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 PA Curves....................................103
5.2.2 M0 Curves .................................... 113
5.2.3 PM 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 PA Curve (4)........105
5.2 Brick Element PA 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 PA
Curve (18) ..............................................109
5.5 Collar Joint Element PA Curve............................111
5.6 Experimental Source of Concrete Block Element PA
Curve (4)......................................... 112
5.7 Block Element PA Curve..................................114
5.8 Experimental Source of Brick Element M9 Curves (4).......115
5.9 Relationships Used to Obtain Desired Data From
Experimental Results For M9 Curves.......................117
5.10 Brick Element M9 Curves..................................125
5.11 Collar Joint Element M9 Curve...........................126
5.12 Experimental Source of Concrete Block Element M9
Curves (4)..................................... ......127
5.13 Block Element M9 Curves................................... 31
5.14 Experimental Source of Brick Element PM Interaction
Diagram (4)...............................................132
5.15 Brick Element PM Interaction Diagram......................134
5.16 Experimental Source of Concrete Block Element PM
Interaction Diagram (4)....................................135
5.17 Block Element PM 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 twodimensional 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 loadbearing 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 loadbearing 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
LoadBearing Concrete Masonry" (14). A third standard on concrete
masonry was recently published by the American Concrete Institute (ACI)
entitled ACI 53179R, "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 loaddeformation 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 stressstrain 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 twodimensional finite element model will take into
account the different nonlinear loaddeformation 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 fixedend 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 fixedend forces
measured in structure coordinates
[K] = N x N structure stiffness matrix measured in structure
coordinates.
Again, fixedend 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 CROSSSECTION
E = MODULUS OF ELASTICITY
L = ELEMENT LENGTH
I = MOMENT OF INERTIA OF ELEMENT CROSSSECTION
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,
crosssectional, 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 crosssection. 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
AHs
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 crosssectional 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
Sf6 1+ f
w6 )
T=0
S(2 ) E .
(1 + 4)1
If the deflection of the lefthand 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 CROSSSECTION
MODULUS OF ELASTICITY
ELEMENT LENGTH
MOMENT OF INERTIA OF ELEMENT CROSSSECTION
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
fwH
(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 righthand 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 loaddeformation 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 CROSSSECTION
MODULUS OF ELASTICITY
ELEMENT LENGTH
MOMENT OF INERTIA OF ELEMENT CROSSSECTION
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 CROSSSECTION
E = MODULUS OF ELASTICITY
L = ELEMENT LENGTH
I = MOMENT OF INERTIA OF ELEMENT CROSSSECTION
= 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 loaddeformation 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
loaddeformation 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
Iz
cZ
0
0J
'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
steadystate 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
W4 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'liStio*
*****SfnsMt* 0S
(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 GaussElimination procedure to the number required by the
modified GaussElimination 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
 irri 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
Ijj
az
wj
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 1014, this solution
converged better than one which produced a value in [ERROR] equal to
9 x 101
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 loaddeformation 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 = ONEHALF THE THICKNESS OF
THE BRICK WYTHE (= 2")
2 = ONEHALF 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 loaddeformation 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
piecewiselinear 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
At
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 setup 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
CROSSSECTION
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
CROSSSECTION 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
