• TABLE OF CONTENTS
HIDE
 Title Page
 Dedication
 Acknowledgement
 Table of Contents
 List of Tables
 List of Figures
 Abstract
 Introduction
 Development of the finite element...
 Special considerations
 Using the model to analyze composite...
 Numerical examples
 Results of analysis
 Conclusions and recommendation...
 Computer program
 Subroutines
 User's manual
 Data files for numerical examp...
 Reference
 Biographical sketch
 Copyright














Title: Finite element model for composite masonry walls
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Permanent Link: http://ufdc.ufl.edu/UF00090209/00001
 Material Information
Title: Finite element model for composite masonry walls
Series Title: Finite element model for composite masonry walls
Physical Description: Book
Creator: Boulton, George X.,
 Record Information
Bibliographic ID: UF00090209
Volume ID: VID00001
Source Institution: University of Florida
Holding Location: University of Florida
Rights Management: All rights reserved by the source institution and holding location.
Resource Identifier: alephbibnum - 000487104
oclc - 11902053

Table of Contents
    Title Page
        Page i
    Dedication
        Page ii
    Acknowledgement
        Page iii
        Page iv
    Table of Contents
        Page v
        Page vi
        Page vii
    List of Tables
        Page viii
        Page ix
    List of Figures
        Page x
        Page xi
        Page xii
        Page xiii
        Page xiv
        Page xv
        Page xvi
    Abstract
        Page xvii
        Page xviii
    Introduction
        Page 1
        Page 2
        Page 3
        Page 4
        Page 5
        Page 6
    Development of the finite element model
        Page 7
        Page 8
        Page 9
        Page 10
        Page 11
        Page 12
        Page 13
        Page 14
        Page 15
        Page 16
        Page 17
        Page 18
        Page 19
        Page 20
        Page 21
    Special considerations
        Page 22
        Page 23
        Page 24
        Page 25
        Page 26
        Page 27
        Page 28
        Page 29
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        Page 64
        Page 65
        Page 66
        Page 67
    Using the model to analyze composite masonry walls
        Page 68
        Page 69
        Page 70
        Page 71
        Page 72
        Page 73
        Page 74
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    Numerical examples
        Page 102
        Page 103
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        Page 149
    Results of analysis
        Page 150
        Page 151
        Page 152
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    Conclusions and recommendations
        Page 200
        Page 201
    Computer program
        Page 202
        Page 203
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    Subroutines
        Page 282
        Page 283
        Page 284
        Page 285
        Page 286
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    User's manual
        Page 368
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    Data files for numerical examples
        Page 384
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        Page 408
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    Reference
        Page 410
        Page 411
    Biographical sketch
        Page 412
        Page 413
    Copyright
        Copyright
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




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