Structural Dynamics - Theory and Computation 4th Ed (1997)

~~~'~;~'~~;~~~~;.~ Dyn;t.mics, 4th programs In 5trUcturai edition features a ti7:~~:f~::~;,::,,~:~~'~r~~~1!11 '~;

~ an 'introduction to the dynamtc analysis of s'U'Uctu;es

Method

• a new addition to the chaptet on Random Vibration' ~ "~i~nse of strtu:tuni~.c·.... •. ,', modeled as 11 multi degrea-of.freedom 5},st.em,5ubje
,

Structural Dynamics Theory and Computation Fourth Edition

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Structural Dynamics Theory and Computation Fourth Edition

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Mario Paz Speed Scientific School University of Louisville Louisville, KY

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CONTENTS

PREFACE TO THE FOURTH EDITION I xv PREFACE TO THE FIRST EDITION I PART I

xxi

STRUCTURES MODELED AS A SINGLE-DEGREE-OF-FREEDOM SYSTEM 1

UN;)I\MPE;) SINGLE-DEGREE-OF-FREEDOM SYSTEM

U U L3 1.4

LS 1.6 1.7 1.8 1.9 1.!O

U I

3

Degrees of Freedom I 3 Undamped System I 5 Springs in Parallel or in Series I 6

Newton's Law of Motion i 8 Free Body Diagram I 9 D' Alembert'S Plinciple I 10 Solution of the Differential Equation of Motion I II Frequency and Period I 13 Amplitude of Motion I 15 Undamped Single-Degree-of-Freedom Systems Using COSMOS I 20 Summary I 22 Problems i 23

Conten:s viii

Contents

2

DAMPED SINGLE.DEGREE·OF·FREEDOM SYSTEM

2.1 2.2

2.3 2.4 2.5 2.6 2.7

3

3J

5

5.5 5.6

5.7 5.8

RESPONSE OF ONE·DEGREE·OF·FREEDOM SYSTElY, TO HARMONIC LOADING 47

6

6.3 6.4 6.5 6.6

6.7 6.8 6.9 6.10

7 4

Principle of Virtonl Work I 162 GeneraLlzed Slngle~Degree-of-Freedom System-Rigid Body I 164 Generalized Single-Degree-of-Freedom System-Distlibu~ed Elasticity I ,67 Shear Fo,ces and Bending Moments I 172 Generalized Equation of Motion for a Multistory Building / 177 Shape Function I 180 Rayleigh's Method i 185 Improved Rayleigh's Method! 192 Shear Walls I 195 Summary I 199 Problems I 200

NONLINEAR STRUCTURAL RESPONSE

7.1 7.2

Impulsive Loading and Duhamel's Integral I 96 Numerical Evaluation of Duhamel's Integral-Undamped

73

4.3

System i 105 Numerical Evaluation of Duhamel's Integral-Damped

7.5

System I 109 Response by Direct Integration I 11 0 Program 2-·Response by Direct Integration I 116 Program 3-Response to Impulsive Excitation f 119 Response to General DynamiC Loading Using COSMOS I 124 Summary I 131

162

205

96

4.1 4.2

4.4 4.5 4.6 4.7 4.8

Fourier Analys" I 139 Response to a Loading Represented by Fourier Series I 140 Fourier Coefficients for Piecewise Linear Functions / 143 Exponential Form of Fourier Series I 144 Discrete Fourier Analysis I ]45 Fast Fourier Transform I 148 Program 4-Response in the Frequency Domain i 150 Summary I 156 Problems I 156

GENERALlZED COORDINATES AND RA YLEtGH'S METHOD 6.1 6.2

Undamped System: Harmonic Excitation / 47 Damped System: Hannonic Excitation / 50 3.2 Evaluation of Damping at Resonance I 58 3.3 Bandwidth Method (Half·Power) to Evaluate Damping, / 59 3.4 Energy Dissipated by Viscous Damping I 61 3.5 Equivalent Viscous Damping I 63 3.6 Response to Suppon Motion / 66 3.7 Force Transmitted to the Foundation I 76 3.8 Seismic InstrUments I 79 3.9 :·3.10 Response of One·Degree-of·Freedom System to Harmonic Loading Using COSMOS i 81 3.1l Summary I 88 Problems I 92 3.1

RESPONSE TO GE':-lERAL DYNAMIC LOADING

FOURIER ANALYSIS AND RESPO':-lSE IN THE FREQUENCY DOMAIN 139

S.I 5.2 5.3 5.4

Viscous Damping i 31 Equation of Motion i 32 Critically Damped System I 33 Overdamped System / 34 Underdamped System / 35 Logarithmic Decrement I 37 Summary / 43 Problems i 44

ix

7,4

7.6 7.7

7.8

7.9 7.10

Nonlinear Single Degree·of·Freedom Model I 206 Integrmion of [he Nonlinear Equation of Motion I 208 Constant Acceleration MeLood I 208 Linear Acceleration Step-by·Step Method I 2l! The Newmark Beta Method I 2 14 Elastoplasric Behavior I 215 Algorithm for the Step·by-Step Solution for Elastoplastic Single-Degree-of-Freedom System I 217 Program S-Response for ElastopJastic Behavioa, System I 221 Nonlinear Stn.lClUra; Response Using COSMOS I 224 Summary I 228 Problems I 229

x

s

Contents

RESPONSE SPECTRA

Contents

11,4 Ha:monic Forced Excitation I 326

233

11.5

8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9 8,10

PART II

Construction of Response Spectrum I 233

Response Spectrum for Support Excitation! 237 Tripartite Response Spectra I 238 Response Spectra for Elastic Design I 241 Influence of Local Soil Conditions! 245 Response Spectra for Inelastic Systems I 247 Response Spectra for Inelastic Design ! 250 Program 6-Seismic Response Spectra .I 257 Response Spectra Using COSMOS I 260 Summary I 265 Problems I 266

12

Equations for Damped Shear Building I 353 Uncoupled Damped Equations I 354 123 Conditions for Damping Uncoupling J 355 12.4 Program ll~-Absolute Damping From Damping Ratios I 362 12,5 Summary I 364 Problems I 364

271

REDUCTION OF DYNAMIC MATRICES 13.1

Sta:ic Condensation ! 367

Static Condensation Applied to Dynamic Problems J 370 Dynamic Condensation J 380 Modified Dynamic Condensation! 387 Program 12-Reduction of the Dynamic Problem! 391 Summary I 393 Problems I 393

133

Stiffness Equations for the Shear Building J 272 P-Ll Effect on a Plane Shear Building! 275 Flexibility Equations for the Shear Buildip.g J 278 Relationship Between Stiffness and Flexibility Mame""s J 280 Program 7-Modeling Structures as Shear Buildings I 281 Summa,), I 283 Problems ! 283

FREE VIBRATION OF A SHEAR BL1LDING

10.1 10,2 10.3 lOA 10.5 lO.6

]3.4 13.5

13.6

II

FORCED MOTION OF SHEAR BUILDING

ILl 11.2

287

Natural Frequencies and Noonal Modes! 287 Orthogonality Property of Ute Noonal Modes I 294 Rayleigh's Quotient J 298 Program 8-Natural Frequencies and Normal Modes I 300 Free Vibration of a Shear Building Using COSMOS! 301 Summary I 304 Problems J 305 310

Modal Superposition Ylethod i 310 Response of a Shear Building to Base Ylotlon I 317

14

366

13.2

PART III

10

352

STRUCTURES MODELED AS SHEAR BUILDINGS 271

THE MULTISTORY SHEAR BUILDING

9.1 9.2 9.3 9,4 9.5 9.6

DAMPED MOTION OF SHEAR BUILDIt-;GS

12.1 12,2

13

9

Program lO·~Harmonic Response I 331

1 1.6 Combining Maximum Values of Modal Response I 334 11.7 Forced Motion of a Shear Building Using COSMOS .I 335 11.8 Summary J 346 Prob;ems ! 348

STRUCTURES MODELED AS DISCRETE MULTIDEGREE,OF-FREEDOM SYSTEMS

DYNAMIC ANALYSIS OF BEAMS

14.1 14.2

14.3 14.4

14.5 14.6 14.7 14,8 14.9 14.10 14.1 I

397

399

Static Properties fnr a Beam Segment J 400 System Stiffness Matrix J 405 Inertial Properties-.Lumped Mass ! 408 Inel1iai Properties-Consistent Mass I 41O Damping Properties ! 414 External Loads ! 414 Geometric Stiffness ! 416 Equations of Motion I 420 Element Forces at Nodal Coordinates I 427 Program 13----Modeling Structures as Beams J 430 Dynamic Analysis of Beams Using COSYlOS I 433

Xl

XII

Conten:s

Contents

14.12

15

DYNAMIC ANALYSIS OF PLA;-!E FRAMES

15.1 15,2

15.3 15.4

15.5 15,6

16

16,2

16.3 16,4

16.5 16.6 16.7

16,8 16.9

442

Eleme"' Sliffness Matrix for Axial Effects I 443 Element Mass Motrlx for Axial Effects I 444 Coordinate Transformation I 449 Prooram 14-Modeiinc:r~ Structures as Plane Frames! 458 ~ Dynamic Analysis of Frumes Using COSMOS I 460 Summary / 465 Problems I 466

DYNAMIC ANALYSIS OF GRIDS

16.1

19

Local and Global Coordinate Systems f 470 Torsional Effects I 471 Stiffness :vIatrix for a Grid Element I 472 Consistent Mass Matrix for a Grid Element I 473 Lumped Mass Matrix for a Grid Element f 473 Transformation of Coordinates I 474 Program IS-Modeling Structures as Grid Frames I 480 Dynamic Analysis of Grids Using COSMOS f 483 Summary I 487 Problems f 488

19, I

19.2

19.3

20

17.6 17,7 17 ,8 17.9

Element Stiffness Matrix f 49l Element Mass :vIatrix f 493 Element Damping :vIatrix I 494 Transformation of Coordinates f 494 Diffetenrial Equation of Motion f 503 Dynamic Response I 504 Ptogtam 16-Modeling Structures as Space Frames I 504 Dynamic Response of Three-Dimensional Frames Using COSMOS f 507 Summary I 510 Problems I 5\0

DYNAMIC ANALYSIS OF TRUSSES

18,1

511

Stiffness and Mass Matrices for ttie Plane Truss f 512 I

.:;'1·1

Plane Elasticity Problems f 539 19, J.l Triangular Plate Element for Plane EJastlcitv Problems I 540 19, L2 Library of Plane ElosticilY Elements (2D Eleoents) I 552 Plute Bending I 555 j9.2.1 Rectangular Finite EleT:1ent for Plate Bending I 556 19.2.2 COSMOS Library of Plate and Shell Elements f 565 Summary I 573 Problems I 575

T1ME HISTORY RESPO;-!SE OF :VIULTiDEGREE-OF·FREEDOM SYSTEMS 577 20,1

17.1 17.2 17.3 17.4 17.5

Program 17-Modeling Structures as Plane Trusses f 520 Stiffness and Mass Matrices for Space Trusses I 522 Equation of Motion for Space Trusses I 525 Program l8-Modeling Str'Jctures as Space Trusses I 526 Dynamic Analysis of Tmsses Using COSMOS f 528 Summary I 536 Problems I 536

DYI'\AM!C ANALYSIS OF STRUCTURES USING THE Fll'\ITE ELEMEI'\T METHOD 538

469

17

18

18.3 18.4 18.5 18.6 187 18.8

Summary I 437 Problems I 438

xlii

Incremental Equations of Motion I 578 The WIlson-8 Method I 579 Algorithm for Step-by-Step Solution of a Linear System Using the Wilson-8 Method f 582 20.31 Initialization f 582 20.3.2 For Each Time Step f 582 20A Program 19-Response by Step Integration f 587 205 Newmark Beta Method f 588 20,6 Elastoplastic Behavior of Framed Structures f 589 20.7 :vIembet Stiffness Matrix I 590 20.8 Membet :vIass Matrix I 593 20.9 Rotation of Plastic Hinges I 595 20.10 Calculation of Member Ductility Ralio ! 596 20.11 Time-History Response of Mul(idegree-of-Preedom Syster.rs Using COSMOS f 597 20.12 Summary I 602 Probler.rs f 604

202 20.3

xiv

PART IV 2!

STRUCTURES MODELED WITH DISTRIBUTED PROPERTIES

607

DYNA.\1IC ANALYSIS OF SYSTEMS WlTH DtSTRIBUTED PROPERTIES

609

2LI 21.2 2 L3

21,4 2 L5

21.6 21.7

22

Flexural Vibration of Uniform Beams J 610 Solution of the Equation of MociOf: in Free Vibration J 6 i ! Nftrura~ Frequencies and Mode Shapes for Uniform Beams / 613 21.3.1 Both Ends Simply Supported I 613 21.3.2 Both Ends Free (Free Beam) ! 617 21.3.3 Both Ends Fixed I 6 i 8 2L3.4 One End Fixed and [he other End Free (Cantilever Beam) I 620 21.3.5 One End Fixed and the other End Simply S"pponed I 622 Orthogonallty Condition Be[weer. Normal Modes I 622 Forced Vibration of Beams J 624 Dynamic Stresses in Beams I 630 Summary I 632 Problems I 633

DISCRETIZATION OF CONTINUOUS SYSTEMS

22.1 22,2 ,,22.3

22.4 22.5 22.6 22.7 22.8

PAHT V 23

Contents

Contents

Dynamic Matrix for Flexural Effects I 636 Dynamic Matrix for Axial Effects I 638 Dynamic Matrix for Torsional Effec[S I 641 Beam Flexure Including Axial-Force Effect J 642 Power Series Expansion of ::,e Dynamic Matrix for Flexural Effects I 646 Power Series Expans~on of the Dynamic MatrIX for Axial and for Torsional Effects j 648 Power Series Expansion of the Dynamic Matrix Including the Effect of Axial Forces I 649 Summary I 650

RANDOM VIBRATION

RA,'fDOM VIBRATION

23.1

23.2 23.3

635

651

653

Statlsrical Description of Random Functions I 654 Probabiliry Dens;ry Function I 657 The Normal Distribution I 659 f

r:.r.;n

23.7 23.8

13.9 23.10 23.11 23.12 23 13

PART VI 24

xv

Correlation I 662 The Fourier Transform I 666 Spectral AnalysIS I 668 Spectral Density FU .. otlon I 672 Narrow-Band and Wide-Band Random Processes I 675 Response::o Random ExcI(3[ion: Single·Degree-of~Freedom System I 679 Response to R2.ndom Excitation: Mulr!ple-Degree~of-Freedom SyStem! 685 Random Vibration Using COSMOS! 696 S ummay ! 700

EAHTHQUAKE ENG:NEERING

705

t:NIFORM BUILDIC'lC CODE 1994; EQUIVALENT STATIC LATERAL FORCE METHOD 707

24.1 24,2 24.3 24.4

Earlhquake Ground Motion! 708 Ec;:..tivalenr Seismic Lateral Force J 7:2 Eanhquake-ResiSlam 8esign Methods I 712 Static Lateral Force Mcrhod I 713 145 Distribution of Lateral Forces j 718 24.6 Story Shear Force I 718 24.7 Horizontal Torsional :Vloment I 719 24.8 Oveflureing Moment I 720 24.9 Srory 8rift Limitm'on I 720 24.10 P-Delta Effect (P-Ll) I 721 24.11 Diaphragm Design Force I 723 24.12 Program 23 UBC·94 Equivalent Static Lateral Force Method I 732 24.13 Simplified Three Dimensional Earthquake Resistant Design of B uilcings I 739 24.l3.l Ylodeiing the Building I 739 24.13.2 Transfomla[jon of Stiffness Coefficients j 740 2413.3 Center of Rigidity I 742 24.13.4 Story Eccentricity I 743 24.13.5 Rotational Stiffness! 744 24.13.6 Fundamental Period I 745 24.13.7 Seismic Factors I 745 24. I 3.8 Base Shear Force! 746 24.13.9 Equivalent Lateral Seismic Forces! 746 24.13.10 Overturning Moments I 747

xvi

Contents

24.13.10 Story Shear Force / 747 24.13.12 Torsional Moments / 747 24.13.13 Story Drift and Lateral Displacements / 748 24.13.14 Forces and Moments on Structural Elements I 749 24.13.15 Computer Program / 750 24.14 Equivalent Static Lateral Froce Method Using COSMOS / 756 24.15 Summary I 761 25

UN1FORIvl BUlLDl.t'!G CODE 1994: DYNAMIC :vlETHOD 25.1

25.2

25.3 25.4 25.5 25.6

Preface to the Fourth Edition

766

Modal Seismic Response of Buildings / 766 25. Ll Modal Equation and Participation Factor / 767 25.1.2 Modal Shear Force I 768 25. L3 Effective Modal Weight / 770 25.1.4 Modal Lateral Forces / 771 25.1.5 Modal Displacements / 771 25.1.6 Modal Drift I 772 25.1.7 Modal Overturning Moment / 772 25.1.8 Modal Torsional Moment / 772 Total Design Values I 773 Provisions of UBC-94: Dynamic Method I 774 Scaling of Results I 776 Program 24-UBC 1994 Dynamic Lateral Force Method i 783 Summary i 787 Problems I 788

APPEI"1DlCES I 789 Appendix I: Answers to Problems in Part I I 79 I Appendix II: Computer Programs I 80 I Appendix Ill: Organization and their Acronyms / 804 Glossary i 807 Selected Bibliograpby I 815 Index / 819 Diskette Order Fonn I 825

I

The basic Structure of the three previous editions is maintained in this fourth edition, although numerous revisions and additions have been introduced. A new chapter to serve as an introduction for the dynamic analysis of structures using the Finite Element Method has been incorporated in Part 1lI, Structures Modeled as Discrete Multidegree,of-Freedom Systems. The chapter on Random Vibration has been extended to include the response of structures modeled as multidegree-of-freedom systems, subjected to several random forces or to a random motion at the base of ~he structure. The concept of damping ~nc1uding the evaluation of equivalent viscous damping is thoroughly discussed. The constant acceleratlon method to determine the response of nonlinear dynamic systems is presented in addition to the Hnear acceleration method presented in past editions, Chapter 8, Response Spectra now includes the development of seismic response spectra with consideration of local soil conditions at the site of the structure. The secondary effect resulting from the lateral displacements of the building, commonly known as the P-!l effect. is explicitly considered through the calculation of the geometric stiffness matrix. Finally, a greater number of iHustrative examples have been incorporated in the various chapters of the Dook using the educational computer programs developed by the author or tbe professional program COSMOS.

xvm

Preface to the Fourth Ed:tion

The use of COSMOS for the analysis and solution of structural dynamics problems is introduced in this new edition. The COSMOS program was selected from among the various professional programs available because It has the capabiiity of solving complex problems in structures, as well as in other engineering fields such as Heat Transfer, Fluid F:ow, and Electromagnetic Phenomena. COSMOS includes routines for Structural Analysis. Static, or Dynamks with linear or nonlinear behavior (materlu! nonUnearlty or large displacements), and can be used most efficiently in the microcomputer. The larger version of COSMOS has the capacity for the analysis of structures modeled up to 64,000 nodes. This fourth edition uses an introductory version thot has a capabiiity limited to 50 nodes or 50 elements. This version is induded in the supplement, STRUCTURAL DYNAMICS USING COSMOS '. The sets of educational programs in Structural Dynamics and Earthquake Engineering that accompanied the third edition have now been ex.tended and updated. These sets include programs to determine the response in the time or frequency domain using the foB (Fast Fourier Transform) of structures modeled as a single oscillator. Also included is a program to deler:nine the response of an inelastic system with e!astopiastic behavior and a program for the development of seismic response spectral charts. A set of seven computer programs is included for modeling structures as two-dimensional and threedimensional frames and trusses. Other programs. incorporating modal superposition or a step-by~step time-history solution, are provided for calculation of the responses to forces or motions exciting the structure. In addition. in this fourth edition, a new program is provided to detennine the response of singleor muItidegree-of-freedom systems subjected to random excitations. The computer programs for earthquake-resistant deSign have been updated using the latest published seismic codes. The book IS organized into six parts. Part I deals with st:1lctures modeled as single-degree-of-freedom systems. It introduces basic concepts and presents methods for the solution of such dynamic systems, Part II introduces conccpts and methodology for solving multidegree-of-freedom systems through the use of structures modeled as shear buildings. Part III describes methods for the dynamic analysis of skeletal structures (beams, frames. and trusses) and of continuous structures such as plates and shells modeled as discrete systems with many degrees of freedom, Part IV presents the mathematical solution for some simple structures modeled as systems with distributed properties. thus baving an infinite number of degrees of freedom. Part V introduces tbe reader to the fascinating topic of random vibrations, which is now extended to multidegree~of-freedom systems, Finally, Part VI presents the current topic of earthquake engineering with applkations for the design of earthquake-resistant

; A cQnVen:erH form lo order this supplemelH is provided in lhe back of the book.

Preface

to the Fourth Edilion

xix

buildings following the provisions of the Cnifonn Building Code In use in the United States. There is a detailed presentation of the seismic analysis of buildings modeled as three-dimensional structures with two independent hori~ zontal motions and one rotational motion about a vertical axis for each story of the building. A computer program for the implementation of this simplified method for seismic analys~s of buildings is jncluded in the set of educational programs. Scientific knowledge may be presented from a general all-encompassing theory from which particular or simple situations are obtained by introducing restricting conditions. Alternativeiy, the presentation may begin by considering particular or simple situations that are progressively extended. The author has adopted this latter approach in which the presentation begins with particular or simple cases that are extended to more general and complex situations. Funhennore, the author believes that a combination of knowledge of applied mathematics, theory of structures. and the use of computer programs is needed today for the succe$.').fu! profes;;;iollal pfflctice of engineering. To provide the reader with such a combination of knowledge has been the primary objective of this book. The reader is encouraged to inform the author on the extent to which this objective has been fulfined. Many of my students. colieagues, and practidng professionals have suggested improvements, identified typographical errorS. and recommended additional topics for inclusion. Ali these suggestions were carefully considered and have been included in this fourth edition whenever possible. I was fortunate to have received valuable assistance and insight from many individuals to whom I wish to express my appreciation. I am grateful to Jeffrey S. Janover, a consulting engineer from New Jersey, who shared his expertise in the implementation of professional computer programs for the solution of complex engineering probtems. I appreciate the discussions and comments offered by my colleagues Drs. Michael A. Cassaro and Julius Wong who helped me in refining my exposition. I am also grateful to my friend Dr, Farzad Naeim who has coIlabornted with me on Seismic Response Spe<:rra in the International Handbook of Earthquake Engineering: Codes, Programs and Examples (Paz. 1994) of which I am the editor. I have incorporated some of the material from the Handbook in updating the chapter on Response Spectra, I also wish to acknowledge Dr. Luis E. Suarez from the University of Puerto Rico in Mayaguez, who provided me wiLi copies of his work in random vibrations and of his class notes on the Finite Element Method. It is with great satisfaction that I acknowledge the help re<:eived from four of my fonner students: Christopher Biles, who carefully studied and commented on Chapter 23, Random Vibrations, as he worked on his Masters' theSIS on that subject; Mahomet Sharif for providing me with actual cases of random vibration problems selected from his professional practice; Zair HiJlal, who made skillful use of the computer in preparing some of the new figures

xx

Preface to the Fourth Edllion

in the book; and Cleryl Hoskins who most carefully checked the solution of the problems for some chapters of the book A special acknowledgement of gratitude is extended to Dr. Edwin A. Tuttle, emeritus professor of educatlon, who provided many suggestions that helped to improve the clarity of my presentation. I aiso wish to express my sincere gratitude to my friend Jack Bension for his professional help in editing the revised sections of the book. My thanks also go to Ms. Debbie Jones for her competent typing sieills in the revisions. To those people whom I recognized in the prefaces to the previous editions for their help, I again express my wholehearted appreciation. To my wife Jean a special thanks for carefully checking the structure of the book and for most graciously aUowing me time to prepare this new edition, particularlY during sev.eral "working vacations." As with the third edition, this volume is dedicated to the everlasting memory of my parents.

Preface to the First Edition

March, 1997

Natural phenomena and human activities impose forces of time~dependent variability on structures as simple as a concrete beam or a steel pile, or as complex as a multistory building or a nuclear power plant constructed from different materiais. Analysis and design of such Stn.lctures subjected to dynawic loads involve consideration of time-dependent inertial forces. The res islance to displacemenl exhibited by a struclure may include forces which are functions of the displacement and the velocity. As a consequence, the governing equations of motion of the dynamic system are generaUy nonlinear partial differential equations which are extremely difficult to solve in mathematical terms. Nevertheless. recent developments in the field of structural dynamks enable such analYSls and design to be accomplished in a practical and efficient manner. This work is facilitated through the use of simplifying assumptions and mathematical models, and of matrix methods and modem computarionai techniques, In the process of teaching courses on the SUbject of structural dynamics, the author came to the realization thai there was a definite need for a text which would be suitable for the advanced undergr'J.duar:e or the beginning graduate engineering stJdent being introduced to this subject. The author is familiar with tJ}e existence of several excellent lexts of an advanced nature but genxxi

xxli

Preface to the First Edmon

eraUy these texts are. in his view, beyond the expected comprehension of the student Consequently, it was his principal aim in writir.g this book to incorporate modem methods of analysis and lechniques adaptable to computer programming in a manner as clear and easy as the subject permits. He felt that computer programs should be induded in the book in order to assist the student in the application of modem methods associated with computer usage. In addition, the author hopes that thIs text will serve the practicing engineer for purposes of self-study and as a reference source. In writing this text. the author also had in mind the use of the book as a possible source for research topics in structurdl dynamics for students working toward an advanced degree in engineering who are required [0 write a thesis, At Speed Scientific School, University of Louisville, most engineering students complete a fifth year of study with n thesis requirement leading to n Master in Engineering degree. The author'S experience as a thesis ndvisor Jeads him to believe that this book may weB serve the students in their senrch and selection of topics in subjects cun·ently under investigation in structural dynamics. Should the text fulfill the expectations of the nuthor in some measure, par~ ticuiariy the elucidation of this subject, he will [hen feel rewarded for his efforts in the preparation and development of the material in this book. MARIO PAZ

December, 1979

PART I Structures Modeled as a SingleDegree-of-Freedom System

1 Undamped Single Degree-of-Freedom System

It is not always possible to obtain rigorous mathematical solutions for engineering problems, In bct, analYlical solutions can be obtaine<;i only fo: certain simplified situations. For problems involving complex malerial properties, loading, and boundary conditions, the engineer introduces assumptions and idealizalions deemed necessary to make [he problem mathematically manageable but still capable of providing sufficiently approximate solutions and sat 4

lsfactory results from the potot of view of safety and economy. The link between the real physical system and the mathemaricaHy feasible solutloil js provided by the mathematical model which is the symbolic designation for the substitute ideaHzed system including a.1I the assumptions imposed on the physi~ cal prublem.

1,1

DEGREES OF FREEDOM

In structural dynamiCs the number of independent coordinates necessary to specify lhe configuration or position of 2. system at any time is referred to as the number of degrees of freedom. If:, ge:1eral, a continuous structure has an 3

4

Slructures Modeled as a Single"Degree-ol-Freedom System

Undamped

Fjrl

~F~t=~ , ,.,

~

pm

• llwnlllHl 1ill1lj

'"

1

lfJ-' Ft.-_-==~---;

n

Fig. 1.1 Examples of Structures modeled as

one-degree~ofMfreedom

Sjng!e~Oegree"of~Freedom

System

5

on the behavior of the real physical system. Nevertheless, from a practical point of view, the information acquired from the analysis of the mathematicai model may very well be sufficient for an adequate understanding of the dynamic behavior of the physical system, including design and safety requirements.

1.2 UNDAMPED SYSTEM

systems,

infinite number of degrees offreedom, Nevertheless. the process of idealization or selection of an appropriate mathematical model permits the reduction in the number of degrees of freedom to a discrete number and in some cases to just a single degree of freedom. Figure I.! shows some examples of structures that may be represented for dynamic analysis as one-degree-offreedom systems, that is. structures modeled as systems with a single displacement coordlnate. These one-degree~of~freedom systems may be described conveniently by the mathematical mode} shown in Fig. 1,2 which has the foHowing elements: (1) a mass element m representing the mass and inertial characteristic of the structure; (2) a spring element k representing the eiastic restO£ing force and potential energy storage of the stnlcture~ (3) a damping element c representing the frictional characteristics and energy losses of the structure; and (4) an excitation force F(t) representing the external forces acting on the structural system. The force F(t) is written this way to indicate that it is a function of time. In adopting the mathematical model shown in Fig. 1.2, it is assumed that each element in the system represents a Single property; that is, the mass m represents only the property of inertia and not elasticity or energy diSSIpation, whereas [he spring k represents exclusively elasticity and not inertia or energy dissipation. FinaIly, the damper c only dissipates energy. The reader certainly realizes thar such "pure" elements do not exist 1n our physical world and that mathematical models are only conceptual idealizations of real structures. As such, mathematical models may provide complete and accurate knowledge of the behavior of the model itself, but only limited or approximate information

\Ve start our study of structural dynamics with the analysis of a fundamental and simpIe system, the one~degree-of-freedom system in which we disregard or "neglect" frictional forces or damping. In addition, we consider the system, during its motion or Vibration, to be free from external actions or forces, Under these conditIons, the system is in motion governed only by the influence of the so-caBed inith.l conditions, that is, the given displacement and velocity at time I = 0 when the study of the system is lnitiated. This undamped, onedegree~of-freedom system is often referred to as the simple undamped oscillator, It is usually represented as shown in Fig, 1.3(0) or Fig, L3(b) or nny similar arrangements. These two figures represent mathematical models that are dynamically equivalent It is only a matter of personal preference to adopt one or the other. In these models the mass m is restrained by the spring k and is limited to rectilinear motion along one coordinate axis. :Ine mechanical characteristic of a spring is described by the relation between the magnitude of the force Fs applied to Its free end and the resulting end displacement y, as shown graphically in Fig. 1A for three different springs.

~

>v

tj~J/ Ib)

fa)

Fig. 1.3 Alternate representations of mathematical models for systems,

, 1

one~degreeMof~freedom

'.

~~---------------,

Fig. 1.2 Mathematical mode! for one-degree-Qf-freedom systems,

Fig, 1.4 Force displacement relation. (a) Hard spring, (b) Linear spring. (c) Soft spring.

6

Structures Modeled as a

Sing;e-Degree~of-Ffeedom

System'

Undamped Slngle.-Degree-or·Freedom System

The curve labeled (a) in Fjg, 1.4 represents the behavior of a "hard spring," in which the force required to produce a given displacement becomes increasingly greater as the spring is defonned. The second spring (b) is designated a linear spring because the deformation is directly proportional to the force and the graphicaJ representation of its characteristic is a straight 1ine. The constant of proportionality between the force and displacement (slope of line (b)] of a linear spring is referred to as the spn'ng constalll, usually designated by the lelter k" Consequently, we may write the foUowing relatlon between force and displacement for a linear spring.

For two springs in parallel the total force requJred to produce a relative displacement of their ends of one unit is equal to the sum of their spring constants. This total (orce is by definition the equivalent spring constant k( and (s given by (12)

In general for n springs 1n paralic.!

,

K, = F,=ky

2: k,

(1.3)

;"'1

(l.l)

A spring with characteristics shown by curve (c) in Fjg. 1.4 is known as a "soft spring," For such a spring the incremental force required to produce additional deformation decreases as the spring deformation increases. Undoubtedly, the reader is aware from his previous exposure to mathematical modeling of physical systems that the linear spring is tne simplest type to manage analytically. It should not come as a surprise [0 learn that most of !be technical literature on structural dynamics deals wHh mode's using linear springs_ In other words, either because the elastic characterisrics of the structural system are, in fact, essentially linear, or simply because of analytical expediency, it is usually assumed that the force-deformation pmperties of the system are linear. In support of [his practice, it should be noted that in many cases the displacements produced in the. structure by the action of external forces or disturbances are small in magnitude (Zone E in Fig. 1.4), thus rendering lbe linear approximation close to tbe actual structural behavior.

7

For two springs assembled in series as s.hown in Fig_ 1.5(b), the force P produces the relative displacements in (he springs

LlYI

and Lj., 1.

=~ k2

Then. (he total displacement y of the free end of the spring assembly is equal to y = LiYI -;. Ll)'2. or substituting LiYI and LlY2,

y

(1.4)

Consequently, the force necessary 1.0 produce one unit dis.placement {equivalent spring constant) is given by

1.3

SPRINGS IN PARALLEL OR IN SERIES

Sometimes it 1s necessary to determine the equivalent spring constant for a system in which two or more springs are arranged in parallel as shown in Fig. 1.5(.) or in series as in Fig. L5(b).

k=P , y Substituting y from this last relation lnto eq, (1.4), we may conveniently express the reciprocal value of :.he equivalent spring constant as

k,

1 +_.

k,

(1.5)

In general for n springs in series the equivalent spring constant m.ay be oblained from

Fig. 1.5 Combination of springs. (3) Springs in parallel. (b) Springs in series.

k,

k,

( 1.6)

8

Struc\ures Modeled as a

Slng\e~Oegree-of-Freedom

System

Undamped ~jngle'Oegree'of"Freedom Sys!em

1,4 NEWTON'S LAW OF MOTION We continue now with the study of the simple oscillator depicted in Fig. L3.

The objective is to describe its motion. that is, to predict the

displaceme~t

or

velocity of the mass In at any time f. for a given set of initial conditions at time f = O. The analytical relation between ~he displacement. y, and time, (, is given by Newton's Second Law of Motion, which in mod~rn notation may be , expressed as

F=ma

(1.8a)

(I,8b) (L8c) The acceleration 1S defined as the second derivative of the position vector with respect to time; it foHows that eqs. (1.8) are indeed differentiar equations. The reader should also be reminded that these equations as stated by Newton are directly applicable only to bodies idealized as panides, that is, bodies that possess mass but no volume, However. as is proved in elementary mechanics, Newton's Law of Motion is also directly applicable to bodies of finite dimensions undergoing translatory motion. For plane motion of a rigid body that is symmetric with respect to the reference piane of motion (x-y plane), Newton's Law of Motion yields the following equations:

IFy mead)' IMr; = fcCt

in which the mass moment of inertia 10 and the moment of the forces Mo are determined with respeet to the fixed axis of rotation. The general motioo of a rigid body is described by two vector equations, one expressing the relation between tbe forces and the acceleration of the mass center and anorher relatioothe moments of the forces and the angular motion of' the body. This Ias~ equation expressed in its scalar components is rather complicated, but seldom needed in structural dynamics.

(1.7)

where F is the resultant force acting on a particle of mass m and a is its resultant tlcce1eration_ The reader should recognize rhat eq. (1.7) is a vector relation and as such it can be written in equivalent fonn in terms of irs components along the coordinate axes .x, y, and z, namely

L F, = m(ad,

9

(1.93) (L9b)

1.5 FREE BODY DIAGRAM At this point, it is advisable to follow a method conducive to an organized and systematic analysis In the solution of dynamics problems, The first and probably the most important practice to follow in any dynamic analysis IS to draw a free body diagram of the system, prior to writing a mathematical description of the system. The free body diagram (FBD). as the student may recall, is a sketch of the body isolated from all other bodies, in which all the forces external to the body are shown, For the case at hand, Fig, 1,6(b) depicts the FBD of the mass In of the oscillator, displaced in the positive direction with reference to coor~ dinate y, aqd acted upon by the spring force F~ ky (assuming a Jjnear spI1ng). The weight of the bod) mg and the nonnal reaction N of tbe sUp'Porting surface are also shown for compieteness, though these forces, acting in the vertical direction, do not enter into the equation of motion written for the y direction. The appHcation of Newton's Law of Motion gives -.. Icy = mji

(LlO)

where the spring force acting in the negative direction has a minus sign, and where the acceleration has been indicated by y. Ir. this notation, double overdots denote the second derivative with respect to time and obviously a single overdot denotes the first derivative with respect to time. that is, the velocity.

(J ,9c)

In the above equations (aG);~ and (aG)" are the acceleration components, along the x and v axes, of the center of mass G of the body; Ct is the angular accelerati{}J;; lr; is the mass moment of inertia of the body with respect to an axis through G, the center of mass; and 'kMc; is the sum of the moments of all the forces acting on the body with respect to an axis through G, perpendicular to the x-y plane, Equations (1.9) are certainly a}so applicable to the motion of a rigid body in pure rotation about a fixed axis. For this particular type of plane motion, altematively, eq. (L9c) may be replaced by ( 1.9d)

,.)

(0)

'c)

Fig. 1.6 Alternate free body diagrams: (a) Singie degree-of-freedom system. {b) Show~ jng only external forces. (c) Showing external and inertial forces,

Sj;uclu,es Modeled as a Slngle·Degree·o!·Freedom System

1.6

Undamped

Single~Deg(ee·ot-Freedom

System

11

O'ALEMBERT'S PRINCIPLE

An alternative approach to obtain eq. (LlO) is to make use of D'Alembert's Principle which states that a system may be sei in a state of dynamic equilibrium by adding to the external forces a fictitious force thai is commonly known as the inertial force. Figure L6(c) shows the FED with inclusion of the inert:al force my. This force is equal (0 the mass multiplied by the acceleration, and should always be directed negatively with respect to the corresponding coordinate. The application of D'Alembert's Prineiple allows us to use equations of equilibrium in obtainJng the equation of motion. For example, in Fig. 1.6(c), the summation of forces in the y direclion gives directly mji + J. :y = 0

SOlution: The FEDs for these two representations of the simple oscillator 1.7(c) and 1.7(e), where the inertial forces are inchIded. are shown in Equating to zero the sum of the forces in Fig. L7(c), we obtain (a)

When the body in Fig. 1.7(d) is in the static equilibrium posit;on, lhe spring is stretched Yo units and exerts a force kyo = W upward on the body, where W is the weight of the body. 'When {he body is displaced a distance y downward from this position of equilibrium the magnitude of the spring force is given W + ky, since kyo = W. Using this result and applying by F. =. k(yo + y) or it to the body in Fig. L7(e), we obtain from Newton's Second Law of Motion

+ W=my

w '
Exam ple 1.1. Show that the same differential equat:on is obtained for a spring-supported body moving verlically as for the same body vibrating along a horizontal axis, as shown in Figs. 1.7(0) and 1.7(b).

- (W+Ay)

(0)

(LlI)

which obviously is equivalent to eg. (1.10). The use of D' Alembert's Principle in this case appears to be triviaL This will not be the case for a more complex problem, in which the application of D'Alembert's Principle, in conjunction with the Principle of Virtual Work, constitutes a powerful tool of analysis. As wUi be explained later, the Principle of Virtual Work is directly applicable [0 any system in equHibriuffi. ft follows then that this principle may also be applied to the solution of dynamic problems, provided that D' Alembert' s Principle is used to establish lhe dynamic equilibrium of the system.

m'j+ i..:y;;; 0

,

(b)

N

fd

Fig. 1.7 Two representations of the simple oscillator and corresponding free body diagrams. or

my+ky=O which is identical to eq. (a).

1.7

SOLUTION OF THE DIFFERENTIAL EQUATION OF MOTION

The next step toward our objective is to Dnd the solution of the differenlial equalion {Lll). \Ve should again adopt a systematic approach and proceed to first classify this differential equation. Since the dependeot variable y and its second derivative y appear in the first degree in eq. (1.11), this equation is classified as linear and of second orGeL The facl that the coefficients of y and y (k and m, respectively) are conslants and the second member (rigbt~hand side) of the equalion is zero further classifies the equation as homogeneous with conslant coemcien~s. We shodd recall, probably with a certain degree of satisfaction, that a general procedure exists for the solution of linear differential equations (homogeneous or nonhomogeneous) of any ordeL For this simple, second~order djfferentja~ equation we may proceed direclly by assuming a trial solution given by y=A cos WI

( 1.12)

12

structures Modeled as a Sing!e-Oegree-of-Freedom System

Undamped Single .. Degree·of-Freedom Sysiem

or

y=B sin

(Ll3)

W!

where A and B are constantS depending on the initia~ion of the motion whlle w is a quantity denoting a physical characteristic of the systert: as it wiE be shown next. The substitution of eq. (1.12) into eq. (1.1 j) gives (- In,,}

+ k) A cos

w!;;;:;

0

(1.14)

13

Next, we should determine the constants of integration ,4, and B. These constants are determined from known values for the motion of the system which almost invariably are the displacement Yo and the velocity Va at the jojtiation of the motion. that is, at time t = O. Tbese two conditions are referred to as initial conditionJ, and the problem of solving the differential equation for the lnltial conditions is caUed an initial value problem. After substituting. for l = 0, y ~ Yo, and }' = Uo into eqs. (Ll?) and (!.IS) we find that

If this equation is to be satisfied at any time, the factor in parentheses must

be equal to z.ero or

Yo=A Vo=

(1.15)

(LJ9a)

Bw

(l.!9b)

Finally, the substitution· of A and B from eqs. (1.l9) into eq. (Ll?) gives The reader should verify that eq. (1. 13) is also a solution of the differential equation (LlI), with OJ also satisfying eg. (US). The positive roO! of eq. (LlS), (1.160)

is known as the natural frequency of the system for reasons that will soon be apparent. Equation (1.16a) may be expressed in tem.s of the static displacement resulting from the weight W = mg. The substitution into eq. (1. i6) of In = Wlg results in

w=

(kg

'8

which is the expression of the displacement y of the simple oscillator as a function of the time variable 1; thus we have accomplished our objective of describing the motion of the simple undamped oscillator mOdeling structures with a single degree of freedom.

1.8 FREQUENCY AND PERIOD

wT= 211'

where Y$! = W Ik is the static displacement due to the weight W. Since eilher eq, (Ll2) or eq. (Ll3) is a solution of eq, (l.ll), and since this differentiaI equation 1$ linear, the superposition of these two solutions, indicated by eg, (1.l7) below, is also a solution. FurthemlOre, eq. (Ll?), having two constants 0: integration, A and B, is, in fact, the general solution for this second~order differential equation, y=A cos wt-B SIn wI

(1.17)

The expression for velocity, ):, is found simply by differentiating eq. (1, 17) with respect to time; that is,

+ Bw cos

( 1.20)

(LJ6b)

w=;V YSl

Aw sin wi

w

An examination of eq. (1.20) shows that the motion described by this equation is harmonic and, therefore, periodic; that is, it can be expressed by a sine or cosine function of the same frequency w. The period may easily be found since the functions sine and cosine both have a period of 211', The period T of the motion is determined from

i-

yW

Hence

y=

Vo .

y=yocos wt+-sm wt

wt

(US)

(1.21 )

or

T= 21T W

The period is usually expressed in seconds per cycle or simply in seconds, whh the tacit understanding,that it is '
1

w

T

21T

f=-=-

( 1.22)

·

14·

:"".'

",:i;

Structures Modeled as a Single-Oegree·o{·Freedom System

Undamped

System

15

Substituting corresponding numericai values, we obtain

l~l;n-ll

M~T 10.691~lit\,

Single~Oegree,of-Freedom

I] \1' 1 1= - X I X = 12 \4J

114 in.

3 X 30

1

. 4 (JIl)

10'

X

k, = (12.5)' X 768 = 60 !blin Fig. 1.8 System for Example 1.2_

Tne natural frequency f is usually expressed in hertz or cycles per second (cps). Because the quantity w d!ffers from the natural frequency f only by the constant factor, 2'lT, w also is sometimes referred to as the natural frequency. To distinguish between these two expressjons for natural frequency, w may be called the circular or an8ular natural frequency. Most often, the distinction j5 understood from the context or from the units. The natural frequency f is measured in cps (IS indicated. while the circular frequency w should be given in radians per second (radfsec). Example 1.2. Detennine the natural frequency of the system shown in Fig. 1.8 consisting of a weight of W = 50,7 lb attached to a horizontal camilever beam through the coil spring ka, The cantilever beam has a thickness t ~ in, a width b = 1 in modulus of eiasticity E= 30 X 106 psi, and a length I. = ! 2.5 in. The coil spring has a stiffness, k, = 10.69 (Ib lin). Solution: The deflection L1 at the free end of a uniform car-tilever beam acted UpOrl by a static force P at the free end is given by

and I 1 =-+--60 ;0.69 k, = 9.07 lb/in

The natural frequency for this system is then given by eq. (1.16a) as

Vligandg=386 in/sec 2)

w=Jkt:!m (m

w = /"9ih X 386/50,7 w

= 831 rad/sec

or using eq. (1.22) 1.32 cps

(Ans.)

1.9 AMPLITUDE OF MOTION Let us now examine in more detail eq. (1,20), the solution de.scribing the free vibratory motion of the undamped osciHator. A simple trigonometric transformatJon may show us (hat we can rewrite this equation in the equivalent foons, namely

PI.' 3EI

y""'-C sin(wl+ a)

(123)

y = C cos (",1- fl)

(1.24)

c = j y~ + (uo}w)J.

(1.25)

or Tne corresponding spring constant k, is then P

3EI

j

I.

where

k!=-;:;;o-)-

J

where I -?7.bt (for rectangular section). Now, the cantilever and the coil spring of this system are connected as springs in series. Consequently, the equivalent spring constant as given from eq. (1.5) is

ran

a

y,

(1.26)

and uo}w

tan fl=-y

( 1.27)

16

I

Structures Modeled as a Single-Degree-of·Freedom System

Fig. 1.9 Definition of angle

I

I

Ck'.

The simplest way to obtain eq. (1.23) or eq. (1.24) is: to :nultip;y and divide eg, (1.20) by the factor C defined in 0'1. (L25) and to defme a (or (3) by eq, (1.26) [or 0'1, (1.27)J, Thus

Y=

I Yo wi + C\CCOS

(1.28)

With the assistance of Fig. L9, we recognize that

.sin ex = Yo C

I

Undamped

~

Solution:

I

W=200 X 25 1

(1.30)

82.5 in

12 X 30 X 10' X 165 (15x 12)'

k

1o

1O,185Ibli"

Note: A unit displflcement oftlte top of afixed column requires aforce equal 12EUI}, Therefore, the Mruralfrequency from eqs, (U6) and (1.22) is

The substitution of eqs, (1,29) and (1.30) into eg, (1.28) gives y :;;;: C (sin a cos wI

+ cos

0'

sin wt)

= 5000 lb

4

E = 30 X 10'psi

(L29)

uo!w

17

The parameters of this model may be computed as follows:

k

0: =

System

Example 1.3. Consider the frame shown in Fig. 1. 11 (a). ThIS is a rigid steel frame to which a horizontal dynamic force is applied at the upper level. As part of the overall structurai design it is required to determine the natural frequency of the frame. Two assumptions are made: (1) the masses of the columns and walls are negJigible; and (2) the horizontal members are sufficiently rigid to prevent rotation at the tops of the columns. These assumptions are not mandatory for the solution of the problem, but they serve to simplify the analysis. Under lhese conditions, the frame may be modeled by the springmass system shown in Fig. 1. 11 (b).

and

cos

Sing!€l-Degree-of~Freedom

=

(1.31)

tv

The expression within the parentheses of eq. (1.3 J) is identical to sin (wt + a), which yields eq. (1.23), Similarly, the reader should verify, without difficulty, the fo:m of solution given by eq, (1.24). The value of C in eq_ (Ll3) [or eq, (L24)] is referred to as the ampliLOde of motion and the angle 0' (or f3) as the phase The solution for the motioo of the simple oscillator is shown graphically in Fig, 1. [0.

_1_1 10,185 21T~

386

X

5000

4,46 cps

y

L '" 15'

I .L

we x 24 -~

-----

(Ans_)

r-'

lj t:: m

FII)

11///////5J~ tbl

Fig. 1,10 Undamped free-vibralion response.

Fig. 1.11 One-degree~of~freedom frar:le and corresponding mathematical model for Example 1.3.

18

Structures Modeled as a Single-Degfee·of·Freedom System

U,'ldamped ,

w

r

Sing~e·DegTee-of~Freedom

System

19

y

F ' ' t:::k-~-J ~

k

k

...........-

,,'

M,

~

",ml,}"""

~b)

Fig. 1.12 (a) Water lower tank of Example 1.4; (b) Mathemutical modeL

~a)

'Ir

Fig. 1.13 (a) Fru:ne of Example 1.5; (b) Mathematical model.

Example 1.4. The elevated water tower tank with capacity for 5000 gallons of water shown in Fig. 1.12(a) has a natural period in laterai vibration of 1.0 sec when empty_ When the tank is fuiI of water. its period lengthens to 2.2 sec. Determine the lateral stiffness k of the tower and [he weight \.11 of the tank. Neglect the mass of the s~lpporting columns (one gallon of water weighs approximately 8.34 Ib) Solution: In its iatem! motion, the water tower is modeled by [he simple oscillator shown in Fig. L12(b) in which k is the lateral stiffness of tile tower and m is the vibrating mass of the tank. fa) Natural frequency We (tank empty):

(a)

(b) Natural frequency Wy: (tank full of water) Weight of water W",:

w, ~ 5000 x

8.34 = 41,700 Ib

_ 2r. _ 21T _

I--kg-

Wt- - - - . - - ) - _.. _ - -

T,

2.2

\ 1'1 + 41.700

and solving for W

1'1= 10,860 ib Subs~ituling

(2.2)'

W + 41,700 W

10,860 10 and g = 386 inlsec', yields

:11

:~

I

tl

k= lila Ibiin

!: ,r

211' 1,0

k386 I-~­ \ 10,860

and (Ans.)

Example 1.5. The steel frame shown in Fig l.13(a) is fixed at the base and has a rigid top that weighs 1000 lb. Experimentally) it has been found that its naturai period in iateral vibration, is equal to 1/10 of a second. It IS required to shorten or lengthen its period by 20% by adding weight or strengthening the columns. Determine needed additional weight or additional stiffness (neglect the weight of the COlumns). Solurion: The frame is modeled by the spring-mass system shown in Fig, 1.13(b). Its stiffness is calculated from

(b)

2r. _

[T

-Y-V In

Squaring eqs. (a) and (b) and dividing correspondingly the left and right sides of these equations, results in

(1.0)' =.

in10 eq. (a), VI

(Ans.)

217 0.:

~L 1000

as

k = 10,228 In lin

,

20

undamp'ect Sing\e-Degree-of-Freedom System

Structures MOdeled as a Sing!e-Degfee-of-Freedom System

(a) Lengthen the period

lO

T,

= L2 X 0.10 = 0.12 sec by adding weight LlW:

2 Ti"

w

!

10,228 x 386

= -0-.'-2 = Y-;-;,OcoOO"'+-:-iLlC::CW';-

Solution: The analysis is performed using a single spring elemem with one concentrated mass element. The foBowing com
Solve for LlW: LlW=440 lb

(Ans.)

(b) Shorten the peliod to T, = O.S X 0 1 = O.oS sec by strengthening columns in Llk:

DISPLAY VIEW, 0,

> 0,

0.08

=

i -,(_10~.2_2_8-,.~.,.·"M,....:..)(",3",8",,-6) Y

>

GEOMETRY

GEOHE'l'RY

Solve for M:

1.10

°

> VJ EW

GRID

>

PLANE

0, 1

(3) Establish a grid with two diVIsions in the X and Y directions, then use the scale command;

1000

Llk= 5753 Ib/in

vn:w_PAR 1,

(2) Define the XY plane at Z ~ 0: PL>.NE, Z,

co =2..".

21

(Ans.)

UNDAMPED SINGLE-DEGREE-OF-FR!::EDOM SYSTEMS USING COSMOS

> C,

0, 1,

2, 2, 2

DISPLAY SCALE, 0

>

DISP_PAR

>

Example 1.6. An instrument of mass m = 0.026 (lb . secL/in) is mOl;.nted on isolation springs of total spring constant k = 29.30 (lb/in). Model the system as an undamped single-degree-of-freedom system (Fig. 1.14) and de;tennine its natural frequer:cy.

SCALE

(4) Generate a curve from 1, 0, O. to I, 1, 0: GEOMETRY

>

CRPCORD,

The foi;owing example is presented to illustrate the use of the program COSMOS in the analysis of structures modeled as single degree-of-freedom systems. Detailed explanations for the llse of COSMOS including numerous examples with data preparation and results are presented in the supplement STRUCTURAL DYNAMICS USING COSMOS .•

GlUDON 1.,

GRIDON,

CiJRVES > CRPCORD , \.I, 0, :, ~, C-

(5) Define element group two nodes:

using the SPRING element formulation with

PROPSETS > EGROUP EGROUP, 1, SPRING, 0, 2, 1, 0, 0, 0, 0

(6) Define real constant for spling element: k PROPSETS

ReOM'ST,

>

29.3 (ib/in):

=

RCONST

1, I, }, 1,

29.3

(7) Generate one spring element along curve 1: MESHING ~CR,

L

> 1f

>

PAR..b.Y"...J,!:ESH

L

2,

1,.

}CCR

1

(8) Define element group 2 using the MASS element fonnulation: PROPSETS

Fig. 1.14 Mathematical model for Example 1.6,

>

EGROUP

EGHOUP, 2. MASS, 0,

0, 0, 0, 0,

C,

°

(9) Define real constant for mass element: m = 0.026 (lb . sec2 /in): PROPSETS • A tonVenlelll foffi'l ror or>Jering ~his supplement is provided :in the lacs! page of this volume,

~CON$'l',

>

RCONS1'

2, 2, 1, 7,

0,

0.026,

0,

0,

0,

0, Q

22

Structures Modeled as a Single-Deg.ee-ol-Free:dom Syslem

Undamped Single-Oegree-of-Freedom System

(10) Gene,rate one mass element at point 2: MESHING

>

>

PARhlCMESH

l"CFT .. 2, 2,

(5)

!'CPT

The differential equalion of the undamped simpie oscHlator in free motion is

~

(11) Merge nodes; MESH ING

and its general solution is

>

NODES.

NMERG£

)' = A cos

NMEHGE, 1, 3. 1, O.aOOL O. 0, 0

(12) App!y constraints in an degrees of freedom at node 1. and an degrees of freedom exce.pt UY at node 2: LOADS-Be > DP1', 1, l\L,

STRUCTURAL 0, 1, 1

>

;)ISPLHKT'S

>

DPT

FREQ!BlJCK

>

>

FREQ/BUCK

>

w=Jkim is the r.atural frequency in rad/sec w

f = ...-...~ is the nalural frequency in cps 2.".

A_FREQUENCY ~

T=

(6)

The equation of motion may be written in the alternate fonns:

(14) List the natural frequency of the system:

>

LIST

>

y

1

1.11

= C sin (UJI + a)

FREQLIST

or

F:~E:QLIST

FREQUENCY#

71 is the natural period in seconds

R_FREQUENCY

~FREQUENCY

RESULTS

FREQU£K'CY (2.AD / SEC)

FREQUENCY (CYCLES/SEC)

{SECONDS)

3,35597E+Ol

5.34278£:+:):)

:.87:68£-:)1

PERIOD

)' """ C eos((ul - f3) where

c

SUMMARY

Several '::>as1c concepts were inLroduced in this chapter.

and

(1)

The mathematical model of a structure is an idealized representation for its analysis.

tan a

{2)

The number of degrees of freedom of a system is equal to the number of independent coordinates necessary to describe its ,Position. The free body diagram (FBD) for dynamic equilibrium (to allow application of D' A~embert's Principle) is a diagram of the system isolated from all orher bodies. showing the external forces on the system. including the inertial force.

llnfJ

(3)

an

(4)

mt

B =v(jiw

A-FP£QUCNCY, 1, 5, 16, 0, 0, 0, 0, IE-OS, 0, 1£-06, 0, 0, 0, ANALYSIS

+ B sin

A; Yo

(13) Set the options for the frequency analysis to extract one frequency using the Subspace Iteration Method with a maximum of 16 iterations, and run the frequency analysis:

>

fJJJ

where A and Bare conslants of integration determined from initial condiiions:

VPT, 2, UX, 0, 2, 1, UZ, RX, RY. RZ

A,NAI.''iSIS

23

The stiffness or spring constant of a iinear system is the force necessary to produce a unir displacement

PROBLEMS L 1 Determine the natural period [or the system in Fig. P U. Assume Lhllt the, bellm and springs supporting the weight Ware massless. 1.2 The foHowing numerical values are given in Problem 1.1; L = 100 in, E! lOlI(lb_inl), W=3000 Ib, and k 2000!biin. If the weight W has an initial

24

Struclures Modeled as a Single-Degrce-Ql-Freedom SysLer

Undampeij

1.S

EI

T

,

System

25

Detennine the natural frequency of the fixed beam in Fig. PLS carrying a concentrated weight W at its center. Neglect the mass of the beam.

~.

..

~

fl

I

ml../2

Y

Sjngle~Degree-Qj-Freedom

····-1.12

=I ~

Y

Fig. Pl.1.

Fig. P1.5.

OJ'=' 20 in/sec, cetermlne the displacement and the veloctry 1 sec later. 1.3 Determjne the natural frequency for norlzontaf motion of the sreel frame in Fjg.

displacement of Yo'= 1.0 in and an initial velocity

1.6

The numerical values for Problem 1,5 are given as; L = 120 in, El = 10 9(Ib . in2), and W 5000 lb. If [he initial displacement and (he initial velociry of the weight are, respectiveiy, Yo = 0.5 in and Vo = 15 in/sec, determine the displacement, vel~ odty, and acceleration of W when time ( = 2 sec,

1.1

Consider the simple pendulum of weight W illustrated in Fig. Pl.?, A simple pendulum is a particle or concentrated weight that oscillates in a vertical arc and is supported by a weightless cord. The only forces acting are those of gravity and cord tension (I.e., frictional resistance is neglected). If the cord length is L, determine the motion ir the maximum 9sdllation angle B is small and the lnllial displacement and' velocity are 80 and 80 • respectively.

P1.3. Assume the horizontal girder to be Infinitely rigid and neglect the mass of the columns.

~ I\~

Fig, Pl.3.

1.4 Calculate the natural frequency in the horizontal mode of the steel frame in Fig. PL4 for the following cases: (a) me horiz,ontal member is assumed:o be infmitely rigid; (v) the horizontal member is flexible and made of steel~WJ:O X 33.

I I

\

I

:-' ;

W"" 25 Kips

;

\

~

W

Fig. Pl.7. 15'

W10

;:

II

il 'I

11

x 33

"

I 1.8 A diver standing at the end of a. diving board that cantilevers 2 ft osciIlates at a frequency 2 cps. Determine the flexural rigidity EI of the diving board. The

_L

wejght of the dIver is 180 lb. (Neglect the mass of the diving board).

1---'5'

1.9

Fig. PlA.

,,j; ..

A bullet weigh1ng 0.2 Ib is fired at a. speed of 100 ft/sec into a wooden block weighing 50!b and supported by a spring of stiffness 300 IbJin (Fig, PL9). Determine the displacement y{t) and velocity v(t) of the block after r sec,

26

Structures Modeled as a Single-Degrae-o!·Freedom System

Undamped S:ngle'Degree-of-Freedom System

1.12

An c;evator weighing 500 Ib is suspended from" spring having a stiffness of 600 IbJin. A weight of 300 Ib is suspended through a cable to the elevator as shown schematically in Fig. PLIO, Determine the equation of motion of the elevator if the ca:::'Je of the suspended weight suddenly breaks.

where w mg, ¥.It, is the critical budding weigbt, and 10 is the natlJral frequency neglecting the effect of gravity. 1.13

=p,

A vertical pole of length L and DexuaJ rigidity EI carries a mass m at ~ts top. as shown in Fig. P LJ 3. Neglecting the weight of the pole, derive the differential equation for small horizontal vibrations of the maSs, and find the natural frequency, Assume tba( Lhe effect of grav!:y is small and nonlinear effects may be neglected,

t--r

CPT~

Fig. F1.l0.

1.11

Sbow thilt lhc nalural frequency for the system of Problem L 11 may be expressed as

f= ie

Fig. P1.9.

1.10

27

I

I

Write the differential equation of motion fo~ lhe inverted pendulum shown in Fig, P Lll and determine its natur
I

'I . ddr-i-

Fig. P1.l3. Equihbril1m POS:tiOf!

I

m

1.14

Show the natural frequency for the system in Probler:> 1,13 may be expressed as i--~¥

/ = fn

y1 -

W~,

where 10 is the natura! frequency calculated neglecting the effect of gravity and W<;Y is the critical buckling wejght. 1.15

Fig. Fl.l1.

Determine an el
28

Struc1ures Modeled as a Slngle,Degree·o/-Freedom System

Undamped Single-Degree·ol-Freedom Syslem

29

Rigid Bearo

w L_-.l

k

I

2

~k ,,,

•

~

I

JA,;.

I

0 C

8

4

{ 1:

~Hinge

•

a

a

I

a

I

Fig. P1.17,

(bi

!---'-t-"r

b--l

[WJi

~

Ie)

lei

£

1.18

Determine the natural freqt:e::cy of v;bration in t,1.e vertical direction for t.ie rigid foundation (Fig. PLlS) transmitting a uniformly distributed pressure on Ihe soil having a resuliant force Q = 2COO kN, The area of the foot of the foundation is 3 A = 10 mJ • The coefficient of elastic compression of the soil is k "'" 25,000 kN 1m • (Problem contributed by Professors Vladimir N, Alekhin and Aleksey A Antipin of the Urals Siate Technicai University, Russia.)

Fig. P1.15.

1.16

A system (see Fig. Pl.16) 15 modeled by two freely v;brating masses m; and!n2 interconnected by a spring having a constant k. Detennine for this system the differential equation of mo!ion for the relative displacement u = Y2 - Yl between the two masses. Also determine the corresponding natural frequency of the system.

I

~.nl,

k ~.------

'1J'l

/=#/d')"'W/A/d/,Pflil.awV~$pp1bh'fliIP Fig. P1.16. 1.17 Calculate ~he natura! frequency for the vibration of the mass In shown in Fig. PL17. Member AE 15 rigid with a hinge at C and a supporting spring of stiffness k at D~ (Problem contributed by Professors Vladimir N. Alekhio and A!eksey A Antipin of t::,e Urals State Technical University, Russia.)

Fig. PL18.

1.19 Calculate the natural frequency of free vibration of the chimney on elastic foundation (Fig. PL 19), pennitting the rotation of {he structure as a rigid body about the horizontal axis x-x. The total weight of the structure is W with its center of gravity at a height h from the base of the foundation. The mass moment of inertia of the structure about the axis x-x is 1 and the rotational stIffness of soil is k (resisting moment of the soil per unit rotation). (Problem contributed by Professors Vladimir N. Alekhin and AIeksey A. Antipin of the Urals State Technical University, Russia.)

30

Struct:1r'eS Modeled as a Single-Degree-of-Freedorn System

2

n

Damped Single Degree-of-Freedom System

~T./Jw rI

:

b

~--- .• +j

Fig. Pl.19.

We have seen in the preceding chapter that the simple oscillator under idealized conditions of no damping, once excited, will oscillate indefinitely with a constant amplitude at its natural frequency_ Experience indicates, how~ ever, that it is not possihie to have a device that vibrates under lhese ideal conditions. Forces designatec as frictional or dampIng forces are always pres~ ent in any physical system undergoing motion. These forces dissipate energy; more precisely, the unavoidable presence of these frictional forces constitutes a mechanism through wh:ch the mechanical energy of the system, kinetic or potential energy> is transformed to other forms of energy such as heaL The, mechanism of this energy transformation or CissJpation is qujle compiex and is not completely :mderstood a! this time. In order to account for these dissipative forces in the una:ysis of dynamic systems, it is necessary to make some assumptions about these forces, on the basis of experience.

2.1

r,

In considering damping forces ;n the dynamic analysis of strJctures. it is usually assumed (hal these forces are proportionaJ to the magnitude of the

"

iJ .."

~

VISCOUS DAMPING

,

.~ "

32

Structures Modeled as a Single-Deg:rse-or-Freedom System

Damped SingJe-Oegree-of-Freedom System

velocity> and opposite to the direction of motion. This type of damping is known as viscous damping; it is the type of damping force that could be developed in a body restramed in its motion by a surrounding viscous fluid. There are situations in which the assumption of viscous damping is realistic and in which the dissipative mechanism is approximately viscous. Nevertheless, the assumption of viscous damping is often made regardless of the actual dissipative characteristics of the system, The primary reason for such wide use of this method is that it leads to a relatively simple mathematical analysis.

2.2

which, after cance1!aiion of the common factors, reduces to an equation called the characteristic equation: for the system, namely

mp2 + cp + k == 0

Let us assume that we have modeled a str"JclUral system as a simple oscillator with viscous damping, as shown in Fig. 2.1(a). In this figure, In and k are, respectively. the mass and spring constant of the osci!!ator and c is the viscous damping coefficIent We proceed. as in the case of the undamped oscillator, to draw the free body diagram (FED) and apply Newton's Law to obtain the differential equation of motion, Figure 2.1(b) shows the FED of the camped oscillator in which the inertial force my is also shown, so that we can use D' Alernbert's Principle. The summation of forces in the y direction gives the differential equation of motion,

my+cy+ky

(2,1)

0

The reader may verify that a trial solution y ;;;; A sin wt or y = B cos t:J! will not satisfy eg. (2.l). However, the exponential function yo:: Cef'i coes satisfy this equation. Substitution of this function into eg. (2.1) results in the equation

The roots of this ;:;uadratic equation are

P,= P2

~--~-... :t It~r-~ 2m'~

+ kCe"

=0

y(l)

= C,e'"

l2m)

m

(2,3)

+ C,e"

(2,4)

where C\ and C1 are constants of integration to be determined from the initial conditions, The final fonn of eq. (2.4} depends on the sign of the expression under the radical in eq. (2.3). Three distinct cases may occur: the quantity under the racical may either be zero, positive, or negative. The limiting case in which the quantity under the radical is zero is treated first The damping present in this case is called crilical damping,

2,3 CRITICALLY DAMPED SYSTEM For a system osciUating with critical camping, as defined above, the expression under the radical jn eq" (2.3) is e<;iual to zero; that is,

(~) __ k 2m

0

m

(2,5)

or ~

err:::::

2 J1m

(2.6)

where Ccr designates the critical damping value. Since the natural frequency of the undamped system is designated by w ,{ki';;, the critical damping coefficient given by eq. (2.6) may also be expressed in alternative notation as

(,j

2k

CCf

(0)

(2,2)

Thus the general solution of eq. (2.1) 1s given by the superposition of the two possible solutions, namely

EQUATION OF MOTION

mCp',,' + cCpe"

33

I ~~"

,

"'---Ic~j

A,,', :

Fig. 2.1 (a) Viscous damped OSCj1Jl'HOf. (b) Free body diagram.

= 2mw = w

(2,7)

In a critically damped system the roots of the characteristic equation are equal. ane, from eq, (2.3), they are PI =P2= -

(2,8)

34

. Structures Modeled as a Single·Degree-of~Freedo'11 System

Dam;JeQ Single-Degree-of-Freedom System

Since the two roots are equal, the generai solution given by eq. (2.4) would provide only one independent constant of integration, hence, one independent solution, namely (2.9)

2.5

UNDERDAMPED SYSTEM

When the value of the damping coefficient is iess [han the critical value (c < c.: r), which occurs when the expression under the radical is negal~ve, the fOots of the characteristic eq. (2,3) are complex conjugates, so that

p,

Another independent solution may be found by using the function

This equation, as the reader may verify, also satisfies. the differential equalion (2.1). The general solution for a critically damped system is. then given by the superposition of these two solurions,

('2m, C'lT

Ik - - - - c· + i 1

p, (2.10)

35

~ m

2m -

(2.13)

where i = J~ is the jmaginary unit For this case, i[ is convenient to :nake use of Eu!er's equations which relate exponential and t.:-igonometric functions. namely ei~ = cos

e - H = cos

(2.11 )

x

+i

X -

sin x

(2.14)

i s;n x

T1H~ substitution of the roots

2.4

Pi and pz from eq. (2.13) into eq, (2.4) together with the use of eq. (2.14) gives lhe foilowing convenient fonn for the general Solullon of the underdamped system;

OVERDAMPED SYSTEM

In an overdamped system, the damping coefficient is greater than the value for critical damping, namely (2.12) Therefore, the expression under the radical of eq. (2.3) is positive; thus the two iools of the characteristic equation are real and distinct, and consequently the solution is given direcdy by eq. (2.4). It should be nOled that, for the overdamped or the critjcally damped system, the resulting motion is not oscil~ latory; the magnitude of the osciHations decays exponentially with time to ~ero. Fjgure 2.2 depicts grapbknJJy the response for the simple oscjHator with c'rilicat damping. The response of the overdamped system is similar to the motion of the critically damped system of Fig. 2.2, but the return toward the neutral position requires more time as the damping is increased,

Y (l)

=e

-{,.;.'2>:.-jl

(A cos woe

+ n sin

(/Jot)

where A and B are redefined conStants of integration and frequency of the system, is given by

(2.15) WD,

the damped

(2.16)

or (uv=

(2.17)

W

This lost result is obtained after substituting, in eq, (2.16), the expression for tbe undamped natural frequency

(2.18)

and defining the damping ratio of the system as

c

Fig. 2.2 Free-vibraLion response with critical damping,

where the critical damping coefficient c'" JS given by eq. (2.6).

(2.19)

36

structures Modeled as a

Singje~Degree·ol-Freedom

System

Damped

Finally, when the initial conditions of dispLacement and veiodty, Yo and VOl are introduced, the constants of integration can be evaluated and substituted ilito eq. (2,15), giving y(t):;:; e

'~I ' Yo cos


' "0

T

The value of the damping coefficient for rea; stmctures is much Jess than the critical damping coefficieJ1t and usuaUy ranges between 2 and 10% of the critical damping value. Substituting for the extreme value g= 0.10 into eq.

(2.20)

It can be seen that the frequency of vibration for a system with as much as a

Alternatively, thiS expression can be written as (2,21 ) where

(2.22)

and

+ Yu~w)

Uma=---"'-

(223)

UJDYO

A graphical record of the response of an underdamped system with inilial displacement Yo but start:ng with zero velocity (uo:;;; 0) is shown in Fig, 2.3. It may be seen in this figure that the motion is oscillatory, but not periodic. The amplitude of vibration is not constant dcring the motion but decreases for succeSSIve cycles; nevertheless. the oscillations occur at equal intervals of time. This time !ntervul is designated as the damped period of vibration and is given from eq. (2.17) by (2.24)

y\tl

(225)

Wo= Q,995w

WI))

(VD

37

(2.17),

+ yogw, wpt,) _·····--sm

\

Sjngle~Degree-of-FreedomSystem

10% damping ratio is essentially equai to the undamped naturai frequency, Thus, in practice, me natural frequency for a damped system may be taken to be equal to the undamped natural frequency_

2,6

LOGARITHMIC DECREMENT

A practical method for determining experimentally the damping coefficienl of a system is to initiate free vibration, obtain a record of the osclIJatory motion, such as the one shown in Fig, 2.4, and measure the rate of decay of the amplitude of motion, The decay may be conveniently expressed by the logarithmic decrement (] which is defined as the natural logarithm of the ratio of any two successive peak amplitudes, y: and Y2, jn free vibration, that is,

y, 8=ln-

(2.26)

The evaluation of damping from the logarithmic decrement foHows. Consider the damped vibration motion represented graphically in Fig, 2.4 and given analytically oy eq, (2.21) as y(t) = Ce -~w.. cos (wn1- a)

rangcnt pcrint~ Ices (wol - q) "" 11

Yo

Peak

Fig. 2.3 Free vibration response for underdamped system.

Fig. 2.4 Curve showing j::eak displacements and displacements at the points of genc),.

tan~

.... "',,

38

Structures Modeled as a Single·Oegree-of-Freedom System

Damped Sing!e-Degree-ofrFreedom System

We note from this equation that, when the cosine factor is unity, the displacement is on points of the exponential curve y (I) = Ce ~f"" as shown in Fig. 2A However, these points are near but not equal to [he positions of maximum displacement. The points on the exponential curve api?car slightly to the right of the points of maximum amplitude. For most practical problems, the discrepancy is negligible and the displacement curve may be assumed to coincide at the peak amplitude. with the curve y (1) = Ce .. §wi so that we may write, for two consecutive peaks, Yl at time '1 and h at To seconds later, Yl

39

and at time 1'2 ::;::c 1J + Tv, corresponding to a period later, whcn again the cosine function is equal to one and the sine function is equal to zero

The ratio of the accelerations at times

(I

and

[1

is then

Ce- f "", and laking nalural logarithmic results in the logarithmic decrement in term of accelerations as

and

Dividing these two peak amplitudes and taking the natural logarithm, we obtain (2.27) or by substituting TD , the damped period, from eq. (2.24), 8=

(2.28)

As we can see, the damping ratio {can be calculated from eg. (2,28) after detennlning experimentally the amplitudes of two successive peaks of the system in free vibration. For small values of the damping ratio, eq. (2.28) can be approximated by (2.29) Alternatively, the logarithmic decrement may be cak:l1ated as the ratio of two consecutive peak accelerations, wbich are easier to measure experimentaty than dispJaceme:1ls. In this case, raking the first and the second derivatives in eq. (2.20, we obtain y(l)

Y(I)

Ce 1~[-§WCOs(wDI-a)-Wosin(wd-a)]

= Ce -,~ [[ -

(Ua sin (wo! - a)] (- {w) + [§WWD sin (UJol - a) -~ wb cos

which is identical to the expression for the logarithmic decrement given by eq. (2.27) in terms of displacements.

Example 2.1. A vibrating syslem consisting of a weight of W = 10 jb and a spring with stiffness k;;::. 20 lb/in is viscously damped so that the ratio of two consecutive amplitudes is LOO to 0.85. Delennine: (a) the natural frequency of the undamped system, (b) the logarithmic decrement, (c) the damp~ lng ratio, (d) the damping coefficient, and (e) the damped natural frequency. Solution: second is

(a) The undamped natural frequency of lhe system in radians per

w= Jklm = )(20 !blin X 386 io/sec')110 lb

or in cycles per second

J=

((J)vt

w

4.42 cps

(b) The logarithmic decremel1t is given by eq. (2.26) as . y, 0= In ~

{w cos (Wol- (X)

= 27.78 radlsec

y,

~ a)J)

= In

100 0.85

= 0.163

(e) The damping ratio from eq. (2.29) is approximately equal to §=

8

0.163 =--=0.026 2,,-

,"

40

St!uctures Modeled as a Single-Degree-of·Freedorn System

Single~Degree·of·Freedom

Damped

(d) The damping coefficient is obrained from eqs. (2,6) and (2.19) as

41

(c) Approximately, the logarithmic decrement is

0.037 Jb . sec

(e) The natural frequency of the damped system is given by 0<;. (2. J7),

System

277(0.05) = 0.314 and the ratio of two consecutive amplitudes

so that

WD=w.:~tj tub

= 27.78; I

m

(0.026)'

27.77 rad/sec

Example 2.2. A platfonn of weight W 4000 Ib is being supported by four equal columns. that are clamped to the foundation as weH as to the platform, Experimentally it has been determined that a static force of F i 000 lb applied horizontally to the platform produces a displacement of Li = 0-10 in. It is estimated that damping in the structure::. is of the order of 5% of the critical damping. Determine for this strocture the following: (a) undamped natural frequency, (b) absolute damping coefficient, (c) logarithmic decrement, and (d) the number of cycles and the time requlre.d for the <1mplitude of motion to be reduced from an initial value of 0.1 in to O.Ol in.

(d) The ratio between the first amplitude Yo and the amplitude YI: after k cycles may be expressed as

F

1000 0.1

Ll

. 10000 Ibltn '

=

110,000 X 386

4000

8+8+ ... +8=k8

In Y;

In 0,1 In 10 0.314

k

The dam pee frequency

Wa

0.314k

7.33 => 8 cycles

is given by

3 1.06J I -- (0.05)'

1 - =..- - -

Y

31.02 rad Isec

and the damped period T[) by

211'

31.02

(b) The critical damping is lb'sec In

and the absolute damping 32.19 lb· sec

in

= 0.2025 sec

Then the time for eight cycles is

64 3.8-.-

fie" = 0.05 X 643.8

=

31.06rad/sec ""'~

c=

Y"w, Y:.:

Yl

Then taking the natural logarithm, we obt<1in

and the undamped natural frequency ! k w= ; - V Wig

Yt

YI<

Solution: (a) The stiffness coefficient (force per unit displacement) is computed as k=~=- ..·......·=

1.37

y,

t (8 cycles)

= 8TD

1.62 sec

Example 2.3, A machine weighing 1000 Ib is mounted through springs having a total stiffness k 2000 lb lin to a simple supported beam as shown in Fig. 2.5(a). Detennine using the mathematical model shown in Fig. 2.5(b), the equivalent mass mE. the equivaient spring constant kE' and equivalent damping

!:

il Ii

,_ f·

42

Structures Modeled as a Single-Degree-oJ·Freedorr. System

Dampec Single-Oegree-ol-Freedom System

43

The equivalent mass is: L = 4(\itl

I

~.~.=.~=

f

1

The critical damp}ng is calculated from eq. (2.6):

(3)

c" (lb'sec\ 12792 [~~~~-! \

10

(Ans)

!

The damping is then calculated from etl (2.19):

(\»

Fig, 2,5 (a) System for Ex:affiplc 2.3; (b) Mathematica! modeL

Cs

£Ccr=O.10X 12792

Ilb.seC) 12.79,-.\

coefficient c£ for the system assumed to have 10% of the criticai damping. Neglect the mass of the beam.

So/mion: The spring constant k.b for a uniform simple supported beam is obtained from the deflection 3 resulting from a force P applied at the center of the beam:

Hence, P

kb~5

48£1

IT

48 X 10'

= - - - = 7500 Jb/in

40'

The equivalent spring constant is then calculated using eq. (1.5) for two springs in series:

I

I

2000

7500

=.---+---kF= 15791blin

(Ans)

2.7

(Ans)

10

SUMMARY

Real stmcturcs dissipate energy while undergoing vibratory motion. The most common and practical method for considering this dissipation of energy is to assume that it is due [0 viscous camping forces, These forces are assumed to be proportional to the magnitude of the veiocilY but acting in the direction opposite to the motion. The ftletar of proportionality is called the viscous dampiflg coefficient. It is expedient to express tbis coefficient as a fraction of the crifical damping in the system (the damping ratio {= c ICe,)' The critical damping may be defined as the least value of the damping coefficient for which the system will ;'lot oscll1ate whe!': disturbed initially, but it simply will return to the equilibrium position. The differential equation ef motion fer the free vibration of a damped single degree-or-freedom system is given by

my + C)' + k)l =

0

The analytical expression for the solution of this equation depends on the magnitude of the damping ratio. Three cases are possible: (l) eriticaUy damped system 1), For the underdamped system (1;< J) the solution of the differential equation of motion may be written as

44

StnJClures Modeled as a Srngle~Degree~of-Freedom System

in which

Dar:;ped 2.5

(jJ;;;

{ki;;;

is the undamped frequency

Y"" e -

';= clccr is the damping ratio where

~

C rr

2.,! km is the critical damping

and Yo and tlo are, respectively, the jnitial displacement and velocity. A common method of detennining the damping present in a sys~em is to

evaluate experimentally the logartthmic decrement, which is defined as the natural logarithm of the ratio of two consecutive ueaks for dis!)iacement or for acceleration. in free vibration, that is, .

;Yo cosh L

Repeat Problem L2 assuming that the system has 15% of crit:cal damping.

In 2.4 it is indicated that the tangent points to the displacement curve correspond to cos (wot- a)"" L Therefore the difference in We( between any twO consecutive tangent points is 211". Show lhat the difference in Wpf between any two consecutive peaks of the curve is- also 211", Show that for an underdamped system in free vibration the logarithmic decrement may be written as 1

k

2.10

decrease 5% on each consecucive cycle of motion. Determine the damping coefficient c of the system~ k = 200 Ib lin and m = 10 lb· sec1 l!n.

IT , "'T I

v'

In -'-'YiH

k

.

It has been estimated that damping in the system of Problem 1.11 is IO% of the

critical va~ue. Determine the damped frequency In of the system and the absolute value of the damping coefficient c. A single degree-of-freedom system consj.5ts of a mass with a weight of 386 lb and a spring of stiffness k 3000 !b/in. By testing the system it was found that a force of JOO Ib produces a relative velocity i2 in/sec. Find (a) the damping ratio §, (b) the damped frequency of vibration If), (c) logarithmic decrement 8, and (d) the ratio of two consecutive amplitudes.

2.11

Solve Problem 2.10 when the damping coefficient is c=2Ib, sec fin.

2.12

For each the systems considered in Problem L15, determine the equivalent spring COnstant k! and (he equivalent damping coefficient CE in the mathematical model shown in Fig. P2.12. Assume that the damping in these systems is equal to lO% of the critical damping.

0:

,i I

v

Fig. 1'2.3. 2.4

1 for

wi; "" wj fj2 - 1

(5: = .. ~

2.2 Repeat Problem 1.6 assuming that the system has 10% of critical damping, 23 The ampiitude of vibration of the system shown jn Fig. P2.3 is observed to

m

!

sinh wi;!

(;)D

a,

2.7

2.9

,

wVr + -VI) + yofjf!J (

where k IS the number of cycles separating two measured peak amplitudes Yi and

PROBLEMS

,

{=1

element Find: (a) rhe damping ratio fj, (b) [he damped period T D, (e) the logarand (d) the ratio between two consecutive amplitudes. ithmic decrement

mately equal to the undamped frequency_

:

+ wr)+ uor] for

EM. r

The damping ratio in structural systems is usually less than 10% of the critical damping (f< 0,]). For such systems, 6e damped frequency is approxi-

_C

45

2.6 A structure is modeled as a damped osciIlntor with spring constant k "" 30 Kips! in and undamped natura! frequency w= 25 rad/sec. Experimentally il was found that a force 1 Kip p;>oduced a relative velocity of 1.0 in/sec in the damping

2,8

2.1

System

Show that the displacement for critical and overcritical damped systems with init:al displacement Yo and velocity Vo may be written as y=e-
is the damped ff"'.....quency

Single·Degree~of·Freedom

It is observed experimentally that the ampIitude of free vibration of a certain structure, modeled as a single degree-of· freedom system, decreases from 1 to 0.4 in 10 cycles. What is the percentage of critical damping?

Fig, P2.12.

I.

Sing!e-Degree-of~Freadom

46

Structures Modeled as a

System

2.13

A vibration generator with two weights each of 30 lb with an eccentricity of lOin rotating about vertical axes in apposiLe direclions is mouo!ed on the
2.14

A syslem is modeled by two freely vibrating masses ntj aud nt2 interconnected by a spring and a damper element as shown In Fig. P2, 14. Determine for this system the differential equation of motion in terms of tbe reln.rlve mOtion of '.he masses U=Yl-Yl

Response of OneDegree-of-Freedom System to Harmonic Loading

Fig. P2.14. 2.15

Determine the relative motion t/ = y~ ~Yl for the syslem shown in Rg. P2.14 it~ terms of the natural frequency fl), damped frequency W.o, and relative damping. Hint: Define equivalenr mass as M = mj1Ji21{ml + 111,).

In this chapter. we will study the motion of slructures idealized as single~ degree~of-freedoIr. systems excited harmonkaHy, lhat is, structures subjected to forces or displacemenl<; whose magnitudes may be represented by a sine or cosine funclion of time. This type of excitation results in one of the most important molions in the study of mechanical vibrations as weB as in applications to structural dynamics. Structures are very often subjected to the dynamic action of rotating machinery which produces harmonic excitations due to the unavoidable presence of mass eccentricities in the rotating parts of such machinery. Furthermore, even in those cases when the excitation is not a harmonic function, tbe response of the structure may be obtained using the Fourier Method, as the superposition of individual responses to the harmonjc components of the external excitation. TIltS approach will De dealt wjth in Chapter 5.

3.1

UNDAMPED SYSTEM: HARMONIC EXCITATION

The impressed force F(t) aCling on the simple oscilhHor in Fig. 3.1 is assumed to be harmonic u:1d equal to Fo sin w/, where Fo js the peak amplitude and iiJ 47

I

48

Structures Modeled as a. Sing!e~Degree-of·Freooom System

Respo-:)Se of One-Degree"of-Freedom System 10 Harmonic Loading

49

or y=

(3.5)

in which r represents the ratio (frequency ratio) of the applied forced frequency to the natural frequency of vibration of the system, that is,

f.-y...j )b)

Fo k-m&/

-:-t-·--'h

>y ...- - - , [ -...

iii r=-

fo 5ir.;:;r

'"

Fig. 3~1 (a) Undamped osclllator harmonically excited. (b) Free body diagrf'.!:1.

is ~"e frequency o~ the force in radi,ans per second. The differential equation ob.amed by summrng all the forces m the free body diagram of Fig. 3, I (b) is

my + ky = Fo sin

?Ix

(3.6)

(3.1 )

Combining eqs. (3.3) through (3..5) with eq. (3.2) yields y(l) ==.1 cos

.

wt+ B Sin

Folk. S10 col 1- r

If the initlal conditions at time t:=: 0 are taken as zero (Yo";;:; 0, constants of integration determined from eq. (3.7) are

The solution of eq. (3.I) can be expressed as

A=O,

(3.7)

wt+~

Of)

0), the

rFolk

B=~17

(3.2)

which, upon substitution in eq. (33). gives ,,:here YC(,I) is the complementary solution satisfying the homogeneous equatlOn~ that IS, eq: (3.1) wIth the left-hand side set equal to zero; and Yp(l) is the partlc~lar solu~Jon based on the solution satisfying the nonhomogeneous differentJal equatIOn (3,1). The complementary solution, v (!) is 2:iven by eg (l.I7) as •< , •

y .. (1) =A cos where w

Wl-:-

B sin

(ut

(3.3)

[kim"

The nature of the forcing function in eq. (3.1) suggests that the partIcular solution be taken as

y (I)

~

Folk.

(3.4)

wher~ Y 15 the peak value of the particular solution. The subsritut10n of eq. (3.4) Into eq. (3.1) followed by canceHation of common factors gives

?naif"'" kY= Fo

(3.8)

As we can see from eg. (3.8), the response is given by the superposition of two hannonic tenns of different frequencies. The reSUlting motion is not hannonic; however, b the practlcal case, damping forces wiH always be present in the system ar:d will cause the last tenn, i.e., the free frequency term in eq. (3.8), to vanish eventually. For this reason, this term is said to represent the transient response. The fordng frequency term in eq. (3.8). namely y{t) =

yp(t) = Y sin iiJt

.)

- - - , (sm iilt - r sm WI 1 - r~

F,lk 1

Si;1

wt

(3.9)

is referred to as the steady-state response. It 1s clear from eq. (3,8) that in the case of no damp1ng in the system, the transient will not vanish and the response is then given by eg. (3.8). It can also be seen from eg. (3.8) or eg. (3.9) that when the forcing frequency ~s equal to the natural frequency (r == LO), the amplitude of the motion becomes infinitely large. A system acted

50

Structures Modeled as a SingJe-Oegree-of-Fre€dom System

Response of One--Oegree-of·Freedom System to Harmonic Loading

For this purpose, t:le reader should realize that we can write eq. (3,10) as

upon by an external excitation of frequency cOinciding with the natural frequency is said to be at resonance, In this circumstance, the ampljtude will' increase gradualiy to infinite. However, materials that are commonly used jn practice are subjected to strength limitations and in actua; structures failures occur tong before extremely large amplitudes can be auained.

3.2

(3.13) with the understanding that only the imaginary component of Fo etOJ<, i.e., the force component of Fo sin WI, lS acting and, consequently, the response will then consist only of the imaginary pan of Ihe total solution of eg, 13.13). In other words, we obtain the solution of eq. (3.13) which has real and imaginary components, and disregard the real component. It is reasonable to expect that t)le parlic"Jlar solution of eq. 13.13) wllJ be in the fOfm of

DAMPED SYSTEM: HARMONIC EXCITATION

Now consider the case of the one¥degree-of-freedom system in Fig. 3.2{a) vibrating under the jnfluence of viscous damping. The differentia! equation of motion is obtained by equating to zerO the sum of the forces in the free body diagram of Fig. 3.2(b). Hence

my + cy + ky = Fo sin

(3.1~)

Substitution of eq. (3.14) into eq. (3.13) and cancellation of the factor e'1:.Jt: gives

(3.10)

[ol

- m{j,}C + iciiJC + kC = Fa

The complete solution of this equation ag;::jn consists of the complementary solution Ye(r) and the panicular solution yp(t). The complementary solution is given for (he underdamped case (c < C,.) by eqs. (2.15) and (2.19) as Y,,{t) = e-~(A cos

(,tJo(+

B sin wni)

The particular solution may be foand by substituting lo be of the fonn

or

(3.11 )

c

k

lIl&l""t" fe&!

Yr, in this case assumed ano (3.12)

(3.15)

into eq. (3.10) and equating the coefficients of the sine and cosine functions, Here we follow a more elegant approach using Eu}er's reJat:on, namely

e;:"

51

By using polar coordinate form, the complex denominator in

eg~

(3.15) may

if

,'f

be written as

cos wI + i sin WI

)'p

I I

or

I

(3,16)

Ii

I~

J

where (bl

(3.17)

Fig. 3.2 (a) Damped oscitlator harmonically exciled. (b) Free body diagram.

I ,

.i

52

Structures Modeled as a

Sir,gle~Degfee..j)f~f(eedom

~esponse of One~De:gree·of·Freedom System to Harmo[',ic Loading

System

The response to the force in Fo sin WI (the imaginary component of Foe i'''') is then the imaginary component of eq. (3.;6), na:nely,

r-O'" ._....

t, 0

(3.18)

,i

o.

,

or

Yp= Y sin (Wi ... Il)

(3.19)

l

:

1

i i i 1

I;;'

!

,

i

,

<125

!

, 'e

where Ys, = Folk is seen to be the static deflection of the spring acted upon by the force F0; g= c lew the damping ratio; and r iJJl lV, the frequency ratio. The total response is then obtained by summing the complementary solution (tran-

sient response) from eq. (3.11) and the particular solution sponse) from eq. (3.20), that is,

(steady~state

re-

i

I

i

I

II

I

I

I

i 1\

II I

(3.20)

(3.21)

r--i-

I

~

II.

tan B=

,

,

,

and

i i

!

where

is the amplitude of the steady~s"'te motion. Equations (3.18) and (3.17) may conveniently be wnuen in terms of dimensionless ratios as

,

I I

0.125

4

53

i\,,\

""I'\.

tI~~

-t'-, t '" 0.1

o

i

i1: 1 t-,

'\. ,\ 1\.\

, r<~'t- ~

1,\ "I

~

'\,

,/

r-:0

~ ;::,.

I I

2.

I.

3.

(3.22)

:F'ig. 3.3 Dynamic magnificatiou factor as a function of the frequency ratio for various amounts of damping.

The reader should be warned that the constants of integration A and B must be evaluated from initial conditions using the total response given by eg. (3_22) and not from just the transient component of the response given in eq, (3.11). By examining the transie
It may be seen from eq. (3.23) that the dynamic magnification factor varies with the frequency ratio r and the damping ratio f Parametric plots of the dynamic magnification factor are shown in Fig, 3.3. The phase angle 8, given in eq. (3.21), also varies with the same quantities as it is shown in the plots of Fig. 3.4. We note in Fig. 3.3 lhat for a lightly damped system, the peak amplitude occurs at a frequency ratio very close to 1~ that is, the dynamic maQnification factor has its maximum value virtually at resonance (r = 1). It can'" aiso be seen from eq. (3.23) that at resonance the dynamic magnification factor is inversely proportionai to the damping ratio, that is,

Y

D= -

y"

= """E==;c;==.c:r

(3.23)

1

D(r=I)=21;

(3.24)

54

Structures Modeled as a

Single~Degree~of-Freedom

•

180

······~··I

!

l

I

..

b- ±iiz !

~

~.- r-' !

o

/'

--

t=

fj

'

!

--ri

' ,2::

!

..

,

,

I

W I

,

large mass of the machine. Figures 3.5 and 3.6 show, respectively, the schematic diagram of a beam~machine system and the adapled model. The force at the center of a slmply supported beam necessary to deflect this point one unit (i.e., the stiffness coefflclent) is given by the formuia

-

k=~= L'

i

+_ i

48X30XIO'X 128,4 =61920 Ib/in •

w

rT

1_

Y III

=

161~92() V 16,000/386

I-~·-- =

38.65 rad/sec

3

i i

the force frequency

as a function of tbe frequency ratio for various amounls of

300 X 27f W = --.. -~ = 31 A I rad/sec

60

Although the dynamic magnification factor evaluated al resonance is close to its maximum value, it ~s not exactly lhe max.imum response for a damped system. However, for moderate amounts of damping, the difference between the approximate value of eq. (3,24) and the exact maximum is negligible.

Example

3~1.

I I I

The natural frequency of the system (neglecting the mass of the beam) lS

I

I 2 r '" w/w

Fig, 3.5 Dlagram [or beam-machine. system of Example 3, I.

i

,

I

I

::::

~- i

0

~

L-···~----{I

4 -

! •

El

--

1

i

!

FreQuency ratio

Fig. 3.4 Phase angle damping.

/,

!

04 / 1

/oi/ Va. .// 1

.~

'

V l =- ~f:::- r-.

if

l'

I V

, V

IIV

~?, I '-44 I. '( V / , !: 2 / L' ,II i /

/

I

-

,

/,

/ 0,'1

0

55

,-

i..,..- f-j ,

·/:02

f

Response of One-Degreo-of-Freedom System 10 Harmonic Loading

I.

...-r

0.1

i

~O

Syslern

+

A simple beam supports at its center a machine having a

1in i:;,

t

weight W 16,000 lb. The beam is made of two standard S8 X 23 sections with a clear span L = 12 ft and total cross~sectiona! moment of inertia 1= 2 X 64,2 = 128A in4. The motor runs at 300 rpm, and its rotor is out of balance: to the extent of W' = 40 lb at a radius of eo = 10 in. What will be the amplitude of the steady-state response if the equivalent viscous damping for the system is assumed 10% of the critical?

TT !m-m'}~;

1- ...

, m

~,

~~Td.

/##»,$#/////.&7/.,w///)..w~ 1,1

SoIUlion: This dynamic system may be modeled by the damped osciHato/ The distributed mass of the beam will be neglected in comparison with the

y

'y

'"

Fig. 3.6 (a) Mathematical model for Example 3.L (b) Free body diagram.

I

'I

56

Response of OnewDegreewof~Freedom System to Harmonic Loading

Structures ModeJed as a Single-Oegree-of-Free-dorn System

t§K;P~$=iS1 T

and the frequency ratio

57

y

Ftr<:---.:;=;s':;S·

I

W8X 20

f'5'

Referring to Fig, 3,6, let m be the total mass of the motor and In the unbalanced rotating mass. Tnen, if y is the vertical displacement of the nonrotating mass (m - m ') from the equilibrium position, the displacement YI of m' as shown in Fig. 3.6 is J

Yt = Y

+ eo sin

(b)

(a)

Wi

Fig. 3.7 (a) Diagram of frame for Example 3.2. (b) Mathematical model.

The equation of motion is then obtained by summing forces along the vertical direction in the free body diagram of Fig. 3.6(b), where the inertial forces of both the nonrotating mass and the unbalanced mass are also showll. This summation yields (m - m')y + m),

+ cj + ky = 0

(b)

Example 3.2. The steel frame shown in Fig. 3.7 supports a rotating machine that exerts a horizontal force at the girder level, F(t) = 200 sin 53! lb. Assuming 5% of critical dampir.g, determine: (a; the steady-state amplitude of vibration and (b) the maximum dynamic stress in the columns, Assume that the girder is rigid.

Substitution of YI obtained from eg. (a) gives

(m-m')Y+m'(Y-eoli.l' sin &;f)-:-cj+ky

Solution: Thi~ structure may be modeled for dynamic analysis as the damped QsciHator shown in Fig" 3,7(b), The parameters in this model are computed 'as follows:

0

and with a rearrangement of terms

mj.i+ C); +ky = m'eo{i} sin

Wl

(c)

3 x 30 x 10' x 2 x 69<2 ' , = 2136 fb/in (12 XL)

(a)

0<05

This last equation 1S of the same form as the equation of motion (3. I 0) for the damped oscillator excited hannonically by a force of amptjrude

w

Ym

r=

F,=(40)(1O)(31.41)'!386= 1022lb The amplitude of the steady-state resulting motion from eqs< (3< 19) and (320) :5 then

y

1022!6L920

J(l -- 0<813')' + (2 x 0<813 x 0< I)' y= 0<044 in

.

= 0.0936 m

II; = 12i36 x 386 =7AI rad/sec

(d)

Substjtuting in this equation the numerical values for this example. we obtain

200

YM

Y

15,000

53 =--=0<715 w 7AI

iii

Tbe steady-state amplitude from e<;s< (3.19) and (320) is

y

,?~~~y~<,~,~~~ = 0<189 in J(I- r'l'+ (2r$'

(Ans<)

'. A coit dls.ptacement at (he top of a pir.ned supported column requires a force equal to 3El JL).

58

SL'lJclures Modeled as a Single-Degree-of·Freedom System

Response of One·Degree-ol-Freedom System to Harmonic loading

59

(b) Then, the maximum shear force in the columns is

3EIY

V~, =

= 2018 Ib

";;

the maximum bending moment

l.1 \l\lU

=V

mIlA

""=, , 0

L = 36,324 Ib . in

~ ·c

and the maximum stress

:r

Mm."

crm~~ = - -

lie

=

36,324

=

2136

.,EE psi

(Ans.)

,.~

"

in whrch J Ie is the section modulus.

3.3

F re<;ucnzy nniQ (r .. {-:i/wi

EVALUATION OF DAMPING AT RESONANCE

We have seen in Chapter 2 that the free~vibration deeay curve oermirs the ev~iuation .of d~mping of a single-degree~of-freedom system by si~pjy calcu~ latmg ~h~ loga:l{hr:n d~crernent as shown in eq. (2.28). Another fechnique for detennlOlOg dampmg 1$ based on observations of steadv~st3,te harmonic re~ sponse, :hi~h requir~s barmor.ic excitations of the st~cture in a range of ,.freq~encles In lb~ neighborhood of resonance. With the application of a harmOnIC force Fo S10 iVt at cJoseJy spaced values of frequencies, the response curve f?r the structure can be plotted, re,sulting in displacement amplitudes as a functiOn of the applied frequencies. A typical response curve for such a moderately damped structure is shown in Fig. 3.8. h is seen from eq, (3.24) that, at resonance, the damping ratio is given by

t=

j

2D(r- 1)

(3.25)

in

whe:-~ D(r= 1) is the dynamic magnification factor evaluated at resonance, pracuce. the damping ratio is determined from the dynamic magnification factor evaluated at the maximum amplitude, namely,

t

I

~=--

2Dm

where

D=!::. }"

(3.26)

Fjg. 3.8 Frequency re:,ponse curve for rr.oderalely damped system.

and Y", is the maximum amplitude. The error involvec in evalualing the damp~ ing ratio using the approximate eq. (3.26} is not significant in ordinary Structures, This me.thod of determining the dampir.g ratio requires only some simple equipment to vibrate the sln;c~ure in a range of frequencies that span the resonance fre,quency and a transducer for measuring umpHtudes; nevertheless, the evaluation of the static displace:r.ent YH = Folk may present a probJem since, frequently, it is difficult to apply a statk lateral load to the structure.

t

3.4 BANDWIDTH METHOD (HALF-POWER) TO EVALUATE DAMPING An examination of [he response curves in Fig, 3.3 shows that the shape of these curves is controlled by the amount of damping present in the system; in particuJar. the bandwidth, that is, the difference between two frequencies co:-responding to the same response amplitude, is related to the damping ];1 the system. A typical frequency amplitude curve obtained experi:r.entaUy fo:- a mOderately damped structure is shown 1n Fig. 3,9. In the evaluation of damp~ ing, it is convenient to measure the bandwidth at 1/[2 of the peak amplitude as shown in this figure, TJle frequencies corresponding in this bandwidth II and Ix are also referred to as hollpolVu po;nts and are shown in Fig, 3.9. The values of the frequencies for this bandwidth can be determined by setting !he

60

Response o~ One~Degre6-of~F'reedom Systei"1 to Harmonic Loadir.g

Struclures Modeled as a Single-Degree-01·Freedom System .12

kp'~k ~ 0.11,34 io

I ,

,

.10

I

I

i

,

AV' ' I

f. ""

'\

f,

1'7.05: 18.97

i . 02

oB

, i

10

-12

/'

V'

14

~

i

20

I

]

r=

~_ fZ - ( l ~ i.87

,

••

18

,

(3.27)

j,+/,

'"

/

and

:1

- /,+1,

/=~2-

-r;:f7;"' ~= 5.2%

+t, '" 17.99,"" 2

16

.....-= I,-/i ..-

1 iiJ,-'W,

2 since

I

I, fr"'~""'1"

i

0,0802

I

f- H

or i

g=~-

I ,,';'Pt'ilk

" i

!

,

I

I

,

:

61

"-

22

, I:

i

I',.

i i

26

2.

Example 3.3. degree-of~freedom

Experimental data for the frequency response of a single system are plotted in 3.9. Detennine the damping ratio

of this system.

Solution: From Fig. 3.9 the peak amplitude is 0.1134 in; hence the amplitude at half-power is equal to

Exciting frequency. f r.ps

0.1134/[2

Fig. 3.9 Experimental frequency response curve of Exa:np!e 3.3.

0.0802 in

The frequencies at this ampUtude obtained from Fig, 3.9 are

response amplitude in eg. (3.20) equal to 1 fj2 times the resonant ampUtude given by eq. (3.24), that is,

Squaring both sides and solving for the frequency rat;o results in

/1 17.05 /,= l8.92 The damping ratio is then calculated from eq. (327) as

f=l, - 11 or by neglecting

j,+/1

e in the square root terms ?, =

1-

r~= J

2g' -

18.92 - 17.05 Ij= i8.92';' 17.05

2§

(Ans.)

5.2%

2£2 + 2~

r,=I-Ij-e r2= 1 + ~~

t·

Finally, (he damping ratio is given approximately by half the diffe:ence between these half-power frequency ratios, namely

3.5

ENERGY DISSIPATED BY VISCOUS DAMPING

The energy, Eo, dissipated by viscous damping during one cycle of harmonic vibration of frequency (i; is equal to the work done by the damping force cy during a differential displacement dy integrated over one ;;>eriod of vibration T = 2.,,-lw. Heoce,

Eo

-'"I" (cy)dy _0

'

r "1"1f/';'

Jo

[2 . . /,;

dy (ey) - , dt = d. ..0

ej'dt

(3.28)

62

Structures Modeled as a Sing!e-Oegree-of-F{eedom System

Response of One-Oegree·of·Freedom System to Harmonic Loading

The velocity, y;;;;; y(t), for damped oscillator acted upon by the harmonic force, F= FQ sin wr, is given by the derivative of eq, (3.19) as y(l)= Ywcos ((I;r- (J)

a?L

and using eq, (3,20) sin8=

(3.29)

2(rY

which substiWled in (3,31) yields

which sUDstituted in eq. (3.28) gives ('2"./w

Eo=cy

2

63

1

cos (Wf- (l)d(=

1TCliJy2

= 21T!;rky1

! "i

2(r

(3,30)

1:

Folk

where

"

21T(rkY'

c

Tv

(=-,

w

r=-. w

c~

It =1 V JJJ

and

Cee

= 2jk;;:;

which is equal to the energy, ED, dissipated per cycle by (he viscous dampjng force as given by eq, (3.30).

Equation (3.30) shows that the energy dissipated by visco:Js dnmping is proportional to the square of the amplit"Jde of the motion Y. The energy dissipated by viscous damping is provided by the work Wp performed by the external force. During one cycle. the work of the external force F:::::: Fo sin WI is WF =

1",· . - 1"'" . Fo

()

~

3m lIJt dy

1,

Po

=

()

SUI

dy wI ··--dr dr

r1,,-j::'

[1", sin iillJj(l}d!

in which j(l) given by eq, (3.29); Heoce WF ~

J,

r2f.'r:.

iF. sin iVI] [iVY cos (wi - 0)] dr

= 1TFOY sin e

(331)

To demonstrate that work, W p , of the exciting force in eq. (3.3 1) is equal to the energy dissipated, ED> by the viscous force in eq. (3JO), we need to substitute the sine of the phase angle 8 intO eq. (3.31). Thus, from eq. (3.2i), we have sin 8 2fT cos8 =1=7 sin'

e

··si·~rfT+ cos 2 0 = (1 -

3.6

EQUIVALENT VISCOUS DAMPING

As mentioned in the introductory sectio:1s of Chapter 2, the mechanism by which structures dissjpr.:e energy during vibratory motion is usually assumed to be viscous_ This assumption provides the enormous advantage that the differentiai equation of motion remains linear for damped dynamiC syslems vibrating in the elastic range, Only for some exceptional situations such as the use of frictional devices installed in buildings to ameliorate damage rest.;,lting from strong motion earthquakes, viscous damping is usually used [0 account for frictional or damping forces in structurr.l dynamics. TIle numerical v<:'.lce aSSigned to the damping coefficient is bil;sed on values obtained experimentally and the determination of an equivaknl YISCOUS damping. The concept of equiva:ent viscous damping is based on test results obtained llsing harmonic forces. Thcs, in reference (0 the experimemal frequency response ptot in Fig. 3.9, equiva:ent dampi!1g may be based on the maximum relative ampli:.ude of motlo!1, D,,, = Yn.lYs1> or on the band with corresponding to

freque:lcies 11 and A at amplitude equal to Thus, the eqUivalent viscous damping (3.26) as

D"Jfi. .f.,q may

Y. 2Y."

be calculated from eq.

(3.32)

(2~r)' rZ)l

+ (2r§)z

or from eq, (3.27) as (3.33)

+/'

64

Structures Modeled as a

S;ng!e-Degree-o~-Freedom

Response of One-Degree-or·Freedom System to Harmon;c Loadlng

S:lsteI":"J

65

Restoring Force

Alternatively, the equivalent viscous damping ~eq may also be evaluated experimentally using the expression for the logarithrn1c decrement fj from eg. (2.28) 0' app
R~ ~ ____ _ E: "" Maximllm st.'"ai:!

(3,34)

energy sto;ec

e· "" Energy dissipated in one cycle

y

The most common definition of equivalent viscous damping is based On equating the energy dissipated, in a vibration cycle of the actual structure, to the energy dissipated in an equivalent viscous system. Hence, equating the energy, EO, dissipated in a cycle of harmonic vibration detennined from experiment to the energy, ED, dissipated by an equivalent viscous system given by eg. (3.3D), we have

Fig. 3.10 Restoring ferce vs. displacement during a cycle of vibration showing the energy dissipated e" (area within the loop) and the maximum energy stored (triangular

area under tpax.:mum displacement).

Example 3.4. Laboratory tests on u structure modeled by the damped spring-mass system (Fig, 3.1 1a;, are conducted to evaluate equivalent viscous damping usjng (3) peak amplitude and (b) energy dissipated. The experimental restoring force-displacement plot at resonance is shown in Fig, 3,11 (b).

and

or k

100 ~ , -_ _ _,

(3,35)

F(l)

W"'" 38.6lh

LS.sinwr(Jb)

c

in which Es, the strain energy stored at maximum displacement if the system were eIastic, is given by 1 , E , ~-kY 2

(,)

(3.36) £~

In the determination of the energy. EO, dissipated per cycle and the elastic energy, E" stored at the maximum displacement, an experimer:t is conduced by vibrating the structure at the resonant frequency for which r= wlw= L At th:s frequency, damping in the system has a maximum effect. With appropriate test equipment and measuring instrumentation. the restoring force and displacement during a cyde of vibration are measured to obtain a plot of the type shown in Fig. 3.10. The area endosed in the loop during one cycle of vibration is equal to energy dissipated. E', and the triangular area corresponding the amplitude Y is equaJ to the maxImum strain energy E., Consequently, the equivalent viscous damping is evaluated by eq. (3.35) from the experimental results E~ and En with r= 1, obtained from resisting force-displacement plot

i1845(lb,in)

Mwmrnn strain energy stored

Oisp!a¢em(1!\l

(b)

Fig. 3.11 (a} ~iatherr.atical mode! for Ex.ample 3.4. (b) Ex.perimental restoring forCe-amplitude plor.

Structures Modeled as a Single,Degree-cf-Freedorn System

66

Solution:

Response 01 One-Degree-of·Freedom

The stalic deflection, mass, and natural frequency are: W 38.6 =~.. = 0.1 (lb· sec'lin) g 386

and

Ik =,1f\OO =3L62radlsec

w= 1-

¥m

¥

OJ

(a) Equivalent viscous damping ca.lculated from peak amplitude:

Sys~em

10 Harmonic Loading

67

where the support of the simple oscillator :nodding the structure is subjected to a hannonic motion given by the expression y~ (r) =

Yo sin

wt

(3.37)

where Yv is the maximum amplltude and liJ is the frequency of the support motion. The differential equation of motion is obtained by setting equal to zero lhe sum of the forces (includIng the inertia! force) in the corresponding free body diagram shown jn Fjg. 3.12(b}. The summation of the forces in the horizontal direction gives (3,38)

0.015 \ (0 = 0,0576 = 5,76% (, .13) 2

by eg. (3,32)

The substitution of eq. (3.37} into eq. (3.38) and the rearrangement of terms result in

(b) F,..quivalent viscous damping calculated from energy dissipated E" = 0.66

my + cy + ky

(lb' in) and E, = O.845(lb, in) shown in Fig, 3.11:

kyo sin

w( + cWyo cos

wI

(3.39)

lWO harmonic terms of frequency w in the right-hand side of this equation may be combined and eq. (3.39) rewritten as [similarly 10 eqs, (1.20) and (L23)]

The

0.66

=0,0621 = 6,21%

4'11'(1.0) 0,845

by eg. (3.35)

my + cj + ky ~ Fo 51n 3,7

RESPONSE TO SUPPORT MOTION

There are many actual cases where the foundation or support of a structure is SUbjected to time varying motion. Structures subjected to ground rnot~on by earthquakes or other excitations such as explosions or dynamic action of machinery are examples in which support mottOl)S may have to be considered in the analysis of dynamic response. Let us consider in Fig, 3.12 [he case

(WI

+ f3)

(3.40)

where

(3.41) and tat, f3 = cwlk = 2Ft,

(3.42)

h is apparent that eq. (3.40) is l.~e differential equation for the oscillalor excited by the harmonic force Fo sin (fur + 13) and is of the same fo:m as' eq. (3.10). Consequently, the steady-state solution of eq. (3.40) is given as before by egs, (3.19) and (3.20), except for ,he addition of the angle {3 in tile argument of the sjne function. that is, Folk

y(Il

=

sin (WI + {3-1!)

(3,43)

jO': I'j'+ (2r[,)'

or substituting Fa from eq. (3.41)

y, Fig. 3.12 (a) Damped simple oscillator harmonically excited through its support. (b) Free body diagram including inertial force.

...

/1+. .(2r[,)'. ,

[(1 - I'l' + (21/)'

sin (WI + {3

I!)

(344)

Equalio::t (3.44) JS the exp:-ession for the relative transmission of the support motion to the oscillator. This is an important problem in vibration isolation in

68

Response of One-Oegree"ot"Freetiom System to Harmonic Loading

Structures Modeled as a Single-Degree··ol-Freedom Syslem

which equipment roust be protected from harmful vibrations of the supporting structure. The degree of relative isolation is known as transmissibility and is defined as the ratio of the amplitude of motion Y of the oscillator to (he amplitude Yo, the motion of the support From eq. (3.44), transmissibility Tr is then given by

69

2~-------7~~k-~~~------~----------~

(3.45)

Analogously to the motion transmitted, we may find the acceleration transmitted from the foundation to the mass. The acceleration transmilted to the mass is given by the second derivative of y(l) in eq. (3.44) as ji (I)

~ ~.?YcF(2·~ff sin (Wt + f3 - fJ)

/(1::: r\{ + {2rf/

(3A6)

while the acceleration )Ii(1) of the foundation is obtained from eq. (3.37) (3.47)

The acceleration transirnissibillty, Tn is then given by the ratio of the amplitudes of the acceleration in eqs. (3.46) and (3.47). Hence,

fr~uer.cy

Fig. 3.13 TransmissibiHty versuS frequency ratio for vibration isolation.

(l/sec) for the type of t:lbber pad material used in the isolation. 'What is the stiffness of the isolation requ:red to reduce the acceleration transmitted to the instrument to 0.0 I g? . Solution:

Setting the acceleration transmissibHlty given by eq. (3.48) equal

. to 0.1 and beginning with an assumed value ~;;;; 0.10 for the damping ratio, we have

J1 + (2r~J' 2 -+(2rl)

(3.48)

T,=~-

J(1

It may be seen that the transmissibility of acceleration given by eq. (3.48) is identical to eq. (3.45), the transmissibility of displacements, Hence, the same

/2.

Exampie 3.5. A delicate instrument weighing 100 lb is to be mounted on a rubber pad to the floor of a test laboratory where the vertical acceleration is 0.1 g at a frequency of f= 10 cps. [t has been determined experimentally that the ratio of the stiffness, k, to the damping coefficient, c, IS equal to 100

-

r)2gSS

=0.1

Squaring both sides

expression wHJ give either displacement or acceleration trar.smissibility.

A pJot of transmissibility as a function of the frequency ratio and damping ratio is shown in Fig. 3.13. The curves in this figure are simHar to curves in Fig. 3.3. representing the frequency response of the damped oscillator, the major difference being that all of the curves in Fig. 3.13 pass through the same It can be Seen in Fig. 3,13 that damping point at a frequency ratio r = tends to reduce the effectiveness of vibratlon isola:ion for frequencies greater than this ratio, that is. for r greater than

ratio If "" w!wl

+ 0.04/ - ....! ~~~,~= = 0.01 (I

and solving this quadratic equation yields T'

= 13.346

r;;;;

wlw;;;' 3.653

w=27if=27T·1O w

62.83radlsec

wlr = 17.20rad/sec

m = 1001386 = 0_2591b· sec'lin k = mw'

0.259 x 17.20' = 76.641blin

(a)

70

Resp'onse of One-Degree-o(·Freedom System to Harmonic Loading

StructureS Modeled as a Single-Degree-of·Freedom SysLem

71

we obtain

Now, we check value of damping:

u

klc= 100

j (l

Yo

c = kll 00 = 76.221100 = 0.766(lb· sec lin)

r')' + (2rEj'

(353)

where 0 is given in eq. (3.2:). Critical dampi:1g: Example 3,6, If the frame of Example 3.2 (Fig. 3.7) is subjected to a sinusoidal ground motion y .. (t) = 0.2 sin 5.3[, determine: (a) the transmissibility of motion to the girder, (b) [he maximum shearing force in the supporting columns, and (c) maximum stresses in the columns.

c" = 2 j'k;; = 2 }715.154 X 0259 - 8.91 (!b . sec lin) Then, the calculated damping ratio is (=clccr=O.76618.91 =0.086

whIch is somewhat less than the assumed value , 10. If desired, an itera!ive cycle could be. performed introducing § = 0.086 in eq, (a) and repeating the calculations. Equation (3.43) provides the absolute response of the damped oscillator to a harmonic morion of its base. Alternatively, we can solve the differential equation (3.38) in terms of the relative motion between the mass m and the support given by

Sotwion: 3.2 as

(a) The parameters for this system are calculated in Example

k~21361blin

0.05

Yo = 0.2jn

Y. = 0.0936 in tv

(3.49)

tv = 5.3 rad/sec

which substituted into eq. (3.39) gives

mu

r

+ eLl + ku =

F~rr(t)

(3.50)

= 7A1 rad/sec

0.715

The transmissibility from eo., (3..15) is

where Fef'r(t) = - my~ may be interpreted as lhe effecti~e force acting on lhe mass of the oscillator, and its displacement is indicated by coordinate 11. Using eg. (3.37) to obtain y, and substituting ;n eq. (3.50) results in

mil + cu + kH

=

myo iii sin ii;t

(35 J)

Again, eg. (3.51) is of the same form as eg. (3.10) with Fo myod. Then, from egs. (3.19) and (3.20), the steady-state response in tenns of relative motion is given by

(Ans.)

(b) The maximum relative displacement U from eo.. (3.53) is

u

j (!

- r')'

+ (2rtl'

0.206in

Then the maximum shear force in each column is (3.52) (Ans.) or substituting (c) The ,maximum bendIng moment iiJ'

kim

.....is/'".= r2 oJ'

l1es';)or.se of

72

and the corresponding stress

Of:e~Degree-of·Freedom

System to Harmor.ic Loading

73

Tee su1;)stitution of these vatues into eq. (3.45) gives 2327 ?si

y J1 + [2 (OA) (L755)1' -y~ ~ 'j=(l=-~~<--~L'='75O=S;';,)cf2 ;"'+~[2;}c(~0~A':")(~I";;.7';"55",,)';fJ'

T,

(Ans<)

in which lle is ::l:e section modulus,

0<6869

and

Example 3.7. A bumpy roadway is modeled as sinusoidal curve having a vertical displacement from peak to peak of 5 in at a distance of L = 25 ft, between peaks as shown in Fig, 3.14, An automobi;e ideaIized as a singledegree-of-freedom system, appropriate for a first approxima~ion. weighs 3500I'O fully loaded. The stiffr:ess of the sllsper.sion system of the automobile is 1000 lbl in and its viscous damping system provides 40% of the critical dampIng. Determine: (a) the amplitude of vertical vibration of the automobile when traveling at 50 mph and corresponding transmissibility, and (b) the speed of (he automobile that wm result in maximum vertical motion. Assume that the vehicle remains in contact with the road and that the tires are rigid in comparison w:th the suspension system of the automobile.

y~

0<6869 X 25

~

L72in

(Ans<)

(b) Speed for maximum vertical amplitude: At relatively small values of damping. the maximum ampiitude occurs close to the resonance condition of r= L However, in this problem. the value of the damping, 40%. is large zoe it is necessary to determine the value of r that wiH result in maximum amplitude. Sqnaring eq, 0.45) and substituting the damping ratio 0.4 gives

1 + 0<64r'

1 + (0<8r)'

1; = U?j2 + (0<8r)' then

The vertical displacement of the tires is y{t) = Y{J sin W!, where Yo = 2.5 in. The forcing frequency is iiJ = 211ft, where the forcing period T=' Llv is the time taken by the automobile with velocity u to traver the distance L over one cycle of curved roadway. Hence,

W

27TU IL

d(T;) ~ 0

dr

(a) 'Ylaximum displacement Y: v

~

w=

50 mph

~

r;;-

w= / \ m

!

Y

w 18A3 r~-~-w

18A3rad/sec

{""iooo X 386 =.

1050

1050 ~ 937rad/sec<

Maximum vertical amplitude occurs at this value of the forcing frequency, which by eq. (a) corresponds to the velocity

7334 ft/sec

2"X 7334

w= O<893w= 0<893 X

and

(a)

gives r = OJ = 0.893 w

u

fiJL

(937) (25)

2".

2"

3731 ftlsec = 25A4 mph

(Ans)

At this velocity, the transmissibility and ampIitude of vibration of the automobile would be calculated by eq. (3A5) as

= 1O.50rad/sec

1.755 T,

5 in.

y

j 1 + [(2)(OA)(O<893)]'

Yo

j 6= 0<893')' + [(2) (CAl (0<893)]' y=

L6Sy,~

L65 X 25 ~ 4J3in

(Ans)

Example 3.8~ A machine having a total weight of 1800 lb, including its foundation. is to be isoLated from the vibration of the ground, which is 900 cps

Fig. 3.14 Idealized bed-road for Example 3.7.

. i

; 5:

!

I I,

I, I!

L6S

and

- -...-;

,

74

Response of One-Degree-of·Freedom System to Harmonjc Loading

Structures Modeled as a Single-Degree-of-Freedom System

owing to other machines operating near. Determine the stiffness of rubber isolation spring (0 limit the transmitted vibration to lJ 10: (a) neglect damplng and (b) consider damping given by the exp:-ession c = k 1170 obtain experimentaHy [units of c (!b sec lin) and of k Oblin)]_

and

w

75

143.24

w= - = - - - = 39.244 (rad Isec) I' 3.65

r-;;-

I-~ = 39.244 (rac./sec)

\ 4.663

m

SolutioJt:

then

(a) §= 0

k=7!8L381b/in

o

By eq. (3.45) with

The damping is then determined from

Y

T,

I

~..,.,-----,-

:!:o-/)

Yo

= 0.1

c=k/170=718L31l70

as

-(1._/)= !O

c = 42.241b· sec/in

-1+,.'=10 1"=11

w

and the damping ratio as

1'=3.3166=w

900

c"

rad

2~

w = - = 143.24-217 sec

m= w=

W

-~I'

Then

k = w 2m

(b) Assume §

2j(7181.244)(4.663) = 348.34 (1b·seclin)

42.24

~=--=0.1212=

348.34

1800

-- = 4.663 Jb· sec l 1m.

Repeating the calculations for

;= 0.11

12%

gives

386

143_24 3.3166

r' - 6.7916 - 99 =0

= - - - = 43 lS8rad/sec

.

r'

(43.188)'(4.663) = 8698Jb/JO

(Ans.)

13.909

W

1'= -

w

= 3.73 ~

14324

w=--=38.40= / 3.73 \ m

0.10

k

Squaring eq. (3.45) and substituting §= 0.10 gives I + (0.21T)' _ I I \' , - (l - 1")' + (0.21')'

T' _

-\1O)

which results in the following quadratic equation in /-:

r' - 5.96/ - 99 = 0 r' = 13.366

w 1'=-=3.65 w

k

6876 1b lin and c = 170

= 2/ 6876X4.663

= =

40.45117 . sec lin 358.14]b· sec lin

and 40.45 358.14

~-=0.113=

[1%

This calculated value, 11 % for lhe damping ratio, agrees with the last value tried. Therefore, the requked sprjng conswnt for the isolation is k 6876 tb/in as calculated above,

7$

Structures Modeled as a Shgle-Oegree-ol-Freedom System

3.8

FORCE TRANSMITTED TO THE FOUNDATION

Response of One.Degree-of-Freedom System 10 H;:lrmO:lfC Loading

and

In the preceding section, we determined the response of the structure to a ham10nic motion of its foundation. In this section we shall consider a sjmilar problem of vibration isolation; the problem now, however, is ~o find the force transmitted to the foundation. Consider again the damped osci;Iator with a hannonic force F(t)·= Fo sin wt acting on its mass as shown in Fig, 3.2. The differential equation of motion is mji

with the

stead)'~srate

e,

(359)

I

Then from eqs. (3.54) and (3.57), 'he maximum force AT transmitted to 'he foundation is

I

+ (.'); + ky = Fc sin wi

are given, respectively, by eqs. (3.20) and (3.21) as

T,

(3.54) and ()=

2fr

1=7

AT I 1 + (2fr)'-Fo = Y (1 - r')' + (2rfJ'

tan ¢

(3.55)

(3.61 )

It is interesting to note that both the transmissibility of motion from the foundation to the structure, eq. (3.45), and the transmissibility of force from the structure to tbe foundation, eq, (3.61). are given by exactiy the same function. Hence the curves of transmissibiHty in Fig. 3.13 represent either type of transmissibility, An expression for the total phase angle 4J in eq, (3,57) may be determined by taking the tangent function to both members of eq. (3.59), so that

The force transmitted to tbe support through the spr:ng is ky and tbrough the damping element is c~v. Hence the total force transmitted F T is

Fr=k:y+C)'

(3.60)

In this case, the transmissibHity T~ is cefined as the ratio between the amplitude of the force transmitted to the foundation and the amplitude of the applied force. Hence from eq. (3.60)

solution. eq. (3.]9),

tan

1 + (2fr)'

A 'F = F" . -:-;--T'i-"-'o-::--;. 'Y (l - ?)' + (2~r)'

y= Y sin (w{- fl) where Y and

77

'an e - tan p 1 + tan Btanp

Then substituting tan 0 and tanp, respectively, from eqs. (3.21) and (3.58), we obtain

Differentlatjng eq. (3.19) and substituting 1n eq. (3.55) yield Fr= Y[k sin (iiJr- 8)+cw cos (WI- 8)]

(3.62)

or ". 0 '" F' T= Y ~'k'-+cw" s;n(i5Jl-IJ+,..,J

(3.56) (3.57)

in which (3.58)

Example 3.9. A machine of weight W.= 3860 Ib is mounted on a sImple supported steel beam as shown in Fig, 3.15(a). A piston that moves up and down in the machine produces a harmonic force of magnitude Fo = 7000 lb and frequency liJ:=: 60 radlsec. Neglecting the weight of the beam and assuming 10% of the critical damping, determine: (a) the amplitude of the motion of the machine, (b) the force transmitted to the beam supports. and (c) the corresponding phase angie.

78

StrJclures Modeled as a Singte-Degn.e-of·Freedom System

Response o!

One·Oeg;ee~of·Freedom

Hence the amplitude of the force transmiued

w

E

System to Harmonic LoadL'1g [0

the foundation is

Ar=F,T,= lO,827lb

30,000 ksl

79

(Ans.)

"1"'12010,"

(c) The corresponding phase angle fror:1 eq. (3.62) is y

(Ans,) (,I

Ib)

Fig. 3.15 (;1) Beam-machine system for Example 3.5. (b) !V1alhematlcal model.

Solution: The damped oscil1ator in Fig~ 3.1S(b) is used to model the system. The following parameters are cakula:ted:

48£1 ,. k=-;r= IO'lb/m

L

rT

/ - = 100 rnd Isec V In

3,9

SEISMIC INSTRUMENTS

When a system of the type shown in Fig, 3J6 is used for the purpose of vibration measurement, the relative displacement between the mass and the base is ordinarily recorded. Such ar. instrument is called a seismograph and it can be designed to measure either the displacement or the acceleration of the base, The peak relative response U/Yo of lhe seismograph depicted in Fig, 3.16, for hannonlc mOlion of the base, ;$ given from eq. (3.53) by

{=OJ

u

/'

W

y,

.f (l :'~')' + (2r,fj'

r=-

w

=0,6

Po 00-' T= . lin (al From eq, (3.20), Ihe amplitude of motion ;s

y= (7;==if~"""",'F= 0.,075 in

(Ans,)

A plot of this equalion as a funcllon of the frequency ratio and damping rat10 is shown in Fig, 3,17. It may be seen from this figure that the response is essentially constant fo. frequency rattos r> 1 and damping raLio ~;;;:: O.S. Consequently, the response of a properly damped instrument of this type is essentially proportional to the base-displacement amplitude for high frequencies of motion of the base. The instrument will thus serve as a displacement :neter for measuring such motions, TIle range of applicability of the instrumenl is 1n-

with a phase angle from eq. (3,21)

B= tan .. !

(3.63)

= 10,6'

(b) From eq. (3.61), the transmissibility is

1547 Fig, 3,16 Model of seismograph,

80

Response of OnewDegree-of-Freedom System to Harmonic Loading

Slructures Modeled as a Single-Degree-of-Freectom System

3.10 t

~

0.15

81

RESPONSE OF ONE-DEGRE-OF-FREEDOM SYSTEM TO HARMONIC LOADING USING COSMOS

Example 3.10. Determine the steady-state response of the system in Example 1.6 subjected to a hannonic force, F 1000 sin wI (lb) applied at tbe mass In in the y direction, The force frequency w varies between 2 and 10 cps, Neglect damping. t '" 0.4

Salurian: The analysiS ~s a continuation of Example 1.5, A harmonic analysis is performed over the frequency runge 2 to 10 Hz. (15) Denne analysis type as hannonic using one natural frequency, units of exciting frequency tn Hz, frequency range from 2 to 10 Hz. 1(0) frequency points in the frequency range, linear interp0lation. and request printout of relative displacements and reiative velocities:

2

ANALYSIS

Fig. 3.17 Response of seismograph

lO

harmonic motio;) of the base.

>'

PD_ATYPE,

POST-.DYN

>'

?D_A'fY?E

5, 1, 1, 2, 10, 1000, 1,

°

(16) Define dynamic forcing function us frequency depended force over full frequency range: creased by. redudng the natural frequency, i.e., by reducing the spdnoa stiffness . or mcreasmg the mass. New consider the response of the same instrument to a harmonic acceleration of the base Y$;;;; Yo sin Wt. The equation of motion of this system is obtained from eq. (3.50) as

mil + cu + ku;;;; - myo sin

{i;t

ANA::"YSIS POS"l'_DYN;. PD_CURVES PD_elJR'!'YP!' 1, 1, 0 ~

AKA:"YSIS

POST~DYN

PD_G.iRTYP

:>

> PD·-CURVES

;. PD-CURDEF

1000, 11, 1000

PD_CURDEF, 1, 1, 2,

(17) Apply dynamic force function at node 2 in the Y direction: (3.64)

The steady-state response of this system expressed as the dynamic magnification factor is then given from eg. (3.23) by

CONTROL )< ACTIVATE )- AC'I'SET ACTSET, TC, 1 L.()p.D- EC

STRUCTUR.L.~

>'

;. FORCE ;. ?ND

FN'b, 2, FY, 1, 2, I, 2, 1, 0

(3.65)

This equation is represented graphically in Fig. 3.3. In this case, it can be seen 0.7, the value of the response is from this figure that for a damping ratio nearly constant in the frequency range 0 < r< OA Thus it is clear from eq. (~.65) that the response indicated by this instrument wi!: be directly proportlOnal to the base-acceleration ampIitude for frequencies up to about six .tenths of the natural frequency. Its range of appHcabiHty wUl be :ncreased by increas~ ing the natural frequency, that is, by increasing the stiffness of the spring or by decreasing the mass of the oscillator. Such an instrument 1S an accelerometer.

(18) Request output for the amplitude of the displacement and of the acceleration in function of the frequency: ANALYS:S

)<

PD~PR::;:N7,

POS_DYN ;. PD_OUTPUT 1, Q, 1, 0, 0, 10, L

>'

PD_PRINT 1000, 1

(19) Request plots for displacement and for accelerarion at node 2 in function of the frequency: P.NA!,YSIS

PD_NRESP, fu~ALY$!$

PD_PLO~,

:>

POS_DYN

:>

PD.. _OUTPUT

:>

P::LNRESP

1, 2, 0 :> P05_DYN > PD 1, 1000, 10, 0

OC~PU~

) ?D_PL07

:

82

Structures Modeled as

a.

Single-Degree-of-Freedom SYSlem

.' ..""

Response of One-Degree-oj·Freedom System 10 Harmonic Loading

83

(20) Execute harmonic analysis: RESULTS

;I>

POS_DYN

R.....!)YNAMIC

:>

R_DYNAMIC

,--1--11

(21) Activate XY plot infonnarion for Y displacement at node 2 as a func~ tion of frequency, and pJot the displacement YS, frequency for node 2: orSPI,AY

XV_PLOTS

:>

>

XYPLOT,

v y

In

the range set

to

, ,,

,I

1

(22) Replot the displacements repajnt the SCreen:

- '-+----\~U '" 1 I I I -~ i '"

ACTXYPLOT

." ~'"'-

1

--"

,

1

I,

I

,I

f--! -

1

,'II

'-i'\

ACTXYPLOT, l, FREQ, OY, 2, C, 12, 1, 0, 2K DISPLAY > XY_PLOTS ? XYPLOT

H

,--

,i

i

\+1 b-I:::t:, f-~I 1 J==t=bk-i

1000 maximum and

'"

i

,

I

,

I,

7 : \

DISPLAY ~ XY_PLOTS :> XYRANGE XYRAAGE, 1, L 2, 10, 0, 1000

l---J,

( L

-'-~,--

:1

DISPLAY REPAINT

LI1

L~

J.'6

'.l.l

\i,\!:

""I.~

8.4

9.2

j$

F!l(C Ifll)

(23) Activate XY plot !nformation for Yacceleration at node 2 as a function

(a)

,of frequenc)', and generate plot of the acceleration vs. frequency for node 2, Then replot with acceleration range from 6000 to lE6: lE'S£t-,,~-~-···~

DISP:0AY ;> XY_PLOTS ACTXYPOST, 1, FREQ,

::>

AC?XYPOS'I' AY, 2, 0, 12,

1,

0,

s.!'i&S€<S!

2N

1

DISPLAY REPAINT

,:':::::::1=-;~~

XY_PLOTS;> XYRANGE 1, l, 2, 10, 6000, lEG ::>

t

VIEW_PAR ~ REPAIKT

I

(24) Obtain a hard copy of the frequency response for displacement and acceleration: CONTROL j LASERJET,

y s

,

, !I

-iT :

1_ _

,

i

I, '

~l

-"~"

. •

t -t2 ~!

i

,

\

I

,

I

,

{l3~'{lSt---+-1'~~ I-

C{ll5t'!i~

'

]

,

~r--

!:I {l12(+$$ ,

O:i:SPLAY > XY_PLOTS > XYPLOT XYPLOT, 1 DISPLAY XYRANGE.

- - '_ _

4~

I,

,!.

II'

1

'

•

,

,

!+---'--\ j - I L., : ~

, _ I __

~'

..

1

'~

,. I, . ,"=,.~,

:

DEVICE > LASERJET 150,0

FRSI)!'U)

Figure 3.18 shows the frequency response in terms of the displacement and of the acceleration,

Example 3.11. Solve Example 3,10 assuming that the damping in the system is equal to 10% of the critical damping.

(b)

, \ I

i

i

t. ';',

" !;

~'

~

Fig. 3,18 Frequency response for Exnnlp1e 3.10. (il) Amplllude of the (b) Amplitude of tbe acceleratiol:.

d~splacemem;

i' u

Ii

84

Response of One-Degree-of-Freedom System to Harmonic Loading

Struc1ures Modeled as a Single-Degree-of-Freedom System

Solution: The analysis is a continuation of Examples 1.6 and 3.10: A harmonic analysis is performed over the frequency range of 2 to to Hz. 10% of critical damping is assumed. The following commands in COSMOS follow those of Examples 1.6 and 3.10:

'"

_~ALYSIS

) POST_DYN

PD_MDAMP I L L

1,

>

PD_DN1P!GAP

>

i

I \ I \

'" , '"

.1

ANALYSIS > POST_DYN ) R_DYNAMIC R_DYNAMIC

, ,

(27) Activate XY plot information for X displacement at node 2 as a function of frequency, and plot the displacement vs. frequency for node 2. Then replot with displacement range set to 200 maximum:

'\

i

,"

PD_MDAMP

(26) Execute hannonic analysis:

\

!I II

V

I

1\

.\

IL

,

'------

,

DISPLAY ) XY_PLOTS ) ACTXYPLOT ACTXYPOST, 1, FREQ, UY, 2, 0, 12, 1, 0, 2N DISPLAY XYPLOT XYPLOT, 1 DISPLAY ) XV_PLOT XYRANGE XYRk~GE, I, 1, 2, 10, 0, 200 DISPLAY ) VIEvCPAR ) REPAINT REPAINT

6.6

-

1$

9.2

FRE01H,1

(a)

2E·('l~

m

I

1.8E+G5

(\

I

1.6E+GS

I --\

! .4E+.:IlS

0, 2N

I I II

1.2E+GS

,,

,

lE+~S

1 8G.:Il.:ll!l

---- .--.. ---- --- ---

1\

\

."

6SGGS

DISPLAY REPAINT

6.4

7.6

(28) Activate XY plot infonnation for X acceleration at node 2 as a function of frequency, and plot the acceleration vs. frequency for node 2. Then replot with acceleration range set to 2E5 maximum: DISPLAY ) XV_PLOTS ) ACTXYPLOT ACTXYPLOT, 1, FREQ, AY, 2, 0, 12, I, DISPLAY XYPLOT XYPLOT, 1 DISPLAY ) XY PLOT XYRANGE XYRANGE, 1, 1, 2, 10, 0, 2E5

'"

'" '"

(25) Define modal damping for mode 1 as 10% of critical damping (0.10):

85

I'---

/

4GGGS

-=-

2GSn

Figure 3.19 shows the frequency response plot for Example 3.11 in terms of the amplitudes of the displacement and of the acceleration.

Example 3.12. Detennine the steady-state response of the system of Example 1.6 when a vertical harmonic acceleration of magnitude 0.3 g and frequency UJ in the 2 to 10 cps is applied at the support. Neglect damping. Solution: The analysis is a continuation of Example 1.6. A harmonic analysis is perfonned over the frequency of 2 to 10Hz. The following commands in COSMOS follow those of Example 1.6:

,~ 2

2.8

3.6 ~

.4

6.8

5.2

7.6

8.4

9.2

lG

FREOIH,)

(b)

Fig. 3.19 Frequency response for Example 3.11. (a) Amplitude of the displacement; (b) Amplitude of the acceleration.

86

Structures Modeled as a Slngle·Oeg:ee-of-Freedom Syslem

Response of One-Degree-of· Freedom System to HarmoniC Loading

.,

(15) Define analysis type as hannonic using one natural frequency, units of ex.citing frequency as Hz, frequency range from 2 to [0 Hz, 1000 frequency points in frequency range, linear inrerpolalion, and reque..')t relative displacement and relative velocity response printout: ANALYS:::S

J

POST_DYN )-

:

(16) Define dynamic forcing function as frequency dependent base acceleration in the Y direction: ANALYSIS

,

,

,

POST_DYN PD_CURVES L 1, 1 POST_uYN PD_CURVES PD ... CUR::>EF, 1. 1, 2, .3, 10, O,}

PD_CURTYP, M1"ALYSIS

,

ANALYSIS ) POST_JYN

,

?D_CURTY? PD_CURDEF

PD_BEXC'!) PD_BASE

PD_BASE, 1, 0, 386.4, 0,

, ,, "

0

(17) Request response at node 2: k~ALYSI$

PO_NRESP, 1, 2, 0 l-.NALYS'1S > POS~DYN

~

PD~PRIN'I',

0,

1, 0, L

:>

!

,.

./

,.,

'-

I

..

J.

,

•••

"

PD._OUTPUT » PD_PRINT 0, 10, 1, 1000, 1

ANALYSIS ) POS_DYM > PD OUTPUT PD_PI,OT, 1, 1COe, 10. 0 (18) Execute harmonic analysis; AJ,"VALYSIS

I

•

,

)- POST_DYN )- Pu_OUTPUT ) PD_NRESP

fJ~1

i

", ,

;

=-t-=

~

J.

, •

"

•

, ,

?D_ATYPE

PDyTYPE, 5, 1, 1, :;L 10, 1000, 1, 0, 0

87

>

(a)

PO_PLOT

I

POST_OYN' ) R_DYNAHIC

FLJYNAMIC

(19) Activate XY plot information for Y displacement at node 2 as a fune·· tion of frequency, and plot the displacement vs. frequency for node 2. Then replot with displacement range set to 10 maximum: DISPLAY ) XY~PLOTS ) ACTXYPOST ACTXYPOST, 1, FREQ, UY, 2, 0, 12, 1, 0, 2N DISPLAY ) XY_PLOTS ) XY~~GE XYRANGE, 1, 1, 2, 10, 0, 10

, ,,

REPAINT

(20) Activate XY plot information for Yacceleration at node 2 as a function of frequency. and piot the acceleration vs. frequency for node 2. Then replot with acceleration range set LO 10,000 maximum: DISPLAY ) XY_?LOTS ) ACTXYPOST ACTXYPOST, 1, FREQ, AY, 2, 0, 12. 1, 0, 2N DISP~AY , XV_PLOTS > XY~~GE XYRANGE, 1, 1, 2, 10, 0, 10000 REPAINT

Fig:lre 3,20 shows the frequency response for Example 3.12 in terms of the amplitudes of the cisplacement and of the acceleration,

). I !

(b)

Fig. 3.20 Frequency response. fer Example 3,12, (a) Amplilude of the displacement;

(b) Amplitude of the acceleration,

Structures Modeled as a Single-Degree-of-Freedom SySlem

88

Response of

One~Degree-of·Freedom

System 10 Harmonic Loading

89

Example 3.13. Solve Example 3.12 assuming that the damping in the system is equal to 15% of the critical damping.

Solution: The analysis IS a continuation of ExampJe 3.12. Damping is increased to 15% of critical damping. (21) Define modal damping for mode I as 15% of critical damping (0.15): O.J:O;

]\l\AL'fSIS > POST_DYN > P;:}_DAMP iGAP ?D_HDA.,'\J?, 1, 1, 1, 0.15

"y, ,

{22) Execute hannonic analysis: m,ALYSIS

>

POST_DYN

>

'.

R_D~NAMIC

R~DYNA1HC

0,

L

/1

:

,

I

\

I

.\

!

,,1--

2N

I

'" I---I

III.SS

, .!"

,

I

, ., 9.1S'

(23) Activate XY P!ot information for Y displacement at node 2 as a function of frequency, and plot the displacement vs. frequency for node 2. Then replot with displacement range set to 0.5 maximum: DISPLAY> XY_?[,OTS ) AC'rXYP['OT ACTXYPIJOT, _, FREQ, UY, 2, 0, 12, DISPLAY > XV_PLOT ) XYRA~GE XY?.ANGE, 1., 1, 2, 10, 0, 0.:) DISPLAY > VIEW~PAR > REPp_IN·1' REPAINT

II

1.. 25

,

'\

/

,.,

., , ,.,

,

! I

i

'.'

'iL7

"

FfI€QHhl

(a) .......~.~.----.:~--------,

(24) Activate XY plot information for Yacceleration at node 2 as a function of frequency, and plot the acceleration vs. frequency for node 2. Then repiot with acceleration range set to 500 maximum: DISPLAY:> XY_PLOTS > ACTXY?LO'f ACTXYPLOL l, FREQ, AY, 2, 0, 12, 1, 0, 2N DISPLAY :> XY_P:01' :> XY?JL~GE XYRANGS, L L 2, 10, 0, 500 DIS?LAY :. VIE~CPAR > REPA:)lT REPAINT

"y

,,

,

Figure 3.21 shows the frequency response for Example 3.13 in terms of (he amplitudes of the displacement and of the acceleration.

3.11

SUMMARY

In this chapter, we have determined the response of a single degree-of~freedom system SUbjected to harmonic loading. This type of loading is expressed as a sine, cosme, or exponential function and can be handled mathematically with minimum difficulty for the undamped Or damped structure. The dIfferential equation of motion fo:- a linear single degree'of-freedom system is the secoodo:-der diffe:-e:1tJai equation

my + cy + ky = Fo sin w!

eq. (3.10)

(b)

Fig. 3.21 Frequency response for Example 3,13. (a) Amplitude of the displacement; (b) Amplitude of the acceleration.

90

Structures Model€ld as a Sing!e-Degree-of"Freedom Syt')tem

Response 01 One"Degree-of·Freedom System 10 Harmonic Loading

or

with respect to the support, In this latter case, tbe equation assumes a much simpler and more convenient form, namely F m

._

y + 2~wY + w Y = -~a Slfi 2

in which

91

w to

wt

+ cu + ku.

mil

Fef! (l)

eq. (3.50)

in whkh

is the forced frequency,

F.;.:r(t)

=

m)'J(t)

is the effective force

is the damping ratio

" u

and and

it =}' _.}',-

r;;W

is the natural frequency

= /~ It!

The ueneral solution of eq, (3.10) is obtained as the summalion of the com-

plem~n\ary

Ii

(transient) and y

= e -,~ (A ,~

cos

~he

particular

WDI

_____

~

+B

. Slfl

(steady~slate)

soh1l10ns. namely

. Folk sin (WI - 0) wvt) ~ r-:--:.I' " j'

J (1

____

~J

transient solution

, r

r

+ (lrt;)

~~--'

steady-state solulion

in which A and B are constaf'lt.5 of integration,

r

w"

€v

=-

w

is the relative displacement

[i,

For harmonic excitation of tbe foundation, the solution of eq, (3.s0) in terms of the relative motion is of the same form as the solution of eq, (3.10) in which the force lS acting On the mass, In this chapter, we have also ShOWli that the equivalent damping in the system may be evaluated experimentally either from the peak amplitude or from the bandwidth obtained from a plot of the ,1mpJitude-frequency curve when the system is forced to harmonk vibnuion. Most commonly, equivaient viscous damping is evaluated by equating the experimentally measured energy dissipated in the system during a vibratory cycle at the resonant fret:;uency to the theoretically calculated energy that the system, assumed viscously damped, would dissipate in a cycle_ This approach leads to the following expression for the equivalent viscous damping:

is the frequency ratio

wJ 1 -

47Tr

/

is the damped natural frequency

and

,I 2rt \ 0= tan - '--,1 \ 1 - r'l

is the phase angle

The transient part of the solution vanishes rapIdly to zer~ because ~f lhe negati ve exponential factor leaving only the steady-state SOIU1l0D: Of partlcui~r significance is the condition of resonance (r = iiJlw =:. 1) for whlch the ampltttldes of motion become very large for the damped system and lend 10 become .. infinity for the undamped system. The response of the structure 10 support or foundaHo.n motl~n can, be obtained in terms of the absolute motion of the mass or of tis relatlve motJon

E~

in which

It = energy dissipated in the system during a cycle of harmonic vibration at reSOnance E~ strain energy stored at max;mum displacement jf the system were elastic r = ralio of forced vibration frequency to the natural frequency of the system.

'i

I I

Two related problems of vibrating isolation were discussed in this chapter: (1) the motion transmissibility, that is, the relative motion transmitted from the foundation to the structure; and (2) the force transmissibility which is the relative magnitude of the force transmitted from the structure to the foundation" For both of these prob!ems, tbe transmissibility js given by

T< -

1 (1

~1+(2rt;)'-­

\

r'l' + (:irt;)'

I

III

,I

I

i

92

Response of One·Degree~or-Freedom System to Harmonic Loading

Str.;clures Modefed as a Singie-Degree-o:-Freedom System

the stiffness of the isolation springs required to reduce the vertical motion amplitude of the instrument to 0,01 in. Neglect damping,

PROBLEMS 3.1

93

An elec(;-ic motor of total weight W = 1000 Ib is mounted at (he cemer of a slm;!ly supported beam as shown in rig. P3.L The unbalance in the rolOr is W' e "" I lb· in, Determine the ste<1dy·state amplitude of vertical motion of rhe motor for a speed of 900 rpm, Assume that the damping in the system lS 10% of the critical damping. Neglect the mass of the supporting beam.

3.7

Consider the water tower: shown in Fig. P3.7 which is subjected to ground motion produced by a passing train in the vicinity of the tower. The ground motion is idealized as a harmonic acceleration of the foundation of the tower with an amplitude of 0.1 g af a frequency of 10 cps. Determine the motion of the tower relative to {he motion of its foundation, Assume an effective damping coefficient of 10% of the cri~icai damping in the system.

Fig. P3.I.

3.2

Delenn~:)e

the maximum force transmitted to the supports of the beam .in

Prob~

lem 3.1.

3,3

Detennine the steady-state amplitude fOf the horizontal motion of the steel frame in Fig. P3.3. Assume the horizontal girder to be infinitely rlg:d and neglec: both the mass of the columns and damping.

3A Solve for Problem 33 assuming that the damping in the system is 8% of the Fig, P3,7,

critical damping,

3.8 3.9 15'

W10X 33

I

Determine the transm;ssibiUty in Problem 3.7, An electric moror of total weight W"'" 3330 lb is mounted on a simple supported beam with overhang as shown jn Fig. P3,9, The unbalance of the rotor is W' e 50 lb ' in. (a) Find the amplitudes of forced vertical vibration of the motor for speeds 800, 1000, and 1200 rpm, (b) Draw a rough plot of the amplitude versus rpm. Assume damping equal to 10% of the critical damping_

~ Fig. P3.3.

3,5

For Probiem 3.4 determine; (a) the maximum force transmitted to (he founda!iotl and (b) the transmissibility,

3.6

A delicate instrument is to be spring mounted to the noor of a test laboratory where it hes been determined that the floor vibrates vertically with harmonic motion of amplitude 0, I in at 10 cps. [f the instrument weighs 100 lb, determine

~

~'----L

;:g;;: = 10'-",

Fig, 1'3,9,

:

--,c-2.5'~---2.5'~

I ·,1

94

Structures Modeled as a Single·Degree-of-Freedom System

ResponSf! 01 One'Degree·ot~Freedom System to Harmonic loading

single~degree-of·[reedor:) system that lS e:xdted by a harmonic force, The peak displacement ampiitude at re.<;;onance was measured equal to 3 in and equal to 0.2 in at one~tenth of the natural frequency of the

3.10 Estimate the damping in a

ratio, (e) the amplitude of the exciting force when the peak amp:itude of the vibrating mass :s measured to be 037 In, and (d) fhe amplitude of the exciting force when the ampJimtle measured is at lhe peak frequency assumed to be the resonant frequency.

system,

3.11

95

Determine the damplng in a system in which during a vibration lest under a harmonic force it was observed that at a frequency 10% higher than the resonant frequency, the displacement amplitude was exactly one·hil.if of the resonant amplitude. r-4>flt)

w

Fig. P3_15.

Fig. P3.12. 3.12

3.16

A s;:ruclmal systeCl modeied as 11 damped oscillatOr is )';ubjecled to the harmonic excit..tion produced by an eccentric rOLOr. The spring eon;;!ant kane the mass m are known but nO( the damping and :he amount of unbalance in the rotor. From measured amplitudes Y, al resonance and Yl at a frequency ralio rl "'" 1, determine expressions to calculate the damping ratio £ and the amplitude of the e:r.:citb& force F, at resonance.

3,17

A system is modeJed by two vibrating masses nil and III; interconnec(cd by it spring k and damper element c (Fig. P3.!f). FOr nannonic force F= Fa sin w( acting on mass /Il2 determine: (a) equation of motion in [crms of the reJative motion of the (wo masses, U )'l - y:; (b) the slcacYwstate solution of the relative motion,

Detennine Ihe natural f,equency, amplitude of vibration, and maxilnum normal stress in [he simple supported beam carrying an engine of weight W = 30 xN (Fig. P3.! 2). The engine rotates at 400 rpm and induces a vertical force

F(:)=Ssinwr(&>""c;400/60(21t) 41.9rad/sec). (E=210X10 9 (N/m\ /= 8950 x W- s (m4) Sect. f!lOdulus. S = 597 X 1O-6m J.} (Problem contributed by Professors Vladimir N. Alekhin and Aleksey A. Amipin of the Urols State Technical Universfty, Russia,) 3.13 A machine of mass m rests on an elastic floor as shown in Fjg. P3, l3. In order to find the natural frequency of the vertlcal motion, a mechanical shaker of mass nt: is boIled to the machine and run at various speeds until the resonant frequency II is found. Detennine the natura! frequency I~ of the floor-machine system in tenns of fr ar:d the given daHL

~r--' y,

Fig. P3.17.

Fig. P3.13.

3.~4

Detenn:::e the frequency at which the peak amplitude of a dumped oscilblor will occur. Also. determine the peak llmplitude and corresponding phase angle.

3.15

A sLructure modeled as a damped spring-mass sys,em (Fig, P3.IS} with mg=2520Ib, k~·'89,OOOlblin> and c=lJ2Ib·sec/in is subjected to a harmonic exclting force. Detennine: (a) the natural frequency, (b) the damping

Response 10 General Dynamic Load:ng

4

97

Fir)

!

Response to General Dynamic Loading

~------- .... ~,~,~;~----~ ..",,---~~-----,

Fig. 4.1

Genera! load function as impulsive loading.

m produces a change in veIocity that can oe determined from Newton's Law of Motion, namely dv

=F(r)

Rearrangement yieJds dr = .:...cc":,:,:,:,,

du

m

In the preceding chapter we studied the response of a single degree~of~freedom system with harmonic loading, Though this tYpe of loading is important, real structures are often subjected to loads that are not harmonic. In the present chapter we shall study the response of the single degree-of-freedom system to a general type of force. We shall see that the response can be obtained in tenns of an integral that for some simple load functions can be eva1uated analyticaUy. For the general case, however, it will be necessa.'1' to resort to a numerical integration procedure.

4.1

IMPULSIVE LOADING AND DUHAMEL'S INTEGRAL

An impulsive loading is a load which is applied during a short duration of time. The corresponding impulse of this type of load is defined as the product of the force and the time of its duration. For example, the impulse of the force F(r) depicted in Fig, 4.1 at time T during the interval d-r is represented by the shaded area and it is equal ~o F('T) dr, This impulse acting on a body of mass 96

(4.1)

where F(r) d1' is the impulse and du is the incremental velocity. This incremental velocity may oe considered to be an initial velocity of the mass at lime r. Now let us consider this impulse F(r) dr acting on the structure represented by the undamped oscillator. At the time r the oscHlalor will expedence a change of velocity given by eq. (4.1), Thi_$ change in veiocity is then introduced in eq. (1,20) as the initial velocity DiJ together with the initial displacement Yo;::;: 0 at time r producing a displacement at a later time t given by dyer)

F(r) dr .

_

....- - sm "'(I - r)

mw

(4.2)

The loading fl;nction may then be regarded as a serres of short impulses at successive incremental times dr, each producing its own differential response at time I of the form given by eq. (4.2). Therefore, we conclude that the total displacement at time t due to the continuous action of the force F (r) is given by the summation or integral of the differential displacements dy (t) from time t = 0 to time t, that is. yCt) =1-

I'

mw ,

F(T) sin w(t - r)d~

(4.3)

98

StruClures Modeled as a

Sjngle~De9fee-of·Freedorr.

System

Response 10 Gar-eral Dynamic Loading

The integral in this equation is known as DHhamei's inregrai. Equation (4,3) represents the total displacement prod'Jced by the exciting force F (1') acting on the undamped oscmatOr~ it includes both the steady-state and the transient

99

2------- -----------I

components of the motion corresponding to zero initial conditions, Yo = 0 and = O. If the function F(T) cannot be expressed analytically, the integral of eq. (4.3) can always be evaluated approximately by suitable numerical methods.

1)0

I

-- ------- -------

------

-------

To include the effect of initial displacement )'0 and initial velocity Dc at ~ime t 0, it is only necessary to add to eq, (4.3) the solution given by e.q, 0·20) for the effects due to the initial conditions, Thus the total displacement of an undamped sing!e degree-of·freedom system with an arbitrary load is given by y(t).,,-;yocos

I

"

wt+~-sin (JJ{+--~-J

mw ()

w

F(r) sin w(f-r)d-r

Constant Force

Consider the case of a constant force of magnitude Fo applied sudc!enly to the undamped oscillator at time f:::':" 0 as shown in Fig. 4.2. For both initial displacement and initial velocity equal to zero, the application of eg. (4.4) to this case gives y(/)=-~ I

mw

f'

degree-of~freedom

system

(0

a suddenly

(4.4)

Applications of eq, (4.4) for some simple forcing functions for wbich it is possible to obla;n the explicit integration of eg. (4.4) are presented below.

4.1.1

Fig. 4.3 Response of an undamped Single applied constant force.

where y~! = Folk. The response for such a suddenly appUed constant load is shown In Fig. 4.3. Il will be observed that this solution is very similar to the solution for the free vibration of the undamped oscillator. The major difference is that the coordinate axis! has been shifted by an amount equal to YSI Folk. Also, it should be noted that the maximum displacement 2Y~l is exactly twice the displacement that the force Fe wou:d produce If it were applied statically. We have found an elementary but important TesUIt: the maximum displacement of a linear elastic system for a constant force applied soddenly is twice the displacement caused by the same force applied statically (slowly). This result for displacement is a;so true for the internal forces and stresses in the structure.

4.1.2

Fes!n W(l-T)dT

Rectangular Load

0

Let us consider a second case, that of a constant force Fo suddenly applied but only during a lim~ted time duration Iff as shown in Fig. 4.4. Up to the time l", eg. (4.5) applies and at that time the displacement and velocity are

and integratlon yields y(l) ~

F, . cos --,-!

w(t -

I'

7) 0

InW

Y (/)

Fe

-(1k

cos

Wf)= y,,(l-

cos

(4.5)

Wi)

and

F(ll j

~c-----------------------

For the response after time ld we apply eq. (1,20) for free vibration, taking as the initial conditions the displacement and velOCity at td. After replacing I by t - Id, and Yo and Uo by Yd and u", respectively, we obtain

1'1

Fig. 4.2 Undamped oscillator actec upOn by a constant force,

y(t) =

100

Response to General Dynamic Loading

Slruclures Modeled as a Slngle·Oegree·of-F:eedom System

2.0:-,n-n"---'-I'Ir-i,7'TTlT,r-il, 1

L6H+t't--+:

in I i i'''ITn I

or uS1ng the trigonometnc identity

~!-/+-I-HH+l-i-+-;-I-'i-li-,Hi"m '

cos a-cos

HH~~-r.~+i4+H+-~L-~~itH

.

.J ~

Li

u'-t+tl't-~Irl/'-t-rtti ffi F1') 1

i~H+~~+--r'i-~++H ,i I '

s o'!.JHftttHiH--t-l-I-'-t'ttI' :/ 1 J4+J,f--+,-+-'!-+

I .

O'tuLL_+-'l '_il-+H+1H1-++ i_++I+t+Wi oI II 0.05

Fig. 4A

iI ' 0.2i L.J..JLLc-LUlic-: -!-::_"-i' -'-::-,-~i';',' 0.5 \.0 2.0 5 lO

0.10

Maximum dynamic load factor for the undamped oscillator aCled upon by a

rectangular force.

which can be reduced to Y (I)

-Fe {cos w(t k

tJ ) -

cos wt }

(4.6)

If the dynamic load factor (DLF) is defined as the displacement at any time t divided by the static displacement y" ~ Folk, we may write eqs. (4.5) and (46) as DLF=l-cos(.o[,

[$(d

and DLF:::: cos w(t -

[d) -

cos

.

a+{3 ,

2

"'4 \ . [ (,

4.1.3

Triangular Load

We consider now a system represented by the undamped oscillator, jnitially at rest and subjected to a force F(t) that has an initial value Fo acd that decreases linearly to zero at time td (J-olg. 4.5). The response may be computed by eq. (4.4) two intervals. For the first interval, 7'~ la, the force lS given by

DLF= I and

DLF ~ cos 2

'\r

and the initial conditions by ~I-

Tf

(4.8b)

The use of dimensionless parameters in eq. (4.8) serves to emphasize the fact that the ratio of duration of the time. that the constant force is applied, to the natural period rather than the actual value of either quantity' is the important parameter. The maximum dynamic load factor (DI.F)m"-H obtained by maxizlng eq, (4.8), is plotted in Fig, 4.4. It is observed from this figure that the max.imum dynamic load factor for loads of duration (.1 IT'?::.05 is the same as if the load duration had been infinite, In general, the maximum response occurs during the application of the load, except for loadings of very shor~ duration (t.1IT
sn

Ii \

2

(4.7)

Wl,

It is often convenient lo express time as a dimensionless parameter by simply using the natural period instead of the natural frequency (w:::: 2r./T). Hence eq, (4.7) may be written as

II

a-{3

f3= -2 Sin - - - sm - - -

DLF'= 2 sm -'-I sm 27T:\ T J \T

I

I

'-I

~01

I

cos 27T-, T

(4.8a)

Yo 0:;:;0,

uo=O

102

Structures Modeled as a 2.0

L

1

6

' ! TTl

I II i

I III I

1. 2

0:

I

!!

I

9. o. 8

1

Y

1

lL Xl

o. ,

J.+

I I

II i

I

J

U4

J

I

System

I

H' I : I ! II 1 I 1.1 I II I I 'I Ll:::: Ull I

~

on

Sjng!e-Degree~of-Freed()m

-1_• .

I I

I I

I

Fir!

IJ F'~

-,

I

.ill

I,

J 0.5 Jill1.0 ; I I

I 0.2

2.0

Response to General Dynamic Loading

'!O3

These values may be considered as the initial conditions at time r = Id for this second interval. Replacing in eq. (L20) t by t td and Yo and Vc. respectively, by Yet and ()d ,md no:ing that FC,) = 0 in this interval we obtain the response as

•I

UL!

1

y = -Fc - - {' SIO

I

. WI - SIn

we;

kWlet

r~ '~

and upon dividing by y,;

·U' : i L1 ,o 5

= Folk I.

DLF = ~[S1O WI,

Fig. 4.5 Mllxiffium dynamic loud fnctor for {he u;tdamped oscillntor acted upon by triangular force,

II

Fo

- f J )'J

Teos

uJt

gives

.

tvl ~ S10

(jJ

(I -

t d)

}

cos wt

(4.12)

In terms of the dimensionless time pantmeter, this last equation may be written as

DLF =

'r ---I 2'l11,IT L

"

'10

'

' .

2,,--·- SIO T

-}i-

1 ( ,'d \'

2,,1\T

T.

cos

I

2,,T

(4.13)

The substitution of these values in eq. (4.4) and integration gives

Fo y=--(1- cos wr) k

+ -Fo (Sin - -wr kttf

(4.9)

UJ

or in terms of the dynamic load factor and dimensionless parameters

DLF

L= I-cos (2m 11) + sin (2mlT) 2m,dr

YSl

which defines the response before time

ld_

For the second interval (t2'!

(4.10)

I d),

we

obtain from eq. (4.9) the displacement and velocity at time Id as

Eo (Sin k

(;,)1,1 -_. cos

Wid

wlt!

'I !

The plot of the maximutn dynamic load factor as a function of the relative time duratiot'. (dlT for the undamped oscillator is given in Fig. 4.5. As would be expected, the maximum va!ue of the dynamic load factor approaches 2 as lalT becomes large; that is, the effect of the decay of the force is negligible for the rime required for the system to reach the maximum peak. We have slUdied the response of the undamped oscillator for two simple impUlse loadings: the rectangular pulse and the triangular pulse. Extensive charts have beer, prepared by the U.S. Army Corps of Engineers I, and are available for a variery of other loading ;Jll]ses.

Example 4.1. A one-slUry building, shown in Fig. 4.6. is idealized as a IS-ft high frame with twO steel columns fixed at the base and a rigid beam supporting a weight of W 5000 lb. Each COlumn has a :nOment of inertia Ix;::; 69.2 inA and a section modulus S = IJ:1c = 17 in) (E = 30 X 106 psi). Determine the m.2.ximum response of the frame to a rectangular impulse of ampli~ tude 3000 Ib and dUHltior: id = 0.1 sec appIJed horizontally on the top member of the frame. The response of interest is displacement at the top of the frame at'.d the bending stress in the COlumns.

and

Fo I'.

Vrl=-

k \

wsm

COS

wt,:

1\

wl,'+-----I Cd

1,1 1

(4.11)

j U.S. Army Corps of Engitleers, Dcsig" oj SUilcfw,:s fO ResiSf the EffeCf:! Of ATomic Weapons, Manunls 415, 41$, 3nd 416. Maret: !5, 1957: Mnnuals 417 lind 419, Janury IS, 1958; Manuals 41$,420.421, January 15, 1960.

104

Response to General Dynamic Loading

Structures Modeled as a Sin!Jle-Degree·of~F{eedom System

4.2

W::::: 5000 l!l

F-"fZ'Z===Zj

105

NUMERICAL EVALUATION OF DUHAMEL'S INTEGRAL-UNDAMPED SYSTEM

In many pract:caj cases the appUed loading function is known on]y from

W Sx20

experimental data as in the case of seismic motion and the response must be evaluated by a numer:cal method. For this purpose we use the trigonometric identity sin <»(/- T) = sin wt cos WT- cos wt sin (;)7, in Duhamel's integraL Then, assuming zero initial conditions, we obtain Duhamel's integral, eq. (4.4), in the form

..-;;---

1 y(t) =sin wr--

Idealized frame

Fig. 4.6

fOf

mco

ExampJe 4.1.

f' F(T) cos wTd,~cos wt··------~ I J;) mw

f'

Fer) sin wTdT

0

or

Solwio": Natural period:

y (I) = (A (I) sin 12X30X IO"X2X69.2

8544 Ib/in

(IS X 12)'

A (I)

~

J:

B(I)

=

L

386

r;; = 217 i112.9534

---~

y

V 8544

= 0.2446 sec

=~ .. = OA08 0.2446

= 1.9

(DLF)~,

Fa

(from

4.4)

A(I)=

y" =

Y•."

F(r) cos wrdT

3000

k = 8544 0.3511 in = (1.9)(0.3511) = 0.667 in

F(r) sin wrdr

(Ans.)

f'

-:-r Ym"'

'.

The elementary operation required for the trapezoidal rule is

(4.16)

= 6 X 30 X

\0' x69.2 0.667 = 256 420(lb' in) (15 X 12l" '

and the maximum s:-ress

0-m~~

(1'=

M s

and for Simpson's rule A (I) = L!rj(l, + 4/,

by 256,420 = --,~= ,I

(4.15)

I(T) dT

Maximum bending StreSS: The bending moment M in the columns is given by

l>f = 6El

(4.14)

The calculation of Duhamel's integral thus requires the evaluation of the integrals A (0 and B (I) numerically. Several numerical integration techniques have been used for this evaluation. In these techniques the integrals are replaced by a suitable summation of the function under the integral and evaluated for convenience at n equal time increments, liT. The most popular of these methods are the trapezoidal rule and the SImpson's rule. Consider the integation of a general function J(T)

J\zfaximum displacement:

:i. T

B (I) cos WI} Imw

where

m= 5000 = 12.95341b.sec'lin T= 211' '

wi -

.

15,083 pSI

(Ans.)

+ 21, + ... + 4/"_, + I")

(4.17)

where n ~ l/1J'T must be an even number for Simpson's rule. The implementation of these rules is straightforward. The response obtajned will be approxi-

'. ':.;

106

Structures Modeled as a Sing!e-Deg;ee,o(-Freedom System

Response 10

F(r)

General Dynamic Loading

107

where

I

t1F, = F(I,) - F(lI_ ,) F(r'>1 1

and

--------"---~---~--~----

L1I'=(;-[;_1

; I I I

F(t,l -.----"'~ ;

1

I

I

"

I

F!r)

L

The substitution of eg. (4.20) jnlo eq_ {4.18) and integration yield

I I

£IF;

I I I

_-:-'-~_._L~_"--~ _~

r,

f,_ 1

t"

________

t1F· +---2j' {cos wt,-COS (ot;_I+w(t;s;nW(,-I;_1 sin wt i - r»)

T

Cd Ll.t,

!

i (4.2])

-:-£l(,~-----l

I

Analogously from eq. (4.19).

Fig. 4.7 Segmentally linenr loading function.

II

B(lI)

mate since these rules arc based on the substitution of the function 1 (r) for a piecewise linear function for the trapezoidal rule, or piecew~se parabolic function for Simpson's rule. An alternative approach to the evalualion of Duhamel's integral is based on obtaining the exact analytical solution of the integral for the loading function assumed to be given by a succession of linear segments. This method does n?t introduce numerical approximations for the integration olher than lhose inherent in the round off error, so in this sense it is an exact method. In using this method. it is assumed that F (r), the forcing function. may be approx1mated by a segmentaHy linear function as shown in Fig. 4,7. To provide a complete response history, i~ is more convenient to express the integrations in eg. (4.15) in incremental form, namely A(r,)=A(I,_,)+

r

F(r) cos ""dr

f.,

+ ~ {sin we; - sio wrj_! U?

w(!, cos Wi j

-

(i_I

cos

Wi_I)}

Equations (4.21) and (4.22) are recurrent formulas for the evaluatlOO of the integrals in eq, (4,15) at any lime!:.:-.;; I;.

Exnmple 4,2. Determine the dynamic response of a tower subjected to a blaslloading. The idealization of lhe stucture and (he blaSlloadJog are shown

in Fig. 4.8. Neglecl damping.

F(r)

F(r) sin wrdr

(4.19) K'" 100 kim.

where A (l,J and B(t,) represent the values of the integals in eq. (4,15) at time tl_ Assuming that the forcing function F(T) is approxin;ated by a piecewise linear function as shown in Fig. 4.6, we may write

(4.22)

JJ.t;

(4.18)

li.1

B(t,)=B(IH)+

i

L1F;

-Y, {"

(4.W) Fig. 4.8

'"

Idealiz.ing srructure and Joadmg for Exan'lple 4. L

!

108

Structures Modeled as a Single-Degree-ci-Freedom SYSlem

Response to General Dynamic Loading

For this system, the natural frequency is

Solurion:

4.3

Since the loading is given as a segmented linear function, the response obtained using Duhamel's integral, eq, (i!.14), with the coefficients A (/) and B (t) detennined from eqs. (4.2]) and (.:!.22), will be exact The necessary calculations are presented in a convenient tabular fonnar in Table 4.f for a few time steps. The integrals in eqs. (4. J 8) and (4.19) are labeled L1A (t) and LlB (I) in this table, since

109

NUMERICAL EVALUATION OF DUHAMEL'S INTEGRAL-DAMPED SYSTEM

The response of a damped system expressed by the Duhamel's integral is obtained in a manner entirely equivalent to the undamped analysis except that the impulse F( r) dr producing an intral velocity dv = F(r) drlm is substituted into the corresponding damped free-vibration equation. Setting Yo = 0, Vo = F(T) dTlm, and substituting I for t - r in eq. (2.20), we obtain the differ-ential displacement at time t as -,f",(I-T)

dy (/) = e '

F(r) dr

.

. ---'-- sm 'Yo (t - r mwo

)

(4.23)

,.[;

L1A(ti)=A(ti)

A(ti~:)=) ~

1,-_,

Surruning these differential response terms over the entire roading interval results in

F(r) cos urrdT

and

y(i) =_1_

LlB(li)

B(I,)

"

B(li~l)

{'., F(r) sin wrdr

Since the blast terminates at t = 0,060 seCt the values of A and B remain constant after this time. Consequently) the free vibration that foHoVots is obtained by substituting these values of A and B evaluated at t = 0.060 sec into eq. (4.14), that is,

mwo

r

Jo

F(T)e--!<4(I-i)sinwD{l

(2571 sin 3 ).621

AD(ti)=AD(t;~I)~'

3585 cos 3 ),621) 13162

BD(t;)=BD(I;~I)+

or Y(I)

0.8130 sin 31.621

(4.24)

which is the response for a damped system in terms of the Duhamel's integraL For numerical evaluation, we proceed as in the undamped case and obtain from eq. (4.24) e' "" (4.25) y(t) ~ (AD(/) sin WD' BD(t) cos WDI}--'-mwo where

y (I)

r)d7

r.,

r;.,

F(r)e'-cos wDrdr

(4.26)

F(.,-)e'··sin wordr

(4.27)

For a linear piecewise loading functIon, F(r) given by eq. (4.20) is substituted into eqs. (4.26) and (4.27) which requires the evaluation of the following integrals:

Ll338 cos 31.621

for I"" 0.060 sec.

(4.28) TABLE 4.1

Numerical Calculation of the Response for Example 4.1

I (sec)

Fer)

WI

L1A (I)

A (I)

LlB (t)

B(I)

Y (I)(in)

0.000

0 120,000 ]20,000 0 0 0

a

0 1082

0 1082 2458 2571 2571 2571

0 486 1918 ]181 0 0

0 486

0 0.078 0.512

omo 0.040 0.060 0.080 0.100

---~----

0.6324 1.2649 :.8974 2.5298 3,1623

1376 113 0 0

2404 3585

3585 3585

(4.29)

(430)

1.134 1.395

1.117

(4.31)

:

ii;

·'i

110

Struclures Modeled as

a.

Single-Degree-of·Freedom System

Response to General Dynamic Loading

where It and I~ are the integrals iIldicated jn eq,s.. (4.28) and (4.29) before their evaluation at the limits. In terms of these integrals, Ao(fj) and Bn(ll) may be evaluated after substituting eq. (4.20) into eqs. (4.26) and (4.27) as

.

tlF, \

A D (,,) = Ao(c,~ ,) + (F(c,.,) - c,_, '~/'

+

111

J~i-'- _.. yl')

tlF,

tl', [,

(4.32) (3) (4.33)

Finally, the substitution of eqs, (432) and (4.33) into eq. (4.25) gives the displacement at time Ii as e -(~i Y Ct i ) = - - {Ad!;) sin

Waf, -~

m(liD

Bp(t i ) COS wolJ

(4.34)

(bl

4.4

RESPONSE BY DIRECT INTEGRATION

The differential equation of motion for a one.degree~of-freedom system repre~ scored by the damped simple oscillator, as shown in Fig. 4.9(a), is obtained by establishing l:he dynamic equiHbrium of the forces iIi the free body diagram, Flg. 4.9(b):

my + cj + ky = F (I)

k

(4.38)

rill

Fig. 4.10 (a) Datnped Simple oscillmor excited by the displacement y, (1) nl itS support. (b) Free body diagram.

in which the function Fet) represenlS the force applied to the mass of the oscillator. When the structure, modeled by the simple oscilhHor, is excited by a motion at its support, as is shown in Fig. 4. t O(a), the equation of motion obtained using the free body diagram in Fig. 4.9(b) is

-I m

:;;;;;J)~ffi/ffi/

my + C (Y - y,) + k(y'" y,) = 0

(4.36)

In this case, i: is convenient lo express the displacement of the mass relative Lo rhe displacement y.. of toe support, namely

u=y-y, The substitution

Ii

(4.31)

and its derivatives frorn eq. (4.37) into eq. (4.36) results in

(4.38)

bl

Fig. 4.9 diagram.

Comparison of eqs, (4.35) and (4.38) reveals that both equations are mathematically equivalent if (he r:ght-hand side of eq. (4,38) is interpreted as the effective force

(a) Damped simpJe oscillator excited by Ihc force F(1). (b) Free body

(4.39)

112

Structures Modeled as a

Shgle-Deg~ee-of-F::"eedom

System

Response to Genera! DynamIC Loacing

On the other hand, the panicular solution of eq. (4.42) takes the form

Equation (4.38) may then be written as (4.40)

Consequently, the solution of the second order differential equation (4.35) or eq. (4.40) gives the response in terms of the absolute motion y, for the case in which the mass is excited by a force> or in terms of the relative motion u =)' -)'1> for the strucrure excited at its support.

4.4.1.

113

Solution of the Equation of Motion

The method of solution for the differential equation of motion implemented in the computer program presented in this chapter is exact for an excitation function described by linear segments, The process of solution requires for convenience that the excitation function be calculated at equal time intervals LIt. This result is accomplished by a lineer interpolation between points defining the excitation, Thus, the time duration of the excitation, including a suitable extension of time after cessation of the excitation, is divided into N equal time intervals of duration For each interval .1t, the response is caiculated by considering the initial conditions at the beginning of that time interval and the linear excita~ion during the intervaL The initial conditions are, in thjs case, the displacement and velocity at the end of the preceding time interval. Assuming that the excitation function F(r) is approximated by a piecewise linear function as shown in Fig, 4.7, we may express this function by

(4.45)

which upon its substitution into eg, (4.42) gjves I-I)

11-1 '\

-~' F+l~-' F I Lli ' Lli i ,+ where A, and BI are constants for the interval t,.:s t S ti + L1t and where we use the notation F; F (t/) and FJ + I = F(t, + Llt). Establishing the identity of terms between the left- and right-hand sides, that is, between the constant terms and the tenns with a factor (t - t;) and solving tbe resulting equations, we obtain

(4.46)

Fi-cA; ··~C-

.at.

The substitution into eq. (4.43) of li}e complementary solution Yc from eq. (4.44) and of the particular solution YP from eq. (4.45) gives the total solution as (4.47)

The velocity is then given by the derivative of eq, (4.47) as (4.41) in which ti = i· t3t for equal intervals of duration .11 and i 1,2,3" .. , N. The differential equation of motion, eq. (4.35), is then given by

my+ <-:v+ J..:y

i

t- " ! 1<,

.

,

f t Ii \ F+.~·-IF . , \ Lli ! d ,

(4.42)

The solut:on of eq. (4A2) may be expressed as the sum of complementary solution YC' for which the second member of eq. (4.42) is set equal to zero, and the particular solution )'Pl that is, }' =}'~ +}'p

UJ[)(t -

t,)

e ......" .. 'i'[(WpOi-,;'WC;) - (WDC i

+ ,;'wD i )

COs

WD(t-'i)

sin wD(t-li)] + A,

(4.48)

The constar.(S of integration Ci and Di are obtained from eqs. (4.47) and (4.48) introducing the initial conditions for the displacement Yf and for the velocity :If at the beginning of the interval .dt, that is, at time ti. Thus, introducing into eqs. (4.47) and (4.48) the initial conditions and solving the resulting relations yields

c,.= y; -Bi

(4.49)

(4.43)

The complementary solution is given in general by eq. (2.15), which for the interval liSt:S:C;+ Lit is

Yc = e - {",(I-r,lle, cos

y

+ D; sin

WD(t - (,)]

(4.44)

The evaluation of eqs. (4.47) and (4.48) at time Ii + ilt results in the displacement Yi+1 and the velocity Yi+l at time !j,..!. Namely, (4.50)

114

Structures Modeled as a

Single~Degree·oIMFreedom

System

Response io General Dynamic Loading

115

and Yi+1

e-fwd'(D,(wocos wD..1t-t'wsin

c,. (~w cos

Fitl

waLll)

waLl! + (.Un sin (UD.1t)J - A,

(4.51)

Finally, L'1e acceleration at time 1; '1'1 t/ + Lit is obtained directly after subs.tituting and J',., from eqs. (4.50) and (4.51) into the differential eq. (4.35) and letting t""'" t/ + ..1r. Specifically.

y,.,

120

Yi+ 1;;;;: A'y, + S'y; + elF, + D/F.+!

(4.53)

Yi+ I = A "Yi + BUYi+ C"Fi+ D"Fi+ I Y;+l -U}Y,+i- 2 gw;h.-I+Fii :/m

(4.54) (4.55)

in which the coefficients A'. B/" .. D'l are:

.,

l-- ___ 7· ~---"

I

(4.52) The substitution of the coefficient Ai, B" C i and D, from eqs (4.46) and (4.49), together with c ~ 2gklw into eqs. (4.50) and (4.5]), results in the following formulas to calculate the displacement, velocity and accelaration al the time step l{+ I = ti + Lk

I

K

o

!, I

,

I

,,

I

I

I

0,02

0.04

.

....;;,;,,---- thee)

0.06

Ibi

Fig. 4,11

Ideail.l.ed structure and loading for Example 4.3,

Equations (453), (4.54), and (4.55) are recurrence formulas to calculate, respectively. the displacement, velocity, and acceleration at the next lime step li+ I = Ii + Lli from the previously calculaled values fOf the these quantities at the proceeding time step f i • Because these recurrence formulas are exact, the only restriction in selecting the length of (he lime step, Ll.r, is that it allows a close approximation to the excitation function and that equally spaced time intervals do not miss the peaks of this function. This numericai procedure is highly efficient because the coefficients in eqs. (4.53), (4.54), and (4.55) need to be calculated only once. Example 4.3. Detennine the dynamic response of a tower subjected 10 a blast loading. The idealization of !he structure and dhe blast loading are shown in Fig.4.1 J. Assume damping equal to 20% of the critical damping.

.

Sin wDClf

2f + - - cos wAr

Wodl

1+ 11/ ~

\

2/; \ 1 --J I

Solu.tion: Since the. loadir:g is given as a segmental linear function, the response obtained using the direct metbod will be exnct The necessary calculations are presented in a convenient tabular format in TabJe 4.2. For this system, tbe natural frequency is

wilt}

UJ

Jk/m

)100,000/100 -31.623 fad/sec

and damped natural frequency WD=

wfl-:::f! ~ 3 1.623 FO.2'

~ 30.984 lad/sec

Hence, the natural period ;s

2". 2". T= ~'- = - _ . - = 0,20 sec w

Recommended practice is 0.02 sec.

,1,

31.623

to select Ll.r::;;; T Ii 0, Specifically, we seiect

I I

Ii

!;

116

Structu~es

TABLE 4.2

Modeled as a

Si~gle-Oegree·:)H~reedcm

Response 10 General Dynamic loading

System

Calculation of the Response for Example 4.2

t, sec

Yi in

Yt in/sec

0 0.02 0.04 0.06 0.08 0.]0

0 0.074 0.451 0.926 1.044 0.778

0 1O.692 25.155

17.096 .- 4.821

.- 20.191

Yi in Isec 2

F,lb

0

0

990.754 430.768 - 1142.511 982.581 - 522.555

120000 120000 0

0 0

We then calculate the coefficients of eqs. (4.53) and (4.54):

A' 0.82180 B' A" = - 16.5170 BO

0.16517 C' = 1.16755 X 10-' D' = 6.1439 X 10-' 0.61286 C"=7.60730X 10. 5 D'=8.9097X 10- 5

with initial conditions Yo = 0 and (4.55) y,

= 6.1439 X lO -, X

1.2 X 10'

Uu =

0, we obtain from cqs, (4,53), (4.54), and

0.074 in

problem, stores the data in a file, and then calculates and prints the response at equal increments of time. Maximum values for the response are also calculated and printed. A description of the programs used in this book and a fonn to order them is provided in Appendix II. Example 4.4.

From Fig. 4.11, we have the following data:

Solution: Mass:

In

Solve Example 4.3 Csing Program 2.

= wig = (38.6 X 1000)/386 = 100 (lb, sec'iin)

Spring constant: k

100 X 1000 = 100,000 (lb . in)

Damping coefficient: c = 2< J kin

= 1265 (lb· sec fln)

Natural period: T = 2"1T ~ :;:; 0.20 sec Select hme step for integration: Lit = T 11 0:;:; 0,02 sec

Input Data and Output Results

lU?:JT DA71"

y, 8.9J24x IO-'x 1.2 X 10'= lO.692 in/sec y, = - 31.623'XO.074-2x0.2X31.623x 10.692+ l.2X 10' 1100 = 990.754 in /sec'· Thus, completing the first cycle of calculations in the direct method of solution, Introducing the calculated values Yh Yb and y! into the recurrer::ce fommlas eqs. (4.53), (4.54), and (4.55), we obtain the response at time = 0.04 sec. The continuation of this process results in the response of this system as showr. in Table 4.2 up to 0.10 sec.

'2

~n¥.st:R

Of' ro:l'';S Ol':f:m'llG THt:

e\CJ':A~'lON

MASS

S?IHNG CQNSTAM'l'

c"

l.~A?

1'II1E STEP cO' :NTEGRA1'!ON

Ii "

.1)2

l>:OEX (;>.CCELEPJI!tlOO C1" G"G>Vl1'Y C'R ZSROj

C " (;

0.150

'tItE

Excr'!',",?lON

TIME

;,;xcrT,~':'XC"

O.l)1C

120000.0:)

() OISC

O.OC

:).00

OL"!'l'vT RBsm.:rs:

4.5

117

PROGRAM 2-RESPONSE BY DIRECT INTEGRATION

The computer program described in this section calculates the response of the simple oscillator excited by a time~dependent external force acting on the mass or by an acceleration applied to the support. The excitation is assumed to be piecewise linear between defining points. The response consislS of a table giving at equal increments of time the displacement. the velocity, and the acceleration of the mass for the case of the oscillator excited by a force applied to the mass. For the case of the osciHator excited at its support, the response is given in terms of the displacement and the velocity of the mass relative respectively to the djsp]ac,ement and velocity of the support ar:.d of the absolute acceieratJon of the mass. The program implements the direct integration method to calculate the response. The program has been written in an interactive mode with the user. It starts by requesting information about the data needed in the solution of the

(1.000

0.000

0.0:'0

0.074

to.692

0.0;\0

0.451 O. 92~

2'),,155

43(;. 7~()

17.096

-t1~:z.SU

(L

aco

9!n.:;:n

L044

-4.821

~1Jl2.581

O. ,-CJ

J.ns

<:O.Bl

-'On.55S

0.120

;).l06

-25.215

13.248

-21.511

42L'S~

c .l40

-, m

C.H;:)

-O.·PS

-;J.!HO

59!1.09S

O •• 1)0

"0.S53

1.()DS

5:<9. MIG i;:<4.160

O.HlQ

-0.424

to. 2~"

'L220 O.HO 0.::£0

-0. :80

11.260

1:.5'>'2

C .072 0.242

:;'1.10$

-l1LJU

5.6n

"3D.4?7

.

D=SP"~ACl':.'{£N"l'

)'AX. VE;t.OCl'J'Y

Ace.

DIS?!...

C .Ji:O C ,060

w.x.

VC".JX:

7::ft:

o.c,jO

>
~

1. 04

;rs .22 ~

!H2.S1

Example 4.5. Consider the tower shown in Fig. 4.11, but now subjected to a constant impulsive acceleration of magnitude y~ = 0.5 g during 0.5 sec applied at the foundation of the tower. Determine the response of the tower in terms of the displacement and velocity of the mass relative to the motion of the foundation. Also, determine the maximum acceleration of the mass, Solution: Mass: m

The following data are obtained from Fig. 4.11:

= IOO(Jb - sec 1 /in)

Spring constant: k

100,000 (lb/in)

Damping coefficient: c

~

1265 (lb· sec/in)

Acceleration of gravity: [; = 386 (in Isec 1)

Select time step:,d{

0.02 sec

4.6

PROGRAM 3-RESPONSE TO IMPULSIVE EXCITATION

Program 3, the same as Program 2, gives the response of the simple osdliator to an excitation specified as linear segments between pOJn;s defining the loading function. Program 2 may be used for any excilation function. while Program 3 is restricted to certain specific excitation functions predefined in the program, A total of eight possible excitation functions are implemented 1n the program. There is also a provision for a ninth fUllction labeled as "new function" which may be imp:emented by the user The excitation functions in the program are s!~own graphically :n Fig. 4.l2 and given analytically by the following expressions: E, (T); P(l)' sin fP(2)*T- P (3))

Te:O

(4.56)

E,(T)~P(l)

T?::O

(4.57)

E,(T)~P(l)*T

0'" To;; 1'(2)

(4.58)

;0

T> 1'(2)

Time (se:~c-,-)_ _+-..:S"pport Acceleration (g)

Excitation:

a

0.5

0.5

0.5

119

Response: to Genera! DynamiC Loading

Structures Modeled as a Slogle-Degree-oI-F:eedom Sys1em

118

p(1If-~----

Input Data and Output Results

"'--~---T

(e) Graph eq (>i.51!.)

(a! Graph eq f'tSS) INPUT D/lTh:

};W,SER or POINTS r-iPHHNG "'tHE: 'tXC17}.:-rO!;

nc "

}'.ASS

A.'! '" lOO

SPRING CONS':h.N7

AK " looeoo c- ~ lHS

OAM'fll
H

~

.G2

"'£lex

G

~

366

\ACccU:AAn~

or \7AAVI'tY OR ?E:RO)

f 4 !Tl

0.500

i

.~~, ,

:1.'>00

ki: GrJph eq. (4,59)

I

P"l~

T

1'(21'---"-'

(n J

p\ll~

Z:XCI1'ATXOfl

0.50

E6

Esln

1 P(1)

E/cn;>,-:!OO

Ii!

~

T

1'(2)

{el Graph eQ. {H;O)

pml2 P(21

i I T

I

(lj Gr~ph ilq. (>i.51)

PRINT TIl'IC-RZSPOflS't TJiBI.f: YIN? N

VM. PlSH.ACZKENT 1 "

".AX.

V£l.OC:1''i:l ~

MAX. ;>,e;::eLE:AA':'!CN '"

Ir; Graph en. {>i.63j

(9) Graph eQ. f4.S2i

2 For excitation at the supper:: Displacement :tnd velocity ue n.~~ative 10 the sup]Xlrt. whereas

the acceleration

)$

absoiute vall;.e.

Fig.4_12

Graphs for excitation

func~ions

grven by eqs. ('L56) through (4.63).

II

120

StructureS Modeled as a Sing'e-Degree-oi·Frcedom System

E,(T)

= P(l}*T

Os T,;; P (2)

=P(l)*P(2)

T>P(2)

E,(T)=P(1)*(I

TIP (2»)

=0

(4.59)

=2*P(I),[l

TIP (2)],

o

OST5P(2)12

Natural frequency: w=~ =3l.623 rad/sec Natural pedod; T

(460)

T> P (2)

E,(T) = P (I) * [2 *T IP(2)J

o

= 0.20 sec

Imput Datn and Output Results (4.6,)

P(2)12<1'5P(2)

051'';;1'(2)

Di?l)"T CATll;

(4.62)

1'",,0

:-v.ss

," ""

SPRING CONS'!.' .....""!' PAJ;fING CO£i"f""lC!£NT

T> 1'(2)

E, (T) = P(l)IEXP [I' (2) * 71

2'lf/w

Select time step: ..1T = 0.01 sec

T> 1'(2)

E,(T)=P(l)*sin [P(3) *71

E, (T)

Os T,;; P(2)

Response 10 Genera! Dynamic Loading

Tn!e S"!'£P Of W'l'EGRA'l""l:)M

(4.63)

= !'lEW FU!'ICTION

R£S?OMS£ K1.X. T!M£

(a) Determine the dynamic response of [he tower shown

in Fig. 4.13 subjected to the sinusoidal force F(l) Fa sin WI applied at its top for 030 sec, (b) Check results using the exact solution which in this case is
Mass:

III

(3) The following data is obt.ained from Fig. 4,! 3:

wig = (38.6 X 1000)1386= IOO(lb sec'linl

Spring constant: k

100 X '000 = 100,000 (Ib/in)

ISO£X (GRAVIn OR ZeRO)

7AAAH£":ER POi

""

.0:\

C

m

.:l is

C'll)-SW!P,2}-T-PO)l

T">C

)1) ~

0

O;nPJT RElhIL1'S:

DIS7: ••

VH.rA:.

Ace

o.oue

0,(::)0

0.010

0.0:10 C .00':

o .O?t

0.016

L1SS 5_ ,,$5

52S.-lSa

0.030

(Lt:n

11. $70

';S(L

O.iHO

c,zn

211.229HZ

6~0.10.j

T;"'£

0.000

:l.()SO

O.

4S~

J.060

o.

]~,

0.0;0

1.c:n

o.oan

l.J96

0-.090

1.199

"" '"

" "

"

'~().H9

~8J

$1:1:.513 ~16.Z&~

16$

-).67.611

:89-

~S20. 916 -len] .9$0 -145·1 .He

6 S~

O,Hl::>

1. SS5

O.H{>

1.5.4;

-13.C1<

-17(14

'50

0.120

.HO

-10.~H

-1112

m

o .DC

0.9J7

-17.;;-.'5

-1£:1S ~H

0.1'"

O. JB5

-61.205

~:l56

O.J.!)O

-0.29: -l m 171

-n.2?9 -10.S.n

4l}

~SjI.H5

(1.1,70

-,

::. !}IE

-~2

.(0)

'"

-6B6. Hi )1.0112

BH_732.

::L 1\;0

-, -, nc -,

~)i!.0:;'$

:lS~L~O,

(I.NC

:it}

-lL60,

2SU.269

O.

Tower for Example 4.5.

~

1\l001l0

PARNren:R ... (2) i'AiVt.M£T£R PO)

0.1£0

Fig.4.13

0

THAX

<:(T)

Solwion:

l,OJONI

C

PhRN·:£':'ER$ OF "7,;e £XCITATICII,

In eqs. (4.56) through (4.63) T is the time variob!e and P(I), P(2) and P(3) a;e cons~ants or parameters to be input as data i:l order to complete!y define the specific excitation function selected.

Exampie 4.6.

100

l~O

I~H.l!),):

0_210

~3.U'

18.}g.j

].115_74:4

0.220 0.23:)

-2.7a~

SC .2""

)()94 .181

-2. :31

19.5B

~JC9

m

C . ./4:J

-L2:t

'O}.2$t5

2GO~

.1Si;

0.2"0

-0.09>

HS.EOl

lOS: .496

121

122

Stn;clures Modeled as a

Singla-Degree-of~F(aadom

System

O.21S0

LL?S

l21,lS1l

~126.4116

0,,;10

:L ~:>o 1. )98

US.574

-1:'09.642 -Z5B,161

(). ?:SC 0,300

4.213 4.674

,,$.996 $5.:n5 2$.&1::;

C.310

4.711

-18.7;;5

o.n::;

Response to General Dynamic Loading

Results shown in the above table are sufficiently close to corresponding values given by the computer in part (a) of this problem. .

-354,LSS4 -O.6~,)76

Example 4.7. Solve Example 4.6 selecting the time step 111 equal to and 0.005. Then compare the displacements at time 1= 0.1, 0.2, and 0.3 sec with the response obtained in Example 4.5 using the exact solution of the differential equation,

~45$tI.3;;)

0.02,

.. n

},.AX . or SPLACEMElIlT'

K'X. V£LCCITY ACC;;;LtAA'l'XON

">-X.

1;<12

n

'5$$.)2

"

(b) The exact solution for the response of <1 simple oscmator to the sinusoidal force Fo = sin wt, with zero initial displacement and velocity, from eg. (3.8) is

Y (t)

Fo /. = "'--:::-T ISin

k-mar \

"

wI

W.

\

w

!

-sm wll

where w is the natural frequency in rad/sec, W the forced frequency also in rad/sec, and Fo the amplitude of the sinusojdal force. Substituting corresponding numerical values for this example yields

om,

SolutiOfl: In solving the probiem using Program 3, it is only necessary to modIfy existing file by selecting option 2 (Modify Existing Data File) at the computer request for file information, The only modification necessary i:1 the dala is to change the time step to the values prescribed by this problem, The foHowing table, records the results provided by the computer and their comparison with the exact solution calculated in ExampJe 4.6:

Exact Time Displacement (sec) (in)

30 31. 6231

02 0.3

. \ sin 31.6231, J

0.94868 sin 31.623t)

10 (sin 30,

300 cos 31.623t

Solution: Mass:

and

Spring Y(I) ~

4t

0.005 sec

DlspL

Dispi. (in)

% error

Dispi. (in)

% error

(in)

1.560 3.092 4.570

2.96 2.96 3.62

1.595 -3.163 4.674

0.78 0.73 1.43

1.604 -3.181 4.701

The following data are obtained from Fig, 4.14:

= 10 lb· sec2/in constant k = lO,OOO lb lin

ttl

9000 sin 30t + 948.7 sin 31.623t Fir)

The evaluation of the response at specific values of time results in the ing table: t

(sec) 0.1 0.2 0.3

Y (t) (in)

1.6076 -3.1865 4.7420

follow~

500011;

yet)

HI)

(in Isec)

(in Isec 1)

2.9379 - 11.6917 26.0822

% error

0.22 0.17 0.86

Example 4.8. A structural system modeled in Fig. 4.14(a) by the simple oscilJato( with 10% (~= 0.10) of critical damping is subjected to the impulsive load as shown in Fig. 4.14(b). Determine the response.

The velocity and acceleration functions are then given by

y(t) ~ 300 cos 30t -

1.6076 -3.1865 4.7420

LIt = 0.01 sec

4t = 0.02 sec

- - - - - "..-

0.1 100,000 I . Y (I) ~ ]00,000 100 (30)' \ SIn 30t ~

123

- 1466.51 2907.53 4298.04

~

-" -

-- --

o Fig, 4.14

0.1 Cb)

(a)

Mathematical model and load function

fOT

Example 4,8

Slructures Modeled as a Singte-Degree-o:-Freedom SYSl~1'1')

124

Dampi:1g coefficient:

GC<;, = 2t i km = 63,25 lb' sec/in

C

Select time step:,1t

JISPL.'1Y )- VIEi-CPAR > VIEt'! VIEt'!, 0, D, 1,

°

0,20 sec

Excitation function: F (t)

(2) Define the XY plane at Z = 0:

50,000; Ib, 0 S r :s; O. I sec, o 1> 0, I sec,

GEOMET"aY )- GRID ) PLANE

0.2 sec.

,1A;

PLAN::, Z, 0, 1

(3) Generate a curve from 0, 0, 0 to I, 0, 0:

Input Data and Output Results

GEOMETRY )- CURVES :. CRPC:):SE) CRPCORD, 1. 0, 0, O. 1, 0. a

::-nmT DA'fA:

f1 = 10

Y.ASS SPRING CO:
i(

~

l.CCCC

Dh.'l?!NG C02F'F!Cl£N'f

c

=

53.~"

'tun::

ii

~

.C~

STEP :)2 :,.'7eGAA1tOCl

?:£Sl'o.>.J$E O'_\X.

?tI~l
(4) Define element group I using the SPRING element fonnul"tion with 2 nodes:

'!'lXE

;;';0;;;< 1::::I<.;\\l11'Y ::;"

10('1')

125

(I) Set view to the XY plane:

= 0.20 sec

Natural period: T =

Maximum time: t rn

Response to General Dynamic Loading

~

PO) '1'

e

0

Ptl)

PARAlis:78R P (2;

PROPSETS )- EGROUP

Z!,~1:('l

EGROU?,

L

SPRING, 0,

2,

L

0,

o.

0,

0

(5) Define real constant for spring element: k= 100,000 (Iblin):

(H':'cP (2)

1'>;>(2')

l?ROPSETS )- RCONST RCONST. 1, L l, L

~ao:Jo

,"

1000000

(6) Define element group 2 using the MASS element formulation: Att,

Dl$?L

o. oct! O. no

O.CJO

::0.1>)(

f:"O(;()

C,OO"

J.n"

6:. ;$7

0.040

(l,015

~. 22~

lH.2SS

~

.691

124..894

'Z .110 '0.5-60 usn t?lI -1:'''1 ~H

O. J6C

C,1)S

!).:lee

O,27S

7.922

o .lDO

{j, 4~6

0.120

O,57{i

8.646 L39"l

0.140

0.51'

_8,O!

c· .1&0

0,26$

PROPSE7S )- EGROUP EGROtJP, 2, Ml\SS, 0,

us

(l.2:(0

.iL t&Z

0.:'10

-;; .4~()

HH"*'1 410_ :iliS

0,

0,

0,

(7) Define the real constant for mass element: PROPSETS > RCONST RCONST, 2, 2, L 7,

lOa, 0,

0, m;;;:

0

100 lb· sec 1 /in:

0, 0, 0, '0, 0

(8) Activate element group I (spring) and real constant 1 (k ~ 29,30 Ib/in), and generate one element along curve 1:

"173 ?EO

-c.on

0,

10

CONTRa!... > ,'1CTIVE )- ACTSET l"..':,;>.CEY;;NT ,. P.AX. V£LOC:Ti '"

ACTSET, EG, 1 ACT SET RC 1 M_CR, 1, 1, 1, 2, 1. 1

1,,';>10

1S .le

I

S9l_2C

4,7

(9) Activate e~ement group 2 (mass) and real constant 2 (m = 100 lb' sec!./

RESPONSE TO GENERAL DYNAMIC LOADING USING COSMOS Example 4.9,

I

in), and generate one element at point 2: CONTROL ~ ACTIVE ACTSE'J', EG, 2

Solve Example 4.2 using program COS}'10S.

Solution: The analysis is pre:onned using a single spring element wIth one concentrated mass element The following commands are implemented:

ACTSET, He,

Ii

2

>

ACTSET

126

Structures Mode!ed as a. Singla-Oegree-of·Freedom Syslem

Response to General DynamIc Loading

(10) Merge nodes; MESHING

07) Define modal damping for mode 1 <:s 20% of critical d4mping (0.20); NODES ) NMERGE

:>

NY-ERG?:, 1, 3,

1, 0,0001, 0,

ANALYS:S :;. POS1'_D":'~J > PD~DAMP/GAP PD-HD.r..Mp, L 1, 1, .2

0, 0

(11) Apply constraints in all degrees of freedom <'I.t node I, and <'!oil degrees of freedom except UX at node 2:

,

,

LOAD_Be STRUCTURAL DrSPLMKTS DPT, L AL, 0, L 1 D?':', ", ey, 0, 2, UZ, RX, RY,

" the f::-equency

,

FREQ /BUCK

:>

1,

A_FREQUENCY,

0,

G,

D,

0

ANALYS:S

>

S,

0,

0,

0,

0,

lE-OS,

0,

18··06,

(13) List the natural frequency of the system:

LIST

1-

:R£<;;:,157

?requency

Frequency irad/scc) 3.1622e+G1

1

(cycles/sec} 5.03292e+OO

Period (seconds; 1,98692e-01

(14) Define analysis type as modal time history using one natural fre~ quency, 1000 time steps starting at t = 0 with a time increment of 0,002 sec; use default values for all integration parameters and reques~ printout of relzrive displacement and absolute acceleration: A."'ll,LYSIS

POST__ DYN

:>

?D._ATYPE,

2,

L

:>

1000,

FD_ATYPE

0,

0.002.

C,

0 S,

0.25,

G

(15) Define dynamic forcing function as time-dependent force: ANALYS:S ~ POST_DYN PD_.CUR';'YI?, 1, 0, :) ANALYSIS :. POST~9YN PD_CURCEF,

1,

1,

P~_CURVBS

0,

0,

.02,

120000,

.O~,

120000,

(16) Apply dynamic forcing function to node 2 in the X direction; ~

'":'C,

AC?IVE }

LOADS_&: :.

FP'!',

2, f'X,

ACTSE~

1

STRUCTURAl, 1, 2,

1

>

>

R_DYNAHIC

R_DYNAr~rC

(20) Activate XY piot information for X displacement at node 2 as a function of time, and plot lhe displacement VS. lime for node 2:

D::SPLAY ) XY_PLO?S >ACTXYPOS'T DISPLAY

:>

XYPLOT,

1

1, TIME, OX,

2,

12, 1, 0,

2N

XY_?LOTS ) XYPLCT

CISPLAY ~ AC?XYPOST,

XY~PLOTS

D!SPLAY

:>

XV_PLOTS ) XYPLOT

XYPLO'!',

1

FORCES, FP'I'

> ACTXYPOST

1. TIME, AX,

2,

12, 1, 0, 2N

i. !

,.I'!

Figure 4.15 shows the displacement and acceleralion response for the first 1 sec
(1) Set view to the XY pla.ne:

DISPr,AY VIE\,v,

ACTSET.

i

Solwion; The analysis is performed using a Single sprbg element with One concentrated mass e1emenL Tbe foHowing commands are implemented:

PD_CUR~BF

.06, 0, 1, 0

CONTROL

POS':'_DY~

>

,

,i·

(21) Activate XY plot informalion for X acceleration at node 2 as a functio:1 of time, and plot the acceleration VS. lime for node 2:

FREQr..IST

:>

A;-.JALYSIS

ACTXYPOS?,

PREQ/8UCK ) R_FREQUENCY

R_?REQUENCY

HESUL':'S

·

POST_DYN ~ PC_OUTPUT) PD_KRESF 1, 2, 0

:>

(19) Execcte modai time his:ory analysis

RZ

A_FREQUEKCY

16,

·· ,,

(18) Request response at node 2: ANALYSIS PD__NRESP.

DPT

(12) Set the options for analysis to exlract one frequency using the Subspace Iteration Method with a maximum of 16 iterations, and run the frequency analysis: ANALYS1S

127

VIb"'f,' PAR> v!EW 1. 0

>

0,

0,

(2) Define XY plane at Z = 0: GEOMETRY PLANE,

Z,

>

GRIJ 0,

1

:>

PLANE

II

Structures Modeled as a

128

1.1)2

& .6-1

&.65 J

x

t:l,! :2

)

.~2

-s .S

I

i , ....

•I

,

,

0.0

\.

• .,' '"

s.<

'-'

(al

"

I

i

?ROPSETS EGROUP,

I /'""'

¥2~tl

PROPSETS :> HGROUP RCONST, 1, 1, 1, L

'--'

3.1

2, L

0,

0,

0,

0

lES

(6) Define element group 2 using the ~1ASS element fomulation:

i

V

0,

(5) Define a real constant for spring element: k = lOO,OOOlblin:

.,

-nIJ I--H>'

,

EGROUP

1,. 'SPRING,

'-...--

-~8&'

-12SG .

>

using the SPRING element formulation with

I"-'~

\

-SSi'il

Water tower and excitation ror Example 4,10.

(4) Define element grQup two nodes:

/\

'"

(b)

Fig.4.16

!

no

,,s

--"~'l

aId

i

G.5

I

'50

,

(t2

--+ 0,

(a)

A

(in/aec')

'i"!\----";f;;1

'-

I

•

"0

a(t)"'300 ain/3t

i,

/" ..........

129

2M1

,!

(

=-1\\

-tl,24

Response to General 0ynamic Loao:ng

I ,

I

System

i

,

-S.96

-&

- ....

i

•• 3

L

, ...

I

0.48

R E

,

I

-

t-I

Sjngle·Degree~of~Freedom

,

"

.:,

0.'

.:,

, .1

3.a

PROPSETS > EGROUP EGROUP, 2. ¥~SS, 0, 0, 0, 0, 0,

•. g

TIME

(7) Define a real constant for mass element: m = 15.5 lb· sec1.lin:

(bl Fig. 4.15 Computer plot for rhe response of Exarr:p!e 4.9. (a) Displaceme':'lt. (b) AcceleratioL

PROPSETS > RCONST RCONST, 2, 2, 1, 7, 15.5, 0, 0, 0, 0, 0, 0

(8) Activate element group I (spring) and real constant I (k ~ 100,000 Ibl in)~

(3) Generate a curve from 0, 0, 0 to I, O. 0, GEOMETKY > CURVES > CRPCORD CRPCORD, 1, 0, 0 1 0, 1, 0, C

0, 0

and generate one element along cunte 1:

CONTROL :> ACTIVE > ACTSET AC'ISE'L EG, 1 ACTSE'!, RC, 1 MESHING " PAR!>'LMESH ) M_CR M_CR, L L 1, 2, .3., 1

130

Structures Modeled as a Smgle-[)egree-of-Freedom System

Respon..l;c 10 General Dynamic Loading

(9) Activate element group 2 (mass) and real constant 2 (n:::::: 155 Jb . sec 1 f in), and generate one element at point 2: CONTROL AC7SET. ACTSE'!', MESHING M~PT,

;.. ACTIVE ;.. ACTSE':' EG r 2 Ret 2 ). PARt.l·CH2:SH

131

(16) Define base excitation as acceleration in the X direction: ANl\LYSlS ~ POS':'_DYN } PD_£5EXCIT > PC_BASE PD_BASE, 1, 1, G, 0, 0

(I7) Req"Jest response at node 2: ANALYSIS ) POST_CYN PJ_N RES?, 1, 2, 0

2, 2,

(10) Merge nodes:

~

PD_OU'l'PU'T ,

PD_NRESP

(18) Execute modal time history analysis:

MESHING > NCDED ;.. t4}1ERGE

NMERGE,

1, ), 1, 0,0001,

0, C,

0

ANALYSIS ) POS'l'_DYN > R_DVNAM!C

(11) Apply constraints in aH degrees of freedom at node 1, and all degrees of freedom except UX at node 2: LOADS-Be ;.. STRUCTURAL DPT. 1, ALL, 0, L 1

~

DISPLMNTS ) CP'r

DPT, 2, OY, 0, 2, 1, UZ, RX,

PREQ !BUCK > A .. FREQUENCY A_FREQUENCY, ~, S, 15, 0, 0, O. 0,

Or 0,

>

lE-C5,

0,

>

DISPLAY ;. XY_PLOTS ) ACTXYPQS,=, AC'l'XYPOST, 1, ':'J:ME, UX, 2, 12, 1, DJ:SPLAY ) XY_PLOTS > XYPLO'l' X,{PLOT, 1

(Figure 4.17 shows the plot of the displacement a: node 2 as a function of time.)

R.]REQUENCY

A.:,"JALYS:r:S > POST_DYN ) PD_OUTPUT :> P:U1AXMIN PD_M..Ji.X:-Hl';, L 1. 10, 2, 2. 0, 2 ANALYSIS } POS'T'~ .. DYN > PD_PREP.lJ..RB

(13) List the natural frequency of the system~ RESUL7$ ~ LIST > FREQLIST F'HEQLrS?

PD_PREPARE,

Frequency#-

~requency

Frequency

1

(rad/sec} 8.032:ge+C1

{cycles/sec) L27B36e..-01

7.82250e-02

A,,),

0,25,

1, 2,

1

ANA:'YSIS ;,

POST~DYN

PD__CURD£F,

1,0,

>-

PD_CURVES

PD~Ct:RDEF

.2,:3CO, 100, 0, 0, 0, 0, C

MaxilTlIJm displacement:

Type

Positive Negative

Slep. no.

Time (sec)

Value (in~

70 87

0.l40 0.174

~0.18413

0.18323

{)

(15) Define dynamic forcing function as time-depe!1dent harmonic base excitation: ANALYSIS > POST_DYN

1

ANALYSIS > POST_DYN ;. PC OU'J'PtJ?

Period (seconds)

(14) De5.ne analysis type as modal time history llsing one natural frequency, 1000 time steps starting at t = 0 with a time increment of 0,002 sec; use default values for all integration paramelers: and request pri:1tom of relative displacements and relative velocities:

PD_CURTYP,

2N

(20) Request SCJn for maximum displacement in the X direction at node 2 from t = 0 to 0.2 sec.

R_FREQUENCY

ANALYSIS ) POST_:)YI", :. PD_ATYl?E PD_ATYPE, 2, 1, 1000, 0, .002, 0,

0,

1E-06,

0, C

ANALYSIS :. FREQ !BtXK

(19) Activate Xl' plot infomuHion for X displacement at node 2 as a function of lime, and plot the displacement VS. time for node 2:

RY, RZ

(12) Set the options for the frequency analysis to exJract one frequency using the S'Jbspace Iteration Method witb a maximum of 16 iterations, and n.l!1 the frequency analysis: A..~ALYSIS

R_DYNAMIC

4.8

SUMMARY

In this chapter, we have shown that the differential equation of motioc for a linear system can be solved for any forcing function in terms of Duhamel's integral. The numerical evaluation of this integra! can be accomplished by any standard method sHch as the trapezoidal or the Simpson's rule. We have

Sir,gle·Deg(ee~or~Freedom

S:ructu:es Modeled as a

'32

System

Response :0 Generar Dynamic Loading

,., ~

Q .l2

,

R E l

/\

a.ea

u

i I

•

•

I II

!

0 04

• 'f.lJI4

-e.l.iB

,i

I

/

[sV ·····r

I\ j

i

-$.12

,

I

-iil.2 Q

0.$2

,

I

I

rL~4

~.~&

•

•

1!l.1i8

r

\

\/

.\

!/ J!m~

11

!, _.. ,I

G,!

•

f;

+-V mJ ;

I 3.12

:

Ii I Ii !-I! i ; ,

\

\

\,

m

-oS .IS

!

1\

i

•

\

@.l4

W8X 24

r

,v, :lLlG

~.18

.., 'J

Fig. P4.l

TIME

i<1g. 4.17

Displacement

re~ponse

at node 2 [or Example 4.10.

preferred the use of a numerical integration by simply assumIng that the forcing function is a linear fU:1ctio:1 between defining points, 3:1d, on this basis, we have obtained the exact response for each time increment. The computer programs described in this chapter use the direct integration method. In this method, the differential equation of motIon is solved for each time increment for the conditions existent at the end of the preceding interval (initial conditions for the new interval) and for the action of the excita~ion applied during the interval, which is assumed to be Ii:1eaL Two programs were presented in this chapter: (1) Program 2 to calculate the response of a single degree-of~freedom system excited by a force applied to its mass (or by an acceieration at the support). (2) Program 3 to determine the response of a single degree-of-freedom system excited by one of the impulsive functions specified in the program, These programs anow us to obtain the response in tenns of dIsplacement, velocity, and acceleration as a fU:Jction of rime for any single degree-of-freedom system of linear-elastic behavior when subjected to a general force function of time applied to the mass or to an acceleration applied to the support.

maximum horizontal deflectioIl. Assume the columns massless and the girder rigid, Neglect damping.

4.2 Repeat Problem 4.1 for 10% of critical damping. 4.3

The steel frame shown in Fig. PA 1 is subjected. to a horizontal force applied at the girder leveL The force decreases linearly from 5 Kip at time t = 0 to zero at ! = 0.6 sec. Determine: (a) the horiZOp.lat deflection at 1=0.5 sec and (b) the

For the load-time function in Fig, P4.3. derive the expression for the dynamic load factor for the undamped simple oscillator as a function of t, &J, and (1.

Fig. 1'4.3

4.4 The frame shown in Fig. P4.1 is subjected to a sudden acceleration of 05 g applied to its foundation. Determine the maxImum shear force in the columns. Neglect damping. 4.5

Repeat Problem 4.4 for 10% of critical clamping.

4.6

Use Duhamel's integra! to obtain the response of damped simple oscliiator of stiffness k, mass tn, damping ratio {, subjected to a suddenly appJied force of magnitude Fa- Assume ini~ial displacement and initial velocity equal zerO.

4.1

Establish ~he equation of motion for the system in Problem 4.6 and solve it by superposition of complementary and particular solutions with initial conditions for dispiacemellt and velocity equal to zero,

PROBLEMS 4.1

~Y

Ffti1'-1==============i I

r •

\

I

20Kip~

,N'

(

1

.IS

134

Structures Modeled as a Single·Oegrce-of-Freedom System

f-Icsponse to Gen-eraJ Oynamic Loading

4,8

A trailer being puUed by .. truck moving a: constn.nt speed u is idealized as a mass In connf'.cted to the {ruck by a spring of stiffness k" Determine the governing differential equation and its solution if the truck starts f£Orr. rest.

4.9

Determine the response of an undamped systerr. to a ramp force (Fig. P4.9) of maximum magnitude FiJ Ilnd duration Id starting with 7.ero initial conditions of

135

8 (/)/g

j

displacement and velocily.

Birl/g'" 0.1.1.

U1

o ----+fJ(t)

ia)

Ib'

Fig. P4,14

Fig. P4.9 4.10

Determine the maximum displacement at the top pf the columns and maximum bending stress in the frame of Example 4.1 ass:.;ming that the columns afC pinned at the base. DIscuss the effect of base fixity.

4.11

Delermine the maximum response (displacement and bending stress) for the frame of Example 4.1 subjected to a trinaguJar food of initia! force ['0 == 6000 !b linearly decreasing to zero at (ime IJ == 0.1 sec.

4.12

For the dynamic sySlem shown in Fig. P4.!2, determine and plot the displacemenl as a funCtion of time for the inlerval 0 S 1:S 0.5 sec. Neglect damping.

,

Fit)

,~~

,owt""

~"'----liwc)

0.2

0.4

4.15

Repeal Problem 4.14 for 20% of crj(icai darr.ping.

4.16

Determine for the tower of Problem 4,15, the maximum dispiacement at (he top of the tower relative to !hc ground displacement.

4.17 The frame of Fig. 4,17(3) is subjected to horizontal support motion shown in Pjg. 4J 7(0). Delermine a:e maxireum absolute deflection at the lOp of the frame. Assume no damping,

~~T

~W8XZO~

I

I"

Repeat Problem 4.12 for 10% of eThical camping.

4.14

The lower of Fig. 1'4. (4(a) is subjected (0 horizOnlal grOlmd acce!ef3tion a (0 shown in Fig, P4.14(b). ;)etermine the relative displacement at the top of the tower at lime I 1.0 sec. Neglect damping.

11

~~ 1.0 Ill)

Fig. P4.12 4.13

I ,I

"

1

10J

10

{ !~ecJ

(b)

Fig. P4.17 4.18

RepeaL Problem 4.J7 for 10% of cHkal damping.

4.19

A strucfural systeo modeled by the simple oscillator with 10% {s:= 0, 10) of criticar damping is SUbjected to [he impulsive load as shown in Fig, P4,19. Determine fhe response.

136

Structures Modeled as a Sing'e-Oegree-of+Freedom System

Response to General Dynamic Loading

r;!)

(Ibl

FW""

0.0

Fig. P4.!9

0.23

4.23

The steel frame in Fig. P4.23 JS subjected Lo the ground motion produced by " passing traill in its vicinity, The ground motion is idealized
rhec)

Jolt)

Fig. P4,23 (bl

(,)

Fig. P4.20

4.21 4.22

0.2

Fig, N,22

A water tower modeled rtf> shown in Fig, P4.20(a) is subjected to g:rouod shock given by the function depicted in Fig, P4.20(b) Determine: ~a) the maximum displacement at the top of [he tower and (b) the ma;omum shear force a! tbe base of the tower. Neglect damping. U,\C time step for integration.:11 0.005 sec,

-

1QS\(\Wf

IL\,

(hI

4,20

137

Repeat Problem 4,20 for 20% of the crhical damping. Dete~ine the maxImum response of rhe tower cf Problem 4.20 when subjected to the iC1pulsive g:uuod acceleration depicted in Fig. P4.22.

4.24

Detenninc the maximum stresses in the columns of the frame in P(oblem 4.23 using the maximum response in terms cf relative motion. Also check that the same results may be obtained using the response in terms of the maximum absolute acceieration.

4.25

A machine having a weig.it W= 3000 ib is mounted through coil springs to a steel beam of rectangular cross sect jon as shown in Fig. P4.25(a). Due to malfunctioning, the machine produces a shock force represented in Fig. P4.25(b), Neglecting the mass of the beam and damping in the system, determine the ma:;;imum displaCement of tr.e machine.

138

S~ruclures

Modeled as a

Sjngje~Degree~of~F(eedom

y k~ =

~

18000 Ib/io.

~==============::=======,=e=3=4~8=,="~,==i>

System

I

I---l0'~-'--I0'---1 101

f(d

F(r)

r

6500"'~ -

-

~~

F(n"'F~sio3!.41Sl

5 Fourier Analysis and Response in the Frequency Domain

I 0.0

0.1

;!~cj

Fig. P4.2S

4.26

For Problem 4.25 determine (a) the maximum [ensUe and compre3Sive stresses in the beam ar.d (b) the maximum force experienced by the coil sprin.gs during the sno-:;k.

This chapter presents the applica~lon of Fourier series to determine: the response of a system to periodic forces, and (2) the response of a system to nonperiodic forces in the frequency domain as an alrernate approach to the usual analysis in [he time domain. In either case, the calculations :-equire the evaluation or integrals that, except for some relatively simple loading rU:1Ctions, employ numerical meti:ods for their compu:ation. Thus, in genera!, to make practical use of the Fou:-ier method, it is necessary to repiace the integrations with finite sums.

5.1

FOURIER ANALYSIS

The s"...lbject of Fourier series and FOt~.rier analysis has extensive ramificatJons in its application to many fields of science and mathematics. We begin by considering a single~degree-of-freedom system under the action of a periodic loading, that is, a forcing function [hat repeats itself at equai intervals of rime T (the period of (he function), Foune, has shown that a periodic function may be expressed as the summation of an infinite number of sine and cosine tenns. S:Jch a sum is known as a Fourier series.. 139

140

Structures Modeled as a

Single~Degree~of~Freedom

System

Fourier Anaiysis and Response in the Frequency Domain

F(r)

141

each component of the series. When the transient is omitted, the response of an undamped system to any sine term of the series is given by eq. (3.9) as y" (I)

where r"

= nwlw and

w=

\l

h Ik.

n =- ,SIn

I -

r~

_

(5.4)

nWl

kim. Similarly, the response to any cosine term is

Fig. 5.1 Arbitrary periodic function. a"lk

_

(5.5)

y" (t) = -----O<:OS nwt

I -

For a periodic function, such as the one shown in Fig. 5.1. the Fourier series may be written as F(t) = ao

+ hi

+ al cos wi + a2 cos

2Wl

+ (/3 cos 3&;1 + ... a" cos nwr + .

sin wt + h2 sin 2Wl + h3 sin 3wr + .. h" sin nwl + ...

The [otal response of an undamped, single~degree~of-freedom system may [hen be expressed as the superposition of the responses to aU the force terms of the series, including the response C10lk (steady-state response) to the constant force Qo. Hence we ha ve

(5.1)

or

y (l) = F(!) = Qo

+ )' {an cos nwt + h" sin nwt}

(5.2)

II"'!

where w= 27TIT is the frequency and T the period of the function. The evaluation of the coefficients Qo, a,,, and h" for a given function FCt) is determined from the following expressions:

J""

aD = - 1 T "

2J"" 2J""

a" = -

k

f

~(~os nwt + ~sin nwt)

,,=11

r,,\k

k,

(5.6)

When the damping in the system is considered, the steady~state response for the general sine term of the series is given from eq. (3.20) as b"lk sin (nwr - iJ) y" (r) = ; ? ? ,(I - r;t + (2r",;t ?

(5.7)

or

F(t) cos nWl dr

The substitution of sin F(t) sin nwl dl

e and cos e from

(5.3)

where [i in the limits of the integrals may be any value of time, but is usually equal to either - TI2 or zero. The constant Qo represents the average of the periodic function F(t).

5.2

~+

F(I)dt

T "

b" = T :,

r~

RESPONSE TO A LOADING REPRESENTED BY FOURIER SERIES

The response of a single-degree-of-freedom system to a periodic force represented by its Fourier series is found as the superposition of the response to

h" y,,(I)=T'

eq. (3.21) gives

(l-r~)sinnwt-2rl1;cosnwt

(l-r;)'+ (2r",;)'

(5.8)

Similarly, for a cosine term of the series, we obtain Q"

y" (t) =

T'

(l - r~) cos nwt + 2r,,; sin nwr (1 r;)' + (2r",;)'

(5.9)

Finally, the total response is then given by rhe superposition of the terms expressed by eqs. (5.8) and (5.9) in addition [Q the response to the constant

142

Structures Modeled as a Single-Degree-of-Freedom System

lj t: m

FIr)

J)?&///};)~7

Fourier Analysis and Response in the Frequency Domain

T--/:;::-=J:---~ FO~;?T---o

T

2T

(.)

FIr)

1

Fie;.!)

3T

-----------------

(b)

Fig. 5.2 Undamped oscillator acted upon by a periodic force. Fir,)

t.erm of the series. Therefore, [he total response of a damped singJe-degree-of freedom system may be expressed as

k

+

a" (1 - r;) - b"2r,,f

,,

(1 - r;)'

?

+ (2r,,§)-

1 cos nwt J

(5 [0)

2fT Fo ~I sin nwr dl = ToT

Fo n7T

~-

b" = -

The response of the undamped system is then given from eq. (5.6) as ~ Fo sin nwt 2 ,,_, n1Tk(l - r,,)

yt~?~L

_k

5.3

FOURIER COEFFICIENTS FOR PIECEWISE LINEAR FUNCTIONS

Proceeding as before in the evaluation of Duhamel's integral, we can represent the forcing function by piecewise linear function as shown in Fig. 5.3. The calculation of Fourier coefficients, eq. (5.3), is then obtained as a summation of the integrals evaluated for each linear segment of the forcing function, that is, as

ao

Fo cos nWI dr = 0 a" = -2i ~r ToT

Fo

I

2

T

~

c"

Fig. 5.3 Piecewise linear forcing function.

ao=~rTFordr= Fo

()

r,

I----Clc,-1

Solution: The first step is to determine the Fourier series expansion of F (t). The corresponding coefficients are determined from eqs. (5.3) as follows:

or in expanded form as

Fo

Fo sin wt

Fo sin 2w{

2k

1Tk (l - r;)

21Tk (l - 4r;)

~-

oF,

1,_ I

Example 5.1. As an application of the use of Fourier series in determininothe .response of a system to a periodic loading, consider the undamped simpl~ oscillator in Fig. 5.2(a) which is acted upon by the periodic force shown in Fig. 5.2(b)

T)o T

__ 1 _________ _

F{r,_ I)

+ 1 ~ [a"2r,,f+b,,(I-r;). . k ,,= L I (1 _ Til')' + (0_TIl':-) "" Sin "wI

_ ao y () r -

y(r) =

143

Fa sin 3wt 31Tk (1 - 9r;)

= -[ )'N J" T7 ,- I 2

a"=T~ N

,- I

b" =

F(r) dr

(5. [ [)

F(r) cos aWl dl

(5.! 2)

F(t) sin nw[ dt

(5.13)

li_'

f" 1;_:

T2 I~I f" N

Ii-I

where N is the number of segments of the piecewise forcing function. The forcing function in any interval [,_I :5. t:5. t,- is expressed by eq. (4.20) as

F(r) = F(r,.,)

flF,

+ ~-(r .1r;

r,.,)

(514)

where TI

=wlw,w=[klm, and W=27TIT.

in which .1Fi = F(l,) - F(ti- I) and ..Jt, = {i - {i_I' The integrals required in the expressions of a" and b" have been evaluated in eqs. (4.21) and (4.22) and

144

StructJfeS Modeled as a Single~Oeg(ee-of"Freedom System

Fourier Analysis and Response In the Frequency Domain

designated as A (r,) and B (I,) in the recurrent expressions (4.18) and (4.19). The use of eqs. (4.18) through (4.22) to evaluate the coefficients a" and bl< yields

(i-

+

elF, '>

')

n~w- .:1/;

elF \ 1--') (sjn nwt ill,

ti -

1

(5.15)

!

?""11'

-·iF(r,.J-tl.' elF .,.:'-)\ (C05 nwtH nw \ .:J" I

cos nWI;

elF

5.5

+ ::_1~ ··«sinnWi;-sinnlVt,_,) n w 4.1[; - nw(t/ cos n&t,

[,_!

cos

The interval of integration in eq. (5.20) has been selec(ed from zero to T for the periodic function. It should be noted that the exponential form for the Fourier series i:1 eg. (5,19) has the advantage of simpliciry when compared to [he equivalent trigonometric series. eq. (5.2). The exponential form of the Fourier series can be used as before (0 deter:uine the dynamic response of structural systems, However, a more effective method is available for the determination of the coefficients as well as for the cakulatIon of the response for a single degree of freedom excited by the force expanded as: in eq. (5.; 9). This method, which is based on Fourier nnalysis for the discrete case, is presented in the next sections.

en

sin n(~t-J)}

N

,; ~

sin nWf, 1)

«(cos nWl;-cos nwti __ ;)

+ nw(ti sjn niixi b,,~ T· L

i-

nZ;)/;_I))~

J

(5.16)

1 N-l

a,. ~ - ) F (ti) cos nMj 4t T i"'O

b_=~ .. T

(5.17)

EXPONENTIAL FORM OF FOURIER SERIES

The Fourier series expression (5.2) may also be written i:1 exponential fonn by substituring the trigonometric functions using Euler's relationships:

DISCRETE FOURIER ANALYSIS

When [he periodic function FCr) is supplied only at N equally spaced time intervals (dl TrN) 10, l" [2, ' ' ' , [/1- h where t} = jLlr, the integrals in eqs_ (5.3) may be replaced approxImately by the summations

The integral appearing in the coefficient ac of eg. (5.3) is readily evaluated after substituti;1g F{r) from eq. (5.14) into eg. (5,11), This evaluation yields

5.4

145

n

0, 1,2, ...

(5.21)

where w= 27T11. The above definitions for the Fourier coefficients have been slightly altered by omitting the factor 2 in the expressions for an and h". In this case eq. (5.2) is then wriaen as F(r,)

2 )-~ {a" cos nliJt + On sin nZVt}

(5.22)

11"'1

cos nwt =

e;,Iid,T e -

,""Wi

---:;---2

(5.18)

If we use complex notation, egs, (5.21) can be combined into a single form by defining

C" =0»- ib"

The result of this substitution may be written as

(5.23)

and using Euler's relationship (5.19)

where

(5.24) to

I

C=~Ii

T

JT F([),e -!r:wid' '. . o

(5.20)

I) into

obtain after substituting eg. I

C"=y

eq. (5.23) (5.25)

FOurier A'1alysis and Response in the Frequency Domaln

Struc:ures Modeled as a Single-Deg:ee-of,Freedom System

Substituting

'i =

jfll, T= Nfll, and OJ = 2'JTIT into eq. (5.25). we obtain n~O,

1,2,."

(5.26)

Equation (5.26) may be considered as an approximate formula for calculating the complex Fourier coefficients in eq, CS.20), The discrete coefficients given by eq, (5.26) do nOt provide sufficier.t infonnation to obtain a continuous function for F(r); however, it is a most importanr fact that it does allow us. to obtain all the discrete vaiues of the series [F(IJ)} exactly [Newland, D.E., 1984). This fact leads to the fonnal definition of the discrete Fourier transform of [he series {F(!j)}.j=O, 1,2, N- 1, given by

147

As a matter of interest, Example 5A is presented later in this chapter to illuslrate the importance of choosing the nl.:mber of sampling points N for Lhe excitation function sufficier.tly large Lo avoid spurious resufts due to aliasing. Having represented an arbitrary discrete function by a fwire sum, we may then r..lso obtain as a discrete fl:nction the respor.se of a simple oscillator excited by the ham:onic components of the loading function. Again, only the steady-state respor.se will be considered. The introduction of the unit exponential forcing function E" = e''''nt into the equation of morior:, eq. (3.13), leads ro (531)

which has a

steady~stale

so:utior. of the form

"q

y(r); H (w,,)e'""'

" = O. L 2, .. " (N - !)

(5.27)

ar.d its inverse discrete Fourier transform by

(5.32)

When eq. (5.32) is i:1troduced i:lto eq. (5.3 t), it is found thai :he function H(w,), which will be desigr.mec as the complex frequency response function, takes the form

If- ;

F( [i) --

Y C.~e '"""1'''' ,}'-O!? - , ,~,., ,(N- I)

L-.,

(5.28)

The range of the sumation it: eq. (528) has been limited froal 0 to (N

l) in order to maintain the sym:netr)' of transfonn pair eqs. (5.27) and {5.28). It is important to realize that in the calculation of the summation indIcated in eg. (5.28), the frequencies increase with increasing jndex n up to n N n. It will be shown very shortly that, for n> N 12, the corresponding frequencies are equal to the negative of frequencies of order N n. This fact restricts the harmonic components that may be represe:1ted in the series to a maximum of N 12. The frequency correspondIng to this maximum order WI-In. = (N fl)liJ is known as the Nyquisl frequency or sometimes as [he joldlng frequency. :\1oreover, if [here are harmonjc components above W,v/2 in the original functior.. these higher compor.ents wilt introduce dis.tortions in the lower harmonic corr.ponents of the series, This phenomenon is called aliasing [Newland., D.E., 1984, p. 118). In view of this fact, it is recommended that the number of intervals or sampJed pOints N should be at least twice the highest harmonic compone:1t present in the function, The Nyquist :requency &J y is given in radians per second by

(5.33)

H(w,,)

,,=0

UpO:l ir.troducing t!le frequency :-atio

and the damping ratio

c

['=-= -

C

Cet

eq. (5,33) becomes

H(w,,)="--

k(l

I 1.

+ ro + 2/f,/;)

Therefore, the response)'4 (;j) al tirt.e {j = jiJr to a harmoniC force cOr:1ponent of arr.pliwde en indicated :n eq. (5.28) is given by (5.34)

(5.29) ar.d the total response due to (he N hannonic force components by

and in cycles per second by

f ; 2." = .y

(cps)

(5.30)

(5.35)

148

Structures Modeled as a

Single~Cegree~ct-Freedom

Fourier Ar_alysis ard Response in the Frequency Domain

System

where ell is expressed in discrete form by eg. (5.27). In the determination of the response y (t) using eg. (5.35), it is necessary CO bear in mind that in eg. (5.28) the force component of [he frequency of order n is equal co [he cegative of [he component of (he frequency of orde-r N - n. ThIs fact may be verified by substitcting - (N n) for n i!1 the exponemial factor of eq. (5.28), In this case we obtain (5.36)

since e -2""/:= COS i sin = l for all integer values of j. Equation (536) together wiili eq. (5.28) shows that harmonic components of the force corres~ ponding to frequencies of orders Ii and - (N - n) have the same value. As a consequence of this fact, r'l w,,/w, where (seJecting N as an even number) as bJ,,=niiJ

for

w=.i KIm

(539) where lY! is an imeger. In this case, :he integers j and n can be expressed in binary form. For the pUf'I,?ose of iHustrution, we wi!: consider a very simple case where :he load period is divided ~nto only eight time increments, that 1s. N = 8, M = 3. In this case, rhe icdices in eqs, (5,27) and (5.35) will have the bicary ::eprese:1tatlon

j

=

jo + 2j, + 4[, ~5.40}

and eq, (5.37) may be written as

n::S;Nt2

(SAl)

A(j)

(N-n)w

for

n>NI2

The exponential factor can be written as

where the frequency corresponding 1:0 fi = tV /2, as already mentioned. is the highest frequency [bat can be considered in the discrete Fourier series. The evaluation of [he sums necessary to determine the respo<1se using the discrete Fourier tninSfOffil is greatly simplified by the fact that the ex.ponential functions involVed are harmonic and extend over a range of Hl, as demonstrated in the next secrion,

5,6

The evaluation of the sum wHl be most efficient if the number of time ir.crements N into which [he period T is divided is a power of 2, that is,

should be evaiuared

and

w"

FAST FOURIER TRANSFORM

A numerical technique is available that 1S efficiem for computer detenr.ination of the response in the frequency domain. This mewod is known as the fast Fourier transfonn (FFT) [Cooley. P. M., eL aL (1965)1. The corresponding compurer program is reproduced as a subroutine of computer Program 4. The response it; frequency domain of a single-degree-of-freedom system to a general force is given by eq. (5.35) and the coefficients required are compmed from eq. (5.27). It can be seen (hat either eq. (5.35) or eq. (5.27) may be represented, excep( for sign in the exponent by the exponential functron as

2: A'" (n)W::;

(5.37)

'l '" (>

W~l

= e1 ,..-j\SiaV ;;;; cos 2-rrI + i sin 21il =

1

where 1= j !nZ +- 2j2n: + hnl is an integer. Therefore, only the remaining three factors need to be considered in the summations. These summations may be performed conveniently in sequence by introducing a new noration to indicate the successive steps in the summation process, Thus the first step can be indicated by

,

",=0

and the third step (final step for lv! = 3) is

where

W,v=

We note that tbe first factor on the right-hand side is unity since from eq. (5.38)

~ A(l)v" n n )W:2I!10/1+J<;) A (~(j(I> J' ;, t to );;;; L O,!. 0 B

."'-1 A ()) =

149

(5.38)

150

Fourier Analysis and Response in the Frequency Domain

Structures Modeled as a Single-Degree-ot-Freedom System

151

An

Time in minutes

w ~ 38.61<

F(t)

0.16

k

~

0.64.

t(sec)

100 K/in. -1201<

-

-

-

-

--

Fan method

~~~~~~==~~~==========~;::;Num~Of 1024 2048 4096 8192 point d,ul Fig. 5.4 Time required for Fourier transfonn using convemional and fast method. [from Cooley, 1. W., Lewis, P. A. W., and Welch, P. D. (1969), IEEE Trans. Education, E-12 (I ).]

The final result Am UO,jl>j2) is equal to A U) in eg. (5.37) or (5.41). This process, indicated for N = 8, can readily be extended to any integer N = 2M. The me [hod is particularly efficient because the results of one step are immediately used in the next step, thus reducing storage requirements and also because the exponential takes the value of unity in the first factor of the summation. The reduction in computational time that results from this formulation is significant when the time interval is divided into a large number of increments. The comparative times required for computing the Fourier series by a conventional program and by the fast Fourier transform algorithm are illustrated in Fig. 5.4. It is seen here how, for large values of tV, one can rapidly consume so much computer time as to make the conventional method unfeasible.

5.7

(b)

Fig. 5.5 Idealized structure and loading for Example 5.2.

Example 5.2. Derermine the response of the tower shown in Fig. S.S(a) subjected to the impulsive load of duration 0.64 sec as shown in Fig. S.5(b). Assume damping equal to 10% of [he critical damping.

Solution: Problem Data:

Mass: m ~ 38,6001386 ~ 100 (lb· sec' lin) Spring conSlant: k ~ 100,000(lb/in)

Damping coefficient: c = 2 ;\ km Select M such that 2M = 8: M Ex c itat ion f u nc ti 0 n :

=3

~T~im:::::e~(c:s::e::c,-)-+_~F-=o::r::.ce=-,(,,1b:,)_

0.00 0.16 0.48 0.64

PROGRAM 4-RESPONSE IN THE FREQUENCY DOMAIN

The computer program presemed in this chapter calculates the response in the frequency domain for a damped single-degree-of-freedom system. The excitation is input as a discrete function of time. The computer output consists of two tables: (1) a table giving the first N complex Fourier coefficienls of [he series expansions of the excitation and of the response and (2) a table giving the displacement history of the steady-state motion of the response. This second table also gives [he excitation function as calculated by eq. (5.28), thus providing a check of the computations. The main body of this program performs the tasks of calculating, using the FIT algorithm, the coefficients ell and the function F(l j ) in eqs. (5.27), (5.28), and the response Y (I) in eq. (5.35).

= 632 (lb· sec lin)

o 120,000 - 120,000

o

Input Data and Output Results 2H.. E:D4 !NPUT D':'7.:'· NiJMEE~

OF

POZ~TS

DEF":::l:KG THE

~" S?R!NG CONSTnNT DA.!1?!NG COEFt'"ICI2!IT

E;~,::r;.!'7To!l

A.V; "

100

;'K ;

100000

c "

632

G ;

0

EXPONii:NT OF 7' M fOP. e,T INDEX

(GRAVITY OR ZEi
152 1'~,~t:

:} ,ec'}

Strwctwres Modeled as ill O::
·J.e?

Single~Degree·ot·Freedom

Fourter An'eJysis and Response in the Frequency Domain

System

~XC!TA'!'!ON

'!'!11"

lZ:'O:,,) ,co

O. ;;:~D

o. co

Exponent of N = 2M: /vl = 4 Gravitational index: G = 0 (force on the mass) Excitation function:

Time (sec)

Force (Kip)

D!AC

0.e;:;:,>

-O.tEn

"co c. ';l~~C

OvlS 1':,'5 '1'10 j010

-0.znr:

-0 DOJ l O.

evoo

COl5

o. ~~-~I)

,1:0:)

_0.0337

<), ;;~:. j

o

0.00 O.lO 0.20 OAO 0.45 0.60 :.00

O.C<>JC

[0 8

o 6

o o

Input Data and Output Results ~O_O:?~

:. ",41

l1C;;;;C

C(>~}

5JC:;C 01%

:;<';>co

.J Ccc:;.

-5J-)('0 O:CO

j.onc

joe-£:

:u:os~

-l1J·XC.

,) OeD;)

>;;)},5::::;: OT

0.£:010

?O~:,1"'$

:);';t1ytxC :-"S

::XC1--:;;,:O
",_'d"iS

Example 5.3.

Determine the response of the simple osciIlator shown in 5,6(a) when subjecled to (he forcing function depicted in Fig. 5,6(b). Use lkf = 4 for the exponent i:1 N 2M. Assl.!me 15% of the critical damping,

C,"..>:7!~ :::;:'HT:C:!~:'"

'!:Xl?OC!f':."':'

::msx

m"

2 '1 ::-OR ??? A

iGRAV:'t:.' C? ZZROl

',rHZ

:;:XC!1"'A"t1Ol-:

J.G;!D

0.00

!::<:;-r:;?
lao

-11, CO

20J

G,,;0

O. ,150

Solution.' Problem Data:

-a,OJ

o.,)C

0,600

Mass: m ~ [001386 ~ 0.259(Klp sec' lin) Spring constant: k

= 200 Kip lin

Damping coeftlcienr: c = c ~ 2 X 0.15}200 X 0.259 ~ 2.159 (Kip· sec lin)

0.0000 ~.CC$9

cess

eeq: f\ti

() .1069

000"

lieS!

k '" 200 K!if'l. 15%

O.lH':'

Fit)

0.11-5$

+-+-+--"'-'l-i'--"i--+_ ..--+-- <~ 0.60

LO

~D.C3~}


0021

CC"" 0000

,~

OS'5

~O

0004

_0. )007

~c.

ClS11

-0 (lOC7

O. DOOl

'J D-JO

DOC;'

oooa

-0 -:j('C7

"J.,DS)

10

0.115$

-0 0$75

-e

:.:

o .:a4

Cl :'-314

-0

0,SH3

tb)

C lOS::

-c o

~s;.;;

C043

_l.:$C~

Fig. 5.6 Simple oscillamr and loading for Example 5.3.

~';!

-0.1624

.5,,52

000"

en,

COS1 0,000:: 0.oe31

0.0040;. 0.e<);)4

154

Structu~es Modeled as a Single·Deg(ee~ofwFreedom System !HSF!.. Rl:;I\L

::l!S?:'.

x.v.c.

FORCE RZ,\L

fCRCE UV,O.

0.0(100

O.JGS2

a.ooon

o .·~ooo

0.0675

v.ODOO

-0.00%

-C.\)OGO

wn.2S00

C.12S0 \) 15,5 l'.;(>GO

coooo

-0. aDeD

-5. SC'OC

-0. )OCO

wO.CG}S

-B.?SOO -i$.0000

0.0000 ·(l.OGeo

-LSO\lO GOOl'

(I.oooc

. 0 ,0S-;1

~O.

0000

,,).O;;S{

(1.0(100

-0.0091 ·0.C1<4

~O.OJOO

0.)75-0

C,

HZs

0.l'000

~l.

TABLE 5.1

- _....

;:: .0.000.

G.~n5

G.OO$6

-(l.oOOC

~. ~OO()

O.5COO

O. (4)0

0.,0000

O.OOGO

~.COOO

0.%25 0.62S>J

O. G:)OO

O. ~21D -:).0167

-o.OOGe

:. SOC'O

0,000:)

C.OOOO

O.6S7S

·a.M'"

-c.oooo O.cooC'

-J,OC:i"J

J .COO:)

0.7SCO O.$U:' C. S7S0

o.oeoo

0,0:)00.

0.0116 0.0018

o.eOOO

-O.OCOO

C.oooc

·o.oon

0.00;';0

", GOCl'

O.Ol'OO

0.00':

·~.OCOO

-0.0000

O.OOOD -0,000;) -0,0(0)

(l.H~:;

Fourier Analysis and Response in the Frequency DQmain

Example 5.4. Consider a smgle-degree·of-freedom und<::!mped sysrem in which k 0:::: 200 lblin, m 100 lb· sec 1 /in subjected to a force expressed as

P(I)

= L" ""

lOO cos 2717"

N

0 0.125 0.250 0.375 0.500 0.625 0.750 0.875 1.000

0.8531 0.9357 1.022 1.071 1.089 107! 1.022 0.9357 0.8531

....-

5.4 (Exdtation Having 16

Number of Sampling Poims for the

Time (sec)

-

Examp~e

---_

..

8

N

16

0.4201 0.4698 0.5 i07 0.5358 0.5443 0.5358 05107 0.4698 OA201

-------

Ex~.~~,1tion

N= 64

N=32 - 0.0416 . 0.0153 0.0052 0.Oi78 0.0221 0.0178 00052 - 0.0153 - 0.0416 -....

_----

- 00416 0.0]53 0.0052 0.0178 0.D22! 0.0178 0.0052 -O.O~53

0.0416 ...

--.~--

=

(a)

Determine the steady-sCate response of the system using Program 4 with

lvJ = 3, 4. 5, and 6 corresponding to N = 8, 16, 32, and 64 sampled points. Then discuss (he results in relation to the limitations imposed by the NyqUIst frequency.

The fundamental frequency of the excitation function. eq. (a), is Wj f2-rr= t sec. Since the ~i!ghest component in eq. (a) is of order Wlf) = 16wJ, to avoid aliaSing, the number of sampled poims should be at least twice that order, that is, :he minimum number of sampled points should be N 32. Sotwion:

WI

Displacement Response for Harmonics)

155

= 21/ and its period T= =

With a simple moodification of Program 4, the applied force is calculated in the program, instead of being supplied through a numeriC2.1 [abLe as nor~ many required by the program. The results given by the computer for this example are conveniently arranged in two tables: Table 5.1, giving the dis~ placement response to the excitation having aU 16 harmonic components as prescribed for this problem; and Table 52, Showing the dispJacement response to a reduced number of harmonic terms in (he excitation function. For this example, in which the exciting force is supplied in 16 harmonic components, the response given in Table 5.1 corresponding to N = 32 or N =:;:; 64 may be considered the exact SOlution. A comparison of the response l6 with (he exact solution (N = 32) shown for sample points N = 8 or N dramarically demonstrates the risk of not choosing N sufficient!y large enough so that none of the frequencies of the components in the exciting force exceed

the Nyquist freqnency. The response obtained for N g or N = 16 gives spuriolls numericai results, . Results in Table 5.2. which "vere obtained using N = 8 sampled pomts, also verify t:tat when the exciting force ContamS.hiHTI:onic compone~ts higher ~~atn the Nyquist frequency which corresponds, In Lhls case, to N/ - 4, the re",ul s are again spurious. . . A fir:al comment is in order. The example ptesen[ed, havlflg equal.amplltude for all the components of the exciting force, serves to e:mphasne the importance of choosing tr.e number of snmpling point~ N s~ff:clent,ly large to avoid aliasing. In pn!c~ical siwa~ions r.ormally the higher :larmOnlCS hav~ a much smalier amplitude thar: that of the fundamental Or lower frequencIes.

TABLE 5.2 Time

(sec) 0 0.125 0250 0.375 0.500 0.625 0.750 0.875 1.000

Displacement Response for Example 5.4 (Excitation Force Sampled at N~ 8 pointS) _____ . _____________ N'Jmber of Harmon~.~ Comp0n.~nts in the Exciration Force

IV

4

- 0.0375 -0.0153 0.0048 0.0184 0.0215 0.0184 0.0048 - 0.0153 - 0.0375

N=8

N=16

0.4246 0.L679 0.5J 12 0.5353 05446 Oj353 Oj112 0.4679 0.4246

0.8531 0.9357 1.022 1.071 1.089 1.071 1.022 0.9357 0.8531

Structures M(ldeled as a Single-Oegfee-cf-Freedom Sys~em

156

Fourier Analysis and Response in the Frequency Demain

Consequent:y, the distortion in the response might nOt be as dramatic as shown

157

k - 120 Kfin.

in Tables 5J and 5.2.

~ "" C.10

' W'" 128.66'"

~r---' 5.8

SUMMARY

In general, any periodic function may be expanded into a Fourier ser:es, eq. (5.1), whose terms are sine and cosine functions of successive multiples of the fundamental frequency. The coeffjc;en~s of these functions may be calculated

integrating over a period the proCUCt of [he periodiC function multiplied by a sine 0:- cosine fU:1ction, eq. (5.3). The response of (he dynamic system is then obtained as the s:.lperposiEion of tJ:e response for each (erm of the Fourier series expansior. of (he excitmior! function. The extension of the Fourier series to nonperiodic functions results in integrals which are known as Fourier trans-

Fig. PS.2.

5.3 The spnng-rn2$5 system of Fig, P5,2 is acted upon by the t:me-varyL1g force shown in Fig:. P53. Assume that (he force :5 p-edodic of period T:= 1 sec and determme steady-state Ies?or:se of the system by applying Fourie; series expansion of F(t).

the

forms, The discrete form of these !mnsfonns, eqs. (5.27) and (5.28), penni!s

,clfi

their use in numerical appiications. An eXlremely efficient algOrithm known as the fast Fourier transform (FFT) can save as much as 99% of the computer time otherwise consumed in the evaluation of Fourier complex coefficients for the eXCHattOn function and for the response of a dynamlc system

t

PROBLEMS

'---;. --..f._ ....... r--"';"--T---1

S.l

Determine the first three terms of [he Fourier series expansion for the lime varying force shown in Fig. PS.I.

5".4 The cantilever beam shown in Fig. PS.4(a) carries a concentrated weight at its free end and it is subjected to a periodic acceieration :rt :its support which is the rectified sine f:.mcrion of period 7=OA sec and amplitude )te"'" 180 as shown 1fl Fig. P5.4(b}. Determine: (a) the Fourier series expansl0n of rhe forcing function. and (b) the steady-state response considering oni:; three terms of the series, damping in the system and assume L1e beam massless.

,

0

05

-30xL.

Fig. P5,3.

r- TI2 -+- m-j

1.0

20

.(sec)

U Fig. P5.1.

~(rl

'0

T

T,

Y,

I.)

5.2

De!ermine the steady-state response fo:- the damped spring-mass system shown in Fig. P5,2 rnl,H is acted upon by (he forcing func~ion of Problem 5. L

(hi

FIg. P5.4.

158

Structures Modeled as a Single-Degree-of-Freedom System

Fourier Analysis and Response in

5.5

Solve Problem 5.4 using PlOgram 4. Take 16 Fourier terms. Inpu[ [he values of the excitation functions Ul intervals of 0,025 sec.

5.6

Solve Proble:n 5,4 in the frequency domain using Program 4. Take the exponent of N 2 M , M=4, Input [he effective force, F df = -mYt(!) cZllculated every 0.025 sec. Repear Problem 5.6 assuming 20% of critica: damping.

5.7

5.8

5.12

~he

Frequency Domain

159

Cor.sider the sy:item shown in Fig. P5.12 and its loading with assu:ned period T= 2 sec. Determine: (a) Ihe firs! four rerrns of the Fourier series expansion for [he forcing functior: in terms of Po; (b) Lhe firs! four terms 'Of the Fourier series expansion fer the resp'Onse.

The forcing fur:c~ion shown in Fig. P5.&(0) is assumed w be periodic in the extended interval T= lA sec. Use Prog:rom 4 to determine the fir:i.t eighl: Fourier coefficienrs and the steady-state response of a .mucture modeled by the undar:J~ ped osciHawr shown in Eg. P5.8(b),

F(t)

2.0 (b)

{ill

·Fig. P5.12. 0.'

0.4

0.6

r

0.8

1.0

I 1.2

r(secl

1.4

0

5.13 Repeat Probler:l 5.12 using Program 4, Take 32 terms

1.4 sec

'h,

I,)

Fig, PS,g,

5.9

Solve Problem 5.8 in the frequency domain using Program 4, Take M=4 for the expOl:ent in N = 2M.

5.10

Cse Prograrr. 4 to determine: (1) the Fourier senes expansion of the rorcir:g func(:on shown in Fig. P5.10(a), and (2) the steady·s(ate :esponse caiculmed in rhe freql.:ency domain for the spring-mass system in Fig. P5.1O(b). Assurr:e ! 5% of the critical damping. Take />1 = 3 for the exponent in N 214 compare results with those in the solution of Example 5.3.

the force at intervals of D. iO

0: Fourier series;

input

,~ec.

5.14

Repeat Probiem 5. J 3 as:;uming: that (he systel7'l has 20% of [he critical darr.ping.

5.15

Ob:ro;n the close solution for the system in Proble:n 5.12 by considering (he half-circle sinusoidal excitation as the s"Jperposition of two sinusoidal functions: PI = Pa sin Tt1 smrring ,n'" 0 and P! and Pc sin 17(1 - I) stanir:g at r = 1 sec as shown in Fig. PS.iS.

p~

P 1

1r\;stnw: L:::, _ ,

Fitl

t ft; ""

200 K/in.

Fig, PS.15.

0.1 0.2

I O.S

• r sec

,

1.0

r--~''''''- 10"'~

5.16

i,

(,j

(b)

Fig. PS.IO.

5.11

Solve Problem 5.10 in the frequer:cy domain l'sing Program 4. Take M 5 for the expOJ1eIU in N 2 M.ColT.pare resules with those in the solutIon of Example 5.3.

A single~degree.of-f:eedom system having a natural period of ;).8 sec ar.d stiff" ness of 5000 Iblin subjected to an impulse of dUr 1 sec for which the value of force is zero. Use F'rograrr: 4 to Obtain: (a) the discrete transform of the forcing function and of [he response. (b) the displacerr:e!lt response, and (c) the applied force calculated using the ir:verse discrete trans:·orm. Neglect damping and discreljze (he forcing fl'ncrior: uSing a rime slep Dr = eu SeC.

160

Fourier Analysis and Response in the F~equency DOIT'.ain

S:ructures Mode;ed as a Single-Deg:ee-or-F . .eedom System F{l)

5.19 5.20

tOOClb

Fig. P5.16. 5.17

Solve Problem 5.16 assuming a 10% of the critical damping in tle system,

5.18

The water rower shown in Fig. P5.J:'8{a) is subjec(ec to impu:s:ve acceleradon of its ba$e lilac varies as half the 3ir.e function shown 1n Fig. P5.1'8(b). {)$e Program 4 to determine: (u) :he discrete Fourier coefficients for the excitation and for tile respo:"Jse. (b) (he re:1I!(ive displacement of rhe tower with respect to the ground displacement, and (c) ~he excitation obtained by the inverse discrete transform. Use an extended e-x;ciration of [Otal duration 1.6 sec and time step .::::il 0.1 sec. Negrect damping.

1--1----

W"'" 3S.6K K""

y

lOO~

'"

O.3g O~-----';;O';g'----"1.76----[g~·

(a)

(b)

Fig. PS.1S.

A{r)

1.2

Fig. PS.20.

I.'

I

seo

161

Repeat ProbleCl 5.18 assuming thal: the damping in the system is 5% of the criticaL Solve Problem 5. ('8 for an accele:adon at lhe base of tower (hat varies as a symmenical triangu!ar load as shown i:1 Fig, P5,20,

Generalized Coordinales and Rayleigh's Method

6 Generalized Coordinates and Rayleigh's Method

163

Principle, which establishes dynamic equiHbrh:m by the inclusion of the inertial forces in the system. The principle of virtual work may be stated as foHows: For a system that is in C
+ cj8y + AJ,8y ~ F(r)8y ~ 0

or

{my + cy + ky -

F(I)}8y ~ 0

(6.1)

oy

Since is arbitrarily selected as not equal to zero, the other factOr in eg. (6.1) must equal zero. Hence,

In the preceding chapters we concentrated our efforts in obtaining the response to dynamic loads of structures modeled by the simple oscillator, that is, structures that may be analyzed as a damped or undamped spring-mass system, Our plan ;n the present chapter is 10 discuss the conditions under which £! structural system consisting of multi pie interconnected rigid bodies or having distributed mass and elasticity can stiH be moceled as a one-degree of-freedom system. We begin by presenting an alternative method to the direct .application of Newton's Law of Motion, the principle of virlual work.

6.i

F (I) ~ 0

(6.2)

Thus we obtained in eq. (6.2) the differential equation for the motion of the damped oscillator.

PRINCIPLE OF VIRTUAL WORK

An alternative approach to the direct method employed thus far. for the formulation of the equations of motion is the use of the principle of vinuai work. This principle is particularly usefuL for relatively complex structural systems which contab. many lnterconnected parts. The prir.dple of virtual work was originally stated for a system 1:1 equilibrium. Nevertheless, the principle can readily be appJied to dynamic systems by the simple recourse to D'Alernbert's 162

my + cS' + ky -

T:

KY+-~-'~ , my tb)

cy ,

,!

mg

:

"
N

Fig. 6.1 Damped simple oscillator undergoing virtual displacement &j.

164

6.2

Structures Modeled as a

Sing!e~Degree~of·rreedom

System

Generalized Coordinates and Rayleigh's Method

GENERALIZED SINGLE-DEGREE-OF-FREEDOM SYSTEM-RIGID BODY

Most frequently the configuration of a dynamic sysrem is specified by coordinates indicating the linear or angular positions of elements of t3t system. However. coordinates do not necessarily have to correspond directly to displacements; they may in general be any independent quantities that are sufficient In number to specify the pOSition of all parts of u1e system. These coordinates are usually caIled generalized coordinates and their number is equal to the number of degrees of freedom of the system. The example of the rigid-body system shown in Fig. 6.2 consists of a rigid bar WIth distributed mass supporting a circular plate at one end. The bar is supported by springs and dampers in addition to a singie frictionless support. Dynamic excitation is provided by a transverse load F(x,t) varying linearly on the portion AS of the bar. Our purpose is to obtain the differential equation of motion and to identify the corresponding expressions for the parameters of the simple osci~lator representing this system. Since the bar is rigid, the system In Fig. 6.2 has only One degree of freedom, and, therefore, its dynamic response can be expressed \vith one equation of motion. The generalized coordinate could be selected as the vertical displacement of any point such as A, B, or C along the bar, or may be taken as the angular position of the bar. This last coordinate des.ignated &(t) is selected as the generalized coordinate of the system. The corresponding free body diagram showing all the forces including the inertial forces and the inertial moments is shown in 6,3, In evaluating the displacements of the different forces, it is assumed that the displacements of the system are small and, therefore, vertical displacements are simply equal to the product of the distance to support D multiplied by the angular displacement 8= 8(1). The disptacements resuhing at tbe points of application of [he forces in Fig. 6,3 due to a virtuai displacement 58 are indicated in this figure, By the principle of virtual work~ the total work done by the farces during this virtual displacement is equal to z.ero. Hence

o8[l,8 + 1,8+ 4L'Jj,ij + mL'ij + cL'iJ+ 4kL'8 -lL1(tl] =

-------- --!

165

I,}

'"'

miiL

4LiriLii

~---L

2LM

2LKfi

D

cLB R

Fig. 6.3 Displacements and resultant forces for system in Fig. 6.2. or, since 88 is arbitrarily set not equal to zero, it follows that

(I, + I, + 4L'm.,. mL')8 + cL'iJ + 4kL'iJ - iL'f(t) = 0,

(6.3)

where

10 = n(4ihL)(4L)2 = mass moment of inertia of the rod

I,

lm(~r

mass moment of inertia of the circular piate

The differential equation (6.3) governing the motion of this system may conveniently be written as (6.4)

where J", Coo, K·, and F· (t) are, respectively, the generalized inertia, generali7.ed damping, generalized stiffness, and generalized force for this system. These quantities are given in eq. (6.3) by the factors corresponding to the acceleration. velocity, displacement, and force tenns, namely, r =/0+ 11 + 4ifle + c~

0

= cL

me·

2

K' = 4kL'

x

r 01 length}

C

Rigid OOf

£

AC===~~~====~======~=====i~=i~ Fig, 6.2 Example of

(t) = i L'!(I)

m

lr:'as~!uni!

slngle·degree.of~fteedom

rigid system.

Example 6.1. For the system shown in Fig, 6.4, detennine the generalized physical properties M', C', K' and generalized loading r (f). Let Y(t) at the point A:: in 6.4 be the generaiized coordinate of the system. Solution: The free body diagram for the system is depicted in Fig. 6.5 which shows all the forces on the two bars of the sytem including the inertial

166

Stfuc~tJre$

Modeled as a Single-Degree-of-Freedom System

Generahzed Coordlnates and Rayleigh's Melhod

167

where the generalized quantities are given by

M' C-

I,

3m

+4-

c 3

K' = 3k, and Fig. 6.4 System for Exnmple 6.1

P" (I) =

force and the )nertial moment. The generaHzed coordinate is Y (1) and the displacemenr of my point in the system should be expressed in terms of this coordinate; neveriheJess. fOf convenience, we select also the auxiliary coordinate Y j (I) as indicated jn Fig. 6.5. The summation of the moments about point A I of all the forces acting on bar Ai - Bh and the summation of moments about Bl of the forces on bar Ai. B 1,) give the roUowing equations:

kl (~Y - Y )3a = 2aFo sin iiJr j

h··Y+--maY+-cY+k. 3 .. a · 1'-Y 2

30

4

3

'3

\J Y,2a+3ak,Y=Q

Subs~itljting

(6.5)

(6.6)

YJ from eq. (6.5) into eq. (6.6), we obtain the differential equation for the rnotion of the system in termS of the generalized coordinate Y(t), namely

6.3

3Fc sin Wt.

GENERALIZED SINGLE-DEGREE.OF·FREEDOM SYSTEM-DISTRiBUTED ELASTICITY

The examp1e presented in the preceding section had only one degree of freedom in spite of the compiexicy of the various parts of the system because the two bars were interconnected through a spring and one of rhe bars was massless so that only one coordinate sufficed to completely specify the motion, If the bars were nor rigid, but could defom) \11 flexure, the system would have an infinite number of degrees of freedom However, a single-degree~of-free­ dom analysis could still be made, provided Ihai only a single shape could be developed during motion, that is, provided that the knowledge of the disp)ace~ ment of a single point in t;)e system determines the displacement of the entire system. As an illustration of this method for approximaling the analysis of a system with an infinite number of degrees of freedom with a single degree of freedom, consider lhe cantilever beam shown in Fig. 6.6. In this iHustration, the physi<;aJ properties of the beam are the Dexural stiffness EI(x) and its mass per unit of iength m (x). It is assumed that the beam is subjected to an arbitrary distributed forcing function p (x, t) and to an axial compH'..ssive force N_

y

Fig. 6,5 Displacements and resultant forces for Example 6.1.

Fig. 6.6 Single-degrcc-of-freedom contif\UOUs system.

168

S!rllctures MOdeled as a Single-Degree-of-Freedom System

Generalized Coordinates

In order to approximate the motion of this system with a single coordinate, it is necessary to assume that the beam deflects durjng its motion in a prescribed shape. Let if>(x) be the function describing this shape and, as a generalized coordinate, yet) the function describing the displacement of the motion corresponding to the free end of the beam. Therefore, the displacement at any point x along the beam is y(x, 1)

~

4> (x)y(t)

(6.7)

where 4>(L) I. The equivalent one·degree·oHreedom system [Fig. 6.6(b): ",ay be def:ned simply as the system for which the kinetic energy, potential energy (straIn energy), and work done by the external forces have at all times the same values in the two systems. The kinetic energy T of the beam in Fig. 6.6 vibrating in the pattern indicated by eq. (6.7) is

T~ r."~m(X){4>(X)Y(t)}'dx

0'

~

f

m (X) 4>' (x) dx

(6.9)

The fiexu:al strain energy V of a prismatic beam may be determined as the work dOiH:; by tbe bending moment fof (x) undergoing an angular displacement dO. This angu]ar displacement is obtained from the well-known formula for the flexural curva!ure of a beam. namely (6.l0)

The factor f is required for the correct evaluation of (he work done by tbe flexural moment increasing from zerO to its final value M(x) [average value M(x)!2J. Now, utilizing eqs. (6.10) and (6.11) in eq. (6.12), we obtain v~

' 1 (d'y\' dx f -EJ(x)(-, 2 ,dx ! i

Finally, equating the potenrial energy, eq. (6.13). for the continuous system to the potential energy of the equivalent system and using eg. (6.7) results in

! K'Y (t)'

10'1EJ (x){ 4>" (x)Y (I))' dx

or K·= fEJ(X){4>" (x)}'dx

E1

(6.1l)

since dy /dx e, where B, being assumed smaH, is :aken as the slope of the elastic curve, Consequently, the strain energy is given by

v~ f1M(X)d8

(6.14)

where

The generali7.ed force F"{t) may be found from the virtual displacement OI(t) of the generalized cnordinate Y(l) upon equating the work performed by

the external forces in the structure to the work done by the generalized force in the equivalent single-degree~of~freedom system. The work of the distributed external force p (x, t) due to this virtual displacement is given by

W= Substituting

oy

LLp (X,t)8ydx

4>(x)8Y from eg. (6.7) gives W

f

p (x, 1)4> (x)OI dx

(6.11)

(6.15)

The work of tlle generaUzed force F·(t) in tbe equivalent system corresponding to the virtual displacement OY of tbe generalized coordinate is

W· M(x)

(6.13)

)0

or d8=--dx

169

(6.8)

Equating this expression for the kjne~ic energy of the continuous system to the kinetic energy of the equivalent single~deg:ee-of~freedom system t.M"Y(r)l and solving the resulting equation for the generalized mass, we obtain M·

and Rayleigh's Method

= ret) IlY

(6.16)

Equating eq. (6.15) with eq. (6.16) and canceling the factor oY, which is taken to be different from zero; we obtain tbe generalized force as r(t) ~

1 rL

p(x,t)4>(x)dx

(6.17)

Similarly, to detennine the generalized damping coefficient) assume a virtual d~spiacement and equate the work of the damping forces in the physical

170

Generalized Coor;j:na\es and Rayleigh's Method

Structures MOdeled as a Slnglc-Oegree-of-Freedom System

sysrem with the work of the dampmg force in the eqUivalent of-freedom system. Hence

. J'

C'Y6Y =

0

sing~e-degree

c (x)y8y dx

where c (x) is the distributed damping coefficient per unit length along the beam. Substituting 6y = ¢(x)oY and ]I: ¢(x)'i' from eq. (6.7) and canceling the COITLt'J:lon factors. we obtain

C': fC(X)[¢(X))'dx

expression waS conveniently set equa! to the initiaL length of the beam L instead of to its borizontal component L I. Now we define a new stiffness coeftlcient to be called the generalized geometric stiffness K~ as the stiffness of the equivalent system requlfe.d to store the same potentia! energy as lhe potentia! energy stored by the normal force N, that is, IK~Y(!)' = No(!)

SUbstituting 8(1) fro:n eg. (6.21) and [he derivative dyldx from eq. (6.7),

(6.18)

we have

which is the expression for the generalized damping coefficient To calculate tne porentlal energy of the axial force N which is unchanged during the vibration of the beam and consequenlly is a conservative force., it is necessary to evaluate the horizontal component of the mOlion 8(f) of the free end of the beam as indicated in FIg. 6.6. For tnis purpose, we consider a differential element of length dL along the beam as Sh?wn in Fig. 6.6(.a} ,\he length of this element may be expressed as

or (6,l9) Now, integrating over the horizontal projection of the lenglh of beam (L f) and expanding in series the binomIal expre.sslon, we obtain

r'l,Y(t)d]' dx

I, 2 I -KeY(,) : -N)

2'

2,[

dx

or (6.22) Equations (6.9), (6,14), (6.17), (6.18), and (6.22) give, respectively, the generalized expression for the mass, stiffness, force, damping,_ and geometric stiffness for 3 beam with distribuled properties and load, modellng it 3S a , simple oscillator. For the case of an axial compressive force, the potentiai energy in the beam decreases with a loss of stiffness in tbe beam. The opposite is true for a tensile axial force, which resuhs in an increase of the flexural stiffness of the beam. Customarily, the geometric stiffness IS determined for a compressive axial force. Consequently, the combined generalized stiffness K; is then given by (6.23)

Finally> the differential equation for the equivalent system may be written as (6,24)

Retaining only the first two terms of the series results in

dy " , Irq- f-) dx ~O 2 \d:x,

L=L +

(6.20)

or

The critical buckling load N n is defined as the axial compressive load that reduces the combined stiffness to zero, that is,

K; = g' 0(1)

=L

Y )'x '_J'lld L--d (} 2 \ dx

(6.21)

The reader should realize that eqs. (6.20) and (6.21) ll\voive approximations since lhe series was truncated and the upper limit of the integral in the final

K~ ~O

The substitutian of K' and Ko from eqs. (6.14) and (6.22) gives

172

General!zed'Coordinales and Rayleigh's Melhod

Structures ,\1odeled as a Single-Degree-or·Freedom System

and solving for the critical buckling load, we obtain E1(d'¢ldx')'dx

(6.25)

173

less accurate than the displacemeflls, because the derivatives of approximate shape functions tends to increase errors in the shape function. A better approach to estimate the equivalent forces is to compute the inertial forces that resulted in the calculated displacements. These inertial forces are

given by

(d¢ldx)' dx

Once the generalized stiffness K~ and generalized mass M'" have been determined, the system can be analyzed by any of the methods presented in the

preceding chapters for single-deg~ee-of-freedom systems. In particular, the square of .he natural frequency, 0'-, is given from eqs. (6.9) and (6.14) by

,

p (x, /)

111 (x)ji(x,

I)

(6.30)

or substituting Y(x, t) from eq. (6.7) by p (x, I)

m (x)
(6.31)

E1 (x)¢" (x)' dx

At any time t the equivalent load p(X,I) given by eq. (6.31) is applied to the m(x)¢'(x)dx

The displacement y(t) of the generalized single-degree-of-freedom system is found as the solution of the differential equation M'Y(I)

r

+ C'Y(t) + K'Y(t)

(!)

and the displacement y (x. r) at location x and time t is then calculated by eq.

(6.7).

6.4

structure as a sta~ic load and the internal forces (shear forces and bending moments), as wen as the stresses resulting fron these internal forces, are determined. To provide an example of the determination of the equivalent one degree of freedom for a system with dIstributed mass and stiffness, consider the water tower in Fig. 6,7 to have uniformly distributed mass in and stiffness EI along its length with a concentrated mass lVi mL at the top. The tower is subjected to an earthquake ground motion excitation of acceleration ag(t) and to an axiai compressive load due to the weight of its distributed mass and concentrated mass at the top. Neglect damping in the system. Assume that during the motion the shape of the tower is given by

SHEAR FORCES AND BENDING MOMENTS

TIle internal forces~shear forces and bending moments-associated with th~

(6.32)

¢(x)

displacements y(x, 1) may be obtained by loading the structure with the eqlliv~ alent forces that would produce the dynamic displacements calculated. From elementa~ beam theory, the rate of loading p (x, t) per unjt of length along a beam is related to the lateral displacement y = Y (x, t) by the differential equation p (x, I)

a [E1(x)a;r a'Y]

(6.28)

or subst;:ut:ng y(x, 1) =
i

hi,FI

p (;:, I)

a' [E1 (x}'T. a'<1>: Y(I) f,X

~

i I

(6.29)

The equivalent force p (x, I), calculated using eq. (6.29), depending On derivatives of the shape function, wi[] in general. give internal forces that are

Fig. 6.7 Water towe!;' with distributed properties for Example 6.2.

174

Structures Modeled as a Single~Degree·of·Freedom System

Generalized Coordinates and Rayfcig,l'l'S Method

Seiccting the lateral displacement Y (/) at the top of the tower as the generalized coordinate as shown in Fig, 6,7, we obtain for the displacement at

By setting

K; =

O. we obtain

any point

trig .. ' . - (3".- - 4\

16

(5.33)

1(4

(!hg)~r

and

(17 \'

<

f

Y (I) - y, (I) is

where M' is given by eq. (6.34), K; by eq. (6.38), and the effecrive force by eq. (6.17) for the effective distributed force and by tMAg (I) for the effective concentrated force at the ~op of the lOwer. Hence

11X

£1 1- : cos'-dx o ,)L) 2L

F;" (I) =

".4£1 K ;321!

(6.35) where Pdr(X,t) = -

The axial force is due to the weight of tbe lower above a particular section, including Lhe concentrated weight at the top, and may be expressed as (6.36)

where 8 is the gravitaliona: acceleration. Since the normal force in this case is a function of x, it is necessary in using eg. (6,22} to include N(::::.) under the integral sign, The geometric stiffness coefficient K~ is then given by

{ x'd 17\' sin' 1IXdx fLmLgI2-~!\.2L L 2LJ \

which upon integration yields

r

L p,,,(x, 1)
F;,,;

r

-iiw,(r)1>(x)dx-liILa,(r)

Substitution of ;:P (x.) from eq. (6.32) :nto the last equalion yields upon in~ {egration F;,,; - .....~~('"'I)~L~ ('TT- 1)

Example 6.2.' As a numerical example of c~lculating the response of 2. system wIlh distributed properlies, consider the water tower shown in ~ig, 6.7 excited by a sinusoidal ground acceleration i.l g (t) = 20 sin 6,361 (in Isec~), Model the slIucwre by assuming the shape given by eq, (632) and determine the response, Solution:

The numerical values for this example are: 1

It. = 0.1 KJp' sec 1in per unit of length

Consequently, the combined stiffness from eqs. (6.35) and (6.37) is 4 17 £1 ,ilg t ····---(317

16

EJ L~

4)

(6.38)

(6.41)

71

(637)

32L'

(6.39)

2(317' -
(6.40)

•

dO

E[

(6.34)

271

K~=

Q

The equation of motion in terms of the relative motion u given by eq. (350) for the undamped syslem as

• fi:L_ M ;-()1T-8)

K"

'

which gives the critical load

The generalized mass and the generalized stiffness of the tower arc compuled, respectively, from eqs. (6.9) and (6.14) as

M'

175

L2 1013 Kip· in 1 100 ft; 1200 in

w= 636 rad/sec

176

Structu,es Modeled a8 11 Single-Oegree-of-Freedom System

Generalized 'Coordinates and Rayleigh's Method

177

From eq, (6.34) the generaHzed mass is 0.1 X 1200 21T (51T- 8)

Kip'sec:!

F4'::')..

m

Fl~ 1--..

m

F,(O

ffl.

F,(')

m.

.".

!47.21~'-;.-111

1-

and from eq. {6.38) the generaJized combined stiffness is w4 L2 1013

32 X

0.1 X 386

(i200)"

-'---:-:~··-(31T' - 4)

1-

= 2[,077 Kiplin

The natural frequency is

,

w=-/K;/lvr

,,"

~

('j

11.96rad/sec

and the frequency ratio

r

w

6.36

'"

11.96

-=--

=0.532

Fig. 6.8 (a) Multistory building sllbjected to lateral forces F,Ct). (b) Column mode:nng the building s'bowing assumed lateral displacement function y{x,t)"" tP(x)Y(t), {c) Equivalent stngle~degree·of~freedom system, (d) Free body diagram.

From eq. (6.41) the effective force IS

6.5

GENERALIZED EQUATION OF MOTION FOR A MULTISTORY BUILDING

Consider a multistory building such as the model for the four-story building shown in Fig. 6.8(a) subjected to lateral dynamic forces Fi(t) at the different levels of the building. The mass of the building 15 assumed to be concentrated at the various levels (floors and root) which are assumed to be rig;d in their own planes~ thus only horizontal displacements are possible in such a building" To model this structure as a single degree of freedom [Fig. 6.8(c»), the lateral dIsplacement shape y (x, t) is defined jn terms of a single generalized coordinate Y(I) as

or

F;" =

(
(0)

-3272 sin 6.361 (Kip)

Hence the steady-sta:e response (neglecting damping) in terms of relative motion is given from eq. (3.9) as

y(x,l) = w(x)Y(t)

327212\ ,077 sin 6361 I - (0.032)

The generalized coordinate ye'l in eq. (6.42) is selected as the lateral displacement at the top level of the building which requires that shape function be assigned a unit value at that level; that is. ¢(H) = 1.0 where H is the height of the building.

" - " - .-':;2

= - 0,217 sin 6.36t in

(6.42)

(Ans.)

I I

178

Structures Modeled as a Single-Oegree-ol-Freedom System

Generalized Coordinates and Rayleigh's Method

The equation of motion for the generalized single-degree-of-freedom system in Fig. 6.8(c) is obtained by equating to zero the sum of forces in the corresponding free-body diagram [Fig. 6.8(d)], that is

179

in which w is the natural frequency calculated for the generalized system as

(6.47) (6.43) where MO, C', KO, and FO (t) are, respectively, the generalized mass, generalized damping, generalized stiffness, and generalized force, and are given for a discrete system, modeled in Fig. 6.8(b), by

The substitution of the generized damping coefficient, (, from eq. (6.46) and of K = w 2Mo from eq. (6.47) into the differential equation of motion, eq. (6.43), results in O

N

M· ~

I

m;q,;

(6.44a)

c,Llq,;

(644b)

(648)

;"'1 N

C ~

I

i= I

conveniently measured relative to the motion of the base Yo (I), that is,

N

K· ~

I

k;Llq,;

(6.44c)

;""1

r

U;

(t) ~

I

F;Ct)
(644d)

where the upper index N in the summations is equal to the number of stories or levels in the building,

The various expressions for" the equivalent parameters in eq. (6.44) are obtained by equating the kinetic energy, potential energy, and the virtual work done by the damping forces and by external forces in the actual structure with the corresponding expressions for the generalized single-degree-of-freedom system. In eq. (6.44), the relative displacement d¢, between two consecutive levels of the building is given by . (6.45) with ¢o = 0 at the ground level. As shown in Fig. 6.8(b), Tn, and F;U) are, respectively, the mass and the external force at level i of the building, while k i and Ci are the stiffness and damping coefficients corresponding to the ith story. It is convenient to express the generalized damping coefficient C· of eq. (6.44b) in terrns of generalized damping ratio thils by eg. (2.7)

r;

i= I

~

y; (I) - Yo (I)

(6.49)

Also, in this case, the effective forces are given from eq. (4.39) as F~ff,i

F~Jl,i

at the various levels of the building

= - m,yo (I)

(6.50)

The differential equation for the generalized single-degree-of-freedom system excited at its base is then written as (6.51) where the generalized coordinate for the relative displacement, V(r) is U(I)

~

YCt) - )'0(1)

(6.52)

The generalized effective force F;JJ(t) is calculated by eq. (6.44d) and (6.50) as N

F;ff~ -yo(1)

I

j=

m;
(6.53)

I

The generalized equation of motion may tllen be expressed as

N

I

(I)

N

i= I

c" ~

When the excitation of the building is due to an acceleration function, a(l) = Yo(t), acting at the base of the building, the displacements Ui(t) are

c;Ll¢; ~ f·e;, ~ 2f·M· w

(6.46)

(6.54)

Generalized 'Coordinates and Rayleigh's Method

Structures Modeled as a Slngle-Deg;ee-of-Freedom Syslem

180

to whieh the coefficient

r'

• • • • • • • •I '"... I •I • • • • • • • •I • • • • • • • • •!

I

is the generalized participation factor given by

r'=

181

'"

(6.55)

iiOJ

6.6

SHAPE FUNCTION

I

The use of generalized coordinates transforms a multidegree-of-freedom sysu tem into an equivalent single-degree-of-freedom system. The shape function describing the deformed structure could be any arbitrary function that satisfies the boundary conditions, However, in practical applications, the success of this approach win depend on how close the assumed shape function approximates the actual displacements of the dynamic system. For structural buitdings, selection of the shape function is most appropriate by considering the aspect ratio of the structure, which is defined as the ratio of the building height to the dimension of the base. The recommended shape functions for high-rise. mid-rise, and tow~rise buildings are summarized in Fig. 6,9. Most sejsmic building codes use the straight-line shape which is shown for the mid-rise building. The displacements in the structure are calculated using eq. (6.7) after the dynamic response is obtained in terms of the generalized coordinate.

8@12ft-96ft

J'

(a)

(

I __ J...

iii

~- ---+

1

AW

Example

,~

-----------

I

6.3~

A four-story reinforced concrete framed building has the dimensions shown in Fig. 6.1 O. The sizes of the exterior columns (nine each

A'

24ft

Af

!!

'" N

~

24ft

1~ J

on lines- A and C) are 12 in X 20 in. and the interior columns (nine on line B) are 12 in X 24 in for the bottom two stories, and, respectively, 12 in x 16 in and 12 in X 20 in for the highest two stories, The height between floors is

Fig. 6,10 Plan and elevation for a four-story buHding for Example 6.3; (a) Plan. (b)

12 ft. The dead load per unit area of the floor (floor slab, beam, half the weight

Elevation.

(0)

of columns above and below the floor, partition walls, etc.) is estimated to be 140 pst The design llve load is taken as 25% of an assumed live load of 125 psf. Determine the generalized ma~s, generalized stiffness, generalized =0.1), and the fundamental period for lateral damping (for damping ratio vibration perpendicular to the long axis of the building. Assume the foHowing shape funetio"s: Ca) ¢(x) =xiH and (b) ",(x) sin ('11Xi2H) where H is the height of the building.

1

Ix

I I 82t8 hi orx) ~-D-----j

(a) .l,.oW~RISE

HID.;, 1.5 ¢(x) = sin;rx 12H

r

~!~'"

p--fl.

(b) MID-R1SE

(c) HiGH-RISE

I.5
HJD>3

q,(x)=xfH

¢{:rJ >., 1 -cosnx

12U

Fig. 6.9 Possible shape functions based on aspect ratio (Naeim !989. p. 100).

I 1

Solution: 1. Effective weight at various floors: No live load needs to be considered on the roof. Hence. the effective weigbt at all floors, except at the roof, will be 140 + 0.25 X 125 = 171.25 pst and the

182

Gcneral;zed Coordinates and Rayleigh's Method

Structures Mode!ed as a Single·Deg.ce-al-Freedorn System

effective weight for the roof will be 140 pst. The pbn area is 48 ft x 96 ft 4608 ft~. Hence, the weights of various leve~s are:

w, ~ 1'1, = I'l, = 4608 x 0,17125 = 789,1 W,

183

and for the cobmns 12 in X 20 in, I

1 12x20'

Kips

, 8000 in- ,

12 X 3 X 10' x 8000 k = -144'

= 965 '

Kiplin

Hence, tOlal stiffness for the third or fourLh stories is

4608 x 0, 140 = 645.1 Kips

K,

~

K, = 18 X 49A

+ 9 X 96,5 = 1757,7 Kip!in

The total seismic design weight of the building is then w~

789.1 X3

3, Generalized mass and stiffness;

+ 645.1 = 3012.4 Kips

2. StOI)' lateral sliffness: It wjll be assumed t:'1at horizontal beam-and-fioor diaphragms are rigid compared to the cobmns of the building in order to simplify the hand calculation. In this case, the stiffness belween two consecutive Ievels is given by k=

(a) Assuming "'(x) ~ x IH, Table 6. J shows the necessary calcuiations to obtain using eqs. (6.44a) and (6A4c) the generalized mass M· and the generaliz.ed stiffness K' for this example assuming $ (x) = x/H. The natural frequency is then calculated from eq, (6.47) as

12El

where and the generalized critical damping C;, and [he absolute generalized damping C', respectivaly, by eqs, (2,6) and (2J9) as

L = 12 ft (distance between two floors)

E

3 X 101 ksi (modulus of elasticily of concrete)

C;, = 2[j('M'

= 2[(625250)(3,559) = 9045 (lb' sec lin)

1

I:;: U 12 X 2D 3 = 8000 in4 (moment of inertia for the concrete section for

and

columns 12 in x 20 in) Therefore, for these co:umns,

k=

c=(C'u

12 X3 X 10' X 8000 144'

96.450 Kip!in

TABLE 6.1

Level

Similariy. for columns 12 in X24 in, J = 13824 in'

and

(0,1) (94345) = 9A3 (Ib ' sec lin)

Calculation of MOe and K"' Assuming q,{x} """ xlH k,

Ini

"':

1.671

1.000

(Kip lin) (Kip . sec z fln)

4

The total stiffness for lhe first and second stories is [hen

3

= 18 X 96.45 + 9 x

2

2.044

0,750

2,044

0500

2,044

0,250

3236 I

-'- ,2 X 16' = 4096 in', 12

k=

~-

....~~:-=--

49A Kiplin

109,875 1150

0.250

3236 Similarly, for the columns 12 in X 16 in. of the lOp and third stories.

1.671

0,250

1758 166,67 = 3236 Kiplin

12x3X lO'X4096

k,Ll",; nli$; (Kip' sec' lin) (Kip!in)

---"

k ~ 166,667 Kipfln

1758

K, = K,

L"p,

lOn75 0.610

0250

202250 0,128

0,250

202,250 A{" = 3,559 K*

625250

184

Slrl!ctures Modeled as a

Generalized Coordinates and Ray!eigh's Method

System

Calculation 01 M' and K" Assuming "",x) _ sin (,-,x 1211)

TABLE 6.2

Level

SjngJe-Oegree~of·Freedom

,p,

m, k, (Kip lin) (Kip· sec 2 lin)

1.671

4

k;tJ.l/Ji

1n;¢7

£1,p;

(Kip· sec' fin) (Kip lin)

1.671

1.000

1758

0.076

3

6.7

2.040

0.924

2.040

0.707

2.040

0.383

1758

10.154 1.745

0.217

3236

82.782 1.022

0.324

3236

339.702 0.300

0.383

476.686

M*

= 4.738 K* =

909.324

(b) Assuming (x) = sin (=I2H).

Table 6.2 shows the necessary calculations to obtain using eqs. (6.44a) and (6.44c) the generalized mass M" and the generalized stiffness K- for this example assuming ,ptA) = sin (ml2H). The natural frequency is then calculated from eq, (6.47) as

'i.'

IM'

I

27r ="!909.324 4.738 =13.85 (rod/sec) then Tb =-;;;- =OA5 sec

and the generaljzed critical damping C~.;t and the absolute generalized damping C, respectlvaly, by eqs. (6.48) and (6.49) as

c,,=2[j?M' =2j(909.324)(4.738)

RAYLEIGH'S METHOD

In the preceding sections of this chapter the differential equation for a vibrating system was obtained by application of the principle of virtual work as an alternative method of considering the dynamic equilibrium of the system. However, the differential equation of motion for an undamped system in free vibration may also be obtained with the application of the Principle of Corr~ servation of Energy. This principle may be stated as follows: 1f no external forces are acting on the system and there is no dissipation of energy due to damping, then the total energy of the system must remain constant during motion and consequently its derivative with respect to time must be equal to zero. To illustrate the application of the Principle of Conservation of Energy in obtaining the differentia] equation of motion, consider the spring-mass system shown in Fig. 6,11" The total energy in this case consists of the sum of the kinetic enorgy of the mass and the potential energy of the spring_ In this case the kinetic energy T is given by (6.56) where y is the instantaneous velocity of the mass. The force in the spring. when displaced y units from the equilibrium position, is ky and the work done by this force on the mass for an additional displacement dy IS - ky dy. This work is negative because the force ky acting on the mass is opposite to the incremental displacement dy given In the positive direction of coordinate y. However, by definition. the potentia] energy is the value of this work but with opposite sign. It follows then that the total potential energy V in the spring for a final displacement y will be

v=

= 13L276(1b·seclin)

and

J:

kydy

jky'

(6.57)

Adding eq. (6.56) and (6.57), and setting this sum equal to a constant, will give C =

rC'" = (0.1)(131.276) = 13.13 (lb· sec lin) Co

For this example, either of the, two assumed shape functions results in essentially the same value for the fundamental period, However, ¢ (x) = x III is a slightly better approximation I to the true deflected shape than is (x) sin ('l7X 12ff) because To> T,.

I

185

For an assumed displacement shape closer to the actual dlsplaeemenl shape,

~he

structure

will vibrate elmer to the free condition with less imposed constraints, thus with the stiffness reduced and

,l

longer period.

Fig. 6.11 Spring-mass system in free vjbration.

(6.58)

186

Slructures Modeled as a

Sir;g!e~Degree~oi~Freedo:n

System

Generalized Coordinates and Rayleigh's Method

187

Differentiation with respect to tiu':e yields

myy + lyy =0 Since

y cannot

be zero for all values of 1. it follows

~hat

(659)

Fig. 6,12

This equation is identical with eg. (l.l I) of Chapter I obtained by applicafion of Newton's Law of Motion. Used in this manner, the energy method has no particuiar advantage over the equilibrium method. However, in many practical problems it is only the natural frequency that is desired. Consider again the

(6.60)

Example 6.4. In the previous calculations on the spring-mass system, the mass of the spring was assumed to be so smaH that its effect on the natural frequency could be neglected. A better approximation to the true value of the naturaJ frequency may be obtained using Rayleigh's Method. The distributed mass of the spring could easily be considered in the calculation by simply assuming that the deflection of the spring along its length is iinear. In this case, consider in Fig. 6.12 the spring-mass system for which the spring has a length L and a total mass In r • Use Rayleigh's Method to detemline the fraction of the spring mass that should be added to the vibrating mass.

and velocity

y= wC cos

(WI + a)

system with heavy spring.

which is the natural frequency for the simple oscillator obtained previously from the differential equation of motion. This method, in which the naturai frequency is obtained by equating maximum kinetic energy with maximum potential energy, is known as Rayleigh's ,"-1etltod.

simple oscillator of Pig, 6.1l, and assurr.e that the motion is hannonk. This assumption leads to the equation of motion of the form

)' = C 51n (W! + a)

Spring~mass

(6.61)

where C is the maximum displacement and we the maximum velocity. Then, at the neutral position (y = 0). there will be no force in the spring and the potential energy is zerO. Consequently, the entire energy is then kinetic energy and (6.62)

Solution: The dispiacemen~ of an arbitrary section of the spring at a distance s from the support will now be assumed to be u = sylL. Assuming iliat the motion of the rr.ass 111 is harmonic and given by eq. (6.60), we obtain

At the maximum displacement the velocity of the mass is zero and ail the

energy is then potential energy, thus s

D

The energy in the system changes gradually over one-quarter of the cycle from purely kinetic energy, as given by eq. (6.62), 10 purely potential energy. as given by eq. (6.63). If no energy has been added or lost during the quarter cycle, the two expressions for this energy must be equaL Thus

(6.67)

.1 . ~

(6.65)

(6.66)

The potential energy of the uniformly slretched spring is given by eq. (6.57) and its maximum value is

(6.64) Canceling common factors and solving eq, (6.64) wi!] give

.

-C Sin (wt+ al

(6.63)

~ l

,i .i

;, i, '1, "

,j;'i

:,i :.."

:l

;~I

A differential element of the spring of length ds has mass equal to nlflslL and maximum velocity um~. = &tim~, wsCIL. Consequently the lotal kinetic energy in the system at its maximum value is (6.68)

188

~89

Generalized Coordinales and Rayleigh's Method

S1ructures Modeled as a Sing!e-Degree-of-F,eedom System

After integrating eq. (6.68) and equating it with eg. (6.67}, we obtain

1

2

1

2 2( m+-I m,,\

-kC ~-UJC 2 2

3 !

(6.69)

y

Solving for the natural frequency yields

(i)

=

E=:=~L·······~~r (OJ

(a}

k

,; ----''---~

'\ m+m sl3

Fig. 6.13 (a) Cantilever beam of uniform maSS with a mass concentrated at its tip. (b) (6.70)

Assumed deflection curve.

end as shown in Fig. 6.13(b). For this static load the deflection at a distance x from the support is

or in cycles per second (cps), (6.71)

(6.72)

The application of Rayleigh's Method shows that a better value for the natural frequency may be obtained by adding one-third of the mass of the spring to that of the main vibrating mass.

where y = deflectio71 at the free end of the beam. Upon substitution into eq, (6,72) of y ~ C sin (WI+ a). which is the harmonic deflection of the free end, we obtain

Rayleigh's Method may also be used to determine the natural frequency of a continuous system provided that the deformed shape of the structure is described as a generalized coordinate. The deformed shape of continuous structures and also of discrete structures of multiple degrees of freedom could in general be assumed arbitrarily. However, in practical applications. the success of the method depends on how close the assumed deformed shape will come to match the actual shape of the stmcture during vibration. Once the deformed shape has been specified, the maximum kinetic energy and the maximum potential energy may be determined by application of pertinent equations such as eqs, (6.8) and (6.13). However, if the deformed shape has been defined as the shape resulting from statically applied forces, it would be simpler to calculate the work done by the extema] forces, instead of directly determining the potential energy. Consequently, io this case, the maximum kinetic energy is equated to the work of the forces applied staticaHy. The fonowing examples illustrate the application of Rayleigh's Method to systems

u

3x'L - x' 2L' C sin (WI + a)

(6.73)

Tbe potential energy is equated to the work done by the force F as it gradually increases from zero to the final value F. This work is equal to i Fy, and its maximum value which is equal to the maximum potential energy is then

•

.

3El '"

Ym~'" = iFC= 2L3 L.

(6,74)

since the force F is related to the maximum deflection by the formula from elementary strength of materials,

FL'

(6,75)

Yma~=C= 3EI

with distributed properties.

The kinetic energy due to the distributed mass of the beam is given by

Example 6.5. Determine the natural frequency of VIbration of a cantilever beam with a concentrated mass at its end when the distributed mass of the beam is taken into account. The beam has a total mass mb and length L. The flexural rigidity of the beam is EI and the concentrated mass at its end is m, as shown in Fig. 6.13. Solution: It will be assumed that the shape of deflection curve of the beam is that of the beam acted upon by a concentrated force F applied at the free

T=

lL .!..L~~~)U2 o 2

I.,

dx

(676)

L !

and using eq. (6,73) the maximum value for total kinetic energy will then be (6,77)

190

Generalized Coordina!es and Rayleigh's Method

SlmellJres Modeled as a Single·Degree-of~F(eedom System

1

(6.78)

have durjng vJbra ..ion, A choice of a shape that gives consistently good results is the curve produced by forces proportional to the magnitude of lhe masses acting on lhe structure. For the simple beam. these forces could be assumed to be the weights W! = mig, W2 = tn18,,, .• W N = mNg due to gravitational action on the concentrated masses. The static deflections under these weights may then be designated by y" y,• .. _, YN- The potential energy j5 tben eq~al to work done during toe loading of the beam, :hU5,

1---··-··

(6.79)

(6.81)

After integrating eg. (6.77) and equating it with eq. {6.74), we obtain

3£1 C' 2L'

I "C 'm+ I 33 -w --·11117 2

'\

140,

and the natural frequency becomes &.I

f~-~-

277

191

I

3£1

211 1 ,I \11 L

\

33.\

,m+-

'\

40

II 'b'

i

It is seen, then, thai by concentrating a mass equal to (33/140)

!1lb at the end of the beam, a more accurate value for Lhe natural frequency of the cantilever beam is obtained compared to the resull ob~'iined by simply neglecting its dist.ributed mass. In practice the fraction 331140 is rounded to 114, thus approximating the natural frequency of a cantilever beam by

For harmonic motion in free vibration, the maximum velocities under the weights would be WYh W),?, ... , WYN, and therefore the maximum kinellc energy would be

(6.82)

(6.80)

J=

When the maximum potentia! energy, eq. (6.81), is equated with the maximum kinetic energy, eq. (6,82), the natural frequency is found to be where The approximation given by either eq, (6.79) or eq, (6,80) is a good one even for the case in which m = O. For this case the error given by these formulas is about 1.5% compared 10 the exact solution which will be presented in Chapter 21.

(6.83) or

Example 6.6. Consider in Fig. 6.14 the case of a simple beam carrying several concentrated masses. Neglect the mass of the beam aod determine an expression for the natural frequency by application of Rayleigh's Method.

Solution: In the application of Rayleigh's Method, it lS necessary to choose a suitable cu~e to represent the defonned shape that tbe beam will

)'i

is the deflection at coordinate i and W, the weight al lhis coordinate. This method is directly applicable 10 any beam, but in applying the method,

it must be remembered that these are not gruvity forces at all but substituted

v

y,

v,

y,

Fig. 6.14 Simple: beam tarrying concentrated masses.

forces for the inertial forces. For example, in the case of a simple be.am with overhang (Fig. 6.15) the force at the free end should be proportionai to m)(F1 m3g) but directed upward in order to obtain the proper shape for the defomled heam. In toe application of Rayleigh's Method, the forces producing the deflected shape do not necessarUy have to be produced by gravitational forces. The only requirement is that lhese forces produce the expected deflection shape for the

192

Structures Modelad as a Single-Oegree-of-Freedom System

Generalized Coordinates and Rayleigh's Method

193

~K? '" 44300 I

i

m1

-""

135!b seci /in.

_Y,

K \ "" 30700 Iblin

Fig. 6.15 Overhanging massless beam carrying concentrated masses, Fig. 6.16 TWQ-stQfy ftame for Example 6.6,

fundarr:ental mode, For example, if the deflected shape for the beam shown in Fig. 6.14 is produced by forces designated by f,. j" .... f" instead of the gravitation forces WI! W~, . " W,\', we wiU oblain, as in eq. (6,81), the maxjmum potential energy (6.84) which equated to the maximum kinetic energy. eq, (6.82), will result in the following formula for the fundamental frequency:

(6.85)

Then an improved value for the natural :-requency may be obtained by loading the strocture with the inertial loads associated with the assumed deflection. This load results in a new deformed shape which is used in calculating the maximum potential energy. 'nle method is better explained with the aid of numerical examples. Example 6.7. By Rayleigh's Method, determine the natural frequency (iower or fundamental frequency) of the two~story fr.!rne shown in Fig. 6.16. Assume that the horiz.ontal members are very rigid compared to the columns of the frame. This assumption reduces the system to only two degrees of freedom, indicated by· coordinates YI and Y2 in the figure. The mass of the structure, which is lumped at the floor levels, has values mj = 136 lb· sec 1 lin and!tl2 = 66 Jb· sec1-lin. The total stiffness of the first Story is k j = 30,700 Iblin and of the second story k2 = 44,300 ib lin, as indicated in Fig. 6,16,

Consequently, the fundamental period could be calculated as

Solution: This structure may be modeled by the two mass systems shown in Fig. 6.17. In applying Rayleigh's Method, let us assume a deformed shape for which Y1 1 and Y1- = 2. The maximum potentia! energy is then (6.86)

~ !(30,700)(1)' + 1(44,300)(1)'

6.8

~

IMPROVED RAYLEIGH'S METHOD

The concept of applying inertial forces as static loads in determining the deformed shape for Rayleigh's Method may be used in developing an improved scheme for the method. In the application of the improved Rayleigh's Method, one would start from an assumed deformation curve followed by the calculation of the maximum values for the kinetic energy and for the potential energy of the system. An approximate va!ue for natural frequency is calculated by equating maximum kinetic energy with the maximum potential energy.

37,500 lb· in

(a)

and the maximum kinetic energy

~ I (136)",'

=200w'

+ 1(66) (2w)' (b)

194

Generalized Coordinates and Rayleigh's Method

Structures Modeled as a Slng\e~Degree"or-Freedom Syslem

195

or in the ratio

y, = 1.00

y,

= 1.34

(c)

Introducing these improved values for the displacements YI and h into eqs. (a) and (b) to recalculate the maximum potential energy and maximum klnetic energy results in (d) (e)

and upon equating V ffi.4>. and T mm we obtain (v

Fig. 6.17 Mathematical mode! for structure of Example 6,'),

= 12,57 rad/sec

Of

Equating maximum pOlential energy with max.imum kinetic energy and solving for the naturai frequency gives w~

13.69 fadlsec

Of

w 2".

~.• ~

2.1& cps

The natural frequency calculated as f;;;:; 2,18 cps is only an approximation to the exact value, since the deformed shape was assumed for the purpose of apply:ng Rayleigh's Method. To improve this calculated value the ~atu~al frequency. :et us load the mathematical model in Fig. 6. 17(a) with the mertlal load calculated as

.of

= (136)( 13.69), (l) = 25,489 F, = m"iy, = (66) (13.69)' (2) = 24.739 F, ~

Ill,

w'y,

The equilibrium equations obtained by equatjng to zero tbe sum of the forces in the free body diagram shown in Fig. 6,l7(b) gives 30,700y,

44,300 (y, - y,)

= 25,489

44,300(y, - y,)

24,739

and solving

f= 2.00 cps This last calculated value for the natural frequency f= 2.00 cps could be further improved by applying a new inertial load in the system based on the last value of the natural frequency and repealing a new cycle of calculations. Table 6,3 shows results obi.ained for four cycles. The exact natural frequency and deformed shape, which are calculated for this system in Chapter 10, Example 10.1, as a twowdegrees~of-freedorn system, checks with the values obtained in tl1e last cycle of the calculations shown in Tab!e 6.3.

6.9

Horizontal forces in buildings, such as those produced by eanhquake molion or wind action, are often resisted by structural elements called shear walls. TABLE 6,3

y,

= 1.64 =

2.19

Improved Rayleigh's Method Applied to Example 6.6

Deformed Cycle

Shape

Inertial Load

Fl

'-~-'~~~-~-~

2 YI

SHEAR WALLS

1:2.00 1 :1.34

J

I: 1.32

4

1.1.27

25,489 21,489 19,091

Natura]

F2

Frequency

"-~-'--

24,739 ;8,725 12,230

2.18 cps 2.00 1.88 1.88

i

. j

:1 !

I

I

196

Structures MOdeied as a

Slng\e~Degree-of·Freedom

Sysler-,

Generalized Coordinates and Raylefgh's MeL'lod

197

x

---+. x Floor mass

L y

r

----L

Fig. 6.19 Assumed deflection curve for Example 6.7,

The de.formed shape equation is assumed as the deflection curve produced on a cantdever beam supporting three concentrated weights lV, as shown in Fig. 6.19. The static deflections Yit Y:h and Y3 cafculated hy using basic knowledge of strength of materials are

----/

i-··o----j Fig. 6.18 Mathematical model for shear wall and rigid floors

These structural elements are generally designed as reinforced concrete walls fixed at the foundation. A single cantilever shea! wall can be expected to behave as an ord;nary flexural member if its length-to-depth ratio (LlD) is greater than about 2. For short shear walls (LID < 2), the shear strength assumes preeminence and both flexural and shear deformations should be considered in the analysis. When the floor syscem of a multistory buildIng is rigid, the structure's weights or masses at each floor may be treated as concentrated loads, as shown in Fig. 6.18 for a three-story building. The response of the structure is then a function of these masses and of the stiffness of the shear walL In practice a mathematical model is developed in which the mass as weI! as the stiffness of the structure are combined at each fioor ieveL The fU.I1damental frequency (lowest natural freqoency) for such a structure can then be obtained using Rayleigh's Method. as shown in the following illustrative example.

y,

IS WL' WL' -::ccc -:::- ~ 0.0926 El

49 V(L J

WL 3

El

0.3025 EI

92 WL'

WL'

y, ~ 162 y, ~ W

El ~ 0.5679

EI

(b)

The natural frequency hy eq. (a) js then cak'uiated as

_ {386 (0.0926 + 0.3025 + 05679)----;;.r· w - nO.092-6-'-+··O;;-.;;-30;;;2"'5""-:+"'O"'.s::-6C:7;;;9-l:i):..·· wL'

I

El

29.66~ WL' radlsec

or Example 6.8. Determine> using Rayleigh's Method, the natural period of the three-story building shown in Fig. 6.18. All the floors have equal weight W. Assume the mass of the wall negligible compared to the floor masses and consider only flexural deformations (1... fD > 2). Solution: The naturai frequency can be calculated using eq. (6,83), which is repeated here fOf convenience:

,---,,-,

i= 1

Y ~WDI

lEI

Example 6.9. For the mathematical model of a three-storv buHdincr shown in Fig. 6.18 determine the total deflections at the floor levels "considering both fleXUf2.1 and shear deformations. j

Solution: The lateral deflection Lly,,> conSidering only shear deformation for a beam segment of length L1x, is given by

iIg"Wv. L.. UI

w= 1-,,-··-

w

f= 21T =4.72 y We cps

(a)

VLIx .dy,~ aAG

(a)

i ~

,

'",."". ,

198

Generalized Coordinates arid

Structures Modeled as a Single·Degree-of-Freedom System

where

TABLE 6.4

v=

shear force

A

cross~sectional

Yl" (in)

y, " (in)

w·" (radisec)

0,09259 0,)3600

0.30247 0.37483

0~56790

0.65465

29,66 27.67

O~OO

G = shear modulus of elasticity

O~50

. At the first story, V

3W. Therefore, by eq. (a) the shear deflection at lhe

first story is 3W(Ll3) WL y" = ~~ aAG -- - aAG

(b)

1.00 1.50 2.00 2.50 3~00

•

O~26620

O~59193

O~91489

23.30

0.48322

O~95376

19~O5

O~78704

1.46032 2.11161

\.34862 1.95585 2~736S8

13~21

3.69079

11.33

Ll77GS 1.65509

2~90764

~-----~

At the second floor the shear deflection is equal to the first floor deflection plus the relative deflection between floors, that is Y~I

YJl

2W(U3)

+

aAG

6WL 3aAG

£i '

49 WL'

Y,

~

5WL

92 wL' 6WL y, = -[:6=2=~~~~ +-3aA~G-

(e)

We can see better the relative importance of the shear contribution to the total deflection by factoring the first terms in eqs. (e). Considering a rectangular waH for which A = D X 1, E 10 2~S, 1= ID'1i2, a = 1.2 (I = thickness of the wall), we obtain IS WL ' [ 1+1.8751-~1 ID\'] y,=-~-

y,

49

'D)"

WL) [

~--I+O~9S7l· 162 El L i 92 WL' r1 +O~611\{D)'1,

y,~~-

162 El ,

L

Example 6.10. For lhe structure modeled as shown in Fig. 6.18, study the relative imporlance of shear deformation in calculatin.g the natural frequency. SolWton: In {his study we will consider, for the wall, a range of values from 0 to 3.0 for the ratio D IL (deplh-to~length ratio). The deflections Yh Yl, y} al the floor levels are given by eqs. (f) of Example 6.9 and the natural frequency by eq. (6.83). The necessary calculations are conveniently shown in. Table 6.4. It may be seen from the last column of Table 6.4, that for this example the natural frequency neglecting shear deformation (D /L = 0) . is

= 162 Ei~ + 3~G

;L j

fE/iwI.',

(d)

WL aAG

162 EI

~~~--~-~~-~~~--~~--~~--

The next iHUSlrative example presents a table showing the relative import~ ance that shear deformation has in calculating the natural frequency for a series of values of the ratio D i L

The total deflection is then obtained by adding the flexural deflection determined in Example 6.8 to the above shear deflections. Hence,

IS WL)

L50 2.10 180

(c)

since the shear force of the second story is V= 2W, and at the third floor

Y ,162

15.71

4.72 4.40 3.71 3.03

'" DIL-O is equivalent. to neglecL shear defoml<1~ions. •• Factor of WefEf.

••• Factor of

5WL 3aAO

roo (cps) ~~~~~-

~~---~~~----~~

a= shope constant (u· = 1.2 for rectflnguJar sections)

199

Calculation of the Natural Frequency for the Shear Wall Modeled as Shown in Fig. 6.18

y, •• (in)

DIL

area

Rayleigh's Method

j

(f)

4~72[EIjWL' cps, For short walls (DII. mation becomes increasingly important

6.7

> O~S)

the effecl of shear defor-

SUMMARY

The concept of generalized coordJnate presented in (his Chapter permits the analysis of multiple interconnected rigid or elastic bodies with distributed ProPM erties as single·degrce-of-freedom systems. The analysis as one.,-degree~of freedom systems can be made provided that by the specification of a single coordi~ nate (the generalized coordinate) the configuration of the whole system is determined. Such a system may then be modeled as the simple osciilntor with its various parameters of mass, stiffness, damping, and load, calculaLed to be

200

Structures MOdeled as a

Single-Degree~of·Freedom

System

Generalized Coordinates and Rayleigh's Method

dynamica!ly equivalent to the actual system to be analyzed. The sOlution of thIs model provides the response in terms of the generalized coordinate. The principle of virtual work which is applicable to systems in static or dynamic equHibdum is a powerful method for obtaining the equations of motIon as an alternative to the direct appiication of Newton's law. The principle of virtual work states that for a system In equil!brium the summation of the work done by ail its forces during any displacement compatible with the constraints of the system is equal to zero. Rayleigh's Method for detennining the natural frequency of a vibrating system is based on the principle of conservation of energy_ In practice, it is applied by equating the maximum potential energy with the maximum kinetic energy of the system. To use Ray[eigh's Method for the detennining of the natural frequency of a discrete or a continuous system, it is necessary to assume a defonned shape. Often, this shape is selected as the one produced by gravitational loads acting in the direction of the expected dispJaceme,lts. This approach leads to the following formula for calculating the nmuml freqc:ency:

,Y. Un!fOf/l'l disk

TOldl mass'" m

: !

V

Rigid beaM Total m
Fig. P6.2. 6.2 6.3

I~'~···-

:

!~e:(tMsibleCilble

1

I i 82.: WiY, w=

201

Determine the generalized quantities M", C', K", and F· (t) for the structure shown i:~ Fig. P6.2. Select yet) as the generalized coordinate. Determine the generalized quantities M·, Co., KA, and F" (I) for the structure shown in Fig. P6.3. Select OCt) as the generalized coordinate.

eq. (6.83)

2.: W,y;

where }; is the deflection at coordinate i and W, concentrated weight at this coordinate. Shear walls are structural walJs designed to resist lateral forces in buildings. For short walls (LID:::; 2) shear defonnations are important and should be considered in the analysis in addition to the flexural deformations.

PROBLEMS 6.1

For the system shown in Fig. P6.1 determine the generalized mass M', damping C·, stiffness K', and the generalized load F* (r). Select Y(r) as the generalized coordinate.

Fig. P6.3.

Y(I)

1

FOI (he elastic cantilever beam shown in Fig. P6A, determine the generaiized quantitIes M", K·, and F~(r). Neglect damping, Assume that the deflected shape is givea by ¢(x) = 1 cos('rrxI2L) and select Y(l) as the generalized coordinate as shown in Fig, P6A. The beam is excited by it concentrated force F(f) Frf(r) at midspan.

6.5 ~'ig.

P6.J.

Defermine the generalized geometric stiffness KG for the system in Fig. P6A if an axia! tensile force N is applied at the free end of the beam aIong the x direction. What is the combined generalized stiffness K;?

'·202

Slnlctures Modeled as a Single-Degree-of·Freedom System Uoiiorm beam TOlal mau m

FltJ

As.ru'fled

man ""m =:0

EI

~hape

Fig. po.g.

y

6.9

Fig. PM. 6.6

203

ConCI!n\r;;1ted

Flexurz:1 rigidity

Generalized Coofdlnates and Rayleign's Me!hod

A concrete conical post of djameter d at the base and height L is shown in Pig. P6.e;, It is assumed lhar the wind produces a dynamic pressu:e poet) per unit of projected area along a vertical plane. Determine the generalized quantities M', K', and F" (1). (Take modulus of elasticity E{ "'" 3 X 10 6 psi; specific weight y= ISO lb/ft) for concrete.)

Determine the natural frequency of the simply supported beam shown in Fig. P6.9 using Rayleigh's Method. Assume the deflection curve given by 2 ¢(x) = Y sin 1'fX IL The total mass of the beam is IItb = 10 th· sec /in, flexural rigidity £1 lOt lb· inl, and length L = 100 in, The beam. carries a concentnncd mass at the center m = 5 lb ' sec" lin. m,

~Y(/}

T I

Fig. PG.9.

6.10 A two·story building ls modeled as the frame shown in Fig, P6,lO. Use Rayleigh's Method to determine the natural frequency of vibration for the case in which only flexural deformation needs to be considered. ~eglect the maSS of the columns and assume rigid floors. (Hint: Use cq. {6,83)l.

i

-~Lll

~W===1IY'

,.--d---j

+,

!\Total)

Fig. P6.6. w

6,7

6.8

A simply supported beam of total unifomlly distributed mass mb, flexural rigidity Ef, and length L, carries a concentrated mass m at its center. Assume the dellection curve to be the defleclion curve due to a concentrated force at the cerner of [he beam and determine the natural frequency using Rayleigh's Method. Determine the natural frequency of a simply supported beam which has a [otaluniformly distributed mass ma. flexural rigidity sions shown in Fig, P6.8. Assume that during vibration the beam is of the shape produced by a concentrated force applied at the beam.

with overhang £1, and dimen· deflected curve free end of [he

{ (Toul)

L

I

~-L... Fig. P6.10.

6.11

Solve Problem 6.lD for the case ir:. which the columns are short and only shear deformation needs to be considered. (The lateral force V for a fw~.d column of

204

Structures Modeled as a

Singl$-Degree~o~~Freedom

System

length L. cross~sec(!onal area A, is approximately given by V = AG,d IL, where the lateral deflection,) G is the shear modulus of eiasticiry and

a

6.12

Calculate the natuml frequency of rhe shear waH carrying concentrated masses at the floor levels of a [cree-s.tory belJding as shown in Fig, P6.12. AssurTle that the deflec~ion shape of the shear wall is rhat resulting from a conceO[ra~ed larera! force applied a~ its rip_ Take :rex-ural rigidity, £1 3,0 X 101< lb - in 2; length L 36 ft; concentrated masses, m = 100 Ib - sec! li n and mass pe:- unit of length along the wull, m= 10 lb· sec 2 /inz, y

7 Nonlinear Structural Response

Fig. P6.12. 6.13

Solve Problem 5.12 on the assumption that the deflection shape of the shear waH IS (hat resulting from a lateral unifonn load applied along irs lengrh.

In discussing the dynamic behavior of single-degree~of-freedom systems, we assumed that in the model representing the structure, the restoring force was proportional to the d:spiacemeor. We also assumed the diSSipation of energy ~hrough a viscous damping mechanism in which the damping force was proportional IO the velocity_ In addition, the mass in the model was always considered [0 be unchanging with time. As a consequence of these assump~ tions, the equation of motion for such a system resulted in a linear, second~

order ordinary differential equation with constant coefficients, namely> m)'+cy+ky

F(t)

(7. I)

In the previous Chapters it was iHustrated that for particular forcing functions such as harmonic functions, jt was relatively simple to solve this equation (7.1) and that a general solution always existed in tenus of Duhamel's integraL Equation OJ) thus represents Ihe dynamic behavior of many structures modeled as a single-degree-of·freedom system, There are, however, physical situations for wh:ch this linear model does not adequately represent the dynamic characteristics of the structure, The analysis in such cases requires the 20:5

2GB

Structures Modeled as a

Slngle·Oegree-of~Freedom

Syslern

Nonlinear Structural Response

introduction of a model in which the spring force or the damping force may not remain proportional, respectively, to the displacement or to the velocity. Consequentiy, the resulting equation of motion will no longer be linear and its mathematical so]mjon. in genemJ, wi1l have a much greater complexity. often requiring a numerical procedure for its integration.

207

F,(yl

I

siope tangen, ~

slope: secant

--,

:- 1:,.:;, .....

7.1

NONLINEAR SINGLE-DEGREE·OF-FREEDOM MODEL

Figure 7. J (a) sho'w's the model for a sing\e-degree-of-freedom system and in Fig. 7,1 (b) the corresponding free body diagram. The dynamic equilibrium in the syslem is esmbllshed by equating (0 zero the sum of the inertial force F t (l) tbe damping force F D (1), the spring fo:-ce F, (t), and the extema! force F (f). Hence, at time t, the equilibrium of these forces is exp:-essed as (7.2)

Considering the case that in (his equation, the mass is constant, F, (r) = my; that the damping force is proportional to [he velocity with the damping coefficient also constant, F b (r) = cj; and the resistance force or spring force is a fUnction of the dispiacernem, Fj(f) = F;(y), we may then express eq. (7.2) as

(7.3) Hence, at time t,-. using the notation Yi = Y (().

my, + cy; + F, (Yi) and at short time later.

'ii =

y(t,) und

Yi = Y (t

Fit,)

~Y ~~~~

Spdng

.-~~

Damper

~!

f----+- F(c}

I

~--lE

(7.6)

Ft:rthermore, we ass:Jme that t:le incremental resisting or spring force is proportional to the incremental displacement, thar is i1F J (yj) = k,Lly"

The coefficient k i is cefined as the c"JITem evaluation for the resisting force per unit displacement (st1ffness coefficient) whlch may be taken as the slope of the tangent of the force~disp!acemem function, at the initiation of ~he time step .::1; for ~hc imcrval L1t or as the slope of the secant Jir:e as s.hown in 7.'2 fOl the plot of the resisting force F$lV). The value of (he coeft1cjenl Xi is calculated at a displacement corresponding to time l; ar:d assumed to remain constant during t~e increment of time Lk Since, in general, this coefficient does not rerr.ain constant during thar time increment, eq. (7.7) lS an approximale equation. The incremental displacement .1)", incremental velocity Lly, and incremental acceleration Jy are given by

: I -_.'5(~}. f---..... F(d itl">---L___J

F,. (tl F(}

(7,7)

(7.4)

(7.5)

,4

j

Subtracting eq. (7.4) from eq. (7.5) results in the difference equmion of motion in terms of increments, na,T',ely

m4j\ + c4Yi + k,4y, = 4F,

l ),

+ .c.1;

li+ I=!r

.--.~'--~----

(7.8) =

Fig. 7.1 (a) Model for a singJe~degree~of·freedom system. (b) Free body ding:am showing {he inertial force, the damping force, the spring force, and the eXtern::>l force.

Y(Ii + Lll) - Y (ti)

(7.9)

+ 41) - Y (Ii)

(7.10)

J,v, = 'Y(I,

208

7<2

Nonlinear Structural Response

Sfruc:ures Modeled as a Single-Oegree·of·Freedom System

209

INTEGRATION OF THE NONLINEAR EQUATION OF MOTION

Among [he many methods available for (he solution of the nonlinear equation of motion. probably one of the most effective is the step-by-step inregrntion method. 1:1 this method, the response is evaluated at successive increments ,dt of time, usually taken of equa~ lengths of time for computational convenience. At the beginniilg of each interval, the condition of dynamic equilibrium is established. Then, the response for a rime increment .1I is eva:uated approximarely 00 the basis that (he coefficients k (y) and c CY) remain constant during the intervai ,jt. The nonlinear characterisrics of rhese coefficiems are considered in rhe analysis by reevaluating these coefficients at (be beginning of each tIme increment. The res.ponse is then obtained using the displacement and velocity calculated at the end of the time interval as the initial conditions for the next time step. As we have said for each time interval, the stiffness coefficient k(y) and the damping coefficient c (j) are evaluated at the initiation of the interval but are assumed to remain constant until the ne.~t seep; thus (he nonEnear behavior of the system ~s ap?roximated by a sequence of successively changing linear systems. It should also be obv:ous that the nssumptlon of constant mass is unnecessary; it could just as weil also be represented by a var:able coefficient There are many procedures available for performing the step-by-step imegrarion of eq. (7.12). Two of the most popular methods are the constant acceleration method and the linear acceleration method. As the names of these methods imply, ;n the first method the acceleration is assumed to remain constant during the (ime interval .1r, while in the second merhod, the acceleration is assumed to vary linearly during the ir::tervaL As may be expected, the constant acceleration method is simpler but less accurme when compared with the linear acceleration method for the same value of rhe time incremem, We shall present here in detail both methods.

ti .......... t I f f ' :

A,

Fig. 73

--l

CO:1stam acceleration assumed during time interval Llr.

Integrating this equation twice with respect to the time between the limits and! ::esults in y(r)=:i<~

j

<

I,'

(7.J 2)

20';+),,+J(I-I,)

and y(t) =y, +y«I-I,)

+

+

The evaluation of eqs. (7" 12) ar.d (7.13) at tir:1e

(7J3)

l, +

I

= I,

+ ill gives (n4)

and (7.15)

7.3

CONSTANT ACCELERATION METHOD

In the constarnt acceleration method, it is assumes that acceleration remains t; and t, ~ 1 = [, + .1l as shown in 73.

constant for lhe time step between limes

at

The value of L'1e constant acceieration during the interval is taken as the average of ~he values of the acceleration )I, at the inii:iation of [he time step and y,.;. :, the acceleration at the end of the time step. Thus, the acceleration y (r) at any time I during rhe time interval Lit is given by

where ay; and Lly; are respectively the incremental displacement and incremental velocity defined by egs (7.8) and (7<9). To use the incremental displacement in the analysis, eq. (7.15) is solved for y;.; and substituted 'nto eq< (7.14) to obtain: (H6)

and (7< II)

2

ar

-2y,

(7<17)

210

SUuctures Modeled as a

Nnw subtracting

'ii

Sjng!e~Degree~of·F(eedom

System

Nonlinear Structural Response

on both sides of eq (7.16) results in

7.4 (7.13)

The substitution into eq. (7.7) of fly, and flY" respectively. from eqs, (7.17) and (7.18) gives

4

-y'.- Oy"

ill

I

-

\

r

2

'J + el,-Lly'" \ .at

\

2y·'

+ k fly· =

Ij'

LlF

f

,

(7.19)

Equation (7.19) is then solved for the incremer.cal displacement ~Yi to ob~ain Lly,

LINEAR ACCELERATION STEP-BY-STEP METHOD

Ir. the Unear accelera(ion method. it is assumed that the acceleration may be expressed by a linear function of rime during the time interval LlI. Let to' ar.d (,;. I = II' + ill be, respectively, the desigr.ation for the time a~ the beginning and at the end of the time interval ,cll. [0 this type of analysis. the material properties of the system C; and k; may include any form of nonlinearity. Thus it is not necessary for the spring force to be only a function of displacement or for the damping force (0 be specified on~y as a function of veiocity. The only restr1ction in the analysis is that we evaluate these coefficients at an instant of :ime t, and then assume [hat they rerr.ain constant during the ir.crement of time ill. When the acceleration is assllmec to be a linear funcllon of time for the interval of time Ii or f, {-! = f; + .1r as depicted in Fig. 7.4, we may express the acceieratjon as

(7.20) \'.. ri ) "

in which the effective sllffness

ki

2c

,

I

.

ilt

(7.25)

'

+ -Llr + k·'

(7.21)

where LlYi is given by eq. (7.10). integrating eq. (7.25) twice with respect to time berwee!1 the ;imits I, anc I yields 1

iiYi

-,

+ Yilt

- t,) I- C)-;;-(t - til' _ uf

- ri) -:-

j 1 ,dv, "TY;(f -.- fit + 6" AI

y(t) = Y,

and the effective incremental force ::jF, is

+

,}

.. + -~ Yi (t -[-)

=)'

is

4m _ k = ----r , Llr

LlF,

211

(7.26)

and

4m

(7.22) y{t):;;;; Yi

+ y;(l

1

{r- fi)

)

(7.27)

=

The displacementy;+) y(ti + Ll/) at time ti't- ( I; +..1{ is obtained fron eq. (7.8) after solving for incremental displacement LlYi in eq. (7.20). The incrementnl velocity is calculated by eq. (7.17) and the velocity atI time Ii + i = t; + ill from eq. (7.9; as

The eva1uation of eqs (7.26) and (7.27) at time [

.vi+-l

at the end of [he time step,

ti+!

+.1l gives (7.28)

(7.23)

Finally, the acceleration

!i

y

[j+,dt, is

obtained directly from the differencial equation of motion, eq. 0.3), rather than using eq. (7.16). Hence, from eq. (7.3j: (7.24)

Fig, 7.4

Assumed linear varialion of acceler<:.tion during time Interval.

212

Nonlinear Str~ctural Response

Structures Modeled as a Single·Degree~cfwFreec.om System

and (7.29)

where Lly, and .6Yi are defined in eqs. (7,8) and (7.9), respectively. Now to use the incremental disp!flCement .:1y as the basic variable in !he analysis, eq. (7.29) is solved for the incremental acceleration Lly, and then substi~uted inro eq. (7.28) to obmin

6

Llj,

6

Lly,--;;-y,-3j, 61

To obtain t,1e displacement y;...-; = yeti + .de) at time is substituted into eq. (7.8) yielding Yi.' = y,

1 kLly· " = LIT, ~

in which

ki

='

(7.31)

(7.32)

(7.33)

6m 3c; le, + -L1t2· -+ £it

(7.34)

and ifF, is the effective iI1c~emental force. expressed by (7.35)

It ShOUld be noted thar eq. is equivalent to the static incrementaIequilibrium equation, and may be solved for the in~remental displacement by simply d~viding th.e effective incremental force tJF,- by the effective spring Constant k" that is, Lly

fJF

,

=-~-'

k,

(7.37)

Finally, the acceleration y,.,. 1 at the end of the lime step is obtained directly from (he differential equation of motion, eq. (72). where the equation is written for tlIT.e t,"'- i ;;;;; Lit. Hence, after setting F; s: my,' + 1 in eq. (7.2), it follows [lim

is the effective spring constant, given by k,

+ .:1!, this value of ilYi

. '.' . Yi+!=Y;-.!J.Yi

FinaHy, transferring in eq. (7,32) all the terms containing the unknown incremental displacement LlYi to the left-hand side gives k,L1y,

!i

Then the incremental velocity LlYi is obtained from eq. (7.31) and the velocity at time [;+; = t; +.cit from eq. (7,9) as

{7J6)

(7.38)

'i .:-

(739)

The substitution of eqs. (7.30) and (7.31) imo eq. (7.7) leads to the following form of the equation of motion: L1r -v 2 .. '

Lly,

=

(7.30)

and

6

T

ti", !

213

where that damping force F DCt,.;. I) and the spring force Fs(t l + I) are now evaluated := rj + Llt. After the displacement, velocity, and acceleration have been detennined at time t; .. 1:= t, - L1£, the outlined procedure is repeated to calculate L1ese quantities at the following time step ti + '2 = t, ... I + .dt, and the process is continued to any desired finai value of time. The reader should, however, realize that this numerical procedure invotves two significant approximations: (l) the acceleration is 3>sumed to vary linearly during (he time increment ilt; and (2) the damping and stiffness properties of the system are evaluated at the initiation of each time increment and assumed to remain constant during the rime interval. In general, these two assumptions introduce errors that are small if the time step is short. However, these errors generally might tend to aCcumuiate from step to Step. This accumulation of errors should be avoided by imposing a total dynamic equilibrium condition at each step in the analysis. This is accomplisned by expressing the acceleration at each step using the differential equation of motion in \vhich the displacement and velocity as well as the stiffness and damping forces are evaluated at that time step. There still remains rhe problem of [he selection of rhe proper time increment ,dt. As in any numerical method. the accuracy of the step-by-step integra::ion method depends upon the magnitude of the time increr:1eot selected. The following factors should be considered in the selection of Llt: (1) the natural period of the structure; (2) the rate of variation of the loading function; and (3) the complexity of the stiffness and damping functions. In general, it has been found that sufficiently aCCurate resulrs can be obtained if the time interval is taken 10 be no longer than one-tenth of the at time !,,,.!

214

Nonlinear Siructural Response

Structures Modeled as a Siogle-Degree·of-Freedorr. System

natura! period of the structure. The second consideration is that the interval should be s:nal! enough to represent properly the variation of [he load with respect to time. The third point that shouid be considered is any abrupt variation it: the rate of change of tbe stiffness or damping fUJ1CliofL For ex.ample. in the usual assumption of elastoplastic materials, the stiffr.ess sud-

denly changes from linear elastic 10 a yielding plastic phase. In this case, tc obtain the best accuracy. it would be desirable to select smaller time steps in the neighborhood of such drastic changes.

Now, the substitution of eqs, (7.42) and (7.43) into the incremental equatlon of motion, eq erl), results in an eqUaliOD to calculate the incremental displacement .dy;, namelY, (7,44)

where ~he effec:ive stiffness k" und the effective incremental force !JF; ure given respective;y by _

m

k,=kl+--"

7.5

215

(3!Jr'

THE NEWMARK BETA METHOD

C

+-1

(7,45)

2{3!JJ

and Tlle Newmark Beta Method includes, in jtS formulation, several time-step methods used ~or the Solulion of iinear or nonlinear equations. It uses a numerical parameter designated as {1 The method, as originally proposed by Newmark (1959), contained in addition to /3, a second parameter y. Particular numerical values for these parameters leads (0 weB-known methods for the solution of the differential equation of motion. the constant acceleration method, and the linear acceleration methcd. The Newmark equation can be written in incrementa] quantities for a COnstant time step Llt. as

!JF,. = .dF, +

In

c

mIl \

+-' y-+--y -(",.:1(11--ly·· 2{3'

2{3'

.

\

4/3 J '

(7.46)

10 these equations, C; and k, are respec;:ively tbe damping and sriffness coefficients evaluated at the :nitia: time I, cf the time step Lit Ii":" I - I;. 1n the implemen;ation of t~e :'\ewm2.rk Beta Me6od, a numerical value for the parameter {3 is selected. Newmark suggested a vabe in the range 1/ 6.::f f3:S: 112. For f3 1 14. the method corresponds (0 the constan~ accelera~ion method and for f3 i 16 tc {he linear acceleration method.

(7.40)

and

7.6 (7.41)

in which the incremental displacement ny: and incremental velocity LlYi 3re defined, respectively, by eq (7,8) and (7,9), It has been found that for values of y different than in, the method introduced a superfluous damping in this system. For this reason, this parameter is generally set as '1= 1/2, The solution of eq, (7.41) for .1.9; and its subsequent substitution j nto eq. (7.40) after setting y =. ! 12 yield

(7.42)

and

ELASTOPLASTIC BEHAVIOR

If any stmcture modeled as a 5ingle-degree··of~freedom system (sp:ing-mass system) is aUowed to yield plustically, then the restoring force exerted is likely to be of the fonn shown in 7.5(a). There IS a porrion of the curve in which linear ei.astic behavior occurs, Whereupon, for any fu::ther defonnation, plastic yielding takes place. When the strucUJ.re is unloaded. the behavior is .again elastic untiJ further reverse loading produces compressive plasric yleldir.g. The stmcture may be subjected to cyclic loading and unloading in this manner. Energy is dissipated during each cycle by an amount thnt is proportional to the area under ~be curve (hysteresis ;oop) as indicated in Fig_ 7.5(a). This behavior is after'. simplified by assul'TJing a definire yield point beyond which additional displacemem takes place ar a constant value for the reswring force without any further increase in the load. Such behavior is known as elastoplasIic behavior: the corresponding force-displacement curve is shown in Fig. 7.S{b)

(1.43)

For the structure moceJec as a spring-mass system, expressions of the restoring force for a system with elastoplastic behavior are eaSily w:inen,

216

Nonlinear Structural Response

Structures Modeled as a Single-Oegree-of-Freedom System i'hmoriN} force

in which

R/!!$,Or;P9

y=O

loree

R

R

y~~x

217

is the maximum displacement along curve T, which occurs when

Conversely. if y decreases to y~, the system begins a plastic behaVIor tn compression along curve C and it remains on this curve as long as y < O. The sysrem returns to an elastic behavior when the velocity again changes direction and y > O. In this case, the new yielding limits are given by

y,

Yrr,in

y, = Yxi'

Fig, 7.5 Elasric~pl,l$(ic :structural models. {a) GenerJi plustic behavior. (b) E!asmpiastic behuvoiL

These expressions depend on the magnitude of the restoring force as well as upon whether the morion is such that the displacement is increasing l": > 0) or decreasing (y <0). Referring to Fig. 75(b) in which a general elastoplastic cyc!e is represented, we assume that the initial conditions are zero (yo ~ 0, Yo 0) for the unloaded structure. Hence, initially, as the load !s appJied, rhe system behaves elastically along curve The displacement y" at which plastic behavior in tension may be initiated, and the displacement Yn at which plastic behavior in compression may be initiated. are calculated, respectively, from

+ (R, - RJlk

(7.50)

in which Ymin is the minimum displacement along curve C, which occurs when .y = 0. The Same condition given by eq. (7.48) is valid for the system [0 remain operaung along any elastic segment such as Eo, E" £];.".as shown 1n Hg. 7.5(b). We are now interesled in calcuiatlng the restoring force at each of the possible segments of the elasrop[astrc cycle. The restoring force on an elastic phase of the cycle (E" E,. E,» .. ) may be calculated as

R=R,

(y,-y)k

(7.51)

on a plastic phase in tension as

R=R,

(752)

and On the plastic compressive phase as

(7.S3)

y, = R,Ik

(7.47)

and

where Rr and Rc are the respective values of the forces that produce yielding in tension and compression and k is the elastic stiffness of the structure. The system will remain on curve Eo as long as the disp:acemem y satisfies


(7.48)

If the displacement y increases to y" the system begins to behave plastically in tension along curve T on Fig. 7.5(b); it remains on curve T as long as the velocity y > O. When y < 0, the system reverses to elastic behavior on a curve such as EI with new yielding points given by

Yr =

The algOrithm for the step-by-step linear acceleration method of a single degree-of-freedom system assuming an elasroplastic behavior is outlined in the fonowing section.

7.7

ALGORITHM FOR STEP-BY-STEP SOLUTION FOR ELASTOPLASTIC SINGLE-DEGREE-OF-FREEDOM SYSTEM

Initialize and input d(1{a:

(1)

Input values for k, m, c, R" R,., and a table giving the time magnitude of [he excimtion Fl'.

(2) (3)

Set Yo and Yo;= O. Calculate initial acceleration:

=

(7.49)

and

°

F(r = 0)

)'ma1

Ii

m

(7.54)

218

(4)

Select time step Lil and calculate constants: 0,

(5)

= 31Llt, a, ~ 61Ll1,

OJ

dli2, a,

(4)

Calculate the incremental effective force:

6/d:'

(7.59)

Calculate initial yield points: YI

(5)

Rilk

y, = R,lk

Solve for the incrementa! displacement:

(7,55)

For each lime step: (l)

219

Nonlinear Structural Response

Structures Modeled as a Single-Degree-of-Freedom System

Lly;

(6)

=

(760)

Calculate rhe incremental velocity:

Use the following code to establish (he elastic or plastic state of the (7.6~)

system: KEY

= 0 (elastic

KEY =

behavior)

(7)

KEY,:,:,;" I (piastic behavior iii tension) (2)

(8)

When the system is behaving elastically at the beginning of the iime step and

+ Lly,

(7.62)

Yi )= y, T i.lYi

(7,63)

Y lot' 1

(7.56)

Calcui:lle Ihe displacement y and '¥eJocity j at the end of the time step and set the value of KEY according to the following conditions; {a)

Calculate displacement and velocity at (he end of time intervat:

I (plastic behavior in compression) y,

Calculate acceie:-ation Yi~ I at the end of time interval using the dynamic equmion of equilib:i.urr.: (7,64)

y,
at which

When (he system is behaving plastically in (cosion a( the begin, fling of the time step and

j>O KEY = 1 When the system is behaving plastknlly in compression at the beginning of the time step and

y <0

KEY=-j

;;>0 KEY (3)

0

CalcUlate the effective stiffness:

k for elastic behavior (KEY

k" = 0 for plastic behavior (KEY

R=R"

if KEY = - 1

(7,65)

Example 7.1 To illustrate the hand calculations in applying the step-bystep integration method described above, consider the single~degree-of-freedom system in Fig. 7.6 w:th elasroplastic behavior subjected to the loading histOry ;1S shown. For this example, we assume that the damping coefEcient remains constant (~= 0.087). Hence the only noniinearities in the system arise from the char,ges in stiffness as yielding occurs.

Solution:

where

kp

if KEY = I

or

y
R=R,

=

0)

= 1 Or

The stiffness of the system during elastic behavior is

12E(2!)

- 1)

12 x 30 X 10'

x 2 X 100

k=~~--=---~····.

(7.58)

Ll

(15 X 12)-'

12,35 Kiplin

220

Strc.lctures Modeled as a Single-Degree-of-Freedom System

Nonlinear Structural Response

numerical convenience, we select Llr (7.57) is

20'

221

0.1 sec. The effective stiffness from eq.

or

k = k, +

128,22

where ibl

k" = k kp

;.:!;;!

1235 (elastic behavior)

0 (plasric behavior)

The effective incremental loading from eq. (7,59) is ---r--~Y

f 6

\

I

L.V

.1F + [-m + 3c)Y' +13m -r-";!Y" '\Lir i 2 j

lit, =

l
Fig. 7.6 Frame with elaStop!l!stic behavoir subjected (b) Loading. (c) Elastoplastic behavoir.

(0

dynamic loading. (a) Frame.

dF,

\

I

+ 12.822Y; + 0,6137y,

The velocity increment given by eq" (7.61) becomes

lIy; ~ 30lly, - 3y; - O.05y, and

~he

damping coefficient

c = tc" = (0.087) (2)

X 12.35 = 0,274 Kip·

Initial displacement and inil:iaI veJocity are Yo = Yo

yo=

=

sec lin

0, Initinl acceleration is

F(O) =0

k Yield. displacements are

The necessary cakulz.tions may be conveniently arranged as illusrrared in Table 7.1. In this example with elasroplastic behavior, the response changes abruptly as the yielding StartS and stops. To obtain better accuracy, it would be desirable to subdivide the time step in the ceighborhood of the change of state; however, an iterative procedure would be required to establish the length of the subintervals. This refinement has not been used in the present analysis or in the computer program describec in [he next section. The stiffness computee at the initiation of the time step has been assumed to remain constant during the entire time increment. The reader is again cautIoned that a significant error may arise during phase transitions uniass [he time step is selected

relatively small, R, v =-~ I k

7.8 and y~. =

1.215 in

The natural period is T = 21T r;;ik = 0,8 sec (for the elastic system). For

PROGRAM 5-RESPONSE FOR ELASTOPLASTIC BEHAVIOUR

The same as for ~he oilier programs presented tn this book, Program 5 initially requests information about the data flle: r.ew, modify, or use existent data file. After the user has selecIed one of these options. the program requests the name of the file and the necessary input dam. The program continues by reading and

\1")0;

Nonlinear Struct\,;fal Response

("4-

00
lI";

....... ( ' j

00 NVi

223

a:N~g~g~$~?1j~;g~~~~

'NMMN-:N~0N

dC'4N"":o6

I

>-'2

~v

::S!;:; '" '" MO "

r- r- 0> r0

" '" (i':;5 "" r00 00 0 00 ON

""

'" " N

I N

OI'X>OONa

~("o,)«")("")

.........

'"

r-oooo

("") 0 0\ ...... Vi <.n.<:::!'OON -.:t-N\Or-C-N""-T 0

"" ooddoood I

!

I

I

~

.-.: ~

c. E ~

"V "'""': .,."" ".,..,. '" r--r-0 r--r--0 r--r--0 r--r-- 0r--r-"~ '" ~ '"ci '" '" '" '" ~

k. c:.

"l

:;;?

"

'" '"

M

I

I

I

""

t-

r--

0

CO 0

I

'" '"

0

'"

OJ

.2 "0 0

,~

~

"

::E c.

I,I')I,I')I,I')V-,>n 1,1') f"".M«)t"'!~ M OOO~NN(,
" iii

>.

.0

6. <1)

iii

,,,c

;; il

N

~"

:>..~ ,S

,..,

e- N <': .,.., 0 od N 0-

'"

~

c: 0

a. ~

<1)

'"" ~

'"

c0" z

~

.-.: OJ ..J Ql

.......

222

.,..,

0- 0>

00

~,

'-0

Example 7.2 Using the computer Program 5, find the response of the structure io Example 7, J" Then repea( the calculation assuming elastic beht-voir. P:o~ and compare results for the elaswpiasric behnvoir with the elas[ic response. Solution: Problem Data (from Example 7.1).' Spring constant: k = 12.35 Kiplin Damping coefficient: c = 0,274 KIp' sec lin

<::O---~~--

..." J,

00 >n

&:. r--

~1T'l>no.riV-)"'-;""';

u

:.

'D

.,..,

0 oc 0 ,... '" '"~ '"""""oZ N =r--0'" '" '" "" NI '"'"I "'"I '" i '-"I r-'"
""" '"'" '"r-- '"'"e- 0-

~~ggg88

-.;

'c:" ::J

printing the dara and by setting the mitial values to the varioLls constants and variables in the equarions. Then by linear interpolation, values of the forcing function are computed at time increments equal to the selected time step Lll for the integration process, In the main body of the program, the displacement. velocity, and acceleratIon are compLted <:::t each time step. The nor.lir:ear behavior of the restoring force is nppropr.meJy cons:Gered in the calculation by the variable KEY which is tesred through a series of condllional statements in order to determine the correct expressions for the yield poinrs nod the magnitude of the restor:ng force in the system. The output consists of II table givlng the displacement, ve]ocity, aod accel~ eration at time increments Lit, The last caluma of the tD.ble shows the value of the tndex KEY which provIdes Information about the stale of rhe elastop. lastic system. As indicated before, KEY = 0 for elastic behavior and KEY = 1 or KEY = : for plastic behavior, respectiveiy, in [eosior: or in compression"

Mass: m = 0.2 (Kip· sec' /in) I,I')vOOOOQ'\...-.Qr--O:-

Max. restoring force (tension): Rt = 15 Kip

O\-«)MOOI,I')OOO\OOOOOOMN

Max. restorir.g force (compresslon): R" = ~ 15 Kip

~~~~::::=:~~ti6~~~~~

oci~""';-.Dr-..:r-..:.nNci-ir-..:
1

;;; ""co '" '" '"ci 0"i ""'"<:5 "i"" 0: ;:;;

N

00 >n 'D


I

Natural period: T=27TFmlk =217/0.2/12.35 -0.8 sec

'" '" '" - 0

'0

I

v-,

"" '"

'"'"f:! '" r. r. N

'Voo{"0t-010000

0

0

0-

" N"" "" '"

;:;; 00

'"

0

'" "" ""

'" N '"r-...,,.., '" "" '"MN'" "" "': "" 00

N N

vt-

~oonr--\Q>nO{"0Vir-­ ~<::O«)t-~{"0{"0N-O

o-i
-----

-I

N «) -q- l.(j ~ t- 00 0\ 0 -: C"J. 0000000000-":

~

-.::t:

V',

Selecl time step: Lit = 0, I Sec Gravitational index: G = 0 (force or. the nass) The same computer program is used [0 obtain the response for a completely elastic be!iavioL J[ is only necessary to assign;) large value to the maximum restoring forces in tension ~nd in compression, These assigned values should be large e:1011gl: in order for tl1e structure to remain in the elastic range. For the present example. R/ 100 Kip 3:lG R," = ~ 100 Kip were deemed adequ&te in this case. In order to visualize and faciljtate a comparison be{ween the elastic and inelasric responses for this example, the displacements are plotted in Fig, 7.7 w1[h [he response during yielding shown as dashed lines.

224

Structures Modeled as a

Single~OegTee·of-Freedom

System

Nonlinear Struc:ural Response

225

Displacement !In.!

Input Data and Output Results

!

4.D~

tiS

0

AX

0

AK OA.:~?I":G :'::~::;:

0

CCS?F!cr€::"

S':'!i:?

o.r

H

nr::'SCftN!:'!CN

orr.;\

RSS~"=;''G

?;)!l.CS IS TS:iS:O"

::;.'!,

:11';

RSS?C~!~:G

?8?.cZ :::m??:::ss~ml

"

:;::;Z;': (G1-1X::"'':' OR ZERO: ':'I!~£

, ,

S;;'C:C:;:,:C;i

,

". <:J'J L .:;::

2X::::T.;T:(',d

'!'!~'S

0;;

.0~

ZG, CO

4SJ

.>::0'1

0

T::',>;

,

:')\'

. '",

35

~3

-:,

0

Z:KCt'!'.::;,:ox 1.

,t

~";~

"

no

:-:YlS

(N

"/S:'-:>:.

;..ce.

300C

::.DOI;O

C.OCO'i :a. 9714 25 .7n.\

:n:6

c.i4$S

0.2325

:>. ~83',

JOel

6,,;;:8 '-,2679 9154

::"::3:' ;")49

:3 lH:

o .~o::

0: 7 3

65CC

0:15

.,

?: ?J

,.

~), 2'~C

SO()

C. CDC C.7C0

soc so:;

-+

no'!

S"7/S

on. ;'(l~:

l.SS?;>

-31 . ?S5e

$:;10

-0.,,1;64

.{1 )1$90

GEOME~RY

BC52

_:4 936:

?:"?-..ol\lE,

lCO

n258 1122

oS ]77';

:leo

120Z

400 l. ,c:;

PH

:!ss:.

;

6~ZS

n57

5:1,12$5

50C

6911

\,. 2lsa

-i.3aM

7-:;0

59/1

.'1471

-s~. 5~6~

,.

,

,sec: ooc-

7.9

~9';;()

630'<

.,

st:i

Fig. 7.7 Comparison of elastoplastic behavoir with elasTic response for Example 7.2.

s. dSJ

~Ja,

"as

on]

0'

,

~;56

B:.2?

(2) Define the XY plane ar Z = 0:

~

. :C2S

il"-3e

~7?_

, c

liiO 3

$753

-,

~;;J.S

-59 441{

Ol';' .,

-'t ga97

·s . .ns~

DISPL.?,';( VI~\\!,

0,

the XY plane; :>

V:::E.:'i1 J?AR )

Dt

1,

C

PLk~B

CEOME-~RY

). C0RVE$

CRPCORD.

1. C, 0, 0, 1, G, 0

:>

capCORD

(4) Define element group 1 using the TRL'SS2D element formulation with etastoplastic behavior PROPS 2'rS ). EGROH? GROUP, 1, 'fRUSS2C, 0,

0, 0,

C, 1, 0, 0

(5) Define real consta!1ts for truss element: A = 1.0 in::!; PRO?SE':'S ). ?CC:TST RCONS?, 1, k, I, ,;. 1, 0, ::::, 0

(6) Define elastic modulus for truss elernem base OIl 1 in length and area I in' (k = AEIL = 12,346 Iblin). Also define yield 'tress as 15.000 Ibl mZ and tangem modulus as tE~7 Ib/tn 2: ?ROPSETS )

\rIE'd

~

1

(3) Generate a curve from 0, 0, 0, to 1, 0, 0

5051

Solve Example 7.1 uSi3g the program COSMOS

to

G.

56.7'~S

Solution: The analysis is perfonned using a single truss element with one cO:1cemrated mass element. The following commands are implemented: (I) Set view

) GRID

Z,

-9.4756

..

NONLINEAR STRUCTURAL RESPONSE USING COSMOS

Example 7.3

, Yielding/""

.72Y1

,

oe·::;

E1utic (~cnse

1/

01--~+---4---~--+---'.;:+--~---+--~,.c.+- .. ~ Time 1.8 (~cJ 0.4 e.8 0.6

::J~s"'t.

,

•

•

+

1':;-;£

t 20D

CU':?'..:'..' ,,;;:5(,.':"75.

1.CGD

~;//.-

30~

,

.

MPROP

t1J?ROP, 1. EX, ~2346 MP?.CP, 1, SIGYLC, 15000, ETP,_N, 1E-7

226

,\onlinear Structural Response

Structures Modeled as a Single-Oegree-of-Freedcm System

227

(15) Request response at node 2:

(7) Generate one truss element along curve 1:

:CNRES?

(8) Define element group 2 using the MASS element formulation:

(16) Set analysis to start at

I

= 0 and

end at l -= '1 sec, with n time increment

of 0.05 sec: PROPSETS EG~OU?,

j

2,

EGROUP K~SS,

0, 0, O. O. 0, 0, 0

LOADS-BC > LOAD_OPS > l'IHES TIMES, 0, 2, .OS

(9) Define real constant for mass element: m ;:;;; 200 !b· sec:? lin:

PROP$ETS ReOl'1ST,

2.

~

RCONST

I

2,

7, 200, 0,

1,

0, 0,

0,

0I

(7) Set the options for [he nonlinear dynamic analysis using default values for toierance parameters and run analysls:

0

t>.NALYSIS ~ NOKL::NE:A::\ > A~K0NLINS:l-_)=<. A NONL1N2AR, :J, l , 1, 20, 0.801, 0,

(0) Generate one mass element at point 2: MESHING ~CPT,

PARA."CMSSH .. 11 ?T

:>

MESHING ~ NODES NMERGE, 1, 3, L

;>

NM2RGE 0.000;", 0,

(18) Activate XY plot info,wation for X displacement at node 2 as a function of time, and plot the displacement vs. time for node 2: 0,

0

(12) Apply constraints in all degrees of freedom at node 1, and all degrees of freedom except UX at node 2: $'l'RUC'TURAL .. DISPLMN'T'S

DPT, 1, AL, 0, 1, 1 DPT, 2, Uy, 0, 2. 1, UZ, RX, RY,

;>

CPT

([3) De:'ine cyr.amic fOfcIng function and apply as force to node 2 in the

LO_!;'DS~8C

> FUNe_cuRvES

1,

1,

>-

0,

CUB-DEF

0,

.45,

20000,

1.1,

0,

L2,

-:'0000, 2.0,0 LQ.lWS·· Be .)

FPT, 2,

STRCC':'URAL FX, 1, 2, 1

>-

~O:.qCES

)

FPT

(14) Define Rayleigb damping coefficients [C = ak + pm]. C" = 2(km),n = 3143 10' sec lin; C 0.087 (C,,) = 273.4 lb· sec lin: Using a= 0.01, fJ [273.4-(0.01)(12346)11200 0.75 ANALiSIS-NONSINEA:R NL-ROf..Hi? NL_RDAMP, 1, 0.01, 0.75

C:S?LAY ~ XY_?LOTS ~ ACTXYPOST ACTXYPOST, 1, TH!E, UX. 2, 12 • .\., DISPLAY XY~?=-O'T'S:> XYPLO'!' XYPLO?, 1

v,

2N

(19) Activate XY plot information for X velocity at node 2 as a functio;i of lirr:e, and plot the veiociry vs. time for ;iode 2:

HZ

d~rection:

CuRC2F,TIME.

0,

R_NOtt,;L'::~SAR

(11) Merge nodes:

X

Y,

,;';"lALYS!S > NOl'lLINEAR

2, 2,. 1

LOADS-Be )

N,

0.001, 0.01. 0

DISPwAY > XY P~OTS } l-CTXiPOST ACTXYPOST, 1, TIME, VX, 2, 12, .\., DIS!?LAY > XY ... PLor{~S } xY2LOT XY?LCYr, 1

0,

2N

(20) Activare XY plo~ informatio:l for X acceleration at node 2 as a function of time, and plOl the acceleration vs. ~~me for :lode 2: DISPL.!;'Y ) XV_PLOTS } ACTXYf'O$T ,l:!,C'rXYPOS'L L 'rIME, AX, 2, 12, 1, DISP.!:.AY XY ...,?LCTS > XYP:"'OI' x YPL01' , 1

Figure 7.8 shows the response for Example 7.3 in (b) velocity, and (c) aeeelerarion.

0,

[elmS

2N

of (:1)

djspJaceJ:len~,

Nonlinear Structural Response

Structures Modeled as a Single-Degree-ol-Freedom System

228

~

i

!

[

,I

I

u

I

I

x

I

I

2

II

,

1.5 I

....r,

,,

I

I '\ I I \!

!

;

I/bl II

'~ ___ ~ '\

/ -f~ I

~

!

I

i-~t

I

I

, .4

.

I

,

I

I I~', .8

-

=~

I

!

I

~

[

'ti

I I

,,

I

I

!

:r , /

'.5

lit

i

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J.5

,i

,

.

~

1.4

I

48

I

!

12 16

/

,

A X

-! 5

·31

-qs

-5, -go

:?

I

I .8

1.2

I 64 I

i ,

,

EJ

,

OJ.2

1.8

i I

,, ,

,i ~ I, ~ i

-~r-

1

,,

~"KI ~

I

. I

I I

I

,

, ,

I

.4 !;!

i

! I I

.5

I

\

\

!

II I , ,

r-

1\ ,i

I

I .2

2N

\

II

i

,I

? '\

,7

i

I~

I

7

! i

i I

)

I

I

I

[

,I

229

I ,I I.,

I, ! I .6

\ II ,

1-1, 1.8

T!HE

,,' i

i

j

I

./

/ i

v

~

I

I

" '\ ,

I

I

-4

I

-8

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L-

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I \ I \

II

I

,

, .8

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,

,

, .4

'.6

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I

l

\

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IT

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"-...

I '"I

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I ! \ I \ I/ \

,

-, -co

i

i\

/

x

Fig. 7 ..Be

,

,

iol

\

I

I ! 1.\

i l.6

I I

I I

\

I

1.8

T!HE Ib)

F;g.7.8

7.10

Response for Example 7.3. (a) Displacement; (b) Velocity; (c) Accelenuion.

motion of an eanhquake or the effects of nuclear explosion. In these cases, it is not realistic [0 assume that the structure will remain linearly elastic and it is then necessary [0 design the Structure to withstand deformation beyond the elastic limiL The simplest and most accepted assumption for the design beyond the elastic limit is [0 assume an elaswpi
SUMMARY PROBLEMS

Structures are usually designed on [he assumption that the structure is linearly elastic and [hat it remains linearly elastic when subjected [0 any expected dynamic excitation. However, [here are siwations in which the structure has [Q be designed for an evemual excitation of large magnitude such as the Slrong

7.1

The single-degree-of-freedom of Fig. P7.1 (a) is subjected to the foundmiol1 acceleration history in Fig. P7.I(b) Delermine [he maximum relative displacement of the columns. Assume eiastoplas[ic behavior of Fig. P7.I(c).

230

Str.JCtures Modeled as a Sjngle~Degree~of~Freedom System

m"'O,5~W¢

Nont!near Struc!ura, Response

7.4

.,I()

231

Repeat Problem 7.2 for the acceleration history show:. in Fig. P7.4 applied horizontally to the fO:Jndation.

50

Ibl

50

Fig. P7,4

7.5

7.6 Ie)

7.7

Fig. P7.1 7.2

Determine the displacement history for the structure In Fig. P7.1 when it is SUbjected to the impulse loading of Fig. P7.2 applied hOIlzomaHy at the mass"

7.8

Solve Prcb!em 7.1 as;;.uCiing elastic behavior of t~e strJctu:e. (Hint Use :::omputer Program 5 with R,o= 200 Kip and l?", = _. 200 Kip.) Solve Problem 72 :or elastic behavior of the struL'ture. Plot the rime-cisplacement rcsponsc and compare with results :rom Problem 7.2. Determine the ductility rado from the results of Probler:l 7.2 {DuctHity ralio is defined as the ;ario cof the maximum displacement to the displacement at {he yield point}. A stru~ture mOdeled as spring-mass shown in Fig. P7.8(b) is subjected to t:'e loading fcree depicted in Fig. P7.8(a). Assume elastoplastic behavior of Fii!. ~. P7.8(c). Determine the response.

Fit}

L _ _-'-_ _-::'::-_ ..... ({sec)

o

0.1

0.2

R

Fig. P7.2 7.3

Repeat Problem 7.2 for the impulse loading s~own in Fig. P7.3 applied hClr1Z0ntally at the mass. ,cUI

200K~

t

1°l7--7

r"

~".;',

··~l

w . 38$1<:

__ -

~F({)

-~

L21= :c)

o Fig. P7.3

Fig. P7.8

232

7_9 1.10

1.11 7.12

$liuctu:es Modeled as a Single-Degree-oH'ii:edom Sys!em

Re.peal Problem 1.& assuming dn,-nping in [lie system equal

10

20% of lhe cri!icul

damping. So~ve Problem 7,& assuming elastic be\)uviOl of Ihe $)'$ll~m. (Hint: Use Program 5 with ff,= 1000 Kip and Rc~ -, 1000 Kip.)

Solve Problem ?,9 assuming c!Jsiic behavior of lhe sys:em. Fig, P?I2(a) is A suucwle modeled as [hI:': damped spling-m:ls.s 5yslem shown subjecteu to the !im~-acctleriH;on ex.citation (luing at ils suppOrt. -rne: excitation (unCI!On is expreised:)5 a{l}::::: (Ju!(J). Whl!feJ(f) is depicted in Fig. P7.12(b). Determine the maxlmUln v,llne (hUI lhe f:lclOr (In nJ£ly have for the S(fllCIlIH! 10 remain e!J:.;{ic. AS~ume !h;)( the SlruclUre has.: nil elns[op!a$!ic behavior of FIg. PI. !1(c}.

il,

8 Response Spectra

A fir'" ;!)OJ _ ,_ _-,

O.~

0.3

(6

-4-.1--+-1 ;! I -t--- I 10.5 0.6 0.7 0.3 0,9 1.Q 0.45

id

101

fig. P?t2

In Ihis Chapter, we inirodllce (he concept of response SpeClrl.tffl, v{hich in recenr years has gained wtde: nccepwnce in SlfUCI'Jfl11 dynamic ptuctlce, particularly in eanhqu"Ke englliteril1,£ design. Slated brieny, the response spec:nlm is a plot of lhe Glilximur:! respof'.sc (maximum displacement, veloci,,!, acceleration, or any olher qu
freedom system, the mJxirJlum d:sp\Oicemenl

or

[he mass

til

relative to the

displacement at the support, TI!\JS, to dClcfmine Ihe response from an nvailab!e speclra! chan, fOf (. specified excitJlion, We need only iO know the :\
8,1

CONSTRUCTION OF RESPONSE SPECTRUM

;I!u$u~[e (he construction of J rCspof\:;e SpecI(;.}1 cllan, consider in Fig. 8.2(a) the unda~Ylped osci!lam( subjcC:;:d :0 one·half period of lhe sinusoidal

To

2)3

8 Response Spectra

j~t1Dduce Ilk. t:lJllI.':CPl or re.f!1oWiC .'ipeClmm, which in recent yents has £:lintu wide ,H..'CCP\;lllC(: iI) ~((Ut:lllf;1l uyn:\m!c prl.lcljc~. p::lniculady iii eanhqui1kc tngi:1Ccfillg th!:i!t;11 St;l\ed hriefly. lilt: response spectrum 1$ il plot or tht.O l11ilXll1HlIn 1!:;;ij)0!lSC (1l"IXllillilli ci:-;placcH\clH, vdacilY, pccdeficll;, :;;)(~ tIll! oH.!;n;He tht 111aximum response, A plot of this type is showll In (:ig. 8.!, j;, wll!cli ;t ene-story building is subjected (0 a grcunu Ji:;p:acCIl1l.::di i:I(:ic,llCJ oy the [uHc\lon Y,(t)_ The n> spcn~e spectral curve sliewn ill i:,g. l:L i(:\) gi"'I~S, [or ;:,\ly singk-degrce-o[freedom system, (he flHlXinHlnl 0J!>p!:'ce.:nerH cr the:. muss III relaflve to lhe displi1cemen( O[ ill:! suppert Tho::., (8 dc(cnni,ie (hi,! ;e.spomc (;om ao available spectra; chan, [or a srecjfi~d l:-xcit:too:\, wt: need only to kllow the oalUral fre.qucilcy of lhc s'yslem

In this chapler, we

8.1

CONSTRUCTiON OF RESPONSE SPECTRUM

To il!u:ar;Hc lh< cO:1s(n.:ctioJ: of it (C:'PCIl$C SjltCHJI ch:irt, consider in pig. 8.2(a) dIe Und~!nlpcJ 8-sci!LLIO( 5uuj';Cld 10 cn(.:-ll:!i[ pcriDd o{ (he. :;inusoid;)i

234

Response Spectra

Structures Modeled as a Slngle>Oegree-cf-Freedom System

Malt felatWII dispt Iy~y.i,.,..,.~

I

'

/////,r),ddddN&&dL/d _ _ _ _ _ _ _ _ _ _-=~_!NatlJr~! frequent;:y. f

~---...

{a) Typical response spectrum. (b) Single·degree-of~freedom system subjected

Fig. 8.1 [0

The Solulion of eg. (8.1) may be found by any of the methods studied in the preceding chapters such as the use of Duhamel's integral (Chapter 4) or the step-by-step linear acceleration method (Chapter 7). However. in this example, owing to the simplicity of the exciting force, we can obtain the solution of eq. (8.1) by the direct method of integration of a linear differential equation, that is, the superposition of the complementary solution Yc and the particular solution yp

(8.41

Y=Yc+Yr

¥,(tl

{b)

The complementary solution of eq. (8.1) (right-hand side equals zero) is given byeq. 0.17) as

ground excitation.

y",=A. cos wt+B sin

exciting force shown in Fig. 82(b). The system is assumed to be inrtially at rest. The duration of the sinusoidal impulse is denoted by la. The differential equation of motion is obtained by equating to zero the sum of the forces in the corresponding free body diagram shown in Fig. 8,2(c), that is, (8.1 )

my+ky=F(t)

235

(85)

UJt

in which w=jklm' is the natural frequency. The particular solution fo( the time interval O:i l:i I, is suggested by the right-hand side of eg. (8.1) to be of the form

)'p=Csin

wt

(8.6)

The substitution of eq. (8.6) into eq. (8.1) and solution of the resulting identity gives

in which

F

r Fo sin F(t) = 1 ,0

wt

C=_·~o~

for 0 s: t ::s; ttl for t > ta

(8.7)

k-mni

(8.2)

Combining (8.4) through (8.7), we obtain the response for 0 s. t ~ ta as

and

, FQsin iiJt y=Acoswt+Bsm W+ k-

(8.8)

°

(8.3)

Introducing the initial conditions y (0) = 0 and y (0) = into eq. (8.8) and calculating the constants of integration A and B. we obtain Folk '. y=.;'-(_, ·)·T[Slll iiJ/

Fit) k

m

~F(I)

+~~

1 -

(wlw) sin UJt]

(8.9)

It is convenient to introduce the foHowing notation: _

7T

w=-

2-rr

t/

,

I )

(e)

W W

Then eg. (8.9) becomes

(b)

(S.lOa) (a) Undamped simple osclllatcr subjected to load F(r). (b) Loading function Fe sin (i)(OSo('5;ld). (c) Free body diagram.

Fig.8.2 F(t)

','" ':,."

236

Structures Modeled as a Slngle·Degree-of-freedom System

237

Response Spectra

After a time lti, the external force becomes zero and the system is then in free vibration. Therefore, the response for {> (d is of the fo~ given by eG:. (8,5) wIth the constants of integration determined from known values of displacement and velocity calculated from eg. (8.lOa) at time 1 ld_ The final expression for the response is then given by ttl _ ~~.;:..- COS 1}-- Sin

)'

)'.s:

2 7T

I'

-I

{i (

~ -

1.1

\1'

.for

{:.?:: t.1

(S.IOb)

It may be seen from eq, (S.lO) that the response in terms of yly" is a function of the ratio of the pulse duration 10 the natural perIod of the system (talT) and of time expressed as liT. Hence for any fixed value of the parameter t,Jr, we can obtain the maximum response from eq. (8.1 0). The plot in Fig. 8.3 of these maximum values as a function of I"IT is the response spectrum for the half~siflusoidal force duration considered in thjs case. It can be seen from the response s;>ectrum in Fig. 8.3 that lhe maximum vaiue of the response (amplification factor) }'IY~I = 1.76 occurs for this particular pulse when (uIT~ 0.8. Owing to the simplicity of the ip.put force. it was possible in this case to obtain a closed soiution and to plot the response spectrum in teons of dimen~ ;:;ion!ess ratios, thus making this plot valid for any impulsive force described by one-half of the sine cycle. However, in general, for an arbitrary input load, we cannot expect to obtain such a general plot of the response spectrum and we normally have to be satisfied with the response spectrum plotted for a completely specified input excitation,

fj--klY -y,~ t}-+ C{y~ Yl1~,-_ _ __ Fig. 8A (a} Damped si'TIple oscillator subjected to support excit
8.2

RESPONSE SPECTRUM FOR SUPPORT EXCITATION

An importanl problem in structural dynamics is the analysis of a syStem subjected' to excitation applied to (he base or support of the structure. An example of such input excitation of the base acting on a d,lInped oscillator which serves to model certain structures is shown in Fig, 8.4, The excitation jn this case is given as an acceleration function which is represented in Fig. 8.S. Tile equation of motion lhat IS obtai!1ed by equaling to zero the sum of the forces in ~he corresponding free, body diagram in Fig. 8.4(b) is m), + cU

----

2.

°1-'- 1. 5

ki i

Fig. 8,3

11--'

~F(r)

'!

0. 0

i

Of)

with the usual substitution

r",

~

~

•

I

Response spectrum for

15

Y"

°

"

,

1.0

ha!f~sinusoidal

force duration (,'

I,

-

(S.l! )

J kim and t = c/c", (cu = 2 Jkm), ~"

(8.12)

Y,ll)

~V M/\~

FoIl<.

F,~

2.,.;p

0'

(tJ;;:

y + 2~wy + w'y = ,.,'y, (I) - 2(,"y, (I)

F(tJ

o / o

r:.-:--

:

-~

:/

.Y,) + k(y - J',):= 0

:

i

V~'·

I

4.0

Fig. 8.5

Acce;e,a~ion

function exei':log the suppOrt of the oscillator in Fig. 8.4.

238

S~ruC!Ufes Modeled as a Single-Degree-of-Freedom System

Response Spectra

Equation (8J2) is the differential equation of motion for the damped oscillator in terms of its absolute motion. A Hlort useful formulation of this o[ob1em is to express eg. (8.12) in terms of the relative motion of the mass w'ith respect to the motion of the support, that is, in terms of the spring defonnation. The relative displacement u js defined as

u= y- Ys

(8.13)

Substitution into eg. (8. (2) yields it

+ 2gwu + uiu =

Ys (I)

(8.14)

The formulation of the equation of motion in eq, (8, 14) as a fur.ction of the relative motion between the mass and the support is particularly important since in design It is the deformation or stress in the "sprIng element" that is required. Besides, the input motion at the base is usually specified by means of an acceieration function (e.g .• earthquake accelerograph record); thus eq. (8.14) containing in the right-hand side the acceleration of the excitation is a more convenient fonn than eq, (S.12) which in the right-hand side has the support displacement and the velocity, The solution of the differential equation, eq. (8.14). may be obtained by any of the methods presented in previous chapters for the solution of one-degreeof-freedom systems. In particular, the solution is readily expressed using Duhamel's integral as

239

and for Ule undamped system

my+ku=O

(8.17)

We observe from eq. (8.17) tbat the absolute acceleration is at aU times proportional to the relative displacement. In particular, at maximum values, the spectral acceleration is proportional to the spectral displacement, that is, from eq. (8.17) (8.18)

where UJ = ik)m is the naturaJ frequency of the system, So. }'mau and SD =: Um,n' When damping is considered jn the system. it may be rationalized that the maximum relative displacement occurs when the relative velocity is zero eu = 0). Hence we again obtain eq. (S.18) relating spectral acceieration and spectra! dispracement. However, for a damped system, the spectral acceleration Sa, is not exact[y equal to the maximum acceleration, although in general, it provides a good approximation. Equation (8.18) is by mere coincidence ~he same as the relationship between acceleration and displacement for a simple harmonic motion, The fictitious velocity associated with the apparent harmonk motion is the pseudovelocity and, for convenience, its maximum value S~ is defined as the spectral velocity, that is (8.19)

(8.15)

8.3 TRIPARTITE RESPONSE SPECTRA It is possible to plot jn a single chart using logarithmic scales the maximum response in terms of the acceleration, the relative disp!acement, and a third quantity known as the relative pseudovelocity, The pseudovelodty is not exact:y the same as the actual velocity, but it is closely related and provides for a convenjent substitute for the true veiocity. These three quantities-the maximum absolute acceleration, the maximum relative displacemeut, and the maximum relative pseudovcJocity-are known, respectively. as the spectral acceleration, spectra! displacemer..t. and spectral velocity, It is significant that the spectral displacement SDz that is, the maximum relative displacement, is proportional to the spectral acceleration Sa: the maximum absolute acceleratior:. To demonstrate this fact. consider the equation of motion, eq. (8.1 I), which, after uSing eq. (8.13), becomes for the damped system

mY+Cli+ku=O

(8.16)

Dynamic response spectra for sjngle-degree-of-freedom e]astjc systems have been computed for a number of input motions. A typical example of displacement response spectrum for a sing!e-degree-of~freedom system subjected to support motion is shown in Fig, S,6. This plot is the response for the input motion given by the recorded ground acceleration of the 1940 E1 Centro earthquake. The acceleration record of this earthquake has been used extensively in earthquake engineering investigations. A plot of the acceleration record for this eartbquake is shown in Fig. 8.7. Until the time of the San Fernando, California earthquake of 1971, the EI Centro record was One of the few records available for tong and strong earthquake mot10ns. In Fig. 8.8, the same type of data that were used to obtain the displacement response spectrum in Fig. 8,6 are plotted in tenus of the spectral velocity, for several values of the damping coefficient, with the difference that the abscissa as well as the ordinate are in these cases plotted on a logarithmic scale. In this type of plot, because of eqs. (8.18) and (8.1 9), it is pOSSible to draw diagonal scales for the displacement sloping 1350 with the abscissa, and for the acceleration 45"" so that we can read from a Single plot values: of spectral acceleration, spectral velocity, and spectral displacements.

240

Structures Modeled as a Single-Degree-of-Fieedom System

] LI 0

,

U

AI

\~\,

/"

,

0

2. 0

~

1.0

<

E

~

''yfv.,

,

•• x

~

o.5,

!

'u~

I

>

, 1

0,02

O.OS 0.1

tal

\8 \J]

V

!

i

,

·5

2O

i

___ M

o. 1

1.S ~

!

o.2 -

0.05

I

\

i i

!

i

! '~I i '

I

0.2

Fig.8.8 Response spectra for elastic system for :he t940 EI Centro earthquake {f:oom Blume et al. 1961.)

Natural frequency, I:pS

Fig. 8.6 Displacement response spectrum for elastk system, subjected to the ground motion of 1940 El Cemro earthquake, (from Design of Multisfory Reinforced Building for Earthquake MOlions by J. A. Blum, N. M. Newmark, and L R Corning, Portland

For constant values of So, eq, (8.20) is the equation of a straight !jne of iog versus log/with a slope of 45", Analogously, from eq. (8,19)

To demonstrate the construction of a tripartite diagram such as the one of Fig, 8.8, we write eq. (8.19) in terms of the natural frequency / jn cycles per second (cps) and take the logarithm of the tenus, so that

S, log S,

coSo = 2'iT/SD log! ~ Jog (2OTSD)

(8.20)

Y/9 (

c

_:~h~I\~httl~tJ.i~i~/v.,~I/Wii~~f,\JlrM~'N*/~"~ ~q o

11

:

5

"I,

10

'~'~-!'~'~~"-';;:'~'~'.L-J....-.i.~~'~-'-Tj(m:.~i: IS

M

25

Fig. 8.7 Ground acceleration record for El Centro, California earthquake of May 1g, 1940 north-south compOnent.

S~

5,,= S" ~~_ -

Cement Association 1961.)

"i

241

~-'

I

i

~

'"~

I

i

,

I

i .~

I

Response Spectra

Jog S"

w

2r.f

So

- log! ~ Jog27r

(8.21)

For a constant value of SI>' eq. (S.2i) is the equation of a straight Hne of log 5<1 versus log f with a slope of 135'-'.

8A

RESPONSE SPECTRA FOR ELASTIC DESIGN

In general, response spectra are prepared by calculating the response to a specified excitation of single-degree-of-freedom systems with various amounts of damping. Numerical inlegration with short lime intervals are applied to caJcuiate the re~ sponse of the system. The step~by~step process is continued until the total earthquake record has been completed. Tne largest vatue of the fcnclion of jnteresl is recorded and becomes the response of the system to that excitation. Changing the parameters of the system to change the natural frequency, we repeat the: process and record a new r:laxirnum response. This process is repeated until all frequencies of interest have been covered and the results ploued. Since nO two earthquakes are alike, this process must be repeated for all earthquakes of interest

242

Response Spectra

Struclure$ Modeled as a Single-Oegree-or-Freedom System

2.
Until the San Fernando, California earthquake of 1971, rJlerc were few recorded strong earthquake motioT.;s because there were few accelerometers emplaced to measure them; the El Centro, California earthquake of 1940 was the most severe earthquake recorded and was used as the basis for much analytical work. Recently, however, other strong earthquakes have been recorded. Maximum values of ground acceleration of about 032 g for the El Centro earthquake to values of more than 0.5 g for other earthquakes have been recorded. It can be expected that even larger vaJues will be recorded as more instruments are placed doser to the epicenters of earthquakes, Earthquakes consist of a series of essentially random ground :notions, esually the north-south, east-west. and vertical components of the ground acceleration are measured. Currently, no accurate method is available to pre~ diet the particular motion that a site can be expected to experience in future earthquakes. Thus it is reasonable to use a design response spectrum which incorporates the spectra for several earthquakes and which represents a kind of "average'; response spectrum for design. Such il design response spectrum is shown in Fig. 8.9 normalized for a maximum ground acceleration of 1.0 g.

freque!'cy, cps

Fig. 8.10 Elastic design spectrum normalized to peak ground acc~leration of LO g fer 5% damping. (Kewrnark and Han 1973,)

This figure shows the design maximum ground motion and a series of response spectral plots corresponding to various values of the damping ratio in the system. Details for the construction of the basic spectrum for design purposes are given by Newmark and Hall (973), who have shown that smooth response spectra of idealIzed ground motion may be obtained by amplifying the ground motion by factors depending on the damping in the system. 1n general, for any given site, estimates might be made of the maximum ground acceleration. maximum ground velocity, and :naxjmum ground displacement. The lines representing these maximum values are drawn on a tripartite logarithmic paper of which Fig. 8.10 is an example. The lines in this figure are shown for a maxJmum ground acceleration of 1.0 g. a velocity of 48 in/sec. and a dlsplace ment of 36 in. These values correspond to motions that are more intense than N

Fig. 8.9 Basle design spectra normalized to 1.0 g. (From Newmark and Hal! 1973.)

244

Response Spectra

Siructures Modeled as a Slngle-Degree-of-Freadom System

TABLE 8.1

Relative Values Spectrum AmpliHcaUon Factors ....

-~

--

..

Amplification Factors

Percent Damping

Displacement

Velocity

o

2.5

0.5

2.2 2.0 1.8 L4 1.2 Ll

4.0 3.6 3.2 2.8 1.9

I 2

5 7

10 20

---

AcceleratJon

6.4 5.8 5.2

43 2.6 1.9

1.5 L3 Ll

LO

1.5

those gene~al!y expected in seismic design, They are, however, of proportional magnitudes which are generally correct for most practical applicarions. These maximum values norrnnlized for a ground acceleration of LO g are simply scaled down fOf olher than 1.0 g acceleration of the grouild, Recommended amp! ificalion factors to obtain the response spectra from maximum values of the ground motion are given in Table 8. L For each value of the damping coefficient, Lhe amplified displacement lines are drawn a1 the left, the ampli~ fled velocities at the top, nnd the amplified acceleration at lhe right of the chart. At a frequency of approximately 6 cps (Fig. 8.9), the amplified accel~ eration regron line intersects a line sloping down toward the maximum ground acceleration value at a frequency of abot:.l 30 cps for a system with 2% damping. The lines corresponding to other values of damping are drawn para;lel to the 2% danlping line as shown in Fig. 8.9. The amplification factors in Table 8.1 were developed 00 the basis of earthquake records available at the time. As new records of more recent earthquakes become available, these amplification factors have been reca[~ culated, Table 8.2 shows the results of a statistical study based on a selection TABLE 8.2 - "..

-

.. -"~""

..

.. -"

Displacement -~

Factor 2

1.691 1465 L234 ..

~-~.-.~-~.

"~

..

-~.-"~

Velocity

Example 8.1 Example 8.1. A Slructure rr:ode~ed as a single-deg;:ee-of~ freedom system has a natural period, T = 1 sec Use l~e response spectral method to determine the maximum absolute acceleration, the maximum relative displacement, and the maximum relative pseudoveJocity for: (3) iJ. foundation tT'.Otion equal to the EJ Centro earthquake of 1940, nne (b) the design earthquake with a maximum ground acceleration equal to 0.32 g. Assume 10% of the critical damping,

Solution: (a) Fro:r. the response spectra 1n Fig. 8.8 with 1 IT == 1.0 cps, corresponding to the Cilrve labelec ~ = O. J 0, we read on the three scales the following values:

SD Su

--~

Factor

0.828 0.630

2.032 1.552

0481

1.201

- -..

..

3.3 in 18.5 in Isec

(b) From the basic design spectra in Fig. 8.9 with frequency f= 1 cps and 10% critical damping. we obtain after correcting for 0.32 g maximum ground acceleration in lhe following resu;is: S" = 9.5 X 0.32

3.04 in

So = 60 X 0.32

= 19.2 in Isec

So

= 0.95 X 0.32 = 0.304 g

..

~~-

Standard Deviatjon

=

Acceleration

.. -"-"~ .. -"

Percent

5 10

of 10 strong motion earthquakes. The tab:e gives recommended amplification factors as well as the corresponding standard deviation val'Jes obtained in the study. Tbe relatively large values shown in Table 8.2 for the standard deviation of the amplification factors ;::rov:des f'Jither evidence on the uncertainties surrounding earthquake prediction and analysis. The response spectra for designs presented in Fig. 8.9 bas been constructed using the amplification factors shown in Table 8,1,

Spectral Amplification Factors and Standard Deviation Values" -"-"~"""-~

245

..

~--~--

Standard Deviation

Factor

Standard Deviation

0.853 0.605 0,432

3.075

0.738

2.281 1.784

0.502 0.321

"--"--"--"--

~ Newmark, N. M., and Riddell, R., Inelastic Spectra for Seismic Design: Seventh World

Eartbquake Conference, Instanbul, Turkey, Vol. 4. PI'. 129-136. 1980.

8.5

INFLUENCE OF LOCAL SOIL CONDITIONS

Before the San Fernando earthquake of 1971, earthquake accelerograms were llrnited in number, and the majonty had been recorded on BIluvium, Therefore, it is only natural that the design spectra based on those data, such as those suggested by Housner (1959) and Newmark~Hali (1973), mainly represent alluvial sites. Since 1973. the wenith of information obtained from earthquakes worldwide and from scbseqt.:ent studies have shown the very significant effect that the Iocnl site conditions have on spec~ral shapes.

Response Spectra

S!ructures Modeled as a S:ngle,Degrcc·of-Freedom System

246

247

'.--~

J

~

NO .... S(R or Rf(.:OROS AIIALY1(O; SPECTRA FO~ 5~ OA~PING

10TAL

A

, 1"" -

I

, ~

/ S O f ' T TO MeO IUr' CLAY ANO

\

I

104

sAl.O - !5 RECOReS

,

"

-.........,./OEEP COH£SION~£SS SOILS /~ (>250 FT) ~ 30 il.HOiH!S

'-,,> . ."" '~ -. .... '

STl FF 50 J LS «200 FT) - 31

~ECOROS:

PERI DO, sec

Fig. 8.11 Average acce),i:f
Soil Type I (Rock and Sliff Solis):

S, = I + 10 T

for

O
SA =2.5

for

0,15
S.

for

T> 0.39 sec

(8.22)

Fig, 8.12 Normaltz.ed design $pectHI shapes contained in Uniform Building Code. (ICBO 1994.)

Soil Type III (Soft to Medium Clays and Sands):

SA

=i

+75 T

for

SA =2.5

for

0.2
5, = 2.28SIT

for

T>O,915 sec

(8.24)

where SA in the spectral acceleration for 5% damping nonnalized to a peak ground acceleration of one g, and T is the fundamental period of the building. It should be noted that values obtained from the UBC speclraj chart of Fig. 8,12, Or alLematively, calculated with eqs. 8.22 to 8.24. are too conservative. In actual design practice, these values are scaled down by lhe structufil1 fac!or Rw (Chapter 25) with values between 4 and 12, depending on the type of building. However the UBC establishes limitations for tbee resultant base shear force obtained by rhe dynam~c method reiative to the base shear given by the static method of a:Jalysis, as explained in Chapter 25. j

= O.9751T

Sail Type II (Deep CohesionJess or Stiff Clay Soils): for

0
for

o 15 < T;;; 0.585 sec

for

T> 0.585 sec

\ From Re{;ommented Lateral Force Requiremenl

36-C

~t1d

(8.23)

Tentative Commentary SEAOL·OO, P.

8.6

RESPONSE SPECTRA FOR INELASTIC SYSTEMS

For cenain types of extreme events such as nuclear bla.st explosions or strong motion earthquakes, it is sometimes necessa.ry to design structures to withstand strains beyond the elastic limit. For example, in seismic design for an ear1hquake of moderate intensity, it is reasonable to assume elastic behavior for a weU·designed and -cons~ructed structure. However, for very strong motions.

248

Slructures Modeled as a Sing!e~Degree-of-Ffeedom System

Response Spectra

249

A.MtOfHlg

lQrce

R

?l,m,e

Y'lrl~Y

"-r---k'.,

k

Y, Y",.,

Y

Elastie p.~ ?l~~tic

fbi

Fig. 8.13 Force-dispJacement relationship for an elastoplastic singJe~degree-of-free­ dom system.

this is not a realistic assumption even for a well-designed strJcture, Although stnlctures can be designed to resist severe earthquakes, it is not feasible economically to design buildings to elastically withstand earthquakes of the greatest foreseeable intensity. In order to design structures for strain levels beyond the hnear range, the response spectrum has been extended to include the inelastic range (Kewmark and Hall 1973), Generally, 'he elastoplastic relation between force and displacement, which was discussed in detail in Chapter 7, is used In structural dynamics. Such a force-displacement relationship is shown in Fig. 8.13. Because of the assumption of elasloplastk behavior, jf the force lS removed prior to the occurrence of yielding. the materia! wE] return along its loading Hne to the origin. However, when yielding occurs at a displacement y/. the restoring force remains constant at a magnitude R" If the displacement is not reversed, the displacement may reach a maximurr: value Ym;!:;- if, however. the displacement is reversed, the elastic recovery foHows along a Hne parallel to the initi2J line and the recovery proceeds elastically until a negative yje!d value R<; is reached in the opposite direction. The preparation of response spectra for such an inelastic system is more difficujt than that for elastic systems, However, response spectra have been prepared for several kinds of input disturbances. These spectra are usually plotted as a series of curves corresponding to definite values of the ductility ratio }Jr. The ductility ratio fL is defined as the ratio of the maximum displacement of the structure in the ihelastic range to the displacement corresponding to the yield point yy, that is,

N~tV(~llJef;od

Fig. 8.14 Response spectra for undamped eiastoplutic system for the 1940 EI Cc.ntro earthquake. (From mume et aL t 961.)

(K25)

The response spectra for an undamped singie-degree-of-freedom system subw jected to a support motion equal to the EI Centro 1940 earthquake IS shown

T, Sl:e

.f

J

.~

i (.

t, riC,

"

in Fig. 8.14 for several values of the ductility ratio. The tripartite logarithmic scales used to plot these spectra give simultaneously for any singk-degree-offreedom system of natural period T and specified ductility ratio fL, the spectral values of displacement~ velocity, and acceleration. Similarly, in Fig, 8.~5 are shown the response spectra for an elastoplastic system with 10% of critical damping. The spectral velocity and the spectral acceleration are read directly from [he plots in Figs. 8.14 and 8.15, whereas the values obtained for the spectral displacement must be multiplied by the ductility ratio in order to obtain the correct va!ue for the spectral dispJacement. The concept of ductility ratio has been associated mainly with steel struc~ tures. These structures have a load-deflection curve that is often approximated 3S the elastoplastic curve shown in Fig. 8.I3(b). For otber types of structures, such as reinforced concrete structures or masonry shear walls, still. conveniently. the load-
Structures Modeled as a Sing!e-Degree-of-Freedom Systetr.

250

~

Response Spectra

251

2

! .!1 <0' J';

l

i 0.1

Fig. 8.15

0.2

0.3 0.4 0,5

2

3.

4

5

Response spectra for elastop:atic system with 10% critical damping for tbe

1940 El Centro earthquak.e, (From Blume et aL 196L)

masonry shear walk The selection of ductility values for seismic design must also be based on the design objectives and the loading criteria as well as the risk level acceptable for the structure as it relates to its use. For reinforced concrete structures or masonry walls, a ductility factor of to to 1.5 seems appropriate for earthquake design where the objective is to limit damage, In other words, the objective of Emit damage requires that structural members should be designed to undergo little if any yielding. When the design objective is to prevent cOllapse of the structure, ductilities of 2 to :} are appropriate in this case.

8.7

RESPONSE SPECTRA FOR INELASTIC DESIGN

In the preceding section of this chapter. we discussed the procedure for calculating the seismic response spectra for elastic design. Figure 8,9 shows the elastic design spectra for severa] values of the damping ratio. The same procedure of constructing a basic response spectrum that consoHdates the "average" effect of

Frequency, cps

Fig. 8.16 Inelastic desjgn spectrum nonnaHzed to peak ground acceleration of 1.0 g for 5% damping and ductility factor j.J.. = 2,0, (NOTE: Chart gives directly Su and Sit; however, displacement va[nes obrainec from the chart should be multiplied by ).J. to obtain

SD')

several earthquake records may also be applied to design in the inelastic range, The spectra for elastoplastic systems have the same appearance as the spectra for elastic systems, but the curves are displaced downward by an amount that is related to the ductility factor JL. Figure 8.16 shows the construction of a typical design spectrum currently recommended (Newmark and Hall 1973) for use when inelastk action is anticipated. The elastic spectrum for design from Fig. 8.9 corresponding to the desired damping ratio is copied in a tripartite loga.rithmic paper as shown in Fig. 8.16 for the spectra corresponding to 5% damping, Then lines reduced by the specified ductility factor are drawn parallel to elastic spectral lines in the displacement region (the left region) and in the velocity region (the central region), However, in the acceleration region (the right region) the recommen-

252

Structures rv10deled a.s a SJngle-Degre8-of-Freedom Sys':em

Response Spectra

ded reducing factor is . This last line is extended up to a frequency of about 6 cps (point P' in Fig. 8,16). Then the inelastic design spectrum is compieted by drawing a line from this last point P' to point Q, where lhe descending Hne from point P of the elastic spectrum intersects the line of constant acceleration as shown in Fig. 8.16. The development of the reduction factors in the displaceme,nt and velocity regions is explained with the aid of Fig. 8.17(a). This figure shows the force--displacement curves for elastic and for elastoplastic behavior. At equal maximum displacement Ymn for the two curves, we obtain from Fig. 8.l7(a) the following relationship:

However, in the acceleration region of the (esponse speclrum, a ductility reduction factor f.L does not result in dose agreement with experimental data a.~ does the recommended reduction factor !2J.L - 1 . This factor may be rationaily obtained by establishing the equivalence of the energy' between the elastic and the jnelastic system, In reference :0 Fig. 8.17(b) this equivalence is eSlab.lished by equating the area under elastie curve "Oab" with the area under the jnelastic curve '
F,.yy

=_._. I" - } ' ) 2 +FyVp..),

.'

= FE

(8.26)

fJ.

(8.28)

I

where FE and Fy tire loe elastic .md inelastic forces corresponding respectively to the maximum elastic displacement)lF. tinct 10 the maximum inelastic displace~ ment Yp. and wbere y., is {he yield displacement. The substitution of }'£ = Ff)k, )'J = FJk, and YP- = p,.y;- into eq. (8.2&) gives

or

F

253

where FE is the force corresponding to the maximum displacement J'm~. in the e!astic CUrve and F" is the force at the yield condition. Equation (8.26) then shows that the for~e and consequently the acceieration in the dastoplastic system is equa! to the corresponding value in the elastic system reduced by the ductility factor. Therefore, the spectral acceleration S" for elasloplastic behavior is related to the elastic spectral acceleration SaC as

Or

Thus,

(8.27)

Fe Fy = .."'('= ... ~= \' 2J-t .--

and, consequenlly,

-;1 /

.L1k

,

---;1 Iflelulic respmlSe

F,

a

,., Fig. 8.17

!nela~lk ~sponll¢

/

.L1k

I

/'

/

-

-

I

c / 1 d 7-,f---+-_.!.--\,'-

I

I

I

I

b

Thus, the inelastic spectral acceleration io the acceleration region is obtained by reducing the elastic spectral by I;,e factor fij;.~_ The inelastic response spectrum thus constructed and shown in Fig. 8.16 gives directly (he 'values for the spectral acceleration S(I and spectral velocity SUo However, the values read from this chan in a displacement scale must be multipiied by the ductility factor J-t to obtain the spectral displacement Sf)' Inelastic response spectral charls for design developed by the procedure eX.plained are plesented in Figs. 8.18, 8.19, and 8.20 correspondingly to damping factors ~;. 0,5%, and 10% and for ductility ratios J-I- = I, 2, 5, and 10.

,e

Yc

Elastic and inelastic force-displacement curves. (a) Dispiacemem and vel-

ocity regions. (b) Acceleration region.

i

., i l

::

254

Structures Modeled as a

Sjngle-Degfee~ol~Freedorn

Syslem

Response Spect:.3

255

.' Frequ(!ncy. Cj}:1

FI4;quency. cp!j

,

f;-

Example 8.2 Calculate the response of the

f,

sjngle-degree~of-freedom sys~

tern of Examp]e 8.: > assuming that the struchlre 1s designed to withstand seismic motions with ar; eiastopimaic behavior having 11 ductility ratio jJ- = 4.0. Assume damping equal to 10% of the cri[ical damping. (a) Use the :"esponse spectra of lhe EI Centro earthquake. (b) Use the design response spectra.

I,

Fig. K 19 Inelastic design spectra normalized to peak ground acceieraJion of 1.0 g for 5% damping. (Spectral y"lues Sa for acceleration and S" for velocity art direcdy obtained from Ihe graph. However, values of SD for spectra! displacement should be amplified by the ductility rutio ,u.)

Fig. 8.18 Undamped inelastic oe:;.ign spectra normalized to peak grounJ acce;eralion of 1.0 g. (Spectra! values 5<1 [or acceleration and S~ (or velocity are obtained direcUY frolD the graph. P.owever, values of S9 for spectral displacement should be amplified by the ductility ~atio jL.)

.

t,

The factor 4.0 IS required in the calculation of So since as previously noted the spectra plotted in Flg. 8.15 me correct for acceleration and for pseudovelocilY, but for displacements it is necessary to amplify tile values read from the chart by the ductiH[y ratio.

i

I t

,I

(b) Using the inelastic design spectra with 10% damping for design in Fig, 8,20, corresponding to 1 cps, we obtain the following maximum values fo: the response:

Solution: (a) From the response spectrum corresponding to 10% of the critical damping (Fig. 8.15), we read for T = l sec and [he curve labeled

S[J

= 3.84 in

:1

'" = 4.0

So = 15.6 X 032 X 5,00 in/sec

I

Slj = 6.2 in/sec and

So

3.0 X 0.32 X 4.0

0.3 x 0.32

= 0.096 g

'I

I

I'

I

[I I

255

S!ruclures Modeled as a Single·Degree·Q!·Freedom Syslmr.

Response Spectra

8,8

257

PROGRAM 6-SEISMIC RESPONSE SPECTRA

The computer program described in this chapler serves LO calculate elastic response spectra in ierms or spectral displacement (maximum reiafive displacement). spect:-al velocity (maximum relative pseudo velocity), and spectral acceleral~on (maximum absolute acceleration) for any prescribed time-acceleration-selsrr.ic excitation. The response is calculated in the specified r
Solution.:

"'n · '""

!-IUXet;R DF ron,Ts C<:FlIHNC THf: EXCtTh1'lill' INITl"!.; fllEQUENCY

\C.I'.".)

. . MAL FREQUEtlCY

(C.

0;'

r. S_)

ON1PlNC RA1'lC

"

" ·· ,"

'nNe STLI" IN'tt(;i!I\T10,",

,'CCf:LER1.Ti0l1 GAAV:;1"(

~'II\;O,

(CI".S

Fig. 8.20 Inelastic: design spectra norm;Jlh:cd to pcnk ground acc:e!cnlliou of 1.0 g' for 10% damping. (Spectral values S" fo:- acceieralion and Sv for velQcity are directly obtained from the graph. Howcycr, vah.;es or Sf) for spr.ctral displacement should be ali'.plified by lhc ductility Hillo p..)

P1SI'"

(IN)

$.ttcT

24.65

, tJ

O. I')

""

2~

G JS

O. '0 O.

~

S

(/.50

" u " " " ." "n

11.t)2

1>. 1~

VEt-X

( INfSEC]

:2:\ .86

c. }O

SI'CCT, ACC.

'>0

" "" ""

20 .31

25. SS

1.41 e.54 jJ.

$:

H,

"

n n.n

"

26. $I' lc. ;)6

';;!i .02

""

6'1.25

1S.17

LaO

,

1S

127.4()

2.41

n.n

lU_S7

.t~

25.H

)16.111

".H

"

"

~:).

H

&1.

$~

]. CO

2

L.50

1.17

H, U

H9.~()

) .oc

O. 7~

1 J. 96

261.1.\

J. 'jU

O.

$)

1l.H

~ ...

coo

\lA, 0.31

lo,a

15'1.

9.1>&

lSS.

$.0

to.oo

".JO I). OS

294.

15.00

0.02

L,--'

" '" "

20.CO

1).l)l

". SOl

120.01'

4. ~o

1."0

,HI>

(,lOsee/SECl

1. 50

S _flO

"

Sf>E:(:T

1)'1!)

O.

somewhat different than those obtair:ed from the respo:lse spectrum of the EI Centro earthouake of J940, Also, if we compare these rest.;lls ror the elastoplaSlic benavi~r Wilh the results in Example 8.1 for the elastic slructure, we observe that the maximum relative. disp!aceme:lt has essentially the same magdtude whereas the acceleration and the reli'ltive pseuGovelocity are appreciably Jess. This obSe;vtllion is in general true for «roy structure when inelas~k response is compared with the response based On elastic behavior.

G

lJ,05

0.20

As can be seen, these spectfJl values based On the design spectrl·.m are

,

·

, · ," ", · "

nlEQU!':1ICY INCREN;;;l4i IC.f.".l

\. )!i

""

liS .0.4

I

III

I

N

~

~.

TABLE 8.3

Digitized Values of the Accel~ration Recorded 10r the First Ten Seconds for the EI Centro Earthquake 011940 ~~"

.~~-~~

Ace,

Time (sec)

(ace. g ')

0.0000

O.OlOS

0.2210 0.3740 06230 0.7890 0.9410 10760 1.3840 1.5090 1.8550 2.2150 2.4500 2.7080 3.0680 3.3860 3.6680 4.0140 4.3140 4.6650 5.0390 5.3020 5.5100 5.8000

0.0189 0.0200 0.0094 - 0.0387 - 0.0402 .• 0.0381 - 0.0828 - 0.1080 0,1428 0.2952 0.2865 0.1087 0.0520 0.1927 0.0365 0.0227 0.1762 - 0.2045 0.0301 0.1290 -0.1021 - 0.0050

TABLE 8.3

Time (sec)

Ace.

Time

(ace. g ')

(se.c)

0.0420 0.2630 0.4290 0.6650 0.8290 0.9610 10940 1.4120 1.5370 1.8800 2.2700 2.5190 2.7690 3.1290 34190 3.7380 4.0560 44160 4.7560 5.1080 5.3300 5.6060 5.8090

0.0020 0.0001 .. 0.0237 0.0138 - 0.0568 0.0603 ~. 0.0429 - 0.0828 - 0.1280 0.l777 0.2634 - 0.0469 - 0.0325 0.1547 - 0.0937 -0.0736 - 0.0435 0.1460 0.060S 0.2183 0.1089 0.0141 - 0.0275

0.0970 0.2910 0.4710 0.7200 0.8720 09970 1.1680 14400 1.6280 1.9240 2.3200 2.5750 2.R930 3.2120 3.5300 3.8350 4.1060 4.4710 4.S31O 5.1990 5.3430 5.6900 5.S690

Ace. (ace. g ')

Time (sec)

-

- ------

Time (sec)

Ace. (ace. g ')

0.0159 0.0059 00076 0.0088 0.0232 - 0.0789 0.OR97 - 0.0945 0.1144 0.2610 - 0.2984 0.1516 0.1033 0.0065 0.1708 0.031 I il.0216 - 0.0047 - 0.2733 0.0267 0.0239 -0.)949 - 0.0573

0.1610 0.3320 0.5810 0.7400 0.9020

Ace, (ace, 8 ')

Time (sec)

]0660 1.3150 1.4810 1.7030 2.0070 2.3950 2.6520 2.9760 3.2530 3.5990 3.904il 42:>20 4.6180 4.9700 5.2330 5.4540 5.7730 5.8830

Ace. (ace. g '; ~

00001 0.0012 0.0425 - 0.0256 - 0.0343 - 0.0666 0.1696 0.0885 0.2355 - 0.3194 0.0054 0.2077 - 0.0803 0.2060 0.0359 0.1833 - 0,1972 0.2572 0.1779 0.1252 0.1723 0.2420 .- 0.0327

(continuation} ~~~-.---

~

v

~

Time

Ace.

Time

(sec)

(ace. g ')

(sec)

5.9250 61320 62290 6.3820 6.5200 6.6030 67280 0.8520 71210 72260 'i4250 76000 7.7520 7.9600 8.1260 8.2780 85330 8.8180 8.9560 9.1500 9.4410 9.8150 )0.0200 )0.1500

0.0216 0.0014 - 0.0381 - 0.0162 0.0043 - 0.0170 0.0009 0.0022 0.0078 0.0576 0.0186 - 0.0628 - 0.0054 - 0.0140 0.0260 0.Q305 - 0.0344 .. 0.0028 OJ 849 0.1246 - 0.1657 - 0.0881 - 00713 0.0024

5.9800 6.1740 6.2790 6.4090 65340 6.6450 6.7490 6.9080 7.1430 7,2950 74610 76;10 7.7940 7.98'10 8.1660 8.3340 8.5960 8.8600 9.0530

92530 9.5100 9.8980 10.0500 10.1900

0.0108 0.0493 0.0207 0.0200 00040 00373 0.0288 0.0092 - 0.0277 - 0.0492 - 0.2530 0.0280 0.0603 - 0.0056 - 0.0335 0.D2.46 -0.0104 0.0233 0.1260 - 00328 0.0419 0.0064 .. 0.0448 0.0510

6.0130 6.1880

6.31.60 6.4590 6.5620 6.6860 6.7690 6.9910 7.1490 7.3700 7.5250 7.6690 7.8350 8.0010 8.1950 8.4030 8.6380 8.8820

9,0950 9.2890 9.6350 9.9390 10.0800

0.0235 0.0149 - 0.0058 - 0.1760 .~ 0.0099 0.0457 0.0016 - 00996 0'{1026 0.0297 0.0347 0.0196 - 0.0357 0.0222 - 0.0128 0,0347 - 0.0260 0.0261 0.0320 - 0.0451 - 0.0936 - 0.0006 .. 0.0221

6.0850 6.1980 6.3680 6.4780 6.5750 6.7140 6.8110 7.0740 7.1710 7.4060 7.5720 7.6910 7.8770 8JI700 8.2230 8.4580 8.7350 8.9150 9.1230 9.4270 9,7040 9.9950 10.1000

Ace.

(ace. g ') -0.0665 - 0 0200 - 0.0603 0.0033 0.0017 0.0385 0.0113 0.0360 0.0272 00109 0.0036 0.0068 0.0716 00468 0.066) - 0.0369 0.1534 - 0.0022 0.0955 0.)301 0.0816 0.0586 0.0093

------_...

~-----

---

---.-.,.=.~~.

Structures Mode!ed as a Single-Oegree-o1-Frcedom System

260

8.9

Response Speclra

RESPONSE SPECTRA USING COSMOS

"'f-:'- r

JIS

Example 8.4 Use the Program COSMOS to genenHe response spectra for :.elastic system subjected to the 1940 EI Centro earhqunke. Assume damping equal (Q 10% of the crillcsl dampir.g.

I

I

I

..

,,-1 \ I (\ , i \/j

I

'\

i

I

....

......

Ii .)2

,1·

.$ .IS

&.tB

, -&

i

'

i

,

I

-9.15

,

<

, I,

,..

,I

I I

:,VY

...............

,

,

i ,

SL:i

I

15 87.:;

I 12 • ~

1?S

[0

the XY plane: >

VIEW

(2) Define the XY plane at Z ~ 0: G~OHE~RY

~ GRID PLANE, Z, 0, 1

>

PLANE

(3) Generate, a curve from 0, 0, 0, to 1. 0, 0,". i

GEOMETRY ) CURVES ) CRPCORD CRPCORD, 1, 0, 0, 0, 1, 0,

~

.......

•------J

i

II

"

I

1'1

Solution: The analysis is performed ~!iing n single spri:Jg element wHh one concentrated mass element The following commands are implemented in COSMOS:

I

• E! '- vr~'._",

I

..

:

Fig. S.22 Ac::elel:likm response spcctwrn ;){ Ihe base of an e!;istic syste:n subjected 10 ground nlOllOO of 1940 E! Cenlro carthquilke,

,

!

i ~

:

DISP::.. AY j VIEW_PAR VIEi'<, 0, 0, 1, 0

':

,A • ¥J .Ia

i

"-.. :

,

FR[O lR.d/S'!)cl

~rv~ I

,

!L'2~

v

I

I

-~

12. ::;

(l) Set view

,.. 1-

····r·--1

'=IH=-:. I

IN

!

lSG ,

Solafion: In generating response spectra, the program, COSMOS bas tbe capability to calculate the spectra at any specified coordinate of [he structure, Because of this capability. it is necessary to model the structure and calculate the natUfal frequencies before requesting the generMior; of ~he response spectra at a specified coordinate. Example 8.~ is implemented in COSMOS as a spring. mass system with spring constant I and mass value 1. Then the natural frequency is calculaled and the response spectra at the base of this s),slem (coordin:He i) generated. Time-acceleration values for EI Centro earhquake (Table 8.3) are inpot in COSMOS through a previously prepared file that 'contains 186 digiH7-ed point of acceleration curve recorded for El Centro earhquake of 1940 for the first 10 seconds. Figure 8.2i shows the plot of this record obtained using COSMOS. The acceleration response spec:ra ge.nerl!;ed by COSMOS for frequencies :n the range 0.314 rad/sec (OJ cps) to 125.7 rad/sec (20 cps) in shown in Fig. 8,12.

,

26i

(4) Define element group J using the SPRING element formula::on with

-

two nodes:

S.S

"

lillE

Fig. 8".21 Ground acceleration for the nest lO scc recorded for the po!.ent of {he 1940 El Centro earthquake.

°

PROPSETS } BGROUP EGROO:?, 1 SPRING, 0, 2,

1,

0,

0,

0,

0

(5) Define real COnstant for spring clement: .1::= I (lb/in) nQl1h~sou!h

com·

PROPSETS ) RCONST RCONST, 1, 1, 1, 1, 1

262

SlrUClures Modeled as a Single"Degree-of-Freedom Systerr.

use default values for all integra~jon parameters; and request printout of relative disp:acemem and relative velocities:

(6) Generate one spring element along curve 1: MESHniG ),

l'CCi{,

L

PARAM_MESH :> l-CCR 1, 1. 2, 1, 1

PROPSS'I'S '

EGROUP

MASS,

0,

J,

(8) Define renl constant for mass elemenr: m

=l

2,

?ROPSETS

0,

0,

0,

C,

)

POST~DYN

ANALYSIS ) PD~ATYPE,

(7) Define element group 2 using the MASS elemenr formulation: EGROUP,

263

Response Speclra

2.

1,

2000,

?D_A?YPE

Ot

.01,

0,

0.5,

C.25,

0

(15) Prepare file ELCENTRO,GEO in a formal suilable for input in COSMOS, cont<'iining the values of the ground acceleration for 1940 EL CE~TRO EARTHQUAKE shown in Table 8.3, aod lisl the firsl five lines of lhis file

0

(lb" sec2 /in):

P!)_Cl:RD£F,

RCONST RC:ONST', 2, 2, 1. 7, 1, 0, 0, 0, 0, 0, 0 >

.J10S,

.o,n .002, .0)7, .015'S,

.H1.

-.0001

.221,

.018S . . 263,

.0001, .291, .0059, .3J2,

S,

. }?4,

.02,

0::37,

PD_CUKDEF,

(9) Generate one mass element at point 2:

G,

1,

PD_C::K::n::? . 1. S, PD_CCRDEF, PD_CURDEfl,

"

.6:n,

13,

"P • .

.42S,

0;)94,

769, -,0331,

,665,

,471, .0076, .Sill,

.013S,

.72. -.00S6,

-.O~12

.0425

.71, -.0256

,S29. -.0568 . . 312. -.0232 • . 902,

-.tlJ'D

MESHING } PAR.l..,"LMESH

:1_?T, 2, 2,

1

(16) Define dynamic forcing function as time-dependent hase acceleration in the X direction, Read forCing function curve from file "ELCENTRO.GEO":

(10) Merge nodes: MESiUNG ,. NODES, 1"mERGE NMERGE. 1, 3, L 0.0001,

o.

C,

(l:) Apply constraints in all degrees of freedom at node I, and aU degrees

of freedom except UX at node 2:

(l2) Set the options for the frequency analysis to extract ~ frcquency using the Subspace Iteration Method with .t: maximum of 16 jterations. and [un tte freqJency analysis:

0,

0,

U'l'ILI'1'Y ) "!?ILS

l',

386.4, 0,

0,

0

DISPLAY )- XY._PL07S :'> ACTXYPRE l, ;., TIt-!£, L 12, L DISPLAY X':CPLO?S) XYPLOT ACTXYPRE,

XYP:"OT,

p~ALysrs

C,

~

PD~C()R?YP

(17) Activate XY plot information for acceleration input at node 1 as a function of time and generate plot (see Fig. 8.20:

DPT, 2, UY, 0, 2, 1, UZ, RX, RY, HZ

1E--05,

0,

1£-06,

0

ANALYSIS > FREQ/BUCK ) R_FREQUSKCY

CONTRCL

>

l, 0, 1

1?l.LE, ELeEN'PRO.GEO, L 1, 1, ::. ANALYSIS > POST_DYN > PD_BECIT

PO_BASE,

:OADS-BC ) S'I'RUCTURAL ;. DISP::"'Yl.JTS ,. DP}' DPT, 1, AL, 0, 1, .:

> FREQ/B~CK > A_FREQUENCY A_FREQUENCY, L S, 15. 0, 0, 0, 0,

POST_D'iN ,. ::URVF::S

ANALYSIS> PD_CURTYP,

0

R_FREQUENCY

0

1

(18) Define modal damping for mode 1 as. 10% of critical damping (0.10): ANA!. 'ISIS ~ ?OST. __ DYN > Pi:LDhliP;'GAP ) PD_MDAN?, 1, 1, 1, .1

PD_HDA.."1P

(3) LiSl the natural frequency of the system:

(l9) Request response at nodes 1 and 2: RES0L'l'S )

LIST) FREQLIST

FREQUENCY

FREQUSNCY

FREQUENCY

PERIOD

imMBER

(RJ:i.D/SEC)

{CYC!~ES!SEC)

(SECONDS)

C 1000000E+01

0.1591549E+00

0.6283185£+01

(14) Define analysis type as modal time history using one natural frequency, 2000 time steps starting at 1=0 with a time increment of 0.01 sec;

ANA~YSIS

>

PD...:.Nf:£sP,

POS'I'_:nfN ) OUTPUT) PD_!'lRESP 1,

1,

2,

0

(20) Execute modal lime history analysis: ANALYSIS

R_DYNM{IC

;>

POST_DYN > R_DYN1\MIC

264

Response Spectra

Stn.lclures Modeled as a Single-Degree-o!-Freedorn System

(21) Define analysis type as response spectrum generation using one natural frequency, starting at 0.314 fad/sec (0.05 Hz). ending at :25.7 TJdJsec (20 Hz), linear scale, with 200 points in frequency range, 10% of critical damping, for node 1 in the X direction:

Use an editor to display the output file (EX8w4.0UT) and print the firsl nine lines of the response spt"~[ra: RISSfOl'lS~

SIT.c7I\lJIf

~N

ANALYSIS ,

?OST_DYN )

3,

PD.J,':''::.'PE. Q,

Q,

1,

PD~ATYPE

.314,

125.7,

1,

1,

200 ..

0.1,

I,

0,

0, 0

'OR

!)~l\gCT;OI

I"REQ\jH1CY fREQUENCY

SEC!

>

NOD:;

,

POST~CYN

) R_DYNAM:C

R_DYN~J"GC

(23) Activate XY plot information for X acceleration at ~?de 1 as a fUfictlor: of frequency (response spectrum), and generate plo~ (see Fig. 8.20):

PElno!)

(S!tCl

D::;:SPLAY

1, FREQ, AX,

1, 12, 1,

0,

The following table reproduces the first len lines of the list;ng shown in the screen:

, 3

5

6 7

8

9 10

'tRUE

AU$

Rtt.

m

Dr$p

\itt.

;£c

VEL

Ace

'tRUE

".:n10E.n~

Q.H11t,01 1.20HE.C)

'.1J7S)SE.ry,

C.';6!HE.Oj!

.Q .2H2~JE.'.t

O.1"f)f..:j1

~.,>~;.;:.c'

J. ,tf"lUE.t2

Q_::tlln;,~:

-0. BO~S;;t.1)2 -$.191SllalS.O'

:';.1Il00E.01 1).1592t,00 C.i.10jE.Ol

:.lun~E.;JZ

C.~~6?Ht.Ol

-O.tUi1:;E.Ol .Q.\%Oilf.llt ·0.))1»9E.01

c. ts.,£.(l. 0.1"(>$E.1)0 t.)SSlt.nl

b.lsn48E4': .n.1)a%tE,ol ·jj_il~J1n:.;n

C.~2SH(f.'n

~_:17H-SE'01

C.no{,>ol

O.)S081'.\j0 O.zaOSE.:l -O.! i'l41St.(>?

-O.2S~SHE.(,:

O.nH~.Oi

8.4";\

;E'O~

O.HnB'Cl

o.a~5L9£.Cl

O_l)~S?lE.Ol

O.29snBl:'Q -0,]2505£.02 -0. );]&3):E,02 O.1?SSOGE.C2 ·O.67501/E.()2 Q.)HUle,l}"l -~ _6n,es!':.o)

1),}HH~(t\

O.5~Ht.CI}

O.HU!:.O:

". ,'64,1,.0:

O.'n~9J!t.h

-().S;;$MS!:.Ol

-q.lO~lH!t.Q:;

OAC94C.Ol ",o6I1t.un (l.lS)$E.¢!

o.n~9'n>~,

J.;1S2{$<=+Ol -O.19:11t .. £.O,

O.H'1)~E.OJ

-o.$P~Ht.u<

Q.nlSE.n o.1Sln.01 O,lHilt.O'

OAOn)e,H

o.

O.20a)"E'~1

-].,.:I$S6'H.:>OZ

B,10

DISPLAY ) XY_PLOTS > XYf''TLIST XY?'TLIS7, 1, 201. 1

2

PSEU:xl

I\Et.

;:;nS.E'~l

·O.~lOjnE.jjZ

J.;'?n~7'1£.Ol

-O.n~678E.n

dJ.6i1.06;;:.02

I

XY_PLOTS) XYPLOT

(24) List values plotted of the response spectrum:

,

('S""'UX

ktt.

IN

XY?LOT, 1

pOINT

TRUE

jj.'H~t"~;;

DISPLAY > XY_PLOTS > ,t..CTXYPOST

ACTXYPOST,

,

01.AC/SEC) iCYCt.Il:/

(22) Execute response spectrum generation: ANALySIS

265

FR8Qt:ENCY (BAD/SEC)

SPECTRAL

NO::;:E 1 U.!1/ sec')

3 .140e-Ol 9 441e-01 1. aOOe-Ol 1.S74e-Ol 2.204e-01 2.834e-Ol 3.464e--01 4.09Se-01 4.72Se-Ol S_3SSe·-Ol

(25) LIst the response spectrum from the output file: CONTROL ) UTILITY :- SySTEM

ACe.

2 .390e+00 1 .353e+01 1 .313e+01 3 .282e+OJ 5, 9S5e+01

6, 925 •.,:+01 6 ,085e+Ol

8 .17ge+01 9 .2S6e401 9.413e+01

SUMMARY

Response spectra are pl0iS tha~ give the maximum response for a single~ degree-of-freedom system subjecied LO a specified excitation, The construction of these ploes requires tbe solution of single-degree-of~f:-eedom systems for a sequence of values of the natural frequency and of the damping ratio in the range of interest. Every so;ut:on provides only one point (the maximum value) of the response spectrum. In solving the single~degree-of-freedom systems, use is made of Duhamel's integrak or of t!:e direct method (Chapter 4) for elastic systems and of the step~by~step linear accelerJ.£ion method for ir,elastk behavior (Chapter 7). Since a la:-ge number of systems must be analyzed to fully plot each response spectmm, the task is Jengthy and time consuming eveu with the use of a computer. However, once these curves 3re constructed and are availabJe for Lhe excitation' of interest, the analysis for the design of sl.rlJctures subjected to dynamic loading is reduced to a simpJe calculation of the natural frequency of the system and the use of the response spectrum. In the following chapters dealing (hat structures which are modeled as systems with many degrees of freedom, it will be shown that the dynamic analysis of 3 system with Jt degrees of freedom can be transformed to the problem of solving n systems in which ead1 one 1S a single·degree~of·freedom system. ConsequentlY, this transformation ex.tends the usefulness of response spectra for single-degree-of-freedom systems to the solution of systems of any number of degrees of freedom. The reader should thus reaiize the full importance of a thorough understanding and mastery of the cor.cepts and methods of solutions for single~degree-of~

:~

i

l

I

I

1

Response SpeClr3

Sln.:ctures Moceled a,s a Singha·Oegree·o(·Freedom System 8.4

freecom SVSlems, since tbese S'lme. methods "fE also llpplicabk to systems of many deg;ees of freedom afler {he prob~eflj has been transformed to if\depcn~ dent s;nglt>degree·of-rreedom systems,

8.5

267

Derennine the lll~iiCirnum S((CSS in (!.e COhOllllS or the frame of Proble:':) g.] The frame sbowll ;;~ Fig. Pil.t i;; sub;ee\O!d to Ihe excll:!.liol\ prowLced by the El Centro eanhqu~ke of i940. A~$ume 10% camping ano from. lhe appropnllle chnrt de:ermine lilt specllal v;)ttJe~ fOf di~;):aCtmeIH, velocity. and :,cceleftIlJon . ..,\s· $.urne e)-..Slic b.;ll,,"VlOr.

8.G

Repeal Pmbkm 8 .5 using thl: b;\sic Gcsign spC<.:lra g,iven In Fig. 8.9 to delermine tile ;;pec!fal vahres for acceli:w;jon, ve:OCily, ;u1d displacement. (Scale down speciml Y
8,7

A slructure morJded as t:~e s:>ring'll:ass systcm sJ,own in fig. ?8.7 is ;:,ssum¢d 10 be. subjcc{c\J to ;:, suppon mOllon :1fouuced by (hI! El Cenlfo eanhquake of

DROBLElVS 8.1

The steel frame show!' in Fig. P8.1 is. subjeclcd ~o horizontal fo.:-ce al the girder level of (1000 sin 101 lb) for a lime clllf~lion of half a cycle or lhe-fofciilg sine funcliol"l. Use !!Je appropriate r..::spons.e spectral chtlf{ to obtaIn {\)¢ rnr.ximum Jisp!acemem. Neg.lec{ d:l.rnping. I

F

l\OCC WI '" 0

1940. Assuming ei~lstic bch:\.vior and using IliC ;,prropria!c response spect{;l; ch::.tn find thi!: !\)3Xj;Y;lHll fcia:ive displacement between lhe mass and Ihe suPPOrt. Also co:nplle :hc m,\f:m:um foro..:..: .:c!ing on Ihe spring.. Negkcl dan\ping.

11 K,p.

'0, tl

<: ~tlOj --+GSSS'S"SSSS\\\SSSS5S'i

\,>~ilCl

'l~

W{;

x 20

\

~.'~ · 1'

l

~1

Fig. P8.1

S.2

Determine the maximum Stfcsses in [be COI...,lnlIS or the (Hune of P(():.lle:n 8. L

S.]

Cons',der the rralne shown in F:g. \)8.3(;;) suojeCied to (l. foundillion excitation

l produced by a h:!!f cycle or Lhe runction til =::; '200 sin WI inise:: as ~bown in ]~ig, pg,3(b). DeLermi.:'!c tht m<1x~murn hOri:r.Olltlll disl,lacemem or the gU'tkf relative to the moCon of Ihe founu:>riofl. Neglect daruping.

Ro::pe(l.l Problem 8"7

8.9

D::!ermin;: the force lr;,Hlsmiul.'c Pwblem 8.8,

8.10

dlt: system lW.$ 10% Df !he ClltiC:!\ tbmping,.

(0

the rOundaliOl1 (or /he sySleti, .,::,,,,1 y-z.eJ In

C(}ns~der Ihe '1jJring-m>lss .'YSIClll of Problem 8.7 ,\I\J tiS$UI11¢ lila! the spring dement follOws, :;,;1.1 tlun c!aslOp!:I~!ic b;:hnvior with a l\l.J.:;.iwutll valu~ fOr (he f~SIOfL!g fo(c~ ill tc.cSIOIl
8.11

Repeal Problc!'o1 B.l0 [Of H}Cf(.l d:-tllipwl::,.

itU

A Mfucture modtk~ as <.l sl:lg!::·ucgn:.e·of·fn:.:eJclll system has a natural period -r"" 0.5 sec, Use (he fCSponsi! spcclc:ll me~hod tQ dCicnnine ill the clastic range lhc m:t,:im..;:n :"!bsohne 'Hxderatl<);). ~ilc ;:,::ximuli\ rdauvc displacemem. and lhe n:l.aximurn p:'l;.'uJcve\oci:y for: (aJ j\ found;:.rion JllOtion ec;ual LO :he EI Centro earlilqu:.tK-e of ;940 and (b) Ihe Gi:!lign spectrum wilh a maxinlllffi ground :!.cceler
ILl3

SOlve Problem 8.12 (l.ssUl.\ing elaslop!a$I;C .alio jJ. 4.

8.14

Use Program 6 10 de.,.dop a lutile :l:.tving: the $p
i·

l f,

JS~U!ll;llg :lm

E,g

!

bdlJ'<'\O{

of (he 5y5.lem with ullctility

r

268

Structures Modeled as a Sirgle·Degree·ol-Freedom System

The excitation is a constant acceleration of rr.('lg:1ilutie O.O! g applled for ;0 sec. Neglect domping. 8.15 Use Program 6 to deve!op the response spectra for elastic syStems subjected to the first 10 sec of {he 1940 El Cen{fO em1hquake. Neglecl damping. The digitized values corresponding [0 (he accelernrions :ecorded ror the firS( to sec of the EI Centro earthquake ue given in Table 8.3. 8.16

Repeal Problerr. 8.! 5 corresponding to a system with JO% of the critical dam?ing.

PART II

Structures Modeled as Shear Buildings

,

9 The Multistory Shear Building

In Part I we analyzed and obtained the dynamic response for structures modeled as a sing;e-degree-of-freedom system. Only if the structure can assume a unique shape during its motion will the single-degree model provide the exact dynamic response. 0!herwise, when the structure takes more than one possible shape during motion, the solution obtained from a single-degree model will be an approximation to the true dynamic behavior. Structures cannot always be described by a single-degree model and. in ge:1erai, have to be represented by multiple-degree models. In fact. structures are continuous systems and as such possess an jnfinite number of degrees of freedom. There are analytical methods to describe tie dynamic behavior of continuous structures thal have uniform material properties and regular ge~ ometry, These methods of aT1a1ysis, though interesting in revealing information for the disc:-ete modeling of structures, are rather complex a:ld are apphcable only ro relatively simpie actual s.tructures. They require considerable mathematical analysis• .including [he solution of partial differential equations. which will be presenred in Chapter 20. For ~he present. we shall consider one of the most instructive and practical types of stmcture which involve many degrees of freedom, the multistory shear building. ,~

27l

272

9.1

273

The Multistory Sheaf Building

StrJC!'Jres Modeled as Shear ouild:r.gs

STIFFNESS EQUATIONS FOR THE SHEAR BUILDING

A shear buildng may be defined as a structure in which there is no rotation of a horizontal section ac the level of 6e floors, In (his respect, the deflec(ed building will have many of [he features of a cantilever beam tha[ 1s c.eflec!tc by shear forces only, hence the name shear building. To accomplish such deflection in a building, we muSt assume that: (1) the towl mass of the StrJCt:.ue is concentrated at the levels of the f1oors~ (2) the girders on the floors are lnfmilely rigid as compared to the columl1$; and (3) the defofF.lation of the structure is independent of the axial forces present in (he coiumns. These assumptions transform the problem from a structure with an infinite number of degrees of freedom (due to the distributed mass) to a structure that has only as many ·:::egrees as it has lumped masses at the floor levels, A three-story struCf:;re modeled as a shear building [Fig. 9. ita)} will have three degrees of freedom, that is, (he three horizontal displacements at the floor :evels. Th.e second assumption introduces the requirement that the joints between girders and columns are fixed against rotation. The third assumption leads to the condition that [he rigid girders will remain horizontal dunng motion. lc should be noted that the building may l:ave any number of bays and that it is only as a matter of convenience that we represem the shear building solely in terms of a single bay. Actually, we can further idealize tne shear building as a single column [Fig. 9.2(a)], having concentrated masses at the floor levels with the understanding that only horizontal displacements of these masses are possible. Another alternative is to adopt a multimass spring system shown in

ta.)

Ib)

Fig. 9.2 Single-COlumn model represerllullon of shear building.

Fig. 9.3(aj to represent the shear building. [n any of the three representations depicted in these figures, the s:~ffness coefficiect or spring conStant Ki showr. between any two consecutive masses is (he force required to produce a re:2t!Ve l.:nit displacement of the two adjacent floor leveis, For a uniform column wilh tne tWO ends fixed against rotation, the spring constar.t is given by

k=

m,

I-r,..j

--, ---1

A,

m,

K1(Yl

AI .,

'-;

I

ro,

I')

~k~ (Yl-y , l

F,t!)--+!

I

k,

II

J-

+--m~

F*){Y:.t-Yl)

Y'I I

Yi t

r--+ '"'J{YJ-y?l--...! F~it)~

I

k,

(9.1 a)

_m'i-4

F'll)~

I

12El

F

_!

4---m~ X I >'1

", i

l

(b)

Fig. 9.1 Single-bay model representation of a shear building.

.

,~

I

i

,

rYz~ Fl

:()--+! *!Y;

.J I")

Fig. 9.3 Multimass spring model

repreSenla~fOn

of a sh
274

The Multistory Shear Building

Structures Modeled as Shear Buildings

and for a column with one end fixed and the other pinned by

3EI k=-L'

(9.1b)

where E is the material modulus of elasticity, I [he cross-sectional moment of inertia. and L the length of the column. It should be clear that all three representations shown in Figs. 9.1 to 9.3 for the shear building are equivalent. Consequently, the following equations of mmion for the three-story shear building are obtained from any of [he corresponding free body diagrams shown in these figures by equating to zero the

275

It should be noted that the mass matrix, eq. (9.4), corresponding to the shear building is a diagonal matrix (the nonzero elements are only in the main diagonal). The elements of the stiffness mmrix, eq. (9.5), are designated stiffness coefficients. In general, the stiffness coefficient kij is defined as the force at coordinate i when a unit displacement is given at j, all other coordinates being fixed. For example, the coefficient in the second row and second column of eq. (9.5) k n = k2 + k3 is the force required at the second floor when a unit displacement is given to this floor.

9.2

P-d EFFECT ON A PLANE SHEAR BUILDING

sum of the forces acting on each mass. Hence mill

+ klYI

m,y, + k, (y, -

- k,(y, - YI) - FI (r) = 0

YI) - k, (yo - yo) - F, (I) = 0

m,l' + k, (y, -

(9.2)

y,) - F, (r) = 0

This system of equations constitutes the stiffness formulation of the equations of motion for a three-swry shear building. It may conveniently be written in matrix notation as [M] {Y}

+ [K] {y}

=

(9.3)

[F]

where [M] and [K] are, respectively, the mass and stiffness matrices given, respectively, by 0 [M]

[ml

= ~

m2

0

[ [K]

= -

kl k, 0

u

+ k, - k2 k2 + k3 - k,

(9.4)

-~j]

(9.5)

k,

and {y}, {)i}, and {F} are, respectively, the displacement, accelermion, and force vectors given by

FI

(r)}

{F}= Fo(t) [ F, (I)

(9.6)

Buildings are generally designed to resist two types of forces: (l) vertical (gravitational) forces originating from the dead weight, occupancy, and equivalent loads; and (2) lateral forces caused by earthquake or wind loading. These two types of forces induce at each story of the building axial vertical forces, horizontal shear forces, overturning moments, and torsional moments. The overturning moments and also The torsional moments of the columns or other lareral resisting elements (structural walls) of the buildings have two contributing components: (1) the primary moments resulting from the vertical and lareral loads applied with their respeC[ive moment anns measured from the undefonned building configurmion; and (2) the second-order moments resulting from the vertical forces acting over the incremental moment anns caused by the lateral deflection of the building. This latter second-order comribution [Q the overturning moment is commonly known as the P-L1 effect; P designating the vertical force and L1 the lateral displacement of this force. Generally, the P-L1 effects in low-rise buildings are negligible because the total lateral displacements of low buildings are kept relatively small by the story drift limitations specified in the design criteria. However, in taller midrise and high-rise buildings, where lareral displacements may be much larger, the maximum interstory drift limitations does not ensure that P-iJ. effects will be negligibly smalL Since these second-order effects are manifested after the building has experienced lateral displacements, an iterative process may be used to determine P-/J effects. However. the P-L1 effect can be introduced directly in the stiffness matrix when (he axial vertical forces are consmnts (weight), thus making it unnecessary to conduct an iterative process. In Fig. 9.4(a), the P-J effect is illustrated for a three-story shear building with lateral displacements y,- subjected to constant vertical forces Wi at the various levels of the building. As shown in Fig. 9.4(b), due to these lateral displacements, second-order overturning moments Mi given by eq. (9.7), are developed at the various stories of the building. (9.7)

276

Structures Modeled as Shear Buildings

The MUltistory Shear Building

E,

M,

277

The application of eq. (9. i 1) to the three-story shear building in Fig. 9.4 results in

h,

I'<

F, F,

w

I'<

Ail

(9.12)

h,

~:

and using eg. (9.J 0) in

h,

..... M 1

-,,/ q ,-

E,

\ hi

(a)

(1))

Fig. 9.4 Three-story shear buHding showing (a) lateral displacement )'" (b) story oyer~ tumiog momentS Ml and equivalent couples ,.0; X hi.

(9.13)

in which PI is the total weight above level i, that is

P;=

"

IW

j

Equations (9. J 3) may in genera! be written in matrix notations Z,s (9,8)

j""i

{q} = [KG]{Y}

with n = 3 for the three-story building in Fig. 9.4. The second*order moments i'VIi of eg. (9.7) can be substitured by equivalent lateral force couples of magnitude equal to the overturning moments M t has shown tn Fig, 9A(b), that is

(9.14)

where [KG] can be referred to as the lateral geometric stiffness matrix which for n~story shear building is given by

r

P,

P , P,

-T-

hl

h:

(9.9)

P,

P;

h,

h,

+

-,

O

Pj

h)

0

0

0

0

0

0

Solving eq. (9.9) for F,' and substituting MI from eq. (9.7) yields

KG

(9.15)

(9.10)

0 with Yo= O. Therefore, the force qi at level i of the building is given by (9.11)

-

0

0

0

0 0

"

~

1\ -

h,,_

0

i I

P,,-I h,,_1

--+ P"

p" hf; P~

hr.

I

J

The consideration of the P-tJ. effect in [he shea:: bl:ilding will rhen be equival e m

278

SUuctures Modeled as Shear Buildings

to adding forces {q}

= [Ke]

The Multistory Shear Suikijng

279

{y} to [he right-hand side of eq, (9.3), Hence,

[M] {Y}

+ [K] {y} = (F) + [Kd {y}

or [lvf] (y) + [[K] - [Kc] (Yl = {F}

(9,16) {f)lY',

+-

From eg. (9.16), it is seen that the P~!J. effect may be intmduced in the dynamic as well as in static analysis by modifying the stiffness ma:rix [K] which is decressed by the geometric matrix lKG], In the sQlution of eq, (9.16), the latera! d!splacement will be larger man those calculated with eq. (9.3) in which p·LJ effects are not conside:-ed. These increased lateral displacements will then result in greater values for the shear forces at [he varioL
9.3

F,

It'H-1=:::'::==I

Fig, 9.6 Forces acting on a rhree-stOry she;:;r building,

stfHe that the displacement aC any coordinate is equal to the sum of the producrs ot" fiexibHiry coefficients ar that coordinare mUltiplied by the corresponding forces. The forces acting On [he three-story snear building (including [he inertial forces) are shown in Fig, 9,6. Therefore, the dispJncemenrs for the three-story building may be expressed in tenns of the flexibility coefficiems as

FLEXIBILITY EQUATIONS FOR THE SHEAR BUILDING

An aHernative approach in developmg the equation of motion of a S[rUClUre is the flexibility forrnularion. In this approach, the elns[ic p!:'operties of the structure are described by flexibility coefficients, which are defined as deflections produced by a unit load applied at one of the coordinates. Specifically, the flexibility coefficient ~~j is defined as the displacement at coordinate i when it unit static force IS applied at coordinate j. Figure 9.5 depicts the flexibility coefflciems corresponding to unit force applied at one of the srory levels of a shear building. Using these coefficients and applying superposition, we may

y,

= (F (t) - mij)/" + (F, (r)

m,y,)f12 ~ (F, (1) -

miN/"

Rearranging the tenns in these equations and using maulx notation, we obtain 1

IJ.J.

c----'---l-"+

(yl

;

I

I

t---+IU....lI

=

(f] {F} -- (rl [M] {y},

(9,17)

where [M] is the mnss ",.trix, eq. (9.4), [tJ is the flexibility matrix given by

I

I

I

r---~!J:~l,

,

I

I

(9,18)

I

I.)

Ib)

tel

and {y}, {y}, and {F) nre, respectively, the dlsplacement, acceleration, and force vectors given by eq. (9.6).

Fig. 9.5 FleXibility coeftlcients for a three-story shear building,

',.

280

9.4

The Multistory Shear .8uHoiog

Structures Modeled as Shear Buildings

Inserting these expressior.s fo, (he flexibllity coefficients into mlHrix, eq. (9.18). resuits in

RELATIONSHIP BETWEEN STiffNESS AND FLEXIBILITY MATRICES

The definitions given for either stiffness Of flexibility coefficients are based on static considerations in which the displacements are produced by static forces. The relationship between stat~c forces and displacements may be obrained by equating to zero the acceleration vectOr O'} 10 either eq. (9.3) or eg.

[j]=

k,

(9, (7), Hence

k, [K]{y)~{F)

(9" 19)

[fJ {F}

(920)

{y}

k

k;

[~e

281

flexibility

k

j

-+k,

k,

l

t

k,

k,

-+-

I

1

k,

kJ

-+i

1

(9.22) I

--+-+k,

k!

k) , ~

The extension of the tlexibility matrix for a rhree~s[Ory shear :,uilding to any number of stories is obviolls from the pattern of eq. (9.22).

From these :-elatlons it follows that the stiffness matrix (K} and [he flexibility matrix [t1 arc inverse manices, that is

9.5

PROGRAM 7-MODELli'iG STRUCTURES AS SHEAR BUILDIi'iGS

[KJ = [fJ or

[fJ = [K]

The computer program presented in this chaprer serves to model a S(t"..!cture as a shear building. Such modeling requires the development of the s{iffness and mass matrices of the sys~em" In Program 7. the elements of these matrices are stored in a file in preparation for dynamic analysis such as the calculation of natural frequencies and the determination of the reSponse of the structure when subjected w dynamic forces or to seismic exciratio"n. Program 7, stiffness and mass matrices of a shear building. contains the option of iocludino- the p,jj effect, thus of calcu;ating the combined stiffness marrix (KcJ = [K} :;, [Kd. Smce the stiffr.ess of the mass matrices are symmetric, only the elements in the upper triangular portIon of these malrices need to be stored in files. The notation used, in this case, consis(s of numbe,ing consecutively those elements of the matrix located in the upper triangular part. For a matrix of order 5, the numbering of elements is as follows:

(921}

I

Consequently, the flexibility matrix Ul for a shear building ;nay be o~tained either by calculating the inverse of the stiffness matrix or directly f:-om lhe definition of the flexibility coefficients. Taking this last approach, we obtain, for :he t~ree-story shear building shown in Fig. 9.5(n),

kdu = 1 and

1(1)

·l

Analogous:y from Figs. 9.5{b) and 9.5(c) we obtain

(3) (6) (10) (IS);

(2)

~!~ :~\ ::iij·

symmeu.c

and and

(II)

Using this notation, we need (0 store only about half of [he total number of coefficients in the stiffness matrix. As a matter of fact, we could save more memory in {he computer :;y no[ slo.-ing the zero coefficients at the wp of each columr. of the m2trix. The implementation of such saving in computer memory

since [he flexibility coefficients for springs in series are given by the summation of the reciprocal values of the spring constants. ,

,

282

The Muttistory Shear Building

Structures Modeled as Sheaf 8v!ldings

requires special coding to locate each entry in the diagonal of the ~t~x and, therefore, each coefficlen! of the matrix. Such coding has not been tmp{emenred in the computer programs presented in this volume.

5.S4';Z.. :U -5.2";:n;::+cz

;;.SA7n~C2

G.ooon_()Q

-3.2'?F£~CZ

283

c.oOD:n::.. JD c .OOJ"E+CO

I),::;{)O();::.OJ

Example 9.1. Detennine the stiffness and mass matrices for the four-story 6 shear building shown in Fig. 9.7. The modulus of elasticity is E = 2 X 10 psi. Neglect the P·Ll effect. Solution:

:.OOG),,:'00

o. COOCf>CO

0,OGDO;;:~0()

c. "c:::n~JC

.

1)

C JOCJS~OO 0. Jih;)OS. 30

,

c.:sn~>o~

1.C~O

0;;)

o .1"':lH:~C$

:.;;00

!SCl.GO

(].:S9~r->-O$

l.CO::

:$:). all

O.l':l~:£~j~

l.OCO

,C0

~eQ,

M'j~S·CO

£.Sq59<:-02

-l.Z73SS.CZ ,,~"n~Cl

,

0<:;00;;; .. 0,]

: _CONS-GO

,

OOCO;;;~OG

, ,

~)"GOS~;;O

O.0C:JO?:+;;iC OJY:2~0Q

1.0"J0('c;;.co

,.

c,

0. :::::OOS_co ,.)

~-;H!>C2

(). 0;0C2~OC

.)

27J6I'+Cl

~}

nl",:.n

:.

.:rn~S.02

J _JC%E""CO

~'. jOCCS_~C

O.OOJC,,;-:)(l

l, OOCC;:;~,jC

o,;:oc::t".oc

0, COvO£ __J{;

O.JOO:;'C.oo

0 QOO')£_(;O C. m')oc-£",c;;

QOOt,;);>-(;O

o .C¢O~::·¢O

effect,

:: .OOCCE+QO 273fS .. Cl2

:s~nS->-G2

. OC.O;;S~:;'C

())C1£~t)D

P~d

o.coo-n_cc

o.ocon~Oj

0

9.6

000(;::;.0<:-

.cooo;;.oo

Solve Example 9.1 considering the

c) .~:H::>02

,

l~C

(l . DC M)1,;' C0

Example 9.2.

Input Data and Output Results

00')0:;:.00

~.

OOD'):;:"CC

"-. CO()CE~')C

SUMMARY

The shear building idealization of structures provides a simple and useful mathematical model for the analysis of dynamic systems. This model permits the representation of the srrucmre by lumped rigid musses interconnected by elastic springs, In obtaining the equations of motion, two different fonnulal:tons are possible: (1) [he s{iffness method in which the equations of equilibrium are expressed in terms of stIffness coefficients and (2) the fleXibility method in which the equations of compatibiEty are written in terms of flexibility coeffi· cients, The stirn,ess matrix and the tlexibillty matrix of a system, in reference to tne same coordinates, are inverse matrices. J""79,55in:~"

,

t I

180 10

!

totl)! In'" 1

PROBLEMS I

r~/

-'--/JW

....

10079 total

55!~ 4~ 1

iZ7T

Fig. 9.7 Modeled structure for Example 9.L

For the three-story shear building verify that the stiffness matrix, eq, (95), and the flexibility matrix. eq. (9.22), are iJ1Verse matrices. 9.2 For r.~e two-Story shear buiJding shown in Fig. P9.2 derennine the stiffness and flexibility matrices ar.d then verify that these matrices are inverse matrices,

9.1

284

Siructures Modeled as Shear Buildings

The Multistory Shear Building

...

lBOO !b/ft

;:'1

([)--1========~========:j-t WlO

x 21

285

Y,

W10 X 21

•....-v

1.'

<---25'--+---3IJ'---i

Fig, P9.1.

9.3

----+

For the rhret>.-srory sheaf building shown in Fig, N.3 obtain the $[lffness and flexibility matrices and show thaI rhese matrices are inverse mmrices. AU the columns are steel members W 10 X 2 L tOOO lb/ft

FJW---.

r

tt)

Fig, P9.6.

9.8

De[ermine the stiffness and moss matrices for the shear building shown in Fig. P9.8. The ::nodulus of elnslicllY is £ "" 30 X 10° Pi>l.

)

15CX) Jb/h

F1irl

Y

)2'

2000 IbJh Y

30'

"""w

.a &

.a;,&

J

Fig. P9.3.

9.4

Write the differential equation of motion using the stiffness formulation for (he shear building in Fig. P9.2. Model the structure by a

multimass~spring

system.

9.5

Write the differential equation for the motion of the shear building In Fig. P9,],

9.6

The three-story shear building in Fig. P9.5 IS subjected t;::; a foundation motion which is given as an accelennion function Y.(r). Obtain the stiffness differemjllJ: equation of motion. Ex?ress the disj:llacement of the floors relative to the foun~ dllticn displacement (i.e .• u, "" )" Yf)

9.9

Model \he stmclure as a shear column with lumped masses as the floor levels.

9.7

Generalize the results of Problem 9.6 and obtain the equations of motion for a shear building of n. slories.

Use Program 7 Ie de,ermine the stiffness and mass matrices for (he she;:;r buliding shown in Fig. P9.9. The modulus of elasticity is f. "" 30 X !OI> pst. Neglec( the p. Ll dfecL

9.10

Solve Problem 9.9 inclUding lhe P-::l

9.11

Use Program 7 to de!er:nlne the stiffness i:.nd mass matrices for a uni form (en-srory shear bui!ding: m which the mass :u each level it: 20 {lb· sec'llin}; ~he imerstory heigh, is 12 ft; rhe mcdulus of elasticity is 3.0 X 106 psi; the ~()tol 9 flexur<:.l slifrncss is: :fEf 24 x 10 (lb in;') for e.!! the stories of {he buildinC" Neglecl the P-.J effect. e'

286

Structures Modeled as Shear Buildings n))

Of.

10 lib ~ec~!;n,l

-r- rim==""""'1 10'

-lI

,

12"

fl'<

10

~if2mrr'3"3'5~~ '""," ,. 400 in. ~

Free Vibration of a Shear Building

Fig. P9.9.

9,12

Soive Problem 9,: I including [he p~tl effecL

\Vnen free vibration :5 u:lder considerarion, the structure is not subjected to any external exciUlllon (force or suppor:: motion) and irs motion is governed only by the initi~1 conditions. There are occasionally circumstances for which it is necessary to detennine (he mmion of the structure under conditions of free v:bra!ion, but ~his is seldom [he case. Nevertheless, (he analysis of the Structure in free morion provides the most impor::um dynamic properties of the structure which are [he natural frequencies and tbe corresponding modal sbapes. \Ve begin by considering both formulations for the equations of motion, namely, tne stiffness and the tlexlbiliry equarlons.

10.1

NATURAL FREQUENCIES AND NORMAL MODES

The prOblem of f'ree vibration req~ires that the force vector {F} be equal to zero in either lhe stiffness, eg. (9.3), or fiexibiiilY, eq. (9.17), formulations of the equations of motion, For the stiffness equation with {F} to}, we have

[M] (Y] + [K] (y) = {OJ

( 10.1)

287

288

Free Vibration of a Srear BuHdif'lg

Structures Modeled as Shear S:lildings

where fD] is known as the dynamic matrix and is defined as

For free vibrations of the undamped structure, we seek solutions of eq. (10.1) in the fonn

Y,=a, sin (wt- a).

289

[DJ

= [f] [,M]

(10,8)

i= I, 2, "" n

Equation (l0.7) may also be written as or 1:1 veclQr notation

(yj

=

{a} sin

eM - a)

[[D} - I Iw'[IJJ {a}; 0

(102)

where [f] is the unit matrix with ones in the ma~r. diagonal and zeros every* where else. For a r.ontrivial solution of eq, (iO.9), it 15 required that the detennioant of the coefficient matrix of {a} be equal to zero, thz.t 15,

where at' is the amplitude of motion of the ith coordinate tnd n is the number of degrees of freedom. The substitution of eq. (to.2) into eq. (10.1) gives - w'[M; {a} sin (we - a) ~ IK] {a) 'in (we - a) or rez.rranging

=0

[DJ - I/w' [I] I

~erms

w' [M]] {a} =

lIK]

{OJ

ExampJe 10.1. The building to be analyzed is tr.e sjmple steel rigid frame lOJ, The weights of the floors and walls are indicated in the shown in f;gure and 3re assumed to include the structural weight as well. The building consists of a series of frames spaced 15 ft apart. It is further assumed that the stmctural properties are aniform along the length of the building and, therefore, the analysis :0 be made of an inte.:ior frame yields the response of the entire building.

0

"''!-

+ Ifl [M]

m

(10,10)

one of the n solutions for (I IW') of eq, 00,10), we can obtain from eg, (lO,9) corresponding solutions for the amplitudes a, in tenus of an arbitrary constant. The necessary caJcuiatior,s are better explained with the use of a numer~cal example,

In general, eq. (lOA) results in a polynomial equation of degree n In u} which 2 should be satisfied for n values of w . This polynomia: is known as the characteristic equation of the system. For each of these values of ul' satisfying the characteristic (lOA), we car. solve eg, (10.3) for a" a2. ,,,,0" in tetros of an arbitrary constant. Analogously, for tr,e flexibillty formulation. we have for free vibra::on from eq, (9,17) with {F} 0,

{y}

0

Equat~o;1 (to.to) is a poiynornial of degree n in I Ivi. This polynomial is the characteristic equation of tne system for the fleXibility formulation. For each

(10.3)

whkh, for the general case, is set for n homogeneo~s (right~hand side equal to zero) algebraic system of linear equations with n unknown displacements a, ar.d an unknown parameter (,1/", The fonnulation of eq. (lO.?) is an impof:.ar.t mathematical problem known as an eigenproblem. Its nontrivial solution, that is, the solution for which not all at = 0, requires that the determinan! of the matrix factor of {a} be equal to zero; in this case,

1[K] - w'[Mj

(10,9)

~:::W~,;=:50=='b=l=ft::::~.-. ~ Y, 20 psf W10 X 21

10' Wj "'" tOO lb/ft2

~

{OJ

r=======H....

(l05)

Y,

W",f

Vole again assume hanuonic motion as given by eq. (10.2) and substitu!c eq, (10.2) into eq, (10.5) to obtain {al

= w' [f] [;11l{a}

WIO X 45

(1M) >

~~

Or

l/w'{a} = [DJ {a}

(107)

"t

I.)

Fig. 10.1 Two~story shear building for Example 10.1.

290

Free Vibration of a Shear SUi!ding

S1Jucture5 Modeled as Shear Buildings

291

In the usual manner, these equatjons of motion are solved for free vibration by substituting

YI = Cl; sin (WI

a)

Y1 =

a)

a2

sin (M -

(l0.1l)

for the displacements and - a,w'Y. sin (wt - a)

YI (h;

h Fig. 10,2 Mult!mass~spring moce! for a two-story sheor building. (u) Model (b) frcc

=

- G.7.W2

sin (<-:Jt - a)

for the accelerations. In matrix nQration, we obtain

body diagram.

co

fk:+k 2 - m l wZ

Solution: The building is modeled as a shear building and, under the assumptions stated, the entire building may be represented by the spring-mass system shown in Fig. 10.2. The concentrated weights, which are each taken as the total floor weight plus that of the tributary wi1i1s, are computed as follows:

l

- k,

(lO.I2)

lo

For a nontrivial SO,'utl'on, we requl're thot ~ ... the d e t ennmant 0 f i he COeftlclent matrix be equai to zero, thar is, <

IVI~ IOOX30x 15+20X 12.5X ISx2=52,5001b

m·

(l O. 13)

1361b . sec' lin

IV, ~ 50 X 30 X 15 + 20 X 5 X 15 X 2 = 25.500 lb

The expansion of this detenninant gives a quadratic equarion in

(t,/", name!y, (lO.l4)

m, = 661b· sec'lin Since the girders are assumed to be rigid. the stiffness {spring constant) of each story is gi ven by

k

12£(21)

-;;;--

or by introducing the numerical values for this example, we obtain 8976&>' - lO,974,800w' + 1.36 X 10' =

The roots of this quadratic are

u/ wi =

and the individual values for the steel column sections indicated are thus k;

140

lO82

12 X 30 X 10' X 248.6 X 2

=_.~_..C':-:-.",,=,,--=30,7001b/in

X

Therefore, the Gatural frequencies of the structure are

12 x 30 X lOb X 106.3 X 3 k,. = ..:::....:.c:.......__._.- ' - ~ 44 300 Ib /in (lOX 12)" , The eqUlltions of mm!on for [he system~ which are obtained by considering In Fig. 1O,2{b) the dynamic equlHbrium of each mass in free vibration, are m,YI

T

k'YI - k?(y~

y,)

=0

m;.,V2 + k2(Y~ - Yl} =

0

w: = 11.83 radJsec Wl

= 32,89 radisec

Or in cycles per second

I;::;;: wJ'21i= 1.88 cps I, = w,I2T1'= 5.24 cps

a

(10 IS)

292

Free Vibration 01 a Shear Building

Structures Modeled as Shear Buildings

293

and the corresponding natural periods

T,

= -1 = 0.532 sec

t,

-

1

tl

= 0.191 sec

To solve eq. (10-12) for the ampHtudes at and Q1. we note that by eqcating [he determinant (0 zero in eq. (l0.l3), the number of indeper:denr equations is one less. Thus in the present case, the sY3tem of two equations is reduced to one independent equation. Considering the nrst equation in eq. (10.12) and

SUbstituting the first natural frequency, 55,960a"

Wj

Fig. 10.3 Norm;,l modes for Example 10.1 ~a} Firs. mode. (b) Second

44,300a" = 0

= L263

a"

It is customary to describe the normal modes ~y assigning a unil value ro one G:: equal to unity so that

of the amplin.:.des; thes, for (he first mode we set att

lb,

(10.16)

We have introduced a second subindex in o! and Q2 [0 indicate (hat the va;ue Wi has been used in this equation. Since in the present case there are two unknowns and onty one equarion, we can solve eq. (10.16) only for Lhe relarive value 0:' all to all' This relative value is known as the nonnal mode or modal shape corresponding to the first frequency. For this example eq. (to. i6) gives Q2'

(.j

= 11.8 rad/sec, \>.,'e obtain

is called a norma! or natura! made of vibrafiofL The shapes (for this example fall and O:fliad
LOOO (\0.17)

SimilarlY, substitlltir.g the second natural frequency, (10.12), \ve obtain the second nonnal mode as

UlZ

an = 1.000

a" = - 1.629

(10.19)

= 32.9 rad/sec into eq.

(10.18)

H should be noted that aJrhough we obu:.ined only ratios, the amplitudes of motion could, of course, be found fro:n initial conditions. We have now arrived at two possible simple hannonic motions of the slructure which can take place in such a way that all [he maSses move in phase at the same frequency, either w, Of W:J.. Such g motion 0: an undt.rnped system

Here C( ar:d as well as Ct\ and ((1 are four COnstants of integration [0 be detennined from fO'Jr initia: conditions which are the initial disptacemer.[ and velocity for each mass in the system, For a [wo-degree-o~-freedom system, these initial conditions are y, (0;

=Yo" y, (0) = y" 00.20)

For computationnl purposes,

j(

is convenient

[Q

eliminate the phase angles

294

Free Vibration of a Shear

StfiJctures Mcdeted as Shear Bui;dings

sln

+

Clall

h(/)

C 1t.l11 sin wjt+ C,P21 cos

Wit

ell/ il

cos Wit Wit

sin

w}}

+ C4Gn

cos ~l

+ C;.a·~"2 sin

(r)i

+ C 4G!Z

COS WJI

'T

C~au

(10.21)

in which C 1, C 2, C)., and C", are new consrunrs of integration, From [he first twO in:tiul eoocirion;:; in eq. (l0.20), we have (he followIng two equations:

00.22) Since the modes are independent, these equmions can always be soived for C z and Similarly. by expressing in eq. (10.2 I) the velocities at time equal to

zero, we find

fo:-ces

(0.23)

The solurion of these

twO

System II:

sets of equatior.s aHows us to express the motlon of

the system in tenns of the two modal vibrations, each proceeciog at its own frequency, completely independem of the other, the amplitudes and phases being detennined by the initial conditions.

,

forces

~ai?mL

displacemegts

ar:.

,

UJ~anm?

an

The application of Bet:i's theorem for these two systems yields 1

afiJIlJG\lG['L

10.2

295

These equations are exactly rhe same as eq. (lO.i2) but written in Lilis form they may be given l.l stade incerpre(arion as [he equilibrium equations for the system acted on by forces of magnitude ffltw"J,a, and mzviaz appJied to masses ml and m;b respectively. The modal shapes may then be conSidered as the static defleclions resulring frorn rhe fo:-ces on [he right-hand side of eq, (025) for any of the twO modes. Tnts interpretation, as a static problem, alIows liS to use the resu:(s of the general smtic ~heory of linear s{ruc(Ures. In partiCUlar, we may make use of Betti's (hecrem, which states: For a structure acted upon by two systems of loads and correspond:r.g displacements, {he wor.i< done by the first system of loads rr:oving [hrough the displacements of the seccnd system is equal (Q [he work done by this second system of loads undergoing the displacements produced by the first load system. The two systems of loading and corresponding displacements which we shall consider are as fot~ lows: System I:

(Ct! and U'2 in eq, (lO.l9) in faver of other cor:stanrs, Expanding the trigonometric functions and renaming the constants, we obtain

YI (/)

Buifding

ORTHOGONALITY PROPERTY 01" THE NORMAL MODES

+

~

~

1

wlm~a~ia2~ = w~m!ar:;(lH ... W2JIl10n(l2:!

or (10.26)

We shaH now introduce an important property of [he normal modes, :he

orthogonality propeny. This property constitutes the basis of one of the rr:.OS[ a.ttractive methods for solving dynamjc problems of mulridegree-of-freedom systems, We begin by rewriting rhe equations of motion in free vibration, eq. (103), as

[.'c {a}

~

w'

,M] {a}

If the nat1Jral frequencies are differem (WI

it follows from eq. (10.26) that

which is the so~cal!ed orthogonality rela:ionshlp between modal shapes of a two degree~of~freedom system. For an n-degree-of~freedom sysre::u in which the mass marrix is diagonal, the orthogonality condition between any tWO modes i and j may be expressed as

( 10.24)

For the two-degree-of-freedom sysrem, we obmin from eq. 00.12) (k , + k2)Clr - k2iJ-;,. = miw::(lj

~ k2C1 1+ k2a2 = m:J.ll Q"J,

r= UJ2),

'"

) ' n1:.-G""Gt; = k-;;-'I

(10.25)

,

0,

for i ')Ai j

(10.27)

296

Free Vibration of a Shear Building

Structures Modeled as Shear 8uildings

and in general for any n-degree-of-freedom system {a,}T[M]{a,j = 0,

ilS

for iY
(10.28)

in which {a,} and {aJ are any two modal vectors and [tvt] is the mass mat:ix of the system. As mentioned before, the amplitudes of vibration in a normal mode are only rdative values which may be scaled or nonnalized to some extent as a :-natter of choice. The foilowlng is an especially convenient nonnalization for a general system:

Solution: The substitution of eqs, (IO,17) and (10,18) together with the values of the masses from Example 10.1 ieto the normalization factor required in eq. (1O.30) gives ,j(136) (100),

+ (66)(1.263)'

/(136)(joO)' + (66)( -

1.629)'

1.00

0,06437,

¢"

(.p};~M]{.p); =

°

fori for i = j

c: 0.31)

311.08

=

LOa v

1.263 1.31

in which ¢ij is the nonnaIized i component of [he j modal vector. For normalIzed eigenvectors, the orthogonality condition is given by

= "

4;11=

(10.29)

(10,30)

,31

=

Consequently. the nor.nalized modes ale

4>" = which, for a system having a diagonal mass matrix. may be written as

297

= 00813,

=0.0567

311

- 1.6287

cf>n

;31L08··~=

- 0.0924

The normal modes may be conveniently arranged in the columns of a matrix :':r.own as the IT'..odal matrix of the system, For lbe general case of n degrees of freedom. the modal matrix is written as

l.p"

¢12"

[
L¢n~ dJa1 "

•

4>'"1

(10.34)

¢?" ,

rp'm:

The orthogonality cor-dition may then be e_xpressed in general as

(10,35) Ar.other orthogonality condition is obtained by writ~ng eq, (10.24) for the nonnaHzed j mode as

[K] {.p}; =

w; [MJ {.p};

where [4>] T is the matrix transpose of [4» and [.1'11] the mass matrix of the system. For this example of two degrees of freedom, the modal matrix is

(I0,32) [0.06437


Then premultiplying eq, (10.32) by {cf;}[ we obtain, in view of the orthogonality condition of eq. (10.31), the followIng orthogonality condition between eigenvectors:

{4>}TrK] {.p}; =

°

fOf i

=

wJ

fori

[ J = l00813

0.05671

(10.36)

- 0.0924j

To check the orthogonality condition, we simply substitute the normal modes from eq. (10.36) into eq. (10.35) and obtain

'" j

(10.33)

Example 10.2. For the tWO-degree shear building of illc.strative Example 10.1, determine the normalized modal shapes and verify the orthogonality condition between modes,

01

[0,06437 008131 [136 [0,06437 :.0,0567 - 0.0924: 0 66: [0.0813

l

005671

I 0j'

-00924(lo :

We have seen (hat to determine the natural frequencies and nonnal modes of vibration of a structural system. we have to solve an eigenvalc.e problem.

29E

Free Vibration of a Shear Building

Structures Modeled as Shear Buildings

299

The direct method of solution based on :he expansion of the determinant and the solution of [he characteristic equm::on is limited in practIce to systems having only a few degrees of freedom. For a system of many degrees of freedom, the algebraic and numerical work required for the solution of an eigenproblem becomes so immense as to make the direcr method impossible. However, there are many numerical methods available for the calcu!ation of eigenvalues and eigenvectors of an eigenprobtem. The discussion of these methods belongs in a mathematical text on numerical methods rather than in a tex! such as this On srructural dynamics. For our purpose we have selected, among :he various methods available for a m;merical solution of an eigenproblern, the Jacobi Method, which is an iterative method to calculate [he elgen~ values and eigenvectors of the system. The basic Jacobi solution merhod has been developed for the solution of standard eigenproblerns (i.e., [IV( being [he identity matrix), The method was proposed over a century ago and 3as been used extensively. This method can be applied to aU symmetric matrices (K] with no reslricdon on the eigenvalues. It is possible to transfom1 the generalized eigenproblem, [[K] - w' [M]] = {
The ratio given by eq. (10.37) is known as the Rayleigh's qumiem. This quotient has the foHowiing properties: 0) It provides the eigenvalue Wj when the eigenvector {
10.3

and into the denominator

RAYLEIGH'S QUOTIENT

Several iterative methods for the solution of an eigenprob!em make use of Rayleigh's quotienL Rayleigh's quotient may be obtained by premultiplying eq. (10.32) by the transpose of the modal vector {4>}J. Hence,

Example 10..3. Use Rayleigh's quotiem to calculate an approximate value for the first eigenvalue of the structure in Example 10.1 beglning with t!le approximate eigenvector for the first mode {¢}T = {LOO L50}, then iterating using eqs. (1025) and (10.37) to converge to the eigenvalue and eigenvectOr for the first mode. Solution: The substitution of the given vector {4>i' {LOO 1.50} and the matrices [K] and [M] into :he nume",:or of eq. (l0.37) results in

I'

I

f 75,000 - 44,300j' .00 (l.00 1.50} 'I , - 44,300 44300 l 150 J

{1.00 LSO}

42.025

136 0 - (1.00) [o 66,• 11.50 = 284.5

!

which substituted into eq. (10.37) yields ,

w;:

, {} w

= {4>}'[,o/f}{P

(10.37)

in which for convenience the subindex. j has been omitted.

42,025

uF = .............-

The property of the mass matrix [AI] of being positive definitely; renders the product {4>};[M] {4>h'" 0, thus, it is permissible to solve for

284.5

The use of the calculate vall:e, eq. (10.25) yields Q;

(J. = 147.9 together wlth a I = LOO into the first

= 1.00

and

A second iteration of eq. (10.37) with I A f7umix [A] is defined as positive definite if ~( satisties the condhior, ,hat fer any tttbi!rary nonzero vector (u}, the product {v}T(AJ {v} >

o.

= 147.9

Q:;.

= 1,24

{4>} T = {LOO

1.24} yields

Free Vibratior. of a Shear BUi!dir.[,;

Structures Modeled as Shear Buildings

300

This value of fJ} is virtuaily equal to the solution w~ = i40 calculated in. Example 10.1. Anothe:- popular iterative method to solve an eigenproblern, that is, for structural dynamics, to calculate nmura! frequencies and modal shapes, is. [he 'subspace ireration method. Computer programs for the solution of an eigenproblem using the Jacobi method or the s:.lbspace iteration method are included as alternative methods in [he set of educat.:onal programs accompanYlng thi.s textbook. Commercial computer program for the solurion of an eigenproblem problem such as COSMOS usually provide severa! other methods in addition to the Jacobi and subspace iteration method. The computer program COSMOS gives the user the option of solving for natural frequencies and modal shapes by any of the following methods·, Jacobi, subspace iteratio:l, Lanczos, inverse iteration power, and ::omptex eigenvalue analysis.

Input Data and Output Results

O. GOCoo,:_co

O~OC');:.'N;

;},' S;)'J'Jt>C;)

C.

:'.10000a.01

:;,Mt
C'. CODCOf:.O:'

e:;~s"'-·.';'.tvC5 ?~.;

10.4

301

: :n);:,,;

l.

.;s;;,:

PROGRAM 8-NATURAL FREQUENCIES AND NORMAL MODES

The program presented in this section uses the generalized Jacobi method to .determine the natural frequencies and corresponding modal shapes for a struc~ ture modeled as a discrete system. Example 10.4.

Use Program 8 to solve the eigenproblem cOtTesponding

to a system having the following stiffness and mass matrices:

[

[KJ =

(M] =

3000 - 150~

[~

0 I

0

- 1500

01

3000 - 1500 1500 )500

j

~l

Solution: The execution of Program 8 to calcuiate natural frequencies and modal shapes requires the previous preparation of a file containing the stiffness a;)d :nass matrices of t:Je system. This me is crez.ted during the execution of one of [he programs to model the structure or by execution of the auxiliary Program Xl. Thls program accepts as input the stiffness a:ld mass matrices of the s~rucrure -and creates the flIe required to execute Program 8 In the solmion of Example 10.4, the required nIe was created by executing Program XL

(l.S;E0:

C.") )ii-a

n.1Hn

i),)2'lH

C.S9'Cl

O.sno!.

fL[3-$SS

10.5

FREE VIBRATON OF A SHEAR BUILDING USING COSMOS

Example 10.5, DeLennine the nan:ral frequencies and modal shapes for the three-slory shear building modeled as shown in Fig, tQ,4. So!wion: The analysis is perfonned using three spring elements and three concentrr..ted mass elements.

0) Set view to the XY plane: :OISPL~Y

VIEW,

~

Cl,

V:S\\ C,

L

P'~R

~

V.:EH

0,

(2) Define the XY plane at Z ~ 0: GEOt1ETRY ) GRID > PLANE

PLANE, Z,

0, 1

(3) Generate th,ee curve segme:HS from 0, 0, 0 to 3, 0, 0: GEO~£TRY

>

CRPCOR;),

:i,

CURV2S 0,

0,

>

0,

CR?COR~

L

C,

;),

2,

G,

0,

:3,

0,

',j,

3,

0,

0

3C2

Structures Modeled as Shear Buildings

Free Vibrat"lon of a Shear Building

40 Kilo.'

50 KJiO.j 10'

l

L)

6'

:1() Ktin.

,

303

CON7ROL ;,. ACTIVS AC':'SET ACTSEl', EX:;, 2 AC'TS£T, ?-C, 4 MESHING > PARAlC1·H::SH: f'CPT, 2, 2, 1

!

~

ACT-S2'! AC'fSET, RC, S MESHING

Fig. lOA Mu(he;nr.nical model for a three-story shear building of Example W5.

:-CPT,

SPR=~G,

1,

D,

0,

2, 1, 0, 0, 0, 0

0,

0,

0,

0.

0,

tESHING ) NOSl2S NM"2RGE. L 9, :i,

, RCONST RCCNST, " , 1, 1, 1, 5000 l , l. 4000 L, RCONS'!'. RCC>lST t 1, 3, 1, 1, 30QO

LCADS~BC

RCONS'1'

j

tCC~,

2,

2,

CONTROL

?

.f..CTSE?,

Re,

MESHING

:>

::-CCFL

3,

1, 2,

3

M_CR

?Af
3,. L

2, L

1

1, UZ,

FREQ/B~CX

A_FREQUENCY,

8.X,

RY,

RZ

0,

il

0

maximu:ll of 16 iterations, and rJn

3,

S,

A_FREQU E:KY

16.

0, 0, 0, 0, 0,0,\),0 !....~ALYSIS ~ f'REQJ2UCK > R.1REQU:::NCY

lE-{)S,

c,

lE-·06,

R_FR£QUENCY

!

ACTIVE ) ]>.CTSE:T

;;,

OUT?:JT_OPS » PRINT__ QPS 1, 0, 0, :., 0, 1, 0, O.

>-

lu\lALYSIS )

;

1, 1

and in all

(he frequency anaJysis:

j

PARA."CHESH

t.

(11) Se[ :he optIons for {he frequency analysis to extract rhree frequencies

CONTROL ;. ACTIVE ) P·.CTSET MESHr~G

C,

usiog the Subspace Iteration Met:lOd with

1

AC'!'SET, RC, 2

0, 1, 1

D.?'l' , 2, t;y,

!?fHN1'~O?S,

CON'I'ROL ~ ACTIVE ;,. ACTSET AC':'SE':' , EG, 1 AC'!'SET, RC, 1 XESHING > ?AR3M.MSSH fo M_CR l, 2, 1,

0

) STRCC":'URAL ) D!SPLr-0-!T$ > DP'I'

DP7, 1, AL,

ANALYSIS

{7) Generate spring and mass elements with appropriate e!ement gro~p and real constants actlv
l,

0,

(£0) Requesr printout of mode shapes:

0 Q, 0, 0, 0 aeCNS'!', 2, 4, 1, 7, 25.88, 0, ", RCONST, 2, 5, 1, 7, 15.53, 0, 0, 0, 0, Q, 0 RCC>lST, 2, 6, 1, 7, 10.35, C, Q, 0, Q, 0, 0

1,

N1";ER.GE 0,0001, 0,

(9) Apply conStrtmts in all degrees of freedom at node degrees of freedom except UX at nodes 2, 3, and 4:

(6) Defi:le real constnnlS for mass eJements: m ~ 25,88 lb· sec:: lin, 15.53 lb· sec'lin, and 10.35 lb· sec' /in:

M_CR,

>

XE':SHI:JG ) NO::JES ) NCOM?RSSS NCOV.PRESS, 1, 0:

?RO?S£'1'S

,

1

(8) Merge find compress nodes:

0

(5) Defi:lt real conSi.ftnts for spring eiemems: k = 50,000 Ib lin. 40,000 Iblil1, and 30,000 Ib/in:

?RO?SE'1'S

P.\RP."M_~lESn

::;,

> ACTLVE: ACTSE:7 Po.C'!'SE'I', Re. 5 !1ESHING > PAR.A'!'LM£SH M_I?'T·, 4., 4, 1

PROPS2TS ' EGR00? SCRO~P,

~

CONTR.OL

(4) Define element group I using the SPRING element formulmion with two- nodes and element group 2 using (he MASS etemem fomlUlation:

EGROUf:'. 2, HASS,

3,

I

(12) List (he natural frequencies of the system: RESULTS > LIST ) FREQLXS':'

Frequency l 2 J

Frequen.cy (Raciisec)

Ft'eC;:".lency

Period

(cycle/sec)

(seconds)

2.S1B60e+Ol

4.00848ef-OO

::.49471e-Ol

S.€9871e .... Ol 8.J6760e,..Cl

9:.Q697ge+OO 1.J3174e+Ol

1.10256e-01 7,50595e-02

304

Free Vl!Jratlon of a Shear Building

Structures Modeled as Shear Buildings

constit~tes the ~odal matrix [
(13) List the ,he mode shapes of the system: 'E\ESU:'TS >

LIST

>

DISLIS'r

, 1, 1, 1, , , 1, 0 2, c, 1, 4, ", 0 , DISL!ST, 3, 1, l, " 1. 0

DIS1,:31', D:;::S;:'IS'l',

)~o,-;:Q

,

1 . SBe-:>"

?;':

',>?

~.(j0e~()t'

'J. :;:;"" Oi}

,

;) .::::;e~v7

:}~"'~oc

S.5s",~i)~

:>;e'(;'

0. JGe~";; ,;;;;",~:)C

:).00>[<·00

0De-cO ~".;;~e-(',:

-2.';'0",·02 :,91;,.··n

,f:[

,,1-

OOe·CC

,()C'?';::;

C .C(j"TCO

0. ::H;e~00 . O(\e.~0

. ::Je~C;;

:;.OO~<-C

.0;:e·00

O.~Q"·:»)

D.OO.:~{)D

G.CJe~;'J0

>Iode sn??'"

,

G.00e C:;

:;:,:

;IX

llY

?-2

1

,

.OCe~a0

,).OOe~C::;

";y

FK

{¢}TrM](¢};

~h",?'"

'.:0:

',)X

M

'L(;~~·O:;

O.0)",·JC

:LME"~O

;;

G.OQM00

(; .:C""C·(j

"'.C::;o-;)<:

J,

OOe~DC

D ~:e·(() () .:;:;..,.::;0

3D"!->-OO

J.CCe+CC

:;. C·:,.,-":; C .»()ii~,";; .; ,;:;e*O;;

t"::>;),,, Co:.

,.

OCe~C,;

0.00'11-00 ,'!o:le sic"p'"

,w

0%

;;":'

:';1.

rtx

0.309-00

C _0Cq~CS

O. CDe-JC

(; . (;C",,,,C;)

j.GJe w ;)0

,

-3.66.;\-:)2

C. G()e .. Ct

·?0]f~C()

c.rte-C]

;;. DOe.O:::

C .C~",·CO

.,

0.:;':2-00

:.OiJe~OC

G. aCe'CU

C.CC?-Co

C,;)0e~:·!)

Ooe~DO

G.00".]C

C. ;;C",.JJ

c. :)O~~~~

.). :;Q!!-_()J

L}i?e~:::

}%,n

,

L

;=1,2, .. ,n

where the normalized modal vec~ors {
.~()"'~S'c:

(loe~C(,

;;,00>1->00

().

305

{¢};(M]{¢h=O

for i

{¢h;M]{¢},=!

fori=J

{d>};[X]{¢};=O

for iFj

ar.d

l\Z C~",-Oj

{¢};[X] {¢}j

wi

for i=j

The above relations are equivalent to

10.6

r
SUMMARY

The motion of an undamped dynamic system 1n free VIbration is governed by a homogeneous system of differential equations which in matrix notation is [M) {y}

+ [X] (y)

= {OJ

The process of solving this system of equations leads to a homogeneous system of linear algebraic equations of the fonn (lK] - ,} [,11]) (o}

{OJ

which mathematically ~s known as an eigenproblem. For a nontrivial solution of this problem, it is required that the determinant of the coefficients of the unknown {a} be equal to zero, that. is,

I. [Kj-

wr

w1 (MJ

[1]

and

in which [
I= 0

The roots of t,1)is equation provide the natural frequencies Wi It is then possible to solve for the unknowns {aJ, in terms of relative values. The vectors {all corresponding to the roots w~ are the modal shapes (eigenvectors) of the dynamic system, The arrangement in matrix format of the modal shapes

PROBLEMS 1O~1

Determine the natura: frequencies and nOf":llai modes for the two-stor}' shear buiIding sho\."itn i:l Fig. PI O. L

306

Free Vibration of a Shear 8uild:r.g

Structures Modeled as St:ea: Buildings

307

k:] "" 500 lotin. '"1 '"

FI,,,,,., "" 5

i

Ib

k<;.l

lin..

t 1 '" 5GO lb/in.

\5"

!

Y,

/).7j>y)/,// k I ""

Fig. P]O.!. 10.2

1

~

X 10" 11) in?

-----.:

5{]

A certain SlruC!UrC has been mou
500 Iblll'L

Fig. PIOA.

10.5

Consider (he uniform shear building in which rhe mass of each Door is m and ~he

stiffness of each slOry is k. Determine :he general fonn of the system of differentia! equations for a uniform shear building of N stories,

fil

KI;n.l~ 4() KI;o r K

~

I..

10

II

:.:0 KIln.

!D.G

~._.

10.7

10.3

Wrile a main computer program which will use Subrouiine JACOBi. of Program 8 to caku;me rhe natural frcque:,.cie!> <.tnd r:iodnl shapes for a uniform shear blli!dir.g as described in Prob:em 10.5. Express (he natural frequencies in fu:,xlion of lhe quantity kIll). Modify the computer program required in Problem 10.6 to incorporate lhe

option of seHing: Ihe ratio kIm to a value that will result in the uniform shear bu:lding having a prescr:bed fundar:,en!al natural period T sec/cycle.

Assume
10.8

Write a main compute, program thnl will use Subroutine JACOSr of Prog~am 8 to calcu!ate lhe natural frequencies and modal shapes for a shear buildir;g of N stories in which the stiffness of each story changes linearly from k at the first !;!ory to k,'l rH [he Nih slOry. Assume equal mass at ench floor.

10.9

10.10

lQA

the shear buijdi~g having a prescribed fundamental natun!: period T sec/cycle.

j

10.11

~ i

Find the naturaf frequencies and modal shapes :or the three·degree-of-freedom shear building tn Fig. P! 0, 1 I.

10.12

Use the results of Problem 10.11 to write Ihe expressions for the free vibration displacements)'\o Y2, and)'1 of ,he shear building in Fig. PIO.II :n terms of constants of integra(ion.

10.13

The stiffness matrix (Xl nne the milSS matrix (MJ of u s{ruc~ure modeled .lIS a three-degreewof-freedom SYStem are given respectively by

A movab~e strucmrtt! frame is supported on rollees as shown in Fig, PIDA.

Dete-:mine natura! periods and co::esponding normal modes. Assume shear building model and check answers usi!\g Program 8.

Modify the compu(er program req:;ested in ?roblem !D.S to inco:porate the

option of seuing: :he m.:lgnirude of the Ooor mass to a value that will result in

,

Fig. PIO.3.

Modify the computer prograr:i requested in Problem to.a for the case rhat the mass in each floor vnries linearly from m at the first floor to m,. at the Nth floor.

30B

Structures Modeled as Shear 8uildings

Free VibraHon ot a Shear Bulicing

10.15

Solve the eigenproblem corresponding to a structure for which [he stif!'ness and ma:;s matrices are: 1263,4

[K]

0 i,052,800

=r

~

631.68

15.192

0

0

15,792

263.200

0

0

263.4

0

15.192 15.792 256.300

631.68

1.052.800

ISymmetric

~

263.4

L

1.05~.800J

and

r

3.7143

;

Fig. PIO.Il. [M]= ;210

[K]=

lot~ "

.'

o

5

210

5

5

J' (Ib!in)

and

23.070 [M]

0

o 23.070

=

1886

1~86

1886: 1886\ (lb, sec1 /in)

187.200,

Oeterr::tioe the natural frequencies and corresponding modal shapes of the smJc-

ture. 10.14

Solve the eigenvalue problem co;responding ness and mass matrices are

r [K]

l

18:500 ~ 12.0000 51430

~

12.0000 15.0000

~

12.0000

LO

a srructure for which

:ne stjff-

! :~~~~~j 5.1430

and

[.0\1) =

l

0.8169 0.1286

~ 0.0740]

0.1286 OJ1571

0.1286 0.8169

0.0740 0.1286

I ISymme;ric '-

0.333

,. ~.

309

0

0.6429

238.10 7.7381

37143

7.7381 ~

89.286 0

138.1

0

0 0.6429 7.7381 3.7143

~7~3J 8~2861 238.1

!

~

Forced Motion of Shear Buildings

11

normal mo~es of vibration. We retun to the equations or marion, eq, (9.3), which for the partIcular case of a two-degree-of·rreedom shear buiiding are mlYj + (k ,

Forced Motion of Shear Building

311

+ kl)y,

fl/l,V'l -

k'JJl

k:tYl = FI (r)

+ klYl

F2 (I)

(11.1 )

. We seek to transform this coupled system of equation into a system of mdependenl or. uncoupled equations in which each equation contains only One unknow.1 f
(1 1.2)

Upon substitution into eq. (l 1.1), we obtain

In the preceding chapter, we have shown lhat the Free motion of a dynamic system may be expressed in terms of free modal vibrallons. OUf prese;1t interest is to demonstrate that the forced motion of such a system may also be

expressed in terms of the normal modes of vibration and thal (he total response may be obtained as the superposition of the solution of independent modal

To ~et~rr:ltne the appropriate functions 2, (t) and z'J (t) that wiH uncouple eg. Lt 15 advantageous to make use of the orthogonality relations to separate the modes. The orthogonnJity relations are used by multiplying the first of eq. (,11.3) by ~II a~~ rh~ second by ali. Addition of these equations after multiplica-

OJ ..J),

equations, In other words, our aim in this chapter is to show that [he normal r:1odes may be used to transfo:Tn the system of coupled differential equations into a set of uncoupled differential equations in which each equation contains only one dependent variable" Thus the modal superposition method reduces the

tion and Sl:nphf,eutlOn by using eq (1025) and (10.27) yields

problem of finding the response of n multidegree~of-f[eedom system to the determination of the response of slngle-degree
Simi.larly, multiplying (he first or eqs. (j 13) by

(IlAa) ~~'n

aIL

ar.d the second by

a

~

we

(I lAb)

1 L1

MODAL SUPERPOSITION METHOD

We have shown that any free motion of a rr.ultidegree-of-freedom system may be expressed in terms of normal modes of vibration. h witi now be demonstrated that the forced motion of such a system may also be expressed in i,erms of the 310

The results. obtained in eqs. (11A) pennir a simple physical interpretation. The force (hat IS effective in exciting a mode is equal [0 the work done b t" t 1 f " y "e ~x erna. or~e ?Isplaced by the modal shape in question. From the mathemm~ leal pOint or vle~, what we h~v~ accomplished is to separate or uncouple, by a change of vnnables, (he onglnal system of differential equations. Conse-

312

Forced Motion 0' Shear Buildings

Structures Modeled as Shear Bu(1dngs

quently, each of these equations, eq. (1 1.4a) or eq: (l L4b), corresponds to a slngle-degree~of~freedom

system which may be wntten as

3,3

root of the sum of the squares of tte modal contributions (SRSS method). Thus the maximum dis.placements may be approximated by

MIZI+K1ZI=Pl(t)

(! LS)

M?Zl + K z::::) = P1 (I) where Ml

. . ~1' = rnltlil') + m2(/ZI2nne. ;1"1 -

2 _ 2 .h~ In:Cln ' mlQ12
m'"o' ............,..

rr"~c.s· ~,

KI :::; w MI

and Kz = w~M 2> the modal spring constan[s~ and Pi (r) :::; a nF I (I) + (/2fF:. (t) and P? (I) = tl12F! (f) + <122F2 {t), the modal forces. Alternarivejy, if we r.ad u~ the previous normalizatior., eqs. ([0.29) or (1030), these equ[trions may be wntten

simply as 21 + l2

Wlzl : ;: ; PI (r)

+ W;Z1

=

Pl

(I ;6)

(r)

where P 1 and P2 are now given by

P,

~

(11.7)

¢"F, (I) + ¢"F, (I)

The solution for the uncoupled differential equations, eqs. (1 LS) or eqs. (! 1,6), may now be found by any of the methods presented in the previous chapters, In particular, Duhamel's integral provides a general solution for these equations regardless of the functions describing the forces acting on t~e struc~ ture. Also, maximum values of the response for each modai equation may readdy be obtained using available response spectra, However, the superposi~ tion of modal maximum responses presents a problem. The fact is that these modal maximum values will in general not occur Simultaneously as the transformation of coordinates, eq. (I 1.2), requires. To obviate the difficulty, it is necessary to use an approximate method. An upper limit for the max~mum response may be obtained by adding the absolute values of (he maXlmum modal conl;ibutions, that is, by SUbstituting ZI and Zl in eqs. (I L2) for the maximum modal responses [:1:\"", and Zlm,) and adding the absolute vaiuts of the terms in these equations, so lha( YlmlV'

= ;
+ !
Y2m~~ ';:;; !
I

¢2ZZ2mu

I

and

7 I

(11.9)

The resutts obmined by appiication of the SRSS merhod (square root of t::e sum of [he squares of modal contributions) may substantially underestimate or overestimare the rowl response when two or more modes are closety spaced, In (his case, anot::er method known as the Complete Quadratic Combination for combining modal responses to obtain the total response is recommended. The disct,;ssion of such a method is presented in Section 11.6. The rransformation ftom a system of two coupled cifferential equations, eq. (Il.l), to a set of two uncoupled dif:eremial equations, eq. {l1.6), may be extended to a system of N degrees: of freedom. For such a system, it is particularly convenient to use matrix notation. Wi~h such notation, rhe equation of motion for a h'ne,Ur system of N degrees of freedom is given by eq. (9.3; as

[.111 {Yl + [K] (y) ~ {FUll where [Ai] and [K} are respectively the mass and the stiffness matrix of the system, {FCr)} [he vector of extemal forces, and [y} rhe vector of unk.'iown cisplacements a~ the nodal coordinates. Introducing into eq. (l L! 0) (he linear transformation of coordinates {y} = [c.b]{z}

(I U 1)

in which (
In + [K] [c.b]{zl = (F(I)}

(ll.l2)

The premultip!icalion of eq. (11.12) by the transpose of the ith modal vector,. {
(11 8)

The results oblained by this method will overestimate the maximum response, Another method, which is widely accepted ar.d which gives 2. reasonable estima~e of the maximum response from these spectral vaiues, is the square

'TIle onhogonality conditions between normalized modes, eqs. (103 nand (10,33), imply that (l ];4)

StrlJ~lJres

Forced Motion of Shea( Buildings

Modeled as Shear BuildLlgs

315

and (! 1.15) 20'

CQnsequently, eq. (I Ll3) may be wr.tteo as

1.2,3,· " N

10"

(I U6)

where the modal force P; (I) is giver. by

Equation (11.16) constitutes a set of N uncoupled or independent equations of motion in terms of the modal coordinates Zi. These uncoupled equations, as may be observed, may readily be written afrer the narural frequencies to and the modal vector {¢}; have been determined in the solution of the correspond~ Lng eigenpmblem as presented :0 Chapter 10.

,OJ

(,)

(11.17)

Fig. 11.1 Shear building wich impulsive loadings. (a} TWO-!jWfY shear buUding. (b) Impulsive load:J)gs.

The 5ubstitu(ton of :hese values: into the uncoupled equations of motion, eqs. (i 1.6),

:, .;- 140" = 2270/(1) Exampie 11.1. The two-story fram~ of Example 10.1 is acted upon at the floor levels by triangular impulsive forces shown in Fig. 1l.1. For this frame. detemline tht! maximum floor displacements and the max!mum shear forces in the columns.

Solution: The results obtained in Examples iD.I and 10.2 for the free vibration of this frame gave [he following values for the natural frequencies and normalized modes:

w = I 1.83 rad /sec, ¢"

~

0.06437,

¢"

~

0.08130,

W:t = 32.89 rad /sec

= 0.0567 t/>n = - 0.0924

1281/(1) in which /(1) ~ 1 - III" for IS; 0.1 "nd fU) ~ 0 for I> 0.1. The maximum values for Zt and z~ are (hen obtained from available spec~ral charts such .as the one shown in 4.5. For [his exam?le, I,

= lO,OOO (l

F, (I) = 20,000 (l - I lid) Ib,

Id

0.1

TJ

O.l91

- = - - .. =0524

1>"

- Ill d ) Ib

0.188

and

in which the modal natural periods are calculated as

The forces acting on the frame which are shown in Fig, 1 LI(b) may be expressed by

F, (I)

0.1

T,

211' T\ = - ..- = 0.532 sec w, From

and

2IT T, =-- = 0.191 sec

w,

4.5, we obtain

for I s; 0.1 sec 0.590

in which

fd

= 0. 1 sec and F,(t) = F,(I) =0,

fOfi>O,1 sec

122

316

Structures Modeled as Shear Buildings

Forced Motion cf Shear Buildings

where in this cas.e the static deflections are calculated as

Fo.. z!q

=

?'770

The max.:mum shear force

Vi!

at story i corresponding to mode j [s given by

- 1281

(l U9)

_ . _ = -1.18

16.3 wf =----= 140 '

1082.41

where Z:;,.,..u is tl':e m~mum modal response, (,; -
Then the maximum modal response is

" = 12£1, _ 12 X 30 X 10' X 248.6 - "--·(15 X IZ)j

15,3451b/in

""I

As indicated above these maximum modal values do not occur simultaneously and therefore cannot simply be superimposed to obtain the maximum response of the system. However, an upper limit for the absolute maximum displacement may be calculated with eqs. (11.8) as Y,~,= 10.06437 X 9.621

v" = Vim"

I

I

9502 Ib

= ,9502' -;.

1253' - 95841b

k, = :2El, _. 12 X 30 X ]0' X 106.3 , (lOx 12)'

A second acceptable estimate of the maxir.lUm response ~s obtained by taking the square root of the sum of the squared moda! contributions as indicated by eqs. (U.9). For this example, we have

22,146Ib/in

V" = 9,62 X (0.08130 - 0.06437) X 22,]46

V" Y'm" = .) (O.o8i3·O XT62)':;' (

~

and for a column in the second story

0.092" x 1.44 i = 0.92 in

Y'm,., = /(0.06437 X 962)' + (0.0567 X 1.441'

9.62 X 0.06437 X 15.345

V" = 1.44 X 0.0567 X 15345 = 1253 lb

+ iO.0567 X 1.44! =0.70 in

Y'm" = 0.08130 X 9.62[ +

317

0.62 in

=

= 3607

Ib

1.44 X ( - 0.0924 - 0.0567) X 22.146 = 4755

V'm~ = ,:3607' + 4755' = 5968 lb

(al

0.0924 X 1.44) = 079 in

112

The maximum shear force Vrna" 1n the columns is given by

RESPONSE OF A SHEAR BUILDING TO BASE MOTION

(IUS)

in which k is the stiffness of the story and .1y the difference between the displacements at the two ends of the column. Since the maximum displacements calculated as jn eq. (a) may have positive or negative values, the relative

displacement Liy cannot be deterrr.ined as the difference of the absolute

dis~

The response of a shear building to the base or foundation mot1on is conveniently obtained in terms of floor displacements relative to the base motion. For the two-story shear building of Fig. 1L2(a), which is modeled as shown in 1 L2(b). the equations of motlor: obtained by equati~g: to zero ri:e sum of forces in the free body diagrams of Fig. 11.2(c) are the folJowing:

placements of the two ends of the column. The maximum possible value for Jy could be estimated as tr.e sum of the absolute maximum dj5placements at

the ends of the columns. However, this procedure will in most cases greatly overestimate the actual forces in the columns. The recommended procedure is first to calculate the shear forces in the columns for each mode separately and then combine these modal forces by a SUItable metbod, such as the square root of the sum of the squares of modal contributions. This procedure is based on the fact that ::nodal displacements are known with their correct reLative sign and not as absolme values.

(11.20) where Yfi = ys(t) is the displacement imposed at the foundatiot. of the structure. Expressing ~he floor displacements relative to the base motion, we have

liZ =)1, -

)1..

(1121)

318

St,uctures Modeled as Shea: 8uildhgs

Forced Molion o! Shear 8u,!dings

where F, and

319

are cniled parricipQriort factors and are given by mrCt~i

+m 1Q l1

mIG;I"+- mla il

,

,

m:G,!! -r m}a~l

Fig. 11,2 Shear building wilh bn.;;e motion. (a) Two~story

S!1e.1l

buiidi:lg. (b) M~\lh·

Tr:e relation between rhe modal displacements ment U I, !f; is given from eqs, (l J .2) DS

£" z~

(l126)

and the relative displace-

emu(jcnl m:)deL (c) Free body diagram.

(1127)

Then dlfferenti3tioJ) yields

(I L22)

Substitution of eqs. (lUI) and (I i.22) into eqs. (I :20) results

In practice I( is convenient to introduce a change of variables in eqs. (l1.25) such thnt the second members of these equiHions equal )11 (t), The required change of variables to accomplish this simplification is

In

(1128)

(11.23) We no:'e that the right-hand sides of eqs. [11,23) are proportional to the snme function of time, y:(!). This fnct leads [0 a somewhat sirn;?ler solution Com~ pared to the solution of eqs. (1\.6) which rr;r.y contain different functions of time if! each equation. For the base motion of the shear building, eqs. (11.4)

may be written as

which when introduced into egs, (I! .25) gives

gl l w~g) =)1: (I)

81 + w~g:! = Y,{t)

(! 1.29)

Finally, solving for g( (/)
(l L30) ( 1124)

or

(1125)

When the maximum modal re.sPO:1se glrnJ< and &:m;u are obtained from s,pectral charts, we may estimate the maxi;-r:'Jm va!ues II rm~~ and !fJm~~ by the SRSS combir:atlon method as

(lUI)

Forced Motion of Shear Bu,'ldi:1gs

Structures Modeled as Shear Buildings

320

m,;

[

321

where y~ y, (r) ;$ the acceierarion function exciting the base of the struCture. Equations (11.34) may conveciemly be written in matrix notation as

--YIf

~i

.~.~

[NfJ {ii} + [K] {u} ~ - (M] (I) LeI)

~;';'..-~'-- ---y"",

,.

in whtch It,,n, the rr:ass matrix:, is a diagonal matrix (for a shear building), [KJ, the stiffness n1:2..lrix, is a symmetric matrix, { I} is ;l Vector with all its elements equa! to 1, Y. = j.\ (£) is the applied acceleration at the foundation of the bui:d~ ing, and {u} and {it} are, respectively, the displacement and acceleration vectors relative to the mOtion of (he foundatico. As has been demonstrated, the system of differential equaiol1s (i 1.35) can be uncoupled [hrcugh the rransformmlon given by eq. (11.1 J) as

~

(11.36)

Fig. 11.3 MU!listory shear bl1.ilding excited Free body diagram.

ilt

where [cP~ is the modal matrix obtained in the solution of cor.esponding eigenproblem l[K] -. w' [I,'f}] {"'} = :O}. The substitution of eq. (11.36) into eq. (I 1.35) followed by premultip!icalion by the transpose of {he hh eigenvector, {¢};' (the ith modal shnpe),

the foundation. (3) $Iruclural mooeL (b)

results in (11.37)

The equations of motion for a;) N-srory s~enr building .(Fig. 11.3(a)) sub~ lected to excitation motion at its base are obtawed by equatmg to zerO the sum

which upon introduction of onhogonalily property of the nonnalized eigenyec~ tors feqs. (11.14) and (1r.:5)1 resu![s in the modal equations

-of fcrces shown in the free body diagrams of Fig. 11.3(b), namely

m,y, "-k,(y,-y,)-k,(y,-y,)

0

mJ, + k,(y, - y;) - k)(y) - Yo)

=0

" mN-l'iN-1

-

+ k N·-1 i" vN-i -

ml'/YN + k,v{Js - YN-I)

,. ') .rN-2

(11.38)

where the modal tJartidpation factor 1; is given i:) generai by

kN ,(y , v - y .v-I ) = 0 (11.32)

=0

Introducing ir.to eq. (Il.32)

(11.39) (i

!tl=Y"-y;

(1l.33)

L 2, "', ,"I)

results in mlii;+kILl!-kl(tlz-lIl) m?ii2

+ k:.u'1

K.) Cu) -

f(1)

and for normalized eigen¥ecrors by

-mlys =

- mij,

T,,= mtl_1iiN_1+kH- (UtI_l-lIt1_·J mN!tH+ kN(fI;,,,-UN-l) =

-m,"/Y$

kN(U",

U,y_:)

=

m,,,,,,

(i = 1, 2, 3, .. , N

(11.40)

-mN-IY5

(l134)

,

The maximum response in terms of maximur:l values for dispJacemenls Or for

322

Structures Modeled as Shea: Build:ngs

Forced Motion of Shear Build:ngs

accelerarions at me modal coordinates calculated by the SRSS method given, respectively. by

~s

then

323

The modal equations (11.29) are

8, + 140g, = 108,47

ii, + lO8ig,

(l1.4 i)

108,47

(bl

and their solution, assuming ZerO initial conditions for velocity and displace~ ment, 1s given by eqs. {4.S) as

and

108.47 8, (I) =·---(1

( 1142)

140

where So/and SA, are, respectively, the spectral displacement and spectral accelera[ion for the jth mode. The participation facto;s Jj indicated in eqs. (11.39) or (11.40) are the factors of the excitation function y, (!) in eo... {ll.38). As presented in Chapter 8, response spectral charts are prepared as the so!ution eq. (11.38) (wlrh for vfllues of ine natuml frequency w, in the range of interest Thecefore, the spectral values obtained from these charts Sm or SA; should be multiplied as indicated in eqs. (11.41) and (I L42) by the participation factor 1~·, which was omitted in the calculation of spectral vaiues.

r, ')

Example 11.2. Determine the response of the frame of Example 1 L 1 shown in Fig. 11.2 when it is subjected to a suddenly applied constant accel~ eration ]1 = 0.28 g at its base.

Solution: The nat'Jral frequencies and cOITesponding r:oIT.:J<11 modes from cc.lc'Jiations in Examples 10. J a~d 10.2 are WI

= 11.83 rad /sec,

f.!J?

= 32.89

q,,, = 0.0567

<{;" = 0.08130.

q,,,

-

386 = 108.47 inlsec'

The participation facto:·s (Ire calc':Jiated from eqs. (11.39) with the denornina~ ~ors set equal to unity since (he modes are normalized. These factors are then

F,

(136 X 0.06437 + 66 X 0.08130) = -14.120

F, =

(136 X 0.0567 - 66 X 0.092<)

- 1.613

(c)

-14.12GXO.06437XO.775(J -cos 11.83£) - 1.613 X 0.0567 X 0.100(1 -- cos 32.891)

ti, (I) ~ - 14. ,20 X 0,08130 X 0.775 (l - cos 11.83i)

+ J.61"'XO.0924XO.lOO(l-cos 32.891) or, upon simplification, as

=

0.7135;- 0.704 cos J 1.831 + 0.009 cos 32.891

- 0.874 '" 0.900 cos J 1.831- 0.015 cos 32.891

(dl

In this example, owing to [he simple excitation function (a constant acceler~ a[ion), it was possible to obtain a dosed SOfU(lon of the problem as a functIon

The acceleraliorl acting at the base of this Structure is

y, = 0.28 X

U,(I)~

U, =

0.0924

cos 32.891)

The response !n terms of the relative motion of the stones .<:.t ~he floor levels with respect to the displacement of {he base is giver. as a f1JTIction of tlme by eq. (i 1.27), and rhe appropriare consideration of (he participation factors, as

i<,

rad Isec

.p" = 0.06437.

i08.47

g,(r) = 1()il2(1

cos 1 1.831)

of lime. For
(n)

glrnJ<

L5S 0.20

(e)

324

Forced MOlion 01

Structures Modeled as Shear Bui,dings

and the maximum response, calculated from the approximate formulas: (I 1.31), is

Sneer 8uild:t1£s

325

The mass matrix is

.

1M]

01 = [136 0 66

(f)

The possible maximum valves for 'he response calculaled from eqs. (d) by setting the cosine functions to their maximum value resuit in U;i1\~"

1.426in

ti;m4>-

1.789 in

(g)

which for this particular exampie certainly compares very well with the approximate results obtained in egs. ([) above.

11-3

PROGRAM 9-RESPONSE BY MODAL SUPERPOSITION

Previous Lo the execution of Program 9, it is necessary (0 execute one of [he programs to model the S;fuCWre (for example, Program 7 [Q model the structure as
Nl.I!<.Ss."( OF Dr:GR2Z$ OF FR£ZOCX <::G}:J>'V~,-u£s

Program 9 calcvlates the response of a linear system by S-UperposiE~on of the Solulions of the moda! equations. Before one Can use this program. it is necessary to so! ve an eigenproblem to determine the nat:Jral frequencies tnd modal shapes of the structure. The prog:am determines the response of the structure excited either by time-dependent forces .appJied at nodal coordin.ates or a time-dependent acceleration at the support of the structure.

?lODE SH;>stS RY RGVS' C.O~:;~?

C.OiEJG

0,0%70

-.O~2~:)

n•

..-r.

OAT",

;';UI1&2R .?F Cac.R£F$ CF I""!'EEC(:11

Example n.3. Use computer Program 9 to find the response of the frame analyzed i:1 Example 11,2. Solmion," The natural freqaencies and the modal matrix for this struC[Ure as calculated in Example 10.1 are

!'I.iMlJ.SR

or

2XT!;;:?;)",;,,,, 701
TIro;<:- $1'£;> 01"" :Nif:CAATtov GRA":':';;'7!;)..""!.. WDEX

FORCt #. COORD

,

,., ;"-<'I£R<: FOr!':;:;;: 1$ A?JiI\.:EP. (HJ;t. \:;1" (>(',:H'tt CB7H1::;,,-c 'the fORe"

(VIIi .83 rad Jsec Wz

32.89 tad Isec

and

[ (ill

[0.0644 0.0567] ,0.0813 -0.0924

G.OO

O.2Si)

l.IJO

O.2!lO

Forced Monon of Shear Suildings

Structures Mode-led as Shear Buildings

326

~,.

COORD

Example 11.4.

elSE.

1"";'1;.

'laoc,

1.405

a.

1. na

,0. oS"

i"..AoA.

327

Ace.

19~

m,

Solve Example 1 i. I using Program 9.

Soiwion:

1

Input Data and Output Results

>,

Fig. 11.4 Shear bl1llding \vich harmonic loading. I:;PUT

~A'1'.A:

mw.EER Of DEeR:;':" Of 'fREE:OC+J

rm~~

~;'Jl'!eER

NE'~2

O' £x7tf'.riA!.. fOi'..cES

'l'!M£ 5:-:;:1> OF'

U • . OJ,

r~:'l'Ii:::;flA1'!Otj

C~C

GRAV!1'h1'IOtl';L HiQ£;{

force F F'J sla Wr which is i.:pplied at ~he level of the second floor. In this = Fo s:n Wf become case eqs. (l L I) With FI (I) 0 and

r;

(11.43) Fo: the steady-sli).te response we seek a solution of the form I)

OQOO"v(lO~

.OOM

I). tO~o

c.oooo

J.4000

{l.jCCOl

3.0000

2.000C.:HiCO

J.!OOO

v.coco

C ••WOO

:l.onno

Y::

= y.~ sin WI

(11.44)

A:rer substitution of eqs. (l1.44) inro eqs_ (11,43) and cancellation of [he commOn factor si;1 wr. we then obttin

!;:XC!1'>.rtON

(k, + k, -

m,w') Y,

k,Y,

=0 (11.45)

tv.X. V£i.OC.

H.:'>.x. Ace,

c.s-nn

?64.33

l1>'7.{10C

o. i5t\$~

~J. ~851

:'!AX.

11.4

!HS?t...

HARMONIC FORCED EXCITATION

When :he excitation, that is, (he external forces or base motion, is ha:r.uook (sine or cosine function), the analysis is quite simple and [he response can readHy be found without the use of modd analysis. Let us consider [he two'-s£Ory shear building as shown in 11.4 subjected £0 a single harmonic

which is a sysrem of two equations in two unknowns, Y I > and Yl - This system always has a un:que SOlutiOn except in [he case when the determinant formed by the coe:ficiems of [he unknowns is equal :0 zerO, The reader should remember [hat in this case the forced frequency w would equal one of the natu!'al frequencfes, since this de(erminam when equated co zero is precisely [he condition used for determining t:-te natural frequencies. In other words, unless the structure is forced to vibra(e at one of the resonam frequencies, the algebraic system 0: eqs. (l L43) has ~ unique sol:Jtion for YJ, and Y::. Example 11.5. De(ermine the steady-state respor:se of the two-Story shear LO,OOO sin 201 is appHed to [:te buildir:g of Example 10.1 when a force F?(r) second story as shown in Fig_ [lA,

328

Fo:"ced Motion of Shear Buildings

Structures Modeled as Shear Buildings

Solution: The natural frequencies for this frame were dete:mined in Example 10, I to be fiJi

= II,83

fJ)2

= 32.89 rad/sec

~(tl

~

_____

m-,'~_-;>--+

329

V,

rad/sec F;{tl

Since the forcing frequency is 20 rad/sec, the system is not at reSonance. The steady-state response is then given by solving eqs. (11 A5) for Y! and Y::. Substituting numerical values in this system of equations, we have (75,000 - 136 X 20') Y, - 44,300 Y, = 0 - 44,300 Y

+ (44,300 - 66 X 20') Y,

10,000

So[v:r.g these equations simultaneously resr.;.lts in 0.13 in

Therefore, according to eqs, 0

~

44), the

steady~state

response is (b)

y,

=

0,28 sin 20l in

Y2 = ~ 0,]3 sin 20t in

(Ans,)

Damping may be considered in the analysis by simply including dampjr,g elements in the ~nodel as is shO\~ln in Fig, 11.5 for a two-story shear building. The equations of motion which a:-e obtained by equatir.g to zero the sum of the forces in the free body diagram sbown in Fig. j L.5(c) are

:t

m,Y, + (0,

+ (k, + k,)y,

-

= F; (t) + klY2 = F, (t) - k,y2

(ll ,46)

Ie)

Fig, 11.5 (a) Damped shear Ol!ilding with harmonic load. (b) Multidegree model, eel Free body diagram,

Now, considering [be general case of applied forces of the fo!T.l given by

Real {(Fe

iF;)e'Wt}

(l1.48)

with the tacit understanding tbat oniy the real part of the eq. (11.48) is the applied force. Vile show that [he real part of the complex force in eq. (11.48)

=

Real {(F"

= F"

(11.47) we, conveniently, express such force in complex fonn as

= cos W[

is precisely the force in (11.47). Using Euler's formuia we obtain

cos

!F;)(cos

wt + Fs sin

[;J{ -7

mass~$pring

+ i sin

rut,

i sin Wi)}

(11.49)

lu(

wbet, is the expression in eq. (l1.47). Assuming that the forces FI (I) and F~(t) in eg. (J 1.46) are in the fonn giver. by eg, (11.47), we substitute eg, (l L48) into eq, (l 1.46) :0 obtain mlil

+ (c: + c:) Yl + (k + k:) Y: j

mlh - c:!J'; - k:YI

~ cJ: - k;:)'-;

(Fd - iF,I) e'i;,!

+ c:!.-y: + ki:ll = (Fd

i

iFde ''''

(11.50)

330

f"orced Motion of Shear Buildings

Slfuctures Modeled as Shear BUi!dings

The solution o~ the complex system of eqs. (11.50) will, in general, be of the form Y{i) = (Yo: +- iY,;)e''&':. Since on!y the re.al parts cf::he fcrces in eqs. (1 LSO) are applied, the sOlution is also only the real part of y (i). Then, analogously to ego (1 1.49). (1151) For the stea.dy~s~ate response of eqs. (11.50), we seek so;utlor!s 1!1 the form of

Solwiofl:

The su::,stitution of egs. (11 Yz, namely

The damping constantS are calculated as c, = a01< ,

3071b·seclin (a)

The substitution of numerical values for this example into eqs. (i J .53) gives the following system of equations: (20,600 + 15.000i) Y, - (44,300 + 8860,) Y, = 0

Y\ = Yle'~ Y2::::

331

yl/,j.;

- (44,300

(11.52)

aed the first and second derivatives of YI and

~

8860i) Y,

+ ( 17,900 + 8860£) Y, = - lQ,OOOi

(b)

The solution of this sys[em of equations is

y, = 00006814 + 0.26865; Y,

into eqs. (l: .50) and leteing Yl Yd + i Y,I and Yz ;;;;: Yc:; + i Yib reselts 1:1 the foEowing syS(em of complex algeb:-aic equfI[ions:

=

0.06309

+ O. 137775i

Therefo::e, by (he re:2.tio:1 established in eq. (j [.54), [he steady-stare response is given by

= 0.0006814 cos 201 y, (f) = - 0.06309 cos 201

y, (r)

(l1.53)

Ytt cos fur - Yt • sin

Y: (fJ

Yd cos

wr -

wt

Ys1 sin Wi

(11.54)

in wpjch YI = Yt ' l + iYs !, Y2 Yel + iYs:; is the solution of the complex equations. (11.53), The necessary calculations are better explained lhrough the use of a !1ur:1ericai example.

Example 11.6. Determine [he steady-scale response for the tWQ-story shear building of Example 11.5 in which damping is considered in me analysis (Fig. 11.5). Assume for this example that the damping constants Cl and C2 are, respectively, proportional to the magnitude of spring constants k; and k'l in which the f.acmr of proportionality, ao = 0.01.

0.137775 sin 201

which also may be wrtnen as

y,

AS aiready stated, the response is (hen ~ound 'JY solving !he complex system of equations (11,53) and considering only the real part of the solution. Hence, from eq. (11.51), we have

YI (t)

0.26865 si, 201

0.2686 sin (20t

+ 3.144) in

y,=O.1516 sin (20t+3.57])in

(Ans.)

Wher: these rest.:lrs a;e compa;ed with those obmined fo~ the undamped Struc~ lure in Exampie 11.5, we nOie only a smail change in the amplitL.de of motion. This is at ways the case for systems ligh(1y damped and subjected to harmonic excitation of a frequency [hal is not close to one of the natt.:ftll frequencies of the system. For (his example. the forced frequency Wj = 20 rod/sec is relativelY far from the naU!;al frequenctes WI ! l "83 rad lsec Or UJz = 32.89 rad Isec which were calculated in Example : 0.1.

11.5

PROGRAM 10-HARMONIC RESPONSE

Prog:am 10 calculates the ;esponse 10 hannonic excitations of a strucrurru sys:em for wh:ch the s:if::1ess and mass matrices have been determined by one

332

Forced Motion :::f Shear BuHdings

Structures Modeled as Shear BuHdings

y, (I)

of the programs modeling the structure (Program 7 to model the structure as a shear building). DampIng in the system is assumed to be proportional to the stiffness and lor mass coefficients, that is, the damping matrix is calculated as (11.55)

[C]=ao[M]+a,[K]

b which ao and at are constants specified in the input data. The program calculates the steady-state response for structures subjected ro harmonic forces applied at the nodal coordinates or a harmonic acceleration applied at the base of the structure. Example 11.7, Obtain tbe response of tbe damped two-degree-of-freedom shear building of Example 1 L6 using computer Program 10. Solution:

333

0.00068 cos 20, - 0.2686 sin 20t

y, (I)

0.0631 eos 20r - 0.1378 sin 20,

or y:(t) =0.2686

y,(t)

s;n (20,+3.'44)

0,1516 s'n (20,.,-3.571)

The results given by the computer, as expected. are the same as the values calcdated in Example I 1.6.

Example 11.8. For the structure modeled as a four~story shear bullding shown in Fig. 1L6, determine the steady-state response when subjected to the force FCt) = 10,000 sin 20r (1b) appl:cd at the top floor of the building. Assume dampIng proportional to stiffness (factor of proportionality al = 0.01. Modulus of elasticity E 30 X 10'" pSI.

Input D.t. and Output Results

m?""

OA'!.''; OAMr:NG S1"U'?Ne:S$ :t.;C':'OZ:

K?AC '"

.c:.

O;;''1?!NO ¥Jl.$S FACTO"

Solution.." The solution of this prOblem is initialed by executing Program 7 to model the structure as a shear building followed by the executior:, of Program 10 to obtain the response to hannonic forces. Modeling of this strucmre has been undertnken in the soiution of Example 9.1. '"

rCRC:;:p tlREQ!;;;:OIC:" l:AAP / $"C::

-~ .4300S~G{

Fltl

~ .4~CO.>o~

V

-1 I '" 7955 in.4 .......... 180 in. m"'1

V

,

,

I "" 79.55 in.

4

180 in.

-........

180 in.

m-
lDC:;:

!

V' '" 79.55 "n.

t t

..........

4

m'" \

r:. <;"S:~£<;~

w5,

':c»!?O;·'!!J7

-2 ES$S;::MC:

VI'" 79.55 in:" ____

:;;:93-~2

~

As given by the output results. of com?uter Progrz,w 10, the response for this two-degree-of-freedom system is

W

Fig. 11.6 Modeled structure

fOf

Ex<.:.mple 1:.8

334

Structures Modeled as Shear BUIldings

Forced Motion 01 Shear Buildings

Input Data and Output Results 9;>,'I'}, flLe,

D';'''i~t:'lC

Applic,:rion of the SRSS method for combining modal respo:1se generally provides a;1 acceptable eSlimation of the tOlat maximum response. However. when some of the modes are closely s~O!ced. the use of tne SRSS method :-nay result in grossly underestimating or overescim::nlng [he maximum response. In 9ar::c:.!lar, large e:Tors have been founG ir: (he analysis of rhree~dimetl$ional struct'Jres in which tors:onal effects are signifk-o.oL The tenn "closely spaced" refe:-ring to modes, intiy arbi:rarily define [he case when the difference between two no.cGral frequencieS is within lC% of the smnlles[ of [he twO frequencies. A formGlmiun known as the Complete Quadratic Combinurlofl (CQC). which is based on the lheory of random vibrations, has been proposed by KiHreghian {l98D) anc by Wilson e{ nl. (l9St). The CQC method, which may be conSidered as an eXlension of dIe SRSS method, \S give.n by the foHowing

OIO.S

S1';rrOiESS F;>CTOR

DM?LtlG h,l;SS fACWR

FOI\Ct:O fit£CUEi'K'Y

jR.~o!sec)

G.S47H:.O~

_).27)7£.01

o.OOOO1l;,QC

"1.2717£.02

i:.'H))(,O'l

-).:n31r;:.0-;:

D.OOOC;:'O()

O.OOQOR.:)()

<, ,:n}7£~02


C. )(!(;05,00

o. OCOCi;~OC

5. S1n:::.J~ .;:.:037<:,01

O.I)()002,CQ

1.C()ON:-~:'!O

J.O(ji)"£~CG

o.cuooe.oo

J .(lOOC';,OO

0.0000:::.00

!..tooo.>oo

0.0'0,,02.00

(;.000)£.00

iLoCCOE.QO

0.3000:;:.00 0.0000<:.00

'.0000£.i)(;

o.ocoo£.oo

0.0000£.30

1. ()OOOi.:.OQ

D.CI}(HH:'CO

335

equation:

(l157) COORD.

FC COVol'Otl£:4T

in whiCh the cross-medal coemCten: P,; may be approximated by

,

(1 1.58)

ljOCll

where r;;: r.vi!w, is lhe rntio of the narural frequencies or order i and j and ~i and §; the corresponding damping «1::0:; for modes j and j. For constao[ modal d4tmping [', eq. (: 1.58) reduces to

Fe COKf'CNZNT

11.6

:'LO)5".(11

2.6qn£:'OI

2 .lO{t:,c'l

1.1"70,,.01.

~.4qlE.Oj

.i.7ll211;.Ol

·"S'Jse.Ol

··3.215H:.Ol

o

,')

COMBINING MAXIMUM VALUES OF MODAL RESPONSE

S§'(; +r)r,n ..---.--C - r-f ,. r {I -+- rY!

~--..,,~~

4f

(l1.59)

It is important to note (ba(, for i eq. 01.58) or eq. (1 i.59) yields Pij= 1 for any value of [he damping ratio, including ~::::: O. Thus, for an undamped strJcture, the CQC melhod (eq, (l 1.57)) ~s idemicaJ to {he SRSS method (eq. (1;.56)]

The sqJare rOOt of the sum of squared comribctioflS (SRSS), to estimate rhe totnl response from calculated maximum modal values, may be expressed in general, from eq. (J j.41) or eq. (l : .42), as

11.7

(: 156) where R is the estima{ed response (force, displaceme:lt, etc.) 3r n specified coordinate and R; is the conesponding maximum response of the i[h mode at that coordtnate

=

I i

•

FORCED MOTION OF SHEAR BUILDINGS USING COSMOS

Exnmple 11.9. De,ermtne ~he In
Forced Motion 0:' Shea.r Bu1ld~"9$

Structures Modeled as Shear 8ulidings F

337

(7) Activate set 1 and generate three spring elements along curves I, 2, and 3: ~

CON'f?OL

ACTIVE ;. ACTSET

ACI'SET, EG, 1 AC~T'SET,

1

ReF

YlESH LNG ~ ?ARJ:IJ.:_MESH !>LCR, l, 3, 1, 2, 1, l

(8) Activate set 2 and generate three mass elements at points 2, 3, and 4: Fig. 11,7 (a) functions"

Three~story

shear building for Example 11.9, (b) Impulsive loading

k = 1500 lb lin and the mass at each fioor level is m Neglect damping in the structure,

= 0.3886

lb· sec'lin).

Solution: The analysis is performed using three spring elements and three concentrated :nass elements. (1) Set view to the XY plane:

i\C1'SE':', EG, 2. AC'T'SE'Y', Re, 2 MESHING > PARitl-:_MESH > r-:: ?T f'LPT, 2, 4, 1 (9) Merge and compress nodes:

N>:3RGE,

L

MESHIN~

>

9, :,

NOJES

L

NCO~!PRESS,

VIEVJ, 0, 8, L

~

GRID

;),

:),

0

CCMPRESS

6

0

(10) Apply constraints in all degrees of freedom at node L and in all degrees of freedom except [IX at nodes 2, 3, and 4:

(2) Define the XY plane Z = 0:

GEOMETRY

:;:,GOO1, >

>

?L?$E

PLANS, Z, :J, 1

(3) Generate three curve segments from 0, 0, 0 to 3, 0, 0: GEOME':'RY ). CURv"E'S

C?J?CORD, L

:>

DPT, 2,

4, 1, UZ,

RX,

RY,

EGROD?

EGROUP, 1, S?RING, 0, 2, l, 0, 0, EGF:OUP, 2, M1\SS, G, C, 0, C, J, 0,

c,

AKP-.L'!SIS ) o,:'j'=-'?u7~C?S " ?RINT_O?S P::=\:NT._OPS, :., 0, 0, 1, 0, L 0, Ot

°

A..."1ALY$rs ~ FE:..E<:::; /BUC3': :> Zl"_r~rtEQU2NCY A_FREQ:.J3NCY, 3, s, :6, 0, 0, 0, ;),

0,

C.

(}, 0

ANALYSIS > r:'?\EQ /8UCK

= 0.3886 Ib

R_FREQUSNCY

sec: lin: (13) List the natuml frequenCies of the system:

RCONST,

;>

RCONST

2 , 2 , 1 , 7,

3886,

(L

0,

°

0

?E\O?SETS ;. RCONS'fRCONST, L L 1, 1, 1500

(6) Define the reat constant for mass eiements: m

0,

(12) Set the options for the frequency analysis to extract three frequencies using the Subspace Iteration Method with a maximum of 16 iter2.tions, and run the frequency analysis:

(5) Define the real COnstant for spring elements: k = 1500 lb 1m:

pRQ?SSTS

!<..Z

0, 0, 2, 0, C,- 3, C, 0, 3, C, 0

(4) Define element group using the SPRl:.~G element fOnT,alation whh two nodes and elemem group 2 using the ~1ASS element formulation: PROPSETS ;.

0,

(l ;) Request ?rintout of mode sha;;es:

CR?CC2..:J

0, :::, 0, L

'J':{,

0,

0,

0,

0

~ESULTS

:>

LIST

:>

?:KEQLIS':'

1S-05,

0,

12-06,

338

f

Structures Modeled as Shear 8uildings

Frequency#

1 2 3

Frequency \ Rad Isec; 2,76500t'HOl

fr.s-quency {cycles /.secl

7.7473Ge+C1

1.23303e+Cl L78178e1'Ol

<1.10C63e+OC

]." 11953e.;.02

Period !$econdSl 2.27240e-Cl

?OST~~)yN

>

PD ..},T'::'PE,

2.

3.

"

500,

.001,

I J .33")

5.512J6e-02 5.54$$

R R R E E£ t. l l

ti[l)-iII_nt~l

?O_ATY?S

0,

0,

0.5,

0.25,

+--_+_

0 ·3

time~dependent

(15) Define the dynarnic forcing funclion as a

339

S.11010e-:J2

(4) Define analysis lype as rnodnl time history using three flt!.turai frequencies, 500 time steps s[aning at 1=0 with a time increment of 0.00 I sec; iJSe default values for a:l integration parameters; and reques: printout of relative disp1acemen!s and re!ative velociries: ANALYSIS

Forced MOlion of Shear 8uHdings

14<1 G

t---+--'

force: -5 568"2"

ANALYSIS)

POS1' __ J'tN

PO_CURTY?,

1,

AJ.'1A.LYSrS ' PD_CURDEF,

PD_CURVSS

L

C, L

)

PO_CURV£S

>

PD_CUF{'l'YP ~1,?9~,+--_~

0 0,

PD_CUROEF

>

lCOO,

.2, 0,

,0

I eo .4\ S

~I--+- .--l----

C.:\lW!

O,JSO)

(16) Apply the dynamic forcing functio(', !o nodes 2, 3, and 4: CONTROL

) ACTIVE )

ACT SST

ACTSS':', 7C <' 1 LOADS-SC ) STRUCTURAL ;. FP'T'r;-

FPT, .:L FPT,

(17)

4,

Reques~

FX, FX, FX,

1, 2, 2, ), :L 4,

?D ..NRESP,

?ORCES

~

2, ), 4,

TiME {sec]

(20) Activate XY plot infoIT:1mion for X displacement at nodes 2, 3, and 4 as a func[;on of time, and plot the displacement vs. time for these modes (see

Fig. 1:.8);

PO_OUTPUT

°

DISPLAY XY_PLO';'S) AC':'iCYPOST ACTXY?OST, 1, 'i11'-1E, UX, 2, 12, L

(18) Execute modal time history a."I<11],5is:

AC':'XYPOST. ACTXYPOS':',

ANALYSIS

:>

POST_OYN > R_OYNAMIC

DISPLAY

R ;)YNAtlIC

~

2, TINE, ), '":'XME,

XY PLOTS

>

UX,

),

0, 10. 1, 0,

UX,

4,

8,

L

0,

2N 3N 4~

XYPLO?S

XYPLOT, 1, 1, 1

(19) Request Scan for m
POST OYK > PD ...OUTPUT 1, 1, 10, 2, 4, 0, .5

ANALYSIS > ?05'T'_DY!\AM:CC

PO_PREPARE, ANALYS~S

~5GI

fPT

>

1 1 1

POS':'_DYN

1,

&

Fig. 11.8 Latera! disploce.ment funcdons at the three-story building of Example J 1.9.

response at nodes 2, 3, and 4:

ANALYSIS )

~.5

iIl.q01J;2"

JGG:'

"

~.es.s'.;!

1 ;. POST_DYNAMIC

PD_MAXLlS1'

~

PD_HAXMIN

PD~PREPARE

(2!) Activule XY plot information for X velocity at nodes 2, 3, and 4 as a function of time, and viOl the velocity vs. time fOf these modes (see Fig. 11.9); DISPLAY

~

XY_?~CTS

~

ACTXYPCST

L TI1·1E, VX, 2, 12, L 0, 2N ACTXYPOS?, 2, ':'!ME, VX, 3. 10, 1, 0, IN ACTXYPOST, ), l'It1E, VX. 4, 8, I, 0, 4N ACTXYPOST,

DISPLAY ) XV_PLOTS ) XY2LOT, l, 1,

XYPLO~

34Q

t

Structures Modeled as Shear Bulldings

Forced Motion of Shear 8 .•dldings 829.2-2 667'1.3 SaSE.

4

3Q36.S

, ", ,

•••

RRR

A AA

Bas

2S2.S1

L L L

" s (inJsecl )

-14 ! S • 2

~

E"

(i'1isec)

X X

162i'1.Ei

-3@33.' -«65l

.2"

."

-3¢5,31

+---l. -+-+f'-'+,~--+ -+-+-1--+,1-[..... ; L \.

$1 llGl!

!:' , I ~1'WS $. 2&ISJS (il Gl.QSI!I9 (i,!SS? (il,2S!]:;

,3~i94

(\) , (il,J5QJ

-62$9

4~G,

e


T1ME (sec)

TIME (sec)

Fig. 11.10 La{ual acceleration fur;doos at the three-story building of Example 11.9 Fig. 11.9 Lateral veloci[y functions at the three-5tory builc.:ng of Example i 1.9.

(I) Set view to the XY plane: (22) Activate XY plot information for X acceleration at nodes 2, 3, rond 4 ns a function of time, and plot the acceleration YS. time for these modes (see Fig. 11.1 0): DISPLAY > ACTXY?OST, ACTXYi?OST, ACTXYPOST,

XY_PLOTS > ACTXYPCST 1, TIME, AX, 2, 12, 1, Of 2N 2, TIME, AX, 3, 10, L 0, 3N 3, TIME, AX, 4, 8, L 0, 4N

DISPLAY ) XY_PLOTS > XYPLO'::' XYPLOl', 1, L 1

Example 11.10. For the structure modeled as a four-story shear building shown in Fig. 11.11 a. determine the response when it is subjected to the force shown in Fig. ll.llb applied at the top level of the building. The modulus of elnSlicity is E::;.; 2.0 X 10 6 psi. Assume dampi!1g in the system proportionaJ ro the stiffness coeffiCIent (Co 0.01). The cross-sectional area of the columns is 0.5 in X 5.0 in.

DISPLAY ,. VIE_PAR ) VIE VIE..), 0, 0, 1, 0

(2} Define [he XY plane Z = 0:

,,

GECMETRY PL,b_I\lE,

0,

) PLANE

1

(3) Gene;:ate four curve segments from 0, 0, 0 to 4. 0, 0:

GEOMETRY ) CURVES

~

CRPCORD

CRPCORD, 1, 360, 0, 0, 360, lac, 0, 360, 180, 0 CRPCORD, 2, 360, 18C, 0, 360, 360, 0, 360, 360, C, CRPCORD, 3, 360, 36G, 0, 360, 540. 0, 360, 540, 0, CRPCORD. 4, 360, 540, 0, 360, 720, 0, 360, 728,

°

(4) Define element group i using the BEAM2D elemerHs, tT.2.;:enal properties, and cross~sectiDnal consrants and genera[e mesh with two nodes and one element along each Cl!rve: ?ROPSE?S EGROIJP,

Solution: The analysis is perfonned using four beam eiemenrs and four concer.trated mass elemems.

> G~ID

Z,

>

L

EGROUP BEAM2D,

PROPSETS :;. HPROP 11?ROP,

1,

EX,

2E6,

0, 0, 0,

0, 0,

0,

0,

t! ,

Struclures Modeled as Sheaf Buildings

342

Piti

m" ,

I

! .. 79551f'!'

ANALYSIS

F(t)(:b}

180'm,

T

m" 1

--+

ANALYSIS)

y,

0,

Ib]

PROPS E'I'S

H_CR,

L

1,

1,

lS9,~,

5,

}

L

5,

0,

0,

G,

0,

0,

M~CR

1

(5) Define group 2 using mass eiemems, real constants, and assign a mass PROPSE':'S

>

EGROUP

e,

2, HASS,

0

v,

0, 8,

0,

0,

L

12,

0,

0,

Q,

0,

0,

0,

MESHING

>

NODES

>

1,

0,

OPT, 2, Uy, 0,

~

A~fREQUENCY

16,

0,

0,

0,

0,

iE-OS,

:),

1E-C6,

R... FREQUENCY

}

FREQLIST

>

Period ( seconds}

1 2

0, 8886530£:+01 C " 2S5877f~£~02 0, 3920269£+02 0 .4808922£:+02

0.1414335E+01 0.4072416£+01 0.6239303£+-01 0 7653638£+0::'

a .707046CE+OO

4

0,

0,

~

POST_DYN )

2,

~

DISPL}-l)l'TS

RX,

RY,

JOG,-

PD_ATYPE

O. . Cl,

0,

,05,

0.25,

0

:n

the X direction: ~

)

POST_CVN "-.

~

PD_CURVES

0, 0

POST_DYN > PD_GJRVES

1,

L

0,

1000,

.25,

PO COROE?

1000,

.5,

0,

5,

0

CONTRO:" ) ACTIVE ,. ACTSST

NCOV.PRESS

L 1 5, 1, UZ,

4,

(12; Define the dynamic forcing function as a time-depender.t force, and

PO_CURDEF,

0

O.245S5~5£ .. OO 0.:602743£:+00 0.:'306568£+00

(IJ) Define analYSIS type as modal time history using four natural frequen~ cies, 300 time steps st2.ning at l 0 wirh a rime increment of 0.01 sec; use default values for all inl:egration parameters; and request printout of relative displacements and reliHive velocities;

ANA;"·YSrS

0,0:)01.

STRUCTUR..A.L

ALL,

0,

Frequency (cyc:::'es Isecl

PD~CURTYP,

(7) Apply constraints in all degrees of freedom at node 1, and in a!i degrees of freedom except UX at nodes, 2, 3, 4, and S: DP':',

G,

{radl5
ANA:..rYSIS

NCOHPRESS, 1. 8

:"OADS--BC )

0,

Frequency

apply to node 5

M_P7

> tH1ERGE

1,

0,

P;:;-i?quency number'

PD_Al'YP£:,

(6) Merge and compress nodes: NMERG£,

5,

FRQ /SUCl'(

>

ANALYSIS

0,

PROPSETS , RCONS':' RCONST, 2, 2, L 0' , L MESHING > PARAM_MESH H-PT, 2, 5, 1

MESHING). KO;)ES

::.,

0, 0

3

8,

1. 2,

P::\l:NT _OPS

0,

FRSQLIS'1'

.1 poinls 2, 3, 4, and 5: EGROUP,

4,

RESULTS ) LIST

PARA~_MSSH

<1.

L

(10) List the naturtt! frequencies of (he syslem:

budding for Ex')Jr.p:e i 1.10. (b) LoaJir;g funclion.

L

0,

R~FR£QUENCY

> :KeONS'}'

MESHING '

0,

f\u\lA:VSIS

Illl shea~

0,

FREQ ISUCK

A_FREQ:.JENCY,

'l~~' SU RCONS'C,

L

(9) Set (he options for [he frequency analysis to extract four natural frequencies using the Subspace Iteration Method with a maximum of 16 iterations, ar.d 11m the frequency analysis:

I'" 79.55 in.4

Fig. 11.11 (0) Four-swry

~

> OUTPUT _OPS

PRINT_0PS,

m<' K\SSSSSSSSS""i - - 4 - ' Yl , .. 79.55rn· J 180in. :

t

343

(8) Request printout of mode shapes:

, t i

Forced MoHon 01 Shear Buildings

~

RZ

AC':'SET,

TC,

CPT,

ex,!,

5,

1

5,

1

(13) Define Rayleigh proportional damping b41sed on stiffness (a i3~OOl):

D?T ANALYSIS > ?OST_DYN ~ PD_RDA.HP, 0, .01

PD_DA.NP lOA? > PD_RDA.HP

= 0,

344

Forced Motion of Shear Bulldiflgs

Struc:ures Moaeled as Snear Build:pgs

345

lIllUt) X; XI

R R R R: E I: E E l L '- :..

'\AAA S ;, Ril

(in)

(inJsec~)

5 SSS

-'.6 i-----.+---+--+-+----'--.... 1-+················· --+-----1

-".~---i-----I---+---+--t----:..C-_ ..- -1---':+ " .~.:---+ ---!" @.S 1.2 1\.I! &.1

$.5

l.S

2,\

2.7

,.,

•. 3

! .5

TfME (sec)

Fig. 11.12 :"'ateral tEsplacemems functions al (he

four~story

building of Example II 10.

'

PD_NRESP,

PD_OUT?D'r 1,

2,

3,

PD_NR£S?

>

4,

5,

\)

(15) Execute modal time history analysis: Al.JALYSIS ~ R_DYNAi1IC

POS'f~;)YN

:>

D:::SPLAY

>

ACTXYPOS? DISP::"AY )

XYPLOT,

_,

XV_PLOTS

>

ACTXYPO$'I

TIME. UX, 2, 'rIME, UX, 3, TIME, :lX,

2, 12, 1, C;, 2N 3, 10, ::., 0, 3N 4, 8, 1, 0, 4N

4, TIME, UX. 5, 6. 1, 0, 5N XL_PLO'TS ;. XYPL07 1,

::.,

w

(1) Define ana:ysis tYpe as frequer:cy respor.se for an excited freque:1cy, 20 (radlsec):

~

.!
PC_CURTY?,

(17) Activate XY plot information for X accelero.don at nodes 2, 3, 4. and 5 as a funCtlon of time. and piot [he response c;;rves (see Fig. l L 13): 2N

ACTXYPOST, 2, TIME, "~X, 3, :0, 1, C, 3N ACTXYPOS'I\ 3, TIME, AX, '2, 8, 1, v, 4N ACTXY?CST, 4, TIME, ~ XY~I?LOTS } XY?::'OT XYPI.,CT, 1, :::.., 1, 1

C

{l2) Define (he dyr:amic force F (1) ;;;;; 1000 sin &t using the following commands: l\1.'\l)'.LYS:S ,.

:.

CrSPLAY :> XY_?LCTS :> ACTXYPOS'T' ACTXYPOST,. TIME, AX, 2, 12, 1, O.

Flg, 11.13 Luteral accelenltlon functions at the four-story building of Example I UO,

Solution: The following commands are implemented in COSMOS after the comma:1ds listed in Example 11.1O up to poir:t lO:

!LDYNAlJ;IC

(16) Activate XY plot information for X displacement at nodes 2, 3, 4, and 5 as a functior: of time. and plot the response curves (see Fig. II, 12): .!l.C'I'XY?OST, AC'T'XYPCST, ACTXYPOST,

, .1

Example 11.11. Determine using COSMOS the steady-stale respOf1se for the building of Example I L 10 wheT'. it is subjected ro a harmonic force F(;) ~ 1000 sin 20t (lb) applied horizontally a the top leveL

(14) Request response at nodes 2, 3, 4, and 5: &~ALYSIS

2.1

TIME (sec)

.k.NALYSIS } PD_CJS\DEF,

?OS?_DYN > PD_CiJRVES

POST_DYN ) PJ._CURVSS > PD_CURDEF 1, 1, 0, 1, 30, 1

I..OA;)S·,BC } STR:JCTUR.h.L > FORCES

Fl'i'D.

5,

FX,

) PD... CURTYP

0

10000,

5,

1,

~

~ND

C

(13) Request values for [he ar:lplitudes of the response in terms of displacemenr, velocity, all::! accelen:tion at the four levels of the buildmg:

346

Struclures Modeled as Shear 8uildHigs PD~PRINT,

L

1,

1,

0,

POS'r~DYN

ANALYSIS ).

PD_NRESP, L

0,

1,

i,

L

> PD_OUTPUT

2. 3, 4.

1 j

TABLE 11.1

(Continued)

PD_NR£SP

5 Nod"

(14) Define Rayleigh proportional damping ~ased on the stiffness (a DO!):

fi

347

Forced MoUon of Shear Buildings

= 0,

,,·t:":I$1.

y"e::i'nsi.

!-q'~'s~,

""1"Uti'tiOfl

:.,I>i'SI<·'

\pl>i'se1

(ph"liel

IphilH.,1

y-~oti'! .. on (ph
~·r"ti'tio"

;) OOOCOC>(Hl

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(I.OOOOCE.OO

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J.CCCOOE,C(l

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O.NlOCCE .. OC

(O,OOOCE·OO)

;O."OOC<:.OOI

10.(I,j<30o;.OOI

{O,OCOOE.:)O)

i!LCOOD<:~CO)

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IC.OCOO£-ODI

(D.OO()-CE.CO)

{"M~e:l

O.OO:)O(}>:.O(l

POS'CD'iN .';~56

(lj) Execute dynamic analysis:

ANALYSIS fCDYl\AMIC

(16) Use an ediror to !1St !.he results from the OUTPUT FILE. (see Table Il,"i for the steady-s:ate response in lerms of dis;::iacement. velocl!y. and

.31~00t,(n

acceleration):

~O.OOOO£.oC)

ACC£t.£RA,tO"S

Hc4'i!

11,8

SUMMARY

For the solution of linear equations of matior:, we may employ either the modal superposition method of dynamic analysis or a step-oy-step numerical TABLE 11,1 Steady~State Response (Displacement, Velocity and Acceler~ alian) ror Examp~e 11.11

,,-t.cdnsL

/·(C"",,1.

z-

I"h"s~j

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(O.OODO".O()-)

iOJ:OO·,:t.COl

e.U170E.OS

O.OOOCO':.OO

O.OGOOO1\:.CO

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CAeoo!);:.cO

1'.)'1%

10.00on.oO)

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(0 tGOOr.CO)

(O.COCOE.OO; 0.000;10,;"00

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(ph~~<»)

o OOOOOE~{lJ

y~<:ot-,,~,

lph"5e!

'In

(p'rM''') O.CCOOO(~OO

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C.OJOOOE~OO

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tC.CQOOE.OQ)

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0 OOOODt.CO

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:C.jOOO;:.O~l

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iO.OO{)Oi:.OOI

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integration procedure" The modal superposition method is restricced to the anatysis of structures governed by linear systems of equations whereas the s(ep~by~step methods of nc:rr:erical integration are equaily applicable to systems With linear or nonlinear behavior. We have deferred the presentation of the ste;:;~by-s~ep integration method to Chapter 20 on time history response of multidegree-of-freedom systems, In the present chapter, we have introduced fhe modal superposition method in ob~aining the-response of 0 sheor building subjected to either force excita~ [ion or to base motton <:!nd have demonstrated that the use of the nonnal modes of free vibration for Iransforming the coordinates leads to a Set of uncoupled

348

Strudures Modeled as Shear Buildings

ForCed Motion ot Shoar Buildings

differential equations. The solution of these equations may then be obtained by any of the methods presented in Part I for the singJe-degree-of-freedom system. When use lS made of response spectra to determine maximum values for modal response, these valUes are usually combi ned by the square roOl of the

W 2 '" 26,500!b 500010

sum of squares
20psi W11 X45

closely spaced, A more precise method of combining maximum valJes of the

modal response is the Complete Quadratic Combination (CQC).

Th~s

349

n:.erhod

has been strongly recommended in lieu of the SRSS method. In the particular case of harmonic excitation. the response may be obtained in closed fann by simply SOlving a system of algebraic equations in wh~ch the unknowns are the amplitudes of the response at the various coordinates.

F;g. PILI.

, F

PROBLEMS 11.1

Determine the response as a function of UC)e for the two-story shear building of Problem 10.1 when a cons!.o'lot fo:-ce of 5000 lb is suddenly applied at the level of the second floor as shown in Fig, PI L L Bays are 15 f: apan.

m

11.2 Repeat Problem 11" 1 if the excitatlon is applied to the base of the structJre jn the form of u suddenty applied acceleration of mugnitude 05 g, 11.3

F,

Delermine [he maximum displacement at the floor levels of the (hree~s{ory shear building [Fig, Pl1.3(a)) subjecLed to impulsive triangular loads as shown in Fig. Pi L3(b). The lOlal stiffness of the columns of each story is k = 1500 Ib/in and 2 the muss at each floor level is m "'" 0.386 [b . sec lin.

11.4

Solve Problem 11.3 using Program 9. Set time increment Lir=O.Ol sec and toeal time of integration T:'en "'" 0.3 sec,

11.5

Detern:.lne (he maximum shear force in the cobmns of the second story of Problem I 1.3. (Hint: Calculflte modal shear fon::es and combine concribmions using method of square root SUC1 of squares.)

L_-.::J",1,2_ Til"fle twx.:)

Fig. PIU.

1l.9

PIL6(b), Determine the response for a toeal rime of LO sec using time slep Lit"'" 0.05 sec, and modnl damping coefficient of 10% for alt [he modes (£ = 30 x 10' psi).

For the Structure (shear building) shown tn Fig. P 11.8 determine (he stead>'.stare motion for the following load systems (loads in pounds): (a)

FI (r) = lOOO sin

(b)

PI (I)

= 2000 cos

f,

r,

F2 (r) = 2000 sin c,

F) (l) ,.. 1500 sin I

F 2 U) "" 3000 COS

F,(r) = 4000 cos

I,

0.

f

A:so load the :;;[rucn.:re simultaneously with load sy!>Lems (n) and (b) and verify :he $upCr90si(ion of reSC;1:s.

Find the steadY-Staie response of the shear building shOwn in Fig. PI! 7 subjected to the harmonic forces indicated In the figure. Neglecr damping,

11.8 Solve Problem 11,7 assuming damping coefficients proponional to the slory stiff ness, C; 0.05 K;

Y,

I"

11.6 Use computer Program 9 to obtain {he time history response of the (hree¥story building in Pl1.6(<.I) subjected lO (he suppOn acceleratior:. ploned in Fig.

1L7

m

11.10

For {he str,Jc[ure modeled as a focr-story shear building shown in P1UO. delermine the s:eacY-SLUte respor;se when i[ is subjected to the force F 10000 sir: 20: Cb) applied at lhe tOP floor of the building. The modc!us of

350

Slruclur$"S Modeled as Shear Buildings

Forced Motion of Shear Buildings

t Ef"""l "5 X lOs Ib ;iI! I 1930 i:;.

"'1 ".

_1 1£ • 1'aU' -'X'~'b'" 'll-. :n. ~:I

351

jO'

t

Ac:cele:ration

~ 9 C.5

CA

elasticity is £ 2.0:x tO~ psi. AS;iume domping in the syste:-n propo;--:ionat to the stiffness coef(1cienl (c)) co- O.O!}.

0.2

0.2

_.-L--;!o-,,',.v, c.• ;fJ\ o.a 0,2 0.3

O. J

OA:

0.9

1

f(d(lhj

lac·jil.

0.5

0.1

m~

fitl

0_2

Ttm~ (!s~C)

1.0

~O_!

!

1000 /----_ _""

I

lS0.ln.

+

- 0.2

,b,

Fig. Pl1.6.

_v, o

--+¥\

EI,~", '" 5 X lOll !lb ;n. 1)

F, '"

2000

WI .. 3860 In

5;n

I

12'

I

lrJI!!b_~z~:;Z:z~~a~z~a*-f""" Y,

F:g. PIL7.

Fig. Pll.lO,

C .25

Ib,

0.5

r[rec)

Damped Motion of Shear Buiidings

12 Damped Motion of Shear Buildings

12.1

353

EQUATIONS FOR DAMPED SHEAR BUILDING

~or ~ viscously damped shear building, such as the three.. s[ory building shown lfl Fig. 12: t, [he equations of motion obtained by suer.ming forces in the correspondlflg free body di&grarr.s are m;'y;

m?"'h + c: Cy~

-

+ CtY! + kIll

- C}(j1 - .vI) - k:(y~ - )'1)

)1) + k3 (y; _.- 11) - CJ (Y) - 'y2) - k; (yJ

Y2)

FI

(I)

= F2 (f) (i21l

F~ It)

In the previous chapters, we de[ermlned the natural frequencies and modal shapes for undamped structures when modeled as shear buildings. We also

determined the response of these structures using the modal superposition method. In chis method, as we have seen, the differential equations of motion are uncoupled by means of a transformation of coordinates that incorporates the orthogonality property of the modal shapes. The consideration of damping in the dynamic analysis of structures complicates the problem ~ot only will the differenti::::1 equations of motion have additional terms due to damj)ing :orces, but the uncoupling of ihe equations

,.)

will be possible only by imposing some restrictions or conditions on the functional expression for the damping coefficients. Tne damping normally present in structures ~s relatively srr.all and practi. caUy does not affect tbe ca1cuh:::.tion of natural frequencies and modal shz?es of the system. Hence, the effect of damping is neglected in determining rhe natural frequencies and modai shapes of the srfucturai systems. Therefore, in practice, the eigenproblem for the damped structure is solved by using the same me~hods employed for undampe~ structures. 352

Fig. 12.1 (u) Damped shear bui:dir._!l, _f,~l . Marhem",;c'; .. ~ .., medeJ' _ (c ) F ree b 0 d y diagram.

354

Damped Motion of Shear Buildings

SI,uc!ures Modeled as Shear Buildings

then (he coefficient of the damping term in eq. ([ 2.6) will :-educe to

These equations may be conveniently written in matrix notation as

[MJ(Y) + rC]{y} + [Kl(y)

~

{F(rll

355

( 12"2)

where the matrices and vec(Qrs are as previously defined, except for [he damping matr.x (C], which is given ::;y

In this case eq 02.6) may be wriaen as

or alternatively as (12"3)

z" + 2§"w"z'1

[n the next seedon, we shall esmblish ~he conditions under which the

M" ~ (¢);'[M]{4»)" K" = {¢};[K] {"oL

C"

UNCOUPLED DAMPED EQUATIONS

(12"lOa)

= w;M"

= {,p};[C] ("oj" =

F.,(r)

To solve ~he d:fferential equations of motion, eq. {l2.2), we seek to uncouple these equatlor.s. We, therefore. inlroduce the trnnSfOli:lation of coordinates

(12.9)

in which case

damped equations of monon may be t:-;:ms:omH!d to an uncoup:ed set of independent equations.

122

+ w~z,,::::: F" (I) 1M"

2(w",M",

(¢};{F(I)}

(12.too)

(l2.lOc) (12.lOd)

The norrnallzntion discussed previously (section 10.2) (! 2.11)

( 12.4)

{y) = [$! (z)

wi1! resu:t in where [~] is the modal matrIx obtai!1ed in the solution of lhe undamped free~vibration syste:n. The substitu~ion of eq, (12.4) and ~ts derivatives ir.ro eq,

M,,=l so that eq. (12.9) reduces to

(12"2) lends to

(12"12)

[M1 [
(12.5)

which

:5

a Se[ of N uncoupled d!fferential equations (n:;;: I, 2, .. > .'1/).

PremultiplYlng eg. (l25) by the transpose of the nth modal vector {¢};, yields

12.3

In the derlvat;on of ~he uncoupled damped equation (12.12), it has been assumed that the nonnal coordinate transformation, eq. 02.4), that serves !o

It ~s noted that the orthogonality property of the mouai shapes,

{¢};[MJ{¢}", =0, {¢};'[KH¢} ... ~ 0

CONDITiONS FOR DAMPING UNCOUPLING

~nco~ple the inertial and elastic :orces also uncouples the damping forces. It

m"n (IV)

causes all components except the nth mode in the first and lhird terms of eq. (12.6) to vanish. A similar reduction is asslIlned to apply to the dampi!1g te!m in eq. (i2.6), thal is, if it is assumed :hat ( 12.8)

:s of mterest to consider the conditions under whiCh this uncoupling wi:! occur, ,hat is, the fO!"T:l of the damping matrix [C] to which eq" (12"8) applies" When the damping IT',atrix is of the form

[C]

~

00[,\4] + a, [K]

(12" 13)

i~ which Q" and at are arbirrary proportionaliry factors. the Orthogo!1
356

Structures Modeled as Shear Buildings

Damped Motion of Shear Buildings

ity condition to eq. (12.13). that is, ?remultiplying both sides of this equation by the transpose of the nth mode {¢'}~ ar.d postmultlplying: by the modal matrix

357

fro;n which

[
(12.18)

(¢};[C][4'J

ao[¢};[M][tP]

+ a,{¢J;[1Q[4'l

(12,14)

Equation (12,i8) may be used to determine the C0nstams 0i for any desired values of modal damping ratios corresponding to any specified numbers of modes. Fo;- example. to eva!uate these constants specifying the first fcur modal damping ratios 5j, [,' £.." we may choose i J, 2. 3, 4. In tn,s case eq. (12.18) gives the following system of equations:

The orthogonality cenditions, eqs. 02.7), then redt.:ce eq. (l2.14) to (1>l; [C] [l"

or, by eqs, (12,10), to

t

= arJvf" + a }t1nw~ {¢};[C] [4'1. = (a, + a,w;)M.

{ ¢};,'[ C) [tJ>1."

6

;,

which shows that, when the damping matrix is of the form of eq. (12.:3), the damping forces are also uncoupled with the transfonnation eq. (12.4). How~ ever, it can be shown that there are other matrices formed frorn the m.ass and stiffness m;;.!rices which also satisfy the orrhogonality condition. In general, the damping matrix may be of the fonn (12,15)

where i can be anywhere in the range :x: < i < 00 and the summation may include as many terms as desired. The damping matrix, eq. (12.13), can obviously be obtained as a speclaJ case of eq. (l2.l5). By raking twO terms correspondbg to i 0 and i = 1 in eq. (12, i5). we obwin the damping matrix. expressed by eq. (12.13), With this fOnT! of rhe damping matrix it is possible to compute the damping coefficients necessary to provide uncoupEng of a system having any desired da;nping rarios in any specified number of modes. For any mode n, the modal camping is given by eq, (12.lOcJ, that is

If [C] as given by eq. (12.15) is subs[ituted in the expression for C" we obtain (12.16)

~,

~

f ,", W:;

21

,

w, OJ,

w,

J

"', "', 5

.

7 ,

w~ ~ w, ,

(;)~ w_~

,

7 '

l w)

(iJ~

l ,

"'; w, J

"a,a, )

(12.19)

a,

In generaJ, eq. 02.19) may be written symboEcally as

W

~HQ]{a}

( 1220)

where fQ] is a squflre matrix: having different powers of the naturaL freque:l~ cies. The solution of eq. (12.20) gives the constants {a} as (12,21)

Finally the damping matrix: 1s obtained after the substitution of eq, (12.2l) into

eq.

(I2, 15),

It is interesting to oeserve from eq. (12, !8) that in the special CaSe when the damping ;nar::1X is proportional to [~e mass {C} = Go [tl1](i = O}, the damping ratios are :nversely proportional to the natural frequencies; rhus the hjgher modes of the structures win be given very little dampir.g. Anaiogously, when the damping is proportionai t:c rhe stiffness matrix ([C) = {II {K}), the damping ratios are directly proponion as can be seen from eq. (12.18) evaJumed for i 1; and in this case the higher modes of the strJcture will be very heavily damped.

Example 12.1. Determine (he absolute d
Now, using eg, (10.24) (K [ U ond perfonning several algebraic operations, we can show (Clough and Penzien 1975, p, 195) that the damping coefficienr associated with any mode n may be written as

SOlution:

From Example 10.1, we have [he fo1!owir.g infcrmarion.

Nawml frequencies: w,

1 i ,83 rad Isoc

W~ =

32.89 r2.dlsec

(12,17) (a)

358

Structures Modeled as Shear Buildings

Damped Motion of Shear Buildings

Modal matrix:

359

Flna:!y, substi:ut:ng this :natrix into eq. (12.15) yields the damping matrix as

[100 I: .26

(b)

Mass malr]x;

r

=

1136 0]1

l0

42172

573.5

- 1.4654]

66,' _. 3.0193

4.7556

199.3

- 19931 31391

There is yet a second met.1od for evaleatlog the damping matrix. corres. por:ding to any set of specified medal damping ratiO. The method may be explained stnning with the relationship

136 [M)=.

I ,C]

°

Sliffness matrix:

(12.22)

_I

75.000 - 44,300j [Kj-l-4".300 44,300j Using eqs. (12.18) and (l2.19) wi~h i =0, i: to calculate the constants needed in eq. (12,15), we obtain the fo)iowir.g system of equJ.tioris:

Where the mutrix [AJ is defined as

{I,

[A]=

r

,

2f,~,ld'

o

0

2f,w,M ,

0

o

26w,u,

C

02.23)

L

in which ,he modal masses Mj, M" M} .... are equal to one if the modal matrix {
Solving this system of equations gives II,

= 0.0:85J

"1=

-O.C{)OOI146 [C]

= [4>]-T[A1 [IP] -,

We also calculate

1

-: _ [0007353 0 [M] -[ 0 0.01515; and

'iY -, ,f)

_[ 551475 - 325.738; [Kj - I - 671.145 671.:45J

02.24)

Therefore, for any specified set of modal damping ;atlos {fl. matrix. [A] can be evaluated from eq. (12.23) and the damping mwix [C] from eq. (12.24). However. in practice, lhe inversion of the modal matrix is a large computational effort, Ir.stend, lakmg advantage of onl,ogonality properties of the. mode shapes, we cun deduce the following expression for the slstem damping matrix:

(12.25)

Then

,

2: a, ([Ur' [Kj)'

0.01851 [

551.475

l-67U45

{1" !

- 325.7381 . I

671.145]

55:.475 - 325738r -0.00001146 [ -671.145 671145J =

l-

1

4.2172 - 1. 4654 3.0!93 4,7556 J

EqUlltion (12.25) muy be obtained from the condition of orthogonality of the norma! modes given by eq. 00.35) ;!s

[I]

= [(p), [.'>f] UP)

Postmultiptying eq. (} 2.26) by UPJ -

I,

(12.26)

we obtuin [
(12.27) .

360

Structures MOdeled as Shear Buildings

Applying the transpose operation

to

Damped Motion of Shear Buildings O.OB.

eq. (12.27) results in

•

0,07

(12.28)

Cl STEEL

~

{

•

REINF. CON CR.

{o

REINF. CONCR.

{

REiNF. CONCR.

0 :=: [M']T since the mUss rr:atrix [A1J IS a symmetric matrix. Finally, the substitution into eq. (l1.24} of eqs. (12.27) and (i2.2S) gives

0.06

in which [A1]

[e] ~ [M1

361

[
which results 1n eg. 02.25) aher substttuting matrix fA) frorr. eg. (12.23). The damping matrix (C] obtained from eq. (l2.25) will satisfy the orthogona!ity propeny and, therefore, the damping term in the differen(ial equallon (12.2) will be uncoupled with the s[.me transformation. eq. (12.4), which serves to uncouple the inertial and etasric forces. h is of interest to note in eq. (l2.25} thal the contribution to the damping m,ltrix of each mode is proportional to the modal damping rario: thus any undamped mode wiH contribute nothing to the G.::mping ma,rix. We should mention at this point the circuf!lstances under which it will be desirable to evaiuate me elements of the camping matrix, &s eq, (12,15) or eq. (12.25). It has been stated that absolute structural damping i.s a rather difficult quantity to determine or even to estimate. However, modal damping ratios may be estim2.ted on the basis of past experience. This past experience includes laboratory determination of damping in different matenals, as well as damping values obtained from vibration tests in exisri:1g buildings and other srructures. Numerical va;ues for damping <'2.110S in structures are generz.Hy in the range of 1% to 10%. These values depend on the type of structure, marerials utiliz.ed, nonsm.lctu[.[,l e:ements, elc. They also depend on the period and [he mz.gnimde of vibratiO!1. It has also been observed that damping ratios corresponding to higher modes have increasing values. Figure l2,2 shows the values of camping ratios measured in existing buildings as reported by H. Aoyama (1980) [in Wakabayashi (1986)). It may be observed from this llgufe that experimental values for damping ScD:trer over a wide range, and that it is difficult !O give definite recommendations. The scatter observed in Fig. 11.2 is typical for ex-periments conducted to determine damping. On this basis, the obvious Gon~ elusion should be [hat the assumption of viscous darr.ping to represent damping does not de$cribe the real mech<.:nism 0: energy dissipation in structural dy· namics. However, for analytical expediency and also because of the uncertain~ ties involved in atterr.pting other formulation of damping, we still accept the assumption of viscous damping. At this time, the best recommenda(10n that can be given in regard to d2.mpil1g is [Q use conservative values. 1% to 2% for steel buildings and 3% to 5% for reinforced concrete buildings for the fundamental frequency. and co r.ssume damping ratios for the higher modes to increase in proPOft!O!l to the natural frequencies. Thus, on the basis of giving

2nd mod~

~

'"z ;;: '~"

•

0,05

•

O,04~..

3ld

•

.,

0.03

"

h. STEEL

•

~:{-o

0.02 0.01

mod~

t:. o ._

!

STEEL

--1

--+0-1:------'-.....

• Cl

0.1)0-'- -... 0

~

Co

1

8

o

... - -...~--~.- _ - - - ' -_ _--l 2 3 .( 5 PERIOD T lurel

Fig. 12.2 Dam~;ng ratios measured in existing buildings. [H. Aoyama in Wakabayashi (1986).J

some coosiderarion to the type of s~ructure, we assig:i numerical values to the damping ratios in aU the modes of interest. These values are then used directly in the modal equations or [hey are used to determine the dampi:1g matrix that is needed when the dynamic response is obtained by SOme analytical method other than modal superposition, e.g., the time history response of a linear or nonlinear s.ystem. Example 12.2. Determine the dump;r.g matrix of Example ILl using the method based on eq. (12.22).

Solution:

From Exampie 10.2 we have :he natural frequencies Wl

= 11.83 rad/sec

w, = 32.89 fad Isec aod the mass matrix r

1"'6

[M; = 1. '" .0

0

1

66.

362

Structures Modeled as Shear 8:;i!dings

Oamped Motion of Shear Buildi.;gs

To determine [CJ. we could uSe eitr.er eq. {l2.24) or eq. (12.25). From Example 10.2, the nOnTIalized mocal matrix 1s [>P) =

CO.064J7

l 0.0813

363

The modal and mass matrices are

r04330 (>P]

00567]

~

-0.0924

: 0.7967 lO.4330

and its :nverse by eq. 02.27) is

0.7228 "!

- 0.7421

0.0000 -0.7719j 0,7228 0.7421

and

'l

S.370 C 6.097

j

[M] =

0.8169 C.1286 - 0.0740

0.1286

- 0.0740 I

0.8571

C.1286] 0.8169

0.1286

Substituting imo eq. {l2.23), we obtain Assume damping ratio of 10% in all ;:he modes.

2f,w,:H, = (2) (0.1)(1 1.83) (I) = 2.366

2f,W1M,

Solwion: The execution of Program [l requires a file that IS prepared during t;,e execution of Program 8 to solve the eige;1problem of a structure previously modeled, Altematively, the required file is prepared by the auxiliary', Program Xl. In the solullon of Sxamp!e 12.3, the auxiliary Progrnm X2 is executed, followed by the execution of Program t 1,

= (2) (0.1) (32.89) (:) = 6.578

Then by eq. (12.24)

(Cj

1[2.366

8.752 7700 [ 5.370 - 6.097 L 572

[C]=l-198

0

o j' [8.752 5.J70j 6518 7700 - 6097

Input Data and Output Results ONH'tfl!G HATiHX

i'ROORAH U

- 198]

(Ans.)

m

NATlJRAi.. l"i<£Q0FW';:ES

0.221

which in this case of eql:al damping ratios in all the models checks with the damping matrix obtained in Example 12,1 for the same St::UCture using eq.

0.514

L241

:HOO>':' SHAPES 8'1' ROW

0.4)10

(12.15).

0,1,>67

PitTA 1'!l.£

tC.?5.l:

l~rG:;N'\ftcroRS:,

D.H);}

~0.1H~

0.0000

D.H21

o.'n~

-0.171;;

0.n25

Hil'lJ'l' OATA,

12.4

PROGRAM 11-ABSOLUTE JAMplNG FROM MODAL DAMPING R.ATIOS

Program 1 I, "DAMPING," serves to calculate the absolUlc dumping coefficients, that is, the elements of the damping matrix from known values of modal damping r2:tios. Example 12.3. Use Program : 1 to calculale the damping matrix for a structure with three degrees of freedom for which t;,e squares of the natural frequencies (eigenvall..es) are

wi

L96!8,

~ = IS 3927,

wj = 60.7968

S,tl;jOS>Cl

L2860Er~,

1.1S&0(>0

~.S710E-~1

L:tt!€();;>Ol

?lJ"n-o,

L ,005>01

-7 AOOCl>CZ

,.'

.,

, I

I ;

1.

·7.10N)E:-c2

OtlH'Vf I
·"SYSTe¥. O;;''!?WG fo'.,I,l'aty,···

4.921-1;:>01 ':<.)0»)<>01

·2.l03J"-Dl

f.pJSZ-C}

·L)Ol'E~'):

5.29SB~-,J;

.PH'!>O ,':;310::<

Oil

364

12.5

Structures f>lodeled as Shear Buildings

Darr.ped .\Aotion of Shear Buildings

SUMMARY

5 Ksec 1lin.

:'11 J =

The most common method of taking into account the dissipation of energy in structural dynamics is to assume in the mathematical model the presence of damping forces of magnitudes that are proportional ro the relative velocity and of directions opposite to the motioo. This type of damping is known as viscous damping because it is the kind of damping that will be developed by motion in an ideal viscous fluid. The inclusion of this type of damping in the equations does not al~er the linearity of the differential equations of motion. Since the amount of darnp:ng commonly present in structural systems is relatively smali, iiS effect IS neglected in the calculation of mllurnl frequencies and mode shapes. However. [0 uncouple the damped differential equations of metlon, it is necessary to impose some restrictions on [he vaJues of damping coefficients in the sys(em, These resrrictions are of no consequence owing to the fac~ thm in practice it is easier w determine or to estimate modal damping ratios rather than absolute damping coefficients. In addition, when solving (he e~u(Hions of motion by the modal superpositIon method, only damping ratios are required. When the solution is sought by orher methods, the absolute value of the damping coefficients may be calculated from modal damping ratios by one of the twO me:hcds presenred ln this chapter.

K 1 = 2000 KJin.

I

fl', " lO K~c' lin.

1

~ Y,

~ K1 = Kiio. 3OC()

y,

Fig_ PI2.3. 12.6

Use Pro_gmm 7 (lnd S (Q model a.:l.d de[erf:")ine rhe natural frequencies and nomal modes tor th~ (ive~s[ory shear building shown in Fig. PI2.6: then use Program ! 1 to de:erf:")me the darr,ping ma(rix c:orresponding to an 8% damping ruti!) In ail {he modeS. Repeat Problem 12.6 fo::- rhe follewing values of Ihe GWdal dampiGg ratios: (,*0.20,

.;,~O.15,

6~O.IO,

;,,~O.05,

PROBLEMS 12.1

The stiffness and mass matrices for a certain [wo-degree~of~freedom Sll'.lcmre are

400 _. 200J 200 '

[K] ~ [ _ 200

12.2 12..3

Determine the dampir.g matrix for this sysrem corresponding to 20% of the critical damping fer the first mode and [0% for the second mode. Use the method :-,ased on eqs. (J2.i6) and (l2.17). Repeat Problem 12.! using the merhod bosed on eqs. (12.22) and 02.25). The natural frequencies and corresponding nonnal modes (arranged in the modal matrix) for the three-slory shear buUdlng show;) ~n Fig. PI1.3 are WI = 9.3! radl sec, W2 = 20.94 rad/sec, w) = 29.00 rnd/sec. and

r

O.l 114

12,4 12.5

- 0.1968 -··0.12451

[
- 0.0277

lO.2703

0.2868

0.2333

1

--0.2114]

De:.ermme the dampl:>g r.Hmix for :he system correspor.ding to damping railos of 10% for all the modes. Repeat Problem 12.3 for dacnping ratios 0: 20% :or aU the modes. Repeat Proble[J1 12.3 for [he foUowing value of modal dampi:"l:g rar:os: (, ~ 0.2.

<, .", O.l.

(,

0.0

365

Y,

K,

10.000 Kiln.

Fig. P12.6.

<,=0

Reouclion of Dynamic Malnces

13 Reduction of Dynamic Matrices

13.1

387

STATIC CONDENSATION

A pracrical merhoc of accomplishing [he reduction of [he stiffness matrix is to identify chose degrees of freedom to be concensed as dependent Of second~ ary degrees of feedom, and express them in terms of the remaining indepenM den! Or primary degrees of freedom. The relationsh!p between [he secondary and primary degrees of freecom is found by establishing the static relmion between the:n, hence [~e name Sialic Condensation lWelhoa (Guy an, RJ., 1965). Tr.is relationship provides the :f1eans to reduce the stiffness matrix. This method is also used in sta(~c problems to eli:ninate unwanted degrees of freedom such as the imernal degrees of freedo:n of an element used with the Finite Element Method. In order to describe lhe Swtic CondenSation Method, let us assume that thos.e (secondary) degrees of freedom to be reduced Or condensed ue arranged as the first s coord!nates, and rhe remaining (primary) degrees of freedoD are rhe hl.';;( p coordinates. With [his arrangement, the sriffness equation for (he struCture may be written using partition matrices as (13,1)

In the discretization process, it is sometimes necessary to divide a structure into a large number of elements because of changes in geometry. loading, or material properties. When the elements are assembled for the entire structure: the number of unknown displacements, that is, the number of degrees or freedom, may be quite large. As a cop-sequence, the stiffness, mass, and damplog matrices will be of large dimensions, The solmlon of the corresponding eigenproblem ro determine narural frequencies and modal shapes will be difficnit and, in addition, expensive, In such cases it is desirable to reduce the size of rhese matrices in order to make the solution of tr.e eiger.problem more manageable and economical. Such reduction is referred to as condensation .. A papular method of reductiOn is the SUllie Condensation Merilod" ThIS method, t!:lough s.imple to apply, is oniy approximate and may produce relative!y large errors in the res.ults when appiied to dynamic problems. An improved method for condensatior. of dynamic problerns, which gives virtually exacl results, has recently been proposed. This method, caned the Dynamic Condensarion Merhod, will be presemed in this chapter afier the introduction of me Static Condensation Method. 366

where {YI} is the displacement vector corresponding to the s degrees of :reedom to be reduced and {Yp} ~s lhe vector corresponding to the remaining p incependem degrees of freedom. In eq, (13.1), it was assumed that the external forces were zero at lhe dependent (i.e., secondary) degrees of freedom; this assumption is no( mandatory (Gallagher, RH .• 1975), but serves to simpHfy explanations withom affecting the final results. A simple multiplication of the matrices on the left side of eq. {l3.!) expands this equation into twO matrix equations, namely, [Ku] {yJ + [K,,] {y,} = {OJ

(13.2)

[K,'] {y,} + (K,,] {y,} = {F,}

( 13.3)

Equation (13.2) is equivalent to (13.4)

where

en is the

transformation mlUrf;c given by

(7] = - [X"l- I [X,,]

(13.5)

Substitming eq. (13.4) and using eq. (13,5) i:i eq. (13.3) results in the reduced

stt(fness equarion relating forces and displacements at Ihe primary coordinates, [,1at is,

(]3,6)

368

Reduction of Dynamic Matrices

Structures Modeled as Shear Buildings

369

rfJ and [K] are precisely the transformarion matrix and the reduced stiffness matrix defined by eqs. (13.4) and (13.6). respectively. In this way, the Gauss-Jordan and the reduced elimination process yields both the transformation matrix s;Hfncs$ rr:atrix [~. Tnere is thus no need to calculate [Ks~] - t in order to reduce L~e secondary coordinates of the system, r~ :nay be seen by expanding eg. (13.11) that the partition matrices

where [ie] is the reduced stiffness matrix given by

en

Equation (13.4), which expresses the static relationship between the secondary coordinates {tTl and primary coordinates {yp}. may also be written using the identity {ypl = [I] {yp} as

Example 13.1. Consider [he two-deg:-ee-of.·freedom system represented by the rr.odel shown in Fg. :3.1 and use static condensation to reduce the first coordinate. Or

{y}

= [TJ b

Solution: (13.8)

p}

For this system [he equcc;ons of equilibrium are readily obtained

ns

where (13.9)

Substituting eqs. (13.8) and (13.9) inlo eq. (13.1) end premultiplying by the transpose of tIl results jn [TJT [K]

[TJ {y,}

[[7)'[1]1 I{O}

The reduction of Jl using Gauss elirnl:1ation 1eads to

II

[0

1

"l{Fp };

-4]IY\ /0) \y,J = IF,

(13.13)

kl2

Comparing eq. (13.13) with eq. (13.11), we identify in [his example

or [TJT[K]

[71=&

[TJ {y,} = {FT}

[K] = kl2

and using eq. (13.6)

[K] = [7l T (K] [TJ

(l3.1O)

thus showing that the reduced stiffness matrix [KJ can be expressed as a t~nsformation of the system stiffness matrix [K]" It may appear that the calculation of the reduced stiffness matrix (K] given by eq. (13.7) requires the inconvenient calculation of the inverse matrix [KssJ I. Flowever, tb.e practical application of the Static Condensation Method does not :-equire a matrix inversion, Instead the standard Gauss-J Qrdan elimination process is applied systematically on the system's stiffness matrix [K] up to the elimination of the secondary coo:-dinates {y,}, At this stage of the elimination process the stiffness equation (13.1) has been reduced to

!{~}

\ {F,,}

)

(13.1\)

(13.14)

Consequently, from eq. (13.9) the transfomlation matrix is

~x,

~x,

t:l:,£:i2l:: Fig. 13.1 Mathematical model for a tWO-degree-of-freedom system.

370

,

Reduction of Dynar'1ic t,,1a!rices

,

Structures Modeled as Shear Buildings

We can now check eq. (13 iO) by simply performing the indicated multiplica~ (Ions, namely

This method of reducing the mass and damping matrices may be justified as follows: The potential elastic energy V and the kinetic energy KE of the s.tructure may be written, respectively, as V

(13.16)

j(yjT[K] [y)

KE=1/Y}r[M](j}

oW"

STATIC CONDENSATION APPLIED TO DYNAMIC PROBLEMS

Introduction of

In order [0 reduce the mass and the damping matrices, it is assumed that ;:he same static relationship berween the secondary and primary degrees of fr-eedom remains valid tn the dynamic probiem. Hence the S2-me rransfonnation based on s(Uric condensation for the reduction of the stiffness matrix is also used in reducing the mass and damping matrices. 1n general this method of reducing the dynamic problem is not exact and imwduces errors in (he results. The magnitude of these e!Tors depends on the relative number of degrees of freedom reduced as well as on [he specific selection of these degrees of freedom for a given structure. We consider first [he case in which the discretization of the mass results in a number of massless degrees of f;eedom_ In :his case it !$ oniy necessary to car.; out :he static condensation of the stiffness matrix and to delete from rhe mass rr.arrix the rows and COlumns corresponding to the maSSless degrees of freedo:n" The Static Condensarion Method in this case does not alter the original proble~n, and ::bus res:Jlts in an equiv:llem eigenprob1em without introducing any error. In the general case, that is, the case involving the condensation of degrees of freedom to which the discretization process has allocated mass, the reduced :nass and damping mmrices are obtained using transformations analogous to eq. (13.10). Specifi.cally, if [M] is the m:lSS marrix of the system, then the reduced mass matrix is given by

V;{]

= [1l T [M][1l

(13.17)

where (T] is the transformation matrix defined by eq. 03.9)_ Ana:ogousiy, for a damped system, the reduced damping matrix is given by

[Cl=[1l T [C][1l where (C] is the damping matrix of the system.

(13.19) (13.20)

Ana~ogous)y, the virtual work EWo! done by the damping forces Fd ;;;;. (C] {j} cOITesponding to tl1e virtual dlsplacemenr {8y} may be expressed as

which agrees with the result given in egs. (13.!4).

13.2

371

(13.18)

~he

{/iy/T[C](y}

(13.21)

transformation eq. (13.8) in the above equations results in T

11l {Yp] KE. = !(Yp)'"[1l' [M] [1l {]i,} oW" {oYp}TI1l T IC] 11l {Yp} V= l{ypJ'[1l [K]

([322)

03.23) (13.24)

The respective substitution of [kJ, [Nfl. and It] from eqs. (D.IO), (l3!7j, and (13. IS) for the product of 'he 'hree cennal matrices in eqs. (13.22), (13.23), and (13.24) yield

V=; {yp}r[Al {yp}

(13.25)

KE = HYp}'[,~f1 Up}

( [3.26)

{Syr}frtJ {Yp}

( 13.27)

oWe Thes~

last three equatio:l.s express the potential energy, rhe kinetIC energy, and the vlrwal work of the damping forces in terms of the independent coordinates {Yp]. Hence the matrices [i;.;, [1\1], and lCl may be inrerp,eted, respectively, as the stiffness. mass, and damping matrices of the structure corresponding to lhe independent of freedom {yp}. Find lhe natural frequencies and modal shapes for the shear building shown in Fig. i3.2; then condense the first of f:-eedom and compare the reslJiling values obtained for natural frequencies and modal shapes. The stiffness of each story and the mass at each floor level are indicated i!1 [he figure.

Example 13.2.

L.~reewdegree-ofwfreedom

SoiuriofL-

Calculation of Naturnl Frequencies and Modal Shapes: The equation of morion in free vibration for this structure is given by eq. (9.3) with the force vectOr {n = (O). namely,

(Ml {y} + [K] {y} = (0)

(a)

372

Reduction of Dynamic Matrices

Structures Modeled as Shear Buildings

The natural frequencies are calculated as

m"3 '" 100 Ib sec in.

373

W /21T, so that

j, = 0,96 cps k, "" 10,000 lb/in.

f; = 3.18

26~,8

j,

k1

okr Fig. 13.2 Shear building for Example 13 2.

where the matrices lMJ and [I<] are given, respectively, by eqs, (9,A) and (9,5), Substitt:ting the corresponding numericai values in eq. (a) yields

Q,QooJ'

y, Yo

1

-

10,000

o

l Y'J' 0

= 3.91,

Y" = 3,00,

Y2J = 3.338

Y:lj:= - LOO.

Yn= - 2.025

Y~:

The sriffness matrix for this structure is

l

40.000 - 10,000

- 10,000

y,. f°'

1

Yn= 1.00

Condensation of Coordinate

0,

Upon substituting Yi = Y; sin wt and canceling the factor sin wt, we obtain

°

Y" = LOa,

Y" =6 It,

~ 'lo r

rI ' - [0,000 20,000 - 50",' 10,000, i - 10,000 10,000 100w lY, LOJ

1.00,

Y"

01 iY'} r01

10,000 _ 20,000 -; o 10,000 ,0,000

r40,000 - 25",'

(b)

o

o

20,000

SOw'

10,000

10,000

- l~oooJ 10.000

Gauss elimination of the first unknown gives - 0.25

o

40,000 - 25",' - 1O,OOO

1

20,000

- 10,000

wh:ch for a nontrivial solution requires that the determinant of the coefficients be equal to zero. that is,

, - [0,000

cps

The modal shapes are then dete!TTI~ned by s'Jbstituting in tum each of the values for the natural frequencies into eq, (b), deleting a redundant equation. and solving the remaining two equations for two of the unknowns in terms of the third, As we mentioned previously, in solving for these unknow:1s it is expedient to set [he first nonzero unknown equal to one. Performing these operations. we obtain from egs. (b) and (c) the following values for the modal shapes:

30,000 lb!ln

40,000 -, 10,000

cps

o.

17.500 - lO.OOOJ' :0.000 10.000

(d)

A comparison of eq, (d) with eq, (l3J 1) indicates that ~O

[T] = [0.25 OJ

10,000 - 100w'

__ [

II<] -

The expansion of this determinant rest:lts in a third degree eqeation in terms of u.i having the following roots:

17,500 - 10,0001

_ 10,000

10,000j

(e)

and from eq, (13,9) (.V~=36.1

wi =4000 wi = 1664,0

(c)

1n =

.0.25 O~]

l~

(f)

374

Reduction of Dynamic Matrices

Structu;es Modeled as Shear 3uHdtngs

(KJ.

To check, we use eq. (13.10) to compule

I 40,000 I - 10,000

rk' = [0,25

I 01.

"

0 IJ :

,

,

0

l

r

[K] =

! 7,500

l- 10,000

0

Corresponding modal shapes are obtained from eq. {g) after substituting the numerical vaiues for w~ or w~ and. soiving (he first equalior: for Y3 with Yl :=; 1. We then obtain

Hence

o 1 ,0,25 0 1

10,000

20,000

!O,OOO!

- 10,000

10,00C

Ii

J .0

0 IJ

- 10,0001

,

o

Yn

=-

0.33

'I =,1r°.25 01· 1 .i0,25} '11.00 0 1'6 =11.00 ly)J, lo I J o ,156

r2~ 5~ ~i r~25 ~'jl d Lao 100) La !

or, nfter r..ormaiizing so that the tJrSl comr:onen[ is L

Y" = ! ,00,

Y" = 4.00,

Y" = 6.24

and analogously for the second mode

, _ r51 ,6 [041]-, 0 l

Y"

0'1' 100

:516 .

0:, 1 [,)" + [

0 IOOj h:

17,500 - iO,OOO1

- 10,000

10,000

j

[Y'] = [OJ' Yj,

:.0

The r.aturnl frequencies and modal shapes are cecemlined from the solUTion of the following reduced eigenproblem:

II7,500-5J.6w'

l 1£(0

10,000

-10,000: lO.OOO-lOOw')

fy,]_'Ol ,y)i-\oi

the determinant of the coefficient matrix if'. eq. (g) and w~ = 36,1

wi = 403.3

Example 13.3. figure 13,3 shows a uniform four-story shear building. For [his structu::e, determine (he foHowing: (a) the natural frequencies and corres· ponding modal shapes as a fou:--degree-of-freedom sysrem, (b) the na~ural frequencies and modal shapes after static condensation of coorc.ina:es Yl and YJ.

(g)

solVIng the resulting quadratic equation in (j}' gives

(h)

1.00,

For this system of only three degrees of freedom, the reduction of one coordinate gives natural frequencies lha( compare rather well for tje first two modes (eqs. (h) and (c)]. However, experience shows that static condensation may produce large errors in the calculation of eigenvalues and eiger..vec:ors obtained from [~e reducec system" A generai :-ecommendadon given by users of this method is (0 aSS-Ur:1e that the static condensation process results in an eigenproblem that 9rovides acceptable approxima(ions of only about a thLrd of the calculated eigenvalues (natt:ral frequencies) ap.d eigenvectors (modal shapes.).

j

The condensed dynamic problem is then

Equaring to

Y:H = 1.56,

I,' Y,

which reselts in

l

Y" = 1.00

. jy,t

[0.25 I 0 1

= lO ~

Y" = 1.00,

Application of eq, (! 3.8) for t:-te firsl mode gives

IomO,

which checks with eqs. (e). The reduced mass matrix is calculated by 5ubsli· wI:ng matrix [Tj and irs transpose from eq. (f) into eq. (13.17), so that

(M!

375

;;.

Solutioli.' (a) Natural Frequencies and r"v1odaJ Shapes as a FourwDegree-of.Freedom System; The stiffness and the mass r:1atrices fe:- this StflJCWre are :-espectively

from which

2 -J

/1

j 36.1 1171 = 0.95

cps

h = J403-:J 121T= 3.20 cps

(K] = 32735

0 I

-I

2

a

I

0

0 -I

~~

2

01

-:1

(al

376

Reduction of Dynamic Matrices

Structures Modeled as Shear Buildings m

F or nil flOOrS: m=1

For all

EJ m

s;ori~a:

and the normalized modal matrix (see Section 102)

---+Y.

----1

k

f----- ---1

r0.2280 ---+

-lo.d

285 11>j - 0.5774

Y,

k

k"327.35~ I m' ~m~.~

r--i

0.42851

0.5774

0.2280

0.6565 .

0

0.5774

0.5774

To reduce coordinates y, and y~, we first, for convenience, rearrange the ness matrix in eq. (a) to have the coordinates in order Yi' Yi-, Yz, y,,:

m

E~rH!lp!e

[~-m"[ : - 1

:3.3.

and

(d)

°

0 1J

Substituting eqs. (a) and (b) into eq. 00.3) and solving the correspondlng eigenvalue problem (using Program 8) yields

(J)~ = 768.3,

-1

1

2

0

0 -I

0

1:

and

- 0.5

0

(bl

1M] = 0 0 1 0

o0

0 -I 2 -1

stiff~

(e)

"

Applying Gauss-Jordan elin:ination to the first two rows of the matrix in eq. (e) results in

o1 010001

327.35,

0.6565

(b) Natural Frequencies and Modal Shapes after Reduction to Two·De· gree-of Freedom System:

k

Fig. 13.3 Uniform four-story shear building for

WT = 39.48, w~

,

0.5774

0.6565 -0.5774 - 0.4285 - 0.2280]

k

[

377

0

0.5

0 1

- 0.5

0 0

327.35

- 163.70

0 0

163,70

163.70

(f)

A comparison of eg. (f) with eg. (13.11) reveals that

(J)~ =: 1156.00

.

[11 ~

,0.5 0 . lO.5 05j

corresponding to the natura] :requencies and

__ [

1,00 cps

w,

I, = - = 2.88 cps 2tr

i~

J.,>

= W:; Wi!.

(e) 441 .•.

cps

1,=-=5.41 cps

21T

327.35 - 163701

IK] - _ 16370 Use of eg. (13.9) gives

r1o.5

0 0.5 0.5 [71=1 0

o

163.70 J

(h)

378

Reduclion 01 Dynamic Matrices

Structures Modeled as Shear Buifdings

The recuced mass matrix can now be calculated by eq. (13.17) as [mi;',; J

'7]T,,,,,, 'Tl; [LS 0,25 L

1 rJ j

~

l

0,25]

(i)

1.25

The condensed eigenprob!em is then 327.35 ~ L5u? [ - 163.70 - 0.15w'

- 163,70

U)

;63.70

Exump1e 13.4. The snear building of Example l33 is subjected to an earthquake mod on at its foundation. For design purposes, use the response spectrum of Fig, 8. to (Secllon 8.4) and delermine the maximum horizonrai displacements of the strucmre at the !eve! of the floors. Solution: The panicipatior, factOr of a shear buiic.ing given by eq. (I ! AD) as

; 36598

{k)

_ [0.4380 0.7056] [YjP-l06723 - 0.6128 wnere

[119

is the mod;)! matrix co;responding

!O

(I)

the primary degrees of free~

dGm. The eiger,yeClOrs for ehe four-degree·of-freedom system are calculated for the first mode by eq. 03.8) l'.S

Y,

05 0

Y,

0,5

y,

1

o

l Y4

°

i

1

0,5

wil~

N stories is

N

T, ~ - :>..: (Jnj
and jrs solution is U)~ = 40-39,

rO,,~ ,

J 1°.4380'

05552

I

(a)

where nl J is the mass at jth ncor and

;

(a) Response Considering Four Degrees of Freedom: The subslitution into eq. (a) of !he corresponding numerical results from Example I),) gives

r, =

r,

- 1.890,

05775,

r,=

-0.2797,

and

r,=

-0,1213

(b)

The spectra! displacemems corresponding to the values of t~e natllra] frequen~ cies of th!s butldmg [eq. (c) of Example 13,3] are obtained from the response spec
• 0.4380

06723)

379

l 0,6723 ;

Sm

~

14.)2,

So,

3240,

SD)

=

1.433,

and

SCA

~

0.969

(e)

Or

Y'1

I' 0,2]90

y; f ~ j y,

y~ ,:

j

0,4380 0.5552

l

The maximum displacements at the floor levels relatIve to [he displacement at the base of the bt.:i:ding are ca!CUlaled using eq. (11A1), namely,

1 (m)

(d)

0.6723 J to obtain

and for the second mode

r

~

~: 1

0.5 0.5

: y, ':

I, r:

or

r

~: 1

1 y; . Y,

I

J,

J,

O k

~'I

07056'

r

, ,

0,6128 1

l

l/ll1)a~·~

0,3528

~

0.0464 - 0.6128

illm;.v. =

0,0464

£t.1/h:lJ;

0.7056

11.65 in,

")"m

= I 5,64 in,

and

= ! 7.81 tn

(b) Response Considering the System Reduced to Two Degrees of Freedom: The natural frequencies, calcu!med from eq. (k) in Example 13.3, are

- 0,6128

0.3528

0,7056

6.274 rn.

/, (n)

~

!21T~

1.011 cps

and

f,

~,

12 7T = 3.044 cps

(e)

380

Structures Modeled as Shear Buildings

Reduction of Dynamic Matrices

Upon introducing, into eq. (a), the corresponding eigenvectors given in eqs. (m) and (n) of Example 13.3, we obtain [he participation factors

T,

=-

1.884

T,

=

0.492

-

case, the equations of free motion may be wriuen in partitioned matrix form as (13.28) The substitution of {y} = {Y} sin eigenproblem

The values of spectral displacements corresponding (e) can be read from Fig. 8.10:

[0

Wi!

in eq. (13.28) results in the generalized

the frequencies in eq. (13.29)

SD' = 14.16 SD2

381

= 2.913

Use of eq. (d) gives the relative maximum displacements at the level of the floors as

where w~ is the approximation of (he ith eigenvalue which was calculated in the preceding step of the process. To start [he process one takes an approximate or zero value for the first eigenvalue wT. The following three steps are executed to calculate the ith eigenvalue and the corresponding eigenve,cror {y}, as well as an approximation of the eigenvalue of [he next order w;+r: Step l. The approximation of is inrroduced in eq. (13.29); Gauss-Jordan elimination of the secondary coordinates {Y,,} is then used to reduce eq. (13.29) to

w;

w;

"'"''' = j (1.884 x ",." =

j (1.884 x

14.16 x 0.2190)' + (0.4920 x 2.913 x 0.3528)'

= 5.864 in

14.16 x 0.4380)' + (0.4920 X 2.913 X 0.7056)' = 11.73 in

",." = j (1.884 X

14.16 X 0.5552)' + (0.4920

X

2.913

X

0.0464)'

",=, =

14.16

0.6723)' + (0.4920

X

2.913

X

0.6128)' = 17.97 in.

)(1.884

X

X

= 14.81

(13.30)

in The tirst equation in eq. (13.30) can be written as {Y,}

13.3

= [T.]{ Y,}

(13.31)

Consequently, the ith modal shape {Y}i can be expressed as

DYNAMIC CONDENSATION

{y}, = [T,]{Y,} A method of reduction that may be considered an extension of the Static Condensation Method has been proposed (Paz 1984). The algorithm for this method Starts by assigning an approximate value (e.g., zero) to the first eigenvalue w~, applying dynamic condensation to the dynamic matrix of the system [Dr] = [K] - w~ [MJ, and then solving the reduced eigenproblem to detennine the fIrst and second eigenvalues wi and ~. Next, dynamic condensation [""1] to reduce the problem and is applied to the dynamic matrix [D 2 ] = [K] calculate the second and third eigenvalues, w~ and wj. The process continues in this manner, with one virtually exact eigenvalue and an approximation of the next order eigenvalue calculated at each step. The Dynamic Condensation Method requires neither matrix inversion nor series expansion. To demonstrate this fact, consider the eigenvalue probkm of a discrete strucLUral system for which it is desired [Q reduce the secondary degrees of freedom {yJ and retain the primary degrees of freedom {yp}. In [his

wi

(13.32)

where

IT)

.

= [it.]] l

[I]

and

{y},=[{~'} \ {Y,} J.

(13.33)

Step 2. The reduced mass matrix [MiJ and the reduced stiffness matrix [k;] are calculated as

= [T,]T[M] [T,}

(13.34)

= [D.] + w; [M.]

( 13.35)

[M,]

and

[K,]

where the transformation matrix [T,] is given by eq. (13.33) and the reduced dynamic matrix [D,] is defined in eq. 03.30).

.382

RedUClion of Dynamic Malrices

Structures Modeled as Shear BuHdlngs from which, by eqs. (13.30) and (13.33)

Step 3. The reduced eigenproblem

· l

0.5 0.01

(13.36)

w' [,w,]]{l',} = {O}

[[Kil

383

0.5 0.5

1S solved [0 obtain an improved eigenvalue wi, its corresponding eigenvector {Yp};, and also an approximation for rhe ne)\;[ order eigenvalue W;i-l_ This three-step process may be applied iteratively, That is, (he value of wi obruined in step 3 may be used as an improvec approximate value in step i to cowip. a further improved value of in SEep 3. Experience has shown thilt one or twO such iterations will produce vInually exact eigensolU!!ons. Once an eigenvec[Qr {YpL for the reduced system is found, the ith modal shape of {he system is determined as {Y}i = [T,J{ l'ph using eq. (13.32).

wi

IT 1= 1 • I

0

o

I

1

:

nne

.

[D,]=

r 327.35 l- 16367

- 163.67] J63.67;

Slep 2, The recuced maS$ and stiffness matrices, eqs. (13.34) Dod (13.35),

Example 13.5. Repeat Example 13.3 of Section 13.2 using the Dynamic Condensarion Method.

are : 1.5

Soiwion: The stiffness ma[rix and the mass matrix with the coordinates in the order Yb Y::" Y2, Y~ are given, respecrively, by eqs. (e) and (b) of Example i 3.3. SubsricU[ion of these matrices Into eq. {t 3.29) results in the dynamic matrix for the sys[em:

[Dj ) 1

I

65437~_~7'~"3"W5"'~" 6547~ w; ,;~~;~~wi .. =3~7.3S. ..

327.35

: 654.70 -

- 32735

0

o

r,

654.70

0

0

654.70

lD,l ~ 1:327:35

~

327.35

- 327.35

0

w?

6)4.,0 0

-0.5

H;'~~J

0

I~

0

327.35

-163.61.

0

i - 163.67

163.67 J

.0

-0.5

... " ..... "." ...

-

w; = 4039

327.35

(b)

r

327.35 163.67

163. 67 1 163.67j

and

=

{OJ

wi = 365.98

These values for w~ and ~ may be improved by iterating the calcularl0ns. that is, by totroducing w;:::::: 40.39 imo eq. (13.29), This substitution results tn

I

O!

.,..

= [D,l + wi [M ,] = l -

Sup 3. The ,olulion of the reduced eigenproblem [[K,] - w' [M,l] {Y,} yields

1

Application of the Gauss-Jordan elimination process to the first twO rows :1

(a)

we Start step

- 327.35 0 - 327.35 - 327.35 .:.- .. ".:.;.... -.., ...... -- .. ,""'"'" .. ,

_

wi

327.35 -

Step 1. Assuming we have no ini[ial approximation of by setting w~:::::: 0 and substituting this value inm eq, (a):

and [K,]

0

0.25':,

~O.25 1.25j

r (D.]

=

614.31

0

- 327.35

0

614.31

- 327.35

- 327.35

·,327.35

614.31

0

0

- 327,35

0

286.96

0

Application of Gauss-Jordan elim~nation to the firSt two rows gives

0

[j

- 0.533

1

0.533

0

265.44

0

- 174.44

~~i~l:~l 112.53 '

384

RedL-clion of Dynamic Matrices

Structures Modeled as Shear Buildings

385

Gauss-Jordan elim:nation of (he first two rows yieldS

from which

0533 r.0.533

[T,]

= l~

0

Li12

I

- !.I 12

0 0

1

00 - LlI2

433.27

- 363.88'

36388

- 396.74 j

from which

and 265.44 - 174.44i

_ : [D,] =

l- 174.44

112.53 J

The reduced mass and stiffness matrices are then and

[M ,] =

[T,] 7 [Ail [T,J

'.568 0.284]

-

[D.:d

[0.284 1.284

=

r - 433.27 - 36388-

:

l - 363.88

- 396.74,

and t~e reduced mass and stiffness matrices are

and ."

..

2 • ..

IKJ = D, + w, lM ,J

_.

328.76 - 162.97 ' 164.39

-l _162.67

-

j

[M,J

..

The solution of the reduced eigenproblem,

[K,]

= [T,] 7 [M][T,) = [3.471 1.236 -

,..

= [D,] + wi [M,] =

1.236, 2.236,

[817.12 81.21 '~ .. 8:._1 408.06.

1 :

The solution of the reduced elgenprohlem [[1(,] .lhe second mode

y:elds the eIgenvalues

wi = 39.48, wi = 360.21

An iteration is perfonned by introducing w~ he following:

r

0.69 35 1

{Yp j,=l-0.6171j

(d)

D-,i =

The same process is now applied to me second mode, starting by substitudng into eq. (13.29) the approximate eigenvalue = 360.21 calculated far the second mode in eq. (c). In this case we obtain

[--

[ n'., 0

-

3~7.35

wi

I

- 327.35 0 - 327,35 327.35 l [DJ = :I'''_'''''''"~'_'''' ..3............-··~:·327.3-" .. ·,··::·· ..<:+·~:_.4 ..9·4:9 :..· ........ ·····0'.....'.1. .)L.., • .)

I c

294.49 0

0

0 294.49

- 327.35

c

0

- 32.86

{Of yields :or

= 328.61

(e)

and corresponding eigenvectors

w' [M,] I (Yp} =

= 328.61

0

- 327.35

326.09

- 327.35

327.35 327.35

into eq. (13.29) to obtain

0 -

326.09 0

0 - 126

Applying Gauss-Jordan elimina:ion to the flrst two rows yields

[

~ ~ o

0

o

0

2.004 - 1004

0.0 - 1004

i I

~~~.~~~~!::~j

386

Slruct'J:'es Modeled as Shear 8ui:dings

I l

and

from which

1.004 0,0 ., T

[

r Y,Y,

_ 1.004 1.004j ,1- i 0

[

o

Y,

i

Y, ,

and [I),]

:~331.14

Hence

" 328,62]

Reduction of Dynamic Ma!rlces

, T [3,015 1.008' [M,I = [1',1 [MI [T,] = 1,008 2008J

and

[t, ) = [0,] + ",,(M,l =

2,54: 2,54 329,89]

1659,78

l

The solution of the reduced eigenproblem [[K 2} -

ui (M z1) {fp} = {OJ now gives

for the second mode

r

05766)

\ - 05766,

(e)

Therefore, from eqs. (c), (d). and (e) we have obtained for the first two eIgenvalues

wi = 39.48

and

wi = 327.35

05766)

. Y, j,Y"Y,

1

-0533 00 = I0533 0,533' (04283) =

II°

0

Hence ( Y

Y, Y,

,y"

I =

\0,6562

i

l05780 0,6562 J

- 0,5766

1

0.0

!

- 0.5766

J

(i)

then by eq, (l3,34)

and since by eq. (13.32)

{f}, = (n {Yp) (g)

we see that {Y}:[M] {Y};= I

thus demonstradng that (he eigenvector {Y}; is nonnalized with respect to the mass matrix of the system [M] if {Yp} is normalized wirh respect to [M].

j' 02283 'l 05780

I 04283 I 13.4

,06562)

f 0,2283 f 0,4283

,

The eigenvalues and eige:1VeclOrs (t). (h), and (i)l caicula[ed for (he first twO modes in this example US!:1g dynamic condensation are virtually identka! to the exact solution derermined in eqs. (c) and (d) of Example 13.3. It should be noted lhat normalizarion of the eigenvectors is not needed in eqs. (h) and (i) if the reduced vectors art normalized with respect to [he reduced mass of [he system, that (s, if a reduced eigenvector {Yp} satisfies the normalizing equation

The eige!lveccors of the sysrem are then computed using eq. (l3.32) as foUmvs:

'Y']

0,5766

(f)

and corresponding eigeEvectOrs 0,5766\

0,0

- 0,)766,

5789 0. 0,5766

and the reduced mass and s[iffness matrices are

,"

[

0.5789 :

r 0 ~7661 = \

lO

329,88

328,62

I

=:• 004 0I 0(Y1 [1.004 0,0

387

(h)

MODIFIED DYNAMIC CONDENSATION

The dynamic condensation method requires the application of elementary operations, as it [s routinely done £0 solve a Unear system of algebraic equations, using [he Gauss-Jordan elimination process. The elementa..ry operations are required (0 [fansform eq, (13.29) to rhe form given by eq. (13.30). However, the method also requires the calcuiation of the reduced mass matrix by eq, (13,34). This last equation involves the mulriplication of three matrices of dimensions

388

Structures Modeled as Shear Buildings

Reduction of Dynamic Matrices

equal to the toral number of coordinates 1n the system, Thus, for a system defined with many degrees of freedom, the calculation of the reduced mass matrix [M] requires a large number of numerical operations, A modification (Paz 1989), recently proposed, obviates such large number of numerical operations. This modificatlo:l consists of calculating the reduced stiffness matrix [k} only once by simple elimination of s dispJacemer.{s in eq. (13.29) after seuir.g wi, = 0, thus making unnecessary the repented calculation of [k} for each mode using eq. (1335), Furthennore, :t also eliminates [he time consumed in calculating the reduced maSS matrix V~fj using eq. (13.34). Ir. the rnodified method, the reduced mass matrix for any mode i is calculated from eq. (1),)5) as _

[M,J

I

_

= -, [[K] w,

-

[D;]J

Now, the reduced mass matrix (ki;] )s calculated from eq. (D.37), after substitution in this equation of [K,J from eg. (a) and [D,] from eg. (d). as 1M,J

=

I

. ]K,] - [D,ll

1.530 0.267] [0.267 1.266

(I)

Ther. the reduced stiffness and mass ma!rix from egs. (a) and (f) are used ro SOlve the reduced eigenprobJem

to obtain eigenvalues (13.37)

wi

where [.K] 1s [he reduced stiffness multi;.;., already calculated, and [i\] is the dynamic matrix given in the partitioned matrix of eq, (1330). As can be seen, the modified algOrithm essentially requires, for each eigenvalue calculated. only the application of the Gauss-Jordan process to eliminate s unknowns in a linear system of equations such as the system in eq. 03.29). Example 13.6. sation method,

389

Repeat Example 13.5 using the modified dynamic conden-

w~

39.46,

= 363,67

(g)

and the eigenvector for rhe first mode

IY'] = 10433591 I Y,

(h)

,0.66424)

The eigenvector for the first mode, in tenns of rhe original four coordinates. is then obtained from eg. (13.3 i) as

0 ' ·0.43359~ _1°.23110) [0..533 0533J 66424) - 058514j 0.533

Solution: The initial calculations of the modified method are the same as ,hose in Example 13.5. Thus, from Example 13.5 we have

to

(i)

The combination of eqs. (h) ar.d (i) gives the eigenvecwr for the first mode as 327.35 - 163.67] 163.67 163.67

=. _ [L5 [M,J = 0.25 [it,]

0.25' 1.25 J

wi = 40.39. wi = 36598 [D,]

=.l

265.44 - 174441 :74.44 t 12.53 J

(a)

[, l

(b)

Y}

I

0.58514

y.

J,

0.66424

y;

(c)

(d)

[0.533

°1

.0533 0533j

0.43359

wi

r- 445.43 l 368.20 (e)

(')

Analogously, for the second mode, we substitute = 363.67 from eq. (g) into eq. (a) of Example 13.5, to obtain the fonowing matrices after reducing the first two coordina~es;

and

[1\1

0.23110

[1',]

=

1.1248 1.1248

- 368201

- 40454 J

°1

1248

390

Reduction oj Dynamic Mauices

S!ructures Modeied as Shear 9wildiogs

13.5

The reduced mass matrix (l~l:d is then calc'Jlated from eq. (13.37) as

.

--;::-;;::;-;' r 327 .J 5 It - :63.67

163.671

/'112 = "

_ (M,J

16J67j

c - 445.43

l-

- 368.2°]1

368.20

404.54

I

'2.1250 0.5624', 1.5624

j

= lO.5624

Then, for the second mode, [he solution of the corresponding reduced eigenproblem give::

319.41, 1.1248

iY')=i 0.61894) l-

~ y~

o 1[

[ 1.1248 1.1248

0.63352

0.61894)_' - 0.63352J

0.696181

and

1

0.61894

I

12~REDIJCTION

OF THE DYNAMIC

The compurer program described in this section reduces by SUllie condensation or by dynamic condens:'Hion the stiffness and mass matrices of a structurat system and solves {he reduced eigenpcoblem. The user has the option of selecting eiLher of these twO :nelhods, Example 13.7. Fo: tht four~5rory shear building shown in Fig. 13.3, use 12 (0 de(errnine the nawral frequencies and modal shapes after reducing the sYi;tem :0 coordinate:; Yl Gnu Y4 using [he following methods: (a) Sta[ic condensmion, (b) Dynamic condensi.\t:on, (c) Exact soiu~io.:l as <1 system wi(h four degrees of freedom. P~ogram

SOhlfl'On.: The execu:ion or "Program 12: Condensation," requires the previous preparation of :1 file modeling Ihe structure, that is, a file storing the sliffness and mass matrix of the sys{em. For this ~xarnple. this file can be prepared by either cxccudng Progr:.m 7 to Madej {he structure as a shear building or execUling [he Auxiliary Program Xl, which directly Stores in a file the stiffness and mass mallict';s fram inpur data.

-1- 0.01640)

0.69618

PROGRAM PROBLEM

Input Datu and Output Results

- 0.01640 - 0.63352 ;

,Q~~")orTDl

127 lSE.Ol

The results obrained for rh:s example using the modified method, although [Q the ex-aCt solution as those obtained in Example D,S by the direct application of the dynamic condensation method. Table 13.1 shows and compares eigenvalues calculated in Examples 13.5 3!ld 13.6 with the exact solution obcained previously in Example DJ.

suffic!er.rly apprOxlrntHe, are not as close

;)S·.j)CE.t:1

,

XltQJ,:'?C

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;171S~.(n

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0,00000£.(;0

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H7}'>E.C;

l!.lOOO(.0[

• .COCOO£.(lO

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' . U;;OOO".-,)[

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• .00002£.OC • .OOC(HH':.OC

'.

~OOj)Ct ,.)t

OtCOflC.OO

.,\10000;:;.010

\flOOJ£.Q1

.-,00000£.00

C1.'rA fll£, 011

TABLE 13.1

Comparison of Results in Examples 13.5 and 13.6 Using Dynamic Condensation and Modified Dynamic Condensation "VIISER 0. 02GR,:ns Ot SUMlER

Eigenvalue

os

tRZE:X:~

i'f\l¥.J;"Y Coo?tHN;'.1'SS

CCIWEtGATW:. :NO;;;<;:

Mode

Exact Solu6on

Dynamic CondenSation

Error %

39,48 327.35

39.43

0.00

Modified Method ...

I

2

327.35

0.00

~~-

39.46

319.41

391

i'f(ll1.;R!tS COO/WIn,:;,;::;:

Error %

...- - " " " - -

CUT!'VT

0.05 2.42

FHEQ.

;I

R:;S'j~T::;:

m:...

2

392

Reduction of DynamiC Matrices

StructU(CS Moceled as Shear Buildings

As expected, results given by the computer for Example 13.7 J.gree with calculations for the same structure obtained previously using static condensa[ion ir: Example 13.3 and dynamic condensatlon in Example U.S.

2IG2tlV£tW& # •

:i<2Ql:ENC'{ Ie ".5.'

J;:tQZtNALllE

L

). Q4~

~6{)C:.Q2

13.6

~UHeER

0.'2

c:sc~e2S

tI(,~

'I I.

n;;csx"

1

ND~

Ot- fRze:::OM

NDM!lSk OF ?,Ur.Afl.Y COORO!t'A72S CO~l'i~NS.;':":ZON

INOU

?/l.lt-<..... R I 5;$ '!:OORPIN;'1'£S;

Oll1'ptr, RSSUt.,-S, fR!SQUENCY

(C.?S.)

LGJO

(;.517311

(L22B219

393

0.6%167

SUMMARY

The reduction of unwanted or secondary degrees of freedom is usually accom~ plished in practice by the Stalic Condensation Method. This method consists of determining, by a parliH! Ga'Jss~Jordan elimInation, [he reduced stiffness marr:x correspondbg to the primary degrees of freedom and the transformatio:1 matrix relaring the secondary and primary degrees of freedom. The same transformation matrix IS ased in an orthogonal transformation [0 reduce the mass and damping matrices of the system. Stmic condensation introduces errors when applied 1O the so1:Jtlon s(rJctural dynamiCs problems. However, as is shown in this chapter, the appiication of the Dynamic Condensation Method s;Jbs!antiaUy reduces or eliminates these errors. FunhemlOre, the Dynamic Conde:1sarion MeLhod cO:1verges rapidly ro the exact solution when iteration is applied.

PROBLEMS 13.1 The stiffness and m:::.ss matrices of a cenain Slructu:e are given by

10 0 OOOOOE_OO

D.

Oi)CD()r~CC

O.65n3E~C)

>;n:nE~O~

- .J21::r/2+0) 'J,OC:)OCE+OO

O.6S47}:,:+03

. :n;nE~in

O . OOQ{)Ce.JO

~,);"';;'::wC:'

O.SS4iJ-lkJ)

_)1n'll;'o~

O.oCOOC£+OO

OO:)OOlt.CQ

-.32731!;'0}

C, J:t7J'l;:+O)

·~·~ss

O.lOOOOF:-Ol 0.00000::",o0 (;, 00000£ .. Of) O. ~COOCt: .. OC

[K]

0,(::10002.00

OOOOO£AlO

O.HjOOCE~O!

j . Q('HlOCE.OO

C.o:)OOO£.OO

O. COCCJf: .. OO

0.100002+01

,).OOO()~E~O~

G.CO:liJ02~OO

O.OOOC(:\:>OO :).lCOOD£ .. rH

13.2

0.22801 D.nSS} 0.51"135 0

···3

J2

2 -1

~l

-J

8J

'I>f< = l,

OOOOI o 0 (J 0 a0 3 a

rcO 0 0 2

(a) "{.Is\: the S(a~i(: Concensation Method w determine the transformation matrix ur.d the reduced stIffness ,lOd mass matrices correspcnding [0 lhe e1iminatior. of lhe first ~wo degrees of freedon: (the massless deglees of freedom). (b) Determine the naturr;.l frequencies and ccrresponding normal modes for 1he

EIC<:"VALi.!ES,

656S~

v.O(HDO 0.S'(7)5

-C.55-tS4 O.22Wl 0

- 1

6 - 3

HA7?:::X¥"

0.0:;00:: ..>00

-O.:'773~"O.5")nS

~

=[ =

2

:::nJs-c .• ns)

OA:::a~n--:L6%S4 O.S77);'~().22~O!

reduced system. Repe
394

13,3

Structures Modeled as Shear BUildings

RedUCtion of Oynamic Matrices

"tn in/sec?

Determi::e the rtllH.:ra) frequencies and modal shape 0:· the system in Problem 13.2 in terms of irs four original coordinates; find the errors \0 the rwo modes Obtained in Part (b) of Problem 13.2.

i

400k

13.4 CO!1sjeer the shear building shown in Fig. P 13.4.

'\

L~__ .____

Fet all floon:

m'" 1!'2 sec

1

!:J '"

1

1949{\~) \'~

Fig. PlJ.9. 13.9

Consicer the five-slO:Y shear building of Fig. P13,4 subjected at it:> foundation £0 !he [ime.·acce!er::.tion excit<:tion depicted in Fig. P13,9. C'se static conden:,'alIon of the coordinate:; YI, )':1. and )'... and determine: (a) The two natumi frequenc"I\';$ and corresponding mocal shapes of the reduced $ysrem. (b} The displacements at the tloor levels considering twO modes. [c) The shear forces in the coh.:mns 0; the struCture also cOns:ldering two modes.

13.1Q

Solve Problem ]3,9 using (he Dynamic Condens
13.11

"The stiffne>.$ and mass

ma~rices

r 0.906 0.294

-

I

288 - 3

(a) Deleff:)ine the transformut;on mfnrix and (he reduced stiffness and mass matrices corresponding to the star:c condensation of the coordln;:tes YI, YJ, ;;!:ld )'4 as indicated in the figure. (b) Determine the natt;ral frequencies and normal moces for tile reduced sys:em Obtained in Part (a} (c) Use rhe results of Pa~ (b) to determine the modal shapes, des.;ribed in Ihe frve original coordinates, corresponding to the two lowest frequencies.

ruo

13,6

ese rhe results In Problem 13.4 to determine the maximurr. shear forces in the sl:)nes of the building i:1 Fig. Pl3A when subjec~d to an earthquake for whIch the response spectrum is giYen in Fig. 8.10 of Secrion 8.4.

13.7

Repeat Problem 13.2 using the Dynamic Condensation Method,

13,8

Repeat Problem 13.4 using the DynamiC Condensation Method and compare results with the eXllC! solution.

0.424] 0.176

l0.424 0.176 80.000

{AI] "'"

Use the results obtained in Problem 13-4 to deterrnine the maximum displace" ments relative [0 tbe foundation for the struc~ure shown in Fig. Pl3A when subjected £0 an earthquake for which the response spectrum is given in Fig. of Section 8.4.

for u ce!1ain structure ::ire

~K] ~ 10';0.2940.318

Fig. PlJA.

13,5

Tim, (,"O'

0.25

\ in. j For all !;.lories: k"

395

r

- 8

304

1556'.\ 6>14

1»6 644 SO,OOO,

Calcu!ate the fundamental nml;(a! frequency of the system after redcction of the first coordiJ1<.Uc by the fo:low~ng methods: (a} Static condensa,ion. (b) Dynamic condensa:.ion. Also obtain (he ~.mural t'requelicles as a [hree~degrees-of·freedom system and compare results for the fundamental trequency. 13.12

Repeat Problem 13,11 using the Modified Dynamic Condensation Method aod com pure results wi[h [he SOlution obtained with no con:knsatlon.

PART III Framed Structures Modeled as Discrete Multidegree-ofFreedom Systems

14 Dynamic Analysis of Beams

.,

?

In this chapter. we shall swdy the dynamic behavior of struCtures designated as beams, that is, struclures that cJ.:TY loads thar nre mainly transverse to the longitudinal d~rectjon, thus producing nexural stresses and lateral displacemems, We begin by establishing the static characterisrics for a beam segment~ we ~he:l inrroduce (he dynamic effects produced by [he inertial forces, Two approximate methods are presented to take into account the inercial effect in the Structure: (1) the iumped mass method in which the distributed mass is assigr.ed to poine masses, and (2) the consistent mass method in which the assignment to point masses includes rotational effects, The latter method is consistent Wilh [he stadc elastic deflecrfons of the benm, In Chap(ers 21 and 22. rhe eXact theory for dynamics of beams considering (he elastic and inertial distriblHed properties will be presented. In these chapters, [he mathemanca: relationship belween [he exact solution and the stiffness and the ccnsisrem mass coefficients will be shown. 399

400

14.1

Framed Structures Modeled as Discrete Multidegfee-of-Freedom Systems

Dynamic Analysis of Beams

401

STATe PROPERTIES FOR A BEAM SEGMENT

Consider a '.mifonn beam segment of cross-sectional moment of inertia 1, length L. and mererial modulus of elasticity E as shown in Fig. 14.1. We shall establish the relation between static forces and moments designated as Ph P2, P;, and V! and the corresponding linear and angular displacements 8), B:!. 0:.. and 0.1 at the ends of the beam segrr.ent as indicated in Fig. 14.1. The relation thus obtained is the stiffness matrix for a beam segment The forces P; and the displacements 0; are said to he at the nodal coordinates defined for the heam

{.,

I~

segmenl The differential equation for small transverse displacements of a beam, which is well known from elementary studies of strength of materials., is given by (bl

M(x)

in which iv1 (x) is the bending moment at a section of the beam and Y IS the transverse displacement. The differential equation (]4. 1) for a unifonn !Jearn segment is equivalent to pix)

(14.2)

Vex)

(14.3)

since dM(x)

dx

Idl

(14.1)

Fig. 14.2 Static deflection cu::ves due to a t:11!r displacement at one of the nodal coordlnaxes

ment at noda1 coordinate j whHe 3:1 other nodal coordinates are maintained at zero displacement Figure 14.2 shows the displacement curves corresponding to a unit displacement at each one of the four nodal coordinates :or a beam segment indicating the cOlresponding stiffness coefficients. To determine the expressions for the stiffness coefficients kif> we begin by finding the equations for displaced curves shown in Fjg. 14.2. We consider the beam segment in F1g. 14. I free of loads [p (x) = 0], except for the forces P" P,. P" P, applied at the nodal coordinates. In this case, eq. (14.2) IS reduced to

and

=p(x) in which p{x) is the beam load per unit length and V(x) is the shear force. We state :1rst the general definition of stiffness coefficient which is designated by k,}, that is, k'j is the force at nodal coordinate i due to 3 unit displace-

(14.4)

Successive integrations of eq. (14.4) y:elds

1

,

,Ix =."" C,x- -1- C,x -1- C, Fig. 14.1 Beam segment showing forces and displacements al the nodal coordinates.

(14.5)

(14.6)

i

402

I Dynamic Analysis at Beams

Framed $rfuctures Modeled as Discrete Mu:tidegree-oi-Freeaom Syslems

in which C:, Ch CJ • and C~ art coostanrs of inregrcrion ro be evaluated using boundilfy conditions. For example, to de(ennine {he function ifil (x) for the curve shown in Fig. 14.2(a), we make est of the following boundary conditions:

at x:;;;;;

0 yeO)

!

.nd

(H }:

L y (Ll = 0

:lnd

~

dy(O) = 0

( 14.7)

dx dy(L)

eLt

( 148)

0

Use of these conditions in eq::L (14 ..5) and (14.6) resulrs in an algebraic sysc:~m of four equations to determIne the conSLants C j , C2 ('3. and C4 _ The 5u'JSrlmtton of [hese consw.tits lOW eq. (i4.6) resulls in the equation of the detlecled curve for the beam segme;j( in Fig. 14.1(a) U£

403

be l:.sed to detef!TIlne expressions for the stiffness coefficients. For example, consider lr.e bean~ in Fig. i4.2(b) \vo;ch is in equilibrium with the forces producing the displacement 02 = ; .0. For (hiS beam in the equilibrium position, we assume tho.t
!

This work. as sUltcd a:;ove. is eqL:al to ,he work performed by the elastic forces during the vlr;~a! displacernenL Considering [he work perfol!1":ed by [he ':Jendirl.g m0f!lea:, we ob:ain for the Jnremal work

('4.90)

["'M(.r)de

IV,

(14.12)

j()

in which t/JI (x) is used instead of y (x) to correspond to the condilion 0 1 = 1 imposed on {he beam segment. Proceeding irl analogous fushioe, we obtain for the equatio.'1s of the deflected curves in the other cases depicted in Fig. 14,2 the following eqt:ations;

(14.9b)

ill which M(x) is the bending moment Jl ~ec[ion x of the beam and dB is the relative angldar disp!aceme:it of this section.

For the virtual displacement t.:.nder consideralion, [he transverse deflection (0 the bending momejl( through (he clifferen{ial equation (14.1). Substitution of rhe second deriva::ive ,p;' (x) of eq (14.9b) Imo eq. (;4.1) results in

of (he beam is given by eq. (l4.9b) wbich is related

£1 $;' (x) = /yj (x)

04.13)

{ l4.9c)

(l4.9d)

The angular dcf1ectio,1 dO produced during (his virtual displacement lS relnted to the resulring transverse det1ectio:i of (he beam l/II (x) ':::y

(;4.iO) The deflection equations which are given by eqs. (l4,9) and which CO!Tespond to unit displacements at the r,odal coordinates of a beam segment may

cI',p, (.rl

dO d,

at

Since tj;j (x) is (he de!lection corresponding to u unit displacement I, [he displacement resuhlog from an arbitrary displacement o\> is tJ;; (x) 0,_ Analogously. the deflection resulting from nodal dispiacements ~, 0), and 04 are, respect" ively, if2 {x) Oz, if1 (x) 8) and t/14 (x) Dol' Therefore, the total deflectlon y (x) at coor~ dinare x due ~o arbitrary displacements at lhe nodal coordinates of the beam segment :s given by superposition .us

...~

ljJ;' (x)

or dB =

,p;' (x) dx

(14.14)

Equating lhe external virtual work w'r from eq. (t 4.11) with the internal virtual work W, from eq. (14.12) of(Of ""ng M(x) and de from eqs. (14.13) and 04.(4), respec,ively, finally gives the stiffness caefficiem as k"

= i' £1 ,W (x) ,pi' (x)dx

J,

(14.15)

404

Framed Structures Modeled as Discrete M".Jitidegree-of-Freedorr: Sys;ems

In general. any stiffness coefficient associated with beam t1exure, therefore, may be expressed as

t £J ¢i'

Dynamic Analysis of Beams Analogous~y,

the forces at the other nodal coordinates are

= k21 (51 + k·Z2 o~ + k2) 0) + k74 o.~ P, :;; k31 01 + k n 02 + k33 o} + k;4 04

P,:

('L

kif

=

It may be seen from eq. (14.16) that

(x)

¢i' (x) d.x

(14.16)

since the interchange of indices requires only an interchange of the :wo factors ¢i' (x) and ¢!' (x) in eq. (14.16). The equivalence of kij :;;; kji is a panicular case of Betti's theorem. but it is better known as Ma:tweU's reciprocal theorem. It should be pointed out that althoogh the deflection functions, eqs. (14.9), were obtained for a uniform bear:1, in practice they are nevertheless also used in detennining the stiffness coefficients for non!;niform beams. Considering the case of a uniform beam segment of length L and crosssectional momen[ of inertia I. we ;nay calculate any stiffness coefficjent from eq. (14.16) and the use of eqs. (l4.9). In particular, the stiff"ess coefficient k" is calculated as follows. Fro;n eq. (l4_9a), we obtain tf"(' (x)

P:,=k:.IO i +k~101+k430;+k.w(j!l.

kij;::; kp

and from eq. (l4.9b)

4

6x

--+L L'

The above equations are written convenie:1tiy in matrix notation as

p,l

6x 1

and ir.tegration gives

{P}

n

,

1

k2~ k y , i k33

~" I K~J

1s, 1

l:

(14.18)

J

= [k]{o}

£1

P,

l~:

J

I

6L -12 4L' 6L : - 12 -6L 12 2L' -6L 6L l ' ,k

6L

l

(14.19)

6EI = ...... , L-

Since the stiffness coefficient k f; is defined as the force at the nodal coordinate 1 due to unit displacement at the coordinate j, the for~es at coordinate l due to successive displacement Ob 02, 6:» and 04 at the four nodal coordinates of the ream segment are given, respectively, by k, J 01> kr:. fh., k); ~> and k!4 0,_ Therefore, the total force PI at coordinate 1 resulting from these nodal displacements is obtained by the superposition of the resulting forces, that is,

6L

0,

-~~: j

0,

4L-

Ii" 0,

f

(14.20)

or in condensed notation IP)

.

kl~ k;4

or symbolicaJ!y as

'

+ L'~ ,I i\ -L + --Idx L~I

k,~

,

k"

P;.=LkJ.i k-l~ k"

, 12x11-4

_I k"

P2: - i k:, k~2 P}J~:k)i kn

(

Substitution in eq. (14.15) gives

-6

(14.17)

in whiC!1 {P} ar:d {8} are, respectively. the force and [he displacement vectors at the nodal coordin~tes of the beam element and LkJ is the bearT: element stiffness r:1lltriX. The use of eq. 04.16) in the manner shown above to determine the coefficient kl2 will result in the evaluation of all the coefficients of the stiffness ;natrix. This result for a uniform beam segment is

=

¢!'(x)

405

= (kJ{o}

(14.21 )

14.2 SYSTEM STIFFNESS MATRIX Thus far we have established in eq, (14.20) the stiffness equation for a unifonn beam segmer.t. that is, we have obtained the relation between nodal displacements (li:1ear anc ar.guiar) a:1d nodal forces (forces and moments). Our next objective is to obtain the same type of relationship between the nodal displacements and the nodal forces, but now for the entire structure (system stiffness equation). Furthermore, our air:1 is to obtain the system stiffness matrlX from the stiffness matrix of each element of the system. The procedure is perhaps better explained through a specific example such as the cantilever beam shown in Fig. 14.3.

406

Dyna:nic Analysis 01 Seams

Framed Structures Modeled as ;Jiscre:e Mulll'degree..of-Freedom Syste:-ns

,

7

7~

~..

;'i'j)

,&

,+ .+ 3

5

4~0

&:,

G)

0

Ei"O'N-m' L = 1m

Fig, Vt3 Cantilever beam divided imo three beam segments with numbered sy"tem nodal coordiflU(e::.

The first step in obraining the system stiffness matrix is to divide (he structure into elements. The beam in Fig. 14.3 has been divided into three elements which are numbered sequemially for idenLificarion. Tne second step is to identify the nodes or joints between elements and to number consecutlve· ly those nodal coordinates [har are not constrained. The conscra:ned or fixed ncdal coordinates are [he laS[ to be labeled, All {he fixed coordinates may be given (he same label as shown in Fig. 14.3. In the presen[ case, we consider only two possible displacemems at each node, a vertical defiecrion nod an angular displaceme:1t. The cantilever beam in 14,3 with its three elements ,esults in a total of six ffee nodal coordinates and twO fixed nodal coordinates, :he latter being labeled with {he number seven, The third step is to obtain sys[ematically [he stiffness marrix for each element in the system and to add [he element stiffness coefficients appropriately to obtain the system sriff:1ess ma[rix. This method of assembling the system stiffness matrix is called [he direct method. In effect, Z!.:ly stiffness coefficient k'j of ~he system may be obtained by adding togerher the corresponding stiffness coefficients associmed with [hose nodal coordinates. Thus, for example, to obtain the system stiffness coefficient k:n, ir is necessary to add [he stiffness coefficients of beam segments In and 11::. corresponding to node three. These coeffjcients afe designated as kW and kW, respectively, The upper i:1dices serve to identify the beam segrnenl, and [he lower indices to !ocme the appro?riare stiffness coefficients in the COITes~ ponding ele:nent stiffness matrices. Proceeding with (}le example in Fig. 14.3 and using eq, (14.20), we obtain the following expression for the sIiffness matrix of beam segmem &, namely

2 12

3

6 - [2

4

- 6 2

Cc

6

12 -6

(422)

407

For rhe Decm segme:1t &,. the element nodal coordinc[es numbered one to fout coincide With the assignmen: of system nodal coordinates also numbered] to 4 as may be seen in Fig. 14.3. However. for the beam segments .& and & of this beam, rhe assignmem of element nodal coordinates numbered 1 to 4 does not coincide wirh the assigned system coordinates. For exam?le, for element & (he assigned system coordinates ~s seen in Fig. 14.3
m.

rk(!)]

=

i 0'

2 6

7

7

12

6

12

6

4

6

- 12 -6

12

2

6

5 12

6 6

6

2

4

12 -6

5

6

7 7

:J

(14.23)

2

and

,:'''1 -

'"~ [

3

4

12

6

•

4

-12 -6

6

2

6

4

3 (l4c24)

6

Proceeding systematically [0 assemble the system stiffness matrix, we [ra05late each entry in (he element stiffness matrices, eqs. (l~.22), (14.23), and (14,24), to the appropriate location in the system stiffness matrix. For instance, l2 X 10" should be translated to the stiffness coefficienr for element .ill, location at row 3 and column 5 since these are the coordinares indicated at right and top of matrix eq. (14.24) for this entry. Every element stiffness coefficient translated [0 its appropriate location in the system stiffness mmrix IS added to the mher coeHiciems accumu~rHed at that loca!ton. The stiffness coefficients corresponding w columns or rows carrying a labei of a fixed system nodal coordinate {seven in the present example) are simply disregarded since :he constrained nodal coordinates are not unknown quantities. The as~

kW

408

Framed Structures Modeled as Discrete MuJtidegree"of"Freedom Systems

sernblage of the system matrix in the manner described results in a 6 X 6 matrix, namely

fOf

DynamiC Analysis of Beams

this example

Mass distribution

Lumped mass.

8

-6

12 -6

14

0

[kJ = 10'

0 - 12

6

0

0

0 -12 0

0

2

6

2

0

8

-6

6 2

0

0

- 12

12

-6

0

0

6

6 2

-6

4

nil =-,-

Uniform

(14.25)

Equation (14.25) is rhus the system stiffness matrix for the cantilever beam shown in Fig, 143 which has been segmented into three elements. As su:::h, the system stiffness matrix relates the forces and the displacements at the nodal system coordinates in the same manner as the element stiffness matrix relates forces and displacements at the element nodal coordinates.

~I

Ai

Im,xl

'8

I

r-'---i 14.3

I

!-----'----'----'---'---·.:........--1 i11A

24

409

INERTIAL PROPERTIES-LUMPED MASS

The simplest method for considering the inertial properties for a dynamic system is to assume that the mass of the structure is lumped at the noda1 coordinates where translational displacements are defined, hence the name lumped mass method. The usual procedure is to distribute the mass of each eJement to the nodes of the element. This distribution of the mass is determined by statics. Figure 14.4 shows, for beam segments of length Land distributed mass m(x) per unit of length, the nodal aHocation for uniform, triangular. and general mass distribution along the beam segment. The assemblage of the mass mauix for the entire structure wUl be a simple matter of adding the contributions of lumped masses at the nodal coordinates defined as translations. In this method, the inertial effect associated with any :-otational degree of freedom is usually assumed to be zero, although a finite value may be associated with rotational degrees of freedom by calculating the mass moment of inertia of a fraction of the beam segment about the nodai points" For exampie, for a uniform beam, this calculation would result in determining the mass :nomem of inertia of half of the beam segment about each node, that is

J

'

L

Genera!

Fig. 14..4 Lumped masses for beam segments with distributed mass. shown in Fjg. 14.3 1n which onlY translational mass effects are consIdered, the mess matrix of the system would be the diagooaJ matrix, namely

2

[M]

=

rm

3

4

5

a

1

2

my

I

0

m, in which ml

=

InLl ,m~

2 -:--2-

mL1 fhL} m· = -....::. +

;)

2

inLJ

where in is the mass per unit length along the beam. For the cantilever beam

6

mS=2

2

3 4

I5

aJ6

(14.26)

410

Framed Structures Modeled as Discrete i'v',lIltidegree-of·Freedom Systems

Using a special symbol (14.26) as

tr J)

iM] ~

rm,

for diagonal matrices, we may write eq.

0

,",0m,

0J

Dynamic Atl;:Hysis of Bearrts

411

The inertial force 11 (x) per unit of length along the beam due to this acceleration is then given by

II (x) ;

(14.27)

Ii, (x) y, (x)

or using eq. (:4.28) by

14.4

iNERTIAL PROPERTIES-CONSISTENT MASS

It is possible to evuluu[e [he mD,Si> coefficients corresponding w the nodal coordinates of a beam element by a procedure similar to the determination of element stiffness coefficients. First, we defJ ne [he mass coefficient rn!; as rhe force at nodal coordinate i due to a unit acceleration at nodal coordinate j w!1ile all Other nodal coordinates are maintained at zero acceleration. Consider the beam segment shown in Fig. 14.5(a) which !las distribuled mass m(x) per uoir of length. In the consisrent mass method, it is assumed that the deflections resulting from unit dynamic displacements a[ the nodaJ coordinares of the beam elemen[ are giVen by the same functions 1/11 (x), ifi2 (x), 1/11 (x), and ifJ4(X) of eqs, (14,9) which were obtained from static consideratlons_ If the beam segment is subjected to a unit nodal acceleration at one of the nodal coordinates, say '~> the transverSe acceleration developed along the length of the beam is given by the second derivative with respect [0 time of eq. (14.10). In this case, with 8; = 8) = 8,,::::; 0, we obmin

y, (x) ~ >/I, (x) iJ,

, ~

t=

or, since

8:: = 1, j, (x)

= iii (x) .p, (x)

(14.29)

Now to determi:1e the mass coefficient {fin, we give to (he beam in Fig. 14.5(b) a virtual displacement cOITesponding to a unit displacement a( coordinate I, 0, = 1 and proceed (Q apply the principle of v!rtual work for an elastic system (extema! work equal to imemal virtual work). The virtual work of the external force is simply (:'4.30) since (he only external force undergoing virtual displacement is the inerrial force reaction !finD, wirh 01 = 1. The virtual work of the internal forces per unit of leng;h along [he beam segment is

(14.28)

oW, = j, (x) t/tl (x)

3

mi.>;} L

1

,-..,

or by eo. (14.29),

t,

(.J

and for The entire beam

('

W, = Join (x) >/I, (x) >/I, (x) dx

(14.3 I)

Equating the external and internal vinuai work given, respectively, by eqs. (14.30) and 04.31) rescirs in

L "

ll/IZ=

m(x) 1jJ?;(.r:.) t/t!{x)dx

(14.32)

(bJ

Fig. 14.5 (a) Beam element wirh dimiDuled mass shOWing four aod,1 coordinates. (b) Beam element supporting inenial toad due to ;)cceleration 5z "'" t. undergoing vinuai displacement OJ = L

which is the expression for (he consistent mass coefficient mn. In general, a consistent mass coefficiem may be caiculated from m" =

L""

(x) >/I,(x) >/I, (x) dx

(14.33)

412

Framed Structures Modeled as Discrete Multidegree-of-Fieedom Syslems

It may be seen from eq, (14.33) that mij = mj; since the interchange of the subindices only results in an interchange of the order of the factors ;,fI, (x) and if'; (x) under the integral. In practice, the cubic equations {14.9) are used ir. calcu!a:ir.g the mass coefficients of any straight beam elemenL For the special case of the beam with uniformly distributed mass, the use of eq. (I433) gives the foHow:ng relation betweer: inertial forces and acceleration at the nodal coordinates:

r P,

I )

22L

156

P2

mL

P, P,

420

27..£ 54

4L' i3L

54

8,

13L

3:2

156

0;.

3L' .- 22L

- !3L

4L'

Dynamic Analysis of Beams 7

3

7~ 210 Kg

When (he mass matrix, eq. (14.34), has been evaluated for eaer. beam element of the structure, the mass matrix for the entire system is assembled by exactly the same procedure (direct method) as described in developing the stiffness matrix for the system. The resulting mass matrix \"lin in general have the same arran2:ement of nonzero terms as the stiffness matrix. Th~ dynamic analysis using [he lumped mass matrix. requlres considerably ~ess computational effort than the analysis usjng the consistent mass method for the following reasons. The h.:.mped mass matrix for the system results in a diagonal mass matrix whereas the consistent mass matrix has many off diagonal terms whkh are called mass coupling. Also, the lumped mass matrix contains zeros in its main diagonal due to assumed zero rotational inertial forces. This fact permits the elimination by static condensation (Chapter 13) of the rotatior.al degrees of freedom. thus redUCIng the dimension of the dynamic problem. Nevertheiess, the dynamic analysjs using the consistent mass matrix gives results which approxlmate bette:- to the exact solution compared to the IUr.1ped mass method for the same element discretization.

Example 14.1. Determine the lumped r.12.SS ar.:d the cor.sis;:ent mass matrices for the cantilever beam in Fig. 14.6. Assume uniform mass. iii = 420

.& 210 _._="--_ ... Kg

2HJ Kg

210 Kg

•

•

21~.K.,-._",&=2,,-_2_'..~ Kg

( 14.34)

8,

Fig. 14.6 Lumped masses for Example 14.1.

example LIm, m::;:;: 420 kg/rn into eq. (14.34) gives the CO:ls:stent mas.s matrix [M~2)1 for any of the three beam segments as

[M:."J =

0

420

0

210

0j

(b) Consistent Mass Matrix, The consistent mass matrix for a unifonn beam segment is given by eq. (14.34). The substitution of numerical values for this

4

22

4

13

5L

13 ..··3

156

22

:,- 13

-22

,12 1

(a)

3

"J 4

The assemblage of the system mass :natrix from the element mass matrices is car:ied out in exactly the same manner as the assemblage of the system stiffness matrix from the element stiffness matrices, that is, the element mass matrices are allocated to appropriate entries in the system mass matrix. For tl:e second beam segment, this allocation corresponds to the first four coordinates as indicated above and on the right of eq. (a). For the beam segment &, the appropriate ailocation is 3, 4, 5, 6 and for the beam segment 8, 7,7, 1. 2 since these are the system nodal coordinates for these beam segments as indicated in Fig. i4,6. The consistent mass matrix [Me] for this example obtained in this manner is given by 3

4

5

6

312

0

54

13

0

01

1

0

8

13

3

0

0

2

54

312

0

54

-- 13

13 -- 3

0

8

13

0

0

156

0

54 -13

13

0

-3

- 22

4

r420

3 54

- 13 1 1 -3

156

0

[M d =

4

22

0

kg!m.

Solution: (a) Lumped Mass Matrix. The lumped mass at each node of any of the three beam segments, into which the cantilever beam has been divided, is simply half of :h; mass of the segr.1cnt. In the present case, the lumped mass at each node is 210 kg as shown in Fig. 14.6. The lumped mass matrix ~Md for this structt:re is a diagonal matrix of d:mension 6 X 6, namelY.

413

[Mol

=

L

I3

!

3

(b)

- 3; 4 -22 4

J5

6

We :'lote that the mass matrix [iWe ] is symmetric and also banded as jn the case of rhe s:iffness matrix for this system. These facc:s are of great importance

414

Framed Structures ModeJed as Discrete Muitidegree·of·Freedorr. Systems

Dynamic Analysis of Beams

in developing computer prog!"nms for strucClral analysis, since it is possible to perform the necessary caicu!.2.lions storing in the computer only the diagonal elements ::md lhe elements (0 one of the sides of the main diagonal. The maximum m:.mber of no:1ZUO elements in any row which are required to be stored is referred to us the bandwidth of the matrix. For lhe ma~rix eq. (b) the bandwidth is equat ~o four (NBW 4). In this case, it is necessary to Store a total of 6 x 4 = 24 coefficier.rs, whereas if the square matrix were [0 be srored, it would require 6 X 6 36 storage spaces. This economy in swring spaces becomes more drar:1:lttC fOf structures with a large number of noda! coordinates. The dir:leoSloli of rhe bandwidth is di:-ectly related to :he largest difference of the nodal coordinates assigned to any 0: the elemc:nts of the structure. There:ore, It is imporrnn{ to number the system nodal coordinates so as to r.'1inimize (his differenCe.

14.5

DAMPING PROPERTIES

Damping coefficients are defined in a manner entirely parallel to the deflni[Lon of [he stiffness coefficient or the mass coefficier:r. Specifically, the damping coefficient Ci' is defined as the force developed at coordinate i due to a unit velocity at j. If [he damping forces cisrributed in [he StfUcmre :::ould be determined, the damping coefficients of the various structural elements would then be used in obtaining the damping coefficient correspor.dir.g to the system. For example, Ihe damping coefficient C;j for an element might be of the form c,=

r'

Jo C(x) "'i(X) ifi;(x)dx

Pix. ()

Fig. 14.1 3
supporti '.g ;!rbilr"ry di$[ribulco load unc.ergoing 'Iltlual

dis~

placem:!Hl 01 = 1.

wrirten direcl:y. In ger.eral, however, loads are app:ied at poims other than r.oda! coordinate:>. Ir. addition, the external load may :r:clude [he action of distributed forces. !n {his case, [he load vectOr corresponding (0 the nodal coordinntes consists of the eql,ivaiem or generalized forces. The procedure to de{emline the equivalem r.odaJ forces which is consistent with the derivation of the sciffness matrix and ~he consistent nU1SS malrix is to assume (he validity of the smtic deflection functions, eqs. (14.9). for the dyr'.amic problem and use the principle of virrua! work. Consider rhe beam elemem in Fig. 14.7 when subjected to an arbitrary distributed force p{x,t) which is a function of position along the bearr as weI! as a :unction of ;::me. The equivalent force PI llt coordinate 1 may be found by giving a virtual cisplacement 0] = 1 at this coordinate and equating the resulrir:.g exrernal work and imerr;a! work during this virtual displacement. !n (his case, the external work is

t;

( 14.35)

where C (x) represems the dis(ributed damping coefficier:[ per unl[ length, If the element damping matrix could be cakulated, the darr.ping matrix for the entire S[f1.lCfUre could be assembled by a supe:position process equivalent to rhe direct stiffness matrix.. In practice, the evaluarion of the dnmpjng property c(x) is impracticaole. For this reason" the damping is generaHy expressed 1n terms of damping ratios obtained experimentally rmher than by a direct evaluation of the damping :natrix usir.g eq. (14.35). These dampir:.g ratios are evaluated or estimatec for each natural mode of vibration. If (he explicit expression of the e.amping matrix [C] is needed, it may be computed from the specified reiative damping coefficienLs by any of the methods described in Chapter l2.

Wc~P,8,

EXTERNAL LOADS

\Vhen the dynamic loads acting on the StruCture consist of concentrmed forces and mOments applied at defined nodal coordinates, the loae. vector Clln be

=P,

(14.36)

since 8, !. The in;:ernal work per unit of length along {he beam is p (x, t) if;1 (x) and (~e total internal work is then

W, =

r

p(X,I) ifi, (x)cL,

(14.37)

Equating exterr;al work, eq, (14.36), and ir.(ernal work, eq. (14.37), gives the equivalent nodal force as P,(l)=

14.6

415

J,r'

p(x,l)ifi,(x)dx

Thus the element equlva:ent nodal forces can be expressed in general P,(r)

f

p(x.l) >j;,(x)dx

(14.38) 2.S

(14.39)

416

Frame-d Structures Modeled as Discrete

MuJtideg:ee-of~Freedom

Systems

Dyr.amic Analysis of Beams

f;

417

Nixl

PVt, r) "'" 2'00 sin 10r N/m

p,(t}1? t t trt1 "1,: t t tnt t~P'1rl Pitt!

Pjltl

[,l

Fig. 14.8 Beam segment subjected to external distributed load showing equivalent nodal forces,

y

14.8 and detennine the Example 14.2. Consider the beam segmenr in element nodal forces for .1 uniform disuibmed force along the Jeng:h of !he beam given by

p (x, t) = 200 sin 101 N 1m lei

Solution: Introduction of numerical values into the displacemenrs functions, eqs, (14,9), and substitution in eq, (14.39) yield P,(f)=200C (1-3x'-

-,

dx sin 101 = lOa sin 101

P,(I)=200('X(1 '-x)'dxsin 101

16,675io 101

00

p) (Il

200

L"

(3x' - 2x'l dx sin 101

= 100 sin

Fig. 14.9 (a) Beam element loaded with arbitrary distrib<.::(ed axial force. (b) Beam element acted on by noda: forces resulting from displacement 02 I, undergoing a

virtual disphcement

o! = 1.

examp.le, kc;r:; 1s the vertical force at the left end. If we now to this deformed beam a unit displacement 0 1 = 1. the resulting ex[ernal work is

:Ot Or

P,(r)

=200fo'x'~t-

I)dx sio 10,= - 16,67 sin lOt

14,7 GEOMETRIC STIFFNESS When a beam element is subjected to an axial force in addition to flexural loading, the stiffness coefficients are modified by the presence of the axial force. The modification corresponding ro the stiffness coefficient ky is known as the geometric stiffness k Gi;, which is defi:led as the force corresponding to the nodal coordi:1ate i due [0 a 'unit displacement at coordinate j and resulting from the axi2.1 forces in the struct:..He. These coefficients may be evaluated by application of the principk of virtllal work. Consider a beam e]ement as used previously bi,lt now subjected to a distributed axial force per unit of length N(x), as depicted in Fig. 14.9(a). In the sketch in Fig. 14.9(b), the beam segment is subjected to a i,lnit rotation of the left end, ~ = 1. By definition, ~he nodal forcC'-s due to this displacer:1ent are the corresponding geometric stiffness coefficients; for

(14.40) since 0= 1. The internal work during this virtual displacement is found by con.~idering a differential element of length dx taken from the beam in Fig. 14.9(b) and shown enlarged in Fig. 14.10. The work done by the axial force N(x) during the virtual displacement is

(14.41) where Dc represents the relative displacement experienced by the nonnal force N (x) acting on the differentia! eiement during tbe virtual displacemen:. From Fig. 14.10, by similar triangles (triangies I and H). we have ~=d~(x) diff, (x) dx

418

Framed Structures Modeled as Discrete Multidegl'ee~ol~Freedom Systems

Dynamic Analys;s of Beams

419

in eq. (14.44) to calcuia(e the geometric sriffness coefficients, the result is

called the consistent geometric s(i.f/ness matrix, In the special case where the

~'I"-+-T"-N'xl

aXial force is constant along the length of the beam, use of eqs, 04.44) and 04.9) gives rhe geometric stiffness equation as

i

--'--

~~~f

i

P, P,

t

p)

36

1i

l-

.y.

3'OL

P,

3L

3L

4C' - 3L

36 3L

- L'

- 36

3LJ -L" :

-3L 36

-3L - 3L 4L'

J

Ii,

Ii, Ii, ~ 15,

j

(14.45)

The assemblage of the sys.tem geometric stiffness matrix car: be carried out exacrly in the same manner as for rhe assemblage of the elastic stiffness matrix. The result:ng geometric Stiffness matrix will have the same configurallon as the elastic stiffness rr.mnx. It is customary to define the geometric stiffness matrix for a compressive axial force. In this case, the combined stiffness maLrix (K"J for :he structure is gi\'en by

w,lxl

--1-~==~======:j

(14.46)

Fig. 14.10 Differemial segmem of deflected beam in Fig, 14.9.

in which (K]ls [he assembled eifistic sdffness mauix for the structure and [KcI (he corresponding geometnc stiffness matrix"

or

Example 14.3. For the cantilever beam in Fig. 14" II, detennine the system geometric matrix when an axial force of magnitude 30 N is applied at the free end as shown. ir. this figure.

15 = d.p, . d.p, dx ¢

15. = .p,

dx (x) .

dx

.p; (x) t:b:

in which t/I{ (x) and I/I~ (x) are the derivatives with respect to x of the corres~ ponding displacement functions defined in eqs. (14.9), Now, substiruting 8~ into eq, (}4.41), we have

Solurion,' The subs[irutlOn of numerical vatues into eq. (l4AS) for any of the three beam segments in which the cantilever beam has been divided gives the element geometric matrix

( 14.42) [Kcl

Then integrating this expression and equating (he result to the external work. eq. (14.40), finally give

-n

3 -36 36 3 -3 -36 -3 36 - 3 ' -3 41 3 -1

"

"

kGn

In

r'N(x).p: (x).p; (x)dx

(14.43)

" general, any geometric stiffness coefficient may be expressed as kc,j= r'N(x).p!Cx)if!j(x)t:b:

7

,~ (14,44)

Jc

3

1

6::,

)~

&,

)~

5

6::,

6 N =. 30 Nl"WtQflS

---lm----{-----lm

1m

------j

In the denvatior. of eq. (14.44), if ;s assumed that rhe normal force N(J;) is icdependent of time. \-Vhen the displacement functions, eqs. (14.9), are used

Fig. 14,11 Cantilever beam subjec(cd

iO

conSlam axial fo(ce (Examp!e 143).

420

Dynamic Analysis of SeaIT's

Framed Structures Modeled as Discrete Multjdegree·of~Freedom Sys:ems

since in this eX2mple L = 1m and N= 30 N. Use of the direct method gives the assembled system geometric matrix as

r! [Kc'

14.8

=

72 0

l-'j

0 -36 8

3

-3 -1

-3

72

0

3 -3 0 - 36 0

3

0

0 0

36

n

3 - 1 4

In practice, the solution of eg. (14.51) or eg. (!4.52) is accomplished by standard methods of analysis and the assistance of appropriate computer programs as those described in this and the following chZl:Jters. \Ve illustrate these methods by presenting here some simple problems fo~ hand calculation. Example 14.4. Consider In 14.12 a :miform beam with the ends fixed against translation Of rotation. In preparatJOn for analysis, the beam has beer. divided into four equal segments. Determine the first three natural frequencies and corresponding modal shapes. Use the lumped rr:ass method in order to simplify l;,e numerical calculations.

!

36 -3

-3

421

j

EQUATIONS OF MOTION

So[urion: We begin by numbering sequentially the nodal coordinates starting with the rotational coord:nates which have to be condensed in the lumped ffi8.SS method (no inert:a! effect associated with rorationai coordinates), con tinuing to numbe, the coordinates associated with translation, and as~igning the dummy last D'Jmber 7 to any fixed nodaJ coordinates as shown in Fig. 14.12. The stiffness matrfx for any of the beam segments for this examo:e is obtained from eq. (\4,20) as . w

In the previous sections of this chapter, the distributed properties of a beam and its lot:d were exp:essed in tenns 0: discrete quantities at the nodal coor~ dina[es. The eq:lJtions of motion as funct:ons of these coordinates may then be established by imposing conditions of dynamic equilibrium between the inertinl forces {FI(r)}, damping forces {FD(I)}, elastic forces {F,(I)}, and the external forces {F(t)}, th2.t is,

7 (14.47) [K]

The forces on the left-hand side of eq, (14.47) are expressed in terms of the system mass matrix, the system damping matrix, and the system stiffness matrix as (14.48)

{FI (In = eM]

= [C] {y}

(14.49)

{F,(I)} = [K] {y)

(14.50)

{FD(t)}

SubstItution of lhese equations into eq. (14A7) gives the differential equation of motion for a linear system as [lvl]{y)C-[C]{j}7[K]{y}

{F(t))

{F(I)}

(1452)

in \-vhich

[Kol = [K] - [Kcl

(j ~.53)

7

\

6

12

4

-6

-6

!2

2

-6

n~

- 6

!

4

( 14.54)

5 2

4

4 J 1

at the top and on the right of the stiffness matrix, eg. (14.54), we proceed to assemble the system stiffness matrix using the direct method. For the beam segment lA, the corresponding labelS are 7. 7, 4, L Since the label 7 which corresponds to fixed coordinates should be ignored, we need to trans]2.te only 7

7

[M) (Y}+[C]{j} +[KJ{y}

2

4

With the aid of the system nodal coordinates for eact: beam segment written

(14.51)

fn addition, if the effect of axial forces is considered in the analysis, eq. 04.51) is modified so that

5

4

~

4 I'

"-

&

EI,iil/

i---lm

(1) I

5

2'

.&.

""

6

&

@ 1m

,I

1m

3' "

0) I

7

&

7~

1m~~

Fig. 14.12 Fixed beam divlded in fo~r elements with indication of system nodal coordinates (Exampie 14.4).

422

Framed Structures

f>.~.odeJed

as Oiscrete tv1ultidegrc€-ol-Freedorn Systems

the lowest 2 X 2 submatrix on ~he cighr to locarions given by the combinatio:;'L of row indices 4, 1 and column i:1dices 4, I; for the beam segment 11::., we mmslnte [he 4 X 4 elements of matrix eq. (14.54) to the system stirfness mu!rix to rows and columns designated by comb~r.atton of indkes 4. r. 5, 2 as labeled for this element 2nd so forth for the other two beam segments. The assembled system stiffness matrix obtained in (his m
r~ [K]

EI

l-j

2

0

0

8 2 6 0

2

6

-6

Oynamic AnalySis of Beams

The generr:i traosforrna(ion matrix, eq. (13,9), is [hen

r

8

0

6

24

-12

6

- 12

24

~2

0

0

- 12

24

0"750 -0.2Ia 0 0.858 0.214 - 0.750 - 0.214

[T] =

l

-6

0

0.214

- 0.858

01

6 0

0 0

{ 14.55)

I

0

0

0 0

I

0

[A]

~

f1

0

0

- 0.214

0

t

0

0.858

!O

0

0

0

0

18.86EI

- 12.00EI

5.14EI

1 ,0

0

0

- 12.00EI

15.00E1

12.00EI

l..O 0 0

S.l4El

12.00EI

18.86EI

- 0.750

As an exercise, [he reader may check eq. (13.7) for this example by simply performi:1g the mutrix multip:icurions

- 0.858

0.750

-0.214

...... " ..

0.214

l4.:3. Therefore, [he reduced lumped mass matrix is

[

18"86 [K] ~ EI - 12.00

{K]

- 12.00

is.OO

5.14- - 12.00 znd

1

The natural frequencies and modal shapes are found by solving the undamped

(14.56)

free vibration problem, that 15,

[,i1) {y} + [K] {y}

= {a}

(14.6l)

Ass:im:ng the harmonic solu(jon ({y} ~ {a} sin Wi), we obtain ([K] -

w' [,i1)) {a}

=

{OJ

(14.62)

requiring for fi nontrivial so~utiO!1 that .he determinant

(\4.63)

514J ;2.00

(14.57)

"'12

•

18.86

",12_ 0.214 0.858

L

0.214 - 0.750 - 0.214J

0.750 - 0.214 0

( 14"60)

~o 0 I j

and the transformation matrix

I

[TJ = -

01

rIO 1 0

[i4J =mlo

Comparison of eq. (\4.56) in partition form with eq" (13.1!) permits the identlfi:::ation of the reduced stiffness matrix [7:, so that

three equal masses

of mag:li[Ude iii at each of [he three translatory coordinates as indicated in Fig.

0.214

0

(14.59)

0

The lumped mass method applied to this example The reduction or condensation of eq. (14.55) is accomplished as explained in Chapter 13 by simply performing (he Gauss-Jordan elimi:1a[ion of the first three rows since, in this Cfise, we should condense these first ~hree coordiames. This elimlrHlrion reduces eq. 04.55) to the foHowing mmrix:

423

1

0.858 1

m(2 inn

inn m/2

ro/2 iiia

m

iii

m

•• •

•(d•

•

•• •

iiil2

•

~m2

(bl

(14.58)

Fig. 14.13 (a) Lumped mas<;e:; for unifcrm beam segmentf>_ (b) Lumped masses at system nod:\! coordinate:).

424

Dynamic Analysis of Beams

Framed Structures Modeled as Discele Multidegree-of·Freedom Systems

SubstitutlOn in this last equation and (14.60) yields 18.86 - "-

[K] and [,11], respectively, from eqs. (14.57) 12.00

r

5.14

The normalized modal shapes which are obtained by division of the elements of eqs. (l4.69~ by corresponding values of (m,-= 100 kg for this example) are arranged In the columns of the modal matrix are

/Y;'-;;;;;

r

II

1=0

15.00 -,I - 12.00

- 12.00

(H.64)

I

18.86

- 12.00

5.14

425

A:

0.0431 [
0.0707

LOQ431

0.0707

0

0.0562 i - 006071

(I ;1..70a)

00562 J

in which (14.65)

EI

!he modal shapes in terms of the six origmal coordinates are then obtained ~y eq, (J 3.8) as

[
The roots of the cubic equation (14.64) are found to t>e

A, = 13.720.

A, = 1.943,

and

,13

= 37.057

r

(14,66)

Then from eg. ([4,65)

[
w:: = 3.704~;

(14,67)

The first three natural frequencie.s for a uniform fixed beam of length L = 4m determined by the exact analysis (Chapter 21) are

0.0301

-0.1212

0.0594

00301

0.0431 0.0793

0.0707

0

0.043 [

-:""l 0.0457 0.0562

(14.70b)

- 0.0607:

°

0.0562"

Fo, sin

wt

F2 = Fez sin

wt

F[

a
i

F;; ; ; ; F03 sin Wt

(l4.68)

WJ (exact) = 7556

The first two natural frequencies determined using the three~degrees-of-freedom reduced system compare very well with the exact values. A practical rule in condensing degrees of freedom is to condense those nodal coordinates which have the least inertial effect; i:l this problem these are the rotational coordinates. The modal shapes are determined by solving two of the equations in eq. (14,62) after substituti:1g successively values of Wit lVz, w} from eqs, (14.67) and conveniently setting the first element ror each modal shape be equal to one. The resulting modal shapes are

[ 1.84J' 1.00

0.0594

0

Example 14.5. Determine the steady~state respOIl$e for the beam of Example 14.4 when subjected to the harmonic forces

(exact) ;;;;;; 1398 ~i £llm

"" (exact) = 3.854 Elfin

1.001

I

L

Ellm

w) = 6.087,;

WI

1-

= [71[ \tiJ.,

1.0°1 LOS

i

LOO"

Sol~tian:

The modal equations {uncoupled equations) can readUy be Wrltusmg the results of Ex~mple 111.4. In general the nth normal equation is given by t~n

(l4.71)

in which ( 14.69)

N

P" = )'
426

Framed Structures MOCleted as Discrete MUltldegree-ol-Freedom Systems

The steady-state solution of eq. (14.71) is given by eq. (3.4) as t" =

2.,

. ~ S;fl I.l/l

Prsin

w[

= "~~~ W"

Dynamic Analysis of Beams

Therefore, the motto:1 at these nodal coordinates is gi Yen by

y, = - 2.616 X 10 -, sin 3OO0lfad

(14.72)

(v

y,= -L818x 10-' sin 3000lrad

The calculaticns required in eq. (14.72) a:e conveniently arranged in Table 14. L The deflections at the modal coordinates are found from the tr2.flsformation {y}

~

[IPJ {z}

3.524 X 10- 6 sin 3000!rad y, =

- L207 X 10 y, = ., 4,094 X 10 -

(l4.73)

in which [IP] is the modal matrix and {y}

{Y} sin

v

r

YJ

I" 00594

Y, Y, , Y,

0.0431

0.0707

0.0793

0

J

or

l

0.0594

- 0.0457

0.0301 - 0.1212

0

0.0431

0

O.OJOI

0.0457 0.0562 - 0.0607

- 0.0707

- 2.616 X 10'" tad

Y,

- 1.818 X IO-'rad

O.Odh ~

=-

TABLE 14.1

Mode n

2 3

w,

1.943 X 10' \3.720 X la' 37.057 X 10 6 ... _ - - - _ ...

sin 3000, m (14.74)

ELEMENT FORCES AT NODAL COORDINATES

The central problem to be solved using the dynamic stiffness method is to dercnnl:1e the d;splacements at the nodal coordinates. Once these displacements have been determined, it is a simpk mauer of sUbstitt:ting the appropriate displacements in rhe condition of dynamic equilibrium fer each element to calculate the forces at the nodal coordinates. The nodal element forces {P} may be obrained by adding the tnen:ial force {PI}, the damping force {P D }, the e:astic force {P s}. and subtracring the nodal equiva1ent forces (P E }. Therefore, we may write

T

[el (ill

+ [k]{B)

(Pel

(14.75)

1.207 X IO-'m In eq. (14.75) me inertial force, (he damping force, and the elasdc fcrce are. respectively,

-3.329x 10-'m

(PI)

Modal Response for Example 14.5

2

14.9

{Pi ~ [m] (8)

Ys= - 4.094 X lO -b m

Y,

6

or

3.524 X 1O- 6 rad

Y3 = Y~

{ - 5.200 ) 5 L500 f 10-

0.0562

Y,=

sin 3000r m

The minus sign in the resulting amplitudes of motion simply indicates that [he motion is 180 0 Ou~ of phase with (he applied harmonic forces.

The substitution of the modal matrix from eq. (14.70) and the vulaes of {Z} from the last celumn of Table !4.1 into eq. (14.73) gives the amplitudes at nodal coordinates as

', Y2

-6

Y6~ -3.329x 10-' sin 3000rm

w(

{ll = {Z} sin WI

,

427

P" =

>- 4>" FOi

367.2 70.7

- 135

. . _ - _..

= [mliB)

(P D )

~

{P,}

= [kl {B}

[el (B) (14.76)

Z,,= - 5.200 10 ,5 l.500 10 -, -0.048 10-'

where [m] is the element mass matrix.; [e] (he element damping matrix; (kJ the element stiffness matrix~ and {8}., {8}, {8} represer.ts, respecti ve)y, lhe displacemer:t, veiocity, and acceleration VeClOrs ar the nodal coordinates of the element Aitho:lgh the deterrnin2:ion of element end-forces at nodal coordinates is

42B

Dynamic Analysis of Beams

FraQed Strllctures Modeled as Discrete Muilldegree·of-Freedom Systems

given by eq, (14,75), commercial computer programs onlY use the forces due to the elastic displacements, that is, the elew.ent end forces are calculmec as

= [k]{8}

(P}

TABLE 14,2 Element Nodal Forces (Amplitudes) for Example 14.6

Bea:n Segment

ExampJe 14.6. Determine the element nodal forces and moment'l for the four beam segments of Example 14.5.

P, P, P,

p,

1347.;' 322.2

121.2 201.0 664.3 - 322.2

,038.3 481.4 ----~

= [m] {S} + [k]{8}

(oj,

J~: l , , I j'

f

8}

)'s

y,

{S';;; =<, y,

Yf

:"YI

{O}"

i Y5

y, , y,

l y, ,

0

(14.78)

0

, 1

12

+£1

6

0

0 0 0

y,

0

Y: J

0

6

12

6

4

-6

2

-12 -6

6

(- ui)

2

12 -6

-6

4

sin Wt

j:l~: \"'.

- 382.3 - 5870 1880.4 1292.6

6

12

6

4

-6

2

6 2

7.5

-6

6 4

0

I

0 - 120.7

I, - 261.6

which then gives

N N·m N

N·m w 3000 fadl

\'"-

121.2 sin JOOOI N

Pi

P,

Un:ts

=

P, =

201.0 sin JOaOt N· m 664.3 sin 30001 N

P, = - 322.2 sin 30001 N· m

where {o} i is the vector of nodal displacement for i beam segment The substitution of appropriate quantities into eq. (14.77) for the frrst beam segment results in

: P, i 00 . P: = in 0 0 0 P, 2 .0 0 I P, , L0 0 0

1947.9 481.4 1392.4 587.0 ....--------

(ld.77)

The displacement functions for tbe six nodal coordinates of the beam in Fig. 14.12 are given by eg. (14.74). These displacements are cenainly also the displacements of the dement nodal coordinates. The identiflcation for this example of corresponding nodal coordinates between beam segments and syste:n nod2.: coordinates is

0 0 y,

4

To cornpiete this example, we substitute the numerical valees of sec, m = i 00 kg 1m, and E! = lOs (N - m J ) and obtain

Solution: Since in this exampie damping ;s neglected and there are no external forces apphed on the beam dements except those at the nodal coor~ dinotes, eg. (14.75) reduces to {P}

3

2

Force

This simpiified approach may be justified by the mathematical model in which the inertial forces. damping forces, and external forces are assumed to be acung directly at the nodes. Therefore, it is important in the process of digitizalion :0 se!ect beam elements relatively short as to approximate the distributed propenies with discrete concentrated values at the nodes between the elements. The detennination of the element nodal forces is :llustrated :n the follOWing example.

429

The nodal element forces found in this manner for all of the four beam segments in this example are given in Table 14.2_ The results in Table 14.2 may be used to check that the dynamic conditions of equilibrium are satisfied in each beam segment. The free body diagrams of the four elements of this hearr, are sbown in Fig. 14.14 with inclusion ofnccal inertial forces. These forces are computed by multiplying the nodal mass by the corresponding nodal acceleration .

MMM~

-121.2

Fig. 14.14

664.3 1347

Dyn~!nic

1038.3 1948

1392.4 -:3a2.3

equi:ibrlum for beam segments of Example 14.6.

1880A

43C

Framed Structures Modeled as Discre!e

14.10

PROGRA~

Mu:llidegree·of~Freedom

Sys:ems

Dy:lamic Ana!ysis of 8eams

431

13-MODELING STRUCTURES

AS BEAMS 100 00

The computer program presented in this section calculates the stiffness and mass matrices for a beam and stores the coeft1cients of these macrices in a fiie, for future use. Since the sriffness and mass matrices are symmetric, only [he upper triangular porrion of [hese mat:ices needs to be stored, The progrrtm also stores in another fi!e, named by rhe user, the general information on the beam, The :nformation s[Ored in these tiles is needed by programs which perform dynamic analysis such as calculation of natural frequencies or de[ennination of [he response of the strUCtLlre subjected to exremal excitation. Example 14.7. Determine the stiffness and mass matrices for the uniform beam shown in Fig. 14.lS, The following are the properties of the beam: Length: L = 200 in Cross-sec~ion moment of inertia: 1 = 100 in ~ Modulus of elasticity: £ ~ 6.58 E6 Iblin' Mass per unit length: fit = O. I 0 Ob" sec:! lin lin)

10C.OO

o .to !)

1"0.00 iOO.M

10

fixed~ended

Solwion:

The beam is divided in four segments of equal length as shown 14,15. This division results in a total of five joints of which two are fixed, thus giving a tota1 of four tixed coordinates (the two ends of the beam are fixed for translation and for rot3lion).

lfi

Input Data and Output R.. ults

?~:H:::.05

O.OOOOi!>OO

·:>.H!i.<1!:.O·;

o. GOOC5;~GO

1.OS2EZ.O$

-1 ,'i;]j:t!,;;6

.,

~l.

}:6as.o~

1.5J;;20:.C5

S192i>G5 1.$):::(;5;.0

O,vQOC£_CC

C.;)O'.i~Ii':.ct;

C,~Qcc:::.eo

D00n~C0

(,1.6),::_05

., :.;

.O{)0~'t.:/) .il.~S't,t;..;

,

c.uocce.co c.occce.oc 5"7no:.o$

l,~n:ie~o~

0.0000=:.00

6noz~:p

v.coooe,,:n

O. Ot:l:tl:;:. 00

-;;.Jl5SE.Ct.

.0<;28"2.n

-1.~7ne:dl6

2,5):20£.01

-:\7$2Z.36

1.26H:::.OS

G,OOOGJ!:.OQ

.6120£.0,

O. COOO£.OO

.0523:;.OS

·7.73S12.0C S106:;.Ot

G. OOOOg.oo

.,

c,5nn.:;5

3.71no:.oo

O.00COE.OO

C .-0000:::.00

1. '81-Ol';.C2

€ .42S.sr:··)1 1.7)$,,,.00

7.73B1';:'00

L1t~3!i>N'

0.00000::.00

6. ... 2362-
lJaH:~CC

.7331S.00

~8.923i>£+Ol

".000::'0:.00

.15LO£.L'2

0.00002+00 0.0000::.;)0

0.0000;:;:.00 O. CjOCS.oO

7.7Hlc+00 3. 7:'~H:.:.10

-5 sa60:~Cl ().CC()()£.CO

.j,1Hil't."'C

.,

42e6:::~Ot

.

6,

~?g:;;<:>Ol

·s

.,

. nal£.GO

o.oooos.OO

ooocz.:.CQ

O.

QCCO£~CC

S2S6E.Gl

O,,1'A P!r.. E, Dll

GJ,:m:iU;l.

Example 14.8.

DA7A:

NUMa.:!!. Of JOINTS !.f.JMflER Of '$:;"0".'1 t:U:'!;{SN'l'5

Ne,,~

h~altA

l>'C.o4

Of nx£o COO)l.D1NA'l'ES

t;,,6Sl>r;OOO

v.cC()L(J$ Of 5;1..,>,5",;:::::1':::-,>

E'or the fixed beam modeled in Example 14.7 determine: (a)

[he natural frequenctes and modal shapes, (0) the response to concentrated force of WOO lb suddenly applied at the center of the beam for 0.1 sec and removed 14.16. Use time step of integration .1: = 0.01 sec. linearly as shown in F{t)

C .00

O.GO

SO .00

" .GO

,OC o co

lUG.OO

'1 5D .CC;

1ooo!):)

c---------..

0,00

~(i)

G)

ffi ~

61

fA

&,

(3)

CD

0

f-5D"-+-50"-\--50"-+-50"~

Fig. 14.15 Fixed benm mocielec for Example 14.7.

a

D.1

Fig. 14.16 Force applied to rhe beam of Examp!e f4.8.

432

Framed Structures Modeled as D!screte Multidegree-?f-Freedcm Systems

DynamiC Analysis of Beams

433

Solution: (a) During the execution of Program to r:1odeI the beam in Problem 14.7, a named "SK" has been created to store the data necess.::.ry to execute Prograa: 8 to calculate natural frequencies and corresponding modal shapes. The execution of Program 8 gives the fonowing output:

me

l':~CJ7

C .554

Output Results

c. ):s

t7""i.N

1.. 1;'<3

'H"S.Sii

().5';.4 :;

14.11

7.:n

.,-,

1) .l.?3S1

c. 0C5.:;2

C.J56Dii

~LC;;CQv

1?JSJ

.o.nne

~C.(j0:;55

LOC::;~O

C.C~2~?

n'ila:

;)::1:

-:;,00030

~?CSl

[j

,0:3::3:;

~S~S,

0.02$,,:

.OCC·JC

~J.0"';;52

c. :; .. %5

c .O7~;'7

." . ~$:;H

C, :/47$8

, nest

-0.

~5l57

,

~S:"S'

:;.06£52

-D.%7S.-

:.D2Z9:

.

,

-(

-C.et:);)o

°J;B£5

::; . .11~3S

C .04758

:: .cc:;-){)

~;:

.,

:Wf42 :;)155

0)"''''

.C~ '1

DYNAMIC ANALYSIS OF BEAMS USING COSMOS

Example 14.8. The cantilever beam shown in Fig. 14.17(a) is subjected to the time·forcing function shown in Fig. 14,17(b). applied at the free e!1d of the beam in the y direction. Use the program COSMOS to: (1) generate a finite element modei of [he beam with len two-dimensional beam elements, (2) calculate the ten lowest narural frequencies of the beam. and (3) detenni:1e the response in tenns of displacements, internal forces, moments. and stresses z.t time = 0.009 sec. Use 50 time steps at interval Lil 0.006 sec .

(b) The execution of Program 8 "Jacobi" creates a file named "EA" in

£",so;o:\d'~ '1=0..3

w;

which the eigenvalues and the eigenvectors {¢}; are stored for subsequent use by other programs such as Program 9 which uses the modal superposition rr.ethod to calculate the response of the structure to externally appUed forces. The following is the computer output for execution of Program 9:

P:::: O.!l ":>-sd-/i,;'

10m

Input Data and Output Results

Fit)

(al

Gfu"\'VI7;':'ZO:';;'~

:::l'DE..X

??::::N? ':"!KS :O::STCRY NP:t';'",:; eN;:'''- «AL VA~d;;;S ::l':rr,,:;

o~----------~----~ D 0.\

(b) o .i.e:;o

:CiJiJO.CC

iJ .20;)()

J.G)

Fig. 14.17 (a) Cam:lever bea;n for Example 14.8. {b) Loading function.

Dynamic Analysis of Beams

Framed Structures Mooc1ed as Dis0rete Multidegrec·of-F"ree:;om Systems

Soituion: COSMOS:

The anaiysis is performed using the following commands in

(I) Set view to the XY plane:

435

(10) Set options for frequency calcuLation to eXfract 10 frequencies using the Subspace Iteration Method wirh a r:1uximuf:i of 16 irerations, and run the frequency analysis; A.."1ALYSIS > fREQ /BUCK > f.._FREQ;JSNCY

DISP::'A't' > V:C:W_PAE{'l' > VIS:''';

JLFREQUENCY, 0, 0, 0" !)

V:EW, 0, ), 1, D (2) Define the XY plane", Z = 0:

ANALYSIS

0,

1

(4) Define element group ?RO~SETS

RESUU;:"S

CURVES ) CR!?CORO

CRPCORD, 1, 0, 0, 0, 10, O.

SGROUP,

f.::eq'lency Q,

10 i

1

using BEAM2D:

j

2

, 3

> EGROU? 1. BEA.M2D, 0, 0, 0, C, 0, 0, 0

l?ROPSETS

5 5 7 8

MPROP

>

1,

EX,

30E6,

DE:::.JS,

0 3

9 10

(6) Define real consranrs: PROPSETS )

(7) Generate mesh of 10 beam elements with two nodes along curve (Fig. 14.18 shows the beam modeled with 10 bea:n segments, 11 nodes): MESH:::NG > PARAM_!4.ESH > I'CCR

L

L

1, 2, 10, :

~

DK!),

P-.L,

1,

STR~~rUR~L

0,

L

;>

NNERGE,

1,

2

NODES 11.

;>

OISPL~u~TS

O.

0,

18-05,

0,

:'E-06,

R~FREQUENCY

?recr.:ency

Frequency

Period

(Kad !sec}

(cycles /sec) 1. 50770e: + 001 9.96317e+OOl

(seconds}

::.. OlOISe T 002 6 . 26005e + 002 1 "56918e + ::;83 .73528e+ 003 3 . 36'.)02e -j- 003 .66891e + 003 5 _SCO",Se -1- 003 7. 65376e + :::03 B" 10530e + 003 1 Q4S00e + 004

2. 497<13e -+- 002

,

C.OOOl,

:>

PD_.ATYPE,

2 .76178e

-r

5" 35560e

+ CO2 + 002

7 .43080e

002

8, 1547Ge + 002 L 21812e -+- 003 1.29016e-+ 003 1. 65316e + CO]

C;.22006e-002 I. ::;0370e-002 4,Q0412e-003 3,62085e-003 ;. 85720e-CC3 1.34575e-003

1.l4224e-C03 8,20938e-004 7.7509ge-004 o.C1263e-004

POST_QYN > PDj.. TYPE

2,

lQ, 50, :::, 0.005, 0,

j

>

POST~DYN

!?C~Cl1RVES)

1, 0, 0 POST_DYN

L

... , 0,

CURVES:>

0, 0,1,

0, 7

PD_CtJR'IYP

P:::'LCURDEF

1000(}, 1, 10000

CO~?I'ROL :> ACeYVE ) AC1'S2T ACTSE'r, 'Ie, 1 LOAD- Be > STRDCTUR.P.L ) FORCES F.ND, 11, FY, L :1, 1

0,

0,5. 0.25, 2

(13) Define dynamic forcing function and apply as force at node II in the Y direction:

PD_CtJRDEF,

1

5

At\ALYSIS

;'.N.!;r~YSIS

~ND

;>

NHERGE 1,

;#

PD,~cu:rI'Yp,

(9) Merge nodes: MESHING

0.

L,IS'I' > FREQLIST

>

AKALYSIS

(8) Apply constraints in all degrees of freedom at node 1. LOAD-Be

>

0,

(12) Define type of dynamic annJysis:

RCONST

RCONST, 1, 1. 1. S, O.l., 0,00833, 1, 0, 0, 0, 8, 0

M_,CR,

FREQ/BGCK

0, 0

(5) Define material properties for EGROUP 1: M?ROP,

15,

(11) List the nlltlicu! frequencies of the system:

(3) Generale line from 0, 0, 0, to 10,0, 0: GEOMETRy ,

S,

R_FREQUENCY

GEOMETR't ) GftID > PLANE:

PLANE, Z,

l,

>

FND

(14) Set options for pr:nting and plotting results:

:)

•

s

10

Fig. 14.18 Cantilever beam modeled with 10 beam segments (11 nodes).

P-l-iAL'(s:::S ) PO:S?~DYN ) PD_PRlr-;:T, 1, 0, 0, 0, J:..l"f.~I.,ySIS ) ?OS'f_DY~ > ?D_?LOT, 1, SO, 1, 0 ANft.:':iSIS ~ POST_DY~ >

OCTPUT > PD_OUTPUT

0,

2.,

OL:Tl?UT

OUTPUT

1,

50, 1 PO_PLOT

> NR2SP

I-t

t~.: t,

! 2.S U

v

Dynamic Analysis of Beams

Framed Structures Modeled as Discrete Mumdegree~of~Freedom Sys:ems

436

11, .2

i

I,

J c5.' c--I/ ,.. y '.2

1.,

-;U '\ I V \ 1/

,

i

9.S

-

I

/ '\

/

\IJ

\/

i

Ii

\L-'t . \

I

\

i

,

,

,I

,i

I

,

I G.le

9,12

GAiS

TABLE 14.3 Forces and Stresses at the Nodes of the First Three Elements of the Cantilever Beam Modeled in Fig. 14.18 for Time Step 15 (t= 0.09 sec)

i

i

&.1')$

i

,

).-/

/

, iLl •

i

437

, G.2l

,

I ! 1$,24

:

;

v;;.~-.'7~'-2·;:;';

c.7i).i.E;.. j~

C~$~V.)C)vS~C)

0,::)Cn.cc

\M$;S$)~Q.::DJ()S~G0

C.vi)OCE;'JG

'J:;"C.DO;;CE.(D

I';.C:;;O£~jr.:

:~:~~.8154ii>CS

C,7%1;:;:')~

W~/$~)~').50?OZ.C7

C.C;5S",>!)7

(~~

, j (LV \'''~''

~'J, OOOQ::>O~

{) .OCJGJS,·()O

\'t~0.Gj"C2<;')

0

(\)C~2'C!,;

H~~·,1Snt~O~

0,5';46£.05

(~;t/Stl",O."5S6S*07

G.4(j~'fZ.07

':'(.~'. %o~";.o~

CJ. 8"O~r::~()~

}~,,~C. OOCCt~OO

'I,

(i{v ,. $",)

Vt~C.oQDCS~Q(;

0.(lD0Q:;:·C~

(::~~.SH6O:~05

o.seMs.~o~

$150E-04 0

.iH5n~c~

Hs~C

.,)000,,")0 0,

C(l()C:;;~CO

iSS;

Fig. 14.19 Time-displacement response at node II. OOOO::~CO

~G.

U)"02·C'} (). COi;'J';:~~J

(Xt:S~!~').1':~':!E.;)-;-

(;.;~::':2.~~

S;;v:;.x~C.1c~n_~7

C,E)~2":7

S;t,,i,.r."~,404~S47

-.~S);.t:.c;,

(15) Periorm dynamic analysis: .::'_NA;:/iS:S ~ POSl'_JY:;; R_DYl,Pol1:C

tCDYNl\.NXC

14.12 SUMMARY

(t 6) Calculate forces and stresses at tbe nodes of the beam elements:

ANALYSIS ) STATIC R_STRESS

>-

R..-STRESS

(17) Activate Xl' plot information for Y displacement at node 11 as a function of time, and plut this function (see Fig. 14.19): D:=SPLAY

>-

ACTXYPOST,

XypLOTS

DISPLAY ) XY?LOTS XYPLO'L

:>

1, T.n1E, >

[M] {j} + [C] {y} + [E] {y}

AC'TXYPOS'I' \JY,

1:, :;'2,

In this chapter, we have formulated the dynamlc equations for beams in reference to a discrete number of nodal coordinates, These coordinates are translational and rotational displacements defined at joints hetween structural elemer::ts of the beam (beam segments). The dynamic equations for a linear system are conveniently written in matrix notation as

1,

{F(!)}

:;:. liN

XY?LOT

-.L

COl>rT'RoL > DEVICES } LAsERJET

LASER...J:::::r, 150, 0

(18) Cse an editor to list from tbe output file the forces and stresses at the nodes of the beam elements: TabJe 14.3 shows the internal forces and the stresses at the nodal coordinates for the first three beam segments calculated at time step 15 (t 0.09 sec).

where F(t) is the force vectOr and [M], aod [E] are, respectively, the mass, damping, and stiffness matrices of the structure. These matrices are assembled by the appropriate superposition (direct method) of the matrices determined for each beam segment of the stmcture. The solution of the dynamic equations {i.e., the response) of a Unear system may be found by tbe modal superposition method, This method requires the determination of the natural frequencies w.~(n::: 1, 2. 3, '"', IV) and the cones¥ ponding norma! modes \vhich are conveniently written as the columns of the nodal matrix [
"

438

I,

Framed Structures Modeled as Discrete Multldegfee-of-Fcsedom Systems

Dyna.mic Anaiysis of Beams

numie equations reduces them to a set of independent equat;ons (uncoupled equations) of the form

439

143 For {he beam in Problems 14.1 and l4,3 use slacie condensation to eliminale the [Ota~ional degrees of freedom. Find the transformation matrix nnd the reduced stiffness l!.nd mass matricc.:>. De:err:line the ::.a:uml frequenci~$ and corresponding normal modes using the reduced stiffness anc mass matrices obta)ned in Problem 14.4,

where (" is the modal damping ratio and P,,(t) :i,¢;"F,(;) is the modal force. An alternate method for determining the response of Hnear systems (also valid for nonlinear systems) is the numerical integration of the dynnmic equa(ions. Chapter 20 presents {he step-by-step linear acceleration method (with a modificarion introdttced by Wilson) which is an efficient method for solving :he dyr.amic equ<1rions. A computer program is r:lso cescribed for the dynamic analysis of beams. Tnis progrr-:m performs the task of assembling and storing in a flIe the stiffness anc mass matrices of the system, T:1.ese matrices are subsequemly used by olher programs to calculate nawoal frequer.cies Of the response of the beam to external excitation.

14.7

De(ermi::.e [he nalt;:al frequencies and corresponding normal modes using the :-educed $[~tbes:; U:1C mass matrices obtained in Problem 14.3.

J.l.8

Delermine the geometr~c s:iffness matrix for the beam of Problem 14J when ;( ~s subjected :0 a co:::;(.:::( lensile force of 10,000 Ib as shown in Fig. PI4,g, y

4

El "'" ,cf{!b~in,71 Q ..

3~

Fig. P14.8.

A uniform benm of flexural sriffncss £1 10it (lb· in 1) and length 300 in h.D.s one end fixed and the other Simply sl,pponed, Determine lhe system sriffness r.,atrix considering rhree beam segu:.ems and rhe nodal coordinaces indicatec in Fig. P14.1. y

3.86 {lbJif'l.)

~~~~~~~x zt,o,€XX) !b

PROBLEMS 14.1

$

t

5

4

I

EI'1<

14.9

Perform .:>tatic condens
14.lU

Use results from Problem 14.4 and 14.9 and determine the n:Hural frequencies a"d corresponding normal modes for the beam shown in Fig, P14,8.

14,11

Use (he results of Problems 14.5 and 14,9 and determme the natural frequencies and corresponct:-,g norma! modes for [he beam shown in Fig, P!4.8.

14.12

Deren:"'J:ne the stiffness matrix for;) beam segment in which the flexural stiffness has a !inear varimion as shown in Fig. Pi4.12.

toa /II) in..l)

3'-'"

~~~~~~_x

tOOin.--i--,oo,n.+,oo,n....,,-1

Cil======i,2EI

Fig. PI4.!.

,'1'

I-~~~-L

14.2

Ass:Jming thar the beam show;) ;n Fig, P14,1 carries a uniform weight per unic 1e:1gth q = 3.86 Ibfln, determine the system mass matrix corresponding to the lumped Class form:.tlal:Ofl.

14.3

Determine the system mass m,t(rix fOf ProbleF.l 14,2 using the consistent mass method.

14.4

For the beam in Problems 14,1 :lOO 14.2, use static conde:1sation !O eliminme the massless degree,) of :reedom. Find the transformation matrix and me: reduc-

ed sriffLes;; aLe mass m;.nnc:es,

3

Fig. P14.12 .

~

.

14.13

Determine the lumped mass m:.itrix for a beam segment in which [he mass has a linear disrriburion as shown tn Fig. P i4,13.

14.14

Determine [he consistem mass matrix for the beam segment shown in Fig, PI4,13.

440

Framed Structures Modeled as Qiscrete MUitidegtee-of-Freedom Systems

Oy!1arnk Analysis of Beams

441

figure. Eliminate t.1e fm
2mo (lb/inJ

2

14,19

Determine the natural frequencies and corresponding normal modes for the beam shown in F!g. P!4.1: (<1) condensing the three rotational nodul coordinates: (b) no condensing coordinates. (Use lhe consistent mass method.)

14.20

Derermine the response for the beam shown in fig. PI4.15. Neglect camping. (a) Do not condense coo::-dinates: (b) condense the three romtionaj coordinates. Repeat Problem l4.20 assuming 1.0% damping in all the modes.

4

f-----L----

14.21

3

14,22

Determine the steady-state response for [he beam show::; In Fig. P14.18 whe::; SUbjected to a harmonic force as shown in the figure. Do not condense coordinates, and neglect damping in the system.

14.23

Repeat Problem 14.22 assuming that the damping is prGponionai [0 stiffness of the sys:em where the constant of proponionalir), Cle = 0.2.

14.24 14.:25

Solve Problem 14.22 after condensing the three rotational coordi::ares. Repeat Problem ]':'.24 assurr.:ng 15% damping in all the modes.

14.26

Derermine the steady-state response for the beam shown in Fig. P14. i5. Do no: condense coordinates. 'and neglect damping in the system.

Fig. P14.13. 14.15

The uniform beam shown in Fig. P 14.15 is subjected (0
y

r 100 11"1.--"1t---100 in.--"1I--

5

4

Fig. P14.15. 14.16

Solve Problem 14.15 using the results obtained in hoblem 14.7 which are based on the cor:sistent mass :·orrnulauon.

14.17 Solve Problem ]4.15 using the results obtained in Probletr. 14.9 which includes the effect of the axial force in the stiffness of the system, 14.18

Detennine the steady-state response for the beam shown in Fig. P14.1S which is acted upon by a harmonic force F{f} 0;.7 5000 s;n 30t (lb) as shown in the

y

5000 sin 3Otilb}

i

/

/ 100 in. _ _I--·100 4

in.~--;

__

s Fig. P14.18.

EI '" 1()9llb-in) q >tt 3.86 Obiln,)

Dynamic Analysis of Plane Frames

15 Dynamic Analysis of Plane Frames

15.1

443

ELEMENT STIFFNESS MATRIX FOR AXIAL EFFECTS

The inciusion of axial forces in the stiffness macrix of a flexural beam segment requires the determination of the stiffness coefficients for axial tonds. To derive the sriffness matrix for an axiaHy loaded member, consider in Fig. 15.1 a beam segmem acted on by the axial forces PI and p~ producing axial displacemems o! and 8::. at the nodes of the element. For a prismatic and Gniform beam segmem of length L and cross-sectional A, it is relativeiy simple ro obtain the stiffness. :-e!ation for axiai effects ":Jy [he apptication of Hooke's law. in relation to [he beam shown in Fig. lS.l. the displacements 0 1 produced by the force PI acting ::.t node i while node 2 is maimained fixed (Ol "" 0) is given by

,

P,L

!;i~­

,

(15.1)

AE

From eq. (15.1) and [he defiol:ion of ~he stiffness coefficient ktl (force at node 1 to produce a unit displacemer:t, oJ), we obtain

k

. II

P! AE =-= Ot L

(15.2a)

The equilibrium of the beam segment acted GPon by [he force ko requires a force at the other end, namely The dynamic u:1alysis using lhe stl ffness marrix melhod for structures modeled as beams was presented in Chapter 14. This method of analysis when applied to beams re(!uires the calculatio:1 of element matrices (stiffness, mass. and damping matrices), the assemblage frorr: these matrices of the corresponding system matrices, the formation of the force vector, and the SOlution of the resultant equations of motion. Thes.e equations, as we have seen, may be solved in general by the modal superposition method or by numerical integrat~on of the differential equations of motioo. l!J (his chapter and in the following chaprers, [he dynamic analysis of structures mode~ed as fromes is presemed, \Ve begin in this chapter with the analysis of structureS modeied as plane frames and with the loads acting in the plane of the frame. The dynamic analysis of such strucwres reqllires the inclusion of the axial effec[s in the s:iffr:ess and mass matrices. It aiso requ~res a coordinate transfonnation of the nodal coordinates from eiement or local coordinates to system or global ::oordinates. Except for the consideration of axial effects and the need [0 transform these coordinates, the dynamic analysis by the sdfffless method when applied to frames is identical ra the analysis of beams as dis::ussed in Chapter 14. 442

AE L

(l5.2b)

Analogously, the ocher stiffness coefficients ure

( 15.2c)

and

AE L

(lj.2d)

Fig. 15.1 Beam elemenr with nod3.1 aXial loads Pl. Pl , and corresponding nodal displacemems 0:, 02.

Framed Structures Modeled as Discrete Multidegree-of-Freedom Systems

444

The stiffness coefficients as given by eqs. (15.2) are the elements of the stiffness matrix relating axial forces and displacements for a prismatic beam segment, that is,

gives half of the total mass of the beam segment allocated at each node. Then for a prismatic beam segment, the relation between modal axial forces and modal accelerations is given by ( 15.6)

(15.3)

The stiffness matrix corresponding to [he modal coordinates for the beam segment shown in Fig. 15.2 is obrained by combining in a single matrix the stiffness matrix for axial effects, eg. (15.3), and the stiffness matrix for flexural effects, eg. (14.20). The matrix resulting from this combination relates [he forces P; and the displacements 0; at the coordinates indicated in Fig. 15.2 as

P, P, P,

EI

p,

- L3

P5 P6

symmetric

AL'II 0 0 - AL' II 0 0

12 6L 0 -12 6L

4L2

0

AL'II

- 6L 2L'

0 0

12 - 6L 4Ll

0, 0, 0, 0,

or, in concise notation,

15.2

where m is the mass per unit of length. The combination of the flexural lumped mass coefficient and axial mass coefficients gives, in reference to the modal coordinates in Fig. 15.2, the following diagonal matrix:

P, P, Po P, P5 P,

(15.4)

Os 86

(Pi = [K] (o)

445

Dynamic Analysis of Plane Frames

(15.5)

2 0

du

P (x)

dx

AE

(15.8)

Integration with respect to x yields P (x) u=--x+C AE

(15.9)

52 = 0

P, k"~j:======KO::"'~"';S"~========~J--P, ~~--=J.-d:~ I=

+

I

Fig. 15.2 Beam element showing flexural and axial nodal forces and displacements.

(15.7)

To calculate the coefficients for the consistent mass matrix. it is necessary first to detennine the displacement functions corresponding to a unit axial displacement at one of the modal coordinates. Consider in Fig. 15.3 an axial unit displacement 01 = 1 of node 1 while the other node 2 is kept fixed so that 02 = O. If u = u (x) is the displacement at section x, the displacement at section x + dx will be u + duo It is evident then that the element dx in the new position has changed in length by an amount du, and thus, the strain is duldx. Since from Hooke's law, the ratio of stress to strain is equal to the modulus of elasticity E, we can write

ELEMENT MASS MATRIX FOR AXIAL EFFECTS

The detennination of mass influence coefficients for axial effects of a beam element may be carried out by any of two methods indicated previously for the flexural effects: (1) the lumped mass method and (2) the consistent mass method. In the lumped mass method, the mass allocation to the nodes of the beam element is found from static considerations which for a unifonn beam

0

filL

81 82 8, 8, 85 86

du

b\§\"'J f----+-dx

+ du

Fig. 15.3 Displacement at node 1 (8 1 = 1) of a beam element.

"k"

446

Dynamic Analysis of Plane Frames

Framed Slructures Modeled as Discrete Mu!t:degrse-of-Freedom Systems

in which C is a constar.t of integration. Introducing the boundary cor.ditions, = 0 and H = 0 ar ;c ;;;:;: L. we obtain [he dispiacement funcrion U I (x) corresponding to .'! unit displacement OJ as u = ~ a: x

u, (x) = l -

x

L

(! 5.LO)

as shown in Fig. 15.4. Hence the inlcmal work for element dx!s obtained fro!":l eqs. (tS_l3) and (l5. :4) as dW,

me::1t

HZ

(X) cor.esponding to u unir displace-

Finc:!ly, equating coefficient

B1 = I is

AEut (x)u{ {x.)dx

and for the beam segment of length L

W,= Analogously, the displacement funCtion

447

WI~=

f' AE!i((x),,;Gr)dx

(15.l5)

J,

W! from eqs, (15.12) ar.d (15.15) glves (he sriffoess

I) k"

= CAE,,; (x),,; (x)dx

(15.16)

~D

The application of the principle of virtual work results if: a general express~ ion for ~he calculation of [he stiffness coefficients. For example, consider the beam in Fig. i5.3, which is in equilibrium with [he forces PI k 1j ar.d P2 = k;n at its two ends, Assume that a virtual Jispiacement S: 1 takes place. Then, according to the principle of virtual work, during this virtual displacement, the work of the ex'[ernul anc imemal forces are equal. The exterr.al force k21

In general, [he stiffness coefflcie:1l k'j foc axial effects may be obto'ined from k" =

r'

AE,,: (x),,; (x)d"

(lS.l?)

'C

Using eq. (1).17), the reader may check the results obtained in eq. (15.3) for

perfonns [he work

a unlfolm beam. Howeve:, eq. (lS.J7) could as welt be used for r:onuniform beams 1n which AE would in general be a function of x. Ir. practice, the same dis;:>1acement U (x)
or

(l5J7) for a nonumform member. The displncemem y(x,t) at any section x of a beam element c:.te (0 dynamic modal displacements, 8 1 (r) and D-z ([), is obtained by superpos:tioo. Hence

j

05.12)

slnce 8z ;;:;; I. The internal force P (x) a( any sec[ion x is obtained from eq. (i5.8) as P (x)

AEu! (x)

([5.l3)

( lS.l8) in which '" (x) and ",(x) are given by eqs. (l5.10) and (IS.ll).

Now consider the beam of F:g. 15.5 while ur:.dergoing a unit acceleration, 51 (t) ;;:: 1 which by eq. (15. J 8) ;-esults in an acceleration at x given by iii (x,r);;:;;

in which !<: (x) = dUt Idx.. The relative displacement of elemem dx during [his virtual displacement is duz dx

du..,=--dx -

HI

(X)SI (I)

or

(15.l4) ~ll{t)"1

============:;;Ji----+

i=:;=·:::·

m\l ...............

CD

®i

p

.. ""

Mll

,,- '-i I-

= Fig. 15.4 Displacement a[ node 2 (~ '.--~ l) of a beam element S:.lbjected to axial dis~ placement at node 1 (0 1 = I).

m n---+L.. __ _

Fig. 15.5 Displacemem at .l.ode 2 (0; erarion llt node (81 (:) = 1].

I) of a beam eiemefU undergoing axial

aC(;el~

448

Dynamic Analysis of Plane Frames

Frame': Structures Modeled as Discrete Multidegree-of-Freedom Systems

since 81 (t) = 1. The inertial force per unit length along t::'e beam resulting f:::-om this unit acceleration is

and m I 2=m:!=

(15.19) where tn-ex) is the mass per unit length along the beam, Now, to determine the mass coefficlent m!h we give to the beam shown in Fig. 15.5 a virtuai disp!acemeot ~ = l. The only external force doing work during this: virtual displacement is (he reaction m2> This work is then

or (15.20)

\L,

(15.25)

6

symmetric

fl40

P,

1 p, r=

P, , P, ,

niL

420

0 0

]56

70

0

0

54

0

( ~,

8;: 8)

22L 140

0

0

13L

!3L - 3L

2

156

0 -22L 4L2..;

3, 3,

(15.27)

Ii,

or, in condensed notation,

J, m "C

(x)u; (x)u, (x) d.x:

Wi =

mL

'-jdx=-

Finally., combining the mass matrix eq. (14.34) for flexural effects with r.q. 05.26) for the axial effects, we obtain the consistent mass matrix for a uniform beam element in reference Lo the modaJ coordinates (t) shown in Fig. 15.2 as

aw, = /' (x)u, (x)

Hence the totai internal work is

x\/x\

(15.26)

P,

.tn (X)LlI (X)U2 (x)

Jo

m 1-

In matrix form. the axial inertial force relationship for a unifoml beam may

P,

SW,

"L-

be written as

since 02 L The internal work pe:- unit length along the beam pei'fonned by t;,e inertial force Ii du:ing this virtual displacement 1S

or. from eq. ([5.19),

449

(15.21)

{PI = [M,] {5}

Finally, equating eqs. (l5.20) and (15.21) yields m"

=

in which [lWcJ is the consistent mass matrix for a beam element

Im "£

(x)", (x)", (x) dx

(15.22)

"'

or, in general,

it

In,!

15.3 COORDINATE TRANSFORMATION fit (x)u; (x)uj(x) dx

(15.23)

The application of eq, (15.23) to the SPecial case of a unifonn beam :-esults in

,2 _

_ (L _ { X ihL mll-TJ dx - 3

mll-:

j"J

\

"

Similarly, m::1;=

lfiL

( 15,24)

The stif:;Jess matrl;¥; for the beam element in eq, (l5.'~) as \veIl as the mass matrix in eq. (15.27) are in reference to nodal coordinates defined by coordinate axes fixed on [he beam eiement. These axes are called local or element axes while the coordinate axes for the whole st:11cture are known as global or system axes. Figure 15.6 shows a beam element wim nodal forces P j, P1",", P (, referred to the local coordinate axes x, y. Z, and p;. Pl, .... P(, re:·erred to giobal coordinate set of axes X, Y. Z. T..'1e objective is to transfonn t;,e element matrices (stj:fness, mass, etc.) from the reference of local coordinate axes to the global coordinate axes. This transfonnation is required in order that the matrices for ail the elements refer to the same set of coordinates; hence, the

450

Ftamed Structures Modeled as Discrete M;;!tidegree-of·Freedom Systems

Dy!1amlc Analysis of Plane Frames

451

Equations (15.28) and (lS.30) may conveniently be arranged in matrix form as ( Pi

I

y

P,

I~:

y

r

P, P,

r cos e : -sin

e

0

0 0 0

0 cos 8 0 sin

f)

0 0

0

0 0

0

0

cos

0

0

PJ

0

p~

0

- sin

0

0

0

cos fJ

0

{PJ

= [1] (10,)

sin

f)

(J

P,

e

0

0 0 0

,, 1 J\ .

L })6

0

(IS.3I)

)

or in condensed notmion

r.Z

Fig. 15.6 Beam element showing nodal fOr(:e~ Pi in loc.)! (;.r, y, z) .)nd nodal forces ia globaI coordino:e axes (X. Y, 2;.

t,;

marrices become compatible for assemblage imo the system ma[rices for the

structure. We beglo by expressing [he forces (P I' P?. PJ ) in termS of the forces (Ph Pz> i\). Since (hese (Wo sets of for:::es are equivalent. we obtain from Fig. 15.6 the fol1owing relationships:

in which {P} and {.o} nre, respectively, [he vectOrs of (he element nodal forces in local and global coordinates and [1] is the transformation matrix given by [he square matrix in eq (15.31). Repea~lng {he same procedure, we obtain the relarion between nodal dis~ pla::.:emems (Oh Oh ".,66) in local ::.:oordinates and the componems of the nodal displacements in global coordin::ues (51) 82, ... , 86), namely

sin 8

cos p,

=

p cos 8 +

o o o o

sin 0

(15.28) The Erst two equations of eq. {[5.28) may be wrinen in matrix omarion as

(15.32)

e

0

o

0

o

o o o

o 0 0 0

e

sir.

e

- 51_""1 {j

cos

e

cos

o

o

u

o o~-1 ~,

8,

l

(1533)

JI

8~

or (15.34)

(15.29)

the substirution of (P) from eq. (15.32) and (0) from eq. (15.34) into the snffness equation referred to ioeal axes iP} = [Kj {o} results in

"OW,

Analogously, we obtain for rhe forces on the other node [he relationships.

or

Ps=

-P

4

sin 8+

0')

cos I) ( 15.30)

= [Tj-' [K] [1] {S}

(15.35)

where [7]-! is the inverse of matrix {T]- However, as the reader may verify, the

452

Framed Structures Modelcd as .)jscrete

M!Jltjdegree~:)f~Freedom

Systems

')ynamic Analysis of Plane Frames

transformation matrix [1] in eq. (15.31) is an orthogonal matrix, that is, [11-' ~ [1]7. Hence

453

2

3

(15.36)

or. in a more convenient notation,

6

05_37)

'"

in which

~

A -.: 6in.2 '''''1001n.4 m'" 4,20 Ib secz/ln.J E~ 10 1 Ib/ln. 2

( 1538) Fig. 15.1 Plane frame for Example i5.:.

is the stiffness m2.trix for a beur:l segment in reference to the global system of coordinates. Repeating the procedure of transfonnation as applied to the stiffness matrix for the lumped mass, eq. 05.7), or consistent mass matrix, eq. (15.27), we obtain )0 a similar manner

The transfonnarion matrix for element

[T.J

~'

Solution: The stiffness matrix for element by eq. (15.4) is

• [K,1

~

[K,l

~

1000

0

h

Symmetric

l

12 600

40.000

0

0

12

-600

600

20,000

600 0 0

12

-600

40,QOOJ

1

0

0

a:ld for element

&.

with

0°

e= if IS

0

0

0

0

0

0 0

0 0

0

0 0 0 0 0 0

2

L

0

-I

0

0

01

0

0

,/2

the identitv , matrix . 'T,] = [I]

The mass matrix in local coordinates for eitber of the two elements of this frame from eq. (15.27) is

140

& or ,£ in local coordinates

r 600

-I ~

(1539)

Example 15.1~ Consider in Fig. 15.7 a pJane frame having two prIsmatic beam elements and three degrees of freedom as indicated in the figure. Using the conSistent mass formulation, determine the three natural frequencies and normal modes corresponding to this discrete model of the frame.

by eq. (15.3l) with 8= 450) is

1

in which

is the r:lass matrix for a beam segment referred to global coordinates and [1] is [he matrix of the transfonnat~on given by the square matrix in eq. (15.33).

&

[Mil = [M,J

°

Symmetric 156

2200

40,000

0 0 54 c 0 -1300

1300

0 : 70

0 140

30.000

°

°

156 - 2200 40,OOOJ

The element stiffness and mass matrices in reference to the global system of coordinates are. respectively. calcu;.red by egs. (15.38) and (15.39) For el-

454

Dynamic Analysis 01 Plane Frames

Framed Structures Modeled as Discrete Mullidegree-of-Freedom Systems

The system stiffness and mass matrices are assembled by the direct method. As was mentioned before, j( is expedient for hand calculation of these matrices [0 indicate [he corresponding system nodal coordinates ac the top and right of rhe e;c:rnc:m matrices. We thus obtain the system stiffness matrix as

emenr 11::. the stiffness matrix is

4

r

4

4

2

3

Symmet:ic

0306 0,294

0306

- 0,424

0,424

1I :

40,000

_ - 0.424

0.424

f0906 0,294

0306 0,424 40000J

20000 0,424

~

and

~he

system

:TIaSS

matrix

148

For elemen'

~

8

148

1556

1556

40,000

62 8

8 62

- 919

148

919

8

148

919 -919 -30,000 1556

1556

l288 -

-8

]556

8 304

644

1556 644 80,000.1 The natural frequencies are found as ~he roots of (he characteristic equation

4

3

2

1

'[K] -

2 ,~O.O{)O _

3

4

0

which, upon substituting the values given for this example, yields

4

Symmerric

0,600

w' [MJ'

·906 - 0,288w'

294 + 0,008",'

424 - L556w~1

10' '294 + O,008,,}

318 - 0.304",'

176 - o.644w~1

:424 - 1.556u/

176 - 0,644w'

80,000 -

w; = 976.6,

~ = 4211.6

0,012

10'

~ 0600 °- °

iii

40,000

°

-0,600

0,600

ar.d

o

0,012

°

0,600

",'; 6385,

w, ~ 25,26 radlsec,

1140 U;1,J ~·I ~

3

40,000

, 70

o

o

54

1300

0

0 -[300

30,000

0

~

4

C 140

156 2200 40,000

W, ~

31.24 rad/sec,

and

w,; 64,90 rodlsec

Or

2

156

2200

SOu.ll

and the nmural freqt.:..encies are

4 4 Symmetric

3

2

a

The roots then are

0,012 - 0,600 0,600 20,000

o

J

£:\$

[,\1] ~

Lb.,

I [1(,]

3

2 Symmetric

4

4

1

,0424 0.176 80,000

and :he mass matrix 4

424 0. 0,;76

[k] ~ IO'! 0,294 0,}18

1

0,424 0.306

- 0,306 - 0294 'i - 0294 - 0,306 - OA24 0,294

455

4

I,

~

4,02 cps, J,

~

4.97 cps,

and

I, =

10.33 cps

The normal modes are given as [he nontrivial solution of the eigenp:"oblem ([K] - ",' [M]) (al = (0)

456

:J'jnar.1ic A'1alysls of Plane Frames

Frarr.ed Structures Modeled as Discrete Multideg.ree·of~F(eedom Sysler.1S

Subst.ituting w~ = 638"5 and setting all = LO. we obtain the first mode shape as

457

The modal eqcations have the form of (a)

l.00}

D8

where

o P,=

(b)

which is normalized with the factor in this example, the nodal applied forces are

F;

F, = 0,

100.000 lb,

=

F; = 0

The normalized eigenvector is then Vie thus obtain the modal equations as

¢H1 {¢,}=

't, + 638.Sz,

f
z) + 4211.6:::; = 5830

Analogously, for the other two modes, we obtain ( .;1>"

0.0583 1;

f
0.0241 ~

l¢33J

0.0218 0.00498 l'

00.0527 0.00206 0.0034l

Z,

['"] =

I - 0.0527 L

0

000206 0.00341

1

00241 J' - 0.0016

P,

(1- cos Wit)

yields =

z, =

- 0.0016

0. 0583

Wi

z,

0.0583 -Ji 0.0241

Solmion: From Example 15.1, the natural frequencies are Wi = 25.26 fad/ sec, W:: ~ 31.24 rad/sec, and W;; &:k90 fad/sec; and the modal matrix is 0.0218 0.00498

Zf=

Substitution for P; and

Example 15.2. Determine the maximum displacement at the moda~ coor~ dinaEes of the frame in Fig. is.? when a force of magnitude 100,000 tb 1S suddenly applied at nodal coordir.me 1. Neglect damping.

(c)

The solutions of these equations are of rhe form

- 0.0016!

From these vectors, we obtain the modal matrix

=

2180

i, + 976.6z, = 498

, ¢"j

[¢]

=

3.414 (1 - cos 25.268St)

cos 31.25061)

05(0 (1 -

= 1.384(1

~.

cos 64.8970t)

(d)

The nocal displacements are obtained from {y}

[
which results in

y, = 0.1577 - 0.0744 cos 25.26, - 0.00254 cos 31.2St- 0.0807 cos 64.9t 0.1455

y,

y,

=-

+ 0,1800 cos

0.000475

25.26r - 0.00105 cos 3L25t - 0,0333 cos 64.9r

+ 0 cos 25.26r

0.00174 cos 31.25, + 0.0022

COS

64.91

(e)

F:amed Slft.-ctures Modeled as Discrete Mui!idegree·of-freedom Systems

458

Dynamic Analysis or Plane frames

The maxiroum possible displacemenrs at the nodal coordinates may then be estimated as [he summation of the absolute va!ues of the coerficienL<; in [he above expressions. Hence

:"_>'SS/LEH7i.h

4.Z0Q

459

SECT. :::W.::r<.T:;>,

too.co

(; .OC

100.00

0: .00

}'.'IT.;,., =03177 in y,",,1'1'; = 03598 in

1I

(l)

15.4

PROGRAM 14-MODELING STRUCTURES AS PLANE FRAMES

Program 14 serves to determine [he stiffness and [he mass matrices for the plane frame and to store the coefficients of these matrices in a file. Since the stiffness anc masS marr.ces are symmetric matrices, only the upper triangular ponion of these matrices needs to be stored. The program also stOres in another file, named by the user, the general information on the f~ame. The informarion srored in these files is needed by programs which the user may call to ?erform dynamic analysis of the frame, such as determination of natura! frequencies or calculation of the response of the structure subject to external eXCitation, Example 15.3. Use Program 14 to determine the stiffness and mass trices for the plane frame show:! in Fig. 15.7, Soli/rion:

ma~

'LO$002.0~

2:. S4cn.cs

• ;;;0('.:.05

._ !BOGE.(:S

4.?4~H:~¢~

. SS::;;;2 .. ;)?

-7. "9?;;E.')::

S:.>SI;2.0J

<1.9999E,J
1 J400;;;-1);I

H40E:~O}

5S3~2,J:l

!f9H2,C.:.

Example 15.4. For the slruc:ure shown in 15.7 which was modeled in Example 15.3, determine: (a) naCJral frequencies and modal shapes: (b) the response to a force of magnit'Jde [00,000 Ib sudcenly applied at nodal coordinate 2, The solurion reqtdres the USB of files prepared during rhe exeCUlion of Program 14 lO model the StnIC!l/re as a plane frame followed by rhe execmion of Programs 8 and 9 ,'0 de!ennine natural frequencies and to obtain the response using che modal superposition merhod.

Input Data and Output Results Solution:

(a) Output Results ~'t.~BEF

cr

WM82R Of'

.JO:rtrS

NJ: ]

seAl~

NE.t

ELEHEN'tS

NC~G

m;,l:iJZR or \<{)iE:J COORtllNA1'2S !
JO!~

,

O.

£~lE.i)7

EW,STICITY

X~COO!l?!:;ATE

Y·COOP.o!t:;:rE:

':ON<:ttl'!'PA,20

.;onn

)o'-~5S

0.00

C _GO

o .ona

7C-.7t

70.71

c. Goe

tLOC~%

nIL II

7:1.11

c coo

1.0S:H:' 0.:124.!5

-c

OllB)

D.DS27J

G_

ilOl0~

, (JonC'

,

G;;~ .. :

·G.

:}0l,,~

Dynamic Analysis of Plane Frames

Framed Structures Modeled as Discrete Multidegn:;)e~o!wFreedoM Systems

460

461

(3) Define keypoints at the end of the two beam elements in Fig, 15.7:

(b) Input Data and Output Results

GECM:STRY :;

POI!\'1"S

> Pi:'

PT,' 0,0,0 PT, 2, 70.7, 78.7, 0 ?1', 3, :70,;,70.7, 0

(4) GC:lerate lines between keypoints: GEOHE'I'?,Y :> CURVES eRLINE, 1, 1, 2, ::::R,;'/~NE, .2, 2, 3

) C:S.LINE

(5) Define eiemen', group "1;)"';

l;!.,,·mEf: 0, O;:C;-i1?:ES 0< ,,,t;Z;C:X:

fl''''l

c; 2.TIS"NA:, FORCES s':';;:? o. r~7~!'t"'''!:C!>

tl~1\S$.R

'?1~S

O1~AV:':N"!ONAL

INo)ZX

IJsing BEAM2D:

PRO?SS~S > 3G20U? EGROtJP, 2, BEAM2:), 0, 0, 0, 0, '-', 0, 0

0.7 (lb· sec 2 lin"')

(6) Define material properties, (E = 10' psi,!5 P20PSETS MPRO.?,

M??.O?

j-

EX,

1,

lS7,

DENS,

Q. 7

(7) Define real const:nts: :).(lC

PROPS~S

:'O:;OCC.'}O

(L2:)

RCONST,

lOGOCJ.OC

ReOKS?

~

2-,

L

L

S,

6,.

lO{l,

3.

0,

0,

C,

0,

::;

(8) Generate mesh of one bez.m element along curves 1 and 2: CODE::>.

!'.A:~.C!S"L.

1'.A..'<; • Ace

(J.~J:!

7.cH

;:$C.!O

~3.;;

6.,!75

:: .001

tl,~ll!l

233.7" 10.:'9

O.

15.5

~:.AX:. '.'1::(.\:1;

.

DYNAMIC ANALYSIS OF PLANE FRAMES USING COSMOS

Example 15.5. Use program COSMOS to model the plane frame showa in Fig, 15.7 with two beam elements and calculate the fi:-st six natural fre~ quencies. Use the opdon for consistent mass matrix formulation. Solution: MOS:

T!1e following commands are implemented in the program COS~

(1) Set view to the Xl' plane: DISPLAY :; VIE';'Ct'Z,.R " VIE""; VIEW, O. C. 1, 0

(2) Defme the XY plane at z ~ 0: GEOMETRY ;. GR=~ ) PLANS i?Lflo..l'\;E, z, G, :

:f:ESHING ;.

lee2.

1,

?ARFY._MBSH > M_CR

2.

1. 2,

1. :

(9) App!y constraints in all degrees of freedom at nodes 1 and 3 and at node 2 for transiation Z and mtation X and Y: L,OAJ_BC

:>

STRUCT(J!<'Jl.L

:>

DISF=.KN'TS

:>

:JND

DND, :, AL, 0, 2, 3

L (lO) Merge and compress nodes ON;:),

2,

UZ,

0,

2,

EX RY

MESHniG > NOuSS > NMSitGE N'KERGE, 1, 4, 1, D :;001, 0, NESHING > 1';01)8S NCOM?RESS, 1, 4

~

0,

0

.0JC:.)MPR=:SS

(11) Set options for f:-equency calculation to ex.tract three frequencies using Subspace Iteration Method with a maAjnum of 16 iterations and the use of consistent mass formuiation, then run the frequency analysis: .';';~ALYSIS

> FE2Q!B:)CK

P.._FRSQ:r:SNCY, 1,

3,

S,

:>

16,

0

Al~A:"YSIS > FREQ!DUCK R_F'REQt:ENCY

P,,-FREQ
0,

C,

C,

0,

1E-5,

1..2>6,

0,

0,

Framed Structures Modeled as Discrete Mu!tidegree~of-Freedom Systems

462

Oynami,; Analysis ot Plane F:ames

(4) Defi!~e [he lype of dynamic analysis as modal superposition:

(12) List the natural frequencies in the system: RESULTS

>

p,f':A.L'!S!S )

LIST >FREQLI5T

PD_A~t'?E,

F'R2QLIST

?OST_.J'!N > PD __!..TYPE 2, 5, 10C, 0, 0.0\)5, 0,

ri<.EQU£.\lCY

:REQl:ENCY

?ER!OD

t-;UHBER

(RA:l/SEC)

(CYCLES/SEC)

(SECONGS)

1 2 3

C "2193783£+02

0.34915:42+01

0.2864:::87£+00

ANALYSIS

0 .2306635£+02

O.36?1124E+Ol

0 .4110362£+02 0 . 5588167£+02 D. 1117692£+03 D. 13527972+03

(;.6541844E+Ol

0.2723962£+00 O. lS28621E+OQ 0 . 1124.373E+00 ,u. 5621573E-Ol 0 .4644590£-01

ANAL.YSIS ),

?OST_0Y:>1

CCR'JES

!?D_CUFl:DEF,

1,

100008,

5 5

8.8893843£+01

o. 177S86H>,.02 ~

.2153(:4,312+02

£i

>

CON'.:RO~

1,

;.. .",.CTIVB

AC'!'S£'I'z 1'C,

i..:ASERJE'T

0,

0

PO_CURDEF

0.2,100000,

0.2l,

.:<.CTS£T

~

1

LOADS-BC ). STRUCTURAL E'ND,

(Fig. 15.8 shows the p!o' for the structure modeled as a plane frame wi[h tOml of four beam elements.)

3,

FX,

L

3,

:>

FCRCES ) F'ND

1

(! 6) Activate and plor the forcic.g func~;on: DISPLAY ~ XY_PLOl'S ~ ;"C1:'Y.,{PRE .:;CTXYPRE, :, 1, TIt,jE, L 12, L

Example 15.6. Use [he program COSMOS w obtain th::: displacemenr resoonse at node 3 of the frame IT.odeled in Exampte 15.5 when If is st!bjected to ; force of magnitude 100,000 lb applied suddenly for 0.2 sec at node 3 ~n [he X direction. Also determine ::he forces and s[resses on the beam elements modeling this frame.

J

DLSPL:\Y ::(Y?:::"OT I

1

CONTRe:.. ) LPtS2RJET,

DEVICES

150,

(Figure 158 Shows

Solurion: The following commands are implemented in COSMOS after the execu~ion of the commands described in Example 155:

~he

> Lfo,$2RJc.:T

0

plot

ob~a:ned

for the forcing function,)

E>osS

)7t-..

.

I

i

Sill9Gfil

15&80

• ,,,, ,, 5 . . . .

,

'r~'

-r

BIJS~"

v

,

0,25,

8.5, 0

C3) Obtain a plot of (he modeled structure shown in the screen: CONTROL > DEVICES LASBR.JET, 150, 0

0.5,

(15) Define the fo:'cing function and apply it as a force at node 3 b the X direc!:on:

FREQUE;/\,CY

4

463

I

i

!

:-

4{j(ih<)&

i

,",. 2&flli& Jg$$iJl

, ,: I

IL 15

Fig. I5.S Modeled pla:;e frame structure of Example 15.5.

,

,:,

Fig. 15.9 Lond funcrion for Example 15.6.

, :<

'.5

464

Framed Sttuc~ures Modeled as Discrete Muitidegree-oP=reedom Systems

Dynamic Analysis of Plane Frames

(20) Detennine the stresses on the beam elements and use an editor to obtain from the output file (EXlS.OUT) a print of the forces and stresses at seep #40 corresponding to time ( = 0.20 sec,

(17) Set options for printing and plotting results: .;"'N".;:'.LYSIS ,.

POS'='_DYN ,.

?D_PRINT,

:, 0,

P;)_P;:;!NT

OUTPU'T' ,.

0, 1, 1, :"00, 1

0, 0,

465

.A..:.'1AZ"YSIS

tCSTRESS ANALYSIS

lr

'?D~PLO'l',

1,

CONTKOI., ) uTILITY

O:.TI'?'J'I'" P"J~ .. 1?LO'Z'

POS't_DY:i

lCe, L

>

SYSTE.>1

:SDIT EX:.5.0UT

'J

(Table 15.1 reproduces t!1e stress

(18) Execute dynamic analysIs:

OU!put

for step # 40.)

.J..;'l£..LYSIS ,. PiJS'T'_DYN ,. R_!J'!)lAH:C:

15.5 SUMMARY

R_DYNA-I'1::L :::

(19) Ob,ain a plot of the displacement response at node 3 in 'he X and Y directions: XY_?LO~S

DIS2LAY ,.

L

-".CTXY?LOT, ACTXY?LGT,

,. ACTXYPLOT

TIME, liX, TIME, UY,

2,

3, 3,

P 1, 10, ..!.-,

0, 0,

3X 3y

:JIS!?LAY ;; Xy_p:c.,01'S } XY?I.CT X'il?LOT, XY?LOT,

TABLE 15.1

1 1

2,

The dynamic analysis of plane frames by the stiffness me~hod requires the inclusion of the axial effects in the system matrices {stiffness, mass. etc). It aiso requires a transfonnation of coordinates rn order to refer all the element matrices to the same coordinate system, so that the appropriate superposition can be applied to assemble the system matrices. Forces and Stresses at the beam elements of Example 15.6 at step "# 40 corresponding to time t=O.20 sec.

CONTROL » DE'l:tCES ~ LASERJ:S~ :U~.SF.RJE'T',

::..:; 0

I

.J

(Fig. 15.10 shows the plot for the dispJacement response at node 3 in the X and Y directions.)

'.4

i'-

~.J2

\

,u,

S 24

.!S

, ~~~~:::

,.,a

(L~fl

, -0.4

r==:-t\ / -j-

/

"'..

'\

[/

r----

.:,

~/

l~ 1 --_.

i--'

/

'\

;--/

,:,

0.!;2e~£~r;~

')'o;)

~C:)7;;;+C'\

·.4S)7<:~Q4

'/~~O

8Y;;)::~OQ

JUlC~:'Ot

?".~,:, <.7S9::~JS

-.<'759£·0£

"

i,D,

V

/

V "

;: i ~28<)e·04

~

?!-:"O.:2~9£~O"i

0,!;41S~~""

:~SfSS:

"C.()%G">cs v.3coc1>?c "2'nn;.o.: C.;J2)E··J::'

(M;:i5~i~

;

Vs,,-j

>;~22-04

·_=--H2::;~~"

""r }c&c::~~a J.cci;tl"";.;;D xs,CC03Ct';;C o.co:cs; .... ro

:Ms!Sn"J.(j00~2:_t)

~.~OCC;::~OD

\r::.~G

JCGC::~:;O

o.:);;{:c;::.:::o

x;:~

(7~~C7~/.:.,)~t

",~

'"

!(S",r,.OOQC">00 ?Gcro;:,t;!,l

:?!Ai

~~35n:;;_:~

'.~:;'S;;;~:::~

.}5~S::+?!

~r76E'~S

(Y:.j$;:)~~."t~l;1i:~C}

c.·.Hn·~",

J¢C'J2~GC

?JtCG;::.Q,'

SN'.x "~.~(:';"E:<)~

.ns.;::.:;q

S"l:1 "·.511tS-O{

5211t:.t4

" j. :. 1

, .3 (LiS

.: 4

I'L3S

,:; Q.45

T! "E

.3!!J;a"~4

C.J!!W;'>J·;

y.sz~

:'O;;CB-'-;¢

('00JS~C:)

(YsiS';; .. C-.CCOC;;
:).~cccc.)C;

'1~,,).:rcJ2~().O

r.:)C;;GF;.(lI)

X:;;~-.3~9i::;;~;JS

t~3~S""Ji

(~~~/s:;;;",¢.S1seS"'O;

_.L4{J1i:~:;4

·:ccn;'-'}:J

-)CC}2·C;;i

~.$""

i'i"r'C70;:t_~"I~C

s:~~X~"

~9SES·;;S

~:;i:-; ~-. 2~50"LJS

Fig. 15.10 Displacement response in the X and Y directions at node 3 for Examp:e 15.5.

.-------

~

·.-ii~<:-:)5

-. ~~S"'2';<,

466

Dynamic Analysis 01 Plane F:arnes

framed StructJres Modeled as Discrete Mul:idegree-ot-Freedom Systems

Use results of Problem is.3 the system.

The requjred mmrices for consideration of axial effects as well as the matrix reouired for the trilnsformatlon of coordinates are developed in this c~aptef. A

co~puter pro~ram for modeling structures as plane frames is also presented.

15.5

Th:s program is organized foilowing the p,mero of the BEAM program of {he preceding chapter.

~o

467

obtain the modal equations, Neglect domping in

Deterrntne the max.~mum response of the frame shown in Fig, PIS.; when subjected (0 ihe triangulnr impulsive [ood (Fig. PL5,5) along (he nodal cootdi~ naie 2. Use results of Problem 15.3 to ocrain rhe r:",odal equarions and use the appropriate response spectrum fO find maximum mOGtll response (Fig. 4,5)_ Neglect da:nping in {he system.

PROBLEMS F(tl !

T~e following problems are imended for hand calculation, though it is recom-

1.0'1: i

mended that whenever possible solutions should also be obtajned using Pro2ram 14 w model [he structure as a plane frame, Program 8 to determ:ne ~atural frequencies and modal sh:pes, and Program 9 to calcuiate the response using modal superposi[ion me(hod. 15.1

For the plane ffame shown in Fig. P1.5.: determine the system stiffness and mass matrices. Base the analysis on the four nodal coordinates indicated in the figure. Use consiSlem mass methoc. Fig. P1S.5. 3

2 ....,..:4'+ict__ Fltl

15.6

Derermine the !lreooY-S((lle response of the fn:tme shown in Fig. PlS.l when subjected to hnrmonic forcl':: F(I) = 10 sIn 30f(Kip) along nodal coordinate 2. Neglect damping in the system.

15.7

Repeat Problem J5.6 ass:Jming thm the damping is proponio:1ul to the stiffness of the system, (C1 = au (KJ, where (iij = 0.2.

15.8

The frame shown b Fig. PIS.S i:> acted upon by (he dynamic forces shown b ~he figure. Determine {he equivalent nodal forces corresponding to each mem~ ber of (he frame.

15.9 Assemble [he system equivalent nodal forces {FJ from eqUivalent membet ncdal forces which were ca!cu!,Hed io Problem ] 5.8. 15.10

Determine the narurflJ ftequencies and corresponding normal modes frame shown in Fig. PIS.S.

15.11

Determine [he ~esponse for [he fro me shown in Fig, P15.t l{a) wilen subjected to Ihe force FJ(r) (Fig. 15.Il(b)] aCling along nodal coordinate I. Assume S% damping in all the modes,

15.12

Determine the steady-state response of the fra:ne in Fig. P15.LO acred upon harmonic fotce F; (!) 10 cos SOf(Kip) as indicated in the tlgure. Negle-:::t damfJing in the system

15.13

Solve Problem 15.1 t usmg 19), !Xeglect damping.

15.14

Determine rhe response of [he frame shown in Fig. PIS,} whet: acted upon by the force F(f) {depicted in Fig. P1S.14 applied ar nodal coordinate 2. Assume 10% damping in all ~he modes. t,'$e modal superposition method.

Fig. PlS.I. 15.2

Use the results obtained in Problem 15.: in perfor:-ning the Srlltlc condensation to eli:ninate ~he rotational degree of freedom al the suppOrt to de~ennlne the ~ransforma[ion matrix and the reduced stiffness and mass ml:ltrices.

15.3

Determine the natural frequencies and corresponding normal modes for the reduced system in Problem 15.2.

15.4

Determine the response of the fl1lme shown in Fig. P15.1 when it is acted upon by a force F (f) 1.0 Kip sudde:lly applied at nodal coordinate 2 for 0.05 sec.

; • >

, ~" , !p

step~byM!-;[ep

fOf

the

linear accelenltion method (Program

Framed Structures Modeled as Discrete MulCdegree-ol-Freedom systems

468

16

J

Dynamic Analysis of Grids

iO f(rHKip)

50 in.

12 '" 104 iksH q"" 386 X 10- 5 Klin. t"" 100 in4

A"'10in 2

45"

,3

Fig. PIS.S. 5

2

100 Ib/ft

w "" HXlib/ft E""

I -1 "

1.1)

X loJ Ksi

A"" 8 in

2

1

t--- 20'----"1, " (,'

f-

(.,

Fig, PIS,IL

Fig. P15.14. 15.15

Find the response in Problem i5,l4 using step-by-step linear acceleration method (Program 19). Assume damping propodonal .to stiffness by a facror an = 0,01.

In Chapter 15 consideration was gi yen to the dynamic analYSIS of the plane frame when subjected to forces acting on the plane of the structure. When the phwar strc.lctural system is subjected to loads applied normally to its plane, ~he structure is referred to as a grid. This S[fUcture can also be treated as a special case of the three-dimensional frame to be presented in Chapter 17. The reason for considering the planar frame, whether loaded in its plane or norma! to its plane, as a special case, is the jnunediate reduction of unknown nodal coordinates for a beam element, hence a conside,rable reduction in 13e number 0: unknown displacements for the struclUral system. Vl~en analyzing the planar frame under action 0: loads in the plane. the possible components 0: joint displacements that had to be considered were translations in the X and Y directions and rotation about the Z axis. However. if a plane frame is loaded nomal to the plane of the structure, the components of joint displacements required ~o describe the displacements of a joint are a translation in the Z direction and rotations about the X and Y axes, Thus treating the planar grid struCture as a special case, it wiH be necessary to consider only three components of nodal displacements at each end 0: a typical grid member. 469

470

16.1

Framed Structures Modeled as Discrete Multidegree-of-Freedom Systems

LOCAL AND GLOBAL COORDINATE SYSTEMS

For a beam element of a grid, the local orthogonal axes WIll be established

s-:.lcn that rhe x detines the longitudinal centroidal axis of the member and the x-y plane wi:1 coincide with rhe plane of the structural system, which wijj be defined by rhe X~Y plane. In [his case, the l axis will define the minor principal cuis of the cross section w~lle (he y axis wiU define {he major axis of the cross section. It wiil be assumed ~hat the s!:ear center of (he cross section coincides with rhe centroid of the CiO$$ secrioo. The grid member may have either a vunnble or constant crOSS section along its length. The possible nodal displacements with respect to the local or to the global systems of coordinates are identified in Fig. 16.1. It can be seen that the linear displacements along the z direction for local axes and along the Z direction for the global sys:em are idenricnl since the tWO axes coincide. However, in general, rmational components at the modal coordinates differ from these twO coordinate systems. Hence, a mmsformarion of coordinates will be required [0 m:msfonn the element rr.nrrices from [he local w rhe global coordinates,

Dynamic Analysis of Grids

16.2 TORSIONAL EFFECTS The dynamic: a;Jalysis by [he st:ffness method for grid srructures, that js, for plane frames subjecred [0 nor.nai loads, requires the derermination of [he tor~ion~1 s[iffness and mnss coefficier.rs for a typical member of the grid. The denvatlon of these coefficier.[s IS essentially tdenrical to the derivation of the stiffness and mass coefficients for axial effee[s on a beam elemem. SimiJilr~ty ,between t~ese two derivations occurs because fhe differentlal equations for b?,rh pro.blems h
-=

P

dx

Likewise, the differential egt:a:ion tor torsional dispiacemem is

dS

T ( 16.1)

ax i,

y

\

.....--' "

6.

?

/

in which 8 is the angular displacement, T is (he tOrsional moment, G IS -:he modulus of elasticl(Y in shear, and j is the torsional constant of l:.:le cross section (polar moment of inertia for circular sections). ,As a consequence of the analogy between eqs. (15,8) and (16.1)., we can nte . the fOHowing results already obtained for axial effects. The displacement r:n~nor.s for :he ~orSlonQl effec:s are the same as the corresponding functions gJVlng _the dispt
:v

~ 5,

0,

471

0, \a)

8, (x) = 1

x\

-[I

(16.2)

'I

"

and X

'.

" 5,

0

x

5,

0, (x)

in which (he angular displacement func~ion 8 1 (x) corresponds to a unit anoular cisplacemem 8 1 = 1 at nodal coordin
It;, z.,

Ito)

Fig. 16.1 Components of nodal displ;."lcemcnts for a grid member. (a) Local coordin:ue system, (b) Global coordinare system,

(16.3)

Fig. 16.2 Nodal torsional coordinates (or a bcilm element.

472

Framed Structures Modeled as Discrete

MuJtjdegre~of-Freedom

Systems

Dynamic Analysis of Grids

mem function resulting from a unit angular displacement 8;. = 1 at nodal coordinate 2. Analogous to eq. (15.17), the stiffness influence coefficients for torsional effects may be calculated from k;;

r'

fG8:(x)e;(x)dx

(16.4)

Jo

in which 8; (x) and tt.;. (x) are the derivatives with respect to x of the displacement functions 81 (x) and (J? (x). Also analogous to eq. (15.23), the consrs!ent mass matrix coefficients for torsional effects are given by

m;;~

r'

radius of gyr.,;.tion may. jn tum, be calculated as the ratio lolA. Therefore 1 the mass polur moment of inertia per unit length 1m is given by ( 16.6)

in which 10 is the polar morne:1t of jnert~a of the cross-sectional area and A the cross-sectional area. The application of eqs. (16A) and (16.5) for a uniform beam yields the stiffness and mass matrices such that

)r"

L

,

r P, P,

I -llf~'1 - I

1jl

o,!

{

.TGL'IE!

(16.7)

E!

P,

-6L

12

0

0 0

,

P6 )

4L'

0 -- .TOL'IE!

- LJ

(

0

JGL'IE!

2L'

- 6L

0

4L'

6L

-12

0

6L

Jl!:l ' lo. ·

12)

(16.9)

,,;J

or in condensed fonn

{Pi

[KI{o}

(16.]0)

16.4 CONSISTENT MASS MATRIX FOR A GRID ELEMENT The combination of the consistent mass matrix for f1exural effects (14.34) with the consistent mass matrix for torsionaJ effects (16,8) results in the consistent mass matrix for a typicaJ member of a grid, namely

r P!

P,

1

140!o1A

Symmetric

4L'

P" p,

,=420

filL

P,

I

P, )

and

Symmetric

0

(16.5)

on in which 1m is the polar mass moment of inertia, per unit length along tbe beam etement. This moment of inertia may conveniently be expressed z.s the product of the mass in per unit Jength times the radius of gyration squared, e". The

fr,) = fG

a grid structure. In reference to [he local coordinate system indicated in FlO'. 16,I(a), the stiffness equation for z: unifonn member is then e

~

!,,8,(x)8;(x)dx

473

o 70lc iA

o o

22L

156

o

0

-3L' - DL 13L 54

140loiA

0

o

or in concise notation

(: 6.8) in which llit is given by eq, (16,6). and T;, are torsional moments at the ends of the beam segments shown in Fig. 16.2 as Pi and P2-

(16.12)

~n which [.iW,J is the mass matrix for a typical unifonn member of a grid structure,

16.5 16.3 STIFFNESS MATRIX FOR A GRID ELEMENT The torsionai stiffness r:1atrix, eq. (l6.7). is combined with the flexural stiffness matnx, eq. (14.20), to obtain the stiffness matrix for a typical member of

LUMPED MASS MATRIX FOR A GRID ELEMENT

;me lu~ped mass aUocatio!l to the nodal coordinates of a typical grid member IS ob£a!:led from static consideratiol1s_ For a '.lfliform member having a distributed mass along its lengtb, the nodal mass is simply one-half of the total

474

Framed

Struc~!Jre:s

Mode!ed as Discrete

~Jl!_liidegree~of·Freedom

Systems

Dynamic AnalYSIS of Grids

475

rotational mass 1",L. The marrix equatIon for the lumped mass matrix corres~ to [he torsional effects is Ll-Jen

p~nding

y

([6.13) y

The combination of the lumped [orsional m2.SS matrix from eq. (l6,D) wi[h [he flexural mass matrix for a typical member of a grid results in the diagonal matrix which is the lumped mass mnrrix for the grid elemem. This matrix, relating forces and accelerations at nodai coordinates, is given by the following

1

equation: rhJA)

o

Fig. 16,3 Compmreuts of the noual moments in tocnl and globrtl coordinates.

( 16.[4)

o

and for node

p~ Of

briefly

Pi cOS 8+P s sin 8

P. . sio

?j=

(16.15)

in which rM J is, in this case, the diagonal lumped mass matrix for a gnd

p(:,=

fJ+P S C05 8

Pf,

(l6.16b)

member.

The identical fomI of these equadons with those derived for the rransformarion of coordinates for nodal forces of un element of a plane frame, eqs. (15,28) and (15.30), s~ould be nored. Equations (16.16) may be wriHen in matrix

16.6

nO(a[ion as

TRANSFORMATION OF COORDINATES

The stiffness matrix, eq. (16.9), as well as the consistent and [he lumped mass matrix in eqs. (l6.11) and (16.14), respec"vely, are in reference [Q the local sysrem of coordinates, Therefore, it is tle:::essary to transform the reference of these matrices to the global system of coordinates before their assemblage in the correspondi!1g marrices for the structure. As has been indicated, the z axis for the loca~ coordinate system coincides with [he Z axis for the global system. Therefore, &..e only step left to perform is a rotarion of the coordinates in the x-y plane. The corresponding ma(rix for this transformation may be obtained by establishing the relationship between components of the moment at the nodes expressed in these two systems of coordinates. In reference [0 Fig. 16.3, these relations when written for nDde (i) are

P2 = --- P! sin 8+ P1 cos ()

P:;

0

e

0

o o

I

o

cos

o o o

0 0

- sin 8

o

0

o

cos $

(16.16a)

o o o sin f)

cos

(l6.L7)

f}

o

or in shon notation {F} = (T] (P)

P1=P 1 cos 8+P2sin 0 P,

sin if

(16.18)

in which {P} ana {P} are, respectively, the vectors of the nodal :orces of a typical grid member in iocal and global coordinates and [11 the Transforml!rion

476

Frcuned Structures Modeled as Discrete Multidegree-of·Freedom Systems

Dyr.am:c Analysis of Grids

matrix. The same transfonnation matrix [71 serves also to transform the noaal components of the displacements from a global to a local system of coo:x:linDres. In condensed notation, this :elation is given by

[1] (iii

{o}

h",,125hJ. 10 [n. 2

A

L =-60 in. 1== 100 lr..4 J"'"200 in.4 m"'"' 10 lb·sec"2/ 1n ,2 c""30x lOOps? G 12X 10 6 psi liff"" 125 lb sec z

(16.19)

vlhere {a} and {8} are. respectively, the component'i of nodal disp{acement in local and globai coordInates. The substitution of eqs. (16.18) and (16.19) in the stiffness relation eq, (16.10) yields the element stiffness mat:ix in reference to the global coordinate system, that is,

111

;;_'4 /

50001b

F3

4

"

is an orthogonal tnat:ix, it follows that

{,o}

If we define

4

[K][1]{/J}

[1]{P'

or, since

477

[kJ

(1]T[K][1]{8)

as

[KJ = [TfiK]

(16.20)

Fig. 16.4.

we obtain {16.21)

The transformation matrix for element fl]. Hence

(T,] =

e=(f' is simply th

fA with

: . e un.' matnx

Analogously, for the mass matrix, we find

{P} =

[,Wi {S}

and fo, element

in which

= [T,j'[Kj [T,]

[.1(,]

(16.22)

J.\ with

e= 90"

( 16.23)

is the transfonned mass matrix, Example 16.1« Figure [6.4 shows a grid structure in a horizontal plane consisting of two prismatic beam elements with a total of three degrees of freedom as lndicated. Detennine the namral frequencies and corresponding mode shapes. Use the consis[enr mass formulation. The stiffness matrix for elements I or 2 of the grid in local coordinates by eq. (16.9) is

r [K,J = [K,J = 10'

I L

40

0

0

-40

0 0 -40

200

-5

-5

[K,J

[T,]

=

[-1 i

0 L 0

I 0 0 0 0

0

0

0 I 0 0 0

0 0 0

0 -I 0

0

0 0 I

11

0

O.

0

lj

so that

[K,l = fT:1T[K:]

[T,]

0

0 100

5

200

0.167

0

-s

- 0.167

0

0

40

0

0

0

100

-5

0

200

5

0

5

- 0.167

0

5

0.167

0 5 100 0

-40

3 5 0 0.167 5 0

-5

0

-0167

0

= iOti

2 0 40 0 0

4 100 0

-5 200 0

-5

4 0

4

-sl

-40 0 0.167 0 5 40 o 0 0.167

I

3 4

T 4

4

478

Framed Slructu~es Modeled as Discrete Multidegree-of·F',sedcm Systems

The. system matrix [KsJ assemhled from

l c

[KJ

= 10'

fK t ]

5

[K1 J is

0

240 0

and

Dynamic Ana!ys!s of Grids

which gives the eigenvalues {squ::tres of the nl(Uml frequencIes}

w; = 396.35,

5

-5 240 5 0.333

[M,]

= [M,J =

'

1

l

UJI

0

0

1250

0

oj

20,570

:886

0

- 15,430

11 i4;

0

1886

223

0

1114

1250

0

°

I!

0

0

- 15,430

1114

2500 0

0

1114

77

0

0 20,570

1886

=

19.91 rad/sec,

! 0,402,

and

L

--1886j

213

w" =:

[a J

101.99 rad/sec,

LOOO

1.000

LOOO

1.000

1.000 i

0

7,765

54,285

!S4A9 rad!sec

- 1.0001

J

214.81

,{a,}' [/,1,] {tl,}

= (I]

,f {a,}T[M'] {all = 120.20

and analogously

The no!"maJized eigenvectors ale arranged :n columns of the modal matrix,

2

3

4

0

- 1886

15.430

0

0 - 1886

2500

0

0

1250

0

223

1114

0

15,430

0 1250

!Il4

20,570

0

0

77

20,570

0 lli4

lM,]

Wj

i{a,}T[Mj{:;'J ~ 97475

since

[M,J =

and

The eigenVectors are conveniently no:malized by divldir.g [he columns of the :nodal matrix, respectively, by [he factors

[M,)=[M,J

[T,)

= 23,866

and the eige:wecw;:$ (modal matrix)

7~ I

Vole then calculate uSing eq. (16.23)

From

~

[hen

Anaiogously, for the mass, we have from eq. (l6.11) ~2500

479

4

0 2500

0

1886

0

so that

4

1114l

Q, 2

"j

188~

223

[
3

I

l

4

4 4

,,I

Example 16.2. when subjected

[0

0.1026

0,4655

0.1026

0.4655

0.8320

5.5691

0

64603

- 083201

Detemline rhe response of the gJld shown in Fig. 16.4 a suddenly applied force F) = 50001";) as indic8.(ed in the

figure.

and (lYf.l ] we assemble the system mass matrix and obtain

[M,]

=

23,070

o

0

23,070

- 1886

1886

- 1886

446

[

:886]

The natural frequencies and mode shapes are obtained hom the solution of [he eigenproblem ([K,]

w' [M,i! {a} = {OJ

SOiUlio,,: The r:aturaI frequencies and modal shapes for this Structure were calculated ie Example 16,1. The modal equation is given i!1 genera! as

z" + w~z" = P" where

q,,,,F,

480

Framed Str.Jctures Mcdelec as :)iscrete Multidegree~of·Freedorr: Systems

and Fi the external forces at the nodal coordinates which for this example are F, = F, 0 and F3 ~ 5000 lb. Hence. we obtain

z, + 396.352, Zl ~ 23,86623

27846

~

Dynam'c Analysis of G'ids

analyslS, such as determination of natural frequencies or caIcu1ations of the response of the structure subjected to external excitation.

Example 16.3. For the grid frame shown in Fig. 16.4 and analyzed in the previous examples, (a) use Program 15 to mode! this structure, (b) use Program 8 to calculate :he natura: frequencies and mode s:,apes, and (c) use Program 9 to determine the response to a constant force of 5000 lb suddenly applied for 0,1 sec as indkated i
323.01

The solu71on of these eqt.:.2.(ions for zero initial conditions is

So/ulion,' 27846 " = - - - ( 1 - cos 19.911) 396.35

323.01

9

=,~ 23.866 (I-cos 154.4 I)

The displacements at (he nodal coordinates are calculated from

{y) - 0.1026

0.4655

0.1026

0,4655

5.5691

o

~

[
- 0 8320J 07026(1- cos 19.911) 08320 o 64603 [00135(1 cos 154,49t)"

Problem Data: Modulus of elasticity:

E = 30 X 10' psi

Modulus of rigidIty:

G ~ 12 X ! 0' psi

Distributed mass:

ih

Cross~sect;o:1al polar moment of :nertia:

I, = 125 'n'

Cross-sectional torsional COnstant:

1=200 in-l

Cross-sectional moment of inertia:

{~

Cross-secuonai area:

A = 10 in'

= 10 lb· sec' !inlin

100 in'

(a) Input Data and Output Resu]ts: Modeling the Structure

and finally y, = 10 -3 (

.-

0.8332 + 0.7209 cos 19.91t + 0.1123 cos 154.49t) rad

y,= 10-'( -0.8332-0.7209 cos

19.91t

0.1123 cos IS4A9t) rad

y, = 10 -,(40 - 39.13 cos 19.911- 0.87 cos 154.491) in

16.7

481

PROGRAM 15-MODELING STRUCTURES AS GRID FRAMES

Program 15 calculates the stiffness and mass matrices for a grid frame and stores the coefficients of these matrices in a file. Since the stiffness and the mass m2.~rices are symmetric, only the upper triangular portion of these matrices :leeds to be stored, The program also stores in another file, named by the user~ the general information 0:1 the grid frame. The information stored in these files is needed by programs that rhe user may call to perform 2. dynamic

",l/:~33ll:

OF ,;O:":'S

NL!"~3:;:R

OF 33-'\.11 CC-:;l\DIV,,"SS

Nt111SER ;)7 FlX2;:J CQORO!NA7E$

t:,;>.snc:::'Y'{

~!?O:J;'U$

01"

:«)t:{)t,t.!'s

Cl' "W10,>"(

8 " 1;;: .. C' GR "

~.

2E:-r,7

>:':;C'O:i<;)IK::'7S

,

J.Cj !>J

,co

3.00 0.::0

:;,00

.cc

0,00

6~

S"lUt

l'IRS:r v0;:~"7

,

stttm~

,

:0;;-:';'

,

0, ·)c

!-'~".SSi

5SC':":-OX

i'iSC7rCN

7J"S!Ce:;L

L:::t.~,r;

:;"r;l',TZA

Ai\]';",

COt':S?"''''':"

:0. CCj

::;:; ,;;;:

::.;'c ,.){o

10. c:::

:r:;: .YO

: c .CO :0 J»

ZC;;,00

Dynamic Analysis cf Grids

Framed Structures Modeled as Discrele Mullidegree-o(-Freedom Systems

482

:l!S?:".

~-AX_

o.c}Crif

o

0.0)09

COil:

o.c%~oc

~Oi)o!!:.oa

O,JCC:;£~O(l

S, {JJ0Ge.C6

:)0002.0

~:1!1
-s. COOCE~C6

5 COOOE.06

-S.00002.%

;1-1lS,05

,

G

. .;CC"

" "

sn~

);):,J?

Tne results given by (he compurer :;ompare very closely to those obtair::ed in Examples 16.1 and 16.2 by hand calculations. For example, the displace" men~ given by [De computer (not printed) ror nodal coordinate 3, a rime 0.1 sec, is

"'SYSl"S:<: S':'tffN;:SS ;<..;\'';'l!.!0;''

,

,~"X

."AX. V=='-C.'C.

0.001.1;1

463

?l(r=0.1)=0.0568 in while the hand cakulation from Example 16,2 is

0.00;)(,,;;'00 Ci)CCS·CO

J.)OIlE.04

.~asn;.o}

-1..seS7£+i}j

y, (r = 0.1)

16_8

2~.

) .li3

S8,

stCi2NV€CTORS B~ ROO1S

wO,0010;'

0.0-n01

C.OO~S6

Q.OOMii

DYNAMIC ANALYSIS OF GRIDS USING COSMOS

Example 16.4. Use t:,e p.ogram COSMOS to determine the first SlX natural frequencies for the grid frame shown in Fig. 16.4. Model the strucwre using BE;\yi3D with a mesh of six elements.

0.0$5;;9~O"G::J)OC (LC6~6C

-c.eeen

(c) Input Data and Output Results: Response by Modal Superposition

Solution: COSMOS:

The analysis is ;:::erformed using the foUowing commands in

(I) Set view to the XY plane: DISP:';'.Y VIEld, 0,

:>

VIE1C!?AR " 0, 1,

°

tJUMS£R Cf DECER:$ Of f'R€SOOH Jiv'HBER Of S:>:1:;;;l't!;.J1\L

0.03913 cos 1.991 - 0.00087 cos 15.45

0.0568 in

(b) Output Results: Natural Frequencies and Modes

f.lC::Nl.lAU)2S,

= 0.04 -

VIE'I~

(1) Define the XY plane at Z = 0:

fC~C8S

1'll'!E !iT;;;? Gr <:!lTEGM710N

GEO~E'TRY

PL;'.N2, roSlCE :, -:::OO:U:. ; ht1:t:I!$

;>-c:o:n:::

,

:5 ;l;f?!..);£D, H0K. Of -"OUIT:> OEPlN1W: '!'iii: ?'ORC£'

:ClY,;::

cxcn,;'!'tON

0.0:) 0_10

50:10.;)0 5000.';::;

Z,

" GRIJ > v, 1

2LA..~E

(3) Gene:ate keypoints at tr_e three joints of the grid frame and a fourth keypcinr at the principal plar.e of the c~oss section of the beam elements: GEON2TRY :> POINTS Fr, 1, 0, 0, 0 ?T, 2, 60, 0, 0 P'l', 3, 0, 60, 0 PT', 4, 0, 0, 10

> ?'1'

464

Dyramlc Analysis of Grlds

Framed Structures Modeled as Discrete Multidegree-of-Freedom Systems

z

(d) Generate lines between keypoints l., 2 and 1. 3:

GEOMETRY )- CURVES

CRL::>rE

f

)

CRL:;:::H::, 1, 1, 2 ::~RLINE, 2, 1, 3

(5) Defme etement group

using the element BEA1\'13D:

)- EGROU?

PRCPS~~S

EGROU?,

485

B~f!;113D,

l,

0,

0,

0,

G,

0,

0,

0

(6) Define material properties for group I; PROPSET$ :- !{PROP

MPROP,

::, ,SX,

3086, GXY,

12.E6,

DEI,,'S, 1.0

, 4:

(7) Deft"e real constants;

IJ.

PROPSETS )- RCONST RCONS?,

1,

10,

1.,

10,

100,

leO,

3.36,

},36,

5 .~

0,0, zOOrc', 0

(8) Generate a mesh of six beam elements with two nodes along curves and 2: NESHING )-

PARk"L~ESH

peeR, 1, 2 ( 1

r

:3

6,

t

;. l!__CR

1.,

Fig. 16.5 Mathematical model for the grid frame of Example J6.4.

~

(9) Merge nodes and compress nodes: XBSnING )- NODES, I"1~EHGE,

1, 80,

MESfL!NG

:>

~~ERGE

1, 0.C001,

(12) Run frequency analysis: 0,

0,

:J

ANALYSIS NO;)ES ,. r-;COH?RESS

KCOMPRESS, 1,

DND, 14, .;".:.."

0,

7,

1

0, :4., 1

DNJ, 1, UX, Q, 5. 1, UY, RZ uN!),

9,

UX,

:;,

14,

1,

UY,

QZ

(11) Set options for frequency calculation to extract SIX natura! frequencies using the Subspace Iteration Metbod and lumped mass formulation: ANAL'iS!S ,. F~EQ/BUC?( ,.,M,~fR£QUENCY ,'\._FREQU:::.}ZCY, 3, S, :'5, 0, 0, 0, 0,

(;,

0,

G

FREQ/BUC::: ;. :ECfRSQUE:NCY

15

(10) Apply constrainl' in all degrees of freedom at nodes 7 and 14, and transiations along the axes X, Y and rotation about the Z axis for all other frame of nodes of the beam elements, (Fig, 16.5 shows the model for the Example 16.4): DND, 7, .;:.."

>

?_?::.EQGE!-JCy'

0,

H>Q05,

lS-0:Jo,

(13) List three natural frequencies:

FREQUENCY

?:REQUE~CY

PERIOD (SECONDS)

0 1963908E+C2

(C'fC:::":SS/SECi O.312S555E+Cl

0.3199328£+8C

O.7779C73E+D2

O.123S078~+02

I}.

O.108592SE+D3

C.17:283092+02

C.5786004E-Ol

Kt:r1BER

FREQUENCY [RA.D/SEC}

1 2 3

S077035£-01

It should be obserled that only the first natural frequency calculated by COSIV10S using stx elements per beam provides a value dose to that determined in Exampie 16.3 uSlng a single element per beam. Example 16.5. Use the program COSMOS to obtain the response of me grid frame of Example 16.4 subjected to a suddenly applied force F = 5000 lb z:s shown in 16.4.

486

.,

Solution.' The following commands are implemented in COSMOS following the commands to solve Exarnple 16.4; ([4) Define the type of dynamic analys~s: }

pcs'r._.oy~

:;

PD_A_TYPE,

2,

6, 50,

C,

0.5,

,,

(15) Defi;;e the dynamic function and appiy it as a force at node

>ll,~15

j

if. the

Z direclion:

I

-&.C:6

I

,:,

}:...NALySIS ). POS'I_DYN > CURVES ;} PD_GJfWEF CURDEF', 1, l, 0, 5:)00, C.2, 5000

COKTROL > ACTIVE AC'l'SE'I', TC,. 1

LOAD_Be

i

,'L!5

iL2S

,

/.

.

.:,• " f :,r

!

.:i

S'1'RUCTURAL , FORCES :; FND

>

1, FZ,

1,

1,

1

(16) Sel options for prir.[~ng and p!auing reslilts: ANALYS!S )

}l-.NA:::'YS!S )

Q,

0,

a,

0,

1, I, SQ, 1

?OST_DYN ~ OUfPUT

PD_PLO~,

1, SO, 1,

ANALYSIS

:>

?D_~RESP,

j-

PD_PLO'!'

a :>

1,

XYPLOTS

L

0,

>

XYRA.;~GE

0,5,

-0.08,

0.08

(21) Obtain a print of the displacement piOL (Fig. 16.6 shows rhe displace-

NRESP

K_DYNA~lC

R~DYNAMrC

(18) Calculate forces and stresses

:>

ZR.A.NGS,

ment response at node l in [he Z direction.)

P05T_DYl'J > OUTPU':' 1, 1, a

POS~~DYN ~

D1S:'LAY

REPRINT

(17) Perform dynamic anaI)isis: :>

(20) Set range between 0 and 0.5 sec for (he time axis and between - 0,08 ar:d 0.08 for the UZ axis:

POS'='_DYN ;, OUTPlJCi' > PD_PRINT'

PD_PRnn, 1,

?~ALYSIS

, .,

i

\

I

Fig. 16.6 Dispb,cemem response in [he z-direction foc node I for the grid frame of Example 16.5.

AC!'SE'I

>

I

\

"rJ .el:?

-,

'"

7i-

\

-/9.(148

CUR':'YP, 1, 0, 0,

/

I

\

-G.GG~

FNC,

!

\

/ /

• /'

A

0.25, 2.

\

/

e,$lZ

?D_,:',TYP£

0, 0.01,

"-

/'

~"$64

:L&48

p,~ALYS1S

487

Dynamic Analysis of Grids

Framed Struclw:es Modeled as Discrete Muaide:gree-o/·Preedom Systems

at

the nodes of the beam element

CONTROL:> DEVICES;. LASERJET l

LASERJE>r,

50" :::

(22) Use an editor to list from [he ompUl me, EX16.0lJr, the forces and stresses at the nodes of the fi::-s[ three beam elemems at [he time for srep No. 15 (r 0.15 sec). Table 16,1 reproduces the output forces and stresses for the first three beam eieme:1ts of (he grid frame for Example 16.5.

ANALYSIS :;. STATIC ) R_s'rRESS ~STRESS

(19) Activate XY plOt :nformatloo for Z displacement at node 1 as a functiO:1 of rime, and plot (his function:

16.9 SUMMARY This chapte:- has presemed the dynamic analysis of pJane frames (grids) supporting loads app:ied normally [Q lts plane. The dynamic aoalys:s of grids

DlS?GAY > XYPLO~S > ACTXYPOST ACTXYP8ST, 1, TIME, ~Z, 1, 12, 1, 0, iN DISPLAY XYPLOT5 ~ X:PLOT XY?LOT, 1

, The

p!ogr
the s.ystem.

386CO!'\F.EXE

$hOuld be :::xec\.itcC J.nd configclred Ie tht available primer in

488

Framed Structures Modeled as Oiscrete Multidegree,of~Freedorn Systems

TABLE 16.1

Dynamic Anaiysis of Grids

489

Forces and Stresses at Step 15 ((=0.15 sec) for Beam EI· ements 1, 2 and 3 of the Modeled Grid Frame for Example 16.5 S":'?ZSSRS ~;o0::::

'is,,-C. H ?$r~N

:;:";5£+:;'';

~:$~Q. CC()~S~iJ:

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Y$,,::;.:G00E>OC ?OCOO:=:'OG Yt,:,2Y73S.05 C,23;HE.~·;5

L:::n:::.J4

as(s·o:;

;y":&:),,J.i::>QCf:
>::<;~S<:

fig. PI6.l . . GJQn:~()O

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::r.61r.9z>OS

'Js,c. c:)Sr~c,;

~99SS·N

':5~:

J.CJC~E.cn

)~
;; .\lClO"S··)C

}'.~ ~-

. OC()C,>OO .2:>2,;0;. OS

r~~t

S't,,)

.:322£.,.!)S

~:),C0002.()G

CYDCOS'C:,

:Ys!Ss;,,().:lOC~E:.Oti

{?/A)

(: {l{lO:·;::·C}

:1:tfS~)~:::

39t::J:>J1

:~.")S~Q4

requires the inclusion of torsional effects in the element stiffness and mass matrices. It atso requires a transformation of coordlnates of the element matrices previous to the assembling of [he system matrix, The required matrices for torsional effects are developed and a computer program for the dynamic analysis of grids is presented. This program is also organized along the same pattern of the programs in the two preceding chapters for the dynamic ana:ysis of beams and plane frames

PROBLEMS The following p:obiems are intended :or hand calculation, though it is recommended chil[ whenever possible solutions 5hould also be obtained using Program 1.5 to model the structure as a grid frame and Prog,ams 8 and 9, respectively. to cz.!cuiate natural frequencies aGd to solve for the response. 16.1

For the grid shown rn Fig. P16.1 determine the system stiffness and mass matrkes. Base the am:.:ysis 0::> the three nodal coordinates indicated in the figure. Use consistent mass method.

16.2

Use static condensf:lrion to eliminate the ~otational degrees of freedom a:1d detem:ine rhe transformaricn matrix and the reduced stiffness and mass rr.atrices :n Problem l6.1.

16.3

Dere~mine

the natural frequency for the reduced system in Problem 16.2.

16.4

Determine the natural frequencies and corresponding normal modes for the grid anHlyzed in Problem 16. L -

16.5

Determine the response of the grid shown in Fig. P;6.1 when acted upon by a force F(t) = 10 Kip sudden:y applied for one second at the nodal coordinate 3 as show;: 1:1 the figure. Use results of Problem 16,2 to obtain the equation of thot:o:; for the :::ondensed system. Assume 10% modnl damping.

16.6

Use results from Problem i 6.4 to solve ?robiem 16.5 on the basis of the three nodal coordinah:;s as indicated in Fig. PI6.L

16.7

Determine the steady-state response of the grid shown i:1 Fig. P 16.: when subjected [0 harmonic force F(l) = 10 sin SOl (Kip) along noda: coordinate 3. Neglect damping in the system.

16.8 Repeat Problem 16.7 assuming that the damping is proporrional to the s:iffness of the system, [C] =oo[K], where 00=0.3. 16.9

Determine the equivalent nodal forces for a member of a grid loaded with a dynamic force, P (t) = Pr/(c), unifor:nly disrributed along ils length.

16.10

Deter;nlne the equivalent :1odal forces for a member of a grid supporting a cOGcentroted dynamic force F(:) as shown in Fig. PHilO.

Fig. P16.10.

490

Framed StructureS MOdeled as Discrete

Mu!tideg(ee·Crf~Freedom

Sys!ems

The following problems are intended for computer sollaion using Program 15, to model the Structure and Program 8 :0 determine natural frequencies and nodal shapes and Program 9 to caiculate the response. 16.U Determine the nawrol frequencies and corresponding nonnal modes fer [he grid shown in Fig. Pl.6.L 16.12 De(ermine {he response of :he grid shewn in Fig. PI6.! when acted upon by the force depicted in rig. P16.12 flc(ing along nodal cooroinate 3. Neglec[ damping in (he system, 16.13 RepeJ.[ Problem 16.12 for 15% damping in all the modes, Use modal super· position me[hod, 16.14 Repem Problem 16.12. Use ~tep·by·:>(ep lineur accelera[ion method, Program 19. Neg:ec! damping,

17 Three-dimensional Frames

Fitl {Kipli

'~i 10

9

a

,

5

2

: : , ~!!!

:

a

0.1

0.2 0.3 OA

0.5 O.S

0.7 0.8 0,9

l.0 '[(sec)

Fig. P16.12.

The stif:ness method for dyncmic analysis of frames presented in Chapter 15 for plane frames and in Chaprer 16 for grids can readily be expanded for the analysis of three~dimensional space frames. Although for the plane frame or for the grid there were only chree nodai coordinares at each joint, the threedimensional frame has a total of six possible nodal dispiacements at each unconstrained joint three translmions along the x, y. Z axes and three rotations about these axes. Consequently. a beam element of a three-dimensional frame or a space frame has for its two joints a total of 12 nodal coordinates; hence the resulting element matrices will be of dimension 12 x i2. The dynamic a:1alysis of rhrce·dimensional frames results in a comparative· ty :onger computer program in geneml, requiring more input data as well as substantially more computational time. However, except for size, [he analysis of three~dirnensional frames by the sliffiless method of dynamic anaiysis is identical to [he analysis of ?lane frames or plane grids,

17.1

El.EMENT STIFFNESS MATRIX

Figure 17.1 shows a beam segment of n space frame with "itS 12 nodai coordinates numbered consecutively. The convention adopted is (0 label first 491

492

Framed Structures Modeled as Discrete Mult;d$9ree·of~Freedom Systems

Three-dimensional Fral'nes

the three translatory displacements of the first joint fonowed by the three rotational displacements of the same joint. then to continue with the three translatory displacements of the second joinl and finaHy the three rotational displacements of this second joint The double arrows used in 17.1 serve to indicate rotational nodal coordinates; hence, these are distinguished from translational nodal coordinates for which single arrows are used. The stiffness matrix for a three-dimensional uniform beam segment is readily written by the superposition of the axial stiffness matrix from eq. (15.3). the torsional stiffness matri, from eq. (16.6), and the flexural stiffness matrix from eq. (14.20). The flexural stiffness matrix is used twice in forming the stiffness matrix of a three-dimensional beam segment to account for the flexural effects in ~he two principal planes of the cross section. Proceeding to combine in an appropriate manner these matrices, we obtain in eq. (i 7.1) the stiffness equation for a uniform beam segment of a three-dimensional frame, namely P,

E' Syffiffie:rie

P,

0

P,

0

0

p,

0

0

0

P,

0

0

-6£1,

P, ...

0

J2 £1"

l2

,. P,

0

0

0

0

6El"

0

J2£1, L'

p,

0

0

P"

0

0

:P,"

0

0

0 f)El,

,.

17.2

0 FA

0

0

0

0

a

0

0

7--

0

0

0

0

0

0

0

0

0

0

0

- Gl L

P"

at 1he

in which II. and It are, respectively, the cross-sectional moments of inertia with respect to the principal axes labeled as y and:: in Fig. 17, Land L, A, and J are respectively the length, cross-sectional area, and torsional constant of the beam element.

GI

-,y-

T -EA

di:;;:l~;}cemer:ts

-----y;-

L

p,

z Fig. 17.1 Beam segment of a space frame showing forces <:.r:ct nodal coordinates:.

L

493

-6EJ,

-,y-

0

0

0

bE(,

2EI, [.

0

ELEMENT MASS MATRIX

The lumped mass matrix for the uniform beam segment of a three-dimensional frame is simply a diagonal matrix in which the coefficients corresponding to translatory and tOrsional displacements are equal to one-half of the total inenia of the beam segment while the coefficients corresponding to flexurnJ rotations are assumed to be zero. The diagonal lumped mass matrix for the unifonn bearr. of distributed mass in and poiar mass moment 1m = ihh/A of inertia !.Xr unit of length may be written conveniently a.~

L 0

0 0

IQ/A 0

0

0

0

(17.1)

or in condensed notation {P} =

[AC {5}

0

0

lolA

0

0 J (17.3)

0

([7.2)

In whjch lois the polar moment of inertia of the cross-sectional area A. The consistent mass matrix for a unifonn be.s.m segment of a three-dimensional frame is readily obta:ned comb~r,jr,g the consistent mass matrices_ eq. (15.26) for axial effects, eq. (16.8) for torsional effects, and eq. (14.34) for Hexuc'al effects. The appropriate combination of t;,ese matrices results jo the

494

Prarr.ed StH.ictures Modeled as Discrete Mllltidegree-of·Freedom Systems

Three·dimensional Frames

consistent mass marrix for the uniforrn beam segment of a [hree~dimensional frame, nnmely,

: P'l

i"O

P,

0

I

P, P.

i

f,

I

p.

I

lilL

'5,-: 156

8,

Symrn<::!(ie

s;

0

0

Jj6

0

0

0

0

C

-21L

0

-ILl

C

l'2L

C

0

0

;So

a~ ilL"

t, &,

P,

Ie

0

0

0

0

0

l40

P,

0

54

0

0

0

I3L

0

156

P.

0

0

- I)L

0

0

J:36

P"

0

0

"

0

,

C

0

0

0

0

lp"J

0

0

J3L

0

3t>

0

0

0

'IlL

0

4£ ~

- lJL

0

0

0

- JL~

0

-22t

0

0

0

0

70/»

A

P"

lc

"

"

140/~

.,

in condensed

.T

=X cos xX';- Y cos xY+Z cos xZ

(17.60)

in which cos xY is the cosine of the ar.gle between axes x and Y ilnd corresponding definitions for other cosines. Similarly, the y and Z cOffi;>onents of A are

8", 4Llj

SnJ

y = X cos yX + Y cos )'y + Z cos yZ

(17,6b)

zX ~ Y cos zy + Z cos zZ

(17.6c)

:: :::: X cos

These equations are conveniently wrinen in marrix r.otarion as

00(;)t100

{Pi ; [/vI] {8}

17,3

the elements of rhese matrices corresponding [0 the same nodal coordina[es of the structure should be added to obtain rhe system stiffness and mass matrices, il is necessary first to transform these matrices to (he same reference system, [he global system of coordinates. Figure 17.2 shows these two reference sysrems, the x, y, Z axes representing the local sys[em of coordir:ates and the X, Y, Z axes representing the global system of coordinates, Also shown in this figure is a general vector A with its compenems X, Y, Z aler.g the global coordioi:ue. This vector A may represent any force or ct;splacement at the nodal coordinates of one of (he joints of the structure. To ob!
8!)

(17.4) Of

495

Jc 1

(17.5)

Y[

cos xX cos xY cos cos yX cos yY cos

k -

ELEMENT DAMPING MATRIX

xzrXI yZ tY[

07.7)

cos zX cos zy cos ,ZJ Z; y

•I

The damping marnx for a uil:lform beam segment of a three-dimensional fmme may be obtained 10 a manner entirely parallel to those of the stiffness, eq.

(17.1), and mass, eq. (17.4), matrices, Nevertheless, as was discussed in Sec-

y

tion 14.5, in practice, damping is generally expressed in terms of damping ratios for each mode of vibration. Therefore, if (he response :s sought using rhe modal superposirlon method> these damping ratios are directly iritroduced in the modal equations. \Vhen the damping marr.x is required explicitly, it may be determined from given values of damping ratios by the methods presented in Chapter 12,

17.4 TRANSFORMATION OF COORDINATES The stiffness and the mass matrices giver. by eqs. (17.1) and (17.4), respec[ively, are refe:Ted (0 local coordinate axes fixed on [he beam segmenL Inasmuch as

{\A

x

--+-x

0

z

z

/

---.-£

Fig. 17.2 Compon.;:nts 0f a g:1!MfOJ veCfor A in locol and global coordinntes.

496

Framed Structures Modeled as Discrete Multlde;;;ree"of·Freedom Systems

Three~dit.)ensional

or in short notation {A}

lTd {A}

(17 .8)

COS xX

cos x Y Cos

-XI

cos xY =

~-::--,

L

cos xZ=

cos

cos zf

cos zX

u=

(:7.9)

= (Yi Zy = ('I Z.,

L

(17.10)

y,) (Zp

Z.)

Z,) (xp

x,) - (x, - x,) (Zp - z,)

Z~"= {Xl - Xi)

(y,'>

y;)

Z,) (yp - y,)

(Xi

.v,) {xl' -

(yj -

( 17.15)

XI)

and

iZI

(17.16)

Ana[ogously. tbe direction cosines of the local axis yare calculated from the condition of orthogonatlty between a veCtor Y along the y axis and the unit vectors Xl and Zl along the x and z axes. respectively. Hence,

or in expanded notation j

where [ is the length of the beam element gi ven by

Ii

i

cos xX cos xY cos xZl, cos

(17.11)

The direction cosines of the z axis can be calculated from the condition that any vector Z along the z axis must be perpendicular to the plane fonned by any two vectors in the local x--y plane. These two vectors simply could be the vector X from point (]) to point (2) along the x axis and the vectOr P from poim Q to pOlnt P. The orthogonality condition is expressed by [he cross product between vectors X and P as

Z ZXP

(17. ;4)

where

cos yX cos yf cos yZ [ cos:eX cos zy cos

The cosines required in the transforrrmtjon matrix IT,], are usually calculated in computer codes from the global coordinates of three points. The two points defining the two ends of the beam element along {he local x axis and any third point ~;cated in x-y local plane in whicb y is one of the principal axes of tbe cross-sectional ZlreZl of tbe membec The input data containing the global coordlnates of these three points are sufficient for the evaiuarion of all the cosine terms in eq. 07.9). To demonstrate this fact let uS designate by x" y;, z,. and xi' Yj. Zj the coordinates of pOint Q nnd (j; at the two ends of a beam element and by x p• yp. Z" the coordinates of a point P placed 011 the iocal x-y plane. Then the direction cosines of local axis x along the beam element are given by cos xX=

497

where i, ], and k are unit vectors along the global coordinate axes X. Y, and Z, respectively. Consequently the direction cosines of axis z are given by

in which {A} and {A} are. respectively, the components in the ~ocal and globai systems of the general vector A and (T I ] the transfoffilation matrix given by

lTJ

Frames

Therefore, cos yZ=

where

Yx = cos xY cos zZ

or substituting the componenrs of these vectors as

07.D)

cos yY

cos yX

(17.12)

J

zX cos zy cosul

and

cos xZ cos zy

iX cos

yy

cos xZ cos

xX cos

zZ

Yt

cos xX cos zy -- cos xY cos

iX

(17.17)

498

Framed SUuctures Modeled as Discrete Multidegree-of·Freedom Systems

Three-dimens;onal Frames

We have, therefore, shown that know!edge of (he coordinates of points at the two ends of an element of tI paim P on the local plane x-y suffices to calculate [he dL:ectlon cosines of the tnmsforrnation matrix [T;J in eq. 07.9). The choice of point P is generaUy governed by [he geometry of [he structure and the orientation of the principal directions of the cross section of the membeL Quire ofren pOInt P is selected as a known point in the structure which is placed on the local axis y, although, as it has been shown, the point P could be any point in [he plane fanned by the local x-y axes. Altema~iyely the direction cosines for the local axis x, CI::;: cos xX, C2 cos xY, and C3 = cos xZ are given directly by [he coordinates of [he two end points (Xi> YI , 2 1) and (X:,!, Yh 2 2) or the beam e!emenr as c~

, .. i i

k

For ~he particular case in which [:,e plane defined by the local axes x-y is vertical, [he direction cosines for the other two local axes, y and l, can also be expressed in terms of the coordinates of the two points at the ends of [he beam element. Since in this particular case the local axis z is perpendicdar [0 the ver.rcal plane defined by the local axis x, and an axis Y' parallel to the global axis Y as shown in 17.3, a vector z along the local axis z may be (ounc as

t=X*]=

Cj

o

C'Z

"3'

I

-

z

I Vertical I Pian;, J(I l--

-

y

Finally, a

unj~

•X

J

coordina~es

(x, y, z)

vector J along rhe local axis y is given by I if

'="'"=Ic

k

j

, -c

0 l

( 1725)

d

c·,

C1

or

1:1 I

P,(X p Y 1' Z ,} I

Fig. 17.3 Global system of coordinates (X, y, Z) and local system of with x-y plane vertical.

(17.20)

J

I I I I

i

I

in which

I I I xl

P, (X:z, Y,.2'l l

I I I I

I

z L=

_--y-

_---

(17.19)

=

_--1

y

~Y

499

= -

'":, c)

i

l T CI K.

07.21) (17.26)

01

in which

Therefore. from eqs. (17.22), (17.23), and (1726), the rransformation in eq. C7.7) jn the particular case considered is given by

(17.22)

is a unit vec~or along the local axis x, anc i, j, .~ are res/=ectively the unit vectors. a!ong t:,e axes X, y, Z of the global system. A unit vector i along the local axis z is then c&lculated from eq. (17.21) as

rei

>1 -

\Y =

lZj

(! 7.23)

where

d = jC5 + d

c,e, --d-

l-dC1

c,

c)

0

r 1

d

d

c, d

Yr

(17.27)

2,

where d is given by eq. 07.24). Equation (17.27) is rhe transformation matrix between local coordinares (x, y. z) and global coordinates (X, Y, Z; for the particular case in which the local plane x-y is vertical. If this plane is not

(17.24)

·1I

"

I 500

Framed Structures Modeled as Discrete Mulfidegree-of·Freooom Systems

501

ertical, the required transformation may be formulated ~n the following three steps: Step 1.- Rotate the local coordinate system (x, y, z) around the local axis x until the iocal y axis and the local axis x form a vertical plane, Let (Xl, yl, z') denote the auxiliary coordinate system resulting from this rotation of the system (x, y. z) about axis x. The transformation matrix between these two local systems, (x', y', z') and (x. y. z), is given by

rx} ~] F ~ lo l=

°

o

1'x'l

cos r!> sin (I, sin 1> cas ¢

0

!Y'f

(1728)

lz'

in which ¢ !s the angle of roiling from the axis y to the axis y' _ This angle is posl(ive for a clockwise rotation around axis x observing the ~otadon :rom the second end point to the first end point of the beam eIement.

It should be noted that transformation matrix [Td is not defined if the local axis x is parallel to the global axis Yo In this case, eqs. (17. J 9) and (17.24) result: in C; 0, CJ = 0, and d= 0, If !he centroidai axis of the element is vertical. that is, the local axis x and the global axis Y ;:>,re parallel. the angle of rollIng is defined as :he angle of rotation about the local axis x to have the local axis z parallei to the global axis Z. Let t;:s designate by (x', yt, Zl) this auxiliary system of coordir.ates in which [he local axis z' is paralle! to the global axis Z as shown in Fig. ;7.4. The transformation of coordinates between this auxiliary system (x',.v'. z') and the global system (X, Y, Z) obtained from Fig. 17.4 is given by

,

Step 2: Transfonn from the global system (X, Y, Z) to the auxiliary local system (Xi, y', z') of coordinates. Since this auxiliary system h2s its x'--~;/ plane in a vertical plane, the necessary transformation matrix is given by eq. (I7.27) as

: '1

,;:r

c, c,

Cj

- clc}.

-

d

d -CJ

(2C3

d

f~}

F ,z'

y'xl r

I

,2"

0

1

='0

cos

sin oS I

~O

sin

cos ¢j

,

0

c,

01 fXf ]! lZi,

0 0: Y 0

(17.32)

"

y

+

(17.29)

xt x p~ 'X~, y~,

Z;l

d

Step 3: Obtain the transformation from the global system (X, Y, Z) [0 the local system (x, y, 2:), by the combination of the transformations indicated in steps 1 and 2. Hence, the ~equjred transformation is obtained by substituting eq. (17.29) into eq. (17.18), namely,

0

j -c~

Z

C,

0

,

r

c,

c, -- C\C;;

C,

d

0

c, -

c,

C2(3

[Xl

gJ

}--------J--------------~x

(17.30)

d

Therefore, the general transformation matrix [T,] from the global system (X, y, Z) to the local system (x, y, z) is given by eq. (17,3]) as the product of the two matrices in eq. (IBO).

z z· Fig. 17.4. (a) Global axes (X, Y,4. (b) Local axes (x, y, z) will) local axis x paral~e! to global axis Y. (c) Auxiliary local axis (x'. y', z ') witt local axis z' parallel to giobal axjs Z.

502

Three·di:-nensionai frames

Framed Structures Moae!ed as Discrete Muliidegree-or-Freedcm Systems

where c! = 1 wJen local axis ;r has (he seose of (he global axis Y; other\vise, c,! = L The simple rransformation of coordinmes from (he (x', y', z') system to the (x, y, z) lS

V yt = rlC

C

i

I,,-

)

0

cos ¢ ~

o

i

10 1 10,

sin

sin cP cos

in which

IX' I

C ,.J

o

LX}

,in¢iF cos

¢J

~z

(1733)

[T,)

=f

l

c,

C2

cos :/>

0

Analogously, the transforma:ion from nodal forces {P} in g!obal coordinates to noda: forces {P} in local coo:dinates is given by

{P}; [Tl (F}

C1

sin ¢

0

(17.34;

[Tl iP)

= ITl {S}

= [K] rT] (iJ)

or ( 17.39) since [TJ

:5

an orthogonal macri,,;. From eq. (17.39), we may write

(l735)

(17.40) in which

where if; is the angie of rolling between the local axis z and (he direction of the global axis Z (C1 = + I w:len loca~ axis x and global axis Y have the same sense and C2 = - 1 for the oppOsite sense of these axes), We have, therefo:-e, shown that the knowledge of the coordbares at the two ends of an element, together with the knowledge of either a third point located in the x-y local pla:1e in which y is one of the principal axes of the cross~ sectional area of the member, or altemati vel y, the knowledge of the angie of rolling. suffices to calculate the direction cosines of the transformalon matrix [Td in eg. (17.7). For the beam segmem of a three~dime:1siona~ fmme, [he transfomtution of [he nodal displacement vectors involve (he transformation of linear and angular displacement vectors at each joint of the segment. Therefore> a be£!.m elemen: of a. space frame reqJ.ires, for the two joints, [he wmsformacion of a total of four displacement vectors, This transformation of the 12 nodal displacements {8} in global coordinates to the displacements {o} in local coordinates may be written in abbreviated form as (o)

( 17.38)

Finally, to obtain fhe stiffness marrix (K] and (he mass matrix rt~'!J in reference to the global system of coordinates, we slmpiy substitute, into eq. (17.2), {o} from eq. (17.36) and [P' from eg. (17.38) to obtain

a : sin 10j cos ¢

( 17.37)

L

Therefore, whe:1 the foe£!.: axis x is parallel IO [he global axis Y, the transformation matrix [T;] between the global coordinates (X, Y. Z) a~d the local coordinates lX, j\ Z) :s gi yen by

0

(T]

(T] =

where the angle ¢ is the angle of roiling around the local uxis x measured from local axis l 1O the aux.iliury axis z'. This angle is posltlve fOf a clockwise rotation observing rhe x axis from the second end toward [he first end of the beam element The substitution of eq (17.32) into eq. ~I7.33) yields

o o

1

: [T,]

,

~

503

(17 .36)

[k] is defined as (17.41)

Ana:ogously, the mass matrix in eq (i 7.5) is transfor.ned from tocal to global coordinates by

[,\1]

=

iTl' [M][T)

(17.42)

[TJ'[C) [T]

(17.43)

and the damping marrix [C] by [C]

17.5

:I

DIFFERENTIAL EQUATiON OF MOTION

The direct merhod which was explained in detail in Chapter l4 may also be used to assemble the stiffness, mass, and damping matrices from (he cOrres-

504

Three~d!mersiona!

Framed Structures Modeled as Discrete Muttidegree-of-Freedom Systems

{F(r:}

505

y

ponding matrices for a three-dimensional beam segment, eqs, (17.41), (17.42), and ([7.43), woich are referred to the global system of coordinates. The differential equations of motion which are obtained by establishing the dynamic equiEbrium among the inertial, damping, and elastic forces w~t~ the externa! forces may be expressed in matrix notation as

[,\1] (y}'" [C] U) + [K] {y}

Frames

(! 7.44)

in w~icn [M]. ~C]. and (K] are. respectively, the sysiem mass, damping, and stiffness matrices, {}'-}, Lv}. and {y} are the system acceleration, velocity, and d1splaceme:1t vectors, and {F(!)! is the force vector which includes the forces app!ied directly to the joints of the structure and the equivalent nodal forces for the forces not applied at the joints.

=-------------------~. x i7.6

DYNAMIC RESPONSE

The integration of the differenrial equations of morion, eq. (17.44), may be accomplished by any of the methods presented in previous chapters to obtain the response of Structures modeled as beams, plane fmmes, or grids. The seiec~ tion of the panicular method of solution depends, as discussed previously, on :he linearity of the differential equatior:, that is, whether the stiffness matrix [K] or any other coefficient matrix is constant, and also depends on the complexity of the excitation as a function of time. When the differential equations of motion, eg. (17.44), are linear, the modal superposition method is applicable. This methOd, as we have seen 1n the preceding chapters, requires the solution of an eigenproblem to uncouple the differential equations resuiting in the modal equations of motion. If the structure is assumed to follow an elastoplastic behavior or any other form of nonlinearity, it is necessary to resort to some kind of numericai integration in order to solve the differential equatiop.s of motion, eq. (17,34). In Chapter 20, the linear acceleration method with a modiflcation known as the Wilson-8 method is presented together \.\'ith a computer program for analysis of Enear structures with an elastic behavior.

17.7

PROGRAM 16-MODELING STRUCTURES AS SPACE FRAMES

Program 16 calculates the stiffness and mass matrices for a &.ree-dimeosional frame and stores the coefficients of t:-tese matrices in a file. Since the stiffness and mass matrices are symmetric matrices, it is only necessary to store the

z Fig. 17.5 Space Frame of Example 17.1.

coefficients ir: the upper triangle of these matrices. The program also stores in another file the general data of the space frame. The infonnation stored in these files is needed by the programs to undertake the dynamic analysis of the structure, such as the detenninatio:l of natural frequencies Or the response of the Structure subjected to ex~emal excitation, Example 17.1 For the L:"ree~dimensional frame shown jn Fig. 175. deter~ mine (a) the stiffness and mass matrices, and (b) the natural frequencies and corresponding modal shapes. Mode! the structure with four beam e:ernents and use consistent mass matrix formulation. Solution: As the first step in the analysis, the frame is divided in four elements, as indicared in 17.:5. This division of the Structure results in five nodes with a total of 30 nodal coon:Hna(es. of which 24 are fixed. T~e numerical values needed for analysis of this structure are given in Table 17.l.

SOB

Framed Structures Modeled as Discreje rv.ult:aeg:ee.-of·Freeoom Systems

Table 17.1

ihree-dimefisional Frar.1es OI}1'I't,l1' RSSt,I[. 'l'5.

Input Data for Example 17.1

Members 1,3

QUfllHl(Y

Members 2,4 ..

'-'~-~'-'-'--~-"--.- ~--

30 X 10' 12X 10'

Modulus of elasticity (psi) Modu Ius of rigidi~y (psi) Distributed mass (lb· sec z lioJ) Cross~sec(ional y moment of inertia (in") Cross-sectionai z moment of inertia (in") Torsional conStant (in") Cross-sectional area (in?) -.-.~.---~~.-.--

O~

t;(Jt13ER OF

0.2

0.1

200

64

200

64

40.0 50

,-C)OOOU.OO

S4!St:dJ

oc::o;;:~oo

0.0000.>00

o[
W;

I,)Q

co

0.00

C fl:)

-2CO.OO

"oe

200,CJ

0.00 0.00

C .;;0

(0.

;)

~OOC::+O&

,
tjOOE~Ol

C,lC())F. .. )l

).ooo!.n

.24:51;.05

~.lOCOlt~O~

v. eoen.Ol

!0002.iH

lSl92E.u,>

. :\12:>0~

.OODO!':':):)

-.~.90<:'C)

OC(lOE.CO

C.:~S!2-06

Q.~:9CE:.C3

C(;oOF..OC

1190);>0)

t 9C,:"CJ

o ,OOO(E~OC

o

\L~J:J'$E.':;l

"19~S-fH

to.

OCOOE~l)Q

Ij

4!.9CE:.C~

O(;OS2.00

o. }(;SS;:¥os

C(l;)C2.0(l

CCCO;;.CC

QiJJOS.')C

(;'~l.~OE;'C)

;:; OOOf)S.CO

3054£.0$

OCO(£.oo

;:;Ot:02.00

O.I:OOCO:;.OO

C.)0':>5lt+)$

Of)OCS ... O0

,n;ce:.n

:.::.>01

~20J

;}

"

:)Q

COt~:;

COOOC

O(h6?

O.O;:;OOO~C.DO!OS

OCOOJ ilJldi>

OOO,,! O.}(lOCC·~ OC~6:; 0000-) :.LOO%7 o.t)Q\)OQ

·0

(lC~8S

K<\SS!

JO~:';7#

20e.OO

HC1'-t

1'01'<3:0..'1

!NtR71A

CONS"?

103.0;;

4G [<0

,,1.

:;\$88 0

ooooc-o

00000

1779;) 0.000000.002420

12. eo

:wo.ce

20:1, CO

10 GO

so

H.C:J

6COO

:2 so

2&. :10

64,00

C 200. (I. ~G3

~'100s

Y~2.

D!R.

J

tZ588 "'.OCOQO

I..1)D£

i(-aC'!',,~.

OHe

'{""it:r.~s,

NCiCS

0:(.;.

CO

or?:.

*'0::

orR,

o.ov:>~8"

(1) Set view to the XY ?Iane: VIEftl,

1:-RQT~C;

NO:;£

;:Joe::;)

ooono

The following commands are implemented in COSMOS:

DISPLAY "1;~1.

O.CQ)5~

DYNAMIC RESPONSE OF THREE-DIMENSIONAL FRAME USING COSMOS

Solurian:

0[<

0.1*0

:::'.~'l'A·

(X~:,

:locae

aeOM 0 O:)(l90

Example 17.2. Use the p:-ogram COSMOS to calculate rhe natural freM quencies of the threekdimensio!lul frame of Example 17. L Model the structural using BEA;.v1JD elemen:s wiLh consistent mass formulation and compare results with those obmined in Example 17J,

" 000

l't":(D coca;;t;;;<"I'!\S

0

C .Ctl

O~

SECOND

0

10,,7 C.OOCOC J,to,")7 0.30(:00-0 000:;0 (l.OOCOO

000

C~R.

gOO~z.06

0, 131 )t .. o~

lHn·o"

o

2-1.

J.OCC C 000

~on£

?l!1l2-0> . 'H)o;;.>06

:I.CCOO£.O(;

17.8

O!R

O.$OOOE.DJO.

o.

coooo

NOes

0 llJGcc.ao

;OOC2,Cl,s

'5l5£:.07 Goon-co

E ~ ~2YC?

cO'! '"

DIREC'!'lON ¥

~OOOE.O;;;

78].]£>;)2

})£

MOOOWS OF lUCIDITY

~200

k.

9::IIJO;:.06

C

ILOOOO£.OO

j.

Z!.AS'1'lC!'l'Y

"

ooooZ~OO

OOOO">~<'

.0000£.01} •. 1l
J'or!'!,:s .a;:.\M E(,.L>!snrs

or

")~15,>01

OOJO£.oo

,oocc;:.c~

12.8 28

.•- - . - - -

Ntil
MOOUl.,VS

0,

30 X 10' 12X 10'

Input Data and Output Results

NUOI.,r:R

507

~::;::

DIR.

VIE_PAR .. V!2Vi

0,

0,

1,

0

(2) Define the XY p:ane at Z = 0:

~

G28METRY PLk'\lE,

Z,

~

GRID ) D.

1

?L~N£

SOB

Framed Structures Modeled as Disc:,ete Mu1tidegree-of~Freedom Systems

Three-dlmensional Frames

(3) Defme key points at the nodes of beam elements:

(2) Merge nodes and compress nodes:

::;EOMETRY )- ?O!K'?S ;. l?T 0,

P'I'r

t.:t-tERGE. 1, 20, 1, 0.C001, C, C, 0 NCOM?RESS, ::., 11

0, :) 0, ~2CC

0, 3 , 0, 20a, 0 ,, 0 4, 200, PT, 5, 0, -200, 0

p~

0

<,

P'" PT,

(13) Apply constraints ill all degrees of freedom at nodes 2 through 5:

,

DK;:),

j.

2, ALL,

0,

5-,

D!s?LMN?S ;. DND

:'-

(14) Set options for frequency calculation to extract six natural frequencies using [he Subspace iteration Method and consistent mass farmulatio;};

CURVSS ) CRLINE

,, 2 2, 3 4 CR:;:"INE, 3, 1 eRLINE, 4 .L, 5 CRLI!'':-E, CR~'.. !NB,

~

LOA.JS-BC > $'TRUC?UR.i\L

(4) Define lines between key points along the beam elements: GEOMETRY

509

l,

ANE.LYS:S > ?E-(EQ/SUCK ) P>._FREQUE:.1CY A_P£Qu'ENC{, 6, S, :'0, 0, 0, C, 0, 1E-05,

t:::::-06, 0,

(l,

!, 0

(15) Run freque>!cy analysis:

(5) Defme element group

ANALYSIS ) ?REQJSGCK

using the BEAM3D:

:>

R_FREQUEl'-!C'!

K ... FREQUE;";CY

PROP SETS

>

EGROUP

EGROUf', 2.

r

BSr-u'13;Jt-::,

0,

0,

0, 0,

0, 0 z

(16) List natural frequencies: :::

RESULTS ) LiST> FEEQL:ST

(6) Define material properties far members It 3: PROPSETS ;. MPROP MPROP, 1, EX, 3CE6, SKY, 12:So, DE:.t\S,

FEmQUE};C! Nut{BER

2

3 )0

RCCNST, L

N?ROP

4

l , 1, 10, 50, 20C, 200, 5, 5, G, 0, 40, {), 0

5 0

(8) Generate a mesh of beam elements for members I, 3: MS$HING ;. ?AR.tlJ1_MESH

M_CR. 1, 3, 2,

)0

3, 1, 1, 3

Frequency

.?ROJ?SE"I'S ,. RCCNST 64,

5,

5, O.

0,

(11) Generate a mesh of beam elements for members 2, 4; V,,-C?,

2,

:.

P.tl,Rl:0CMESE :. M_CF. 4, 2, 3, L 1. :;;

O.80S0547:S+J2 C 8066980E+02 C:.88SB077E+C2 O.4178062E-C3 C·.48936
0.1281284::>-82 (:.1283900.8-.-02 O.1<109807E""C2

0.778877:::"5:-01

O.6649S93E~02

:::.77884lSE+02 C,82306531;:;':.. 02

C.7804669S~C::'

0.70931712-01 O. :'S03852E-01 8 . 12839581£-01 0.12l.~970E-Cl

Natuml Frequency (CPS)

(0) Define rea( constants for members 2, 4:

~f:ESE:::~G

(SEC:O.:JDS)

TABLE 17,2 Natural Frequencies Calculated by COSMOS Compared with Values Obtained in Example 17,1

PRC.?SE1'S :;. Y-PROP MPROP, EX, 3GE6, GXY, ::'2E6, DENS, O.OCJ57

1, 2, 1, 10 , 28, 64,

PERTOJ

(CYCLES/SEC)

Table 17"2 shows the comparison of the natural frequency vaJues obtained using COSMOS with those obtained in Example 17.1.

:-'CCR

(9) Define material properties for members 2, 4:

RCONST,

FREQUS[(CY

(R]\"D/SSC)

(;.084

(7) Define real constants for members 1. 3: PRO]?SETS

FREQUEN2Y

.... ".8,

0,

°

2 3 4 5 6

~o.

COSMOS

Example 17.1

12.81 12.84 14.10 66.50 77.88 82.31

12.75 12.83 14.09 66,50

77.90 82.32

510

17.9

Framed Structures Modeled as Discrete Mu!i'ldegree-o!-Freedom Sys:elT'.s

SUMMARY

This chap[er presel1:ed the formulation of [he stiffness and mass :naulces for an element of a space frame, as well as the transformation 0: coordinates required to refer these :natrices [0 the global system of coordinates. Except for the larger dimensions of the matrices resulting from modeling a space frame, [he procedure is :c.emical :0 the case of a beam, plane frame, or grid frame described in ~he preceding chapters.

PROBLEMS 17.1

18 Dynamic Analysis of Trusses

Use Program 16 to detennine [he s(iffne:::>$ and mass m;;!,rices of the (:-tree dimensiollul frame modeled
Problem Daw (for all members): Modulus of eias(ici(y: Modulus of rigidity: Distribu[ed mc.ss: Concemmted ma:;ses: Cross-sectional y momem of inertia: Cross-s~c(ional

z moment

E.=;30X lei' psi G=12XW{,psi in = 0.2 !b· sec' iin:; m=!O Ib-scc 2 /in = 300 in~ It = 400 in 4 J=500 in~ ;1=20 in~

'I

of inertia:

T crsi.onal COnstant: Cross-sectional area:

CD 10. 120. 01

&

(j) {120. 120, OJ

&, :0.120.1201

Y

6 @

10

0f---t-'=...::~

(0 ';;;i",;r-~ X

I '

® 1120,

(0,0,0)

z

~

0(0,0,120}

®

(120,0, 12m

Fig. PI1.1.

D.Oj

The sTaric analysis of trusses whose members are pin-connected reduces to the problem of determining the bar :orces due to a ser of loads ap?Hed at the joints, When [he s.ame [:llsses are subjected w the action of dynamics forces. the simple simution of only .axial stresses in the members is no longer present. The inertial forces developed .::.long the members of the (russ wlll, in genera!, p!"oduce t1exurat bending in addition [0 axial forces. The bending moments at the ends of (he truss members will stiB remain zero in the absence of external joInt moments. The dynamic stiffness method for the analysis of trusses is developed as in the case of framed strucwres by establishing (he basic reJatior:.s between external forces, elastic forces, damping forces, inertial and the resulting displacemems, velocities, and accelerations at the nodal coordinates, that is, by detennining the s[iffness, dam?lng, and mass matrices for a member of (he truss. The assemblage of system stiffness, damplng. and mass matrices of the truss as well as the solution for [he displacemems at [he nodal coordinates fonows aiong (he standard mel.hod presented in [he preceding chapters for framed srn:.cwreS. 51t

Dynamic Analysis of Trusses

Framed Structures Modeled as DIscrete Multidegree-ot-Freooom Systems

512

18.1

or in condensed notation

STIFFNESS AND MASS MATRICES FOR THE PLANE TRUSS

A member of a plane truss has two nodal coordinates at each joint, that is, a total of four nodal coordinates (Fig. 18.i). For small deflections:. it may be

assumed that the force-displacement relationship for the nodal coordinates along the axis of the :nember (coordinates I and 3 in 18 1) are independent of the transverse displacements along nodal coordinates 2 and 4. This assumption is equivalent to stating that a displacement nodal coordinates 1 or 3 does not produce forces along nodal coordinates 2 or 4 and vice versa. The stiffness and r:1.ass cocfficlents corresponding to the axial nodal coor· dinates were derived in Chapler i5 and are given, in by eg. (15.17) for the stiffness coeffidents and by eg, (15.23) for consistent mass coefficients, Applying ~hese equations to a uniform beam element, we obtain, using the ;lotation of Fig. 18.1, the fonowing coefncients:

k:,

=

.4E

=-

mil;;;; m), =

mL

AE

k,}

L

mu = m3:

mL

(

l

J p,

P, P,

0 0 0 - 1 0 0 0 1

PI

AE

I

0:"

0

01

0

~J

0: 0, 0, 0,

{Pi

= {K}{8}

(18.4)

in whi::h [K] is the e!emen~ s~iffness matrix. The :::onsistent mass matrix is obtained, as previously demonstrated, using expressions for s:atic displacement f'Jnctions in the application of the principle of virtual work. T~le displacemem functions corresponding to a unit deflection at nodal coordinates 2 and 4 ind:cated in Fig. 18.2 are given by

x

L

u: = 1 -

(185)

and

x

u':=L

(l8.1)

(;8.6)

The consistent mass coefficients are given by the genera] expression, er;.. (18.2)

in which m is the mass per unit length, A is the cross~sectio.,al area" a::1d L is the length of the element The stiffness coefficients, for pin-ended elements, corresponding to the nodal coordinates 2 and 4 are all equal to zero. since a force IS not required to produce displacements at these coordinates. Therefore, arranging the coefficients given by eg. (18.1). we obtain the stiffness equation for a uniform member of a truss as I

513

(j ),23), which is repeated here for convenience, namely, mij =

rL

10 m(x)ui(x)uAx)dx

( 18.7)

For a uniform member of mass in per unit length, the substitution of eqs. (18.5) acd (18.6) into eq. (18.7) yields

(18.8)

(18.3) Finany, the combination of the mass coefficients from eqs. (I8.2) and (I8.S)

x

{ai

Fig. 18.1 Member of a plane truss show:ng nodal displacements and forces.

Fig. 18.2 Displacement fU:lctions. (a) For a djsplacement 8~ "'" L

l):lj~

Ciliplacement 02 = 1. {b) For a un:t

514

Dynarr.ic Analysis of Tn.:sses

Framed S!fuctt:res Modeled as Di~fete MuJtidegfee-of~Freedom Syslems

forms (he consisrent mass matrix. relating forces to accelerations at the nodal coordinates for a unifonn member of a plane truss, :lamely,

r P, 1 . p.,

!

mL

r2

10

0,1 r 8, ~

o 2

0

,8,

a

2

0'

')=-.

, P,

l P,

J

6 II

La

2J

a

I

S!rJ.ce (71 is an on:hogonal matrix ([71

!

[7]r), it follows that

or ( 18.9)

~~J' u,

(18.14)

in which

or in cO:1.cise :lorarion

(1&.15) ( ,810)

(P) = [M] {S}

18.2

is the e!;!ment s:iffntss rnlltnx :n the global coord1nare system. Analogous!y> substituting eqs. (13.12) and (18.13) into eq. (lS.IO) results in

TRANSFORMATION OF COORDINATES

The stiffness matrix, eq. (18.3), and the muss matrix. eq. (18.9), were derived in reference to nodal coordinates associated with the local or element system of coordinates. As discussed before in the chapters on framed strucmres, it is ne('~ssary (Q (ransform rhese matrices to a Cop.1mon system of reference, the globa] coordinate system. The transformation of displacements and forces at the nodal coordmates is accomplished, as was demonstrated in C~ap(er 15, perfomiing a rotation of coordinates. Deleting the angular coordlnates in eq. (:5.31) and relabeling the remaining coordinates resulr in rhe foHowing tranSformarion for (he nodal forces:

r P, 1

r

{ P,.,=

1-

l ;: J

_

i

cos {)

sin {)

0

sin 8

cos 0

0

~

0 0

0

cos 8

sin B

0

-sin {)

cos 8

r- ,

I:: j

{P} ~ [T([1I1J (7]

(i'l

= [7] {,o}

{8} e~s.

[7] {S}

~

(18.12)

(l8.13)

(18.12) and (18.13) into the stiffness equation (18.4)

[NI] (ii)

(18.17) (18.18)

!fi which [i~ll is the element mass matrix referred to the global system of coordinates, HO\'lever, [here is no need (0 use eq. (l8.18) to calculate matrix [.~1]. This :natrix is e~ual to the mass matrix [At] in reference to lo:;al axes of coorciinmes. To verify this fact, we substiwre into eq, (18,18) mar:ices (M] and [7]. respectively, from eqs. (18.9) and (l8.11) to obtain

(18.11)

in which {P} ane {P} are the nodal forces in reference to local and global coordinates. respectively. und (TJ the ~ransfonnation matrix defined 1n eq. (18.11) The sa:ne transformation matrix [71 also serves to transform the nodal displacement vector {6) 1n the global coordi:,uue system to tbe nodal displace~ ment vector {o} in local coordir:ates:

(18.16)

[M] ~ [71 T [1\1] [7]

[NI] {P}

{il}

or

or in condensed notation

The substitution of

515

c

s

0

0

5

c

0

0

0

c -s

0

s

c

2

;1

[M]

0

2J

l~

f2 [M) =

r~

:!

ihL;

filL ,0

l~

0

I

c

s

0

0

s

c

0

0

0

c

0

0

s

0

2

0

i] [

:1 c

I

2 0 0

l~

0 2

(18.19)

e

in which we use [~e notation c::::;; cos 0, S = sin and the fact that cos 2 f) + l sin 8= 1. A simUar relationship is also obtained for rhe element damping :natrix, namely,

gives

(7]

(PI

~ [K]

[11 {81

(el

[TlT[C}[TJ

(1&.20)

516

Dynamic Analysis of Trusses

Framed Structures Modeled as Discrete Multjdegree~ot~Freedom Systems

!o which [C] and [CJ are the damping matrices referred, respectively, to the global and the local systems of coordinates.

517

Then from eqs. (18.15)

Example IS.l. The plane truSS shown in Fig. 18.3 which has only three members is used to illustrate the application of the stiffness method for tn.;:sses. For this [russ determine the system stiffness and the system consistent mass matrices.

[1?,] ~ [T,f[K.] [T,]

!Er~ La

0

10

-1

a

0

0' ! -I.

0

01

0 0

Jj

For member Solution: The stiffness matrix. ego (18.3). the mass motrix, eg. (18.9), and the transformation matrix. eq. nS.1I), are applied to the three members of this truss. For member 6, = 90'\

e

M[ :

[K]~-

.

0

- I

0

0

L-I

0

0

0

°1

0,

I.

0: 0

2 0 ~

2 0 0 2

inL 0 (Md = ,M" ~ •I -

r

r

oj

I

:K,]

AE

I O. I

0

- I

0

0

0 0

0

0

I

0

0

~J

L

01

0

~

[T ,]

0

0 0

0

~l

- I

OJ

[1

A

I'

P,

L

~

"

- 1

0

0

!

- IJ

2,

30 x 1\f Ib/ln

For member

ill,

-1

-:1

-I

- 1

' - 1

10 in 2 2

I

- I

l

1

- 1

0

0 0

I

0

1

0 0

0

- 1

:

Ll AE I,

&

4

f.---

&. L • 6tT

~ I

fig. 18.3 Example of a plane truss_

(,- ,

=

0

_

f2

0

1

filL 10 6: 1

2

0

M'j=[M,'~-, )

• 3

1I

e~ 0°,

~

(j)

~J

I,

I -1

° AE

[K,] = [T,f[K,]

_

I

!

~l

0

r

[K,) = [K,]

,1;

~ [M"

0

1 •

®

1,

2 0 In ~ -"~6-l~ 0 2

Then from eqs. (18.15)

m = 0.1 Ib-se
I

01,

0

0

0

For all members:

2

[M,]

-I

I

and

0

f2J~

;

-1

[Tol

0

-

1

and

0 ?' -J

,0

,

I

4

>"

lo

0

2 0

~l

0, 0'J

;]

substituting the proper numerical vaJues for this exampie; L = 60 in, A = lOin 2, 6 fil = O. i lb, sec:; lin:!, E = 30 X 10 ib /in~, ~nd following the rules of the direct

518

Dynamic Analysis of Trusses

Framed Structures Mode!ed as Discrete Multidegree-of-FreedOr.1 Syslems

method of assembling [he system stiff;)ess and mass matrix from the above element matrices, we obtain r

1.768

- 1.768

10'1 ~. 1.768

6.768

- I 768j 1.768

- 1.768

1.768

6.768

!

[Ksl

r4.828 [Ms~ =

0 l:A14

4.828

0

0

4.828.

whic:t may be normalized using the factors

This r:ormalizatlor. ,esults in

1.414!

0

{4>2} =

;,vhere [Kd and (M)'l are, respectively-, the system stiffness and mass matrices for the truss shown in Fig. 1&,3. Example 18_2. Determine lhe natural frequencies and normal modes for ~hc truss of Example 18.1.

Soiwion:

Substituting {y}

= {a}

sin

WI,

+ [Ks]{Y}

0

These :lormai;zed eigenvecwrs form the modal matrix:

[
([Ksj-

Example 18.3.

(b)

For the nonrrivial solurion, we requite u;' (MsJ' = 0

or

Wi

= 415 rod Isec

O.02t4

or

Ul-:!

= 1034 fad/sec

A, = 0.0466

or

w)

= 1526 rad/sec

Substituting in tum w" Wz. and and G3 give rhe r:1odal vectors

LOaD} {a,}= 0.216. { 0.274

1iJ,

into eq. (b), seuing

G1

= L and solving for.:J2

l.0001 {a,}

=1

5.488 f.

, -4.000j

0.246 : - 0.2461

- 0.272 - 0.375 J

Determi:1e ,he response of the truss in Examples 18.1 a;)d

T~e

modfll equations are given in general [eq. {I 1.6)] by

(e)

= 0.00344

,.1.2 =

0.373

I as shown in Fig. 18.3. SOl~lfiOtL'

Substituting from Examp!e 18.1 {Ksl and (MsJ and expu:1ding tbe above determinant give a cubic equation in A = u.:?mL2/6AE, which has ~he following roots Ai

0.068

18.2 when a constam force Fl ::;; 5000 lb is suddenly applied along coordinate

w' [,11,]) {a} = {OJ

I [Ksl -

,0.402

: 0.087 lOl10

(a)

we obtain

0068 1 OJ73\ - 0172.1

The differential equations ot morion for (his s),sreo":. are

[Ms]{ji)

5i g

l.000' {a)}=

{

LOOOr' - L524

in whIch the modal force

Hence using the resclts that were cakulated in Example 18,2 we obtain

520

Framed Structures Modeled as Discrete Mujtidegree~of·Freedorr:: Sys~err:.s

The solution of the above equations is given by eqs, (4.5) as

fOf

2010 340 Z;:=

zero initial conditlons (Zl! = 0, it! = 0)

respDnse to a constant force of FJ nate 1 as shown in the figure.

~

5000 Ib suddenly applied at nodal

cos 1034;)

Input Data and Output Results 1230 2:)=

(l

cos 1526t)

The response at the nodaI coordinates is then calcuiated from {.v}

Iv'

Y" -

l

)'3,

[0.402

[
0.246j{Jz,

0.373

0.246

0.1I0 - 0.272

0.375

0.087

J

or YI

= 10- 3 [4.843

4.692 cos 41St

)/,= 10-'[1.004-1.015 cos 41St

0.022 cos 1034t

0.130 cos 1526tJ

0.119 cos 1034t';'0.130 cos 1526
)/, = 10 -; [0.999 ... 1.284 cos 415r + 0.087 cos 1034t + 0.198 cos 1526r]

18.3

521 coordi~

Solution: In the solution of [his problem, it is necessary to modeJ the structure as a p)ane truss (Program 17), to solve the eigenproblem to determine natural frequencies ar.d modal shapes (Program 8), and finally to calculate the response (Program 9).

cos 41St) (!

Dynamic Analysis of Trusses

PROGRAM 17-MODELING STRUCTURES AS PLANE TRUSSES

Program 17 calculates the stiffness and mass matrices for the plane truss and stores the coefficients of these matrices in a file. Since the stiffness and the mass matrices are symmetric, onty lhe upper triangular portion of these matrices needs to be stored. The program also stores in another file, which is named by the user, the genera! information of the plane tf',JSS. The information stored in these files is needed by programs; which the user may call, [0 perform dynamic analysis of tbe truss, such as determination of natural frequencies Dr calculatior. of the response of the structure subjected to external excitation Example 18.4. Use Program 17 for (he dynamic analysis of the truss shown in Fig. 18.3. Determine the natural frequencies, modal shapes. and the

:. 7""e¥;~05 q.,'7PiJ~.0"

;:.32"">C5

522

Framed Structures Mode!ed as D;'screte Mu!ddegree·of·Freedom 8yslems

DynamIC Analysis

ot

Trusses

523

164 51

I,

I

,

IIUl>i!2R Df S;C::'£RNAL fO?C2S

(.(

Ss'

"ftME $1'<:7 Of !t>.1'SCR.!'.'!'!CN CRA'll'A1'!O!.'';';'' !!1tlEX ?R.H,':' 'ftxs HtS'i'(:R't S?RT " 1; O~l.:t "'.AX.

>/I..L0£S tli'RT

y

r

'.

0 t;

?xrrr".:"ON

o.co

SJ:)O,OO

:2.0C

SOOQ,CO

CD

, Z

e::::oa\:l

"'..AX,

DIS?L.

!"_;x. Vl::!"OC.

".;,;.r. ACC. 1:')2.,(;

0.0095$

"2.o:n

(LOOHl

C.6H

517.11

(U;013
0.8G

.sSl. 70

18.4 STIFFNESS AND MASS MATRICES FOR SPACE TRUSSES The stiffness matrix and the mass matrix. for a space truss can be obtained as an extension of the corresponding matrices for (he piane [fUSS. Figure l8.4 shows the nodal coordina~es in the local system (unbarred) and in the global system (barred) for a member of a space truss. The local x axis is directed along the longitudinal axes of the member while the}' and z axes arc set to agree with the principal directions of the c-;oss section of the member. The following mmrices may then be wrinen for a uflifoffil member of a space truss

"

/ "

-x

{h)

Fig, 18.4 :V1errber o~ a space ~ru$S showing nodal coordinates. (a) In the local system (unbarred). (b) In (he g!obn! system (barrtd).

as an extension of the scifr:Jes$. eq. 08.3), u:Jd the mass, eq. (18.9), matrices for a member of a plane [russ, Stiffness matrix: c-

AE

[K] = - -

L

~1

-I

0

0

0

0

0

0

0

0

0

0

0

-I

0

0

1

0

0 0 0 0 0

0

0

o· 01

0

0

OJ

O 0

0

(18.21)

524

Framed Str.Jctuies Modeled as Discrete

Muitldegree~of.Freedom Systerrs

Dynamic Analysis of Trt,;sses

525

and

Consistent mass matrix:

nur~

[MJ=(f 1

l~

0 2

0 0

°l °0 0 ° 1 0

1 0 2 0 0 2 0 0 1 0 0 2 0 0 0 2 0

[,til = [n T 1M] [11 ( 18.22)

Lumped mass matrix:

mL [M,l=2 .~,

r1

1 1 I 1 Ij

rcos xX

cos xY

cos xZi

IT,] = cos yX

cos yf

cos y2 cos zz...,

I

lcos zX

I'

cos zy

(18.25) with

I;)'

+ (Z, ..- z,l'

lk}

(18.24)

The direction cosines in [he first rOW of eq. (1824), CJ = cos xX, C2;;:;:: cos xY, and Ct cos xl., may be calculated from the coordinates of the two points Pi (XI. Y;, Z;) and P, (X" f 2,) at the ends of the truss element, that is "

L = f(X,-X,l'+ (Il

In the case of an element of a space truss, it is only necessary to calculate the direction cosines of the centroidal axis of the element which are given by eq. (!8.25)~ The other direction cosines in eq. (18.24) do not appear in the final expression for the element stiffness matrix (k] as may be verified by substituting eqs. (18.24) and (18.27) into eq. (18.28) and proceeding to multi~ ply the matrices indicated in this last equation. The final result of this operation may be written as follows;

(1823)

The transfonnation matrix [Til corresponding to three nodal coordinates at a joint is given by eq. (17.9). It is repeated here for convenience.

(18.26)

_ [[T] [0]

I

CIC)

c;;:c,

c:;

C2C}

C;}Cr

CJC:;

C;

,

,

-d

-Ct Cl

-

C:.C,

-d

:... -

C;;C;

-

C3C~

-

c~

-

C(C}

-c;c}

'2CI

-

Ci

---

C;CI

-

CJC:,

-Cy,

,

C~C5

2

08.30)

Cr C )

C~

Ct C }

C;C}

-

Cle;;

CZCI

d

CZ C 3

-

C;

Clei

C3 C2

d

-~

,

[M]

(MJ

(18.31)

(18.27)

18.5 EQUATION OF MOTION FOR SPACE TRUSSES

in which [Ta is given by eq. (18.24), The fonowing transformadons are then required to obtain the member stiffness matrix lKl and the member mass maLix :M) in reference to the global system of coordinates:

[k} = In T [KJ In

AE

C ,cz

Consequently, the detennimnion of the stiffness matrix for an element of a space truss., in reference to the global system of coordinates by eq_ (18.30), requires the evaluation by eq. (18.25) of only the direction cosines of the local axis x alo:Jg [he element. Also t analogously, to eg. (18.19) for an element of a plane truSS, the mass matrix vii] for an element of a space truss in reference to gJobal coordinates is equal to the mass matrix of the element [M] in local coordinates. Thus. there is no need to perform the operations indicated in eq. (18.29). The substitution into this equation of [M] from eq. (18.22) and In from eqs. (18.24) and (827) results, after perfonning the multiplications established by eg. (18.29), in the sa:ne matrix ~A1], that is,

The transformation matrix for the nodal coordinates at the two ends of a truss member is then given by

In - JO] [T,]J

(1829)

(18.28)

The dynamic equilibrium conditions at the nodes of the space truss resuit in the differential equations of motion which in matrix notation may be written

as follows:

:t\1J {y~ + [C] {]i} + (k} {y) = {F(I)}

(18.32)

Framed Structures Mode!ed as Discreto Multldegree-of-Freedom Systems

526

i)'1oamic Analysis of Trusses

y

in which (y}, Ii}, anc {y} are, respectively, the dispiacement, velocity, and acceleration vectors at the noeal coordinates, {F(t)} is [he vector of external are the system maSs, damping, and stiffness nodal forces, ane {Jf], [(:], and matrices. In the stiffness method of 2nalysis, the system mmrices in eq. (18.32) are obtained by appropriate superposition of [he corresponding member matrices using the direct method 2.S we have shown previously for the framed structures. As was discussed in the preced:ng chap(ers, the practical way of evaluating damping IS to prescribe damping ratios relative to the critical damping fOf each :node. Consequently, when eq. (18.32) is solved using the modal superposition method, the specified modal damping ratios are introduced directlY into the modal equutiOf'.S. In this case, there is no need for explicitly obtain:ng the system damping matrix (C). However, th~s matrix is required when the solution of eq, (} 832) is sought by other methods of solution, such as the step~by~step integration merhod. In this case, (he system damping marrix [C} can be obtained from rhe specified modal damping ratios by any of (he merhods discussed in Chapter t2.

t

[R1

f-,---lOO in,~----l

T

100 in.

50in'---r

1S.6 PROGRAM is-MODELING STRUCTURES AS SPACE TRUSSES Program 18 calculates the stiffness and rhe mass marrices for the space truss anG stores the coefficients in a file. Since the stiffness and mass matrices are symmetric, only the upper triangular portion of these matrices neecs to be stored. The prograr:t also stores in another file, which is named by the user. the gene:al information on the space truss. The infom1ation stored in these tiles is needed by programs. which the user may caU, to perform dynamic analysis of the truss, such as determination of nurural frequencies Of calculation of the response of the Structure 5ubjecred to external excitation.

100 to.

z Fig. 18.5 Space

[111S$

x-ceOR!:>.

Example 18.5. Detennine the stiffness nod mass mauices for the space trusS shown in Fig. 18.5. The mass per unir length of any member is iil 0.1 (lb· sec 2fin 1). Also determine the first three natural frequencies of the truss.

so. cc

for Example (8.5_

Z·CCQ;(lJ.

O.

cc

100.CO

leo .00

.oc

100.\10

:00.00 iLOO

oJ.00

000

C. (')0

100.00

l::~

JO!NT

~";$$

0.00

SolutiDn'

Input Data and Output Results

EreCt;:> JOiXl'

1'.l..$Sf:'ZM.)1'1I ;; ~()()

.

:c.co

o .H)e

10.00 10.0" 10.00

",.s

C. !'1)0

,a.ma!:'R 01" a!t)..l1 £L£Y<:N'!'S

Ill:",?

o _~OO

:-'\J~ar;R

l'-'C,,::

0.100 (\ .100

OF FI;('ZO

Co;n~:;rw;':,,$

07 !!:L,>,ST::CCfY

e",H:~?

10. JO

'J .100 ;'
~ODOLIJS

SEC'!'. AAi!.A

lO. :Ie Hl.OO

527

Dynamic Ana!ys:s of TrU$ses

Framed Structures Modeled as Discrete Multidegree-of·Freedcm Systems

528

,

t!:
orR.;:::,:CN

~

:1,,,1.

'{~1,

,

::,,~:

1

~I .. '

OL'1'l?1J, Y£SU:"7S'

.:

529

II ;:.:v ~l

" j,,*J:.

.4t,;~!>0S

l:l.3$S,"r:~(}"~

, - 24' i ,5::;OS1£:+0:

:).c;WOS-.JO

::;.CO%:;:'::0

(;.

(j:)OC£~C0

",

'-"

-.>11--'- - 24' --..1'f-·~-24'

~I'

I

\\,1l

.-.-.~

24' ~

.."

'1'" , ! i

""

,

seQi'<

Sll.'E'&

I

I

i

'&01) i

3&00

Zl\lflG

,

•

'

,

,

\

<

' 's

l. 7S

2.25

d

(bi

18.7

DYNAMIC ANALYSIS OF TRUSSES USING COSMOS Fig. Ul6 (8) Plane :russ and (b) load for Example 18.6.

Example 18.6. The steel plane truss shown in Fig. 18.6(a) is subjected to the impulsive force depicted in Fig. lS.6(b) applied in the vertical ctitection at joint 3. Determine the first five natural frequencies and the maximum member 3 ~forces. Cross-sectional area A = 6 in 2 ; mass density /).;:: 0.3 (lb, sec?!in fin ): 6 modulus of elasticity E 30 X 10 psi (for all members). Solution,- The following commands are irnpiemented 10 COSMOS to model, extract natural frequencies, and determine the response for L1lS plane truss: (I)

DISPLAY ;. VIEW_?AR

0, 0,

1, 0

GEOEETRY ?~AKE.

>

0:

GR!::; > PLANE

Z, 0, 1-

(3) Define lines along the members of the truss: GSOM2:TRY

:>

C:URV.sS

>

C??CORD

CRPCOR:;, L 0,0, 0, 288,0 , 0, 576, 0,0, 864, 0,::;,1152, C, 0, :152, 0,0,

Set view to the XY plane: 'v"IEw,

(2) Define the XY plane at Z

CRPC:)?D, 5,288,0,0,288,384,0,288, 38'L 0,

V! Ev·J

Ci:\?COR:J,5,576,C,O,S76,38:J:,0,575,384,D, C?\PCC?D, /,864,0,0,864,384,0,85'1,384,0,

530

Framed Structures Modeled as Oiscrete MU!lIdegr88-of-F reedom Systems

CRLINE, 8,7,.4, CRi?CORD, '3,0, C, 0, 288,384,0,570. 0,0, 86-'L 384, 0.1152, 0, 0,1152, C, 0, CRPCORD,:3,288,384,O,576r3S4,0.864,384,:),864,38~,O,

(4) Defir.e element group 1 using TRUS2D:

?RO?SE?S EGROJ?,

TRU5S20,

0,

0,

0,

0,

0,

0,

0

Set optiOns . for frequency calculation to ex·r,ic'. u _ l'2 natural freque!)cies: F-..!"1ALYSrS ;. PREQ/8UCK "f'___ FREQ'JENCY A_FRSQUENCY, 12, $, H;, 0, 0, 0, 0,

0,

1,

(14) Run frequency caicuiarion:

1,

1. 2,

6.

°

FRE:Q:.!£NCY

Fi<.EQUENCY

fRE:Jl!ENC'!

?~!UOD

NUMB2R

{R?.D/SC:C)

{CYCLEs/sSe)

(SECONCS}

O.48C7G03F.:-Ol

O.7E50583E+OO ~ .11200962 ... 01 o .193:)B5H>Ol 0.2471774E+01 C,11661}lE+01 0.3927617E+01 o .Ii.0120:)]E:+Ol

0.1307090£+01

{7) Generate a mesh of one element along each c:.Jrve: ?ARA..Y._MESH 14, :, 2, L

>

M_CR

(8) Merge and compress nodes; MESHING ~

~fEKGE,

NODES

L

~E:SEING >

:>-

J:lft-tERGE

20, I, O.08CL 0, 0,

°

)lCOES. NCOHPRESS

~COMPRESS,

I.

14,

(9) Apply constraints in rhe X, Y, Z axes at node and In the Y,. Z axes at node 5. Also apply constraints for rotations about X, y, Z axes for aU (he nodes of truss; LO":;'0S~BC } STEvCTOR~L

:>-

2

).703777~2tOl

J

0.12131-392 ... (;2 0.15530622 ... 02

5

0.19893392+02

6

O.2~5779SE-02

7

o

2520S16E+02

o. 4517391E+~1

~ND.2zUZ,8,4,l,~X,Ry,RZ

DND,6,l'Z,Q,g,:,RX,RY,RZ

o

Q.1878521E+00

11

5323336E:+:Jl 0.615933)£+01

12

O.406~_S2E+02

0.6463572£+01

0.154.7132£>.. 00

Q.5L193)72+01

ANALYSIS PDJ\TYPE,

j

PO$t_DYN 2, 12, 50,

>

C 195B78E,OO

Q,:62J552E,OO

PC_ATY?.2

0,

C.05, 0, 0 " 5

EGROUP

10. lO, 0, 0, 0, 0, 0

(12) Mesh concentrat PT

0 ,~, 2~ 0

. C~) Define forcing functlon and app Iy It as a forCe at node 3 dJrectlOn: in the Y

(11) Define real constraints for- group 2:

P'I'l,3,1

0,2213667E-;.CO

(16) Defi:le dynamic analysis:

EGROUP, 2, H..J..SS, C,O,C,C,O,O,C,

PROPSE':'S :> RCONST RCC)JST, 2, 2, L 7,

O,4345677E+OO 0,3158,29E+00 0.25460738 ... 00 C ,24925212+0C

8.28383612 ... :)2 Cl 3216574£+02 0.3344751£+02 0,3870023:::+:12

8 9 18

(lO) Detlne group 2 using concentrated mass element: >

0.89278072700 0.5:790652+00

DISPLM.'\!'r

:)ND, L UY , 0" 1 t 1 t GX, UZ, RX, RY, RZ DNO,5,UY,O,5,:,I)Z,RX,RY,RZ

PROPS2rs

0,

RESU':"'!-S

PROPSSTS ) RCONST

>

lE-OJ6,

0, 0

(15) List nataral frequencies:

(6) Define real constants;

HESHING t-LC~, L

0,

~tPRO!?

>

M?KOP, :, EX, 30B6, DE~S, 0.3

RCONST,

531

.'1.NALYSIS ~ ?REQ/aUCK >a_?RsQuENCY ZLFREQYJENCY

(5) Denne material propeI1ies: PROPSETS

( l,C) .-

lE'006,

£GROCP

~

1,

DYl"am:c Ana!ysls af Trusses

CONTRCL ACTSE~,

AC?iVE ~ 1'..C':'S2'1'

TC. 1

LO}I..DS_BC > STRVCTURP.L ) F'ORCSS, FN.J, 3, FY, :., 3. 1

FND

Dynamic Analysis of Trusses

Framed Structures Modeled as Discrete Mu!\idegree-of-Freedom Systems

532

TABLE 18.2 Stresses at Step No. 25 Corresponding to Time t= 1.25 sec lor Example 18.6

(18) Set options for printing and plotting results: _~~ALYSIS

POST_DYN

>

PD_PRINT, Al"i.;.LYSIS

0,

1,

0,

POST_LlYN

>

PD_?LOT, I, 50,

OUTPUT

;>

0,

0,

;>

1,

PD_PRINT

;>

50,

1,

OUTPUT

o

1, 0

POST_LlYN

;>

PD_NRESF, 1, 3,

2NSRGY Cl::::-;"S:TY

.1678~2:::·05

0.2797372:+04

0.2254S+0]

.1599J~":+05

o.

26;;556::'~O"

0.20~5S+03

0.95959lE-O';

G.735?S~02

.~253:::-~·:'

;;2582E>O~

O.3:4.3S ... 02

.181.9::-0"

G.

Al"\)ALVSIS

5,;\255

FORCS

1

Pu_PLOT

;>

533

O.25407?~>04

0

-~

o. 3064,}OS.G~

(19) Merge nodes:

~g0'?62=>03

0.58%;:;.0'-

.2939:::-02

0.1257:; ... 01

0.5458S·03

.5:'G8~7S.03

r;.922.3S-04

-O.~q5345:::+03 -O.2G360j::~0~

HESEING ;> NODES ) Nl1ERGE Nl1ERGE, 1, 20, 1, 0.0001,

O.58:8;:;+G~

-0.2088862+0';

0,

0,

0

.2020::-07.

:J.1554S+02

265):::-0:

(20) Perform dynamic analysis: .Zl.NALYSIS

-0.

270~25:::+0!'

_.J

4S6995s .. ~~

5850S ... 0l Q

.31:36S-02

.1897::: ... 02

POST_LlYN >R_DYNi\.,.'1IC

;>

R_DYN,?.}lIC

(21) Calculate forces and stresses at the nodes of the beam elements: ~~ALYSIS

;>

STATIC

>

, .1

R_STRESS

G.~8

R_STRESS

~U16

G.G4

Output Results The output of the program COSMOS includes the displacements of the joints or nodes of the truss and the stresses in the members determined at each time

step, at intervals Llt = 0.05 sec, for all 50 step requested. However, to save space, values for displacements and stresses are reproduced, respectively, in Tables 18.1 and 18.2 only for step No. 25 corresponding to the time t= 1.25 sec. A plot of the displacemem response in the Y direction at node 3 where the excitation is applied is shown in Fig. 18.7 as a function of time. The deformed shape of the truss at step No. 25 corresponding to time t = 1.25 sec is shown in

R E L

," I ,

17 V

/l / i

-G.G2

~ ,

I

I i ,i

1

-Ill. ill 8 -(L!

,

ill .25

I

i

i

,

\! ~

1\

i

1

1\

,I

1 \

.. ,

i i

i i

i

,

1 1

I \,,

-al.~4

-(LG6

i

I

,,

/" ~

!

/

1/ 1

II /1

I

I

}

i\ i i \ i

I

1

I

\1

1

1

:

i

G.IS

I

\1

I

~

1

i

1 1

1

I

/

"1,

I., 1.25

1. 75

2

, i

2. , 2.25

TIME

Fig. 18.8. Fig. 18.7 Response displacemem in the Y direction at node 3 for Example 18.6.

TABLE 18.1

Displacement at step No. 25 Corresponding to Time t=1.25 sec for Example 18.6

"

NODS

"

"

,,1

"

,,3

o.ooooO::' .. OQ

00000:0:+00

0.00000.:+00

00000£+00

o .ooooos .. oo

c.ooooo£.oc-

0.00000:: ... 00

00000=: .. 00

0.00000::+00

o. oooon:o>oO

0.00000::: .. 00

268SS::-0l

6021U'>Ol

00000:;: .. 00

S244~:;:-Ol

56,99£-0~

00000£+00

00000S+00

5U38S-0:

0.000002 ... 00

0.00000£+00

00000£+00

0.00000:::+00

57798£-01

0.00000::: .. 00

00000::: .. 00

00000.:: ... 00

ooooos+oo

0.00000:0:+00

0 }"'827:::-02

o. 6~00S::-01

5~79lE-Ol

0.00000:::"'00

0.00000::' .. 00

00000:: .. 00

0.000.00::·00

000008+00

O.OOOOOE+OO

0.0000C::;:+00

0.00000::: .. 00

OOOOO:::~OO

ocooo~>co

0.00000:: .. 00

O.OOO·JO::'·OO

,.

0.59679::'-0:

5S815E-Ol

O.51sa7S-0.l.

.4~700E-Ol

O.

Example 18.7. Use program COSMOS to obtain the natural frequencies for the three-dimensional of Example 18.5. Solution: COSMOS:

The following commands are implemented in the program

(1) Set view to the XY plane: DISPLAY > VIEW_~AR VIEW, 0, 0, 1, 0

>

VIEW

534

Dynam;c Analysis

Framed Structures Modeled as Oiscre:e Multidegree-of-Freedom SysterP-li

PON S':ap :15

0:

Trusses

535

(7) Define real constants;

:;:'1.25

?ROPSE'TS ~ RCONST, 1,

8.CO~ST

1,

1,

2,

:'0,

0

(8) Generate a mesh of one elemenr along defined Hnes: t,:ESEING ). P.;\,RN·\_H2SH H<_CR r.!_C8., 1, 7, ::, 2, 1, 1

(9) Merge and compress :lades: t-1ESHING ) KODES

NMERGE,

GRL,) ) 0, 1

Z,

POINT'S

0, 1:)0,

,

>

t:-U$S

elements:

eRLINE, 1, CRLI::-J2, 2, eRLINE:. 3, eRLINE, 4, CRLIt\E, 5, eRLINE, D, CRLn~E, 7,

ANALYSIS

,

,

5,

0,

A_?REQUENCY 0, 0, 0, 1E-O:)6,

0

1;:;-006,

C,

AN,!..LYSIS > FP.EQ/BUCK >R_FREQUENCY R_FREQLJENCY

(13) List (he natural frequencies:

element TRUSS3D: 0,

(6) Define material properties {E

0, 0, 0,

= 306 psi,

p;{.OPSETS ) MPRO?

30£6,

),

(12) Run frequency "nalysis:

CRLINE

EGROUP

LEX,

fREQ/3l!CK ;.

0, 1, 0

3, 5

EGROl1P, 1. 'rRl1SS3~,

MPROP,

;

A_FREQUENCY,

(5) Define eler:1em group PROPSE'T'S

0

{I !) Sec options for frequency calcu Imion to extract three natural frequencies using ~he Subspace Iteration Method with a maximum of 16 iteratIons and the use of consistent mass fonnula[ion:

?T

:)

CURveS 2, 1 J 2, 3 2, 5 3, 4 2, 4

0,

1, ;o.I,L, 0, 2, "'" DND, 4. ALL, 0, 5, 1

(4) Generate lines between key points: GEO:1E'TRY

0,

DND,

P?, L 50, 0, Cl PT, 2, 0, 100, 100 !?T, 3, 100, 100, :..00 ?'r, lCiC, 100, :) PT,

0.08(;1,

PLA!\'E

(3) Define key points at the nOGes of the GEONE'1'RY ;,

) NMERG£

1,

(0) Apply constrat:1ts in ali degrees of freedom at all [he nodes except at node 3:

(2) Derine the XY piane at Z = 0: PLANE,

2:.L

MES;-.!:NG ;, NCDES ) KCOHPRESS NCOM?RESS, 1, 10

Fig. IS.8 Deformed tru;i$ of Example 13.6 at time t.: : .2.5 SeC.

GEOMETRY )

L

D,SNS,

0.01

0,

0, 0 2

8 = 0.01 (tb· sec lin4):

FREQCENCY

FREQU2NCY

FREQUENCY

PSrHOD

:.rJMBER

(R2\-D/SEC)

(CYCLES/SEC)

[SECONDS}

1

2 .06357E+02 4. 34476E+01

3.284282.01

2 3

6 .217!'3E:'.. 02

9.89487E,Ol

3 ,04481£-02 1 .44615E-32 1 01062£-02

6. 9 J.. 49 OE:'i-O 1

The natural frequencies caiculaled by COSMOS are, as expected, equal :hose given by [he educational program in Example 18,5,

(0

536

Dynamic Ana!ysis of Trusses

Framed Structures Modeled as Discrete Multidegree-of-Freedom Systems 4

18.8 SUMMARY

Concentrated

ma~

10K ~:?/jn.

The dynamic analysis of trusses by the stiffness matrix method was presented in this chapter. As in the case of framed structures, discussed in the preceding chapters, the stiffness and ma<;s. l:1atnces fo~ a member of a truss were developed. T~e system matrices for a truss are assembled as ex?lained for framed structures by the appropriate superposition of the coefficients in the matrices of the eleoents.

T ~~~I

PROBLEMS

ConC1!ntrated 10 K se2/in.

18,1

537

For ~he plane truss shown in Fig. PIE.l detennine (he syster:l stiffness and mass matrices corresponding to the two nodal coordinates indicated jn the figure.

---... Flll)

fl'llti

=

1

E= 10 4 ksi iii =- 0,$ K sec 2fjn. 2 A'" 101n. 2

, t

Fig. P18.1. 18.2 18.3

18.4

18.5

18,6

18.7

Determine the natural frequencies and corresponding nonnal modes for t!'le L'Uss shown in Fig. P18. lDe(ermine the response of t:'e truss shown in Fig. PtS.l when sub~ected to a force F(t) 10 Kip suddenly applied for 2 sec at nodal coordinate!' Use the results of Problem l8.2 to obtain the modal equations. Neglect damping in the system, Solve Problem; 8.3 assuming W% damping in all the modes. Detennine the maximum response of ihe truss shown in Fig. PI~U when subjected to a rectangular pulse of magnitude F Ij = 10 Kip and duration td"'" 0.1 sec. Use [he appropriate response spectrum to detcmrine the r.1a:dmum modal response (Fig. 4.4). Neglect damping in the system, Determine the dynamic tesponse of the frame shown in Fig. P1S.! when subjected to a hanno:1ic force F(i) = 10 sin lOr (Kips) along nodal coordinate L Neglect damping in the system. Repeat Probler:l 18.6 assuming that the damping in [he system is proportional to the stiffness, (C]=aa[K] whereac=O.l.

18.8

Determine the response of the truss shown in Fig. P Uta when f!.cted upon by the fOlees F; (I) = lOt and F, (I) '"" 5t~ during I sec. Neglect damping.

18.9

Solve ProbJem 18.8 aS5umir;g iO% modal damping in all the modes.

Dynamic Analysis or' Sttuch..lteS Using the Finhe Element Method

539

19 Dynamic Analysis of Structures Using the Finite Element Method

In the preceding chapters, we considered [he dynamic analysis of structures modeled as beams, frames, or trusses. The elemencs of all these types of strucLUres are c.escribed by a single coordinate along their longitudinal axis; that is, h'1ese are structures with unidirectional elements, calied "skeletal Struc-

tures." In general, these skeletal Structures consist of individJal members or elemems connected at points designated as "nodal points" or "joints." For these types of structures, the behavior of each eiement is first cOl1sldered independently through ;:he calculation of the element sliifness matrix and the element masS 17'..arrix. These marrices are then assembled into the sysrem stiffness marrix and the sysIem mass matrix in ~l!ch a way that the equillbrium of forces and the comparibility of displacements are smisfied ar each noda~ point, The analysis of such structures is commonlY knowr. as the Mcurb: Structura.l Method and could be appJied equally [0 static and dynamic problems. T~e s(rucmres presented in this chapter are continuous structures which Ure convenlen~:y idealized as consisting of rwo-dimensio:1al elements connected only at the seiecLed nodal poir;ts. For example, Fig. 19.1 shows a th~n plate idealized with plane triangular elements. The static or dynamic analYSIS of such idealized structures is known as the Flnite Element Method (FEM). This :s a powerful method for the analysis of structures including those with complex geometrical confIgurations. material properties, or loading conditions, This method ~.s entirely analogous to Matox Structural Ana!ysis for skeletal 538

Fig, 19_1 Finite dem~n( modelir:g of ;) thin plate with [riangular e!emems.

structures (beams, frnmes, and trusses) presen~eC. in the preceding chaplers. The Finite Elemem Method differs from the Matrix S~ructural Mer.Jod only in two respects: (I) (he selection of elements and nodal points is not nmuraHy Or c:early established by the geomerry as it is for skeletal strucrures and (2) the displacernems a: interior points of an element are expressed by approximate inrerpotariflg juncrions and :'lot by an exaCt analytical relationship as it is in the Matrix Srrucmral Method" F:Jrthermore. for skdetaJ structures, (he displacement of an interior point of an e!emem ;s governed by an ordinary differential equation and for a continuous two~dimensional element (he displacement IS governed by a partial differential equation of much greater comp!exity"

19,1

PLANE ELASTICITY PROBLEMS

Plane elasticity prOblems refer ro plates that are Ioaded in thet; own Dlanes_ Out-oF-plane displacements are induced when plates are loaded by ~onnaJ forces that are perpendicular to the plane of the plate; such problems are generally referred to as plate bending. (Piate bending is considered in Section 19.2.) There are two differem types of plane elasticity problems: (1) plane stress and (2) plane srrain. In the plane stress problems, the plate is lhin relative ro [he other dimensions and the stresses r.onnal to the plane of the plate are nm considered, Figure 19.2 shows a perforated strip-plate in tension as an examp:e of a plane stress problem. For plane strain problems. the strain

--.~

0

-~f-..

r

Fig. 19.2 Perfo,ated plate t~nsio:l e!cmer.t a::; an example of a slnlctural member loaded in plane stress.

540

Framed Structures MOdeled as Discrete

Mumdegree~of~Freedom

Systems

.i Dynamic Analysis of Structures Usjng the Finite Element MethOd

541

p q 6,

6

y !

p q 5,

5

p q 3,

(j)-pq 1,

3

1

~g. 19.4 Triangular plate clcr:1em showing nodal forces (l)Spiace::1cnts qi at the three nodes.

,0,

and correspondir.g nodal

fig. 19,3 Retaining waH showing a plate sliee as an example of plane strain conditions.

p q 6,

normal to the plane of loading is suppressed and assumed to be zero, Figure 19.3 shows a transverse slke of a retaining wall as an example of a plane strain problem. In the analysis of plane elasticity problems, the continuous plate is idealized as finite eler.1ents interconnected at their nodal points. The displacements at these nodal points are the basic unknowns as are the displacements at the joints in the analysis of beams, frames, or trusses. Consequently, the first step in the apphcation of the FEM is to model the continuous system into discre:e elements. The most common geometric dements used for plane elasticity probIems are trianguJar, rectangular, or quadrilateral, although other geometrical

6

y

p q

1

5,

p q

,

4,

3,

p q 1,

Triangular Plate Element for Plane Elasticity Problems

The foHowing steps are used in the application of the FEM for the analysis of structu:al problems:

Step I: Modeling the structure Figure 19.4 shows a triangular plate element with nodal forces {P}~ and corresponding nodal displacements {q} with components in the x anci y djrections at the three nodes of this element. In dynamic analysis, the plate element t'

4

p q

shapes could be used.

19.1.1

5

' - -..

3

1

~-~~-

----_X

Fig. 19.5 Tr.angular plate element showing the external :orces.

15 assumed to be loaded by external forces distributed in the plane of the plate (Fig. 19.5). These external forces are: (1) body forces with components of forces per unit of volume in the x and y directions, hx and by, conveniemly an'anged ill the vector {b}.. and (2) inertial forces, p {q}, per unit of volume in which p is the mass density of the piate and {q} is the acceleration of a

:: 542

Ffar:1~d Structures Moceled as

Discre(e Multidegree-of-Freedom Systems

Dynamic Aoejysis of Structures Using

point in the plate element with components til; and ti.y along rhe coordinates x and y and du is a differential element of volume, In addition to [hese lWO rype of forces, other external forces :rH1Y also be considered in the analysis. In a plune elasticity problem, rhe triangular element with three nodes has

two degrees of freedom at each node, as shown in Fig. 19.5. resulting in a total of six degrees of freedom. Thus, the nodal displacement vector {qL and con"esponding nodal force vector {P},,, fOf the plune rr:<:::ngulur element have six terms. These vectOrs may be written as: r

'1,

c]-;.

{q},.

q) q) Cfs eft,

,

fP},J

[

P, P, P,

p)

I P,P,

f

1

C2_t'

and

(195) The subsequent introduction of (e) from eq. (19.5) into eq. (l93) gives ;he

d:splacemenrs at any inrerior point of {he element in ierms of the nodal displacements (q!.- as , [u(x,y)] , . _, , {q(x'YJl=\ f=[g,x,Y)llA] '{qj, v (x,y»)

j

{ q(x,y)) =

Srep If: Selection of a suilClble displacement jllncrion The displacements, u = U (x,y) and v ~ U (x,y), respectiveiy in [he x and y directions at any interior point P(x,y) of [he triangular element, are expressed approximately by polynomial functions with a total of six coefficients equal [Q the number of possible nodal displacements. in this case, the s~mplest expressions for the displacement functions, u(x,y) and u (x,y), at an imeri.or pOlnt of (he triangular element tue:

+

+ C)y (19.2)

= [g(x, y)] (e)

d

The evaluation of eq. (193) for the displacements of the three nodal joints of the triangular element followed by [~e soiUlIon of the coefficients Ci (i I, 2 6) results in

=

{q},; [A]{c}

([9/1)

~~

!

[f(x,y)] {q},

tx,y)}

(19.7)

. Srep IV: Relationship benveen srrain,

€

{(x,y)} at any paim within the

:lemen: to [he dispiacemems {q (x,y)} and hence to the nodal displacemenrs ,q). , In th~ Theory of Elasticity (S. Timoshenko and 1. K Goodier, ! 970), it is s:Jow~ tna.t the linear stram vector, {€(x,y)}, with axial components along (he x, y direclions, c<, cy and shearing s[rain )'.:" is given in terms of denvacives of the displacement functions:

au

(l93)

pOint ill the element are expressed in

(II (x,y)!

i

dlsplacemel~:S.

{.(x,y)}

where {c} IS a vecwr containing (:--te six coefficients Cj, [g(x,y)] a mtHrix function of 6e position coordina~es (.l,y), and {q(x,y)} (l vector with the disptacement components u(x,y) and V (x,Y) at an interior POInt along /: and y directions, respectiVely.

Srep Ill.· Displacemerus {q(x,y)} at terms of the nodal displacemems {~, 1~

(l9,6)

where [he matrix [f(x, y)] = ~g (c,y)] rAJ - I is oniy a fU:lCIion of the coordinates x, y of a poi:lt :n rhe ~riangular elemem. Because the displacement functions u(_: .. y~ and v (x,y) in eq. (19.7) are borh linear in x and y, displacement conti~ n~HY !s ensured along the interface adjoining elements for any value of :1odal

Or (q(x,y)}

543

or

relating nodal forces and nodal. displacements ar.d the element mass matrix [Nl1~ relating nodal forces and nodal acceienuions, are of dimension 6 X 6,

(:1

Finite Element Method

(I9.l)

Therefore, for (his plane triangular element, [he element stiffness marrix (1\1 ..

u (x,y) =

tr,e

r

= :E,

~

oy

l U

l

ax

au

~y,,-.

in which tt == u (x,y) and u = and y direction, respec~:vely.

i

au

au ax

oy+ -

(x,y) refer

IO

(19.8)

J

displacement functions in the x

T~e sttain comp~ne~(s may be expressed i:l terms of the nodal displace:nen.s tq} .. by subsmu~tng from eq (J9.7) (he deriv<1!;ves of u(x,y) and v (x,y) 10to eq. (19.8); r

{E~t,y)}

= [8] {q},

(19.9)

514

Dynamic Analysis of

Frame<:: Structures Modeted as Discrete Multdegree-of-Freedom Systems

in wh:ch the matrix [B) is solely a function of the coordinates (x,y) at an interior point of the element. Seep V: Relationship between internal srresses {a-(x,y)} to strains {€(x,y)} and hence to the nodal displacements {q,..} For plane elasticity problems, the relationship between the normal stresses a:n if? and shearing stress T(,. to the corresponding strains E,<. €,\" and y,." may be expressed, in general, as follows: (19.10)

Structures Using the Finite E!ement Method

Step VI.' Element stiffness and mass mmrices Use is made of the Pri:1c1?ie of Virtual Work to establish the expressions for the element stiffness matrix [K]( and the element mass matrix [M],," This principle states that for structures in dynamic equilibrium subjected to small, compatible virmal displacements, {8q}, the virtual work, OWE, of the external forces 1S equal to the virtual work of internal stresses. OlVf' that is (l9. :5)

In applying this principle to a finite element. we assume a vector of displacements &j {(x.y)). Hence, by eq. (19.7) [f(x,yJ:B{q}~

O{q(x,y)}

or in a short matrix notarion as: {cr(x,y)} =

IDJ {«x,y)}

(19.11)

The substitution in eq. (19.11) of {«x,y)} from eq. (19.9) gives the desired relationship between stresseS {cr(;c y)} at a point in the element and the displacements {q},. at the nodes as: {iT(X,y)} = [DJ [El {q),

(19.121

The terms dij of the matrix D in eq. (19.10) have different expressions for plane stress problem than for plane strain problems. These expressions as glven by the Theory of Elasticity are: For plane stress problems:

E

d" = d,,= I

vE

The change of the internal work over the volume of the element during the virtual displacement is then given by the product of the virtual strain and the stresses integrated over the volume of the element:

oW, Ir "

8{
"E

E

(19.14)

d yJ ;;;:-.2..-

in which E is the modulus of elasticity and

'V

r

is the Poisson's ratio.

J" 8{q}'p{q)dV+o{q};{Pje

(1919)

The second tenn of eq. (19,19) is negative because the inertial forces act in directions that are opposite to the positive sense of the accelerations. The substitution of eqs. (19.18) and (19.19) into eq. (19.15) results in

"

+ v)(l-

(19.18)

The external vlrtual work W! includes the work of the applied body forces {b}"dV; the work of inertial forces p{q}dV; and ~he work of the nodal forces P,(i= 1,2, .. 6). shown in Fig. 19.5. The total external vi:1:ual work is equal to the product of these forces times the corresponding virtual displacements integrated over the volume of the element, namely

J

EO dll=d,,= (1+")(1-2")

d"=d,, = (I

o{e)T{lT(x,y))dV

(

For plane strain problems:

(19.16)

(19.17)

oWe = J, o(q)'{b},dV(19.13)

virtua~

lising the strain-displacement relationships, eq. (19.9). we obtain

1E d--=--....·.... "" 2(l+v)

5..:15

r o{q)'{b}i dV - r 8{q)Tp(q}dV-co{q;'{P}, Jv

Jo

(19.20)

By substituting into eq. (19.20) O'(x,y) (Dl [EJ {q}, from eq 09.12), wd the (ranspose vectorS 6{q}T and 8{ef from eqs. (19.16) and (19.17) respectively, we obtain, after cancellation of the corr;mon factor () {q} T, the equation of motion for the element:

(l9.2!)

546

Dynarnic Analysis

Frameo Structures Modeled as Discrete Multideg:ee-of-F(eedom Systems

where [K], = [Mt

J,

[BJ'ID] [B] dV

r

{J"(I»),

547

coordinates {y}, including equivalent forces for the body forces and for tmy other forces distributed over (he structural element,

p[f(x,y)],'f(A,y)]dV

Slep VllJ: SD/WiD'! of (he differential equations of mOfion

(1923)

J"

and

ot S:n,c(ures Using the Finlte Eh~men! Method

L

[f(x.yll (b},dV

(19.24)

Matrix [K]", in eq. (\9.22) is the element stiffness matrix with the terms expressing the force for unil disptacemem at a nodal coordinate of the elemenL Equation (19.23) gives the consistent mass matrix for [he eiement with components expressing forces due to a unit value of the acceleration at a nodal coordlnate. Finally, [he vector {P}r contains the external forces applied to the r.odes of the element and the vector {P b }( in eq. 09.24) is the consistent nodal forces due to the body forces {b} r on {he element. The element sriffness matrix, [K]~, and the element mass matrix, lll1L. as well as [he equivalent veCtOr of the applied body forces, {Ph}'" may be readily

obtained explicitly from eqs. (19.22). (19.23). and (19.24) respectively, for the simple plane elasticity element However, compt.:.ter codes ate generally written

The so!urion of the system of differential equations of mo6on in [erms of [he nodal displacements {y} is usually obtained by the modal $uperposirior. method presented in Chapter 11. Damping in (he syscem can readily be included in the analysis by (he simple addition of the modal damping term to the modal equa[ion. For systems with nonlinear behavior, the modal superposition is TIm valid and [he solulion miJst be obtained using a numerical method such as the step-bys:ep integ;a:ion melhod presented for a single~degree-of~freedom system In Chapter 7 and ro: a multiple-degree-of freedom system in Chapter 20. Sfep jX: De(ermina.rion of nodal stresses The fi(m! step is (he calculation of srre.sses at the nodal poims. These stresses can be calculated from the element nodal displacemenLs {q}~ selected from [he system nodal displacements {y} al:eady delermined. The element nodal Slresses, {O'(xpYj)}" for node j of an element, are given from eq.( 19.12) by

( 1929)

in which (he matrices [Dt and fBl; are evaluated for the coordinates of the

1:0

node j of the element.

polynomials.

Example 19.1. Consider the steel deep beam of rectangular cross section shown in Fig. 19_6, fixed at the left end as a cantilever beam carrying a static

calculate these matrices by numerkai methods, particularly when the structure is modeled using more advanced elements developed from higher order

Step VII: Assemblage of the sysrem stiffness matrix [K], [he system rrlil.Ss malrix [M], and lhe system equivalent nodal force veClOr {Pl>}~ due to [he body

forces The system s(iffness matrix [K] and the system mass matrix l!Yl} are assembled from (he appropriate summations of Ihe corresponding element mat:ices by exacdy the same process used in the previous chapters to assemble these matrices for skele(al~type structures: the system force vector is assembled from the element equivalent nodal forces, Hence, we may symbolically write

[K]

I[K],

([9.25)

[Ail

IlM],

(19.26)

and ([9.27)

p= 10,000 #

T

~

....

/1 /1

/1

4 in

1

~

The system of differential equations of motion IS then given by

[M] {y}

+ (K}{y)

= {F}

6in

(19.28)

in which {F} is the system vector of the external forces at the sys(em nodal

H

0.5 in Fig. 19.6 Deep

c(lJ,titi!v~r

beum for Example \9.1.

Dynamic AnalYSIS of Structlires Jslng the Finite Element Method

Framed Stn.:ctures Modeled as Discrete Muttidegree-of~Freedom Systems

548

549

concentrated force of 10,000 (Jb) as shown. This simple example was chosen because an exact SOlution has been obtained by using the Theory of Elas~icity (Tlmoshenko and Goodier, 1970.) It is, therefore, possible to compare the finite element solution with the exact solution for different finite element modeling. Solution:

Figure 19.7 shows !:hree possible idealizations of this deep beam

with 48, 108, and 432 elements 1 using triangular plane elements. The fix.ed supported condition is implemented by specifying zero vertical, horizontaL and

rotarional displacements at an the nodes at the left end of the beam. Figure 19,8 shows the venicai deflection shapes at the center Ene of the beam. for the three idealizations of the beam that have been obtained by the use of the program COSMOS. This figure also shows the deflected shapes 0: the beam calculated using the Theory of Elasticity (Timoshenko and Goodier. 1970} and by using simple bending theory which ignores shear defomnHions. The following commands are implemented in the program COSMOS to perform the analysis of this deep beam using 48 triangular elements with three nodes: (,)

A. GEOMETRY (I) Set view to the XY plane: GEOHE:KT'::'

GRID PLANE )

>

P!:]V";E, 2,

0,

PLAJ.'\!E

1

(2) Define the XY plane at Z ~ 0: DISPLAY;. VIE/CPAf{ ,. VIEW

VIEW,

C,

0,

1, D

(3) Generate curve lines at the periphery of the plate: GEO~ETRY

>

CRPCORD,

L

CURVES 0,

>

C,

CRPCORD 0"

6,

C,

0,

6,

,;,

0,

0,

4,

v,

0,

0, 0

(4) Define the sutface of the plate: GBOMEi"f'R::- ) S?2CR,

L

SDR:?ACES ,. SF2CK 1,

3,

0

B, MATERIAL PROPERTIES (b)

(5) Define element group 1 using triangular elements:

Fig. 19.7 Idealizations used for Example 19.1. (a) 4S elements. (b) 1Q8 eiements, (c)

?RCPSETS ) EGROU? EGROUP>

1,

'TRIANG,

0,

~;

0,

0,

C,

C,

0

432 elements.

(6) Pick material (steel): The In:rQduclory v¢rsior. of COSMOS limited to 50 elements C:l.O be used for ,he benr:; modeled wi~h 48 elements: other larger versior.s of COSMOS must be used to obrain the sol:nion with J08 or 432 elements. I

PROPSE'I'S ;. PICK )·i1;'! PIC}Ct'lAT I 1, A_S':'E:E:::"',

P?S

Dynamic Analysis 01 Structures Using the Finite Element Method

Framed Structures Modeled as Ciscrete ML.J:idegree-of~F(eedom Systems

550

551

C MESHING

~VI/I/I/I/l/i/I//I/I/I//[7177)

)~~~/VI//I/i//I/l//I//I/I/I/I/

(8) Mesh the surface of the plate with six ele:-nents a!ong the X axis and four elements a~ong the Y ax.is:

'V Vi/VIL 1/1/1/1/1/1/ /1/ 1/l7l7r/i)

/1/1///1/1/1//./1//1/1/,) /1/1/' >* Vl/1/\7 / /\/1/./1/1/1/ /1/ 1717/1/ , Vi/V./I/I/I/V:l/I/I/ 1/ 1/./1) 1//1/ , VV /1/ / /1/1/ /1/1/1/1717VI/I//, >*VI/I//I/I/I/I/I// /Vl/V~I/I/ /1 >* VI/\/I/I/I)I/I//I/ :)1/1/1/ ,/1/1/ 1 }» VI/ 1/1/'/I/I/I/I/I/I//I/I/1//./ 7i /,171/1//1/1/1/1/1/1/1/1/1/1/1/ /1/.

MES}r;:)!G > PARAN_H£SH

M_St,

L

:-CSF

>

1. 3, .5 >

1,

,:;,

L

1

I

~/j//I~IAI)1//I/I/I/1/1/ ./1/1/\/17 I>';:

(9) Merge nodes: NNERCE,

1, 4.8,

1. 0"0::::01,

0,

0,

°

D, BOUNDARY CONDITIONS AND LOAD (lO) Apply constraints in ali degrees of freedom on the nodes of the left side of (he plate: LO?CS_BC

DND,

>

STRUCTUR..-\:"

1, AL,

0, 29,

~

DlSf'L!-'f.NTS )

DND

7

(It) Define a constant s:mlc force of - 10,000 (lb) in the Y direction

applied

(c)

a(

the mp-righ{ node of (he p:ate:

LOADS~BC

)

F:-;JD,

F'Y,

35,

S1'RDC'!'L";:i,';L > FORCES ;.

10000,

35,

F~·lD

1

E. RESPONSE

Fig. 19.7 (co:uinucd).

(12) Pertorm (he s:atic analysis: t. ...NALYSIS

>

ST.!..'T'IC

R~ST.s.TIC

R_S'IAT!C

F. POSTPROCESSING (13) Print a table of displacemem along the center line of '{he plate:

?(ssuurs ) LIS':' )0 DISL:::ST DISLISl', 1, :, IS, 21, 1, n ....~t

1" .... /..
~~~4-;' tk~"' .. I}

0

(fable 19.1 shows [he displacement values at the center line of the plate.)

1('11<1.,1<1:111\
TABLE 19.1

msplacements at the Nodes on the Center Line of the Plate for Example 19.1

~10D2

Fig. 19.8 Deflection of (he cantilever deep b
(7) Define real constants t?!ate thicku0SS, d::;; 0.5 (in)]: PROPSF.'IS > RC:)KST

RCONST, 1, 1, 1, 2. 0.5, 0

15 16 17 18 19 20 21

x-DISPL. (in:

Y-D::SPL, (in)

O,OO~OO£+CC

O.OOOOOE .. CO

4.,405}6E:-07 -:.2;:'342£-06 -5.Q334.5E>06 -1.01584E-(;5

-6.386JCE-05

-l.2B7lo£··C5 -4,325888-06

-3,54295E:-C'; -5,53207E-C4 -] .69722E-04 ~9. 7909JE~04

Dynamic Analysis of Structures Gsing the Ffr..lte E!ement Method

Framed Structures Modeled as Discrete Multidegree~of-Freedom Systems

552

l'klodlt Plane Quadrilateral

([4) Plot the defonned plaw: L

L

48, l, 53.2105

( 15) Superimpose the original rnesh of the plate to the deformed p;ot: M2:S:-!:::NG )

SPLO";::,

ELE}llS!~':' ~

Ei~r:'HM'\t

EP=.,Q?

NaMe; PLANE20

Noaes; 4

t:J \ \ 0

E"m,'" N,m. PLANEZ" Nooes:S

1. 48, :.

Element Nama: ?LAN!:.':.2D Nodsl>: 4

(16) Obtain hard copy of deformed plate: 1k10d9 Axh)'mmelrl<;

CO~!?ROL ~ CEVICES ) LASERJ!'::? ~ASER~ST,

3-node ?1a<'19 ir1an.gle

6-hod& PlanEt Trlangllt

Ouadrllakull!

ISQ, 0

(A plot of the deformed plate is: shown in

19.. 12

4-nooe Axisymmetric

Ouadrllataral

R3.SUL,TS ) PLOTS > DE??LOCr

DSFPL01,

553

\9.9)

Elem01\ Na:me:TRIA"IG Nodes: 3

Library of Plane Elasticity Elements (2D Elements)

'l"i computer procrams fo: the solution of problems using ~ . + I .-l Genera II v, comm erc (. Finite El~ment Method (FE:vi) provide several geometric types o~ e.emeDrs an!")

Elem ..m Nanc: TR1ANG Nodes: S

E!e:r.&nt Nar:1a: Pl.ANE2D

Nodn:9 •

Fig. 19.10 Plane elasticity elements available in progn:m COSMOS,

related number of nodes. The most common of these plane elasticity elements are triangular with three or six nodes or quadrjiateral etemer:rs wlth four or nodes. Figure 19.10 shows a pictorial representation of plane eJasticity elemen;: available in the program COSMOS. The development of the stiffness and mass :natrices as well as the vector of nodal forces for djfferenr types of elements can he obtained by mer.hods analogous to the appro2.ch presented in Section :9.1.1 for the simple triangular eJement. The following example uses the four-node quadrilateral element for the dynaT:1ic a:1alysis of a simple supponed deep-beam.

,. I

"

IJ.

i\

I

L::

/

I

,

~

/

I

Fig, 19.9 P!Q( of the deformed plate superimposed on the undeformec original mesh for the plate of Example 19. L

Example 19.2. A simple supported deep steel beam shown in Fig. 19.II(a) with a distributed mass of 0.03 (lb· see'lin') is loaded at its cen:er by the impulsive force depicted in Fig. 19.11(b}. Detennine the first five na~t:rai freql1encies and the b:1e-dispi2cemem response at :he center of the bea:n~ also determine the stress pattem in this structure. Use rectangular plane elements (PLANE2D with four nodes) available in COSMOS element library. Solution: Th:s structure is modeled with a mesh of 10 rectangular elements as shown in Fig, 19,12, The si:mpje supported condition at the ends of the beam is implemented in the program by requiring that the y components of the displacement at all the nodes on these ends be equal to zero.

Dynamic Analysis of Structures Using the Finite Element Method

Framed. $truclu.-es Modeled as Oiscrete Mullidegree-of-Freedom Systems

554

TABLE 19,2

F

First Five Natural Frequencies for the Deep Beam of Example 19.2

i ~r I

~.

______________________________

-;;:;0

555

I

~

;g

f.------

10 in (a) Q. :

F(#)

, "I

{LQS

t

f"..

{"~3

'~l!\

U

y

(in)

~

/

.In

.~

.!lEi

0.3

0.4

$,$1

,;

Fig, 19.11 (a) simple supported deep be:.>m, and (b) load for Example 19.2.

\ \ \

./

!LS2

(b)

\

I

IL~3

(,«)

"--"

../

Ii.H

0,2

\ ;

/

~.1$5

time

0.'

$

/

:

/

i S .t2~

""

~.G75

.:,

. S. : i'S

" .,

i\ i \ :

tL! 15

,:

~

., "

/""".

T1KE (sec)

,

Fig. 19.13 Tirr:e-·di$placemem function. for the vertical compon.ent at the center (r.ode 17) of the ceep beam of Exarr.ple 19.2.

!

'.~--I" .---,;,.'"-TT'i Sis..

l ..

~..

~

·t _l_,-i'

12

\'1 '

.b_......_____ 9 -

I"

..

Fig. 19.12 Simple s!.lpporred deep beam of Example 19.2 modeled with four-nodes pian.e rectangular eiements.

The first five natural frequencies calculated by COSMOS are listed in Table 19.2. The rime--disvlacement function for lhe center of [he beam (node 17) is show~ in Fig. 19.13. The contour plot of the stress distrjbution [hroughout the beam for step 15 corresponding to t:me 1 = 0.075 sec, which COSMOS pro-

step specified. The results presemed for (his example are only a portion of the solution contained in the output tile given by the program COSlvfOS. 2

19,2

PLATE BENDING

The application of the :mite element method is now considered for the analysis of plate bending. Plate bending refers [Q plates loaded by forces that are perpendicu;ar to the plane of [he plate. The presentation (hal foHows is based On tWO assumptions: (I) the thickness of the plate is assumed to be small

duces in color code, is given in Fig. 19.14.

The computer provides detailed results consisting o~· listings and piots of componems and resultant stresses at ttH;~ nodes of the elements for every time

, Data prepara:io{~ for lhis enmple aad for Other ilh.ts,riltive ~xamp!es
556

Framed Structures Modeled as Discrete \o1ulfcegree-of-Freedorn SYStems

Dynamic Analysis of Structures Usi:1g the Fi:"lite Elemcpt Method

von Mises

557

y

y

I @/ - - - - - - - - : ; ,(i)

z Fig. 19,14 Contour-stress plot for the deep beam of Example j9.2 for slep 15 corres· pordlng 10 time t 0.075 (sec).

compared to other dimensions of the plate and (2) the detlections of the loaded plate under the fond are assumed to be smaD relative to its thickness. These assumptions are not particular to the application of finite element method; they are also made in the classical theory of elasticity for tiending of thlo plates. These two assumptions are necessary because if the thickness of the plate is large, then the plate has to be analyzed as a three-dimensional problel.! and if the deflections are also then in-plane membrane forces are developed and should be accounted for in rhe analysis. The analysis of plates can be undertaken by the finite element method without applying these two assump~ tlons. The program COSMOS used In this book is genera; and may be used for the analysis of either thin plates undergoing small deflections or to [hick plates in which these two assumptions are not required. However. the presentation in this section is limited to thin plates that undergo smaH deflections.

19,2,1

Rectangular Finite Element for P'ate Bending

The same steps used for derivation presented for a triangular element SUbjected to in-plane loads in tbe preceding section can be used for the derivation of the stiffness and mass .r.1anices as well as the vector of equlvalent nodal forces for body forces, inertial forces, or any other forces dist:ibuted on the plate elements.

Step I: .\1odeUng the structure A suitable system of coordinates and node numbering is defined in FigJ9.15(a) with x,y ax.es along the continuous sides of the rectanguIar plate element, and the z axis normal to the plane of the plate, completing a righthand system of Cartesian coordinates. This rectangular element has three nodal

(a)

q',Pl-

1

q"p, Ff.g. 19.15 Rectangula.r plate bending element. {a) Coordinate system and node numbenng" (0) ~odal coordinates q, and corresponding nodal forces p,.

coordinates a[ each of its four nodes, a rotation about the x axis (O.r.), a rotation about the y axis (Oy), and a nonnal displacemenr (w) along the z axis transverse to u1e plane of the plate. These nodal displacements are sbown in their positive sense a~d iabeled qi (i = 1,2,.. 12) in 19.15(b). Corresponding to the three nodal displacements at each node, two moments and a force. which are labeled Pi, are also shown in this figure. These 12 element nodal displacements and 12 nodal forces are conveniently arranged in [wo vectors of 12 components. {q}" and {P}e> Therefore, the stiffness matrix and the mass matrix for this rectangular plate bendir:..g eiemem with four nodes are of d~mension 12 X l2. The angular displacements {)t and fj" at any point (x, y) of plate element are related to the nom:al displacement }9 by the foHowing expressions: (19.30)

Framed Structures J~v1adeled as Discrete Multlcegree~Qf~Freedom Systems

558

Dynamic Analysis of Structures Using the Finite Element Melhod

The positive directions of e~ and O,y are chosen to agree with me angular nodal 19.15(b). Therefore, <::tfrer a displacement qlo qz, q.h '11, etc, selecred in function, w = w (x, y), lS chosen for (he lateral displacement the angular displacerner.ts are derennined (hrough rhe relatIOns in eq. (19.30).

Step ll: SelecIlon 0/ a suilabie displacement /uncrlon Since the rectangular elemer.t in plate bending has 12 degrees of freedom, the polynomial expression chosen for the norrm:d dispiacements, Hi, must con* tain 12 constants. A suita~le polynomial funcrlon is given by

Sup til: Displacements {q (x,Y)} Gl a poim within fhe elemem are expressed in terms of lhe nodal ciispiacemenrs {q}. Wricing eqs. (19.31), (19.32), and (19.33) in matrix oO[atton, evaluating the displacements at the nodal coordinates, and solving for {he unknown constants results in (

1

8 ,

" !

{q (x,})} = t~:I; [g (x,y)]{c)

fA.i {e}

{q},

w (19.31)

ow

--=

oy

(! 9.32)

and

(19.34)

(19.35)

and (e); fAj -, {'ll,

The displacement function for the rotations 0., and Ov are then oi:Hained from eqs. (:9.30) and (19.31) as

559

(19.36)

where [AJ -, is ,:,e inverse of the matrix fAJ in eq. (19.35), [g (x,y)] is a function of the coord:nates x, y at a poim in the element, and {q}~ is the vector of the 12 displacements J.( the nodal coordinates of the ele:nent. The sub$.ti~u(iol1 of the vector of conSUlntS {c} from eq. (19.36) intO eq. (19.34) provides ,he required reia,iooship for displacements (q(x,y») 2r aoy interior point ir. [he reclat.gular eiemem ~nd the displacements {q}~ at the nodes: (19.37)

8y

ow = c" + 2c.~x + c:;.y + ax -

+ 2caxy + cui + 3C11

+ CI1/

(19.33)

By considering the displacements at the edge of one elemenr, that is, tn the bounc.a:-y between adjacent elemems., It may be demo!lstrated that {here is continui[y of normal latera! disphlcemems ().nd rotational displacement :n the direction 0: the boundary line. but not in the direc~ion transverse to tt:is line, a3 Ir is shown graphically- in 19.16. The displacement function in eq. 09.30 is called a ;'nonconforming function" because It does not satisfy [he concition of continuity at the boundaries between elements for all rhree displacements w, 8 and Or p

Deflected Elements

\ Slopes normal to edge are disconlint1ous Fi.g. 19.16 Detlecred coruinuous recrangular elements.

or using eq. (l9.30): UW

1

0"'" aw . ; U(x,y)J{q}.

a"

( 1938)

J'

IV

in which [j(x,y)] fB (x,y)] (A]-: is SOlely a function of the coordinares x, y at a pOint within [he element.

Srep IV: Relalionship herp/un :i!raills (..;(x,y) (jf any poin{ within rile el~ ement ro displacemenfS {q(x, v)} and hence!O rhe nodal displacemenrs {q},. For bending, Lhe state of strain at any pobt of the element may be represented by three components: (he curvature in the x direction, the curvature in the y direction. and a componem represeming tarsior. in the plate, The curvature in {he x direction is equal {Q the rate of chnr.ge of the slope in that direction, [hac is, to rhe derivarive of the slope,

c~{ aWi = dX\(!X/

(19.39)

560

.Jynamic Anaiys·is of

Framed Structures Modeled as Discrete Multidegree-of-Freedom Systems

Structures USing the Finite Element Method

Vy

SimUarly; the curvature in the y direction is

Ny

Vx

),

Myx

(19.40)

Nx

Finally. ihe torsionaj strain component is equal to the rate of change, with respect [0 }', of the slope in the x direcrion, that is (19.41) Mxy

The bendjng moments: M, and lvl,. and the torsionat moments ll{n and Iv1 each act on two opposites sides of the element, but since A{(> is :;1UmericaJl),: equal to M)~n one of these torsional moments. MxY' can be considered [0 act on all four sides of the element thus allowing for simply doubling the torsional strain component. Hence, the '·strain" vector, {e(x,y)}, for a plate bending element \'A

can be expressed by

{«x,y))

o.~:~

and moment corr.poncnts per unit length as defined by .

r

(19.42)

;)y'

M"

a1 w

,

The substitution in eq, (19.42) of the second derivatives obt2.ined by differen~ tiation of eq, (19,38) yieldS {«X,y)}

for~e

element of the plate bending, t~ese internal moments are shown jn Fig. 19.17. Tr.e mome:1[-curvature relationships obtained from plate bending th~ory (5. Tirr:oshenko and N. Goodier, 1970) are: '

t,tZw a=
=

Fig. 19.17 Direction of

COSMOSJ:vf for rhin shells.

= [BJ {q},

1'11,. =

(I9A5)

(l9A3)

in which [BJ is a function of the coordinates (x,y) only,

Step V: Relationship between internal stresses {cr (x, y)} and fmernai srrains {e(x,y)~ and hence nodal displacements {q:e In a plate bending, the internal "stresses" are expressed as bending and tWlsting moments. and the dstrains" are the curvatures and the twist. Thl1s, for plate bending, the state of Internal "stresses" can be represented by

rM ,1

{,,(x,y))

= \M,

~

(19M)

These rela~ionshlps are written in general for an orthotropic pi ate. i.e" a plate that has different eLastic properties in two perper.dicular directior.s, in which Dx and Dy are ftexural rigidities in the x and y directior:s. respectively' D· is a "'coupling" rigIdity coefficien~ representing the Poisson's ratio rype of effect: and D~} is tt.e torsional rigidity. For an isotropic plate that has the same properties in all directions, the t1exural and twisting rigidities are given "::::oy

D\.=D=-,~

,

lM",1 where Mr and M" are intemal bending marr-ents per unit of length and lvI.ty is the lr:temal twisting moment per 1I:1l1 of length, For a smaU rectangular

D,= vD

12(1

Dx:

v')

(1 - v)

~~D

2

(19.46)

552

Dynamic Analysis of Structures USing the flnite Element Method

Framed Structures Modeled as Discrete Mull:degree-of-Freedom Systems

in which the matrices [DJj and [B]) are evalu(!ted for the coordinates of node j of [he element.

Equa[ions (l9.45) may be written in marrix form as

{"()e,y)}

t,l [ ) , ~ ',~ ,tv[,

)\t[",-J

D, 0 D, D, 0 0

563

Example 19.3. A square steel pIme assumed to be fi.xed at [he supports on its four sides [Fig. 19.18(a)] is loaded by a uniform!y distributed dynamic force applied in the Z direction with the time variation shown in Fig. 19.i8(b). Use program COS'MOS [0 determine (n) the first 10 natLlrei frequencies, (b) the time-displacemem function a[ the center of the plate. (c) the !lme-s:ress

(19.47)

0 D.ty ";

or symbolically as {U(X,Y)} ~ [[)]{«x,y))

(19.48)

y

where {he marrix [D] is defined in eq. (19.47), The subsri[Utlon of {e(x,y)} from eq. (19.43) gives the required relationship between element stresses and r:.odaI displacements as {cr(x,y)}

[D)[B] {qL

1 ,i"

(19.49)

/

/

""

Step VI: Element stiffness and mass matrices The stiffness matrix and the mass matrix for an element of plate bending obtained by applying the principle of virtual work results in eqs. (19.12) and (19.23), and for the equivalent forces due to the applied body forces on the e!emeot, by eq, (19,24), The marrices (f(x,y)j. [BI. and ID] required in these equations are defined, respectively, in egs, (19.38), (19,43), and (19,47), The calcu:ation of these matrices 2nd a!so of the integral .. indicated in eqs. (19.22), (!9,23), and (19,24), is usually undertaken by numerical methods implemented in the coding of compuler programs.

Slep IX: DeterminctIion of nodal stresses The element nodal S[[esses, {u(Xj,Yj)}t> for node j of an element are giver, from eq, (19.49) as (19,50)

/

/

I I

,

,

Step VII: Assemblage 0/ the system stiffn.ess matrix (K], [he sysi€m mass matrix [M], and the vector of (he external forces {F} Q[ the system nodal coordinateS {y}, which includes the eqtlivalent forces for the body jorces {P b} and for any Olher forces distributed over the structwai element. The system stiffness and mass matrices, as well as the nodai force vector, due to appEed body forces, are assembled from the corresponding elemenr marrices and vector as indicared in eqs, (19,25), (19,26), and (19,27), Srep VllJ: Solution of the differential equafion of motion The system differential equation of mOtion is given by eq. (19.28) in which [,W] and [K] are the assembled ma.ss and stiffness matrices, respectively and {FJ is the system vector of the external fotces, The solution of eg. 09.18) thus provides the system vector of nodal displacemenrs {y}.

/

/

I I ,! I \ I I , I, I i '1 t ! , ,

(a)

P(lblin.2)

!

, i

"V\ 0.<

0.3

0.2

DA

!ime (s.::c)

(b)

Fig. 19.18 (a) Square plate for E.umpk 19.3 supponing a nonnal uniform pressure. (b) Time variation of the normal pre~.~ure.

se4

Dynamic Ana!ys's of Structures Using the Finite Element Method

Framed Structures Modeled as Discrete MuJtjdegree~of·Freedom Systems

50S

TABLE 19.3 Ten Lowest Frequencies for the Plate of Example 19.3 No,

Frequency (cycles Isec)

2

3.78674e+OQ U7632e+01

3

1.3835ge~OI

4

2.25472e.;.0 I

5 6

3.1917!e+01 3.1 9532e+0 I 3.93305e-01

7

3.95 i38e-O I

8 9

5.4 I448e+0 1 5.645280+01

10

",

,

S.[lSS I Ij.l a8;

Fig. 19.19 A c.uarter of the plate for Example 19.3 showing the pressure load and boundary conditions.

function for a point at the center of plate. a:1d (d) the contour stress plot at time I = 0.1 sec. Use time step ilt 0.005 sec.

IJ G.:.a73

Cal Table 19.3 shows the values obtained for the first 10 natural frequencies.

,

G.1l1ll2£

-G.~::;S6

, , , , ,

,'.'"

"

/':

1 fLr.£5, 5

-!L,.!8~

Solution: Due to the symmetry of geometry and loading of this plate, only a quarter of the plate needs to be modeled. The boundary cO!lditions along the symmetry lines requires that the slope transverse to these lines be set equal to zero. The quarter of the plate is modeled with ames" 6 x 6 SHELL4 eleme~ts (Fig. 19.19). Detailed explanation for data preparation and the necessary com· mands for analysis and listing or plotting 0: desired output are presented in the supplement "Structural Dynamics Using COSMOS,"3 Results:

y

ISPL· .

/1

/

\

\

/

\

V

\

\

-til.asc)?

Q.1!1295

, .'54

'.·os

, . ,.m

(\1.1..;7 ...

(I,

!7SS

G.;"

til" 2255

.. ,', \

Tl/it:

Fig. 19.20 Time--displacement (normal) at the center of the plate :or Example 19.),

(c) The time-stress function in the .'t d!rection for an element at the center of the plate j5 shown in Fig. 19.2]" Cd) The COntour-stress plot for a quarter of the plate, which the computer SCreen provides in color coding. is shO)vTI in black and white in Fig. 1922.

Cb) A plot of the time-displacement function for the point at the center of the plate is given in Fig. 19.20.

19.2.2 COSMOS Library of Plate and Shell Elements 3 A convenient form to order the sllpplement Structural Dynamics Using COSMOS is provided in the last page of this Yolume.

The program COSr,,10S provides a library of elements for thin or thick p]ar.es and for shell elements in addition to the rectangular thjn plate element.

555

Oynamic Anatysis of Sltuctl,.:res Using the Finile Element Melhod

F"mmed Structures MOdeled as Discrete Multldegree-of-freedom Systems

567

3-Node Thin Shell

D88fJ

l'

I

,,

,J'

S S

5

[9522 I 4 ]}6

.;I

9950

,~

J Je

I ,.;"

\,

•..... _-

!

I I

L."

i

,. ;03

0.1275

..

i

I ;"

_ ,.i"

E!emen! Name: SHE!..L3

i Eisman; Name; SHELL4

~es, 3:......~_ _~~_-'-i:.:N.:.o.:.d•.:.':.:'_<~_

,

i

-9:94

-13'3S<'.i

"\

L

b-"

5[64

L

J~.

""-"\

/

,4S>SB

5

r

'"1'l

y/'

I

29994

iN

1\---,

0.2255

Bement Name: ShELLS

..~.?
Fig. 19.23 Element;; ava!!::.b:e In fhe progr:'-.nt COSMOS for [hi:: plates und shells.

.-'5

of an element at the cen:er of the plate of Fig. 19 ..'d T·llne-::iU ess plo! [.-.~ "'·'.eC{'ion) "" Example i9.3.

E,,,merif Name: Sj-(E~L3r Noo\ls:3

E1J¥rom NarM: SHELl-tT Kodu: 4

III

ElomGfl! Name: SHELL9L ~II~:

illS

£lnm"llI Namo: SHEll:!L Noc!G$;3

I

..

_

. .- - - .

~--.-

8a.'TIQfl! Nama; SHELL4L Noo..$: 4-

Fig. 19.24 Ekme:nt$ avuilable in (he program COSMOS for thick plates and sheHs..

SHELL4, used in Example 19.3, Figures i9.23 and 19.24 show, respectively, me ty?e of thin and :hick shet: eieer.ems tiyallable in COSN!OS. Several lllustradve examples in strJcrural dynamics modeled by thin and thick shell elerr:ents are presented in derai: in th¢ su;::piement '''Structural Dynamics Using COSMOS.") Example 19.4. A caorilevered steel beam. 100 (in) lenglh with a semicircular cress section of radius !O (ic) and thickness l.0 (in) (Fig. ! 9.25) is 2 subjected to a uniform normal ;:ressure of to \lblin ) suddenty ap;::lied. Use the program COSMOS to derermine (a) [he first 10 natural frequencies, and (b) [he response in renns of displacements and stresses. FIg. 19.22 Con:our~uess plot (Von Mises stresses) at time 1=0.1 (sec) for of (he ylate of Example !9.3.

« quarter

Sollilion: COSMOS.

The fol,owing commucds are lrnplememed in me program

568

Fra(t1ed Structures Modeled as Disc;8!e Mu1tidegree·of~Freedom Systems

Dynamic Analysis of Structures Using the Rnil€'! Element Method

569

(6) Remove grid and repaint: ~EOM:STRY

)

202:::-:1"'r > GRIDO??

Grt::::::OFz.~

RE?AIN'T (7) Generate an arc passing thmugh the three key points defined. Pick the

end poin!.s as I and 3 with yoint 2 1n the midcle: GBC~B'?RY

;, CURVES

Ct',.:::'.RC3?'L

(8) View the

l,

l,

CIRC!...ZS

:>

3,

three~dimen5ional

C:::S?L;',Y

»

V::::EW, L

). C?-ARC3 PT

]

V::;:SS
geometry in perspective: Vlr:w

(9) Extrude the curves on the arc (1 and 2) in the Z direction by 100 units and generate the surfaces; Fig. 19.25 Sketch for the beam of E.>:ample 19.4.

GEOl>1ETRY > S:JRFACSS " SFGSKR " SFEXTR Sf2X:"FL 1. 2, 1, Z, 100

(10) Scale the new geometry:

A. GEO:vIETRY (I) Define the XY plane at Z ~ 0:

(Fig I 9.25 shows a sketch of the beam "ovith labels for the keypoints at the PLAt"iE,

Z,

0,

two end of the beam.)

1

B. MATERIAL PROPERTIES

(2) Set view to the XY plane:

(11) Define element group 1 with SHELLA elements: P?,C?SE'!'S BGROO?,

(3) Set the grid from the onglo of 20 X 20 d1\':8ioos witt increments of

one unit: SEO~E'rRY

GR:::DO:N,

0,

1, 1

20,

20, 2

SHSLL4,

(12) Pick the aHoy steel material from the library: (Its ?roperties will be taken from existing database,) 2~OFSST$

" GRID )- GRICCN 0,

). EGROUP J.,

PIC?C:-LZ\'!',

)

?ICK~~.T

1, A_STEEL,

FPS

(13) Assign thickness of 1.0 (in) for the shell model:

(4) Scale the currer.t view: DI$?L}I.Y

:.>

?ROP$ET$

DTSP~?AP. .,.

SCALE

(5) Define key points:

2'1.

P'T', PT,

.,. ;:O:NTS D 0 2, ::'0, C, 0 3, 20, ,0

1, D,

1.,

6,

l.O,

0

C. MESHING

seALS, 0

~EOMS'T:?Y

) RCCNT

RCONST, 1, if

1

PI'

(14) Mesh the surfaces (1 and 2) us~ng four node SHELL4 elements creating four ele:nents in the X direction and 20 elements in the Z dIrection for each surface: M:t::SHING ;. }'1~SF,

1,

PARllJC:'USSr: ) M.. _S?

2, 1, 4,

4,

2C,

1.

Dynam:c Araiysis of Structures Using the Finite Element Methcd

:=raf1leC 3!ructures Modeled as Discrete M;/tjdegre€·ol~FreedQm Systems

570

(15) Merge (he nodes of the twO surfaces together to create one continuous model: NOTE: If [he <'PIC" icon is activa:cd, press ESC to obtain [he prompts for beginning and ending node numbers. t>lESHING ) NODES > NY.£RGE NMERGE, 1, lOO, L 0.0001,

0,

OCR,

STR()Cr:JR,.t.L )

:, ALL,

0,

2,

DIS?LMN't'S )

PREQUE>JCY 1RJl..D I sse}

FREQUENCY :CYCLES ISEC)

1 2

C .255015:£ .. 02

5

0.16023078+03 0.2184975E,,03 0.5997459E-rO) 0.63606281::,..03 o .1106SGlF>Ot;

0

O.1173S24£~04

7 8 S

O.lJ90516£+04

0 95452528+02 0 , :012325£+03 0,17610512+03 0.1867722£+03 0,2213234£-..03

0.1390822£-<-04

G.22l)S61S+03

O.~649;J06Et04.

10

"},220924:0£ ... 04

0.26244742+03 O. 3516114£+03

PER roe

(S2CC~mS)

0.39213378-·01 0.28757648-01 C .104764:E··Cl 0.98782482-02 C .56784298-C2 0 " 5354117£-02 0 .45182752-02 0 .4517607E-02 0 38102878-02

O. 3477)372+02

.--""~--.-----.--

OCR

----

FREQUENCY N:);-'18ER

4

(16) Constrain all the nodes on (he back end of [he beam (all (he degrees of freedom) by issuing the DCR command for curveS I !ll'.d 2. This commiJnd will constrain ail [he nodes of these curves: >

Natural Frequencies fo( Examp!e 19.4

-~.--------~.--.

J

0, 0

D. BOUNDARY CONDITIONS

LOADS_Be

TABLE 19.4

Q.2844049£-02

..

1

(Fig. 19.26 shows a view of (he mesh and boundary support for the beam

ANALYSIS

).

;'~RSQ

/seCK

>

fCFREQU£NCY

R~XR2QUSNC'::'

of Example 19.4.)

(18) US( i1a~ural frequencies.:

E. NATURAL FREQUENCIES

RESULT'S

(17) Set parameters to request and calculate the lowest 10 r.amra! frequencies.:

ANALYSIS) FREQ/SUCK } A_FREQUENCY A_F'REQ:.JENCY, :0 S, 16, 0, 0, 0, C, 0, :), 0, 0

1.2-5,

0,

1,2-6,

;. :"lS'r ).

F'REQL!ST

(Tab!e 19.4 shows the values of the first 10 natural frequencies.)

F. PRESSURE LOAD

",,,.e funct;on:

(19) Activ",e the

CON'l'ROL , ACTIVE ). .)..CTSS1'

•.t"CTSET,

're

1

I

(20) DeDne the calculation for 50 steps of 0.01 (seo): ANALYSIS ?D_ATYP£,

:>

POST_:JY;l > PD_ATYPE 2, 3,50,0,0.01. G,

0.5,

8.25, 0

(21) Def'ne the load curve type: ].,.NALYSIS ~ ?D_CURTYP,

?OST_lJY::-J > ?D_CURVES > PO_CUR'I'YP _, C, 0

(22) Specify the toad Curve as a functiOn. of time: ANALYSIS } POS,?_DYN )

?D_CURCE?,

L

:, 0,

PD~_GJRVES

-10,

0.25,

)

P:LCURDEP

-10,

26,

(23) Apply norma; p!'essure or. {he $hell elements:

Fig. 19.26 Mesh and support for the beam of Example 19.4

571

LOADS_9C ). STRUC'I'UFZ.A.L ) !?RES.s:JRE > PSF PSF, 1. L 2, i, ::., L 4

0,

0.5, 0

572

Dynamic Analysis of Structures Using the Finite E;€ment Method

Framed Structures Modeled as Discrete Multidegree-of-Ffeedom Systems

S73

G. SPECIFY DESIRED RESPONSE VALUES (24) Request pnr,t of displacements for 50 time steps: ANA'LYS:;:S

>

PD_J?RI~lT,

POST __D"fN :;., 0,

PD_OUTPC7 > ?D_?RINT

>

0, 0,

(:,

2.,

L

50,

1

(25) Request stresses calculated at a node on the free end of the beam (node 101): ANA:"YSIS ) :?oS'1'_:::riN N~ES?, :, 101, 8

> ?D_OJ'IPU'1' )

?D_NRSS?

(26) Request a plot of stresses at step 25 corresponding to time {sec): hNAI..YS!S

>

PD'M}'!_::Y:',

25, 25, ::., 0

0.25

I

Fig. 19.27 COnlour--displacement plot at ttme t

POS'T_DYN > PD_vi)'I'P;JT

H. RESPOlSSE

0.25 (sec) for the beam of EX3.mple 19.4.

(32) Activare the Von Mises stresses:

(27) Execute postdynamic analysis: (33) Plot contour stresses: RESUI..,'TS

(28) Calculate clemem stresses:

>

:::CSTRESS

(6) Obtain a hare. copy for the stress plotted on the screen:

1. POSTPROCESSING

C:O~TRO:'

:>

L.?:.SERJE'1',

(29) Activate displacement for plotting: RESU:/I'S > AC'!'IVAT2 ACTDIS?,1;

:>

PLG?

D:::::S?LO'I, 0,

>

:,

~

:::.ASERJE'I'

= 0.25 sec.)

19.3 SUMMARY In this chapter we have ;;resented ar. introduction to the Flnite Element Method (FEM) for the analysis of problems in Structurat Dynamics The theory of FE:..1 In structural dynamics was fonnulated througJ the following steps:

C!SPLOT

0, l6C,

DEVI:SS 150, 0

(Fig. 19.28 shows the contour-stress plot at step 25 correspondi:1g to lime f

.!:..CT:::S?

:>

(30) Plot contour displacements: R~SCLTS

PLOT

STRPLO'Ti

'1.

(31) Obtain a hard copy of the displacement plotted on the screer.:

(1) CON':'ROI.t > DE\':ICES ). :L.ASER"::-E7'

(A contour-..cHplacement plot at step 25 corresponding to time (sec) 1S showr. in Fig. 1927.)

f

= 0.025

(2)

Modeling of the entire structure into one-dimensional, two-dil7lensional, or three-dimensional beam, rod, triangular, quad:ilateral, rec~ tangular, or other types of structural elements. Identifying nodes and nodal coordinates at joints betweer. structural elements.

574

Dynamic Analysis of Structures Using the Fini!e Eleml$ot Method

FraMed Suuelutes Modeled as Discrete MuJtideg(ee~ol-F(eedom Systems

575

(9)

Salving [he s),s(em differe:1rial equation of motion, eq. 09.28), to obtain the sys[em nodal displacements [y}.

(to)

Determining the clemem nodal stresses {u(x;,y)} uSlng "(he calculated element noda; displacements {q}.-_

In eng:neering pructice, [he solution of p:ub!ems by [he Finite Element Me[hod ;$ ob{t.in~d with a computer and appropriate software, such as the program COSMOS. In (he implemema[ion of [his program, items are selected from menus presented by (he program and dum are supplied in response at the prompts i:l the progran:.. Tr.e fonowing are bnsic steps in (he Impleme:1tation of a FEM p:ogram such as COSMOS: (1)

GEOMETRY: Keypoin(S, Lnes, corves, surfaces, regions, etc. ore used [Q define the geomerrical characterisr:c of [he srructure.

(2)

Selecting un inlerpol
PROPERTIES: Sees of properties a:-e given to define the type of eleme:1ts .1nd (heir cha:-ac[eristics as well as numerical values for materia! properties and geon~errical consrams.

(3)

of [he displacements at its noda! coordulatcs" Establishing the relattons-hips at the nodal coordinates of a s(r~~:uml element between fo:ces and displacements (the element stltrness

MESHING: A :nesh using selected types of elemcr.~s is established [0 model [he Sl:"uCture into s[ructural elements and to idennfy the sysrem :lodal coord:nmes.

(4)

POSTDYNAMrCS: Instructions are implemented [Q establish the type of dynamic anaiysis and to define [he exciracion functions.

(5)

LOADS AND BOUNDARY COND!TlONS: Loads a:e applied at specified points, lines, or surfnces, and boundary conditions are implemented at specified locations in the Structure.

(6)

ANALYSIS: Commands are given ro perform desired cnlculatlons for the determinn:ion of nUtural frequencies and the calculation of the response in :erms of dispbceme:1ts. velocities, acceleration, for::::es, tind stresses.

(7)

POSTPROCESSING: Lisls, graphs, and plots are obtained for desired functions such as displacemems, deformations, forCeS. and stresses.

o Con'o"r~!:;trcs.':i plot tll lime Fig. 19,20 '"

(3 )

(4)

f""

0,25 (:-.ec) for ~he beam of Example 19A

matrix) and between forces and accelc:-arions (the elemen~ mass

matrix). Obtaining the vecror of the equivalent nodal forces for the body or (5) other external forces acting on the element. (6) Assembliilg the system stiffness rr.auix, the sys(em masS ~atriX, and the system vector of [he equivalent nodal forces, r~speCt1velY, from the element stiffness matrjc~$, element mass mamces, and [he element vectors of the equivalent nodal forces. (7) Establlshing the dynur:1ic equillbr~um at the sys[em nodal coorcinates, amon o the elastic forces, inc:tial forces, J:1d tl':e external forces to c;, " • obtain the system differential equation or monon: ~

[At]

iJJ

+-[K] (y)

= (F(l)}

(19.28)

in which (/1-'1] and [K] are, respec(lvely, the system mass .matrix a.nd the system stiffness matrix. {F (I)} is [he system :-ector of th~ equlvalent nodal forces, und {y} and {y} are, respectively, the dtsplncemtnl and the acceleration vectors at the system nodal coordinates.

(8)

Lntrociucing 1n the sYStem differemial equruion the bo~ndary conditions resrricti:1g displacements at specified nodal coordmates.

PROBLEMS 19.1

The SIeel pio:e showl'! ir. fig. P19.! of diffi¢nsiolis 20 in X 20 in and thickness 0,10 in wirh a circular hole of radius r = 0.5 in is subjeCted to 0. sucdenly applied in-plane lateral compressiVe pressure along (he edges AD llnd Be of magnitude p == 100 pSi. Mode! the PltH¢ wi~h eler.'iefHs PLANE2D on !l 6 X 6 mesh in e:ach qU~1f1er SeCltOll of [he place. Deremllne the fiw five n3tural frequencies anc plms of the response for displacements and $ireSSes on [he elements.

576

Framed Structures Modeled

as

Discre~e Multidegree-of-Freeccm Systerrs

D~"' _ _ _~----'IC

-! ~:I

p= 100 psi

~3

r.-/j

~ ~_/

i=

i

li

AL:::==:::::::w::=====; 8 ~.

20

!~

20'

1

Time History Response of Multidegree-ofFreedom Systems

Fig. P19.1.

Solve Problem 19.1 taking advant:looe of the double syrn:Tletry of geometry and load introducing appropriate bcunda.',)' conditions and analyzing only a quarter of the structure. 19.3 Solve Problem 19. i assuming that applied pressure reaches its maximum value in 0.1 sec and then decays to z.cro in next 0.;: sec as shown in Fig. ?19J.

19.2

P (psi)

Ie;)

L~~-

Fig. P19.3. Pressure load for Problem 19.3.

rn Chapter 7 the nonlinear analysis of a sing!e~degree-of-freedom system using the step~by-step linear acceleration method was presented. The extension of this method, with a modification known as the Wilson-e method, for the solution of structures modeled as multidegree-of-fretdom systems is developed in this chapter. The modification introduced in the method by Wilson et aL 1973 serves to assure the numerical stability of the solution process regardless of the magnitude selected for :he time step: for this reason. such a method is said to be unconditionally stable. On the other hand, without Wilson's modification. the step~by-.s:tep linear acceleration method is only conditionally stable and for numerical stability of the solution may require such an extremely small time step as [0 make the method impractical jf not impossible. The deve!opment of Li.e necessary algorithm for the linear and nonlinear :nu~tidegree-of­ freedom systems by the step¥by ~step linear acceleration method parallels the presentation for the s!ngie-degree-of~freedom system in Chapter 7. Another well-known method for step-by-step numerical integration of ~he equations of motion of a discrete system is the Newmark beta method. This met30d which also may be considered a generalization of t.he linear acce!erSi7

Framed Slrl.;.ctures Modeled as Discr€H~: M.litidegree<:f-Freedom Systems

578

Time HiSleI)' Response of Multidegree-of·Freecom Systems

allan method is presented later in this chapter after discussing in detail the Wilscn-f) methcd.

Fs;

1

r i

F',U+d

20.1

579

.

iNCREMENTAL EQUATIONS OF MOTION

··-----Af~r_­ Tangent $tiHneu

a

The basic assumption of the Wilson method is that [he acceleration varies linearly over the time interval from I W {+ edt where e 2':: 1.0. The value of [he filcwr I) IS de[ermined [Q obtain optimum stability of the numerical process and accuracy of the solution. It has been shown by Wilson that, for e?::. 1.38, [he method becomes unconditionally srable. Tbe eq:.Jations expressing the incrementai equ;libr:um conditions for a muhidegree~of~freedom system car. be derived as the matrix I equivalent of [!Ie incremental equation of motion for the single degree~of-freedom system, eq. (7.6;. Thus raking (he difference between dynamic equilibrium cOl1diljons defined at [i~,e I, and at I; + T, where "T= Btl/, we obtain the Incremental equations

r-6y.W--: i

Yitr}

+ 7) -

Li y , ~ Y(Ii +

y(tJ

(20.2)

T)·~ y{I,)

(20.3)

dYi= j(l; + T) -

y{t;)

(20.4)

and

ill';

F{r,

+ T) -

1'(1;)

(20.5)

In writing eq. (20.1), we assumed, as explained in Chapter t for single~degree­ of~freedom systems, [hat the stiffness and damping are obtained for each rime step as the inir;al values of the rangem tc the corresponding curves as shown in Fig. 20.1 rather {han the slope of the secant line which requires iteration. Hence the stiffness coefficiem is defined as (20.6)

Cij'

(b)

and the damping coefficient as. dF£,

=

(20.7)

dy;

in which F.i and Fbi are, respectively, ~he elastic and damping forces at modal coordinate i and Yj and Yj are, respecrively, the displacement and velocity ar nodal coordinate I

20.2

THE WtLSON-8 METHOD

The integration of the non:inear equations of motion by the step .. by-step linear acceleration method with [he extended time step imroduced by Wilson is based, as has already been mentioned, on the assumption that the acceleration may be represented by a :.inear function during the time step T:::;:;; eLll as shown in Fig. 20.2. From this figure we can write the linear expression for the acceleration during the extended rime step as .1" ,

y(r) =j,+ .... y, (r

[I)

T

(20,8)

in which .aji is given by eq. (20.4). (nregrating ee;, (20.8) twice between limits [,- ar.d r yieldS j(r)~y,+j,(1

I Ma:rices and Vectors ore defloted with boldface lene:-ing lh.oughollt this chupru.

Yi''''

Fig. 20.1 Definition of idl~encc ccefficitnt$. (a) ~
c'J

.dy;= Y(li

:

·--,.'--"'---'i d

Ibl

(20.l) in which the circumflex over d indicates that the increments are associ:ued with the extended (ime step T = eLl:. Thus

'

"'-----

I,)

1 d Yi ~ + ,---(1-1,)" _ T

(20.9)

580

Time History Response of Multidegree-of-Freedom Systems

Framed Structures Modeled as Discrete Multidegree-of-Freedom Systems

where

Yltl

J

1

581

T ..I : y(r;l

I.

(20.16) and

I

I

I

I

I

I

-,-,

Y(t,+Ob.t)

: y(ti+b.tl r----b.t~ I I I

\ T

I I

I

L ___L..L____,.J-:-:-.L---:-~;-;----'--_ ~ _ _-~~8l\[- _ _~

Fig. 20.2 Linear acceleration assumption in the extended time interval.

and y(t) ~ y,

I

0

I

Liy,

,

+ y,(I-I,) + lY' (t -I,t + 6-;;-(1 -I,)'

\)

I

7

\

2!

+C,[3y,+-Y,i

(20.17)

Equation (20.15) has the same form as the static incremental equilibrium equalion and may be solved for the incremental displacements L1Yi by simply solving a system of linear equations. To obtain the incremental accelerations ..1Yi for the extended time interval the value of ..1y; obtained from [he solution of eg. (20.15) is substituted int~ eg. (20.13). The incremental acceleration Lly; for the normal time interval ..1t is then obtained by a simple linear interpolmion. Hence

time

ti+Li.t

t,

1/6

jjF,~jjF,+M..-y,+3Y,

i

Li

(20.10)

.... = -YLly

e

Evaluation of eqs. (20.9) and (20.10) at the end of the extended interval gives

t= lj+ T

(20.11)

(20.18)

To calculate the incremental velocity LlYi and incremental displacement Lly, corresponding to the normal interval L1 t, use is made of eqs. (20.11) and (20.12) with the extended time interval parameter T substituted for Ll t, that is,

and

(20.19)

(20.12) and in which Li y , and Li y, are defined by eqs. (20.2) and (20.3), respectively. Now eg. (20.12) is solved for the incremental acceleration .:1y; and substituted in eg. (20.11). We obtain o

6Ll'

LlYi=--"

-r

6.

Yi~-Yi-

T

3" Yi

(20.20) Finally, the displacement Y;';'I and velocity Y;';'I at the end of the normal time interval are calculated by

(20.13) (20.21 )

and

and (20.14)

(20.22)

Finally, substituting eqs. (20.13) and (20.14) into the incremental equatio~ of motion, eg. (20.1), results in an equation for the incremental displacement L1y;. which may be conveniently written as

As mentioned in Chapter 7 for the single degree-of-freedom system, the initial acceleration for the next step should be calculated from the condition of dynamic equilibrium at the time t + Ll t; thus

(20.15)

(20.23)

Framec Structures Modeled as Discrete

582

Mu~tideGree·ot~Freedom

Systems

to which F D (Y i ... I) and F s (y ,'+ I) ;epresent, respectively,. the daC1pt:1g force 2.nd stiffness force vectors evaluated 2.[ the end of [he time step to ... ! :;;:; Ii + LlL Once the displacemenl, ve!ocity, and acceleration vectors have been determined a( tiC1e l;_! = 'I + Lir, (he ou:.Hned procedure is repeated to calculate these quaotlcles at the next time step !, ... '1 = t i + I + .d t and rhe process is cominued to any desired final rime. The step-by-step linear accelern:;on. as indicated in :he discussion for the single degree-of-freedom system, involves (WO basic approximations: (i) the 2cceleration is assumed to vary linearly during the time step, and (2) the damping and stif~hess cha!"acreristlcs of the Slf\jcrure are evaluated at rhe initiation of the time step and are assumed to remain constant during {his rime intervaL The algorithm for the integration process of a linear system by :he Wilson-8 method is outlined in the next section. The upplication of this method to lir.ear structures is then developed in the foHowing section.

Time !-iisto:), Response 01

(2)

Multidegree"ol~Freedom

Systems

Calculate the effective incremental load iiFi for (he time interval + T, from eq. (20,17) as

583 !; to

!,

F, ~ LlFj...L. (a:;!YJ + 3 C)J;

+ (3iH + flJC)Yi

(3)

Solve for the incrernemal displacemem AYi from eq. (20.15) as

(4)

Calculate the incrementa: acceleration for [he extended lime intertal T, from the relation eq. (20.13) as

(5)

CalcUlate 'he incrcmenul acce:eration for [he nomtr.i interval from (20.18) as

kJviiF,

JY,=a 4.dy; a2iJ-Jy, e~,

dy

20,3 20,3,1 (l)

(2) (J)

ALGORITHM FOR STEp·BY·STEP SOLUTION OF A LINEAR SYSTEM USING THE WILSON-S METHOD

.:lY=e (6)

Initialization Assemble the system stiffness matrix K, mass man1x M, ""d dampir:g matrix C. Set initial values for displc.cement Yo. velocity Yo, a:1c forces Fo, Calculate iniIial accelenuion Yo from

Ca;cula[e the incremental velocity £1y; and the incrementa! disp!ace~ men: .:::1y, from lime f, to !, -{-..1! from eqs. (20.19) and (20.20) as .:::1Y'=Yi.1I+~£1.Y;L1t

.:::1y, = j, Lit + ty, .112 + i.1y,df 1 (7)

Calcult::e (he displacement and velocity at time YI-;-I

(4)

8 DI,

411=

3 .,

,

(1)

(8)

6

lt2=·-, ·1

T

ill using

.:1Yi

Calculate (he acceleration .Yi t I at time [i+ equilibrium equation of motion. namely,

I

= Ii + L.\!

directly from the

For.n the effective stiffness matrix K [eq. (20.16)], namely, K=K+-a4 Nf

20,3,2

ri

Select a {ime step .:1!, the factor 8 (usually raken as 1.4), and calculate the CD:1stants r, 4'1, 0:2, aJ, and a .. from the relatlonships 7

(5)

=], +-

fir 1=

+ aI C

For Each Time Step Calcuiarc by linear Interpolation the incremental load dF; for the time interval Ii [Q t; + T, from [he relationship

Example 20.1. Calculate the displacement response for a two-story shear building of Fig, 20.3 subjected [0 a suddenly applied force of 10 Kip a[ the leve! of the second floor. Neglect damping and assume elastic behavior.

Soimion: The equarions of mOlion for [his strucLUre are [0 J36 0

75.0

:0

44.3

0,066

584

Time History Response of

Framed Structures Modeled as Discrete Multidegree·of-Freedorr: Systems

and

~he

Muitideg(ee-of~Freedom

Systems

585

natural periods are

The initial acceleratIOn is calculated from E"" 30,000 Kipfin:

h I;.

1I !'0 .1.,'6 0 ,0 0.066:

497.21n.4 212.6 in.4

r

75.0 • - 44.3

giving

YlO = 0 }:w;;:;;; 151.51 infsec::

Fig. 20,3 Two-story shear building for Examples 20.l and 20.2.

If we seJect Lli = 0.02 and stants. we obtain

e= 1.4.

= 8ill = 0.028, and calculate the con·

T

which, for free vibmtion. become

3

lj'Y,] [ 'J+

;0.1360 i

lO

Substitution of

}i =

0.066,

.Y,]

-

75.0· -

44.3~

44.3

44.3 j

~

a,

= -,.

a,

= -,. = 214.28>

=1'[00]

a, sin wt results in the eigenproblem

a>.'

= 107.14.

6.

=..:.2 = 0.014

a, =

6

= 7653

The effective stiffness is then

it = K + a,M ~ Q, C (C = 0

I

which reaquires for a nontrivial solution 75.0

~

44.3 - 0.066 w'

44.3

~

!1:;:: 0

0.136 w' -44 .-
a:lC lle

Expa:Jsion of thIS determinant yields

w' - 1222.68 w'

effec~ive

3F

wi = 139.94

wi = 1082.0

and

- 44.31 44.3 •

443

-

44.3

= 11.83 rad fsec,

(02

= 32,90 fad fsec

force

ro';+2J4.28 [0.136

1

lOi

>0

f, = 1.883

cps,

f, = 5.237

cps

0

]' {[OJj-3 rO.136

0.066 iO;

l

0

° '.: 0.066J

JP = [3~1 Solving for

Lly

from

k Jy;;:;;; L1F

, 1115.8

or

1

549.4,

Hence, [he natural frequencies are Wi

+ 7653,[0 136 [0

F= 3F + Ca,M + 3 C)j + (3M +a,C)jI

+ 1515 16 ~ 0

which has the roots

f 1115.8

K=!

I

75.0 - 44.3

undamped system)

-44.3

yields

fO.002175l

549AJ

Ll~·

.

; ~ [0054780,

S86

Framed Sln:ctures Modeled

Solving for , Llji

3jl

3S

Time History Response 01 Muilidegree-of-Freecom Systems

Discrete Mu!lidegree-ot-Freedom Systems

integration process_ The cominua[ion in determining the response for this struCture is given in Ex.ample 20.2 with the use of the computer program described in [he next section.

from eq. (20.13), we obtain

= 7653 (0.002175') 0.054780

214.28

3; _

rO)

\0

y,

i

l-

16.6451

35299)

20.4

Then

16.647) 35.2991

=I

18891] 1-2521

1(0.02)4"--, 0.02 ( 11.8911

1I51.51)

for a :inear system using lhe linear accelermion method with the

FrO~l

:0.11891 ; [2.7781,

structure. The program peri'orms a linear interpolation between the load data points, which result in a table giving the magnitude of [he applied forces at each nodal coordinate calculated at increments of time equal to the rime step At. The output consists of a tab;e giving (he response for each nodaf coordinate in (erms of displacement, velocity, and acceleration at increments of time .1t up to the maximum time specit1ed by (he duration of the forces including, if desired, an e,'uenston wir;, forces set to zero.

eq. (20.20), . , (0.02)' f 0 ] 4"-'--,,.(0.02)' Ll v = 0) '0.02) T , ..... , [0 \ 2 :~51.51

I

q_

11.89 1000081 25.2 j : -100286)

From eqs, (20.21) and (20.22).

,

(0) , [O.OOOS: _ fO,00081

\y}

=:0

T

l00286) -(00286)

(a)

and

Example 20.2 Use Program 19 (0 detennine the response of the [WO~story shear buiiding shown in Fig. 20'}. The first cycle of the integration process for this structure has been hand calculated in Example 20.1.

Solurion.'

. _ fO] ~ [0.1189) = (0.1189\

(Y) - ,0 '12.7781

2.778lj

(D)

Program Input Data and Output Results

From eg. (20.23)

[0.136 0 'I '. r 01 [ 75.0 - 44.3) [0.00081 lO 0066jlYl=[IO;- -44.3 44.3 [0.0286)

IN?CT O,;?;;,:

Nu;:eBH Of" CEGS225 Of [,«2200" t{:JM9£? C?

,"01011'$ DEF!!HN"C 71iS ;;;xcr:,,:r-:cII

CCORtl- Xt EXC:r" ... 'l'ro:~

(?OK $\.lt-O'crrr M01':C~<~O)

'l'!ME: S'rE? Ct 'N"'tSCAA'l'lON

which gives :

{y}

8.875]

= [132.85

Wilson~e

modification. The program requ:res previous modeling of the sLructure to determine the stiffness matrix and the mass matrix of the system. Input dam include (he time step 11:, [be maximum time computation, TMAX, nnd a tab!e of the time-force values for each ioad applied a( the nodal coordinates of the

1=1

-2).21 J

2

PROGRAM 19-RESPONSE BY STEP INTEGRATION

Progra:Y, 19 performs [he step-by-step imegration of the equa[lons of motion

From eq. (20.19), it follows that

Lly=, o

587

(e)

The results given in eqs. (a), (b). and (c) far the dispiacement, velocity, and acceleration, respectively, at rime 1\;;0· '0 + L.1! complete a I1rst cycle of the

;j.1S()OE~02

- .HJ;:lE.OZ

~,H)OE?02

: .H10E.C2

0,11602.0(;

O.00CO;;:':3

() ,ON01:'OO

(;. 661;01':-C1

NO",.

!4lWbJ N(");,,2 H~.

04

588

Tlrr:e History Response of Multidegree·of~Freedom Systems

f'ramed StruClures Modeled as Discrete Mullidegree~of·Freedom Systems

589

results In an eq:.mtion to calculate the incremental displacement Lly;, name!y, G.OOOO".OO

(}.

oooo.:~oo

(20.29) where the effective stiffness matrix vector .1Fi are given respectively by

K:

and the effective incremental force

(20.30) VA~.

CIS.?S

0.6?2

~A:<.

vn:::c_

itA;,.

Ace.

$')<

5.375>

and

1',,:

Hi .5:: f, C Llt I 1 . \

20.5

--

1'\." 4{3} i

-'Y

(2031)

THE NEWMARK BETA METHOD

The Newmark beta method may be considered a generalization of the linea: acceleration method. It uses a numerical parameter designated as /3. The method, as originaily proposed by Newmark (1959), contained in addition to ,8, a second parameter y, These parameters replace the numerical coefficients ~ and ~ of the terms containing the incremental acceleration LiYi in eqs. (20.19) and (20.20). respectively. Thus. replacing y the coefficient t of Lly, in eg. (20.19) and by {3 the coefficient ~ aiso of LlYi in eg. (20.20), we have (20,24)

In these equations C,. and Ki are respectively the dumping and stiffness matrices evaluated at the initial time r,. of the time step.at [1_ i - [ i ' In the imp!ementation of the Newmark beta method. the process begins by selecting a numerical value for the parameter {i Newmark suggested a value in the range i:s: f3 s t. For f3!::: t the method is exactly equal to the linear acceleration me [hod and is only conditionally stable. For f3 = ~ the me [hod is equivalent to assuming that the velocity varied linearly during the time step, which would require that the mean acceleration is maintained for the intervaL In this last case, that is, for f3 = L the Newmark beta method is unconditionally stable and it provides the satisfactory accuracy.

and (20,25)

It has been found that for values of !' different than t, the method introduces a superl1uous damping in the system. For this reason this parameter is generally set as y= t, The solution of eg, (20.25) for ;lYi and its substitution into eq. (20.24) after setting y = ~ yield

1

Lly;

1

LlYi

1 , {3LlI y;

(20,26)

I,.' 2pYI4' 1\ 1 -

(20.27)

= pLlI' Lly; -

= 2{3Llt LlYi -

Then the substitution of eqs. (20.26) and (20.27) into the incremental equatlon of motion (20,28)

20,6

ELASTOPLASTIC BEHAVIOR OF FRAMED STRUCTURES

The dynamic analysis of beams and frames having linear elastic behavior was presented in the preceding chapters. To extend this analysis to structures whose members may be strained beyond the yield point of the material. it is necessary to develop the member stiffness matrix for the assumed elastopiastic behavior, The analysis is then carried out by a step-by~step numerical integration of the differential equations of motion_ Within each short time interval ill. the s~ruc­ ture is assumed to behave in a linear elastic manner. but the elastic properties of the structure are changed from one interval to another as dictated by the response. Consequently, the nonlinear response is obtained as a sequence 0: linear responses of different elastic systems. For each successive interva:, the stiffness of the structure is evaluated based on the moments in the members at the beginning of the time increment.

590

Framed Structures Modeled as Discrete Multidegree-of-Freedom Systerr,s

The changes in displacemems of the linear system are computed by inregration of the differentia] equations of !Dotion over the finite interval and (he total displacements by addition of the incremental displacement to the displacements calcuJared 1ft the previous ti!1le step. The incremental displacements are also used to calculate the increment in member end forces and moments from the member stiffness equation. The magnitude of these end moments relative to the yield conditions (p!astic moments) determine the characteristics of the stiffness and mass matrices to be used in the next time step.

Time History Response of M1JJtidegree·of-Freedom Systems Beflding moment

- --,'----,;f----+---M;.

20.7

591

Angular displacement

Plastic

MEMBER STIFFNESS MATRIX

[f only bending defannations are considered, the force-displacement reladonship for a unifonn beam segment (Fig. 20A) with dasric behavior (no hinges) is given by eq. (1420). This equation may be written in incremental quantities as foHows:

lI" )= j

12

£1

6L

6L - 12 4L2 -6L

6L (20.32)

-12 -6L 12 6L 2L z - 6L

in which tJ. Pi and Lloi are, respectively, the incremental forces and the incremental displacements at the nodal coordinates of the beam segmenL When the moment at one end of the beam reaches the value of the plastic moment Mp. a hinge is fonned at that end. Under the assumption of an elastoplastic relation-between L.fte bending moment and the angular dispiacement as depicted in Fig. 20.5, the section that has been tra:1sformed into a hinge cannot suppOrt a moment higher than the plastic moment A(, but it may continue to defonn plastically at a constant moment M;)" The relationship reverses to an elastic behavior when the angular displacement begins to decrease as shOwn in Fig. 20.5. We note the complete similarity for the behavior between an eiastoplastic spring (Fig. 7,5) in a singie degree-of-freedom system and an elastoplastic section of a beam (Fig. 20.5).

Fig. 20.5 Ela~toplastic relationship betwee:! bending rcoment and ang::!ar displrrcemem al a section of a beam.

The stiff:less matrix for a beam segment with a hinge at one end (Fig. 20.6) may be obmioed by appiication of eq. (l4.16) which is repeated here for convenience, namely (L

k,,= Jo n",;' (x) "'/,(x)dx

where if.!;(x) and if.!;(x) are displacement functions. For a uniform beam in which the formation of the plastic hinge takes place at end Q as shown in Fig. 20,6, the deflection functions corresponding to unit displacement at one of the nodal coordinates 8.• 8z, 83 , Or 6... are given respectively by "', (x) = 1 ",,(x)

3 -=+ 2L

x'

=0

(20.34a) (20.34b)

3x


x' x'

"', (x) =

AP., A&.

""'Q)

L·----Il

APl,AS:!

Fig. 20.4 Beam segment indicating incremental end forces llnd corresponding incrementa! displacements.

(20.33)

Fig. 20.6 Beam geometry with a plastic hinge at joint (J).

(20.340) (20.34d)

SS2

Framed Siwctures Modeled as Discrete Multidegree"of-Freedon Systems

For examp!e, to calculate K!i, we subs:itute the second derivative

if;:"

Time History Response of

(x) from

ego (20.}4.) into eg. (20.}3) aod obtain k

"

3£1 ~£JJ"(3-'" _Id.<; -,-

0.L')

J

resulting stiffness equation in incremental fom"; IS

1

LlP, ~p)

tJ P,

<:[

·3

3

0

0

0

3

0

3L

0 - J[

° 3

l tJP, J 20.8

~8, ° .f ~81

3L '

I : Llb} !' 3L?j l tJ8, j 3L

(20.36)

may

1

ItJP,,= EJ tJP, I l dP i j

1

l r

3 3L

-]

0

]L 3L!

-]

···3L

-3L -3 0 0

1 r ~8,

1

0 0.

~8,

0c

tJo,;

01[ ~:'

(2037) !

iO 0

c

(20.38)

0,

60,)

The relationship between forces nnd acceler;::t;ons al the nodal coordinares of an elastic uniform rr..e;nber considering fiexura: deformation is given by eq. (14.34). This equation wrinen in incrementa! q1..lantiiies is

r 156

JilL

I

~p) r 4201

Ll P,

l

22L

22L

54

4e

i3L

13L

54 13L

156

~ 3L'

22L

"T

- 3L' - 22L

M]

tJo,

tJo,

(20.39)

2

4L "i ,LioJ

where d Pi anG. t1;.t are, respectively, the incremental forces a:1d the incrementa! accele:-altO;1S at the nodal coordinaLes, L is (he length of (he member, and in is its mass per unit length, Assuming elastoplastic behavior, when the moment at an end of the beam segment reaches the magnitude of {he plastiC momem ir'lp and a hi:1ge is formed, the consistent mt.l.SS coefficients are de. termined from eq. (14.33) using lile ~ppropriMe deflectlor. Curves. For a uniform beam in which the forma~ion of the p!llSllC hinge develops at end J as shown in Fig. 206, the def1ection functions corresponding to a unit displacement of the nodal coordinates are given by eqs. (20.34). Analogously, Jhe deflection functions of a beam segment with .a plastic hinge at end ] as shown in Fig. 20.7 for unit displacement at nodal coordinates 8 h ~, 8;, 0:- S~ art! respectively given by

x'


Fig. 203 Be:rm geometry 'NIlh a pias:lc hinge at join! Q)

593

MEMBER MASS MATRIX

[ U, ~P,

It should be pointed out lhM .101 is the incremental rOlation of joint CD at the frar:1e and n'Ot the increase in rotation at end Q of the beam under consider~ ation. The incremental rotation of t~e plastic hinge is given by the difference between .,101 and the increase in rotation of the end (J) of tl;e member. Hinge rolatlO:1 be calculated for the various cases with formulas developed in the next section. Analogous to eg. (20.36), the following equation gives the rela~ionship between incremcatzl forces and incremental displacements for a uniform beam with 0. hinge at end CD (Fig. 20.7), namely,

r tJP,

~;; 1J~ ~ ~ ~l [~::tJo, I

I~p,(-iooco

SimilarlY, illl ihe other stiffness coefficients for {he CDse 1;1 whic~ {he [ofIT.alion of the plastic hinge takes pince at end CD of a be.lm segmenr urc determined using eq. (20.33) and [he deneclion ft1r.ciions given by eqs. (20.34-). The

r LlP.

Systems

Findiy, if hinges me formed a~ both ends of the bellm, the stiffness matrix becomes null. Hence in lhis case the stiffness equatior: is

(20.35)

L

Mu!:idegree-o!-F~eedof71

2L

+x

(20AO)

594

Time History Response of

Framed Siruc;ures Modeled as ;)iscrete Multidegreewot-Freedom Systems

The mass coefficients for a beam segment with a hinge at one end are then ob,ained by application of eq. (l4.33) which is repeared here, namely, m., =

where

i/I; (x)

J,'

MuJtidegroe--of~F(oodom

595

and the corresponding relationship in incremental form by

(20.41)

in fi (x) f;!x) dx

Systems

(20.45)

and f; (x) are the corresponding displacement functions from eqs.

(20.34) or (20.40) . AppEca~lon of eg. (20.4l) and the use of displacement functions (20.34) results in the mass marrix for a beam segment with a hinge at the CU end. Tne resulting mass matrix relates incremental forces and accelerations at Ihe nodal coord;naces, namely,

58.5 0 0 0 ,hl [ ~ 420 99 0 - 16 5L 0 -36L 204

LIP, LiPl

LIP, LIP,

- 16.5L

1fLIB, 1

o :

0 585

-36L

lLI~

(20.42)

LIS, ,r

Analogously to eq. (20.42), the following equation the relationship between incrememal forces and incremental accelerations for a uniform beam segment with a hinge at the CD end:

LIP, LIP, LIP, lLlP,

99 inL 36L =-420 58.5

0

36L 8L'

585 16.5L

16.5L 204 0 0

T" o o

I\ LIB, I

0:,

,:

(20.43)

Fi::1ally> if hinges are formed at both ends of the beam segment, Ihe deflection curves are given by

20,9

ROTATION OF PLASTIC HINGES

fn the soh.Hion process, at the end of each step imerval ir is necessary to calculate the end momems of every beam segment to check whether or not a plasric hinge has been formed. The calculation is done using the element incremental moment~displacemen( relationship. It is also necessary [0 check if the plastic deformation associated with a hinge is compmible wilh the sign of the moment The plastic hinge lS free to rorate in one direction only. and in the other direction the section recums [Q an elas~ic behavior. The assumed moment rotation charactertstics of the member are of the type iHustrated 10 Fig. 20.S. The condltions implied by (his model are: (1) [he moment cannot exceed the plastic moment; (2Y if the moment is less than the plastic moment, the hinge cannot rotate; (3) if the moment is equal to the plastic moment, then the hinge may rotate in the direction consistent with the sign of the moment~ and (4) jf the hinge starts to rotate in a directIon inconsistent with the sign of the moment. the hinge is removed. The incremental rotation of a plastic hinge is given by the difference between the incremental joint rotation of the frame and the increase i;t rotation of the end of the member at that joint. For example, with a hinge at end CO only (Fig. 20.6), the incremental joint rotation is .18:: and the increase in rotation of this end due to rotation 48,~ is - d~/2 and that due to the displace~ ments do! and .:18) is 1.5 (.:183 - 40!) IL Hence the increment in rotation dpi of a hinge fanned at end G is given by

x

.'-, (x) = - -L + 1

(20.46)

'P. .

1;, (x) = 0

Similarly, with a hinge formed at e!ld (]) only (Fig. 20.7), the increment in rotation of this hinge is given by

(20.44)

(20.47)

~.~,I' 596

~

,

Framed Structures Modeled as Discrete Multidegree-of-Freedom Systems

--------.\CD

Ci) ~

Ao, I

....

Time History Response o!

Mumdegfee~of~Freedom

---------------;., ~ r.s,

Y ""

II

597

(2052)

:. where

mm.=-J i

i'

Systems

which from eq, (20,50) becomes

Fig. 20.8 Beam geometry with plataic hinges at both ends.

20.11

Finaliy, with hinges fOfIT'.ed at both ends of a bearr. segment (Fig. 20Js), the rottltions of the hir.ges ure given by (20.48)

(20.49)

20,10

,

CALCULATION OF MEMBER DUCTILITY RATIO

Nonlinear beam deforrnat~on5 are expressed lr: terms of [he member ductility ratio, which lS deflned as the ratio of the maximum total end roranon of the member to the end rotation at the elastic limit. The elastic limit rotation is the angle deveioped when the member is subjecred to antisymmetric yield moments My as shown in Fig. 20.9. Ir. thIS case the relationship between the er.d rotation and end moment is given by

M,L

6E1

(20.50)

Pm~~

is the maximum rotation of the plastiC hinge.

TI~v1E-HJSTORY

RESPONSE OF MULTJDEGREE-OF-FREEDOM SYSTEMS USING COSMOS

The corr.purer p:-ogram COSivl0S in::.:!udes rhe step-by-step Wilson-8 Method, as well as, several other methods for the ar.a!ysis of structures with linear or nonlinear behavior (:naterinl or large displacements nonJineariries). The following example illustrates the use of the prog:-am COSMOS to obtain. the respor.se of a Structure with linea:- or norJinear behavior: Example 20.3. ese the program COSMOS to calculate the natura! frequen:.:ies ar.d the response of the twO-story shear building of Examples 20.! and 20.2. (l) Assume elastic behavior and, (2) assume elastoplastic behavior with yield stress valUe O:v = 10,000 psi. So{urion: The Structure is modeled using BEAM3D elements and concen~ truted ma'ises. The following commands are implemented in COSMOS:

L Elastic Behavior (l) Set viev.,t

~o

CISPLAY VIEW, 0,

The member du:.:tHity ratio J..L is then defined as

:>

the XY plane: ·,iIEW_P.:J..R ;. VIEW 0, L 0

(2) Define the XY piane at Z (2051)

GEotl;:E'"::'RY }

GRID

~

Oc

?LAt.:E

?::'AKE, 2, C, 1

(3) Define three key points at the levels of [he StruCture and a fourth point in the principal plane of the cross~sectionaI areas of the beam elements: GEOHETE<.Y

PT

Fig. 20.9 Definir:or. of yield rOtation for beam segmenL

,

POINTS

l. :C20, 0, 0 PT, 2, l20. 180. 0 p ... :; , 12O, 3CO, 0 PT 4, 240, 300, 0

,

PT

598

Time History Response of

Frarr,ed S:(Ijc~ure$ Modelea as Discrete Multidegree·of·Freedcm S),stems

G20r1E'IRY

:>

~,

(5) Define group havior (option 5 = 0): PR02SSTS BORG:]!?,

>

1,

M. __ PT . .

1, 2

Cc
3

r';:?RQP,

EX,

?RO?SE7S ) RCONST

2GrtOUP B£P,M3D,

iL

0,

0,

0,

0,

9,

1:.42,

0, 0,

'"h,

0,

v,

0,

D,

1,

0,

0, C

"3, 1

HESH:::NG

NODES

~

9,

1,

:>

XCO~?RESS

1., 4

(!7) Apply constraims at all degrees of freedom at node 1 and at all degrees of freedom at nodes 2, 3, 4 except for the X-direction:

DND, 2, UY,

4,

Bt1SECDEF

BHSSCDEF, 1,. L

Q,

1:~_?'::

?ARl-..H_HZSH )

>

LO?DS_BC ;. STRCCTUR:!t.L DK!], 1, AI..,L, 0, 1, :..

P.ll.?A:'CMESn: > :'LCR

1, 1, 3, 1,

>

1.fESBING i-LP'I, 3,

NC'OMPRESS,

(9) Define real constants for tbe second srory with a rectangular cross section of height H = 9.47 in and IOta! width b 3,0 in: P:rlO?S2TS

0,

C.>

30::.:6, :--~UXY,

(8) Mesh ane beam element on curve I: :>

0,

(16) Merge 2.!1d compress modes:

P?,OPSETS > 3t<:SBCDEF 1, 1, L L

""

66,

(1.5) Gene::af.e concentrated mass at key point 3:

C

1,1PROP

8MSECDSF', J

M_CR,

1

RC0NST, 2, 4, 10 7,

(7) Define real conStanlS for the first story with a rectangu;ar cross section of height H = 11.42 in and [Q(al wtd[h b = 4,0 tn:

HESEING

2,

(t 4) Define real conStants for mass at key point 3:

Poisson's ra-c!o ;.t = 0.3: :>

2,

using BEA.\13D elements and requesting elasLic be-

(6) Define material properties. modulus of elasticity, E = 30E6 psi and

?RO?SETS

599

1>':3SHING > PAE...L,,t.CMESH

CRLINE

CC::WES

Systems

(13) Generate concentrated mass at key point 2:

(4) Define lines between key points: CRLIN:::::,

MultJdegree·of~Fteedom

9.47,

3,

0,

0,

0,

0,

C,

0,

>

DISPLM.NTS

>

DND

4, 1, UZ, RX, RY, RZ

(18) Set parameters for frequency analysis: .~"\.t.,.I..,YS=S ). FR£Q iB:JCK " J.LFREQUENCY A_?REQUEJ;JCY, 2, 5, 16. Q, 0, 0, 8, lB-CC5,

0, iE-OOO',

0,

(19) Run frequency calculation:

(0) Mesh one beam element on curve 2.: ~..sS:::::NG :> PARA.iCME5H :> M_CR, 2, 2, 1, 3, 1, l,

At\l}4..r..YSIS

M~::R

4

EGRO(]?,

>

RESULTS ) LIST)

0,

0,

0,

0,

Of 0, 0

(12) Define real consrants for mass at key point 2: ?ROP$STS RCONST,

>

2,

RCONST 3,

FREQLIST

FREQLISl'

EGROD?

2, :-1...:;SS,

FIl.EQ /3UCK '" R_FREQuENCY

(20) List natural frequency:

(II) Define group 2 using mass elements: PRCPSE~S

>

R_FR3QU3NCY

1, 7,

:'36,

0,

G,

;:;,

0,

0,

C

FRE:QU=:NC"t"

FREQUENCY

NtJ}1BER

(RAD/SEC;

PREQUSNCY (CYC;:'2S ISEC)

1

0.7922314£+01

0.126:Ja7SE+Cl

2

O,32v7882E+C2

O.5iG5S03E+Ol

PERIOD (SECONDS)

O.793099SE+QO 8.1958671:;;+00

600

,

(21) Define dynamic forcing functLon and apply a force in [he X-d1rection LOADS-BC ~ FUNC CURVES > CURDS;;T:~H:;:, 1, 1, 0, 10001), 8,(;, LOA~S-BC ) STRUC?JRAL ~

FPT,

3,

FX, 1,

3,

.,

18000,

0.4.1,

0,

, , ,., ' 1...---;/ uu / " ., , ~ , :

.8

i

/V I (/'1 'j

.

., ., . ,.6

ANALYSIS )0 NONLINEAR ". NL~N~ESP NL_NESP, l, 2, 4, 0

LJ!

start at r = 0 and end at

to

1

I,

with increments of

,:,

<.; Q.5

8.3

Q. \

'\

:f

;f i~

4

' - :.-/

'.6

.j

'.1

.6

..,

I

rlME {sec)

0,0; sec: LOADS_BC TIMES, 0,

LOAD~C?S

>

) TIV-ES

1, 0.01

(24) Set options for nonlir.ear dynamic »n11:Y5i5 using default values fer roiernnce and run nonlinear analysis: A~AL YSIS

NOi;E. . n..;2AR > A_NONL:: NEAR A_NONLINeAR, D, J., 1, 20, 0.0C:, 0, 0.00:i,

/

\

,

.,

= 1.0 sec

\

i\

·s .8

2N

•

\

.

1.

{22) Request response at nodes 2 and 4:

(23) Set analysis

~ u, '\.

(in)

fORCES ) fPT

:

U:

.6

at key point 3: Ct;~;)EF,

601

rime History Response of Multidegree-ot·Freedorr, Systems

Framed Structures Modeled as Discrete Multidegree-of-Free
(28) ODtai;: 11 hard copy of the displayed displacements:

>

0,01,

A..~ALYSrS

Fig, 20.10 Lin:!ar response dispincement in lhe X direCTion aL the fino;( and second levels (nodes 2 and 4) of the il!..i!ding in Example 203.

)0

N,

0,

0,

_E+OIO,

8

NONLIKEAR } fCNONL!!'-lEAR

CON'1'RO:"" ;.

DE\/ICES

:"'ASERJET,

150, 0

> ::"'ASERJE'T

Figure 20. i 0 reproduces the plot obtained for the displacements at the two levels of [he buildir.g, nodes 2 and 4 in the modeL

R_Nor-:L!NEAR (25) Activate and Plot displacemeilts at the two levelS of the building. nodes :2 and

4~

OrsPLAY > XY PLOTS > ACTXYPOST ACTXYPOST, 1, 'tH12:i UX, 2, 12, L 0, ACTXYPOST, 2, TIME. ex, 4, 10, ". 0,

l ~

(29) Use an editOr to print the response at step 30 corresponding to time 0.30 sec:

2:-1

CONTROL

4N

EDIT'

(26) Plot displacement function for the horizont
UT!LITY '

>

SYSTEM

Ex20.0UT

Table 20.1 reproduces tr.e porrion of the output file EX20.0UT for step 30. II. Nonlinear Re-sponse

DISPLAY , XV_PLOTS ) XY?LOT XYPLOT, L 1

(27) Plot the displacements in the range 0:$ t -s; 1,0 ar.d ar.d repaint the screen: :):;:SPLAY > XY_PLO'I'S j. XYRlU"SGE XY({.A..l"'lGE,:, 1, G, 1.0, ~2.0, 2.0 D1SPLAY > VIEvCPAR > REPAZK':'

0.8 < UX <1.2

Nonllr.ear response is obtained by executing the above commands used to obtain the Imear response except that in the command EGROUP option #5 is set equal to 1 for nO:1:inear response and [he value for the yield stress is set m irs specified value, uy =:. 10,000 pSI, Hence the following are the new commands for EGROU? and fo. M?RO?: PROPSE'T'S

EGROUP,

>

l,

EGROUP

BEA.H3D,

0,

C,

0,

0, 1,

0,

0

602

Framed Strt.!ctures Modeled as Discrete

TABLE 20.1

Multidegree-Qf~Freedom

Systems

Linear Response for Example 20.3 at Step 30 (1=0.30 sec)

Time History Response ot Multidegree·of-Freedom Systems

TABLE 20.2

:c

::

'jC:C:':~:J

).oo,»:;:.~)

::;;<",;<",,1

'~;;S5;.:,,>

-: JoSS£.;:;

:0

/1 o. ~t;::GaE~C:.l

. :J0D0C:;;.O(l

::;. y;o.;i0E:.:J-O

~,OOC~(}s.('0

II.O(lCCC8.~tI

(}. :iOcn03-0J

;:.Q()CM"2.0n

) ·:s~-

-:"-C70"I.;,,,,0

Nonlinear Response for Example 20.3 at Step 30 (/=0.30 sec)

c. :OGOJilE.OC

J.CC()CC2-OC :. )'}C,J02.CO

:6;JS~05

:;-.:6:~Z-CS

X$"c.oaQC~>()O

~'.~JO";£~~O

if"f$$"~.)CJO~.O(l

?~01it:;:.,~(;

s~>~

.";1,,0£_:3

-.;;:SS.r:~C5

;"; 'C-;Of( /:t;;J;. COO~i'>CC-

~,':nQ(l;'>~C

:-!s.' $~~v .0CCO.s~C:i

'J. {)~acc.QC

5:::;:<;<.0,:5:1£.3;

C.2S:2£~'J~

Sm:,~-.Z;;ia;:;~OS

,25112~()~

;!:.'$:"t.o:sa£_,~?

-.)3ES.OS

t$~"O. :"':2:;:.~5

-. :S:t::>C$

%:

and

603

cr

.()CC,,(lS~C()

C. J()()()0£~CD

(1 ()()ooos.co

J _::':'130:<>01

cC·j(l()s~::':():;

_O~"OOE~J0

(). OCOO(i£-OC

C .{)DCCO;:.O(1

J. OOQCCE .oc

O. ()ODO(l1i .. CO

o(:;)Cn.C0

(}. ¢JOOQ".<:j·Q

. ;)O()\)02.06

O. GO~D(H';.O'J

0 . cocoa ;:.if!)

C".:' 7a!E~:.O:'

0, i)C0C(]S~JC

:'J, O}C¢02~O'J

ODc~n.o,;

0 .i)()000!:~jO

t. C"JO<:ji)~~~0

CDtlCOE.\l-o

?,.~"

2523:2··:1

2S23~·'.:

-::",C-.

oeoos.:·t;

0.0CI)C:::·;;O

0

-::;,;

":::5::.J~

~E~Q

~47J::-l:l

, ,-

,

o. C::>C)"E.C
\~;:;T04

~'t.-,,-

~ci$E-!4

!6'2~":-:4

1S.2S '.2

x~~-

$2D4\';'0';-

32040:'%

cu

""J

PRC?SE{IS M?R:)? ~PRC?,

1, EX,

30~6,

NUXY, 0,3, SIGYLD, 10QOO

Table 20.2 reproduces the porrioo of the output file for the nonIinear response for step 30 corresponding to r = 030 sec and Fig. 20.11 reproduces the plm for the dispiacements at the two levels of the building, nodes 2 and 4 in the computer modeL

20.12

SUMMARY

The determination of the nonlinear response of multidegree-of-freedom structures requires the numerical lntegra(ion of the governing equations of motion. There are many methods available for (he solution of these equations. The step-by-step linear acceleration method with a modification known as the ViUson-B method was presented in this chapter. This method is unconditionally

Fig. 20.11 !\l'onlinear response displacement in {he X direction .at :he first and second levels (nodes 2 and 4) of the building in Example 20.3.

stable, that is, numerical errors do not tend to accumulzte during the imegration process regardless of the magnitude selecced for the :irne step. The basic assumption of the Wilson-B method is rhat the acceleration varies linearly over the extended interval r= Bill in which 82:. 1.38 for unconditional stability.

504

Time History Response of Mu!tidegree~ot·Freedom Systems

Framed Structures Modeled as Discrete Mullidegree-of-Freedorn Systems

In the final sections of this ch2.pter, stiffness and mUSS matrices for elastoplastic behavior of framed structures ure presented, Formulas to determine the plasTic rotation of hinges and to calculate the corresponding ductility ratios ure

20.3

605

Use Program 19 [Q dete;mme the response of the three-sLory s~ear buildi:1g subjected to the force F, (1; as deplcred in Fig. no.) applied at the third level of the building, Neglect darr.ping in the system.

also presented in this chaprcc Y)~--;-;-2Kip~ FlU)

~ PROBLEMS 20.1

i

The stiffness and [he mns!; murrices for

it

certain struclure modeled as n twO-

Y1

___-1

K,'" 60 K'p1i{"'. fTI}"'3

degree-of.freedom system are K2 ..

[K]

r

100 - 501

""!

50

Use Program ! 9

(0

[M)'"

1 (Klp/in),

50;

eo Kip/in.

[~ ~l {Kip· sec] Ii:s) Kl c:",loOJKbJinl

determine the response whc:1 the structure is acted upon by

the forces

30

./f o~~Lj-+--+-+--""OI I

.o!

.02

.03

.0'

.05

I

,,' ItFdr)J

'F (,)'I ~

20.2

30

!

(,I

1772) f(!) (Kip)

10)

Fig. P20.J.

)86

where f(r) is gIven graphically in Fig. P20.1. Neglect darn ping in the system.

20A

Solve Prcblem 20.3 considering damping in the sys(err. of W% in all the modes.

Sclve Problem 20.1 considering [h.H [he damping prese:1t in the $ysf~m resI.J.lts in

20.5

0se Program 19 to oowin the response in the elastic range for (he st;ucture of Problem 20. J subjected <0 a:) acceleration at its fcundation given by [he function !\i) shown in Fig. ?20.1.

20.6

Solve Probler:s 20.5 ccnsidenng darr.ping in the syslem as indicated 1:1 Problem 20.2.

10.7

Use Progr;;m i9 to obtain (he response in the elastic r.ange of the she,ar building shewn in Fig, P203 when subjected to &:1 acceleraticn of its foundarion g:ven by the functionj{t) depic:ed in Fig. no.!. Neglect dampir:g in the system.

the following damping malrix;

[(.1

~!

10

,-5

:]tKip.seClin)

f(tl

0.:31

1

L---+----1-I-.' 0.6

0,27

Fig. P20.I.

O,g

1.0

r{sec)

; '1

PART IV Structures Modeled with Distributed Properties

21 Dynamic Analysis of Systems with Distributed Properties

The dynamic ar.alysis of structures. modeled as lumped parameter syslems with discrete coordinates, was presented in Part I for singie-degree-of"freedom systems and in Parts II and TIl for multidegree~of-freedom systems. Modeling struc[Ures with discrete coordinates provides a pracrical approach for lhe ana!y~ sis of structures snbjected to dynamic loads. However, the results obtained from these discrete models can only give approximate solutions to the actual behavior of dynamic systems which have continuous distribured propenies and, consequently, an infinite number of degrees of freedom, The present chapter considers (he dynamic theory of beams and rods having distributed mass and elasticity for which the governing equations of mOllon are partial differential equations. The !ntegratio:l of lhese equations is in genera! more complicatec than 6e solulion of ordinary differeiltial equat~ons governing ciscrete dynamic systems. Due to this mathematical complexity. the dynamic ana1ysls of structures as continuous systems has limited use in practice. Nevertheless, (he analysis. as cominuous systems, of some simple structures provides. without much effort. results which afe of great importance in assessing approximme methods based on discrete models. 609

610

21.1

Dyramic Analysis of Systems with Distdbuted Properties

Structures Modeled wi.th Distributed Properties

and

FLEXURAL VIBRATION OF UNIFORM BEAMS

The treatment of beam flexure developed in this section is based on the simple bending theory as it is commonly used for engineering purposes. The method of analysis is known as the Bemoulli-Euler theory which assumes that a plane cross s~ction of a beam remains a plane during flexure. Consider in Fig, 21. 1 the free body djagraGl of a short segment Qf a beam, It is of length dx and is bounded by plane faces that are perpendicular ,to its axis, The forces and moments which act on :he element are also shown 10 the figure: they are [he shear forces V and V + (av lax) dx; the bending mo~men~s A1 and /v! -:- (aM lax) {Lr; the lateral ioad p dr.:; and the inertia force (in dt.:)iry Idt~, In this notation ih is (he muss per unit length and p p (x, t) is the load per unit length, Partial derivatives are used to express accel~ration, and var1.a:ions of shear and moment because these quantiries are functIons ot two vanaoles, position x along [he beam and time f. If the deflecrion of the beam is smaU, as the cheory presupposes, the inclination of the beam segment from ,he unloaded position is also smalL Under these conditions. the equ3[ion of motion perpendicular to the x axis of the deflected beam obt~ined by eq~ating to zero the sum of the forces in the free body diagram of Flg. 21.l(b) lS f av \ _ fly v-! v+ ~dx! + p(x,r)dx - mdx -,=0

\

ax

iiI

J

611

which, upon simplification, becomes (21.1)

aM

V=-

ax

(2L3)

where E is Young's modUlUS of elasticity and I is the moment of inertia of the eoss-sectional area with respect to the neutral axis through the centroid" For a uniform beam, the combination of eqs. (21.1), (2L2;, and (21.3) results in

a'y ax

V=EI~l

(21.4)

and

EI

a' a' = p (x, l) -Z ax· + fit ---;'. ar .

(2L5)

It is seen that eq, (21.5) is a yartiaJ differentinl equation of fourth order. Ir is an approximate equarion, Only lateral flexural deflections were considered while the deflections due to shear forces and the inertial forces caused by the romtion of the cross seclion (rotary inenia) were neglected. The inclusion of shear deformations and rotary inertia in the differential equation of motion considerably increases its complexity. The equation taking into consideration shear deformation and rotary inertia is known as Timoshenko's equation. The differential equation {21.5) also does not include [he flexural effects dt
From s1mple bending theory, we have the re1ationships

M = El

21.2

a'y

(212)

--:-:r oX

SOLUTION OF THE EQUATION OF MOTION IN FREE VIBRATION

For free vibration [p (x:, t) ential equat10a

=

0], eq, (215) reduces to [he homogeneous differ-

(21.6)

lb)

Fig. 21.1 Simple beam with distributed mass and load.

The solution of eq. (21.6) can be found by ,he method of separation of variables. In this method, it is assumed that the solution may be expressed as the product of a function of position $(x) and a function of dme f(t), that is, y(x, t) = tP(x)f(i)

C2U)

Dynamic Analysis of Systems with Distributed Properties

Structures Modeled with Dislrlbuted Properties

612

The substitution of eq. (2 i.1) in [he differential equation (21.6) le2.ds to _

EIJ[ll

1i'
i'!U)

_

+ m
o

(2L8)

The subStitution of eq. (21.15) imo eq. (21.10) results in

which, for r. nor::rivial solution, requires lh2t

a';

s~

=0

(21.16)

This last equation may be written as The rOOlS of eq. (21.16) a,e

EI ..rJfv (x) . -_

__

in

J(I)

(21.9)

f(l)


s:

Ln this notation Roman indices indicate derivarives with respect to x and overdots ~ndicate derivatives with respect to time. Since [he left-hand side of eq. (21.9) is a function only of x whiie the :ight-hand side is a function only

of I, each side of the equation must equal the same constant value~ otherwise. the identity of eg. {2L9) cannot eXIst. \Ve designate the constant by w' which equated separareJy [0 each side of eq. (21.9) results 10 the twO (oHewing differential equations:

rJfV (x) -

a'¢(x)

(2 J.J 0)

0

~Ci,S",=

-(Ii

(21.17)

The subs[jrution of ead: of these roots into eg. (21.15) provides a solution of eq. (21.10). The general solution is then given by [he superposI[ion of these four possible solutions, namely, (2 L 18) Where C h C2 > C;> and C~ are conSlams of lntegration. The expor.er.~iat functions in eq. (21.1&) may be expressed in terms of trigonometric and hyperbolic functions by means of the re;ationships

and

e ~,n:;:::: cosh ax::t sinh ax

J(I) + w'!(t) ~ 0

(21.11)

where

± {sin cu:

e:::"":::;; cos ax

(21.19)

Subs~irution of these rehltiol1ships into eq. (2l, 18) yields

a'

(21.12)

EI

It is particularly convenient to soive eg. (2t 12) for wand to use the foHowing notation, namely,

W(x)

A sin ax

+B

cos ax

+C

sinh

t/,.t

+ D cosh

CL,(

(21.20)

where A. B, C, and J) are new constar.ts of integration. These four constants of integration defme the shape and the ampUtude of the beam in free vibfarlon; they are evaluated by considering the boundary conditions at [he ends of the beam as illustrated in the examples presented in the following section.

(21.l3)

21.3

in which C = (aL)'. Equation (21.11) is (he familiar free-vibration equation for the undamped

sing:e

degree~of-freedom

system and its solution from eq. (L 17) is f(f) = A

(:05

Wi

+ B sin

WI

21.3.1 (2 J.l 4)

where A and B arC constantS of integration. Equation (21 10) can be solved by letting (2 l.l 5)

NATURAL FREQUENCIES AND MODE SHAPES FOR UNIFORM BEAMS Both Ends Simply SlIpported

In [his case the displacements and bending mOments must be zero at both ends of [he beam; hence the boundary conditions for the simply supported beams are y(O,r) ~ 0,

lv1(0,r) = 0

0,

M(L,!)=O

y(L,t)

614

Dynamic AnalySIS of Systems with Distributed Properties

Structt;res Modeied with Distributed PrOperties

In view of eqs, (21.2) and (21.7). these boundary condjtions imply the foHnwing conditions on the shape function (x). At x O. (21.21)
At x

=

L.


= 0,


=

°

(21.22)

Substitution of the roots, eq. (21.26). into eq. (21.13) yields

(2127)

where the subscript n serves to indicate the order of ~he natur~l Since B = C = D = 0, it fnl:ows thal eq. (21.20) reduces to

The substitution of [he first two of these boundary condirions into eq. (21.20) yields

Bl + CO+Dll


frequenc~es.

Ln11X

or simp:y


615

rp

0

. nm

,,(X)=51O

L

(21.28)

which reduce to Vole note that in ee;. (21.28) the constam A is absorbed by the other constams

B+D=O

in the modal response given below by eg. (21.29), From eq. (21.7) a modal shape or normal mode of vibration is given by

-B+D=O Hence

y" (x, I) = " (x) III (t) or from eqs. (21.14) and (2U8) by

Similarly, substituting the last two boundary conditions into eq. (21.20) and setting B ;;;;; D 0 leads to (2129)
T

C sinh aL = 0


(21.23)

which, when added. give

The general solution of the equ2.ticn of motion in free vibration [hat satisfies (he bounda..ry conditions, eqs. (21.21) and (21.22), is the sum of all the nonnal modes of vibration, eq. (21.29), m:mely,

2C sinh aL= 0 From this last relation, C = 0 since the hyperbolic sine function cannot vanish except for a zero argument Thas e<;.s. (21.23) reduce to A sin aL=O

Excluding the trivial soiution (A

(2L24)

The constants A" and B" are determined, as usual, from the initial conditions. If at t = 0, the shape of the beam is gjven by

0), we obtain the frequency equation sin aL

=

0

(21.25)

which will be satisfied for (21.26)

(21.30)

Y (x,!)

y (x, 0)

= p(X)

and the velocity by 0\ or • = tjJ(x)

6":6

Dynamic Ana!ysis or Systems with Distributed Properties

Structures Mode!ed wIth ")is\fibuted Properties

21,3,2

for 0::5 x ::5 L, It faHows from eq. (2! 30) that

617

Both Ends Free (Free Beam)

The boundary conditions for a Jearn with ooth ends free are as follows.

At x=O, M (0,1)

and

= 0 or

YeO, I) = n'7TX

= L'" B"w" sin L

Ij;(x)

Therefo;-e, as shown in Chapter 5, Fourier coefficients are expressed ns

, fL p(x) sin -·····dx n'i'lX

"

L

(p'" (0)

0

M (L, r) = 0

or

(p" (L) = 0

0

or

'P'" (£) = 0

V (L, I)

(2132)

=

(21.33)

The substimLions of tllese cond;rions in eq. (21.20) Yield

L

{j

or

=0

Arx=L,

,,=1

A = ~

°



a' ( - B + D) = 0

(/J'''(O)=a'(

(21.3 I)

A+(');;;;O

and The fIrst five values for (he natural frequencies

a

J

( -

A cas aL + B sin tiL + C cosh aL + D sinh flL)

Normnl Modes

u)" =

c"fEI---_. 1,;

n

4/11'

o

2 :I

4/3",

4

o

5

* IN =

25.r f~ p,,(x)d.x! I~ (p\x)dx

4/57t

@" .....-

= sin

nm L

D=B,

C=A

(21.34)

which, 5ubstlruced into the l:iSr twO equarions, result in

(sinh aL

sin aL)A

.. ------~-~--

Shnpe

=0

From the first two equations we obtain

.---~---

t-:atural Frequencies

+ C sinh aL -r D cos.h tlL)::= 0

(cosh oL - cos eL)A

+ (cosh aL

cos aL)B

=

0

+ (sinh aL + sin aLlB = 0

(21.35)

For nontrivwl solution of eqs. (21,35), it is required thar the determinant of the unknown coefficients A and B be equal to zero; hence

Isinh aL ~ sin (lL I cosh aL -

cosh oL cos oL cos aL sinh aL + sin aL

'I'

= 0

(2l.36)

The expansion of this dererminant provides the frequency equation for (he free beam, namely cos aL cosh aL - I

=0

(21.37)

The first ftve natural frequencies which rue obtained by substituring the roots of eg. (21.37) into eq. (21.13) are presented in Table 21 2. The corresponding normal modes are obrair.ed by Jeuing A = 1 (normal modes are decer-

618

Dynamic Analysis of Systems with Distributed Properties

Structures Modeled with Djstributed PropertIes

At x;L,

TABLE 21.2 Natural Frequencies and Normal Modes for Free Beams

Nanna! Modes

y (L,;) ; 0

or

4>(L}; 0

(sinh a~ + sin a,,x)

y' (L, I) ; 0

or

4>' (Ll ; 0

Natural Frequencies tP,,(x) = cosh a,,): -:- cos a,.x

C)

=c

,0

'I

, £1 in!;"'

cosh

i_~ ~

Va

=

oJ. -

cos

C,,; (a"L)'

2

The use of the boundary conditions> eqs. (21.40). imo eq. (21.20) gives

~----

r"

(T"

Shape

22.3733

0,982502

0,8308

61.6728

1.000777

a

120,9034

0,999967

0.3640

B""D;O

and

A-C;O

while condidons, eqs. (21 A I), yield the homogeneous system (cos aL - cosh aLlB + (sin aL - sinh aL)A ; 0 ." (sin aL

3

(21.41)

L

sinh a"L - sin anL ............

n

i7 l1

O'it

619

+ sinh aL)B + (cos aL -

cosh aL)A; 0

(21.42)

Equating to zero the determinant of the coefflciems of thIS system results in

199,8594

4

the frequency equatlon

1.000001

0

1.000000

0,2323

(21.43) 2985555

5

From the first of eqs. (21.42), i( follows that

~ [" = J~ W" (x)dx Ifi; q;;. (x) dx

A; mined only to a relative magnitude), substituting in eqs" {21.35) the roots of a_~ of eq. (21.37), solving one of rnese equations for B, and finally introducing intO eq, (21.20) me constants C, D from eq. (21.34) together with B. Performi~g

cos aL - cosh aL B sin aL ..~ sinh aL

(21.44)

where B is arbitrary, To each value of [he natural frequency

:hese operations, we obtain


=

cosh a..,x + coS a,,x - ern (sinh a~ + sin a~)

(21.45)

(21.38)

where J,,;

cosh a"L - cos a,~ . h L . L s~n an - sm ali

obtained by the substitution o[ me roOts of eq. (21.43) into eq. (2U3), there (21.39)

corresponds a normal mode


21 .3.3

cosh a"x

cos

a~ -

(ff1{sinh a"x - sin anX)

(21.46)

Both Ends Fixed

The boundary conditions for a beam with both ends fixed are as follows: At x; 0, y(O.I);O

or

4>(0);0

y' (0,;); 0

Or

4>' (0) = 0

(21.40)

(21.47)

The first five natural frequencies calculated i'rom eqs. (21.43) and (21.45) and the corresponding normal rr:odes obtained from eq. (21.46) are presented in Table 2L3,

620

Oyr,arr.ic Ana;ysis 01 Systems with Distributed Properties

Stn.lctures Modeled with Distributed Properties

These boundary conditions \vhen substituted inro the shape equation (2~.20) lead to the frequency equation

TABLE 21,3 Natural Frequencies and Normal Modes for Fixed Beams

Natural Frequencies

cP" (x)

621

Nonnal Modes

cosh a"x - cos £l,,x - aft (si:1h

(/".X

cos G"L cosh (l"L

sin (I,,x)

+ 1=0

(2150)

To each root of eq. (21.50) corresponds a na:ura! freqiJency

cos: lI~L - cosh (l"L sin a"L ~ sinh {I'lL

-----~--

C" = (a"L)"

n

2

w"

Shape

, i EI

= (a"L)-.1 ~_~

22.3733

0,982502

0.8308

61.6728

1.00()777

0

(21.51)

\' mL

~-.-~~----~---.-.--

and a normal shape C/),~ (x) = (cosh

Ci'rt

~ cos (lnx) - v" (sinh

Q".X -

sin a"x)

(21.52)

where 120~9034

3

Y~~~f

0.3640

0.999967

f.J'"

cos a"L + cosh a"L = --.---------;----Sl n {I"L

4

199,8594

1.000001

0

1~

5

298.5555

i ,000000

0.2323

i ~<)n'h\}4091..

=O:f)IL

+ Sinh

The. first Eve na{u;al frequencks and the corresponding mode shapes for ca:-wiever beams are presented in Table 21.4.

offi'Li

TA!:.~-=~~4 Nat~raJ Freq uendes and Normal Modes for Cantilever Beams

Natural FreGJencies

CP" = (cosh

21.3,4

One End Fixed and the Other End Free (Cantilever Beam)

[r -.-

At the fixed end (x =: 0) of [he cami1ever beam, the deflecrion and the slope mus( be zero, and at the free end (x L) the bending moment and the shear force must be zero. Hence the boundary condilions for this beam are as

At x

a

y' (0,1) =0

or


°

0

or


V(L,r)

0

or


0

(J'"

(sinh a,,.t

sin a,,x)

a"

1,7

-~--

Shape - - ~-

3.5160

0,734096

0,7830

2

220345

10!8466

0.4340

~'.u:'

3

61.6972

0999225

0.2589

}L'~ -,,58!

~ t

4

1200902

1.000033

0,0017

~~~"

.:t

5

1998600

LOOOOOO

0,0707

'* I" ~ J~ (jJ"

(x)dx I Jt ffr" (x) dx

(2148)

= L, M(L,I)

cos a,.~{) -

-----~.-------.------.-.------.--

follows. At x=O,


CI"X'-

cos a"L + cosh tt"L Sm (!,,1- + sinh Cl"L

C" = ((I"L) '.:

=

or

Normal Modes

~-----------~--

fl

y(O,I)=O

(2 L53)

(I"L

I, (2149)

~

,

*

~yn:5L

___

!~./.

Dynamic Analysis of Systems with Distributed Properties

Structures Modeled with Distributed Properties

622

21.3.5

TABLE 21.5 Natural Frequencies and Normal Modes for Fixed Simply Supported Beams

One End Fixed and the Other Simply Supported

The boundary conditions for a beam with one end fixed and the other simply supported are as follows. At x=O, y (0, t) = 0

or

prO) = 0

y' (0, t) = 0

or

p' (0) = 0

Natural Frequencies ~(x) =

w

(21.54)

I!

11

At x = L.

y(L,l) = 0

or

peL) = 0

M(L, r) = 0

or

p" (L) = 0

!£[

=c l __ "¥ me

Normal Modes

cosh a.,x - cos anX + (T" (sinh a,,x - sin a,,x)

0",,=

cos a"L - cosh a"L sin a"L - sinh a"L

C" = (a"L)2

0',

I,:

15.4118

1.000777

0.8600

(21.55)

The substitU[ion of these boundary conditions into the shape equation (21.20) results in the frequency equation tan a"L - tanh a"L = 0

623

(21.56)

To each root of this last equation corresponds a nmural frequency (21.57)

Shape

,

1-----==

------

I

~

..

2

49.9648

1.00000 I

0.0826

~~O!£IJL

3

104.2477

1.000000

0.3345

1.....-=::-4?-~6.~"

4

178.2697

1.000000

0.0435

5

272.0309

1.000000

0.2076

,....--0294£ ---..O.7SSL ~ ---=--='''

1

Q23SL -6.'l2"9L

Q6\9L _ -tiSIOL ...

* i, = Ii
and a normal mode

P, (x) = (cosh a.,x - cos a.,x) + 0', (sinh a.,x - sin a.,x)

(21.58)

where

cos QIIL - cosh a"L sin a"L - sinh Q"L

(21.59)

The first five natural frequencies for the fixed simply supporred beam and corresponding mode shapes are presented in Table 21.5.

Consider in Fig. 21.2 a beam subjected to the inertial forces resulting from the vibrations of two different modes, ifJm (x) and ifJ" (x). The deflection curves for these IWO modes and the corresponding inertial forces are depicted in the same figure. Betti's law is applied to mese two deflection patterns. Accordingly, the work done by the inertial force, If", acting on the displacements of mode m is equal to the work of the inertial forces, I'm, acting on the displacements of mode n, that is,

(21.60)

21.4

ORTHOGONALITY CONDITION BETWEEN NORMAL MODES

The most important property of the normal modes is that of orthogonality. It is this property which makes possible the uncoupling of the equations of mmion as it has previously been shown for discrete systems. The orthogonality property for continuous systems can be demonstrated in essentially the same way as for discrete parameter systems.

The inertial force Ill! per unit length along the beam is equal to the mass per unit length times the acceleration. Inasmuch as the vibratory motion in a nonnal mode is harmonic, the amplitude of the acceleration is given by w~ifJ" (x). Hence the inertial force per unit length along the beam for the nth mode is

If" = w~m (x)cP" (x)

624

Structures Modeled with Disl.:ibuted Properties

Dynamic Analysis of Systems with Dis!nbuted PrOi='el1ies

625

in which p ()::, r) is the external loed per unt! length along the beao.. We assume that the general solution of [his equation may be expressed by the summation of the products of the norrr.al r:1odes
cfl"

y(x,r) =

(x) z" (I)

(21.65)

The nonn<:.J modes rP,,(x) sat;sfy the differential equation (2l 10), which by eq. (21.12) may be wr!:[en £~s

(2J.66)

Fig. 21.2 Beam showing two modes of vibrution and (b) Ine.'1iai forces.

i!\e;~ial

forces. {a) ;)jsplacements.

The nonnal modes should also satisfy the specific force bO'Jndary conditions m the ends of the beam. SubSliturion of eq. (21.65) in eq. (21.64) gives

p(x,r)~m:;
ElI
and for the mth mode

j,,,, = w;,m(x) ",(x)

(21.61)

In view of eq. (21.66), we can

wn:e eq. (21.67) as

,'-

Subst!l:Jtlng these expressions in eq. (21.60), we obtain

L

(21.67)

' - ' ' ' () mW"YJ..\Xj2:" r

') =P\.X,l

"

m:;
(21.68)

Mul:[plying both sides of eq. (2L68) by tP", (;t) dx and integrating between 0 and L result in

which may be writter, as

(w;~w~)r;.

(21.62)

(2169)

Ir follows that. for two differenr frequencies w" ~ w"" the nonnal modes must satisfy the relationship

We nore that aH the tenTIS that cont;;.in products of differem indices (n # m) vanish from the st.:r.lmations in eq. (21.68) in view of the orthogonality conditions, eq. (21.63), berween noma! modes, Equation (21.69) may conveniently be wriaen as

<1>," (x)
JC

"L

I

(21.63) ,,(x)m(x)dx=C Jc which is equivalent to the oahogonal condition between normal modes for dIscrete parameter systems, eq. (lO.27).

(21.70)

where n

21,5

/v/"

FORCED VIBRATION OF BEAMS

For a unifom:. beam acted on by lateral forces p (x, I), the equation of motion, eq. (21.5), may be written as

a\ ax

EI --'.f = p (x, /)

I mr:fr" (xld,

(lUI)

}o

is the modal mass. and n

F. (I) (2164)

is {he modal force.

i

h


(21.72)

626

Structures Modeled with Distributed Properties

Dynamic Analysis of Systems with Distributed Properties

The equatlor. of motion for the nth nOffilaI mode, eq. (21.70), is completely

Solution:

627

The modal s'lapes of a simply supported beam by eg. (2 L28) are

analogous to the modal equation, eq. (12.9), for discrete systems. Modal damping could certainly be introduced by simply adding the damping tenn in eg. (2 L70); hence we would obtain

(2 L73)

m. '¥"

,n;rx

n= 1,2,3,.

= SIn

(a)

and the modal force by eq, (21.72)

J, (/)" CL

which, upon dividing by M", gives

F, er) = (2L74)

[n this problem p(X,I)

Po at x=x,; otherwise. p(x,/)=O. Hence F" (r)

where §II = c,,/C.,,(r is the modal damping ratio and Kn AI/"w;, is the modal stiffr.ess. The rotal response is rhen obtained from eq. (21.65) as (he superposition of the solution of the modal equation (21.74; for as mar.y modes as desired. Though the summation in eg. (21.65) is over an infinite number of £ems, in most structural problems only the first few modes have any significant contriburion to the total response and in some cases !he response is given essentially by the con[dbutron of the first mode alone. The modal equation (21.74) is completely general and a?plies to beams with any type of load distribution. If t.he ~oads are concentrated rather than distribmed, the integral in eq. (21.72) merely becomes a summarion having one term for each load. The computation of the integral in eqs. (21.71) and (21.72) becomes tedious except for the simply supported beam because the normal shapes are rather complicated functions. Values of the ratios of these integrals needed for problems wirh unifonn distributed load are presented in the last columns of Tables 21.l through 2l.5 for some common types of beams.

(x) P (x, I) dx

Po(/)" (x,)

or using eq. (a), we obtain n11X,

= Po sin --.-

F,,(l)

(h)

D

The modal mass by eq. (21.71) is C'

M"

)0- ;;r
I

t..

o

.:< niTX

fhL

msm - - d x = D

2

(c)

Substituting the modal force, eq. (b), and the modal mass, eq" (c), into the modal equation (2] .70) resulrs in

Example 21.1. Consider in Fig. 21.3 a simply supported unifonn beam subjected to a concentrated constant force suddenly applied at a section x! units from the left support. Determine the response using modal analysis. Neglect damping.

(d)

For initial conditions of zero displacement and zero velocity, the solution of eq. (d) from egs. (4.5) is

m.EI

in which (Z'()II = 2Po sin .,(n'TlX1IL) Flg. 21.3 Simply supported beam subjected IO a suddenly applied force.

w;mL

(I)

628

Structt.:res Mode!ed with Distrib\Jled Properties

Dynamic Analysis 01 Systems with O[stribl;led Properties

629

so that £1

z"

=

2P, sin (n11X,IL) 1

W"m

m."

c

L ····~tl - cos w,,t)

(g)

30 x loa Ib~if'l? O. I Ib·'sec1/,n?

Pc) '" 200 lb/i:••

W "" 300 radhec

The modal deflection at ar:y section of the beam is

L "" 24Qin,

Fig. 21A Fixed beam with uniform harmonIc load.

(h)

which, upon substitution of eqs. (a) and (g), becomes ) ' (, i1

"."

2PD sin (nr.x,

=~-'-'"

iLl (

w!mL

1

cos w"t) sin

rr7TX

L

Soimion: {2L13) as

The na:ura! frequencies for unifonn beams are given by eq,

(i)

w"~c,,J ~~

By eq. (21.65), the lotai deflection is then

'll

2Pt; mL"

y (x, f) = -_-

--y

w,!

sin

n1T.X,

L

' (1 - cos wut) sin

As n special case, let us cO!lsider the force appUed Hence eq. U) becomes in this case

r-2·1 sa:.

D y(x,J) = 2P _

mL

II

~ W"

nIT 2 (1 -

at

n-rrx l --j L

midspnn, i.e., Xl

cos 1)),,1) sin

Lflm]

U)

or, substituring nJmerical v.dues for this exnmple, we get

Li2.

(a)

(2l.75)

whe:-e .the values of C" ue giver: for the first five modes 1:1 Table 21.3. The defiectron of the beam is given by eq. (21.65) as

From the latter (due to the presence of the factor sin n1T12) it is apparent that all the even modes do not contribute to the deflection at ar.y point. This is I.rue because such modes are anrisymmetrical (shapes in Table 2 L 1) and are not excited by a symmetrical load. It is aiso of interest to compare [he contributior. of the various modes to the de:Iection at midspan, This comparison will be done on the basis of maximum modal displacemer.t d;sregarding the manner in which these dis~ placements combine. The amplitudes will indicate the relative tr:iportance of the modes. The dynamiC load fac(Or (1 - cos Wi/f) in eq. (21.75) has a maximum va~ue of 2 for all the modes. Funhermore, since all sines are unity for odd modes and zero for even modes, the modal contributions are simply in proportion to l/w~. Hence, from Table 21.l the maximum modal deflections are in proportion <0 1, 1/81, and 11625 for the first, tbird, and fifth modes, respectively. It is appilrent. in this example, that the higher modes conD"ibute very llnle ~o the midspan deflection.

y(x,r)

=

(b;

in which <1>" (x) is 'he roodal shape defined for a [i"ed beam by eq (21.46) and (r) is the modal response. "

?;fI

The modal equation by eq, (21.70) (neg!ecting damping) may be written as ..

>

z" (t) ~ W~Z([)

fp(X,l)¢,,(X)dx

= ~··~-L-·----.----( ih¢;;(x)dx )0

Then, substituti:1g numerical values

to this

example, we obtah'.

Ii

200)0 ¢,,(x)dx ~--.--

Example 21.2. Determine the maximum deflecdon at the midpoir.t of the fixed beam shown in Fig. 21.4 subjected (0 a harmonic load p(x,t)=posin 3001 lb/in uniformly distribLHed along the span, Consider in the analysis the first three modes contributing [Q the response.

01)0 ¢;(x)d< or

Zn (r) -+- w;,z" (/):= 20001" sin 300r

(e)

630

Structures Modeled with Distributed Properties

Dynamic Analysis of Systems with Distributed Properties

TABLE 21.6 Modal Response at Midspan for the Beam in Fig. 21.4 .-.~-.--~---.

Ylode 67.28 185.45 363.56 600.98 897.76

2 3 4

5

20001. . W,,-w

Z" =---:r"-_T(m)

{I,l,

I"

4.730 7.853 lO.996 14.137 17.279

0.8380 0 03640 0 0.2323

- 0.0:94 0 0.0173 0 0.00065

I L\
Therefore, the calculation of the bending mOment or (he shear force requires onlY differentiation of the deflection function y = y (x, l) with respect to x. For example, in the case of the simple supported beam with a cor:centrated load suddenly applied at its center, difterentiation of the deflection function, eq. (21.75), gives

M=

(21.76)

1.414

2rr'P,E! "\'

L

rn'

.

nr,

-y Sill )(1 -

,,~WII

in which

r .--I ¢;(x)dx L

cp,,(x)dx

1,,""'- ~:L ~O

is given for the first five modes in Table 21.3. The modal steady-state response is

20001. . 0 - ' - - , 5m 30 I «>;- (300,-

-

y,

'L \ y I--n = .- 0.0541 sin 3001 (in) \ 2' }

JYNAMIC STRESSES IN BEAMS

To deteGTIine stresses in beams, we apply the following well-known ships for bending moment M and shear force V, namely

a2 v dX-

i!:VJ = ax

Wilt)

n=

I

cos - - , L j

(2177)

3' = 81

V,

3

This tendency in which higher modes have increasing iInportance in moment and shear calculation is generaHy true of beam response. In those cases in which the first mode dominates the response, it is possible to obrain approximate deflections and srresses from static values of these quantities amplified by (he dynamic load factor. For example. Lie maximum det]ec:;on of a si:nple supported beam with a concemraced force at midspan may be closely approximated by

relation~

If we consider only the first mode, the corresponding value given by eq, (21.75) Is

M = £! ---=:.,.,

V=

cos

We note that the higher modes are ir.creasingly more important for mOments thar; for deflections nnd even more so for shear force, as indicated by the 2 1 factors 1. n , and n , respectively, in eqs, (2L75), (21.76), and (2L77). To illustrate, we compare the amplitudes for the first and third modes at (heir maximum values. Noting that U!~ is proportional to n4- [eq" (21.27)]. we obtain from eqs. (21.75), (2:'76), and (21.77) the fOllowing rados:

The numerical calculations are conveniently presented in Table 21.6. The deflections ut midspan of the beam are then calculated from eq. (b) and values in Table 2L6 as

21.6

_

(d)

Y( 2L, I) ~ [(1.588)( -0.0194) + (- 1.410)(0.0173) + (1.414)(0.00065)] sin 30G,

631

l:.1 3'~.

r}x

l

I

L'

2P,

\

2}

mLw;

yl.<:=-b-_--,(I-c05 «>,1)

Dynamic Analysis of S,slems with Distributed Propertfes

633

Structures Mode;ed with Distributed Properties

632

Since, by eq. (2112),

wi = rr' £ilmL", i

L\

,

2,l

in which

it follows that F" (I)

2P"L'

Y lx = -I = ------{l 4 17

£1

cos Wit)

=

J,'

and

p"L' ,

,;;;;; _ _ ~\! -cos Wjt)

A4" =

48.7£1 .

The close agr<.!emem between these (wO computations is due to the fact that static deflections can also be expressed in terms of modal components, and for a beam supporting a concentrated load at midspan tbe first mode dominates

" (x) p C<, I) dx

The be:"'dl' .l no ~ mome:H

Joe· IYltjJ;Cr)d.r.

1>'U", an d t he 51'car ' for V al any section of a beam are

calculated from :he we;l-k:lOwn relations

both static and dynamic response.

21.7

SUMMARY

The dynamic an
discretization of the structure. The natural frequeocies and corresponding normal modes of single-span beams with different supportS are determined by solving the differential equa~ tion of motior:. and imposing the corresponding boundary conditions. The normal modes satisfy the ortilogonaiity condidon betweeo any ~wo modes m and n, name]y.

f

,"(x) ¢,,(x)mdx=O

PROBLEMS 21.1

(WQ

ends tixed.

21.4

Determ:_ne the maximum deflection
21.5

~ ==im p:y su~~o.::ed ~ea~, is prismalic and has the followi:;g propenies: m 0.3 " . b" lb d- se . . . Itn per mC:1 of span, EI = 1O~ lb· in~ ' and 1..-t = 1....'0'In. TI le bell,;) :s su Jcc:e to a uniform distributed static lo;:.d po which is suddenly rem ed the s:ries expression for the resulting free vibr
Wri!~

d~~rmi~~v[h~

;~e ~eam

0:. Pr:bler:: 2L~, is .acl~d l.~on by a concenlmted force given by

,I)

lO~O.:.m

Solve

~roble:n

:OOr ,.0 applled at ITS midspan. Determine the amplitude of [he :tea.dY-S[o
¢,,(x)z,,(1) 21.7

where z" (r) is the solution of n modal equation

nutural frequencies and corresponding nodal shapes of reJOrorced concrete beam huving tl cro:;s section to ~:1 wide ~~ ~410 dee~ ~ll~ ~ $p~n_o~ 36 fL Assume the flexurul stiffness of the be~m. ~.l- 3:5 x 10 .b· m~ ;:.nd wt:ght per unit volume W = 1:50 Ib/ft). (NegJec: shear dlstortlOn and rotary inertia.) Solve ?::-oblern 2 L1 for the beam wt{h one end fixed 'n" [de • other simply • «'..I supported_

21.6

I

tr.~ee

21.2 Solve Problem 21.1 for :he beum with its

The respDilse of a continuous system may be deterrni:1ed as the superposi-

y(x,') =

The firSf

s\Jppor1~a

21.3

(m ;in)

tion of modal contributions, that is,

De:er~ine

u slmp:y

det~rmt.:!e

mc",es,

21.6 asswming 10% of criticnl damping in each mode. Also lhe ste~dy-$!nte rr:otion ;),1: the quarter point consideri.1g the firs~ (wo

634 21.8

Structures Modeled with Distributed Properties The cantilever beam shown in Fig. P2L8 is prismatic and has the following properties. ih = 0.5 lb, sec2- lin per inch of span, £ = 30 X lOt, psi, L = toO in, and 1= 120 in4. Considering only t;,e first mode, compute the maximum deflection and the maximum dynumic bending moment in the beam due to (he load tine function of Fig. P2L8(b). (ChllXt in Fig, 4.5 may be used.,

,,)

22 Discretization of Continuous Systems

IbJ

Fig. P2LS.

21.9

A prismaric simply supported beam of the following prcperties~ L 120 in. El = Wi lb. in!, and iiI = 0.5 lb· sec: lin per inch of span is loaded as shown in Fig. P2L9. Write the series expression for :he deflection at the midsection of [he beam.

P(tl

P(r)

1 ! .

PlrJ

t :l)OO Ib f

1_ _ _ _ _ _ • '''''

Fig. P2L9.

21.10

21.U

21.12

Assuming rtllit the forces on the beam of Problem 21.9 are applied for only a time dnration I,} = O.l sec, and considering only the first mode, de{ermine the maximum deflection at each of (he load poi.ms of the beam. (Chart in Fig, 4.4 may be used.) A prismatic bear:'! wirh itS tWO ends fixed has the following properties: L = 1gO In. Ef = 30 x 10$ lb· in1 , Ii! = : Ib, sec? lin per inch of span. The beam is acted upon by a uniforrn:y distributed irr.pulsive force p (x, t) = 2000 sin 400t lb during a time interval equcJ to half of the period of the sinusoidal load function (rd = 1T1400 sec). Considering only the first mode, determine the maximum deflection ot {he midsection, (Chart in Fig. 8.3 may be used.) Solve Problem 21. il considering the first two modes,

Tne modal superposition method of analysis was applied in the preceding chapter to some simple strJctures having distributed properties. The determination of the response by this method requires the evaluation of several narural frequencies and corresponding mode shapes. The calculation of these dynamic properties is rather laborious, as we have seen, even for simple structures such as one-span unifonn beams, The problem becomes increasingly more complicated and unmanageable as this method of solution is applied to more complex structures. However, the analysis of such structures becomes relatively simple if for each segment or elemem o~ the structure the properties are expressed in tenns of dynamic coefficients much in the same manner as done previously when s:atic deflection functions were used as an approximation to dYnamic deflections in determining stiffness, mass, and other coefficients. . In this chapter the dynamic coefficients relating harmonic forces and displacemems at the nodal coordinates of a beam segment are obtained from dynamic deflection functions. These coefficients can then be used [0 assemble the dynamic matrix for the whole structure by the direct method as shown 1n the preceding chapters for assembling the system stiffness and mass mamces, 635

636

S1'L:~tures

Modeled with Distributed Properties

Djscfet;za~jon of COnHrtuous Systems

Also,. lr. the present chapter, [he mathemati:::al relationship between the dy-

namIc coefficients based on dynamic dlsplacement funcrions and the coefficients of the stiffness and consistent muss matrices derived from sratic displacement functions is established"

22.1

DYNAMIC MATRIX FOR FLEXURAL EFFECTS

As in the Case of statIc influence cOefficients (stiffness coefficients, for eX2.mple). the dynamk influence coefficients also relate forces and displacemems at the noda~ coordinates of a beam element. The difference between the dynamic and static coefficientS IS that the dynamic coefficients refer to nodal forces and displacements that v8ry harmonically while the static coefficients relate static forces and displacements or the nodtll coordinates, The dynam:c influer.ce coefficient 5/i is then defined as the hcrmonic force of freqJer.cy w a~ nodal coordinates i, due to a haml0nic displacement 0: r: unit ampliwde nr.d of the same frequency ii{ nodal coordinale i, To determine the expressions for the various dynamic coefncients for a uniform beam segmenc as shown in Fig, 22.1, we refer to the differential equation of mOtion, eq. (2 \.5), which io the absence of ex.ternal loads in the span, thar js, p (x, r) ::=:: 0, is

w~ note that eq. (22.3) is equiva:ent to eq. (2LW), Which is the differenttal e~ua\.JOn for rile shape funclior. of a beam segment in free vibration. The d;fference ~etw.een ,these (Wo equations is that eq, (22.3) is a function of ~he parameter a Which, 1n tum, is a function of the forcing frequency w while "a" in (2LIO) depends on the natural frequency w. The soli.Hion o"f e~. (22,3) is of the same form r;s the solution of eq. (21.l0). Thus by analogy with eq. (21.20), we can write

Cos ax + C) sinh ax + C~ cosh ax

i]>(O) ~

wt

(22.2)

Substitution of eq, (22.2) into eq. (21.1) yields

cp'v (x) -

a'
,

8,

i]>' (L) ~ 8,

8"

(22.6)

El '

,D , (22.i)

E1 '

I~ eqs, (22.6), 0\, &:., OJ, and 04 are amplitudes of linear and angular harmonic chspincements at. the nodal coordinates wble in eqs (')27' P!> P2, P d P4 are . ••.. .) 3, an the c?rr~spondJng harmOnIC forces nnd moments as shown in Fig. 22.1. The SUbstltu.tlOn of the oounda;-y conditions, eqs, (22.6) and (22.7), into eq, (22.5' resuJ ts 10 )

(22,3)

r8,1 8,

1°,0

0

ii

8,

s

c

S

-l lO'J

(214)

lie

0

~

I

Cis liC

1I'c'l

~~m:j

(22.8)

and

r;: ~ Fig. 22.1 Nodal coordinates of a flexural becm segment.


cP m '0) ~ P,

where

flJn

8,.

Also

i]>" (0) =

y (x, ,)

(22.5)

Now,. [0 obtain the dynamic coefEcient for [he be'am secm'r.t boundary conditions indlcmed by egs. (22.6; ~lOd (22.7) are imposed: ;:> 1;;.,

(22.1 )

For harmonic bOl.:nctary displacements of frequency W, we introduce in eq. (22.1) the trial solution

637

r

aJ

0

ii'

P;j EI[_,a c

(il

0

... ills

-aJc

as

-' a-c

LP~

. 0

2

8 1S

0 ~

i C ,r

til

I

- ;o'S

I~:

li-C

~ C4..,

_,I

(22.9)

638

Struc1ures Modeled wilh Distrfbuted Properties

:)iscre:ization of Continuous Systems

639

in which

s c

~

sin aL,

S

sinh ilL

cos iiL, C = cosh ilL

(22,10) lbi

Next, eq. (22,8) is sOlved for [he constants of integration C h Cz> C3 , and C~, which are subsequently substituted into eGo (22.9). 'vVe thus obtain [he dynamic matrix relating hamlo:1ic displacements and hannonic forces at the nodal coordinate of the beam segme;1t. namely

rP,,[

I

B

l;:J

Symme:ric

a' (cS+ sC) sC- cS ass

- (i2 (s - S)

l

arc - cJ

il Ie - (.j

c? (cS + sf)

S-s

asS

differential element.

"::1& r5'1 ,

!( : i

sc

(22,Il)

where u is the displacemenr at x. The displacement at x -;- dx will then be

0:;;

" 8~J cS ~:..

u + (au/lJx)dx< It is evident that the element dx in the new pOSition has changed leng"h by an amount {3u/ux)dx. and thus (he strain is au/ax. Since

from Hooke's law the tatio of stress to strain is equal to the modulus of elasticity £, we can wrl[e

where

B=

Fig. 22.2 Axiai erfeelS 0:1 a beam. (a) Nodal axial cccroll1ates. (b) Forces Jetiflg on a

aEf 1-

{22.l2)

We require the denomina[or to be different frorr, zero, that is,

au

P

ax

AE

(22.15)

(22.13)

where A is the cross-sectional area of the beam. Differemiating with respect to x resuJts in

T.'1e element dynamic matrix in eq. (22.1 1) can then be used to assemble the system dynamic matrix for a continuous beam or a plane frame in a manner entirely analogous to the assemblage of the system stiffness matrix from element stiffness matrices"

(22.16)

1 - cos aL cosh aL7*O

and combining eqs. (22.14) and (22.16) yields the differential equation for axial vibration of a beam segment, namely,

22.2

DYNAMIC MATRIX FOR AXIAL EFFECTS (22.17)

The governing equation for axinl vibration of a beam elemen: is obtained by establishing t:te dynamic equilibrium of a differential element dx of the beam, as shown jn Fig. 22.2. Thus

ap ,

a2u

aX ,

ar

P +-.-dxl\ .. p - (mdx)-,

o

A solution of eg. (22.17) of (he form u (x, I) = V (x) sin WI

(22.18)

will result in a haITilonic motion of amplitude

v (x) ~ C,

sin bx + C, cos bx

(22.\9)

D!scretization of Continuous Systems

641

Structures Modeied wl1h Distributed Properties

640

22.3

DYNAMIC MATRIX FOR TORSIONAL EFFECTS

where !m(j}

b=

I-I AE

(22.20)

The equation of motion of a beam segment lr. wrsional vibration is si:nilar to that of fhe ax:al vibration of beams discussed in the preceding section, Let x (Fig. 22.3) be measured along ~he length of the beam. Then the aoale of twist fo;- any element of length dx of the beam due to a torque T is D

and eland C:; are consmnts of inregrmio o.

To obtain the dynamic r:1utrix for the axially vlbrrtting beam segment,

boundary condirlons indicated by eqs. (22.2!) and

ore imposed,

dB=

Tdx

namely, U(L)

U(O) = 8"

= 5,

(22.21}

U' (L) = P,

U' (0)

(22.22)

AE

where 01 and ~ are the dlsplacements and PI and p} are the forces coordinates of the beam segment ~s shown i:1 Fig. 22.2.

at

~ors:onal

Substitution of the bou:1dary conditions, eqs> (22.2l) ar.d (22.22), lnto eq.

:rq

stiffness given by the product of [he torsional constani

aT ax

the :1odal

(22.19) results in

[0

where IrG is the

Ir (If is the polar moment of inertia for circular sections) and the shear modulus of elasricity G, The torque applied on the faces of the elemerH are T and T + (oT lax) dr as shown t:1 Fig. 22.3. From eq. (22"28), ~he net torque is then

-d.( = Ire

eq:.lmion of mottor.

a'e

and

'IrC,]

0

(22.29)

(22.23) fTC ax: dx

-1

dx

Equatiilg this torque ro the product of the :nass moment of inertia f;, dx of :he element (I.x and rr.e angular acce1era[ion ;i 8/o;z, we obtain the differential

lSlo bL cosbLJlC"

AEb [

d'e

(22.24)

cos bL - sin bLh C1

a'8 1"

dx

Of i)

2

8

1,;,

--0 -

ax·

iTG

o~

()

0

(22.30)

Then, solving eg, (22.23) for the constants of lotegratior;, we obtain -cot bL coseC bLl [ I 0

[l8 ,1 oJ;

(22.25)

subject to the condition (22.26)

sin bL# 0

Finally, the substitution of eq. 02.lS) iow eq. (22.24) resuirs in eq. (22.27)

relating harmonic forces and displacemeo. 2t the nodal coordlnr,tes through the dynar.1lC ffif:.tr1X for an axialiy vibrating beam segment. Th\.1S we have

P11 r cot bL [p~J =EAb ~ - cosec bL

- cosec bLl[ 0,1 cOt

bL ,.0,]

(2227)

Fig. 223 Torsional effects on a beam. (a) Nodal IOfsiona! coordin.ates. (b) Mome:1fs acting on II differemial elemenL

642

SlructUf&; Mod~led with Oistributed Properties

where

1m

Discretization of Contjnuous Systems

is the mass moment of inertia per unit length about the longitudinal

axis x given by (22.31)

in which 10 is the polar moment of inertia of the cross-sectional area A. We seek a solution of eq, (22.30) in the form

the beam is affected by the presence of this force. Consider the beam shown in Fig. 22A in which the axial force is assumed to remain constant during flexure w1th respect to both magnitude and direction. The dynamic equilibrium for a differential element dx of the beam [Fig, 224(b)] is established by equating to zero boch the sum of the forces tind the sum of the moments, Summing forces in the y direc[io[l, we obtain f &V \ a2v V + p(x, r)dx -! V + -dx! - (ihdx)-';-

\.

O(x, I) = O(x)

643

ax

ae

=a

(22,36)

son WI which, upon reduction, yields

which, upon substitution into eq. {22.30), resui.ts in a harmonic torsional motion

av +m~=D(X[) a'v

of amplitude fJ(x) = C , sin ex

+ C2 cos

(22.32)

ex

dX

dl""

The summation of momenrs about point 0 gives

in which

(2233)

I i 2 \

_ a'y \

,,

, iJy

+ -,p(X,I) - m-, lax" -- N - dx= 0 (jt- J

ax

(2238)

For a circuiar section, the torsional constant J, IS equal to the polar moment

of inertia If)' Thus eq, (2233) reduces

to r:-::T"

'mOl

(22.34)

c= I - V fIG

sir:ce 1m = IDmfA as indicated by eq. (22.31). We tlote thal eq. (2230) for wrSi'onal vibration is analogous to eq, (22,}7) for axial vibraIion of beam segments. It follows thal by analogy to eq. (22.27)

-we can write ;he dynamic relation benveen torsional moments and rotations in a beam segmen.t. Hence

[~:l 22.4

J Gc: T

cot cL

l - cosec cL

- cosec eLf e,l cot cL

Jl e,;

(22.35)

BEAM FLEXURE INCLUDING AXIAL-FORCE EFFECT

When a beam is subjected to a force along iIs longitudinal axis in addition to lateral loading, the dynamic equilibrium equation for a differentiai element of

Fig. 22.4 Beam supporting (onstam axial force and lateral dynamic load. (a) Loaded beam. (b) Forces acting on a differential element

644

OiscreHzation of Continuous Systems

Structures Modeled with Distributed Properties

Where A, B, C, and D are constants of integration and

Discarding higher order (ems, we obtain for [he shear force the expression

.

ay. aM

(2239)

V~N-'-T-

ax

645

iJx

Then using ihe familiar relationship from bending theory, p, =

(22.40)

and combining eqs. (22.37), (22.39), and (22.40). we obtain (he equation of motion of a beam segmenr including [he effect of rhe axial forces, that is,

Ct

-+ 2

(22.46)

tV a;;;;-~

(22.47)

EI (22.4 I)

(22.48) A comparison of eqs. (22.41) and {215) reveals that the presence of the axial force gives rise to an additional transverse force acting on the bea;1). As indicated previously in Secrion 21.1, In the derivation of eq. (22.41) it has been assumed that the deflections are small and that the defiectior.s due to shear forces or rotary inertia nre negligible, In tne absence of external1000s applied to tr.e span of me beam, eq. (12.41)

Tc o~tain the dynamic matrix (which in this case incluces the effect of axial vibration of [he beam elpm"'n' [he boua·" ry . cor. d: orees) or '"Y" the transverse t; _ '-' '-' ~, • 1,oos, eqs. t_-".49), are lmposed, n''1mely

f

I.,;. ....

prO) = Ii,.

reduces to

d
{22.42)

-._={). ~

d'

The solution of eq. (22.42) is fOlind as before by substituting

-dx]-= (22.43)

y (x, r) = 4>(x} sin w1 We ~hereby obtain the ordinary differenti2.1 equation

d?CP(L}

p~

-·d:/~--

£1

(22.49)

~-

(22.44)

0 j, S1 ana'!52, u~ < are, respective~y, the transverse and angUlar dIn eqs I . 12? , 49) r' lSP acemeo.s at the ends of the benm > wh;le P J, p,~ and P2 ,P. t are eorrespondJntl . foreps . ' ?? <:; ana momeniS at [heS~, nodal, coordinnres, The substitution i:HO e ~ (__ condltlons ";Ven by eqs , '(2249) q f .4_) .(ih of1 the boundary • • ,"> ;.. resuI ts '10 a system ~ ell;> : algebraic equalJons whlcr. upon elimination of the four Consrar:'s of ,nregraoon ABC ~ d D Yle, . 'ds LI",e dynamIC " matr1X ., " of , • , ll.n (meluding the effect N_

v

The solution of eq. (22.44) is
+ Bees p],x + C sinh

{J;x

+D

cosh

PIX

(22.45)

Discretization of Continuous Systems

646

647

Structures Modeled with Distributed Properties

axial forces) relating harmonic forces and displacements at the nodal coordinates of a beam segment. The final result is

n["

Symmetric

p)

52t 522 5 J ! 532 5"

P4

S~l

Pz =

5"'2 543

5"

[;]

..,~ -1' 4+32)C] S,,=S,,=B[(piP;+P,P,)cST(p,p, p,p,s

=

-

543 = B [(p I p~ - p~ P2) + (p~ P2 -

PI p~)cC +

2p~p~sS1

5" = 5" = B [(pip, + p;)sC - (p; + p; p,)eS] 5,,= -S,,=B[(p,pl+p;p,)(C-c)]

0'

~)S - (3 s,,=B[(-PiP,-P;P, p,p,2.

+ p,p,~)s]

5" = B [(pi p, + pl)s - (p, pi + p;)s]

(22.51)

In the above, the letters s, c, S, and C denote s = sin P2L,

5 = sinh PIL

= cos P2L,

C = cosh PIL

c

5"

(22.50)

where

52]

Taylor's series (Paz 1973). For the sake of the discussion, we consider the dynamic coefticient from the second [OW and tirst column of the dynamic matrix, eg. (22.11),

a1 EI sin ZlL sinh aL I - cos aL cosh aL

= -----0---,----0-

In the following derivation, operations with power series, including addi[ion, subtraction, multiplication, and division, are employed. The validity of these operations and convergence of the resulting series is proved in Knopp.! In general, convergent power series may be added, subtraC[ed, or multiplied and the resulting series will converge at least in the common interval of convergence of the two original series. The operation of division of two power series may be carried out formally; however, the determination of the radius of convergence of the resulting series is more complicated. It requires the use of theorems in the tield of complex variables and it is related to analytical continuation. Very briefly, it can be said that the power series obtained by division of twO convergent power series about a complex point 20 will be convergent in a circle with center 20 and of radius given by the closest singularity to Zo of the functions represented by the series in the numerator and denominator. The known expansions in power series about the origin of trigonometric and hyperbolic functions are used in the intermediate steps in expanding the function in eq. (22.54), namely, X

cosx coshx = 1 - and the letter B denotes El

B

I

4

X

+-- -

6

2520

[Q

22.5

X

6 1 85 X4 +- + +. x' 70 2,910,600 lO

.. "x x smx smhx=x- --+-~= 90 113,400

the condition

2p, p, - 2p, p,eC + (p; - pi)ss" 0

11

-=-=:-0= + ... 7,484,400

(I - cosx cosh x) - = -

(22.52)

6

Furthermore, eq. (22.50) is subject

(22.54)

(22.53)

POWER SERIES EXPANSION OF THE DYNAMIC MATRIX FOR FLEXURAL EFFECTS

where x = aL. Substitution of these series equations in the dynamic coefficient, eq. (22.54), yields

521

=

Z/ E1 sin aL sinh aL :-1:---c-o-s-=Z;-:L-c-o-s7h:-a:-:L:-

6El L' -

llmL'",' 210

223m' L6 ",'

2,910,600 El

Ie is of interest to demonstrate [hat the influence coe~ficients of the stiffness

matrix, eq. (14.20), and of the consistent mass matnx, eq. (l~.34), ~a~ be obtained by expanding the influence coefticients of the dynamlc matnx 10 a

I

Knopp, K., Theory and AppEicalioll of Infinice Series, Biackie, London, 1963.

(22.55)

649

Discretiza1ion of ContinuolJs Systems

Structures Motieled with Oisttibltted Pr;;p€rties

The first ternl on the righ\~hand side of eq. (22.55) is the stiffness .coefficient kZI in the stiffness matrix, eq. (14.20), and the second .lerm, the ~onslstent?mass 'fi' t . the mo'S matrix eq (1434). The senes expanSion, eq. (2_.55), coe: [Clen in21 In ... ~ ,.,

axial and for torsional effects, The Taylor's series expansions, up to three (erms, of the coefficienls of tbe dynam~c matrix in eq, (22.27) {axial effects) are

is convergent in the positive real field for

o<

aL

A£ mZL;:: L ,L:.,'::ih:.,',::w:.,' AEb cot bL = - - - --

(12.56)

< 4.73

or from eq. (22.4)

t


56) the numencal value 473

slngulanty to the angio of the

(22.57)

< (4.73),

W

. n eq. --

IS

an approximation of the closest

funCtions in (he quotient expanded in eq.

13 Lflt &i

12 £J

:0 each ser!es of eq. {22,59) is equa.! to the corresponding stiffness coefficient of :he m,n:-ix in eg, (!53), iind the second terrn to the consistent mass coe:ficient of the matrix in (15.26), Similarly, the Taylor's series expansions of the coefficients of the dyn'lmic matrix for tor . slona! e~Tects. eq. {22.35), a:-e

, , I1L-m&;-

Sn :::: --;:2 -

JTG L

Ll.;,&} 3

L3 j~,w~ 45 Gl r

=-- - - - - --(12.60)

S~

Comparing the firs: twO terms of rhe ::bove series with the stiffness and mass influence coefficients of the matr:ces in eqs. (J6.7) and 06.8), we find that for torsional effects (he first term is also equal to the stiffness coefficient, and me second term to the consisren: mass coefficient

20_3L6 m u'.!-4 w 2,9;0.600EI

420

~

[}fhZtl-

22.7

71L';P?6/

= -,- - - ,0-5 _. 4 365 900 E1 Lc

12£1

= .- -i:i--

I

9 LTn;;..? 70

"

5

POWER SERIES EXPANSION OF THE DYNAMIC MATRIX INCLUDING THE EFFECT OF AXIAL FORCES

1

1279 L "t 'f.t;4 - 3,880.860EI

series expansions of the coefficients of [he dynnmic matrix, eq. (with axial effects), are Obral:1ea by the method described in the last tWO sections. Detailed derivation of these expansions are given by Paz and Dung {l975). The series expansion of the dynamic matrix, eq. (22.50), is

The

(22.58)

[Sl = [ill

22.6

L

c

I

l r Gc cosec cL =

13L1tn&l...;.., 168i L6 mlr;,/

6E1

4£1

S)I

·

J .. Gr. cor

59 L) tn?- ZI./

-50'=7---210---

S~I:;;;; -

5" =

(22.59)

S" ="""l:' - --35-- -"161 ,7ooi'j 6£1

S,,=

45AE

It may be seen that (he first te:T:l

(22.54). "" . h d ." 't' eq The series expansions for all the coeffICIentS In ( ~ . ynarm" rna r,.x, ~ (22.ll), are obtained by the method explained In obtammg the expanSiOn o. the coefficient S~I_ These series ex.pansions are:

S)J =

3

L

AE iiIZ,l'L 7 L 3 /h z ;;/ - AEb cosec bL = - - - - - L 6 30004£

o<

649

POWER SERIES EXPANSION OF THE DYNAMIC MATRIX FOR AXIA,- AND FOR TORSIONA,- EFFECTS

Proceeding in a manner entireiy analogous to expansion of ~he dyn~n:ic coef[tcicms for flexural effects, we can also expand the dynamiC coeftJctenrs for

(Gol N - [Mol w' - [A ,] Nw' -IG,]N' - IM,l w' - ...

(22.6l)

where the first three matrices in this expJ;}sion [K]. [G o1. and Uvic) are, respectively, the stiffness, geometric. and mass mmrlces which were obtained in previous chapters 0;) the basis of s(a.~ic displacement functions. These matrices nre given, respectively, by eqs. 04.20), (14,45), aod (1434). The other rna,

trices in eq, (22,61) cO:Tesponding

to

higher order terms are represented as

650

Structures MODeled with Distributed propertieS

PART V

follows. The second~order mass¥geometrical matrix:

[A,l=

mL3

EI~

Syrnmetric

3150 L

L'

1260

3150

I

3150

1

L 1680 (1

L 1680 3600 ~

3150 L 1260

I

I i I

Random Vibration

L'I

3i50J

The second-order geometrical matrix:

17~0

I

L'

.] =~-I-I400 [G , £1 _~ _ 700

C

Symmetric 11 l' 6300" __I-I} 1400 13L3

700 ~1~_L2 II L' ~6300 1400

-~-.-

c

1400

1

L

12600

I

J

The second~order masS matrix:

m"! L' [M,]='-IOOOEJ

c

22.S

59 161.7 223L 291;16 1279 3880.8 - 1681 L 23284.8

Symmetric 71 L' 4365.9 ,681 L ---23284.8 1097 L' -69854.4

--~

59 161.7

223L ----~

2910.6

71 L'

43659

I,

J

SUMMARY

The dynamic coefficients relating harmonic forces and displacements at the nodal coordinates of a beam segment were obtained from dynamic deflection equations. These coefficients can then be used in assembling the dynamic matrix for rhe entire structure by the same procedure (direct method) employed in assembling the stiffness and mass matrices for discrete systems. 10 this cbapter ir has been demonstrated that the stiffness, consistent mass, and other influence coefficients may be obtained by expanding rhe dynamic influence coefficients in Taylor's series. This mathematical approach also provides higher order influence coefficients and the determination of (he radius of convergence of Ihe series expansion,

23 Random Vibration

,

p.

The preVlOl:S chapters of this book have dealt with rhe dynnmic analysis of structures sUbjected to excitat:ons which were known as a function of time. Such an analysis is said to be detemlinistic. When an excitation function applied (0 a Structure has an irregular shape that is described indirectly by statistical means, we speak of a random vlonlfion. Such a ft'IlCtlon is usually described as a continuous or discrete function of the exciting frequencies, in a manner similar to the description of f!. function by Fourier series. In stnJctural dynamics, the random excitations most often encountered are either motion rransmmed through ~he foundmion or acoustic pressure. Both of these types of load:ng are usually generated by explosions occurring in :he vicinity of the structure. Common sources of these explosions are COi1structlOf'. work and mining. O:her types of loading, such as earthquake excitation, may also be considered a random functicr. of rime. In these cases the structural response is obrair.ed in probabilistic terms using random vibrntion theory. A record of random vibratton is a time functlon such as shown in Fig. 23, L The main characteristic of such a randor:1 function is that its instantaneous value cannot be predicted in a deterministic sense. The description and ana;:ysis or random processes are established in n probabiiistic sense for which it is necessary to use ~ools provided 'Jy the theory of s[at!stics. 653

Random Vibration

655

Random Vibration

654

assume that all random processes considered are stationary and ergodic. The :andom function of time shown in Fig. 23.1 has been recorded during an mterval of time T, Several averages are useful in describing such a random function The most common are the mean value i: which is defined as

r T

x= Fig. 23.1 Record of a modam function of time.

23.1 [0

1

T ),

x(I)dl

and the mean~square value .~2 defined as

STATISTICAL DESCRIPTION OF RANDOM FUNCTIONS

any statistical method

2.

large number of responses 1S needed to describe a

n:md~m function. For example, to establiSh the statistics of the foundation

excitation due to explosions in the vicinity of a st"nlcture, many records of the type shown in Fig. 232 may be needed. Each record is called a sample, and the !Otal collection of samples an ensembie. To describe an ensemble statisticalty, we can compute at any time t; the average vajue of the insrantaneous

If such averages do not differ as we select diff~~ent ~~iues of l". then [he random process is said to be stationary. 1-n addItIon, It the aver'~ge obtained with respect to time for any member of the ensemble is equal

displacements

(23.2)

Both the mean and the mean~square values provjde measurements for the average value of the random function x(t). The measure of how widely the funcrion x(t) differs from the average is given by its variance, namely,

,

1

r

T

&:=-, , r)o

Xi'

to the average acrosS the ensemble at an arbitrary time t l , the random process is called ergodic. TIlliS in a stationary, ergodic process, a sIngle record may be used to obtain the smtistical description of a random function. We shall

(231)

, [x(I)~XJ-dl

.

.

(23.3)

When the expression under the integral is: expanded and then integrated. we find that (23.4 )

which means that the variance can be calculated as the mean~square minus the square of the mean. Quite often the n:ean value is zero, in which case variance is equal to the mean square value. The root mean~square RNfS" of the random function x(t) is defined as (23.5)

The standard deviation. eg. (23A)

"

I A

•

'

IV~.

ft." ~ ....

1\1\

1\"

Ii/',

t;pI

iWdyrIP>lJfwvvrv'\i 'ljJ Vi

1

O'r

of xU) is the S<1uare root of the variance; hence from

(23.6)

v

[, i

Fig. 23.2 An ensemble of random functions of time.

Example 23~1~ Determine me mean value F, the mean-square value F2, the variance ~> and :he root mean square values R,.\1Sp of the forCing function F(t) shown in Fig. 23.3.

656

657

Random Vibration

Random Vlbrallon Xld

1

I~' - ...

X",I'~_

x~

-~

---~. - -

I

___ _

-------f-, -------

ifr-:-"-'I :;~" ,

,,:

!

Fig. 23.3 Forcing function for Exampie 23, L

SOlution: Since the force F(lJ is periodic with period T, we can take the duration of the force equal to T; hence by eq. (23. t) we have I F=T

JT F(I)dl=-T F=< 0

I Fig. 23.4 Por-ion of random record showing determination of probabilities.

-

23.2 and noting that

PROBABILITY DENSITY FUNCTION

Figure 23.4 shows X2), we may draw horizontal lines through the values Xi and X2, and [hen measure :he corresponding time intervals at;. The ratio indicated by (23.7)

we obtain by eg:, ('23.2)

Tne variance may now be caiculated from eg:. (23.4) as

oi= F' -

and calculated for the e.nt:re record !cngrh T, is the probability of x having a value between Xl and x; at any selected time l, during the random ,?rocess. Similarly, [he probability of x (I) being sma:ter than a value x cail be expressed as P(x) = PIX{l)

(FI' 4

Finally the root mean square value of F(t) JS

< xJ = ;im

~I

T- '"

,

where the rirr.e interv::ds nt, are. now those for which value smalier than the specified .Y. The funCtion P(x) in eq. {23,S} is known as the fUf,ction of ihe random functio:1 xC!). 1n:s function is a function of x The cumulative distribution function creasing function for which P(

00)=0.

O~P(x)S

I,

(23.8)

Liti

the function X(I) has a cumulative disrriburlon ploned in 23.5a as is a monotonica:ly in-

P(x)~

I

658

Rartdorr. Vibration

Ra"ldom Vlbra:icn

659

represented as the [wo shaded "ta:1" areas in 23.5(b). Since every real x lies in the interval ( - 00, co). the area under Ene entire probability density function is equal to L that is. ('p(x)dx=1

(23.1 i)

~-,

Thus as x tends ish (0 zero,

(Q

infinity in either direction. p (x) must asymptotically dimin-

(3) p(x)

23,3 THE NORMAL DISTRIBUTION The most commonly used probability density function is the nonnat disrribution. also referred to as the Gaussian distribution, expressed by p(x) Ax

pix) =

(23.12)

-x", (b) Fig. 23.5 (a) Cut:mlative probability function P(x) and (b) Probability density function p(X) of the random variable x"'" XCI).

Now, the probability that the value of the random variable is srnaUer than the value x.,. ili; is denoted by P(x + ili;) and that x(r) takes a value between x and x + ili; is PIx + Llx) - P(x), This allows us to define 'he probability

Figure 23.6 shows the shape of this function. It may be obser.red that the normal distribudon function is symmetric about the mean value x. In Fig. 23.7 the standard normal distr.bU(ion is plotted nondimensionally in terms of (x - J:)!(]'. Values of P( ox; , x,)lx, - ox; in eq, (23.10)] are tabulated in many sources including mathematical handbooks.! The probability of x being be~ P(x)

density function as P(x

+ ili;) -

P(x)

dP(x)

ili;

(23,9)

dx

Thus, the probability density function p(x) is represen,ed geome:rically by the slope of the cumulative probability function P(x). Tne functiO,ns P(x) and p(x)

are shown in Figs. 23.5(a) and (b), respectlvely, From eq. (2",.9) we conclude that the probability that a random variable xU) has a value between ,X and x·f- dx is given by pix) dx, where p(x) is the probability density function. Having prescribed p(x), for ex.ample, as the function plotted ,in P~g. ~3.5(b). the probability of x being in the range (X11 X2) at any selected tIme is glVen by P(x,";XSx,)

= tp(X)dx

x

Fig. 23.6 Nomlul probability density function,

(23,10)

and is equal to the shaded area shown between x\ and X2 .in Fig. ?3.5(b). Similarly, the probability of x being greater than X m• that is, P(;x~ >xm} can be

1 5;andard Mathematical Tables, T:!e Chemical Rubber Co. (CRe) 20th Ed., 1972. pp. 566-575.

660

Random Vlbra:ion

Random Vibration

661

,

pix)

o.st' o.~

O.4~ O.3~ I

0.2:-

~

0.1 ':

Fig. 23.1 Stnm1ard normal probabili,y density function.

o -..-----'----,..~.. la

tween i-ACT and

x + AU,

where A is any positive number, ~s g~ven by the

20

A

30'

4"

Fig. 23.8 Ray:eigh probability density function.

equation

r i" l' + P[x-Acr
J21TCT

(23.13)

The [;lean and :near. square valL:es for the Rayleigh distribution function can be found as

J-A(f

Equacion (23.13) represents the probability that x lies within A standard deviulions from x. The probability of x lying more rhzn A sta:1dard deviations from x IS [he probability of lx xl exceeding Au, which is 1.0 mlnus [he value gIven by eq. C23.13), The following wble presents numencLI va;ues for the nonnal distribution associated wirh ;\ = 1> 2, and 3:

A -~~

I 2 3

..- -

p~-Aa
.-~

... --~

A

'--I« -'0

A'p(A,)dA

=

-.";:.<;,,,., dA = 2cr 1'< /\) (J":::e

Jo

Therefore, the ROOf Mean Square va:ue of the random variable A. (RMS.1) is

p[lx-xi >Aaj

..

68.3% 95.4% 99.7%

31.7% 4.6% 0.3%

(23.15) The varis.nce associ.u:ed with the Rayleigh distribution function is, by eq. (23.4),

23.4

THE RAYLEIGH DISTRIBUTION

(23.16)

Variables that are positive, such as the absolute value A of the pe2.ks of vibration of 2. random function xC!), often tend to follow the Rayleigh distribution, which is defined by the equation (23.14)

The probabmty of A exceedin bo a specified val"e ) "'" P (A>'f\U;,.lS de fime d as • '--' /lv,

P(A> Au) = (" p(A)dA J

A!T

which after substituting p(A) from eq. {23.14) results in where (T is tr.e standard deviation of the function X(f). The probability density p(A) is zerO for A < 0 and has tbe shape shown in Fig. 23.8 for positive values A.

P(A>Acr)=j('"

'a

A

e,~";'r.!q-:d4

662

Random Vibration Random Vibration

which may be written as

r·

P(A>AIT)= "

663

e('A'M)d(-A'/21T')

Au

Thus,

Also, the p::obability of A exceeding a specified value AVA, P(A > AVA) is given by

Fig. 23.9 Correlation between

Xl (()

and x.!(t).

Substituting from eq. (23,16) v~ = 17~/0.429 and integraring resu:ts in (23,18)

The folluwing table presents values for the probability of A, calculated from eqs. {23.17) and (23.18)~ of exceeding Av, or At7A for A = 1, 2, 3, and 4, ,\

P[A>AIT]

PtA > AlTA]

%

%

60,65

80,69

13,53

4240

1.11

14.51 3.23

~.~--.~---

----~-.--.

2 3 4

0-03

Fig. 23.10 Autocorrelation between x(t) and xC< + T)<

'".

~he correlation between x(t) and x!(t) = xU + 1") is known as the auwcorrela-

23,5

CORRELATION

Correlation is a measure of the dependence between two random processes. Conslcer the two records shown in Fig. 23.9. The correlation between them is calculated by multiplying the coordinates of these two records at each time t, and computing the average over aU values of L It is evident that the correlationship so found will be iarger when the two records are simiizy, For dissimilar records with mean zero, some products will be positive and others negative, Hence LlJeir average product will approach zero. We consider now the two records shown in Fig. 23.10 in which XI(I) is identical to X(l) but shifted to the left In (he amount 7, that is, XI (J) = x(t + 1").

!Jon R(T) and

IS given by

R(7) = lim -I T--ott T

When

';-=

iT

x(r)x(r+ 7)dt

0

(23.19)

0, the autocorrelation reduces to the meat!. square value, that is,

R(O) = lim T_",

. [r [xl!)]' dt =

~ ~

JO

(23,20)

Since the second record of Fig. 23.10 Can be considered to be delayed with

8andom Vibration

664

Random Vibration

665

I

fAN~'vL' Fig. 23.13 Periodic function x(r) of Example 23.2.

(bl

(oj

Fig. 23.11 Wide-band random process x(r) and its aUlocorreiation function R(T).

(a; respect to the first record, or the first record advanced Wilh respect to the second record. it is evident that R{'T) = R( - T) is symmerric about the R axis and that R(7) is always less than R(O). Highly random or wide-bar.d functions such as the one shown tn Fig. 23.1 {(a) lose their similarity within a short time shifL The autocorrelation of such a function, therefore. is a sharp spike at 7= 0 to
Example 23.2. For the function depicted in Fig. 23.13, de:ennine: (a) the mean, (b) [he mean square value, and {c) the flu(Ocorrelarion function. Also,

plot the autocorreiatlon function.

XCI) =

T

_ I (~ Xv x'!!;;xodt:=:-

T )0

(b)

Xc

=0

Tn
,

1 q ') Xu TJ xod!= 2 •

To c.a.lcula,te (he autocorrelation function. It is necessary to distinguish b~tween the (:me shift 0 < T < T 12 and T 12 < 7' < T as shown, respectively in FIguces 23.14(a) and 23.14(b),

'

Usmg eq. (23.19) and considering Fig. 23.14(a), we obtain for 0<7
J J::(I;X(t+T)d;= T1 (I:: -xo·xodr

, rr

r

)0

Q

(a)

(c)

0

AutOcorrelation fU:1c(lor::

R(T) = T'

O
T

,

(c}

(b)

2

Mean square value: by eq. (23.2)

x~=

fd) ,

Solurion: Because the function X(l) is periodic, ave:-ages calculated over a long time approach those calclllated over a single period. Considering the period between 0 and T. the ft:nction is described analytically by

Mean value: by eg. (23.1)

x5

T

T\2

\

~!!

T\

\.L

TJ

TI=xoi~i

for

0<7<

T (d)

in which the limits of integration are given by {he overlapping portio!lS of xU) and x(t+ T) as it.dicared by the shaded areil in Fig. 23.14(a). When T 12 < r < T, we obtain from Fig. 23,14(b)

,.) Fig. :13.12 Narrow-band random process x(t) and its autocorrelation function R(,j.

2

+~l

T

for

~-
2

(e)

ceo

Random Vibration

Raf"dom V;bratlon X(t),X(t + .)

t

_

_-.JII-.JI_-1il-1I_....,.r-+IULIIilw.!I'u,.c'i -"..'!'}____ , 1 ; ~!nlOI! -:-1_ _.,'1.1,._ '_. t

r

-.

T

T

,.,..

-,

T I 'f-'l 1:

(al

,1"'

r

Vle begin by showing that the Fourier integral can be viewed as a limiting case of the Fourier series as the period goes to infinity. Toward [his objective we consider the Fourier series in exponential form and substitute the coem~ cient en g:ven by eq. (520) into eq. (5,19):

11"

,

(23.21)

X{t),X (r + .)

'~--, ,. , ITT--l I ! !

t-' ~

Iii I

(b)

Fig. 23.14 plot of

667

;r. (I)

und

;r.(t

+ 7);

(0.) for 0

~I

, .

~T

-: T

I

In eq. (23.21) we have selected the integration period from - Tl2 to TI2 and substituted the symbol T for r as the dummy vuriable of integration. The frequency w =:::: nw is specified here at discrete, equally spaced values separated by the increment

2"

Llw = (n -,. 1)& - n& = &

T

< 7< Tn, (b) for Ti2 < 7< T.

We $ubs~:tute w for nw and /J.w/2" for 1 IT in eq. (232U and notice that as T ~ 00, Llw-) dw. Thus, in the limi;:, eq. (23.21) becomes F(t) =

I

0_ _II

r

I

-.

J

-00

'r" ,

I

LJ

,.

F,,)e

.' ,

-,«rt.

10;<

aT> e dw

J

-;>!

(23.22)

which is the Fourier integral of F{t)_ Sir:.ce the function within the !nner braces 1S a function of only w. we can \vrite this equation in two parts as (23.23)

Fig. 23.15 Autocorrelurron function R(,) for the function of Exumpl.e 23.2.

and From exarnination of Fig. 23,14, it can be conduded tha~ the autocorrelation function R C~) for the function x (t) must be periodic in t with period T. Tnerefore, from eq. (d), (e), and rhe facr 'hat R(7) is periodic, we can plor rhe autocorrelation function as shown in Fig, 23. J5.

23.6

THE FOURIER TRANSFORM

1n Chapter 5, we Ilsed Fourier series to obtain the frequency components of periodic functions of time. In general, random vibrations are not periodic, and the frequency analysis requires the extension of Fourier series to the Fourier integral for nonperiodic functions, Fourier transfonr.s, which result from Fourier integrals, enable a more extensive treatment of the random vibration problem.

F(t)

r'-. C(W)Ci~

,

The validity of these relations, according is subject [0 the condition that

r'

, -.

to

dw

(23.24)

classical Fourier transfonn theory,

IF(rlidr<

00

(23.25)

The funcrion C(w) in eq. (23.23) is 'he Fourier rrans/orm of F(r) and the functio:1 F(;) in eq. (23.24) is the inverse Fourier transform of C(w). The pair of functions Fet) and C(w) is referred to as a Fourier transfonn pair. Equation (23.23) resolves the function F(r) ioro harmonic components c(,o). whereas eq. (23.24) synrhesizes F(t) from these harmonic components, In practice it is more convenient to use frequency f in cps rather than (he angUlar frequency

668

Random VIbration

Random Vib:ation

win rad/sec. Mathematically, si:1(;e (;:) 211fand dw=2rrdJ, ~his also has the advan:age of reducing the Fourier transform pair into a more sj'mmelric form,

subs~:t;Jtion

of x(r) from eq {23.28) for One of the factors in the definition of mean square value gives

namely, F(t) =

r~ e(f)e""" elf

(23.26)

669

, I x' ="-

T

iT X"(l) ., clf

J,

and (23.27) fro [,.

=-:;:.

~JO

23.7

SPECTRAL ANALYSIS

We have seen in Chapter 5 that the application of Fourier analysis to a periodic function yields the frequency components of the function given by eicher trigonometric tenns (eq. (5.2)] or exponential termS [eq. (5.19)}. When the periodic function is known at N discrete, equally spaced times, the frequency components nre then given by rhe termS in eq. (5.28). Our purpose in this section is to relate the Fourier analysjs for a given function x(r) to its meiln

-,

square value x . The cO:1tributions of the frequency componems of x(r) to the value x:' are referred to as the spectral jUflction of x (t). Hence spectral analysis consists 1n expressing? in terms of either the coefficientS of the Fourier series (i.e., an and b" or equivaJentiy Cn) when x(!) is periodic, or in terms of the Fourier transform [i.e., C(w)] when x(r) is not a periodic f:mction. We begin by performing (he spectral analysis of a periodic function X(l) expressed [n Fourier series, eg. (5,2), WhiCh is

',"

x(r)dr+;>

I.: I f {/

b, '--r

'--T

1

x(f)COStlWtrlr+-=-1

.<(l)sinr.wfr/fi TJv J

""'1,1.,0

FinaEy subsritutlng the integral exp!."essions frorr. eqs. (23.29) we get ehe desired formula as x" = a~

+

The speClrum of the f'Jnc:ion xU) is then given by the terms of the series in eq. (23"30). Each term of th:s series is the contribution of the corresDondinf! frequency :0 the mean square value of xCf). • We now consider a discrete [in:e function F(/f) expressed as a discrete Fourier tmnsform [eq. (5.28)], that is. as ,v-I

')' C"e! m'''Jf,Yj

(13.3 J)

,,"';"0

where C" IS given by eq. (5.27) as xU) =

+

Go

{a" cos nwf

+ b" sin !1wt}

(23.28) (23.32)

where the coefficientS are given by eq. (5.3) as

00

I ('

= -;:c

,0

?

J

a -::::'.-..::. " T

17

"

I

t

x(t)dt

rf

x(t) cos flwdt

1)

? IT X(I) sin nwdt =.::.. T 0

In eq. (23.29). T is the period of the function and IV

(23.29)

2'1T/T its frf'4uency. The

As indicated io Chapter 5 reg. (5.29) or (5.30)]. harmonic companeots of tile function F(ri) higher than the Nyquist freque:icy, WNll = If/ijl rad/sec or /''',1 = 1nilt cps, are nct i:tduded l;l the discrete Fourier transform [eq. (23.31)), Also as noted in Chapter 5, if there are harmonic components in F(l;) higher than this limiting vaLle, these higher frequencies introduce distorting contributions to the lower harmonic frequencies. Hence it is imperative that the value of N be selected sufficiently large to include the frequencies rhat contribute s:gnificandy to the onginal function, To be certain thnt this condition is satisfied, one may filter the signa! of the function electron~ca!ly tc remOve aU frequencies higher than the Nyquisr frequency.

Random Vibration

E70

$71

Aa:1dom Vibration ~J\lps)

Fir)

1:1e mean square value of a discrete funcdon F{t})(j = 0, 1, 2, ... , N ~ 1) is

t

obtained from eg. (23.2) as 1 8 - 1

_

F'

= ~ :>

T j"';;O

F'(t.)Llr

(2333)

J

Substituting LIllY for liN and using egs. (23.31) and (23.32) for one factor F(r) in oS. (23.33), we obtain

-120'

Fig. 23.16 Forcing funcrion for Example 23.3. )'i-l

=

:>

j70

C"C:

Table 23, I IS ?,5~OO E 10. We check this value by calculating the mean square value of F(l) dlfecrly from the definition eq. (23.2), namely,

(23.34)

where C,; is the complex conjugate of Cn .? Hence /'i-I

= I ic"I'= ico!'+ !cJ+ Icd"'''·''' I

F' =

T

in which we substitute

The terms of the summation in eg. (23.35) ate the required spectrum of rhe discrete function F(lj); that is, these terms are the frequency contributions to the mean square value F2. As we -can see in eq. (23.35), the contribution of each frequency is equal to the square of the modulus of the corresponding complex coefficient Cn whic!i is given by eq. (23.32).

F(t) = !20,OOOriO.16

F ' = _4_ 1'°·16 0.64 ), TABLE 23,1

vrus, Use the spectrum to estimate the mean square value F:' and compare this result with the mean square value of F(t) calculated directly from the defini-

We use compmer Program 4 with N = 8 to determine the discrete Fourier coeffidents C". The values thus obmined are shown in Table 23, 1 together with the calculation needed to obtain the speCtrllm using eq. (23.35). The summation of rhe spectral values for F(r) shown in 'he last column of

n

Re (C,)

0

0 0 0 0 0 0 0 0

2

3 4 5

6 7

I[S

conjugate F(ll)' = F(!)
e~"'("I!NI,

j[

0.16

Fourier Coefficients C"

rion of

Solution:

[ 120,000 'I:

dl

0.4800 E 10

Spectral Analysis for the Function F(t) in Fig, 23,16

Example 23~3. Determine the spectrum of the function F(t) shown in Fig. 23.16. Assume that the function is defined at eight equally spaced time inler~

: Since F(I!) is a rca! function.

iT F'(t)dr jc

(23.35)

Jl"'[j

follows :.... a: tiN ~N,,-(.I F(t:)i''''"!I.>-'.". C.

~

1m CC) 0 - 51,210 0

Spectrum of F (tl

IC!

i

en 12

51,210

0 0.2623 E 10

8787

8787

0.7721 E 8

0 8787

8787

0.7721 E 8

51,210

0 0.2623 E !O

0

0

0 51,2lO

0.5400 E 10

672

Random Vibration Random Vibration

TABLE 23.2

Mean Square Value for Function F(t) in Fig. 23.16

ExponelH

Time Intervals

M

JV= 2M

0.5400 0.4950 0.4838 0.4809 04802

32

6

64 !28

7

and

Mean Square Value pl

8 16

3 4 5

E E E E E

(23.36b)

In eq (?3 36 , S . ). . . . . . -, a" .t\W IS ca.led the spectral aeflsuy fum..·lion of x(/) Th 1mp0.r{;)nt properly of SAw) becomes apparent by leulno- T= 0 ZO e~. (2; ;~~s)t hi thiS case ¢ > . •

10 10 10 10 !O

R,(O) =

.~--.~

-_.__.

..----..

Exac, p' --.~

..

= 0,4800 E

r:

S,(w)dw

(23.37)

which by eq. (23,10) is equal to mean squa,e vniue, that is,

0.4800 E 10

Cons.idering that we have used In this examp!e a relatively small number of intervals or sampled pOint!> (/1;'::;. 8), the value of p!. = 0.5400 E 10 obtained from the s?ectium or F(!) in Table 23.1 compares fairly well with the ex~c[ mean square va}ue F1 = 0.4800 E W. As discussed before, errors are introduced in :he caiculations when the number of time intervals Iv" is not Jarge enough :0 include the higher frequency components of F(E). In order !O improve the ca:culat:on of the spectmm, it is necessary to use more time intervals in the discrete Fourier Series. Table 23.2 shows results obtained using Program 4 with M = 3, 4, 5, 6. 7, corresponding to N = 2M = 8. l6, 32, 64, 128 time intervals for the function F(I) shown in Fig. 23.16. It may be observed that values displayed in the last column of the table are converging to the exact value

.073

(23.38) The mean square value of n rnncom process is therefore given by the area

~nder the graph of (h~ spectra! density function as shown in Fig. 23.17. Consequently~ the contnbutIOn of an incremental frequency .dw to the me

square value

an

IS

(23.39)

~f11e ~ourier tr.ansform pair for the discrete amocorrelarion function R. = R( T) and tor rhe discrete specmd density f'Jnclion S = ('(w) of 'I ' . " ...,,, x ( f), J$ definec from eqs. (527) and (5.28) as

x

'f .' Ime unctIOn

0 "

10 as the number of time intervals, N, is increased.

(23.400)

23.8

and

SPECTRAL DENSITY FUNCTION

If a random process x(t) is normalized (or adjusted) so that the mear. value of the process is zero, then, provided [hat x{t) has no periodic components, the autocorrelation function R.cC r) approaches zero as '7 increases, that IS,

lim Rir) ,~"

(23.40b)

=0

Vle therefore expccr that R,kr} shouJd sat:isfy the condition in eq. (23.25). We

can now apply eqs. (23.23) and (23.24) to obtain the Founer transform SJw) of the autoco::-relation function RAT) ane. its inverse as (23.36a) Fig. 23.17 Spectral density funcricn showicg area equal to mean-square value.

674

Random Vibration

Random Vibration

Equarion (23.40b) provides the contribution, [he frequency w'" as

R, to mean square

value

2, at

in eq. (23,36a) yields the cosine trnnsfomJ:

C:.

&1

=::

C"C:

277

R,(r) cos wrdT

(23.47)

It is clear from eg. (23.47) {ha~ SAw) :s also nn even function of w; hence eq. (23.36b) may be written as

(23.42)

Therefore, it follows from eqs. (23.41) and (23,42) 'hot

=

1-. r·

1

SAw)

(23.41) The contribution to the mean square value may also be expressed, in (enns of the Fourier transform coeft1cient C" and ![S conjugate using eg. (23.34) as

675

R,(T);;';;;

Jr'

-t>!

S,{w) cos wrdw

(23.48)

Alternatively, eqs, (23.47) and (23A8) may be wrItten as

(23.43) The spectral density of a given record can be obtained electronically by an Instrument ca\!ed afrequency analyzer or specrraf density analyzer. The ourput of an accelerometer or other vibration transducer is fed into the instrument, which is essentially a variable frequency narrow-band filEer with a spectral meter to display [he fiitered output. With this instrument [he experintenter searches for the predominant frequencies present in a vibration signal. The output of the spectral density analyzer IS l1e contribution lo the mean square value ilx? of the input signai X(f) for a smnll range Llw around the set frequency. \Vhen dealing with theory, the natural unit for the frequency is rad Isec. However, in most practicnl problems the frequency is expressed in cycles per second or Hertz (abbreviated Hz), In the laner case, we rewrite eg. (23.39) as

= S,(j)t!.j

(23.44)

where f is the freque:1cy in Hertz. Since ilw =: 27TLlj, it foHows from eqs_ (23.39) and (23,44) that

(23,45)

(23.49)

and r· R,(T)=2 Jo S,(w) cos w-cdw

These are the celebrated Fliener--Kinchin equations, which describe how the spectral density funn:on can be detenni:1td from the autocorrelation function and vice versa.

23.9

NARROW-BAND AND WIDE-BAND RANDOM PROCESSES

A process whose spectral density function has nonzero values oniy in a !larrow frequency range as shown in Fig. 23.18 is called a narrow band process. In contrast, a wide-hand process is one whose spec(ra~ density function is nOnzero Over a broad range of frequencies. The time history of such a process is then

When the spectrni density function for the excitmion is known, irs mean-square value may be determined from eq. (23.44) as:

';:' = [

(23.46)

SAJ)dj

The spectral density function SA]) is expressed in square units of x per Hertz, Since rhe autocorrelation funcrJon Rx{ T) is real and even. the use of Euler's relationship

i""'" = cos

Wi

+ i sin

wr

(2350)

Fig. 23.18 Spec!ral density function for a natrow-band random process.

Random Vibraticn

677

Random Vibration

676

S..,{wl

xitl I

•

~J ,,'

10)

Fig. 23,19 Wide-band process. (a) Time history. (b) Spectral density funCtlon.

fig. 23.20 A:.J.tocorrelaiior; for

0:

made up of the superposltion of the whole band f;-equencies as shown in Fig. 23,1 9(a). In the limit, whe;,1 the frequency band extends from WI = 0 to WJ. = 00, this spectrum is called white noise. From eq. (23.38) the mean square value of a white noise process must be infinite; therefore, the white noise process is only a theoretical concept. In practice a process is called white noise when the bandwidth of its frequencies extends wei! beyond all the frequeacies of interest.

Example 23.4~ Determine the mean square value and the autocorrelation function for the na.rrow-band ra:ldom process .:tCt) whose spectral density function is shown in Fig. 23.18.

Solution,'

II nllrrow~btlnd

mndom process.

The autocorrelation function for a narrow-band random process given by has the forrT'. shown in Fig. 23.20. where the predominant frequency of R,.vr; 15 th~ average value (w~ + 6wl2). The autocorreJation for such a process has a maXJmum of 250 au) when T= 0 and decreases like a cosine graph as T moves from T= O. The autocorrelation function R,( r) for a wIde-band random process whose spectral ~en~ity function extends in. the range w: to Wz as shown i~ Fig. 23,19(b) ~an be o:)tamed from the result of Example 23.4. In t.his case, letting the lower trequency Wi = 0 and the upper frequency .dw,;:;: W:h we obtain from eq. (b) of Example 23.4 . eq~ (?~

(23.51)

From eq. (23.38) which has the fom: shown in Fig. 23.22(a).

and from eq. (2350)

RAT)

L

= 2 '"

250

S,(w)

(a)

cos wrdw

.

Exam,ple 23.5~ Determine the autocorrelation function for constant spec~ tral densIty fUnctIOn S(Q);:;;; Sa (white noise) for aH value of the frequency w as shown in Fig. 23,21. ~ ~e ~utoco~eiation function for white noise may be obtained from eq. (23,) 1) oy lettmg Wz -7 cc" In this case adjacent cycles come closer together,

~Slw)

""..w,ju

=--Ism w'l'i..., 7

s, R(Cr) ,

4S

=- r 0

coS

(

\

w\

Llw\ 2

.

Llw -

+--)7' SIn - ? '

(b)

Fig. 23.21 Spectral density function for Example 23.5.

678

Random Vibratlon

RaMom VlbrailOn R~

679

Hence

Ir) "" 2 ... Sc 5{7j

c

1

S,~-

(2355)

21T

Finally, so!ving for C 8,od substituting into eq, (23.53), we obtain rhe at.:tocor· relat;on function for white noise shown in Fig_ 2322(b), as (23.56)

23.10

(0)

I.)

Fig. 23.22 Autocorrelation for" wide-band fandor., process becomes a delta function

for white noise"

resulting in a high peak at r= 0 and zero value elsewhere as shown in Fig. 23.22(b)_ This high peak will be of infinite height, zero widrh, bur of a finite area. Such behavior may be described mathematically using Dirac's delta june/ion 0(-.-). The delca function B( T) is defined as having zero value everywhere except at T= 0 in such a way that

r< ~

8(7)/(T)dT~ 1(0)

RESPONSE TO RANDOM EXCITATION: SYSTEM

SINGLE~DEGREE·OF·FREEDOM

(23.52)

~<

for any function of time f( T) defined at T = 0. Using the deltz. function o(r) we can express [he autocorrelation function for white noise as

(23.53)

To detennine the response of a strucrura: system subjected (0 a random exdtattO:1, we need w examine :he freque:1cy content of the excitmion function. We are mostly interested in estimating the spectral function or the spectral densiEY function of the excitation. Until recently, the procedure for estlmarlng the spectrum of a discrete time series has been [0 first determine the aurocorrelarion function reg. (23.19)] and {hen apply the Fourier transform (0 chis funcrion [0 obrain the required spec[rum (eq. (23.36a) or eq. (23.49)]. However, the method of calculation has changed since the developrr.ent of fast Fourier transform (usualJy abbrev!flted FFT). As has beea indicated in Chapter 5, the FFT is a remarkably efficient method for calcdaring the Founer transform of a time series. Rather than escimate the specrrum by firsc determining the autocorrelation function and then calculating the Fouri.er transfonn, it is nOw more efficient and more accurate [0 calculate spectra directly from the original time series, Consider the damped, single-degree-of-freedom system shown in Fig. 23_23(a) subjected to a random force F(l), a sample of which is shown in Fig. 23.23(b). We will assume that this force is known at N discrete equally spaced

where C must be determined from eg. (23.36a) using the fact that S,(w) must be the conw.mt SQ. Subsdtuting eq. (23.53) into [he eg. (23.36a) gives

and "sing eq. (23.52) yields (al

(23.54)

Ih)

Fig. 23,23 Single-degree-of ·freedom $ystem .:>ubject to a random force sampled ut reaular time lntervnls. .:;

680

Random Vibration

Random Vibration

. - ..1 (. = t1rnest--)t)

0 , 1,_, 7

... ,

N - 1). Fourier analysis of F(c) results in..the

681

Now, using egs. (23.63) and (23.64), we may write eg. (23.62) as

freque~cy components as given by eqs. (23.31) and (23.32). By su~erposltlOn

i

the response of the single degree-of-freedom system to the hanTIome components of F(t) is given by eq. (5.35) as

N-I

I

=

IH" I' S,(wo)Llw

(23.65)

,,=0

Or as

(23.57)

N-I

--,

:>

y

~

.

(23.66)

for

n

w" = - (N- n)&

for

n >N!2

!k

2",

w= , -

w=~

T '

(23.67)

N /2

w,,:::::nw

- UJ"

S,.(w")Llw

where

in which, as discussed in Section 5.6,

(" - -;;;'

I~O

Ym

(23.58)

When [he frequency is expressed in cps. we may write eq. (23.66) as

i

(23.59)

,v-I

= ) ' S,.(j,,)Llj

(23.68)

" =0

N the number hof I n t hese f onnu la S, T ·IS the time duration of the excitation, . ' equal intervals of the excitation, and §n the damping ratiO correspondmg to t e • frequency W n • Equation (23.57) may conveniently be wntten as (23.60)

where (23.69) Equation (23.69) states the importam result that, when the frequency response H" is known, the spectral density Sy(Jn) for the response can be calculated from the spectral density SF(f,,) of the excitation.

Example 23.6. Detennine the mean square value of the response for the single-degree-of-freedom system in which k = 100,000 lb/in, m = 100 lb· sec 2 I in, c = 632 lb· sec lin subjected to the F(t) shown in Fig. 23.16. Choose N = 8 for the number of intervals.

where the Frequency response, Hr; is given by (23.61)

Solution: The mean square value

]2 of the

The mean square value? of the response is given by eq. (23.62)

as response can be obtained from eg. (23.35) as iV-I

,v-

,e-I'

"I y'=" L IH"I-

(23.62)

. I· in terms of the spectral Alternanve 'y, eq. (73 - .67) - may be expressed _ ----"2 density . function. The frequency contributions /J.F2 to the mean square value F are gIVen

by eg. (23.35) as (23.63) which. by eg. (23.39), may be expressed as

= SF(w,,)Llw

,,=0

iH"I'IC,,I'

From eg. (23.61)

n=O

LlF'

i= I

I

;H

i"

I'=.!.... 2 k

1

') 2 (1 - T"2)2...l..,(_T"(,,)

where the frequency ratio is Til = W" Iw and the damping ratio is g" = C n Ic er . The natural frequency is w = J kim = 31.62 rad Isec and the frequency components by eq. (23.58) are given as W,,=nWI,

(23.64)

w" = - (N - n)w"

n> N 12

682

Random Vibration

Random Vibration

it. which Wt 21TJT= 21TIO.64 9.8175 rad/sec since the duration of the applied force is T ~ 0.64 sec. Vaiues of le"i, for the function F(t) shown in Fig. 23.16, have been de:ermined in Exampte 23.3 and are shown in Table 23. L The necessary computations to determine? are conveniently shown in Table 23.3. From this table the mean square value of the response is 0.9352. A.nalogous to eq. (23.65), when the excitarion is a random acceleration applied to the support of the srructure, the mean-square acceleration response at a point P of the srructu;e: is given by

€83

e::;!

B{r)

~.A~" i, "" ..

?:;;

(a)

S;/)

a;

__

Sew,)

,,"";;'0

iH.,i'Jw

g
.osL

N-:

a;= ')

i

.06~

.04ri

(23.70)

.03t-

-:~~orilll_ J-LLLJ_L ,

where H'I is now the frequency response in terms of the acceleration at point P resulting from a unit harmonic acceleration at {he support of the stru::ture. The response of a single-degree-of-freedom SITucture subjected ~o a single point random excitation can be determined by a simple numerical calculation provided that the spectrol function or the spectral density function of the excitation and the frequency response of the structure are known, The frequency response Hr. may be obtained experimentally by applying a sinusoidal excitation of varying frequency at the fOl
1

2

3

4

5

6

1

a

9

10

l!

';"2--"3-'~~- -

Jicps_l

(n)

Example 23.7# Determine the respo:Ise at a point P of the structu re modeled as a shear building shown schematically in Fig. 23.24(a) when $ub~ jec~ed (Q a random acceleration at itS foundation. The spectral density function of the excitation is k..TJ.own and ;;hO\vn in Fig. 23"24(b), The frequency response TABLE 23.3

Calculation of

11

w" (rad Isec)

0

0

2 3

9.8!75 19.6350 29.4524 39.2699

r=::

W,,/(:)

Y' for Example 23.6 iff";

IC,I

(in/lb)

(I b)

£17

(in')

----~--.~--.--~--

4 5

6 7

- 29.4524

-- 0.9314

- 19.63S0 - 9.8175

- 0.6209

1.000 l.l04 1.595 4.376 1.677 4.376 : .595

- 0.3:05

U04 E-S

0 0.3105

06209 0.9314

12418

--.-~~-

---~--~---~--~-'

E-5 E-S E~5

0 0.5121ES 0 08787 E 4 0

E-S E·, E-S

0.8787 E 4

E~5

0

0.5121 E 5

lei

lH"I'IC"i' 0 0.3197 C 0.1479

0 0.1479 0

03197 .~-~---~-~.-

? =0.9352 ---~-~~-~-~-~

~ig.. 23.24 (~) S!rucmre sebjecred to random acceleration at the base. (b) Soectral aer>Slty functlon of the excitation. (c) Relative frequency response at point P. • QpJao of the structure at point P, obtained experimentally when the 'oum!anOn . •

•

'

4

by_ a s~nusOidal accelerarion of amplitude 00 and varying frequency shown m FIg. 23.24(c).

IS :xclied

j;,

lS

Solution:

The mean square value a'p of the reSponse at th ' P is e pomt

calcUlated from eq. (23.70) as

N--'

;;;= ')

,,--;;'"0

S(f,')IH":'Jf

(a)

684

Handom Vibration

Random Vibration

23.;;

where

iH"i

RESPONSE TO RANDOM EXCITATION: MULTIPLE-OEGREE-OF-FREEOOM SYSTEM

ap/oo

Table 23.4 summarizes [he computiHiooal procedure, By eq. (a) we obrair. from the sum in the last column of this table the mean square value of the :-esponse = 1.8100g' and by eq. (23.3) (assuming mean value a = 0)

The extension for determining the response to random vibrmion from a sing!edegree-of-freedom system to a rnu!riple-degree-of·freedom systerr: can readily be accomplished using the modal superposition method. This method, t.s w~ have seen in previous chaplers, transforms a system of coupled differential equatiolis into a set of independeol 0: uncoupled differentin.i equations of only ont dependent variZlb!e in each eqoation. Thus, each equation of this set is equivalent w [he differentia! equation for a single-degree-of-::-eedom system and consequently can be solved by (he melhod presented in the preceding section for fl. slngle-degree-of~ffeedom syslem. Vie presem first, ~he relationship between the complex frequency response and the llnir impulse resj)onse followed by the development of random vibra~ tion for a system :)f two degrees or freedom and the generalization to multip!edegree-oF-freedom systems.

a;

~

r:r= {1.8100g

1.3458

The probability of exceed~ng specified accelemrions can now be found assuming the normal distribution for jarl > u: 1.345 g and for !opt> 30-= 4.041 g, respectively, as

P[ia,i > 1.345 g)

31.7%

P[ia,i>4.04Igj

0.3%

685

and

Similarly, the probability thaI the peak acceleration f1p at point P will exceed a specified value (10' or 3u) is found using the Rayleigh distribution <:s

PlAp> 1345g\

60.65%

P[A,,>4.04Ig)

Lll%

23.11

....... _ - -

IHe i'S U~,)dl

S(f,,)

.H"

(cps)

(cps)


(in /l~)

(g' units)

0 1.0

LO LO

0 O.OiD

2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 13.0 14.0

1.0

0.020 0.015 0.030 0.050 0.040 0.020 0.0l5 0.020

0.5 1.0 1.8 lJ 1.4

0 0.0100 00648

I"

1.0

LO l.0 1.0 1.0 l.0 1.0 1.0 1.0 1.0 1.0 1.0

3.0

0.0253 0.0588 0.2420 1.0816 0.1800

1.8

0.0486

1.7

0.0578 0.0361 0.0050

22 5.2

O.OJO

1.9

0.005 0 0

1.0 0.4

0

c'

FU) ::::

C\w) =

sum

= 1.8100

-:.{

oy

I " 2-1 "

0

0

J_'"

where the coefficient C(w) is given

0

0 0

Relationship Between Complex Frequency Response and Unit Impulse Response

Considet a linear dynamic system of a single~degree-of~freedom represented by the damped simple oscillator shown in 23.25. The mndom force F(l) acting on this system ma.y be expressed in terms of its frequency COntent by mea.ns of the inverse Fourier trnnsform given by eq. (23.24) as

TABLE 23,4 Calculations of the Response for Example 23.7

ill

1

C(w)e"" dw

eq. (23.14)

Fourier transform. eq (2323) as

J~ ..

F(,)e-'~dl

eq, (23.23)

686

Random Vibration

Random Vibration

The diffel;"ential equation of motion for the system shown in Fig. 23.25 is

my-cy + ky = F{t)

687

23.11.2 Response to Random Excitation: Two-Degree-of~Freedom System

(23.71)

which for an excitarion of a un!\: exponential function, F(r) = e"", as presented in Chapter 3, has a steady-srate solution, YI(t) or the form (23.72) in which

The concepts and methodology for determining the respOI"ose of mult:pledegree-of~rreedom systems can be adequately presented through the ((ealment of n two-degree-of-freedom system. We begin by presenting a two-degree-offreedom system subjected to a single random excitation. This case is then extended by subjecting this two~degree-of~freedorn system to a second excitation, tn nddit:on to the first rando:n excitation. Finally. we present the general case of a multjp!e-degree-of-freedom system subjected to multiple random

excitations.

Consequently, the response yff) due to the force expressed by :he inverse of Fourier transform in eq. (23.2.4) is given by y(r) =

r'

C(w)H(w)ei~ dl

Vie consider in 23.26(a) (he structural model of a two-degree-offreedom system subjected to a mndom sta(ionary force F 1 (t) applied at coordinate Yl- The equations obtained by establishing the dynamic equilibrium in the free-body-diagram in Fig. 23.26(b) are:

(23.73) (23.78)

.J -~'"

Consider now the special case where the total excitation consists of a single impulse of magnimde 1 applied at time t = O. In this case, C(w) in eq. (23.23) becomes

The natural frequencies a:ld nodal shapes are then found as presented in

(23.74)

because e -

,'ld

=

1 at t

=

0 and

r"

t:

I

(23.75)

F(t)dt=!

.J -,,"

rOl;" a unit impulse. The expl;"esion given by eq. (23.75) is known as the Dirac's 3 function. Thus, from eq. (23,73) and (23.74), the response, h(t), to unit

k, m:

1111

_________

F:.{t)

:1d:m//Q/:"'wnnmnJ,L,//J,):""mn (a)

impulse is given by

r" H(w)ei~

h(r) =_1_ 27f J-1><

dt

J

(23.76)

_ y;}

which is recognized as the inverse Fourier tra.»sfonn of H(w)= ,

r'

h(l)e~i~dw

(23.77)

_ m_,y_'1_...J1

~____

F,(r)

(b)

-~

We observe that the unit impulse response il(r) of eq. (23.76) and the comp,ex frequency response H(w) of eq, (23.77) fonn a Fourier transform pair.

Fig. 23.26 Two-degree-of-fn,:cdom system. {a) Mmhemmical rr.odeL (b) diagram.

Free~bady­

688

Random Vlbration

Random Vibration

Chapler 11-, by setting F 2(I);;;;:: 0 in eq. (23.78) and substituting a trial solUtion y

= a 1 sin

Wi

and h

= a 2 sir. (k ,

(jJ[.

Hence, we obtain

+ k2 ~ m,o/){!I -

689

Damping in the sysrem may readily be induded by adding in eqs. (23.84) damping terms expressed in function 0: the modai damping ratios ~I and Hence,

k2az = 0

- k. 2Cl I + (k2 - m,,!(2)a:o;;; 0

(23.79) (23,85)

For a nomrivlal solution of eq. ('2.3.79), we set equal to zero the determinant of the coefficients:

The solution of eqs. (23.85) :nay be expressed by Duhamel"s integral which for a damped system :s given from eq. (4.24) as

(23,80)

"U);

- k,

r',

¢"F , (t - A,)h,(,\,)d'\,

(23,86)

and

The expans;on of [he determinant in eq, (22.80) re~;ul~s in the following quadranc equation in w2 : (23,81)

Ec;uation (23,8 1) prov~des the rootS (;)7 and w~ which are the squa;e values of the natural frequencies for this two-degree-of.freedom system. The subs[ltution of w1 into One of the equations in eq, (23.79) :-esult3 in the firs: mode a I b a2:, and the subsequent substitution of w~, i:1 the secor.d mode a il, an. These modal shapes are conveniently norml:llized by dividing the components ail and ali of the first mode by ./ mJa~1 T m1a~1 and the components, a 12 and tJ:n, of the second mode by Jm\a\ +;;;;;~r; to obtain the modal matrix given by eq. (23.82) ii1 wn:ch the :lOrmaiiz.ed modes are arranged in the columns of this matrix.

("

:,(t);

in Which

~he unit

J

_%

¢"F,(, - A, )h , (A,) d)',

irnpalse functions hl(') and

y,(t)

4>"z,(,) + ¢"z,(r)

y,(t); ,p,.,z,(r)

+ ¢"z,(r)

(23.83)

are given for ;;::0 by (23.87)

ane

e

\1"'.:'

h,(t) = -,.. _sin W1,,/ I ._. ~?:

tl

(23,88)

and for (< 0 by h,(,)

The uncoupled modal equation, as presented in Chapter 1 i, is obtained by introducing in eq. (23,78) the linear transfonnatlOD

h~(r)

h,(t) = 0

The response is ~hen obta:r:ed in ~errn3 of Ihe displacemer.ts y,(!) and Y2(t) by SUbstituting [he solution of [he modai equations, Z I (I) and z:~(t), into eq. (23,83). We proceed nOw to the determination of the mean square value of the response, YJ (t) in order to estimate its standard deviarion and its cor.fidence interval, doe to nmdom excitmion. The mean square value of rhe response may be calculated from the autocorrelation function R,.! using eqs. (23. i 9) and (23.20) as

resulting in

z, +

4>"F,(t)

l2 +- wit: = (PtzF 2(0

in which ZJ and

Z2

are lhe modal displacement functions.

(13.84)

or using a more convenient nota rioI'. as R",(') = E[y,(t)y, (t

+

7)]

(23,89)

690

Random Vibration

Random Vibration

in which the letter E denotes expected Or mean value of the function in the bracket The substitution into (23.89) of y,(I) and y,(IT T) from eq. (23.83) resultS in

691

The spectral density function S.,Aw) for the response is then given by eq. (23.36a) as

(23.92) The multiplication of eq. (23.91) by

or

(23.90) The substitution of eq. (23.86) into [he expected value for the product of Zj (!)Zl (r +- i) in eq. (23,90) results in

r""

E[Zi(I):l(r+llJ=Ei ¢21F-:(t-A;)hr(A;)d'\1 . J-~

f'"

J-~


Interchanging the symbols for expected vaiue and integration. we obtain E[,.(t)z,(/+ T)J=

.

I~

followed by integration yields

r' E[F,(t-A,)F,(t~T-A,)]h,(A,)h,('\')1>;,dA,dA,

J -"")-.,,

+ 1>" ¢"4>,, 4>"h, (A,)h,(A,) T ¢111>" 4>"1>,,h, (kJh,(A,) "- 4>i,¢i,h,(A,)h,(,I,)]e ,w"'-"'] dTdA, dA, After using eq. (23.77), Syl(W) may be exptessed as

or

,

S,,(w) ~

:> :>

4>,,4>,,4>'j
(23.93)

j'" I ~"" I

since the expected value, E[F;(r ~ )';)F2(t -1- 7 - A2)J is equal to the autocorrelation of the function F 2(1- A2 + AI)' Analogously, for tbe other terms ~n eq. (22.88), we obtajn E(z,(t)z,(I+r)J=

r'

IX R2(T-A,+A,)h,(~\tlh,(.\,)1>"4>,,dA,dA,

J -'" J-':'>

E [Zl (I)Z, (I + T)J

=

J' r' _>:

R,(T - A, + A,) h,(A ,)11,(,1,)

EquatIo~ (23.93) provides the spectral density function, Sjl(W), for the response at coordInate 1 of a two-degree~of-freedom system excited by a random force at coordinate 2 expressed by the spectral density function S2(W). Now we use eq. (23.43) to obtain the discrete spectral function, Syj(w,,), evaluated at frequency WrI as

1>i, d,\, d,l,

J-!I:

which substituted into eq. (23.90) yields R" (T)

~

r: L:

1n which Hk~ (w) is the complex conjugate of H;,(w) and

(23.94)

R,(T- A, T .\.,)[4>;,1>i,h, (A,)h,(A,) + 4>"4>"4>,,q,,.:;h, (A,)h,(A,)

+ 1>"4>,,4>,,4>,,h, (A,)h,(A ,) T w7,1>ih(,I ,)h, (.\,)]d,l, dA,

(23.91)

wbere C:t{w,,) is the discrete Fourier transform coefficient of the random force at coordinate 2 and C; (cu,,) j(5 complex conjugate.

692

Random Vibration

Random Vibration

If the single excitation is the random force FI (r) at coordimue 1 instead of F 2 (l} at coordinate 2, the expression for the spectra! density funalon for the

response at coordinate 1 win be

(Klp)

toO

(23.95)

o

The generalization to a system of N degrees of freedom subjected to multiple random excitations leads to the foHowing equation:

t(sec)

0.125

0.375

0.500

-tOO N

N

S,,(w,) =

I

i"

1>,1>,,), !

N

I

¢4¢",C,(w,)C,;(w m)Hi (w,)H;(w,)

(23.96)

L "'! ",,," I

F,(r)

where

S,,{w,,) is the discrete spectra! density functjon of the response at coordinate i Cr(W/l) is rhe Discrete Fourier Transform coefficient for the force at coordi· nate L C:;(w,,) is the complex conjugate of the coefficient Cm(W.~)

(Kip)

!

t(sec)

23.11.3

Computer Program for an N Degree-ot-Freedom System Subjected to Rar,dom Excitations Fig. 23.27 Excitalion P:.mcdons for Exa;npk 23.8.

A computer program written in BASIC has been developed to evaluate USing eq. (23.96) the spectral density function of the response Sy,(w) at coordinate; of a muitidegree-of~freedom system subjected to multiple random stntionary excitations, The program assemble the stiffness and mass matrices of the system, calculate the natural frequencies and corresponding modal shapes and spectral density functions for the applied random exdtations. It also calculates the mean square value of the response, the standard deviation (RSM for a funclion having a mean equal to zero), and the confidence intervals corresponding to one, two, and three values of the standard deviations for each noeal coordinate of the system.

Solution:

INPUT DATA AND OUTPUT RESULTS FOR EXAMPLE 23.8

Nrn,G'!ea 07 n£G%SZS Of f?S2::x:nl !lUl'{ Of EX'!'

"'ORe,,;;

'!'1Y.E .0l!RA":':O:: FOR 7)1£ fC?::25

!7.ro:ltN'l' 02 K~:"'l

Example 23.8. Determine the response for a two~degree-of~freedom struc~ ture modeled by system shown in Fig. 23.26 in which k! = 300 lb/in, k-z = 100 lb/ln, m; = mJ = 1 lb· sec::: fin" The random excitation forces F, (t) and F 2 (r) are depicted in Fig. 23.27. Use Program 22.

693

Random Vibration

R.andom Vibra:ion

694

695

EXCl1',\':'tCN DL'fu.

I),DCOe~0C

Cc}!?,

Jooe . 'i3StQ7!h)S S.g~tJ4aE>05

~.

~J,

on:.:o

,,
iii: ,S4::;';"

5(;.OCC;:

sa. ::·O(J 2500

~50

0000

-Sf)

co·~o

:::;:;:(;:x

.) .0C)') ;'.:>~C55?">0S

-J

.~~:5;'H;>06

<9 8,,956

-:2.56637 .;; .2S:1135

For this example, the standard deviation for the random vibration displacements of the response at either coordinates 1 or 2, given by the computer results. has essentially the same value, v, ~ 0.034 in (i = L Therefore, [he probability for the dl-splacements Yi at coordinates l or 2 of exceeding 10-, 2u, or 30- (normal distribution) are:

1:.6771 -17.0')77

0.0000

C, !)OCC p, srn

.0000

P

C. ?()QO

-n

?:CAL

::c,c

-,

O.(:JOO

-, nu " . )80:: -37

o.u;)O;)

5$~e

l:'L 75JO 12 :;'C30 -1\1 75~:J

. ilJOO

. GOGO

- iJH2 .500e

18 .750a

)' ~F

. SSMi

-La . 1580

-2

·1

-,

r: y; I > 0.068 in] =

4.6%

5777

, .,

p[iyd > 0.034 in] = 31.7%

.0000

sooo

and the :?robabili[y for the peak values of [he displacements Yl or h of exceeding 10-, 2u, Or 30-, (Rayleigh distribution) are:

pry;> 0 034 in] = 60_7% P [y, > 0.068 in] = 13-5%

pry, > 0_;02 in] = Lll%

696

R.andom Vibration

Random Vibrahon

T9~YJ 180 in

b===~:====

-

Y1

I

+

m, :-. 1,

180 in

+

+

180 ill

A :;;;; 10 in'!.

1i

1= 1600 in4

77777

(a)

(b)

Fig. 23.28 (a) Three-sto!y frame of Example 23.9. (b) Mmhemntical modeL

23.12

RANDOM VIBRATION USING COSMOS

Example 23.9. The three-story frame s.hown in Fig. 23.28(2.) is subjected to ground random vibration motion represented by a constant spectral density functIon So = 0.12 (in!sec 1 )l(radian!sec) from zero w w lOO(r2.dlsec) as shown in Fig. 23.29. :Y1odel the structure 2.S a shear building and determine the RMS (standard deviation) of the response for the horizontal displacements

a[ the Lhree levels of the building. Also determine the probability for the horizoma~ displflcemems a~ the three levels of [he buildmg or for their peak amplitudes of exceeding j, 2, or 3 rimes the calculated standard deviation. The following numerical data apply to this problem: Modulus of elasticity: E = 3.0 x 10' (psi) m l = ml = m) = 10.0 (lb, sec 2 fin) Concentrated masses; Column's cross-sectional area: AI = /\2 =: 5.0(0 1) Cross~Sectional momen~ of inertia: 11 ""'! fJ. = 800,0 (in 4) Column's depth: d, =d, 10,0 (in) Modai damping: q= 0.05 (all modes) Story height: L, = L, = L, 180.0 (in) Seisrr.ic spectral densiTY: So = 0.12 (in !secl)~ !(rad !sec) Solution: The structure is modeled as a massless column with concentrated masses (.It the three levels of the budding as shown in Fig. 23.28{b), The following commands are implemented in COSMOS:

(I) Set view to the XY plane: DISPU.. Y > VIE:\\_P.;?~ > \i::EI'i V:EW, 0, C, 1, J

(2) Define ,he XY p,ane at Z = 0: GEOMETRY :. GRID:. PLAl"'¥E ?LA..NE. Z, O. 1

(3) Generate keypoims at levels of the building: GEOHET'R'i , PO!NTS~ PT 1, 360, A", G.

?T

PT 2, 360, ::'8C, J, PT J, PT 4,

(in/sec)' s-""""-·""·~ ( rad/sec)

697

360. 36C 360, 540,

J,

3.

(4) Generate lines between keypoints: GEOMETRY

~

CUR'./2S

eRLINE, 1, .1.., eRLINE, 2, 2, eRLINE, 3, 3,

>

CRLINE

2, J, 4,

(5) Define elements of group 1 :Js:ng BEA:V12D, pro;;eny (£ and real cons tams {A;o:; 10.0 :n", f= 1600.0 in~, depth ~ 10 in):

0.12

PROPSETS )

EGROt;P

EGROtJP 1, BEAA2n,

:00 Fig. 23.29 Spectral density functioCl of the ground acceler..tt:on in Ex<:mple 23.9.

3"OE6 psi),

0,

0. 0, 0,

0, 0,

0,

PROPSS'I'S ) MPROP

MPROP, 1, EX, 3E6, PROPSETS ~ RCONST RCONST, 1, L ~.. 8, 10, l6DO,

lO,

Q,

0, ;),

0,

"

698

Random Vibration

(6) Generate one beam element along each defined curve: MSSHING M_CR.

1,

3,

L

2,

(13) Define the type of analysts, random vibration using three frequencies in the range fro:TI 0 [0 100 rad/sec:

>M_CR

PA~_HESH

j

L

natura~

1,

(7) Define group 2, concentrated maSSeS at the three levels of the building

and [heir value:

ANALYSIS

~

Pu_ATY?E,

?OST_CYN 4.

3,

0,

0,

a,

100,

L.,

3,

lEl:),

EGRO:.JP

PROPSSTS EGRCUP..

RCONS'?,

:J, Or 0, 0, 0,

t'<,-~.ss,

2,

PHOPSE't'S

(14) Define modal da:-r.ping equal 5% for aU modes:

C, O.

RCONST

~

2,

2,

1. 7, 10

0,

O.

I

0,

0,

0,

P-.NALYSIS ) pu_MD.A_'.1P <

0,

(8) Gener8.te mass elements at the three levels of the building: 2,

4,

1

(9) Merge and compress nodes: MESHING

,.

["ODES ,.

NES~::lG

,. NODES

DAMP IGA? ;;.

i\•.Nl\LYS:S ~ POS1'... DYN PD_CURTYP, 1, 1, 1,

NM£RGE

.:1HSRGE, 1, 10, :,

POS'1'~DYN

0"0::;01,

0,

.. , 0

A-t:IA!...YSIS

) NCOHPRESS

)

?D~CURDEF,

?D __l.I!DAMP

1, 1, 3, 0.05;

([5) Define dynamic excinuJon as wIth value 0.12

MESHING ,. PARAtLMESI-! ,. t-CP':'

(>CPT,

699

Random Vibration

frequency~dependent

C:.JRVES

POST_DYN CC'RVES 1, ~, 0, 0.12, 100,

specrral density

?D_.CURTYP PC_CURDE?

0.12,

NCOMPRSSS, l, 6

(16) Apply base acceleranao:

(l0) Apply constraints in all degrees of freedom at node 1, and all of freedom except UX at nodes 2. 3, and 4:

Al\!,';LYSIS > FOST' DYN > !?D_BEXCIT ) :OADS-BC ;). STR()C?URA::'"

DND, 1 A!.., ONe, 2, UY,

C, I, 0, 4,

L L

~

DISPLt-fNTS

>

UZ,

RX,

RY,

DND

pu_a~SE,

RZ,

;:.

F~EQ

A,.t;}>,LYSIS

;:'-J."JALYSrS

~

l?OST~;:)YN

> B._DYNAMIC

R_DYNAMIC

run the frequency analysis: j.I._I\!.~LYSlS

PD~BASE

0,

Q,

(17) Execute dynamic analysis:

(l r) Set the options of the frequency analysis to extract rhree frequencies using Subspace Itera[ion Method wim a maximum of 32 iterations, and then

A_FREQUENCY, 0, I), 0, 0,

1, 1, 0.

/BlJCK > F._!?REQUENCY

3,

S,

32,

0,

0,

0,

0,

(18) Execute stress calculations: lE-0,5,

0,

lE-E, .hl~ALYSIS

;. FREQ /EUCK )

R_FR£QUENCY

;;. ST}I..T!C ) R._STRESS

R_STRESS

R_FREQUE::K'{

(19) Exit COSMOS and use editor stresses from o"tput file:

(12) List the natural frequencies of the system: RESULTS ,. LIST' ,.

F R E Q

u

~

N C Y

FREQLIST

A N A L Y S I

S

to

print R.V1S for displacements and for

CON'rROi... :;. EXIT

by S U B 5 P A C E FREQUENCY

NUMBER 1

2

3

l\ :0 G 0

R I

l' H 14

FREQUENCY (RM ISSC)

FREQUENCY

PER!OD

(CYCLES j SEC)

(SECONDS]

0,1398632£... 02 0 .3918879E+02 0 5662944E+02

0.2225991£+01 0.6237089£+01 0.9012855E+01

G . 4492380E ... 80 ::::.1603312£+0::::

0.E09525£·CO

Tables 23.5 and 23.6 present resul!:S from (he OUtput fiJe for the Rlv1S values for Im:eral displacemen!s at the three levels of the building and ~he RSi'vi for stresses at the nodes of the three column elements modeling the building, The probability for the magniwde of the instantaneous displacements IYI I, Iy')!, :Y3 I, or their peaks values, Yj, YJ , YJ of exceeding aI, 2, or 3 standard deviation va;ue. 0-= RMS is given respectively in Tables 23.7 and 23.8.

700

Random Vibratlon

Random Vibration

TABLE 23.7 Probability for the Displacements 3" (Normal Distribution)

TABLE 23.5 R,MS, RESPONSE

:;.Qn:}0()s+c:J

0"~{)c0:r:;+!)(l

1),0'.:GG"n.,.oc

::J.CGCC'Qs"t'·?C

Q,ccccor.",CG

O.O(l.J0CS"'~0

C.~';~lS::-;;;~ 2~':'6S-0: :::.32067:':-01

D.C~QCOE+~O p _OOOCO::"" C~ O,OO~COS ... t)O

C.~Oi)CGE+J0

C.:)0C-GOS+GO

0.00(;00,.,,"0<:;

(;.0(';00(;2+:::0

C. 00CO,,8

co \), ~OOOOE"' 00

0.000002';' 00

I).

+ O{)

C.OO()OCS+OC

<).

OOOCO<: +

O.OOOOOC+~O

10-

Prob,

(in)

(in)

%

20(;n)

0,0144 0.0257 0.0321

3[7 3[7 31.7

0,0288 0.0514 0.0642

y,

R2

:<1

Displ.

(i,COC'OOE+Cr;

I

y, I

Y3!

IYil

701

of Exceeding 1fT, 20"', or

Prob.

3(J'

Prob,

%

On)

%

4,6

0,0432 0,0771 00963

03

4.6 4.6

0,3

0.3

O.OOOOOS··O:::

TABLE 23.8 ProbabUity for the Peak Displacements Y j of Exceeding 1(7, 20', or 3" (Rayleigh Distribution)

y"",C.CC:lO!;:+C{l

C,OCCO:::.,.00 "tnO 00':.\:;:;.00

-)ccn",c::

';l;"'-C ~42'2~C3

t.:4242'O;;

1-1:.,,{).;;:;')01:+00

~0CO">CC

\'t~O.jD0n~OO C.O~C(l!1>OO Mttf::.:1SE~05

O.1.nl:;>OS

(?Y' (;TORiJpl

,,0.00005>00

"'::~C~.00C-C2-C~

C.::'J5E~C~ }l$~Q.CJQ):::+:)O O.C:;;COE>C<' Ht~O.1;::.n;;:.J5

,;:;.c:;c.c:;;.oc 0.0000£.00

:'004;;:~02

O.4001,2~O"

(?

iOl.::'!St:j<>o

l?lAl"O.OJOC);+s>C

J.CJcce->-oC

C.:lOCCt-0C

;~$!S$l"J.C:J()C<:+CC' ::J.~C002+[l0

0.1023,,:"-0$

;!>:It. ISt.)".:L319SS.Cl $t:;,1;XoO.3Hii::>;}?

0.31$5=>0"

Sn;i:t"O ::;: 96;;:;,,02

(}. }lS52~il2

0.3:;.)6);+;")2

F;;"D. CC0C::<:~OO vsd . 57EZ .. C::

4 ' [l _CCCOS;.CO 7t "C. OOCL2 .. CC :;;.6713£ .. 02 Y;s",'). ococo:~oc

C _CCJC;:>Nl '} .OCC';l:2:+CJ

{?! r.) -d). ·);;;en-co !NS (5s) .. (;. C')CC!:>CG

L. Jocn~CJ O. ~~oc<:·co

\'~"c.ocnC2 .. nQ

c,occ:):,,;,.;;1

::1~~O.&C"::;S+'~4

Q.f0422-04

\H-:./$t)£O.1SB8E .. 02 Srnax"O. ;'!3*5E.~l $;\\il1~O, 1Il95,>02

).1888=:-02 (). -;'S9SB+O'

23.13

(in)

Prob, %

20' {ir.)

Frob. %

Y, Y, Y,

0.0144 0.0257 0,0321

60.7 60,7 60,7

0,0288 0.0514 0.0642

135

Iff

13.5 13.5

3,;r (in)

Frob. %

0,0432 0,0771 0.0963

1. 11 1.11 1.11

D. OODOE~OO

, I .f:-~C.OCOO">;)O C,ocoos~D::: 'i::~O.GJOCE>JO c.oC'cno.-co '';s~C l':'352~03

·Af~O.01C()f;.O:)

WS(S$l#(l,OCO(<:~OG

Peak D;spL (in)

(} .1895::+02

SUMMARY

The objective of this chapter was to introduce the fundamentals of the theory of random vibrations for applicarion in structural dynamics. In structural dynamics the most common source of random vibration is due to explosions occurring in the viCinity of the structure. The response of a structure to earthquakes may also be predicted using random vibration theory. A random process is described by a function of time whose value at any time is !mown oniv as a set of sample records known as an ensemble. Such a function can only ~be described in probabilistic terms llsing the tools of statistics. The most important statistics of a random process xV) are its mean value X, its mean square value;?, and its variance given, respectivelY, by egs. (23.1).

u;

(232). and (233). The most commonly used probability distribution for a random process is the normal distribution. However, when the random variab:e can only assume positive values (e.g" the absolute values of the peaks of vibration), the process tends to follow the Rayleigh distribution. The autocorrelation RAT) of a random variable xU) is defined by eg, (23,19), The spectral density function SAw) is defined as the Fourier transform of the autocorrelation function R(T) reg, (23.363)J. Although the spectrum of x(l) can be obtained from RIC T), it is now mOre efficient to determine the spectrum of a random function from its discrete Fourier series [eq. (23.35)] using the fast Fourier transform (FFT), If the spectrum of the exciration function and the frequency response of a dynamlc system are known, it is a simple matter to caku1ate the mean-square value of the response using eq. (23.62). Knowing the mean-square value of the response and using standard probabiHty functions (such as the normal or the Ray~eigh distributions). we can predict the response in probabilistic terms. In this introductory chapter on random vibration, the presentation has been given in detail for single~degree-of-freedom systems and extended to multidegree-of-freedom systems using the modal superposition method. This method. as we have seen in previous chapters, transforrr.s a system of differential equations into a set of independent or uncoupled differential equations. Each equation of this set ;s equivalent to the differential equation for a smgle degree-of~ freedom system ar:d consequently Can be solved for random vibration excitation by the methods presented in this chapter.

702

Raodom Vibration

RaCldorn Vibration

I x Itl ~ 15m <-'II

PROBLEMS 2),1-2}.5

703

~Vvv 1

Detenniae !ne mean and mean square values for Ihe functions shown in

Figs, P23.l to P23S 23,6<13.10

De{errnine Ihe Fourier series expansions fer The periodic functions shown

23.11-23.1S

in Figs. P2T I !O P23,5. Deler:nine and plm lilt spectral fuonio))s for Ihc functlcl1s shown in Pigs,

~211/w

-l1i"",

7l!W

P2J.! to P23.). X(rl

xld

I

___ -2 /wj ~.

Jr

.

X(n for

l[---i-d-

. ' -~l---··' -----, -l'f~ 0: IfJWI21l/W: L~.~l L_

~

0

Xlrl ~ e~"' 10. r:;;" C

{< C

------t--.-::::",-..- - - ,

Fig, P23S Fjg. P2},1.

23.16

A s:ne wave with

23.17

Dcturnine the Fourier coeEiciefHS C~ rind !ile spectral fOflCtiOn for rhe penodic fUllc(lon shown in rig, ?23. :7.

:1

"teady-S::He cc-mponem is given by

Xlti

Delerminc the me;w value j: and the mean-square vulue

7,

fig. P23.2, 1.0 -050

D

0_5

10

15

2.0

2.5

3.0

3.5

xlI]

2,

Fig.1'23,},

23.18

~ random ~o-rc:~ has:) :llcan value r = 2 Kips and spectral density funcBOO shown In Fig. ?23. I 8. Dt:lennlnt: ilS standard dev:rHion O"f ~nd ils ro:::( lYIe:ln square R.\·iSr.

704

Random Vibration

PART VI f cps

1200

Fig. P23.18.

23.19

23.20

Calculare the autocorrelation function for un ergodic random process x(:). Each sz,mple function 1s a square wave of amplitude a and pe..iod T, Deterrn·\ne Fourler transform and Fourier integral representations for the function shown in Fig. P23.20 .

.. x(rl

I

.drJ '" 0 ~)(U! '" q-ft? !oU
--.~-_

f

I Fig. P23.20.

23.21

A stngie degree-of~freedom system with mass LO 1b, se.} (in. stiffness 100 lb lin, and damping t= 0.20 is excited by the force F(t)::::::; 1000 coS 5; + WOO cos lOt + 1000 cos 15t Cb)

Detennlne the spectral function and me mean square valt:.e of the res;Jonse. Also plOt the spectra for the input force and the output response displacement

Earthquake Engineering

24 Uniform Building Code-1994 Equivalent Static Lateral

Force Method

Several seisr.1lc building codes are cc.rrently 10 use in ciffereM regions of the United $[ates. The Umform Bul1ding Cod!:: (CBC) published by [he International Conference of Building Officials (t 994), is [he building code most ex.tensively used, panicula;ly in (he western par: of the ccontry. In addition to (he UBC, two Olhe; major bt.:lld:ng code;; are used: (1) The BOCA, or Basic Bt.ilding Code [Building Oftlcials u:1d Code Administrators international 1996]: and (2} The Standard Building Code, of the Southern Building Code Congress International (1994), In addition [0 the preceding codes, there is the ASCE Standard: Minimum Design Loads in Buildings. and Other Stn.JCLUres, ASCE7-95 (995), (ReviSIon of ANSI (ASCE7-93). Severa! orgar.lzacions involved in eanhquake-resisram design have pL:blished recommendations thaI fom'! [he basis for requireP.1enc in the offlcial codes. These organizations include (1) the SlrucLUrai E:lgineers Associatio.1 of

California [SEAOC (1990)], (2) [he Applied Technology Council [ATC3·06 (1978), (3) the Building Seismic Safety Council [BSSC (1994)], and (4) The

Federal Emergency Agency [FE;vtA (l90-\')J which le.ads lhe National Earthquake Hazard Redt.:ction Program {NEHRP) wilh publications issued by L~e Building Seismic Safety Council. These orguniZ.:llions periodica!ly issue recommendations

Un!form Building Coce-1994 Equivalent Slatic lateral Force Method

709

Earthquake Engineering

708

measu~e

The safety of the strJcture is not assured in the event of a major earthquake. The objective of seismic codes is that a minor or moderate earthquake will not damage the structure and that a major earhquake will no produce coUapse of the slructure. To understand how structures are affected by earthq'Jake, it is necess"

ary to understand the ground motion produced by earthquake.

24.",

where M is the magnitude of the earthquake on the Richter scale. Equation (24,2) reveals increases by a factor of about ~- f or an .Increa.se ~ fl. ' hthat the energy ' o. ~nlt to t e m~gmtude M and by a factor of 1000 for an increase of 2 units 10 the magmtude of M. Although the Richter mao-nitude provides of the total energy released by an earthql1ake, it d;es not describe th: dnma~l~g e.ffects ~aused by an earthquake at a parricu~ar location. Such a descnpllon 15 prOVIded by the modified Mercalli intensitv scaie (-M~,t11) w"'h" to' If" ' . ./ ! > 1 .:; ~a 0," ._ lntens:ty values \,,;,suaHy expressed in Roman numerals, Table 24.1 s~ows ,he Mi\H scale. The MMI value assigned to an earthquake at a particular ~lte base~ on the observation of r~sultant damage and is useful in lieu of .nstrJmen. records of ground motion. IntensitIes up 10 VI uS:JnUy do not produce dam~g~ whHe intensities from VI to XII result in progressively greater . ~ danage to building and other structures. Graphical records or time histories of earthquakes are obtained with lr.stru:nents cal1~~ strong-molion ~ccelerographs. These instruments are ir.sta:led on ,he grounc in basements" or :n other locations of bui:dings or other struct'Jres. They are commor:ly, desl~~ed to register the three onhogor.al cornponer.ts of the ground acceleration. f
,?

and requirements for earthquake-resistant design of structures that are based on a combination of theory, experiment. and practical observations. Building codes are intended to provide guidellnes and formulas that constitute minimum legal requirements for design and construction within a particular region" These requirements are intended to achieve satisfactory performance of the structure when subjected to seismic excitation. althQugh they are not optimal:

EARTHQUAKE GROUND MOTION

y!ost earthquakes are caused by energy release at a dislocation or rupture in cn.:sta! plates generated at a point in the interior of the earth known as the focus or hypocenter. The point on the earth's surface directly above the focus is the epicenter. The magnitude of an earthquake is commonly measured by the Richter magnitude (M).l which is commonly defined as the reading regis tered by an instrument called a Wood-Anderson seismograph at a specified distance of 100 km from the epicenter of the earthquake. Specifically, lhe Richter magnitude (A1) for an earthquake is calculated as H

H

(24.1)

where A is the maximum amplitude registered by a Woody~Anderson instrument located at 100 km from the epicenter and An is the reference ampHtude of one thousandth of a millimeter (for A expressed in mmimeters~ Ao ~ 0,001), When there is no instrument loca.ted at this distance, an estimation of the Richter magnitude based on readings. registerd at seismographs in the region affected by the earthquake. Eanhquakes of magnitude 5.0 or greater generate motion sufficiently severe to be potentially damaging to struct'Jres. The e:lergy E in Joules released by an earthquake can be estimated by the fonnuIa

,',

, I\~ t,}" "I ","' >,,,, M

-1$ ,

c

•< E • .E 0

E=

10'''' ,'M (Joules)

(24.2)

Several definitions for the magnimoe of earthquakes are in use in different countries, ali emanating from the original formulation by C. L Richler, A brief description of :>eJ;:mlc magnitude scales )$ given in the {ntemariOllol Handbook oj Earthquake Engineering.· Codes. Programs, and !

Examples. edited by M. Paz, 1994,

lit..

"~ Q

W

20

75, I

I

'v\

t\ !', ! ~i\jl\fV

...d

,I

'"

Co

v

0'

io

,

\

~J.5·1

"

\J~'\l' 'J7"--. i,

'0'

Fig. 24.1. North-:>cuth component of the EI Accelerat1On. (b) Velocity, (c) Displacement.

JO

$0

"

---./">

v-

=-

"

"

,," Centro e"~,.·nquake

,

"

"

CaIifcrnia, 1940. (a) '

Uniform Building Coda--1994 Equivalen! Static La:eral Force Melnod

710

711

Eat1hqi..ake Enginee!ing

TABLE 24.1 (ConUnued)

TABLE 24.1 Modified Mercalli Intensity Scale (1956 Version)'

I:ue:lsiry Intensity

Value

Value

L II. llI.

V.

VI.

VII.

Not felL Marginnl and long-period effects of large earthquakes. Fe!( by person at rest, on upper floors. or favorably pl~ced. . Felt indoors. Hanging objects swing. Vibration like passmg of llght trucks. May liOT be recognized as an earthquake. Hanging objects swing, Vibration like passing of heavy trUCKS:. or sensation of a jo][ !ike a heavy ball srnking the walls. Srantllng cars rock. Windows; dishes, doors rattle. Glasses clink. Crockery ci~lshes. In the upper range of IV, wooden walls and frame creack. Felt outdoors; direction estimmed, Sleepers wakened. Liquids distu:-bed, some spilled, Small unstable objects displaced or upset. Doors swing, close. open. Sl;unerS, pictures move. Pendulum clocks swp, stan:, change rate. Felt by alL Many frightened and run outdoors. Persons walk unsteadily. Windows, dishes. glassware broken. Knickknacks. books, e~c. off shelves. Pictures off wails. Furnit'Jre moved or overt'Jr:1ed. Weak plaster and masonry D*'" cracked. Small bens ring (church, school). Trees, bushes shaken visibly, or heard to

XL

XII. ~

rustle. Difficult to stand. Noticed by drivers. Hanging objects quiver. Furniture broken. Damage to masonry D"'*, inclUding cracks. V/eak chimneys broken at roof line. Fall of plaster, loose bricks. stOnes, tiles, coniees, also unbraced parapets and architect'Jral omarr.ems. Some cracks in masonry C**, Waves on ponds, water turbid with mud. Small slides and caving in along sand or bens ring. Concrete irrigation ditches gravel banks, damaged. ~ Steerino of carS affe::ted. Damage to masonry C; partIal collapse. Som~ dama'2e to m:lsonry B; none to masonry A. Fall of stucco and some ;;asonry walls. Twisting, faU of chimneys, factory stacks, monuments, towers, elevated tanks. Frame houses moved on foundations if not bolted down; loose panel walls thrown our. Decayed piling broken off. Branches broken from tree~. Changes in flow or remperarure of springs and wells. Cracks 10 wet ground and on steep slopes Generai panic. Masonry D;Y* destroyed: masonry C** heavily

VllL

IX.

•........... -

X.

....

--

**

Description damaged, somelimes with comple{e collapse; masonry B se:-iously damaged. General damage [Q foundations, Frame structu:-es, if not bolted, shifted off foundations. Frames racked. Serious damage to ;eservoiL Underground pipes broken. Conspicuous cracks in ground. In aUuviated area and mud ejected, earthquake fountulns. sand craters. Most masonry and f;ume structures destroyed with their foundations, Some wel!,bui~r wooden structures nod bridges destroyed. Serious damage to dams, dikes, embankments. Large L:mdslides. Water thrown on banks of canal. rivers, lakes, etc. Sand and mud shifred horiz.onraHy on beaches and flat land. Rails bent sligh[iy. Rails bem greatly, Underground pipelines complece!y out of service. Damage near~y [otaL Large rock masses displaced. Lines of sight and level dis[oned, Objecrs thrown in [he air.

Original 1931 version in Wood, H, 0., and Neumann, F .. 1931, Modified ~ercalli imensity scale of 19"31: St!ismorogical SOctt!!}' oj America Bullt!!in, v. 53, no, 5, pp. 979~987. 1956 version prepared by Charles F. Richter, in Elemetrlary Seismology, 1958, pp. t37-! 38, W. !-L Freeman and Company. Masonry A, B, C, D. To avoid ambiguity of language, the quality of masonry, brick or otherwise, is specified by the foHowing lenering. Masonry A. Good workmanship. mOHar, and design: reinforced, especially laterally, and bound together by using steel, concrete. etc.; designed !o resist la(e:aJ forces. Masonry B. Good work.manshi? and monar; reinforced, but not designed in detail iO resist luteral forces. Masonry C Ordir:ary workmanship a:!1d monur; no e;meme weaknesses like failing to {he in at COwers, bt.:{ neither reinforced nor designed against horizontal forces, Masonry D. Wea.1.c materials, such as adobe; poor mon:ar; low standards of workmanShip; weak hOrizontally.

:,:

It was shown ~n Chapter 8 that the earthquake response spectrum provides meaningfu1 information for use in structural design. The spectrum, as explained in thac Chapter, provides values for (he maximum absolute acceleration (spec~ tral acceJerarion). the maximum relative pseudove!ociry {spec!ral veloclty), and the maximum relative displacement (spectral dispLacement) of the single~de­ gree-of-freedom sysrem fa; various damping values. Response spectral diagram 8.8, are usually prepared for a specific earthquake, as is lhe one shown in

712

Or are contrucced to be us.ed in design on the ba'i:is of several past earthquakes. as are the spectml plots shown in 8.9.

24.2

EQUIVALENT SEISMIC LATERAL FORCE

The earthquake resistant design regulations of the J994 Uniform Building Code [UBC-94 (International Conference of Building OfOcials 1994)) are based mainly on the publication by the Structural Engineers Association of California 1990 (SEAOC-1990) entitled Recommended Laler"l Force Reqwrc· ments and Tentative Commentary, which is based [0 some extent on the provisions of the Applied Technology Council (l978) recommendations (ATC3-06-i978) and of the Building Seismic Safety Council (1994) guidelines (BSSC-1994). The key UBC-i994 provisions for earthquake resistant design are presented in this chapter for the equivalent Static Lateral Force Method and in the followi:1g chapter for the Dynamic Method.

24.3

Uniform Building Code-1994 EqUivalent Static Latera.! Force Mel"lod

Earthquake Engineering

two methods for earthquake resistant design: (1) The Slali:: lateral force method and (2) 'he dynamic method. The first method is applicable all structures in Zone 1 and those in Zone 2 designed for Occupancy Importance Factor IV as described in Table 24.2. It is also applicable, in any seismic lone, to regular s~ructu(es under 240 ft in height and to irregUlar s~ructures:1 of not more than five s~ories nor over 65 ft in height. The dynamic method mal' be used fo:- any structure but must be used for structures over 240 ft in he1ghi, irregular structures of over five stories or 65 ft in height, and structures located in seismic Zones 3 and 4 with dissimila:- structural systems. The dynamic method must aiso be used for structures which have a period greater than 0.7 sec and are locatee on Soil P:-ofile Type 54 as described in Table 24.3,

;0

24.4 STATIC LATERAL FORCE METHOD The UBC·94 stipulates that the structure should be designed foc a total base shear force given by the following forrnu;a:

EARTHQUAKE-RESISTANT DESIGN METHODS

The seismic zone map of Fig. 24.2 shows: the continental United States classified in the following seismic zones: 0, 1, 2A, 2B, 3, and 4. The code provides

713

V= Zle

w

(24.3)

in which (24.4)

The code also s:tipu~ares a minimum value for the ratio CIRW of 0.075 except where the code prescribes scaling of forces by 3Rw!8 The various factors in eqs. (24,3 and (24.4) are defined as follows: Z is the seismic zone factor related to the seismic zones {Fig. (24,2), It is equal to 0.075 for Zone 1,0.15 for Zme 2A, 0.20 for Zone 2B, 0.30 for Zone 3, and OAO for Zone 4. The val'Jes of [his coefficient can be considered to represent the effective peak groilnd accelemtion (associated with an earthq'Jake that has a 10% probability of being exceeded in 50 years) expressed as a fraction of the acceleration c.ae to gravity. I is the occupancy importance jacroT related to the anticipated use of the structures as classified in Table 24.2. The importance factor I is equal to 1.25 ror essential and hazardous. facilities. This value is less than the maximum Fig. 24.2 Seismic map of the United States (UBC-94). {Reproduced from the 1994 Uniform Building Code, copyright 1994, wirh permission of the publishers, the Internat:onal Conference of Building Ofncials.)

): Ir:egulur struCl:Jres are those thc: hcve significant pbysical discontinuities in configuration Or ir. toetr lateral foree-resi,sting sYS(erM. Specific fcatures fer vertical structural irregularities are de~Gribed in Table 16--L and for horizontal irregularities in Table J6·M of UbC·94.

714

Unifo(m Building Cod&-·1994 Eqt,;jvalent Static Lateral Force Method

Eal1hquake Engineer;ng

TABLE 24.2 (Continued)

TABLE 24.2 Occupancy Importance Factor

- - -....

--~

...•..

---~--

Seismic Occupancy Category

Occupancy

Of

Functions of Structure

Esse:1riai

Group LDivision 1 Occupancies having

facilities:!

surgery and emergency treatment areas

Importance

Occupancy

Factor, 1

Category

1. Divisions i and 2 Occupancies with 5D or more resident incapacitated patients, but nor induded tn Category 1 Group I, Division 3 Occupancies

vehicles and emergency aircraft

Struct:Jres and shelters in ernergency-prepared:1ess center

Ali s~rnctures with an occupancy greater

A viation control towers

than 5.000 persons

Structures and equIpment in government

Srrecu.:.res and equipmem in

communication centers and other facilities required for emergency response

power-generating statlons: and other publiC utlli(y facilities :1ot included in Category I or Category II above, and required for continued opermion

Standby power-generating equiprnem for

Category I facilities

125

structures)

Occcprmcies BuHdings housing Group E, Divisions 1 and 3 Occupancies with a capacity greater than 3eO students

l.00

-----

(Reprinted from the 1994 Uniform Building Code, 1994, with permission of the publishers, the k,emational Conference of Building Officials.) The timitlulon of II' for panel COfltlcc[lons in Secrion 1631.2.4 shat! be LO for rhe emire connector. ! Structural observation requirements are given in Sections 108, 1701 and 1702. .. For anchorage of machinery and equipment required for life-safety sysrems [he value of Ip shall be taken as L5. !

~~-----------

Group A, Division 1,2, and 2.1

Group U Occupancies except for towers

structures

NonbuUding struClures housing, supponing or containing quantities of toxic or explosi ve substances which, jf contained w:thin a building. wocld cause that building to be classified as a Group H, Division 1, 2 or 7 Occupancy

----------~----

5. Miscellaneous

occupancy

housing or supporting water or other fire-suppression material or equipment required for tbe protection of Category I, ]] or 1lI structures

Group H, Divisions l, 2, 6 and 7 Occupancies ft:1d structures. therein housing or supporling toxic or explosive chemicals or subs[ances

LOa

structcres"

All StruCtUre housing occupancies or having functions not listed in Caregory 1. II or III and Group U Occupancy towers

4. Standard

Tanks or others structures conraining

3. S;Jecial occupancy

---------~

Group

Garages and shelters for emergency

facilities

Occupancy or Functions of Structure

Seismic Importance Factor, J

Buildings housing Groep B Occupancies used for college or adt:.lt education with a capacity greater than 500 students

1.25

Fire and police stations

2. Hazardous

715

LaO

value of 1= 1.50 in the pas~ code of 1985, UBC-85, but additional requlrements are included in UBC~94 which should further jncrease (he margin of

safety.

S is the site coefficient depending 0;1 ~he characterisrics of the soil at [he site as described in Table (243). This coeffiCIent is no longer specified on the basis of the ratio of the building penod to the soil period as in UBC-85. Rw is [he sfrtlclural focror ranging from 4 (0 12 as given in Table 24.4. Analogoes [Q [he coefficient K in UBC-85, the coefficient Rw is a measure of

716

Unifol1YJ Bu!ldir::g Code--1994 EqUivalent Static Lateral Force Method

Earthquake Engineerirg

TABLE 24.3 Site Coefficient'

Type

5!

s,

s,

Description

S Factor

A soil profile w~th eltber: (a) A rock-like material cbaracterized by the shear-wave velocity greater than 2,500 feet per second (762 m/s) or by otber suitable means of dassification, or (0) Medium-dense to dense or medium-stiff to stiff soil conditions. where soil depth is less than 200 feet (60960 mm).

LO

A soil profile with predominantly medium-dense to dense or medium-stiff to stiff soil conditIons, where the soil depth exceeds 200 feet (60960 mm).

1.2

A soil profile containing more than 20 feet (6096 mm) of soft to medium-stiff clay but more than 4() feet (12 192 mm) of soft clay. A soil profile containing more tban 40 feet (12192 mm) of soft clay characterized by a shear wave velocity less than 500 feet per second (142.4 m/s).

717

design snow load is greater thar. 30 psf, tbe snow load sbou!d be included in W, although it could be reduced up to 75% when approved by the building officiaL The seismic weight W should also include the weight of permanent equipment and partitions. T is the f"mdamemal period of the building which may be approximated from the following form",a (Method A):

T= C,rh;~'l

(24.5)

where

hI\'

towl height of the building in feet

Cr

0.035 for steel moment-resisting frames 3

C1

0.030 for reinforced concrete moment-resisting frames and eccentrically orz.ced frames

C.

0.020 for all other boildings. ' Alternatively, the fundamen:a! period of d:e structure may be determined from Rayleigh's formula (Method B) as ,v

2.0

(Reprinted from the 1994 Uniform Building Code 1994, with permission of the publishers, the International Conference of Building Officials.) ! The site factor shall be established from properly substantiated geotechnicai data. In locations where the soil properties are not known in suff:cient detail to detennine the soil profile type, soil profile S;. shall be used. Soil profile S~ need not be assumed unless the building official de(ennines that soil profile S4 may be present at the slte. or in the evenr rhat soU prOfile S!, is established by geotechnical data"

[he capacjty of tr.e structural system to absorb energy in the inelastic range through ducrlHty and redundancy. It is based primarily on the performance of similar systems in past earthquakes. The ap;:;roximate relationship between the two coefficients is Rw=8/K. W is the seismic weight whicb includes the dead weight of the bull ding and a minimum 25% of the floor live load for storage or warehouse occupancy. For office buildings 10 psf is added to the dead load, For cases in wbich the

ti

T:;::2,,~' ~ l~(,o;/g,::'/i8,

1.5

(24.6)

where the values of Ii represent any lateral force distribution approximately consistent with resu~ts obtained using eqs. (2&',7) and (24.8), or any other rational force distribution~ and the values of DI are the elastic lateral deflections produced by the lateral forces;; In the determination of the coefficient C ;n eg. (24.4) the value of T calculated from eg. (24.6) shall not be over 30% gre.:er than the value of T obtained from the use of eq. (24.5) in Seismk Zone 4 and not over 40% in Seismic Zone I! 2, and 3. This provision of the code is to avoid the possibility of L:sing an excessively long calculated period to justify an unreasonable low base shear.

of

1 A mCf:1Cl1I-resisling frame is a s(nJc:~ral frame if! which 1,'1e rnernber,~ and joints a:e capable forces primarily by f1e:<;ure. ~ Alternlltively. :he value of C, for ..::;ucmres with concrete or masonry structural walis (shear

rest~!ing

wall;:) may be taken as 0.1 ,,' Ac. The vaiue of A,

~hall

be determined from itt fo;!cw:ng: :orr,u:;a:

A~ = 2A~ [0.2";' {d~!h.v):J

where .4.~ is :he mini:tlum cross-sectional area in any horizontal plane in the fits( Sl.Ory of a 5trtlc[Urul wall (in square feet), D~ the length. in feet, of a structural wall in rhe first s(Ory in the direction parallel to :je applied forces. ar:.d ii,v ,he heigh. of the bullding in feet

Umiortf'l Su;ldi;1g Code--: 994 Equivalent Static Lateral FOrce Me:hod

24.5

DISTRIBUTION OF lATERAL FORCES N

C

The totai base shear force V calculated from eq. (24.3) is distribured over the height of the structure as a iareraJ force at each level, Fro plus aD additional force Fr (at [he top of the building), given by

i

!

FN ,--'-'--,/----, Ft

r----!j

J

I

7,9

I

:

i

(24.7)

I and F,

~

0.07 TV

< 0.25 V fo:' T > 0.7 sec for

1'50.7 sec

,

(24.8)

i

consequemly F,

m W

F.,. F i , F,"f FI

=

panioo of (he base force V at [he rop of the struCture in addition La F N

= seismic

weighc of ihe: xth or iIh ievet

The distribution of rhe lateral forces F.r and FI are shown in Fig, 24.3 for a multistory budding, The code stipulates thut the force Fl>' ar level x, be applied over the area of the building according to the mass distribution at {hat leveL

24.6

,

/m?7

k/7.

?77

Fig. 24.3 Di::mibucion of Jacerai forces in muhistory building.

force applied at level x, i, or N

h,(, h,;;;; height of level x or i above (he base

W,(. Wi

,

X

v

total number of stor:es above [he base of the building

= lateral

:

(24.9)

where N

:

..

.~

calculared center of mass. in each direction, a dismnce equal to 5% of the building dimension at that level perpendicular to the direction of the force under consideration. The effect of (his displacement shall be considered on the story shear dismbutiofl. Diaphragms are considered f1exible when the max:murn latera! defonnation of the diaphragm is mare than lWO :imes the average Story driE! of the associated Story. Th~s may be determined by comparing the computed midpoim-ir:-plar:e deflection of [he diaphragm itself under lateral load with the story drih of adjoining verrjcal~resisting elements under eqUivalent tributary lateral lead.

STORY SHEAR FORCE

The shear force V-" a~ any story x is gi "en by the sum of (he lareral seismic forces at that S(ory and above, rna, is,

Fi

(24.10)

Where the horizonta1 diaphragms at the floor levels of the building are not nexible, the mass at each level shall be assumed [0 be displaced from the

24.7

HORiZONTAL TORSIONAL MOMENT

The UBC-i994 Stales chat provisions should be made for me increased shear force resulting from horizontai (orsion whece diaphragms are not flexible. As already stared. diaphragms are considered flexible when the maximum lateral deformation of the diaphragm is more than twice the average story drift of the associared stones.

720

Uniform Building Code---1994 EqUivalent Static Latera! Force Method

Earthquake Engineering

The torsional moment at a given scory results from the eccentricity between the applied seismic lateral forces at the levels above that story and (he center of stiffness of the resisting elements of the story, The code also requires that an accidental eccentricity be added by displacing the center of r.JaSS in five percent of the bonding dimension at that level as it was described in Section 24.6. Further provisions in the code account for torsional irregularities in the buildir:g by increasing the accidental torsion by an amplification factor A,r; determined as

(24.11) where OmllA 8~vt

24.8

= the

maximum displacement at level x

the average displacement at the extreme points of the structure at level x.

detemIlne the calculated drift may be derived using a value or the coefficient C (eq. (24.4) based on the period determined by eq. (24.5), neglecting the lower bound ratio for C IRw of 0.075 and the limitations mentioned for the value of the period T used in the caIcu1ations of the coefficient C by eq. (24.4).

24.10

P-DELT A EFFECTS (P-il)

The so-called P-t1 effect refers to the additionaj moment produced by the vertical loads and the lateral deflection of columns, Or other elements of the building resisting lateral forces. F:gure 24.4 shows a column supportir:g an axial compressive force p, a shear force V, and bending moments 11;(-', and ,tin at the two ends. Due to this load, the column ur.dergoes a rejative lateral displacement or drift .;1. In this case, the P-!J. effect results in a secondary moment ,11$ P.1, which is resisted by an additional shear force PfllL in the column, It should be realized that this simple calculation of the P~.1 effect involves an approximation. since the secondary momen.t !'vfJ = P.1 will further increase

OVERTURNING MOMENT

The code requires that the overturning moments be detemIined at each level of the builcing. The overturning moment is calculated by static using the seismlc design forces F$. and FI regs. (24.7) and (24.8)] which act on levels of (he building above the level :mder consideration. Hence. the overturning moment M$. at level x of the buildtng is given by (24.12)

24.9 t \

f

STORY DRIFT LIMITATION

Story drift, the relati'v"e displacement between consecutive floor levels produced by tbe design lateral forces. shan include calculated transiational and torsional deflections. The calculated story drift shaH not exceed 0,04 IR w times the story height, nor 0.005 times the story height fOI buildings less than 65 ft in height. POI buildings greater ir. height than 65 ft, the calculated dnft shall not exceec. 0.031Rw times the story height nor 0.004 times the story height. These drift limitations may be exceeded when it js demonstrated that a greater drift can be tolerated by bmh structural elements and ~on structuraJ elements that could affect. life safety. Furthermore, the design lateral forces used to

721

v

Fig. 24.4 Deflected coiurnn showing the

P~d

effect.

:.Jniform Building Coce-1994 Equivalent Static Lateral Force Method

Earthquake Engineering

722

me drift: in the column and consequently will produce an increment of the secondary moment and shear fo;-ce in the column. An acceptable method

to

estimate the final drift is to add, for each Story,

the incremental drifts: .,1 .. due to the primary overturing moment M,(p; j:: . tt,{ due to the secondary momem wtn = M"p' 8~, Ll.1 • tJ.. due lO the next incremental moment M..s' Ot = U!lI,p' at) , B.o etc,. where the primary and secondary moments M"I' and M.m as weI[ as their ratio Ox> tire defined in eq. (24.13), Hence. the total story drift Ll''rcrAI. including the P-!J effect is given by the geometric series

24.11

DIAPHRAGM DESIGN FORCE

The code sripulates :hat floor and roof diaphragms should be designed [he forces de~er.nined by the following formula:

Fpt

= ~-.~v"'·"~'-- Wp,<

Iw,

i '"

(

=£1"

!

\

!-I-" I \

- Vd

Consequently. the P~d effect may be considered by mUltiplying, fOf each Story, ~he calculated srory drift and lhe calculated stOry shear force by [he amp!ifi~ cation factor 1/(1 - B.~) and [hen proceed to recalculate the overtumi:1g moments, and other seismic effects for these amplified story shear forces. The code specifies that (he resulting member forces and moments, as well as rhe Story drift induced by the P-Ll effect, shaH be considered in the evaluation of overall srruc[ural frame stability, The P-::l effect need not be considered when the ratio of secondary moment (resulting from the story doft) to the primary moment (due to the seismic lateral forces), for any story. does not exceed 0.10 or in Seismic Zones 3 and 4 where the stOry drift ratio nO( exceed O.02IR w• This ratlo may be evaluated at each StOry as [he product of the tota: dead and live louds above the story, times the seism~c drift divided by the producr of [he seismic shear force times rhe height of the Story. That IS, the ratio 0" at level x belween the secondary moment Mn resul[ing from P~tJ effect and the prima!"y moment M"p, due to the seismic lateral forces, may be calculated from the following formula:

me

(2413) where

Hx

=

(l)

J

Seismic Weighr al Various Floors:

For a warehouse, the design load should include a minimum 0; 25% of the live load. No live load needs to be considered in the roof. Hence, the effective weight at all floorS, except at the roof, will be 140 + 0,25 X 125 171.25 psf, and rhe effec~ive weigh[ for the roof be 140 psf. The plan area is 48 ft x 96 ft = 4608 fe. Hence. the seismJC weights of various levels are:

W,; W,

W,=4608XO.17i25=789.IKip'

IV, = 4608 X O. i40 = 645.1 Kip

of story x

heighr of story

'x.

(24.14)

Solution:

= drift of story x

= shear force

resist

Example 24.1. A four-stOry reinforced concrete framed building hus (he dimensions shown in Fig. 24.5. The sizes of the exrerior columns (nine each on lines A nnd C) are 12 in X 20 in, and [he interior columns (nine on line B) are 12 in X 24 in for [he bottom (WO s[Qries. and, respectively, 12 in x 16 tn and 12 in X 20 in for the two highes( stories. The height between floors is 12 fe The d.ead load per unit area of floor is escimured to be l40 psf; it consists of floor slab. beam, half the weight of cOlumns above and below the floor, partition walls, etc, The normal live load is assumed as 125 psf. The soil below the foundation IS assumed [0 be hard rock. The building site is located in Seismic Zone 3. Tf:.e building is imended to be used as 11 warehouse. Perform the seismic analysis for th is struCture (in [he direction normal to lines A, B, C) in accordance with the Uniform Building Code of 1994_

p.. == toml weight at leveJ x and above

V~

[0

in which Fi is the lo[erai seismic force, Wi is [he seismic weight ar each level, and Wp.{ [he weight of the diaphragm and auached parts of the building. The code States that the force FJH calculated by eg. (24.14) need not exceed 0.75 ZIW,,, but it shail not be less 'han 0.35 ZIW",.

which is equal to

Lim::JT,\L.

723

S

I Kip"" 1000 pounds.

724

Earthquake Engineerlng

Unirorrr:: Building Code--199.d. Equiva!ent Static Lateral Foree Method

The total seismic weight of tne building is then

where

W=789.l X3 + 6"-5.1 =3,012.4 Kip (2)

C( = 0,030 (for reinforced concrete moment-resisting frame)

h n = 48 ft (total height of the building)

Fundamental Period:

T = 0.030 X 48';' = 0.55 sec

by eq. (24.5)

•• ._

725

(3)

Base Shear Force:

V= ZiC IV

---.

....

by eg. (24.3)

Z = 0.3 [site in Zone 3 (Fig. 23.2)J

L. .I--I.I--I.I--I.I--I.I--I.I---I........ -~

1= 1.0 [warehouse (Table 24.2)J Rw= 12 [special

moment~:-esist;ng

space frame (Table 24.4)J

S = 1.0 [rock (Table 24.3»)

·c

,25S

~:52.75

C

7""

by eg, (2L.4)

1.25 X LO C = ~(O'5' ,,1IJ = 1.862

"J

and C IRw

0.155 > 0.075

Therefore. 0.3 X 1.0 X 1.862 V=-_·---·-30124 = 1402' 12 . . - Kip

(4)

Lateral Forces: F,

ib)

F,) W/h,

by eg. (24.7)

W"h i

E.e....oon

F"ig. 24.5 Plan all.:! elevation for a four-story boilding of Exampie 24.1.

= (V -

F,=O

for T=0.55 < 07

by

eg.

(24.8)

726

EarthqClake Engineering

Unjform BUild!!g Code--t994 Equ!valenl Static latera! Force Method

TABLE 24.4 Structural Factol

~-

..~ ..

i. Bearing

wall sys[em

Lateral-Force-Resist.ing System--Description

Light-framed walls with shear panels Wood structural panel wails for srrllctures three stories or less b. All other light-framed walls 2. Shear walls u. Concrete b. Masonry 3. Light steel-frameo bearing walls with

~-------

1. Steei eccentrically braced frame (EBF) 2. Light-framed wails with shear panels a, Wood s(ructural panel walls for structures three stories or less O. All other Ught-framed walls 3. Shear wails u. Concrete b. Masonry 4. Ordinary braced frames a, S,eel 4 b. Concrete c, Heavy timber S. Special concentrically braced frames

a, Steel resisring frame system

Basic Structural System l

L Special moment-resisting frames (SMRF) t!.. Steel b. Concrete "- Nfasonry moment~resisting waH frame 3. Concrete mtermedlate momem-resisring frames (lMRl')'

6

65 65

6 6

160 160

8

4

4. [leal systems

65

6

i60

4 4

65

Ie

240

9

65

7

65

8 8

240

La[eral ~ Force, Resisti ng Sys tem,"-Descri ptio n 4. Ordinary moment*rcsisting: frames (OMRF) a. SteerS b. Concrete L :>

[I,

gmviry loads a, Sreei b. Concrete'" c. Heavy timber

3. Momeot-

H'

------------------

lenslon"only bracing 4. Braced frames whcte bracing carries

2. Buliding frame sysrem

TABLE 24.4 (Conlinuea)

------

Basic Structural System'

727

160

1. Sheu wal!s d. Masonry wi,h SMRl' e. Maso.1ry with steel OMRF f. Maso:lry with conCrere lMRF~ 2. Steel EBl' a. With s(eel SMRF b. Wi,h Sleel OMRF 3. Ordinary braced frames a. S'eel wi,h sreel SMRF b. Steel wlrh steel OMRF c. Conccete with conCrete SMRF'; d. Cor,crete wi:h concrete IMRF 4. Special concentrically braced frames a. Steel with sreel SMRF b. S,eel wi,h steel OMRF

H)

6 5

160

12

NL

6

160 160 160 160

9 8

6 7

12 6

i':.L

10 6

NL

160

160

9

6 II 6

N.L. 160

----~------------------------

8 8 8

1'60

9

240

12 12

NL. N.L

9

160

8

65

), Undefined systems

See Sections 1627,8.3 and 1627.9.2

(Reprinted from the 1994 Uniform Building Code 1994, with permission of the publishers, the Imemmional Conference of BuHding Officials.) N ,L ~ No :imit. I Basic s(rucr.. ral syslem$ are defined in Section 1627.6. See Section :628.3 for combiMtion of s:rucn.:ral S'lstem. j H-Hcight limit applicable (0 Seismic ZOrles 3 ;1;"d 4 . See Sec!ion 1627.7. Prohibited in Seismic Zones 3 and 4, $ ProhibIted in Seismic Zones 3 and 4. except as permiucd in Section 1632.2, 6 Ordinary rcoment-resisting frames in Seismic Zone I meeting rhe :-equirerr.etHs of Section 221 1.6 may ;,lSI! an R,., value of 12. 7 Pro::ibiled in Seismic Zones 1, 3 and 4, S Prohibired in Seismic Zones 2A, 2B. J :1nd 4, See Seerle;; 1631.2.7. J

728

Eartr,q:;ake EngIneering

t:nlforrr BUild'r;g Code--1994 Equivalent Static Lalera! Force Method

TABLE 24.5 lateral Forces lor Example 24.1

Level

Wi (Kip)

--~--~----

4 3 2 I

64S 12 789.12 789.12 798.12

h, (ft)

Wth.< (Kip-ft)

Therefore, for these columns,

F, (Kip)

V, (Kip)

12 X3 X 10' X 8000 k=

.~-~--~-~~~~--~

48

30965 28408 18939 9469

36

24 12

49.5 45.4 30.3 15.1

49.5 94.9

Similarly, for COlumn 12 in X 24 in,

140.3

J = 13824 in',

K, = K, = 18 X 96.45 + 9 X 166.67 = 3236 Kip lin SimiJarly. for the third or fourth stories,

Shear Force at Story x:

Lateral Displacement: 0.>: and Story Drift LlE :

12£1 L'

k= 12x3x IO'X4096 =49.4 Kiplin

y, 12 X 20' = 8000 in',

12X3X IO X8000 k =----144 6' ' - - - . = 96.5 KIp/in

Hence, totat stjffness for the third or fourth stcries is

The lateral displacements of the buildlfl,g m~y be de~nni~ed using a ,computer program \'I,.hich impiements the statIC stlffness ,methOd o~ analysIS f~r plane frames. Ii However, to simplify the hand calculatiOns for thlS e,x_~ple, It is assumed that the building is a shear building, In this case. the strtt;1ess for a column between two consecutive floors is gIven by the formula k=-~­

f=;!;12X 16'=4096 in'.

1=

Calculated shear forces are shown ;n Table 24.5.

K) = K, = 18 x 49.4 + 9 X 96.5 = 1757.7 Kiplin The drift L1 for story x is caIcu:ated by

(24.15)

where

L = 12 ft (distance between two floors) E = 3 X 10 3 ksi (modulus elasticity concrete)

I

k = 166.667 Kiplin

The tOtal stiffness for the first a:1d second stories is then

Calculated lateral forces are show:'! in Table 24.5.

(6)

= 96.450 Kiplin

125,2

1:= 87781

(5)

729

= '" 12 X 20' = 8000 in'

(moment of inertia f~r the concrete section for columns 12 in X 20 in.;.

6 A comp~Her pro.l}t'.un for (he analysis of plane orthogonal frzme! is induded
where V", is the shear force of the story (Table 24.l) an:::: K~ is the stiffness of the story already calculated using eq. (24.15). 'The lateral displacement at any level is given by the Sum of the drifts of the overlying stories, Results of the calculations to obtain the story-drifts and lateral displacements are shown In Table 24.6. The code stipulates that the stot}' drift Ll,: should not exceed (0,04!R w) times

t~e story height or 0.005 times the story height. Hence, the maximum penulssible drift

LIma>.:

is given by the smallest of the following results: LI,,,, = (0.04112) [44

0.48 in

730

Earthquake Engineering

TAB LE 24.6

Uniform Building Code-1994 EqUivalent Static Lateral Force Method

Story Drift and Lateral Displacement for Example 24.1

Level 4

3 2

(8)

Scory Shear V, (Kip)

Story Stiffness K; (Kip/in)

Slory Drift .1.., {in)

Lateral Displacement 0, (in)

49.5 94.9 125.2 140.3

1757.7 1757.7 3236.0 3236.0

0.028 0.054 0.039 0.043

0.!36 0.082

731

Ovenurning Moments:

The overturning moment at each ievel of the building is given by eq. (2412) as

0.164

lvi,

F,(hs-h,)

+

s

L, F,(h,-h,) ;"" I

0.043

.....- - - -

Table 24.8 shows the resu[[s of the necessary calculations to determine {he

ovenurning moments. We observe in Table 24.6 that vaiues of [he drift Ll_r for all stories are well below [he maximum permissible drift.

(9)

Accidental Torsion,,! Momems: T, :::::: 0.05 *- D *' V,

(7)

Natural Period Using Rayleigh's Fonnu!a:

0.05 * 96* V,

0,

by eq. (24.6)

(10)

!--3ii~

T = 27(yl 386 = 0.48 sec X 17.42

The results of the necessary calculations are shown in Table 24.7. Because RayJeigh's formula yields a value for the fundamenmJ period (T= OA8 sec) somewhar lower [han the approximate value used in the calculations (T ~ 0.55 sec), [he seismic analysis could be recalculated using T= 0.48 sec. TABLE 24.7

Displacement

-----_.... 645.12 789.12 789.12 789.12

...

4

3 2

TABLE 24.8

--

0; (in)

Lateral Force F, (Kip)

W,8;

0.164 0.136 0.082 0.043

49.5 45.4 30.3 15.1

1735 14.60 531 L46

I= 38.72

F,o;

8.12 6.17

2A8 _ ..

I

0.65 _---

1742

The p·LJ Effect;

The code stipulates [hut [he P~!J effece need not be e\ta~uated when the rario of [he secondary moment Mn (0 [he primary moment My. at each level of the building, is less (hun 0.10. The results of the necessary calculations to evaluace this ratio [eq. (24,i3)] are shown in Table 24,9, This table shows that the largesr moment rario is 0.008, which is well below the code stipulation of 0.1. Consequently, there is no need (0 account for the P-iJ effecL

Ca[culations for Rayleigh's Formula for Example 24,1

Weight W,(Kip)

Leve!

Table 24"8 also shows the necessary calculations for the [orsiona! moments at the various levels of the building considering only accidental torsion.

Level x 4 3 2 Base

Overturning Moments and Torsional Moments for Example 24.1 Lateral Force F., (Kip)

SlOry Height H,(fl)

49.5 45.4 30.3 15.1

12 12 12

11

Overturning Moment M,(Kip·ft)

594 1724 3235 4919

Shear Force V, (Kip)

Torsional Moment T.(Kip"ft)

49.5 94.9 125.2 140.3

238 456 601

673

Earthquake Engineering

732

TABLE 24.9

Level x

4 3 2

Uniform Building Code----1994 Equivalent Static Lateral Force Method

Ratio of Secondary Moment to Primary Momen! for Example 24.1

P.,(Kip)

Story Drift .1, (in)

Story Shear V,(Kip)

Story Height H,(in)

M,IM,,= PxLl.o:IVxHx

645 !434 2223 3012

0.028 0,054 0,039 0043

49.5 94.9 125,2 140.3

144 144 144 144

0.002 0.006 0.005 0,008

Story Weight (Kip)

Overlying Weight

645.1 789.1 789.1 789.1

TABLE 24,10

Example 24.2. Use Program 22 to solve Example 24.1 afcer mode:ing the building as a pJane orthogonal frame (Fig. 24.6). Assume All r.otizon:al :r.e:r.bers to be 12 in by 24 in of reinforced concrete be<"ms = 3000 ksi)i.

Diaphragm Forces for Example 24.1

Fi (Kip)

W. (Kip)

:LF. (Kip)

4

49.5 45.4 30.3 15,1

645.1 789.1 789.1 789,1

49.5 94.9

( 11)

(24.5), for calculating the total base shear force V from (243) and the lateral forces F, and F, from eqs. (24.7) and (24.8). The program also calculates at each level of the building the shear force V.., [he overturning moment M,,;, the torsional moment the lateral displacement Ou the diaphragm design force and the story drift ~,. The seconda.ry mOment Mxn the primary moment M;p, and their ratio (j.f using eq. (24.13) are also caIculated. Finally, the fundamental period T is recalculated using Ray:eigh's fonnula, eg. (24,6).

rE

Level x 3 2

125,2 140.3

:LW; (Kip)

645J 1434.2 22232 3012.4

FpAeq. (24.14)] F p.
67.7 82.8 82.8 82.8

SolUlion: Program 29. which is included in the set of programs for this chapter, is des:gned [0 model the StruCture as a plane frame. The program uses tte stiffness method and starlc condensmion method to determine [he reduced stiffness matrix that corresponds ~o the Iate::-a! displacement coordinates at the various leve!s of the builciing. Next, ?;ogram 22 is executed to implement tbe

Diaphragm Design Force:

@

The code requlres that horizontal diaph;agm (floors and root) be designed to resist the force 7

F,+

F. by eg, (24.14)

I I,,.

" which need not to be greater than 0.75 ZlW"", but shall not be less than 0.35 ZIWp.o.-. Table 24.10 contains the results of the necessary calculations to determine the diaph,agm design forces at the various levels of the building. The minimum values of Fp"" shown in the last cdumn, in this case, are the design forces for the diaphragm of the building.

PROGRAM 22 UBC-94: EQUIVALENT STATIC LATERAL FORCE METHOD

Program 22 implements the provisions of the Uniform Building Code of 1994 usin£ the static lateral force method. The impJementation of the program provides for estimating the fundamental period T of the building from eq.

&& ® && ® M,

&&

, &ffi 0

24.12

733

ill

I

0. &ffi " ®

M,

@ ~

#,

24'

•

-'"

&~ CD &ffi 0 @

~ 7-

&~

& ".~-

x

24'

Fig. 24.6 Plane frame modeling the four-srory buHdbg of Example 24"2,

734

Earthquake Engineering

Uniform Building Code-1994 Equivalent Static Lateral Force Method

provisions of UBC-94. The implementation of this program requires numbering me joints of the frame consecutively excluding the fixed joints at the foundmion (labeled zero) as shown in Fig. 14.6. In this model each vertical or horizonral member represents all nine elements or members along the building. The flexural stiffness values El for me members of the plane fmme are calculaEed (lS follows: Horizontal members:

'"EIGHT i;;,?1

789.

'?

E/=9X3000X 12X24'112

oL:::1..

STrfF.

(£11

f:;:P-!W"2

= 0.37325 X 10' (Kip· in')

O. }1)S.09 }, 3';:'O~

Exterior columns:

.31)E.03

First and second srories: El

= 0.116 X 109

Third and fourth stories: £1 = 0.111 X 10

(Kip· in~)

C.3nE.05

,

9

'.). j

G.

:0

73E.O';

3nE~Oj

O.3"E~09

Interior columns: First and second stories: El = 0.373 X 10 9 (Kip. in 1 ) Third and fourth stories: El = O.? 16 X 10

Sc.;;T" fOR ·."ERT!e.;;!.. EL£:'12!~T$' ELEolENT ...

S'!'ORY

LO\o:ER JG!~:T ... 'J'??ER ':;O!NT

NO.

9

NO.

fLE1.UR,;.!.. ST!::"FNSSS

sr

KI?-n.·"2

O.E62.09

II

\I Ii \1 I:

37JS.09

The numerical results given by the computer for lhis example are the same as the values obtained by hand calculmion in Exnmple 14.1, except for the values of the smry-dri[[s and lateral displacemems. For the building that is modeled as a pinne frame, the story-drifts and lateral displacements are Slightly more than twice the values obtained in Example 14.1 for the struc(Ure modeled as a shear building.

216£.09

" l)

216£.09

0.37)E.09 0.21610.09

15

J 111S.09

!.6

G.216£.09 a.HIS.09

Input Data and Output Results ?ROGR;,.'{ 29,

HODEl..!NG ELD

:0

,

0.l1UO:.09 0.H6E.09

s

AS I'Ll-.NE fR ......,E:

o .lEE.09

'OU',?IJT R£SIJ!..'!'S·

GENERAL O,\TA

tilST

NUMBER Of S'!'ORISS NUHBE:R 02 JOINTS

NJ

(<::XCLUDING FlKEC')

!-If-:

SLE~ENTS

NUKESR Of KOR!Z-OtI?;"L

IN

NUMBER OF V2RTIC,;.L ELSHENT5

JOINT ..

X-COORDINA.T£

Y-CCO!WINA'fS

~

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0.00

24.00 ~6

00

0.00 2-1.00 ~a.

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4.5796:::.02

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~5S2ETOl

686'iE.02

~79S:::.02

.. \

86~'.'E.(J)

. ()4.·l·1E.OO

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0.0000£+00

OOOOE_OO 0000': ... 00

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2.68692 ... 02 . i.6782 ... 0J

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IMPORTM';CE :".;eTO;l.

STRUCTUR..;L SYSTE~1 ::".>.CTOR SOIl.. cos:"::"rGSN,

roo

1

73S

UniforM Building Code-1g94 Equivalent Static Lateral Force Method

Earthquake Englneertng

736

EiUH.. O:NG CAT;:', 2u:;..crw;; '::;:iJ£N$!Og ~8R:..'{,\L :'0 :.---,:rSAA> NRC;;:S
737

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:1<

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stORY ',11:::::::;;<,

,

lU?

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<

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,es

12 DC :: ;;0

?SOl

362

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,

.0-'

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67 _74 .0C

S;Uj6 32.5£

3.<. S"

2. USE

C7~J,(;}O

?OR RS!<::OORC:C;D CCl'}..."?.:;rE

DISC'=-.

~!Ol".EN1' RESIS,I¥G i"?_>"'V,ES

t...'1D ;;:CC7~'tR:tC B".A(:!:~ ".";.~.M.:':S

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00(;

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11'.').75 "'Ii

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45 S5 JO

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:?$

Example 243~ SOlve Example 24.2 for the period T= D_nA calculated by Rayleigh's fonnula in that example.

'?'C

,.

: :,;;

Solution: Program 23 is executed after modeling the building as a p~ane frame using Program 29. Option 4 is selected in the "Estimated Namral Frequency Nfenu" and the value T= 0.714 is input. From the results that follow, the base shear force is nOw equal to 117,84 Kip. Tbis value is greater than 80% of the base shear force lO.80X 1"0.75 = 112.8) obtained in Example 24.2 using the empirical formu!a to estimate the period of the building" Therefore, the reduced value for the shear force (V= 117.84 Kip) ;s accepta;,le by UBC-94<

:n

Input nata and Output Results

'78;:<510,,;..:. MOM?::T

Kt?hf7

S7SC'C?U;o,;,.:' S';lS'tEX ?;,r.;rOR

51,:0; C:C3fT:;;::Sir.

!LCO

Hi

555,30

45" 51)1

J. 712

,

"'

~l4S Cl

4,;;1

"

"

"

1~

5 7 $ .;;0

,:}. ·)0

s "

Uniform Building Code---i994 Equivalent Stalk Lateral Force Method

Earthquake Engmeering

738

S'rORY wr:Oi-iT

fI~rG;;7

$1'O,\'{

,,!?

E££1'

l:<:

';L !.lSi:

1a9.t2

12.CO

71l\'. :'1

~2

C'":

LEV'SL 6~S.U

" "on

n

,

739

7SS, t2

O,OJO ,'OR ;:;2!~IFC!!:CZD C~NCl'lETE !'ICI'.EK1' RESlS't!~G FHA:<£$ ;c.'1D :tC::£Y"'~.lC g!I.AC£:J FRA.'!fS

"" ~~

z,;

32 .. d

"

.~~

AA'!:'W OF

LEVEL

:~mlO:llTS

00;1'>'\.11' ?£S:;LTS:

o .oo~ () .26 ~

7. • G.

S2!S!'!!C ZON;; F.;CTC";

(LOLl:

l"r.;}lOAt!Z}/l';':'" PERIOD

SOIt. fAC1"C?:

o:m,;,'C(;

L:;G ..

PACTO? 'i

':'O'!'AL SASE SEtA"

111,31 KIf

24.13 S!MPLIFIED THREE-DIMENSIONAL EARTHQUAKE-RESISTANT DESIGN OF BUILDINGS ~:.,

";(;

24,13.1

::S

"

The general provisions of most seismic codes allow the mOdeling of a building by concentrating its weight at the various horizomal levels (floor diaphtilgms) which are assumed :0 be rigid in their planes. At each ievel of (he building three degrees of freedom are considered, twO translational degrees of freedom on a horizontal plane along [he main axes X and Y of [he building

lS

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('cw,!;:?

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and one torsional degree of freedom along the vertical axis Z. The main lateral resisring srructural elements are pro\dded by cobmns, structural walls (shear walls), and bracing elements. The horizonral diaphragns at the various levels of the building may provide different degrees of fixity at the ends of columns and sflucrural walls depending on the relative stiffness of the srructural walls and coL..!mns and [he horizontal diaphragms. In genera!, the laren:d stiffness, K, for a lateral resisting element may be expressed by [he following formula:

,n."

"

:.s:n. e9

SO,. CS

.50

27il:>.04

565.63

:: .GO

Modeling of Buildings

0.00

K

(L J / A£1) + (1.2 UGA)

(24.16)

Earthquake Engineering

740

Uniform Bullding Code-1 994 Equivalent Static Lateral Force Method

741

y

with

where E = modulus .of elasticity G

= modulus

of rigidity

cross-sectional moment of lnertia

I

A ;;;;;; cross-sectional area

= length of the eJeme.,t v = POIsson' s ratio,

L

Fig. 24.7 Schema;ic diagram of

The coefficient A in eq. (24.16) has a theorellcal value of 12 for a structural element [hat is completely fixed against rotation about both enes, and 3 for a cantilever elerr.ent fixed in one end and free to rotate in the other end. Investigating studies (Blume, J.A. et aJ. 1961; MacLeod, l.A. ,971) on the seismic response behavior of multistory reinforced-concre:e buildings have shown that the coet1iciem A in eq. (24.16) has a value from 1.30 to 6.91 depending on the relative rigidity of co1umns and the floor slab, In practice, a numerical value of 3 is usually assumed for the coefficient A.

24.13.2 Transformation of Stiffness Coefficients

linear biaxial stiffness elemem.

wher~ k.~ and ky are the necessary forces, respectively, in the x a~d v directions resu:tm g fro~ a uni.t dI3~lacerr~ent. in the corresponding directions; and n (k", = k". by lvlaxw~11 ~ rec!procl:y pnnclple) js the force in the x direction resulting from a uOlt dIsplacement in the v direction.

k.

The force components Ix and/ ;-equired to produce the displacements hs.vin o components Oy and i5, are {hen the sums from eqs. (24.17) (lnd (24.18): 1:0

(24.! 9)

or

The use of eq. (24.16) provides the values of the stiffness coefficients of the structural elements in :eference to principal cross-sectionai axes (x, y) of the element. Given a general layout of structural elements {columns. walls) at any story of the buildings, it is then necessary to transform the stiffness coefficients reterred to local or eiement axes (x, y) to the global system axes (X,

11

{f}

[kl{ oj

in which (24.21)

n.

Figure 24.7 shows a schemattc diagram of linear biaxial stiffness element

acted upon by a force F witt; (..'Ornponent.~ F;: and FY' ir.: the local axes, producing a displacement 8 with components 0;; and (\ along these axes. The forces f and and y directions, required to produce the displacement 0;; are given by Of

/'1). ' in the x

The transformation of the st:ffness coefficients k r , k.,. and k;u = k,'I. from ;ocai axes x, y to the global axes X, Y (Fig. 24.8), requires th~ su~titution into eg. (24.20) of

{f} = [I1 {f1

(24.l7j

Analogously, the forcesi'! and/,,- required to produce the displacement

oJ are

(24.18)

(24.20)

and

(oj = [11 {~j

which gives (2".22)

742

Uniform Building Code--1994 Equivalent Static lateral Force Method

Earthquake Engineering

y

y

'"

y

ky

"

(x~k~

y

Kyx A

resis':iog

i{x

e!eme."lt

i

~C.R.

iXR,YR) .~~~-------..... ) X

{b)

(aj

Fig. 24.& Force and displacement diagrams: for a linear btaxial ::tiffnes5 element

with

(24.23)

[K] = [TJ'[kJ [Tl and

Fxl [F] = [

FJ

[KJ = k x

k·j

cos

(j

sin

(j :

Fig. 24.9 Forces al the ceMer of rigidity C.R. and at u resisting element due (u) to a unit displacement ~n t~e Y-direction; (b) due to () ufli( dlsplaceme!1t in rhe X-direcrion.

the structural element.') due to a l!nir displacement in the X direction, or the sum of the mOmems of the forces (k y, k:
Xy >

[k", k y

K>xX,+K,Y,~I(kxY)

sinO cosO!

- KyX, + KxyY,

where

= I(kxr Y)

I(krxX,)

(24.25)

I(k y.\)

(24.26)

In eqs. (24.25) and (24.26) (he summations include all the lateral resisting stll!crura! elemenrs in the Story.

Fx. F y = force components in the X, Y direction Dx, .LI y = displacement componentS in the X, Y direction 8 = angle be[ween the global axis X and the element local axis x. or

ecos e k y = Xx sin'2 () + ky cos} 8 + 2 kJ..v sin 8 cos e

kx == kx

C05

1

8 + k.~ sin 1 f:J - 2 k.(t sin

k n = KYX, = (k.< - kv}

24.13.3

sin 0 cos 0 + k.

f,

(CO$2

8-

sin:' 8)

Center of R"lg·ldity

The center of rigidity, Of center of resistance, at a story of a buHding is defined as the point wl:ere the total stiffness Kx and K y= Ixi' and Kxy = K rx = Z krx of the StOry may be assumed (0 be concentrated; these $urmnations include all the structural elements in the story, The coordinates (X r Yr ) of the cemer of rigidity at a story are detennined as the solution of eq. (24.25) and eq. (24.26) expressing COat the sum of the moments of the forces (k x• k",) in

24.13.4

Story Eccentricity

The story eccentrichy is defined as the distance (measured in the direction normal to [he assumed seismic force direction) between the center of rigiditY and (he cemer of the applied design lateral forces above [he story conside:ed. The cemer of the lateral seismic forces (X n , Yit) of the stories above (he Story i is calCUlated as

(24.27)

744

U~ifo~m BUlldi:lg Coce--1994 Equ'valent StatiC Lateral Force Method

Earthquake engineering

24.13.6

where Fj

= laternl

745

Fundamental Period

(a) Empirical formulas:

force applied at level j

Xi' Y; == coordinates of the mass center for story i (point of application of the seismic force Fj )

IV = number of levels in the building. The eccentricities e,~, and e". at story i, in the X and Y directions are then given by (14.28)

L T=aN

(24.32)

2. T=bffiL

(24.33)

3. T=C.H'

(24.34)

where the va:ue of the coefficients a, b, Cr, ilfId of the exponent x are pre~ scribee by seismic codes, and Hand L a:e respectively the total height and the length of the building as. defined by the seismic code. (b) Rayleigh's fonnula:

(2435) in which X n , rri are the coordinates of the center of rigidity for story i [solution of eqs. and (24.26)]. where

W; = weight at level i

24.13.5

Rotational Stiffness

b,' = Lateral displacement at level i

Fi = lateral force at level i prescribee. by the seismic code

The rotational stiffness at story i is defined as the torsional moment necessary, at this story, to produce a unit rotation of the story. The ro~atlonal stiffness. Ki> at story i is then given by

g = acceleration of gravity

24,13.7 Seismic Factors (2 4 .29) in \<.ihich kxtj , k y1i , and kxY,j are the j~resisting element stiffness coefficients of the i story evaiuated by eq. (24.}4); and

(24.30)

where Xtj and Yo are the centroida! coordin2.tes of the resisting element j of story i. The story rotation B, is then given by

The following factors are generally require m the implementztion seismic codes.: (l)

Z,

= peak

ground acceleration

peak grour:d velocity

(2)

Z"

(3)

J

= occupancy importance factor = sl'1lctural

(4)

K

(5)

R = structural reduction factor

(6)

S = soil profile factor

(7)

C

= selsmic

factor

coefficient given in seismic codes by

(24.31)

(206)

where the torsional moment, J1z11" is calculated by eq. (24-44) and the torsional stiffness, K1j, by eq. (14.29).

in which the numericai values of p, q, and x are prescribed by the seismlc code; T anc S are, respectively, the fundamentai period of the building and the

746

Earthquake Engineering

Uniform Building Code--i994 Equivaler:t Static Lateral Fo.ce MelhOd

soil profile factor whose value is specified by the Code for dIfferent types of soils.

24.13.8

747

in which

T

fundamental period of the building

V = base shea( force [eq. (24.37) or eq. (24,38)J,

Base Shear Force

The bnse shenr force V is determined in different 5leismic codes by formulas of [he foon

24.13.10 (24.37)

Overturning Moments

The overturning momem, /1.,;1;, at (he level i of the building is calculated as

or

Ie

V=

R

where the values of the seismic factors Z", f, K, C, S are prescribed by the seismic code and W is the rotal seismic weight of the building which is calculated by

W=D+AL

F;(h, - h,)

(24.38)

(24.39)

(24.42)

where i=O, 1,2, ... , (N- I) f~ =

equivalem seismic force

fit

level j

F, = addirionnl force at the top of (he building [eq. 24.41)]

where

h" hi> h"

D = dead load L

= design

height

(0

levels i, j, N from the base of (he building

N = tOral number of kvels in the building.

Eve ioad

A = fraction of the live load prescribed by the seismic code.

24.13.9

24.13.11

Equivalent Lateral Seismic Forces

The equivalent Seismic Force, F," acting horizontally at the varioU5l Jeve!s of the building are given by

F,

(V

F.) W;hf N

The stOrj shea: force, V" at level i is given by

, V,~P,+)' F,

(24.43)

1"'1

(24.40) in whkh F! \s given by eq. (24.41) and Fi by eq. (24.40).

L,,"' W·h" ;, J'"'

Story Shear Force

I

where

24.13.12 Torsional Moments

W = seismic weight at level i

hi> hJ = height of level i, j :'rom (he base of the building

Torsional moments, A1;;" resuhing from [he seismic forces in the Y direction are given I'lL each SrofY of the building as the product of the story shear force, V, [eq. (24.43)) and the eccentricity at the story [eq. (24.28)] usuany increased by a faclor a and by an additive (eon eo to account for accidental eccentricity. Hence,

x = ex;;onem prescribed by the seismic code (u5lually. x = 1) F,

additional force at [he top of (he building given by

F,

=O.07TVS; 0.25 V

F,=O

for T> 0.7 sec (24.41)

(24.44)

748

Uniform Building Code-1994 Equivalent Static Lateral Fo!'Ce Methcd

Earthquake Engineering

24.13.13

24.13.14

Story Drift and Lateral Displacements

The elastic story drift or interS tory displacement at story i, Ll,(, due to the lateral shear force only, may be calculated by (2<1.45)

Forces and Moments on Structural Eiements

The story shear force Vi [eq. (24.43)] and the overturning moment Mi reg. (24.42)], as well as the shear force resuhing from the torsional momem Mri [eq. (2.1,44)J. are dIstributed among (he laterai resisting elements in proportion ~o their relat.ive stiffness as presented in the following sectioD,

24.13.14.1 in which K, is the total stiffness of story i and 'Vi is the story shear force. To account for inela.stic deformations, most seismic codes specify an amplification of the story drift by dividing the calculated elastic story drift by the structural type factor K when K < LO or mUltiplying the elastic story drift by the reduction factor R. Hence the amplified or inelastic story drift.

Element shear force resulting from the story shear force

The shear force isY; on elementj, assuming the seismic forces in the Y direction, is given by (24.52)

.1" is calculated by .1 = .1"

,

K

(K < ].0)

749

where (2446)

= stiffness in the Y direction of element j Ll y; == ejastic drift in the Y directIon for story i.

or by (2447)

The lateral displacement, 0" at level i due to the lateral shear force only IS then given by the summation of the story drifts of the above stories. That is

" .11 6,= )'

24.13.14.2

Element shear force resulting from the torsional moment

The shear force 1;(/ in the Y direction on eiementj resulting from the torsional moment is calculated by (24.53)

(24.48)

J""';

The total disp]acements including the effect of story rotation Opx and opy, in the X and Y directions, at a selected poim P(Xp, Yp), at level i, is then given by (24.49) (24.50)

where OiX and Ory are the lateral displacements due to the seismic lateral forces applied either in the X or in the Y direction; and 8 is the story rotation given by eg. (24.31). The total displacement 8, of a selected point, P, is then calculated as (24.51)

where k yj , kYXJ = stiffness coefficients in the Y direction of elementj, due to displacemems of this element, respectively, in the X and Y directions Xli' Y 1/

X

rl ,

= centrojdal coordinates of a resisting element j

y'..; =

in the i story and

coordinates of the center of rotation of story i.

24.13,14.3

Element total shear force

The total shear force in the Y directiQn~ on the element j resulting from the story shear force and from the torsional moment is then given by

ir.,==fr't)'

(24.54)

750

Earthq:.Jake E.ngineering

Uniform 8uilding Code-1994 EqulvGife:ll Static Lateral Force Method

Ana:ogously for the X direction. the element shear forces j,x;, /'Xj, and jXj afe calculated by (24.55) (24.56)

and j'<.j=j,;<.J

24.13.14.4

(2457)

Element moments resulting from the story overturning moment

The smry overtumlng moment M, is distributed among the resisting el~ effiems in lhe story. The overturning mOment ttli; distdbuted to the resisting element j of story i is given by

k Yi

session) or tn a batch mode (with the dara being input from a file previously prepared), The program aigorirhm assumes that all lateralload-resis[ing elements extend the entire height of the bu!tdlng, However, structural systems WiEh discontinuous elerntnts may be analyzed by modeling a discontinuous eltment as a comineous element having zero lateral stiffness in those sraries in which It does nor exist. A lateral load~resisdng element may be oriented in an arbitrary direction of a noor plan. The orlentation of the resisting element is defined by the angle ~hat is measured counterclockwise from the posiLlve global X axis to the POsili'"e !ocal element x ax.is. The lateral stiffnesses of the resisting elements are described in [heir local element coordinate system. These coordinates are specified as input data for the analysis. The program calculates (he seisrr.ic design base sheflr ar,d the corresponding lateral seismic forces distributed ro the stories, The program also determines the centers of rigidity at each of the scones, the distribudon of the lateral shear among the resisting elemen~ at each story, rhe effect of the torsional moments corresponding [0 the design eccentricities specified in the code, the story drift resu~([ng from the lateral Story shear, and the rotations due (0 the story torsional moment. Example 24.4. A 20-story steel building, with a latera! load resisrino system composed of it momen~-resisrlog spac; frame, is shown in Fig. 24.{Q.

where

Mi = overturning momenl

at stery· i

stiffness in of resisting element j in the Y direction

Xi = wral stiffness of story i in the y direction. The components m"., and m'll of the moment m;i along the local coord ina res x,

y of the resisting element are [hen calculated as mji.1

= m;) cos e (24.59)

o

N

where

f)

is the angle between the global axis X and the element local axis x.

.

24.13.15

751

Compuier Program

A general computer program has been developed to implement (he provisions of the Unifonn Building Code 1994 (UBC·94). The program is written in ADY ANCE BASIC for execution on m;crocomp:.lters. It may be executed ei[ner interactively (wiTh [he data supplied and stored in a file during the jmeractive

I

It,,,,,,,,,,nmrl Fig. 24.10 Building modeled for Example 24.4.

Uniform Building Code-1994 Equivalent Static La1eral Force Method

Earthquake Engineering

752

A gravity load of 1320 Kips is attributed [0 each floor level of the building. Lateral stiffness of each column at all floor levels in 250 Kiplin; it is assumed to be the same in both principal directions. The buHding is located in a seismic region for which the design peak ground acceleration ZA = 0.2 g; Occupancy Importance Facror I to; Soil Site Factor S = LO and Structurai Reduction Factor R = 12. The analysis of building using the computer program is to be perfomed in the N-$ direction according to the UBC-J994 Seismic Code of United States of America.

10

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Uniform Building Code--1994 Equivalent Static Late~al Force Mel.'1od

EarthquaKe Engineering

755

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Ea.rthquake Engineering

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Fig. 24,11 Structural model for ::he shear building shown in Fig. 24.5.

n

H6HZ,,0:; -.HHSS:~C2

Solution: The structure is modeled as shear building represented by the column with concentrated masses as shown in Fig. 24.11.

{).2ii'iS}E>C}' t.1i9::1C;;;-C1 13923.C;r.. ('!1

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The following numerical values are calculated, from data provided in Example 24. I, for implementation in COSMOS:

CaJ Story heights: 24.14

757

EQUIVALENT STATIC LATERAL FORCE METHOD USING COSMOS

Example 24.5. Use Program COSMOS to obtain the lateral displacements and shear force at the various !evels of the four-story buiJding of Example 24.1,

h:

(b) Concentrnted masses: m,

, To save space only results for STORY -20 corresponding m design eccentricity (in) are reproduced.

Bee = 80

h,=h,=h,=12X =12=I44m

m,

m, = 789 i386 = 2.036 (Kip· sec'iin)

758

":otform Bui:ding Code--1994 Eqt.:iva!enl Statk Lateral Force MalMo

Eartr,quaxe Engineering

759

(5) Define element group I using BEAM3D:

(c) Cross-sec[lonal :areas:

PROPSE'1'S

A,=A,=240X 18+288x9=6912 to'

A,=A,= 191

x 18 ·c240X9=5616in'

SGRODP,

EGROUP

>

1,

8£f:._"13D,

0,

0,

0,

0,

:::,

(6) Defir.e materia! property for group I {E

0

3000 ksi);

?ROPSEiS } NPROP (d) Ctoss-sectional moment of ~nertit1:

MPRO?,

i, =1,=8000x 18+ 13824X9

268,416 ;n'

i, = J, = 4096 X 18 + 8000 X 9 = 145.728 in"

1.

EX,

3QCC

(7) Define real consrams for beam elements and generate beam elements and 2: RCO~S'l'

PROPS21'S ;.

RCONST, :, 2, ;),O,C,O,O

(e) Lateral forces:

1:::,

0-,

t1E$HING ' ~l_CR rCCR, 1, 2, 1, 3,

From values calculated in Example 24,1:

6912,

1, 1,

268416,

268416,

2C,

12,

16,

12,

6

(8) Define real COnStams und generate beam e!er::1ems 3 and 4: F,

= ! 5.1

Kip

F,

= 30.3

Kip

F3 = 45.4 Kip

F, = 49.5 Kip

The following commands are implemen!ed in COSMOS: (1) Set view

[0

the XY plane: ~

DISPLAY , VIEvCP.!l..R

IJISW,

0,

C, 1,

VIEW

?ROPSE'I'$ , RCONS'!' RCONS?, 1. 2,

}.O, 5616, G,O,O .. O,O ? t-LCR !-l ..,CR, 4, 5, 1, 3, l. L 6

l45728,

(9) Merge il:1d compress nodes:

;) MESHING '

(2) Defi ne the XY plane at Z

= 0,

NODES " !-<1'lERGS

?-tNERGE, L :-iCO!/;PRESS,

GEOMETRY ) GRIu > ?LANE PLP-_\J£. Z, (I, 1

20, L 1, l2

C,OOC1,

0,

1,

?ROPSETS ) 2GROD?

EGROUP,

2,

t>VSS,

0,

0,

0, 0, 0,

(1 J) Define real coostant for mass elements; GEOMETRY > '?OIN'!'S > PT PT, ~ O. O. 0

.

RCONST,

2. O. 144.0 PT. 3, 0, 288, 0 0, 432, 0 PT. PT, e O. 576, 0 , PT, E. 432. 0, " PI' ,

(12) Generare

••

;r.aSS

NESHINC 2,

L

7,

elements

m" m,. m, = 2.036;

2.036, 0 , 0 , 0 , 0 , at

O. O.

14<;;,

0 CRPCOR.D, 2. O. l44, 0, 0, 288, 0 CRPCOfW, 3. O. 288. O. O. 432, 0 CRPCOR;), <• O. 432, O. O. S76, G

RCONS'1',

:'1_PT,

4,

0

M~.PT

4, 1

2,

4,

1, 7,

1.671, 0,

(14) Generate rhe mass element at node 5: MESH:NG

C,

nodes 2, 3, and 4:

PAR.l>l·CM£SH )

>

G

(13) Define u1e real constant for the mass element m4

(4) Generate lines between keypoints:

c.

2, 3,

)1~p'r,

~,

0,

0

(10) Define element group 2 for mass elements:

(3) Generate keypoints at the levels of the buiiding and a point on the XY plane:

CRPCOR;),

145728,

>

P}\R,.llu.'1_HESH S, 1

\-:~PT

0,

0,

0,

= :.671: 0,

G

Uniform Building Code--1994 Equivalent StatIc Lateral Force Method

761

Earthquake Engineering

760

TABLE 24.12

(15) Merge and compress nodes: L

MESHING

:>

20, 1, 0,

NODES

)<

(iOOl,

£>O~CJ:S

S;"E>l£N,

MESHING ) NODES ) :N'MERGE Ki~ERGE,

Forces and Stress Calculated by COSMOS for Example 24.S

W~B~

0, L

1"00;::2

(}

NCOM?RESS

(16) Apply constraints in all degrees of freedom at node 1, and at all degress of freedom at nodes 2 through 6 except for direction UX'

S7?2SSES

,.

NODBl

'F.-"c. cocce~c;)

'.1.0000;;+')0

V$T',

:41.l.n-~c~

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ooco:;~c-c

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f~~) OOOOs~OO

mID, L AL, 0, ::., 1 DNC, 2, UY, 0, 6, 1, UZ, :iX, RY, RZ ([7) Define lateral forces at the various levels of the building (nodes 2, 3,

O
;;.COOO;;;~Oj

(K~ IS~) ~[), ,~5'!,,~on

5;',""><

~C.

33'5aS-I)C

:7 fA;

z:';.

5, and 6) as specifled by the code [eqs. (24.7 and 24.g)]: ~OADS_B:J

FNJ,

2, FX,

15.1,

2,

FX, 30.3, 4, FX, 45,4, 5,

4 S-ND, 5

2ND, ?ND,

S7RJCTCRAL

6,

FX,

49.5,

6,

FORCES

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(':"'r·CT;l':.

"

?,!:",o

(18) Execute static analysis:

,0:;C?E~(:C

!1~'" ~,~a3

OOOOS.OJ

:nCCZ~:;)

C 'COC!i:-CC

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;:-r~c:1:0){/,/p!",~.OCOC'::-CC

O.OOoO:!>';:?

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s.=x S",::.;;

(19) List [atera! displacements at the nodes:

",).1957;<;-OC

~,:9$~:::_0C

~:L~95i£~O·;

O,:9~7E*0C

m_.

;.B7S~r,()

_,

~:;$7E~C{l

RESt;L':.'S > LIST'" DISFL;

(Table 24.11 shows the lateral displacements given by COSMOS)

forces of the building, Other resulrs, such as stresses at the nodes of the beam elements, given in Table 24.12 refer to the dements in the model for this structure shown in Fig, 24. r L

(20) Execute stress analysis:

(2l) Use an editor to list from the output me resulting forces and stresseS in the structural elements (Table 24.12, shows forces and stresses calculated

by COSMOS): Comparison of results given by the program COS:vfOS with those obtained in Example 24.1 by hand calculations shows, as expected, a complete agree~ ment in the lateral displacements at the various leveis as wen as in the story TABLE 24.11

NOLl!:

Lateral Displacements Calculated by COSMOS lor Example 24.5

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C.bH2J

SUMMARY

Th~

Uniform Building Code of 1994 provides two methods for earthquake reSIstant design of buildings, the static lateral force method and the dynamic method. The stanc lateral force method is applicable to (1) s.tructures in Zone I ~nd to those structures in Zone 2 designed for occupancy importance factor IV. (2) regular structures under 240 ft jn height, and (3) irregu]ar structures of no more ~han five stories nor over 65 ft in height. The dvnamic method may be used for any structure including tall structures over ft with !ITegul~r features, as pennitted by the code. The first of these methods is presented in [his chapter. The second method, which is based on the modal superposition method and the use of response spectra, is presented in Chapter 25. Computer Program 23 implements the provisions of UBC-94 using the Static Lateral Force :vfethod.

240

ZZ~?07

,'!-:t07.

24.15

COO-~o£ .. ~C

762

Uniform Building CCde---1994 Equivalent StaLe Lateral Force Method

Earthquake Engineering

The size of the exterior co!UffiftS (nine each on !ifle A and on Hne C) is 12 in X 20 in and the interior columns (nine on line B) nre l2 in X 24 in ror all the Stones or the building. The heighf between £1OOts is 12 fe The bui!djn,g is located in !he Los Angeles. Califomia area and is if'ltended (0 be used as a hospital. The estimated wcigh( at each !evel, inclUding per:nnnent equipment, is 950 Kip. except u( tl1e lOp level (roof), which is 650 Kip. ~1ooel the structure as II shear building (Program 27).

PROBLEMS 24.1

Use Program 23 fer seismic resis(unt design of a lO-slory concrete building (£ "'" 3000 ksi) having [lte plan. configuration and elevation shows in Fig P24 . .(.

--.

c--

._24.2

I

B--

A--

1-

,

1

! J.

Use Program 23 for the seismic resistant design of (he W.story steel building shown in Fig, P24.2. For each story. she flexural rigidity £1 = 8 X 10' Kip in 2 and

!

- - -• ,..

11

".

~

I,

!

II !

II II

I,

I

~I

.• ~

~

,I

I

'"

Fig. P24.1.

763

I W&;,//////I"h\

Fig. P24.2.

764

Earthquake Engineering

Uniform Building Code--1994 Equlvalent Static Lateral Force Method

the story weight at the floor level is W = 500.00 Kip. Model this structure as a shear building (Program 27), 24.3

24.4

Repeat Problem 24.2 modeling L.1e structure as a plane orthogonal frame in which the flexural stiffness of the horizomal members is EI= 6.0 X 10-' Kip· in! (Problem 29;.

A

12~story

moment-resisting steel frame sbwn in Fig. P24.4 serves to model a

forces for the herizor.tal diaphragms. The overall length of the buiiding is

D = 6 x 3C = 180 ft. ~vlodeI :he building as a plane orthogona! frame. (Use E = 30.OCC ksi.)

~

x

". x

j'

moment~reslstlng

24.5

Solve Prob:em 24,4 arre: mode Eng the structure as a stear building with rigid horizontal diaphr&gms af ali :eveis of the bvHding,

24.6

A three-story concrete warehouse building with shear walls as shown in Fig. P24.6 is located in San D~ego, California 0:1 a son ?fOfile of type 2, The stor), heights are 20 ft fer the first floor and !.5 ft for the second and third tloors. The total design load for the first, second, and third levels are res;Jectively 300, 2Si), and 220 Kip. Using the UBC-85, determine: (a) equivalent lateral s.eismic forces, (b) Story shears. and (c) horizontal iateral displacements and story drift. Model the shear walls as deep cantilever beams (Program 28). Assume modulus of elas~jcity for concrete E, = 3.6 X l O~ ksL

b:.tilding for earthqaake anaiysis. The effective weight on all the levels is 90 psf except on the roof where it is 30 psf. The tributary lead for the fmme:; has a 30 ft width. The building is located in Los Angeles, California cn a siLe with a soil profile classified as Type 5,. The imended use of the building 1s for offices, Use the provision of the UBC·94 to determine: (a) equivalent lateral forces, (':1) sLOry shears. (c) horizontal torsional moments (assume accidenral eccentriclty}, (d) overturning moments, (e) lateral displacements and story drifts. and (0 (he design

,-

,",ex

765

I

WjSx4l)

=J -I

••

li

W16x«l

~'

;

~

~

~

x

!

w,Sx.u)

--L [Io,,_tim;

•

W21xU ~

IS

i!

~

x

x

~

W2.1 xU

; "Ii

W21 x So

~. x

• ~.

~

.. x

:;;i

~

~

x·

~i

I

~I x

~

W21 X

so

W24)(

S5

~

<.

x

x

w:t4 x 55

I,

1

wtlxS5

~ d7

14.-

." ,,'. ,.,.

J

Fig. P24.6.

W2.txS5

~

;;;:;',; ~

•

~

l~ , I'

Fig. P24.4.

24.7

Solve Problem 24.6 using the value of the natural period derermined with Rayleigh's formula. Observe the limitations in the value of coefficie:1t C calculated by eq. (24.4),

Uniform 8ui:ding Code~1994; Dynamic MethOd

25

[he sar'ne direction of [he lateral displacement, may be written, neglecting dampi::g. f:om eq. (! 1.35) as

[Nf] (iI} + [K] {uj = - [M){l}Y,(r)

Uniform Building Code-1994 Dynamic Method

767

(25.1)

In eq. (25.1), [,\ef] and [KJ are respectively the masS and the s[iffness matrices of the system, {u} and {ii-} are respectively the displacemem and acceleration vec[ors (relative to :he base), Yr(£) is [he function of [he seismic acceleration at the base of the buiidir.g, and {l} a vector with all its elements equal to L

25 i .1.

Modal Equation and Participation Factor

As presented in Chapter 11, the solucion of eq. (25.1) may be found by solVing the corresponding eigenprob:em

((K]] -w'(MJi (qi) ={O}

(25.2)

determine the natural frequencies Wi, w?,_.~ w" (or natural periods T" T2."., TN) and the modal matrix. (
(25.3)

In Chapter 8, we introduced the concept of response spectrum as a plot of (he maximum response (spectral displaceme!lt, spectral velocity, or spectral acceieratio:::)) versus the nawral frequency or natural period of a single degree~of~ freedom system subjected co a specific excitariofl_ In the present chapter, we will use se~smic response spectra for earthquake resistant desIgn of buildings

\5 introduced in eq" l", in which

F",

1) to yield the modal equations

+ w:,z", = - r mY! (r)

(m

= 1, 2, ... , ,/>/)

(25.4)

is the partiCipation facwr given by eq. (11.39) as

modeled as discrete sYStems wirh concenc-ated masses at each level of the building. It is necessary to perform a transformation of coordinates to obcain the modal equations of motion, then combine (heir spectral responses to obtain the maximum response of the structure because response spectral charts are prepared for single degree-of-freedom systems. In eap.hquake resistant design of buildings, [he maximum responses include displacements, acceleralions, shear forces, overturing moments, and to!-sional moments.

r"=-~

(25.5)

N

L Wi4>~,

i"'l

For normalized

e~genvec[ors,

the participation factor reduces to (25.6)

25.1.

MODAL SEISMIC RESPONSE OF BUILDINGS

The equarions of motion of a shear building modeled with laterdl displacement coordInates ar the N levels and subjected to seismic excitation i1! the base in 766

because in this case

768

Un:form Building Code- ~ 994: Dynamic MethOd

Earthquake Engineering

where g is the acceleration due to gravity. Damping may be introduced in the modal equation {25.4) by simply adding the damping term to this equation, namely,

From eqs, (25.3) and (25.9). tbe maximum acceleration at the level x of the building is given by

a.~",

769

of the mth mode

(25.10) (25.7)

where /;'" is the modal damping ratiD. Equation (25.7) can be written for convenience with omission of the participation factor as

in which Sam and axltl. are usually expressed in units of the gravitational acceleration g. As stated in Chapter 8, the modal values of the spectral acceleration S""" the spect.-al velocity S,w, and the spectral displacement Sa", are related by an apparent harmonic relationship:

with the substitution (25.9)

25.1.1.

or in teillls of the modal period T", = 21r/w", by

Modal Shear Force

...

. ...

':'11

! ,

The value of the maximum response in eq. (25.8) for the modal spectral acceleration, Sam = (qrn):n;H' is found from an appropriate response spectral chart, such as the charts in Chapter 8 or the spectra! chart provided by UBC~94 (Fig. 25.1).

'r----,--

Sum

T Slim

lT

,-;,

= I £7Tr

'I'

\

! SJm

HI I

On the basis of these relations, the spec[ral acceleration S..m in eq, (25.10) may be replaced by the spectral displacement SEn: times w:~ or by the spectra] velocity SVNI times W m _ The modal lateral force F.,m at the level x of the building is then given by Newton's Law as

or by eq. (25.10) as (25.11)

---

in whid: Sam is tr.e modal spectral acceleration in g units and W): is the weight attributed to the level x of the building. The modal shear force V=, at the level x of the building js equal Lo the sum of the seismic forces F,m, above thaI: level, namely, N

V,.~= )' F"

(25.12)

i =:A

The total modal shear force V", at the base of the building is then calculated

as Fig, 25.1 Normalized response spectnl shape:;, (Reproduced from 1994 Unfform Build· ing Code, © 1994. with permission of the pUblishers. the International Conference of

Budding Officials.)

N

V",= )' F.,. ,-"

~

(25.13)

772

Uniform Building Code... 1994- Oynamic Metr.od

Earthquake Engineer:ng

The modal drift .:1(", for the xth story of the bulldir.g. defined as the re!ative displacement of two consecutive levels. is given by

with "em = O. As indicated in Article 14.7, the UBC-94 stipuiates thar the calculated storY drift, which must include translationai und torsio:lal deflections, shaH n~t exceed Q,04iRw times (he story height or 0.005 times [he story height for buildings less than 65 ft in height. For buildings grea:er in height thar. 65 ft. the calculated drift shall not exceed O.03/R" times the story h~ight or 0.004 rimes the story height. 1

25.1.7.

Modal Overturning Moment

The modal overturing moment M.tll'l at the level x of the building which is calculated as the sum of !he moments of the seismic forces Far, above that level is given by N

iWxm =

;;

Fm: (hi

h~)

(25.26)

773

As mentioned 1n Chapter 24, the UBC-94 requires that an accidenta! torsional moment be added to the torsional moment existem at each level. The recommended way to add the accidental torsion is to offset the center of mass at each level by 5% of the dimension of the building normal to the direction under consideration.

25.2.

TOTAL DESIGN VALUES

The design values for the base shear. story shear, lateral deflection, story drift, overtuming moment, and torsional moment are obtained by combining corres~ p~nding modal responses. Such combinatlon has been performed by applica:ion of the technique SRSS. This technique, as mentioned in C:tapter I!, estimates the maximum modal response by calculating the square root of the sum of the squared values of the modal contributions. However, as discussed in Section 11.6 of Chapter l j, the SRSS technique may result in retatively large errors when some of (he nawral frequencies are closely spaced. This siwation generaiIy occurs in the analysis of space structures. At the presem, the more refined technique described in [hat section, CQC (complete quadratic contribution), is becoming the teChnique of choice for implementation in computer programs. For preliminary or hand calculatior., however, the simpler technique SRSS is commonly used. The following formulas may be used to estimate maxi:num design values by application of the SRSS technique:

i"''-:

where hi is the height of level i. The modal overturning moment ,VIm at the base of the building then is given by

(1) Design Base Shear:

V=

(25.29)

,v

kl",= ') Fimh, ;"';;1

(25.27)

where the modal base shear V", is given by eq. (2S.!5). (2) Design Lateral Seismic Force at Level

25.1.8.

x:

Modal Torsional Moment (25.30)

The modal torsional moment Mwr. at level x. which is due to eccentricity ex between the center of the above mass and the center of stiffness at that level (measured normal to the direction considered), is calculated as

where the seismic modal force F.
(25.28) where 'ill'" is the modal shear force at level x.

I R" is. the su....JctIJral facloi given in Table 24A

(3) Design Shear Force at Story x:

(2531) where the modal shear force V'xl" at level x is given by eq. (25.12).

770

Ea;lhquake Engineering

Un;form Building

or using eg. (25.11)

25.1.4.

Code~lee4;

Dynamic Melhod

771

Modal Lateral Forces

N

Vm

;

)'

Tm¢"mW,s.m

(25.14)

i= I

25.1.3.

By corr.bining eg. (25.11) wirh eqs. (25.15) and (25.16), we may express the modal lateral force F.
Effective Modal Weight

(25.19)

The effectlve modal weight W", is defined by the equation

where the modal seismic coefficient C,m at level x )5 given by

(25,15)

(25.20)

Then, from eq. (25.,4). the modal welght is ,v

w.=r. I

j ..

,m W,

(25.i6)

25.1.5.

Modal Displacements

!

Combining eqs. (25.5) and (25.16) results in the foHowing important ion for the effective modal weight:

express~

The modi'll displacement 8(", a;: ::he levei x of the building may be expressed, in view of eqs. (25.3) and (25.9), as (25.2lj

(25,[7)

It can be proven (Clough and Pecz;en 1975, pp. (559-560) analytically tbar the sum of the effective modal weights for aI! the modes of the building is equal to the totai design weight of the building, that is,

(25.18)

Equation (25.1 &) is most convenient i:1 assessing the number of significam modes of vibration to consider in [he design. SpecificaUy, the UBC-94 requires that, in applying the dynamic merhod of analysis, all [he significant modes of vibration be included. This requirement can be satisfied by including a suffi~ dent number of modes such that their total effective modal weight is at least 90% of the total design weight of the building. Thus, this ;equirement can be satisfied by simply adding a sufficient number of effective modal weights [eq. (25.17)! umil their total weight is 90% or more of the seismic design weight of the building.

where F", is the participation factor for the fnth mode, ¢~m is the component of [he modal shape at level x of the bui;ding, and 5j!m ls the spectral displacement for that made. Alternatively. the modal displacement 0;,.. may be calculated from Newton"s Law of Motion in the form

(25.22) because the magnitude of tr.e modal acceleration correspo:1ding to the modal displacement 0,'111 is w~5s",. Hence, from eg. (2522) (25.23) or sUbstituring

W,,,"!;;;;:

2'ifIT",

0"., where T", is the mth namral periDd.

(2524)

774

Uniform Building Code-1994; Dynamic Method

Earthquake Engineering

Zone

(4) Design Lateral Det1ection
2A 28 3

(2532)

where the modal displacement 0(111 at level x is given by eq. (25,23) or eq. (25.24). (5) De$ign Drift

fOf

Story x:

4

where the modal drift £1"", at srory x is given by eq. (25.25)" (6) Design Overturning Moment at Level x:

~evel

~he

design eanhquake.

r

(2) The OCCL!pnncy imporlCIlIce Factor This factor is eqmd to 1.25 for essential or hazardous facili[les and LO for special or standard occupancy structures as described in Table 24,2,

(3) The Sire Coefficieru S. The site coefficiem is def!r:ed by four soH profiles Sf, S2, S), and S:, described in Table 14.3 with numerical values 1,0, 1.2, 1.5, and .2.0, respectively. (25.34)

where the modal overturning moment lY1iC
Coefficient 0.075 0.15 0.20 0.30 0.40

The seismic coefficient indIcates the effccnve peak ground accele:-urion of

(25.33)

775

x IS given by eq.

(4) The Structural FaclOr R,,~ Numerical values for the structural facror R", are given in TubJe 24,4 for various structural systems. As described in Chapter 24, the Rw factor is related to the duc(ility of the structural system.

(2526) (7) Design Torsional Momem at Level

1::

(25.35) where the modal rorsio:lal moment (25.28).

25.3,

M~,m

ar level x is given by eq.

(5) The Seismic Weigh! W. The seismic weigh' W includes the dead weigh, of ~he building, permanent equipment, and par.itions. In specific cases, it also may include some of [he snow load and some of the live load as indicated in Chapter 24. The U!1iform Building Code of i 994 contains ?rovisions for the imptementation of [he dynamic method of ana1ysis, The code stipurates [hat when this method is used it shaH be implememed using accepted principles of dynamics and it shaU conform (0 the criteria established in the code. These criteria include the following provisions:

PROVISIONS OF UBC-94: DYNAMIC METHOD

In Chapter 24, we defined the various facwrs needed to calculate the base shear force given by eqs, (24.1) and (24.2) of the UBC·94 using the eq!1ivalenr laleral force method. The fOllowing is a recapitulation of iliese faclors which are also pertinent to the Dynamic Method: (1) Setsmic ZOILe Factor Z. This coefficient has been defined by classifYIng the United States in five seismic zones on the basis of past historical earthquake activity. These zones are designated as I, 'lA, 2B, 3, and 4 as shown in Fjg. 24.1. The nomerical values assigned to the coeftldent are:

0) The ground motion represenla:ion may be one of the foUo¥."ing:

0) The response spectra 1 given in the code and reproduced in Fig, 25.1.

2 Valuei obrained [rom the UBC·94 ResponJI: Spectra should be multiplied by tr.e modi fica· (1on [aero. 7.1R", where Z I:J the seismic ::;Onl! roelfide/lf aod R", ~hG structural faclor, Tr..e UBC-94

indic(l,cs explicitly :~e moditknicn faCtor Z bu! not R... although such factor was intended ir. !.he code, Also, (he response spccrrum :::l..rve$ provided by (fBC~94 have a j% damping as stated in Ihe Commenlary of RdcOt1lmelVied LOfl!r(l! For,'e Requirement a'ld Tenf(lJive Commenlary (Source; SEAOC-1990).

776

Earthquake Engineering

(ii) An eiastic response spectrum develo:;>ed specifically for the slte. The preparation of such spectra is mandatory for soil type S•. (iii) Ground motion time histories developed for the specific site. (iv) The vertical component of the ground motion may be defined by scaling the corresponding horizontal accelerat!on by a factor of twothirds.

Uniform Building COde-i994: Dynamic Method

777

Solution: Problem Data (from Example 24.1) Seismic weights: lV, ~ 789.1 Kip. lV, = 645.1 Kip

W, = W,

(2) The adopted mathematical model shou Id represent the spatial distribu-

tion of the mass and stiffness of the structure to on extant that is adequate for the calculation of the significant features of its dynamic response, (3) The code indicates that the dynamic analysis may be performed using:

(a)

(b)

25.4.

An elastic response spectrum in which all the significant maximum modal contributions are combined in a statistical manner to obtain approximate total structural response. As indicated in Section 25. L3, this requirement is :-eadily satisfied by including a s:jffi~ dent number of modes with a totai effective modal weight equal to 90% or more of the seismic weight A time history response in which the dynamic response of the structure to a specific ground motion is obtained at each :ncrement of time.

SCALING OF RESULTS

When the base shear force calculated by the dynamic mechod js Jess than that determined by the equivalent lateral force method, for irregular buildings, the base shear calculated by dynamic analysis shall be scaled up to match 100% of the base shear detennined by the static lateral force procedure. For regular buildings. the base shear force shall be scaled up to match 90% of the base shear determined by static lateral force procedure, eXCept that it shall not be less than 80% of l'le value obtained when eq. (24.5) is used to detemline the fundamental period of the structure, All corresponding response parameters, including member forces and moments, shall be adjusted proportionally" The code also stipulates that the base shear for a given direction, detennined using dynamic analysis, need not exceed the vafue obtained by the static !ateral force method. In this case, all corresponding response parameters are adjusted proportionately.

Total seismic weight Seismic zone facto!": Occupancy importance factor: Sire coefficient:

W= 3012.4 Kip Z = 0.3 (for Zone 3) 1= 1.0 (warehouse)

S = i.O (rock)

(1) Modeling the Structure,' This structure has been modeled as a plane

orthogonai frame (using Program 29) in the soiu~!On of Exam{)!e 24.2. The reduced stiffness matrix in reference to the four lateral displacement coordinates of the building, from Example 24.2, is given bv

[1.1 =

5611

2916

458

-45

I

2916

3863 - 1885

269

l

-45

458

]885

Consider again the four~story reir.forced concrete building of Example 24,1. Model this building as a plane orthogonal frame and perfomI the seismic anaiysis by the UBC-94: dynamic method. Use the response spectra provided by the code Fig.(25.1).

I

J

2893 - ]398 (Kipiin)

1168

269 - 1398

and the mass matrix (W;lg) by "2.036

0

0

0

i

[M]

l

,

0

2.036

0

0

0

0

2.036

0

0

0

0

1671J

I(Kip· sec' lin)

(2) Natural Periods and Modal Shapes: The natural frequencies and the normalized modal shapes are obtaIned by solving the eigenproblern [[K] -

w' fM:J] {,p} = {O}

(e)

The fOots of the corresponding cha:a.cteristic equation IrK] - w' [M]I

Example 25.1

,

{O}

(I)

are

wf = 77,22, wi = wi 678.60. w; =

1939.75

4052)8

(g)

778

EanhQu.l. Eng;neer;ng

Unilorrn Building Code-1994: Dynamic Method

resulting in the :!"lacural frequencies (j= wl27T)

I,

= lAO cps,

I, = 7.01

j, = 4.14 cps, I,

TABLE 25.1

= 10.13 cps

Modal Elfecllve Weight and Modal Base Shear

Modal Effective \Veight (%) Wm(Kip)

Mode m

cps

2465

0.715 sec,

T, = 0.241 sec,

T3

Modal Base Shear Vm (Kip)

(h)

or natural periods (T= IlfJ

1',

779

0.143 sec

T, = 0.099 sec

(i)

2 3

366

99

82 t2 3

4

82

3

Total Weight

84 23 6 4

3012 Kips

and the corresponding modal shapes arranged in the columns of tbe

modaJ matrix are

Numerical values of Vt" a!so are given in Table 25.1. The rotal base shear force given by e.q. (25.29) then is calculated from values In the last column of Table 25.1 as

O.IIZ77 - 0.31074 050518] 0.35308 [
r

lO.51540

0.45452

0)

- 034652 - 0.07766

(3) Spectral Accelerations: The spectral accelerations (scaled down by the effecti ve peak groU:1C acceleration of 03 g and divided by [he structural factor R,., = 12) for (he natural periods in eq. (i) obtained from fhe spectral chart for rock Or stiff soils in Fig. 25.1 are

S",

=

0.034 g,

S,) = 0.061 8

5",

=

0.063 g,

S", = 0.0508

(k)

(4) Effective Modal Weigh!s: The effective modal weighr is given by eq, (25.17) as

W. N

I

¢~,Wi

,,,, I

Values obtained for W", (m 1, 2, 3, 4) are shown i:l Table 25,1. This table also shows the effective modal weight as a percentage of the total seismiC weight of the building.

(5) Modal Base Shear: The modal base shear is given by eq. (25.15) as V'l=W",S"",

v ; (84)' + (23)' + (6)' + (4)'

87,4 (Kip)

(6) Scaling Modal Effec!iYe Weighl and Modal Base Shear: As required by (he UBC-1994, the modal values for the effective weight and for the base shear force in Table 25.1 are scaled up by me ratio r equal EO 90% of the base shear determined by the equivalent static lateral force method and Ihe value for the base shear force calcu lated by the dynamic method. The base shear determined in Example 24.2 using me equiv~ alent static tateraI force method was equal to 140.75 Kip, while the value calculated in :his example using the dynamlc method is equ8t to 87A Kip. Therefore, (he scaling ratio is r=

0.90 X 140.75

-1,45

Table 252 shows the resu\[ of scaling (by the faeror r = 1.45) the values in Table 25.1 for the modal effective weight Wm and the modal base shear V"i. (7) Modal Seismic Force: Tr.c numerical values for the seismic coefficients Cx;" and for the seismic lateral forces F"" calculated from eqs. (25.20) and (25.19) are shown respectively in Tables 25.3 and 25.4. The des:gn seismic forces calcuia(ed by eq. (25.30) are shown in the last column of Table 25.4. (8) }lfoda! Shear Force: Values for ihe modal shear force V.om at level x calculared using eq. (25.12) and results from Table 25.4 are given in Table 25.5. Design values for the story shear forces v." calculated by eq.

Uniform Building Code·1994: Dynamic Method

Earthquake Engineering

780

TABLE 25.2

Scaled Values for Modal Effective Weight and Modal Base Shear

Mode m

1<1odal Effective Weight (0/0) Wm(Kip)

82

3S84 531 144 119

2 3 4

.r = 4378 TABLE 25.3

Levei

i22 33 8 5

12

3 3

:Vlode 2

Mode 3

Mode 4

..

~

0.340 0.348 0219 0,091

2

Design

0,780

Ll41 - 2.027 0.463 1.422

0,145 0.982 0.652

-0,28: 0,942 - 1.897 2,235

Modal Seismic Force Fxm(Kip)

-26

9

84

-21

2

III

II 33

-7 -3

3 6

g

5

Mode

Mode 2

Mode 3

Mode 4

4 3

41

26

42

5

9 - 16

2

27

32

4

5 9

II

22

II

11

-1

Fx Values 50 46 43 29

Mode I

:Viode 2

4

0.321 0,268

- ,023 0,003 0,023 0.015

0,169 0,070

(25.31) are given in the la't column of Table 25.5, It should be noticed that the values shown in Table 25.5 for the modal shear force at level x = 1 are precisely the values for the base shear force calculated by eq, (25.15) and shown in Tab[e 25.2.

Modal Story Drift

Story

Mode I

4 J

0.053 0,099 0,099 0.070

2

49 86 III 126

Modal Lateral Displacement oxm(ln)

Level

3 2

V, Values

Mode 4

41

Design Level

Mode 3

4 3

TABLE 25.7 TABLE 25.4

Mode 2

Mode

TABLE 25,6

Modal Seismic Coefficient C xm

4 3

Level x

Modal Seismic Force Vxm{Kip)

122

Kips

Mode I

--.~

Modal Base Shear Vm(Kip)

TABLE 25.5

781

Mode 3 0,002 - 0,005 0,000 0.002

Design Ox Values

Mode 4 0,001 0,000 0,002 0.001

- -.......

.....

0.320 0.266 0,167 0.063 ~--

~xm(in)

Mode 3

Mode 2 - 0,026 - 0,020 0.008

oms

0.007 - 0,005 - 0,002 0,003

Mode 4 - 0001 0,002 -0,003 0,001

Design LI. Values 0.059 0.101

0.099 0.071

eq. (2533) are given in the last column of this table. The maximum story drift permitted by the UBC·1994 should nOl exceed (0,04/ Rw) fl. = [(ON/12) 144 = 0.48 in] or 0,005 HJO.005 X 144 = 0.72 in). where Hx is the story height and R." the structural factor from Table 24,3 T.'1C desigr. values Ll.r calculated in Table 25.7 for this example are weB

below these

limi~

ror stor; drift.

(9) Modal Lateral Displacement: Values for the modal lateral dispi2.ce~ ment D.vl< at level x calculated from eq. (25.24) are shown in Table 25.6. This table also shows in the last column design values for lateral displacements 0. calculated by eq. (25,32),

(11) JWodai Overturning Moments: Table 25.8 shows rhe vaLles for modal overturing moment calculated using eq. (25.26). The last column of this table g;ves the design values for overturning moments M.. calculated by eg. (25,34),

(10) Modal Story Drift: Table 25,7 shows the values for modal story drift LI.•m calculated by eq. (25.25). Design values for story drift obtained from

(12) Modal Torsional Momerus: Table 25.9 shows the values calculated from eq, (25.28) for :.:he I7lodal torsional mOffients assumir.g only

782

Earthqua;';e Engineering

TABLE 25.8

Uniform BUilding Ccde-1994: Dynamic Method

Modal Overturning Moments Mxm(Kip.Jt)

Level

x -~

..

-~

Mode i

..

Mode 2

Mode 3

Mode 4

497

- 310

- 17

1504 2832 4293

- 425 - 30

109 24 17 79

4 3

2 Base

TABLE 25.9

-----------------562

Mode I - - - ..- -.. 4 199 3 402 2 53! 584

22

~605

2864 4293

Level x

..

-

Mode 4

.-_.

..

- 124

43

- lOl

35 -17

54 157

.. _ . _ . - .

__

Mode 3

Mode 2

---~-

TABLE 25.10

596

Modal Torsional Moments M"m(Kip· jt)

x

_._-

Design lvf, Values

- 52 8

Levei

Design .

-7 15 - 30

38 __._---_

23

.. _ .

At,. Values

238

416 534

__606 ._._-

Calcutation of Raflo of Secondary to Primary Moments

Srory Weig!il W, (Kip)

Above Vleight P, (Kip)

Story LlJin)

Drift

Story Shear V, (Kip)

Story Height Hx(in}

below (he code limiI of 0.1. Consequently, [here is for 'he P·ll effecL

M,llv1p = p ,.,:J. ¥ I V.1.H.r

25.5.

Program 24 can be used to caiculare. using modal superposition and response spectra, [he response of bt:ildings subjected to seismic eK:;I{arion a[ the base. The use of this program requires [he previous determination of [he s[iffness and mass matrices of the s[ruCtcre as well as the solution of the correspondlng eigenprobiem to cetermine the naLUral frequencies (or nmural periods) and the modal shapes. The stiffness and the mass matrices may be obtained by the ex.ecution of one of the following programs for modeling buildings: Program 27 for modeling structures as shear bGi!dings; (2) Program 28 fOf modeling s(ruCtures as cantilever buildings; or (3) Program 29 for mOdeling structures as plane onhogonal frames. The solulion of the corresponding eigenproblem may then be obtained by the execution of Program 30 or Program 31 to determine natural frequer.des and normal modes, using the Jacobi Method or using [he subspace itera[!on method. Program 24 uses the fol:owing analytical expressions) of [he response spectra provided by the UBC·94 in Fig. 25.1: Soil Type S,:

S"

1 + lOT

fOf

0< TS 0.15 sec

< TS0.39 sec

645.1

645

0.059

49

S" = '2.5

fOf

0.15

789.1

1434

O.lOl

86

144 144

0.005

3

0.012

2

789.1 789.1

2223

0.099

III

144

S" = 097511'

for

T>0.39 sec

3012

0071

126

l44

0.015 0.013

S" = I + JOT

for

0
= 2.5 S,. = 1.46311'

for

0.15

for

1'>0.585 sec

= 1+ 7ST

for

0
for

0.2 < TS 0.915 sec

for

T>0.915 sec

Soil Type 52:

accidental e:::cemridty e" of 5% at each level x of the building (for this ex.ample e.t =. 0.05 X 96 ft = 4.8 f[), Design values for torsional moments M,z calculated from eq. {25.35) are shown in the iast column of Table 25.9. (13) The P-tJ Effect: The t:BC-94 specitles that the P·ll effect does not need to be considered when [he ratio of the secondary moment MJ '::;;: p .. .:1 .. to !.he overturing or primary moment Mp is less than 0,10 calculated for each level x of the building. The results of the necessary calculations to evaluate this ratio [eq. (24.13)1 are shown in Table 25.10. The largest moment ratio in Table 25.10 is 0.013, which is well

need to account

PROGRAM 24-UBC-1994: DYNAMIC LATERAL FORCE METHOD

4

I

[;0

783

S~

< Ts 0.585 sec

Soil Type S,: Sp

S., = 2.5 Sp

= 2.2881T

--'--'-} From ReCOmmended La/eral Force Rt'quitl!me!ltS and TerUaliw! Commentary. SEAOC~9D.

784

Uniform Building Cade-1994: Dynamic Melhod

Earthquake Engineering

where Sn is the spectral acceleration normalized Co a peak ground acceleration of one g and T 15 the fundamental period of the building. The program calculates the effective modal weightS Wm from eq. (25.17) and the modaJ base shear force Vm from eq. (25.15). Then :t requeSts i"fonnation for scaling these effective values when the base shear calculated by the egu~vaient lateral force method differs from the vaiue obwined fro:Yl [he

(e) Analisys by UBC-94: Dynamic Method:

dY:1amic method. The output from Program 24 includes the modal values for the lowest four modes and the design values for each level of the building as follows:

,r:c?",

,i"ICi-;,

'<:II'

5.;.'1 12

(I) Seismic laterai forces F, from eqs. (25.19), (25,20), and (25.30).

is])

(2) Shear forces V, from eqs, (25,12) and (25,31).

,,2

11. :0

,fig 12

;2,tv

i2

(3) Lateral displacements Ii., from eqs, (25.24) and (25,32). (4) Story drifts 4" from eqs, (25,25) and (25.33).

(5)

Overtum~ng

moments M" from eqs. (25.26) and (2534).

z ,

,"

(6) Torsional moments !VI"" from egs. (25.28) and (25.35).

;<1}

(7) Diaphragm forces Fp , from eq. (24.14),

L"YO

(8) P-4 effect, that is, the secondary moment M", the primary moment Mpp and their ratio e., ~ M"IMp, f,om eq. (24.13), 2. :;:;;,,;> CCH:::S:O:-::'SSS O? 5:::;:"7 CL;W SOiLS

Example 25.2 Solve Example 25, I using: (a) Program 29 to model the structure as a plane orthogonal frame; (b) Program 30 to calculate the natural periods and nonnal modes; (c) Program 24 for earthquake resistant design by means of the Dynamic Method of the UBC-94.

Solution: Ca) Modeling the Building as a Plane Frame: The input data and output results for Program 29 to model the structure as a plane orthogonal frame are cO:1tained in the solution of Example 24.2(b) NatUral Periods and Modal Shapes:

~.

seT: w,,?r::m

j

.\153

"'~:'E;

57 i!? .41

32.:'::

~:)_

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()'50518

::L1,j~5

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~v_:l69C'7

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1,13J'

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:} .()5C

34.:»

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C\..AYS 8" S';"DS

0.21286

:;,15<;),

-C.Yi755

,

, II %

785

Unllorm Bu;lding Ccde-1994: Dynamic MethOd

786 K:)DA)",

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SUMMARY

The provisions of [he Uniform Building Code 0f 1994 for earthquake-resistant desigo by the dynamic method requi re the modeling of the bui~ding as a discrete sysi.em with one coordinate (horizontal displacement) m each level of the building, Possible ;r.odels jnclude: {1) the shear building in which the horizonml diaphragms are assumed absolutely rigid, (2) the cantilever building

788

Earthquake Engineering

in which the horizontal djaphragms are considered absolutely flexible, and (3) the plune orthogonal frame in which the stlffness of horizontal diaphragms is considered as part of the effect of horizontal members of the frame. The implementation of the dynamic method also requires the solution of the cor~ responding eigenproblem for the modeied structure (0 determine its natural periods and modal shapes. The maximum response of the modal equations is then obtained from a spectral chart such as the one provided by the code. Modal seismic forces and moda: response in terms of shear forces, overturning moments, torsional moments, lateral displacemenrs, :l.nd s[Ory drifts are determined at each level of the building, The final design values are calcuJated using the SRSS technique or the CQC technique to combine the modal contribmlons. The modaJ method 1n which the modal responses are combined is valid while the structure remains in linear elastic behavior as expected when sub~ jected to an earthquake of moderate intensity. However. for a strong earthquake, the provisions of the UBC-94 are not intended to maintain the structure VIbrating within its linear elastic behayjor. Wben subjected to a strong earth~ quake, d1e structure will deform in [he inelastic range producing plastic deformations and structura; damage. However, its ductility will provide a rr;echan· ism to absorb energy and structural stability will be maintained. although the structure may continue to undergo large deflections which may create further damage.

PROBLEMS Use Program 24 for the seismic resiStant de~lgn of the lO~story building of Problem 24.1 by the dynamic method, 25.2 Solve Problem 24.2 using the dY:lamic method of the UBC-94. The total length of the building is 120 ft and its width 24 ft. Consider only :he accidental torsional eccentricity. 25.3 Solve Problem 24.3 using .he dynamic method of the UBC-94. The length of the building is 96 ft and its width is 36 ft. Estii.lated eccentricity at each level ex""Gft. 25.4 Solve Problem 25.3 modeling the Structure as a plane orthogonal frame, 25.1

APPENDICES

1

1

I j I

Jf\j lilt!

1111'

Appendix I Answers to ProbJems in Part

J

~

1.1

fw

L

T~2rrL i-'~--,-

3£1 + 2J:L)

Vg

1.2 y(r

=0:

y(t

~

i) =

0.89 in

I) = 22.66 ;nfsec

1.3 f"'" 2.24 cps

lA

(a)f= 2.87 (b~ f"'"

cps

2,74 cps

. 4. F3EIg 1.5 j=-;~LJW

1.6 y(f=2)= - 0.474 in y((=2)= 21.05 in/sec

,,

! M'

.,

.-i'

y (t =

2)

4065 iflIsec 1 rg

1

1.7

B',"" 80 cos I-i!

1,g

k~

\

80

+~-

w

sir.

I

73,Mlblin 791

792

Answers ':0 Problems in Part I

Appendices

2.5 (a) for

C;1~3.39X IO'(lb'in ' ) L9 Y(I)

0.0995 sin 48.031(10)

~= 1,

y

(b) for

t>

where

roo = w~·

l, y

y(r} = 4.78 cos 48.D3t(intsec)

1.10 y(;) = 0.5 cos 27.78t 2.6 l.ll

1.13

t~

I

27.

0.2765 sec

Ii

2.880; 17.816:

Y::

2.9 2.10

I

(b) f = ? _ )'

-"

(e)f=

C-=~~'''''

1

to

0.076

(b) fo

8.69 cps

(c) Ii

048

(d)

].61 y~

2.11

j

f= ....-

(a)f 0.018 (b) VJp= 57.76 red/sec (c)
0.H3

(d)

1.12

2"

w=

1.18 w

05 . \;

(alf

271

(d)f-

1.17

0.4167

(b) TD

(d)

1.15 (a)f-

1.16

(a) §

(e)

f=

-I

2.12

y,

(a) KF,

I-

I m

3Elk

3E;+

C,- 0.2\'

35.018 (rad/sec)

1.19

(b)

K,-

48Elk

unstable for k::; Wh 2.1

Y(I=l) Y(I = D

2.2

C,

0,037 in

- 0,570 in/sec (e) K,

y(I=2)~

4.65 X IO-'io

'1(t -.-., 2) =

4.083 X 10- 5 in/sec

y(r=2)

4.18xlO- 5 inf.s.ec

2.3

c = 0.73 lb sec/in

2.4

f=

15%

0.2 3ElL

! 3E1Lm

CF, = 0.2 ¥ a~b2

1

(d)

3ElkL

Ke==':===

793

794

Ar.5wers to Problems in Part 1

Appendices

3.15

(a) j ~ 13.58 cP'

(b) 2.13

.f =

2.14

m:rn:!u

2.15

i.:

(c) Fr,=4825 1'0

6.7% (ml

+ 2£wu 7

+ m:!)cu + (m,

;. m:<)k.u

(d)

=0

U}U "" 0

Y = 0.0037 In

3.2

Ar""5l.21b

3.3

y= 0.823 in

3.4

Y=O.746 in

),5

(a) AT (b) T,

3.6

minI.

4.1

3.8

Tx :;.;034

(a)

yO = 0.5)

- 0.407 in

4.2

(a)y((=oO.5) (b)

4.3

Ym;tX

-O.!02in

= 1.17 in sin r.vt - - for

!

DLF=/J

3.9

+ !n;

(b) y ....... = 137 in

k = 93 JbJin u=O,O!3in

Tn,F)

!HI +m~

mj

3.16

3.7

,

ell -:- Xli = --._.- sin ltH

(<.I) /\.Iil. -1-

where :H = - - - ' -

15,803 Ib

~

484C Jb

w/t- f l,

3.17 3.1

F,~

3.16

where: u)

~

"~0.0735

In far J, ~

(,) Y,

~

0.064

Y,

~

0.0446 in far j,

800 RPM lOOO RPM

Y; = 0,0302 i:: for h = 1200 RPM (b) Y(r= 1) ~ 0.076 in

I

:0LF

~

4.4

l'!il d

W(d

l+-(sin Wl-sin w(!+([)}

fOr!.~f4

WI,

Vmn = l8,093 Io for left calun:n Vflnx = 1908 !b for right column

3,10

§"'" 3.3%

3.\1

f=

3,12

w = 49.2 (radJsec)

4.5 6.3%

Y·· 0.387 (cm) U~,

V"' ....

,~

15,640 lb for left column

V m...< = 1649 10 for right column

4.8

u

/I{l} = - - cos WfT ur

w

= 110.7 (MPa)

,r - ' - -

3.lJ

j.j;i J +"'2. m

3,14

wp=w.,fl-ie

," (;;! y",r "m 4.9 y(r)=-11--"fd l

W

1 for j

y" [f sin WI, '""- If'f-~--i EJ l\ w! 4.10 Ym'~:= 1.348 (tn) 0"%", = I l,906(ps\)

CC~ u;(t

fJ)+

~ sin W(f-iil)l w

,

for

i>t
795

796

Appendices

Answers to Problems in Part I

4.11

y_, ~ O.72(in) (Tm.l~ = 16,282 (psi)

4.12

y (I ~ 0.5) ~ - 1.903 in

G, ~ -

4.13

y (I ~ 0.5) ~ - 0.060 in

(b) Y (I ~ 0.35) ~ 0.2570 in

4.14

U(I~ I)~ -2.809 in

4.15

l/

4.16

= 6.03 in limn = 4.59 in

=

(t

5,8

4.17

Yma.,

= 1.51

in

4.18

)'''''tX

= 1.42

in

4.19

)'m~~

= 0.58

in

Go ~

G, ~

I) = - 2.397 in

lI"",x

(a)

(undamped sys[em)

0.0350 b, ~ - 0.0361

0.0069

b, ~ - 0.0402

0.0724

5.9

y (t

= 0.35) = 0.2327 in

5,10

Y(!

~

0.5)

~

0.027 in

5.11

y (1 ~ 0.5)

~

(0.02842 - 0.000 II i) in

5.12

(a) ao = -

(with 20% damping)

Po

'" ClIO

= 0,

an

= -

= 1,3.5, ... ,

n

Pn

2

n = 2, 4, 6.

I _ n"'

4.20

Ymax

4.21

Ym .., = 0.76 in

5.13

y (f

~

0.5)

~

0.0731 in

4.22 Ym" ~ 0.66 in

5,14

y(t

~

0.5)

~

0.0543 in

= 0.34 in

5.15

4,23

)'mox

4.24

= 1.00 in

(Tlm.ox

= 3842 psi

(TZma:<

= 6831 psi

4.25

Ym~,=0.7l in

4.26

(a) (T~ ± 6193 psi

F(r)

5.2

sin 6m+ -I sin 10m 1 3 5 "'J

6

F(t)= 10- [357 sin 2711-26 cos 2m cos 6711

- 35 sin 10m - 7 cos 10m

5.4

5.16

Y(I~

5.17

y (I ~ I sec)

5.18

y (t

~

5,19

Y(1

S.20

y (t

Go

720

an=O,

n=1,3,5,.

bo ~ 0,

n ~ I, 2, 3....

= 0.05) =

n> 1

0.75 sin 4.19t)

O$t$l.O sec

~

0.0532 (in)

I sec)

~

0.0035 (in)

~

I sec)

~

0.0034 (In)

=

1 sec)

= 0.0028 (in)

K*~

- 0.2065 in

5.5

1I

(t

5.6

U

(I ~ 0.05) ~ - 0.2064 in

5.7

u(t=0.05)= -0.1295 in

sec)~0.2419(in)

C* = 2250 lb· seclin F* (1)

n = 2,4,6,.

I

for

M* = 4.48 lb· sec 2 lin

6.1

+ ... J

u(I~0.5)~0.3518in

(a)

7Jt -

y(t)=YI(t)+yz(t)

+ 36 sin 6m - 532 5.3

for

yU)=YI(t)

. 2m + -I = -120 [sm 7r

= 0.229 (sin

b" = 0,

Y2 (1) ~ 0.229 [sin "'(1- I) - 0.75 sin 4.19 (1 - I)]

(b) Fm~ ~ 11,3761b 5.1

Yl (t)

'"

bi

6.2

~

45.000 Ibhn 625/(1) Ib

5 M*=-m 6

C*

=c

K'~

k

F* (I) ~ M(t) L

6.3

2 3

•

M*~-,"C

t?:l.Osec

797

798

Appendices

Answers to Problems in Part I

~cL

C'

7.7

I'- ~ 1.7

7.8

Y(I ~ 0.5)

7.9

y (t

K*= kL

6.4

m

M'~~(5,,-8)

EI,,'

K'~~-

32L' ~

P (I)

y(l = 0.5)

7.11

y (I

7.12

ao

Kt

6.6

M*~

y", ..,

= 0.418 in

yd

8.4

O""m:u;

= 12.788

g

8.5

Y"'o.<

ksi

5 0 = 1.9 in

Ec7Jd~

5\. = 22.4 in /sec

128[-'

5"

I

8.6

6.8

riEl w=7.8 ) !~-} rad/sec

6.9

f= 3.51 cps

Y

m,L

_ I gEl

We

w=2.}) ~i

~

0.68 8

5 0 = 1.28 in 5,. = 15.36 in/sec

48EI rad Isec 17'

I.

in

= 2.78

8.3

0.1237-

in

0.1340 in

8L

YC(m +}5m,)

6.10

~

8.2

P(r) ~ -0.1807P,(I)Ld w~

0.5)

N-rr

8.1

K'~~~.

6.7

~

= 0.1423

= 0.374 in if",,,-, = 7 .246 ksi

0.2929FcJ(l)

= ---

6.5

0.4477 in

= 0.5) = 0.2643

7.10

27T

~

5" ~ 0.48 g 8.7

50

= 1l.0 in

(F,)m~ ~

88.0 Kip

8.8

(Fs)= ~ 36.0 Kip

8.9

(F T)m~ ~ 44.0 Kip

8.10

5 o =8.0in

8.11

I'-~

8.12

(a) 5 0 = 40 in

rad/sec

1.8

,~-

6.11

w = 0.62

f gAG

\

i-WL

6.12

f= 0.496 cps

6.13

f~

0.492 cps

7.1

Umax

=-

7.2

Ym:u

= 2.56 in

7.3

Ym:lX

= 6.47 in

7.4

Um,",

= 1.03

in

7.5

U mou

= 5.19

in

7.6

Ymax

= 2.38

in

rad Isec

5 = 50.3 in Isec y

5" = 1.63 g (b) SD~ 4.8 in

So

10.27 in

= 60.0 in Isec

S,,~1.96g

8.13

5 0 = 0.46 in Sy = 5.8 in Isec S,,~0.19g

8.14

SD~

0.78 in

(atf~

So = 2.46 in Isec 5" = 7.72 in Isec

2

0.5 cps)

799

800 8.15

Appendices So'"" 8.03 in S~

= 50.45

(Ill f= 1.00 cps) in/sec

S" = 317.00 in/sec1

8.16

S,~ 3.26 ;n (n< f~ 1,00 cps)

Sf' "" 20.38 in/sec

S"

127.40 in/sec 2

Appendix II Educational Computer Programs

The computer programs used throughout this book are written in the BASIC language. Ttlese programs are organized in two sets: structural dynamics programs and earthquake engineering programs. Table AILl presents a brief description of the educat'lona! computer prograras in these two sets. The programs are written for IBM or compatible microcomputers. They are available directly from the author in a compiled or noncomplled version of BASIC (BASICA or QUICK BASIC). A convenient form for ordering diske[[e, with the programs of interest is provlded in the back of this book.

SOl

802

CompL:ter Programs Edl:caliol".al

Appendices

TABLE AIL 1 (Continued)

TABLE All. 1 Educational Computer Programs

Program

File

Purpose

Chap
25

Structural Dynamks 2

3 4

Menus with

o~t!ons

and commands

MASTER PP2

* 4

Response simple oscill,l[o( by direct

PP3

4 5

PP4

imegradon Response simple oscillator to impulses Response simpie oscil~ator in frequency

27

30

EQ7 EQ8 EQ9 EQIO

31

EQII

PP5

7

6 7

PP6

8

Seismic response spectra

PP7

9

8

PP8 PP9 PPIO PPl! PPI2 PPI3 PPI4 PPI5 PP16

10

15 16 17

PP17

18

PPl8 PPI9 PP20

19

13

14 15 16 17 18

19

20

-,

0'

XI Xl

PP21 XI X2

11

Modeling structures as shear bUliriings Natural frequencies and normal modes Response by modal superposltion Response (0 harmonic excitation

12

Absolute damplng from modal damping

13

Reduction of the dynamic prob!em Modeling s[ructures us beams Modeling structures as pJane frames Modeling structures as grid frames

11

14

18

* 23

* *

Modeling structures as space frames Modeling strucrures as plane trusses Modeling srructures us space trusses Response by step integrBtion Maximum member's end forces and moments Response to random vibrarions File for struCtures File for eigensoiution

Earthquake Engineering

•

EQ5 EQ6

5

12

File

26

domain Response oscillator for elDstoplasEic behavior

9 iO II

Program

Chapter

Purpose

*

ED.rlhquake·resistan< design: NEHRP-94**

----~~-~-----

--~~

803

22

MAIN EQ2

23

23

EQ3

24

Menus with options and commands Earthquake~resistant design: UBC~94 (Statics) Earthquake-resistant desig:1; UBC-94

24

EQ4

25

(Dynamics) Earthquake-resistant design: UBC·94 {Simplified three-dimensional)

28 29

* 9

* 15 JO .'f.

(srarlcs} Earthquake-resistant design: NEHRP-94 (dynamics) :'vlodeling structures as shear buildings Modeling structures as cantilever buildings Modeling structures as plane frames Natural frequencies and modal shapes \Jncobi Method) Nawral frequencies and modal shapes lterado:l Method)

"" Program IlO( described in the book. "'''' Na!ionu! Earthq:..w.ke Hazr:rd Reductiun Program.

Organizations ard Their Acronyms

Appendix III Organizations and Their Acronyms

,, ,

ff h n

Earthquake Engineering Reseaxh Center University of California [30 I South 46th Street Richmond. CA 94304 Phone (415)525·3668

EERI

Earthquake Engineering Research Institute 2620 Telegraph Avenue Berkeley, CA 94704 Phone (415)525·3668

FEMA

Federal Emergency Manageme::1t Agency Administrators of the National Earthquake Hazard Reduction Program (NEHRP) National Emergency Training Center Emmitsburg. MD 21727 Phone (301)447-6771

ICBO

International Conference of BuHding Officials Publishers of Uoiform Building Code (UBC) 5360 South Workman Mill Road Whittier, CA 90601 Phone (310)699·0541

MITA

Micro-Text Association

American Institute of Steel Construction 400 North Michigan A venue Chicago, IL 60GlI Phone (312)670·5432

ANSI

American National Standards Ins(itute 1430 Broadway New York, NY 10018 Phone (212)354-3300

NBS

National Bureau of Standards Department of Commerce Washington, DC 20234 Phone (301)975-2000

BOCA

Building Officials and Code Administrators 4051 West Rossmonr Road Country Club Hills, IL 60478 Phone (708)799·2300

PCA

PortIand Cement Association 5420 Old Orchard Road Skokie, IL 60077 Phone (708)966-6200

H :{

"

II,

EERC

AISC

;!

1) j,

Building Seismic Safety Council 1201 L Street, N.W. Suite 400 Washington. DC 20005 Phone (202)289·8000

American Concrete Institute P.O. Box 19150 Redford Station Detroit, MI 48219 Phone (313)532·2600

i!

i1i!

BSSC

ACl

il l!

n

11

" 804

805

P.O. Box 35101 LouisvHle, KY 40292 Phone (502)895-1369

806

Appendices

SBCCI

SEAOC

Southern Building Code Congress International Publisher of the Slandard Building Code (SBC) 900 Mom Clair Birmingham, AL 35213 Phone (205)591,1853

Glossary

Structural Engineering Association of California, Wilh a

-research group-Applied Technology Council (ATe) 217 Second Stree! San Francisco, CA 94105 Phone (415)974,5147 SEAOW

Structural Engineers Association of Washington p.o, Box 4250

Seat::e, W A 98104 Phone (206)632,6026

i'

Ii \I

"i ,

H:'

II!

;

If: II

II, Ii II l!

II ,1i

I'

I,

Accelerometer. A seismograph for measuring ground accelerarion as a func-

tion of lime" AHasing. The phenomenon in which higher hannonies inrroduce spurious low frequency componenrs. This occurs when the number of sampled points of a function is insufficient [0 describe the function. (See Nyquis[ Frequency,) AmplitudiL Maximum value of a function as it varies with time. If the variation wirh time can be desclibed by either a sine or cosine funcrion, it is said to vary harmonicaHy. Angular Frequency/Circular Frequency. The frequency of periodic fuoc{~on in cycles per second {He::tz) multiplied by 21T, expressed in rad/sec. Autocorrelation of a Random Function x (I). Correlation between the func,ion x (I) and the out,of·phase func!ion x (l + r) as defined by eq, (23.19), Base Shear Force. The total lateral force 00 (he structure equivalent to the earthquake excitation at (he base of the structure, Basic Design Spectra. Smooth or average plots of maximum response of singJe degree-of~freedom syscems used in seIsmic design of srrucIUres, Bearing \Vall System. A structural system without a complete vertical load carrying space frame.

I 8m

808

Appendices

Boundary Condition. A constraint applied to the structure independent of time.

,,

Braced Frame. An essentially vertical truss sys[e:n of the concentric or eccentric type which is provided to resist lateral forces, Building Frame System, An essentially complete space frame which provides support for gravity loads. Characteristic Equation. An equation whose roots are [he natural freq:Jer.cies. Ckcuiar FrequEncy. See Angular Frequency_ Complementary SolutioR The solution or a homogeneous differential equa~ ~1on (no excita:ion). Complete Quadratic Combination (CQC). A method of combining values of modal contributions which is based on random vibratior. theory and ir.eludes cross correlation terms. Concentric Braced F'rame. A braced frame ;n which the members are subjected primanly to axial forces. Consistent Mass. Mass influence coefficients detennined by assuming that the dynamic displacement functions are equal [0 the static displaceI71ent functions, Correlation between Random Variables XI (t) and X 2 (I). The time average of the product of the functions x, (I) and x, (I). Coupled Equations. A systeI71 of differential equations in which the equations are not independent from each other. Critical Damping. Mimmum amount of Y:scous dar:lping for which the system will not vibrate. D' Alembert Principle. This principle states that a dynamk system may be assuoed to be in equilibrium provided that the inertial forces are considered as external forces. Damped lfrequency. The frequency at which a viscously damped system osciIlates in free vibration. Damping. The property of the structure to absorb vibrational energy, Damping RatiQ. The ratio of the viscous damping coefficjent to the critical damping. Degrees of Freedom. The number of independent coordfnates required to compietely define the position of the system at any time, Deterministic Vibration_ A process which can be predicted by an exact mathematical expression. Diraes Delta Function. A generalized function having the properties described in eq. (23.52). Direct Stiffness Method. The method of assembling the system stiffness m2.trix by proper summation of the stiffness coefficients of the elements in the system. Discrete Fourier Transform. A summation of harmonic tenns to express the Fourier transform for a function defined by a finite number Df points.

Giossa:y

809

Dual System. A combinatIon of a special or intermediate moment-resisting space frame and shear walls or braced frames. Ductility Ratio. The ratio between the maximum displacement for eiastoplastic behavior and the displacement corresponding to yield point. Dynamic Condensation. A method of reducing the dimension of the elgen~ problem by establishing the dynamic relatIonship betv,feen p:imary and secondary coonjinate!:L Dynamic l\'lagnification Factor. The ratio of the maximum displacemem of a single degree of freedom excited by a harmon:c force to the deflection that would result if a force of that magnitUde were applied statically. Earthquake. The vibrations of the Earth caused by the pa::.sage of seismic waves radiating from some source of elastic energy. Eigenproblem. The problem of solving a homogeneous sys!em of equations :::ontaining a parameter which should be determir:ed to provide nontrivial solutions. Elastic Rebound Theory. The theory of earthquake generatior: proposing that fauits reoain locked while strain energy slowly accumulates in the SliI~ rounding rock. and then suddenly slip. releasing [his energy. Elastoplastic. A system which behaves elastically for a force that does r.ot exceed a maximum value and plastkal1y above this maximum. Ensemble. A set of samples or records of a random process. Epicenter. The point on the Earth's surface directly above the focus. Ergodic Process. A stationary r2.ndom process for which the time average of any record is equal to the average across the ensemble. Fast Fourier Transform (FFT). A very efficient algorithm impie;nented in a computer program for the calculation of the response in the frequency domain. Flexibility Coefficient /,; is the displaceoent at coordinate i due to a unit force (or moment) applied at coordinate i Forced Vibration. Vibration in which the response is due to external excitation of the systen. Fourier Analysis. bt1ethod of determin;r,g the response by superposition of the responses to the harmonic components of the excitation. Fourier Transform. The Fourier transform C (w) of a function F (1) is defined by eq. (23.23). Fourier Transform Pair. Ir. reference to the function F(t), the Fourier transfonn pair is given by eqs. (23.23) and {23.24). Free Body Diagram. A sketch of the system, isolated from all other bodies, in which all the forces external to the body are shown. Free VibratiQR The vibration of a system il1 absence of external eXCItation. Frequency AnaJyzer or Spectral Density Analyzer, An lnstrumen:: that measures electronically the spectral density fur.ction of a signaL Frequency Ratio. The ratio between the forcing frequency 10 the natural frequency for a system excited by a harmonic load,

810

Append;ces

Fundamental Frequency_ The lowest natural frequency of a multidegree~of freedom vibratl:lg sys[eTn. Gauss-Jordan Reduction or Elimination. A compuc<1lionaJ technique in which elemeomry row operatio:ls are applied systematically LO solve a linear system of equations. Generalized Coordinates. A set of independent quamities which describe (he dynamic system at any time. These quantities are generally functions of the geometric coordinates. Geometric Stiffness Coefficient. k Cij is [he force at coordinate i due to a unit dispLacement at coordinate j and resulting from the axial forces in the Slnlcture.

Harmonic. A sinusoidal function having a frequency (hac is an integral multiple of the fundamental frequency. Harmonic Force, A force expressed by a sine, cosine (or equivalent exponenial) function. Hypocenter or Focus. The point in the imerior of [he earth at which rupture is initiated during an earthquake. Impulsive Load. A load that is applied during a relatively sherr rime interval producing an instantaneous change in velocIty. Initial Conditions. The initial values of specific functions such as dispiacerr.ent, velocity, or acceleration evaluawd at time! = O. Intensity (of Earthquakes). A measure 0: grot.:nd shaking obtained from the damage done to structures bt.:ilt by man, changes :n the Earth's surface, and felt repor1S, Intermediate 1.\1oment~Resistlng Space Frame (l.J.\tlRSF). A concrere space frame designed in corJormance wirh Section 1921.8 (k) of UBC~94. Isolation. The reduct!on of severity of the response, usually auained by proper use of a resillent su ppon:. Lateral Force :\tlethod, A method of analysis in which lateral horizontal forces at various levels of [he stmcmre are equivalent to seismic excitarion at the base cf the structure. Lateral Force-Resisting System. That part of the structural system assigned to resist latera! forces. Linear Acceleration Method. A step-by-step method for the integration of the differential equations of motion in which [he acceleration is assumed ro be a linear function during each (ime step. Linear System. A system of differential equations in which no term comains products (or exponents) of the dependem variables or their derivarives. Logarithmic Decrement. The natural logarithm of the ratio of any two successive amplitudes of the same sign obtained in the decay curve in a free vibration lest. Lumped Mass. A method of discretiza[ion in which [he distributed mass of the elements is lumped at the nodes or joints.

Glossary

811

Nlathematical l'vIedeL The ldeaHzatton of a system i:1cluding an the assumptions imposed 0:1 the phys:cul problerr.. Mean..square Value. The [irr.e average of the square of a random function as denned by eq, (23,2), Mean Value, The (ime average of a random funcrion defined by eq. (23.i). Modal Shapes (also Nannal Modes). The relative displacemems ar the coordinates of a multidegree-of-frccdom system vibrating at one of the natural frequer.cies. ;'.lodal Superposition Method, A method of solution of multidegree-of-freedom systems in which the respcnse is determmed from [he solution of independent modal (cr nonna!) equations, Modified Mercalli Intensity (MMI), A measure of the effect of an eanhquake at a panicular location. NIoment~Resistjng Space Frame. A space frame in which the members and joints are capable of resisting forces primarily by flexure. Narrow~Band Process. A rundorr. process whose spectral density function has nonzero values only in a narrow frequency range. Natural Frequency. The number of cycles per second at which a Single degree of· freedom system vibrates freely or a multidegree~of-freedom sys[em vibrates. in one of the normal mode::;. Natural Period. The time interval for a vibrating system in free vibratjor. to

do one oscillation. Newmark Beta ivlethod. A numerical method to ca1cu!me (he response of a strUCture subjected to external excitarjon (force or rr.otion). Node or Joint A point joining e;emencs of the structure and at which displacements are known 0, to be determined. Normal Distribution or Gaussian Distribution. A function whose probability density function is by eq. (23.l2). Normal Modes, See modal Shu pes, Nyquist Frequency, The maximt1m frequency component that can be detected from a funerion sampled at time spacing Llr[N, lI2LlI(Hz)), Occupancy Factor. A numerical faCtOr in the calculation of the base shear force [har depends on the intended use of the structure. Ordinary Moment-Resisting Space Frame (OMRSF). A moment-resisting space frame not meeting sped-a! detailing requiremencs for ductile behavior. P-Delta Effect The secondary effect on shears, axial forces, and moments of frame members :ndt;ced by the vertical toads on the laterally displaced buildi:lg frame. Periodic Function. A fU:1clion rhat repeats itself at a fixed time inrerval known as the period of the function, Power Spectral Density, A renn used to describe the intensity of random vibrarion at 11 given frequency, measured in g2/Hz.

$12

Glossary

Appendices

Principle of Virtual Work The work done by aJl the forces acting or:: a system in static or dynamic equilibrium, which occurs during a vinuai disp~acement comparible with the constmints of the system that is equal to zero, Probability Density Function. A function p (x) such [hat the probability of x (I) of being in the ,ange (x, x + dx) is p (x) dx, Pseud{)~Velocity, The veiocity calculated by analogy with the apparent har~ monic motion for a system seismically excited. Random Function, A function (as. opposed to a deterministic function) whose value at any time cannot be detennined exacrly, but can only be predicted In probabilistic terms by statistical methods. Random Vibration or Random Process, A process which cannot be predicted in a deterministic sense, but o:1iy probabilistically using the theory of statistics. Rayleigh Distribution, A function whose probability density function is given

by eq. (23.14). Rayleigh's Formula. A formula to estimate the fundamental period of the stwcture. Resonance. The condition in which the frequency of the excitation equals the natural frequency of the vibrating system. Response. The force or motion that results from external excitation on the

stnJcture. Response Speetrum. A plot of maximum response (displacement, velocity, or acceieration) for a single degree-of~freedotT, system defined by its natum! frequency (or period) subjected to a specific excitation. Richter lVlagnitude (M). A measure related to total energy released during an earthquake.

Root Mean Square (RMS). The square root of the mean-square value of a random function [eq. (23.5)]. Sample. A record of random process. Seismic Zone Factor. A numerical factor in the calculation of the base shear force at a given geographic location. Seismograph. An instrument for recording as a function of time the motions of the Earth;s surface that are caused by seismic waves.

Shear WalL A wall designed to resist lateral forces parallel to the plane of the waH (sometimes referred to as a venlcal diaphragm or structural wall),

Shock Spectmm. See Response Spectrum. Simple Hannonic Motion. The motion of a system which may be expressed by a sine or cosine function of time. Site-Structure Resonance Coefficient A numerical factor in the calculation

of the base shear force that depends on the condition of the soil. Space Frame. A three-dimensional structural system, without bearing walls, composed of members interconnected so as to function as a complete

813

self-contained unit with or without the aid of horizontal diaphragms or floor-bracing systems. SpeciaJ Moment~Resisting Space Frame (SMRSF). A moment-resisting space frame speciaHy detailed to provide ductile behavior and comply with the "equiremems given in Chapter 19 0:- 22 of l:BC-94 (bte:-national Conference of Building Officials 1994). Spectral Analysis or Spectrum. A description of contributions of the frequency components to ~he mean-square value of a random function. Spectral Density Function. A function that describes the imensity of random vibration in ter:ns of the mean-square value per unit of frequency,

Square Root Sum of Squares (SRSS).

A

method of combining maximum

values of modal contributions by takng the square root of the sum of the sq;,Jared modal contribuvtions, Standard Deviation. The square root of rhe variance. It may be calculated by

eq. (23.6). Static Condensation. A method of reducing the dimensions of the stiffness and of the mass matrices by establishing the static relation between primary and secondary coordinates, Stationary Process. A random process for which the average across the ensemble has the same value at any selected time. Steady~State Vibration, The motion of the system thm remains after the transient mOlion existing at the initiation of the motion has vanished. Stiffness Coefficient. klj is the force at coordinate i due to a u:lrt displacement at coordinate j, Story Drift. The relaEive lateral displacements of consecutive levels of a building, Spring Constant. The change in load on a linear elastic structure required to produce a unit increment of deflection. Strong :Motion AcceJerograph. An instrument (0 register seismic motions higher than a specified amplit;]de. Structure. An assemblage of framing members designed to support gravity loads and resist lateral forces. Structures may be categorized as b~ilding Structures or nO:lbuilding s[ructures. Strudural Factor. A numerical factor in the calculation of the base shear force that depends on the ~ype of structurai system. Tectonic Earthquakes. Earthquakes resulting from the sudden re~ease of energy stored by a major defonnation of the earth. Time History Response. The response (motior. or force) of the Structure evaluated as a function of time. Transient Vibration. The in:tial portio:) of the r.lotion which vanishes due to the presence of damping forces in the system. Transmissibility. The nondimensional ratio, in the steady~state cor.dition, of the :esponse motion to Ehe input motion. Or the C)ondimensionaJ ratio of

814

Appendices

the amplitude of the force transmitted to [he foundation [0 the amplitude of rhe force exciting the system. Variance of xU,. The average of [he squares of the devi£ltions of x (:) values from the mean value "( [eq. (23.3)J. Viscous Damping. Dissipation of energy s'Jch [hat the morion is resi:aed by a force proportional to the velocIty but in [he opposite direction. White Noise, A wide-band random process for which the spectra! density function is constant over the whole frequency range. \.yide~band Process. A random process whose spectral function has nonzero value over a large range of frequencies. \Viener-Kinchin Equations. These are equations (hat reI ale the autocorrelation function and the spectral density function [eqs" (23.49) and (23.50)]" Wilson's & Method. A modification of the step-by-step lineJ.f accelerution method in which the time step is multipiied by u fuetor necessary [Q render the method unconditionally stable. Wood¥Anderson Seismograph. An instrument used to regIster seismic motions.

Selected Bibliography

EARTHQUAKE ENGINEERING Blume, j, A, Ciewmark, N" tvL, and Coming. L (l96!), Design of Multistory Reinforced Concrete BHildings for Earthquake MOfions, Portland Cement Assodatjon, Chicago. Hart" Gary C, and Englekitk, Robert E (1982), Earthquake Design of Concrete Masonry Buildings.' Response Spectra Analysis and General Earrhw quake Modeling Considerations, Prentice-Hall, Englewood Cliffs, NL Housner, G, W. (1970), Design speCltum, in Earthquake Engineering (R, L Weigel. Ed.), Prentice-Hall, Englewood Cliffs, Nl Housner, G. VI" and Jennings, P. C. (1982), Earthquake Design Crileria, Earthquake Engineering Institute, Berkeley., CA. Hudson, D. E. (l97C}, Dynamic tests of full scale structures, in Earthquake Engineering (R. L Weige!, Ed.), Prentice-Hall, Eng!ewood C!iffs, NJ. Naeim, Farzad (1989). The Seismic Design Handbook, Van Nostrand Reinhold, New York" Newmark, N. M" and Hall. W" J" (1973), Procedures and Criteria for Eanhqu.C!.ke ResisIanl Design: BUIlding Pracfices for Disaster Mitigation, Building Science Series 46, National Bureau of Standards, Washington, DC, pp, 209-237, a[5

t 816

Appendices

Newmark. N. M .. and Hall, W. J. (1982), Earthquake Spectra and Design, Earthquake Engineering Research Institute, Berkeley, CA, Newmark, N. M., and Riddell. R. (1980), in Inelastic Spectra for Seismic Design, Proceedings of 7th 'World Conference on Earthquake Engineer:r:g, Istanbul. T"rkey, Vol. 4, pp. 129·136. Newmark, N. \1., and Rosenbl"eth, E. (1971). FU!ldamentais of Earthrjuake Engbteering, Prentice~Han, Englewood Cliffs. NJ. Paz, Mano (1994) International Handbook of Earthquake Engineering.' Codes, Programs and Examples, Chapman and Hal!, New York. Popov. E P., and Benero. V, V. (1980), Seismic analysis of some steel building frames, 1. Eng. Meeh., ASCE. 106, 75·93. Steinbrugge, Karl V. (970), Earthqullke damage and structural performance in the united States, in Earrhquake Engineering (R, L Vleigel, Ed.), Prentice-Hail, Englewood Cliffs, NJ Wakabayashi, Minoru (l986), Design of Eanhquake-Resistant Buildings, McGraw-HIlL New York.

Selected Bibliogrophy

817

Kjuregbia:l, A. D. (1980), A response spectrum method for random vibration, Report No. UCBlEERC-80115, Earthquake Engineering Research Center, University of California. Berkeley, CA. Nashif, A. D., Jones, D. I. C" and Henderson, J. P. (1985), Vibrarion Damping. Wiley, New York. Newmark, N. M, (l959), A method of computation for structural dynamics. Trans. ASCE. 127. 1406·35. Paz. Mario (1973), ?-1athematical observations in structural dynamics, Int. 1. CompUl. S/rue/.. 3, 385·396. Paz, Mano (l983), Practical reduction of strucl"Jral problems. 1. Struct Eng., ASCE, 109( i I I), 2590·2599. Paz, Mario (1984a), Dynamic condensation, AfAA 1" 22(5), 724·727. Paz, ';"'1ar.o (1984b). in Structural lvlechanics Software Serit!s, The university Press of Virginia, Charlottesville, VoL V. pp. 271 M286. Paz. Mario (1985). Micro~CompUfer Aided Engineering: Structurai Dynamics. Van :-;ostrand Reinhold, New York. Paz, Mario (1989). Modified dynamic condensation method, 1. Str"cr, Eng., ASCE, llS(I). 234·238.

STRUCTURAL DYNAMICS Bathe, K. J. (i 982). Finite Element Procedures in Engineering Analysis, Prentice·Hall, Englewood Cliffs, Nl. Berg, Glen V. (1989), Eiemen/s of Struc/ural Dynamics, Prentice·Hali. Engle· wood Cliffs, ]\'1. Biggs, J. M.(1964), introduction to Structural Dynamics. McGraw-Hill, New York. Blevins, R. D. (1979), Fonnuias for Natural Frequency alld Mode Shape, Van Nostrand Reinhold, New York. Chopra, A. (1981), Dynamics of Structures: A Primer, Earthquake Engineering Research Institute, Berkeley, CA. Chopra, A. K. (1995) Dynamics of Structures: Theory and Applications to Earthquake Engineering, Prentice Hall, Englewood Cliffs, N. J. Clough, R. W., and Penzien, J. (1993), Dynamics of Strucrures, 2nd Edition McGraw·HilL New York. Gallagher, R. H. (1975), Finile Elemeni Analysis, Prentice·Hall, Englewood Cliffs, NJ, p. 115. Guynn. R. J. (1965), Reduction of stiff"ess and mass matrices, AIM J_, 13,380. Hanis. Cyril M. (1987). Shock and Vibration Handbook, 3rd ed., McGraw Hill, New York. Humar, J. L. (1990) Dynamic of Struetures, Prentice Hall, Englewood Cliffs. "i. J.

Paz, M., and Dung, L (975), Power series expansion of the general stiffness matrix for beam elements, Int. ], Numer. Methods Eng., 9, 449M459. Smith, 1. W. (1988) Vibration of Structures: Applications in Civil Engineen'ng Design. Chapman and Hall, New York. Timoshenko, S. P. and Goodier, J. N. (1970) Edition Theory of Elasticity, McGraw·Hill. New York. Wilson, E. L, Farhoomand, L, and Bathe. K. J. (1973). Nonlinear dynamic analysis of complex structures. In!. I Earthquake and Structural Dynamics, VoL I. pp. 241·252. Wilson. E. L. Der Kiureghian, A., and Bayo, E. P. (1981), A replacement for the SRSS method in seismic analysis, In:. J. Earthquake Eng. Struct. Dyn., 9, 187-194.

BUILDING CODES American Insurance Association (] 976), The National Building Code, New York. American Society of CivH Engineerings (1995) ASCE Standard: Minimum Design Loads for Buildings and Other Structures (Revision of ANS I ASCE 7·93) ASCE New York. Applied Technology Council (1978), Tentative Provisions for {he Developmeni of Seismic Regulations for BUildings. ATe 3-06, National Bureau of Standards Special Publication 510, U.S. Government Printing Office, \Vashjng. ton. DC.

818

A;:.;:.endices

Berg, Glen V. (1983), Seismic Design Codes alld Procedtlres, Earthquake Engineering Research Institute, Berkeley, CA. Building Offic;a1s and Code AdministratOrs International (1996), BOCA Basic Building Code, Homewood, IL. Building Seismic Safety Council (BSSC) (1994), Recommel,ded Provisions for the Developmem of Seismic Regulations for New Buildings, NEHRP (~a­ (ionai Earthquake Hazard Reduction Program), Washington, DC. Intem:uiona! Conference of Building Officials (1994), Uniform Building Code (UBC), Whit:ier, CA. Southern Building Code Congress Imern::Hior,al (1994), Standard BuUding Code, Birmingham, AL. S[ructural Engineering Association of California [SEAOC (1990)), Recom~ mended Laleral Foret Requiremenfs and Tentative Commentary, San Fran· CISco, CA.

INDEX

RANDOM VIBRATIONS AND FAST FOURIER TRANSFORM COOley, P. M., and Tukey, J. W. (1965), An algorithm for the machine compL::tation of complex Fourier series, Math. Compul., 19. pp. 297-301. Cooley, J. W., Lewis, P. A. W., and Welch, P. D. (1969), The Fast Fo"rie, Transform and its Applications, IEEE Transacrions on Edu.cation, VoL ;:2, No. l, pp. 27-34.

Newland, D. E. (1993), All Imroduction 10 Random Vibrations and Spectral Analysis, 3,4 ed., Longman, New York. Ramirez, R. W. (1985), The FFT: Fundamentals and Concepls, Prentice-Hall, Englewood Cliffs, Nl Yang, C Y. (1986), Random Vibrmion of SUI.lCIUreS, Wiley, New York

acce!l!fOme!er, 80 (.I1\:1sing, ! 46 ampliwde of motion. i 5 anguiar frequency, 14 ANSI. 707 ATe, 707 autOCOfrdatiOfl, 663

3:xi.::l fcrce effect on flexural s:lffn.es$, 443 force-displacement relation, 444 stiffness codfici~!l!, 447 base shear force, 713 Basic Building Code. 707

""urn deiini!:orr. 399 dynamic s!ress, 6}0 forced ~'ibr.. (;cn, 420. 624 geometric s[i:flless, 416 m"S$ m,l(rix, 412

stiffness m
complementary so[uc:on, 48 complete quadrrnic combination (CQC). 335 complex equD.t!on,~, 330 complex frequency response, 147 conc:;nsatlon dyt1;1mic. 380 modified dynamic, JS7 s(:Hic, 367 consistent mass detini,ion, 410 mamx, 412 819

Index

lndex

constant acceleration rr.e:hod. 208 conswnts of integration continuous system, 613 rr.u;tidegree~of-freedom system, 293. 40? singlewdegrec"of~freedom

r

I,I i!

IiIi

II

q

systerr" 13

conslsten( geometric stiffne!S m:J.trlx. 419 Cooley. lW., 148 coordinate transformatlO:1. 449, 47'::', 494 correlation, 662 com?!ex freque:1cy t:nit :mpube ,elationship. 68S COSMOS library, Elements Plane elasticity, 553 Plate bendi:1g and shells, 567 COSMOS examples

Ell.rthquake·resist;:1g

de~jgn

Eqt:ivalem ;mnic la!eral force method. 756 Mulridegree-of-freedom Systems Free vibration shelli~ building, 301 forced motion sheili bui:ding, 331 Dynamic analysis of beams, 433 Gf-ds, 433 ?Iane frllmes, 460 Three dimensional frames, 507

T:usses, 528 Random vibration, 691 Time history response. 597 Single--degree~of-freedom systems FTee vibration, 20 Nonlinear strJcturu! response, 224 Response spectra. 260 Res.po:1se to gene:aj Ioodir.5. 124 Response co har.nonic loading. 81 cou?led diffetential equation, 3 to critical b11ck:ing load, 171 critical damping, 33 D'Aleml)en's Principle, !O damped £ystem. 31 freqc.ency,33 harmonic exdtation. 47, 50 shear building:, 271 oscillator. 32 period,36 steady-st.1te response, 49 tOtal response, 52 danpmg equivalent ..iscous, 63 coeffit:ic!'lt, 32, 4!4 critical. 33

distribL:ted coefficient, 414 ev:tluation of :;anowidth (half power), 59 logarilhmic decrernenl, 37 resonance. 58 :~(10.

3-5

uncoupling. 355 degrees of freedom. 3 design response S?CCirurn Elastic, 241 Inelastic. 250 de(erministlc anaiysis, 653 diaphragm forces, 723 Dirac's dellR fUllction, 673 direct :nerhod, 406 discrete Fourier !ransforr:!, 146 discretization consistent m:J.S5, 4; 0 h:mp::d mass, 408 dispiucement meter, 70 disrribmed property syslern, 609 ducrility ratio. 243, S96 Duhamel's In!eg=al definition. 96 numericaJ cvajuarion damped syste.71, 109 undamped sys;em, 105 dynamic

condensntion, }eo equilibrium. 1O lond factor, 100 magr:Wcation factOr, 52 macrix, 189 dy narnic maL.';x for axial effect, 638 flexural-axial effec~, 642 flexurJI effect., 638 power series: expansion, 646 torsional cHeer, 648 stresses. 630 earthquake desigr: response spectrum Elastic, 242 inelastic, 25; Ei Cen::-o, CA. (1940), 239 San Fernando, CA (l971), 239 tripartite respoIlse spectra. 238 effective rr.odal weighr~ 770 eigenproolem. 288 EI Centro eanhquake, 239

earthquake (COni,) dastoplastic system frame";' StruCturr.:5, 583 to
resim.nl design. simplified Ihrec-dimensionui center or rigidity, 142 de~ign of bUildings, 739 forces and momem in e!e:nems. 749 row,jonul eccentricity, 743 story eccentricity, 7::"3 equatio:1s of motion discrete system, 3$3

dislribt:led

sy~tem.

61 :

unco'Jp!ed, 355 equilibrium. dynam;cs, 10 equivalent Slll(Jc iate:-a! force, 707 ergodic ?fOCeSS, 65::" Eliler's relation. 50 Fast Fourier Tronsforrn {FIT), 148 FEMA.707 Finile Element 0,-Iethod (FEM). 538 Displacement FunctiOrt rec:angular eleJ:len!, SSg triangular element, 542 Element Stiffness reat~ix place bending, 562 friar-gular, 545 Element mass rnatrix plate ber.d:ng, 562 triangular, .545 Plane elasticiTY problems Plane strain, 540 plane stress. 539 first o~ f:;ndamenla: mode, 293 flexibili:y coefficient. 278 matrix, 279 foc'.!s, 70S folding frequency. 146 forced rr.otion. 310 force tra:lsmiuco to L'1e fOllndat:on, 76 Fourier lmalysis., : 39 coefficJent, J40

discrete. :45 exponential fo;m, 144:megra!. 667 series. ;40 transform pair. 667 fra!nc

p;ane. ~: three dimensior.aL c!91 free body di:lgram, 9 free vibrations critically damped. 33 overdam?Cd, 34 undamped. 5 underdnmpcd. 35 frequency analyzer, 674 cirCUlar 0:- ar,gubr, J4 damped. 35 fundament;ll, 1.515. 293 r:a!l':'r.li. 12 ratio, 49 fundamental period, 7! 7 Galaher, R. H" 367 Gaussian Distriburior.. 659 Gauss-Joronn, 368 generalized coo;dtna!~s, 162 generaliz.ed properties damping, t78

force, 165, 178 geometric $riffnes$, 171 inertia, !65 mass, 168., I stifrness. l6S, 178 Goodier,l.N. 543, 548, 56] geometrical stiffness, 17:,2'1'1,416

is

grids, 469 G~yar" R.L 367

harmonic excitatior:., 47 harmonic motion damped,50 undar.tped, 47 harmonics or h igh::r harmonics. 293 horizontal torsio:1al moment, 719 hypocemer. 708 impulsive load, 96 rectangular, 99 triangu!;;r, lOt inelastic spectra, 2~ tll.ert;al force. 10.54;

821

.r !>Ii II;

.

~

it ~.

~ ~ ~,

822

Index

~ndex

lolli .. ! conditions, 13 initial value problem, lJ

Integruliou

nonlin~ar

equations

cons;"ni (l,e.ceterallon me'hod. 208

linear acceleration method. 21 J irregtJu; Sl(UCWrCS, 713 L:lOlatlon, 67 Jacobi Melhod, 299

shapo!5 for uniform beams, 513 seismic coefficient, 771 $etsmlc response, 766

,;>heilf force, 768 syring constant. 312 $uperpos(lion m~{hod, 3iG (orslonul moment, 172 modif:ed dynamic conCer.:s~uion, 387 Modified MetCal:y Intensity scale (MM1). 70<;

~ ~.

~.

~I

~:

tal· ~I: ~i

~II

.

/I'l, .

~!

~: ~'.

~I:.

pq'

~. ~, )Itt!

J-l: ~:

kinetic energy. 168, 371 Kuopp, K., 6';'7

nlirrOW-O
lateral force. dis:ribu:ion, 7!8

mUltidegree-of-freedom system, 287 single-dcgr.:e-of-t::-eedom system, ! 2 (1a\umJ period. 13 Newh:nd, D. E., !46 NeWmilrf: Bela Me!hod, 2!4. 588

llnear tlcce;C(aliOfl melhod, 211 Ic~ding

genera!, 96 harmonic, 47

imp:.il:::lvc. 96 periocic, t 40 rcc[J,r.gular, 99 trianglblJ, :01 logari:hmic UeCrCfl'.em, 37

magnification [neror, 52 muss mlHrlX consL~lenl

beam,4l0 grid. 473 plane ~rame, 449 plane truss. 513 three dimensional frame. 493 three dimensional ,rus;;, 524 lumped

beam. 408 grid, 47) plane frume. 445 three dimension;)1 frilme, 493 malhe:nat;cal model, 3 Maxwell's R~ciprocal Theorem, 404 mean·square niue, 655 modal camping, 357

dispiacemenLs, 77:

~f

lateral forces, 771

1-1.

"""'1i ~.

r.odal coordinates, 400 elemer,t force, .:27 noalineil~ re::;pOnse incrementa! eq:'HI.(ion of motion. 578 mullidegree-of-freedom system, 577 single degree-of-freedom syStem, 106 unconditionally stablo! solucion, 577 Wilwn·8 method, 577 normal rJ;lIuibw:loo, 659 modes, 194 meae of vibrtHio::l, 293, 615 vibration, 293, 615 numerical integration, 105 pH;ce0wise linear function damped, i09 undemped, 106 SImpson's Rt:i!:, 105 trllpezoidal :ule. 106 Nyquis: freq'Je:1cy, 146,669

mean value, 655

t-/:

i-i

Newmilrk, N, M, 243, 251 Newton's Lnw of Motion, S

drift, 772

torce.312

occupancy impOrlilnCe filctor, 7t3 onhogonulity properly cOnlinuous system, 622 dhcrete sys:em. 294 (lverl'Jming momeR!, 720

mass, 312 m:alrix, 297

participation factor, 3 '.9, 767

overturning moment, 771

Paz. Mario, 380, 647, 649

particular SOI\l!lon, 48

P-do!h:t o!ffect. 275 Pen!ie:!, ;., 356 period, naH.It, &;6 Aexu;al a:'fecB including ::Ix;a!, 649 princiyli: of con;.;erv;..:ion of energy, iSS principle of virmal work, 162 proollbility den;:;l)' function, 65& pseudov
ruoctioo, 654 process, 654 vibration, 653 Rayleigh

qnoliec{, 298 Distribution, 660 717

fOffll ..tla,

Metho(L IRS red"cec

damping motrix, 370 dynnflltC malrix, 3S1 ronss miJLrix, 3Sl stiffness matrix, 367 resonaoce, 50 respon:>e spectrum

chan construction, 233 elastic design, 241 inela$tic design, 250 ioelJ.S:ic sys~em, 247 (ripartite C!-ilrt, 238 ,R"1Opoc1Oe 10 random exdwtlon multiple-d.;gtee-of-frecco:n system, 685 two degree-of-freedom system, 68] restoring force, lIS Richter ,Magnilaee. 708 root mean square (RMS), 655 rOlary inenia, 655 sample, 654

scaling re$uirs, 776 SEAOC, 707 sej}:mic iwmurnenls, 79 se:smic weIght, 716, 115 se:smic zone, 713, 114 s!Jenr Quitding

dornped. 353 definition, 272 dynamic matrix. 289 forc
hig!-ier harL::on ICl(, 293 naturul frequencies. 287 nOrlnal mede:;, 294 [i!SponSe to base motion, :3 t7 shear walls, 195 S,mf'$O(j's nile, 105 single-degree·of-t"reedom sysfem :Jamped fre
nonlinear .c:.Ipcnse. 205 reSponse in (he frequency dO:ll2:n, 139 responso! sp
analy:m. 668 density analyzer, 614 density function. 672 ;ji$place~nt,

238

funcdon, 668 velocity, 238 spectrum, 669

spring COClstnnt, 6 spnogs in parallel. 6 ,in seri;!s, 6 SRSS method, 3D st
Standard BU;lding Code, 707 standard deviullon. 655

Slade lateral force method, 713 st::ady-s!llie response, 49 step-by-step inregratloCl,

eon;w:m:

acce~er(lJion

method. 208

linear n,:celeta.tlcm method, 2\1 uncondition;.!Lly stab:e, 577 Wilson's-9: me!p.od, 579 ,;(iffness coefficient,40r equation (or shear buildings, 272

823

824

Index

stiffness (COni.; mat,i;; axial force effect, 443 axial mass effec(, 444 dynamic. 633flexural. 405 plane trUsS, 512 space lru!>S, 522 three dtmensional frame, -492 torsional, a. 72 ~tory drifr, no slOry ShCOf. 718 mong rr:mion eanhquaKe, 2.39 strucltiral factOr, 715, 726 wpport motion, 66

Tnylor"s ser:es. 647 Timoshek.c, $.,543.54'8,561 three-dimensional frumes. 491 wr;:ional effect, 471 10(31 design v<':.:ues Pase sheil!' force, 773 lateral deflection, 774 la:cral sei&mtc force. 773 overtJming moment, 774 story drift, 774story shea, force, 773 rOfsiOf.a! moment. 774 transformarion miltriX, 451. 369, 475, 496 trantmissibi:ity. 68, 77 tronslent respO:1.$c, 49 triang'Jinr plate elcmenl:., 540

tripartite chart, 138

mSKETTE ORDERING FORM

of, 243 elastic Cesign, 24! inelastiC design. 250 cons~r'.lction

Professor

(russes pln:1c, 511 space, 522

usc response

!'v~ario

Datc _ _ _ _ __

paz

P.O. Bo' 3510! Louisvme, KY 40232 (;SA s;x:cmL 247, 768

\lneoupioo eqmltions damped. 354 u:lCampcci, 310 undamped free vitJrrHion condn'Jou$ system, 6: 0 :nulticlegre~of·freedom

Pieuse se:1::l

to;

NAME _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ sys!em. 287

single degree system • .; undampec. o$cil\mor, 5 Ur.iform Building Code (DEC), 707

variance, 655 vibration isolation. 67. 76 vibration, normal modes. :lS7 virtual work. pri:1ciple oC 10. 162 viscous camping, 31

WaKubnyashi, Mir.ul''U. 360 white noise, 676 w~de·band pr
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