALCULATE STIFFNESS FACTORS (CALL BRPDMT) I PRINT 'ELEMENT/ HAS FAILED / SELECT AND STORE APPROPRIATE STIFFNESS FACTORS (CALL STIFAC) Figure A.1-continued 331 Table B.21 Nomenclature For Subroutine CJPDMT VARIABLE TYPE DEFINITION NCJPDP INTEGER NUMBER OF POINTS IN THE COLLAR JOINT P-DELTA CURVE NCJMTP INTEGER NUMBER OF POINTS IN THE COLLAR JOINT M-THETA CURVE CJDCOR DOUBLE PRECISION MATRIX THAT STORES COLLAR JOINT DELTA COORDINATES CJPCOR DOUBLE PRECISION MATRIX THAT STORES COLLAR JOINT VERTICAL LOAD COORDINATES CJTCOR DOUBLE PRECISION MATRIX THAT STORES COLLAR JOINT THETA COORDINATES CJMCOR DOUBLE PRECISION MATRIX THAT STORES COLLAR JOINT MOMENT COORDINATES CJEF DOUBLE PRECISION COLLAR JOINT ELEMENT FORCE MATRIX SCJPD DOUBLE PRECISION MATRIX THAT STORES THE SLOPES OF THE COLLAR JOINT AXIAL LOAD DELTA CURVE SCJMT DOUBLE PRECISION MATRIX THAT STORES THE SLOPES OF THE COLLAR JOINT MOMENT THETA CURVE CJVERF DOUBLE PRECISION ASOLUTE VALUE OF COLLAR JOINT SHEAR FORCE CJMOM DOUBLE PRECISION ABSOLUTE VALUE OF AVERAGE OF END MOMENTS FOR A COLLAR JOINT ELEMENT CJSHRS DOUBLE PRECISION COLLAR JOINT SHEAR SPRING STIFFNESS CJMOMS DOUBLE PRECISION COLLAR JOINT MOMENT SPRING STIFFNESS ITER INTEGER DO-LOOP PARAMETER 50 Thus, accounting for material nonlinearity in the analysis of a structure by the Direct Stiffness Method involves applying the load in increments and carefully monitoring the selection of modulus of elasticity values used in constructing the stiffness matrix of each element. The remainder of the general procedure outlined in section 2.1 remains the same. 3.6 Equation Solving Techniques Recall from Chapter Two that once the structure stiffness matrix [k] and the structure force matrix [f] have been constructed, solving for the structure displacement matrix [w] entails solving the matrix equation [f] = [k] [w] for [w]. This requires solving N simultaneous equations, where N is the number of structure degrees of freedom. Two efficient techniques for solving simultaneous equations are Gauss Elimination and Static Condensation. Meyer (8) discusses both of these methods and was used as a reference for the following two sections. 5.6.1 Gauss Elimination As previously discussed, the equation [f] = [k] [w] takes the form shown below. K11 K21 KN1 K12 * k1n *1 *r K22 K2N w2 F2 %2 %N % % Gauss Elimination essentially consists of two steps: APPENDIX A COMPUTER PROGRAM A.1 Introduction The finite element program for composite masonry walls was written in the Fortran-77 language and run on a VAX 11/780 computer. A deliberate effort was made to avoid the sophisticated features available in Fortran-77 but not permitted in the WATFIV version of Fortran, in order to increase the program's usability. Customary composite wall failure is characterized by a failure of the block in compression. If loads are placed on the wall at too great an eccentricity towards the block face (off of the cross-section), this may cause the wall to fail by tension induced in the brick wythe. If in the analysis the wall fails because either a brick or block element goes into axial tension, this will be identified by the program and a recommendation to check for unusual loading made, since this should only occur if the vertical load is placed at an eccentricity greater than that at which it could be physically placed in the prototype wall or test wall. Analytically, wall failure is defined by the model as: 1) A brick or block element going into tension as mentioned above 2) A brick or block element exceeding the allowable load and moment combination permitted by the interaction diagram for that element type 3) A collar joint element exceeding its allowable shear load capacity. 371 Table B.23 Nomenclature For Subroutine STIFAC VARIABLE TYPE DEFINITION VERTF DOUBLE PRECISION ELEMENT VERTICAL FORCE MOMF DOUBLE PRECISION ABSOLUTE VALUE OF AVERAGE OF END MOMENTS FOR AN ELEMENT STOVF DOUBLE PRECISION MATRIX THAT STORES THE ABSOLUTE VALUE OF THE VERTICAL FORCE FOR EACH ELEMENT STOMF DOUBLE PRECISION MATRIX THAT STORES THE ABSOLUTE VALUE OF THE AVERAGE OF END MOMENTS FOR EACH ELEMENT VSTIF DOUBLE PRECISION ELEMENT VERTICAL STIFFNESS FACTOR MSTIF DOUBLE PRECISION ELEMENT ROTATIONAL STIFFNESS FACTOR STOVST DOUBLE PRECISION MATRIX THAT STORES THE VERTICAL STIFFNESS FACTOR FOR EACH ELEMENT STOMST DOUBLE PRECISION MATRIX THAT STORES' THE ROTATIONAL STIFFNESS FACTOR FOR EACH ELEMENT TRSTIF INTEGER VARIABLE THAT IDENTIFIES WHETHER OR NOT ANY ELEMENT STIFFNESS FACTORS HAVE CHANGED; 0 = NO; 1 = YES ITER INTEGER DO-LOOP PARAMETER KEY INTEGER VARIABLE THAT IDENTIFIES THE TYPE OF ELEMENT; 1 = BRICK; 2 = COLLAR JOINT; 3 = BLOCK ELEMNO INTEGER ELEMENT NUMBER D.5 Example Number 3 CALL COORD (NBLIDP,BLIDM ,BLIDP) C BEAD Â£ FEINT MAXIMUM BEICK ELEMENT COMPRESSIVE LOAD CAPACITY. CALL TITLE HEAD (5,10) BEHAXP 10 FORMAT (D13. 6) BITE (6,20)ERMAXP 20 FOB BAT ( ',17X,D13.6) C BEAD 6 PRINT BAXIBUM COLLAR JOINT ELEMENT SHEAR LOAD CAPACITY. CALL TITLE REAL (5,30)CJMAXE 30 FORMAT(D13.6) WRITE (6 ,40) CJKAXP 40 EOBMAT(1 ,17X,D13.6) C READ 6 PRINT BAXIBUM ALLOWABLE BLOCK ELEMENT COMPRESSIVE LOAD. CALI TITLE REAE (5,50) BIMAXP 50 FCRMAT(D13.6) WRITE (6,60)ELBAXP 60 FOHMAT(* 17X ,D 13.6/) ^ C READ 6 PRINT FORCE APPLICATION INFORMATION. WRITE(6,65) 65 FCRKAT('I') CALL FORCES (EOF, NDQF,SPIRIT,5FINCÂ£,SFMAX,KSBCF) C READ 6 PRINT INSTRUCTIONS ON INTERMEDIATE PRINTOUTS. CALL TITLE READ (5,70)FfiGUI 70 FORMAT (11) IF (FECUT. EQ. 0) GO TO 110 WRITE (6,80) 80 FORMAT(0*,INTERMEDIAIS RESULTS WILL BE PRINTED FOR , 'STRUCTURE FORCE*) READ (5, SO) NGFN NFORCE, FMU1T 90 F0RMAT(2I3,D13.6) WRITE (6,100) FKULT, NFORCE 100 FORMAT(' ','VALUES WHICH ARE MULTIPLES CF ,1X,D 13.6, 1X , $ 'FCfi DCF NO. ,14) IF (AVGHCM. II. BEMCOR (NOBRPC, 3) ) GO TO 150 140 CGN1INUE 150 BfiMCM=AVGKOM BRVEEE=BREF(1) BE1DRN END C SUBEGUTINE CJPEMT (NCJPDP,NCJMTP,CJDCOB,CJPCOR,CJTCOfi, i CJMCOR,CJEF,SCJPD,SCJMT,CJVERF,CJMOM,CJSfiES,CJKCMS, ITER) C SUBROUTINE CJEEMT WILL CALCULATE I BE SLOPES OF I HE C LINEALLY APPROXIMATED P-DEITA AND E-THETA CURVES FOB C THE COLLAB JOINT ELEMENT THE FIRST TIME IT IS CALLED. C IT ALWAYS FINDS THE SLOPE FOB THE REGION IN WHICH THE C ELEMENT P FALLS AND THEBEFORE GENERATES THE APPROPRIATE C SHEAR SPBING STIFFNESS FACTOR. SIKILAIIY, IT FINES THE C SICEE FCB THE REGION IN WHICH THE AVERAGE MOMENT FALLS C CN THE CURVE, AND GENEEATES THE MOMENT SEEING C STIEFNES3 EACTOR. DOUBLE PRECISION CJDCOR,CJPCCS,CJTCOR,CJKCCR,CJEF,CJSHRS DOUEIE PRECISION 0JM0M3,SCJPE,SCJMI,CJAXF,AVGMOM,CJMCM DOUBLE PRECISION CJVERF DIMENSION CJDCCE(20) CJPCOR (20) ,CJ1C0R (20) ,CJMCOR(20) DIMENSION CJEF (4) ,SCJPD (20) ,SCJMT (20) IF (ITER. NE. 1) GO TG 60 NPLS 1 = NCJPDP-1 DC 20 1= 1 ,N ELS 1 J=I* 1 SCJPD (I) = (CJECCH(J) -CJPCOR (I) )/(CJBCOR (J) -CJDCOR(I) ) 20 CONTINUE NMIS 1= NC JMT P- 1 DC 4C L = 1 MLS 1 R=I+ 1 SCJMT (L) = (CJMCCR (K) -CJMCCR (L) )/ (CJTCCR (K) -CJTCCB(L) ) CONTINUE CJSHBS=SCJPD ( 1) CJ MC ES=SCJ Ml (1) 40 323 Table B.18 Nomenclature For Subroutine PRINT VARIABLE TYPE DEFINITION A DOUBLE PRECISION N x 1 MATRIX WHICH THE SUBROUTINE PRINTS N INTEGER NUMBER OF ROWS IN MATRIX [a] 21 Thus, once the structure displacement matrix values are available, the displacement matrix for each element is easily obtained with the help of the index matrix for each element. 2.8 Solving For the Element Force Matrix In section 2.1, it was mentioned that the force displacement equations for an element i can be expressed as Mi = Mi Mi Thus, the force matrix for each element is calculated simply by multiplying the stiffness matrix for that element by the displacement matrix for that element. The values in the element force matrix for an element will correspond to the element displacements for the element. For example, considering element 1 in the frame of Figure 2.3, the force matrix for element 1 will take the form [f] 1 f f f f f f 1 2 3 4 5 6 A positive value in the element force matrix implies that the resulting force for that DOF acts in the direction shown for that DOF. Conversely, a negative value represents a force which acts opposite to the direction shown. 2C6 Table A.1- -continued. VARIABLE DEFINITION TRCONV VARIABLE THAT IDENTIFIES IF A CONVERGENCE ITERATION IS NECESSARY; 0 = NO, MEANING [ERROR] < TOLER AND SOLUTION IS ACCEPTABLE; 1 = YES, MEANING [ERROR] > TOLER AND AN ATTEMPT WILL BE MADE TO CONVERGE ON THE SOLUTION BY USING THE INCREMENTAL FORCES TO SOLVE FOR THE INCREMENTAL DISPLACEMENTS TRELPR VARIABLE THAT IDENTIFIES WHETHER OR NOT ELEMENT FORCES AND DISPLACEMENTS ARE TO BE PRINTED; 0 = NO; 1 YES TPRINT VARIABLE THAT CONTROLS WHETHER OR NOT A STATEMENT IDENTIFYING EACH ELEMENT THAT HAS FAILED IS TO BE PRINTED; 0 = NO; 2 = YES TRFAIL VARIABLE THAT IDENTIFIES THE FAILURE STATUS OF THE WALL; 0 = HAS NOT FAILED; 1 = HAS FAILED, THEREFORE PRINT STRUCTURE FORCES AND DISPLACEMENTS; 4 = STOP AFTER PRINTING ELEMENT FORCES AND DISPLACEMENTS 344 (start) CONSTRUCT THE STIFFNESS MATRIX FOR THE BLOCK ELEMENT (return) Figure B.26 Algorithm For Subroutine BLESM p Figure 5*15 Brick Element P-M Interaction Diagram 329 Table B.20-continued. VARIABLE TYPE DEFINITION BR3EIL DOUBLE PRECISION BRICK 3EI/L (ROTATIONAL STIFFNESS FACTOR) ITER INTEGER DO-LOOP PARAMETER inspiration and must not go unnoticed. The encouragement of family, friends, and former teachers is also appreciated. Finally, the author extends heartfelt thanks to Ms. Cindy Zimmerman and Ms. Joanne Stevens for typing the manuscript and helping with its preparation. 120 CONTINUE BI3 EIL= LLSM1- l ( (BIPCV (LLP) -BIEF 1) ) / ( BLPC V (ULP) -BLPCV (LLP) ) ) $ *(OLSMl-ILSMT)) GC 1C 150 125 DO 1C0 1 = 1,NMIÂ£ 1 IF (AVGMCM. GE.BIMCOR(NOBLPC, I) ) BL 3KIL=SBLMT (NCBLFC ,1) IF (AVGMOH.II.BIMCOR(NOBIPC,I)) GG 1C 150 140 CC MINUE 150 BLMGM=AVGMGM BIVEbF=ELEF (1) RETURN ENE C SUBROUTINE STIFAC (VERTF,MOME,STOVE,STOMF,VSTIF,MSTIF, $ ST0VS1,S1CMS1,1RSTIF ,11 EÂ£ KE Y E IE H N C) C SUBROUTINE STIFAC ilLL SELECT THE APPROPRIATE STIFFNESS C FACTORS FOR AN ELEMENT BASED CN THE VAIUFS CE THESE C FACTORS AS DETERMINED FROM THE P-CELIA AND B-1BE1A C CURVES, THE CURRENT LEVEL CF ELEMENT LCADING, AND THE C EREVIOUS LEVEL OF ELEMENT LCADING. USE KEY = 1 FOE C BRICK ELEMENTS, KEY = 2 FOR CCILAE JOINT ELEMENTS, AND C KEY = 3 FCfi BICCK ELEMENTS. INT EGER IR STIE,ELEMNO DOUBLE PRECLSLC N VERTF,MOMF,STOVF,STOMF,VSTIF,MSILF,S10VST DUELE PRECISION STOMST DIMENSION STOVE (3,117) ,STOMF (3 1 1 7) S TG VS1 (3 1 1 7) DIMENSION STOMST (3,117) 1 = KEY J =EIEMNG IF (ITFB. NE. 1) GC TC 10 STGVF (I,J) = VSRTF S1CMF (I J) = MCKÂ£ STQ V ST (I J) = V S11F STOMST (I J) = MST IF GO TO 50 10 IF(VERIF.GT.STGVF (I,J)) GO TC 20 34 Similarly, if the bottom end of the element is fixed, as shown in Figure 36b, then by use of the differential equations for the beam deflections or the condition of symmetry it can be demonstrated that k = k 12EI 2,2 5,5 (1 + *)!3 (3.19) -k -6EI 3,2 6,5 (1 + #)li (3.20) -12EI 5,2 2,5 (1 + #)13 (3.21) k = -k -6EI 6,2 3,5 (1 + #)li (3.22) with the remaining coefficients in column 2 equal to zero. In order to determine the stiffness coefficients associated with the rotations wg and w^, the element is subjected to bending moments and the associated shears, as shown in Figure 3.6c and d. The deflections can be determined from Equation (3*6), but the constants C1 and C2 in these equations must now be evaluated from a different set of boundary conditions. With the boundary conditions (Figure 3.6c) v = 0 at x = 0, x = 1 (3.23) and dv dvs dx dx GAc at x = 1 , (3.24) Equation (3*6) becomes Elv _5(X3 A) lT(lx x2) + Â£6{1i . 6 2 2 (3.25) 342 (start) CONSTRUCT THE STIFFNESS MATRIX FOR THE COLLAR JOINT ELEMENT (return) Figure B.25 Algorithm For Subroutine CJESM 362 (start) DETERMINE IF EACH VALUE IN MATRIX [A] IS LESS THAN TOLER (return) Figure B.34 Algorithm For Subroutine CHKTOL WALL HEIGHT, INCHES 182 Figure 6.30 Wall Wythe Vertical Load Versus Height At 0.60 Pmax Example Number 5 For 128 (c) Relationship between vertical compressive load and vertical strain for 6 X 32 X 24-in hollow block prisms at e= t/4. .(d) Relationship between vertical compressive load and vertical strain for 6 X 32 X 24-in hollow block prisms at e= t/3. Figure 512-continued 205 Table A.1 Variables Used in Detailed Program Flowchart VARIABLE DEFINITION [K] STRUCTURE STIFFNESS MATRIX [w] STRUCTURE DISPLACEMENT MATRIX [F] STRUCTURE FORCE MATRIX [aw] INCREMENTAL STRUCTURE DISPLACEMENT MATRIX [ af] INCREMENTAL STRUCTURE FORCE MATRIX [K*] MODIFIED STRUCTURE STIFFNESS MATRIX WHICH AT THE TOP OF THE WALL ONLY CONSIDERS THE TWO PLATE DEGREES OF FREEDOM [V] MODIFIED STRUCTURE DISPLACEMENT MATRIX WHICH AT THE TOP OF THE WALL ONLY CONSIDERS THE TWO PLATE DEGREES OF FREEDOM [F] MODIFIED STRUCTURE FORCE MATRIX WHICH AT THE TOP OF THE WALL ONLY CONSIDERS THE TWO PLATE DEGREES OF FREEDOM [I] ELEMENT INDEX MATRIX 1 1 ELEMENT STIFFNESS MATRIX [w] ELEMENT DISPLACEMENT MATRIX [f] ELEMENT FORCE MATRIX [error] MATRIX THAT MONITORS SOLUTION CONVERGENCE TOLER TOLERANCE OF VALUE AGAINST WHICH THE NUMBERS IN [ERROR] ARE COMPARED NCONV NUMBER OF CURRENT CONVERGENCE ATTEMPT; A MAXIMUM OF 10 ATTEMPTS ARE MADE NSTAT STATUS OF AN ELEMENT AFTER IT IS CHECKED FOR FAILURE; 0 = O.K.; 1 = FAILED; 9 = IN TENSION TRSTIF VARIABLE THAT IDENTIFIES WHETHER OR NOT ANY ELEMENT STIFF NESS FACTORS HAVE CHANGED; 0 = NO; 1 = YES, THEREFORE RUN IS REPEATED WITHOUT INCREMENTING THE STRUCTURE FORCES WALL HEIGHT, INCHES 183 Figure 6.31 Wall Wythe Vertical Load Versus Height At 0.90 Pmax For Example Number 5 210 Figure A.1-continued 201 When a failure of one of the latter two types occurs, the program identifies the wall loads at which this takes place, as well as which element(s) has (have) failed. Once the wall has failed, or all the loads reached their maximum value, the program will output: 1) The structure forces and displacements 2) The lateral wall deflection versus height 3) The vertical wall deflection versus height 4) The element forces and displacements 5) The wall wythe vertical load versus height. This information may also be displayed for intermediate load values as described in the data input section. In listing the structure forces, a subheading of external wall loads will identify the degrees of freedom loaded when the load plate degrees of freedom are used. A subheading of equivalent wall loads will represent the resulting degrees of freedom loaded when the original structure degrees of freedom are considered. The program uses 36 subroutines each of which performs a unique operation. Throughout the program and subroutines, comments have been strategically interspersed to explain how the analytical process is undertaken. Furthermore, extensive documentation in the four appendices provides detailed information on the finite element program and its usage. All program input and output is in basic units of inches, pounds, or radians 3S P Figure 3.8 Moment Magnification 41 V 1 [w]T [f] I1 (y)2 dx e 2 2 0 (3.37) in which the first term represents the work of the [f] forces and the second term the work due to P. Since the ends of the member approach each other during bending, the axial force does positive work when it is a compression force and negative work if it is a tension force. The strain energy stored in the member during stage two is due only to bending. Thus U-S. I1 (y")2dx. 5-38) 2 0 Equating the strain energy to the external work gives 4 MT [f] +4 i1 (y')2 dx = .M. J1 (y)2 dx (3.39) Making use of the relationship [f] = [k] [w], in which [k] is the element stiffness matrix, Equation (3*39) becomes [w]T [k] [w] = El f1 (y")2 dx P f1 (y*)2 dx (3-40) 0 0 To evaluate [k], it is necessary to put the right-hand side of Equation (340) into matrix form. This can be accomplished if the deflection y is assumed to be given by y = A + Bx + Cx^ + Dx-5 (3-41) The choice of a deflection function is an extremely important step. A cubic is chosen in this instance because such a function satisfies the conditions of constant shear and linearly varying bending moment that 9 Usually an n x m transformation matrix [t] is also required to transform structure displacements into element displacements. However, if the element and structure coordinate systems are coincident, a transformation matrix is not required. Such is the case in this model. The solution procedure is generally as follows. For each element, 1) Construct the index matrix [i]. 2) Construct the element stiffness matrix [k]. 3) Insert the element stiffness matrix [k] into the structure stiffness matrix [k] using [i]. Once the structure stiffness matrix has been formed, 4) Construct the structure force matrix [f]. 5) Solve for the structure displacements by solving [f] = [k] [w] for [w]. Finally, for each element 6) Extract the element displacement matrix [w] from the structure displacement matrix [w]. 7) Multiply the element stiffness matrix by the element displacement matrix to yield the matrix of element forces, i.e., [f] = M M. Sections 2.2 through 2.8 discuss these matrices in more detail. 2.2 Construction of the Element Stiffness Matrix Consider the element shown in Figure 2.1a. This is the basic flexural element found in any text on matrix structural analysis. It contains the six degrees of freedom shown which include a rotation, axial displacement, and lateral displacement at each end. Recall that the stiffness coefficient k^j is defined as the force developed at the ith degree of freedom (DOF) due to a unit displacement at the jth DOF of the element while all other nodal displacements are maintained at zero. For example, the stiffness coefficient k^ is the force developed at the first DOF due to a unit displacement at the first Abstract of Dissertation Presented to the Graduate School of the University of Florida in Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy FINITE ELEMENT MODEL FOR COMPOSITE MASONRY WALLS By GEORGE X. BOULTON December 1984 Chairman: Dr. Morris W. Self Major Department: Civil Engineering Composite masonry design standards are at an early stage of development. To improve the understanding of composite masonry wall behavior in response to load application, a two-dimensional finite element model has been developed. The model considers a wall subjected to vertical compression and out of plane bending. It takes into account the different strength- deformation properties of the concrete block, collar joint, and clay brick, as well as the nonlinear nature of these properties for each material. The effects of moment magnification and shear deformation in both the brick wythe and the block wythe of a composite wall are also considered. The primary purpose of this model is to analytically represent typical composite masonry walls that might be tested in a laboratory. Wall tests attempt to duplicate conditions found in prototype walls. By comparing the results of the analytical and experimental tests, needed 238 (start) r~ I I l_L- * FOR EACH ROW) FOR EACH COLUMN) : SET [c] = [a] + [b] (return) Figure B.3 Algorithm For Subroutine ADD 295 (start) r jFOR EACH ROW IN COLUMN VECTOR [c]) I SET [c] = [A] x [B] BUT PERFORM MULTIPLICATION AS IF [a] WERE AN N x N MATRIX (return) Figure B.6 Algorithm For Subroutine BMULT STRUCTURE STIFFNESS MATRIX C C C (EVEEY CIHER TIBE) C 1. RECALL THE INDEX MATRIX. C 2. EXTBACT T BE ELEMENT DISPLACEMENT MATKIX FROM THE C SIBOCTUHE DISPLACEMENT MATEIX. C 3. BECALL THE ELEMENT STIFFNESS MATEIX. C 4. MULTIPLY THE ELEMENT STIFFNESS MATBIX EY THE ELEMENT C DISPLACEMENT MATBIX TO YIELD THE ELEMENT FOBCE MATBIX. C 5. PBINT THE ELEMENT FOBCES WHEN AEEROEEIAIE. C 6. POINT THE ELEMENT DISPLACEMENTS WHEN APPROPRIATE. C 7. CHECK FOB ELEMENT (AND THEREFORE, WALL) FAILURE USING C IKE P-M INTEBACT ION DIAGRAM. C 6. GET STIFFNESS FACTORS BASED CN THE ELEMENT FCBCES FBCM C THE P-DELTA S M-THETA CURVES S COMPARE THEM WITH THCSE C OF THE PREVIOUS CYCLE TO DETECT ANY CHANGES. C 9. CCNSTEUCT TEE ELEMENT STIFFNESS MATRIX. C 10. INSERI THE ELEMENT STIFFNESS MATRIX INTC THE C STRUCTURE SIIEENESS MATRIX. C 11. MULTIPLY THE ELEMENT FORCE MATRIX BY -1.0 & INSERT IT C INTO THE STRUCTURE FORCE MATRIX FOR CONVERGENCE CHECK. 142 CON1=0 DC 2 80 1=1,NEMI,3 IF (IIER.NE.1) GO TO 145 CALI INEXER (IBR,TOTEIN,I,NEMT,BE,NELEM) GO TO 2 5 C 145 CALL PULEOW (IBB TOT EIN, I, UB N ELEM) CALL NULL (BREW,BE,1) IF (I,EQ.N EMT)CALL EXTRAK (STEW,BEEN,2,3,NSDCF,BE,1BB) IF (I.EQ.NEMT) GO TO 170 ' IF(I. EQ. 1) CALL EXTRAK (SIRW,BE EH,2, 1,NSDGF,BR,1BO) IF (I N Â£. 1) CALL EXTRAK (SIRN,BREW ,2,2, NSDCF, BR,.IÂ£fi) 17 CALL NULL (EBEK,BR,BB) CALL EULHAT (BEEK,TOTEK,I,BB,NELEM) CALL MULT (EflEK,BREM,EEEF,BS,BR, 1) CGUNT=CCUNT* 1 i\) V>) 199 elements at a node are not in equilibrium. When the collar joint element end moments decrease at a node due to a decrease in shear stress, this causes the moments in the brick (or block) wythe at that node to be lower than they would have been otherwise. This produces the change in curvature in the moment diagrams observed previously. Eventually, if the collar joint shear stress becomes negative, as it does in Figure 6.43, this causes the moment in the brick and block wythes to actually decrease as shown in Figures 6.35 and 6.39, respectively. Not surprisingly, the maximum moment in each wythe for example 4 occurs at a height of 168" which is the height at which the collar joint shear stress is zero. 287 Table B.3 Nomenclature For Subroutine ADD VARIABLE TYPE DEFINITION A DOUBLE PRECISION N x M MATRIX WHOSE VALUES ARE ADDED TO THOSE OF MATRIX [b] TO OBTAIN MATRIX [c] B C N M DOUBLE PRECISION DOUBLE PRECISION INTEGER INTEGER N x M MATRIX WHOSE VALUES ARE ADDED TO THOSE OF MATRIX [a] TO OBTAIN MATRIX [c] N x M MATRIX WHOSE VALUES ARE THE SUM OF THE CORRESPONDING VALUES IN [a], [b] NUMBER OF ROWS IN MATRICES [a], [b], [c] NUMBER OF COLUMNS IN MATRICES [A], [B], [C] 73 *, = ONE-HALF THE THICKNESS OF 1 THE BRICK WYTHE (= 2") = ONE-HALF THE THICKNESS OF THE BLOCK WYTHE (= 2", 3" or 4") ks = SHEAR SPRING STIFFNESS FACTOR k = MOMENT SPRING STIFFNESS FACTOR m Figure 4.4 Finite Element For Collar Joint 284 (start) r { FOR EACH ~ROw) -|for each column) I L_- SET [A] = 0 (return) Figure B.1 Algorithm For Subroutine NULL ML BIOGRAPHICAL SKETCH George Xavier Boulton was born April 13, 1959, in Havana, Cuba. To escape communist rule, he moved with his parents to the United States in 1961. They lived in New York City until he was five, then moved to Mobile, Alabama. There he attended parochial schools and graduated from McGill-Toolen High School in 1977. He then enrolled at the University of South Alabama, where he received his Bachelor of Science in Civil Engineering degree in 1991, graduating with high honors. In August of 1981, he moved to Gainesville, Florida, to pursue graduate studies at the University of Florida. In August of 1982, he received his Master of Engineering Degree in civil engineering from the University of Florida. He then enrolled in the doctoral program of the structures area of the Civil Engineering Department. George is an American citizen. He is also a member of the American Society of Civil Engineers, the National Society of Professional Engineers, and is registered as an Engineer in Training in the State of Alabama. He speaks fluent Spanish and enjoys most outdoor sports. Since his sophomore year in high school, George has been working part-time while also going to school full-time. He is anxiously anticipating entering professional practice upon graduation, and plans to become registered as a professional engineer as soon as possible thereafter. 326 (start) /PRINT THE DIMENSIONS N AND M OF MATRIX [a] / /print matrix [a]/ (return) Figure B.19 Algorithm For Subroutine WRITE 2 4.1C9D-Q3 1186D+05 3 6.616D-03 1.780+05 4 1.0D-02 2,58179D+05 COLLAR JOINT ELEMENT P-DELTA CORVE 2 PT DELTA COORDINATE P COORDINATE 1 O.OD + OO O.OD+OO 2 4,6 13D-02 9 777D+03 COLLAR JOINT ELEMENT B-THETA CURVE 2 PT THETA COORDINATE F COORDINATE 1 0.004-00 U.D+Q0 2 1. OD4-00 0.0D+00 BLOCK ELEMENT P-DELTA CURVE PT 1 2 3 4 5 3 5 PT 1 2 3 4 5 PT 1 2 3 DELTA COORDINATE P COORDINATE 0,QD+00 4.0D-03 .0D-03 1.20-02 1,60D-02 0.0D+00 2.5125D+04 5.0250+04 7.5375D+04 9,315D+04 BLOCK ELEMENT M-THETA CURVES 2.5E+04 THETA COORDINATE M COORDINATE 0.0D+00 0.0D+00 5.63D-04 2.3325D+04 8,460-04 3.5D+04 1.49D-03 4.6675D+04 3.5D-03 8.3114D+4 5.0D+04 THETA COORDINATE H COORDINATE O.OD+OO O.OD+OO 1.0350-03 4,665D+04 1.674P-03 7.0D+04 vi KO r\> n r. CJEK (2,2) = CJSHRS* (LNHBR**2) +CJMOMS CJEK (3, 2) =-1.G*CJEK (2, 1) CJEK (4,2)=CJSBBS*LNHBR*LNHEL-CJM0MS CJEK (3, 3) =CJEK (1 1) CJEK (4,3) =-1.Q*CJEK (4, 1) CJEK (3, 4) =CJEK (4,3) CJEK (4,4) = CJS HRS* (1NHBL**2) +CJMOMS DO 3C J=1,CJ EC 20 K= 1,CJ CJEK (J K) =C J EK (K, J) TCTEK (J,K,I) =CJEK(J, K) 20 CONTINUE 30 CONTINUE RETURN EKE SUBROUTINE ELESM (BLEK,ELÂ£F,TOTIK,ELAEL,BL3EIL,LN,ELASBL, $ AKSHBL,I,BI,KEIEE) SUBROUTINE ELESM NHL CCNSTBUCT TEE STIFFNESS MATRIX FOR EACH CGNCE ETE BLOCK BLE EE NT. INTEGER BL ECU ELE PRECISION BLEK,BLEF,TCTEK,LK,BLAEI,E13 EIL,ElAS EL DCUELE PRECISICN A ES HBL,GBL,PHIBL,POISBL DIMENSION TOTEK(6,6,NÂ£LEM),BLEK (BL ,BI) ,EIEF(EL) fCISEL=0,15 E 0 0 GBL=ELASBL/(2.C*(1.+POISBL)) PIiIEI=EL3EII*4. 0/ (GBL*ARSHEL*LN) BLEK (1, 1) =ELAEL BIEK (2, 1) =0. 0 ELEK (3, 1) = C. 0 BLEK (4, 1) = -10*ELEK(1, 1) ELEK (5, 1) =0.0 BIEK (6, 1) =0.0 BLEK (2, 2) = (BL3EIL* 4.0/ { (1.0+PHIBL) (LN**2.0) ) ) - $ (ELE.E(1) *6. 0/(5.O+LN)) BLEK (3, 2) = (-BL3EIL*:2.0/ ( (1. O+PHIBl) *LN) ) + (ELEF (1) / 10.0) non lj k: - n n HITE ( 6,45) K,XCOOR (I, 0) ,50000(1,0) 45 FORMAT(1 *,13,3X,Â£13.6,35,013.6) 50 CONTINUE 60 CONTINUE WHITE (6,70) 70 FORMAT (* ) BET DEN ENE C SUBROUTINE FBI NT (A H) SDBfiOUTINE PRINT WILL EfilNI AN NX! MATRIX [A] AND IDENTIFY EACH BOW AS A DEGREE CF EBEEECM. DCOEIE EBECISICK A DIMENSION A(N) WHITE (6,10) 0 FORMAT ( ) DC 30 1=1,N WHITE (6,20) I,A (I) ECEM AT (' ,COE *,I4,2X,D13.6) 0 CONTINUE BETUHN ENE SUEBOUTINE WHITE (A N,M) SUBROUTINE WRITE WILL PRINT OUT THE DIMENSIONS N AND M OF A MATRIX Â£ A J AND THE MATRIX OF CCEEEICIE NTS IN EOWS. DOUEIE E BECISIC N A DIMENSION A (N ,M) WBITE (6,20) K,M DO 1C 1=1,N IE (M. LE. b) THEN WRITE (6,30)1, (A (I,J) ,0=1 M) ELSE WRITE (6,40)I,(A (1,0),0=1,6) WfilTE (6,50) (A ( I, 0) J = 7,M) END IF 260 211 Figure A.1-continued EXAMPLE #5 H/T=2Q PLATE LOAD ECCE NT R.ICI T Â¥= 2 DESCRIPTION CE WALL 200 8.D + 00 2. GD+CO 3.D + 00 24 6 PT 1 2 3 4 5 6 3 4 PT 1 2 3 4 PT 1 2 3 4 PI 1 MATERIAL PROPERTIES BRICK ELEMENT P-DELTA CURVE DELTA COORDINATE P COORDINATE 0,GD+00 4.GD-03 8,D-3 1. 2D-02 1.6D-02 2- C8D-02 0.0D+00 9.2175D+04 1,671750+05 2.32725D+05 2,9 1450D+05 3.3885OD+ 05 BRICK ELEMENT M-IHETA CURVES 1.CC IN El A G 2. 1 9.7 1 1.ED THETA C J. 1 7.8 1 2. CE THETA C + 05 CCCRDIN AT , CD + CO 17D-03 C2D-G3 ,0D-02 1. + 05 COORDINA! .CD+OO 31D -03 t SD-3 .GD-02 1, + 05 COORDINA! .CD+CO M COORDINATE O.GD+OO 5.93D+04 1.1870 + 05 2 1033D+05 E M COORDINATE 0. CD + OO 8.895D+04 1 335D + Q5 5359 1D+05 E M COORDINATE 0. CD + OO IN. 406 317 iil \rt) /read the wall description data/ /print the wall description data/ calculate the modulus of elasticity AND AREA IN SHEAR FOR THE BRICK CALCULATE THE MODULUS OF ELASTICITY AND AREA IN SHEAR FOR THE BLOCK urn) Figure B.15 Algorithm For Subroutine READ LIST OF TABLES Table Page 4.1 Variables Used in Constructing Brick Element Stiffness Matrix 30 4.2 Variables Used in Constructing Block Element Stiffness Matrix 87 6.1 Summary of Wall Failure For Examples Number 1 Through 5 151 A. 1 Variables Used in Detailed Program Flowchart 205 A. 2 Program Nomenclature 215 B. 1 Nomenclature For Subroutine NULL 283 B.2 Nomenclature For Subroutine EQUAL 285 B.3 Nomenclature For Subroutine ADD 287 B.4 Nomenclature For Subroutine MULT .289 B.5 Nomenclature For Subroutine SMULT 291 B.6 Nomenclature For Subroutine BMULT 294 B.7 Nomenclature For Subroutine INSERT 296 B.8 Nomenclature For Subroutine BNSERT 299 B.9 Nomenclature For Subroutine EXTRAK 301 B.10 Nomenclature For Subroutine PULROW 304 B. 11 Nomenclature For Subroutine PULMAT 306 B.12 Nomenclature For Subroutine GAUSS1 3CQ B.13 Nomenclature For Subroutine STACON 311 B.14 Nomenclature For Subroutine TITLE 314 B.15 Nomenclature For Subroutine READ 316 B.16 Nomenclature For Subroutine COORD 319 Table A.2-continued VARIABLE TYPE NBRIDP INTEGER NBRMTP INTEGER NBRPDP INTEGER NBLIDP INTEGER NBLMTP INTEGER NBLPDP INTEGER NCJMTP INTEGER NCJPDP INTEGER NCONV INTEGER NDOF INTEGER NELEM INTEGER NEMO INTEGER NEMT INTEGER NFORCE INTEGER NOBLPC INTEGER DEFINITION NUMBER OF POINTS IN THE BRICK INTERACTION DIAGRAM NUMBER OF POINTS IN EACH BRICK M-THETA CURVE NUMBER OF POINTS IN THE BRICK P-DELTA CURVE NUMBER OF POINTS IN THE BLOCK INTERACTION DIAGRAM NUMBER OF POINTS IN EACH BLOCK M-THETA CURVE NUMBER OF POINTS IN THE BLOCK P-DELTA CURVE NUMBER OF POINTS IN THE COLLAR JOINT M-THETA CURVE NUMBER OF POINTS IN THE COLLAR JOINT P-DELTA CURVE NUMBER OF CURRENT CONVERGENCE ATTEMPT NUMBER OF THE DEGREE OF FREEDOM OF A PARTICULAR FORCE NUMBER OF ELEMENTS NUMBER OF ELEMENTS MINUS ONE NUMBER OF ELEMENTS MINUS TWO NUMBER OF FORCE DEGREE OF FREEDOM, MULTIPLES OF WHICH WILL CAUSE AN INTERMEDIATE PRINTOUT HUMBER OF BLOCK M-THETA CURVES (EACH IS IDENTIFIED BY THE P OR AXIAL LOAD LEVEL FOR WHICH THE MOMENT CURVATURE RELATIONSHIP IS VALID) 217 14 (a) Frame (b) Structure Degrees of Freedom w (c) Element Degrees of Freedom w. w 2 Figure 2.3 Structure and Element Degrees of Freedom For a Frame DC 96 1=1, II=I+K-J C (K,I)=C (K,I) + COEE E*C(J, II) 96 CONTINUE 97 CONTINUE 98 CONTINUE C CCNSTEUCT [K4' ] EEC Â£C]. K=C DC 120 1= NUT P1, N K = K+ 1 DC 110 J=1,M1 K4 (K, J) =C (I, J) 110 CONTINUE 120 CO N1INUE C CCNSIEUCl Â£ JE2 j FECH Â£ E ]. D= 0 DO 140 1= NETP 1, N L=L + 1 2(I) = E(I) 140 CONTINUE C SOLVE Â£ K4 ]Â£ W2 ]=Â£ F2 ] FOE Â£W2J USING GAUSS1 CALL GAUSS1 (W2 ,K4 ,F'2, M 1 NEBC1 1) C CCNSIEUCl Â£ K2' ] Â£C Â£C]. CALL NULL (K2,NRTCP,H1) DC 160 NE=1,N ETCP I=NK1P1-Nfi J=0 N EP1=NR+ 1 IF (NEP1.GI. Ml) GO TO 165 DC 150 L=NBP 1,M1 J=J+1 IF (J .Gl. NBBC1) GO TO 160 K2 (I,)=C (I, L) 150 CONTINUE 160 CONTINUE f\3 OI OI 97 presents no difficulty in executing the previous procedure. In fact, the nature of the transformation matrices [xj and [x^] are such that all multiplications in which they are involved can be performed very efficiently. The model takes advantage of this additional opportunity to keep numerical computations to a minimum. 4.9 Solution Procedure The solution procedure used by this finite element model is basically the general procedure for the Direct Stiffness Method outlined at the end of section 2.1, modified to include all of the special considerations discussed in chapters 3 and 4 in the fashion indicated there. The presence of different materials is handled by having three types of elements as discussed in section 4.2 and using the material properties of each element type in constructing the respective element stiffness matrices. Shear deformation in the brick and block wythes is accounted for by special terms in the stiffness matrices of the brick and block elements. Similarly, moment magnification is considered in the brick and block wythes by special terms in the stiffness matrices of the brick and block elements. Material nonlinearity is accounted for by applying the loads on the structure in increments, obtaining a solution after each load step is applied, and using the information obtained to determine the material properties from the nonlinear curves for the next load step. Static Condensation with Gauss Elimination is used for solving the structure matrix equation while taking advantage of the symmetry and bandedness of the structure stiffness matrix. Solution convergence is monitored and promoted by the method indicated in section 3.6. The application of loads on the wall through the test plate at the 149 7 IN 3 IN f/m////! 4 IN -ni 6 IN TEST WALL Sign Convention H/T = 20 P = P + AP AP = 4000 LB e = 2 IN M = Pe = 2P V' P L K I 8 IN TYPICAL /7T7/7T7 2 INHW3 IN MODEL Figure 5.26 Example Number 5 148 Wi21 vt *, r* 7 V Tv > v+'r'+'' 4 4 VET^ U V47~ 4 V7~/ I f W1 t. 4^ /& 4 4 4 virra vF'Tji 4 '4Tr3# 4 v^rq* 4 iw uThn* +3 A? - 4 4 4_ / t i#-** I rf#P 4 rf** * 4 4 4_ t 4__ * w 2 ~ w, w 3 EXTERNAL WALL LOADS ^121 W122 w 1 * w123 W124 V.#1 V?vf t i V V 4 v. 4 v 4 vi# 4 vp 4 V|# 4 4 VJ# 4 vj* 4 Wi4 w,v 4_ 4_^ t 4 if .4_ if 4 V 4 ir 4__ r 4_^ if 4 if"* 4_^ if 4 If 4_<_ f 4__ if i- t t f W3 w5 3 /77 /77 EQUIVALENT WALL LOADS 2l 2 2 4 4 ^1 2 3 v. 7"v ^124 V >. 9 4 V tLJU 4 v. <# . r 'i '< 4 4 t w Jt 1 4 *4- 4 ^ VUf N. # H V4f s+f 4. 4- sir <4# 4 +- 4- v+# 4 'JT~3i 4 v 4 V 4 4 . f 4 f 4_ < 4_ f 4_ # 4^ 4_ * 4_ 4_ f 4 f T 4 4_ 4 4_ 4_ }"w2 w, w. /7r/rr WALL DISPLACEMENTS Figure 5.25 Structure Degrees of Freedom For Example Numbers 4 and 5 RIGID WYTHE COLLAR JOINT ELEMENT CONCRETE BLOCK ELEMENT Figure 4.1 Finite Element Model of Wall 51 1) Forward elimination to force zeros in all positions below the diagonal of [k] by performing legal row operations on [k] and [F] 2) Backward substitution to solve for [w]. The following simple example helps illustrate how this is done. Assume it is desired to solve the four simultaneous equations: 2W1 + W2 =10 2Â¥1 + 6W2 6W5 =0 - 6W2 + 9W5 7W4 = -28 - 7W3 + 5 W4 = 0 They can be rewritten in matrix form as 2 10 0 *1 10 2 6-60 w2 0 0-6 9-7 Â¥3 -28 0 0-75 *4 0 __ Forward elimination is performed by 2 10 0 10 (-1 x row 1) + row 2 2 6-60 0 0-6 9-7 -28 0 0-75 0 which yields 207 (start) DECLARE MATRICES AND VARIABLES REAL OR INTEGER AS NEEDED DOUBLE PRECISION MATRICES AND VARIABLES DIMENSION MATRICES /READ AND PRINT PROBLEM DESCRIPTION (CALL READ)/ NULL MATRICES AS NEEDED (CALL NULL) /read AND PRINT P-A CURVES FOR brick, collar/ / JOINT, AND BLOCK ELEMENTS (CALL COORD) / /READ AND PRINT M-0 CURVES FOR BRICK, collar/ / JOINT, AND BLOCK ELEMENTS (CALL CURVES) / / READ AND PRINT P-M INTERACTION DIAGRAMS FOR BRICK / AND BLOCK ELEMENTS (CALL COORD) Â¡ ' READ AND PRINT MAXIMUM VERTICAL LOAD CAPACITY FOR BRICK, COLLAR JOINT, AND BLOCK ELEMENTS (CALL TITLE TO READ HEADINGS) /READ AND PRINT FORCE APPLICATION INFORMATION (CALL FORCES)/ r T / /READ, PROCESS, AND PRINT INSTRUCTIONS FOR INTERMEDIATE/ I PRINTOUTS (CALL TITLE TO READ HEADINGS) / DO FOR ITER-1 to 10000) I Figure A.1 Detailed Program Flowchart Table B.20 Nomenclature For Subroutine BRPDMT VARIABLE TYPE DEFINITION NBRPDP INTEGER NUMBER OF POINTS IN THE BRICK P-DELTA CURVE NOBRPC INTEGER NUMBER OF BRICK M-THETA CURVES (EACH IS IDENTIFIED BY THE P OR AXIAL LOAD LEVEL FOR WHICH THE MOMENT CURVATURE RELATION SHIP IS VALID) NBRMTP INTEGER NUMBER OF POINTS IN EACH BRICK M-THETA CURVE BRDCOR DOUBLE PRECISION MATRIX THAT STORES BRICK DELTA COORDINATES BRPCOR DOUBLE PRECISION MATRIX THAT STORES BRICK AXIAL LOAD COORDINATES BRPCV DOUBLE PRECISION MATRIX THAT STORES THE VALUE OF P BY WHICH EACH BRICK M-THETA CURVE IS IDENTIFIED BRTCOR DOUBLE PRECISION MATRIX THAT STORES BRICK THETA COORDINATES FOR EACH M-THETA CURVE BRMCOR DOUBLE PRECISION MATRIX THAT STORES BRICK MOMENT COORDINATES FOR EACH M-THETA CURVE BREF DOUBLE PRECISION BRICK ELEMENT FORCE MATRIX SBRPD DOUBLE PRECISION MATRIX THAT STORES THE SLOPES OF THE BRICK AXIAL LOAD DELTA CURVE SBRMT DOUBLE PRECISION MATRIX THAT STORES THE SLOPES OF THE BRICK MOMENT THETA CURVE BRVERF DOUBLE PRECISION VERTICAL (AXIAL) FORCE AT THE BOTTOM OF A BRICK ELEMENT BRMOM DOUBLE PRECISION ABSOLUTE VALUE OF AVERAGE OF END MOMENTS FOR A BRICK ELEMENT BRAEL DOUBLE PRECISION BRICK AE/L (AXIAL STIFFNESS FACTOR) BLMCOR BLOCK M COORDINATE DOUBLE PRECISION D13.6 67 BLPCV VALUE OF P WHICH IDENTIFIES THE THAT FOLLOWS BLOCK M-THETA CURVE DOUBLE PRECISION D13-6 68 - COMMENT OF 'PT' - - - COMMENT OF 'THETA COORDINATE' - - - COMMENT OF 'M COORDINATE' - - 69-73 K POINT NUMBER INTEGER 13 BLTCOR BLOCK THETA COORDINATE DOUBLE PRECISION D1 3.6 BLMCOR BLOCK M COORDINATE DOUBLE PRECISION D13-6 74 BLPCV VALUE OF P WHICH IDENTIFIES THE THAT FOLLOWS BLOCK M-THETA CURVE DOUBLE PRECISION D13.6 75 - COMMENT OF 'PT' - - - COMMENT OF 'THETA COORDINATE' - - - COMMENT OF 'M COORDINATE - - 76-80 K POINT NUMBER INTEGER 13 BLTCOR BLOCK THETA COORDINATE DOUBLE PRECISION D13*6 BLMCOR BLOCK M COORDINATE DOUBLE PRECISION D13.6 81 COMMENT OF 'BRICK ELEMENT P-M INTERACTION DIAGRAM' 17 1 2 6 24 1 4 17 1 2 6 24 1 4 17 1 oeÂ£ DOF. Similarly, the stiffness coefficient is the force developed at the first DOF due to a unit displacement at the second DOF. The total force F at the first DOF can, therefore, be represented as f1 = k11w1 + k12w2 + k13w3 + k14w4 + k15w5 + kl6w6 where w^ equals the element displacement at the ith DOF. Analogously, the forces at the other degrees of freedom are: f2 = k21w1 + k22w2 + k23w3 + k24w4 + k25w5 + k26w6 f6 = k6lw1 + k62w2 + k63w3 + k64w4 + k65w5 + k66w6 These equations can be written conveniently in matrix form as f1 k11 k1 2 k13 k14 k15 kl6 f2 k21 k22 k23 k24 k25 k26 f3 k31 k32 k33 k34 k35 k36 f4 k41 k42 k43 k44 k45 k46 f5 k51 k52 k53 k54 k55 k56 _f6_ k6l k62 k63 k64 k65 k66 or simply as [f] [k] [w] [w] = element [f] = element [k] = element displacement matrix force matrix stiffness matrix. where 104 loading them axially and noting the vertical strain variation with load. The results of their tests are shown in Figure 5.1* In order to use their results, it is first necessary to convert their data, which is for 4x32x16 in prisms, to equivalent data for 4x24x8 in prisms. This is done as follows. The actual prism height is 15.7 in. The desired prism height is 8.0 in. Let = P on actual prism Pp s P on desired prism p Aqa gross area of actual prism = 4x32 = 128 in O Aqp = gross area of desired prism = 4x24 = 96 in . The vertical stress c is given by a AGA agd therefore (5.1) The vertical displacement A for 15.7 in high prisms is not given directly, instead the vertical strain e is given. Recall that e is given by therefore A = e L (5.2) 86 stiffness factors, as well as the maximum axial compressive load capacity and the axial load versus moment interaction diagram for the block element. Table 4.2 shows all the variables needed to construct the block element stiffness matrix which will be identical to the brick element stiffness matrix shown in Figure 4.8. The block element axial load versus moment interaction diagram should have the same general shape shown in Figure 4.9. It will be approximated by straight line segments and used by the model to determine when a block element has failed. Block prism failure will be defined as the point in the loading process at which the block prism is unable to support additional load. 4.4 Load Application This model is an analytical representation of the test walls that will be built and tested in the experimental phase of the project. In the tests, vertical compressive loads will be transmitted to the top of the wall through a rigid steel plate. When the load is applied eccentrically, it will also cause a moment to be applied to the top of the wall. The external axial load and moment will be applied through the vertical force and moment degrees of freedom at the top of each wythe. The magnitude of these, however, will depend on the relative stiffness of each wythe as well as on the relative magnitude between the axial and rotational stiffness for a wythe. The stiffnesses though, are a function of the level of loading. In short, the problem is that two loads (a force and a moment) are applied to the test plate, but there are four degrees of freedom (a force and a moment for each wythe) at the top of the wall through which these loads can be applied. Since properly converting the two loads into four loads requires consideration 43 In view of (3*44) and (3*45), one can write (y'>2 MT [c]T [c] [v] and (y")2 []T MT [B] [] . Substitution of these relations into (3*40) gives (3.46) (3.47) (3.48) (3.49) H'MM MT (El I1 [D]T[D]dX P /! [c]T[c]dx [] 0 0 / from which [k] = El I1 [D]T[D]dx P f1 [c]T[c]dx . 0 0 (3.50) Using the expressions given in (3-46) and (3-47) for [c] and [d] and carrying out the operations indicated in (3*50), one obtains [k] = El 12 6 12 6 1? l2 l5 l2 6 4 6 2 l2 1 l2 1 _ 12 6 12 6 l3 l2 l2 6 2 6 4 l2 1 l2 1 6 1 6 1 51 10 51 10 1 21 + 1 1 10 15 10 30 _ 6 + 1 6 1 51 10 51 10 1 1 1 21 10 30 10 15 (3.51) APPENDIX D DATA FILES FOR NUMERICAL EXAMPLES Table B.22-continued VARIABLE TYPE DEFINITION BL3EIL DOUBLE PRECISION ITER INTEGER BLOCK 3EI/L (ROTATIONAL STIFFNESS FACTOR) DO-LOOP PARAMETER 311 Table B.13 Nomenclature For Subroutine STACON VARIABLE TYPE DEFINITION X C B K4 F2 W2 K2 K2W2 DOUBLE PRECISION N x 1 MATRIX WHICH STORES THE SOLUTION TO THE MATRIX EQUATION OF THE FORM [A] [X] [B] WHERE THE SOLUTION IS CALCULATED BY THE SUBROUTINE DOUBLE PRECISION N x HI MATRIX WHICH STORES THE UPPER TRIANGULAR PORTION OF A SYMMETRIC N x N MATRIX [a] THAT HAS A BANDWIDTH OF (2 x M1) 1 DOUBLE PRECISION N x 1 MATRIX THAT STORES THE NUMBERS THAT EQUAL [a] [x] DOUBLE PRECISION MATRIX WHICH EQUALS THE PORTION OF THE [c] MATRIX THAT IS EQUIVALENT TO THE BOTTOM LEFT FOURTH OF THE N x N MATRIX [A] DOUBLE PRECISION MATRIX WHICH EQUALS THE BOTTOM HALF OF THE N x 1 MATRIX [b] DOUBLE PRECISION MATRIX WHICH EQUALS THE BOTTOM HALF OF THE N x 1 MATRIX [x] DOUBLE PRECISION MATRIX WHICH EQUALS THE PORTION OF THE [C] MATRIX THAT IS EQUIVALENT TO THE TOP RIGHT FOURTH OF THE N x N MATRIX [a] DOUBLE PRECISION MATRIX WHICH EQUALS [K2] [W2] F1 F1MKW2 K1 W1 M1 N DOUBLE PRECISION MATRIX WHICH EQUALS THE TOP HALF OF THE N x 1 MATRIX [B] DOUBLE PRECISION MATRIX WHICH EQUALS [Fl] ([K2] [W2]) DOUBLE PRECISION MATRIX WHICH EQUALS THE PORTION OF THE [c] MATRIX THAT IS EQUIVALENT TO THE TOP LEFT FOURTH OF THE N x N MATRIX [a] DOUBLE PRECISION MATRIX WHICH EQUALS THE TOP HALF OF THE H x 1 MATRIX [x] INTEGER HALF THE BANDWIDTH (INCLUDING THE DIAGONAL) OF MATRIX [a] INTEGER NUMBER OF ROWS IN MATRICES [x], [c], [b] 3. 3885D + 05 MAXIMUM CO LIAR JOINT ELEMENT SHEAS LOAD CAPACITY 9,777D+03 MAXIMUM BLOCK ELEMENT COMPRESSIVE LOAD CAPACITY 9 3 15D + 04 STRUCTUSE FORCE APPLICATION INFORMATION 1 DOF INITIAL FORCE FORCE INCREMENT MAXIMUM FORCE 71 -4.QD+03 -4.0D+03 -4.0D+05 INSTRUCTIONS ON PRINTING INTERMEDIATE RESULTS: 1 171 -5.6D+04 106. 107. 108. 109. 110. 111. 112. 113. 114. 115. 116. 117. 309 Figure B.12 Algorithm For Subroutine GAUSS1 16 [k]2 - 1 4 2 5 k11 k1 2 k13 k14 k21 k22 K\ C\J k24 k31 k32 k33 k34 k41 k42 k43 k44 1 4 2 5 6 [k] 3 " 0 0 0 2 5 5 k11 k12 k13 k14 k15 kl6 k21 k22 k23 k24 k25 k26 k31 k32 k33 k34 k35 k36 k41 k42 k43 k44 k45 k46 k51 k52 k53 k54 k55 k56 k6l k62 k63 k64 k65 k66 0 0 0 2 3 5 Define LKijJm as the stiffness coefficient in row i, column j of the stiffness matrix for element m. The numbers in the index matrix along the sides of the element stiffness matrix for each element identify the rows and columns in the structure stiffness matrix in which each coefficient belongs. Since the frame has 5 DOF, the structure stiffness matrix will be a 5 x 5 matrix. The term in row 1, column 1 of the structure stiffness matrix is, therefore, 8 In general, [k]^ and [f]j_ can be found from standard cases. Since this model only considers nodal loads, [f0]^ matrices will not exist. The force displacement equations for an element i, therefore, reduce to [f]i = Mi Mi (2.2) Suppose an element i is connected to other elements to form a structure with N structure degrees of freedom. The structure force displacement (or equilibrium) equations can be expressed as !>] = [K] [w] + [F] (2.3) where [w] = N x 1 matrix of independent structure displacements measured in structure coordinates [f] = N x 1 matrix of corresponding structure forces measured in structure coordinates [f] = II x 1 matrix of corresponding structure fixed-end forces measured in structure coordinates [k] = N x N structure stiffness matrix measured in structure coordinates. Again, fixed-end forces are not in the model so these equations reduce to [f] = M M (2.4) If m equals the total number of structure degrees of freedom that are related to the element degrees of freedom for element i, an index matrix [l]j_ can be defined as [l]j_ = m x 1 matrix whose elements are the numbers of the structure degrees of freedom that are related to the element degrees of freedom for element i. 349 (start) CONSTRUCT THE INDEX MATRIX FOR THE COLLAR JOINT ELEMENT (return) Figure B.28 Algorithm For Subroutine INDXCJ 91 r 7^ CD -o a J U a 1 % a16 a17 (e) Test Plate Displacements (f) Test Plate Displacement Geometry Figure 412-continued. 139 Pp = P on desired wall cross-section P^ = P on actual wall cross-section Aqj, = gross area of desired wall cross-section Aqa = gross area of actual wall cross-section. p From the cross-section dimensions given earlier, Aqjj=240 in and Aq^=316 p in Neglecting the difference in end conditions, this means the wall test failure load value to be compared with the model failure load is P = 240 x 18o,000 136,700 lb if the average failure load of the two tests is used. To account for the difference in end conditions, an approximate wall failure load which might have been obtained from the tests if the end conditions in the model had been used, can be calculated using Figure 5*19 taken from the Manual of Steel Construction (7). The end conditions shown in (d) were those used in the test walls. The model considers the end conditions given in (b). Note that K for the test walls equals 1.0 and K for the model wall equals 0.80. The wall test failure load value expected for the end conditions in the model would be P (approx) 136,700 x -1aÂ£_ = 170,900 lb . 0.80 Since the model predicted failure at 132,000 lb, it underestimated the failure load by about 23$. Nonetheless, this is not excessive since a 12% difference in wall failure loads was obtained in the two wall tests. Furthermore, on the basis of the experimental testing program to be conducted in the future, further refinements to the analytical model This dissertation is dedicated to Almighty God in thanksgiving for all of His blessings. WALL HEIGHT, INCHES Figure 6.17 Wall Wythe Vertical Load Versus Height At 0.JO ?max Example Number 2 For 108 (a) Isometric of Typical Assembly. 17 5/8" 15 5/8" L I a. Side View 5/8" 5 5/8" ? 5/S / ' 3/8^ (oj 00 CO m ro b. Front View (b) Geometry of Test Assembly. Figure 5*3 Test Assembly Used By Williams and Geschwinder For Collar Joint Tests (18) 20 30 C c c BBEK (4,2) =0, 0 BREK (5, 2) =-1.C*BHEK (2,2) BREK (6,2) = EBEK (3,2) BREK (3, 3) = (BR3EIL* (4.0+PRIBR)/ (3, 0* (1.0 + EiiIBS) ) ) - $ (2. 0 ER E E (1) IB/15.0) BREK (4, 3) =0.0 BREK (5,3)=-lQ*EREK(3,2) BREK (6, 3) = (BR3EIL* (2.0-PHIBR)/ (3. 0* (1.0+IHIBB) ) ) + $ (EBEE (1) +LN/30.0) BREK (4, 4) = BREK(1, 1) BREK (5,4) =0 .0 BREK (6, 4) =0. C BREK (5,5) = E BEK (2,2) BHEK(5,6)=-1.G*BEEK(3,2) BREK (6,5) =EEÂ£K (5,6) BREK (6, 6) =BHEK (3,3) DC 30 J= 1 ,EE EC 20 K=1,B2 BREK (J,K)=BEER (K, J) 10TEK (J, K,1) =BKKK (J,K) CONTINUE CONTINUE RETUEN END SUBROUTINE CJESM (CJEK ,TCTE K ,C JSH ES, C J f C MS ,L NH Â£E L Nil EX , $ I,CJ,NELEM) SUBROUTINE CJESM WILL CONSTRUCT TflE STIFFNESS MATRIX ECB EACH COLLAR JOINT ELEMENT. INTEGER CJ DGUEIE PRECISION CJEK,TGTEK,IN HER,IN UBI,CJSHRS,CJMOMS DIMENSION TOTEK(6,6, NELEM) ,CJEK (CJ,CJ) CJEK (1, 1)=CJSHES CJEK (2, 1) =CJSIifiS*INHBR CJEK (3,1)=-1.Q*CJÂ£K(1, 1) CJEK (J, 1)=CJSHBS*LNHBL 268 2 4.1C9D-3 1.186D+05 3 6.Â£ 16D-03 1,730+05 4 1.CD-02 2.58179D+05 COLLAR JCIKT ELEMENT P- DELTA CURVE 2 PT DELTA COCEDI NATE Â£ COORDINATE 1 0.0D+00 0.0D+00 2 4.C13D-2 S .777D+03 COLLAR JCINT ELEMENT H-THETA CURVE 2 PT THETA COORDINATE K COORDINATE 1 C. CD + 00 0.GD+00 2 1.0D+00 O.OD+OO BLOCK ELEMENT P-DELTA CURVE PT 1 2 3 4 5 3 5 PT 1 2 3 4 5 PI 1 2 3 DELTA COORDINATE P COORDINATE C.0D+00 4. CD-03 8.0D-3 1.2D-02 1.68D-02 0.0D+00 2.5125D+04 5,025D+04 7.5375D+04 9.315D+04 BLOCK ELEMENT M-THETA CURVES 2,5E+04 THETA COORDINATE M COORDINATE .0D+0 5.63D-04 8.46D-04 1.49D-03 3.5D-03 0.0D+00 2.3325D+04 3.5D+04 4.6675D+04 8.31 14D+04 5.OD + 04 THETA COORDINATE K COORDINATE C.CD+CO C.CD+00 1.035D-03 4.665D+04 1.674D-C3 7.D+04 407 85 k = MOMENT SPRING STIFFNESS FACTOR m Figure 4.11 Determination of Collar Joint Moment Spring Stiffness Factor VALUES [k] = Figure KNOWN: #t E, L, P, AE, 3EI L L LET ka = AE, kr = 3EI L L ka 0 0 -ka 0 0 0 4 x kr 6P -2 x kr A P 0 -4 x kr + 6P -2 x kr + P L2 (1 + $) 5L L( 1 + #) 10 L2 (1 + *) 5L L(1 + ) 10 0 -2 x kr P (4 + *) x kr 2PL 0 -2 x kr P (2 *) x kr + PL L(1 + 10 3(1 + *) 15 l(i + *) 10 3(1 + *) 30 -ka 0 0 ka 0 0 0 -4 x kr A 6P 2 x kr P 0 4 x kr 6P 2 x kr A P L2 (1 + *) 5L L(1 + $) 10 L2 (1 + *) 5L L(1 + *) 10 0 -2 x kr + P (2 $) x kr A PL 0 2 x kr P (4 + ) x kr 2PL L(1 + $) 10 3(1 + $) 30 L(1 + $) 10 3(1 + '*) 15 4.8 Brick Element Stiffness Matrix Formulation Using Experimentally Determined Axial and Rotational Stiffness Factors 306 Table B.11 Nomenclature For Subroutine PULMAT VARIABLE TYPE DEFINITION A DOUBLE PRECISION T x T MATRIX WHOSE VALUES ARE EXTRACTED FROM MATRIX [b] B DOUBLE PRECISION 6 x 6 x N MATRIX FROM WHICH MATRIX [a] IS EXTRACTED I INTEGER NUMBER OF 6 x 6 MATRIX IN MATRIX [b] FROM WHICH T x T MATRIX [a] IS EXTRACTED (MATRIX [A] IS EXTRACTED FROM MATRIX 6 x 6 x I IN MATRIX [b]) T INTEGER NUMBER OF ROWS AND NUMBER OF COLUMNS IN MATRIX [A] N INTEGER NUMBER OF 6 x 6 MATRICES IN MATRIX [b] 294 Table B.6 Nomenclature For Subroutine BMULT VARIABLE TYPE DEFINITION A DOUBLE PRECISION NxHI MATRIX WHICH CONTAINS THE UPPER TRIANGULAR PORTION OF A SYMMETRIC N x N MATRIX THAT HAS A BANDWIDTH OF (2 x M1) 1 B DOUBLE PRECISION N x 1 MATRIX WHOSE VALUES ARE MULTIPLIED BY THOSE IN MATRIX [a] TO OBTAIN MATRIX [c] C DOUBLE PRECISION N x 1 MATRIX WHOSE VALUES ARE THE PRODUCT OF [A] x [b] N INTEGER NUMBER OF ROWS IN MATRIX [a], ALSO NUMBER OF ROWS IN COLUMN VECTORS [b], [c] M1 INTEGER HALF THE BANDWIDTH (INCLUDING THE DIAGONAL) OF MATRIX [a] 361 Table B.34 Nomenclature For Subroutine CHKTOL VARIABLE TYPE DEFINITION A DOUBLE PRECISION N x 1 MATRIX N INTEGER NUMBER OF ROWS IN MATRIX [a] TOLER DOUBLE PRECISION NUMBER WHICH EACH VALUE IN MATRIX [a] IS COMPARED WITH KEY INTEGER VARIABLE WHICH IDENTIFIES WHETHER OR NOT EACH VALUE IN MATRIX [a] IS LESS THAN TOLER; 0 = EACH VALUE IS LESS THAN TOLER; 1 = AT LEAST ONE VALUE IS NOT LESS THAN TOLER 42 exist in the beam element. Taking the coordinate axes in the direction shown in Figure 3-9e, the boundary conditions for the element are and y = -w1 y' = w2 at x = 0 y = -w-j y' = w4 at x = 1 . Substitution of these conditions into Equation (5-41) makes it possible to evaluate the four arbitrary constants and to obtain the following expression for y: 3(w< w,) p 2wp + w4 p y = -w + w x + x - 2- x 1 2 ,2 1 + W4 x3 + 2(w3 ~ w1^ x3 (3.42) Equation (3*42) can be rewritten in matrix form as ',y2 or - 1) (x *Â£ + \(2^ . x? y = [a] [w] . W1 w2 w3 W (3.43) Differentiating the expression in (3.43) gives and y [c] W y" [d] [w] (3.44) (3.45) in which 35 and f 6y 61% . 5 (4 + $)1 (4 + $)1 (3.26) As before, the remaining forces acting on the element can be determined from the equations of equilibrium, i.e., Equations (3.12) and (3.13)* How at x = 0 dv-L dv b _dv_ s_ _ dx dx dx 6 (3.27) so that w = ffi (1 + *)1 Mrn (1 + *)1 6 El (4 + *) El (4 + i>) (3.28) Hence, from Equations (3.12), (3-13), (3*26), and (3-28) 6,6 \ vf r6 \ = (4 + $ )EI ) (1 + *)1 'T0 (3.29) f-U -6EI 26 \w6/ lw6 ) (1 + $)l2 ' 't=o 't=o (3.30) 3,6 lw T=0 + ^ w6 ) (2 0) El . (1 + *)l [=0 (3.31) If the deflection of the left-hand end of the beam is equal to zero, as shown in Figure 3.6d, it is evident from symmetry that k 3,3 k 6,6 (4 + $)EI (1 + *)1 (3-32) 351 (start) CONSTRUCT THE INDEX MATRIX FOR THE BLOCK ELEMENT (return) Figure B.29 Algorithm For Subroutine INDXBL 129 follows. For a given value of P, the end moment M, which equals P x e, was calculated for the different eccentricities in parts (a), (b), (c) and (d) of Figure 5.12. Using Equation (5*13), where L = 23.7 in t = 5.6 in e2, ei read from the curves, the end rotation 9 for the prism was calculated for each value of applied end moment M. This end rotation is for the 6x32x24 in prisms used in the tests. An equation similar to Equation (5.19) which is valid for calculating an equivalent end rotation for 6x24x8 in block prisms is necessary. This is developed in the same way as before. From Figure 5-9e and the first Moment Area Theorem, Equation (5.14) becomes 0 = Z 24 EI24 for the 24 in high prisms, and 4Ma 9 = 8 EIq for the 8 in high prisms. Expressing 9g in terms of 24 98 = x 924 or 8 = (x) _12M24 . ETg EI24 (5.20) (5.21) (5.22) 347 B.28 Subroutine INDXCJ Subroutine INDXCJ will construct the index matrix for each collar joint element. Table B.28 defines the nomenclature used in this subroutine. Figure B.28 is an algorithm for this subroutine. B.29 Subroutine IHDXBL Subroutine IHDXBL will construct the index matrix for each block element. Table B.29 defines the nomenclature used in this subroutine. Figure B.29 is an algorithm for this subroutine. B.30 Subroutine FORCES Subroutine FORCES reads and prints the number of degrees of freedom that are loaded, the initial load at each degree of freedom, the load increment at each degree of freedom, and the maximum load at each degree of freedom. Table B.30 defines the nomenclature used in this subroutine. Figure B.30 is an algorithm for this subroutine. B.31 Subroutine APPLYF Subroutine APPLYF will place the initial loads in the structure force matrix the first time it is called. Thereafter, it will increment each load according to the information read by subroutine FORCES from the data set. It will not increment any load beyond its specified maximum value. Table B.31 defines the nomenclature used in this subroutine. Figure B.31 is an algorithm for this subroutine. B.32 Subroutine PLATEK Subroutine PLATEK will construct a structure stiffness matrix which only considers the two structure degrees of freedom at the top of the wall that correspond to the test plate. It will do this by modifying the original structure stiffness matrix which considers four degrees of 339 Table B.24 Nomenclature For Subroutine BRESM VARIABLE TYPE DEFINITION BREK DOUBLE PRECISION BREF DOUBLE PRECISION TOTEK DOUBLE PRECISION BRAEL DOUBLE PRECISION BR3EIL DOUBLE PRECISION LN DOUBLE PRECISION ELASBR DOUBLE PRECISION ARSHBR DOUBLE PRECISION I INTEGER BR INTEGER NELEM INTEGER BRICK ELEMENT STIFFNESS MATRIX BRICK ELEMENT FORCE MATRIX MATRIX THAT STORES ALL OF THE ELEMENT STIFFNESS MATRICES BRICK AE/L (AXIAL STIFFNESS FACTOR) BRICK 3EI/L (ROTATIONAL STIFFNESS FACTOR) VERTICAL LENGTH OF THE BRICK ELEMENT MODULUS OF ELASTICITY OF THE BRICK AREA IN SHEAR FOR THE BRICK ELEMENT NUMBER NUMBER OF BRICK ELEMENT DEGREES OF FREEDOM NUMBER OF ELEMENTS 15 20 25 30 40 45 50 70 C C C IBB (4) = 1 IBE (Â£) =3 IBB J6) = 4 GC 1C 50 DC 20 J= 1,3 Â£=J + 3 IBB {J ) = IE F< (!) CONTINUE DC 25 K=4,6 IBB (K) =IBB (K) +5 CGK1INUE GO 10 50 IBB (1) = IBB (4) DO 4C M=2, IEB(K) = IÂ£ B(H)+5 CONTINUE IF (I,EC,NEBI) GC TO 45 GO 10 50 I EE (6) = IBB (5) IBB (5) = 0 DC 70 L=1,BE 101EIN (1, L) =IBB (L) CONTINUE BE10BN END 3UBE0T1NE INLXCJ {ICJ,TOTEIN,I,NEDO,CJ,NELEM) SUBROUTINE INDXCJ WILL CCNS1NUCT IBE INDEX EATBIX FOE EACH CCILAE JCINX ELEMENT, INTEGEB TG1EIN,CJ DICE NSION TCTEIN (N ELEM, 6) ICJ (4) IF (I. EQ.NEMO) GO TO 35 IF (I,NE,2) GO TC 5 ICJ {1) =1 ICJ (2) =4 ICJ (3) =2 198 respectively, for the brick wythe, and Figures 6.35 and 6.39 respectively, for the block wythe. Also, at Pmax> the block wythe vertical deflection increases nonlinearly above these heights for each example as shown in Figures 6.9 and 6.11, respectively. Examination of the program output revealed that at these heights and load levels in each example the block element axial force is very close to 75,375 lb and, in fact, above this height it exceeds that amount. At this value, a sharp decrease occurs in the block element axial stiffness as shown in Figure 5*7. A possible explanation of these occurrences is the following. In response to the compressive load applied at the top of each wall, the walls assume the deflected shapes shown in Figure 6.1 for example 2 and Figure 6.3 for example 4. The bending action which produces these shapes tends to put the outer, or brick, wythe of the wall in tension and the inner, or block, wythe of the wall in compression. Similarly, considering each wythe separately, the inner or right side of the brick wythe is subject to compression and the inner or left side of the block wythe is subject to tension. This creates a positive collar joint shear stress. When the block elements near the top of each wall suddenly decrease in axial stiffness, this causes the vertical deflection in the block wythe to increase rapidly at this location. This leads to a sudden reduction in shear stress in the collar joint, producing the discontinuities previously observed. Since all elements must remain in equilibrium, the shear or vertical forces in the collar joint element directly affect the end moments on the left and right ends of the collar joint element. These end moments are also the value by which the end moments due to the other two contributing 142 Ws 6 W57 Wx W4 W56 W5: w56 w 57 w. w 58 v> Wx f Wt " mVs V.> t w2 w a Wb /777 rrn EXTERNAL WALL LOADS EQUIVALENT WALL LOADS WALL DISPLACEMENTS Figure 5*20 Structure Degrees of Freedom For Example Number 1 336 B.23 Subroutine STIFAC Subroutine STIFAC will select the appropriate stiffness factors for an element based on the values of these factors as determined from the P-A and M-9 curves, the current level of element loading, and the previous level of element loading. Table B.23 defines the nomenclature used in this subroutine. Figure B.23 is an algorithm for this subroutine. B.24 Subroutine BRESM Subroutine BRESM will construct the stiffness matrix for each brick element considering shear deformation and moment magnification. Table B.24 defines the nomeclature used in this subroutine. Figure B.24 is an algorithm for this subroutine. B.25 Subroutine CJESM Subroutine CJESM will construct the stiffness matrix for each collar joint element. Table B.25 defines the nomenclature used in this subroutine. Figure B.25 is an algorithm for this subroutine. B.26 Subroutine BLESM Subroutine BLESM will construct the stiffness matrix for each block element considering shear deformation and moment magnification. Table B.26 defines the nomenclature used in this subroutine. Figure B.26 is an algorithm for this subroutine. B.27 Subroutine INDXBR Subroutine INDXBR will construct the index matrix for each brick element. Table B.27 defines the nomenclature used in this subroutine. Figure B.27 is an algorithm for this subroutine. BLIDP BLOCK INTERACTION DIAGRAM P COORDINATE DOUBLE PRECISION D13.6 17 105 - COMMENT OF 'MAXIMUM BRICK ELEMENT COMPRESSIVE LOAD CAPACITY' - - 1 106 BRMAXP BRICK MAXIMUM P DOUBLE PRECISION D13-6 1 107 - COMMENT OF 'MAXIMUM COLLAR JOINT ELEMENT SHEAR LOAD CAPACITY - - 1 108 CJMAXP COLLAR JOINT MAXIMUM P DOUBLE PRECISION D13-6 1 109 - COMMENT OF 'MAXIMUM BLOCK ELEMENT COMPRESSIVE LOAD CAPACITY - - 1 110 BLMAXP BLOCK MAXIMUM P DOUBLE PRECISION D13*6 1 111 - COMMENT OF 'STRUCTURE FORCE APPLICATION INFORMATION' - - 7 112 NOF NUMBER OF FORCES INTEGER 13 1 113 - COMMENT OF 'DOF' - - 1 - COMMENT OF 'INITIAL FORCE' - - 7 - COMMENT OF 'FORCE INCREMENT' - - 23 - COMMENT OF 'MAXIMUM FORCE - - 41 114 NDOF NUMBER OF THE DEGREE OF FREEDOM OF THE FORCE INTEGER 13 1 SFINIT INITIAL FORCE VALUE FOR THAT DEGREE OF FREEDOM DOUBLE PRECISION D13.6 4 63 Thus, Static Condensation uses Gauss Elimination as part of its procedure for solving simultaneous equations. Section 3*6.1 addressed the large storage and computational savings that result from using a modified Gauss Elimination technique which only uses half of the symmetric nonzero terms in the stiffness matrix. The use of Static Condensation in conjunction with the modified Gauss Elimination procedure was explored. This technique was found to be less efficient than just modified Gauss Elimination for small matrices, but for large matrices it was up to 11.5$ more efficient than even modified Gauss Elimination. Figure 3*16 compares the number of operations required by modified Static Condensation to the number required by modified Gauss Elimination for a structure stiffness matrix with a value of 9 for half the bandwidth. The percent savings (or loss) in computational efficiency that results from the use of modified Static Condensation is shown in Figure 3.17. Since both of these methods each require storage of half of the symmetric nonzero values in the stiffness matrix, the storage requirements of each technique are the same. 3.7 Solution Convergence In the standard use of the Direct Stiffness Method, convergence of the solution rarely presents a problem. However, with consideration of the special items discussed in sections 3*1 through 3.5 immediate convergence of the solution is not guaranteed. One method of monitoring the accuracy of the solution is to add a step to the procedure detailed at the end of section 2.1. After the force matrix for each element is calculated, an equilibrium check can be made by multiplying every element force matrix by -1.0 and inserting each one into the structure force matrix. The resulting values should IN-LB P = 25,000 LB 6, RADIANS Figure 5.13 Block Element M-Q Curves 390 $ 420 430 435 440 450 460 470 DC 530 K=3,NELEM, 3 IF (IIEB.NE.1) GO 10 390 CALL INEXBL (IBL,TOTEIN,K,EL,NELEM) GO TO 5CC CALL EULfiCW (IBL,TOTElN,K,BI,NELEM) CALL DULL (BLEW,BL,1) IF (K, EQ NELEM) CALL EXTRAE (STRW,BLÂ£W,2,3,NSDOF,BL , IBL) IF (K. EQ. NELEM) GOTO 420 IF (K.EQ.3) CALL EX1RAK (SIRW,BLEW,2,1,NSDCF,EL,I EL) IF(K. BÂ£.3) CALL EXTRAK (STBW,BLEW,2,2,NSDCF,BL,IBL) CALI NULL (BLEK,BL,BL) CAII 10IB AT (BLEK,TOTEK,K,BL,NELEM) CALL BOLT (BLEK,BLEW,BLEE,BL,BL,1) CCUNT-CCUNT+1 BLWLD(COUNT)=BLEF(1) IF (TBCONV. Â£C> 1) GO TO 525 IF (IRELPR.NE.1) GO TO 440 WRITE (6,430) K FORMAI(* 01/' 'FORCES FCR BLOCK ELEMENT NUMEEE',14) CALI PRINT (BLEF,EL) WRITE (6,4 35)K FORMAT (*0', DISPLACEMENTS FOR BLOCK ELEMENT NUMBER*, 15) CALL ERINT (BLEW,EL) CALL CUKFAI (BLEF NBLID E BLIDM BIIE Â£ ELM A XP 3, NST AT) IF (NSTAT.EQ.O) GO TO SCO IF (NSTAI,EQ.1) GO TO 470 WRITE (6,450) K FORMAT ( 0 1 WARNING*4'* BICCK ELEMENT SC. ',14, IX, 'IS IN AXIAL*) WRITE(6,460) FORMAT (' 14X, 'TENSION, CHECK FOR UNUSUAL LOADING') STOP IF (TRFAIL. IQ, 4) GO TO 480 TRFAIL=1 9.77D+C3 fiAXIflOH EICCK ELEEEKT COMPRESSIVE LOAD CAPACITY 9.315E+4 STRUCTURE FORCE APPLICATION INFCBKATICN 1 DOF INITIAL FORCE FORCE INCREMENT HAXIKUH 121 -4,0D+U3 -4.0D+03 -4.0D+05 INSTRUCTIONS ON PRINTING INTERMEDIATE' RESULTS: 1 1121 -4.8D+04 FORCE 404 367 (start) PRINT THE VERTICAL LOAD CARRIED BY EACH WALL WYTHE AND THE PERCENT 'OF THE TOTAL LOAD WHICH THAT REPRESENTS, (return) Figure B.36 Algorithm For Subroutine WYTHE CHAPTER FIVE NUMERICAL EXAMPLES 5.1 General Comments In this chapter, five numerical examples are presented to illustrate the use of the analytical model and show how the information it generates may be used. Since the results of the experimental phase of this project are not yet available, strength and deformation properties very similar to what might be expected from actual tests of each element type were extracted from similar tests done by Fattal and Cattaneo (4) and Williams and Geschwinder (18). This is dealt with in the next section. Once the experimental phase is complete, and the results of the actual wall tests are compared with those from the analytical model, it may be desirable to perform an extensive parametric analysis by systematically ranging the values of some variables, while keeping the others constant. This should give a good indication of the relative sensitivity of the variables that affect composite masonry wall behavior and isolate which factors are the most important. It would also provide much needed insight into composite masonry wall behavior, and would likely be a source for further experimental research. The current model is based on sound principles and has been developed to consider what are believed to be the most important effects that take place in composite walls. Nonetheless, until actual data are available, it would not be prudent to perform a parameter analysis, since the results might be 102 8.00 FT 6.25 IN II 1-3.75 IN P 4 IN H I16 IN TEST WALL Sign Convention H/T = 9.6 P P + AP AP 4000 LB e 1.25 IN M Pe 1.25P ^ m is 1 8 IN TYPICAL 2 IN /7T7fi77 -I M3 IN MODEL Figure 5*18 Example Number 1 59 KÂ¡a KÂ¡b Ka Kb W f: a = a Wb n Note that [Ka] [WQ] + [Kb] [wb] = [F] but because of the forward eliminations [Kb ] = 0 so this equation reduces to [Rbb] W = ^ (3-52) 3) Solve [K{J [Wb] = [F] for [wfc] . 4) Note that [Ka] [wj + [Kb] [wb] [F] and only [Wfl] is unknown, so solve ka) W G>] Kl>] KB <5-53) for [W ] by backward substitution. 5) Construct j^j _ In performing Static Condensation, best efficiency is obtained if: D [K] is partitioned into four equal quarters and [w] and [f], therefore, are each divided in half, and 2) Gauss Elimination is used to solve Equation (3.52). The numerical example of section 3*6 is solved below using Static Condensation. Recall that [k] [w] = [f] was written in matrix form as 137 5.3 Example Number 1 Finite Element Analysis of a Test Wall Figure 5*18 shows a composite masonry wall which was tested by Fattal and Cattaneo (4) and also analyzed using the finite element model. As shown in the figure, the wall is 8 feet high and 10 inches thick. Both were loaded at the same eccentricity. Two differences between the test wall and the model wall exist. The cross-section of the test wall is 10x31.6 in and the cross-section of the model wall is 10x24 in. The second difference is that the test wall was free to rotate at the top and the bottom, but the model wall is free to rotate at the top but fixed at the bottom. Both of these differences can be taken into account so the failure load of the actual wall and the failure load predicted by the model can be compared. Fattal and Cattaneo (4) tested two wall specimens with the characteristics just described and one failed at 170,000 lb and the other at 190,000 lb. Failure was characterized by vertical splitting in the webs of the block, due to compression, at the top or bottom three courses of the specimen. The analytical model yielded a wall failure load of 132,000 lb characterized by a compression failure in the block 4 in from the top of the wall. In order to make a valid comparison of the failure loads, the differences in cross-section and end conditions must be considered. Recall Equation (5.1), which was developed assuming equal vertical compressive stress and stated P X P. . D Aga A Redefine the variables as 524 (start) PRINT MATRIX [a] IDENTIFYING EACH ROW AS A DEGREE OF FREEDOM (return) Figure B.18 Algorithm For Subroutine PRINT 321 Table B.17 Nomenclature For Subroutine CURVES VARIABLE TYPE DEFINITION NOCURV INTEGER NUMBER OF CURVES NOPPC INTEGER NUMBER OF POINTS PER CURVE CURVAL DOUBLE PRECISION MATRIX WHICH STORES THE VALUE EACH CURVE IS IDENTIFIED BY XCOOR DOUBLE PRECISION MATRIX WHICH STORES THE X COORDINATE OF EACH CURVE; THE FIRST SUBSCRIPT IS THE CURVE NUMBER AND THE SECOND SUBSCRIPT THE POINT NUMBER YCOOR DOUBLE PRECISION MATRIX WHICH STORES THE Y COORDINATE OF EACH CURVE; THE FIRST SUBSCRIPT IS THE CURVE NUMBER AND THE SECOND SUBSCRIPT THE POINT NUMBER 10 w 4 w 3 w. (a) Basic Flexural Element (b) Application of Unit Displacements to Establish Stiffness Coefficients Figure 2.1 Development of Element Stiffness Matrix 94 expressed in matrix form as A16 1 1 1 2 Ap1 a17 1 - 2 0 1 9P1 919 0 1 ^ [<.T] (4.4) From this, a matrix [x^] can be constructed such that [old] [*T] [neJ (4.5) where [W0iJ = original displacement matrix which considers all 4 DOF at the wall top [xT] = transformation matrix used to transform the new displacement matrix with 2 DOF at the wall top, to the old displacement matrix with 4 DOF at the wall top [Wnew] = new displacement matrix which only considers 2 plate DOF at wall top. The matrix [x^] is shown below where this matrix equation is illustrated. 23 other words, it too is accounted for during the construction of the stiffness matrix for each element. As mentioned in section 2.2, the stiffness matrix for an element is constructed by applying unit displacements, one at a time, at each DOF. If a structure is modeled by more than one type of element, this only means that the coefficients and variables in the element stiffness matrix of each element type will be different. Once all of the values in an element stiffness matrix are calculated, the element stiffness matrix is inserted into the structure stiffness matrix in the same fashion as discussed previously in section 2.4. Because the structure stiffness matrix is assembled using the element stiffness matrices, the solution of structure displacements and element displacements and forces will reflect the presence of different types of elements in the model. Consider the frame with the structure degrees of freedom, element degrees of freedom, and property values shown in Figure 3*1 Notice that this frame is similar to the one in Figure 2.3, which was previously discussed, except that: 1) Elements 1 and 3 have different property values 2) Element 2 is of a different type than the previous element 2 and is different from the present elements 1 and 3 Assume it is desired to analyze this structure by the Direct Stiffness Method. From the preceding discussion, it was learned that the procedure is identical to the one outlined at the end of section 2.1, but that the stiffness matrix for each element will be different due to the presence of different materials and different element types. To illustrate, the stiffness matrix for each element will be constructed. 316 Table B.15 Nomenclature For Subroutine READ VARIABLE TYPE DEFINITION WHI INTEGER WALL HEIGHT IN INCHES LNVER DOUBLE PRECISION VERTICAL LENGTH OF THE BRICK ELEMENTS AND BLOCK LNHBR DOUBLE PRECISION HORIZONTAL LENGTH OF THE BRICK HALF OF THE COLLAR JOINT ELEMENT LNHBL DOUBLE PRECISION HORIZONTAL LENGTH OF THE BLOCK HALF OF THE COLLAR JOINT ELEMENT NSTORY INTEGER NUMBER OF STORIES OF WALL WHERE EACH STORY IS 8 INCHES TALL NSDOP INTEGER NUMBER OF STRUCTURE DEGREES OF FREEDOM NELEM INTEGER NUMBER OF ELEMENTS ELASBR DOUBLE PRECISION MODULUS OF ELASTICITY OF THE BRICK ARSHBR DOUBLE PRECISION AREA IN SHEAR FOR THE BRICK ELASBL DOUBLE PRECISION MODULUS OF ELASTICITY OF THE BLOCK ARSHBL DOUBLE PRECISION AREA IN SHEAR FOR THE BLOCK 307 I- I L Figure (start) FOR FIRST T x T VALUES IN 6 x 6 x MATRIX IN MATRIX [b] COPY VALUE INTO MATRIX [a] (return) B.11 Algorithm For Subroutine PULMAT 84 P = O Al P > O k = SHEAR SPRING STIFFNESS FACTOR s Figure 4.10 Experimental Determination of Collar Joint Shear Spring Stiffness Factor 2 4.1C9D-Q3 1.186D+05 3 6.616D-03 1.78D+05 4 1,0D-02 2,58179D+05 COLLAR JOINT ELEMENT P-DELTA CURVE 2 PT DELTA COORDINATE P COORDINATE 1 0.D+00 0.0D+00 2 4.613D-02 Â§.7 77D + 03 COLLAS JCLNT ELEMENT M-THETA CURVE PT 1 2 5 PT 1 2 3 4 5 3 5 PT 1 2 3 4 5 PT 1 2 3 THETA COORDINATE K COORDINATE 0.D+00 .0D + 00 1.D+00 .D+0 BLOCK ELEMENT P-DELTA CURVE DELTA COORDINATE P COORDINATE 0.0D+00 0.D+0 4.0D-03 8,OD-03 1.2D-02 1.68D-02 BLOCK ELEMENT 2.5125D+4 5.025D+04 7.5375D+04 9.3 15D+04 M-THETA CORVES 2.5D+04 THETA COORDINATE M COORDINATE 0.OD + OO 5.63D-04 8.46D-04 1.49D-03 3.5D-03 0,OD+OO 2.3325D+04 3.5D+04 4.6675D+04 8.31 14D+04 5.0D+Q4 THETA COORDINATE K COORDINATE C.OD+OO 0.OD + OO 1.035D-03 4.665D+04 1.674D-03 7.0D+04 402 88 of the stiffnesses, which depend on the load values, an unusual challenge exists. This problem is solved by temporarily reducing the four degrees of freedom at the top of the wall to the two which correspond to the test plate for load application purposes. By taking advantage of the force and displacement relationships between the two plate degrees of freedom and the four nodal degrees of freedom at the wall top, the model will solve for the equivalent four nodal forces that the plate degrees of freedom produce, as well as the structure displacements and element displacements and forces considering all four degrees of freedom at the top of the wall. The technique by which this is done is shown below. Figure 4.12a shows the four nodal degrees of freedom at the top of a wall. The wall wythe axial loads and moments which transmit the vertical load and moment from the test plate to the wall wythes are shown in Figure 4.12b. A freebody diagram of the test plate is illustrated in Figure 4.12c. From equilibrium (ZF=0 and ZM=0), PV = P16 + P17 Â¥ = M18 + Miq + P17 ~ P16 (1) . 2 These force relationships between the two plate degrees of freedom and the four nodal degrees of freedom at the wall top can be expressed in matrix form as Table 4.1 Variables Used in Constructing Brick Element Stiffness Matrix VARIABLE DEFINITION SOURCE 4 FACTOR USED IN ACCOUNTING FOR SHEAR DEFORMATION CALCULATED, $ = 1 2EI GA L2 s E BRICK MODULUS OF ELASTICITY TABATABAI (l5), E = 2,918x106 PSI (ONLY USED FOR SHEAR DEFORMATION) I MOMENT OF INERTIA OF BRICK ELEMENT CROSS-SECTION NOT USED DIRECTLY; 3EI/L FACTOR FROM TESTS USED G BRICK SHEAR MODULUS CALCULATED, G = .E, 2(1+v) V BRICK POISSON'S RATIO GRIM (5), v = 0.15 As BRICK AREA IN SHEAR CALCULATED, As = .84 Anet A = 17-19 IN2 b Anet BRICK NET AREA MEASURED, Anet = 20.47 IN2 L BRICK ELEMENT LENGTH LENGTH DEFINED BY MODEL = 8 IN A AREA OF BRICK ELEMENT CROSS-SECTION NOT USED DIRECTLY; AE/L FACTOR FROM TESTS USED P BRICK ELEMENT AXIAL FORCE ERICK ELEMENT FORCE MATRIX DEGREE OF FREEDOM NUMBER 1; CALCULATED BY PROGRAM FOR EACH LOAD LEVEL DCUEIE PRECISION A,B,C DIMENSION A (N1,N2) ,B (N2 ,N3) #C (N1 ,K3) DC 10 1=1,N1 EC 10 J= 1, N 3 C <1, J)=0,0 DC 10 K=1,N2 10 C(I,J)=C(1,J) + A(I,KJ*B(K,J) RETURN END C SUBROUTINE SMUI1 (A,B,C,N,M) C SUBROUTINE SMDLT WILL MU1TIPIY A SCAIAE A TIBES AN tiXM C MATRIX Â£Ej TO OBTAIN AN NX H MATRIX C. DOUBLE PRECISION A,B,C DIMENSION E (N,M) ,C (N,H) DO 1CC 1=1,N EC 100 J=1,H 100 C(1,J)=A*B(I,J) RETURN ENE C SUBROUTINE BMUIT (A,B,C,N,M1) C SUBROUTINE EMULT RILL MULTIPLY A SYMMETRIC BANDED C MATRIX EY A VECTOR TO OBTAIN ANCTBER VECTCE. IT WILL C I ERF CRM Â£ A ]*Â£ E ]=Â£C ], Â£A] IS AN NXM 1 MATRIX WHICH C CCNTAINS THE UPPER TRIANGULAR PORTION CF A SYMMETRIC C NX N MATRIX THAT HAS A BANDNIDIfc OF (2*M1)-1. Â£B] IS AN C NX 1 MATRIX. Â£ C] IS AN NX1 MATRIX KUCSE VALUES ARE THE C PRODUCT Cf Â£ A ]*Â£ B J. DOUBLE PRECISION A,B,C DIMENSION A (N,B1) ,U(N) C(N) N E Â£ G = 1 ICC 1= M1 N5T0P = N-MH-1 DC 10 1=1, N C (I) =0.CDfCC ro VJI 147 5 IN HH5 IN P t fN rnfTnw 4 IN III 6 IN Sign Convention H/T = 20 P P + AP P = 4000 LB e = 0 IN M = Pe = 0 mis /777/777 2 INHII 3 IN J8 IN TYPICAL TEST WALL MODEL Figure 5*24 Example Number 4 167 Figure 6.15 Plate Load Versus End Rotation For Example Number 4 C Â£C]. [C] CONTAINS THE UPEEB TRIANGULAI PORTION OF AN C NXN NATHI X TH 21 HAS A BANDWIDTH OF (2*H1)-1. THE ACTUAL C NXN MATRIX IS NOT STORED. If IXPE=1, THE FIRST 3 C NUMBERS IN THE [INDEXj MATBIX ABE NOT USED. IF TYFE=2, C ALL THE NUMBEBS IN THE Â£ INDEX ] MATBIX ABE USED. IF C IÂ¥PE=3, THEN THE FIFTH NUMBER IN THE [INDEX] MATBIX IS C SCI USEE. INTEGER TYPE DIMENSION C (N,M1) B (M, M) INDEX (M) DOUELE PRECISION C,B 1=1 IF (TYPE.EQ. 1) L = 4 DO 20 1=1,M II = INDEX (I) IF (I. EC. 5) GC TO 1 GO TO 4 1 IE (TYPE.EQ.3) GO TO 2 4 DO 10 J=L#M *3 J=INDEX (J) IF (J.EQ.5) GO TO 6 GC TO 8 6 IF(1YPE.EQ.3) GO TO 10 8 IF(JJ.LI.II) GO TO 10 JJJ=JJ-II+1 C (II,JJJ)=C (II,JJJ) +B(I,J) 10 CONTINUE 20 CONTINUE fiElUEN END C SUBECUTINE EXTBAK (A,B,KEY,TYPE,N,H,INDEX) C SUBBOUTINE EXTBAK HILL PICK UP A MATRIX [B] CUT OF A C IAEGEE MATBIX Â£ A ]. THE VALUES OF [ A ] ABE NOT AFFECTED. C THE OLD VALUES OF [B], If ANY, ABE BEELACEE EY THE C VALUES FOUND IN Â£ AJ. IF KEÂ¥=1, THEN Â£A] IS AN NXN C MATBIX AND Â£B] IS AN MXM MATBIX. IF KEY=2, THEN [A ] IS CHAPTER ONE INTRODUCTION 1.1 Background A composite masonry wall is a wall which consists of a clay brick wythe, a concrete block wythe, and a collar joint which forms a bond between the two wythes. A section of a typical composite masonry wall is shown in Figure 1.1. The collar joint between the two wythes consists of either masonry mortar or concrete grout. It forces the wall to behave as one structural unit, even though the wall consists of different materials. Composite masonry walls are frequently used as exterior bearing walls with the brick exposed as an architectural surface and the concrete block used as the load-bearing material. Thus, when a floor slab or roof truss bears on a composite masonry wall, it transfers vertical load directly to the block wythe. A typical composite masonry wall load-bearing detail is shown in Figure 1.2. Two design standards have been widely used in the United States as references for the design of engineered masonry construction. These are the Brick Institute of America (BIA) "Building Code Requirements for Engineered Brick Masonry" (2) and the National Concrete Masonry Association (NCMA) "Specifications for the Design and Construction of Load-Bearing Concrete Masonry" (14). A third standard on concrete masonry was recently published by the American Concrete Institute (ACl) entitled ACI 531-79R, "Building Code Requirements for Concrete Masonry 1 292 nomenclature used in this subroutine. Figure B.5 is an algorithm for this subroutine. B.6 Subroutine BMULT Subroutine BMULT will multiply a symmetric banded matrix by a vector to obtain another vector. It will perform the operation [a] x [b] = [c], where [a] is an n x ml matrix which contains the upper triangular portion of a symmetric n x n matrix that has a bandwidth of (2 x ml) 1 and ml equals half the bandwidth (including the diagonal) of matrix [a]. Matrix [b] is an n x 1 matrix whose values are multiplied by those in matrix [a] to obtain the n x 1 matrix [c]. Table B.6 defines the nomenclature used in this subroutine. Figure B.6 is an algorithm for this subroutine. B.7 Subroutine INSERT Subroutine INSERT will insert a matrix [b] into a larger matrix [a]. In other words, it will add the values of [b] to certain values in [a]. If KEY equals 1, then [a] is an n x n matrix and [b] is an m x m matrix. If KEY equals 2, then [a] is an n x 1 matrix and [b] is an m x 1 matrix. The m x 1 matrix [INDEX] contains the numbers which identify the positions in [a] to which the appropriate values in [b] should be added. If TYPE equals 2, then all the numbers in the [INDEX] matrix are used. If TYPE equals 3, then the fifth number in the [INDEX] matrix is not used. Table B.7 defines the nomenclature used in this subroutine. Figure B.7 is an algorithm for this subroutine. B.8 Subroutine BNSERT Subroutine BNSERT will insert the upper triangular portion of an m x m symmetric matrix [b] into a matrix [c]. Matrix [c] contains the upper triangular portion of an n x n matrix that has a bandwidth of 47 Consider the nonlinear load-deformation curve in Figure 3*12 which is approximated by three straight line segments. Notice that three modulus of elasticity (E) values exist, and each is valid only over a certain region of load. Assume it is desired to load a structure to a value in load region 3 First of all, the load would be divided into increments. This is necessary since the solution to the application of one load increment will affect the response to the next load increment, and so forth. How, apply the load to the structure in increments. After the application of each load step, go through the entire process of constructing the stiffness matrix for each element, constructing the structure stiffness matrix, solving for structure displacements, and obtaining element forces and displacements. To decide what value of E to use in constructing the stiffness matrix of an element, determine which load region the element forces fall in, based on the solution to the previous load increment. Since, in this fashion, the modulus of elasticity value is indeed related to the load each element experiences, the solution for the analysis of the structure will reflect the true load-deformation properties of the material from which it is made. Two additional points should be considered. First of all, when an element changes from one load region to the other, say from region 1 to region 2 in Figure 3.12, its modulus of elasticity will decrease from a value of E^ to a value of EÂ£. This means the stiffness of this element has decreased. Loads in the structure are resisted by the elements in accordance with their stiffnesses such that stiffer elements resist a larger part of the load and, therefore, develop larger element forces. After the application of the next load increment, since the modulus of elasticity for this element has gone down from E^ to E2 this element CHAPTER THREE SPECIAL CONSIDERATIONS 31 Different Materials The stiffness matrix for an element is dependent on the geometric, cross-sectional, and material properties of that element. This is evidenced by the nature of the variables in the element stiffness matrix shown in Figure 2.2. A structure which is comprised of different materials can be analyzed by the Direct Stiffness Method. The presence of different materials is accounted for by using the appropriate material property values when constructing the stiffness matrix for each element. Since the structure stiffness matrix is assembled using the stiffness matrix for each element, the solution for the structure will then reflect the presence of material differences among the different elements into which the structure is divided. In short, the presence of different materials in a structure is taken into account during the construction of the stiffness matrix for each element in the structure. 3.2 Different Types of Elements Occasionally, the accurate matrix analysis of a structure involves the use of more than one type of element in the analytical model. This presents no particular difficulty and, in fact, is handled in much the same way as the presence of different materials in the structure. In 165 Figure 6.13 Plate Load Versus End Rotation For Example Number 2 314 Table B.14 Nomenclature For Subroutine TITLE VARIABLE TYPE DEFINITION ALPHA REAL VARIABLE CHARACTER HEADINGS WHICH READS ALPHANUMERIC OR DATA USED TO READ AND PRINT WALL HEIGHT, INCHES Figure 6.9 Block Wythe Vertical Deflection Versus Height For Example Number 2 296 Table B.7 Nomenclature For Subroutine INSERT VARIABLE TYPE DEFINITION A DOUBLE PRECISION MATRIX INTO WHICH MATRIX [b] IS INSERTED; IF KEY = 1, THEN [a] IS AN N x N MATRIX AND IF KEY = 2, THEN [a] IS AN N x 1 MATRIX B DOUBLE PRECISION MATRIX WHICH IS INSERTED INTO MATRIX [A]; IF KEY = 1, THEN [b] IS AN M x M MATRIX AND IF KEY = 2, THEN [b] IS AN M x 1 MATRIX KEY INTEGER VARIABLE THAT KEEPS TRACK OF DIMENSIONS OF MATRICES [a], [b] TYPE INTEGER VARIABLE THAT KEEPS TRACK OF WHICH NUMBERS IN THE INDEX MATRIX ARE USED; IF TYPE = 1, THEN THE FIRST THREE NUMBERS IN THE INDEX MATRIX ARE NOT USED; IF TYPE = 2, THEN ALL THE NUMBERS IN THE INDEX MATRIX ARE USED; IF TYPE = 3, THEN THE FIFTH NUMBER IN THE INDEX MATRIX IS NOT USED N INTEGER NUMBER OF ROWS (AND POSSIBLY NUMBER OF COLUMNS) IN MATRIX [a] M INTEGER NUMBER OF ROWS (AND POSSIBLY NUMBER OF COLUMNS) IN MATRIX [b] INDEX INTEGER M x 1 INDEX MATRIX WHOSE VALUES GIVE THE POSITIONS IN MATRIX [a] INTO WHICH THE VALUES IN MATRIX [b] ARE INSERTED 61 Note that [Ka] = O Solve [Kb] [wb] = [F] for [wb] using Gauss Elimination. 9/5 -7 -40 (^/9 x row 3') + row 4 -7 5 0 gives 9/5 -7 W3 40 0 -200/9 _W4_ -1400/9 From backward substitution W4 = 7 and = 5 therefore Ob] 5 7 Multiplication of K] gives 5 7 0 -6 0 -30 212 Figure A.1-continued x WALL HEIGHT, INCHES 187 Figure 6.34 Brick Wythe Moment Versus Height For Example Number 3 Page 2 of 2 Internet Distribution Consent Agreement In reference to the following dissertation: AUTHOR: Boulton, George TITLE: Finite element model for composite masonry walls / (record number: 487104) PUBLICATION DATE: 1984 I, %, as copyright holder for the aforementioned dissertation, hereby grant specific and limited archive and distribution rights to the Board of Trustees of the University of Florida and its agents. 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Signature of Copyright Holder Georgiy, feoU H~gw Printed or Typed Name of Copyright Holder/Licensee Personal information blurred e-2(-zoe> Date of Signature Please print, sign and return to: Cathleen Martyniak UF Dissertation Project Preservation Department University of Florida Libraries P.O.Box 117007 Gainesville, FL 32611-7007 6/21/2008 Table A.2-continued SUBROUTINE DESCRIPTION BRESM CONSTRUCTS BRICK ELEMENT STIFFNESS MATRIX BRPDMT CALCULATES BRICK AXIAL AND ROTATIONAL STIFFNESS FACTORS CHKFAI CHECKS AN ELEMENT FOR FAILURE CHKTOL CHECKS TO SEE IF ALL THE VALUES IN [ERROR] ARE LESS THAN THE VALUE OF TOLER WHICH IS THE TOLERANCE CJESM CONSTRUCTS COLLAR JOINT ELEMENT STIFFNESS MATRIX CJPDMT CALCULATES COLLAR JOINT AXIAL AND ROTATIONAL STIFFNESS FACTORS COORD READS THE COORDINATES OF A CURVE CURVES READS THE COORDINATES OF UP TO TWENTY CURVES DISPLA CONVERTS THE STRUCTURE DISPLACEMENT MATRIX WITH PLATE DEGREES OF FREEDOM TO THE REGULAR DISPLACEMENT MATRIX EQUAL MAKES TWO MATRICES EQUAL BY COPYING THE VALUES OF THE FIRST MATRIX INTO THE SECOND MATRIX EXTRAK EXTRACTS A MATRIX OUT OF ANOTHER MATRIX FORCES READS AND STORES THE STRUCTURE FORCE APPLICATION INFORMATION GAUSS1 SOLVES A MATRIX EQUATION OF THE FORM [a] [b] = [c] USING GAUSS ELIMINATION WHERE [a] IS A SYMMETRIC BANDED MATRIX INDXBL CONSTRUCTS THE INDEX MATRIX FOR A BLOCK ELEMENT INDXBR CONSTRUCTS THE INDEX MATRIX FOR A BRICK ELEMENT 225 10.00 FT 146 7 IM Wi I3 IN P rN ni'nin// 4 IN H 6 IN TEST WALL Sign Convention H/T = 12 P = P + P P = 4000 LB e = 2 IN M = Pe = 2P mi*. I 8 IN TYPICAL /mnv 2 IN nI 3 IN MODEL Figure 5*23 Example Number 3 BRPCOR BRICK P COORDINATE DOUBLE PRECISION D13.6 18 - COMMENT OF 'BRICK ELEMENT M-THETA CURVES - - 19 NOBRPC NUMBER OF BRICK M-THETA CURVES INTEGER 13 20 NBRMTP NUMBER OF POINTS IN EACH BRICK M-THETA CURVE INTEGER 13 21 BRPCV VALUE OF P WHICH IDENTIFIES THE BRICK M-THETA THAT FOLLOWS CURVE DOUBLE PRECISION D13.6 22 - COMMENT OF 'PT' - - - COMMENT OF THETA COORDINATE' - - - COMMENT OF 'M COORDINATE - - 23-26 K POINT NUMBER INTEGER 13 BRTCOR BRICK THETA COORDINATE DOUBLE PRECISION D13.6 BRMCOR BRICK MOMENT COORDINATE DOUBLE PRECISION D13.6 27 BRPCV VALUE OF P WHICH IDENTIFIES THE BRICK M-THETA THAT FOLLOWS CURVE DOUBLE PRECISION D13.6 28 - COMMENT OF PT - - COMMENT OF 'THETA COORDINATE __ COMMENT OP 'M COORDINATE INDXCJ CONSTRUCTS THE INDEX MATRIX FOR A COLLAR JOINT ELEMENT INSERT INSERTS A MATRIX INTO A SECOND ONE MULT MULTIPLIES TWO MATRICES NULL SETS ALL VALUES OF A MATRIX EQUAL TO ZERO PLATEK TRANSFORMS A REGULAR STRUCTURE STIFFNESS MATRIX INTO A STRUCTURE STIFFNESS MATRIX THAT CONSIDERS PLATE DEGREES OF FREEDOM PRINTS A MATRIX AND IDENTIFIES THE ROWS AS DOF PULMAT PULLS A MATRIX OUT OF ANOTHER MATRIX PULROW PULLS A ROW OUT OF A MATRIX READ READS PROBLEM DESCRIPTION DATA AND PRINTS IT SMULT MULTIPLES A MATRIX BY A SCALAR STACON SOLVES A MATRIX EQUATION OF THE FORM [a] [b] = [c] USING GAUSS ELIMINATION AND STATIC CONDENSATION WHERE [a] IS A SYMMETRIC BANDED MATRIX STIFAC SELECTS AND STORES THE PROPER STIFFNESS FACTORS AND DETECTS ANY CHANGES IN THE STIFFNESS FACTORS FOR AN ELEMENT TITLE READS AND PRINTS A COMMENT CARD WRITE PRINTS A MATRIX AND IDENTIFIES THE DIMENSIONS AND ROWS WYTHE CALCULATES THE VERTICAL LOAD FOR EACH WYTHE AND PRINTS IT 226 25 JO 40 45 50 70 C c c c c c c c 10 15 IEL(K) = IBL(K)+5 CONTINUE GC 1C 50 IBL (1) =IBL (4) DC 40 M=2,6 IBL(M)=IBL (M)+5 CCK1INUE IF(I.EQ.NELEM) GO 10 45 GC 1C 50 IBL (E) =0 IEI (6) = IBL (6) -1 DO 70 L=1#BL 1G1EIN (I,L)=IBL (L) CONTINUE RE1UEN ENE SUEBGOTINE FORCES (NGF,NDOE,SFINI1,5FINCR,SFMAX,NSECF) SUBROUTINE FOECES READS AND PHIH1S 1BF NUMBER OF 3 DEGREES OF FREEDOM 1HAT ARE LCADED, 1 HE INITIAL LOAD AT ^ EACH DCF, THE INCREMENTAL LOAD AT EACH DOF (AMOUN3 BY WHICH LOAD Al 1HAT DOF IS TO BE INCRE EE NT EE) AND THE MAXIMUM LGA E AT EACH DOF. IF NO MAXIMUM LOAD IS DESIRED AT A DOF, PUT A LARGE NUMBER IN THE EA1A SET, SAY 1. 0E + 2, DOUBLE PRECISION SFINIT,SFINCR,SFMAX DIMENSION SFINIT (N5DOF) SFINCE { NS DCF) SFM A X (N SDOF) DIMENSION NDOF(NSDCF) CAII TITLE READ (5, 10) NCF FORMAT (13) WRUE (6,15)NOP FORMAT ('O' NC, GF FORCES ON STRUCTURE =',I4) CALL TITLE DC 30 I=1,NCF READ (5, 20) NDOF (1) ,SFINI1 (I) ,SFINCR (I) ,SFHAX (I) 302 r~ i l l Figure (start) FOR EACH NUMBER IN THE INDEX MATRIX) EXTRACT FROM THE PROPER LOCATION IN MATRIX [A] THE VALUE FOR MATRIX [b] (return) .9 Algorithm For Subroutine EXTRAK H SI Afil=lH1M 1M3 + 1 NSDIi4=KSDGT-4 NS1CE=8 DO 40 K=NSTABl#NSDfl4 DC 30 L=1,flSTOÂ£ NSTBK (K,L) =SIBK (K ,L) 30 CCKTINE 11 Â£10 P=N STOP- 1 40 CGSIIKUE NSTBP1=BSIAHT+1 NSI6 E2= N3T ABT + 2 NSlBE3=NSlABT+3 NHIE= 10 WHICH 1=9 IC2= (OHBE + iWHI:I)/2.0 MLC2=-1.0*L02 NSTBK ( NSTAfiT, rfIDfi 1) =SXBK (NSIABT, 9) 11S1BK (N STAB! N KID) =JSL02* SXRK (NSlABIf9) NHIE=NHID-1 NWICM 1=NHIDH1-1 NSIEK (NSTBP1,NWIE1 1) =ST BK ( N SIS P 1, 8) + SIKK (NSTB.P 1,9) NSIBK (WSTBP1#NHID) =HL02 *STfi K (ESTE E 1,8) +LC2*STfiK (NSTRPl,y) Â£UIE=KUID-1 WHICH 1=NHIDH1-1 BS1EK (NSIRE2, MIDE 1) =STRK (NSIRP2,7) +SIBK (NSTBP2,8) NS1RK (USURP 2, MUID) =HL02*STRK (BSTBÂ£2,7 ) +102*STBK 11 UIC = NHID 1 NICn=NWIDHl-1 H C C L = 6 DC 60 I=NSTBE3,RSÂ£M4 RCQLE 1 = 11C0L+ 1 KCCLÂ£2= KCCI + 2 NCCLP3=RCOI+3 KSIBK (I, MICM 1) =STRK (I,NCC1) +51RK (I, NCCIP 1) NS1RK (I,MID) =KL02*STRK (I ,NCOL) +L02+STBK (I, KCCLE1) 9.777C+03 MAXIMUM BICCK ELEMENT COMPRESSIVE LOAD CAPACITY 9,31513 + 04 STRUCTURE FORCE APPLICATION INFORMATICS 1 DOF INITIAL FORCE FORCE INCREMENT MAXIMUM 71 -4.0D+03 -4,0D+U3 -4.0D+05 INSTRUCTIONS ON PRINTING INTERMEDIATE RESULTS: 1 1 71 -5.6D+04 FORCE VjJ -P WALL HEIGHT, INCHES 186 MOMENT, INCH-POUNDS Figure 6.33 Brick Wythe Moment Versus Height For Example Number 2 52 2 10 0 10 0 5-60 -10 0-6 9-7 -28 0 0-75 0 then 2 1 0 0~ ~10~ 0 5-60 -10 (6/5 x row 2') + row 3 0-6 9-7 -28 0 0-75 0 yields 2 10 0 10 0 5-60 -10 0 0 9/5 _7 -40 I ITv t- 1 o 0 and finally 2 10 0 10 0 5-60 -10 0 0 9/5 -7 -40 x row 3') + row 4 0 0-75 0 produces CHAPTER FOUR USING THE MODEL TO ANALYZE COMPOSITE MASONRY WALLS 4.1 Structural Idealization of Wall The analytical model considers the lowest story of a composite wall, the same portion that will be represented by the laboratory test walls. The wall is divided into a series of three types of elements, one for the brick, one for the block, and one for the collar joint. Each element extends the entire depth of the wall, which for the test walls will be 24 inches. Figure 4.1 shows a wall divided into these elements. As shown, the lowest nodes of both wythes are fixed to the foundation and the top nodes are restrained from lateral displacement as they will be in the laboratory test fixture. Thus, the wall is modeled as a frame with two lines of columns connected at 8 inch intervals by shear beams with rigid ends. Figure 4.2 shows the structure and element degrees of freedom used in the model. The manner of numbering the structure degrees of freedom follows the pattern shown regardless of wall height. Of course, the element degrees of freedom for each element of a given type are as shown. For the highest test wall, which will be 26 feet in height, there are 194 structure degrees of freedom and 117 elements, 59 of each type. 68 145 5.4.2 Example Number 5 Example number 5 considers a wall 10 feet or 120 inches high, 10 inches thick, and 24 inches deep. The plate axial load is applied in increments of 4000 lb at 2 inches of eccentricity toward the block wythe, up to failure. The wall of example 5 is shown in Figure 5.23. The data file for this example is given in Appendix D.3. 5.4.3 Example Number 4 Example number 4 considers a wall 16.67 feet or 200 inches high, 10 inches thick, and 24 inches deep. The plate axial load is applied in increments of 4000 lb at 0 eccentricity up to failure. Figure 5.24 shows the wall of example 4. The degrees of freedom for external wall loads, equivalent-wall loads, and wall displacements for examples 4 and 5 are shown in Figure 5.25. The data deck for this example is given in Appendix D.4. 5.4.4 Example Humber 5 Example number 5 considers a wall with a height of 16.67 feet or 200 inches, a thickness of 10 inches, and a depth of 24 inches. The plate axial load is applied in increments of 4000 lb at 2 inches of eccentricity toward the block wythe up to failure. Figure 5.26 illustrates the wall of example 5. The data file for this example is provided in Appendix D.5. 7C BRICK ELEMENT W/ W 16 W 17 K w w, 18 ^:%rtr-Wl3 W14 W15 wfi i W, oj a M /777 /V77 \ 19 12 W- WÂ£ 10 r w Wo w, V-r Wo * d w6 COLLAR JOINT ELEMENT Cf Wr U- W, W- w. w. w 5 BLOCK ELEMENT Figure 4.2 Structure and Element Degrees of Freedom 26 [k], - [k] 3 ~ A1E1 0 0 'A1E1 0 0 L L 0 12E11! -6E1I1 0 -12E1I1 -6E1I1 L2 l? L2 0 -6E,I, 4E111 0 6E1I1 2ElIl L2 L L2 L -A1E1 0 0 A1E1 0 0 L L 0 -1 2E111 6E,I, 0 12E1I1 6E,I, L3 L2 L2 0 -6E,I, 2E111 0 6E1I1 4EiIi L2 L L2 L (a) Element Stiffness Matrix For Element 1 A 2^*2 0 0 "A 2^2 0 0 L L 0 12E2I2 -6E2I2 0 -12E2I2 -6E2I2 l? L2 L2 0 6E2I2 4E2I2 0 6E2I2 2E2J2 L2 L L2 L A 2 0 0 2^2 0 0 L L 0 -12E2I2 6E2I2 0 12E2I2 6E2I2 1? L2 L2 0 -6E2I2 2E2^2 0 6E2I2 4E2I2 L2 L L2 L (b) Element Stiffness Matrix For Element 3 Figure 3*2 Element Stiffness Matrices For Elements 1 and 3 in Figure 3.1 76 end. It is identical to the brick element which is the same as the basic flexural element previously illustrated and discussed in section 2.2. The effects of shear deformation and moment magnification will also be considered in the block wythe of the model according to the procedures outlined in sections 3.3 and 3*4. This means that the stiffness matrix for a block element, like the stiffness matrix for a brick element, will also take the form shown in Figure 3.11* 4.3 Experimental Determination of Material Properties As mentioned earlier, experimental tests will be done to establish the material properties of each type of element. 4.3.1 Brick Brick prisms, like those shown previously in Figure 4.3, will be tested experimentally to obtain the values of some of the variables that appear in the terms of the brick element stiffness matrix. Two types of tests will be performed to determine strength and deformation properties for the brick element. In the first type of test, prisms will be loaded axially to failure and the axial deformation noted for each level of load. This will establish the relationship between axial load and axial deformation. The axial load will then be plotted against the axial deformation to obtain a curve. This curve will be approximated in a piecewise-linear fashion, i.e., divided into a series of straight line segments. The slope of each straight line segment will be equivalent to the axial stiffness factor AE/L for the region of load values established by the load coordinates of the end points of each line segment. This is shown in Figure 4.6. WALL HEIGHT, INCHES 154 Figure 6.3 Lateral Wall Deflection Versus Height For Example Number 4 44 Equation (3*51) gives the stiffness matrix of a beam column element with the 4 DOF shown in Figure 3*9c. The matrix consists of two parts: the first is the conventional stiffness matrix of a flexural element and the second is a matrix representing the effect of axial loading on the bending stiffness. Figure 3*10 shows the stiffness matrix for the 6 DOF element shown in Figure 2.1a and Figure 3*9a considering moment magnification. Figure 3.11 shows the stiffness matrix for the same element' considering both shear deformation and moment magnification. 3.3 Material Nonlinearity Sometimes it is necessary to analyze a structure which is made up of a material whose load-deformation response is nonlinear. In such a material, the modulus of elasticity will vary and will be a function of the level of loading which the material is subjected to. A stiffness analysis of a structure composed of a nonlinear material can be performed if provision is made for the variation in elastic modulus. This can be done as follows. Recall that the modulus of elasticity is one of the variables required to construct the stiffness matrix for each element in the structure. As mentioned above, a nonlinear material's modulus of elasticity depends on the level the material is loaded to. To properly account for nonlinearity, three things are done. 1) Modulus of elasticity values are made dependent on load level. 2) The load is applied to the structure in increments up to the load for which a solution is desired. 3) The modulus of elasticity value used in constructing the stiffness matrix of an element is based on the element forces resulting from the application of the previous load increment. [k] AE L 0 0 -AE L 0 0 0 1 2EI 6P -6EI + P 0 -12EI + 6P -6EI A P L3 (1 + $) 5L L2 (1 + *) 10 L3 (1 + *) 5L L2 (1 + $) 10 0 -6EI + P (4 + $) El 2PL 0 6EI P (2 $) El + PL L2 (1 + *) 10 L (1 + *) 15 L2 (1 + <*>) 10 L (1 + 30 -AE 0 0 AE 0 0 L L 0 -12EI + 6P 6EI P 0 12EI 6P 6EI P L3 (1 + $) 5L L2 (1 + $) 10 L3 (1 + $) 5L L2 (1 + *) 10 0 -6EI + P (2 *) El + PL 0 6EI P (4 + *) El 2PL L2 (1 + $) 10 L (1 + *) 30 L2 (1 + *) 10 lT+T)~ 15 WHERE: A E L I $ G v As p = AREA OF ELEMENT CROSS-SECTION = MODULUS OF ELASTICITY = ELEMENT LENGTH = MOMENT OF INERTIA OF ELEMENT CROSS-SECTION = FACTOR USED IN ACCOUNTING FOR SHEAR DEFORMATION = -l2!! GAL2 = SHEAR MODULUS = ... E . 2(1+v) = POISSONS RATIO = AREA IN SHEAR = .84 x (NET AREA) = ELEMENT AXIAL FORCE Figure 3.11 Element Stiffness Matrix Considering Shear Deformation and Moment Magnification 20 30 40 45 50 60 70 75 80 90 100 105 READ (5,20) SOI FCHEAT (14) READ (5, 30) LEVER FCEEAT (Â£10,4) READ (5,40) LNHER FCEEAT (D1G. 4) REAL (5,45)IKHEL FORMAT (D1 0.4) BEAE (5,50) WLEETH FORMAT (14) WHF=WHI/12. 0 THICK=2.0*LE1IBR + 2.G*LEHBL BRW YTH=2, Q*INEBB BLWYTH=20*LNHBL NST CÂ£ Y= Will/ 8 NSBGF=(ESTORY* 5)- 1 NEIEE=NSTCBY*3 Â£LEE=LE VER CJIEK=LKHBfi+LNHEL WRITE (6,60) Will, WHE FORMAT ('0*,'WAIL BRIGHT =,14,1X,INC BES* ,3X [ ,F,2,1X, l F E EI ]) WRITE (6,70)THICK FORMAT(O',* HALL THICKNESS =,F6.2,1X MECHES ) WRITE (6,75)WDEETH FORMAT(*0*,'WALL DEPTH =*,I4,1X,*IECHES') WRITE (6,80) ERWYTH FO R M AT ( 'O','BRICK WYTHE THICKNESS =',F5.2,1X,I ECHES ) WRITE (6,90)ELWYTH FORMAT(*0','BLOCK WYTHE THICKNESS =,F5.2,1X, I ECHES') WRITE (6,100)NELFM FOBHATpO* ,'HOUBEH OF ELEMENTS IE FINITE EIEKEKT MODEL =', $ 15) WRITE(6,105)ESDCF FORMAT(* 0 *,NUMBER OF STRUCTURE DEGREES CF FREEDOM =',I4) WRITE(6,110)BLEN 357 Table B.32 Nomenclature For Subroutine PLATEK VARIABLE TYPE DEFINITION STRK DOUBLE PRECISION STRUCTURE STIFFNESS MATRIX NSTRK DOUBLE PRECISION STRUCTURE STIFFNESS MATRIX USED IN SOLVING FOR STRUCTURE DISPLACEMENTS M1 INTEGER HALF THE BANDWIDTH (INCLUDING THE DIAGONAL) OF THE STRUCTURE STIFFNESS MATRIX M1P1 INTEGER HALF THE BANDWIDTH PLUS ONE NSDOF INTEGER NUMBER OF STRUCTURE DEGREES OF FREEDOM NSDH2 INTEGER NUMBER OF STRUCTURE DEGREES OF FREEDOM MINUS TWO LNHBR DOUBLE PRECISION HORIZONTAL LENGTH OF THE BRICK HALF OF THE COLLAR JOINT ELEMENT LNHBL DOUBLE PRECISION HORIZONTAL LENGTH OF THE BLOCK HALF OF THE COLLAR JOINT ELEMENT 116 (c) Relationship between vertical compressive load and vertical strain for 4 X 42 X 16-in brick prisms at e t/4. (d) Relationship between vertical compressive load and vertical strain for 4 x 32 X t6-in brick prisms at e t/3. Figure 5.8-continued " > EXAMPLE #4 H/T=20 PLATE LOAD ECCÂ£NTBICITÂ¥=0 DESCRLPTIGN OF WALL 200 8,OD +00 2. CD + OC 3. OD + OO 24 6 PT 1 2 3 4 5 6 3 4 PT 1 2 3 4 PT 1 2 3 4 PT 1 MATESIAL PHCEERTIES BRICK ELEMENT P-DELTA CURVE CELTA COORDINATE P COORDINATE 0,OD+OO 0.OD+OO 4.OD-03 8.0D-03 1. 2D-02 1.6D-02 2.8D-02 BRICK ELEMENT S.2175D+04 1*67175D+05 2* 3272 5D+ 05 2.9 1450D + 05 3.3685OD+ 05 H-THETA CURVES 1.GC+05 THETA COORDINATE 0. OD + CO 2,117D-03 9.7C2D-3 1.0D-02 1.ED+05 THETA COORDINATE 0.OD+OO 3. 131D-03 7.85D-03 1,OD-02 2.0E+G5 M COORDINATE 0.OD+OO 5 93D+04 1.187D+05 1 2 1033D+0 5 M COORDINATE 0.OD+OO 8.895D+04 1.3350+05 1 > 5359 1D+05 THETA COORDINATE R COORDINATE 0,OD+OO 0.OD+OO IN -p* o n n rÂ¡ n uÂ¡ c o1' C AH NX1 MATRIX AND [] IS AN KX1 MATRIX. IF 1YPE=1, THEN C THE FIESI 3 NUMBERS IN I EE [INDEX] MATRIX A EE NOT USED. C IF TÂ¥PE=2, THEN ALL THE NUMBERS IN TEE INDEX MATRIX ARE C USED. If TÂ¥PÂ£=3, THEN THE FIFTH NUMEEB IN THE INDEX C MATRIX IS NCT USED. INTEGER TYPE DOUBIE PRECISIC N A B DIMENSION A (N ,N) ,B (M,M) ,INDEX (M) L= 1 IF (TYPE.EQ. 1) L = 4 DC 30 I=L,M II=INDEX(I) IF (I. EC. 5) GC TO 1 GO TO 4 1 IF (T YPE. EC. 3) GO TO 30 4 IF (KEY.EQ. 2) GO TO 20 EC 10 0=1,K 00=INDEX (0) IF (J. EQ. 5) GO TO 6 GO TO 8 IF (TYEE. EQ-3) GO TO 10 B (I, J) =A (11,00) CONTINUE GC TO 30 JJ=1 0= 1 B (I,0) = A (11,00) CONTINUE RETUBN END ru *Â£> SUBROUTINE PULRGW (A,B,I,T,N) SUBROUTINE EULROW WILL PULL THE FIRST T NUMBERS IN ROW I OUT OF AN NX6 MATRIX [Bj AND STORE IDEM IK A TX1 MATRIX [Aj. INTEGER A,B,T 341 Table B.25 Nomenclature For Subroutine CJESK VARIABLE TYPE DEFINITION CJEK DOUBLE PRECISION COLLAR JOINT ELEMENT STIFFNESS MATRIX TOTEK DOUBLE PRECISION MATRIX THAT STORES ALL OF THE ELEMENT STIFFNESS MATRICES CJSHRS DOUBLE PRECISION COLLAR JOINT SHEAR SPRING STIFFNESS CJMOMS DOUBLE PRECISION COLLAR JOINT MOMENT SPRING STIFFNESS LNHBR DOUBLE PRECISION HORIZONTAL LENGTH OF THE BRICK HALF OF THE COLLAR JOINT ELEMENT LNHBL DOUBLE PRECISION HORIZONTAL LENGTH OF THE BLOCK HALF OF THE COLLAR JOINT ELEMENT I INTEGER ELEMENT NUMBER CJ INTEGER NUMBER OF COLLAR JOINT ELEMENT DEGREES OF FREEDOM NELEM INTEGER NUMBER OF ELEMENTS 289 Table B.4 Nomenclature For Subroutine MULT VARIABLE TYPE DEFINITION A DOUBLE PRECISION N1 x N2 MATRIX WHOSE VALUES ARE MULTIPLED BY MATRIX [b] TO OBTAIN MATRIX [c] B DOUBLE PRECISION N2 x N3 MATRIX WHOSE VALUES ARE MULTIPLED BY MATRIX [a] TO OBTAIN MATRIX [C] C DOUBLE PRECISION N1 x N3 MATRIX WHOSE VALUES ARE THE PRODUCT OF THE CORRESPONDING VALUES IN [A], [B] N1 INTEGER NUMBER OF ROWS IN MATRICES [a], [c] N2 INTEGER NUMBER OF COLUMNS IN MATRIX [a], ALSO NUMBER OF ROWS IN MATRIX [b] N3 INTEGER NUMBER OF COLUMNS IN MATRICES [b], [c] 105 Relationship between vertical compressive load and vertical strain for 4 X 32 x 16-in brick prisms at e= 0. Figure 5*1 Experimental Source of Brick Element P-A Curve (4) WALL HEIGHT, INCHES 173 Figure 6.21 Wall Wythe Vertical Load Versus Height At 0.30 Pmax For Example Humber 3 130 The moment of inertia for each prism is and ! = 24(6)3 _ 8 12 432 in4 . For equal end moments (M = Mg = if these values are inserted into Equation (5.22), it reduces to (5.23) 8 = 0.444424 Thus, by using Equation (5.23), an equivalent end rotation for 6x24x8 in prisms was obtained. This procedure was carried out for several values of P, and the results illustrated in Figure 5.13 were obtained. For values of P higher than 75,000 lb and lower than 25,000 lb, the M-9 relation did not differ from that shown for these values. 5.2.3 P-M Interaction Diagrams The P-M interaction diagram for the brick element is generated in the manner discussed in section 4.3.1. The P-M interaction diagram for the brick element used in the numerical examples was obtained from tests performed by Fattal and Cattaneo (4) on brick prisms. They tested fifteen 4x32x16 in brick prisms and obtained the results shown in Figure 5.14. By converting their results to equivalent data for 4x24x8 in prisms, their P-M interaction diagram can be used to generate a valid one for the 4x24x8 in prisms. This is done by applying Equation (5.1) to obtain a value of P for the desired prism size. The eccentricity e 290 320 321 322 323 325 330 340 350 360 GO TO 36C CAII FULSCW (ICJ,TOTEIE,J,CJ,NElEK) CALI NULL (CJÂ£W,CJ, 1) CALL EXTBAK (STBW ,CJEW ,2 ,2 NSDOF ,CJ ,ICJ) CAII NULL (CJEK, CJ CJ) CALL PULMAT (CJEK ,TOTEK, J ,C J ,NEIEM) CALI MULT (C J EK, CJ EW C JEF, C J,CJ 1) IF (TfiCGNV.EQ. 1) GO TO 375 IF (TBELEB. NE. 1) GO TO 330 BITE (6,320)J FGBMAT(*0/' ',* F0BCE5 FGB CCLLAB JOINT ELEMENT *, ' HUMI3EB ,14) CAII PHINT (CJEF,CJ) HITE (6,321)J FCBMAT ('0, 'SHEAR STHESSES FGB CCLLAB JOINT ', ELEMENT NUMBER',14) S HE AJB1= CJ EF (1) /19 2. 0 SHEAHB=CJEF (3)/192.0 WHITE (6,322) SHEABL vS FHMA1 ('O' 'BRICK FACE 1 ,2X, D 13. 6) WBITE (6,323)SHEABR FORMAT(' *,'BLOCK FACE ,2X,D13.6) WHITE(6,325)J FORMAT(0*,'DISPLACEME STS FOE CCLLAB JCIKT ELEMENT 'NUMBER', 14) CALL PRINT (CJEW,CJ) CALL CHKFAI (CJEF,NBHIÂ£P,BBIDM,BÂ£IDE,CJMAXP,2,NSTAT) IF(NSTAT.EQ.G) GO TO 36 IF (TBFAIl. Et4) GO TO 340 1RFAIL=1 If (Tf BINT. BE. 2) GO TO 360 WRITE(6,350)J FOBMAT ('0',* *** ELEMENT NO. ',14, 1X,* HAS FAILED ***') CALL CJPDMT (NCJPDP,NCJETE,CJDCCB,CJECCB,CJTCOE, CJMCCB,CJBF,SCJPD,SCJMT,CJVEBF,CJMOM,CJSHRS,CJMOMS ,ITEB) CALL ST IF AC (CJ 7EBF ,C J MOM ,STCVF ,STC IS C JSfi ES C J KCMS , FINITE ELEMENT MODEL FOR COMPOSITE MASONRY WALLS By GEORGE X. BOULTON A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY UNIVERSITY OF FLORIDA 1984 IN-LB 126 150,000 i 120,000 No Test; Collar Joint Moment Stiffness = 0 90,000 3E 60,000 30,000 0 | I I I I | 0 .20 .40 .60 0, RADIANS !' I I I I | .80 1.0 Figure 5-11 Collar Joint Element M-9 Curve WALL HEIGHT, INCHES 176 120.0 112.0 104.0 96.0 88.0 80.0 72.0 64.0 56.0 48.0 40.0 32.0 24.0 16.0 8.0 0 1 Figure 6.24 V/all Wythe Vertical Load Versus Height At P Example Number 3 100 For 312 Table B.13-continued. VARIABLE TYPE DEFINITION NRTP1 INTEGER NUMBER OF ROWS IN THE TOP HALF OF MATRICES [X], [c], [b] PLUS ONE; I.E. NRTP1 = (0.5N) + 1 NRBOT INTEGER NUMBER OF ROWS IN THE BOTTOM HALF OF MATRICES [X], [c], [b]; I.E. NRBOT - N NRTOP NRTOP INTEGER NUMBER OF ROWS IN THE TOP HALF OF MATRICES [X], [c], [b]; I.E. NRTOP = 0.5N EXAMPLE #2 H/T=12 PLATE LOAD ECCENTBICITY=G IK. DESCElPITON Of WALL 120 8.CD+GQ 2. OD + OO 3, D + QC 24 6 PT 1 2 3 4 5 6 3 4 PT 1 2 3 4 PT 1 2 3 4 PT 1 MATERIAL PROPERTIES BRICK ELEMENT P-JCELTA CURVE DELTA COORDINATE P COORDINATE 0, OD + OO 4.0D-03 6.0D-03 1. 2D-02 1.6D-02 2.C8D-2 BRICK ELEMENT 0,00+00 9.2175D+04 1,o7175D+U3 2,32725D+05 2.9145OD+05 3.38850D+05 M-THETA CURVES 1,0D+05 THETA COORDINATE M COORDINATE 0.GD+00 5,93D+04 1.187D+05 1.21033D+05 0.OD+OO 2. 117D-3 9.702D-03 1,00-02 1.5D+05 THETA COORDINATE B COORDINATE C,OD+OO 0,OD+OO 3.131D-03 8.895D+04 7.865D-3 1.335D+05 1.OD-02 1.53591D+05 2. 0E+05 THETA COORDINATE fl COORDINATE 0-OD+OO 0.OD+OO VjJ VO NUMBER OF OPERATIONS 57 Figure 3*14 Comparison of the Equation Solving Operations of Standard Gauss Elimination Versus Modified Gauss Elimination WALL HEIGHT, INCHES 179 Figure 6.27 Wall Wythe Vertical Load Versus Height At 0.00 P For max Example Number 4 290 (start) FOR EACH ROW IN MATRIX [cj) | | -FOR EACH COLUMN IN MATRIX [c]) 1 !' 1 1 *-Z_T'Z_rZ: SET [C] = [A] x [B] (return) Figure B.4 Algorithm For Subroutine MULT WALL HEIGHT, INCHES 159 Figure 6.8 Brick Wythe Vertical Deflection Versus Height For Example Number 5 CHAPTER SEVEN CONCLUSIONS AND RECOMMENDATIONS A two-dimensional finite element model was developed to analyze composite masonry walls subject to compression and out of plane bending. The model considers the factors that most strongly influence composite wall behavior. These factors include the different strength- deformation properties of the concrete block, collar joint, and clay brick, the nonlinear nature of these properties, the load transfer properties of the collar joint, and the effects of shear deformation and moment magnification in the brick and block wythes. The accuracy of the model was verified by comparing the results of a wall analyzed by the model with test results, taken from the literature, of a similar wall. Furthermore, four examples were done in which the slenderness ratio and eccentricity of load application were varied for four walls. The analytical results of these examples were found to conform with accepted patterns of structural behavior. Originally, it was intended to conduct extensive experimental testing and compare the results obtained with those predicted by the model. Due to problems associated with funding, this was not possible. Therefore, it is recommended that physical testing be performed next and that the results generated be correlated with those obtained from the model. Based on the findings acquired from this comparison, it will be possible to make further refinements to the model and consider such effects as temperature changes and shrinkage. Once 200 355 Figure B.31 Algorithm For Subroutine APPLYF WALL HEIGHT, INCHES 171 Figure 6.19 Wall Wythe Vertical Load Versus Height At 0.9C P__, For Example Number 2 20 FORMAT (13,2D 13.6} WR1TE(6,25)J,XCCOR(I),YCCCR (I) 25 F CBM AT ( ,13,3X, D13. 6, 3X D 13. c) 30 CONTINUE WRITE (6 ,40) 40 FORMAT(* ) EETUEN END C SUBBOUTINE CURVES (NOCUBV,NOPPC,CURVAl,XCCCB,YCCCR) C SUBBOUTINE CUEVES WILL BEAD AND PEINT THE X AND Y C COOBDINATES OF UP TO 20 POINTS FOE UP TG 20 DIFFEBENT C CUBVES. NOCUBtf IS THE NUMBER OF CURVES, NOPPC IS THE C NUMBER CF POINTS PEB CUBVE, CUBVAI IS THE VALUE EACH C CUBVE IS IDENTIFIED BY, XCCOR STORES TEE X COORDINATE C OF EACH CURVE WHERE THE FIRST SUBSCRIPT IS THE CUBVE C NUHBEB AND THE SECOND SUESCBIPT IS THE POINT NUMBER, C AND YCOOR STORES THE Y CCOBDINATE CF EACH CUBVE WHERE C THE FIBST SUBSCRIPT IS TEE CURVE NUHBEB AND THE SECOND C SUBSCRIPT IS THE POINT NUMEER. DOUBLE PRECISION CURVAL,XCOOE,YCOOH DIMENSION CUR VAL (2 0) XCCOR (2 0,2 0) ,YCOCfi (20,20) CAII TITLE READ (5, 10) NOCURV 10 FCB CAT (13) READ (Â£, 20) NOPPC 20 F CEE AT (13) DO Â£G 1=1,NOCURV BEAD (5,30) CUBVAI (I) 30 FORMAT(D136) WEITE (6,35) CURV AL (I) 35 FCflM AT('C,9 X,'P = ',013.6) CALL TITLE DC 50 J=1,NOPPC BEAD(5,40)K,XCOOB (I,J) ,YCOOR(I,J) FORMAT(13,2D 13.6) 40 96 By substituting Equation (4.7) into Equation (4.2), one obtains ^Fnew^ ~ W fKold^ fWold^ * (4.8) Then by substituting Equation (4.5) into Equation (4.8), (4.9) the "new" structure stiffness matrix considering only the two plate DOF at the wall top can be constructed from the original or "old" structure stiffness matrix which considers all 4 DOF at the top of the wall. The procedure then, to solve for the original forces and displacements in the model while only considering the plate degrees of freedom at the wall top for load application purposes, is as follows: 1) Construct the original structure stiffness matrix [Kq^] as described in section 2.4. 2) Obtain the new structure stiffness matrix by [^new] = [l] [*T] 3) Construct the new structure force matrix [Fnew] considering only the axial force and moment delivered to the test plate at the top of the wall (plus other forces elsewhere, if any). 4) Solve Equation (4.9) for the new structure displacement matrix [w,! L newJ 5) Calculate the original structure displacement matrix from Equation (4.5). 6) Calculate the original structure force matrix [F0i[] from Equation (4-7). 7) Solve for element displacements and forces as usual (described in sections 2.7 and 2.8). Efficient handling of the structure stiffness matrix by taking advantage of its symmetry and banding as discussed previously in section 3.6 MATRIX TYPE DESCRIPTION BLDCOR DOUBLE PRECISION BLEF DOUBLE PRECISION BLEK DOUBLE PRECISION BLEW DOUBLE PRECISION BLIDM DOUBLE PRECISION BLIDP DOUBLE PRECISION BLMCOR DOUBLE PRECISION BLPCOR DOUBLE PRECISION BLPCV DOUBLE PRECISION BLTCOR DOUBLE PRECISION BLWYLD DOUBLE PRECISION BRDCOR DOUBLE PRECISION BREF DOUBLE PRECISION BREK DOUBLE PRECISION BREW DOUBLE PRECISION BRIDM DOUBLE PRECISION STORES BLOCK DELTA COORDINATES BLOCK ELEMENT FORCE MATRIX BLOCK ELEMENT STIFFNESS MATRIX BLOCK ELEMENT DISPLACEMENT MATRIX STORES BLOCK INTERACTION DIAGRAM MOMENT COORDINATES STORES BLOCK INTERACTION DIAGRAM AXIAL LOAD COORDINATES STORES BLOCK MOMENT COORDINATES FOR EACH M-THETA CURVE STORES BLOCK AXIAL LOAD COORDINATES STORES THE VALUE OF P BY WHICH EACH BLOCK M-THETA CURVE IS IDENTIFIED STORES BLOCK THETA COORDINATES FOR EACH M-THETA CURVE MATRIX THAT STORES THE VERTICAL LOAD CARRIED BY THE BLOCK WYTHE AT EACH WALL LEVEL STORES BRICK DELTA COORDINATES BRICK ELEMENT FORCE MATRIX BRICK ELEMENT STIFFNESS MATRIX BRICK ELEMENT DISPLACEMENT MATRIX STORES BRICK INTERACTION DIAGRAM MOMENT COORDINATES 220 Figure B.22 Algorithm For Subroutine BLPDMT WALL HEIGHT, INCHES 184 Figure 6.22 Wall Wythe Vertical Load Versus Height At Pmax Example Number 5 For 293 i I r~ I I LJr-^ (start) - FOR EACH ROW IN MATRIX [c]) FOR EACH COLUMN IN MATRIX [c] SET [c] A x [b] (return) Figure B.5 Algorithm For Subroutine SMULT COMPRESSIVE IGAD CAPACITY 9.7771*03 MAXIMUM EICCK ELEMENT 9.3 15D + 04 STRUCTURE FORCE APPLICATION INFORMATION 2 DO? INITIAL FORCE FORCE INCREMENT MAXIMUM 121 -4.0D+03 -4. QD*-03 -4.QD+05 122 -fi.OD+03 -8.0D+03 -8.0D+05 INSTEOCTICNS ON PRINTING INTERMEDIATE RESULTS: 1 1121 -8.0D*03 FORCE s 305 (start) r FOR FIRST T NUMBERS IN ROW I OF MATRIX [b]) COPY VALUE INTO MATRIX [a] (return) Figure B.10 Algorithm For Subroutine PULROW 348 Table B.28 Nomenclature For Subroutine INDXCJ VARIABLE TYPE DEFINITION ICJ INTEGER INDEX MATRIX FOR A COLLAR JOINT ELEMENT TOTEIN INTEGER MATRIX THAT STORES ALL OF THE ELEMENT INDEX MATRICES I INTEGER ELEMENT NUMBER NEMO INTEGER NUMBER OF ELEMENTS MINUS ONE CJ INTEGER NUMBER OF COLLAR JOINT ELEMENT DEGREES OF FREEDOM NELEM INTEGER NUMBER OF ELEMENTS 135 can be calculated from (5.24) Then, the moment for the desired prism size is obtained from (5.25) By using this technique for several values of P, Figure 5.15, which was used for the examples, was generated. In the above equations, Pj) = P on desired prism P^ = P on actual prism Mjj = M on desired prism M^ = M on actual prism. The P-M interaction diagram for the block element is generated in the manner discussed in section 4.3*3. Once again, block prism tests performed by Fattal and Cattaneo (4) were used as a basis for the block element P-M interaction diagram used in the numerical examples. They tested fifteen 6x32x24 in block prisms and obtained the results shown in Figure 5.16. As for the brick element, an equivalent P-M interaction diagram for 6x24x8 in prisms was generated by applying Equations (5.1), (5.24) and (5.25) on several values of P. This new P-M interaction diagram is shown in Figure 5.17. Because, for both types of elements, the P-M interaction diagram is a measure of cross-sectional capacity, only the differences in prism cross-section need to be considered when converting results from 6x32x24 in prisms to results for 6x24x8 in prisms. 34b (start) CONSTRUCT THE INDEX MATRIX FOR THE BRICK ELEMENT (return) Figure B.27 Algorithm For Subroutine INDXBR WALL HEIGHT, INCHES 162 Figure 6.10 Block tfythe Vertical Deflection Versus Height For Example Humber 3 5*3 Test Assembly Used By Williams and Geschwinder For Collar Joint Tests (18) 108 5.4 Experimental Source of Collar Joint Element P-A Curve (18) 109 5*5 Collar Joint Element P-A Curve 111 5.6 Experimental Source of Concrete Block Element P-A Curve (4) 112 5*7 Block Element P-A Curve 114 5.8 Experimental Source of Brick Element M-9 Curves (4) 115 5.9 Relationships Used to Obtain Desired Data From Experimental Results For M-9 Curves 117 5.10 Brick Element M-9 Curves 125 5-11 Collar Joint Element M-9 Curve 126 5.12 Experimental Source of Concrete Block Element M-9 Curves (4) 127 5-13 Block Element M-9 Curves 131 5.14 Experimental Source of Brick Element P-M Interaction Diagram (4) 132 5.15 Brick Element P-M Interaction Diagram 134 5.16 Experimental Source of Concrete Block Element P-M Interaction Diagram (4) 135 5.17 Block Element P-M Interaction Diagram 136 5*18 Example Number 1 138 5.19 Effective Column Length Factors Based On End Conditions (7) 140 5.20 Structure Degrees of Freedom For Example Number 1 ...142 5.21 Example Number 2 143 5*22 Structure Degrees of Freedom For Example Numbers 2 and 3 144 5*23 Example Number 3 146 5> 24 Example Number 4 147 320 (start) /read the number of points in the curve/ /read the X and y coordinate of each point/ (return) Figure B.16 Algorithm For Subroutine COORD D.2 Example Number 2 358 (start) CONSTRUCT STRUCTURE STIFFNESS MATRIX THAT ONLY CONSIDERS TWO DEGREES OF FREEDOM AT THE TOP OF THE WALL (CORRESPONDING TO TEST PLATE) FROM THE ORIGINAL STRUCTURE STIFFNESS MATRIX THAT CONSIDERS FOUR DEGREES OF FREEDOM AT THE WALL TOP (return) Figure B.32 Algorithm For Subroutine PLATEK 114 Figure 5.7 Block Element P-A Curve 60 GC 1C 120 CJAXF=DABS(CJEF (1) ) DC 80 I=1,NILS1 IF (CJAXF.GE.CJPCGR (I) ) CJSHRS=SCJPD (I) IF (CJAXF.L1.CJECOR(I) ) GO 1C 90 80 CONTINUE 90 AVGMCM=DABS ( (CJEF (2) +CJEF(4) )/2.0) DO 1GC L-1,NMLS1 IF (AVGMGM. GF. CJMCOfi (L) ) CJMGMS=SCJMT (L) IF (AVGHM.LI.CJMCOR(L) ) GC 1C 120 100 CONTINUE 120 CJMCM=AVGMGM C]V EEF=CJAXF EE1UBN END C SOEEGUTINE ELIDM1' (NBLPDP, NOBLPC, NBLMTP,BLDCOE ,BLPCOB ,BLPC V , $ BLTCCfi,BLMCOB,BLEF,SBLPD,SBLMI,BIVERF,BLi!CM,BIAEL,BI3EIL,ITEE) C SUBBCUTINE ELPDMT HILL CALCULATE 1 EE ELOPES CF THE C LINEARLY APPROXIMATED P-DEI1A AND K-THETA CURVES FCB C THE BLCCK ELEMENT THE FIBBT TIME 31 IS CALLED. IT C ALWAYS FINDS THE SLCPE FCB THE REGION IN WHICH THE C ELEMENT P FALLS AND THEREFORE GENERATES THE APPROPRIATE C AXIAL STIFFNESS FACTOR OF A*E/I. SIMILARLY, IT FINDS C THE SLOPE FCB THE REGION IN WHICH THE AVERAGE MOMENT C FALLS CN THE CURVE AND GENERATES TEE FCTATICNAL C STIFFNESS F ACTOB 3*E*I/I FCB THE ELEMENT P. INTEGER ULP DOUEIE PRECISIC N BLDCOR,BLPCCR,BLTCOR,BLMCCR,BIFF,BLAEL DOUBLE PRECISION BL3EIL,SBLED,SBLEl,AVGMCt,BLECV,LLSMl,ULSMT DOUBLE EBECISIC N ELMOM,ELV ER F DIMENSION BLDCCB (2 C) ,BLPCOR (20) ,BLTCOR (20,20) ,BIMCCR (20, 20) DIMENSION ELEF (6) ,SBLPD (20) ,SBLMT (20,20) ,BIPCV (20) IF (ITER.NE. 1) GC TO 60 NPLS1=NELPDÂ£-1 DC 20 1 = 1,NPL S1 264 n tr. o u> n n n m $ ELASBR ,ARSHBR,ELASBL,ARSHBL) C NULL CUT MATRICES AS NEEDED. CAII NUIL (STRK,NSDOF, 9) CALL NULL (STR 4i ,N SDOF, 1) CAII NULL (SIRE,NS EOF,1) CALL NULL (BREK,6,6) CAII NULL (EREW,6,1) CALL NULL (BEEF,6,1) CAII NULL (ElÂ£K,b,6) CALL NULL (BLE h,6,1) CAII NII (ELE E,6, 1) CALL NULL (CJEW,4,1) CAII NUIL (C JEE 4 1) C BEAD Â£ PRINT BRICK ELEMENT P-DEL1A CURVE DESC E.IFTIC K. C ALI COCED (NEBFDE,BRDCOR,BE ECOS) WRITE (6, 1) 1 FCRM AI (1 11) C READ & PRINT BRICK ELEMENT M-THE1A CURVES DESCEIPTICN. CALI CURVES (NCEBFC,NBRMTP,B'PCV,BfiTCOH,BBMCOR) WRITE (6,2) FORKAT (* 11) SEAL S PRINT COLLAR JOINT ELEMENT P-DELTA CURVE DESCRIPTION. CALI CCCRD (NCJEDP,CJDCOR,CJECOR) READ & PRINT COLLAR JOINT ELEMENT M-THETA CUBVE DESCEIPTICN. CAII CCCRD (NCJ Ml Â£ CJT COR, C J ECCB) BEAD & PRINT BLOCK ELEMENT P-DEITA CUBVE DESCEIPTICN. CAII CCCRD (NELPBE,BLDCCR,BLPCCH) WHITE (6,3) FOB MAT (* 1 *) HEAD 6 PRINT BLCCK ELEMENT M-THEIA CURVES DESCEIPTICN. CAII CURVES (NGELEC,NBLMIP,BLPCV,BLTCOB,BLMCQR) WRITE (6,5) FCBMA1 ( 1 *) READ & PRINT BBICK ELEMENT p-M INTERACTION DIAGRAM DESCRIPTION. CAII CCCRD (N EBIDE BRIDM, BR IEF) C READ Â£ PRINT BLOCK ELEMENT P-M INTERACTION DIAGRAM DESCRIPTION. OZ WALL HEIGHT, INCHES 196 Figure 6.43 Collar Joint Shear Stress Versus Height For Example Number 4 |