Citation
Finite element model for composite masonry walls

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:

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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Full Text











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




Full Text
[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


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=£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 1£0 I=ii£'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 ]£ 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
ks£1
_ks
ks£2
ks£1
ks£f + km
*ks£1
ks£2£1 km
"ks
~ks£1
ks
ks£2
ks£2
ks£1£2 ~ km
"ks£2
ks£2 + km
WHERE: k = SHEAR SPRING STIFFNESS
b
km = MOMENT SPRING STIFFNESS
£1 = LENGTH OF LEFT PART OF ELEMENT
£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 § 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) + £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 *£ + \(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 E£. 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 E£ to E-¡ 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 + \¡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
K¡a
K¡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] [K¡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-
W£
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

00
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 £ ^ *" £ -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.¡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
£2 £] 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 -1a£_ = 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
W V*
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
¡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
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vt F
*, r*
7
V Tv >
v+'r'+''
4
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EXTERNAL WALL LOADS
^121 W122
w 1 *
w123
W124

V.#1
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t

i

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EQUIVALENT WALL LOADS
2l 2 2
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^1 2 3 v. 7"v ^124
V >. 9
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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) ¡
' 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
IL=4?
''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
PRINT
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,B£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,TOL£R,SHEARL,SHEARE
BRHOM,BRVERF,CJMCM,CJVEBF,BLMCK,BLV£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) ,B£EW (6) ,STBF(194)
DELTAE(194), GDELTF (194) ,ERRCE(194) NS1BF (192)
NGSTRF (192) ,NDCF( 192) ,SFINIT(192) ,SFINCR (192)
SFMAX (192) ,BR£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 £ 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
(El£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 £ 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 £ 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 £ PRINT BLOCK ELEMENT P-M INTERACTION DIAGRAM DESCRIPTION.
OZ

CALL COORD (NBLIDP,BLIDM ,BLIDP)
C BEAD £ 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,5FINC£,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
KSEF£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 £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 £. 1) CALL EXTRAK (SIRN,BREW ,2,2, NSDCF, BR,.I£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 £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.£Q.N£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 £ 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 (CJ£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,NBHI£P,BBIDM,B£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,LNH£E,LNHEL,
$ J,CJ, NELEM)
CALL BNSERT (STRK ,C JE K ,2 NSDCF M1 ,C 0 IC J)
I F (IT £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 £. 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,BL£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. B£.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. £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 £ 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,BLMCC£, 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,BL£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 CONTINUE
IF (lEELPfi. N£. 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 £K]*£W]=£F] FCB £N], WHERE £K] IS THE
STBCTUBE STIFFNESS MATRIX, £H] IS TEE STBUCTUBE
DISPLACEMENT MATRIX, AND £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 £ F]=£ 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*£WJ=£F] FOR £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 £ 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,NSIG£,M1)
CALL NULL (DELL AW, NS DCF, 1)
CALL SIACON (DELTA W,GSTRK ,GD£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 £UAL (NGSTRK, NSTRK NSEM2 N IP 1)
CALL NOLL (NSTR ts,NSDM2,1)
T£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 £ 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 £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)+£ (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 £BJ TO 0ETA IN AN N1XN3 MATRIX £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 £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 £ A ]*£ E ]=£C ], £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. £B] IS AN
C NX 1 MATRIX. £ C] IS AN NX1 MATRIX KUCSE VALUES ARE THE
C PRODUCT Cf £ A ]*£ B J.
DOUBLE PRECISION A,B,C
DIMENSION A (N,B1) ,U(N) C(N)
N E £ 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 £5
100
CONTINUE
BE1UEN
EKE
C
SBFCUTINE INSEBT (A,B,KEY,TYPE,N,M, INDEX)
SUBBOU1INE INSEBT BILL INSEBT A MA1BIX [£] INTC A
LAfiGEfi HAIBIX [A]. TEE VALUES CF £B] ABE ADDED 10 1UE
VALUES OF [Aj. IF K£Y=1, THEN [A] IS A K NXN HATBIX AND

non u> to cd a.
C £B] IS AH axil MATRIX, IF KEY=2, THEM £ 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 £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 [£] INTO A MATRIX
247

C £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 £ 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 £ A ]. THE VALUES OF [ A ] ABE NOT AFFECTED.
C THE OLD VALUES OF [B], If ANY, ABE BEELACEE EY THE
C VALUES FOUND IN £ AJ. IF KE¥=1, THEN £A] IS AN NXN
C MATBIX AND £B] IS AN MXM MATBIX. IF KEY=2, THEN [A ] IS

n n r¡ n u¡ 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 T¥P£=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
*£>
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 £ B] AND STORE THE VALUES IN A TXT MATRIX £ 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 £A ]*£X ]=£B ], WHERE £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. £X] IS THE
C MATRIX THAT STORES THE SOLUTION, £C] IS THE MATRIX USEE
C TO STORE THE SYMMETRIC NON-ZERO VALUES IN THE £A]
C MATRIX, £ B] IS THE MATRIX THAT STORES THE NUMBERS
C THAT = £A]*£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 £ 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 £ A ]*£X]=£ B], WHERE £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. £X] IS THE MATRIX THAT STORES
C THE SOLUTION, £C] IS THE MATRIX USED TO STORE THE
C SYMMETRIC NGN-ZERO VALUES IN THE £A] MATRIX, £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 £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) +CC£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 £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 £ JE2 j FECH £ E ].
D= 0
DO 140 1= NETP 1, N
L=L + 1
2(I) = E(I)
140 CONTINUE
C SOLVE £ K4 ]£ W2 ]=£ F2 ] FOE £W2J USING GAUSS1
CALL GAUSS1 (W2 ,K4 ,F'2, M 1 NEBC1 1)
C CCNSIEUCl £ K2' ] £C £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=£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 1£0 I=ii£'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 ]£ 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 £ K 1 ]£ M1 ]=£ E 1 1 J-£ K2* ]£ W2 ] FOB £ 'ill ] BY BACK SUBSTITUTION
C (SINCE FOHWABD ELIMINATION PORTION OF GAUSS 1 A IDEADY DCNE ON
C £ K 1 ] IAET CF £ 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 £ HI 3 £ £W2 j IKTC £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 (£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 C£ Y= Will/ 8
NSBGF=(ESTORY* 5)- 1
NEIEE=NSTCBY*3
£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.E£. 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 EIAS£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 (£, 20) NOPPC
20 F CEE AT (13)
DO £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,£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 £ 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,
$ £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 VGEC£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,K0£W1
L=K+ 1
IF (BBEF (1) GE, EEPCV (K) ) LL£=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=NELPD£-1
DC 20 1 = 1,NPL S1
264

20
J=I+1
SBIJED (I) = (ELECCS (J) -ELPCGL (I) )/ (BLDCC£ (J) -BICCCS (I) )
CONTINUE
Hf5L£ 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,NMI£ 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 E£ 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) = MCK£
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,B£AEI,ES3EIL,ELAS£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))
?HI£E=£B3£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*EB£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) =EE£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 £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*CJ£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,EL£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,N£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) = £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) = EI£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 TCT£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 (£) =3
IBB J6) = 4
GC 1C 50
DC 20 J= 1,3
£=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) = I£ 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 = DAB¡~ (SFMAX (I) )
IF (DSTRF.GT. E5FMAX)STBF (K)=SFMAX (I)
IF (DSTRF.LT.DSFMAX) K£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) = ST£F (J) -SFINCR (I) / 1 0
CONTINUE
HEX 1 = 0
GO 30 80
DC 70 1=1,NCF
I=NDOF(I)
STBF (LJ=STB£ (!) +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 (NS£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,flSTO£
NSTBK (K,L) =SIBK (K ,L)
30 CCKTINE
11 £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)
£UIE=KUID-1
WHICH 1=NHIDH1-1
BS1EK (NSIRE2, MIDE 1) =STRK (NSIRP2,7) +SIBK (NSTBP2,8)
NS1RK (USURP 2, MUID) =HL02*STRK (BSTB£2,7 ) +102*STBK $ +STHK (NSTBP2, 9)
11 UIC = NHID 1
NICn=NWIDHl-1
H C C L = 6
DC 60 I=NSTBE3,RS£M4
RCQLE 1 = 11C0L+ 1
KCCL£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*ST£K(N3EM3, 1) L02*STRK{NSDM3,2)
1 + SIHK (N£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 £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,NSTH£ 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 r¡ 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, KE¥= 1. £ A] IS AN NX1 MATRIX.
DOUBLE PRECISION A,TOLER,ABSCL J
DIMENSION A ( N) 03
K£¥= 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(E2£1.££.£I!S1) GC TO 70
AI£CBH=£lIfl1 + (P1-PIH1)* (MIP1-KIM 1) / (PI P1-PIM 1)
IF(KEY.NE.2) GG TC 75
KAX£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 £ 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*£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+£.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
oe£

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 ECC£NTBICIT¥=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 §.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 £5D* 05
INTERACTION
¡ 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 ¥= 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.£ 16D-03 1,730+05
4 1.CD-02 2.58179D+05
COLLAR JCIKT ELEMENT P- DELTA CURVE
2
PT DELTA COCEDI NATE £ 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
1,6£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 £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
£.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.E£. 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 EIAS£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 £5D* 05
INTERACTION
¡ 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 £5
100
CONTINUE
BE1UEN
EKE
C
SBFCUTINE INSEBT (A,B,KEY,TYPE,N,M, INDEX)
SUBBOU1INE INSEBT BILL INSEBT A MA1BIX [£] INTC A
LAfiGEfi HAIBIX [A]. TEE VALUES CF £B] ABE ADDED 10 1UE
VALUES OF [Aj. IF K£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 £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 § 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*ST£K(N3EM3, 1) L02*STRK{NSDM3,2)
1 + SIHK (N£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 £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,NSTH£ 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 r¡ 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, KE¥= 1. £ A] IS AN NX1 MATRIX.
DOUBLE PRECISION A,TOLER,ABSCL J
DIMENSION A ( N) 03
K£¥= 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
1,6£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 £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
£.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*£WJ=£F] FOR £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 £ 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,B£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,TOL£R,SHEARL,SHEARE
BRHOM,BRVERF,CJMCM,CJVEBF,BLMCK,BLV£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) ,B£EW (6) ,STBF(194)
DELTAE(194), GDELTF (194) ,ERRCE(194) NS1BF (192)
NGSTRF (192) ,NDCF( 192) ,SFINIT(192) ,SFINCR (192)
SFMAX (192) ,BR£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 £ 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) )/ (BLDCC£ (J) -BICCCS (I) )
CONTINUE
Hf5L£ 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
KSEF£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 £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 £B] IS AH axil MATRIX, IF KEY=2, THEM £ 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 £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 [£] 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 £ B] AND STORE THE VALUES IN A TXT MATRIX £ 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 £A ]*£X ]=£B ], WHERE £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. £X] IS THE
C MATRIX THAT STORES THE SOLUTION, £C] IS THE MATRIX USEE
C TO STORE THE SYMMETRIC NON-ZERO VALUES IN THE £A]
C MATRIX, £ B] IS THE MATRIX THAT STORES THE NUMBERS
C THAT = £A]*£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 £ 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) = £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) = EI£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 TCT£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 £ A ]*£X]=£ B], WHERE £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. £X] IS THE MATRIX THAT STORES
C THE SOLUTION, £C] IS THE MATRIX USED TO STORE THE
C SYMMETRIC NGN-ZERO VALUES IN THE £A] MATRIX, £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 £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) +CC£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 £UAL (NGSTRK, NSTRK NSEM2 N IP 1)
CALL NOLL (NSTR ts,NSDM2,1)
T£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 + \¡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,BLMCC£, 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,BL£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 CONTINUE
IF (lEELPfi. N£. 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 £K]*£W]=£F] FCB £N], WHERE £K] IS THE
STBCTUBE STIFFNESS MATRIX, £H] IS TEE STBUCTUBE
DISPLACEMENT MATRIX, AND £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 £ F]=£ EBBOfi ] SO IT CAN BE USED TO INSURE CONVERGENCE.
3


28
ks
ks£1
_ks
ks£2
ks£1
ks£f + km
*ks£1
ks£2£1 km
"ks
~ks£1
ks
ks£2
ks£2
ks£1£2 ~ km
"ks£2
ks£2 + km
WHERE: k = SHEAR SPRING STIFFNESS
b
km = MOMENT SPRING STIFFNESS
£1 = LENGTH OF LEFT PART OF ELEMENT
£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
W V*
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
¡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 VGEC£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,K0£W1
L=K+ 1
IF (BBEF (1) GE, EEPCV (K) ) LL£=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] [K¡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,B£AEI,ES3EIL,ELAS£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))
?HI£E=£B3£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*EB£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,LNH£E,LNHEL,
$ J,CJ, NELEM)
CALL BNSERT (STRK ,C JE K ,2 NSDCF M1 ,C 0 IC J)
I F (IT £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 £. 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,NSIG£,M1)
CALL NULL (DELL AW, NS DCF, 1)
CALL SIACON (DELTA W,GSTRK ,GD£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 E£ to E-¡ 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.¡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
£2 £] 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(E2£1.££.£I!S1) GC TO 70
AI£CBH=£lIfl1 + (P1-PIH1)* (MIP1-KIM 1) / (PI P1-PIM 1)
IF(KEY.NE.2) GG TC 75
KAX£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 £ 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*£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+£.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.£Q.N£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 £ 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 = DAB¡~ (SFMAX (I) )
IF (DSTRF.GT. E5FMAX)STBF (K)=SFMAX (I)
IF (DSTRF.LT.DSFMAX) K£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) = ST£F (J) -SFINCR (I) / 1 0
CONTINUE
HEX 1 = 0
GO 30 80
DC 70 1=1,NCF
I=NDOF(I)
STBF (LJ=STB£ (!) +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 (NS£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 £ K 1 ]£ M1 ]=£ E 1 1 J-£ K2* ]£ W2 ] FOB £ 'ill ] BY BACK SUBSTITUTION
C (SINCE FOHWABD ELIMINATION PORTION OF GAUSS 1 A IDEADY DCNE ON
C £ K 1 ] IAET CF £ 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 £ HI 3 £ £W2 j IKTC £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 £ ^ *" £ -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,
$ £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 £ 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 £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)+£ (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 £BJ TO 0ETA IN AN N1XN3 MATRIX £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
IL=4?
''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 £ 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,5FINC£,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,NMI£ 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 E£ 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) = MCK£
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) + £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 £. 1) CALL EXTRAK (SIRN,BREW ,2,2, NSDCF, BR,.I£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,EL£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,N£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,£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 £ 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 ¥= 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 £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 £ JE2 j FECH £ E ].
D= 0
DO 140 1= NETP 1, N
L=L + 1
2(I) = E(I)
140 CONTINUE
C SOLVE £ K4 ]£ W2 ]=£ F2 ] FOE £W2J USING GAUSS1
CALL GAUSS1 (W2 ,K4 ,F'2, M 1 NEBC1 1)
C CCNSIEUCl £ K2' ] £C £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 F
*, 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) ¡
' 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
oe£


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 -1a£_ = 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) =EE£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 £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*CJ£K(1, 1)
CJEK (J, 1)=CJSHBS*LNHBL
268


2 4.1C9D-3 1.186D+05
3 6.£ 16D-03 1,730+05
4 1.CD-02 2.58179D+05
COLLAR JCIKT ELEMENT P- DELTA CURVE
2
PT DELTA COCEDI NATE £ 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 *£ + \(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 (£) =3
IBB J6) = 4
GC 1C 50
DC 20 J= 1,3
£=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) = I£ 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,BL£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. B£.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. £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 £ 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
K¡a
K¡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

00
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 §.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 £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 £ A ]*£ E ]=£C ], £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. £B] IS AN
C NX 1 MATRIX. £ C] IS AN NX1 MATRIX KUCSE VALUES ARE THE
C PRODUCT Cf £ A ]*£ B J.
DOUBLE PRECISION A,B,C
DIMENSION A (N,B1) ,U(N) C(N)
N E £ 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 £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 £ 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 £ A ]. THE VALUES OF [ A ] ABE NOT AFFECTED.
C THE OLD VALUES OF [B], If ANY, ABE BEELACEE EY THE
C VALUES FOUND IN £ AJ. IF KE¥=1, THEN £A] IS AN NXN
C MATBIX AND £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 E£. 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
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AUTHOR: Boulton, George
TITLE: Finite element model for composite masonry walls / (record number:
487104)
PUBLICATION DATE: 1984
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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
PRINT
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,flSTO£
NSTBK (K,L) =SIBK (K ,L)
30 CCKTINE
11 £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)
£UIE=KUID-1
WHICH 1=NHIDH1-1
BS1EK (NSIRE2, MIDE 1) =STRK (NSIRP2,7) +SIBK (NSTBP2,8)
NS1RK (USURP 2, MUID) =HL02*STRK (BSTB£2,7 ) +102*STBK $ +STHK (NSTBP2, 9)
11 UIC = NHID 1
NICn=NWIDHl-1
H C C L = 6
DC 60 I=NSTBE3,RS£M4
RCQLE 1 = 11C0L+ 1
KCCL£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-
W£
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 (£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 C£ Y= Will/ 8
NSBGF=(ESTORY* 5)- 1
NEIEE=NSTCBY*3
£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 ECC£NTBICIT¥=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 r¡ n u¡ 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 T¥P£=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
*£>
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 (CJ£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,NBHI£P,BBIDM,B£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 (£, 20) NOPPC
20 F CEE AT (13)
DO £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=NELPD£-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
(El£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 £ 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 £ 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 £ 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