Introduction to Structural Motion Control Connor

INTRODUCTIONTO STRUCTURAL

MOTION CONTROL

J. J. Connor

Contents i

Contents

Preface x

Chapter 1 Introduction

1.1 Motivation for structural motion control 1• Limitations of conventional structural design 1• Motion based structural design and motion control 3

1.2 Motion versus strength issues for building type structures 4• Example 1.1: Cantilever shear beam 6• Example 1.2: Cantilever bending beam 7• Example 1.3: Quasi-shear beam frame 9

1.3 Design of a single-degree-of freedom system for dynamic loading 11• Response for periodic excitation 11• Design criteria 16• Methodology for acceleration controlled design 17• Example 1.4: An illustration of acceleration controlled design 20• Methodology for displacement controlled design 21• Example 1.5: An illustration of displacement controlled design 21• Methodology for force controlled design 23• Example 1.6: Force reduction 25

1.4 Design of a single-degree-of-freedom system for support motion 26• Response for periodic support motion 26• Design scenarios 27• Example 1.7: Controlling acceleration due to ground motion 27• Example 1.8: Controlling relative motion due to ground motion 28

1.5 Stiffness distribution for a two degree-of-freedom system 30• Example 1.9: 2DOF system - with equal masses 33

1.6 Control strategies for motion based design 331.7 Scope of text 42Problems 45

ii Contents

Part I: Passive Control

Chapter 2 Optimal stiffness distribution

2.1 Introduction 512.2 Governing equations - planar beam 53

• Planar deformation-displacement relations 54• Optimal deformation and displacement profiles 55• Equilibrium equations 57• Force-deformation relations 58• Example 2.1: Composite sandwich beam 59• Example 2.2: Equivalent rigidities for a discrete truss beam 60• Governing equations for buildings modelled as pseudo shear

beams 642.3 Stiffness distribution for a continuous cantilever beam under static

loading 69• Example 2.3: Cantilever beam - quasi static seismic loading 71• Example 2.4: Truss-beam revisited 72

2.4 Stiffness distribution for a discrete cantilever shear beam - staticloading 75

• Example 2.5: 3DOF shear beam 762.5 Stiffness distribution - truss under static loading 77

• Example 2.6: Application of least square approaches 81• Example 2.7: Comparison of strength vs displacement based

design 892.6 Stiffness distribution for a cantilever beam - dynamic response 93

• Rigidity distribution - undamped response 932.7 Stiffness distribution for a discrete shear beam - dynamic response 98

• Example 2.8: 3DOF shear beam 992.8 Stiffness calibration 100

• Governing equation - fundamental mode response of a discreteshear beam 100

• Example 2.9: 3DOF shear beam revisited 101• Governing equations - fundamental mode response of a

continuous beam 105• Example 2.10: One-dimensional parameters - continuous beam108• Stiffness calibration - periodic excitation 109• Example 2.11: 3DOF shear beam revisited again 111• Example 2.12: Stiffness calibration - continuous beam 113• Example 2.13: Example 2.12 revisited 114

Contents iii

• Stiffness calibration - seismic excitation 116• Seismic response spectra 119• Example 2.14: Example 2.12 revisited for seismic excitation 128• Example 2.15: 5DOF shear beam 129

2.9 Examples of stiffness distribution for buildings 1322.10 Stiffness modification 139

• Iterative procedure - seismic excitation 139• Multiple mode response 140• Building examples - iterated stiffness distribution 143

Problems 149

Chapter 3 Optimal passive damping distribution

3.1 Introduction 1673.2 Viscous, frictional, and hysteretic damping 171

• Viscous damping 171• Example 3.1: Viscous damper 172• Friction damping 174• Hysteretic damping 176• Example 3.2: Stiffness of a rod hysteretic damper 177• Example 3.3: Stiffness of two hysteretic dampers in series 178

3.3 Viscoelastic material damping 179• Example 3.4: Viscoelastic damper 182

3.4 Equivalent viscous damping 184• Example 3.5: Structural and hysteretic damping comparison -

seismic excitation 186• Example 3.6: Determining for 3M ISD110 damping material 193

3.5 Damping parameters - discrete shear beam 193• Damping systems 193• Rigid structural members - Linear viscous behavior 197• Example 3.7: Example 2.15 revisited 198• Rigid structural members - Linear viscoelastic behavior 200• Example 3.8: 201• Example 3.9: Example 3.7 revisited 202• Example 3.10: Viscoelastic damper design 203• Example 3.11: Hysteretic damper design - diagonal element 205• Flexible structural members - linear viscoelastic behavior 206• Example 3.12: Coupled spring-damper model 209

3.6 Damping parameters - truss beam 211• Linear viscous behavior 213• Linear viscoelastic behavior 214

3.7 Damping distribution for MDOF systems 215

αd

iv Contents

• Multi mode free vibration response 216• Example 3.13: Eigenvalue problem - 2DOF 219• Example 3.14: Modal response for nonproportional damping 226• Stiffness proportional viscous damping 229• Examples - low rise buildings 232• Examples - building #4 239

Problems 245

Chapter 4 Tuned mass damper systems

4.1 Introduction 2614.2 An introductory example 262

• Example 4.1: Preliminary design of a TMD for a SDOF system 2644.3 Examples of existing tuned mass damper systems 266

• Translational tuned mass dampers 266• Pendulum tuned mass damper 272

4.4 Tuned mass damper theory for SDOF systems 277• Undamped structure - undamped TMD 277• Undamped structure - damped TMD 280• Example 4.2: Design of a TMD for an undamped SDOF system293• Damped structure - damped TMD 293• Example 4.3: Design of a TMD for a damped SDOF system 299

4.5 Case studies - SDOF systems 3004.6 Tuned mass damper theory for MDOF systems 307

• Example 4.4: Design of a TMD for a damped MDOF system 313• Example 4.5: Design of TMD’s for a simply supported beam 315

4.7 Case studies - MDOF systems 322Problems 331

Chapter 5 Base isolation systems

5.1 Introduction 3395.2 Isolation for SDOF systems 340

• SDOF examples 340• Bearing terminology 344• Modified SDOF model 347• Periodic excitation - modified SDOF model 349• Example 5.1: Stiffness factors for prescribed structure and base

motion 351• Seismic excitation - modified SDOF model 352• Example 5.2: Stiffness parameters - modified SDOF model of

Contents v

Building example #2 3565.3 Design issues for structural isolation systems 357

• Flexibility 357• Rigidity under low level lateral loads 358• Energy dissipation/absorption 359• Modeling of a natural rubber bearing (NRB) 360• Modeling of a lead rubber bearing (LRB) 364• Applicability of base isolation systems 367

5.4 Examples of existing base isolation systems 368• USC University Hospital 369• Fire Department Command and Control Facility 369• Evans and Sutherland Manufacturing Facility 370• Salt Lake City Building 371• The Toushin 24 Ohmori Building 371• Bridgestone Toranomon Building 374• San Francisco City Hall 374• Long Beach V.A. Hospital 375

5.5 Optimal stiffness distribution - discrete shear beam 376• Scaled stiffness distribution 376• Example 5.3: Scaled stiffness for a 4 DOF beam with base isolation

379• Fundamental mode response 380• Example 5.4: Example 5.3 revisited 380• Stiffness calibration for seismic isolation 382• Example 5.5: Stiffness calibration for Example 5.4 382

5.6 Optimal stiffness distribution - continuous cantilever beam 386• Stiffness distribution - undamped response 386• Fundamental mode equilibrium equation 393• Rigidity calibration - seismic excitation 395• Example 5.6: Stiffness calibration - Example Building #2 396

5.7 Building design examples 397• Stiffness distribution on fundamental mode response 397• Stiffness distribution including the contribution of the higher

modes 404Problems 413

vi Contents

Part II: Active Control

Chapter 6 Introduction to active structural motion control

6.1 The nature of active control 423• Active control versus passive control 423• The role of feedback 427• Computational requirements and models for active control 427

6.2 An introductory example of quasi-static feedback control 428• Example 6.1: Shape control for uniform loading 431• Example 6.2: Discrete displacement data 432

6.3 An introductory example of dynamic feedback control 433• Example 6.3: Illustrative example - influence of velocity feedback

4366.4 Actuator technologies 441

• Introduction 441• Force application schemes 442• Large scale linear actuators 446• Large scale adaptive configuration based actuators 450• Small scale adaptive material based actuators 451

6.5 Examples of existing large scale active structural control systems 462• AMD in Kyobashi Seiwa Building 463• AVS Control at Kajima Technical Research Institute 465• DUOX Active-Passive TMD in Ando Nishikicho Building 468• ABS - 600 Ton Full Scale Test Structure 472

Problems 474

Chapter 7 Quasi-static control algorithms

7.1 Introduction to control algorithms 4797.2 Active prestressing of a simply supported beam 480

• Passive prestressing 480• Active prestressing 482• Active prestressing with concentrated forces 488• Example 7.1: A single force actuator 488• Example 7.2: Two force actuators 492• A general active prestressing methodology 499

Contents vii

• Example 7.3: Multiple actuators 5037.3 Quasi-static displacement control of beams 506

• Continuous least square formulation 506• Discrete least square formulation 508• Example 7.4: Cantilever beam - Least square algorithm 510• Extended least square formulation 516• Example 7.5: Example 7.4 revisited 518

7.4 Quasi-static control of MDOF systems 521• Introduction 521• Selection of measures 522• Example 7.6: Illustrative examples of observability 523• Example 7.7: Illustrative examples of controllability 525• Least square control algorithms 527• Example 7.8: Example 7.7 revisited 528• Example 7.9: Example 7.7 revisited with an extended least square

algorithm 531• Problems 534

Chapter 8Dynamic Control Algorithms

8.1 Introduction 5458.2 State-space representation - time invariant SDOF system 546

• Governing equations 546• Free vibration uncontrolled response 548• General solution - time invariant systems 550• Example 8.1: Equivalence of equations (8.18) and (8.24) 551• Stability criterion 552• Linear negative feedback 553• Effect of time delay on feedback control 555• Stability analysis for time delay 560

8.3 Discrete time formulation - SDOF systems 566• Governing equation 566• Linear negative feedback control 568• Stability analysis time invariant linear feedback control 569• Example 8.2: Stability analysis-SDOF system with no time delay

575• Example 8.3: Stability analysis - SDOF system with time delay 584

8.4 Optimal linear feedback - time invariant SDOF systems 586• Quadratic performance index 586• An example - linear quadratic regulator control algorithm 588• The continuous time algebraic Riccati equation 593

viii Contents

• The discrete time algebraic Riccati equation 598• Example 8.4: Expanded form of discrete algebraic Riccati equation

for an undamped SDOF system 600• Finite interval discrete time algebraic Riccati equation 609• Example 8.5: Example revised 610• Continuous time Riccati different equation 611• Variational formulation of the continuous time Riccati equation

612• Example 8.6: Application to scalar case 617

8.5 State-space of formulation for MDOF systems 623• Free vibration reponse-time invariant uncontrolled system 625• Example 8.7: Free vibration solution for proportional damping 629• Example 8.8: General uncoupled damping 630• Orthogonality properties of the state eigenvectors 631• Example 8.9: Initial conditions-free vibration response 632• Determination of W and 633• General solution-time invariant system 634• Modal state space formulation-uncoupled damping 635• Modal state space formulation-arbitrary damping 639• Example 8.10: Modal formulation-undamped case 642• Example 8.11: Modal formulation-uncoupled damping 643• Example 8.12: Modal parameters-4DOF system 644• Example 8.13: Modal response for example 8.12 651• Example 8.14: Modal response response with feedback for

example 8.12 654• Stability analysis-discrete modal formulation 659• Example 8.15: Stability analysis for example 8.14 662• Controllability of a particular modal response 679• Example 8.16: Controllability analysis for a 20 DOF Modal 680• Observability of a particular modal response 682• Example 8.17: Observability analysis for a 20 DOD Modal 684

8.6 Optimal linear feedback-MDOF time invariant systems 686• Continuous time modal formulation 686• Discrete time modal formulation 688• Application studies - LQR control 690• Example 8.18: Control force design studies for a 20 DOF shear

beam 698• Example 8.19: Alternate choice of response measures 711

Problems 714

f j

Contents ix

References 725

Bibliography 735

Index 737

x Introduction to Structural Motion Control

1

Chapter 1

Introduction

1.1 Motivation for structural motion control

Limitations of conventional structural design

The word, design, has two meanings. When used as a verb it is defined as the actof creating a description of an artifact. It is also used as a noun, and in this case, isdefined as the output of the activity, i.e., the description. In this text, structuraldesign is considered to be the activity involved in defining the physical makeupof the structural system. In general, the “designed” structure has to satisfy a set ofrequirements pertaining to safety and serviceability. Safety relates to extremeloadings which are likely to occur no more than once during a structure’s life. Theconcerns here are the collapse of the structure, major damage to the structure andits contents, and loss of life. Serviceability pertains to moderate loadings whichmay occur several times during the structure’s lifetime. For service loadings, thestructure should remain fully operational, i.e. the structure should suffernegligible damage, and furthermore, the motion experienced by the structureshould not exceed specified comfort limits for humans and motion sensitiveequipment mounted on the structure. An example of a human comfort limit is therestriction on the acceleration; humans begin to feel uncomfortable when theacceleration reaches about . A comprehensive discussion of human comfort0.02g

2 Chapter 1: Introduction

criteria is given by Bachmann and Ammann (1987).

Safety concerns are satisfied by requiring the resistance (i.e. strength) of theindividual structural elements to be greater than the demand associated with theextreme loading. The conventional structural design process proportions thestructure based on strength requirements, establishes the corresponding stiffnessproperties, and then checks the various serviceability constraints such as elasticbehavior. Iteration is usually necessary for convergence to an acceptablestructural design. This approach is referred to as strength based design since theelements are proportioned according to strength requirements.

Applying a strength based approach for preliminary design is appropriatewhen strength is the dominant design requirement. In the past, most structuraldesign problems have fallen in this category. However, a number ofdevelopments have occurred recently which have limited the effectiveness of thestrength based approach.

Firstly, the trend toward more flexible structures such as tall buildings andlonger span horizontal structures has resulted in more structural motion underservice loading, thus shifting the emphasis from safety toward serviceability. Forinstance, the wind induced lateral deflection of the Empire State Building in NewYork City, one of the earliest tall buildings in the United States, is several incheswhereas the wind induced lateral deflection of the World Trade Center tower isseveral feet, an order of magnitude increase. This difference is due mainly to theincreased height and slenderness of the World Trade Center in comparison to theEmpire State tower. Furthermore, satisfying the limitation on acceleration is adifficult design problem for tall slender buildings.

Secondly, some of the new types of facilities such as space platforms andsemi-conductor manufacturing centers have more severe design constraints onmotion than the typical civil structure. In the case of microdevice manufacturing,the environment has to be essentially motion free. Space platforms used tosupport mirrors have to maintain a certain shape to a small tolerance in order forthe mirror to properly function. The design strategy for motion sensitive structuresis to proportion the members based on the stiffness needed to satisfy the motionconstraints, and then check if the strength requirements are satisfied.

Thirdly, recent advances in material science and engineering have resultedin significant increases in the strength of traditional civil engineering materials

1.1 Motivation for Structural Motion Control 3

such as steel and concrete, as well as a new generation of composite materials.Although the strength of structural steel has essentially doubled, its elasticmodulus has remained constant. Also, there has been some increase in the elasticmodulus for concrete, but this improvement is still small in comparison to theincrement in strength. The lag in material stiffness versus material strength hasled to a problem with satisfying the serviceability requirements on the variousmotion parameters. Indeed, for very high strength materials, it is possible for theserviceability requirements to be dominant. Some examples presented in thefollowing sections illustrate this point.

Motion based structural design and motion control

Motion based structural design is an alternate design process which ismore effective for the structural design problem described above. This approachtakes as its primary objective the satisfaction of motion related designrequirements, and views strength as a constraint, not as a primary requirement.Motion based structural design employs structural motion control methods todeal with motion issues. Structural motion control is an emerging engineeringdiscipline concerned with the broad range of issues associated with the motion ofstructural systems such as the specification of motion requirements governed byhuman and equipment comfort, and the use of energy storage, dissipation, andabsorption devices to control the motion generated by design loadings. Structuralmotion control provides the conceptional framework for the design of structuralsystems where motion is the dominant design consideration. Generally, one seeksthe optimal deployment of material and motion control mechanisms to achievethe design targets on motion as well as satisfy the constraints on strength.

In what follows, a series of examples which reinforce the need for analternate design paradigm having motion rather than strength as its primaryfocus, and illustrate the application of structural motion control methods tosimple structures is presented. The first three examples deal with the issue ofstrength versus serviceability from a static perspective for building typestructures. The discussion then shifts to the dynamic regime. A single-degree-of-freedom (SDOF) system is used to introduce the strategy for handling motionconstraints for dynamic excitation. The last example extends the discussionfurther to multi-degree-of-freedom (MDOF) systems, and illustrates how to dealwith one of the key issues of structural motion control, determining the optimalstiffness distribution. Following the examples, an overview of structural motioncontrol methodology is presented.


1.2 Motion versus strength issues for building type structures

Building configurations have to simultaneously satisfy the requirements of site(location and geometry), building functionality (occupancy needs), appearance,and economics. These requirements significantly influence the choice of thestructural system and the corresponding design loads. Buildings are subjected totwo types of loadings: gravity loads consisting of the actual weight of the structuralsystem and the material, equipment, and people contained in the building, andlateral loads consisting mainly of wind and earthquake loads. Both wind andearthquake loadings are dynamic in nature and produce significant amplificationover their static counterpart. The relative importance of wind versus earthquakedepends on the site location, building height, and structural makeup. For steelbuildings, the transition from earthquake dominant to wind dominant loading for aseismically active region occurs when the building height reaches approximately

. Concrete buildings, because of their larger mass, are controlled byearthquake loading up to at least a height of , since the additional gravityload increases the seismic forces. In regions where the earthquake action is low(e.g. Chicago in the USA), the transition occurs at a much lower height, and thedesign is governed primarily by wind loading.

When a low rise building is designed for gravity loads, it is very likely thatthe underlying structure can carry most of the lateral loads. As the buildingheight increases, the overturning moment and lateral deflection resulting fromthe lateral loads increase rapidly, requiring additional material over and abovethat needed for the gravity loads alone. Figure 1.1 (Taranath, 1988) illustrates howthe unit weight of the structural steel required for the different loadings varieswith the number of floors. There is a substantial weight cost associated withlateral loading.

100m250m

1.2 Motion Versus Strength Issues for Building Type Structure 5

Fig. 1.1: Structural steel quantities for gravity and wind systems

To illustrate the dominance of motion over strength as the slenderness ofthe structure increases, the uniform cantilever beam shown in Fig. 1.2 isconsidered. The lateral load is taken as a concentrated force applied to the tip ofthe beam, and is assumed to be static. The limiting cases of a pure shear beam anda pure bending beam are examined.

Fig. 1.2: Building modeled as a uniform cantilever beam

0 50 100 150 200 2500

20

40

60

80

100

120

140

Gravity loads Lateral loads

Floor Columns

Num

ber

of fl

oors

500 1000

Structural steel - N/m2

1500 2000 2500

p

H

d

w

a a

p u

d

section a-a


Example 1.1: Cantilever shear beam

The shear stress is given by

(1.1)

where is the cross sectional area over which the shear stress can be consideredto be constant. When the bending rigidity is very large, the displacement, , atthe tip of the beam is due mainly to shear deformation, and can be estimated as

(1.2)

where is the shear modulus and is the height of the beam. This model iscalled a shear beam. The shear area needed to satisfy the strength requirementfollows from eqn (1.1):

(1.3)

where is the allowable stress. Noting eqn (1.2), the shear area needed to satisfythe serviceability requirement on displacement is

(1.4)

where denotes the allowable displacement. The ratio of the area required tosatisfy serviceability to the area required to satisfy strength provides an estimateof the relative importance of the motion design constraints versus the strengthdesign constraints

(1.5)

Figure 1.3 shows the variation of r with . Increasing places

τ

τ pAs------=

Asu

u pHGAs-----------=

G H

As strength

pτ∗-----≥

τ∗

As serviceability

pG---- H

u∗------⋅≥

u∗

rAs serviceability

As strength

--------------------------------------τ∗G----- H

u∗------⋅= =

H u∗⁄ H u∗⁄


more emphasis on the motion constraint since it corresponds to a decrease in the

allowable displacement, . Furthermore, an increase in the allowable shear

stress, , also increases the dominance of the displacement constraint.

Fig. 1.3: Plot of versus for a pure shear beam

Example 1.2: Cantilever bending beam

When the shear rigidity is very large, shear deformation is negligible, and thebeam is called a “bending” beam. The maximum bending moment in thestructure occurs at the base and equals

(1.6)

The resulting maximum stress is

(1.7)

where is the section modulus, is the moment of inertia of the cross-sectionabout the bending axis, and is the depth of the cross-section (see Fig. 1.2). Thecorresponding displacement at the tip of the beam becomes

u∗

τ∗

200 300 400100Hu∗------

r

τ1*

τ2* τ1

*>

r H u∗⁄

M

M pH=

σ

σ MS----- Md

2I--------- pHd

2I-----------= = =

S Id

u


(1.8)

The moment of inertia needed to satisfy the strength requirement is given by

(1.9)

Using eqn (1.8), the moment of inertia needed to satisfy the serviceabilityrequirement is

(1.10)

Here, and denote the allowable displacement and stress respectively. Theratio of the moment of inertia required to satisfy serviceability to the moment ofinertia required to satisfy strength has the form

(1.11)

Figure 1.4 shows the variation of with for a constant value of the

aspect ratio ( for tall buildings). Similar to the case of the shear

beam, an increase in places more emphasis on the displacement since it

corresponds to a decrease in the allowable displacement, , for a constant .

Also, an increase in the allowable stress, , increases the importance of thedisplacement constraint.

For example, consider a standard strength steel beam with an allowable

stress of , a modulus of elasticity of , and an

aspect ratio of . The value of at which a transition from strengthto serviceability occurs is

(1.12)

For , and motion controls the design. On the other hand, if high

u pH3

3EI----------=

IstrengthpHd2σ∗-----------≥

IserviceabilitypH3

3Eu∗-------------≥

u∗ σ∗

rIserviceability

Istrength-------------------------------

pH3

3Eu∗------------- 2σ∗

pHd-----------⋅ 2H

3d-------- σ∗

E------ H

u∗------⋅ ⋅= = =

r H u∗⁄H d⁄ H d⁄ 7≈

H u∗⁄

u∗ H

σ∗

σ∗ 200MPa= E 200,000MPa=

H d⁄ 7= H u∗⁄

Hu∗------

r 1=

32--- E

σ∗------ d

H---- 200≈⋅ ⋅=

H u∗⁄ 200> r 1>


strength steel is utilized ( and )

(1.13)

and motion essentially controls the design for the full range of allowabledisplacement.

Fig. 1.4: Plot of versus for a pure bending beam

Example 1.3: Quasi-shear beam frame

This example compares strength vs. motion based design for a single bay frame ofheight and load (see Fig. 1.5). For simplicity, a very stiff girder is assumed,resulting in a frame that displays quasi-shear beam behavior. Furthermore, thecolumns are considered to be identical, each characterized by a modulus ofelasticity , and a moment of inertia about the bending axis .

The maximum moment, , in each column is equal to

(1.14)

σ∗ 400MPa= E 200,000MPa=

Hu∗------

r 1=

100≈

200 300 400100Hu∗------

r

σ1*

σ2* σ1

*>

r H u∗⁄

H p

Ec Ic

M

M pH4

--------=


The lateral displacement of the frame under the load is expressed as

Fig. 1.5: Quasi-shear beam example

(1.15)

where denotes the equivalent shear rigidity which, for this structure, is givenby

(1.16)

The strength constraint requires that the maximum stress in the column be

less than the allowable stress

(1.17)

where represents the depth of the column in the bending plane. Equation (1.17)is written as

(1.18)

The serviceability requirement constrains the maximum displacement to

be less than the allowable displacement

u

H Ec , IcEc , Ic

Ig ∞=

p2---p

2---

u pHDT--------=

DT

DT24EcIc

H2-----------------=

σ∗

Md2Ic---------

pHd8Ic

----------- σ∗≤=

d

Ic strength

pHd8σ∗-----------≥

u∗

1.3 Design of a Single-degree-of Freedom System for Dynamic Loading 11

(1.19)

The corresponding requirement for is

(1.20)

Forming the ratio of the moment of inertia required to satisfy the serviceabilityrequirement to the moment of inertia required to satisfy the strength requirement,

(1.21)

leads to the value of for which motion dominates the design

(1.22)

1.3 Design of a single-degree-of freedom system for dynamic loading

The previous examples dealt with motion based design for static loading. Asimilar approach applies for dynamic loading once the relationship between theexcitation and the response is established. The procedure is illustrated for thesingle-degree-of-freedom (SDOF) system shown in Fig. 1.6.

Response for periodic excitation

The governing equation of motion of the system has the form

(1.23)

pH3

24EcIc----------------- u∗≤

Ic

Ic serviceability

pH3

24Ecu∗------------------≥

rIc serviceability

Ic strength

-----------------------------------σ∗3Ec--------- H

d---- H

u∗------⋅ ⋅= =

H u∗⁄

Hu∗------

3Ec

σ∗--------- d

H----⋅≥

mu t( ) cu t( ) ku t( )+ + p t( )=


Fig. 1.6: Single-degree-of-freedom system

where , , are the mass, stiffness, and viscous damping parameters of thesystem respectively, is the applied loading, is the displacement, and is theindependent time variable. The dot operator denotes differentiation with respectto time. Of interest is the case where is a periodic function of time. Taking tobe sinusoidal in time with frequency ,

(1.24)

the corresponding forced vibration response is given by

(1.25)

where and characterize the response. They are related to the system andloading parameters as follows:

(1.26)

(1.27)

(1.28)

(1.29)

k

c

m

u

pR

m k cp u t

p pΩ

p t( ) p Ωtsin=

u t( ) u Ωt δ–( )sin=

u δ

upk--- H1=

H11

1 ρ2–[ ]

22ξρ[ ] 2

+

--------------------------------------------------=

ω km----=

ξ c2ωm------------ c

2 km---------------= =


(1.30)

(1.31)

The term is the displacement response that would occur if the loading

were applied statically; represents the effect of the time varying nature of the

response. Figure 1.7 shows the variation of with the frequency ratio, , for

various levels of damping. The maximum value of and corresponding

frequency ratio are related to the damping ratio by

(1.32)

Fig. 1.7: Plot of versus and

(1.33)

ρ Ωω---- Ω m

k----= =

δtan 2ξρ

1 ρ2–

---------------=

p k⁄H1

H1 ρ

H1

ξ

H1 max

1

2ξ 1 ξ2–

-------------------------=

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

ρ Ωω---- Ω m

k----= =

H1

ξ 0.0=

ξ 0.2=

ξ 0.4=

H1 ρ ξ

ρmax 1 2ξ2–=


When ,

(1.34)

(1.35)

Since is usually small, the maximum dynamic response is significantly greaterthan the static response and is close to 1. For example, for whichcorresponds to a high level of damping, the peak response is

(1.36)

(1.37)

When the forcing frequency, , is close to the natural frequency, , theresponse is controlled by adjusting the damping. Outside of this region, dampinghas less influence, and has essentially no effect for and .

Differentiating twice with respect to time leads to the acceleration, ,

(1.38)

Noting eqn (1.26), the magnitude of can be written as

(1.39)

where

(1.40)

The variation of with for different damping ratios is shown in Fig. 1.8. Notethat the behavior of for small and large is opposite to . The maximumvalue of is the same as the maximum value for , but the location (i.e. thecorresponding value of ) is different. They are related to by

ξ2<< 1

ρmax 1≈

H1 max

12ξ------≈

ξρmax ξ 0.2=

H1 max2.55≈

ρmax 0.96=

Ω ω

ρ 0.4< ρ 1.6>

u a

a t( ) u t( ) Ω2u Ωt δ–( )sin– a Ωt δ–( )sin–= = =

a

apk--- Ω2H1

pm---- H2= =

H2 ρ2H1ρ4

1 ρ2–[ ]

22ξρ[ ] 2

+----------------------------------------------= =

H2 ρH2 ρ H1

H2 H1ρ ξ


(1.41)

(1.42)

The ratio is the acceleration the mass would experience if it wereunrestrained and subjected to a constant force of magnitude . One can interpret

as a modification factor which takes into account the time varying nature ofthe loading and the system restraints associated with stiffness and damping.


Once the system and loading are defined (i.e. , , , , and are

specified), one determines and computes the peak amplitudes using the

following relations

(1.43)

(1.44)

ρmax1

1 2ξ2–

----------------------=

H2 max

1

2ξ 1 ξ2–

-------------------------=

p m⁄p

H2

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

ρ Ωω---- Ω m

k----= =

H2

ξ 0.0=

ξ 0.2=

ξ 12

-------=

H2 ρ ξ

m k c p ΩH1

upk--- H1=

a Ω2u=


(1.45)

Note that for periodic response, the acceleration is related to the displacement bythe square of the forcing frequency. One can also work with instead of .

Design criteria

The design problem differs from analysis in that one starts with the mass of

the system, , and the loading characteristics, and , and determines and

such that the motion parameters, and , satisfy the specified criteria. In general,one has limits on both displacement and acceleration

(1.46)

(1.47)

where and are the target design values. In this case, since and arerelated by

(1.48)

one needs to determine which constraint controls. If , the accelerationlimit controls and the optimal solution will be

(1.49)

(1.50)

If, on the other hand, , the displacement limit controls. For this case, theoptimal solution satisfies

(1.51)

(1.52)

In what follows, both cases are illustrated.

H1 H1 Ω m k c, , ,( ) H1 ρ ξ,( )= =

H2 H1

m p Ω k c

u a

u u∗≤

a a∗≤

u∗ a∗ u a

a Ω2u=

a∗ Ω2u∗≤

u a∗

Ω2------- u∗<=

a a∗=

a∗ Ω2u∗≥

u u∗=

a Ω2u∗ a∗<=


Methodology for acceleration controlled design

One works with eqn (1.39). Expressing the target design acceleration as afunction of the gravitational acceleration,

(1.53)

and defining as

(1.54)

the design constraint takes the form

(1.55)

where is the weight of the system.

The totality of possible solutions is contained in the region below .Figure 1.9 illustrates the region for . For low damping, the intersection of

and the curve for a particular value of , , establishes twolimiting values, and . Permissible values of for thedamping ratio , are

a∗ fg=

H2∗

H2∗ a∗

p m⁄-----------=

H2 H2∗≤ W

p----- f=

W

H2 H2∗=

H2∗ 2=

H2 H2∗= H2 ξ ξ∗ρ ρ1 H2

∗ ξ∗,[ ] ρ 2 H2∗ ξ∗,[ ] ρ

ξ∗


Fig. 1.9: Possible values of

(1.56)

The second region does not exist when .

Noting eqn (1.40), the expressions for and are

(1.57)

These functions are plotted in Fig. 1.10 for representative values of .

The limiting values of for reduce to

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

ρ Ωω---- Ω m

k----= =

H2

ξ 0.0=

ξ ξ∗=

ξ 12

-------=

H2∗

possible solutions

ρ1 ρ2

H2 H2∗≤

0 ρ ρ1 H2∗ ξ∗,[ ]≤<

ρ ρ2 H2∗ ξ∗,[ ]≥

H2∗ 1<

ρ1 ρ2

ρ1 2,

1 2ξ∗ 2– 1 2ξ∗ 2

–[ ]2

1–1

H2∗

----------2

++−

1 1H2

∗----------

2–

----------------------------------------------------------------------------------------------=

ξ∗

ρ ξ∗ 0=


(1.58)

Fig. 1.10: Plot of and versus and

Noting eqn (1.29), one can express eqn (1.56) in terms of limiting values ofstiffness. By definition,

(1.59)

Then, letting

(1.60)

and noting that , the allowable ranges for are given by:

(1.61)

ρ1 2,1

1 1H2

∗----------±

--------------------=

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

H2

ξ∗ 0=ξ∗ 0.1=

ξ∗ 0.2=

H2∗

2=

ρ2

ρ1

ρ Ω mk----=

ρ1 ρ2 ξ∗ H2

k Ω2m

ρ2------------=

kjΩ2m

ρ j2

------------=

k2 k1< k

H2∗ 1< k1 k ∞< <


and (1.62)

Given , one specifies a value of , computes with eqn (1.57), andselects a value for which satisfies the above constraints on stiffness. Thedamping parameter is determined from

(1.63)

Example 1.4: An illustration of acceleration controlled design

Suppose and . Applying eqn (1.54) leads to .Figure 1.10 shows that damping has a negligible effect for this value of ; thedesign is essentially controlled by stiffness. Taking , and using eqn (1.58)results in

To illustrate the other extreme, and is considered.

Here, . The two allowable regions for k corresponding to different

values of are obtained by applying equations (1.57), (1.60), and (1.62).

ξ k1/Ω2m k2/Ω2m c1/Ω m c2/Ω m

0 1.5 0.5 0 0

0.1 1.439 0.521 0.24 0.144

0.2 1.231 0.610 0.444 0.312

H2∗ 1> 0 k k2< < k1 k ∞< <

H2∗ ξ ρ j

k

c 2ξωm 2ξ km= =

p 0.1W= a∗ 0.05g= H2∗ 0.5=H2

∗ξ 0=

ρ12 1

3---=

k 3Ω2m>

p 0.1W= a∗ 0.2g=

H2∗ 2.0=

ξ

k k1≥ k k2≤


Methodology for displacement controlled design

The starting point is eqn. (1.26). Noting equations (1.40) and (1.59), eqn(1.26) can be written as

(1.64)

Then, defining as

(1.65)

where is the target displacement, the design constraint is given by

(1.66)

The remaining steps are the same as for the previous case; the only difference is

the definition of . One applies equations (1.57) thru (1.63), using instead

of .

Example 1.5: An illustration of displacement controlled design

Suppose

Then

upk---H1

p

Ω2m------------ρ2H1

p

Ω2m------------H2= = =

H2**

H2** Ω2mu*

p------------------=

u*

H2 H2**<

H2** H2

**

H2∗

p 10kN=

u* 10cm=

m 1000kg=

H2** 1000( ) 0.1( )10000

-----------------------------Ω2 0.01( )Ω2= =


Various values for are considered.

1.

For this value of , the design is controlled by stiffness. Taking , and

using eqn (1.58),

The value of is constrained by

which corresponds to

2.

Since is greater than 1, there are 2 allowable regions for . Also these

regions depend on the damping ratio, . Results for different values of arelisted below. They are generated using equations (1.57), (1.60), and (1.62).

Ω

Ω 2π r s⁄=

H2** 0.394=

H2** ξ 0=

1

ρ12

------ 1 1

H2**

--------+ 3.538= =

k1Ω2m

ρ12

------------ 139.7kN m⁄= =

k

k k1 139.7kN m⁄=≥

ρ ρ1 0.532=≤

Ω 4π r/s=

H2** 1.576=

H2** k

ξ ξ

ρ ρ1< k k1≥


The solution for is radians out of phase with the forcing function.

Decreasing reduces the natural frequency, , and moves the system away from

the resonant zone, . For small , approaches unity, and the response

measures tend toward the following limits

(1.67)

Methodology for force controlled design

The previous examples dealt with limiting the displacement oracceleration of the single degree of freedom mass. Another design scenario isassociated with the concept of isolation, i.e., one wants to limit the internal forcethat is generated by the applied force and transmitted to the support. The reactionforce, , shown in Fig (1.6), is given by

(1.68)

Expressing as

(1.69)

and using equations (1.24) thru (1.31), the magnitude and phase shift are given by

ξ ρ1 k1(kN/m) c1 (kN*s/m) ρ2 k2(kN/m) c2 (kN*s/m)

0 0.782 258.1 0 1.654 57.72 0

0.1 0.795 249.6 3.16 1.627 59.63 1.54

0.2 0.839 224.1 5.98 1.541 66.5 3.26

ρ ρ2> k k2≤

k k2≤ π

k ωρ 1= k H2

apm----→ u

p

Ω2 m⋅----------------→

R

R p ma– ku cu+= =

R

R R Ωt δ δ1+–( )sin=


(1.70)

(1.71)

(1.72)


Figure 1.11 shows the variation of with and . At ,for all values of . When , the minimum value of corresponds to

, which implies that damping magnifies rather than decreases the responsein this region. The strategy for reducing the reaction is to take the stiffness as

(1.73)

Decreasing “softens” the system and reduces the internal force. However, thedisplacement and acceleration “increase”, and approach the limiting values givenby eqn (1.67)

R H3p=

H31 2ξρ[ ] 2

+

1 ρ2–[ ]

22ξρ[ ] 2

+----------------------------------------------=

δ1tan 2ξρ=

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

ξ* 0=

ξ* 0.2=

ξ* 0.4=

ρ Ω mk----=

H3

2

H3 ρ ξ

H3 ρ ξ ρ 2= H3 1=ξ ρ 2> H3

ξ 0=

k Ω2m 2⁄<

k


Example 1.6: Force reduction

Suppose

and the desired magnitude of the reaction is 10% of the applied force. Then,

Taking in eqn (1.71)

results in

Finally, the corresponding stiffness is

p 10kN=

m 1000kg=

Ω 4π r s⁄=

H3 0.10=

ξ 0=

H31

1 ρ2–------------------ 0.1= =

ρ 11 3.317= =

k Ω2m

ρ2------------ 14.36kN m⁄= =


1.4 Design of a single-degree-of-freedom system for support motion

Response for periodic support motion

Fig. 1.12: Single-degree-of-freedom system

The equation of motion for a SDOF system subjected to support motion is writtenas

(1.74)

where is now considered to be the relative displacement with respect to thesupport, and denotes the motion of the support. Assuming periodic excitation

(1.75)

the relative and total displacements are given by

(1.76)

(1.77)

where

(1.78)

(1.79)

The total acceleration is related to the support acceleration by a similar expression

(1.80)

k

c

m

u ug+ug

mu t( ) cu t( ) ku t( )+ + mug t( )–=

uug

ug t( ) ug Ωtsin=

u t( ) u Ωt δ–( )sin=

ut t( ) u t( ) ug t( )+ ut Ωt δ1 δ–+( )sin= =

u H2ug=

ut H3ug=

ut t( ) at t( ) at Ωt δ1 δ–+( )sin= =

1.4 Design of a Single-degree-of Freedom System for Support Motion 27

(1.81)

Design scenarios

There are a number of design scenarios for support motion. If the system issensitive to the total acceleration, one can limit the response by requiring to besmall with respect to unity. In effect, one isolates the mass from the supportmotion. From Fig. 1.11, must be greater than for isolation to be feasible. Thecorresponding constraint on stiffness is .

If the support motion is interpreted as seismic excitation, and the systemrepresents a structure, one wants to control the relative motion of the structureand the total acceleration of the equipment attached to the structure. Oneapproach is to isolate the structure so that the acceleration constraint is satisfied.The relative motion will be essentially equal to the ground motion, so anadditional mechanism is needed to localize the motion. An example of thisapproach is a rigid building supported by low stiffness springs; the motion in thisapproach is confined to the support springs. Another approach would be to workwith so as to satisfy the displacement constraint with stiffness and adjust the

damping to satisfy as close as possible the constraint on acceleration. Theexamples presented below illustrate both approaches. A detailed discussion ofisolation of building structures for seismic excitation is contained in Chapter 5.

Example 1.7: Controlling acceleration due to ground motion

This problem concerns selecting and such that the total acceleration is afraction of the support acceleration, say

(1.82)

Then from eqn (1.81)

(1.83)

As pointed out earlier, yields the minimum value of . Specializing eqn

at H3 Ω2ug–[ ] H3ag= =

H3

ρ 2k mΩ2 2⁄<

H2

ρ ξ

at ν ag= ν << 1

H3atag----- ν= =

ξ 0= H3


(1.71), one obtains

(1.84)

as the lower limit on . The stiffness limit follows from the definition of

(1.85)

For example, suppose . Then, and .

Example 1.8: Controlling relative motion due to ground motion

This example illustrates how to constrain the relative motion due to supportmovement. A typical scenario is a structure subjected to seismic excitation.Defining as the desired relative displacement, the appropriate equationfollows from eqn (1.78).

(1.86)

For this type of problem, is usually greater than unity. The procedure forsolving eqn (1.86) was discussed in the previous section. Considering the solutionzone corresponding to (See Fig. 1.9) and starting with leads to

(1.87)

(1.88)

as the first estimate. The change in and due to including damping can beevaluated using eqn (1.57).

ρ 1 1ν---+≥

ρ ρ

k m Ωρ----

2 mΩ2

1 1ν---+

------------≤=

ν 1 3⁄= ρ 2≥ k mΩ2

4------------≤

u∗

H2u∗ug------≤ H2

∗=

H2∗

ρ ρ2> ξ 0=

ρ2 ρ2 ξ 0=≈ 1

1 1H2

∗----------–

-------------------=

k ξ 0=mΩ2

ρ2 ξ 0=

2---------------------------=

ρ2 k

1.4 Design of a Single-degree-of Freedom System for Support Motion 29

To illustrate the process, the following numerical values are considered.

Applying eqns (1.86) through (1.88), the values of the various parameters are

For no damping, one has to take in order to satisfy the constraint onamplification.

To assess the sensitivity to damping, is taken as . Using eqn (1.57),

the corresponding value of is

The other parameters are determined with

Ω 4π rd/s= T 2πΩ------ 0.5s= =

u∗ 0.1m= ug 0.033m=

H2∗ 0.1

0.033------------- 3= =

ρ2 ξ 0=

1

1 13---–

------------ 1.22= =

T ξ 0=ρ2 ξ 0=

T 0.61s= =

k ξ 0=mΩ2

ρ2 ξ 0=

2--------------------------- m 4π

1.22----------

2105.27m= = =

k k ξ 0=≤

ξ∗ 0.1

ρ2

ρ2 ξ 0.1=1.18=

T ξ 0.1=ρ2 ξ 0.1=

T 0.59s= =


The solution is fairly insensitive to damping. However, if one uses the largerstiffness value, and the damping assumed in the calculation is not realized, theamplification will be greater than the design value, .

1.5 Stiffness distribution for a two degree-of-freedom system

The last two examples were concerned with how to determine the stiffness for aSDOF system subjected to dynamic loading when motion is the controllingdesign criterion. The procedure for a multi-degree-of-freedom (MDOF) system ismore involved since one has now to determine a set of stiffnesses rather than justa single stiffness. A two-degree-of-freedom system is considered here to illustratethe approach. The next chapter extends the approach to deal with continuousbeam type structures and discrete mass systems having both stiffness anddamping.

Consider the two degree-of-freedom system shown in Fig. 1.13. The SDOFnotation is extended to the MDOF case by including a subscript to indicate thecorresponding object. Enforcing equilibrium for the individual masses leads tothe following governing equations

(1.89)

(1.90)

Specializing these equations for free vibration, the solution is expressed as

(1.91)

(1.92)

k ξ 0.1=mΩ2

ρ2 ξ 0.1=

2------------------------------- 112.53m= =

H2∗

m1 u1 t( ) k1 k2+( )u1 t( ) k2u2 t( )–+ p1 t( )=

m2 u2 t( ) k2u2 t( ) k2u1

t( )–+ p2 t( )=

u1 t( ) Φ1 ωtsin=

u2 t( ) Φ2 ωtsin=

1.5 Stiffness Distribution for a Two Degree-of -freedom System 31

(1.93)

where , are the amplitudes and is a frequency measure. Requiring eqns(1.89) and (1.90) to be satisfied for yields the following equation for thefrequency,

(1.94)

where .

Fig. 1.13: Two-degree-of-freedom system

The displacement amplitudes are related by

(1.95)

There are 2 values of , and 2 corresponding sets of ’s. Each set of ’s defines adisplacement pattern which is called a mode shape. A detailed description of freevibration analysis for multi-degree-of-freedom systems is presented later inchapter 2.

When and are specified, one solves eqn (1.94) for and thendetermines the displacement amplitude ratio with eqn (1.95). The actual modeshape is defined only in a relative sense since is an arbitrary constant. Aninverse approach is followed in motion based design. One specifies the relativemode shape based on the desired deformation, and then determines the relativestiffness distribution. The details are presented below.

p1 t( ) p2 t( ) 0= =

Φj ωΦj 0≠

λ2 k1 k2+

m1-----------------–

k2m2-------– λ

k1k2m1m2---------------+ + 0=

λ ω2=

m1 m2

k2k1

u2u1

p1 p2

Φ2

k1 k2 λm1–+

k2---------------------------------- Φ1=

λ Φ Φ

k1 k2 λ

Φ1


The differences in displacement amplitudes can be interpreted as therelative displacements between the masses

(1.96)

(1.97)

For optimal motion behavior, the relative displacements should be equal.This constraint results in a linear mode shape.

(1.98)

(1.99)

Setting in eqn (1.95) and solving for yields

(1.100)

where an asterisk is used to indicate the value associated with a prescribedrelative mode shape. Introducing in eqn (1.94) leads to a relation betweenand

(1.101)

Finally, substituting for in eqn (1.100), results in a relation between thefundamental frequency (lowest) and :

(1.102)

Once is specified, the fundamental frequency and stiffness parameters areuniquely determined. One can also specify the desired fundamental frequencyand then use eqn (1.102) to determine the necessary value of . Given and

, eqn (1.94) can be solved for , and using this value, the second mode shapecan be established with eqn (1.95).

∆u1 Φ1=

∆u2 Φ2 Φ1–=

∆u1 ∆u2=

Φ2 2Φ1=

Φ2 2Φ1= λ

λ1∗

k1∗ k2

∗–

m1----------------------=

λ1∗ k1

∗k2∗

k1∗ 1

m12m2----------+ k2

∗=

k1∗

k2∗

λ1∗ ω1

2 k2∗

2m2----------= =

k2∗

k2∗ k1

k2 λ2

1.6 Control Strategies for Motion Based Design 33

Example 1.9: 2DOF system - with equal masses

The case where is considered to illustrate the procedure. Applyingeqn (1.101), is related to by

(1.103)

Suppose the design objective is to have (i.e. ). Then onesets in eqn (1.102) and solves for

(1.104)

Once is specified, the stiffness parameters can be evaluated. Substituting forthese parameters in eqn (1.94) leads to . The corresponding modeshape is .

1.6 Control strategies for motion based design

The previous sections dealt with the problem of limiting the response of a singledegree of freedom system subjected to periodic excitation. Although the problemconsidered was rather simple, the strategies used to generate appropriate valuesfor the system stiffness and damping are basic paradigms that are also applicableto more complex structures. In what follows, the SDOF strategies are reviewed toprovide the basis for establishing a general motion control methodology forstructural systems.

Consider the linear elastic SDOF system shown in Fig (1.14). Suppose the

design loading is static, and the motion criteria is . Enforcing equilibriumleads to an expression for the required stiffness.

(1.105)

m1 m2 m= =k1∗ k2

∗

k1∗ 3

2--- k2

∗=

T1 1s= ω1 2π rd/s=ω1 2π= k2

∗

k2∗ 2m 2π( )2=

mλ2 3k2

∗ m⁄=Φ2 0.5Φ1–=

u u*<

k pu*----- k*≡≥


From a stiffness perspective, control for static loading is realized by adjusting thestiffness such that k is greater than a certain limiting value which depends on thedesign criteria for displacement.

Fig. 1.14: Single degree of freedom model for passive and active control

Another way of expressing the motion control requirement is in terms ofenergy. In general, the work done by the external forces acting on a system isequal to the sum of the mechanical energy stored in the system and the energytransformed to another form, through either energy dissipation or absorptionmechanisms. This system is elastic, and the loading is static. It follows that thestored energy is equal to the strain energy, and since no energy is dissipated, thework done by the external loading must be balanced by the strain energy.Expanding these quantities leads to eqn (1.106).

(1.106)

From an energy perspective, static control is achieved by providing sufficientenergy storage capacity to satisfy the energy demand.

The design value of stiffness defined by eqn (1.105) is based on thehypothesis that the system properties (k,m) remain constant as the load is beingapplied, and there is no other agent which assists in resisting the load. Control isachieved solely by the action of the stiffness embedded within the system. Thistype of control action is called “passive control” since the system responds in apassive manner.

A different strategy for limiting the displacement is based on using anexternal energy source to satisfy some of the energy demand. In this approach, the

k

c

m

u ug+ ut=ug

p F

External Work Strain Energy≤12---pu* 1

2---k u*( )2≤ ⇒ p

u*----- k≤


energy requirement is taken as:

External work Strain energy + Supplied energy (1.107)

Energy can be supplied in different ways. For the SDOF system, a force actuatorcan be used to generate the force, F, shown in Fig (1.14). Force actuators aredevices which convert energy into a mechanical force, and provide the ability toimplement this control strategy. The work done by F is negative, and itsmagnitude represents the external energy that needs to be supplied.

(1.108)

Equation (1.107) expands to

(1.109)

and is follows that

(1.110)

One can also arrive at this result by enforcing equilibrium. Controlling motion bysupplying external energy is referred to as “active control”.

One of the key decisions in motion based design is the selection of thecontrol strategy. Fully passive static control is simple to implement, since itinvolves only providing stiffness. Active control employs actuator technologywhich is costly and less reliable. However, active control has considerablepotential, particularly for structural applications where weight is a critical issue,such as aerospace vehicles and long span bridges.

The concepts discussed above are also applicable for dynamic loading.When the loading is periodic, the response amplitudes are given by:

(1.111)

≤

WorkF

12---Fu*–=

12---pu* 1

2---k u*( )2 1

2---Fu*+≤

p ku* F+≤

up

Ω2m------------H2=

apm----H2=


where is plotted in Fig (1.15). Since civil structures are generally lightlydamped, a realistic upper limit for is 0.3. Assuming , , and are specified,one needs to select values for stiffness and damping that satisfy the design motioncriteria.

(1.112)

where the value of depends on whether displacement or acceleration is thelimiting motion constraint.

(1.113)

Fig. 1.15: Response function for periodic excitation

H2ξ p Ω m

H2 H2*<

H2*

H2* mΩ2u*

p------------------= displacement( )

H2* ma*

p---------= acceleration( )

0 0.5 1 1.5 2 2.5 3 3.5 40

0.5

1

1.5

2

2.5

3

3.5

4

H2

ρ Ωω---- Ω m

k----= =

Stiffnessdominatesresponse

Stiffnessdominatesresponse

Stiffness and dampingare coupled

ξ 0=

ξ 0.3=

H2*

H2*

H2*


Figure (1.15) shows that is essentially independent of damping forand . If , stiffness is the only parameter that can be

“adjusted” to satisfy the design constraint, and the design process is essentiallyquasi-static. For , both stiffness and damping are available ascontrol variables. One can generate a spectrum of designs by specifying stiffness

, and determining the corresponding damping that satisfies .When , there are 2 possible design zones, one for and the other for

. As increases, these zones merge into a single zone centered at ,the resonant state of the system. Damping dominates the response in theneighborhood of . The strategy employed for can also beapplied for . Given , one selects a stiffness value within the stiffnessrange defined by the specified value of , and the boundary curves and

, and then determines the required value of . For , thesecond stiffness range is outside . Designs for this region are controlled bystiffness, and are characterized by their low stiffness. Design for the first regionare controlled by both stiffness and damping, and have a higher stiffness.

The response for periodic ground excitation is given by

(1.114)

where is plotted in Fig (1.16). This function also relates the applied periodicforce and the corresponding reaction,

(1.115)

These equations are used to establish a control strategy for isolating a systemfrom an external action, either an applied loading or a support motion. In general,one wants for isolation. Defining as the design requirement, theallowable stiffness range is bounded by the curves for and , asindicated in Fig (1.16). The absolute upper limit on stiffness is

(1.116)

Designs are generated by specifying stiffness, and then determining thecorresponding value of . Isolation is achieved by reducing the stiffness belowthe critical level, . If , isolation is not possible even for the casewhere maximum damping is introduced.

H2*

ρ 0.5≈< ρ> 3.0≈ H2* < 0.5≈

0.5 H2* 1.0< <

ρ( ) ξ( ) H2 H2*=

H2* 1> ρ 1<

ρ 1> H2* ρ 1=

ρ 1= 0.5 H2* 1.0< <

H2* 1.0> H2

*

H2* ξ 0=

ξ 0.3= ξ 1.0 H2* 1.5≤ ≤

ρ 3.0=

ut H3ug=

H3

R H3p=

H3 1< H3*

ξ 0= ξ 0.3=

ρ 2 kmax⇒ Ω2m2

------------= =

ξkmax k kmax>

ξ 0.3=( )


Fig. 1.16: Response function for periodic ground excitation

Active control of a SDOF system for dynamic loading involves applying acontrol force, F(t ), and adjusting its magnitude over time in such a manner thatthe resulting motion is constrained within the desired limits. One needs a forceactuator that responds essentially in real time. Specifying the magnitude andsense of F(t ) is the key issue for active control. Various approaches are discussedlater in the text. In general, F is selected to oppose the motion and does negativework on the mass. This work has to be supplied by an external energy source.

Taking F to be proportional to the displacement and velocity,

(1.117)

corresponds to increasing the stiffness and damping of the original system. Thisobservation follows by substituting for F in the equilibrium equation. Then, activecontrol based on eqn (1.117) can be interpreted as introducing virtual stiffness and

0 0.5 1 1.5 2 2.5 3 3.5 40

0.5

1

1.5

2

2.5

3

3.5

4

H3

ρ Ωω---- Ω m

k----= =

ξ 0=

ξ 0=

ξ 0.3=

ξ 0.3=

H3*

F t( ) +kdu t( ) kvu t( )+=


damping.

From an energy perspective, stable active control reduces the energydemand for the system. The energy balance equation for a linear SDOF systeminitially at rest is given by

(1.118)

The last term is the energy dissipated through viscous action. Passive controlprovides energy storage and energy dissipation mechanisms to resist the demand.When F(t ) is taken to have the same sense as , active control decreases theenergy input to the system since the integral is always negative. The effect of thisaction is a lower energy demand that has to be met by passive controlmechanisms.

Selecting a motion control strategy for dynamic excitation involves anumber of decisions. Firstly, should the strategy be purely passive or acombination of passive and active control? Secondly, what percentage of theenergy demand should be met by energy storage (stiffness) vs. energy dissipation(damping)? Thirdly, the type and properties of passive energy dissipation deviceneeds to be specified. There are a number of passive energy dissipation devicesthat are appropriate for structural systems. Fourthly, if active control is used tosupplement passive control, the type and capacity of the force actuator needs tobe established.

Although the discussion has been focussed on a SDOF system, theconcepts introduced here are also applicable for a general multi degree offreedom structural system. The only difference is that now one is dealing withsets of stiffness and damping parameters, and motion criteria involving thedisplacement variables associated with the degrees of freedom. Using vectornotation, the problem can be stated as: given a desired displacement response

vector, , determine k, c, and F, the vectors containing the stiffness, damping,

and active control force variables. These vectors are functions when the system iscontinuous.

The desired response for a structure is related to the nature of the loadingand the critical performance measures chosen for the structure. For service

pu t Fu–( ) td0

t

∫+d0

t

∫ 12---mu2 1

2---ku2 cu2 td

0

t

∫+ +=

u

u* t( )


loading, the damage that non-structural elements, such as wall panels ofbuildings, can experience constrains the magnitude and distribution ofdisplacements, while human and equipment comfort limits the peak acceleration.The controlling criterion for wind dominant design tends to be the peak velocityand acceleration. Displacement is the controlling criterion for earthquakedominant designs. In general, the structure is required to remain elastic underservice loading.

Under extreme loading, structural performance and stability requirementsconstrain the magnitude and distribution of inelastic deformation that thestructural components can experience. Structural deformation is the key measurefor earthquake dominant design. Although design codes allow a structure toexperience inelastic deformation under an extreme earthquake with no collapseor loss of life, the current trend is to reduce the allowable inelastic deformation(Fajfar & Krawinkler 1997). This shift is driven by the need to lower the cost ofrepair. Furthermore, recent studies have shown that the initial cost of a structuredesigned to remain elastic under extreme seismic load may be less than the costcorresponding to the conventional design approach which allows for residualdeformation (Connor et al, 1997). Combining these savings with the decrease ofdamage repair and increased revenue due to maintaining operability can result ina substantial reduction in life cycle cost.

In order to satisfy the complete set of motion requirements, a multi-levelcontrol strategy is required. Figure (1.17) shows the strategy adapted for this text.The first step for passive control is to establish the distribution of stiffness, whichinvolves the choice of material stiffness and cross-sectional properties, byenforcing the requirements on the spatial distribution of displacementscorresponding to the design loading condition. Example 1.9 illustrates thiscomputation for a 2-DOF system. The ideal state is uniform deformationthroughout the structure. However, this state is possible only for a staticallydeterminate structure. The requirement on the magnitude of the response is metwith a combination of scaling the stiffness distribution and incorporating passiveenergy dissipation devices such as viscous dampers, tuned mass and liquidsloshing dampers, and base isolation systems. Depending on the level ofdamping chosen, damping and stiffness may be coupled. In this case, one cannotindependently scale stiffness and damping, and iteration on the scaling processand possibly also on the stiffness distribution may be required. Typical civil

structures are usually lightly damped, with , and coupling is not aξ 0.05=<


problem. However, the current trend is toward more heavily damped structures

with to 0.3. Seismic design for a heavily damped structure requires

iteration between stiffness and damping. The response for extreme loading iscontrolled by supplementing stiffness and damping with mechanisms that absorbenergy. An example of an absorption mechanism is a hysteretic damper. Thesedevices experience residual inelastic deformation, which is a measure of theenergy absorbed.

The first step for active control is to decide on a location for force actuatorswhich provides the capability for controlling the displacement response in thedesired manner. Controllability is a critical issue for active control. It relates to thesensitivity of the response to the control force magnitude. Given the distribution,the desired force magnitudes are established using a control algorithm, andinstructions then are sent to the actuators to deliver these forces. Examples offorce actuators are active mass drivers and active tendon systems. Active controlcan be utilized to supplement passive control for service loading. Since the levelof power, energy, and force required for active control of large scale civilstructures under extreme loading cannot be economically achieved with thepresent technology, passive control is the predominant choice for this loading.

Fig. 1.17: Multilevel control strategy

Load Functional Control Action

Level requirements Passive Active

Service -Control the spatialdistribution of dis-placements

-Control the magni-tude of the response

-Establish the spatialdistribution of stiffness

-Specify a distributionof damping (energy dis-sipation) devices-Scale the stiffness anddamping magnitudes

-Select an appropriatespatial distribution offorce actuators

-Establish the magni-tude of the controlforces-Activate the force actu-ators

Extreme -Control the distribu-tion and magnitudeof the response

-Include energy absorp-tion mechanisms

ξ 0.2≈


1.7 Scope of text

This text presents a systematic treatment of the concepts and computationalprocedures for motion control of civil structures. The material is organizedaccording to the nature of the control process. Part I concerns passive control, andincludes Chapters 2 thru 5. Methodologies for stiffness and damping basedcontrol are presented, and applied to typical building type structures. Part IIconsists of Chapters 6 thru 9. These chapters are intended to provide anintroduction to active control concepts and computational algorithms.Quasi staticcontrol is discussed first, since it is easier to deal with analytically. Classicaldynamic control algorithms are treated next.Their application to civil structures isillustrated with a series of computer simulation studies. A brief description of theindividual chapters is provided below.

Chapter 2 deals with optimizing the distribution of stiffness for beam andframe type structures. The chapter begins with an overview of the governingequations for beam type structures and then addresses the static problem. Amethod for distributing rigidity to attain the desired static deformation profiles ispresented. The method is extended to the dynamic case and an analyticformulation is developed for structural systems characterized by a dominantfundamental mode response. How to distribute stiffness in structures so as togenerate a fundamental mode shape with uniform shear and bendingdeformation profiles is the key issue. An iterative numerical scheme based on theresponse spectrum method is developed to incorporate the effect of the highermodes which get amplified as the slenderness of the structure increases. Resultsfor a set of representative buildings subjected to seismic excitation are presented.They provide an assessment of the effectiveness of controlling displacement withstiffness.

Chapter 3 examines the role of damping for motion control of dynamicallyexcited structures. The chapter starts by examining the response characteristics ofsingle-degree-of-freedom (SDOF) systems with various types of dampingmechanisms. The concept of equivalent viscous damping is introduced and isused to express both structural damping and hysteretic damping in terms of theirviscous counterpart. Numerical results are presented to evaluate the validity ofthis concept for SDOF systems. Multi-degree-of-freedom (MDOF) systems arediscussed next. A formulation is presented for the case where the dampingdistribution is taken proportional to the stiffness distribution generated inChapter 2. The effectiveness of damping under seismic excitation is assessed by

1.7 Scope of Text 43

numerical simulation. Examples illustrating the design of viscoelastic andhysteretic dampers are also included.

Chapter 4 assesses the effectiveness of tuned mass dampers (TMD) forcontrolling the motion of dynamically excited structures. The theory of TMDs forSDOF systems is developed and applied to a range of SDOF systems subjected toharmonic and seismic excitations. The theory is then extended to MDOF systems,and a procedure for designing a TMD to control a specific modal response isdeveloped and illustrated. Descriptions of existing implementations of tunedmass dampers in building structures are also presented.

Chapter 5 considers base isolated structures (BIS). Base isolation is acontrol strategy that modifies the stiffness distribution. A base isolation controlstrategy for SDOF systems is developed, and then extended to the MDOF case.The procedure is similar to the approach followed in Chapter 2. One distributesstiffness along the height of the building so that the fundamental modeexperiences uniform deformation, and selects the stiffness of the base isolationsystem such that the deformation in the isolator is bounded. The iterativeprocedure developed in Chapter 2 to account for the effect of the higher modes, ismodified to incorporate the effect of the base isolation system.

Chapter 6 provides an overview of active control. The architecture of anactive control system is viewed as having 3 main components: a monitoring systemwhich acquires data; a cognitive unit called a controller which processes theacquired date and decides on a course of action; and an actuator system composedof a set of physical devices that execute the instructions from the controller.Various types of controllers are described, and examples illustrating theapplication of particular type of control action called linear feedback control arepresented. Existing established actuator technologies are discussed inconsiderable detail. An overview of emerging actuator technologies based onadaptive materials is also included. Lastly, examples of existing structural controlsystems for large scale structures are presented.

Chapter 7 addresses quasi-static control, i.e., where the structural responseto applied loading can be approximated as static response. Since time dependenteffects are neglected, stiffness is the only quantity available for passive control.Active control combines stiffness with a set of pseudo-static control forces.Fundamental concepts such as observability, controllability, optimal control, andcomputational procedures are introduced in this chapter. Both continuous and


discrete physical systems are treated.

Chapter 8 considers dynamic feedback control of time-invariant multi-degree-of freedom structural systems. A combination of stiffness, damping, andtime dependent forces is used for motion control of dynamic systems. The state-space formulations of the governing equations for SDOF and MDOF systems areused to discuss stability, controllability, and observability aspects of dynamicallycontrolled systems. Continuous and discrete forms of the linear quadraticregulator (LQR) control algorithm are derived, and examples illustrating theirapplication to a set of shear beam type buildings are presented. The effect of timedelay in the stability of LQR control, and several other linear control algorithmsare also discussed.

Problems 45

Problems

Problem 1.1

Refer to example 1.2 and equation 1.11. Construct plots of , as

a function of the aspect ratio for the following ranges:

•

• from to

• from 3 to 8

Recommended design values of for a building are in the region 400 to 500.Comment on when “motion” rather than strength controls the design.

Problem 1.2

Refer to example 1.3. A typical value for the ratio of column depth to storyheight is 0.10. Using eqn (1.22), determine the value of for which theconstraint on displacement controls the design taking the following ranges for

and :

•

• from to

Problem 1.3

Refer to example 1.3. Suppose a mass, , is attached to the “infinitely stiff”girder. Assuming the columns have negligible mass, determine the expression forthe natural frequency, , for lateral vibration in terms of the material andgeometric properties of the frame.

H* u*⁄( )γ 1=

H d⁄( )

E 200 000MPa,=

σ* 200MPa 600MPa

H d⁄( )

H u∗⁄

H u*⁄( )c

σ *

Ec 200 000MPa,=

σ* 200MPa 600MPa

m

ω


Problem 1.4

Consider a SDOF system having and subjected to asinusoidal force with amplitude and frequency . Recommenddesign values for stiffness , and damping, , corresponding to the followinglimiting values for peak acceleration:

•

•

•

Problem 1.5

Consider a SDOF system with mass of . The system is to besubjected to a periodic loading having a magnitude of and frequencyhertz. Recommend design values for stiffness and damping corresponding to thefollowing limiting values of peak displacement:

•

•

•

Problem 1.6

Suppose the mass of a SDOF system is known, but the stiffness, , anddamping, , are unknown. Discuss how you would determine the stiffness byapplying a periodic loading for which both the frequency and magnitude can bevaried, and monitoring the response. How would you determine the dampingparameter, c?

Problem 1.7

Consider a SDOF system having . The system is to besubjected to a periodic force with a magnitude of and frequency .

m 1000kg=

p 500N= 2πr s⁄k c

a* 0.1m s2⁄=

a* 1m s2⁄=

a* 3m s2⁄=

5000kg4kN 2

u* 1mm=

u* 10mm=

u* 20mm=

kc

m 1000kg=

5kN 3Hz

Problems 47

Recommend design values for , and such that the magnitude of the reactionforce is less than:

•

•

Problem 1.8

A SDOF system having is to be subjected to the followingground motion:

meters

Recommend values for and for the following design values of peak relativedisplacement:

•

•

Problem 1.9

Recommend values for and for a SDOF system with mass ofsubjected to the following ground motion:

meters

The motion constraint is . What is the corresponding value for the peak

relative motion, ? Discuss how the response measures, and , vary as the

spring stiffness is reduced from the value you select.

Problem 1.10

Consider a SDOF system subjected to ground excitation. Takeand the ground acceleration as

k c

10kN

3kN

m 1000kg=

ug 0.3 4πtsin=

k c

u* 0.2m=

u* 0.4m=

k c 2000kg

ug 0.5 3πtsin=

ut 0.1m≤u ut u

m 3000kg=


Recommend values for and such that the peak total acceleration is less than

.

Problem 1.11

A machine represented by the mass is to be supported by the spring anddashpot shown in the following figure. The machine is sensitive to totalacceleration and therefore needs to be “isolated” form the ground motion.

Consider the ground acceleration to consist of two dominant components

where

is the gravitational acceleration ( ) and are random phase angles that can range from 0 to

A reasonable approximation for the peak acceleration of the combined response is

where and are the total accelerations due to the individual harmonic

ag 4.0 4πtsin=

k c

1.0m s2⁄

m

m

ck

u+ug

ug

m=2000 kg

agg----- 0.1 2πt δ1+( )sin 0.2 4πt δ2+( )sin+=

g 9.81m s2⁄δ1 δ2, 2π

at at 1,( )2 at 2,( )2+=

at 1, at 2,

Problems 49

excitations with random phasing.

Suppose the desired maximum total acceleration is . Describe how

you would establish design values for and .

Problem 1.12

Consider the 2 DOF system discussed in Section 1.5. Take .Determine the stiffness parameters and corresponding to and

the following range of :

Also determine and the shape of the second mode corresponding to each

stiffness distribution.

Problem 1.13

The derivation in Section 1.5 was based on specifying the fundamentalmode to be linear. Suppose one takes where is a scale factor.

Generalize eqn (1.101) to incorporate . What is the significance of taking ?What happens when is taken to be negative?

0.05g

k c

m2 0.5m1=k k m1 1000kg=

ω1

ω1 π 2π 4π,,=

ω2

ϕ2 αϕ 1= αα α 2≠

α


Problems 149

Problems

Problem 2.1

Determine the transverse shear and bending deformations correspondingto the following displacement distributions

a)

b)

c)

d)

Problem 2.2

Determine the shear and bending rigidity coefficients for the cross sectionshown below. The dimensions are in centimeters.

u nπx2L

----------sin=

β 0=

n 1 3 …, ,=

u x2 a1x3 a2x4+ +=

β 2x 3a1x2 4a2x3+ +=

u a1x a2x2 a3x3++=

β 2a2x 3a3x2+=

u a1 x x2

2L------–

a2x2

2----- x4

12L2------------–

+=

β a2 x x3

3H2----------–

=

150 Chapter 2: Optimal Stiffness Distribution

Problem 2.3

Consider the chevron bracing scheme shown below. Determine anexpression for . Assume the members carry only axial force.

Problem 2.4

The force-displacement relationship for the structures shown below iswritten as .

face

core

E f 210 000 MPa,=

Gc 2 000 MPa,=

25

1

50

1

DT

AdAd

αα

B

h

P ku=

Problems 151

Establish the expression for for each structure. Comment on the relativeefficiency of the structures for resisting lateral loading.

Problem 2.5

Consider a five story five bay rigid frame modeled as a shear beam.Assume all the columns in a story have the same properties, but allow forvariation in column size over the stories. Establish an approximate expression forthe equivalent shear beam stiffness for a typical story.

Problem 2.6

Diagonal bracing is added to a rigid frame as indicated in the figure below.

B

h Ic

Ib

Ad

P u, P u,

B

h

B

h Ic

Ib P u,

a) b) c)

k

hH

i

B


a) Establish an approximate expression for the equivalent shear beamstiffness for a typical story. Assume the column properties are constant in a story,but vary from story to story.

b) Suppose the diagonal bracing system is expected to carry a specifiedfraction of the total shear in a story. Discuss how you would select the cross-sectional area of the diagonal braces and moment of inertia for the columns in atypical story.

Problem 2.7

Determine the expressions for the distribution of shear and bendingrigidities corresponding to constant deformations and .

hH

i

B

γ∗ χ∗

Hx

b0

b b0πxH------sin=

Problems 153

Problem 2.8

Determine the shear and bending rigidity distributions. Take:

Note: these values correspond to (see eqn 2.14).

Problem 2.9

A particular steel having an allowable stress of 600 MPa and Young’smodulus equal to 210,000 MPa has been selected as the “design” material.Assuming the material is to be used in bracing members, for what range oftransverse shear strains will the design be controlled by motion constraints ratherthan strengths?

Problem 2.10

a) Refer to problem 2.5. Consider a uniform lateral loading of 30 kN/m, astory height of 4m, and a design shear strain, , equal to 1/300 for each story.Estimate the value of for each story assuming steel is selected as the material.

b) Refer to problem 2.6. Use the loading defined in part a. Select the

γ∗ 1400---------=

χ∗ 120 000,------------------=

s 0.1=

10m

1kN/m10kN

45°

γ∗Ic


bracing properties such that the bracing carries 25% of the total story shear.

Problem 2.11

Determine the member stiffness factors for the following prescribedloading and displacement quantities.

Problem 2.12

The design objective is to determine the member stiffness factors for the

3m

4m1

P1 u1,

P2 u2,

2

P1 100 kN= P2 50 kN=

u1 0.01 m= u2 0.005 m=

3m

4m1

P1 u1,

P2 u2,

2

4m

3

Problems 155

above truss such that the nodal displacements corresponding to ,are , . Generate solutions using the least

square and mean value least square approaches.

Problem 2.13

Recommend stiffness factors for the truss shown above based in thefollowing requirements

Problem 2.14

Consider two cantilever beams coupled with a rigid link at the top.

P1 25 kN=P2 50 kN= u1 0.01 m= u2 0.01 m=

4m 1

P1 u1,

P2 u2,

2

4m

3

4

5

P3 u3,

4m

u1 0.0087m u2– 0.0012m u3 0.0019m–= = =

P1 100kN P2– 50kN P3 100kN–= = =

1 2

P

H

u∗ u∗

αP 1 α–( )P

Beam 1 is a shear beam.Beam 2 is a bending beam.


Assume beam 1 is a shear beam and beam 2 is a bending beam.

Determine the optimal rigidity distributions that satisfy the displacementrequirement and divide the lateral load between the two beams asindicated.

Problem 2.15

The bending beam model is based on the assumption of no transverseshear strain, i.e. . Theoretically, one sets , which results in .The displacement distributions for periodic vibration with constant curvature,

, and are

Consider uniform mass density. Determine for this prescribedmode shape.

Problem 2.16

Consider a cantilever beam having the mass distribution indicated in thefigure. Assuming the beam acts as a shear beam, determine the transverse shear

u∗ H β⁄=

γ 0= DT ∞= s ∞=

χ∗ γ 0=

u qBxH----

2=

β qB2xH2-------=

qBH2

2-------χ∗ ω1t δ+( )cos=

DB x( )

Hx

m1 aρmH=

ρm ρm 1 bxH------–

=

Problems 157

rigidity distribution required in order that the fundamental mode shape have thefollowing form:

Investigate how shear rigidity varies with , taking .

Problem 2.17

The structure shown below consists of a cantilever beam and a rigidweightless link connecting 2 masses to the top end of the beam. Assume thecantilever beam has uniform mass density and negligible transverse sheardeformation.

a) Determine the bending rigidity distribution required to produce thefundamental mode shape corresponding to constant curvature. The displacementexpressions for this case are:

b) Investigate the sensitivity of the solution to a variation of .

φ x( ) xH----=

a b 0=

ρm

b b

H

m1 m1m1 ρmH( )a=

u qBxH----

2=

β qB2xH2-------=

qBH2

2-------χ∗ ω1t δ+( )cos=

a


c) Suppose outriggers are attached to the rigid links and the beam isextended as shown below. Determine the bending rigidity distribution for thecase where a uniform static loading is applied, and the design objective isconstant curvature.

d) Repeat part a using the structure defined in part c.

Problem 2.18

Consider a cantilever shear beam having uniform mass density andconstant transverse shear rigidity. Allowing for material damping and externaldamping, the equilibrium equation for free vibration has the following form (seeeqns (2.197) and (2.199)):

(1)

The boundary conditions are

(2)

(3)

Substituting for in eqn (3) leads to

(4)

The general solution of eqn (1) which satisfies the boundary conditions at is

(5)

m1 m1

AA

H/2

DTu,xx CTu,xxt ρmu,tt– C ′u,t–+ 0=

u x 0=( ) 0=

V x L=( ) 0=

V

DTu,x L( ) CTu,xt L( )+ 0=

x 0 L,=

u eα t λxsin=

Problems 159

Let

(6)

(7)

a) Derive the expression for in terms of and . Assume . Thesolution corresponding to is the fundamental solution. Comment on thenature of this solution.

b) Compare the mode shapes with the profiles based on constant sheardeformation for the fundamental mode. Consider from 1 to 3.

c) Let be the value of corresponding to , i.e., the ‘th mode.Comment on how varies with .

Problem 2.19

Refer to the figure in problem 2.5. Assume the structure is a discrete shearbeam. Determine the distribution of equivalent shear beam stiffness based on alinear profile for the fundamental mode and the following floor masses:

Relate the shear beam stiffness to the column size for each story.

Problem 2.20

Repeat problem 2.19 using the structure defined in problem 2.6. Assumethe bracing contributes 25% of the story stiffness.

λ 2n 1–2L

---------------π n 1 2 …, ,= =

ω λDTρm--------

1 2/

=

λ2CT C ′+

ρm-------------------------- 2ξω=

α ω ξ ξ 1<n 1=

n

ξn ξ λ λ n= nξn n

m1 m2 m3 m4 10 000 kg,= = = =

m5 5 000 kg,=


Problem 2.21

A single degree of freedom system is to be designed to displace a givenamount, , under a specified seismic excitation. Use the response spectrumfor the pseudo spectral velocity shown in the figure, and take kg.Determine design values for and for a range of damping coefficients( ) and two values of :

.

Also determine the maximum values of the stiffness and damper forces.

m

k c ug

u ug+

umaxm 10 000,=

k cξ 0.05 0.10 0.20, ,= umax

umax = 0.1 m, 0.2 m

0.1 0.6 1.0 10

0.1

1.0

Period

Spectral Velocity

T (sec)

Sv (m/s)

Sv 1.4 ξ, 0.05= =

Sv 1.0 ξ, 0.1= =

Sv 0.8 ξ, 0.2= =

Problems 161

Problem 2.22

The structure consists of 3 shear beam segments and the lumped massesindicated in the figure. Take

.

a) Determine the shear rigidity distribution required for the fundamentalmode vector to have the form

b) Assume a periodic force, is applied at node 3. Calibratethe rigidity for the following conditions:

c) Calibrate the shear rigidity factors for the following conditions:

Use equations (2.249) - (2.251).

d) Assume linear viscous dampers having the same property,, are installed. Determine based on the results of part c.

h

h

h

p3 u3,

p2 u2,

p1 u1,m

m

m 2⁄

h 4 meters= m 1 000 kg,=

Φ 13--- 2

3--- 1, ,

=

p p Ωt( )sin=

p 10 kN= Ω 4π= qmaxh

150---------= ξ 0.05=

qmaxh

100---------= Svmax

1.2 m/s= ξ 0.02=

c1 c2 c3 c= = = c


Problem 2.23

Refer to problem 2.19.

a) Calibrate the stiffness for the following seismic excitation criteria:

b) Assume viscous dampers are installed in the third, fourth, and fifthstories. Take the damper coefficient to be the same for each story. Determine thecoefficient required by the calibration specified in part a.

Problem 2.24


a) Calibrate the structure for the following seismic criteria:

b) Assume a uniform distribution of damping over the height. Determinethe damping coefficient required by part a.

c) Suppose a single damper is connected to the top node (node 5).Determine the value of required by part a.

Problem 2.25

Refer to problem 2.15. Take and .

a) Determine and .

b) Assume the following loading

where

u H( )max 0.125 m=

Svmax1.0 m/s= ξ 0.05=

u H( )max 0.125 m=

Svmax1.2 m/s= ξ 0.02=

c

H 200 m= ρm 10 000 kg/m,=

m Γ

b b0x Ωtsin

H-------------------= b0 6 kN/m=

Problems 163

is applied. Calibrate the stiffness based on limiting to 0.5 m for a frequencyof 1 radian per second.

c) Suppose the design is constrained by the peak acceleration atunder periodic loading. Calibrate the stiffness for where

. Consider to be equal to 0.05 and use the loading defined in partb.

d) Assume the stiffness calibration is dictated by seismic excitation. Take, , . Determine . Assume

where is a constant. Determine corresponding to.

Problem 2.26


a) Determine and in terms of , , and .

b) Assume a uniform periodic loading is applied to the beam. Calibrate theshear rigidity for the following specifications.

c) Calibrate the rigidity for the case of seismic loading. Take

Use eqn (2.249).

u H( )

x H=u H( )max 0.015 g=

g 9.86 m/s2= ξ

u H( )max 1 m= Svmax1 m/s= ξ 0.05= DB x( )

CT x( ) αDB x( )= α αξ 0.05=

m Γ ρm a b

H 10 m= u H( )max H 300⁄=

ρm 100 kg/m=

a 1= b 0.5=

Ω 2π rad/s=

u H( )max H 300⁄=

Svmax1.2 m/s=

ξ 0.02=


Problem 2.27

Refer to problem 2.17. Determine and .

Problem 2.28

Consider eqn (2.249). Let where is a design parameter.Express as where is the typical story height and is the number ofstories. This leads to . Estimate a typical value of .

Problem 2.29

This problem concerns the preliminary design of the lateral stiffnesssystem for a 10 story rectangular rigid frame. The frame properties and designcriteria are as follows:

• Height = 5 m/story• Width = 10 m/bay• Mass/floor = 10,000 kg• Seismic loading: = 1.2 m/s, = 0.02

• Max. Deflection at top = 0.25 m

m Γ

q∗ H α⁄= αH H nh= h n

T fn= f

10 m

5 m

Sv ξ

Problems 165

• = 1/200

Model the frame as a 10 DOF shear beam.

a) Determine an initial estimate for the lateral stiffness at each story levelconsidering a single mode response and . Use the spectrum shownbelow.

b) Consider linear viscous dampers to be installed at each story level. Takefor story . Using the stiffness distribution established in Part a

(frequencies and mode shapes), establish the modal equations for a modaldecomposition in terms of the first 3 modes. Determine such that .Determine and corresponding to this value of . Estimate the maximum‘‘mean square’’ deformation response using the response spectrum shown below,and modify the initial stiffness distribution using eqn (2.252). Carry out thiscomputation for 2 iterations.

c) Suppose the lateral stiffness is maintained constant over 2 story heights,i.e.

, etc.

Discuss how you would determine initial estimates forconsidering a single mode response and . Also discuss how you wouldmodify the stiffness distribution.

γmax

ξ 0.02=

ci αki= i

α ξ 1 0.02=ξ2 ξ3 α

k2 k1=

k4 k3=

k1 k3 …, ,ξ 0.02=

0.1 0.6 1.0 10

0.1

1.0

Period

Spectral Velocity

T (sec)

Sv (m/s)

Sv 0.92 ξ, 0.05= =

Sv 0.8 ξ, 0.1= =

Sv 0.6 ξ, 0.2= =

Sv = 1.2, ξ = 0.02


167

Chapter 3

Optimal passive damping distribution

3.1 Introduction

Damping is the process by which physical systems such as structures dissipateand absorb the energy input from external excitations. Damping reduces thebuild-up of strain energy and the system response, especially near resonanceconditions where damping governs the response. Figure 3.1 illustrates theinfluence of damping on the time history of the strain energy for a system withmass of 1 kg and a period of 1 second subjected to the unscaled El Centroaccelerogram shown in Fig. 2.21. The symbols in Fig. 3.1 refer to the energy input( ), the energy stored ( ), and the energy dissipated ( ). During the earlystage of the response, there is a rapid build-up of the input energy, similar to animpulsive loading. Damping dissipates energy over a response cycle, in this case,1 second. For low damping ratio, the energy dissipated per cycle is small, andmany cycles are required before the input energy is eventually dissipated. As isincreased, the energy dissipated per cycle increases, and the stored energy buildup is reduced. Shifting from = 0.02 to = 0.1 reduces the peak stored energydemand by a factor of 3.7 for this particular system and earthquake excitation. Itshould be noted that seismic accelerograms differ with respect to their frequencycontent and intensity, and therefore one needs to carry out energy time historystudies for individual excitations applied to a specific system. For example, Fig3.2 shows the response of the same system for a typical Northridge accelerogram.The input energy build up for the Northridge loading is quite different than forthe El Centro loading.

EI Es ED

ξ

ξ ξ

168 Chapter 3: Optimal Passive Damping Distribution

Fig. 3.1: Energy Build Up, El Centro (S00E), Imperial Valley 1940

0 2 4 6 8 10 12 14 16 18 20

0.2

0.4

0.6

0.8

1.0

0

0 2 4 6 8 10 12 14

0.2

0.4

0.6

0.8

0

0 2 4 6 8 10 12 14 16 18 20

0.2

0.4

0.6

0.7

0

Time (s)

Ene

rgy

(J)

Ene

rgy

(J)

Ene

rgy

(J)

ξ 2%=

ξ 5%=

ξ 10%=

16 18 20

EI

ED

ES

3.1 Introduction 169

Fig. 3.2: Energy Build Up, Arleta Station (90 DEG), Northridge 1994, = 2%

Dissipation and absorption are attributed to a number of external andinternal mechanisms, including the following:

• Energy dissipation due to the viscosity of the material. This processdepends on the time rate of change of the deformations, and is referredto as material damping. Viscoelastic materials belong to this category.

• Energy dissipation and absorption caused by the material undergoingcyclic inelastic deformation and ending up with some residualdeformation. The cyclic inelastic deformation path forms a hysteresisloop which correspond to energy dissipation; the residual deformationis a measure of the energy absorption. This process is generally termedhysteretic damping.

• Energy dissipation associated with overcoming the friction betweenmoving bodies in contact, such as flexible connections. Coulombdamping refers to the case where the magnitude of the friction force isconstant. Structural damping is a more general friction dampingmechanism which allows for a variable magnitude of the friction force.

• Energy dissipation resulting from the interaction of the structure withits surrounding environment. Relative motion of the structure

EI

ED

ES

0 2 4 6 8 10 12 14 16 18 200

Time (s)

0.2

0.4

0.6

0.8

1.0

Ene

rgy

(J)

ξ


generates forces which oppose the motion and extract energy from thestructure. Fluid-structure interaction is a typical case. The fluid exerts adrag force which depends on the relative velocity and functions as anequivalent viscous damping force.

• Damping devices installed at discrete locations in structures tosupplement their natural energy dissipation/absorption capabilities.These mechanisms may be passive or active. Passive mechanismsrequire no external energy, whereas active mechanisms cannot functionwithout an external source of energy. Passive devices include viscous,friction, tuned mass, and liquid sloshing dampers. Active damping isachieved by applying external control forces to the structure overdiscrete time intervals. The magnitudes of the control forces areadjusted at each time point according to a control algorithm.

• Passive damping removes energy from the response, and therefore cannot cause the response to become unstable. Since active control involvesan external source of energy, there is the potential for introducing aninstability in the system. The term “semi-active” refers to a particularclass of active devices that require a relatively small amount of externalenergy and apply the control force in such a way that the resultingmotion is always stable. Chapter 6 discusses active control devices.

In this chapter, the response characteristics for material, hysteretic, andfriction damping mechanisms are examined for a single degree-of-freedom(SDOF) system. The concept of equivalent viscous damping is introduced and isused to express viscoelastic, structural, and hysteretic damping in terms of theirequivalent viscous counterpart. Numerical simulations are presented todemonstrate the validity of this concept for SDOF systems subjected to seismicexcitation. This introductory material is followed by a discussion of the influenceof distributed viscous damping on the deformation profiles of multi-degree-of-freedom (MDOF) systems. The damping distribution is initially taken to beproportional to the converged stiffness distribution generated in the previouschapter, and then modified to allow for non-proportional damping. Numericalresults and deformation profiles for a range of structures subjected to seismicloading are presented, and the adequacy of this approach for distributingdamping is assessed.

Distributed passive damping can be supplemented with one or morediscrete damping devices to improve the response profile. Discrete viscous

3.2 Viscous,Frictional, and Hysteretic Damping 171

dampers inserted in discrete shear beam type structures are considered in thischapter; the basic theory for tuned mass dampers is presented in the next chapter.Subsequent chapters deal with base isolation, a form of passive stiffness/damping control, and active control.

3.2 Viscous, frictional, and hysteretic damping

Viscous damping

Viscous damping is defined as the energy dissipation mechanism where thedamping force is a function of the time rate of change of the correspondingdisplacement measure:

(3.1)

where is the damping force and is the velocity in the direction of . Thelinearized form is written as:

(3.2)

where , the damping coefficient, is a property of the damping device. Linearviscous damping is convenient to deal with mathematically and therefore is thepreferred way of representing energy dissipation.

In general, the work done on the device during the time interval, is given by

(3.3)

Considering periodic excitation

(3.4)

and evaluating eqn (3.3) for one full cycle under linear viscous damping leads to

(3.5)

This term represents the energy dissipated per cycle by the damping device, asthe system to which it is attached undergoes a periodic motion of amplitudeand frequency . Figure 3.3 shows the force-displacement relationship forperiodic excitation; the enclosed area represents .

F f u( )=

F u F

F cu=

c

Wt1 t2,[ ]

W F ud

u t1( )

u t2( )

∫ Fu td

t1

t2

∫= =

u u Ωtsin=

Wviscous cπΩu2

=

uΩ

W


Fig. 3.3: Viscous response - periodic excitation

Example 3.1: Viscous damper

Figure 3.4 shows a possible design for a viscous damping device. The gapbetween the plunger and external plates is filled with a linear viscous fluidcharacterized by

(3.6)

where and are the shearing stress and strain measures respectively and isthe viscosity coefficient. Assuming no slip between the fluid and plunger, theshear strain is related to the plunger motion by

(3.7)

Fig. 3.4: Viscous damping device

t 0=

t πΩ----=

u u Ωtsin=

uuu–

cΩu

F

τ Gvγ=

τ γ Gv

γ utd-----=

td

td

L

F , u

Side View End View

w

fluid

fluid

plunger


where is the thickness of the viscous layer. Letting and represent theinitial wetted length and width of the plunger respectively, the damping force isequal to

(3.8)

Substituting for results in

(3.9)

Finally, eqn (3.9) is written as

(3.10)

where represents the viscous coefficient of the device,

(3.11)

The design parameters are the geometric measures and the fluid viscosity,.

A schematic diagram of a typical viscous damper employed for structuralapplications is contained in Fig 3.5; an actual damper is shown in Fig 3.6. Fluid isforced through orifices located in the piston head as the piston rod position ischanged, creating a resisting force which depends on the velocity of the rod. Thedamping coefficient can be varied by adjusting the control valve. Variabledamping devices are useful for active control. Section 6.4 contains a description ofa particular variable damping device that is used as a semi-active force actuator.This chapter considers only passive damping, i.e. a fixed damping coefficient.

Fig. 3.5: Schematic diagram - viscous damper

td L w

F 2wLτ=

τ

F2wLGv

td------------------- u=

F cu=

c

c 2wLtd

----------- Gv=

w L td, ,Gv


Fig. 3.6: Viscous damper - 450 kN capacity(Taylor Devices Inc. http://www.taylordevices.com)

Equation (3.5) shows that the energy loss per cycle for viscous dampingdepends on the frequency of the excitation. This dependency is at variance withobservations for real structural systems which indicate that the energy loss percycle tends to be independent of the frequency. In what follows, a number ofdamping models which exhibit the latter property are presented.

Friction damping

Coulomb damping is characterized by a damping force which is in phase with thedeformation rate and has constant magnitude. Mathematically, the force can beexpressed as

(3.12)

where denotes the sign of . Figure 3.7 shows the variation of withfor periodic excitation. The work per cycle is the area enclosed by the responsecurve

F F u( )sgn=

u( )sgn u F u


Fig. 3.7: Coulomb damping force versus displacement

(3.13)

Figure 3.8 shows a coulomb friction damper used with diagonal X bracingin structures. Friction pads are inserted at the bolt-plate connections. Interstorydisplacement results in relative rotation at the connections, and the energydissipated is equal to the work done by the frictional moments during thisrelative rotation.

Fig. 3.8: Friction brace damper

Structural damping removes the restriction on the magnitude of thedamping force, and considers the force to be proportional to the displacementamplitude. The definition equation for this friction model has the form

(3.14)

where is a pseudo-stiffness factor. Figure 3.9 shows the corresponding cyclicresponse path. The energy dissipated per cycle is equal to

(3.15)

uu

u–

F

Fu u Ωtsin=u u=

Wcoulomb 4Fu=

F ks u u( )sgn=

ks

Wstructural 4ksu2

2-----------

2ksu2= =


Fig. 3.9: Structural damping force versus displacement

Hysteretic damping

Hysteretic damping is due to the inelastic deformation of the material composingthe device. The form of the damping force-deformation relationship depends onthe stress-strain relationship for the material and the make-up of the device.Figure 3.10 illustrates the response path for the case where the material force-deformation relationship is elastic-perfectly plastic.

The limiting values are , the yield force, and , the displacement atwhich the material starts to yield; is the elastic damper stiffness. The ratio ofthe maximum displacement to the yield displacement is referred to as theductility ratio and is denoted by . With these definitions, the work per cycle forhysteretic damping has the form

(3.16)

uuu–

Ft π

2Ω-------=

ks

u u Ωtsin=u u=

Fy uykh

µ

Whysteretic 4khuy2 µ 1–[ ] 4Fyu µ 1–

µ------------= =


Fig. 3.10: Hysteretic damping force versus displacement

Figure 3.11 shows a bracing element that functions as an hysteretic damper(Watanabe, 1998). The element is composed of a core member fabricated withhighly ductile low strength steel (yield strength of 200 MPa., maximum percentstrain of 60%), a cylindrical jacket, and mortar placed between the core memberand the jacket. The jacket functions as an additional bending element, and itscross sectional moment of inertia is selected such that the buckling load is equal tothe yield force. This design feature allows the brace to be used for both tensile andcompressive loading. Triangular plate hysteretic dampers that dissipate energythrough bending action have also been used for buildings (Tsai, 1993 and 1998).

Fig. 3.11: Hysteretic damper brace element (Watanabe, 1998)

Example 3.2: Stiffness of a rod hysteretic damper

Consider a damping device consisting of a cylindrical rod of length and area .

u

kh

uuy

u–

Fy

Fu u Ωtsin=u u=

µ uuy-----=

L A


Suppose the material is elastic-perfectly plastic, as shown in Fig. 3.12.

The relevant terms are

(3.17)

(3.18)

Fig. 3.12: Elastic-perfectly plastic damper device.

Then,

(3.19)

(3.20)

(3.21)

Example 3.3: Stiffness of two hysteretic dampers in series

The device treated in Example 3.2 is modified by adding a second rod in series, asshown in Fig. 3.13. The yield force for the second rod is assumed to be greaterthan the yield force for the first rod,

(3.22)

Since the force is the same for both devices, the total elastic displacement is thesum of the individual contributions.

ε uL---=

F Aσ=

εεy

σy

E

σ

F , u

L

F

uy Lεy=

Fy Aσy khuy= =

khAEL

--------=

A2σy 2, A1σy 1, Fy 1,≡>

3.3 Viscoelastic Material Damping 179

Fig. 3.13: Two rod hysteretic damping device.

(3.23)

Specializing eqn (3.23) for the onset of yielding, one obtains

(3.24)

(3.25)

When two elements are used, one can vary both the yield force, , and the elasticyield deformation . The energy dissipation increases with decreasing , for agiven deformation amplitude .

3.3 Viscoelastic material damping

A material is considered to be elastic when the stresses due to an excitation areunique functions of the associated deformation. Similarly, a material is said to beviscous when the stress state depends only on the deformation rates. For simpleshear, these definitions translate to

Elastic

(3.26)

Viscous

(3.27)

The stress-deformation paths for periodic strain are illustrated in Figs 3.14(a) and

F F , u

A1, E1 A2, E2

L1 L2

uL1

A1E1-------------

L2A2E2-------------+ F 1

kh----- F= =

uyFy 1,kh

----------- L1εy 1, 1A1L2E1A2L1E2--------------------+= =

kh

A1E1L1

------------- 1

1A1L2E1A2L1E2--------------------+

-----------------------------=

Fyuy uy

u

τ Geγ=

τ Gvγ=


3.14(b). There is no time lag between stress and strain for elastic behavior,whereas the stress is radians out of phase with the strain for viscousbehavior. If these relations are linearly combined, one obtains the path shown inFig. 3.14(c).

Fig. 3.14: Stress-deformation relations.

Materials that behave similar to Fig. 3.14(c) are called viscoelastic. Theproperties of a linear viscoelastic material are determined by applying a periodicexcitation and observing the response, which involves both an amplification anda phase shift. The basic relations are expressed as

(3.28)

(3.29)

where is the storage modulus and is the loss modulus. The ratio of the lossmodulus to the storage modulus is defined as the loss factor,

(3.30)

An alternate form for eqn (3.29) is

(3.31)

(3.32)

The angle is the phase shift between stress and strain. Delta ranges from forelastic behavior to for pure viscous behavior.

Experimental observations show that the material properties and

π 2⁄

γ γγγ

Geγ

τ ττγ γ Ωtsin=

γ

GvΩγ GvΩγ

(a) Elastic (b) Viscous (c) Viscoelastic

γ γ Ωtsin=

τ γ Gs Ωtsin Gl Ωtcos+[ ]=

Gs Glη

ηGlGs------ δtan= =

τ γG Ωt δ+( )sin=

G Gs2 Gl

2+ Gs 1 η2

+= =

δ 0π 2⁄

Gs η


vary with temperature and the excitation frequency. Figure 3.15 illustrates thesetrends for ISD110, a 3M product. The dependency on frequency makes it difficultto generalize the stress-strain relationships based on periodic excitation to allowfor an arbitrary time varying loading such as seismic excitation. This problem isaddressed in the next section.

To determine the damping properties at the desired temperature and frequency from the data graph shownabove, proceed as follows:

• Locate the desired frequency on the RIGHT vertical scale.• Follow the chosen frequency line to the desired temperature isotherm.• From this intersect, go vertically up and/or down until crossing both the shear (storage)

modulus and loss factor curves .• Read the storage modulus and loss factor values from the appropriate LEFT hand scale.

Fig. 3.15: Variation of 3M viscoelastic material, ISD110, with frequency andtemperature.

The energy dissipated per unit volume of material for one cycle ofdeformation is determined from

(3.33)

Substituting for and using eqns (3.28) and (3.29) results in

(3.34)

Gs η

Wviscoelastic τγ td

0

2π Ω⁄

∫=

τ γ

Wviscoelastic πGlγ2

=


The corresponding expression for a pure viscous material is generated using eqn(3.27)

(3.35)

This expression involves the frequency explicitly whereas the effect of frequencyis embedded in for the viscoelastic case.

Example 3.4: Viscoelastic damper

A damper device is fabricated by bonding thin sheets of a viscoelastic material tosteel plates, as illustrated in Fig. 3.16.

Since the elastic modulus for steel is considerably greater than the shearmodulus for the sheet material, one can consider all the motion to be due toshear deformation of the sheets. Defining as the relative displacement of theends of the damper device, the shearing strain is

(3.36)

Given , one evaluates with the stress-strain relation and then using theequilibrium equation for the system

(3.37)

Applying a periodic excitation

(3.38)

Fig. 3.16: Viscoelastic damper device.

and taking according to eqn (3.29), one obtains

Wviscous πGvΩγ2=

Gl

Gsu

γ utd-----=

γ τ F

F 2wLτ=

u u Ωtsin=

td

L

F , u

Side View

td

LTop View

F , uw

τ


(3.39)

(3.40)

Equation (3.39) can also be written as

(3.41)

(3.42)

Finally, the energy dissipated per cycle is given by

(3.43)

Typical polymer materials, such as Scotchdamp ISD110 (3M Company, 1993) have in the range of and .

Based on the result of the previous example, the expressions defining theresponse of a viscoelastic damper due to periodic excitation can be written in ageneralized form

(3.44)

(3.45)

(3.46)

where depends on the geometric configuration of the device, is the storagemodulus, and is the material loss factor. Figure 3.17 shows the variation ofwith over the loading cycle.

F f dGsu Ωtsin η Ω tcos+[ ]=

f d2wL

td-----------=

F f dGu Ωt δ+( )sin=

G Gs 1 η2+=

W πη f dGsu2

=

Gs 1.5MPa η 1≈

u u Ωtsin=

F f dGsu Ωt η Ω tcos+sin( )=

Wviscoelastic πη f dGsu2

=

f d Gsη F

u


Fig. 3.17: Variation of with for viscoelastic material.

3.4 Equivalent viscous damping

The expression for the damping force corresponding to linear viscous damping isthe most convenient mathematical form, in comparison to the other dampingforce expressions, for deriving approximate analytical solutions to the forceequilibrium equations. Therefore, one way of handling the different dampingmodels is to convert them to equivalent viscous damping models. In what follows, aconversion strategy based on equating the energy dissipated per cycle of periodicexcitation to the corresponding value for linear viscous damping is described.

Linear viscous damping is defined by eqn (3.2):

(3.47)

Specializing eqn (3.47) for periodic excitation

(3.48)

leads to

(3.49)

Also, noting eqn (3.5), the energy loss per cycle is

(3.50)

The force and energy loss for the other models are expressed in terms of anequivalent damping coefficient,

u

F

η f dGsu

f dGs

t πΩ----=

t π2Ω-------=u u Ωtsin=

t 0=

u

Fd ud

F cu=

u u Ωtsin=

F cΩu Ωtcos=

W cπΩu2

=

ceq

3.4 Equivalent Viscous Damping 185

(3.51)

(3.52)

Substituting for a particular damping model in eqn (3.52), and taking, one obtains the equivalent damping coefficient. The coefficients for the

various models are listed below:

Coulomb

(3.53)

Structural

(3.54)

Hysteretic

(3.55)

Viscoelastic

(3.56)

These expressions are valid for periodic excitation of amplitude andfrequency . They can be used to approximate structural and hysteretic dampingas pseudo-viscous damping but require specifying a representative frequency,

, and amplitude, . In this case, eqns (3.54) and (3.55) are written as

Structural

(3.57)

Hysteretic

(3.58)

where

(3.59)

(3.60)

Numerical simulations illustrating the accuracy of this approximation are

F ceqΩu Ωtcos=

W ceqπΩu2

=

Wu u=

ceq4F

πΩu-----------=

ceq2ksπΩ--------=

ceq

4FyπΩu----------- µ 1–

µ------------=

ceqf dηGs

Ω-----------------

f dGl

Ω-------------= =

uΩ

Ωr ur

ceq2ksπΩr----------=

ceq

4FyπΩrur---------------- µ 1–

µ------------=

uyFykh------=

µuruy------=


provided by the following examples.

Example 3.5: Structural and hysteretic damping comparison - seismic excitation

A 1 DOF shear beam having the following properties is considered:

The equivalent structural stiffness is generated using eqn (3.57), takingand equal to the fundamental frequency . The corresponding structuralstiffness is

(3.61)

Results for this model subjected to Taft excitation are compared with thecorresponding results for the linear viscous model in Figs 3.18 and 3.19. Closeagreement is observed.

m 4 6×10 kg= ω1 1.17rd s⁄=

k 5517kN m⁄= T1 5.35s=

c 187.9kNs m⁄= ξ1 2.0%=

c ce=Ωr ω1

ksπω1ce

2--------------- π 1.17( ) 187.9( )

2------------------------------------- 346.6kN m⁄= = =


Fig. 3.18: Response of SDOF with structural damping.

Fig. 3.19: Structural damping force versus deformation.

0 10 20 30 40 50 60−6

−4

−2

0

2

4

6x 10

−3

Time t - s

γ-

m/

mStructural dampingversus Viscous dampingT1 5.35s=ξ1 2%=

Viscous dampingStructural damping

−6 −4 −2 0 2 4 6

x 10−3

−4

−3

−2

−1

0

1

2

3

4x 10

5

γ - m/m

Structural damping

Taft excitation

T1 5.35s=ξ1 2%=

F-

N


The hysteretic model calibration defined by eqn (3.58) is not as straightforward since both the yield force and the ductility are involved. For periodicmotion, the maximum displacement is known. Then, one can specify thedesired ductility and compute the required force level and initial stiffness with

(3.62)

(3.63)

(3.64)

For non-periodic motion, one needs to specify the limiting elasticdisplacement , and estimate the maximum amplitude . This leads toestimates for the ductility ratio and the peak force . Figures 3.20 and 3.21show the results based on taking equal to the peak amplitude observed for pureviscous damping, and a ductility ratio . One can adjust and toobtain closer agreement. Since the energy is dissipated only during this inelasticphase, hysteretic damping is generally less effective than either viscous orstructural damping for low intensity loading.

Fig. 3.20: Response of SDOF with hysteretic damping.

ud ud=µ∗

Fy

πω1uceq4

---------------------- µ∗µ∗ 1–---------------=

uyu

µ∗------=

khFyuy------=

uy uµ∗ Fyu

µ∗ 7.5= µ∗ Fy

0 10 20 30 40 50 60−6

−4

−2

0

2

4

6x 10

−3

Time t - s

γ-

m/

m

Hysteretic dampingversus Viscous dampingT1 5.35s=ξ1 2%=

Viscous dampingHysteretic damping


Fig. 3.21: Hysteretic damping force versus deformation.

The calibration of the equivalent viscous damping coefficient was based onassuming a periodic excitation. As discussed above, non-periodic excitationrequires some assumptions as to the response. An improved estimate of theequivalent damping coefficient can be obtained by evaluating the actual workdone by the damping force. Starting with

(3.65)

and writing

(3.66)

leads to

(3.67)

Equation (3.67) can be used to evaluate the variation over time of the equivalentdamping ratio. Taking = , the total duration of the response, provides anestimate of the effective damping ratio. Figure 3.22 shows results generated for arange of seismic excitations and hysteretic damper yield force levels. As expected,the effective damping increases with increasing excitation and decreases with

−6 −4 −2 0 2 4 6

x 10−3

−3

−2

−1

0

1

2

3

4x 10

5

γ - m/m

Hysteretic damping

Taft excitation

T1 5.35s=ξ1 2%=

F-

N

ED actualFu td

0

t

∫=

ED eq. viscousceq u2 td

0

t

∫=

ceq t( )

Fu td

0

t

∫

u2 td

0

t

∫---------------- 2ξeqωm= =

t tend


increasing yield force.

Fig. 3.22: Equivalent viscous damping ratio vs. yield force

The viscoelastic model calibration is more involved since the materialproperties are also frequency dependent. Referring back to eqn (3.45), thedamping force for periodic excitation

(3.68)

was expressed as

(3.69)

where is a geometric factor defined by the geometry of the device. Ourobjective is to express as

(3.70)

where and are equivalent stiffness and damping terms. Consideringperiodic excitation, eqn (3.70) takes the form

(3.71)

One can obtain estimates for and with a least square approach. Assuming

0 0.01 0.02 0.03 0.04 0.050

0.1

0.2

Fy mg⁄Equ

ival

ent V

isco

us D

ampi

ng R

atio

ξH

T = 4.91sec

Sv 1.2 m/s=

Sv 2.4 m/s=

Sv 0.6 m/s=

ζN 2 %=

u u Ωtsin=

F u f dGs Ωt η Ω tcos+sin( )=

f dF

F kequ cequ+=

keq ceq

F u keq Ωtsin Ωceq Ωtcos+[ ]=

keq ceq


there are material property data sets, and summing the squares of the errors for and over the ensemble results in

(3.72)

(3.73)

Minimizing eqn (3.72) with respect to yields

(3.74)

Similarly, minimizing eqn (3.73) with respect to results in

(3.75)

The form of eqn (3.75) suggests that be expressed as

(3.76)

Substituting for and leads to the definition equation for

(3.77)

Note that depends only on the material, i.e. it is independent of the geometryof the device.

With this notation, the equivalent viscous force-deformation relation for a linearviscoelastic damper is written as

(3.78)

Example 3.6: Determining for 3M ISD110 damping material

This example illustrates how the procedure discussed above can be applied tocompute the parameters for the 3M Scotchdamp ISD110 material. Using Fig. 3.15,data corresponding to five frequencies is generated. Table 3.1 contains this data.

Applying eqns (3.74), (3.75), and (3.77), one obtains

Nkeq ceq

Jk keq f dGs Ωi( )–[ ] 2

i 1=

N

∑=

Jc cef dGs Ωi( ) η Ω i( )⋅

Ωi--------------------------------------------–

2

i 1=

N

∑=

keq

keq f d1N---- Gs Ωi( )

i 1=

N

∑ f d G s= =

)

ceq

ceq f d1N----

Gs Ωi( )η Ω i( )Ωi

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

i 1=

N

∑=

ceq

ceq αdkeq=

keq ceq αd

αd

GsηΩ

----------

ii 1=

N

∑Gs Ωi( )

i 1=

N

∑-----------------------------=

αd

F kequ αd kequ+=

αd

keq 5.7 f d= αd 0.104= ceq 0.593 f d=


Table 3.1: Data for ISD110 Scotchdamp material (from Fig. 3.15).

3.5 Damping parameters - discrete shear beam

Damping systems

This section extends the treatment of discrete shear beams to includedamping devices located between the floors. Figure 3.23 illustrates 2 differentplacement schemes of viscous type dampers for a typical panel. Scheme acombines the damper with a structural element and deploys the compositeelement on the diagonal between floors. Scheme b places the damper on a rollersupport at the floor level, and connects the device to the adjacent floor withstructural elements. An actual installation of a scheme b system is shown in Fig.3.24. The structural elements are modelled as linear springs and therepresentation defined in Fig 3.25 is used.

0.628 1.0 1.0

3.14 2.5 1.0

6.28 3.7 0.93

12.56 5.0 0.85

31.4 9.0 0.65

62.8 13.0 0.55

Ω (r/s) Gs (MPa) η

3.5 Damping Parameters - Discrete Shear Beam 193

Fig. 3.23: Damper placement schemes.

θ

(a)

ui

ui 1–

ui

ui 1–

(b)


Fig. 3.24: Viscous dampers coupled with chevron bracing.


Fig. 3.25: Idealized models of structures with viscous dampers.

A differential story displacement generates a deformation of the damper,resulting in a damper force which produces the story shear, . A subscript isused to denote quantities associated with the damper. The total story shear is thesum of the “elastic” shear force due to elastic frame/brace action and the“damper” shear force. The former was considered in Chapter 2. This contributionis written as

(3.79)

where subscript refers to “elastic” frame/brace action. The damper shear forceis a function of both the relative displacement and the relative velocity. This term

(a)

c

k'

ui 1–

ui Vd i,,

ui 1–

Vd i,

Vd i,

c

k'

ui Vd i,,

(b)

Vd i, d

Ve i, ke i, ui ui 1––( )=

e


is expressed in a form similar to eqn (3.78):

(3.80)

where and are “equivalent” properties that depend on the makeup of thedamping system. Various cases are considered in the following sections.

Rigid structural members - linear viscous behavior

Consider first the case where the stiffness , of the structural memberscontained in the damping system is sufficiently large so that the extension of themember is negligible in comparison to the extension of the damper. Definingas the extension of the damper, and considering scheme a, the damper force forlinear viscous behavior is given by

(3.81)

The corresponding shear force is

(3.82)

The equivalent damping coefficient for story is obtained by summing thecontributions of the dampers present in story .

(3.83)

Equation (3.83) also applies for scheme b; = 0 for this arrangement of structuralmembers and dampers. Scheme b is more effective than scheme a (a factor of 2 for

bracing), and is more frequently adopted.

The general spring-dashpot model shown in Fig 3.26 is useful forrepresenting the different contributions to the story shear force. For this case, thedamper ( ) acts in parallel with the elastic shear stiffness of the frame/bracingsystem ( ) and is equal to the interstory displacement. An extended version ofthis model is used to study other damping systems.

Vd i, kd i, ui ui 1––( ) cd i, ui ui 1––( )+=

kd cd

k'

ed

Fd cded cd ui ui 1––( ) θcos= =

Vd i, Fd θcos cd ui ui 1––( ) θcos2= =

ii

cd i, cd θcos2( )story i∑=

θ

45°

CK e


Fig. 3.26: Spring and dashpot in parallel model.

Example 3.7: Example 2.15 revisited

Consider the 5 DOF shear beam defined in example 2.15. Taking theconstant nodal mass as 10,000 kg, and using the stiffness calibration based on

, results in the following values for the element shearstiffness factors, fundamental frequency, nodal mass, and damping:

, ,

,

(1)

The element damping coefficients are related to by

(2)

• Taking constant leads to

(3)

• Assuming damping is proportioned to the element stiffness, ,and selecting according to , (the basis for this equation isestablished in Section 3.7) one obtains the following:

e

V

K

C

Svmax0.7 m/s= ξ 0.1=

k1 13.59 MN/m= k2 12.68 MN/m= k3 10.87 MN/m=

k4 8.15 MN/m= k5 4.53 MN/m=

ω1 9.52 rad/sec=

m 22 000 kg,=

c 41.9 kN-s/m=

c

c 125------ c1 c2 c3 c4 c5+ + + +( )=

ci

c1 c2 … c5 5c 210 kN-s/m= = = = =

ci αki=α α 2ξ1 ω1⁄=

α 0.0210=


(4)

Suppose the chevron brace scheme (scheme b) is used, and 2 dampers aredeployed per floor. The “design” values for the dampers are obtained by dividingthe above results by a factor 2. For the uniform case, . In order todesign the damper, one also needs to specify the peak value of the damper force.This quantity is determined with where is the maximumrelative velocity of the damper piston. For this damper deployment scheme, therelative damper displacement is equal to the interstory displacement. It followsthat for level is equal to

(5)

The nodal displacements for this 5 DOF model are considered to vary linearlywith height:

where is the modal amplitude. Then

The peak amplitude is determined with

(7)

One can estimate by assuming the response is periodic, with frequency .

(8)

Using the problem data,

(9)

and the peak damper force is estimated as

c1 285 kN-s/m= c2 266 kN-s/m= c3 228 kN-s/m=

c4 171 kN-s/m= c5 95 kN-s/m=

c 105 kN-s/m=

Fmax cvmax= vmax

vmax i

vmax level iui ui 1––( )max hi γi( )max= =

ui ui 1––15---q=

q

vmax15--- q( )max=

qmax1ω----ΓSv ξ1 ω1,( )=

qmax ω1

qmax ΓSv ξ1 ω1,( )≈

qmax 1.36( ) 0.7( )≈ 0.95 m/s=


(10)

Rigid structural members - linear viscoelastic behavior

Fig. 3.27: Spring-dashpot model for viscoelastic damping.

The case where the damping mechanism is viscoelastic is represented by themodel shown in Fig. 3.27. Here, the damping force has an elastic component aswell as a viscous component. Noting eqn (3.78), the damping force is expressed as

(3.84)

where and (or ) are the equivalent stiffness and damping parameters forthe visco-elastic device, and is the interstory displacement. This formulationassumes the viscous device is attached to a rigid element so that all thedeformation occurs in the device. The more general case is treated later. Using eqn(3.84), one obtains

(3.85)

When is small with respect to unity, the contribution of the visco-elasticdamper to the stiffness can be neglected.

Fmax 105( ) 0.955

---------- ≈ 20 kN=

e

V

K

C

K1Vd

Ve

Vd K1e Ce+ K1e αdK1e+= =

K1 C αde

V K K1+( )e Ce+ K K1+( )e αdK1 e+= =

K1 K⁄


Example 3.8:

Consider a SDOF system having an elastic spring and a visco-elasticdamper modeled as shown in the figure. Suppose , , and are specified, andthe objective is to establish values for the spring stiffness and damper properties.

The governing equation has the form

(1)

By definition,

(2)

(3)

Given and , is determined with eqn (3). The stiffness factors are related by

(4)

Our strategy for dealing with a visco-elastic device is based on expressingthe equivalent damper coefficient as (see eqn (3.78)):

(5)

where is a “material” property. Example 3.6 illustrates how to evaluate fora typical visco-elastic material. The procedure followed here is to first determine

, using eqns (3) and (5),

m ω ξ

u

p

k

c

k1

m

mu cu k k1+( )u+ + p=

ω2k k1+

m--------------=

c 2ξωm=

ξ ω c

k1 k+ mω2=

c αdk1=

αd αd

k1


(6)

and then substitute for in eqn (4). This operation results in an equation for k.

(7)

Suppose =10,000 kg, = 2 rad/s, and . Then,

Using a typical value for ,

leads to

For these parameters, the visco-elastic element contributes approximately 20% ofthe stiffness.


Suppose visco-elastic dampers are used for the 5 DOF system consideredin example 3.7. The damper force is taken as

(1)

where and depend on the device, and is the displacement of thedamper. Consider the case where a chevron brace with 2 dampers is installed ineach floor and the damping distribution defined by eqn 4 in example 3.7 is used.The damper coefficients are determined by dividing the values listed in eqn 4 by 2(2 dampers per floor):

k12ξωm

αd--------------- c

αd------≡=

k1

k mω2 2ξωmαd

---------------– mω ω 2ξαd------–

= =

m ω π ξ 0.1=

c 2 0.1( ) 2π( ) 104( ) 12.56 kN-s/m= =

αd

αd 0.15=

k1 84 kN/m=

k 394 84– 310 kN/m= =

Fd kdud cdud+ kdud αdkdud+= =

αd kd ud

cd 1, 142.5= cd 2, 133= cd 3, 114=


(2)

(units of kN-s/m)

The damper stiffness is determined with

(3)

Assuming , the corresponding values of damper stiffness are:

(4)

(units are kN/m)

The total story shear stiffness distribution is given by eqn (1) in example 3.7. Thisvalue is the sum of the elastic stiffness due to frame/brace action and the stiffnessdue to the 2 dampers

(5)

Using (4) and the data from example 3.7, the frame/brace story shear stiffnessfactors for this choice of are

(6)

(units are MN/m)

The contribution of the damper stiffness is about 14% of the total stiffness for thisexample.

Example 3.10: Viscoelastic damper design

Referring back to eqn (3.74), the elastic stiffness of the damper depends on theaverage storage modulus of the viscoelastic material and a geometric parameter

.

cd 4, 85.5= cd 5, 47.5=

kd1

αd------cd=

αd 0.15=

kd 1, 950= kd 2, 887= kd 3, 760=

kd 4, 570= kd 5, 317=

k ke 2 kd⋅+=

αd

ke 1, 11.69= ke 2, 10.93= ke 3, 9.35=

ke 4, 7.01= ke 5, 3.89=

f d


(1)

Given and , one solves for

(2)

To proceed further, one needs to specify the geometry of the device. The figurelisted below shows a system consisting of layers of a viscoelastic materiallocated between metal plates. Considering the metal elements to be rigid withrespect to the viscoelastic elements, the shape factor is given by

(3)

The layer thickness is usually fixed by the material manufacturer, and thereforethe design variables are the number, length, and width of the viscous plates.

As an illustration, suppose

(4)

Taking as the “average” modulus for 3M ISD110, thecorresponding shape factor is

(5)

K1 f d G s=

)

K1 G s

)

f d

f d

K1

G s

--------= )

2n

f d 2nwLd

td----------=

1

2

n

Ld

w

td Ld

kd 1, 10 000 kN/m,=

G s 2.5 MPa=

)

f d10 000 3×10,

2.5 6×10----------------------------- 4.0 m= =


Substituting for in eqn (3), the variables are related by

(6)

Suppose and .

(7)

Taking results in

(8)

Example 3.11: Hysteretic damper design - diagonal element

Equation (3.58) defines the equivalent viscous damping parameter for hystereticdamping. Substituting the extension, , for the displacement measure , andsolving for the yield force, , results in

(1)

where and are representative extension and frequency values, and is givenby

(2)

where is the extension at which the diagonal material yields.

The representative extension is a function of the representative transverseshear deformation . Taking equal to , the design level for , leads to

(3)

and

(4)

f d

2n wLd( ) td f d⋅ 4.0td (meters)= =

n 2= td 10 2– m 1 cm= =

wLd 1.0 2–×10 m2=

w Ld=

w Ld 0.1 m 10 cm= = =

e uFy

Fy

πΩrer4

--------------- µµ 1–------------ C=

er Ωr µ

µerey-----

erLεy---------

er θsin

hεy----------------= = =

ey

γr γr γ∗ γ

er γ∗ h θcos=

µ 2θsin2

--------------- γ∗εy-----=


A typical design value for is . Ideally, one should use a lowstrength material so that the response is essentially inelastic throughout theloading duration, thus maximizing the energy dissipation. One potentialcandidate material is the 200MPa yield strength steel developed by Nippon Steel(Nippon Steel Corp., 1990); the corresponding yield strain is . Using thesevalues and taking provides an upper bound estimate for the ductilityratio

(5)

Flexible structural members - linear viscoelastic behavior

For completeness, the analysis for the refined viscoelastic model shown in Fig.3.28 is presented. The component attached to the damper device is modeled as aspring in series with the damper which is considered to be linear viscoelastic withfrequency dependent properties and . It is convenient to deal first with aperiodic excitation, and then average the properties over the appropriatefrequency range.

Fig. 3.28: General spring-dashpot model.

Letting represent the displacement of the damper and considering tobe periodic, the corresponding damper force, , follows from eqn (3.69)

(3.86)

(3.87)

γ∗ 1 200⁄

1 2000⁄θ 45°=

µ 20002

------------ 1200---------

≈ 5=

Gs η

e

F

K

C

K1

K2

Fe

Fd

Primary elastic element

Secondary elastic element

ed edFd

ed ed Ωtsin=

Fd Gs f d ed Ωtsin η Ω tcos+( ) Gs f d ed Ωt δ+( )sin= =


where is a characteristic geometric parameter for the device and .

Since the force in the secondary element must be equal to the dampingforce, the extensions are related by

(3.88)

Substituting for and leads to the expression for the total displacement

(3.89)

Equation (3.89) can also be written as

(3.90)

where

(3.91)

(3.92)

The force in the primary elastic element depends only on

(3.93)

Combining and , the total force is given by

(3.94)

A more compact form for is

(3.95)

where represents the phase shift between the excitation and the force response,and is the total stiffness measure. The definition equations are

(3.96)

(3.97)

(3.98)

The equations can be expressed in conventional form by shifting the timereference point. Defining as

f d δtan η=

e ed–FdK2------=

ed Fd e

e edFdK2------+ ed 1

Gs f dK2

-------------+

Ωtsin edGs f d

K2-------------

η Ω tcos+= =

e e Ωt δ1+( )sin=

e ed 1Gs f d

K2-------------+

2 ηGs f dK2

-----------------2

+ ℵ ed= =

δ1tan η 1

1K2

Gs f d-------------+

-----------------------=

e

Fe Ke Ke Ωt δ1+( )sin= =

Fe Fd

F Fe Fd+ Ke Ωt δ1+( )sin Gs f ded Ωt δ+( )sin+= =

F

F Ke Ωt δ1 δ2+ +( )sin=

δ2K

K K δ1cosGs f d

ℵ------------- δcos+

2K δ1sin

Gs f dℵ

------------- δsin+2

+=

δ1 δ2+( )tanK δ1sin

Gs f dℵ

------------- δsin+

K δ1cosGs f d

ℵ------------- δcos+

---------------------------------------------------- ℜ= =

δ2tan

ℜδ1tan

--------------- 1–

ℜ 1δ1tan

---------------+--------------------------=

t'


(3.99)

transforms eqns (3.90) and (3.95) to

(3.100)

(3.101)

Finally, expanding eqn (3.101), the result can be expressed in a form similar to theconventional viscoelastic form

(3.102)

where

(3.103)

(3.104)

One can interpret as the storage stiffness and as the loss factor for theassemblage.

Equivalent parameters can be generated following the proceduredescribed in Section 3.4. One writes

(3.105)

and equates eqn (3.105) with (3.102). The error terms for a periodic excitation are

(3.106)

(3.107)

Minimizing the sum of the square of these terms over the frequency range withrespect to and produces the following expressions

(3.108)

(3.109)

where is the number of frequencies composing the data set. If one considersto be very large with respect to the visco-elastic damper stiffness, , thevarious terms simplify to

(3.110)

t t'δ1Ω-----–=

e e Ωt'sin=

F Ke Ωt' δ2+( )sin=

F Kse Ωt'sin ηKs e Ωt'cos+=

Ks K δ2cos=

η δ2tan=

Ks η

F Keqe Ceqe+=

E1 Keq Ks–=

E2 CeqηKsΩ

----------–=

Keq Ceq

Keq Ks average

1N---- Ks Ωi( )

i 1=

N

∑≡=

Ceq1N----

η iKs i, Ωi( )Ωi

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

i 1=

N

∑=

N K2f dGs

ℵ 1≈


(3.111)

(3.112)

(3.113)

Example 3.12: Coupled spring-damper model

This example considers the case where the damper is viscous, and thestructural element connecting the damper to the floors is flexible. This is asimplified version of the model considered above; the stiffness component of thevisco-elastic device is deleted.

The steps are similar to the previous steps. We take

(1)

Then,

(2)

The total force is expressed as

δ1 0≈

K K 12Gs f d

K----------------- δcos

Gs f dK

-------------2

+ +≈

η δ2tan ℜ η 1

1 KGs f d δcos--------------------------+

------------------------------------= = =

eC

Fe

FdK2

K

F

ed ed Ωtsin=

Fd CΩ ed Ωtcos=

e e Ωt δ1+( )sin=

e ed 1 CΩK2--------

2+

1 2/=

δ1tan CΩK2--------=


(3)

where

(4)

(5)

The remaining steps are the same. One expresses as

(6)

and determines and using eqns (3.108) and (3.109).

When , and

(7)

Then,

(8)

and

(9)

3.6 Damping parameters - truss-beam

This section extends the treatment of the truss-beam discussed in Example 2.2 toinclude damping. The typical panel shown in Fig. 3.29 is considered to becomposed of two sets of elements: an elastic system which provides the stiffness(shear and bending) and a second system which functions as a distributed energydissipation/absorption mechanism.

F Ke Ωt δ1 δ2+ +( )sin=

K K δ1 1 δ1tan CΩK

--------+2

+1 2/

cos=

δ2tan CΩK

-------- 1

1 CΩ( )2 1K2------

1K2------ 1

K----+

+-------------------------------------------------------------⋅=

F

F Keqe Ceqe+=

Keq Ceq

K2 ∞= δ1 0=

δ2tan CΩK

--------= δ2cos KK----=

Ks K= nKs CΩ=

Keq K≡ Ceq C≡


Fig. 3.29: Truss-beam structure - geometry and forces.

The various force quantities are determined with

(3.114)

(3.115)

(3.116)

(3.117)

where superscripts ‘c’ and ‘d’ refer to the column and diagonal elements, andsubscripts ‘e’ and ‘d’ denote the elastic and damping force components.

Defining as the extension, i.e. the total change in length of a structuralelement, and noting eqns (2.36) and (2.39), the extensions for the diagonal andchord elements are related to the transverse shear and bending deformations by

(3.118)

(3.119)

The elastic force is a function of and the element properties

(3.120)

where denotes the length, the cross-sectional area, and the Young’sModulus for the elastic element.

h

B

θ

EdEc FcFcFd Fd

V

M

AdAc

Fd Fed Fd

d+=

Fc Fec Fd

c+=

V 2Fd θcos=

M BFc=

e

ed γh θcos=

ec Bh2

------- χ=

e

FeAEL

-------- e K e⋅= =

L A E

3.6 Damping Parameters - Truss Beam 211

Linear viscous behavior

The dissipative force-deformation relation depends on the nature of the dampingdevice. For viscous behavior, is a function of . The linear viscous force isexpressed as

(3.121)

where is a property of the device, which may be an actual viscous damper, oran equivalent viscous coefficient. Summing the elastic and damping forces leads tothe element force-extension relations

(3.122)

(3.123)

One can associate these relations with the parallel spring/dashpot model shownin Fig. 3.26. The member force-deformation relations are obtained by introducingthese expressions into the definition equations for and . They are written as

(3.124)

(3.125)

where

(3.126)

(3.127)

(3.128)

(3.129)

It is convenient to express the damping coefficients in terms of the stiffnessparameters,

(3.130)

(3.131)

where and follow from the previous equations

(3.132)

Fd e

Fd Ce=

C

Fc Kcec Ccec

+=

Fd Kded Cded

+=

V M

DTγ CTγ+=

M DBχ CBχ+=

DT 2hcos2θ[ ] Kd=

CT 2hcos2θ[ ] Cd=

DBB2h

2--------- Kc

=

CBB2h

2--------- Cc

=

CT αTDT=

CB αBDB=

αT αB

αTCd

Kd-------=


(3.133)

Note that and depend on the damping device configuration and stiffnessof the diagonal and chord elements. Expressing in terms of and takingconstant over the height simplifies the analysis of the beam as a continuum usingmodal shape functions by uncoupling the equations for the modal coordinates.

Linear viscoelastic behavior

The case where the damping mechanism is viscoelastic is represented by themodel shown in Fig. 3.27. Here, the damping force has an elastic component aswell as a viscous component, and is expressed as

(3.134)

where and are the equivalent stiffness and damping parameters for thedevice, and is the total extension of the member. This formulation assumes theviscous device is attached to a rigid element so that all the deformation occurs inthe device. Using eqn (3.134), one obtains

(3.135)

(3.136)

Substituting for the forces and deformations leads to

(3.137)

(3.138)

and

(3.139)

(3.140)

The proportionality factors in this case are given by

(3.141)

(3.142)

αBCc

Kc------=

αT αBC D α

Fd K1e Ce+ K1e αdK1 e+= =

K1 αde

Fc Kc K1c

+( )ec αdc K1

c ec

+=

Fd Kd K1d

+( )ed αddK1

d ed

+=

DT 2hcos2θ[ ] Kd K1d

+( )=

CT 2hcos2θ[ ]α ddK1

d=

DBB2h

2--------- Kc K1

c+( )=

CBB2h

2---------αd

c K1c

=

αBCBDB------- αd

cK1

c Kc⁄

1 K1c Kc⁄+

--------------------------= =

αTCTDT------- αd

dK1

d Kd⁄

1 K1d Kd⁄+

----------------------------= =

3.6 Damping Parameters - Truss Beam 213

3.7 Damping distribution for MDOF systems

Chapter 2 dealt with the problem of specifying stiffness distributions for beam-type structures. Numerical simulations for seismic excitation showed that one canestablish stiffness distributions with an iterative procedure that accounted for thecontribution of the higher modes. However, the amplitude of the deformationprofile cannot be totally controlled with just stiffness, and other mechanisms needto be incorporated. An examination of the effectiveness of distributed linearviscous damping to decrease the deformation amplitude and therefore providemore control on the response is presented in this section.

The structure is assumed to be discretized as an ’th order MDOF systemgoverned by the following equation.

(3.143)

where contains the nodal translation and rotation measures, is the systemmass matrix, is the system damping matrix, represents the system stiffness,and represents the load vector. A similar set of equations was considered inSection 2.10, where a restriction was placed on and a procedure for transferringthe coupled matrix equations to an uncoupled set of equations in terms of modalcoordinates was described. The procedure is extended here to allow for anarbitrary form of .

Multi mode free vibration response

The case where is considered first. The general homogeneoussolution can be expressed as

(3.144)

where A and are arbitrary scalars, and is an unknown vector of order n.Substituting for in eqn (3.143) leads to an equation relating and .

(3.145)

Solving eqn (3.145) for and is referred to as the quadratic eigenvalueproblem. A detailed discussion of the mathematical aspects of this problem iscontained in Strang,1993.The discussion presented here is intended to provide anintroduction to this topic which is also revisited in Chapter 8.

n

MU CU KU+ + P=

U MC K

PC

C

P 0=

U Aeλ tΦ=

λ ΦU λ Φ

λ2M λC K+ +( )Φ 0=

λ Φ


When , eqn (3.145) reduces to

(3.146)

Since and are positive definite, must be negative. Then, can beexpressed as

(3.147)

The solution corresponding to the pair of values for is

(3.148)

Since U must be real, the A’s must be complex conjugates,

(3.149)

Substituting for and transforms the solution to

(3.150)

Equation (3.146) has distinct solutions, i.e., n ‘s and corresponding‘s. Each solution satisfies

(3.151)

and the following orthogonality relations

(3.152)

(3.153)

Suppose is a scaled version of .

(3.154)

where is a positive scalar. This condition is called “stiffness proportionaldamping”. Substituting for in eqn (3.145), and rearranging terms leads to

(3.155)

Setting

C 0=

K λ2M+( )Φ 0=

K M λ2 λ

λ iω±= ω 0>

λ

U A1eiωtΦ A2e iωt– Φ+=

A112--- AR iAI+( )=

A212--- AR iAI–( )=

A1 A2

U AR ωtcos AI ωtsin–( )Φ=

n Φω

KΦi ωi2MΦi= i 1 2 … n, , ,=

ΦjTKΦi Φj

TMΦi 0= = i j≠

ΦiTKΦi ωi

2ΦiTMΦi ωi

2mi= =

C K

C αK=

αC

K λ2

1 αλ+----------------M+

Φ 0=

λ2

1 αλ+---------------- ω2–=

3.7 Damping Distribution for MDOF Systems 215

converts eqn (3.155) to the “undamped” form. It follows that the eigenvectors(mode shapes) are the “same”, and the eigenvalues ( ) depend on and .

(3.156)

The eigenvalues are expressed as

(3.157)

where is the damping ratio for the ‘th mode.

(3.158)

The modal damping ratio increases with increasing mode number. Substitutingfor , the typical free vibration modal response is still periodic, but with a timedecaying amplitude.

(3.159)

The case where is an arbitrary symmetric positive definite matrixinvolves an additional complication; the eigenvectors as well as the eigenvaluesare complex quantities. These quantities occur as complex conjugates and areexpressed as

(3.160)

The ‘th solution is generated by combining the corresponding complexsolutions.

(3.161)

Expanding the complex products leads to the following “real” solution:

(3.162)

λ ω α

λ j2 αωj

2λ j ωj2+ + 0=

λ j ξ jωj– iωj 1 ξ j2–( )1 2/± ξ jωj– iω' j±= =

ξ j j

ξ jαωj

2----------=

λ

U e ξ iwit– Φi AR i, ω'icos t AI i, ω'itsin–[ ]=

C

λ j λR j, iλ I j,+=

λ j λR j, iλ I j,–=

Φj ΦR j, iΦI j,+=

Φj ΦR j, iΦI j,–=

j

U Ajeλ jtΦj A jeλ jtΦj+=

eλR j, t AR j, ΦR j, AI j, ΦI j,–( ) λ Itcos A– I j, ΦR j, AR j, ΦI j,–( ) λ Itsin+ =


Equation (3.162) is expressed in a form similar to eqn (3.159)

(3.163)

Combining the time varying terms leads to a simpler form:

(3.164)

The second term vanishes for stiffness proportional damping.

Our strategy for generating an initial estimate for the stiffness distributionis based on using as the modal shape function. The actual modal responseinvolves an additional shape function when is an arbitrary matrix. Theimportance of this second term depends on the amount of damping. Variouscomputational issues related to stiffness are addressed by the following examples.This topic is discussed in more detail later in Chapter 8, where the generalsolution for arbitrary loading is presented.

Example 3.13: Eigenvalue problem - 2DOF

The governing equations for free vibration of the 2DOF system shownabove have the form:

(1)

U e ξ jωj– t AR j, ΦR j,( ) ω' jtcos A– I j, ΦR j,( ) ω' jtsin+ =

e ξ jωj– t AI j, ΦI j,–( ) ω' jtcos AR j, ΦI j,–( ) ω' jtsin+ +

U f jΦR j, gjΦI j,+=

ΦRC

m2

m1

k2 c2,

k1 c1,

u2

u1

m1 u1 c1 c2+( )u1 k1 k2+( )u1 c2u2– k2u2–+ + 0=

m2 u2 c2u2 k2u2 c2u1– k2u1–+ + 0=


Expressing the displacements as

(2)

and substituting in eqn (1) leads to a set of equations for the terms.

(3)

For a non-trivial solution to exist, the determinant of the coefficient matrix mustvanish. This requirement results in the following equation for .

(4)

The elements of are not uniquely defined since eqn. 3 is a homogeneousequation. Solving the first equation leads to a relation between and .

(5)

The undamped solution is obtained by setting .

(6)

(7)

Expanding eqn (6) results in

(8)

Noting that the coefficients are all positive, it follows that must be a negativereal number. Expressing as

(9)

and solving eqn (8) for leads to

(10)

where

u1 φ1eλ t= u2 φ2eλ t=

φ

m1λ2 λ c1 c2+( ) k1 k2++ +( )φ1 λc2 k2+( )φ2– 0=

λc2 k2+( )– φ1 m2λ2 λc2 k2+ +( )φ2+ 0=

λ

m1λ2 λ c1 c2+( ) k1 k2+ + +( ) m2λ2 λc2 k2+ +( ) λc2 k2+( )2– 0=

Φφ1 φ2

φ2 φ1m1λ2 λ c1 c2+( ) k1 k2+ + +

λc2 k2+-------------------------------------------------------------------⋅=

c1 c2 0= =

m1λ2 k1 k2+ +( ) m2λ2 k2+( ) k22– 0=

φ2 φ1m1λ2 k1 k2+ +

k2------------------------------------⋅=

m1m2( )λ4 m1k2 m2 k1 k2+( )+( )λ2 k1k2+ + 0=

λ2

λ

λ i ω±=

ω 2

ω 2 a 1 b±( )=


(11)

(12)

Equation (10) has 2 real positive roots. The elements of are also real quantities inthis case. Each eigenvalue/eigenvector produces a solution of the form

(13)

where and are integration constants that are evaluated using the initialconditions at time .

To illustrate the computations, the system examined in example 1.9 isconsidered. The mass and stiffness are:

(14)

These stiffness values produce a fundamental mode which is linear andcorrespond to a fundamental frequency of rad/s. Evaluating and for thisset of properties results in

(15)

and

(16)

The mode shapes are determined with eqn (7).

Fundamental mode

am1k2 m2 k1 k2+( )+

2m1m2------------------------------------------------=

b 14m1m2k1k2

m1k2 m2 k1 k2+( )+( )2-------------------------------------------------------–

1 2/

=

φ

u1 Φ j( )1 AR j, ωjtcos AI j, ωjtsin–( )=

u2 Φ j( )2 AR j, ωjtcos AI j, ωjtsin–( )=

AR AIt 0=

m1 m2 1 000 kg, 1 kN-s2/m= = =

k2 2( ) 2π( )2 78.88 kN/m= =

k1 3( ) 2π( )2 118.3 kN/m= =

2π a b

a 138.0 (1/sec2)=

b 1 2449------–

1 2/

57---= =

ω1 2, 2π , 2π 2.45( ) 6.28 15.29,= =


(17)

Second mode

(18)

It is convenient to normalize the modal vectors such that the magnitude of themaximum element is unity. For this example, the normalized vectors are

(19)

The case where is proportional to is considered next. Settingand defining as

(20)

reduces eqns (4) and (5) to the “same” form as eqns (6) and (7) with replacedwith .

(21)

(22)

One sets

(23)

and determines with eqn (10). The eigenvector is the same as the undampedvector, and the eigenvalues are determined by combining eqns (20) and (23).

(24)

The solution is written in the following general form:

λ2 ω2– 2π( )2–= =

φ2 2φ1=

λ2 6 2π( )2–=

φ2 φ1 2⁄–=

Φ11 2⁄

1= Φ2

11 2⁄–

=

c k ci αki=λ′

λ′( )2 λ2

1 αλ+----------------=

λλ′

m1 λ′( )2 k1 k2+ +( ) m2 λ′( )2 k2+( ) k22– 0=

φ2 φ1m1 λ′( )2 k1 k2+ +

k2-------------------------------------------=

λ′ ω2–=

ω2

λ2

1 αλ+---------------- ω2–=


(25)

Continuing with the numerical example, suppose the design objective is. The required value of is

(26)

Then,

(27)

The damping ratio for the second mode is proportional to the frequency ratio:

(28)

Lastly, the modified frequencies are:

(29)

The case where is not proportional to either the mass or stiffnessmatrices involves solving a complete 4’th degree equation for . Equation (4) iswritten as

(30)

where the coefficients are

λ j ξ jωj iωj ′±–=

ξ j αωj 2⁄=

ωj ′ ωj 1 ξ j2–[ ] 1 2/=

ui Φ j( )ieξ jωjt– AR j, ωj ′ tcos AI j, ωj ′ tsin–( )=

j 1 2,=

ξ1 0.1= α

α2ξ1

ω1-------- 2 0.1( )

2π--------------- 0.0318 sec= = =

c1 αk1 3.762 kN-s/m= =

c2 αk2 2.508 kN-s/m= =

ξ2

ξ1-----

ω2

ω1------ 2.45= =

ξ2 2.45 0.10( ) 0.245==

ω1 ′ ω1 1 ξ12–[ ] 1 2/ ω1 0.99( )1 2/ 0.995ω1 6.25== = =

ω2 ′ ω2 0.94( )1 2/ 0.97ω2 14.83== =

Cλ

λ4 f 3λ3 f 2λ2 f 1λ f 0+ + + + 0=


(31)

Since all the coefficients are positive, the real part of must be negative. Thefactored form of eqn (30) is expressed as

(32)

where

(33)

With this notation, the 2 pairs of complex conjugate roots are

(34)

The eigenvectors are determined with eqn (5) which is written as

(35)

where is a function of . Each pair of complex conjugate roots generates acorresponding pair of complex conjugates for and . Considering the first pairof eigenvalues,

(36)

and

(37)

f 3 m1m2( ) 1– m1c2 m2 c1 c2+( )+( )=

f 2 m1m2( ) 1– m1k2 m2 k1 k2+( ) c1c2+ +( )=

f 1 m1m2( ) 1– k2c1 k1c2+( )=

f 0 m1m2( ) 1– k1k2( )=

λ

λ2 a1λ a2+ +( ) λ2 b1λ b2+ +( ) 0=

a1 b1 f 3=+

a1b1 a2 b2 f 2=+ +

a1b2 a2b1 f 1=+

a2b2 f 0=

λ1 λ1,( ) 12--- a1– i 4a2 a1

2–( )1 2/±[ ] ξ 1ω1– iω1 ′±==

λ2 λ2,( ) 12--- b1– i 4b2 b1

2–( )1 2/±[ ] ξ 2ω2– iω2 ′±==

φ2 hφ1=

h λh φ

λ1 λ1,( ) h1 h1,( )⇒

φ1 2, h1φ1 1,=

φ1 2, h1φ1 1,=


The modes defined by eqn (37) are represented as:

(38)

Since is not uniquely defined, one can set the magnitude of one element,and scale the other elements using eqn (35). Adopting the normalized formdefined by eqn (19), the definition equations for the elements are:

Mode 1

(39)

Mode 2

(40)

Results based on the mass and stiffness properties defined by eqn (14) and5 different distributions of the damping coefficients are listed below in Table 3.2.The first case is stiffness proportional damping with . The next 3 casescorrespond to uniform, linear, and concentrated distributions with the sum of thedamping coefficients held constant. The last case is a repeat of case 2 with thedamping doubled. Concentrating all the damping in the second element producesthe maximum value of , and does not result in a significant change in the modeshape. However, it increase the damping force, and requires a “larger” damper.Selecting a uniform distribution has the advantage that the damper size is lessthan the sizes for the other choices. To arrive at an optimal distribution, one needsto consider the total damper costs and other constraints on placement of dampers.There are situations where dampers cannot be placed in certain floors. Computerbased simulation is used in this case to “optimize” damper placement.

Φ1 Φ1,( ) ΦR 1, iΦI 1,±=

Φ

φ2 1≡

φ1 φ1, 1h1----- 1

h1-----,=

φ1 1≡

φ2 φ2, h2 h2,=

ξ1 0.1=

ξ2


Note: , ,

Units of c are kN s/m

Example 3.14: Modal response for nonproportional damping

Equation (3.163) defines the general homogeneous solution for non-

Table 3.2: Frequencies, damping ratios, and mode shapes for 2DOF system

6.28 6.286 6.325 6.504 6.305

0.1 0.1 0.0987 0.0926 0.1987

0.5 0.502 0.509 0.528 0.511

1.0 1.0 1.0 1.0 1.0

0 -0.025 -0.066 -0.151 -0.0486

0 0 0 0 0

15.29 15.37 15.276 14.854 15.32

0.245 0.265 0.301 0.385 0.532

1.0 1.0 1.0 1.0 1.0

-0.5 -0.515 -0.544 -0.620 -0.564

0 0 0 0 0

0 0.059 0.161 0.37 0.107

c1 3.762=

c2 2.513=

c1 3.135=

c2 3.135=

c1 2.09=

c2 4.18=

c1 0.0=

c2 6.27=

c1 6.27=

c2 6.27=

ω1

ξ1

φR 1,

φR 2,

φI 1,

φI 2,

ω2

ξ2

φR 1,

φR 2,

φI 1,

φI 2,

m1 m2 1 000 kg,= = k1 118.3 kN/m= k2 78.9 kN/m=


proportional damping. A typical solution is actually the sum of 2 solutions (thesubscript is dropped for convenience):

(1)

where

(2)

(3)

The responses associated with and are out of phase with each other.

This formulation is applied to the 2 DOF system treated in example (3.13).Table 3.2 shows that the mode shapes have the following general form:

Mode 1

(4)

Mode 2

(5)

where and depend on the level of damping. Substituting for the ‘s in eqn(1) leads to:

Mode 1

(6)

Mode 2

(7)

j

U f ΦR gΦI+=

f AR e ξωt– ω′tcos( ) AI e ξωt– ω′tsin–( )+=

g AR e– ξωt– ωsin ′ t( ) AI e ξωt– ωcos ′ t–( )+=

AR AI 90°

ΦRaR

1= ΦI

aI

0=

ΦR1

bR= ΦI

0bI

=

a b Φ

u1 aR f aI g+=

u2 f=

u1 f=

u2 bR f bI g+=


Expanding the terms in eqn (6) shows the phasing effect on thefundamental mode.

(8)

(9)

Comparing there terms shows that the solution for is out of phase with respectto . Defining and as

(10)

(11)

transforms eqn (8) to

(12)

The shear strains in the element are related to the difference indisplacements.

(13)

Using eqns (8) and (9),

(14)

Equation (14) is written in a form similar to eqn (12). The amplitude and phaseterms are:

u1 e ξ1ω1t– AR aR ω1 ′ t aI ω1 ′ tsin–cos( ) AI a– R ω1 ′ t aI ω1cos ′ t–sin( )+[ ]=

u2 e ξ1ω1t– AR ω1 ′ tcos( ) AI ω1 ′ tsin–( )+[ ]=

u1u2 β a

β1tanaI

aR-----=

a1 aR2 aI

2+[ ] 1 2/=

u1 e ξ1ω1t– ARa1 ω1 ′ t β1+( )cos AI– a1 ω1 ′ t β1+( )sin[ ]=

γ11h1-----u1=

γ21h2----- u2 u1–( )=

u2 u1–

e ξ1ω1t– AR 1 a– R( )( ω1 ′ t aI ω1 ′ tsin+cos ) AI 1 a– R( ) ω1 ′ t aI ω1cos ′ t–sin( )–[ ]=


(15)

(16)

With this notation, eqn (14) takes the form

(17)

The peak differential displacements are:

(18)

Table 3.2 lists results for and corresponding to various distributionsof and . The largest deviation from the proportional damping case( ; ) is case 4.

(19)

Using the above values, the various parameters are

(20)

For this example, the peak shear strains in elements 1 and 2 differ by only 10%

Stiffness proportional viscous damping

An introductory treatment of the case where was presented insection 2.10. This section expands upon that analytical treatment and alsocontains some examples which illustrate the influence of damping on theresponse of the higher modes.

Expressing as an expansion in terms of the undamped mode shapevectors,

β2tanaI

1 aR–--------------–=

a2 1 aR–( )2 aI2+[ ] 1 2/=

u2 u1– e ξ1ω1t– ARa2 ω1 ′ t β2+( ) AIa2 ω1 ′ t β2+( )sin–cos[ ]=

u1 max a1 AR2 AI

2+( )1 2/=

u2 u– 1 maxa2 AR

2 AI2+( )1 2/=

aR aIc1 c2

aR 0.5= aI 0=

aR 0.528= aI 0.151–=

β1 15.96°–=

a1 0.549=

β2 17.74°–=

a2 0.495=

C αK=

U

3.7 Damping Distribution for MDOF Systems 227n

(3.165)

and noting the orthogonality relations leads to the following set of equations forthe q’s

(3.166)

where , the damping ratio for the ith mode, is related to and by eqn(3.158). The magnitude of is determined by specifying the damping ratio forthe first mode

(3.167)

Given , one calculates the other damping ratios with

(3.168)

Suppose the initial conditions on are

(3.169)

Expressing as

(3.170)

and noting the orthogonality relation for which follows from eqn (3.151),

(3.171)

one can transform eqn (3.170) to

(3.172)

by premultiplying by . The ‘th term has the form

(3.173)

Applying this transformation to the initial conditions on leads to the initialconditions for .

(3.174)

(3.175)

U qiΦi

i 1=

∑=

qi 2ξ iωiqi ωi2qi+ +

ΦiTP

mi-----------

pimi------= = i = 1, 2, 3, . . . .

ξ i α ωiα

α2ξ1ω1---------=

α

ξ i ξ1

ωiω1------

=

UU 0( ) U∗=

U 0( ) U∗=U

U Φq=

M

ΦTMΦ miδij[ ] m= =

mq ΦTMU=

ΦTM i

miqi φiTMU=

Uqi

qi 0( ) 1mi------φi

TMU∗ qi∗= =

qi 0( ) 1mi------φi

TMU∗ qi∗= =

i 1 2 … n, , ,=


The orthogonality relationships for the coefficient matrices reduces theproblem of solving an ‘th order coupled equation to one of solving uncoupledequations for the modal coordinates. A similar result can be obtained for non-proportional damping, but that formulation is considerably more complicatedthan this formulation. The nonproportional case is discussed again in Chapter 8where a general approach for dealing with arbitrary damping based on the “state-space” formulation is presented. The total response for the i’th modal coordinatecan be expressed as:

(3.176)

(3.177)

(3.178)

The procedure for generating the stiffness distribution presented inChapter 2 assumes the fundamental mode ( ) is the dominate term and basesthe stiffness distribution on the displacement profile defined by . Othermodes may contribute, depending on the frequency of excitation and modaldamping. Ideally, one wants high modal damping for to minimizetheir contribution. With stiffness proportional damping, the modal dampingratios are fixed by the frequency ratios. They increase with mode number, whichis desirable, but one cannot arbitrarily increase a particular modal damping ratioto further reduce the modal response. The only recourse is to increase thefundamental mode damping ratio, . This action increases since ,and requires additional damping. A set of examples illustrating the effect ofincreasing on the response are presented in the following section.

Example 3.15: Low rise buildings.

Results for the shear beam type buildings defined in Table 3.3 arepresented in this section. The building sites are assumed to be in California, andthe reference spectral velocity is taken as for . Threeearthquakes representative of the sites, namely, El Centro, Northridge Station01,and Northridge Station03, are used to generate ensemble averaged responsespectra. Their scaled accelerograms are shown in Figure 3.30 through 3.32.

n n

qi t( ) e ξ iωit– qi* ω′t 1

ω′----- q* ξωq*+( ) ω′tsin+cos +=

1ωi ′-------+

pi

mi------e ξ iωi t τ–( )– ωi ′ t τ–( )[ ]sin τd

0

t

∫qi t( ) e ξ iωit– qi

* ω′t q– *ξ iωi

ωi ′---------- ωi ′2– ξ i

2ωi2–( )q*+

ω′tsin+cos –=

ωi

ω'i------

pi

mi------e ξ iωi t τ–( )– ωi ′ t τ–( ) δi–[ ]sin τd

0

t

∫–δitan1 ξ i

2–( )1 2/

ξ i--------------------------=

i 1 2 … n, , ,=

i 1=q1Φ1

i 2 3 …, ,=

ξ1 α α 2ξ1 ω1⁄=

ξ1

Sv 1.2 m/s= ξ 0.02=


Table 3.3: Example shear buildings

BuildingNumber of

storiesStory height

(meters)Mass/story

(103kg)

1 3 4 10 0.005

2 6 5 10 0.005

3 9 5 10 0.005

γobj


Fig. 3.30: Scaled time history and frequency content of El Centro ( &)

0 10 20 30 40 50 60−3

−2

−1

0

1

2

3

4

time s

accele

ration m

/s2

0 20 40 60 80 100 120 140 1600

50

100

150

ω (rad/s)

Spectr

um

Sv 1.2=ξ 0.02=


Fig. 3.31: Scaled time history and frequency content of Northridge: Station 1component 090 ( & )

0 10 20 30 40 50 60−3

−2

−1

0

1

2

3

4

time s

acce

lera

tio

n

m/s

2

0 20 40 60 80 100 120 140 1600

20

40

60

80

100

120

ω (rad/s)

Sp

ectr

um

Sv 1.2= ξ 0.02=


Fig. 3.32: Scaled time history and frequency content of Northridge: Station 3component 090 ( & )

0 10 20 30 40 50 60−2

−1

0

1

2

3

time s

acce

lera

tio

n

m/s

2

0 20 40 60 80 100 120 140 1600

20

40

60

80

100

120

140

ω (rad/s)

Sp

ectr

um

Sv 1.2= ξ 0.02=


Stiffness and damping distributions are generated for values of ranging from0.02 to 0.3. For each value, the following quantities are determined: ; thestory stiffness coefficients, ; the damping parameter ; and lastly thestory damping coefficients, . Figure 3.33 shows the variation of thefrequency and period for the first three modes with increasing damping of thefundamental mode.

Fig. 3.33: Frequencies and periods of first three modes

ξ1ξ1 ω1

ki α 2ξ1 ω1⁄=ci αki=

0 0.1 0.2 0.3 0.40

20

40

60

ξ − 3 Story Building

ω (

rad

/s)

Mode 1Mode 2Mode 3

0 0.1 0.2 0.3 0.40

0.2

0.4

0.6

0.8

T (

s)


0 0.1 0.2 0.3 0.40

10

20

30

40


ω (

rad

/s)

0 0.1 0.2 0.3 0.40

0.5

1

1.5

2

T (

s)


0 0.1 0.2 0.3 0.40

10

20

30


ω (

rad

/s)

0 0.1 0.2 0.3 0.40

1

2

3

T (

s)


ξ1 ξ1

ξ1 ξ1

ξ1 ξ1


As is increased, the required stiffness decreases, and the correspondingdamping increases. The total stiffness is defined as the sum of the storystiffnesses. A similar definition is used for the story damping coefficients. Thesetwo quantities provide a measure of the “design” cost. The data for the threeexample structures are plotted in Fig 3.34. Total damping varies essentiallylinearly with , whereas total stiffness behaves nonlinearly, and the degree ofnonlinearly increases with the number of stories. Given costs for stiffnesselements and dampers, one can generate data on cost as a function of andestablish an “optimal” damping system for stiffness proportional damping.

Fig. 3.34: Total stiffness (N/m) and total damping (N-s/m) for the three examplebuildings

ξ1

ξ

ξ1

0 0.1 0.2 0.3 0.40

1

2

3x 10

7 Total Stiffness


N/m

0 0.1 0.2 0.3 0.40

2

4

6

8x 10

7


N/m

0 0.1 0.2 0.3 0.40

2

4

6

8x 10

7


N/m

0 0.1 0.2 0.3 0.40

2

4

6

8x 10

5 Total Damping


N/(

m s

)

0 0.1 0.2 0.3 0.40

1

2

3x 10

6


N/(

m s

)

0 0.1 0.2 0.3 0.40

2

4

6x 10

6


N/(

m s

)


Using time-history analysis, the peak shear deformation was generated foreach story of each building, and each earthquake. Figure 3.35 shows the variationof the mean deformation (the average over the building height) and standarddeviation with . As expected, the response depends on the earthquake eventhough the earthquakes are scaled to have the same . The ensemble averageprovides a better estimate of the response. In general, the deformations are lowerthan the target value and the gap widens as is increased. Taking asthe typical value, the results are within 80% of the target. Considering theuncertainty associated with specifying the seismic excitation and modeling thestructure, one needs to apply a conservative design procedure.

Fig. 3.35: Mean story deformation and standard deviation for designearthquakes

ξ1Sv

ξ1 ξ1 0.1=

0 0.1 0.2 0.3 0.40

2

4

6

x 10−3

El C

entro

0 0.1 0.2 0.3 0.40

2

4

6

x 10−3

Nor

thrid

ge01

−090

0 0.1 0.2 0.3 0.40

2

4

6

x 10−3

Nor

thrid

ge03

−090

0 0.1 0.2 0.3 0.40

2

4

6

x 10−3


Ens

embl

e A

vera

ge

0 0.1 0.2 0.3 0.40

2

4

6

x 10−3


0 0.1 0.2 0.3 0.40

2

4

6

x 10−3

0 0.1 0.2 0.3 0.40

2

4

6

x 10−3

0 0.1 0.2 0.3 0.40

2

4

6

x 10−3

(a) 3 story building (b) 6 story building


Fig. 3.35: Mean story deformation and standard deviation for designearthquakes

Example 3.16: Building #4

The building designated as example #4 in section 2.9, Table 2.4 isconsidered here. This building has a height of 200m, which corresponds to about50 stories. The relevant properties are listed below for convenience.

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.40

2

4

6

x 10−3

El C

entr

o

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.40

2

4

6

x 10−3

Nort

hridge01

−090

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.40

2

4

6

x 10−3

Nort

hridge03

−090

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.40

2

4

6

x 10−3


Ense

mble

Ave

rage

(c) 9 story building


Table 3.4: Properties for Building #4

Three values of the damping ratio for the fundamental mode areconsidered. Figure 2.27 is used to obtain the value of Sv corresponding to eachvalue of . Given Sv, one finds T with eqn (2.220), and the rigidity distributionswith eqns (2.119) and (2.220). These rigidity distributions are modified using theapproach described in sections 2.10, and are scaled during the iteration processsuch that the final values of T are equal to the values predicted by eqn (2.220).Table 3.5 contains the design data for Sv and the corresponding periods. The“converged” rigidity distributions for each damping ratio are plotted in Figure3.36, 3.37, and 3.38.

Table 3.5: Design data for Building #4

H = 200m H/B = 5

= 20,000 kg/m

Design earthquake: Sv = 1.2 m/s for = 0.02

Design deformation:= 1/200

= 4s = 0.63

Case Design values Period

Sv (m/s) T1 (s)

1 0.02 1.2 1/200 5.36

2 0.05 0.94 1/200 6.85

3 0.10 0.70 1/200 9.25

ρm

ξ

γ*

f *

ξ

ξ1 γ*


Fig. 3.36: Converged shear rigidity distribution for Building 4, = 0.02


0 2 4 6 8 10 12 14

x 108

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----

Shear rigidity distribution DT - N

Building #4Quadratic BasedT=5.36 s

SRSS

mode 1

mode 2

mode 3

ξ

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x 109

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Building #4Quadratic basedT = 6.85 s


Nor

mal

ized

hei

ght

x H----

ξ



Profiles of the maximum transverse shear deformation due to seismicexcitation scaled such that Sv = 1.2 m/s for = 0.02 are plotted in Figures 3.39,3.40, and 3.41. The trend is similar to what was observed for the shear buildingexamples examined in the previous section. As the damping is increased, themean deformation trends to decrease, even though the design value of thespectral velocity is decreased with increasing . This behavior indicates that thedesign procedure is conservative, at least for this example building. It should benoted that the periods are in the range where the dynamic amplification is not assignificant as for low period buildings. Also, one should consider a number ofexcitations in order to generate a more representative “ensemble” average.

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x 109

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----


Building #4Quadratic basedT = 9.2 s

ξ

ξ

ξ


Fig. 3.39: Maximum shear deformation for Building 4, = 0.02

Fig. 3.40: Maximum shear deformation for Building 4, = 0.05

0 0.002 0.004 0.006 0.008 0.01 0.0120

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1N

orm

aliz

ed h

eigh

tx H----

γ∗

_________ El Centro- - - - - - - Taft

(Sv=1.2 m/s, ξ=0.02)

Maximum shear deformation γ - m/m

Building #4T = 5.36 s

ξ

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----


El CentroTaftγ∗


ξ1


Fig. 3.41: Maximum shear deformation for building 4, = 0.10

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----


El CentroTaft

γ∗


ξ1


Problems 243

Problems

Problem 3.1

Consider a SDOF system. The total stored energy is the sum of the kineticenergy and the strain energy.

Note that and are out of phase for periodic excitation. Suppose there is arapid build-up of strain energy during the initial phase of a seismic excitation,and the design objective is to limit to a target value, . What strategywould you apply? Assume the mass cannot be varied.

Problem 3.2

Give examples of energy dissipation and absorption devices that are usedin structures such as vehicles, crash barriers, machine supports, buildings, andbridges.

Problem 3.3

Consider a SDOF system having a linear spring and linear viscous damper.Suppose the initial conditions at are:

(1)

and there is no external loading.

Stored Energy ES EK+=

12---ku2 1

2---mu2+

=

ES EK

umax u∗

t 0=

u 0( ) 0= u 0( ) u∗=

u

k

c


a) Determine the expression for the free vibration response corresponding to theseinitial conditions.

b) Construct a plot of vs. for ranging from 0 to . Comment on theinfluence of damping on the time history response. Illustrate for ,

.

Problem 3.4

Refer to problem 3.3. Suppose the device is a Coulomb function damper.

a) Determine the free vibration response. Since the sense of the friction force isdetermined by the sense of the velocity, one needs to generate solutions fordifferent time intervals. During the first interval, , the velocity is positive.

b) Construct a plot of vs. for sufficient to include one full cycle.

Problem 3.5

The SDOF system shown below is subjected to a periodic force. Determinethe expressions for the spring and damper forces. Let denote the total internalforce.

(1)

Determine the expression for . Compare the magnitude of with themagnitudes of and .

u u t 4π ω⁄m 1 000 kg,=

k 9 000 kN/m,=

0 t t1≤ ≤

u u t

F

F Fs Fd+=

u

k

c

m

Fs

Fd

F p Ωtsin

F FFs Fd

Problems 245

Problem 3.6

Refer to example 3.3. Consider material “one” to be low strength steel with, and material “two” to have steel. Design a

hysteretic damper for the following criteria:

Problem 3.7

Design a viscoelastic damper for the following conditions:

• Temperature =• Frequency excitation = 1 Hz• Material 3M-ISD110• Maximum displacement = 0.025 m• Maximum force = 10 kN

Problem 3.8

The SDOF system shown above has a linear viscoelastic damper whichproduces the force F. Assume and use given by eqn (3.45).Determine .

σy 200 MPa= σy 500 MPa=

Fy 100 kN=

L1 L2+ 5 m=

kh 6 000 kN/m,=

20°

u

k

F

m p

u u Ωtsin= Fp


Problem 3.9

A convenient way of dealing with periodic excitation is to introducecomplex notation. The basic identity is

(1)

where is an arbitrary scalar. Using this notation, an arbitrary periodic forcingcan be expressed in terms of a complex amplitude,

(2)

One can work with the complex form, and then retain either the real term (forcosine forcing) or the imaginary term (for sine forcing).

a) Consider a 1 DOF system having a spring and linear viscous damper inparallel. Suppose is a general periodic excitation, . Letrepresent the response. Express the relationship between and aswhere is the complex system stiffness. Determine .

b) Consider a linear spring and viscous damper in series. Determine .

c) Refer to part a. Express the relationship between p and u as .Determine . Suppose the loading consists of a set of periodic

eiθ θ i θsin+cos=

θp

p peiΩt peiδ( )eiΩt pei Ωt δ+( )= = =

p Ωt δ+( ) i Ωt δ+( )sin+cos( )=

p p peiΩt= u ueiΩt=p u p Ku=

K K

k

c

m p u,

K

k cm p u,

u Hp=H Heiα

=

Problems 247

excitations,

(3)

Express u in terms of and . Discuss how you would evaluate u at sometime t for a cosine forcing, i.e., the loading corresponding to the real part of eqn(3).

Problem 3.10

Equations (3.28) and (3.29) define the stress-strain response of a linearviscoelastic material to periodic sinusoidal excitation. Using complex notation,one can generalize these equations for arbitrary periodic excitation, ,where is the complex amplitude. The complex shear modulus is defined as

(1)

where

(2)

With these definitions, the shear stress can be written as

(3)

a) Consider a linear viscoelastic damper subjected to periodic excitation,. Express the force as and let . Determine by

generalizing eqn (3.41).

b) Consider a SDOF system having a spring and linear viscoelastic device inparallel. Assume is a general periodic excitation and let . Determine .Equate this result to the result for a linear spring and linear viscous damper inparallel (see problem 3.9a), and determine the equivalent stiffness and dampingfactors, and .

p p1eiΩ1t

p2eiΩ2t

… pneiΩnt

+ + +=

pj H j

γ γeiΩt=γ

G Geiδ=

δtan η the loss factor,=

G Gs2 Gl

2+[ ] 1 2/=

τ τ eiΩt=

τ Gγ=

u ueiΩt= F FeiΩt= F kvu= kv

p p Ku= K

k' c'


c) Consider a linear spring and linear viscoelastic device in series. Determine .

d) Consider a SDOF system having a linear spring in parallel with a linearspring/linear viscoelastic device. Determine

Problem 3.11

Consider a single degree of freedom system having = 1000 kg and =36 kN/m. This problem concerns designing various types of damping devices.

k

m p u,

(Actual)

k'

m

(Equivalent Viscous)c’

K

km p u,

K

k1

m p u,

k

m k

Problems 249

Assume the system is subjected to seismic excitation, and the design objective is adamping ratio of 0.1. Take = 1 m/s as the response for pure viscous dampingwith = 0.1. Propose damper properties for the following cases:

1.Viscous2.Structural3.Coulomb4.Hysteretic with = 5

Discuss the basis for your recommendations.

Problem 3.12

Refer to problem 3.10, part b. Take = 2000 kg and = 25 kN/m.Determine the equivalent stiffness and viscous damping coefficients for the casewhere the material is 3M ISD110, the excitation frequency ranges from 0.5 Hz to 5Hz, and .

Problem 3.13

Refer to problem 3.10, part b. Suppose = 4000 kg and = 50 kN/m.Design a visco-elastic damper to produce an equivalent viscous damping of3 kNs/m at an excitation frequency of 2 Hz. Use 3M ISD110 material. Alsodetermine the corresponding equivalent stiffness.

Problem 3.14

Refer to problem 3.10, part b. The equivalent stiffness and viscousdamping coefficients corresponding to a periodic excitation of frequency are:

(1)

(2)

where and are functions of . Suppose one wants to approximate thesecoefficients with “constant” values over the frequency range of interest.

Svξ

µ

m k

f d 10 2– m=

m k

Ω

k' k f dGs+=

c'η f dGsΩ

-----------------=

Gs η Ω


(3)

(4)

Show that a least square approximation leads to:

(5)

(6)

where and are defined by equations (3.74) and (3.75).

Problem 3.15

Consider a SDOF system having a spring and a viscoelastic dampermodelled as shown below. Take .

a) Determine expressions for the fundamental frequency, , and damping ratio,.

b) Suppose m = 1000 kg and k = 9 kN/m. Determine the damper properties c andk1 that correspond to a damping ratio of 0.1. Take . Also determine thefundamental frequency.

c) Suppose m = 1000 kg and the total stiffness, , is equal to 9kN/m.Determine the stiffness and damping parameters required for . Take

.

Problem 3.16

Refer to example 3.12

k' Ω( ) k*≈

c' Ω( ) c*≈

k* k keq+=

c* ceq=

keq ceq

c αdk1=

ωξ

αd 0.15=

k k1+ξ 0.1=

αd 0.15=

m

uc

k

k1

Problems 251

a) Take C = 600 N s/m. Determine and corresponding to a periodicexcitation, , and the following values for stiffness:

1.

2.

3.

b) Suppose m = 1000kg. Determine and corresponding to the values ofand obtained in part a.

Problem 3.17

Consider the truss beam segment shown below. Assume the diagonalbracing system consists of elastic elements and linear viscous dampers attached torigid links connecting the end nodes. Dimension the elastic element and damperfor the following requirements:

Use steel for the elastic element.

Problem 3.18

Refer to problem 3.17. Assume the diagonal bracing system consists of

Keq CeqΩ 2π r s⁄=

K 9 kN m⁄= K2 4.5 kN m⁄=

K 9 kN m⁄= K2 18 kN m⁄=

K 9 kN m⁄= K2 ∞=

ω ξ KeqCeq

DT 260 103× kN=

T 5 103× kN s/m=

45 0

4 m


elastic elements and linear viscoelastic elements. Dimension the components forthe requirements specified in problem 3.17. Use equivalent stiffness and dampingparameters for the viscoelastic component, and take .

Problem 3.19

Consider a structural system composed of 2 subsystems: the first is linearelastic and provides only stiffness; the second system is linear viscoelastic andprovides both stiffness and damping. Let denote the stiffness matrix for theelastic component and , the stiffness and damping matrices for theviscoelastic component. The equations of motion are:

(1)

where is related to . Specializing eqn(1) for free vibration and expressingthe solution as:

(2)

leads to the following equation:

(3)

The solution for the primary elastic system is expressed as:

(4)

(5)

a) Consider the pure elastic case, and . Determine thecorresponding and for the case where as a scalar multiple of , say

.

b) Consider the pure viscous case and . Take . Determine and . Express as:

(6)

c) Consider the linear viscoelastic case. Take where is a materialparameter. Assume stiffness proportional damping and consider 2 cases:

αd 0.15=

KeK1 C

MU CU Ke K1+( )U+ + P=

C K1

U Aeλ tΦ=

λ2M λC Ke K1+( )+ +[ ]Φ 0=

λ iω±= ω 0>

Ke ω2M–( )Φe 0=

C 0= K1 0≠λ Φ K1 Ke

K1 α1Ke=

C 0≠ K1 0= C α2Ke=λ Φ λ

λ ξω– iω'±=

C αdK1= αd

Problems 253

1.

2.

Determine expressions for the corresponding and .

d) Consider the case where and . Suppose thesolution is expressed as:

(7)

where are the undamped modal vectors, i.e., the solution of eqn (5). Assume are normalized with respect to .

(8)

Derive the governing equation and initial conditions for . Express it in a formsimilar to eqn (3.166), using “equivalent” terms for and . Utilize the resultsgenerated in part a to establish the natural frequency for the case where

. Discuss how damping and frequency vary with , for a givenviscoelastic material. Note that for pure viscous damping.

Problem 3.20

The governing equation for an order system is given by eqn (3.143)

(1)

Expressing the homogeneous solution as

(2)

and specializing eqn (1) for leads to:

(3)

Various cases ( , , and arbitrary) are discussed in section 3.7.

C α2Ke=

C α3 Ke K1+( )=

λ Φ

C α3 Ke K1+( )= C αdK1=

U qjΦe j,j 1=

N

∑=

Φe j,Φe M

Φe d,T MΦe d, m j=

qjωj ξ j

K Ke K1+= α3αd ∞=

n'th

MU CU KU+ + P=

U Aeλ tΦ=

P 0=

λ2M Cλ K+ +( )Φ 0=

C 0= C αK= C


Suppose one defines a new variable, ,

(4)

Differentiating and substituting for using eqn (1), one obtains:

(5)

Denoting the coefficient matrix as , eqn (5) simplifies to

(6)

The homogeneous solution of eqn (6) has the general form:

a) Determine the equation relating and .

b) Show that , i.e., is an eigenvalue of eqn (3).

c) What is the significance of the real part of ?

d) Consider a 2 DOF discrete shear beam having the following properties:

Form and determine the eigenvalues and eigenvectors using the eigen routinein MATLAB.

e) Repeat part d using the MOTIONLAB program.

X

X U

U=

X U

X U

U

0 I

M 1– K– M 1– C–

U

U

0

M 1– P+= =

A

X AX B+=

X eα tΨ=

α Ψ

α λ= α

α

m1 m2 1000 kg 1 kNs2 m⁄= = =

k1 118.3 kN m⁄ k2 78.9 kN/m= =

c1 2.09 kNs/m c2 4.18 kNs/m==

A

Problems 255

Problem 3.21

Consider the 3 DOF system shown above.

a) Determine the magnitudes of , , and such that the first mode hasthe form

and the fundamental frequency is rad/sec. Take = 2000 kg.

b) Determine the viscous damping coefficients , , and such that thedamping ratio, , for the first mode is 0.1. Consider both stiffness proportionaldamping and uniform damping. Use MOTIONLAB to determine the modalproperties for the case of non-proportional damping.

c) Suppose viscoelastic dampers are used. Let and denote theequivalent stiffness and damping coefficients, and take where is amaterial property. Determine and the elastic stiffness for each element.Assume . Consider stiffness proportional damping and uniformdamping. Compare the modal properties for these 2 cases.

Problem 3.22

Refer to example 3.7. Use MOTIONLAB to determine the modal propertiesfor the first 3 modes. Take according to eqn (1) and based on:

u3

u1

u2

k3 c3,

k2 c2,

k1 c1,

m3 2m=

m1 m=

m2 m=

k1 k2 k3

Φ113--- 2

3--- 1, ,

=

2π m

c1 c2 c3ξ1

kd cdcd αdkd= αd

cd i, ke i,αd 0.15=

ki ci


a) damping proportional to stiffness

b) uniform damping

c) and

d) other combinations of the that you believe may be more optimal in the senseof resulting in higher damping ratios for the higher modes.

Problem 3.23

Refer to example 3.7. Suppose viscoelastic dampers are used for elements3, 4, 5 and there is no damping in elements 1 and 2. Assume uniform viscoelasticdamper properties and take . Starting with the data contained in eqn(1) of example 3.7, determine the damper properties and modified elastic stiffnesswhich correspond to a damping ratio of 0.1 for the fundamental mode. UsingMOTIONLAB, assess the effect of non-proportional damping on the mode shapefor this design value of . What would be the effect if is increased to 0.2 andthe stiffness is maintained at the level defined by eqn (1)?

Problem 3.24

This problem concerns the preliminary design of a 10 story rectangularrigid frame for seismic excitation. The frame properties and design criteria are:

• Height = 5 m/story

c1 c2 0= = c3 c4 c5= =

c'

αd 0.15=

ξ1 ξ1

10 m

5 m

Problems 257

• Width = 10 m/bay

• Mass/floor = 10,000 kg

• Max. deflection at top = 0.25 m

• Max. story shear deformation = 1/200

• Response spectrum shown below.

The chevron bracing system is similar to the scheme shown in Fig. 3.23(b); itallows 2 viscous dampers to be placed on each floor.

a) Determine the stiffness distributions based on a linear fundamental modeprofile and equal to 0.02, 0.10, and 0.20. Evaluate the stiffness cost, , foreach distribution.

b) Assume stiffness proportional damping. Determine the correspondingdistributions for the story damping coefficients, and the damping cost, .

c) Repeat b) for “uniform” damping. Using MOTIONLAB, determine theproperties for the fundamental mode. Compare the displacement profiles.

d) Repeat b) for and .Determine the actual damping ratio and profile for the fundamental mode.

e) The damper cost increases nonlinearly with the damper coefficient, i.e., the cost

0.1 0.6 1.0 10

0.1

1.0

Period

Spectral Velocity

T (sec)

Sv (m/s)

Sv 1.2 ξ, 0.02= =

Sv 0.8 ξ, 0.1= =

Sv 0.6 ξ, 0.2= =

ξ ki∑

ci∑

c1 c2 c3 c4 c5 0= = = = = c6 c7 c8 c9 c10 c*= = = = =


for 2c is more than twice the cost for c. Also, the damper force increases with c andplaces more loading on the brace-floor connection. With these limitations, discusshow you would select a damper placement that satisfies the performancerequirement on the maximum transverse shear deformation for each story andminimizes a “cost” function.

Problem 3.25

Consider a 5 DOF shear beam having equal masses and equal nodal forces.Suppose the force consists of a combination of a static component and a periodicexcitation,

where is a random quantity. Take

a) Determine the stiffness distribution such that the interstorydisplacement under the static loading is 0.02m for each story.

b) Assume coincides with the frequency for the fundamental modecorresponding to the stiffness distribution generated in part (a). Suppose thedesign objective is to have the peak acceleration less than 0.02g where g is theacceleration due to gravity and is equal to 9.87m/s2. Suggest various schemes forgenerating the required energy dissipation. Comment on what you consider to bethe optimal solution.

p ps pd ωtsin+=

ω

mi 10 000kg,=

ps 100kN=

pd 1kN=

ω

259

Chapter 4

Tuned mass damper systems

4.1 Introduction

A tuned mass damper (TMD) is a device consisting of a mass, a spring, and adamper that is attached to a structure in order to reduce the dynamic response ofthe structure. The frequency of the damper is tuned to a particular structuralfrequency so that when that frequency is excited, the damper will resonate out ofphase with the structural motion. Energy is dissipated by the damper inertia forceacting on the structure. The TMD concept was first applied by Frahm in 1909(Frahm, 1909) to reduce the rolling motion of ships as well as ship hull vibrations.A theory for the TMD was presented later in the paper by Ormondroyd & DenHartog (1928), followed by a detailed discussion of optimal tuning and dampingparameters in Den Hartog’s book on Mechanical Vibrations (1940). The initialtheory was applicable for an undamped SDOF system subjected to a sinusoidalforce excitation. Extension of the theory to damped SDOF systems has beeninvestigated by numerous researchers. Significant contributions were made byRandall et al. (1981), Warburton (1980,1981,1982), and Tsai & Lin (1993).

This chapter starts with an introductory example of a TMD design and abrief description of some of the implementations of tuned mass dampers inbuilding structures. A rigorous theory of tuned mass dampers for SDOF systemssubjected to harmonic force excitation and harmonic ground motion is discussednext. Various cases including an undamped TMD attached to an undamped

260 Chapter 4: Tuned Mass Damper Systems

SDOF system, a damped TMD attached to an undamped SDOF system, and adamped TMD attached to a damped SDOF system are considered. Time historyresponses for a range of SDOF systems connected to optimally tuned TMD andsubjected to harmonic and seismic excitations are presented. The theory is thenextended to MDOF systems where the TMD is used to dampen out the vibrationsof a specific mode. An assessment of the optimal placement locations of TMDs inbuilding structures is included. Numerous examples are provided to illustrate thelevel of control that can be achieved with such passive devices for both harmonicand seismic excitations.

4.2 An introductory example

In this section, the concept of the tuned mass damper is illustrated using the two-mass system shown in Fig. 4.1. Here, the subscript d refers to the tuned massdamper; the structure is idealized as a single degree of freedom system.Introducing the following notation

(4.1)

(4.2)

(4.3)

(4.4)

and defining as the mass ratio,

(4.5)

Fig. 4.1: SDOF - TMD system.

ω2 km----=

c 2ξωm=

ωd2 kd

md-------=

cd 2ξdωdmd=

m

mmdm

-------=

k kd

c cd

m md

u u ud+

p

4.2 An Introductory Example 261

the governing equations of motion are given by

Primary mass (4.6)

Tuned mass (4.7)

The purpose of adding the mass damper is to limit the motion of thestructure when it is subjected to a particular excitation. The design of the massdamper involves specifying the mass , stiffness , and damping coefficient

. The optimal choice of these quantities is discussed in Section 4.4. In thisexample, the near-optimal approximation for the frequency of the damper,

(4.8)

is used to illustrate the design procedure. The stiffnesses for this frequencycombination are related by

(4.9)

Equation (4.8) corresponds to tuning the damper to the fundamental period of thestructure.

Considering a periodic excitation,

(4.10)

the response is given by

(4.11)

(4.12)

where and denote the displacement amplitude and phase shift respectively.The critical loading scenario is the resonant condition, . The solution forthis case has the following form

(4.13)

(4.14)

1 m+( ) u 2ξωu ω2u+ + pm---- mud–=

ud 2ξdωdud ωd2 ud+ + u–=

md kdcd

ωd ω=

kd mk=

p p Ωtsin=

u u Ωt δ1+( )sin=

ud ud Ωt δ1 δ2+ +( )sin=

u δΩ ω=

up

km------- 1

1 2ξm------ 1

2ξd---------+

2+

----------------------------------------=

ud1

2ξd---------u=


(4.15)

(4.16)

Note that the response of the tuned mass is 900 out of phase with the response ofthe primary mass. This difference in phase produces the energy dissipationcontributed by the damper inertia force.

The response for no damper is given by

(4.17)

(4.18)

To compare these two cases, one can express eqn (4.13) in terms of an equivalentdamping ratio

(4.19)

where

(4.20)

Equation (4.20) shows the relative contribution of the damper parameters to thetotal damping. Increasing the mass ratio magnifies the damping. However, sincethe added mass also increases, there is a practical limit on . Decreasing thedamping coefficient for the damper also increases the damping. Noting eqn (4.14),the relative displacement also increases in this case, and just as for the mass, thereis a practical limit on the relative motion of the damper. Selecting the final designrequires a compromise between these two constraints.

Example 4.1: Preliminary design of a TMD for a SDOF system

Suppose and one wants to add a tuned mass damper such that theequivalent damping ratio is . Using eqn (4.20), and setting , thefollowing relation between and is obtained.

δ1tan 2ξm------ 1

2ξd---------+–=

δ2tan π2---–=

upk--- 1

2ξ------

=

δ1π2---–=

upk--- 1

2ξe--------

=

ξem2---- 1 2ξ

m------ 1

2ξd---------+

2+=

m

ξ 0=10% ξe 0.1=

m ξd

4.2 An Introductory Example 263

(4.21)

The relative displacement constraint is given by eqn (4.14)

(4.22)

Combining eqn (4.21) and eqn (4.22), and setting leads to

(4.23)

Usually, is taken to be an order of magnitude greater than . In this case eqn(4.23) can be approximated as

(4.24)

The generalized form of eqn (4.24) follows from eqn (4.20):

(4.25)

Finally, taking yields an estimate for

(4.26)

This magnitude is typical for . The other parameters are

(4.27)

and from eqn (4.9)

(4.28)

It is important to note that with the addition of only of the primarymass, one obtains an effective damping ratio of . The negative aspect is thelarge relative motion of the damper mass; in this case, times the displacementof the primary mass. How to accommodate this motion in an actual structure is animportant design consideration.

A description of some applications of tuned mass dampers to building

m2---- 1 2ξ

m------ 1

2ξd---------+

2+ 0.1=

ud1

2ξd---------

u=

ξ 0=

m2---- 1

udu

------

2+ 0.1=

ud u

m2----

udu

------ 0.1≈

m 2ξe1

ud u⁄-------------

≈

ud 10u= m

m 2 0.1( )10

--------------- 0.02= =

m

ξd12--- u

ud------

0.05= =

kd mk 0.02k= =

2%10%

10


structures is presented in the following section to provide additional backgroundon this type of device prior to entering into a detailed discussion of theunderlying theory.

4.3 Examples of existing tuned mass damper systems

Although the majority of applications have been for mechanical systems, tunedmass dampers have been used to improve the response of building structuresunder wind excitation. A short description of the various types of dampers andseveral building structures that contain tuned mass dampers follows.

Translational tuned mass dampers

Figure 4.2 illustrates the typical configuration of a unidirectionaltranslational tuned mass damper. The mass rests on bearings that function asrollers and allow the mass to translate laterally relative to the floor. Springs anddampers are inserted between the mass and the adjacent vertical supportmembers which transmit the lateral “out-of-phase” force to the floor level, andthen into the structural frame. Bidirectional translational dampers are configuredwith springs/dampers in 2 orthogonal directions and provide the capability forcontrolling structural motion in 2 orthogonal planes. Some examples of earlyversions of this type of damper are described below.

Fig. 4.2: Schematic diagram of a translational tuned mass damper.

md

Support

Floor Beam

Direction of motion

4.3 Examples of Existing Tuned Mass Damper Systems 265

• John Hancock Tower (Engineering News Record, Oct. 1975)

Two dampers were added to the 60-story John Hancock Tower in Boston toreduce the response to wind gust loading. The dampers are placed at oppositeends of the 58th story, 67m apart, and move to counteract sway as well as twistingdue to the shape of the building. Each damper weighs 2700 kN and consists of alead-filled steel box about 5.2m square and 1m deep that rides on a 9m long steelplate. The lead-filled weight, laterally restrained by stiff springs anchored to theinterior columns of the building and controlled by servo-hydraulic cylinders,slides back and forth on a hydrostatic bearing consisting of a thin layer of oilforced through holes in the steel plate. Whenever the horizontal accelerationexceeds 0.003g for two consecutive cycles, the system is automatically activated.This system was designed and manufactured by LeMessurier Associates/SCI inassociation with MTS System Corp., at a cost of around 3 million dollars, and isexpected to reduce the sway of the building by 40% to 50%.

• Citicorp Center (Engineering News Record Aug. 1975, McNamara1977, Petersen 1980)

The Citicorp (Manhattan) TMD was also designed and manufactured byLeMessurier Associates/SCI in association with MTS System Corp. This buildingis 279m high, has a fundamental period of around 6.5s with an inherent dampingratio of 1% along each axis. The Citicorp TMD, located on the 63rd floor in thecrown of the structure, has a mass of 366 Mg, about 2% of the effective modalmass of the first mode, and was 250 times larger than any existing tuned massdamper at the time of installation. Designed to be biaxially resonant on thebuilding structure with a variable operating period of , adjustablelinear damping from 8% to 14%, and a peak relative displacement of , thedamper is expected to reduce the building sway amplitude by about 50%. Thisreduction corresponds to increasing the basic structural damping by 4%. Theconcrete mass block is about 2.6m high with a plan cross-section of 9.1m by 9.1mand is supported on a series of twelve 60cm diameter hydraulic pressure-balanced bearings. During operation, the bearings are supplied oil from aseparate hydraulic pump which is capable of raising the mass block about 2cm toits operating position in about 3 minutes. The damper system is activatedautomatically whenever the horizontal acceleration exceeds 0.003g for twoconsecutive cycles, and will automatically shut itself down when the building

6.25s 20%±1.4m±


acceleration does not exceed 0.00075g in either axis over a 30 minute interval.LeMessurier estimates Citicorp’s TMD, which cost about 1.5 million dollars,saved 3.5 to 4 million dollars. This sum represents the cost of some 2,800 tons ofstructural steel that would have been required to satisfy the deflection constraints.

• Canadian National Tower (Engineering News Record, 1976)

The 102m steel antenna mast on top of the Canadian National Tower in Toronto(553m high including the antenna) required two lead dampers to prevent theantenna from deflecting excessively when subjected to wind excitation. Thedamper system consists of two doughnut-shaped steel rings, 35cm wide, 30cmdeep, and 2.4m and 3m in diameter, located at elevations 488m and 503m. Eachring holds about 9 metric tons of lead and is supported by three steel beamsattached to the sides of the antenna mast. Four bearing universal joints that pivotin all directions connect the rings to the beams. In addition, four separatehydraulically activated fluid dampers mounted on the side of the mast andattached to the center of each universal joint dissipate energy. As the lead-weighted rings move back and forth, the hydraulic damper system dissipates theinput energy and reduces the tower’s response. The damper system was designedby Nicolet, Carrier, Dressel, and Associates, Ltd, in collaboration with VibronAcoustics, Ltd. The dampers are tuned to the second and fourth modes ofvibration in order to minimize antenna bending loads; the first and third modeshave the same characteristics as the prestressed concrete structure supporting theantenna and did not require additional damping.

• Chiba Port Tower (Kitamura et al. 1988)

Chiba Port Tower (completed in 1986) was the first tower in Japan to be equippedwith a TMD. Chiba Port Tower is a steel structure 125m high weighing 1950metric tons and having a rhombus shaped plan with a side length of 15m. Thefirst and second mode periods are 2.25s and 0.51s respectively for the X directionand 2.7s and 0.57s for the Y direction. Damping for the fundamental mode isestimated at 0.5%. Damping ratios proportional to frequencies were assumed forthe higher modes in the analysis. The purpose of the TMD is to increase dampingof the first mode for both the X and Y directions. Figure 4.3 shows the dampersystem. Manufactured by Mitsubishi Steel Manufacturing Co., Ltd, the damperhas: mass ratios with respect to the modal mass of the first mode of about 1/120 inthe X direction and 1/80 in the Y direction; periods in the X and Y directions of2.24s and 2.72s respectively; and a damper damping ratio of 15%. The maximum


relative displacement of the damper with respect to the tower is about ineach direction. Reductions of around 30% to 40% in the displacement of the topfloor and 30% in the peak bending moments are expected.

Fig. 4.3: Tuned mass damper for Chiba-Port Tower.

The early versions of TMD’s employ complex mechanisms for the bearingand damping elements, have relatively large masses, occupy considerably space,and are quite expensive. Recent versions, such as the scheme shown in Fig 4.4,have been designed to minimize these limitations. This scheme employs a multi-assemblage of elastomeric rubber bearings, which function as shear springs, andbitumen rubber compound (BRC) elements, which provide viscoelastic dampingcapability. The device is compact in size, requires unsophisticated controls, ismultidirectional, and is easily assembled and modified. Figure 4.5 shows a fullscale damper being subjected to dynamic excitation by a shaking table. An actualinstallation is contained in Fig. 4.6.

1m±


Fig. 4.4: Tuned mass damper with spring and damper assemblage.

Fig. 4.5: Deformed position - tuned mass damper.


Fig. 4.6: Tuned mass damper - Huis Ten Bosch Tower, Nagasaki.

The effectiveness of a tuned mass damper can be increased by attaching anauxiliary mass and an actuator to the tuned mass and driving the auxiliary masswith the actuator such that its response is out of phase with the response of thetuned mass. Fig 4.7 illustrates this scheme. The effect of driving the auxiliary massis to produce an additional force which complements the force generated by thetuned mass, and therefore increases the equivalent damping of the TMD (one canobtain the same behavior by attaching the actuator directly to the tuned mass,thereby eliminating the need for an auxiliary mass). Since the actuator requires anexternal energy source, this system is referred to as an active tuned mass damper.The scope of this chapter is restricted to passive TMD’s. Active TMD’s arediscussed in Chapter 6.


Fig. 4.7: An active tuned mass damper configuration.

Pendulum tuned mass damper

The problems associated with the bearings can be eliminated bysupporting the mass with cables which allow the system to behave as apendulum. Fig 4.8a shows a simple pendulum attached to a floor. Movement ofthe floor excites the pendulum. The relative motion of the pendulum produces ahorizontal force which opposes the floor motion. This action can be representedby an equivalent SDOF system which is attached to the floor as indicated in Fig4.8b.

Fig. 4.8: A simple pendulum tuned mass damper.

Support

Floor Beam

Direction of motion

ActuatorAuxiliary mass

md

ud u

Lθ

(a) actual system

keq

md

u+ud(b) equivalent system

t=0t

u


The equation of motion for the horizontal direction is

(4.29)

where T is the tension in the cable. When is small, the followingapproximations apply

(4.30)

Introducing these approximations transforms eqn (4.29) to

(4.31)

and it follows that the equivalent shear spring stiffness is

(4.32)

The natural frequency of the pendulum is related to keq by

(4.33)

Noting eqn (4.33), the natural period of the pendulum is

(4.34)

The simple pendulum tuned mass damper concept has a seriouslimitation. Since the period depends on L, the required length for large Td may begreater than the typical story height. For instance, the length for Td=5 secs is 6.2meters whereas the story height is between 4 to 5 meters. This problem can beeliminated by resorting to the scheme illustrated in Fig 4.9. The interior rigid linkmagnifies the support motion for the pendulum, and results in the followingequilibrium equation

(4.35)

The rigid link moves in phase with the damper, and has the same displacementamplitude. Then, taking u1= ud in eqn (4.35) results in

(4.36)

T θsinWdg

-------- u ud+( )+ 0=

θ

ud L θ Lθ≈sin=

T Wd≈

md ud

WdL

--------ud+ md u–=

keq

WdL

--------=

ωd2

keqmd------- g

L---= =

Td 2π Lg---=

md u u1 u+ d+( )WdL

--------ud+ 0=

md ud

Wd2L--------ud+

md2

------- u–=


The equivalent stiffness is Wd/2L , and it follows that the effective length is equalto 2L. Each additional link increases the effective length by L. An example of apendulum type damper is described below.

Fig. 4.9: Compound pendulum.

• Crystal Tower (Nagase & Hisatoku 1990)

The tower, located in Osaka Japan, is 157m high and 28m by 67m in plan, weighs44,000 metric tons, and has a fundamental period of approximately 4s in thenorth-south direction and 3s in the east-west direction. A tuned pendulum massdamper was included in the early phase of the design to decrease the wind-induced motion of the building by about 50%. Six of the nine air cooling andheating ice thermal storage tanks (each weighing 90 tons) are hung from the toproof girders and used as a pendulum mass. Four tanks have a pendulum length of4m and slide in the north-south direction; the other two tanks have a pendulumlength of about 3m and slide in the east-west direction. Oil dampers connected tothe pendulums dissipate the pendulum energy. Fig 4.10 shows the layout of theice storage tanks that were used as damper masses. Views of the actual buildingand one of the tanks are presented in Fig 4.11. The cost of this tuned mass dampersystem was around $350,000, less than 0.2% of the construction cost.

u+u1

md

u+u1+ud

u

L


Fig. 4.10: Pendulum damper layout - Crystal Tower.


Fig. 4.11: Ice storage tank - Crystal Tower.

A modified version of the pendulum damper is shown in Fig 4.12. Therestoring force provided by the cables is generated by introducing curvature inthe support surface and allowing the mass to roll on this surface. The verticalmotion of the weight requires an energy input. Assuming θ is small, the equationsfor the case where the surface is circular are the same as for the conventionalpendulum with the cable length L, replaced with the surface radius R.

4.4 Tuned Mass Damper Theory for SDOF Systems 275

Fig. 4.12: Rocker pendulum.

4.4 Tuned mass damper theory for SDOF systems

In what follows, various cases ranging from fully undamped to fully dampedconditions are analyzed and design procedures are presented.

Undamped structure - undamped TMD

Figure 4.13 shows a SDOF system having mass and stiffness , subjected toboth external forcing and ground motion. A tuned mass damper with massand stiffness is attached to the primary mass. The various displacementmeasures are: , the absolute ground motion; , the relative motion between the

R θ

md

u (a)

(b)

md

keq

Floor

m kmd

kdug u


primary mass and the ground; and , the relative displacement between thedamper and the primary mass. With this notation, the governing equations takethe form

(4.37)

(4.38)

where is the absolute ground acceleration and is the force loading applied tothe primary mass.

Fig. 4.13: SDOF system coupled with a TMD.

The excitation is considered to be periodic of frequency ,

(4.39)

(4.40)

Expressing the response as

(4.41)

(4.42)

and substituting for these variables, the equilibrium equations are transformed to

(4.43)

(4.44)

ud

md ud u+[ ] kdud+ mdag–=

mu ku kdud–+ mag– p+=

ag p

k kdm md

u ug+ ud u ug+ +

p

ug

Ω

ag ag Ωtsin=

p p Ωtsin=

u u Ωtsin=

ud ud Ωtsin=

mdΩ2– kd+[ ] ud mdΩ2u– mdag–=

kdud– mΩ2– k+[ ] u+ mag– p+=


The solutions for and are given by

(4.45)

(4.46)

where

(4.47)

and the terms are dimensionless frequency ratios,

(4.48)

(4.49)

Selecting the mass ratio and damper frequency ratio such that

(4.50)

reduces the solution to

(4.51)

(4.52)

This choice isolates the primary mass from ground motion and reduces theresponse due to external force to the pseudo-static value, . A typical range for

is to . Then, the optimal damper frequency is very close to the forcingfrequency. The exact relationship follows from eqn (4.50).

(4.53)

One determines the corresponding damper stiffness with

(4.54)

u ud

upk---

1 ρd2

–

D1---------------

mag

k----------

1 m ρd2

–+

D1--------------------------

–=

udp

kd----- mρ2

D1-----------

mag

kd---------- m

D1-------

–=

D1 1 ρ2–[ ] 1 ρd

2–[ ] mρ2

–=

ρ

ρ Ωω---- Ω

k m⁄---------------= =

ρdΩωd------- Ω

kd md⁄----------------------= =

1 ρd2

– m+ 0=

u pk---=

udp

kd-----ρ2

–magkd

----------+=

p k⁄m 0.01 0.1

ωd opt

Ω1 m+

------------------=

kd optωd opt

2md

Ω2mm1 m+

-----------------= =


Finally, substituting for , eqn (4.52) takes the following form

(4.55)

One specifies the amount of relative displacement for the damper anddetermines with eqn (4.55). Given and , the stiffness is found using eqn(4.54). It should be noted that this stiffness applies for a particular forcingfrequency. Once the mass damper properties are defined, eqns (4.45) and (4.46)can be used to determine the response for a different forcing frequency. Theprimary mass will move under ground motion excitation in this case.

Undamped structure - damped TMD

The next level of complexity has damping included in the mass damper, as shownin Fig. 4.14. The equations of motion for this case are

(4.56)

(4.57)

The inclusion of the damping terms in eqns (4.56) and (4.57) produces a phaseshift between the periodic excitation and the response. It is convenient to workinitially with the solution expressed in terms of complex quantities. Oneexpresses the excitation as

(4.58)

(4.59)

where and are real quantities. The response is taken as

(4.60)

(4.61)

kd

ud1 m+

m-------------- p

k---

ag

Ω2-------+

=

m m Ω

md ud cdud kdud md u+ + + mdag–=

mu ku cdud– kdud–+ mag– p+=

ag ageiΩt=

p peiΩt=

ag p

u ueiΩt=

ud udeiΩt=


Fig. 4.14: Undamped SDOF system coupled with a damped TMD system.

where the response amplitudes, and are considered to be complexquantities. The real and imaginary parts of correspond to cosine andsinusoidal input. Then, the corresponding solution is given by either the real (forcosine) or imaginary (for sine) parts of and . Substituting eqns (4.60) and(4.61) in the set of governing equations and cancelling from both sidesresults in

(4.62)

(4.63)

The solution of the governing equations is

(4.64)

(4.65)

where

(4.66)

(4.67)

and was defined earlier as the ratio of to (see eqn (4.48)).

Converting the complex solutions to polar form leads to the following

k

kd

cd

m md

u ug+ ud u ug+ +

p

ug

u udag

u udeiΩt

mdΩ2– icdΩ kd+ +[ ] ud mdΩ2u– mdag–=

icdΩ kd+[ ] ud– mΩ2– k+[ ] u+ mag– p+=

up

kD2---------- f 2 ρ2

– i2ξdρf+[ ]agm

kD2---------- 1 m+( ) f 2 ρ2

– i2ξdρf 1 m+( )+[ ]–=

udpρ2

kD2----------

agm

kD2----------–=

D2 1 ρ2–[ ] f 2 ρ2

–[ ] mρ2 f 2– i2ξdρf 1 ρ2 1 m+( )–[ ]+=

fωdω

-------=

ρ Ω ω


expressions

(4.68)

(4.69)

where the factors define the amplification of the pseudo-static responses, andthe ‘s are the phase angles between the response and the excitation. The variousH and δ terms are listed below

(4.70)

(4.71)

(4.72)

(4.73)

(4.74)

Also

(4.75)

(4.76)

(4.77)

(4.78)

(4.79)

upk---H1e

iδ1 agm

k----------H2e

iδ2–=

udpk---H3e

i– δ3 agm

k----------H4e

i– δ3–=

Hδ

H1f 2 ρ2

–[ ]2

2ξdρf[ ] 2+

D2------------------------------------------------------------=

H21 m+( ) f 2 ρ2

–[ ]2

2ξdρf 1 m+( )[ ] 2+

D2--------------------------------------------------------------------------------------------------=

H3ρ2

D2----------=

H41

D2----------=

D2 1 ρ2–[ ] f 2 ρ2

–[ ] mρ2 f 2–( )

22ξdρf 1 ρ2 1 m+( )–[ ]( )

2+=

δ1 α1 δ3–=

δ2 α2 δ3–=

δ3tan2ξdρf 1 ρ2 1 m+( )–[ ]

1 ρ2–[ ] f 2 ρ2

–[ ] mρ2 f 2–

------------------------------------------------------------------=

α1tan2ξdρf

f 2 ρ2–

------------------=

α2tan2ξdρf 1 m+( )

1 m+( ) f 2 ρ2–

-------------------------------------=


For most applications, the mass ratio is less than about . Then, theamplification factors for external loading and ground motion areessentially equal. A similar conclusion applies for the phase shift. In what follows,the solution corresponding to ground motion is examined and the optimal valuesof the damper properties for this loading condition are established. An in-depthtreatment of the external forcing case is contained in Den Hartog’s text (DenHartog, 1940).

Figure 4.15 shows the variation of with forcing frequency for specificvalues of damper mass and frequency ratio , and various values of thedamper damping ratio, . When , there are two peaks with infiniteamplitude located on each side of . As is increased, the peaks approacheach other and then merge into a single peak located at . The behavior of theamplitudes suggests that there is an optimal value of for a given damperconfiguration ( and , or equivalently, and ). Another key observation isthat all the curves pass through two common points, and . Since these curvescorrespond to different values of , the location of and must depend onlyon and .

Proceeding with this line of reasoning, the expression for can bewritten as

(4.80)

where the ‘a’ terms are functions of , , and . Then, for to be independentof , the following condition must be satisfied

(4.81)

The corresponding values for are

(4.82)

0.05H1( ) H2( )

H2m fξd ξd 0=

ρ 1= ξdρ 1≈

ξdmd kd m f

P Qξd P Q

m f

H2

H2a1

2 ξd2 a2

2+

a32 ξd

2 a42

+-----------------------

a2a4-----

a12 a2

2⁄ ξd2

+

a32 a4

2⁄ ξd2

+---------------------------= =

m ρ f H2ξd

a1a2-----

a3a4-----=

H2

H2 P Q,

a2a4-----=


Fig. 4.15: Plot of versus .

Substituting for the ‘a’ terms in eqn (4.81), one obtains a quadratic equation for

(4.83)

The two positive roots and are the frequency ratios corresponding to points and . Similarly, eqn (4.82) expands to

(4.84)

Figure 4.15 shows different values for at points and . For optimalbehavior, one wants to minimize the maximum amplitude. As a first step, onerequires the values of for and to be equal. This produces a distributionwhich is symmetrical about , as illustrated in Fig. 4.16. Then, byincreasing the damping ratio, , one can lower the peak amplitudes until thepeaks coincide with points and . This state represents the optimalperformance of the TMD system. A further increase in causes the peaks tomerge and the amplitude to increase beyond the optimal value.

0.8 0.85 0.9 0.95 1 1.05 1.1 1.150

5

10

15

20

25

30

ρ Ωω----=

H2

m 0.01=f 1=

P

Q

ξd 0=

ξd 1=

0 ξd 1< <

ρ1 ρ2

H2 ρ

ρ2

ρ4 1 m+( ) f 2 1 0.5m+1 m+

---------------------+ ρ2– f 2

+ 0=

ρ1 ρ2P Q

H2 P Q,

1 m+

1 ρ1 2,2 1 m+( )–

------------------------------------------=

H2 P Q

H2 ρ1 ρ2ρ2 1 1 m+( )⁄=

ξdP Q

ξd


Fig. 4.16: Plot of versus for .

Requiring the amplitudes to be equal at and is equivalent to thefollowing condition on the roots

(4.85)

Then, substituting for and using eqn (4.83), one obtains a relation betweenthe optimal tuning frequency and the mass ratio

(4.86)

(4.87)

The corresponding roots and optimal amplification factors are

(4.88)

(4.89)

0.85 0.9 0.95 1 1.05 1.1 1.150

5

10

15

20

25

30

ρ Ωω----=

H2 P Q

ξd opt

ξd ξdo>

ξd ξdo<

ρ1 optρ2 opt

H2 ρ f opt

P Q

1 ρ12 1 m+( )– 1 ρ2

2 1 m+( )–=

ρ1 ρ2

f opt1 0.5m–1 m+

-------------------------=

ωd optf optω=

ρ1 2, opt

1 0.5m±1 m+

--------------------------=

H2 opt

1 m+

0.5m----------------=


The expression for the optimal damping at the optimal tuning frequency is

(4.90)

Figures 4.17 through 4.20 show the variation of the optimal parameters with themass ratio, .

Fig. 4.17: Optimum tuning frequency ratio, .

ξd opt

m 3 0.5m–( )8 1 m+( ) 1 0.5m–( )-------------------------------------------------=

m

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10.88

0.9

0.92

0.94

0.96

0.98

1

m

f op

t

f opt


Fig. 4.18: Input frequency ratios at which the response is independent ofdamping.

Fig. 4.19: Optimal damping ratio for TMD.

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10.8

0.85

0.9

0.95

1

1.05

1.1

m

ρ 12,

op

tf opt

ρ1 opt

ρ2 opt

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

m

ξ do

pt


Fig. 4.20: Maximum dynamic amplification factor for SDOF system(optimal tuning and damping).

The response of the damper is defined by eqn (4.69). Specializing thisequation for the optimal conditions leads to the plot of amplification versus massratio contained in Fig. 4.21. A comparison of the damper motion with respect tothe motion of the primary mass for optimal conditions is shown in Fig. 4.22.

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

5

10

15

20

25

m

H2

op

t


Fig. 4.21: Maximum dynamic amplification factor for TMD.

Fig. 4.22: Ratio of maximum TMD amplitude to maximum system amplitude.

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

50

100

150

200

250

300

m

H4

op

t

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

2

4

6

8

10

12

14

16

18

20

m

u d u------

H4

H2

-------

=

f opt ρopt ξd opt, ,


Lastly, response curves for a typical mass ratio, , and optimaltuning are plotted in Figs 4.23 and 4.24. The response for no damper is alsoplotted in Fig. 4.23. One observes that the effect of the damper is to limit themotion in a frequency range centered on the natural frequency of the primarymass and extending about . Outside of this range, the motion is notsignificantly influenced by the damper.

Fig. 4.23: Response curves for amplitude of system with optimally tuned TMD.

m 0.01=

0.15ω

0.8 0.85 0.9 0.95 1 1.05 1.1 1.15 1.20

5

10

15

20

25

30

H2

ρ Ωω----=

m 0.01=f opt 0.9876=

ξd 0.0=ξd 0.03=ξd 0.061 (optimal)=ξd 0.1=No damper

P Q

ρ1 optρ2 opt


Fig. 4.24: Response curves for amplitude of optimally tuned TMD.

The maximum amplification for a damped SDOF system without a TMD,undergoing harmonic excitation is given by eqn (1.32)

(4.91)

Since is small, a reasonable approximation is

(4.92)

Expressing the optimal in a similar form provides a measure of the equivalentdamping ratio for the primary mass

(4.93)

Figure 4.25 shows the variation of with the mass ratio. A mass ratio of isequivalent to about damping in the primary system.

0.85 0.9 0.95 1 1.05 1.1 1.150

50

100

150

m 0.01=f opt 0.9876=

ρ Ωω----=

H4

ξd 0.0=ξd 0.03=ξd 0.061 (optimal)=ξd 0.1=ξd 0.2=

H 1

2ξ 1 ξ2–

-------------------------=

ξ

H 12ξ------≈

H2ξe

ξe1

2H2 opt

--------------------=

ξe 0.025%


Fig. 4.25: Equivalent damping ratio for optimally tuned TMD.

The design of a TMD involves the following steps:

• Establish the allowable values of displacement of the primary mass andthe TMD for the design loading. This data provides the design valuesfor and .

• Determine the mass ratios required to satisfy these motion constraintsfrom Figs 4.20 and 4.21. Select the largest value of .

• Determine form Fig. 4.17.

• Compute

(4.94)

• Compute

(4.95)

• Determine from Fig. 4.19.

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

0.02

0.04

0.06

0.08

0.1

0.12

m

ξ ef opt ρopt ξd opt

, ,

H2 optH4 opt

m

f opt

ωd

ωd f optω=

kd

kd mdωd2 mk f opt

2= =

ξd opt


• Compute

(4.96)

Example 4.2: Design of a TMD for an undamped SDOF system

Consider the following motion constraints

(4.97)

(4.98)

Constraint eqn(4.97) requires . For constraint eqn(4.98), one needs to take. Therefore, controls the design. The relevant parameters are:

Then

Damped structure - damped TMD

All real systems contain some damping. Although an absorber is likely to beadded only to a lightly damped system, assessing the effect of damping in the realsystem on the optimal tuning of the absorber is an important designconsideration.

The main system in Fig. 4.26 consists of the mass , spring stiffness , andviscous damping . The TMD system has mass , stiffness , and viscousdamping . Considering the system to be subjected to both external forcing andground excitation, the equations of motion are

(4.99)

cd

cd 2ξd optωdmd m f opt 2ξd opt

ωm= =

H2 opt7<

H4H2 opt

----------------- 6<

m 0.05≥m 0.02≥ m 0.05≥

m 0.05= f opt 0.94= ξd opt0.135=

md 0.05m= ωd 0.94ω= kd m f opt2 k 0.044k= =

m kc md kd

cd

md ud cdud kdud md u+ + + mdag–=


(4.100)

Fig. 4.26: Damped SDOF system coupled with a damped TMD system.

Proceeding in the same way as for the undamped case, the solution due toperiodic excitation (both p and ug) is expressed in polar form:

(4.101)

(4.102)

The various H and δ terms are defined below

(4.103)

(4.104)

(4.105)

(4.106)

mu cu ku cdud– kdud–+ + mag– p+=

kd

cd

m md

u ug+ ud u ug+ +ug

k

c

p

upk---H5 e

iδ5agm

k----------H6 e

iδ6–=

udpk---H7 e

i– δ7agm

k----------H8 e

iδ8–=

H5f 2 ρ2

–[ ]2

2ξdρf[ ] 2+

D3------------------------------------------------------------=

H61 m+( ) f 2 ρ2

–[ ]2

2ξdρf 1 m+( )[ ] 2+

D3--------------------------------------------------------------------------------------------------=

H7ρ2

D3----------=

H81 2ξρ[ ] 2+

D3-------------------------------=


(4.107)

(4.108)

(4.109)

(4.110)

(4.111)

(4.112)

The and terms are defined by eqns 4.78 and 4.79.

In what follows, the case of an external force applied to the primary mass isconsidered. Since involves ξ, one cannot establish analytical expressions forthe optimal tuning frequency and optimal damping ratio in terms of the massratio. In this case, these parameters also depend on . Numerical simulations canbe applied to evaluate and for a range of , given the values for , , ,and . Starting with specific values for and , plots of versus can begenerated for a range of and . Each plot has a peak value of . Theparticular combination of and that correspond to the lowest peak value ofis taken as the optimal state. Repeating this process for different values of andproduces the behavioral data needed to design the damper system.

Figure 4.27 shows the variation of the maximum value of for theoptimal state. The corresponding response of the damper is plotted in Fig. 4.28.Adding damping to the primary mass has an appreciable effect for small .Noting eqns (4.101) and (4.102), the ratio of damper displacement to primarymass displacement is given by

(4.113)

Since is small, this ratio is essentially independent of . Figure 4.29 confirmsthis statement. The optimal values of the frequency and damping ratios generatedthrough simulation are plotted in Figs 4.30 and 4.31. Lastly, using eqn (4.93),

can be converted to an equivalent damping ratio for the primary system.

D3 f 2ρ2m– 1 ρ2–( ) f 2 ρ2–( ) 4ξξ d f ρ2–+[ ] 2=

4+ ξρ f 2 ρ2–( ) ξd fρ 1 ρ2 1 m+( )–( )2+[ ]

δ5 α1 δ7–=

δ6 α2 δ7–=

δ8 α3 δ7–=

δtan 7 2ξρ f 2 ρ2–( ) ξd fρ 1 ρ2 1 m+( )–( )+

f 2ρ2m– 1 ρ2–( ) f 2 ρ2–( ) 4ξξ d f ρ2–+----------------------------------------------------------------------------------------------------=

α3tan 2ξρ=

α1 α2

D3

ξH5 H7 ρ m ξ f

ξd m ξ H5 ρf ξd H5 ρ– H5f ξd H5

m ξ

H5

m

udu

---------H7H5------- ρ2

f 2 ρ2–[ ]

22ξdρf[ ] 2

+

------------------------------------------------------------= =

ξ ξ

H5 opt


(4.114)

Figure 4.32 shows the variation of with and .

Tsai & Lin (1993) suggest equations for the optimal tuning parametersand determined by curve fitting schemes. The equations are listed below forcompleteness

(4.115)

(4.116)

Fig. 4.27: Maximum dynamic amplification factor for damped SDOF system.

ξe1

2H5 opt

--------------------=

ξe m ξ

fξd

f 1 0.5m–1 m+

------------------------- 1 2ξ2– 1–+

=

2.375 1.034 m– 0.426m–[ ]ξ m–

3.730 16.903 m– 20.496m+( )ξ2 m–

ξd3m

8 1 m+( ) 1 0.5m–( )------------------------------------------------- 0.151ξ 0.170ξ2

–( )+=

0.163ξ 4.980ξ2+( )+ m

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

5

10

15

20

25

30

35

40

m

H5

op

t

ξ 0.0=ξ 0.01=ξ 0.02=ξ 0.05=ξ 0.1=


Fig. 4.28: Maximum dynamic amplification factor for TMD.

Fig. 4.29: Ratio of maximum TMD amplitude to maximum system amplitude.

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

50

100

150

200

250

300

m

H7

op

tξ 0.0=ξ 0.01=ξ 0.02=ξ 0.05=ξ 0.1=

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

2

4

6

8

10

12

14

16

18

20

m

u d u------

H7

H5

-------

=

ξ 0.0=ξ 0.01=ξ 0.02=ξ 0.05=ξ 0.1=


Fig. 4.30: Optimum tuning frequency ratio for TMD, .

Fig. 4.31: Optimal damping ratio for TMD.

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10.8

0.82

0.84

0.86

0.88

0.9

0.92

0.94

0.96

0.98

1

m

f op

tξ 0.0=ξ 0.01=ξ 0.02=ξ 0.05=ξ 0.1=

f opt

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

m

ξ do

pt

ξ 0.0=ξ 0.01=ξ 0.02=ξ 0.05=ξ 0.1=


Fig. 4.32: Equivalent damping ratio for optimally tuned TMD.

Example 4.3: Design of a TMD for a damped SDOF system

Example 4.2 is reworked here, allowing for damping in the primary system.The same design motion constraints are considered:

(4.117)

(4.118)

Using Fig. 4.27, the required mass ratio for is . The otheroptimal values are and .

Then

In this case, there is a significant reduction in the damper mass required for this

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

m

ξ e

ξ 0.0=ξ 0.01=ξ 0.02=ξ 0.05=ξ 0.1=

2%

H5 opt7<

H7H5 opt

----------------- 6<

ξ 0.02= m 0.03≈f opt 0.965= ξd opt

0.105=

md 0.03m= ωd 0.955ω= kd m f opt2 k 0.027k= =


set of motion constraints. The choice between including damping in the primarysystem versus incorporating a tuned mass damper depends on the relative costsand reliability of the two alternatives, and the nature of the structural problem. ATMD system is generally more appropriate for upgrading an existing structurewhere access to the structural elements is difficult.

4.5 Case studies - SDOF systems

Figures 4.33 to 4.44 show the time history responses for two SDOF systems withperiods of 0.49s and 5.35s respectively under harmonic (at resonance conditions),El Centro, and Taft ground excitations. All examples have a system damping ratioof 2% and an optimally tuned TMD with a mass ratio of 1%. The excitationmagnitudes have been scaled so that the peak amplitude of the response of thesystem without the TMD is unity. The plots show the response of the systemwithout the TMD (the dotted line) as well as the response of the system with theTMD (the solid line). Figures showing the time history of the relativedisplacement of the TMD with respect to the system are also presented.Significant reduction in the response of the primary system under harmonicexcitation is observed. However, optimally tuned mass dampers are relativelyineffective under seismic excitation, and in some cases produce a negative effect,i.e. they amplify the response slightly. This poor performance is attributed to theineffectiveness of tuned mass dampers for impulsive loadings as well as theirinability to reach a resonant condition and therefore dissipate energy underrandom excitation. These results are in close agreement with the data presentedby Kaynia et al. (1981).

4.5 Cases Studies - SDOF Systems 299

Fig. 4.33: Response of SDOF to harmonic excitation.

Fig. 4.34: Relative displacement of TMD under harmonic excitation.

0 10 20 30 40 50 60

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Time - s

u-

mT 0.49s=ξ 0.02=m 0.01=

without TMDwith TMD

0 10 20 30 40 50 60-6

-4

-2

0

2

4

6

Time - s

u d-

m

T 0.49s=ξ 0.02=m 0.01=


Fig. 4.35: Response of SDOF to El Centro excitation.

Fig. 4.36: Relative displacement of TMD under El Centro excitation.

0 10 20 30 40 50 60

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Time - s

T 0.49s=ξ 0.02=m 0.01=

without TMDwith TMD

u-

m

0 10 20 30 40 50 60-6

-4

-2

0

2

4

6

u d-

m

Time - s

T 0.49s=ξ 0.02=m 0.01=


Fig. 4.37: Response of SDOF to Taft excitation.

Fig. 4.38: Relative displacement of TMD under Taft excitation.

0 10 20 30 40 50 60

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Time - s

T 0.49s=ξ 0.02=m 0.01=

without TMDwith TMD

u-

m

0 10 20 30 40 50 60-6

-4

-2

0

2

4

6

u d-

m

Time - s

T 0.49s=ξ 0.02=m 0.01=


Fig. 4.39: Response of SDOF to harmonic excitation.

Fig. 4.40: Relative displacement of TMD under harmonic excitation.

0 10 20 30 40 50 60

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Time - s

T 5.35s=ξ 0.02=m 0.01=

without TMDwith TMD

u-

m

0 10 20 30 40 50 60-6

-4

-2

0

2

4

6

u d-

m

Time - s

T 5.35s=ξ 0.02=m 0.01=


Fig. 4.41: Response of SDOF to El Centro excitation.

Fig. 4.42: Relative displacement of TMD under El Centro excitation.

0 10 20 30 40 50 60

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Time - s

T 5.35s=ξ 0.02=m 0.01=

without TMDwith TMD

u-

m

0 10 20 30 40 50 60-6

-4

-2

0

2

4

6

u d-

m

Time - s

T 5.35s=ξ 0.02=m 0.01=


Fig. 4.43: Response of SDOF to Taft excitation.

Fig. 4.44: Relative displacement of TMD under Taft excitation.

0 10 20 30 40 50 60

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Time - s

T 5.35s=ξ 0.02=m 0.01=

without TMDwith TMD

u-

m

0 10 20 30 40 50 60-6

-4

-2

0

2

4

6

u d-

m

Time - s

T 5.35s=ξ 0.02=m 0.01=

4.6 Tuned Mass Damper Theory for MDOF Systems 305

4.6 Tuned mass damper theory for MDOF systems

The theory of a SDOF system presented earlier is extended here to deal with aMDOF system having a number of tuned mass dampers located throughout thestructure. Numerical simulations, which illustrate the application of this theory tothe set of example building structures used as the basis for comparison of thedifferent schemes throughout the text are presented in the next section.

Fig. 4.45: 2DOF system with TMD.

A 2DOF system having a damper attached to mass 2 is considered first tointroduce the key ideas. The governing equations for the system shown in Fig.4.45 are:

(4.119)

(4.120)

(4.121)

The key step is to combine eqns (4.119) and (4.120) and express theresulting equation in a form similar to the SDOF case defined by eqn (4.100). Thisoperation reduces the problem to an equivalent SDOF system, for which thetheory of Section 4.4 is applicable. The approach followed here is based ontransforming the original matrix equation to scalar modal equations.

Introducing matrix notation, eqns (4.119) and (4.120) are written as

k2kd

c2 cd

m2 md

u2 ug+ u2 ug ud+ +

p2

k1

c1

m1

u1 ug+

p1

ug

m1 u1 c1u1 k1u1 k2 u2 u1–( )– c2 u2 u1–( )–+ + p1 m1 ug–=

m2 u2 c2 u2 u1–( ) k2 u2 u1–( ) kdud– cdud–+ + p2 m2 ug–=

md ud kdud cdud+ + md u2 ug+( )–=


(4.122)

where the various matrices are

(4.123)

(4.124)

(4.125)

(4.126)

One substitutes for in terms of the modal vectors and coordinates

(4.127)

The modal vectors satisfy the following orthogonality relations (see eqn (2.257))

(4.128)

Defining modal mass, stiffness, and damping terms,

(4.129)

(4.130)

(4.131)

expressing the elements of as

MU CU KU+ +p1 m1ag–

p2 m2ag–

0kdud cdud+

+=

Uu1

u2

=

Mm1

m2

=

Kk1 k2+ k2–

k2– k2

=

Cc1 c2+ c2–

c2– c2

=

U

U Φ1q1 Φ2q2+=

ΦjTKΦi δijωj

2ΦjTMΦi=

m j ΦjTMΦj=

k j ΦjTKΦj ωj

2m j= =

c j ΦjTCΦj=

Φj


(4.132)

and assuming damping is proportional to stiffness

(4.133)

one obtains a set of uncoupled equations for and

(4.134)

With this assumption, the modal damping ratio is given by

(4.135)

Equation (4.134) represents two equations. Each equation defines aparticular SDOF system having mass, stiffness, and damping equal to , , and

. Since a TMD is effective for a narrow frequency range, one has to decide onwhich modal resonant response is to be controlled with the TMD. Once thisdecision is made, the analysis can proceed using the selected modal equation andthe initial equation for the TMD, i.e. eqn (4.121).

Suppose the first modal response is to be controlled. Taking in eqn(4.133) leads to

(4.136)

In general, is obtained by superposing the modal contributions

(4.137)

However, when the external forcing frequency is close to , the first moderesponse will dominate, and it is reasonable to assume

(4.138)

Solving for

ΦjΦj1

Φj2

=

C αK=

q1 q2

m jq j c jq j k jq j+ + Φj1 p1 m1ag–( )= j = 1 , 2

Φj2 p2 m2ag– kdud cdud+ +( )+

ξ jc j

2ωjm j----------------

αωj2

----------= =

m kξ

j 1=

m1 q1 c1q1 k1q1+ + Φ11p1 Φ12p2+=

m1Φ11 m2Φ12+[ ] ag– Φ12 kdud cdud+[ ]+

u2

u2 Φ12q1 Φ22q2+=

ω1

u2 Φ12q1≈

q1


(4.139)

and then substituting in eqn (4.136), one obtains

(4.140)

where , , , , and represent the equivalent SDOF parametersfor the combination of mode 1 and node 2, the node at which the TMD is attached.Their definition equations are

(4.141)

(4.142)

(4.143)

(4.144)

(4.145)

Equations 4.121 and 4.140 are similar in form to the SDOF equationstreated in the previous section. Both set of equations are compared below.

TMD equation

(4.146)

q11

Φ12--------- u2=

m1e u2 c1e u2 k1e u2+ + kdud cdud+=

p1e Γ1e m1e ag–+

m1e c1e k1e p1e Γ1e

m1e1

Φ122

--------- m1=

k1e1

Φ122

--------- k1=

c1e α k1e=

p1eΦ11p1 Φ12p2+

Φ12--------------------------------------=

Γ1e

Φ12m1--------- m1Φ11 m2Φ22+( )=

md ud cd ud kd ud+ + md u ag–( )–=

vs

md ud cd ud kd ud+ + md u2 ag–( )–=


Primary mass equation

(4.147)

Taking

(4.148)

transforms the primary mass equation for the MDOF case to

(4.149)

which differs from the corresponding SDOF equation by the factor Γ. Therefore,the solution for ground excitation generated earlier has to be modified to accountfor the presence of Γ.

The “generalized” solution is written in the same form as the SDOF case.One needs only to modify the terms associated with , i.e., H6, H8 and δ6 , δ8 .Their expanded form is listed below.

(4.150)

(4.151)

(4.152)

(4.153)

(4.154)

(4.155)

mu cu ku+ + cd ud kd ud p mag–+ +=

vs

m1e u2 c1e u2 k1e u2+ + cd ud kd ud p1e Γ1em1eag–+ +=

u2 u≡ m1e m≡ c1e c≡ k1e k≡

p1e p≡ Γ1e Γ≡

mu cu ku+ + cd ud kd ud p Γmag–+ +=

ag

H6Γ m+( ) f 2 Γρ2

–[ ]2

2ξdρf Γ m+( )[ ] 2+

D3-------------------------------------------------------------------------------------------------------=

H81 ρ2 Γ 1–( )+[ ]

22ξρ[ ] 2

+D3

---------------------------------------------------------------------=

a2tan2ξd fρ Γ m+( )

f 2 Γ m+( ) Γρ2–------------------------------------------=

a3tan 2ξρ1 Γ 1–( )ρ2+---------------------------------=

δ6 a2 δ7–=

δ8 a3 δ7–=


where is defined by eqn (4.107), and is given by eqn (4.111).

From this point on, one proceeds as described in Section 4.4. The mass ratiois defined in terms of the equivalent SDOF mass.

(4.156)

Given and , one finds the tuning frequency and damper damping ratiousing Figs 4.30 and 4.31. The damper parameters are determined with

(4.157)

(4.158)

(4.159)

Expanding the expression for the damper mass,

(4.160)

shows that one should select the TMD location to coincide with the maximumamplitude of the mode shape that is being controlled. In this case, the first mode isthe target mode, and is the maximum amplitude for .

This derivation can be readily generalized to allow for tuning on the thmodal frequency. One writes eqn (4.139) as

(4.161)

where is either or . The equivalent parameters are

(4.162)

(4.163)

Given and , one specifies and finds the optimal tuning with

D3 δ7

mmdm1e---------=

m ξ1

md m m1e=

ωd f optω1=

cd 2ξd optωdmd=

md m m1em Φ1

TMΦ1[ ]

Φ122

-------------------------------= =

Φ12 Φ1

i

qi1

Φi2-------- u2≈

i 1 2

mie1

Φi22

-------- mi=

kie ωi2mie=

mie ξ i m


(4.164)

Example 4.4: Design of a TMD for a damped MDOF system

To illustrate the above procedure, a two DOF system having isconsidered. Designing the system for a fundamental period of and auniform deformation fundamental mode profile yields the following stiffnesses(refer to Example 1.6)

Requiring a fundamental mode damping ratio of , and taking dampingproportional to stiffness ( ), the corresponding is

The mass, stiffness, and damping matrices for these design conditions are:

Performing an eigenvalue analysis yields the following frequencies andmode shapes

The corresponding modal mass, stiffness, and damping terms are:

ωd f optωi=

m1 m2 1= =T1 1s=

k1 12π2 118.44= =

k2 8π2 78.96= =

2%C αK= α

α2ξ1ω1--------- 0.02

π---------- 0.0064= = =

M 1 00 1

=

K 197.39 78.96–

78.96– 78.96=

C 1.26 0.51–

0.51– 0.51=

ω1 6.28r/s= ω2 15.39r/s=

Φ10.51.0

= Φ21.00.5–

=

m1 Φ1TMΦ1 1.25= = m2 Φ2

TMΦ2 1.25= =


The optimal parameters for a TMD having a mass ratio of and tunedto a specific mode are listed below.

Mode 1 - optimum location is node 2

Mode 2 - optimum location is node 1

The general case of a MDOF system with a tuned mass damper connectedto the nth degree of freedom is treated in a similar manner. Using the notationdefined above, the jth modal equation can be expressed as

(4.165)

where denotes the modal force due to ground motion and external forcing, andis the element of corresponding to the nth displacement variable. To

control the ith modal response, one sets in eqn (4.165), and introduces theapproximation

(4.166)

This leads to the following equation for

(4.167)

where

k1 Φ1TKΦ1 39.48= = k2 Φ2

TKΦ2 236.87= =

c1 Φ1TCΦ1 0.25= = c2 Φ2

TCΦ2 1.51= =

ξ1c1

2ω1m1----------------- 0.02= = ξ2

c22ω2m2----------------- 0.049= =

0.01

f opt 0.982= ξd opt0.062=

md 0.0125= kd 0.4754= cd 0.0096=

f opt 0.972= ξd opt0.068=

md 0.05= kd 11.1894= cd 0.1017=

m jq j c jq j k jq j+ + p j Φjn kdud cdud+[ ]+= j = 1 , 2 , . . .

p jΦjn Φjj i=

qi1

Φin--------- un≈

un

mie un cie un kie un+ + pie kdud cdud++=


(4.168)

(4.169)

(4.170)

(4.171)

The remaining steps are the same as described above. One specifies and ,determines the optimal tuning and damping values with Figs 4.30 and 4.31, andthen computes and .

(4.172)

(4.173)

The optimal mass damper for mode is obtained by selecting such that isthe maximum element in .

Example 4.5: Design of TMD’s for a simply supported beam

Figure 1

Consider the simply supported beam shown in Figure 1. The modal shapes

mie1

Φin2

--------- Mi1

Φin2

--------- ΦiTMΦi= =

kie ωi2mie=

cie α kie=

pie1

Φin--------- pi=

m ξ i

md ωd

md m miem

Φin2

--------- ΦiTMΦi= =

ωd f optωi=

i n ΦinΦi

x

y,u

x*P*

L

EI constant


and frequencies for the case where the cross sectional properties are constant andthe transverse shear deformation is negligible are:

(1)

(2)

One obtains a set of N equations in terms of N modal coordinates byexpressing the transverse displacement, u(x,t), as

(3)

and substituting for u in the Principle of Virtual Displacements specialized fornegligible transverse shear deformation (see eqn (2.194)),

(4)

Substituting for ,

(5)

and taking

(6)

leads to the following equations

Φn x( ) nπxL

----------sin=

ωn2 EI

ρm------- nπ

L------

4=

n 1 2 …, ,=

u qi t( )Φj x( )

j 1=

N

∑=

Mδχ xd

0

L

∫ bδu xd∫=

δχ

δχx2

2

d

d δu( )–=

δu δqjΦj=


(7)

Lastly, one substitutes for M and b in terms of and q, and evaluates theintegrals. The expressions for M and b are

(8)

(9)

Noting the orthogonality properties of the modal shape functions,

(10)

(11)

the modal equations uncouple and reduce to

(12)

where

(13)

(14)

MΦj xx, xd∫– bΦj xd∫=

j 1 2 … N, , ,=

Φ

M EIχ EI qlΦl xx,

l 1=

N

∑–= =

b ρmu– b x t,( )+ ρm Φl ql b x t,( )+

l 1=

N

∑–= =

ΦjΦk xd

0

L

∫ δjkL2---=

Φj xx, Φk xx, xd

0

L

∫ jπL-----

4δjk

L2---=

m jq j k jq j+ p j=

m jLρm

2-----------=

k j EI jπL-----

4L2---=


(15)

When the external loading consists of a concentrated force applied at thelocation (see Fig 1), the corresponding modal load for the j’th mode is

(16)

In this example, the force is considered to be due to a mass attached to the beamas indicated in Fig 2. The equations for the tuned mass and the force are

(17)

(18)

Figure 2

Suppose one wants to control the i’th modal response with a tuned massdamper attached at . Taking j equal to i in eqns (12) and (13), the i’thmodal equation has the form

(19)

p j b jπxL

---------sin xd

0

L

∫=

x x*=

p j P* jπx*

L-----------sin=

md u* ud+( ) kd+ ud cdud+ 0=

md u* ud+( ) P*–=

kd cd

x*

md

u*+ud

u*

x x*=

miqi kiqi+ kdud cdud+( ) iπx*

L----------sin=


Assuming the response is dominated by the i’th mode, is approximatedby

(20)

and eqn (19) is transformed to an equation relating and .

(21)

where

(22)

The remaining steps utilize the results generated for the SDOF undampedstructure - damped TMD system considered in section 4.3. One uses andas the mass and stiffness parameters for the primary system.

To illustrate the procedure, consider the damper to be located at mid-span,and the first mode is to be controlled. Taking i=1 and , thecorresponding parameters are

(23)

(24)

(25)

One specifies the equivalent damping ratio, , and determines the required massratio from Fig 4.32. For example, taking requires . The otherparameters corresponding to follow from Figs 4.29, 4.30, and 4.31.

(26)

u* x* t,( )

u* x* t,( ) qiiπx*

L----------sin≈

u* ud

mieu* kieu*+ kdud cdud+=

mie1iπx*

L----------sin

2---------------------------- mi=

mie kie

x* L 2⁄=

iπx*

L----------sin 1=

mie m1Lρm

2-----------= =

kie k1EIL

2--------- π

L---

4= =

ξeξe 0.06= m 0.03=

m 0.03=

f opt

ωdω1------ 0.965= =


(27)

(28)

Using these parameters, the corresponding expression for the damper propertiesare

(29)

(30)

(31)

(32)

Once and are specified, the damper properties can be evaluated.For example, consider the beam to be a steel beam having the followingproperties

(33)

The beam parameters are

(34)

Applying eqns (29)-(32), results in

ξd opt0.105=

ud

u*------ 5=

md 0.03m1=

ωd 0.965ω1=

kd ωd2md=

cd 2ξdωdmd=

m1 ω1

L 20m=

ρm 1000kg m⁄=

I 8 10 4– m4×=

E 2 1011N m2⁄×=

m1 10 000kg,=

ω1 9.87r s⁄=


(35)

The total mass of the girder is 20,000kg. Adding 300kg, which is just 1.5% of thetotal mass, produces an effective damping ratio of 0.06 for the first moderesponse.

The mode shape for the second mode has a null point at x=L/2, andtherefore locating a tuned mass at this point would have no effect on the secondmodal response. The optimal locations are and . Taking

and i=2, one obtains

(36)

(37)

(38)

(39)

The procedure from here on is the same as before. One specifies , determinesthe required mass ratio, and then the frequency and damping parameters. It is ofinterest to compare the damper properties corresponding to the same equivalentdamping ratio. Taking , the damper properties for the example steelbeam are

(40)

(41)

(42)

md 300kg=

ωd 9.52r s⁄=

kd 27 215N m⁄,=

cd 599.8Ns m⁄=

x* L 4⁄= x* 3L 4⁄=x* L 4⁄=

iπx*

L----------sin 1=

m2e m2Lρm

2-----------= =

k2e k2 8EIL πL---

4= =

ω22 16EI

ρm------------ π

L---

4=

ξe

ξe 0.06=

md 300kg=

kd 435 440N m⁄,=

cd 2400Ns m⁄=


The required damper stiffness is an order of magnitude greater than thecorresponding value for the first mode response.

4.7 Case studies - MDOF systems

This section presents shear deformation profiles for the standard set of buildingexamples defined in Table 2.4. A single TMD is placed at the top floor and tunedto either the first or second mode. The structures are subjected to harmonicground acceleration with a frequency equal to the fundamental frequency of thebuildings, as well as scaled versions of El Centro and Taft ground accelerations.As expected, significant reduction in the response is observed for the harmonicexcitations (see Figs 4.46 through 4.49). The damper is generally less effective forseismic excitation versus harmonic excitation (see Figs 4.50 through 4.61). Resultsfor the low period structures show more influence of the damper which is to beexpected since the response is primarily due to the first mode. This data indicatesthat a TMD is not the optimal solution for controlling the motion due to seismicexcitation.

4.6 Case Studies - MDOF Systems 321

Fig. 4.46: Maximum shear deformation for Building 1.


0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Building #1Quadratic basedInitial

HarmonicTMD - Mode 1

H 25m=ρm 20000kg/m=s 0.15=Sv 1.2m/s=ξ1 2%=


Nor

mal

ized

hei

ght

x H----

γ∗

m 0%=m 1%=m 2%=m 5%=

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----



H 50m=ρm 20000kg/m=s 0.25=Sv 1.2m/s=ξ1 2%=


γ∗

m 0%=m 1%=m 2%=m 5%=




0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----Building #3Quadratic basedInitial


H 100m=ρ 20000kg/m=s 0.40=Sv 1.2m/s=ξ1 2%=


γ∗

m 0%=m 1%=m 2%=m 5%=

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----



H 200m=ρm 20000kg/m=s 0.63=Sv 1.2m/s=ξ1 2%=


γ∗

m 0%=m 1%=m 2%=m 5%=




0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght


El CentroTMD - Mode 1

H 25m=ρm 20000kg/m=s 0.15=Sv 1.2m/s=ξ1 2%=


γ∗

m 0%=m 1%=m 2%=m 5%=

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----


TaftTMD - Mode 1

H 25m=ρm 20000kg/m=s 0.15=Sv 1.2m/s=ξ1 2%=


γ∗

m 0%=m 1%=m 2%=m 5%=




0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght



H 50m=ρm 20000kg/m=s 0.25=Sv 1.2m/s=ξ1 2%=


γ∗

m 0%=m 1%=m 2%=m 5%=

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----


TaftTMD - Mode 1

H 50m=ρm 20000kg/m=s 0.25=Sv 1.2m/s=ξ1 2%=


γ∗

m 0%=m 1%=m 2%=m 5%=




0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght



H 100m=ρm 20000kg/m=s 0.40=Sv 1.2m/s=ξ1 2%=


γ∗

m 0%=m 1%=m 2%=m 5%=

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----


TaftTMD - Mode 1

H 100m=ρm 20000kg/m=s 0.40=Sv 1.2m/s=ξ1 2%=


γ∗

m 0%=m 1%=m 2%=m 5%=




0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght



H 100m=ρm 20000kg/m=s 0.40=Sv 1.2m/s=ξ1 2%=


γ∗

m 0%=m 1%=m 2%=m 5%=

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----


TaftTMD - Mode 2

H 100m=ρm 20000kg/m=s 0.40=Sv 1.2m/s=ξ1 2%=


γ∗

m 0%=m 1%=m 2%=m 5%=




0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----



H 200m=ρm 20000kg/m=s 0.63=Sv 1.2m/s=ξ1 2%=


γ∗

m 0%=m 1%=m 2%=m 5%=

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----


TaftTMD - Mode 1

H 200m=ρm 20000kg/m=s 0.63=Sv 1.2m/s=ξ1 2%=


γ∗

m 0%=m 1%=m 2%=m 5%=




0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght



H 200m=ρm 20000kg/m=

s 0.63=Sv 1.2m/s=ξ1 2%=


γ∗

m 0%=m 1%=m 2%=m 5%=

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----


TaftTMD - Mode 2

H 200m=ρm 20000kg/m=s 0.63=Sv 1.2m/s=ξ1 2%=


γ∗

m 0%=m 1%=m 2%=m 5%=

Problems 329

Problems

Problem 4.1

Verify eqns (4.13) through (4.17). Hint: express p, u, and ud in complexform

and solve eqns (4.6) and (4.7) for and . Then take

Problem 4.2

Refer to eqns (4.14) and (4.20). Express as a function of , , and. Take , and plot vs. for a representative range of the

magnitude of the displacement ratio, .

Problem 4.3

Figure 4.7 illustrates an active tuned mass damper configuration. Thedamper can be modelled with the 2 DOF system shown below. The various termsare: is the total displacement of the support attached to the floor beam; isthe self equilibrating force provided by the actuator; are parameters forthe damper mass; and are parameters for the auxillary mass.

p peiΩt=

u ueiΩt=

ud udeiΩt=

u ud

u ueiδ1=

ud udei δ1 δ2+( )

=

ω ωd Ω= =

ξe m ξu ud⁄ ξ 0.05= ξe m

u ud⁄

us Famd kd cd, ,

ka ma


a) Derive the governing equation for and . Also determine an expressionfor the resultant force, R, that the system applies to the floor beam.

b) Consider to be several orders of magnitude smaller than , eg,. Also take the actuator force to be a linear function of the relative

velocity of the damper mass.

Specialize the equations for this case. How would you interpret the contributionof the actuator force to the governing equation for the damper mass?

Problem 4.4

Design a pendulum damper system having a natural period of 6 secondsand requiring less than 4 meters of vertical space.

ka

Fa

maus ud ua+ +

kd

cd

us ud+us

md

R

md ma

ma mdma 0.01md=

Fa caud=

Problems 331

Problem 4.5

The pendulum shown above is connected to a second mass which is free tomove horizontally. The connection between mass 1 and mass 2 carries only shear.Derive an equation for the period of the compound pendulum and the length ofan equivalent simple pendulum. Assume the links are rigid.

Problem 4.6

Refer to Fig 4.12. Establish the equations of motion for the mass, ,considering to be small. Verify that the equivalent stiffness is equal to .

Problem 4.7

Refer to Fig 4.15 and eqn (4.84). Derive the corresponding expression forstarting with eqn (4.70) and using the same reasoning strategy.

Considering the mass ratio, , to be less than 0.03, estimate the difference in the“optimal” values for the various parameters.

Problem 4.8

Generate plots of vs. for ranging from 0 to 0.2, , and. Compare the results with the plots shown in Fig 4.23.

m1

m2

L1

L2

mdθ Wd R⁄

H1 P Q,m

H1 ρ ξd m 0.01=f 0.9876=


Problem 4.9

Consider a system composed of an undamped primary mass and a tunedmass damper. The solution for periodic force excitation is given by (see eqns (4.60)to (4.79))

(1)

(2)

(3)

(4)

(5)

(6)

(7)

The formulation for the optimal damper properties carried out in Section4.3 was based on minimizing the peak value of H1 (actually H2 but H1 behaves ina similar way), i.e., on controlling the displacement of the primary mass. Supposethe design objective is to control the acceleration of the primary mass. Notingeqns (1) and (3), the acceleration is given by

(8)

(9)

Substituting for k transforms eqn (9) to

u ueiΩt=

ud udeiΩt=

u pk---H1e

iδ1=

udpk---H3e

iδ3=

H1f 2 ρ2

–[ ]2

2ξdρf[ ] 2+

D2------------------------------------------------------------=

H3ρ2

D2----------=

D2 1 ρ2–[ ] f 2 ρ2

–[ ] mρ2 f 2–( )

22ξdρf 1 ρ2 1 m+( )–[ ]( )

2+=

u a aeiΩt= =

a pΩ2

k----------H1e

i δ1 π+( )=

Problems 333

(10)

where

(11)

Investigate the behavior of with and . If it behaves similar to asshown in Fig 4.15, describe how you would establish the optimal values for thevarious parameters, and also how you would design a tuned mass system when

is specified.

Problem 4.10

Design a TMD for a damped SDOF system having . The designmotion constraints are:

a)

b)

c) Repeat part b considering to be equal to 0.05.

Problem 4.11

This problem concerns the design of a tuned-mass damper for a damped

a pm----H′

1 ei δ1 π+( )

=

H ′1 ρ2H1=

H ′1 ρ f m, , ξd H2

H ′1

ξ 0.02=

H5 opt10<

H7H5 opt

---------------- 5<

H5 opt5<

H7H5 opt

---------------- 5<

ξ


single degree of freedom system. The performance criteria are:

a) Determine the damper properties for a system having = 10,000 kg and for the following values of :

•

•

b) Will the damper be effective for an excitation with frequency? Discuss the basis for you conclusion.

Problem 4.12

Refer to example 3.7. Suppose a tuned mass damper is installed at the toplevel (at mass 5).

a) Determine the damper properties such that the equivalent dampingratio for the fundamental mode is 0.16. Use the values of , , from example3.7. Assume stiffness proportional damping for c.

b)

Consider the tuned mass damper to be a pendulum attached to .Determine and for the damper properties established in part a.

c) Repeat part a for the case where the mass damper is tuned for the secondmode rather than for the first mode. Use the same values of , , and assumestiffness proportional damping.

ξeq 0.1= ud u⁄ 5=

mk 395 kN/m= ξ

ξ 0.02=

ξ 0.05=

2.5π rad/sec

m k c

u5m5

md

Lk5 c5,

m5md L

m k c

Problems 335

Problem 4.13

Consider a cantilever shear beam with the following properties:

•

•

•

a) Model the beam as a 10 DOF discrete shear beam having 5m segments.Determine the first 3 mode shapes and frequencies. Normalize the mode shapessuch that the peak amplitude is unity for each mode.

b) Design tuned mass dampers to provide an effective modal dampingratio of 0.10 for the first and third modes. Take and assume modaldamping is proportional to stiffness.

Note: you need to first establish the ‘‘optimal location’’ of the tuned massdampers for the different modes.

Problem 4.14

Consider a simply supported steel beam having the following properties

a) Design tuned mass damper systems that provide an equivalent dampingof 0.05 for each of the first 3 modes.

b) Repeat part a) with the constraint that an individual damper masscannot exceed 300kg. Hint: utilize symmetry of a particular mode shape to locatea pair of dampers whose function is to control that mode.

H 50 m=

ρm 20 000 kg/m,=

DT 8 5×10 1 0.6xH

----------– kN=

ξ1 0.02=

L 30m=

ρm 1500kg m⁄=

I 1 2–×10 m4=


Problem 4.15

Consider the simply supported beam shown above. The beam has auniform weight of 15 kN/m and a concentrated weight at mid span of 100 kN.The flexural rigidity is constant and equal to 200,000 kN-m2.

a) Assume the first mode can be approximated by:

Determine the governing equation for using the Principle of VirtualDisplacements.

b) Design a tuned mass damper to provide an equivalent damping ratio of0.05 for the first mode. Assume no damping for the beam itself.

c) Will the damper designed in part b be effective for the second mode?Explain your answer.

Problem 4.16

Refer to problem 3.25 part (b). Suggest a tuned mass damper for generatingthe required energy dissipation.

15 m 15 m

W

u1

constant EI

u u1πL---x

sin=

u1

337

Chapter 5

Base isolation systems

5.1 Introduction

The term isolation refers to the degree of interaction between objects. An object issaid to be isolated if it has little interaction with other objects. The act of isolatingan object involves providing an interface between the object and its neighborswhich minimizes interaction. These definitions apply directly to various physicalsystems. For example, one speaks of isolating a piece of equipment from itssupport by mounting the equipment on an isolation system which acts as a bufferbetween the equipment and the support. The design of isolation systems forvibrating machinery is a typical application. The objective here is to minimize theeffect of the machine induced loading on the support. Another application isconcerned with minimizing the effect of support motion on the structure. Thisissue is becoming increasingly more important for structures containing motionsensitive equipment and also for structures located adjacent to railroad tracks orother sources of ground disturbance.

Although isolation as a design strategy for mounting mechanicalequipment has been employed for over seventy years, only recently has theconcept been seriously considered for civil structures, such as buildings andbridges, subjected to ground motion. This type of excitation interacts with thestructure at the foundation level, and is transmitted up through the structure.Therefore, it is logical to isolate the structure at its base, and prevent the ground

338 Chapter 5: Base Isolation Systems

motion from acting on the structure. The idea of seismic isolation dates back tothe late nineteenth century, but the application was delayed by the lack of suitablecommercial isolation components. Substantial development has occurred sincethe mid 1980’s (Naeim and Kelly, 1999), and base isolation for certain types of civilstructures is now considered to be a highly viable design option by the seismicengineering community, particularly in Japan (Wada, 1998), for moderate toextreme seismic excitation.

A set of simple examples are presented in the next section to identify thekey parameters and illustrate the quantitative aspects of base isolation. Thismaterial is followed by a discussion of practical aspects of seismic base isolationand a description of some seismically isolated buildings. The remaining sectionsdeal with the behavioral and design issues for base isolated MDOF structuralsystems. Numerical results illustrating the level of performance feasible withseismic base isolation are included to provide a basis of comparison with theother motion control schemes considered in this text.

5.2 Isolation for SDOF systems

The application of base isolation to control the motion of a SDOF systemsubjected to ground motion was discussed earlier in Section 1.3 as part of ageneral treatment of design for dynamic excitation. The analytical formulationdeveloped in that section provides the basis for designing an isolation system forsimple structures that can be accurately represented with a SDOF model.Examples illustrating the reasoning process one follows are presented below. Theformulation is also extended to deal with a modified version of a SDOF modelthat is appropriate for a low-rise building isolated at its base. This model is usefulfor preliminary design.

SDOF examples

The first example considers external periodic forcing of the SDOF system shownin Fig. 5.1. The solution of this problem is contained in Section 1.3. Forconvenience, the relevant equations are listed below:

5.2 Isolation for SDOF Systems 339

Fig. 5.1: SDOF system.

(5.1)

(5.2)

(5.3)

(5.4)

(5.5)

(5.6)

Given and , one can determine for a specific system having mass ,stiffness , and damping . With known, the forces in the spring and dampercan be evaluated. The reaction can be found by either summing the internalforces, or combining with the inertia force. With the latter approach, one writes

(5.7)

and expands the various terms using eqns (5.1) through (5.6). The result isexpressed as

(5.8)

(5.9)

k

c

m

u

R p

p p Ωtsin=

u u Ωt δ–( )sin=

uH1k

------- p=

H11

1 ρ2–[ ]

22ξρ[ ] 2

+

--------------------------------------------------=

ρ Ωω----=

δtan 2ξρ

1 ρ2–

---------------=

p Ω u mk c u

p

R p mu–=

R R Ωt δr–( )sin=

R H3p=


(5.10)

(5.11)

The function, H3, is referred to as the transmissibility of the system. It is a measureof how much of the load p is transmitted to the support. When , and

reduces to . Figure 5.2 shows the variation of with and .

Fig. 5.2: Plot of versus .

The model presented above can be applied to the problem of designing asupport system for a machine with an eccentric rotating mass. Here, one wants tominimize the reaction force for a given , i.e. one takes . Noting Fig. 5.2,this constraint requires the frequency ratio, , to be greater than , and itfollows that

H31 2ξρ[ ] 2

+

1 ρ2–[ ]

22ξρ[ ] 2

+----------------------------------------------=

δrtanρ2H1 δsin

1 ρ2H1 δcos+------------------------------------–=

ξ 0= δ 0=H3 H1 H3 ρ ξ

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

ρ Ωω----=

H3

2

ξ 0=

ξ 0.2=

ξ 0.4=

H3 ρ

p H3 1<ρ 2


(5.12)

The corresponding periods are related by

(5.13)

where is the forcing period. For example, taking results in, a reduction of from the static value.

The second example illustrates the strategy for isolating a system fromsupport motion. Applying the formulation derived in Section 1.4 to the systemshown in Fig. 5.3, the amplitudes of the relative and total displacement of themass, and , are related to the support displacement by

(5.14)

(5.15)

Taking small with respect to unity reduces the effect of support motion on theposition of the mass. The frequency and period criteria are the same as those ofthe previous example. One takes to reduce . However, since H2approaches unity as increases, the magnitude of the relative motion increasesand approaches the ground motion, . Therefore, this relative motion needs tobe accomodated.

Fig. 5.3: SDOF system subjected to support motion.

These examples show that isolation is obtained by taking the period of theSDOF system to be large in comparison to the forcing (either external or support)

ω Ω2

-------<

T Tf 2> 2 2πΩ------

=

Tf T 3Tf=R 0.125p= 87.5%

u ut

u ρ2H1ug H2ug= =

ut H3ug=

H3

ρ 2> utρ

ug

k

c

m

ut ug u+=ug


period. Expressing this requirement as

(5.16)

where depends on the desired reduction in amplitude, the constraint on thestiffness of the spring is given by

(5.17)

It should be noted that this derivation assumes that a single periodicexcitation is applied. The result is applicable for narrow band excitations whichare characterized by a dominant frequency. A more complex analysis involvingiteration on the stiffness is required to deal with broad band excitations. One hasto ensure that the forcing near the fundamental frequency is adequatelycontrolled by damping in this case.

Bearing terminology

The spring and damper elements connecting the mass to the support areidealizations of physical objects called bearings. They provide a constraint againstmotion relative to a support plane, as illustrated in Fig. 5.4. The bearing in Fig.5.4(a) functions as an axial element and resists the displacement normal to theplane with normal stresses (tension and compression). The bearing shown in Fig.5.4(b) constrains relative tangential motion through shearing action over theheight of the bearing. These elements are usually combined into a singlecompound bearing, but it is more convenient to view them as being uncoupledwhen modeling the system.

Fig. 5.4: Axial and shear bearings.

T ρ∗ Tf≥ ⇒ ω Ωρ*-----<

ρ∗

k m Ωρ∗------

2< m 2π

ρ∗ Tf------------

2=

Fn , un

axialbearing

Ft , ut

shearbearing

(a) (b)


When applying the formulation developed above, one distinguishesbetween normal and tangential support motion. For normal motion, axial typebearings such as springs and rubber cushions are used; the defined by eqn(5.17) is the axial stiffness of the bearing . Shear bearings such as laminatedrubber cushions and inverted pendulum type sliding devices are used when theinduced motion is parallel to the ground surface. In this case, represents therequired shearing stiffness of the bearing, .

Figure 5.5 shows an air spring/damper scheme used for vertical support.Single and multiple stage laminated rubber bearings are illustrated in Fig 5.6.Rubber bearings used for seismic isolation can range up to 1 m in diameter andare usually inserted between the foundation footings and the base of thestructure. A particular installation for a building is shown in Fig 5.7.

Fig. 5.5: Air spring bearing.

kFn un⁄

kFt ut⁄


a) Single stage

b) multiple stage

Fig. 5.6: Laminated rubber bearings.


Fig. 5.7: Rubber bearing seismic isolation system.

Modified SDOF Model

In what follows, the support motion is considered to be due to seismic excitation.Although both normal (vertical) and tangential (horizontal) motions occur duringa seismic event, the horizontal ground motion is generally more significant forstructural systems since it leads to lateral loading. Typical structural systems aredesigned for vertical loading and then modified for lateral loading. Since thevertical motion is equivalent to additional vertical loading, it is not as critical asthe horizontal motion.

The model shown in Fig. 5.3 represents a rigid structure supported onflexible shear bearings. To allow for the flexibility of the structure, the structurecan be modeled as a MDOF system. Figure 5.8 illustrates a SDOF beam typeidealization. One can estimate the equivalent SDOF properties of the structure byassuming that the structural response is dominated by the fundamental mode.The data provided in earlier chapters shows that this assumption is reasonable forlow-rise buildings subjected to seismic excitation.

An in-depth analysis of low rise buildings modeled as MDOF beams ispresented later in this chapter. The objective here is to derive a simple relationshipshowing the effect of the bearing stiffness on the relative displacement of the


structure, , with respect to the base displacement, . The governingequations for the lumped mass model consist of an equilibrium equation for themass, and an equation relating the shear forces in the spring and the bearing.

(5.18)

(5.19)

Fig. 5.8: Base isolation models.

Neglecting damping, eqn (5.19) can be solved for ub in terms of u.

(5.20)

Then, substituting for ub in eqn (5.18) leads to

(5.21)

Equation (5.21) is written in the conventional form for a SDOF system

(5.22)

where Γ is a participation factor,

(5.23)

u ub ug+

mu cu ku+ + m ub ug+( )–=

kbub cbub+ ku cu+=

ug

ub ug+

u ub ug+ +

m

k , c

kb , cb

(a) Actual structure (b) Beam idealization (c) Lumped mass model

ubk

kb----- u=

m 1 kkb-----+ u ku+ mug–=

u ωeq2 u+ Γ ug–=

Γkb

k kb+--------------

kbk-----

1kbk-----+

⁄= =


and ωeq is an equivalent frequency measure

(5.24)

In this case, is the fundamental frequency of the system consisting of thestructure plus bearing. Taking small with respect to decreases the inertialoading on the structure as well as the effective frequency. Consequently, thestructural response is reduced.

Periodic excitation - modified SDOF model

To illustrate the effect of base stiffness on the response, the case of periodicground motion, , is considered. The various response amplitudesare given by

(5.25)

(5.26)

(5.27)

where the brackets indicate absolute values, and ρeq is the frequency ratio

(5.28)

Comparing eqn (5.27) with eqn (5.15) shows that the results are similar. Onereplaces with in the expression for . The limiting cases are and

. The former is the fully isolated case where and ; thelatter corresponds to a fixed support where and .

Suppose the structure is defined, and the problem concerns selecting abearing stiffness such that the total response satisfies

ωeq2 Γk

m------ Γω2

= =

ωeqkb k

ug ug Ωtsin=

uΓρeq

2

1 ρeq2

–--------------------ug=

ubk

kb----- u=

ut u ub ug+ +1

1 ρeq2

–--------------------ug= =

ρeqΩ

ωeq---------=

ω ωeq H3 kb 0=kb ∞= ub ug–≈ ut 0≈

ub 0≈ ut u ug+≈

ut νug≤ ν 1<


One needs to take . Noting eqn (5.27), the required value of is

(5.29)

Substituting for in eqn (5.28) leads to

(5.30)

Finally, using eqn (5.23) and (5.24), the required bearing stiffness is given by

(5.31)

The more general problem is the case where both structural stiffness andthe bearing stiffness need to be established subject to the following constraints onthe magnitudes of and .

(5.32)

The typical design scenario has larger than . Noting eqn (5.26), thestiffness factors are related by

(5.33)

Equation (5.25) provides the second equation relating the stiffness factors. Itreduces to

(5.34)

where

ρeq 2> ρeq

ρeq2 1 1

ν---+=

ρeq

ωeq2 Ω2

1 1ν---+

------------=

kb k 1k

ωeq2 m

-------------- 1–---------------------------- k

k 1 1 ν⁄( )+( )

mΩ2-------------------------------- 1–-----------------------------------------= =

u ub

u νsug=

ub νbug=

νb νs

kb

νsνb-----k=

1– 1ρeq

2--------+ Γ

νs-----=


(5.35)

Solving eqn (5.34) for leads to , and then k.

(5.36)

The following example illustrates the computational steps.

Example 5.1: Stiffness factors for prescribed structure and base motion.

Suppose and . The relative motion of the base withrespect to the ground is allowed to be 10 times greater than the relative motion ofthe structure with respect to the base.

(1)

The stiffness factors are related by

(2)

Evaluating and , using eqns (5.34) and (5.35),

(3)

(4)

leads to

(5)

and finally to k

Γkb k⁄

1 kb k⁄+---------------------

νsνs νb+-----------------= =

ρeq2 ωeq

2

ωeq2 Ω2

ρeq2

-------- Γ km----= =

νs 0.1= νb 1.0=

ub 10u=

kb

νsνb-----k 0.1k= =

Γ ρeq

νsνs νb+----------------- 0.1

1.1------- 0.0909= =

1 1ρeq

2--------– Γ

νs----- 1

1.1------- 0.909= = =

ρeq2 11.0011=

ωeq2 0.0909Ω2=


(6)

Seismic excitation - modified SDOF model

An estimate of the stiffness parameters required to satisfy the motion constraintsunder seismic excitation can be obtained with the response spectra approachdescribed in Chapter 2. Taking to be the seismic excitation, the solution of eqn(5.22) is related to the spectral velocity by

(5.37)

where is a function of the equivalent frequency, , and the equivalentdamping ratio for the structure/bearing system, . Substituting for and ,eqn (5.37) expands to

(5.38)

The relation between the maximum relative displacement of the bearing and themaximum structural motion follows from eqn (5.20)

(5.39)

In this development, the criteria for motion based design of a base isolatedstructure are expressed as limits on the relative motion terms

(5.40)

(5.41)

The values of and required to satisfy these constraints follow by solvingeqns (5.38) and (5.39).

k mΓ----ωeq

2 mΩ2= =

ug

u maxΓSvωeq----------=

Sv ωeqξeq Γ ωeq

u max Sv

mkbk k kb+( )----------------------=

ub maxk

kb----- u max=

u max u∗=

ub maxub

∗=

k kb


(5.42)

(5.43)

One assumes is constant, evaluates and , determines the frequencywith eqn (5.24), and then updates if necessary.

It is of interest to compare the stiffness required by the base isolatedstructure with the stiffness of the corresponding fixed base structure. Taking

reduces eqn (5.38) to

(5.44)

The fixed base structural stiffness follows from eqn (5.44)

(5.45)

Using eqn (5.45) and assuming the value of is the same for both cases, thestiffness ratios reduce to,

(5.46)

(5.47)

The ratio of the isolated period to the fixed base period can be generated with eqn(5.24)

kbku∗ub

∗---------=

kmSv

2

u∗[ ] 2------------- 1

1ub

∗

u∗---------+

-------------------mSv

2

u∗ u∗ ub∗+( )

---------------------------------= =

Sv k kb ωeqSv

kb ∞=

u max Svmk----=

k f

k f k kb ∞=

mSv2

u∗[ ] 2-------------= =

Sv

kk f-----

1

1ub

∗

u∗---------+

-------------------=

kbk f-----

u∗ub∗--------

1ub

∗

u∗---------+

-------------------=


(5.48)

Figures 5.9 and 5.10 show the variation of and with for agiven constant . The increase in the period is plotted in Fig. 5.11. There is asignificant reduction in the structural stiffness required by the seismic excitationwhen the base is allowed to move. For example, taking decreases thedesign stiffness by a factor of . However, one has to ensure that a potentialresonant condition is not created by shifting the period. There may be a problemwith wind gust loading as the period is increased beyond 3 seconds. This problemcan be avoided by providing additional stiffness that functions under windloading but not under seismic loading. Section 5.3 deals with this problem.

Fig. 5.9: Variation of with .

TeqT f---------

ω fωeq--------- 1

ub∗

u∗---------+= =

k k f⁄ kb k f⁄ ub∗ u∗⁄

Sv

ub∗ 2u∗=

3

0 1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

k kf

-----

ub∗

u∗---------

k k f⁄ ub∗ u∗⁄




0 1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1k b k

f----

-

ub∗

u∗---------

kb k f⁄ ub∗ u∗⁄

0 1 2 3 4 5 6 7 8 9 100

2

4

6

8

10

12

ub∗

u∗---------

Teq Tf

--------

-

Teq T f⁄ ub∗ u∗⁄


Example 5.2:Stiffness parameters - modified SDOF model of Building example #2.

The procedure for establishing the appropriate values for and isillustrated using building Example as the reference structure. Table 2.4 lists therelevant design information. The period for the fixed base case is 1.06 sec. Sincebad isolation increases the period, the assumption that is constant is valid.

The relative displacement at the top of the building is estimated aswhere is the height of the structure and is the prescribed shear deformation.Taking and leads to .

The allowable bearing displacement depends on the bearing configurationand response characteristics, as well as the seismic excitation. For the totally softcase, is equal to the ground excitation. Hardening the bearing reducessomewhat, so a reasonable upper limit is the peak ground displacementcorresponding to the design value of for representative earthquakes. A typicaldesign value for is 0.3m. Using and corresponds tothe following stiffness factors

The required structural stiffness is reduced by 55% for this degree of baseisolation.

These scenarios provide an indication of the potential benefit of baseisolation for seismic excitation. However, one should note that the isolatedstructure is less stiff than the fixed base structure, and therefore will experiencelarger displacement under other types of loading such as wind. Also, thesimplified model considered here is based on linear undamped behavior, whereasthe actual bearings have some damping and may behave in a nonlinear manner.More complex models are considered in a later section.

u∗ ub∗

2

Sv

Hγ∗H γ∗H 50m= γ∗ 1 200⁄= u∗ 0.25m=

ub ub

Svub

∗ u∗ 0.25m= ub∗ 0.3m=

k 0.455k f=

kb 0.833k=

T 2.2T f=

5.3 Design Issues for Structural Isolation Systems 355

5.3 Design issues for structural isolation systems

The most important requirements for an isolation system concern flexibility,energy dissipation, and rigidity under low level loading. A number of solutionshave been proposed for civil type structures over the past thirty years. The mostsignificant aspects of these designs is discussed below.

Flexibility

A structural isolation system generally consists of a set of flexible supportelements that are proportioned such that the period of vibration of the isolatedstructure is considerably greater than the dominant period of the excitation.Systems proposed to date employ plates sliding on a curved surface (eg., aninverted pendulum), sleeved piles, and various types of rubber bearings. Themost popular choice at this point in time is the rubber bearing, with about ofthe applications.

Rubber bearings consist of layers of natural rubber sheets bonded to steelplates, as shown in Fig. 5.12. The steel plates constrain the lateral deformation ofthe rubber under vertical loading, resulting in a vertical stiffness several orders ofmagnitude greater than the horizontal stiffness. The lateral stiffness depends onthe number and thickness of the rubber sheets. Increasing either quantitydecreases the stiffness; usually one works with a constant sheet thickness andincreases the number of layers. As the height increases, buckling becomes thecontrolling failure mechanism, and therefore, the height is usually limited toabout half the diameter. Natural rubber is a nonlinear viscoelastic material, and iscapable of deforming up to about without permanent damage. Shear strainon the order of is a common design criterion. Bearing diameters up toand load capacities up to 5 MN are commercially available.

90%

300%100% 1m


Fig. 5.12: Typical natural rubber bearing (NRB).

Rigidity under low level lateral loads

Increasing the lateral flexibility by incorporating a base isolation system providesan effective solution for high level seismic excitation. Although the relativemotion between the structure and the support may be large, the absolutestructural motion is generally small, so that the structure does not feel theearthquake. The effect of other types of lateral loading such as wind is quitedifferent. In this case, the loading is applied directly to the structure, and the lowlateral stiffness can result in substantial lateral displacement of the structurerelative to the fixed support.

To control the motion under service loading, one can incorporate anadditional stiffness system that functions for service loading but is notoperational for high level loading. Systems composed of rods and/or springs thatare designed to behave elastically up to a certain level of service loading and thenyield have been developed and are commercially available. There are a variety ofsteel dampers having the above characteristics that can be combined with therubber bearings. Figure 5.13 illustrate a particular scheme. The steel rod isdimensioned (length and area) such that it provides the initial stiffness and yieldsat the intended force level. The earliest solution and still the most popularapproach is to incorporate a lead rod in the rubber bearing, as illustrated in Fig.5.14. The lead plug is dimensioned according to the force level at which thesystem is intended to yield.

Rubber

Steel shims

D

Mounting plate

h


Fig. 5.13: Steel rod damper combined with a NRB.

Fig. 5.14: Typical lead rubber bearing (LRB).

Energy dissipation/absorption

Rubber bearings behave in a viscoelastic manner and have some energydissipation capacity. Additional damping can be provided by separate devicessuch as viscous, hysteretic, and friction dampers acting in parallel with the rubberbearings. The lead rubber bearing (LRB) is representative of this design approach;the lead plug provides both initial stiffness and hysteretic damping. Sincehysteretic damping action occurs only at high level loading, hysteretic-typesystems require additional viscous damping to control the response for low levelloading. High damping natural rubber with a dissipation capacity about 4 timesthe conventional value is used together with other devices to improve the energydissipation capacity of the isolation system. Figure 5.15 illustrates the deploymentof a combination of NRB’s, steel dampers, and viscous dampers. This schemeallows one to adjust both stiffness and damping for each load level, i.e., for bothlow and high level loading.

Rubber

Steel shims

D

Mounting plate

h Lead plug


Fig. 5.15: Isolation devices of Bridgestone Toranomon Building.

Modeling of a natural rubber bearing (NRB)

For the purpose of preliminary design, a NRB can be modeled as a simple shearelement having a cylindrical shape and composed of a viscoelastic material.Figure 5.16 defines the notation and shows the mode of deformation. The relevantequations are

(5.49)

(5.50)

(5.51)

γ uh---=

F τA=

h ntb=


where is the cross-sectional area, is the thickness of an individual rubbersheet, and is the total number of sheets. Each sheet is assumed to be in simpleshear.

Applying the viscoelastic constitutive relations developed in Section 3.3,the behavior for harmonic shear strain is given by

(5.52)

(5.53)

Fig. 5.16: Natural rubber bearing under horizontal loading.

where is the storage modulus and is the loss factor. In general, andare functions of the forcing frequency and temperature. They are also functions ofthe strain amplitude in the case of high damping rubbers which exhibit nonlinearviscoelastic behavior. Combining the above equations leads to

(5.54)

(5.55)

where

(5.56)

(5.57)

Note that depends on the bearing geometry whereas and are materialproperties.

The standard form of the linearized force-displacement relation is defined

A tbn

γ γ Ωtsin=

τ Gsγ Ωt ηGsγ Ωtcos+sin=

h

tb

uF

τ

Gs η Gs η

u u Ωtsin=

F f dGsu Ωt η Ω tcos+sin[ ]=

u γh γntb= =

f dAh---- A

ntb--------= =

f d η Gs


by eqn (3.70)

(5.58)

where and are the equivalent linear stiffness and viscous damping terms.Estimates for and can be obtained with a least squares approach.Assuming there are material property data sets covering the expected range ofstrain amplitude and frequency, the resulting approximate expressions are eqns(3.74), (3.76), and (3.77) which are listed below for convenience.

(5.59)

(5.60)

(5.61)

Equation (5.58) is used in the MDOF analysis presented in a later section.

Figures 5.17 and 5.18 show that the material properties for natural andfilled rubber are essentially constant for the frequency range of interest. Assuming

and are constant, the equivalent properties reduce to

(5.62)

(5.63)

where Tav is the average period for the excitation and , are the “constant”values.

F kequ cequ+=

keq ceqkeq ceq

N

keq f d1N---- Gs Ωi( )

i 1=

N

∑ f d G s= =)

ceq αkeq=

α

GsηΩ

----------

ii 1=

N

∑Gs Ωi( )

i 1=

N

∑-----------------------------=

Gs η

keq f dGs=

α η2π------Tav=

Gs η


Fig. 5.17: Storage modulus and loss factor for natural rubber vs. frequency(Snowden, 1979)

Gs Pa( )

η


Fig. 5.18: Storage modulus and loss factor for filled natural rubber vs. frequency(Snowden, 1979)

Modeling of a lead rubber bearing (LRB)

As a first approximation, the LRB can be considered to consist of two elements: i)a linear viscoelastic element representing the rubber component, and ii) a linearelastic-perfectly plastic element simulating the lead plug. This model assumesthat the static force response relationship is bilinear, as indicated in Fig. 5.19. Thestiffness defined by eqn (5.62) can be used for the rubber bearing, i.e. for .

(5.64)

Considering lead to behave in a linear elastic manner, the plug stiffness can beexpressed as

Gs Pa( )

η

k1

k rubber( ) k1≡ f dGs=


(5.65)

where , , and denote the cross-sectional area, height, and shearmodulus for the plug. Lastly, the displacement corresponding to the onset ofyielding is related to the yield strain for lead by

(5.66)

Fig. 5.19: Lead rubber bearing model - quasi static response.

Interpreting the behavior of the lead rubber bearing for large deformationas viscoelastic, the response due to harmonic motion is expressed in terms of asecant stiffness, , and equivalent loss factor, ,

(5.67)

(5.68)

where is related to the elastic energy storage capacity and is a measure of theenergy dissipated through hysteretic damping of the rubber and leadcomponents. Defining as the ductility ratio

(5.69)

the secant stiffness is related to the individual stiffness terms by

k lead( ) k2≡ApGp

hp---------------=

Ap hp Gp

uy hpγy=

u

F

k1 rubber( )

k2 lead( )

uy uu

F

Fyk1 k2+

k1

ks

ks η

u u Ωtsin=

F ksu Ωtsin ηksu Ωtcos+=

ks η

µ

µ uuy------

γγy-------= =


(5.70)

The equivalent loss factor is defined as

(5.71)

where is the hysteretic work per cycle and is the maximum strain energy.Evaluating the energy terms

(5.72)

(5.73)

and substituting in eqn (5.71) leads to

(5.74)

Noting that is about and the typical peak response strain isabout , one can estimate as

(5.75)

A typical value for the ratio of to is

(5.76)

Then, reasonable estimates for and are

(5.77)

(5.78)

The loss coefficient for high damping rubber can be as high as . Combining a

ks k1k2µ-----+=

η 12π------ W

ES------=

W ES

W 4 µ 1–( )k2uy2 πηk1µ2uy

2+=

ES12--- ks µuy[ ] 2

=

η4 µ 1–( )k2

πksµ2-------------------------- η

k1ks-----+=

γy 5 3–×100.5 µ

µγγy------- 100≈=

k1 k2

k1 0.1k2≈

ks η

ks 1.1k1=

η 411π---------

0.10.11----------η+ 0.12 0.909η+= =

0.15


high damping rubber bearing with a lead plug provides an effective solution forboth initial stiffness and damping over the range from low to high excitation.

The last step involves transforming eqn (5.68) to the standard form, eqn(5.58). Applying a least square approach and treating and as functions ofboth the strain amplitude and frequency leads to

(5.79)

(5.80)

where N is the number of data sets, i.e., values of and . It is reasonable toassume and are constant, and evaluate these parameters for arepresentative range of the ductility parameter, .

Applicability of base isolation systems

The feasibility of base isolation depends on whether it is needed, whether theproposed structure is suitable for base isolation, and whether it is cost effectivecompared with alternative solutions (Mayes et al. 1990). The need for baseisolation may arise if the location is an area of high seismicity, if increasedbuilding safety and post earthquake operability are required, if reduced lateraldesign forces are desired, or if an existing structure needs upgrading to satisfycurrent safety requirements. A structure is considered suitable if: i) the subsoilconditions do not produce long period input motions to the structure, ii) thestructure is less than about 10 to 15 stories and has a height-to-width ratio thatprevents overturning, iii) the site permits the required level of motion of the basewith respect to ground, and iv) the non-seismic lateral loads (such as wind) areless than approximately 10% of the weight of the structure.

The cost effectiveness of a base isolated structure can be assessed byassigning values to both the initial and life cycle costs and benefits. Examples ofcost items are: the bearings, changes to accommodate the isolation system,maintenance and inspection of the isolation system, and the cost of maintaining

ks η

keq1N---- ks µi Ωi,( )

i 1=

N

∑=

ceq1N----

ks µi Ωi,( )η µi Ωi,( )Ωi

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

i 1=

N

∑=

µ ΩGs η

µ


operability after earthquakes. Examples of savings are: lower initial cost of thestructural system, less construction time, lower insurance premium, reduction inearthquake structural and nonstructural damage, and the reduction in injuries,deaths, and lawsuits from related damages. When disruption costs and the valueof the building contents are important, seismic isolation has a substantialeconomic advantage over other systems provided that such an isolation scheme istechnically feasible. Under such conditions, initial cost savings of up to 5% of thebuilding cost have been noticed. For conventional buildings where disruption ofoperation is not important, there may not be sufficient cost savings in thestructural system to offset the cost of the isolators (Mayes et al. 1990).

The greatest advantage of base isolation is achieved when it is consideredin the early planning stages of the project, since it is possible to take advantage ofthe reduced response due to the isolation system. If the Base Isolation System isselected and added after completion of the structural design, many complicationsmay arise since the construction techniques may have to be altered.

For bridge construction on the other hand, the economic issues are verydifferent from those for buildings. In bridges, the implementation of seismicisolation simply requires the use of a seismic isolation bearing rather than aconventional bearing. Since bearings are only one or two percent of the cost of abridge, an increase in the cost of isolation bearings will have very little impact onthe overall construction cost and consequently, the use of a seismic isolationsystem is expected to reduce the overall construction cost (Billings et al. 1985).

5.4 Examples of existing base isolation systems

The past few years, especially since the Kobe earthquake in Japan, have seen asignificant increase in the number of base isolated structures which suggests thatthe technology is gaining acceptance. A short description of some of the firstimplementations of base isolation systems is presented here to provide anindication of the type of buildings that are being isolated and the cost savings, ifany, achieved by employing this technology. More comprehensive descriptionsare contained in Kelly (1993), the Architectural Institute of Japan Guide to BaseIsolated Buildings in Japan (1993), and various company web sites listed in theElectronic Reference Section of the text.

5.4 Examples of Existing Base Isolation Systems 367

USC University Hospital (Myers 1989, Asher & Van Volkingburg 1989)

This eight-story structure, shown in Fig 5.20, is used as a teaching hospital by theUniversity of Southern California. It resists seismic forces with a steel bracedframe located on the perimeter, and is supported on 68 LRB and 81 NRB isolators.The seismic design was based on a 0.4g response spectrum increased by 20% toaccount for near-fault effects. The decision to incorporate seismic isolation wasmade in the preliminary design phase of the project. Structural cost comparisonsfor conventional and isolated structures were developed and the benefits ofseismic isolation were assessed. It was determined that the cost savings in thestructural frame would be sufficient to pay for the new structural ground floorslab and the isolation system. The additional cost of mechanical and architecturaldetails was 1.3% and there was a 1.4% cost savings in the soil nailed retaining wallused in the isolation design versus the conventional retaining wall. Consequently,there was no net additional cost for incorporating seismic isolation on thishospital project.

Fig. 5.20: USC University Hospital

Fire Department Command and Control Facility (Mayes et al. 1990)

This is a two-story, steel perimeter braced frame structure that utilizes 36 high-damping elastomeric isolation bearings. The decision to utilize seismic isolationon this project was based on a comparison of two designs (conventional andisolation) that required maintaining the functionality of the structure after theextreme design event. This project reflects the first such detailed comparison fortwo designs to meet a performance specification. In the case of this two-story


structure, the isolated structure was found to be 6% less expensive thanconventional design. A reduction in losses by a factor of 40 is expected with theseismic isolation.

Evans and Sutherland Manufacturing Facility (Reaveley et al. 1989)

The building, (see Fig 5.21), is a four-story manufacturing site for flight simulatorslocated near the Warm Springs and East faults in Salt Lake City. The buildingmeasures 280ft x 160ft in plan and rests on 40 LRB and 58 NRB isolators.Preliminary costs for conventional and isolated designs were developed and thebenefits of seismic isolation assessed at the conceptual design phase. Thestructural engineers decided to design the structural framing system for the UBCcode forces for conventional design and, consequently, there were no structuralframing cost savings. The additional structural cost was the basement structuralfloor (versus a slab-on-grade) and the heavy fail safe system used. Based on costdata developed by the contractors, the cost premium for incorporating seismicisolation was 5% or $400,000 on an $8 million project. Important in the decision toemploy seismic isolation was protecting the building contents, including work inprogress, the value of which exceeds $100 million (approximately 12 times thecost of the structure).

Fig. 5.21: Evans and Sutherland Facility


Salt Lake City Building (Mayes et al. 1987, Walters et al. 1986)

This facility, shown in Fig 5.22, is a five-story, Richardson Romanesque Revivalstructure constructed between 1892 and 1894, 265ft x 130ft in plan, and built ofunreinforced brick and sandstone. Its 12 story tower is centrally located and isalso constructed of unreinforced masonry. The building was restored and acombination of 208 LRB and 239 NRB isolators were installed, separating thebuilding from its foundation. The structure is now protected against damage forthe 0.2g design earthquake event. This project was the subject of a detailed studyof several retrofit schemes among which were base isolation and UBCstrengthening. The schemes were developed in sufficient detail to permit costestimates and an evaluation of performance. Although the cost of these twoalternatives was comparable, the decision to use seismic isolation was madebased on the considerably better performance that results from theimplementation of such a scheme. The complete architectural and historicrestoration, and seismic rehabilitation work was estimated to be $24 million. Theapproximate value of the seismic isolation work reported by the contractor was$4,414,000 including the cost of the 447 seismic isolators.

Fig. 5.22: Salt Lake City Building

The Toushin 24 Ohmori Building (Kajima, 1989)

This building has 1 underground story which is used as a parking garage, and 9stories above ground. It is located adjacent to 2 of the busiest railway lines inTokyo, and the isolation system was required to reduce the traffic inducedvibration as well as seismic motion. Figure 5.23 shows a view of the building, a


sectional plan, and the isolation scheme. A combination of laminated naturalrubber bearings and steel rod dampers were deployed. Thick layers of rubberwere used to decrease the vertical stiffness and thus filter out vertical micro-tremors.

a)View of building


b) Section

c) Devices

Fig. 5.23: The Toushin 24 Ohmori Building


Bridgestone Toranomon Building (Shimizu, 1987)

The Bridgestone Toranomon Building (see Fig 5.24) is an office building of theBridgestone Corporation, a major supplier of rubber products such as bearings.The base isolation system consists of 12 laminated rubber bearings, 25 steeldampers, and 8 viscous (oil) dampers. Figure 5.15 shows the layout of the devices.The viscous dampers are intended to dissipate the energy associated with windand low intensity excitations. At this load level, the steel dampers are designed tobehave elastically and provide stiffness. Energy associated with a large seismicexcitation is dissipated/absorbed primarily by the steel dampers.

Fig. 5.24: Bridgestone Toranomon Building

San Francisco City Hall (1994)

San Francisco City Hall is an historic structure that is currently being


retrofitted with a seismic isolation system consisting of 530 lead rubber isolators.The design basis earthquake is 0.50g. Cost of retrofitting the structure is estimatedat $105 million.

Fig. 5.25: San Francisco City Hall

Long Beach V.A. Hospital

The hospital is 12 story concrete structure with shear walls. A combinationof 110 LRB, 18 NRB and 18 sliding bearings were installed in the mechanical crawlspaces below the building to improve the building’s ability to surviveearthquakes up to magnitude 0.32g.

Fig. 5.26: Long Beach V.A. Hospital


5.5 Optimal stiffness distribution - discrete shear beam

The theory developed earlier in this chapter for the SDOF case is extended here todeal with the more general case of a deformable beam-type structure supportedby a base isolation system. Linear behavior is assumed since the objective is togenerate results which are suitable for preliminary design. The approach followedto establish the stiffness distribution for the structure is similar to what waspresented in Chapter 2. The only modification required is to include the effect ofthe stiffness and damping associated with the base isolation system. Most of thenotation and relevant equations have been introduced in Chapter 2.

In what follows, the stiffness distribution corresponding to uniformdeformation for the fundamental mode of the composite system consisting of adiscrete shear beam and isolation system is derived. The theory is extended todeal with continuous beams in the next section.

Scaled stiffness distribution

Figure 5.27 defines the notation used for the base isolated shear beam. Thebearing system is represented by an equivalent linear spring, k1, and linearviscous damper, c1; m1 represents the mass lumped at the foundation level abovethe bearings; ui is the displacement of the mass mi with respect to the ground; kiand ci are the story stiffness and viscous damping coefficients for the actualstructure. The governing equations for free undamped vibration are expressed as

(5.81)

where the various matrices are the same as defined in chapter 2.

MU KU+ 0=

5.5 Optimal Stiffness Distribution - Discrete Shear Beam 375

Fig. 5.27: Notations for base isolated discrete shear beam.

In the previous development, the modal displacement profile was selectedsuch that the interstory displacement was constant over the beam. That strategy ismodified here to allow for a different interstory displacement for the first story,which, in this model, represents the relative displacement of the bearing.

Fig. 5.28: Example displacement profile.

mn

kn,cnmn-1

m2

m1

kn-1,cn-1

k2,c2

k1,c1

u1

u2

un-1

un

ub

usub

H

4

3

2

1ν

ν +14

3

2

1

a) actual b) scaled


Figure 5.28a illustrates the choice of displacement profile; us is the displacementat the top node due to deformation of the beam, and ub is the bearingdisplacement. For equal story height, the bilinear profile corresponds to uniformshear in the beam, . The bearing displacement is expressed as amultiple of the maximum structural displacement,

(5.82)

and the profile is scaled by taking as the independent displacementparameter. Fig 5.28b shows the scaled profile. With this choice of displacementparameter, the displacement vector takes the form

(5.83)

Note that the choice of q as the maximum structural displacement due todeformation of the structure is consistent with the approach followed for the fixedbase case. The modified displacement profile introduced here allows for anadditional story at the bottom of the beam and distinguishes between thedeformation at the base and within the beam.

Generalizing this approach for an n’th order system, the fundamentalmode profile is taken as

(5.84)

The remaining steps are the same as followed in section 2.7. One writes and substitutes for in eqn (5.81). This leads to

(5.85)

Scaling K and rearranging the equations results in

(5.86)

γ us H⁄=

ub ν us=

us

U qΦ=

us ν ν, 13--- ν, 2

3--- ν, 1+ + +

=

U qΦ=

Φ ν ν, 1n 1–------------ ν, 2

n 1–------------ … ν, , 1+ + +

=

U e iωt± Φ= U

KΦ ω2MΦ=

Sk' MΦ=


where and S is defined by eqn (2.161), listed below for convenience.

(5.87)

Given M and Φ, one solves eqn (5.86) for . This procedure is illustrated withthe following example.

Example 5.3:Scaled stiffness for a 4DOF beam with base isolation.

Consider the beam shown in Fig 5.28. The various matrices are

(1)

(2)

(3)

(4)

(5)

When the masses are equal, eqn (5) reduces to

k'i ki ω⁄ 2=

S' i i,( ) Φi Φi 1––=

S' i i 1+,( ) Φi Φi 1+–=

S' i j,( ) 0 for j i i 1+,≠=

k'

Φ ν ν, 13--- ν, 2

3--- ν, 1+ + +

=

MΦ m1ν m2 ν 13---+

m3 ν 23---+

m4 ν 1+( ), , ,

=

S

ν 1 3⁄–

1 3⁄ 1 3⁄–

1 3⁄ 1 3⁄–

1 3⁄

=

k' k'1= k'2 k'3 k'4

k'4 3m4 ν 1+( )=

k'3 3m3 ν 23---+

k'4+=

k'2 3m2 ν 13---+

k'3+=

k'1 m1k'23ν------+=


(6)

Fundamental mode response

Taking U according to eqn (5.84), the response of the fundamental mode isgoverned by

(5.88)

where the modal parameters are defined as

(5.89)

Since Φ now involves the relative displacement factor, , these terms will alsodepend on .

Example 5.4:Example 5.3 revisited.

Modal parameters for the 4DOF shear beam considered in example 5.3 arelisted below

k'1 m* 4 2ν---+

=

k'2 m* 6 9ν+( )=

k'3 m* 5 6ν+( )=

k'4 m* 3 3ν+( )=

mq cq kq+ + p mΓag–=

m ΦTMΦ= c ΦTCΦ= k ΦTKΦ=

p ΦTP= Γ ΦTMEm

------------------= ξ c2ωm------------=

νν


(1)

When the masses are equal, and simplify to

(2)

Values of and for a range of values of are listed below in Table 5.1.There is a significant reduction in with increasing , and this results in areduced response to seismic excitation.

Table 5.1: Modal mass and participation factors for 4DOF shear beam with equalmodal masses.

The modal damping parameter, , depends on both the bearing dampingand the structural damping properties. Incorporating damping in

the bearing is more effective than distributing damping over the structure for the

0 1.556 1.2861 7.556 0.7942 21.556 0.4643 43.556 0.3214 73.556 0.2455 111.556 0.197

m m1ν2 m2 ν2 23---ν 1

9---+ +

+=

m3 ν2 43---ν 4

9---+ +

m4 ν2 2ν 1+ +( )+ +

ΦTME m1ν m2 ν 13---+

m3 ν 23---+

m4 ν 1+( )+ + +=

c ν2c119--- c2 c3 c4+ +( )+=

p νp1 ν 13---+

p2 ν 23---+

p3 ν 1+( )p4+ + +=

m Γ

m m* 4ν2 2ν 149------+ +

=

Γ 2ν 1+

2ν2 ν 79---+ +

----------------------------97--- 1 2ν+

1 97---ν 1 2ν+( )+

-------------------------------------= =

m m*⁄ Γ νΓ ν

ν m m*⁄ Γ

cc1 c2 c3 c4, ,( )


fundamental mode response. Structural damping is needed mainly to control thehigher modes.

Stiffness calibration for seismic isolation

The peak fundamental mode response due to seismic excitation is given by

(5.90)

One specifies , , and , and determines by iterating on eqn (5.90).By definition, is the maximum structural displacement relative to the basemotion due to deformation of the structure. It is evaluated using the design valuefor the maximum transverse shear strain and the structural height,

(5.91)

The peak amplitude of the bearing displacement relative to the ground followsfrom eqn (5.82)

(5.92)

Given and , is determined with eqn (5.92). This approach has tobe modified when the structure is taken to be rigid, i.e., when . In thiscase, the system reduces to a SDOF model, and the formulation presented insection 5.2 is applicable.

Example 5.5:Stiffness calibration for Example 5.4

Returning to the 4DOF example structure, the following data is assumed.

(1)

qmax1ω----ΓSv ω ξ,( )=

qmax ξ Sv ω ξ,( ) ωqmax

qmax γ*H=

ub maxu1 max

≡ νqmax=

qmaxub max

νqmax 0≈

H 15m=

γ* 1 200⁄=

Sv ω ξ,( ) defined by Fig 5.29


Using (1),

(2)

To proceed further, one needs to specify . Various cases are considered below.

Fig. 5.29: Spectral velocity.

Case 1

The parameters corresponding to this bearing displacement are

(3)

Substituting in eqn (5.90) leads to an expression for the period, T.

(4)

Suppose . From Fig 5.29, for T>0.6 sec. No iteration isrequired here, since eqn (4) predicts T>0.6.

qmax 15( ) 1 200⁄( ) 0.075m= =

ν

0.1 0.6 1.0 10

0.1

1.0

Period

Spectral Velocity

T (sec)

Sv (m/s)

Sv 0.92 ξ, 0.05= =

Sv 0.8 ξ, 0.1= =

Sv 0.6 ξ, 0.2= =

Sv = 1.2, ξ = 0.02

ub max0.3m=

ν 0.3 0.075⁄ 4= =

Γ 0.245=

T2πqmax

ΓSv T ξ,( )------------------------- 1.922

Sv T ξ,( )---------------------= =

ξ 0.05= Sv 0.92=


(5)

The stiffness coefficients are generated using the results contained inexample 5.3. For the case of uniform mass, eqns (7) of example 5.3 apply. Taking

and according to eqn (5) above leads to

(6)

Damping is determined with eqn (3) of example 5.4. For and ,

(7)

(8)

The individual damping coefficients are related to by

(9)

One has to decide how to allocate damping to the various components. Forexample, assuming 75% of is contributed by the bearing requires

(10)

Placing damping at the base is an order of magnitude more effective thandistributing the damping throughout the structure for this degree of isolation.

Case 2

For this case, . The various parameters for , , anduniform mass are as follows.

T 1.9220.92------------- 2.13s= =

ω 2πT------ 2.94r s⁄= =

ν 4= ω

k4 129.6m*= k3 244.6m*=

k2 354.3m*= k1 37.9m*=

ν 4= ξ 0.05=

m 111.6m*=

c 2ξωm 2 0.05( ) 2.94( ) 111.6m*( ) 32.8m*= = =

c

c 16c119--- c2 c3 c4+ +( )+ 32.8m*= =

c

c1 1.54m*=

c2 c3 c4+ + 73.8m*=

ub max0.15m=

ν 2= ξ 0.05= ν 2=


(11)

Case 3 Fixed base

The fixed base case is treated in examples 2.9 and 2.10. Specializing theseresults for uniform mass and results in the following parameters andproperties.

(12)

Γ 0.464=

T 1.025s=

ω 6.16r s⁄=

m 21.556m*=

c 13.28m*=

4c119--- c2 c3 c4+ +( )+ 13.28m*=

k4 341m*= k3 644m*=

k2 909m*= k1 189m*=

ξ 0.05=

Γ 1.286=

T 0.5s=

ω 12.56r s⁄=

m 1.556m*=

19--- c2 c3 c4+ +( ) 1.95m*=

k4 473m*=

k3 788m*=

k2 946m*=

k1 ∞=


5.6 Optimal stiffness distribution - continuous cantilever beam

Stiffness distribution - undamped response

The equilibrium equations for undamped motion of the base isolated continuousbeam shown in Fig. 5.30 are

(5.93)

(5.94)

Fig. 5.30: Base isolated continuous beam.

The transverse shear and bending deformation measures for the beam arerelated to the translation and rotation quantities by

V x t,( ) ρm u x t,( ) xd

x

H

∫–=

M x t,( ) V x t,( ) xd

x

H

∫=

V

M

dx

kb

ub

DT DB,

x

H

uT uB

V dV+

M dM+

u

5.6 Optimal Stiffness Distribution - Continuous Cantilever Beam 385

(5.95)

(5.96)

Considering and to be functions only of time, integrating the resultingequations with respect to x, and imposing the boundary conditions at , oneobtains expressions for and in terms of , , and

(5.97)

(5.98)

where denotes the displacement of the base of the structure with respect toground. Taking

(5.99)

(5.100)

(5.101)

produces a periodic motion of the beam. Noting that the deformation measures and are related by (see eqn (2.14) and Fig 5.30)

(5.102)

and expressing in terms of the displacement at due to sheardeformation (see Fig. 5.30)

(5.103)

transforms eqn (5.97) into

(5.104)

The function defines the fundamental mode. The corresponding expression

x∂∂u γ β+=

χx∂

∂β=

γ χx 0=

u β γ t( ) χ t( ) ub t( )

u γx 12---χx2 ub+ +=

β χx=

ub

γ γ∗ eiω1t

=

χ χ∗ eiω1t

=

ub ub∗ e

iω1t=

γ∗ χ∗

suBuT------ χ∗ H

2γ∗-----------= =

ub∗ x H=

ub∗ νuT

∗ νγ∗ H= =

u xH---- sx2

H2-------- ν+ +

γ*Heiω1t

Φ x( ) (γ*Heiω1t ) q t( )Φ x( )= = =

Φ x( )


for the fixed base case is eqn (2.189).

Differentiating with respect to time,

(5.105)

and substituting for in eqn (5.93) leads to

(5.106)

The corresponding relation for the bending moment is

(5.107)

Lastly, the shear and bending rigidity distributions are determined with thedefinition equations

(5.108)

(5.109)

Equation (5.108) is written as

(5.110)

where is the shear rigidity at the base.

(5.111)

The parameter, , can be interpreted as an equivalent shear stiffness

u

u x sx2

H-------- νH+ +

– γ∗ ω12e

iω1t=

u

V ρmω12H2 1

2--- s

3--- ν x2

2H2----------– sx3

3H3----------– νx

H------–+ +

γ∗ eiω1t

=

M ρmω12H4χ∗ e

iω1t 16s----- 1

8--- ν

4s----- 1

4s----- 1

6--- ν

2s-----+ +

xH----–+ +=

νx2

4sH2------------- x3

12sH3---------------- x4

24H4-------------+ ++

DTVγ*----- ρmω1

2H2 12--- s

3--- ν x2

2H2----------– sx3

3H3----------– νx

H------–+ += =

DBMχ*----- ρmω1

2H4 16s----- 1

8--- ν

4s----- 1

4s----- 1

6--- ν

2s-----+ +

xH----– νx2

4sH2------------- x3

12sH3---------------- x4

24H4-------------+ + + + += =

DT

DT 0( )

12--- s

3--- ν+ +

--------------------------- 12--- s

3--- ν x2

2H2----------– sx3

3--------– νx

H------–+ +

=

DT 0( )

DT 0( ) ρmω12H2 1

2--- s

3--- ν+ +

ksH≡=

ks


measure.The shear force at the base of the beam must equal the shear force in thebearing to satisfy the force equilibrium condition for undamped motion. Equatingthese forces

(5.112)

and solving for results in

(5.113)

The fundamental frequency follows from eqn (5.111)

(5.114)

Figures 5.31 and 5.32 shows the mode shapes and shear deformationprofiles for the first five modes of a typical low rise building. The variation in themode shape profiles with the ratio of the stiffness of the isolator, , to the shearbeam stiffness are illustrated by Figures 5.33, 5.34 and 5.35. This ratio is equalto . Figure 5.36 displays the variation in the periods of the highest threefundamental modes. The primary influence is on the period of the fundamentalmode which is significantly increased when the stiffness of the isolator is severalorders of magnitude lower than the beam stiffness. The effect on the periods ofthe second and third modes is relatively insignificant. Figure 5.37 shows thevariation of the participation of the second and third modes relative to the first.The plot shows that the contribution of the second and third modes is alsosignificantly reduced by decreasing the stiffness of the isolator with respect to thebeam stiffness.

DT 0( )γ∗ kbub∗ kbνγ∗ H= =

kb

kbDT 0( )

νH----------------

1ν---ks= =

ω12

DT 0( )

ρmH2 12--- s

3--- ν+ +

-------------------------------------------=

kbks

1 ν⁄


Fig. 5.31: Mode shapes for a typical base isolated structure.

Fig. 5.32: Mode deformation profiles for a typical base isolated structure.

1 2 3 4 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Mode number

Nor

mal

ized

hei

ght

x H----

1 2 3 4 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----

Mode number


Fig. 5.33: Variation of mode 1 shape with relative stiffness of isolator.


−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----

Mode 1

kb ks⁄ 0.001=

kb ks⁄ 0.1=

kb ks⁄ 10.0=

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----

Mode 2

kb ks⁄ 0.001=kb ks⁄ 0.1=kb ks⁄ 10.0=



Fig. 5.36: Variation of periods with relative stiffness of isolator.

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----

Mode 3

kb ks⁄ 0.001=kb ks⁄ 0.1=kb ks⁄ 10.0=

10−3

10−2

10−1

100

101

0

1

2

3

4

5

6

7

8

kb ks⁄

Peri

od

T1T2T3


Fig. 5.37: Variation of relative participation factors with relative stiffness ofisolator.

Fundamental mode equilibrium equation

Incorporating the contribution of the base isolation system, the principle ofvirtual displacements has the form

(5.115)

where is the shear force in the bearing. The equations relating internal forcesto deformations and deformation rates are taken as

(5.116)

(5.117)

(5.118)

10−3

10−2

10−1

100

101

0

0.2

0.4

0.6

0.8

1

kb ks⁄

Γ2 Γ1⁄Γ3 Γ1⁄

Rel

ativ

e pa

rtic

ipat

ion

fact

ors

M δχ⋅ V δγ⋅+( )dx Fb δub⋅+

0

H

∫ b δu⋅( )dx

0

H

∫=

Fb

V DTγ CTγ+=

M DBχ CBχ+=

Fb kbub cbub+=


The form of the modal expansion follows from eqn (5.104).

(5.119)

(5.120)

Assuming external loading and seismic excitation, the loading term is

(5.121)

Finally, introducing the various terms in the principle of virtual displacementsleads to the equilibrium equation for

(5.122)

where

(5.123)

(5.124)

(5.125)

(5.126)

Expressing and as

u qΦ x( ) q xH---- sx2

H2-------- ν+ +

= =

β qΨ x( ) q 2sxH2---------

= =

b ρmag

– ρmu– b x t,( )+=

q

mq cq kq+ + p=

m ρmxH---- sx2

H2-------- ν+ +

2dx

0

H

∫ ρmH 13--- ν ν2 s

2--- 2sν

3---------+ s2

5-----+ + + += =

cCT

H2-------

4s2CB

H4----------------+ dx cbν2+

0

H

∫=

kDT

H2-------

4s2DB

H4----------------+ dx kbν2+

0

H

∫=

p b ρ– mag( ) xH---- sx2

H2-------- ν+ + dx

0

H

∫ ρmH 12--- ν s

3---+ + ag– pe–= =

c k


(5.127)

(5.128)

transforms eqn (5.122) to

(5.129)

where

(5.130)

For a pure shear beam, and the participation factor for the fundamentalmode reduces to

(5.131)

The expression for the modal damping ratio depends on how one specifies thedamping over the beam.

Rigidity calibration - seismic excitation

The calibration procedure presented in Chapter 2 is applied to the base isolatedmodel. Starting with

(5.132)

and substituting for results in

(5.133)

c 2ξωm=

k ω2m=

q 2ξωq ω2q+ + Γag–1m---- pe+=

Γν 1

2--- s

3---+ +

13--- ν2 ν s2

5----- s

2--- 2sν

3---------+ + + + +

--------------------------------------------------------------=

s 0=

Γν 1

2---+

13--- ν2 ν+ +-------------------------=

qmax

ΓSv ω ξ,( )ω

-------------------------=

qmax

ωΓSv

γ∗ H-----------

νΓ Sv

ub∗

-------------= =


One specifies in addition to the other parameters , and solves for .This value is then used to determine and with eqns (5.111) and (5.113).

Example 5.6:Stiffness calibration - Example Building #2.

The stiffness calibration for Building example #2 was considered inChapter 2. In what follows, the calibration procedure is extended to includestiffness and damping components located at the base. The design data are:

(1)

Using (1), the peak relative structural displacement is

(2)

To proceed further, one needs to specify the base displacement and then establishthe value of with eqn (5.103). We take

(3)

Then

(4)

Given , the participation factor follows from eqn (5.130)

(5)

The modal mass is determined with eqn (5.123)

(6)

Assuming CT = constant and CB =0 in eqn (5.124), the modal damping

ν γ* ξ Sv, ,( ) ωDT 0( ) kb

H 50m=

γ* 1 200⁄=

ξ1 0.05=

ρm 20 000kg m⁄,=

s 0.25=

Sv 0.92m s⁄=

qmax γ*H 0.25m= =

ν

ub 0.25m=

νub

qmax------------ 1.0= =

ν

Γ 0.600=

m 2.638 106kg×=

5.6 Building Design Examples 395

coefficient reduces to

(7)

Equation (5.127) relates to .

(8)

Lastly, the frequency is found using eqn (5.133) and the design data for( and )

(9)

With known, the modal damping coefficient follows from eqn (8),

(10)

and the transverse shear rigidity at the base of the beam is determined with eqn(5.111).

(11)

Finally, given , the isolation stiffness is estimated using eqn (5.113), whichis based on neglecting the contribution of the damping force in the bearing.

(12)

5.7 Building design examples

Stiffness distribution based on fundamental mode response

Since base isolation is a potential solution for buildings with less than about 10stories and having an aspect ratio that prevents overturning, only Buildings 1 and

cCTH------- ν2cb+ 0.02CT cb+= =

c ω

c 2ξωm 0.2638 106× ω Ns m⁄= =

Svξ 0.05= Sv 0.92m s⁄=

ωΓSv

γ*H---------- 2.4Sv 2.61r s⁄= = =

T 2.41sec=

ω

c 0.688MNs m⁄=

DT 0( ) 538.6MN=

DT 0( )

kb 10.77MN=


2 of Chapter 2 are considered in the simulation. Tables 2.4 and 2.5 lists the designdata. The damping in the structure is obtained by specifying and computing astiffness proportional damping matrix considering a fixed based structure, ascarried out in Chapter 2. The damping in the bearing for specified bearingdamping ratio is obtained by assuming the structure to be a SDOF systemhaving a mass equal to the total mass of the structure and a stiffness equal to thestiffness of the bearing as obtained from eqn (5.113). Figures 5.38 and 5.39 showthe shear rigidity distributions for Buildings 1 and 2 obtained with eqns (5.108)and (5.109), taking . The corresponding shear deformation profilesfor different combinations of building damping ratios and isolator damping ratioswhen the structures are subjected to scaled versions (to for

) of the El Centro and Taft accelerograms are plotted in Figures 5.40through 5.47. The maximum deformation in the bearing is also indicated on theplots. The plots show that increasing the damping in the bearing tends to reducethe bearing deformation without significantly altering the shape of thedeformation profile along the structure’s height. Furthermore, just as for the fixedbase structures, the contribution of the higher modes becomes more significant asthe structure becomes more slender. The structural parameters, as well as themean and standard deviation results for the deformations of the above buildingsare tabulated in the following section.

ξ1

ξb

Sv 1.2m s⁄=

Sv 1.2m s⁄=ξ 0.02=


Fig. 5.38: Initial shear rigidity distribution for Building 1.

Fig. 5.39: Initial shear rigidity distribution for Building 2.

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x 109

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Building #1Quadratic basedInitialH 25m=ρm 20000kg/m=s 0.15=Sv 1.2m/s=ξ1 2%=kb 6.35 6×10 N/m=


Nor

mal

ized

hei

ght

x H----

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x 109

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----

Building #2Quadratic basedInitialH 50m=ρm 20000kg/m=s 0.25=Sv 1.2m/s=ξ1 2%=kb 9.93 6×10 N/m=





0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----


γ∗

ξ1 2%,= ub 0.18m=ξ1 5%,= ub 0.18m=ξ1 10%,= ub 0.17m=ξ1 20%,= ub 0.17m=


El Centro

H 25m=ρm 20000kg/m=s 0.15=Sv 1.2m/s=ξb 2%=

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----


γ∗

ξb 2%,= ub 0.18m=ξb 5%,= ub 0.16m=ξb 10%,= ub 0.13m=ξb 20%,= ub 0.11m=


El Centro

H 25m=ρm 20000kg/m=s 0.15=Sv 1.2m/s=ξ1 2%=




0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----


γ∗



Taft

H 25m=ρm 20000kg/m=s 0.15=Sv 1.2m/s=ξb 2%=

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----


γ∗



Taft

H 25m=ρm 20000kg/m=s 0.15=Sv 1.2m/s=ξ1 2%=




0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----


γ∗



El Centro

H 50m=ρm 20000kg/m=s 0.25=Sv 1.2m/s=ξb 2%=

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----


γ∗


Building #2Quadratic BasedInitial

El Centro

H 50m=ρm 20000kg/m=s 0.25=Sv 1.2m/s=ξ1 2%=




0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----


γ∗



Taft

H 50m=ρm 20000kg/m=s 0.25=Sv 1.2m/s=ξb 2%=

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----


γ∗



Taft

H 50m=ρm 20000kg/m=s 0.25=Sv 1.2m/s=ξ1 2%=


Stiffness distribution including the contribution of the higher modes

This section extends the iterative procedure developed in Section 2.10 toincorporate iterating over the stiffness for the beam and isolator. For simplicity,only shear deformation in the isolator is considered. The method consists ofincluding the contribution of the higher modes to the transverse shear, thebending deformation, and the base shear, and then updating the shear andbending rigidities and isolation stiffness using

(5.134)

(5.135)

(5.136)

The peak values are found with eqn (2.270).

Rigidity iterations are performed on building examples 1 and 2. Table 5.2lists the parameters of the buildings for the initial rigidity distributions as well asfor a single iteration. For both building examples, one iteration was sufficient toachieve convergence.Tables 5.3 and 5.4 contain the deformations averaged overthe height of the structure and the corresponding standard deviations forBuildings 1 and 2 subjected to scaled versions of El Centro and Taft excitations.

DTi 1+( ) x( ) DT

i( ) γ i( ) x( )[ ] max

γ∗-------------------------------=

DBi 1+( ) x( ) DB

i( ) χ i( ) x( )[ ] max

χ∗--------------------------------=

kbi 1+( ) Vb

i( )[ ] max

ub∗

-------------------------=


Table 5.2: Modal parameters - building examples.

Figures 5.48 through 5.57 show the shear and bending rigiditydistributions resulting from the first iteration, as well as the shear deformationprofiles corresponding to different combinations of building damping ratios andisolator damping ratios under El Centro and Taft excitation.Results from oneiteration provide sufficient convergence accuracy. For Building 2, the iterativescheme tends to pull the top back in, resulting in a more uniform deformationprofile. The effect of damping is similar to that noticed in the examples of theprevious section.

Bldg #1Initial (Q) 2.09 0.62 0.33 1.56 3.55 6.26 0.11 0.03Iteration 1 2.01 0.60 0.32 1.54 3.64 6.37 0.12 0.03

Bldg #2Initial (Q) 2.51 0.90 0.49 1.48 3.70 6.60 0.20 0.06Iteration 1 2.45 0.85 0.46 1.48 3.83 6.89 0.19 0.06

T1 s( ) T2 s( ) T3 s( ) ξ1 %( ) ξ2 %( ) ξ3 %( ) Γ2 Γ1⁄ Γ3 Γ1⁄


Table 5.3: Mean and standard deviation deformation results for Building 1.

Bldg #1 (%) (%) (m) (10-3) (10-4) (10-5) (10-6)

El CentroInitial (Q) 2.00 2.00 0.18 2.80 0.70 3.18 0.56

5.00 2.00 0.18 2.74 0.46 3.11 0.3110.00 2.00 0.17 2.61 0.11 2.93 0.0720.00 2.00 0.17 2.43 0.22 2.72 0.24

2.00 2.00 0.18 2.80 0.70 3.18 0.562.00 5.00 0.16 2.55 1.09 2.93 0.952.00 10.00 0.13 2.36 1.92 2.79 1.312.00 20.00 0.11 2.19 2.35 2.62 1.57

Iteration 1 2.00 2.00 0.17 3.28 1.36 2.86 4.405.00 2.00 0.17 3.09 1.05 2.68 4.3110.00 2.00 0.15 2.79 0.66 2.41 4.1020.00 2.00 0.15 2.37 0.62 1.98 3.63

2.00 2.00 0.17 3.28 1.36 2.86 4.402.00 5.00 0.14 2.55 1.23 2.22 3.082.00 10.00 0.12 2.26 0.92 1.96 3.212.00 20.00 0.10 2.16 1.38 1.90 2.98

TaftInitial (Q) 2.00 2.00 0.09 1.61 3.42 2.04 2.76

5.00 2.00 0.08 1.42 2.22 1.74 1.8610.00 2.00 0.07 1.25 1.32 1.49 0.9620.00 2.00 0.07 1.08 0.49 1.25 0.31

2.00 2.00 0.09 1.61 3.42 2.04 2.762.00 5.00 0.06 1.34 2.78 1.67 2.602.00 10.00 0.06 1.21 1.80 1.45 1.792.00 20.00 0.05 1.17 1.68 1.41 1.54

Iteration 1 2.00 2.00 0.08 1.80 2.26 1.64 2.005.00 2.00 0.08 1.57 1.00 1.38 1.9710.00 2.00 0.08 1.38 0.39 1.19 1.9720.00 2.00 0.07 1.17 0.27 0.99 1.84

2.00 2.00 0.08 1.80 2.26 1.64 2.002.00 5.00 0.07 1.41 1.73 1.27 1.452.00 10.00 0.06 1.25 0.63 1.07 1.432.00 20.00 0.05 1.22 0.93 1.07 1.29

ξ1 ξb ubγm γsd χm χsd


Table 5.4: Mean and standard deviation deformation results for Building 2.

Bldg #2 (%) (%) (m) (10-3) (10-4) (10-5) (10-6)

El CentroInitial (Q) 2.00 2.00 0.24 4.61 7.49 4.90 5.96

5.00 2.00 0.23 4.13 4.17 4.25 3.2810.00 2.00 0.21 3.77 2.49 3.81 1.8320.00 2.00 0.20 3.26 1.22 3.25 0.81

2.00 2.00 0.24 4.61 7.49 4.90 5.962.00 5.00 0.22 4.15 6.25 4.39 4.882.00 10.00 0.19 3.61 4.98 3.79 3.722.00 20.00 0.14 2.99 4.10 3.15 2.78

Iteration 1 2.00 2.00 0.23 4.67 4.36 4.04 5.925.00 2.00 0.22 4.11 1.84 3.44 5.6710.00 2.00 0.21 3.65 1.68 2.98 5.5420.00 2.00 0.20 3.12 2.34 2.50 5.14

2.00 2.00 0.23 4.67 4.36 4.04 5.922.00 5.00 0.21 4.21 3.61 3.62 5.432.00 10.00 0.18 3.65 2.87 3.12 4.822.00 20.00 0.14 2.98 2.43 2.55 4.15

TaftInitial (Q) 2.00 2.00 0.14 2.77 6.24 3.00 5.56

5.00 2.00 0.13 2.30 3.49 2.41 3.1910.00 2.00 0.11 1.89 1.74 1.91 1.7420.00 2.00 0.09 1.49 1.33 1.52 1.17

2.00 2.00 0.14 2.77 6.24 3.00 5.562.00 5.00 0.11 2.34 6.53 2.56 6.052.00 10.00 0.08 1.99 6.03 2.23 4.982.00 20.00 0.07 1.87 5.69 2.16 4.48

Iteration 1 2.00 2.00 0.11 2.45 4.92 2.22 2.845.00 2.00 0.10 2.05 2.68 1.80 2.3810.00 2.00 0.09 1.69 1.22 1.43 2.0320.00 2.00 0.08 1.38 0.75 1.16 2.00

2.00 2.00 0.11 2.45 4.92 2.22 2.842.00 5.00 0.09 2.16 4.57 1.97 2.732.00 10.00 0.08 1.87 3.92 1.69 2.472.00 20.00 0.07 1.74 2.93 1.57 1.72

ξ1 ξb ubγm γsd χm χsd


Fig. 5.48: Converged shear rigidity distribution for Building 1.

Fig. 5.49: Converged shear rigidity distribution for Building 2.

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x 109

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----Building #1Quadratic basedIteration 1H 25m=ρm 20000kg/m=s 0.15=Sv 1.2m/s=ξ1 2%=kb 6.94 6×10 N/m=


0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x 109

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----

Building #2Quadratic basedIteration 1H 50m=ρm 20000kg/m=s 0.25=Sv 1.2m/s=ξ1 2%=kb 1.05 7×10 N/m=





0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----


γ∗


Building #1Quadratic basedIteration 1

El Centro

H 25m=ρm 20000kg/m=s 0.15=Sv 1.2m/s=ξb 2%=

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----


γ∗



El Centro

H 25m=ρm 20000kg/m=s 0.15=Sv 1.2m/s=ξ1 2%=




0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----


γ∗



Taft

H 25m=ρm 20000kg/m=s 0.15=Sv 1.2m/s=ξb 2%=

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----


γ∗



Taft

H 25m=ρm 20000kg/m=s 0.15=Sv 1.2m/s=ξ1 2%=




0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----


γ∗



El Centro

H 50m=ρm 20000kg/m=s 0.25=Sv 1.2m/s=ξb 2%=

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----


γ∗



El Centro

H 50m=ρm 20000kg/m=s 0.25=Sv 1.2m/s=ξ1 2%=




0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----


γ∗



Taft

H 50m=ρm 20000kg/m=s 0.25=Sv 1.2m/s=ξb 2%=

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Nor

mal

ized

hei

ght

x H----


γ∗



Taft

H 50m=ρm 20000kg/m=s 0.25=Sv 1.2m/s=ξ1 2%=

Problems 411

Problems

Problem 5.1

Consider a SDOF system with m=1000kg. The design requirements for thesystem involve 2 nonconcurent loading conditions, a periodic external forcingand a periodic ground motion.Under the external forcing, the relative motion isprescribed; the total motion is prescribed for the case of ground excitation. Thesedesign conditions are summarized below. The asterisk denotes a specifiedquantity.

1) (1)

(2)

2) (3)

(4)

The design problem involves determining k and c such that the aboveperformance requirements are satisfied.

a) Recommand values for k and c corresponding to the following design data:

1)

2)

k

c

m

ut ug u+=ug

p

p p* ΩP* tsin=

u u*<

ug ug* Ωg

* tsin=

ut ut* H3

* ug*≡<

Ωg* 2πr s⁄= ug

* 0.2m= H3* 0.125=

p* 400N= ΩP* π

2---r s⁄= u* 0.2m=

Ωg* 2πr s⁄= H3

* 0.125=

p* 800N= ΩP* π

2---r s⁄= u* 0.2m=


3)

b) It may not be possible to satisfy both constraints with the same value ofstiffness. When this problem arises, one can determine the stiffness correspondingto each design condition and design a spring mechanism that allows the stiffnessto be varied. Suggest design concepts for such a mechanism. This situation mayoccur for lateral loading applied to a seismically isolated structure. Discuss howyou would implement a variable stiffness scheme for non simultaneous wind andseismic loading.

Problem 5.2

Refer to the modified SDOF model defined by Fig 5.8. Take m=10,000 kgand k=400 kN/m. Suppose the system is to be subjected to a ground motion,

, and the total displacement is required to be less than0.2m. Determine the appropriate bearing stiffness.

Problem 5.3

Refer to the modified SDOF model defined by Fig 5.8. Take m=10,000 kgand the ground motion to be . Determine and suchthat

a) and .

b) and

Problem 5.4

Refer to the modified SDOF model defined by Fig 5.8. Take m=10,000 kgand consider the system to be subjected to seismic excitation of intensity

. Determine , and for the following design conditions:

Ωg* 2πr s⁄= H3

* 0.125=

p* 1600N= ΩP* π

2---r s⁄= u* 0.2m=

ug meters( ) 0.5 4πtsin=

ug meters( ) 0.5 4πtsin= k kb

u 0.3m≤ ub 0.3m≤

u 0.1m≤ ub 0.3m≤

Sv 1.2m s⁄= k kb Teq

Problems 413

a)

b)

c)

Problem 5.5

Consider a cylindrical bearing having a diameter of 0.6m, a height of 0.3mand composed of filled natural rubber. Take the rubber properties according to Fig5.18. Assume the frequency range is from 1 to 5 Hz, and the temperature is 20oC.

a) Estimate the equivalent linear stiffness and linear viscous dampingparameters, and .

b) Determine the diameter of a lead plug for the case where the plugstiffness is 10 times the stiffness of the rubber bearing. Assume the lead plug andrubber cylinder have the same height. Take .

c) Assume the bearing experiences periodic excitation resulting in a shearstrain amplitude of 50%. Determine the secant stiffness, , and loss factor, ,using eqns (5.70) and (5.74).

d) Instead of lead, consider using low strength steel as the material for theinitial stiffness element, . Take and for thesteel plug. Repeat parts b and c.

Problem 5.6

Consider an isolation system composed of a NRB, a steel hysteretic

u* 0.1m= ub* 0.1m=

u* 0.1m= ub* 0.2m=

u* 0.1m= ub* 0.3m=

keq ceq

GP 4 3×10 MPa=

ks η

k2 τy 150MPa= G 80 3×10 MPa=


damper, and a viscous damper. Neglect the damping provided by the NRB.

a) For low level loading, one specifies the initial stiffness, damping, and yieldforce level. Describe how you would design the steel damper.

b) For high level loading, one specifies the secant stiffness and equivalent viscousdamping based on a seismic analysis. Describe how you would design the NRB.How would you select the viscous damper?

c) Suppose the isolation system is composed of spring and damper elementswhose properties can be varied instantaneously. Assuming the elements behavelinearly, the force is given by

k1 rubber( )

c

F, uk2 damper( )

F

Fyk1

k1+k2

uy u* u

F k t( )u c t( )u+=

k

c

F, u

Problems 415

where k(t) and c(t) are properties that can be “adjusted”. Describe how you wouldutilize this system for a building subjected (nonsimultaneously) to both wind andseismic excitation. How would you design these devices? Note: a system whichhas the ability to change its properties is said to be adaptive. Adaptive systemsare discussed in Chapter 6.

Problem 5.7

Consider a 6 story building with base isolation modelled as a 7 DOFsystem. Take the floor masses as and

.

a) Find the scaled stiffness for the profile based on eqn (5.84). Take .

b) Evaluate the expressions for , and .

c) Calibrate the stiffness distribution for seismic excitation. Use the spectralvelocity plot contained in Fig 5.29. Take , , and

. Allocate 75% of the damping to the bearing. Use MOTIONLABto determine the modal shapes, frequencies, and damping ratios for the 2nd and3rd modes.

d) Repeat part c assuming the base is fixed. Take the sum of the stiffness factors as

u7m7k7 c7,

u1m1k1 c1,

m1 m2 … m6 10 000kg,= = = =m7 20 000kg,=

ν 2.0=

m c p, , Γ

ξ 0.05= us max0.15m=

ub max0.3m=


a measure of the cost of stiffness. Compare the costs of the fixed base and baseisolation solutions. Also compare the modal properties.

e) Suppose filled rubber bearings having a diameter of 0.5m and a height of 0.25mare to be used for the isolation system. Assume and for therubber. How many bearings are required for the design conditions specified inpart c? Also discuss how you would provide the damping required for theisolation system.

f) Discuss how you would deal with lateral wind loading. Assume the dominantwind gust frequency is 0.2Hz.

Problem 5.8

Consider a base isolated continuous cantilever beam having a uniformmass density of 2000 kg/m and a height of 30m.

a) Generate shear and bending rigidity distributions and isolation stiffness whichcorrespond to a fundamental frequency of 0.33 Hz and various values for s and

. Take and . Comment on the sensitivity of the rigidityparameters to variation in s and .

b) Approximate the continuous beam with 6 shear beam segments plus anadditional segment to simulate the bearing. Determine the first 3 mode shapesand frequencies corresponding to s=0 and , 1.5, and 3. Consider the lumpedmasses to be equal. Comment on the sensitivity of the modal properties to theratio of isolation stiffness, , to the shear beam stiffness measure, .

G 4MPa= η 0.15=

30m ρm constant

x

ν 0 s 0.25≤ ≤ 1 ν 3≤ ≤ν

ν 0=

kb ks

Problems 417

Problem 5.9

Consider a base isolated continuous shear beam having a uniform massdensity and constant transverse shear rigidity. Assume the lateral displacement isapproximated by

a) Establish an equation for q(t) using the Principle of Virtual Displacements.Allow for linear viscous damping in the bearing and uniform material dampingin the beam. Write the result in the same form as eqn (5.122) and determine theexpressions for , and .

b) Suppose H=30m, , and v=2. Calibrate the stiffness distribution( and ) for , , and defined by Fig 5.29. Determinethe damping parameters assuming the bearing contributes 75% to .

c) Approximate the continuous beam with 6 shear beam segments plus anadditional segment to simulate the bearing. Determine the first 2 mode shapesand frequencies, using the design data generated in part b.

H DT and ρm

constant

x

ug

kb cb,

qub

u ug ub q xH----+ + ug q ν x

H----+

+= =

m c k p ω ξ, , , , , Γ

ρm 2000kg m⁄=DT kb ξ 0.05= qmax 0.15= Sυ

c


Problem 5.10

Consider the base isolated shear beam shown above. Assume uniformmass density, , and the shear rigidity to be defined as

a) Derive the equilibrium corresponding to the following approximatedisplacement expansion

Discuss how the undamped free vibration response behaves as , , and arevaried.

b)

Suppose the force, , applied at the top of the structure is generated by a

m1

H

ub ug+

ub u1 ug+ +

p

ρm DT x( ),

cbug

kb

m1 a ρm H( )=

DT x( ) D∗ 1 x2H--------–

=

u u1 t( ) xH---- ub t( )+=

D∗ kb a

ub u1 ud+ +

ub u1+

mdm1

p

Problems 419

tuned mass damper as shown in the figure. How would you estimate theproperties of the tuned mass damper to obtain an effective damping ratio of 0.05for the mode shape approximation considered in part a?

c) Suppose has the form

where are constants and is prescribed. Assume and determinethe equation for . What effect do and have on the response? Illustrate fora periodic excitation.

p

p b1u1– b2u1– p t1( )+=

b1 b2, p cb 0=u1 b1 b2


421

Part II: Active Control

Chapter 6

Introduction to active structural motioncontrol

6.1 The nature of active structural control

Active versus passive control

The design methodologies presented in the previous chapters provide systematicprocedures for distributing passive motion control resources which, by definition,have fixed properties and do not require an external source of energy. Onceinstalled, a passive system cannot be modified instantaneously, and therefore oneneeds a reliable estimate of the design loading and an accurate numerical modelof the physical system for any passive control scheme to be effective. The inabilityto change a passive control system dynamically to compensate for an unexpectedloading tends to result in an over-conservative design. When self-weight is animportant design constraint, one cannot afford to be too conservative. Also,simulation studies on the example building structures show that passive controlis not very effective in fine tuning the response in a local region. Consideringthese limitations, the potential for improving the performance by dynamicallymodifying the loading and system properties exists. An active structural controlsystem is one which has the ability to determine the present state of the structure,decide on a set of actions that will change this state to a more desirable one, andcarry out these actions in a controlled manner and in a short period of time. Suchcontrol systems can theoretically accommodate unpredictable environmental

422 Chapter 6: Introduction to Active Structural Motion Control

changes, meet exacting performance requirements over a wide range of operatingconditions, and compensate for the failure of a limited number of structuralcomponents. In addition, they may be able to offer more efficient solutions for awide range of applications, from both technical and financial points of view.

Active motion control is obtained by integrating within the structure acontrol system consisting of three main components: a) monitor, a dataacquisition system, b) controller, a cognitive module which decides on a course ofaction in an intelligent manner, and c) actuator, a set of physical devices whichexecute the instructions from the controller. Fig 6.1 shows the interaction andfunction of these components; the information processing elements for activecontrol are illustrated in Fig 6.2. This control strategy is now possible due tosignificant recent advances in materials that react to external stimuli in a non-conventional manner, sensor and actuator technologies, real-time informationprocessing, and intelligent decision systems.

Fig. 6.1: Components of an active control system

STRUCTURE RESPONSEEXCITATION

SENSORSSENSORS

ACTUATORAGENTS TO CARRY

OUT INSTRUCTIONS

IDENTIFY THE STATE OF SYSTEM

DECIDE ON COURSE OF ACTION

DEVELOP THE ACTION PLAN(SET OF INSTRUCTIONSTO BE COMMUNICATED

TO THE ACTUATORS)

CONTROLLER

MONITOR

to measureexternalloading

MONITOR

to measureresponse

6.1 The Nature of Active Structural Control 423

Fig. 6.2: Information Processing Elements for an Active Control System

Fig. 6.3: Passive and active feedback diagrams

Physical System

Sensor Sensor Sensor

Data

Processing Transmission channel Processing

Modeling &Analysis

DecisionMaking

Action

Visualization

Archivaland

Access

Fusion

h p( )p u

(a) Passive

h' p ∆pe ∆pf+ +( ) u

(b) Active

observe u

u εu+

decide on ∆pf

+

∆pf

p

observe p

p εp+decide on ∆pe

+

∆pe

decide on changingh to h'


The simple system shown in Fig. 6.3 is useful for comparing active andpassive control. Figure 6.3(a) corresponds to passive control. The input, , istransformed to an output, , by the operation

(6.1)

One can interpret this system as a structure with denoting the loading, thedisplacement, and the flexibility of the structure. The strategy for passivemotion control is to determine such that the estimated output due to theexpected loading is contained within the design limits, and then design thestructure for this specific flexibility.

Active control involves monitoring the input and output, and adjusting theinput and possibly also the system itself, to bring the response closer to thedesired response. Figure 6.3(b) illustrates the full range of possible actions.Assuming the input corrections and system modifications are introducedinstantaneously, the input-output relation for the actively controlled system isgiven by

(6.2)

Monitoring the input, and adjusting the loading is referred to as open-loop control.Observing the response, and using the information to apply a correction to theloading is called feedback control. The terminology closed-loop control issynonymous with feedback control.

In addition to applying a correction to the input, the control system may alsoadjust certain properties of the actual system represented by the transformation

. For example, one can envision changing the geometry, the connectivity,and the properties of structural elements in real time. One can also envisionmodifying the decision system. A system that can adjust its properties andcognitive processes is said to be “adaptive”. The distinguishing characteristic ofan adaptive system is the self-adjustment feature. Non-adaptive active structuralcontrol involves monitoring and applying external forces using an invariantdecision system. The make up of the structure is not changed. Adaptive control isthe highest level of active control.

The role of feedback

Feedback is a key element of the active control process. The importance offeedback can be easily demonstrated by considering a linear static system and

pu h p( )

u h p( )=

p uh

h p( )

u h' p ∆pe ∆pf+ +( )=

h p( )

6.1 The Nature of Active Structural Control 425

taking the input correction to be a linear function of the output. For this case,

(6.3)

(6.4)

where and are constants. Substituting in eqn (6.2) specialized for h’=h andsolving for results in

(6.5)

When is positive, the sensitivity of the system to loading is increased byfeedback, i.e. the response is amplified. Taking negative has the opposite effecton the response. Specializing eqn (6.5) for negative feedback ( ), the responsebecomes

(6.6)

Increasing decreases the effect of external loading. However, the influence of, the noise in the response observation, increases with and, for sufficiently

large , is essentially independent of the feedback parameter. This resultindicates that the accuracy of the monitoring system employed to observe theresponse is an important design issue for a control system.

Computational requirements and models for active control

The monitor component identified in Figs 6.1 and 6.2 employs sensors tomeasure a combination of variables relevant to motion such as strain,acceleration, velocity, displacement, and other physical quantities such aspressure, temperature, and ground motion. This data is usually in the form ofanalog signals which are converted to discrete time sequences, fused with otherdata, and transmitted to the controller module. Data compression is an importantissue for large scale remote sensing systems. Wavelet based data compression(Amaratunga, 1997) is a promising approach for solving the data processingproblem.

The functional requirements of the controller are to compare the observedresponse with the desired response, establish the control action such as the levelof feedback force, and communicate the appropriate commands to the actuator

u hp=

∆pf kf u εu+[ ]=

h kfu

u h1 hkf–---------------- p ∆pe+[ ]

hkf1 hkf–---------------- εu+=

kfkf

kf 0<

u h1 hkf+-------------------- p ∆pe+[ ]

hkf1 hkf+-------------------- εu+=

kfεu kf

kf


which then carries out the actual control actions such as apply force or modify astructural property. The controller unit is composed of a digital computer andsoftware designed to evaluate the input and generate the instructions for theactuators.

There are 2 information processing tasks: state identification and decisionmaking. Given a limited amount of data on the response, one needs to generate amore complete description of the state of the system. Some form of modelcharacterizing the spatial distributions of the response and data analysis arerequired. Once the state has been identified, the corrective actions which bring thepresent state closer to the desired state can be established. In this phase, a modelwhich defines the input - output relationship for the structure is used togetherwith an optimization method to decide upon an appropriate set of actions.

For algorithmic non-adaptive systems, the decision process is based on anumerical procedure that is invariant during the period when the structure isbeing controlled. Time invariant linear feedback is a typical non-adaptive controlalgorithm. An adaptive controller may have, in addition to a numerical controlalgorithm, other symbolic computational models in the form of rule-basedsystems and neural networks which provide the capability of modifying thestructure and control algorithm in an intelligent manner when there is a change inthe environmental conditions. Examples illustrating time invariant linearfeedback control algorithms are presented in the following sections; a detailedtreatment of the algorithms is contained in Chapters 7 and 8.

6.2 An introductory example of quasi-static feedback control

Consider the cantilever beam shown in Fig 6.4. Suppose the beam acts likea bending beam, and the design objective is to control the deflected shape suchthat it has constant curvature. The target displacement distribution correspondingto this constraint has the form

(6.7)

where is the desired curvature. One option is to select the bending rigidityaccording to

(6.8)

u* x( ) 12---χ*x2=

χ*

DB x( ) M x( )χ*

----------------=

6.2 An Introductory Example of Quasi-static Feedback Conrol 427

where M(x) is the moment at location x due to the design loading. This strategy isa stiffness based passive control approach. A second option is to select arepresentative bending rigidity distribution, and apply a set of control forceswhich produce a displacement distribution that, when combined with thedisplacement due to the design loading, results is a displacement profile that isclose to the desired distribution. In what follows, the latter option is discussed.

Fig. 6.4: Cantilever beam with control force

Suppose the control force system consists of a single force applied at x=L.Assuming linear elastic behavior, and using the linear technical theory of beamsas the model for the structure, the displacement due to F is estimated as

(6.9)

where is the bending rigidity, considered constant in this example. Thedisplacement due to the design loading is also determined with the technicalbeam theory. This term is denoted as , and expressed as

(6.10)

Combining the 2 displacement patterns results in the total displacement, u(x).

(6.11)

The expanded form of eqn (6.11) corresponding to the particular choice of controlforce location for this example is

(6.12)

The difference between and is defined as and interpreted

L

u(x)

x

F

uc x( ) F2DB----------- Lx2 x3

3-----–

FDB-------h x( )= =

DB

uo x( )

uo x( ) 1DB-------g x( )=

u x( ) uo x( ) uc x( )+=

u x( ) 1DB------- g x( ) Fh x( )+[ ]=

u x( ) u* x( ) e x( )


as the displacement error.

(6.13)

For this example, is considered to be fixed, and therefore is a functiononly of the single control force magnitude F.

(6.14)

Ideally, one wants for . This goal cannot be achieved, and it isnecessary to work with a relaxed condition.

The simplest choice is collocation, which involves setting equal to zero ata set of prescribed locations. For example, setting at x=L leads to

(6.15)

A more demanding condition is a least square requirement, which involvesfirst forming the sum of evaluated at a set of prescribed points, and thenselecting F such that the sum is a minimum. The continuous least square sum isgiven by the following integral

(6.16)

Taking J(F) as the measure of the square error sum, F is determined with thestationary condition

(6.17)

Differentiating the integral expression for J,

(6.18)

and using eqn (6.14), which defines for this particular example, results in

(6.19)

The value of F defined by eqn (6.19) produces the absolute minimum value of J. Aproof of this statement is presented in Section 7.2 which treats in more detail the

e x( ) uo uc u*–+=

DB e x( )

e x( ) 1DB------- g x( ) Fh x( )+[ ] u* x( )–=

e 0= 0 x L≤ ≤

ee 0=

FDB u* L( ) g L( )–

h L( )------------------------------------------=

e2

J 12--- e2 xd

0

L

∫ J F( )= =

F∂∂J 0=

F∂∂J e

F∂∂e xd

0

L

∫ 0= =

e x( )

F

h x( ) DB u* x( ) g x( )–[ ] xd0

L

∫h x( )( )2 xd

0

L

∫-----------------------------------------------------------------------=

6.2 An Introductory Example of Quasi-static Feedback Conrol 429

least square procedure for quasi-static loading.

Example 6.1: Shape control for uniform loading

This example illustrates the application of the approach described above tothe case where the design loading is a uniform distributed load extending overthe entire length of the beam. The corresponding deflected shape is

(1)

Applying collocation at x=L leads to

(2)

The least square solution is

(3)

Both solutions are approximate since they do not satisfy . One canimprove the performance by taking additional control forces. Selecting the spatialdistribution of the control forces is a key decision for the design of a controlsystem.

The above discussion assumes that there is some initial loading, and onecan determine the corresponding displacement field with the physical model ofthe structural system. This control strategy is similar to the concept ofprestressing. A more general scenario is the case where one is observing theresponse at a set of “observation” points and the loading is being appliedgradually so that there is negligible dynamic amplification and sufficient time to

uo x( ) 1DB-------g x( ) 1

DB------- w

24------ x4 4x3L– 6x2L2+( )= =

F e L( ) 0=32---

DBχ*

L-------------

32---

wL–=

F ls9166------

DBχ*

L------------- 2065

5280------------

wL–=

1.379( )DBχ*

L------------- 0.391( )wL–=

e x( ) 0=


adjust the control forces. Here, one needs to establish using the observeddisplacement data.

Suppose there are observation points located at ( j= 1, 2, ..., s), and attime t the monitoring system produces the data set . This data can beused together with an interpolation scheme to generate an estimate of for theregion adjacent to the observation points. A typical spatial interpolation modelhas the form

(6.20)

where are interpolation functions.

Given , one forms the displacement error,

(6.21)

and determines F(t) with either collocation or a least square method. Thecontinuous least square estimate for F(t) is given by

(6.22)

Example 6.2: Discrete displacement data

Suppose the displacement observation points are located at x=L/2 andx=L. Given these 2 values of displacement, one needs to employ an interpolationscheme in order to estimate . Taking a quadratic expansion,

(1)

uo x( )

s xjuo xj t,( )

uo

uo x t,( ) uo xj t,( )Ψj x( )

j 1=

s

∑=

Ψj x( )

uo x t,( )

e x t,( ) uo x t,( ) u* x( )–1

DB-------h x( )F t( )+=

F t( ) DB

h x( ) u* x( ) uo x t,( )–( ) xd0

L

∫h x( )( )2 xd

0

L

∫---------------------------------------------------------------------=

uo x( )

uo x( ) ao a1x a2x2+ +=

6.3 An Introductory Example of Dynamic Feedback Control 431

and specializing eqn(1) for points 1 and 2 results in the following approximation,

(2)

The expression for the control force is obtained using eqn (6.22) and eqn(2). Evaluating the integrals leads to

(3)

as an estimate for F.

6.3 An introductory example of dynamic feedback control

To gain further insight on the nature of feedback control, the simple SDOF systemshown in Fig. 6.5 is considered. The system is assumed to be subjected to both anexternal force and ground motion, and controlled with the force . Starting withthe governing equation,

(6.23)

and introducing the definitions for frequency and damping ratio leads to the

L/2

u1 u2

L/2

x

uo x( ) u1 4 xL---

4 xL---

2– u2

xL---

– 2 xL---

2++≈

F 9166------

DBχ*

L-------------

DB

L3------- 98

33------u1

13366---------u2+

–≈

F

mu cu ku+ + mag– F p+ +=


Fig. 6.5: Single-degree-of-freedom system.

standardized form of the governing equation

(6.24)

The free vibration response of the uncontrolled system has the generalform

(6.25)

Substituting for in eqn (6.24), one obtains two possible solutions

(6.26)

(6.27)

Considering and to be complex conjugates,

(6.28)

where and are real numbers representing the real and imaginary parts of, the solution takes the form

(6.29)

k

c

m

u ug+

F

ug

p

u 2ξωu ω2u+ + ag– Fm---- p

m----+ +=

u Aeλ t=

u

u A1eλ1t

A2eλ2t

+=

λ1 2, ξω– iω 1 ξ2–± ξω– iω′±= =

A1 A2

A1 2,12--- AR iAI±[ ]=

AR AIA

u e ξωt– AR ωt 1 ξ2–

cos AI ωt 1 ξ2–

sin+=


One determines and with the initial conditions for and . Theresulting expressions are

(6.30)

Considering negative linear feedback, the control force is expressed as alinear combination of velocity and displacement

(6.31)

where the subscripts ‘v’ and ‘d’ refer to the nature of the feedback, i.e. velocity ordisplacement. Feedback is implemented in the actual physical system by:

• observing the response

• determining and

• calculating with eqn (6.31)

• applying with an actuator

Mathematically, one incorporates feedback by substituting for in eqn (6.24). Theresult is

(6.32)

Equation (6.32) can be transformed to the standardized form by definingequivalent damping and frequency parameters as follows:

(6.33)

(6.34)

With this notation, the solution for the free vibration response of the linearfeedback controlled case has the same general form as for no control; one justreplaces and with and respectively in eqn (6.29). It follows that theeffect of linear feedback is to change the fundamental frequency and damping

AR AI u u

AR u 0( )=

AI1ω′----- u 0( ) ξωu 0( )+( )–=

F kvu– kdu–=

u u

F

F

F

u 2ξωkvm-----+

u ω2 kdm-----+

u+ + ag– pm----+=

ωeq2 ω2 kd

m-----+=

2ξeqωeq 2ξωkvm-----+=

ξ ω ξeq ωeq


ratio. Solving eqns (6.33) and (6.34) for and , results in

(6.35)

(6.36)

where is the increment in damping ratio due to active control

(6.37)

Critical damping corresponds to

(6.38)

Equation (6.35) shows that negative displacement feedback increases thefrequency. According to eqn (6.37), the damping ratio is increased by velocityfeedback and decreased by displacement feedback. If the objective of includingactive control is to limit the response amplitude, velocity feedback is theappropriate mechanism. Displacement feedback is destabilizing in the sense thatit reduces the effect of damping. Stability and other issues associated with linearfeedback are discussed in Chapter 8.

Example 6.3: Illustrative example - influence of velocity feedback

This example demonstrates the influence of velocity feedback on the response of 2SDOF systems subjected to seismic excitation. The properties of the systems are

System 1:

ωeq ξeq

ωeq ω 1kdk

-----+=

ξeq ξ ξ a+=

ξa

ξa1

1kdk-----+

1 2⁄-----------------------------

kv2ωm------------ ξ 1

kdk-----+

1 2⁄1–

–=

ξeq 1=

kv2mω------------

ξeq 1=

1kdk

-----+ ξ–=

m 10 000kg,= ω1 6.32rd s⁄=

k 400 000N m⁄,= T1 0.99s=

c 2 500Ns m⁄,=


System 2:

The models are excited with the El Centro and Taft accelerograms scaled to. Figures 6.6 through 6.8 contain plots of the maximum relative

displacement, maximum control force magnitude, and maximum powerrequirement. The power requirement is computed using the following expression

(1)

which assumes the control force is a set of self-equilibrating forces applied asshown in Fig 6.5. Ground motion has no effect on the work done by F with thisforce scheme.

m 10 000kg,= ω1 2rd s⁄=

k 40 000N m⁄,= T1 3.14s=

c 830Ns m⁄= ξ 0.0208=

amax 0.5g=

Power force velocity× Fu= =


Fig. 6.6: Variation of maximum displacement with active damping.

0.05 0.1 0.15 0.2 0.25 0.30.04

0.06

0.08

0.1

0.12

0.14

0.16

ξa

Max

imum

dis

plac

emen

t-

m TaftEl Centro

System 1 T1=0.99s

0.05 0.1 0.15 0.2 0.25 0.30.1

0.15

0.2

0.25

0.3

0.35

Max

imum

dis

plac

emen

t-

m

ξa

El Centro

Taft

System 2 T1=3.14s


Fig. 6.7: Variation of maximum control force level with active damping.

0.05 0.1 0.15 0.2 0.25 0.30.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

2.2

2.4x 10

4

Max

imum

forc

e-

N

ξa

El Centro

Taft

System 1 T1=0.99s

0.05 0.1 0.15 0.2 0.25 0.31000

2000

3000

4000

5000

6000

7000

Max

imum

forc

e-

N

ξa

System 2

El Centro

Taft

T1=3.14s


Fig. 6.8: Variation of maximum power requirement with active damping.

0.05 0.1 0.15 0.2 0.25 0.32000

4000

6000

8000

10000

12000

14000

16000M

axim

um p

ower

- N

.m/

s

ξa

El Centro

Taft

System 1 T1=0.99s

0.05 0.1 0.15 0.2 0.25 0.3500

1000

1500

2000

2500

3000

3500

4000

4500

Max

imum

pow

er-

N.m

/s

ξa

System 2

El Centro

Taft

T1=3.14s

6.4 Actuator Technologies 439

6.4 Actuator technologies

Introduction

The actuator component of the control system generates and applies thecontrol forces at specific locations on the structure according to instructions fromthe controller. Over the past several decades, a number of force generationdevices have been developed for a broad range of motion control applications.These devices can be described in terms of performance parameters such asresponse time, peak force, and operating requirements such as peak power andtotal energy demand. The ideal device is one that can deliver a large force in ashort period of time for a small energy input.

Civil structures generally require large control forces, on the order of ameganewton and, for seismic excitation, response times on the order of milli-seconds. The requirement on peak force coupled with the constraint on energydemand is very difficult to achieve with a fully active force actuator system. Thereare force actuator systems that are capable of delivering large force, but they alsohave a high energy demand. Included in this group are hydraulic,electromechanical, and electromagnetic devices. All these types are based on verymature technology.

There is considerable on-going research and development of new forceactuators that have a low energy demand. One approach is based on modifyingthe physical makeup of the device in such a way that only a small amount ofenergy is required to produce a significant increase in force. Varying the orificearea of a viscous type damper is an example of this strategy. A typical viscousdamper can deliver a force on the order of a meganewton. By adjusting thedamping parameter, the damper can be transformed into a large scale forceactuator.

The second approach employs adaptive materials as the force generatingmechanism. These materials respond to a low energy input by changing theirproperties and their state in a nonconventional manner which results in a force.Although these technologies are promising, the current devices can produce onlylow forces, on the order of a kilonewton, and therefore their applicability for civilstructures is limited at this time.

There are 2 issues that need to be addressed: i) how the force generationmechanism works, and ii) how the forces are applied to the structure. The first


issue is related to the physical make up and underlying physics of the device. Thesecond question is concerned with how the device is attached to the structure soas to produce the “desired” control force. In what follows, the attachment issue isdiscussed first, then the state of the art for linear actuator technologies isreviewed, and lastly some adaptive material based actuators are described.

Force application schemes

The schematic drawing contained in Fig 6.9 shows the typical makeup ofhydraulic, electro-mechanical, and electro-magnetic linear actuators. There are 2basic elements, a piston and a mechanism that translates the piston linearly eitherby applying a force to one end or by moving the end with a gear mechanism. Theinteraction of the piston with an adjacent body produces a pair of contact forces Fat the contact point and a corresponding reaction force at the actuator support. Ifthe body moves under the action of F, the mechanism usually compensates forthis motion such that the force remains constant until instructed by the controllerto change the force magnitude.

Fig. 6.9: Linear actuator

Consider the structural frame shown in Fig 6.10a. Suppose the objective isto apply a horizontal force at point A and there is no adjacent structure whichcould support the actuator. One option (Fig 6.10b) is to fasten a tendon to point A,pass it over a pulley attached to the base, and then connect it to a linear actuatorwhich can generate a tensile force in the tendon. In this scheme, the actuatorreaction force is transmitted directly to the base. A second option would be toplace the actuator directly on the structure. The actuator reaction force is now

F

u

Structure

Piston Mechanism

Actuator Support

F


transmitted to the structure; however, the other end of the piston needs to berestrained in order to generate a control force. If the restraining body is rigidlyconnected to the structure as shown in Fig 6.10c, the force system is self-equilibrating and the structure “feels” no lateral force. Member AC is in tension.In order to have a “non zero” lateral force acting on the structure, the restrainingbody must be allowed to move laterally. This objective can be achieved byattaching an auxiliary mass, ma, to the piston and supporting the mass on rollers.The mass moves with respect to the structure with an absolute acceleration equalto F/ma. One specifies the peak force and magnitude of the auxiliary mass, anddesigns the actuator so that it can provide the required force at that level ofacceleration. Since the force is generated by driving the mass, this scheme isreferred to as an active mass driver.

Fig. 6.10: Control force schemes

The extension of these schemes to a multi-story structure is shown in Fig.6.11. A linear actuator placed on a diagonal produces a set of self-equilibratingforces which impose a shearing action on the particular story to which it isattached. The other stories experience no deformation since the story shear due tothis actuator is zero. It follows that one needs to incorporate active braces

FA

F

Tendon

Pulley

A

(a) Active force (b) Active tendon

(c) Self-equilibrating forces

ma F

a

(d) Active mass driver

FA

C


throughout the structure in order to achieve global displacement control. Forcesgenerated with active mass drivers are not self-equilibrating and consequentlyhave more influence on the global displacement response.

Fig. 6.11: Control force schemes for a multi-story structure

The previous examples relate to shear beam structures which require forcesthat act in the transverse direction. For bending beam problems, control forcesystems that produce bending moments are required. This action can be obtainedwith linear actuators placed on the upper and lower surfaces, as illustrated in Fig6.12. The region between A and B is subjected to a constant moment equal to F d.Another scheme is shown in Fig 6.13. The actuator is attached to the beam withrods that provide the resistance to the piston motion, resulting in the self-equilibrating system that produces a triangular moment distribution over theregion A-B-C. By combing a number of these actuator-rod configurations, one cangenerate a piecewise linear bending moment distribution.

F3

F2

F1

T1

T2

T3

F3

F3

F2

F2

F1

F1

(b) Active mass driver(a) Active brace


Fig. 6.12: Constant moment field

Fig. 6.13: Triangular moment field

F

F

d

a)

A B

F d F db)

A B

F d c)

(-)

L

Fa)

F

F/2F/2

b)

( - )

FL/4

A BC

c)


Linear actuators generate control force systems composed of concentratedforces. For discrete structures such as frames, this type of force distribution isappropriate. However, for continuous structures such as beams and plates, acontinuous force distribution is more desirable. One strategy that has beenexamined is based on using a adaptive material in the form of a thin plate. Fig6.14 illustrates this approach for a continuous beam. Plates are attached by epoxyto the upper and lower surfaces. Applying a voltage to the plate generates alongitudinal strain. Since the plate is attached to the surface, motion of the plate isrestrained and an interfacial shear stress is generated. This stress produces adistributed control moment equal to

(6.39)

where is the width of the plate. Spatial and temporal variation of the controlforce system is achieved by varying the voltage applied to the plate.

Fig. 6.14: Moment generated by strain actuators

Large scale linear actuators

Referring back to Fig 6.9, a linear actuator can be considered to consist of apiston and a mechanism which applies a force to the piston and also controls themotion of the piston. This actuator type is the most widely available andextensively used, particularly for applications requiring a large force and shortresponse time. The descriptors hydraulic, electro-mechanical, and electro-magnetic refer to the nature of the force generation mechanisms. These devicesgenerally have a high energy demand.

τ t( )mc x t,( )

mc x t,( ) τb f d=

b f

d +mo dx beam

film

dx

τ

τ


Hydraulic systems generate the force by applying a pressure on the face ofa piston head contained within a cylinder. Fig 6.15 illustrates this concept. Fluid isforced in or out of the cylinder through the orifice at C1 to compensate for thepiston displacement and maintain a certain pressure. These systems have thehighest force capacity of the linear actuator group, on the order of meganewtons(Dorey et al., 1996). Precise control movement and force can be achieved with asuitable control system. Protection against overload is provided by a pressurerelief value. The disadvantages of this type of system are: the requirements forfluid storage systems, complex valves and pumps to regulate the flow andpressure, seals, and continuous maintenance. Durability of the seals and thepotential for fluid spills are critical issues.

Fig. 6.15: Schematic cross section view - a hydraulic cylinder (Dorey et al, 1996)

Electromechanical linear actuators generate the force by moving the pistonwith a gear mechanism that is driven by an electric motor. The motion, andtherefore the force, is controlled by adjusting the power input to the motor. Thesedevices are compact in size, environmentally safe, and economical. Figure 6.16illustrates the various components of a linear electric actuator system (Raco,www.raco.de). The largest electric actuator that can be ordered “off the shelf” israted for 600kN force.


Fig. 6.16: Components of linear electric actuator systems(Raco, http//:www.raco.de)

Hydraulic and electromechanical actuators are composed of many partsthat are in contact with each other, and therefore have a high risk of breakdown.Since electromagnetic actuators are driven by magnetic forces, which do notrequire mechanical contact, they are theoretically more reliable. Small scaleelectromagnetic actuators with a force capacity ranging from tens of newtons toseveral kilonewtons are commercially available. In addition to their compactnature and low voltage and amperage requirements, their response time is low,on the order of milliseconds. These features are ideal for active force generation,and electromagnetic actuators are a popular choice for small scale structures.Large scale electromagnetic actuator technology is still in the research anddevelopment phase, and there is no commercial product with a force capacity inthe meganewton range available at this time. However, since the force generationmechanism is so simple, it is reasonable to expect that economical designs willeventually emerge, and be viable candidates for controlling large scale structures.

Figure 6.17 shows a schematic cross-sectional view of a design concept fora large scale electromagnetic actuator developed by Chaniotakis et al (99) at thePlasma Science and Fusion Center, Massachusetts Institute of Technology.

Positioning Control Amplifier unitfor motor

Servomotor

Electric Actuator


Fig. 6.17: Schematic view - MIT Electromagnetic actuator design

The unit consists of a cylindrical shell housing and a piston. Two sets of axi-symmetric electromagnets are used. The field coil is embedded in the cylinder,and generates a stationary magnetic field. A driving coil is attached to the pistonwhich translates with respect to the housing. The force mechanism is based on theinteraction between the magnetic field generated by the stationary field magnetand the current in the driving coil. For this particular design, the electromagneticforce is a linear function of the coil currents and is independent of the position ofthe piston.

(6.40)

where are the currents in the field and driving coils, and c is a designparameter. The equation of motion for the piston relates the electromagnetic force,the inertia force for the piston mass, and the external contact force, .

(6.41)

When the device is used as an active mass driver, an auxiliary mass, , is

Fem

Fc

R=Fem

pistonfield coil

driving coil

Fem cI f Id=

I f Id,

Fc

mpu Fem Fc–=

ma


attached to the end of the piston, and is set equal to . The reaction at thebase of the housing is equal to the electromagnetic force which acts on the fieldcoil and is transmitted to the housing.

There are several advantages to this concept: i) the response time is on theorder of milliseconds, ii) there is minimal mechanical contact, and iii) thetechnology for controlling the current is mature. The disadvantages are: i) thecurrent and voltage requirements for a force on the order of meganewton cannotbe satisfied with conventional electrical power supply technology, and ii) there isminimal experience related to the design, fabrication, and operating performanceof large scale electromagnetic actuators.

Large scale adaptive configuration based actuators

This category includes mechanical devices such as dampers, frictionelements, and stiffness elements which have the ability to generate a large forceby changing their physical makeup. Their distinguishing features are their lowratio of energy demand to force output, and their large scale force capacity. Thesefeatures are very desirable for applications to large scale civil structures. Adaptivedevices which have a low energy demand and also function as energy dissipationmechanisms are referred to as semi-active actuators since they behave like passivedevices in the sense that they increase the stability of a structure and require noexternal energy. The key point is stability. A semi-active actuator will neverdestabilize a system whereas an active actuator may destabilize a system eventhough it has a low energy demand.

The use of a variable orifice damper as a force actuator was suggested byFeng et al. (1990), and developed further by Shinozuka et al (1992). Kurata et al.(1994) implemented variable dampers in a large scale 3 story frame structure andSack and Patten (1994) installed a hydraulic actuator with a controllable orifice ona bridge on interstate highway I-35 in Oklahoma. Figure 6.18 illustrates theconcept of a variable orifice damper. The flow control valve is operated by acontroller which in turn responds to the demand for a particular magnitude offorce specified by the feedback control algorithm; the controller adjusts the valveopening according to the force demand. Since the valve motion is perpendicularto the flow, the force required to adjust the valve position is small, and thereforethe energy demand is low. Kurata’s experiment required only 30 Watts to operatea valve.

Letting v denote the velocity of the piston, the damping force, F, can be

Fc mau


expressed as

F= c v (6.42)

where c is a function of v and the geometry of the valve. In the case of a passivedamper, c is a prescribed function of v, and F is determined uniquely byspecifying v. In the case of a variable damper, c is a function of the valve positionas well as v. When used as an actuator, F is specified, v is observed, c is calculatedwith eqn (6.42), and the valve position is determined using the values of c and v.The limitation of this device is the dependency of the sense of F on the sense of v;F is always opposite in sense to v. Therefore, if the force required by the controlalgorithm at a particular time has the same sense as the observed velocity, theforce demand cannot be met, and the actuator needs to be inactivated until a latertime when the phase is reversed.

Fig. 6.18: Variable damping mechanism(Kajima Corporation, http://www.kajima.com)

Small scale adaptive material based actuators

Low force capacity electromechanical and electromagnetic linear actuatorsare standard “off-the-shelf” products offered by a number of suppliers. Ourinterest here is not with these devices but rather with a new generation of small


scale force actuators that utilize the unique properties of adaptive materials togenerate the force. Research and development in this area was initiated by theaerospace industry as a potential solution for shape control of satellite arms andairplane control surfaces. As the technology evolved, other applications related tomotion control of small scale structures such as robot arms and biomedicaldevices have occurred. Although the technology continues to evolve, andreliability is still a major concern, these devices are being seriously considered ascandidates for force control where the required force level is on the order of akilonewton. Brief descriptions of the various adaptive material based actuatorsare presented below.

• Piezoelectric actuators

Piezoelectric materials belong to the electrostrictive material category.When subjected to a voltage, they undergo a molecular transformation whichresults in the material extending (or contracting) in a manner similar to thePoisson effect for applied stress (Ikeda, 1997). Figure 6.19 illustrates thisbehavioral mode; a voltage in the Z direction produces extensional strainsand in the X and Y directions. The opposite behavioral mode occurs when thematerial is stressed in the X-Y plane; a voltage in the Z direction is generatedby and . This behavior was first reported by Pierre and Jacques Curie in1880. Pierre later won a Nobel price in Physics with his wife Marie for their workon radioactivity. Historically, piezoelectric materials have been used as strainsensors. Their use as actuators is more recent and stimulated primarily by theAerospace Industry. (Crawley, 1987)

Vz εxεy

Vz'σx σy


Fig. 6.19: Piezoelectric electrical-mechanical interaction

Piezoelectric actuators are fabricated with piezoceramic block typeelements or piezopolymer films. Lead zirconate titanate (PZT) is the dominantpiezoceramic composite used for sensors and actuators in the frequency range upto Hz. Polyvinyldene fluoride (PVDF) is the most common piezoelectric film.Since it has a relatively low strength, PVDF is used mainly as a sensor,particularly for the high frequency range up to Hz. The underlying principleis the same for both materials. The piezoelectric object is attached to a surfacewhich restrains its motion. When a voltage is applied, the object tends to expandimmediately, and contact forces between the object and the restraining mediumare produced. Two actuator configurations have been developed. The first modelis a conventional linear actuator, such as shown in Fig 6.20. Piezoceramic wafersare stacked vertically, bonded, enclosed in a protective housing, and fitted withelectrical connectors. These devices can deliver large forces, on the order of 20 kN,with a response time of several milliseconds (Kinetic Ceramics Inc.,www.kineticceramics.com).

Z

X

Y

εx

V'ZVZ

σyεy

σx

106

109


Fig. 6.20: Cylindrical piezoceramic linear actuator (Kinetic Ceramics Inc)

The second configuration has the form of a thin plate as illustrated in Fig6.21. Piezoceramic wafers are distributed over the area in a regular pattern. Theymay also be stacked through the thickness. This type of device is bonded to asurface, and applies a pair of self equilibrating tangential forces to the surface.The peak force depends on the applied voltage and degree of restraint. A forcelevel of 500 N at 200 volts, and millisecond response, is the typical upper limit foroff-the-shelf plate type piezoceramic actuators (Active Control Experts,www.acx.com).

Fig. 6.21: Plate type piezoelectric actuator

• Shape memory alloys

Shape memory alloys are metal alloys which, if deformed inelastically at

bond wafer bond

impressed voltage


room temperature, return to their original shape when heated above a certaintemperature. Figure 6.22 illustrates this behavior. The initial straight form isdeformed inelastically at room temperature to the curved form. When thetemperature is elevated, the curved form shifts back to the straight form, andremains in that form when the temperature is lowered to room temperature. If nofurther deformation is introduced, the straight form remains invariant during anysubsequent thermal cycling.

Fig. 6.22: One Way Shape Memory Behavior

The ability to return to its initial shape when heated is due to a phasetransformation from martensite at room temperature to austenite at elevatedtemperature (Wayman et al., 1972). Inelastic deformation introduced during themartensite phase is eliminated when the state passes over to the austenite phase.The phase transitions are illustrated in Fig 6.23: As and Af define the temperaturesfor the start and finish of the transition from martensite to austenite for the casewhen the material is being heated; the corresponding temperatures for thecooling case are Ms and Mf. When T is greater than Af, the phase is austenite but itis possible to convert it back to martensite by applying stress. The quantity Md isthe temperature beyond which austenite cannot be converted to martensite by

a) Initial shape at room temperature

b) Deformed shape at room temperature

c) Shape at elevated temperature

d) Shape after cooling to room temperature


stress, i.e., the phase remains austenite for arbitrary applied stress.

Fig. 6.23: Martensitic transformation on cooling and heating

The stress-strain behavior is strongly dependent on temperature. Figure6.24 shows the limiting stress-strain curves for Nitinol, a nickel-titanium alloy(Jackson et al, 1972). For T<Mf, the lowest temperature for the fully martensiticphase, the material behaves like a typical ductile metal. Yielding occurs at about150 MPa, and inelastic deformation is introduced. The behavior for T>Md iselastic up to 650 MPa, and the modulus, E1, is about 4 times larger than the initialvalue, E0. Between these limiting temperatures, there is a graduation in behaviorbetween fully ductile to fully elastic.

Temperature, T

Vol

ume

frac

tion

mar

tens

ite 1

0.5

0

Mf As

Ms Af Md


Fig. 6.24: Effect of temperature on stress strain behavior of Nitinol (Jackson et al.,1972)

Referring to Fig 6.24 the one way effect can be explained by tracking theresponse as a stress, , is applied and then removed at T<Mf. This actionproduces the path a-b-c. Increasing the temperature at this time to T>Md shifts c

T > Md

T < Mf

ε (%)

σ (MPa)

0 2 4 6 8 10

100

200

300

400

500

600

700

E1

E0

a

σ∗ b’ b

c

c’

c’’

E0

σ*


back to a since the behavior for T>Md is elastic. Further cycling of the temperaturewith no stress applied will not cause the point to shift from a.

A different type of response is obtained when the stress is held constantand the temperature is cycled a number of times between Mf and Md. In this case,the path is b-b’-b. The effect of thermal cycling is to introduce 2 way shapememory. Repeating the scenario described above, starting at b and removing thestress shifts the b position to c. Then, increasing the temperature causes theposition to shift to a. However, when the temperature is now lowered to Mf, theposition shifts back to point c instead of remaining at a. Further thermal cycling atno stress results in the deformation switching between a and c. At hightemperature, the material remembers the initial state; at low temperature itremembers the deformed state. Figure 6.25 is a modified version of Fig 6.22 whichshows this behavioral mode. Applying the stress and subsequent thermal cyclingtrains the material to remember 2 shapes (Fremond et al, 1996).

Fig. 6.25: Two-way shape memory behavior

The two-way shape memory behavior provides the basis for forceactuation. If a trained shape memory alloy is restrained at low temperature so thatit cannot deform, a force is generated when the alloy is heated since it wants toreturn to its initial undeformed shape. Referring back to the stress-strain plots inFig 6.24, suppose the position at low temperature is point c. When heat is applied,the behavior is governed by the curve for T>Md. The material reacts as if it weresubjected to the positive strain a-c, and the position jumps to c’. This behavior is“ideal”; the actual position is lower, such as point c’’, but the induced stress is stilla significant value, on the order of 300 to 400 MPa.

b) Initial shape at elevated temperature

a) Deformed shape at low temperature


Nitinol alloys in the form of small diameter wires ( ) are used toassemble a force actuator. Heating is applied by passing an electric currentthrough the wire. This process limits the response time to seconds vs.milliseconds for piezoelectric materials. Another limitation is the material cost; atypical price is on the order of $300/kg. Most of the applications of shape memoryactuators are for products requiring low capacity, and where cost and responsetime are not critical issues.

• Controllable fluids

Controllable fluids are characterized by their ability to change from a fluidto a semi-solid in milliseconds when subjected to an electric or a magnetic field.This effect was identified by Winslow in 1949. Two materials belonging to thiscategory are electrorheological (ER) and magnetorheological (MR) fluids. In theirinitial state, they behave as viscous fluids. Application of the field (electric ormagnetic) introduces an additional plastic solid type behavior mode, and theresponse is now a combination of plastic and viscous actions. Figure 6.26illustrates this transformation for the case where the material is idealized as aBingham solid, which is defined as an ideal plastic solid in parallel with a linearviscous fluid. The stress-strain relation for a Bingham solid subjected to sheardeformation has the form:

(6.43)

where denotes the yield stress and is the viscosity. Experimental resultsshow that the yield stress increases significantly with the field strength, f, up to alimiting value . However, the viscosity is essentially constant (Carlson, J.D.et al, 1995). Therefore, it is reasonable to take equal to a constant, , andinterpret the material behavior as a combination of variable coulomb hystereticdamping and constant linear viscous damping. The equivalent linear materialviscosity, , is dependent on the amplitude and frequency of the sheardeformation. For the case of periodic excitation, is given by

(6.44)

where is the shear strain amplitude, and is the excitation frequency.

0.4mm≈

τ τ y γ( ) ηγ+sgn=

τy η

τmaxη η o

ηeqηeq

ηeq ηo

4τyπΩγ-----------+=

γ Ω


Fig. 6.26: Effect of electric and magnetic fields on the stress-strain relationship

Devices containing controllable fluids can be used as either variabledampers or as semi-active force actuators. Figure 6.27 shows a schematic view ofa prototype developed by the Lord Corporation (Lord Corp.,www.lordcorp.com). The main cylinder houses the piston, the MR fluid, and themagnetic circuit. A small electromagnet is embedded in the piston head andsupplied with current that generates the magnetic field across the annular orifice.Typical small scale versions have a force capacity of about 3 kN, millisecondresponse, and draw about 10 watts of power. A large scale version with capacityof 200 kN and millisecond response has also been produced (Spencer, B.F. et al,1998).

τ

τy,max

τy f=0

f1

f2 > f1

η

f= field strength measure

Saturation limit

γ


Fig. 6.27: Rheonetics SD-1000-2 MR damper (Pang, L. et al, 1998)

Experimental results for quasi-static loading applied to the small scaleversion are plotted in Figure 6.28. These results show that the Bingham solididealization is a reasonable approximation for the actual behavior, and also thatthe plastic yield force is limited by saturation of the fluid. For this device,saturation occurs in the vicinity of 1.5 amps. When used as a force actuator, theamperage is adjusted such that, for the observed value of velocity, the desired


force magnitude is generated. A controllable fluid type device is more effectivethan a variable orifice damper since the yield force is the primary component, andthis force is independent of velocity. The low external power requirement, rapidresponse, and the potential for large capacity are attractive features. With furtherdevelopment, MR based actuators should be competitive solutions for large scalecivil structures.

Fig. 6.28: Quasi-steady force velocity response for Rheonetics SD-1000-2 MRdamper (Pang, L. et al., 1998)

6.5 Examples of existing large scale active structural control systems

Kajima Corporation has pioneered the research, design, experimentation, andimplementation of active control of large scale building structures (Sakamoto et al.1994). Their work has been concerned with the following active motion controlschemes: active mass driver (AMD), active variable stiffness (AVS), and hybridmass damper (HMD). These schemes have been implemented in the set ofbuildings listed in Table 6.1. Other Japanese organizations such as Shimizu Corp.,Takenaka Corp., and Mitsubishi have also carried out substantial research and

6.5 Examples of Existing Large Scale Active Structural Control Systems 461

implementation in the field of active structural control. In the USA, variousgovernment organizations such as NSF, ARO, AFOSR, ONR are funding researchon active control of structures. A description of some of the implementations isgiven below. A detailed assessment of the performance of these structures duringthe earthquakes and strong winds that they have already encountered is found inSakamoto et al. (1994).

Table 6.1: Implementations of active and hybrid control systems in buildings

(by Kajima Corporation)

Note: BG: Below ground level.ME: Moderate earthquake.LE: Large earthquake.SW: Strong wind.

AMD in Kyobashi Seiwa Building (Kajima 91-63E, 1991a)

The Kyobashi Seiwa Building (see Fig 6.29) is a very slender building with awidth of only 4m, a length of 12m, and a height of 33m (11 floors). It is constructedof rigidly connected steel frames consisting of box columns and H-shaped beams.The total structural weight is about 400 metric tons (1 metric ton equals 104

newtons). An active mass driver (AMD) system is installed on the top floor. Theobjective of the AMD is to reduce the maximum lateral response associated withfrequent earthquakes (i.e. peak ground acceleration of ) and strongwinds (i.e. maximum speed of ) to about one third of the uncontrolledvalue.

Name of Building KyobashiSeiwa

KaTRINo.21

AndoNishikicho

ShinjukuPark

DowaKasai

Phoenix

Completion date Aug. 1989 Nov. 1990 July 1993 April 1994 Jan. 1995

Number of floors 11+1(BG) 3+2(BG) 14+2(BG) 52+5(BG) 29+3(BG)

Building height (m) 33.10 16.30 68.00 232.60 144.45

Total floor area (m2) 423.37 465.00 4,928.30 264,140.91 30,369.66

Typical floor area (m2) 37.32 150.00 324.15 4,523.54 1,072.36

Control system AMD AVS HMD HMD HMD

Type of disturbance ME,SW LE ME,SW ME,SW ME,SW

10cm s2⁄20m s⁄


a. AMD System installation

b.Schematic of AMD

Fig. 6.29: AMD for Kyobashi Seiwa Building


The AMD system is illustrated in Fig 6.29(b). The basic components are:

• Sensors installed at several locations including: ground level, mid-height, and the roof level to detect seismic motions and tremors at theground level and in the building.

• The control computer which analyzes each signal and issues a driveorder. The control algorithm of the system is of the closed-loop controltype where the active control force is determined through linearvelocity feedback of the structure’s response.

• Actuators which execute the control order and drive the masses. Thehydraulic pressure source for the actuator consists of two pumps andan accumulator. One pump is small in comparison to the other; itsfunction is to provide a minimum level of pressure continuously. Thelarger pump is activated when the earthquake occurs.

• Two added masses driven by the actuators. Lateral vibration in thewidth direction is controlled with one mass (weighing about 4 metrictons, about 1% of the building weight) located at the center of thebuilding. Torsional vibration is reduced with a second mass, weighingabout 1 metric ton, located at one end of the structure. These masses aresuspended by steel cables to reduce the frictional effects. The lag in theresponse is about .

AVS Control at Kajima Technical Research Institute (Kajima 91-65E, 1991b)

An active variable stiffness (AVS) device is a non-resonant control type systemwhich is designed to reduce the energy input to buildings from external excitationby continuously altering the building’s stiffness, based on the nature of theexcitation, to avoid a resonant state. The major advantage of an AVS is that it canbe activated and driven with a small amount of power, partially solving theenergy requirement problem of active control.

An AVS system was implemented in a steel structure, three stories high,weighing about 400 metric tons, located at the Kajima Technical Research Institute(see Fig 6.30). The lateral bracing system consists of steel braces placed in thetransverse direction (gable side), and variable stiffness devices (VSD) installedbetween the brace tops and the lateral beams. These devices alter the building’sstiffness by shifting from the locked mode (engage: brace is effective) to the

0.01s


unlocked mode (release: brace is ineffective). The variable stiffness range of thebuilding was designed such that the natural resonant frequency is about 2.5Hz forthe locked condition and about 1Hz for the unlocked condition. Auxiliaryreinforcing braces are installed in the longitudinal direction to increase thebuilding’s stiffness so that the control for the transverse direction is executed withminimum torsion.

a. View of building


b.Schematic of locking device

c. Close up view of locking device

Fig. 6.30: Kajima Technical Research Institute AVS System

The AVS system consists of:

• A measurement and control device consisting of an accelerometerplaced on the first floor of the building which feeds the earthquakeinput into a motion analyzer. The analyzer consists of several specialband-pass filters which approximate the response transfercharacteristics to each stiffness type. Three stiffness types (eightstiffness types are possible) with mutually different resonantfrequencies were selected for the building: all braces locked, all braces


unlocked, and only the braces in the bottom floor locked. Based on thefiltered output, the control computer selects the instantaneous stiffnessthat yields the minimum building response. The control intervalrequired to judge stiffness selection is 0.004s. To maximize thereliability and redundancy of the system, the control computer systemis made up of a host computer which determines the operation of theentire system, and three controllers. As the host computer requiresextremely fast data acquisition and processing, the real-time UNIXoperating system was used. Personal computers utilizing MS-DOSwere selected for the controllers which are placed between the hostcomputer and each VSD.

• The VSD switches over the connection condition between the brace andthe beam in accordance with the signals from the control computer. AVSD consists of a two-ended-type enclosed hydraulic cylinder with aregulator valve inserted in the tube connecting the two cylinderchambers. The open/close function is controlled by oil movements,thus locking or unlocking the connection between the beam and thebraces. Twenty watts of electric power per device is required for thevalve function. The time needed to alter the lock/unlock condition isabout 0.03s.

• The emergency power source counteracts power blackout and enablesthe entire control system to continue to operate for at least minuteseven after sudden termination of the regular power supply. Moreover,in case of sudden power termination, the mechanism will cause thedevices to automatically adopt a locked condition, thus increasing thebuilding’s strength capacity. Under normal conditions, all the bracesare kept in the locked condition.

DUOX Active-Passive TMD in Ando Nishikicho Building (Kajima 93-82E, 1993)

A combination of an AMD and a TMD, the DUOX system is installed in the AndoNishikicho Building show in Fig 6.31. This structure consists of four main steelcolumns located at the corners, 14 stories above ground, and two basements. Thebuilding is located in an area of Tokyo that has mainly small, low-rise buildings,and consequently is susceptible to strong winds. The DUOX system was installednear the center of the top floor (i.e. at the building’s center of gravity) to controlvibration in both horizontal directions.

30


a. View of building


b. Schematic of control system

c. Conceptual diagram of DUOX


d. Bi-directional control

Fig. 6.31: Nishikicho building motion control system

The DUOX system consists of:

• Sensors installed in the basement and on the top floor of the building tomonitor the ground acceleration and building motion. Sensors are alsoinstalled on the AMD and TMD to measure the motion of these devices.

• A digital control computer which receives the feedback signals from thesensors, analyzes them, and determines the optimal control forces thatwill achieve the required control effect and also maintain the stroke ofthe AMD within the allowable range.

• A passive TMD weighing 18 metric tons, about 0.8% of the weight ofthe above ground portion of the building (about 2,600 metric tons). TheTMD is supported by laminated rubber bearings which provide therequired stiffness. Oil dampers provide additional viscous damping inthe system.


• Two AMD units, driven by AC servo motors and ball screws andweighing about 10% of the weight of the TMD (about 0.08% of thebuilding’s weight). The units are mounted one on top of the other in acriss-cross manner to provide control in the two horizontal directions.The mass of the TMD moves out of phase with the building so that thebuilding motion is always being retarded. The active mass is driven inthe direction opposite to that of the TMD so as to magnify the motion ofthe TMD. When the building response falls within the allowable zone,the AMD operates to dampen out the motion of the TMD.

ABS - 600 Ton Full-Scale Test Structure (Soong et al. 1991, Reinhorn et al. 1993)

This 600 metric ton test structure, located in Tokyo, Japan, consists of a symmetrictwo-bay six-story building, 10m x 10m x 21.7m, constructed of rigidly connectedsteel frames composed of box columns and W-shaped beams with reinforcedconcrete slabs at each of the floors. The structure was intentionally designed to bea relatively flexible structure in order to simulate a typical high-rise building. Thefundamental periods are 1s and 1.5s. The spectral properties of the structureobtained both analytically (with a shear beam model) and experimentally, werewithin 3% for all modes. A peak ground acceleration of 0.1g was taken as thedesign loading. The control forces were applied with an Active Bracing System(ABS).

The ABS system consists of:

• Velocity sensors, with an output range of , installed at thebase, the first, the third, and the sixth floor, and an accelerometer, withan output range of , placed at the base of the structure.This data is fed into the controller and used as the state variables forfeedback.

• A PC-Limited 386 microcomputer facilitated with Analog-to-Digital(A/D) and Digital-to-Analog (D/A) boards is used for on-linecomputation. The control logic is implemented on an Intel 80386 25-MHz processor equipped with an Intel 80387 25-MHz coprocessor.

• Two algorithms based on classical linear optimal control theory wereused. One control model works with the complete state vector andrequires about 0.014s to process the data at each time point. The other

100cm s⁄±

1000cm s2⁄±


control law is specialized for a reduced state vector corresponding to a3DOF approximation which considers the translations of the first, third,and sixth floors as variables. The computation time for this algorithm isestimated at around 0.005s.

• A biaxial active bracing system (ABS) consisting of a pair of hydraulicactuators (length 73.5cm, piston diameter 152.4mm, rod diameter63.5mm) aligned in each of the principal directions and located at thefirst floor level. The actuators, monitored by two hydraulic servovalvecontrollers with auxiliary hydraulic accumulators, are capable ofgenerating 685kN of control force in each direction with a time delaybetween the command signal and the actual actuator response of about0.012s. The primary design parameters for the ABS are the requiredcontrol force, the maximum displacement of the actuator ( ), andthe actuator velocity (6.6cm/s). These parameters influence theactuator capacity, cylinder stroke, and flow rate requirement of thehydraulic servovalve. The servo-controlled system was designed topump the hydraulic oil at a constant speed during control action, andthe difference between the required and supplied oil flow is adjustedthrough the hydraulic accumulators. Each actuator is equipped with alinear variable displacement transducer (LVDT) having a range of

, which is used to adjust the length of the brace via theservovalve loop.

• Fail-safe features are incorporated in the hardware and software toprotect the system in the event of a failure of one or more componentsof the system.

5cm±

12mm±

472 Introduction to Structural Motion Control

Problems 473

Problems

Problem 6.1

Consider the cantilever beam shown above. Assume transverse sheardeformation is negligible. The desired state is constant curvature and

.

a) Determine the bending rigidity distribution based on the equilibriummodel,

(1)

b) Determine a “constant” value of using the least square proceduredescribed below. The displacement function for the triangular loading is given by:

(2)

where . One forms the “displacement” error

(3)

and the error functional,

(4)

x

L 10 m=

b0 20 kN/m=

u L( ) L 400⁄≤

DBM x( )

χ∗-------------=

DB

u0

b0L4

6DB------------ x2 1 1

2---x–

120------x3+

DB1– g x( )= =

x x L⁄=

e x( ) u0 x( ) u∗ x( )–=

J e2 xd0

L

∫=


Let . The value of is found by minimizing with respect to .

(5)

Establish the expression for .

c) Assume is specified. The displacement is now fixed. Suppose avertical force, , is now applied at , and the magnitude of can beadjusted. Determine a value for by minimizing with respect to . Theappropriate form of the error is , where

(6)

Evaluate for a representative range of .

d) Let represent the displacement error at point .

(7)

Suppose one uses a “discrete” error functional consisting of terms.

(8)

Derive the general expression for corresponding to requiring to bestationary. Evaluate taking constant = 1/2 the value obtained in part b,

, and = 1/3, 2/3, and 1. Note: it is convenient to introduce matrixnotation and express the error as:

(9)

e) Refer to part d. Consider 2 control forces, at and at

DB1– f B= f B J f B

J∂f B∂

--------- 0=

f B

DB u0F x L= F

F J Fe u0 uc u∗–+=

ucFL3

2DB----------- x2 x3

3-----–

FDB-------h x( )= =

F DB

ei xi

ei u0 xi( ) uc xi( ) u∗ xi( )–+=

N

J 12--- ei

2

i 1=

N

∑=

F JF DB

N 3= xi

e

e1

e2

e3

e f F+= =

J 12---eTe=

F1 x L 2⁄= F2

Problems 475

. Evaluate , for the conditions specified in part d. Compare thedisplacement solutions for part d and part e.

f) Suppose one introduces weights for the error at the discrete points, ,and also introduces a cost associated with the control force applied at x = L. Theexpanded functional has the form,

(10)

where is the weight for and is the cost weight for . Increasing placesmore weight on vs. . Similarly, increasing places more weight on . Derivethe expression for corresponding to requiring J to be stationary. Rework part dusing this formulation, taking and = . Investigate thevariation of with .

Problem 6.2

Consider a cantilever beam subjected to a uniform loading. Takeas constant and neglect transverse shear deformation. The deflected shape isgiven by:

x L= F1 F2

xiF

J 12--- qiei

2

i 1=

N

∑ 12---rF2+=

qi ei r F rF e qi ei

Fq1 q2 q3 1= = = r r∗ DB

2⁄F r∗

x

L

ub0

DB EI=

ub0L4

24EI------------ x2 6 4x– x2+( )[ ]=


where . Suppose the ‘‘desired’’ deflected shape is where is aspecified value.

a)

Consider a concentrated moment, , applied at to ‘‘correct’’ thedifference between the actual and desired deflected shape. How would youdetermine ?

b)

Suppose a uniform distributed moment, , is used as the control forcesystem. How would you determine ?

x x L⁄= u aLx2= a

x

L

M

M x L=

M

x

L

m0

m0m0

Problems 477

Problem 6.3

The structure show above consists of two single spans interconnected atthe interior support with a force actuator system that can instantaneously apply amoment, M(t). The structure also has the capability to observe the transversedisplacement at a limited number of points along both spans. Consider a force, P,which moves with velocity v along the span. Assume the velocity is sufficientlysmall such that the actual response of the structure can be approximated by thecorresponding static response. Describe how you would establish the value of theactuator moment, M , such that the maximum transverse displacement at anytime is less than a specified value, u*.

L L

MP

v


479

Chapter 7

Quasi-static Control Algorithms

7.1 Introduction to control algorithms

Referring back to Fig 6.1, an active structural control system has 3 maincomponents: i) a data acquisition system that collects observations on theexcitation and response, ii) a controller that identifies the state of the structureand decides on a course of action and iii) a set of actuators that apply the actionsspecified by the controller. The decision process utilizes both information abouthow the structure responds to different inputs and optimization techniques toarrive at an “optimal” course of action. When this decision process is based on aspecific procedure involving a set of prespecified operations, the process is said tobe algorithmic, and the procedure is called a “control algorithm”. A non-adaptivecontrol algorithm is time invariant, i.e., the procedure is not changed over thetime period during which the structure is being controlled. Adaptive controlalgorithms have the ability to modify their decision making process over the timeperiod, and can deal more effectively with unanticipated loadings. They also canupgrade their capabilities by incorporating a learning mechanism. This text isconcerned primarily with time invariant control algorithms which are wellestablished in the control literature. Adaptive control is an on-going research areawhich holds considerable promise but is not well defined at this time. A briefdiscussion is included here to provide an introduction to the topic.

The topic addressed in this chapter is quasi-static control, i.e., where the

480 Chapter 7: Quasi-static Control Algorithms

structural response to applied loading can be approximated as static response.Since time dependent effects are neglected, stiffness is the only quantity availablefor passive control. Active control combines stiffness with a set of pseudo-staticcontrol forces. The quasi-static case is useful for introducing fundamentalconcepts such as observability, controllability, and optimal control. Bothcontinuous and discrete physical systems are treated.

The next chapter considers timeinvariant dynamic feedback control ofmulti-degree-of freedom structural systems. A combination of stiffness, damping,and time dependent forces is used for motion control of dynamic systems. Thestate-space formulations of the governing equations for SDOF and MDOFsystems are used to discuss stability, controllability, and observability aspects ofdynamically controlled systems. Continuous and discrete forms of the linearquadratic regulator (LQR) control algorithm are derived, and examplesillustrating their application to a set of shear beam type buildings are presented.The effect of time delay in the stability of LQR control, and several other linearcontrol algorithms are also discussed.

7.2 Active prestressing of a simply supported beam

Passive prestressing

The concept of introducing an initial stress in a structure to offset the stressproduced by the design loading is known as prestressing. This strategy has beenused for over 60 years to improve the performance of concrete structures,particularly beams. The approach is actually a form of quasi-static control, wherethe variables being controlled are the stresses. Figure 7.1 illustrates prestressing ofa single span beam with a single cable. When the cable shape is parabolic, thetension introduced in the cable creates an “upward” uniform loading, wo, that isrelated to the tension by

(7.1)

The initial moment distribution is parabolic, and the moment is negativeaccording to the conventional notation.

woL2

8d------------- T=

7.2 Active Prestressing of a Simple Supported Beam 481

Fig. 7.1: Passive prestressing scheme

Suppose the design loading is a concentrated force that can act at any pointon the span. The maximum positive moment due to the force occurs when theforce acts at mid-span, and the resultant positive moment at mid-span is given by

(7.2)

The initial mid-span moment is negative and equal to

(7.3)

If the prestress level is selected such that

(7.4)

which requires

(7.5)

L

d

wo

(-)M+

woL2/8

(+)PL/4

M+

P

cable

d = distancebetween topand bottomlocation ofthe cable

M L2---

PL4

-------woL2

8-------------–=

M L2---

woL2

8-------------–=

woL2

8-------------

12--- PL

4-------

M*= =

T M*

d-------- T*≡=


then the maximum positive and negative moments are equal. The cross sectioncan now be proportioned for , which is 1/2 the design moment correspondingto the case of no prestress. This reduction is the optimal value; taking willincrease the initial moment beyond and result in the cross-section beingcontrolled by the initial prestress. The limitation of this approach is the need toapply the total prestress loading prior to the application of the actual loading.Since the tension is not adjusted while the loading is being applied, the schemecan be viewed as a form of passive control. The best result that can be obtainedwith prestressing for this example is equal design moment values for theunloaded and loaded states.

Active prestressing

Suppose the cable tension can be adjusted at any time. The equivalentuniform upward loading due to the cable action can now be considered to be anactive loading. Deforming as the equivalent active loading and noting eqn(7.1), the loading is related to the “active” tension force, T(t ), by

(7.6)

The time history of T(t ) can be established using simple static equilibriumrelations. Given the spatial distribution and time history of the loading, T(t ) isdetermined such that the maximum moment at any time is less than the designmoment for the cross-section.

When the applied loading is uniformly distributed, the momentdistributions are similar in form. The expression for the net positive moment hasthe form

(7.7)

Enforcing the constraint on the maximum moment, which occurs at mid-span,

(7.8)

results in the following control algorithm,

(7.9)

M*

T T*>M*

wa

wa8dL2------

T t( )=

M*

M x t,( ) w t( ) wa t( )–( ) 12--- Lx x2–( )=

w t( ) wa t( )–( )L2

8------ M*≤

1. wa t( ) 0 for w t( ) 8L2------M* ws≡≤=

2. wa t( ) w t( ) ws for w t( ) ws>–=


No action needs to be taken until reaches , since the maximum moment isless than . Above this load level, the active loading counteracts the differencebetween and . With active prestressing, the constraint imposed on theinitial prestressing is eliminated. Theoretically, the total applied load can becarried by the active system for this example. This result is due to the fact that themoment distributions for the actual and active loadings have the same form.When these distributions are different, the effectiveness of active prestressingdepends on the difference between the distributions. The following discussionaddresses this point.

Consider the case where the loading is a concentrated force that can act atany point on the span, and the prestressing action is provided by a single cable.The moment diagrams for the individual loadings are shown in Fig 7.2. Whenthese distributions are combined, there is a local positive maximum at point B, thepoint of application of the load, and possibly also at another point, say C.Whether the second local negative maximum occurs depends on the level ofprestressing. As is increased, the positive moment at B decreases, and thenegative moment at C increases. For a given position of the loading, the controlproblem involves establishing whether can be selected such that themagnitudes of both local moment maxima are less than the prescribed targetdesign value, , indicated in Fig 7.2. With passive prestressing, the optimalprestressing scheme produced a 50% reduction in the required design moment,i.e., it resulted in =0.5(PL/4). Whether an additional reduction can beachieved with active prestressing remains to be determined.

w wsM*

w t( ) ws

wa

wa

M*

M*


Fig. 7.2: Active prestressing scheme for a concentrated load

The net moment is given by

Region A-B

(7.10)

(+)

P

PabL-----

a L a–

waL2

8------–

(-) M+

M+

parabola

low prestress

high prestress

M+Mnet

M*+

M*–

A B C D

xc

x

M x t,( ) Px L a–( )L

------------------------ waLx x2–

2------------------

–=


Region B-C-D

(7.11)

Specializing eqn (7.10) for leads to

(7.12)

The location of the second maxima is established by differentiating eqn (7.11) withrespect to x and setting the resulting expression equal to 0. This operation yields

(7.13)

The value for M at has the following form:

(7.14)

When , the maximum negative moment occurs outside the span, andis taken as 0.

The control algorithm is established by requiring

(7.15)

for all . Starting at , no action is required as increases until the momentdue to the force P is equal to . The limiting value is denoted as , anddetermined with

(7.16)

When is greater than , the maximum positive moment, , is set equal to,

(7.17)

Solving eqn (7.17) for leads to

(7.18)

The last step involves checking whether for this value of , the maximumnegative moment, , exceeds .

M x t,( ) Pa L x–( )L

------------------------ waLx x2–

2------------------

–=

x a=

M+ M a t,( )≡ a L a–( ) PL---

12---wa–

=

xcL2--- Pa

Lwa----------+=

x xc=

M- M xc t,( )≡ L xc–( ) PaL

------waxc

2------------– for xc L<=

xc L> M–

M+ M*≤

M- M*≤

a a 0= aM* as

PL---as L as–( ) M*=

a as M+M*

a L a–( ) PL---

12---wa–

M*=

wa

wa 2 PL--- M*

a L a–( )-------------------–

=

waM- M*–


It is convenient to work with dimensionless variables for x and M.

(7.19)

where

(7.20)

The factor, f, can be interpreted as the “reduction” due to prestressing. Noprestress corresponds to f=1; passive prestress for this loading and prestressingscheme corresponds to f=0.5. Using this notation, is given by

(7.21)

The dimensionless form of eqn (7.18) is written as

(7.22)

Lastly, the dimensionless peak negative moment is expressed as

(7.23)

where

(7.24)

The peak negative moment is a function of the position coordinate, , andthe design moment reduction factor, f. Since must be less than 1 for all valuesof between 0 and 0.5, the magnitude of f is constrained to be above a limitingvalue, fmin. Figure 7.3 shows plots of vs. for a range of values f. For thiscase, the limiting value of f is equal to 0.26. Therefore, with active prestressing, thedesign moment can be reduced to 50% of the corresponding value for passiveprestressing. The influence line for the cable tension required for the “optimal”active prestressing algorithm is plotted in Fig 7.4. Also plotted is the requiredtension corresponding to f=0.5, the optimal passive value. As expected, loweringthe cross-sectional design moment results in an increase in the required cabletension. In order to arrive at an optimal design, the costs associated with thematerial (cross-section) and prestressing need to be considered.

x xL---= M M

M*--------=

M* f PL4

------- =

as

as12--- 1 1 f–( )1 2⁄–[ ]=

waL2

M*------------- 8d

M*--------

T≡ 8f--- 2

a 1 a–( )-------------------– g a f,( )= =

M-

M*-------- M- 1 xc–( ) 4a

f------

gxc2

--------–= =

xc12--- 4a

fg------+=

aM-

aM- a


Fig. 7.3: Influence lines for the peak negative moment

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5−1.8

−1.6

−1.4

−1.2

−1

−0.8

−0.6

−0.4

−0.2

0

M–

a L⁄

f=0.2

f=0.3

f=0.4

f=0.5

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5−1.8

−1.6

−1.4

−1.2

−1

−0.8

−0.6

−0.4

−0.2

0

M–

a L⁄

0.270.260.25


Fig. 7.4: Influence lines for the optimal cable tension

Active prestressing with concentrated forces

In this section, the use of concentrated forces to generate prestress momentfields is examined. The design loading is assumed to be a single concentratedforce that can act anywhere along the span.

Example 7.1: A single force actuator

Consider the structure shown in Fig (1). The active prestressing is providedby a single force, F, acting at mid-span. This loading produces 2 local momentmaxima, M1 and M2. The moment at mid-span may be negative for certaincombinations of and F, and therefore it is necessary to check both M1 and M2when selecting a value for the control force. Adopting the strategy discussedearlier, the control algorithm is based on the following requirements

(1)

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50

1

2

3

4

5

6

a L⁄

8dPL 4⁄--------------

T

f=0.26

f=0.5

a

M1 M*≤

M2 M*≤


for all where

(2)

(3)

and is the design moment for the cross-section.

Figure 1

a

M1Pa L a–( )

L----------------------- Fa

2------–=

M2Pa2

------ FL4

-------–=

M*

(+)

P

Pa L a–( )L

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

a L a–

FL4

-------–

(-) M+

M+

L/2F

M+

M1

M2


Shifting to dimensionless variables,

(4)

transforms the equations to

(5)

(6)

No action is required for where is defined by eqn (7.21).

(7)

When , the force is selected such that .

(8)

The corresponding expression for is

(9)

Figures 2 and 3 show the variation of and with and f. The limiting valueof f for active prestressing is 0.345; when f>0.345, the negative moment at mid-span is greater than the design moment, . For passive prestressing, theoptimal solution is f=0.5. Shifting from passive to active control results in anadditional 30 percent reduction in the allowable design moment.

a aL---=

M* f PL4

-------=

MiMi

M*--------=

F FP---=

M14f--- a 1 a–( ) aF

2------–=

M21f--- 2a F–[ ]=

a as≤ as

as 1 as–( )4f--- 1≡

a as> M1 1≡

F 2 1 a– f4a------–

=

M2

M24a 2–

f--------------- 1

2a------+=

M2 F a

M*


Figure 2

Figure 3

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5−4

−3.5

−3

−2.5

−2

−1.5

−1

−0.5

0

0.5

1

0.2

0.25

0.3

0.35

0.4

0.45

0.5

M2

a

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50

0.2

0.4

0.6

0.8

1

1.2

1.4

0.2

0.25

0.3

0.35

0.4

0.45

0.5

F

a


Example 7.2: Two force actuators

The previous example showed that the effectiveness of a prestressingscheme depends on the difference between the moment distributions for theapplied loading and the prestressing forces. In the case of a single control forceapplied at mid-span, the limiting condition occurs when the applied load is nearthe end support, where the difference in the moment distributions is a maximum.Increasing the number of control forces provides the capability to modify the“shape” of the “prestress” moment distribution to conform better with theapplied moment field, and therefore increase the amount of prestressing that canbe applied. This example illustrates the use of self-equilibrating control forcesystems which provide the maximum flexibility for adjusting the moment field.

Consider the self-equilibrating force system shown in Fig. 1. This forcesystem produces a bilinear moment field which is local, i.e., confined to theloaded region. Therefore, perturbing the control force magnitude, F, has no effectoutside of this region.

Figure 1

Applying a set of these self-equilibrating systems results in a piecewizelinear moment distribution. Figure 2 illustrates the case of 2 force systems locatedimmediately adjacent to each other. The corresponding moment field is defined interms of 2 force parameters, F1 and F2.

M+

F F/2F/2

l l

Fl2-----–


Region A-B

(1)

Region B-C

(2)

Region C-D

(3)

where

(4)

Figure 2

Example 7.1 treated the case of a single actuator deployed on a simplysupported beam subjected to a single concentrated force that can act at any pointon the beam. Suppose the control force system now consists of 2 local momentfields centered at the third points of the span. Figure 3 shows the 2 loadingscenarios for this example. The moments at B,C, and D corresponding to thedifferent loading scenarios are:

MM1

l--------– x xA–( )=

MM1

l-------- l xB x–+( )–

M2l

-------- x xB–( )–=

MM2

l-------- l xc x–+( )–=

M1F1l

2--------= M2

F2l

2--------=

M+

F12

------

A B C D

x

F1 F22

------F12

------

F2 F22

------

l ll

(-)M1 M2


Region A-B (Fig 3a):

(5)

(6)

(7)

Region B-C (Fig 3b):

(8)

(9)

(10)

Figure 3a

Mp PL a 1 a–( )( ) 3aM1–=

MB23---PLa M1–=

MC13---PLa M2–=

Mp PL a 1 a–( )( ) M1 2 3a–( )– M2 3a 1–( )–=

MB13---PL 1 a–( ) M1–=

MC13---PLa M2–=

M+

L/3 L/3 L/3

A B C D

Pa aL=

(+)

M1M2

(-)

MP MB MC


Figure 3b

Let represent the design moment. Expressing as a fraction of themaximum moment for the case where there is no prestressing,

(11)

and working with dimensionless moments,

(12)

transforms eqns (5) thru (10) to the following:

Region A-B

(13)

(14)

(15)

M+

L/3 L/3 L/3

A B C D

Pa aL=

(+)

M1M2

(-)

MPMB

MC

M* M*

M* f PL4

-------=

M( )M( )

M*------------=

Mp4f--- a 1 a–( )( ) 3aM1–=

MB8

3 f------a M1–=

MC4

3 f------a M2–=


Region B-C

(16)

(17)

(18)

The control objective for this example is to limit the peak value of each ofthe dimensionless moment variables to be less than unity.

(19)

Since there are 3 constraints and only 2 control parameters, the problem isoverconstrained. The strategy followed here is based on determining and

using 2 of the constraints in eqn (19), and then adjusting f such that the thirdconstraint is also satisfied. Figure 4 shows the variation in the moment measureswith , the coordinate defining the position of the load, corresponding to thefollowing choice of constraints:

Region A-B

(20)

Region B-C

(21)

The minimum value of f is equal to 0.228. This value is controlled by theconstraint on in the region A-B. For the case of a single actuator applied at

Mp4f---a 1 a–( ) M1 2 3a–( )– M2 3a 1–( )–=

MB4

3 f------ 1 a–( ) M1–=

MC4

3 f------a M2–=

MP 1≤ MB 1≤ MC 1≤ 0 a 12---≤ ≤

M1M2

a

MP +1=

MC +1=

MB 1<

MP +1=

MC +1 4 3a 1–( )–=

MB 1<

MB


mid-span, the optimum value for f was found to be 0.345. Applying 2 forceactuators leads to an additional reduction in the “permissible” design moment.

Figure 4a0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

0

0.5

1

1.5

2

2.5

3

3.5

4

M1

a

f 0.228=


Figure 4b

Figure 4c

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50

0.5

1

1.5

2

2.5

3

3.5

4

M2

a

f 0.228=

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5−1.5

−1

−0.5

0

0.5

1

MB

a

f 0.228=


Figure 4d

A general active prestressing methodology

The discussion to this point has been concerned with a specific loading anda specific prestressing scheme. In what follows, a general methodology fordealing with the combination of an arbitrary design loading and prestressingscheme applied to a simply supported beam is described, and the controlalgorithm corresponding to a particular choice of error measure is formulated.This methodology is also applicable for displacement control which is discussedin the following section.

Let denote the moment due to the design loading, themoment generated by the prestress system, and M(x) the net moment. Bydefinition,

(7.25)

The design objective is to limit the magnitude of to be less than , thedesign moment for the cross-section.

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

MC

a

f 0.228=

Md x( ) Mc x( )

M x( ) Md x( ) Mc x( )+=

M x( ) M*


(7.26)

Equation (7.26) imposes a constraint on the magnitude of .

(7.27)

These limits establish the lower and upper bounds for . Given , onegenerates the limiting boundaries and then decides on a “target” distribution for

.

Fig. 7.5: Limiting prestress moment fields

Figure (7.5) illustrates the process of establishing the desired distributionfor the control moment. The curves shown in Fig (7.5b) correspond to

; allowable values of are defined by the shaded area. Thedistribution corresponding to selecting the minimum allowable value ofat each x is plotted in Fig (7.5c). Assuming magnitude is the dominant measure,

Md Mc+ M*≤ 0 x L≤ ≤

Mc

Md– M* Mc M* Md–≤ ≤– 0 x L≤ ≤

Mc x( ) Md x( )

Mc x( )

10

-1-2

a bc d

e

(a)

(b)

01

3

-1-2

10

-1-2

-4

a bc d

e

(c)

upper bound

lower bound

Md

M*--------

Mc

M*--------

Mc*

M*--------

Permissible region

Optimum solution

Mc Md– M*±= McMc x( )


this distribution represents the optimal “target” prestress moment field.

Let denote the desired “target” prestress moment field. Supposethe actual prestress moment field is a linear combination of r individual fields,

(7.28)

where are moment amplitude parameters and are dimensionlessfunctions. The error associated with a specific choice of moment parameters isrepresented by the difference, ,

(7.29)

Ideally, one wants for . However this goal cannot be achievedwhen is an arbitrary function, and it is necessary to work with anapproximate error condition established using collocation, the least squaremethod, or some other weighted residual scheme.

The least square method is based on taking the integral of as ameasure of the accuracy of the approximation represented by eqn (7.28). Thisintegral is denoted as J.

(7.30)

In general, J is a function of the r moment parameters. Equations for theseparameters are generated by requiring J to be stationary.

(7.31)

Expanding eqn (7.31) results in the following linear matrix equation,

(7.32)

where are r’th order matrices, and the elements of and are:

(7.33)

(7.34)

Given ,one determines and then solves for . This solution producesthe least value for J, for a particular set of ‘s. A sense of convergence can be

Mc* x( )

Mc x( ) mjψ j x( )j 1=

r

∑=

mj ψ j x( )

e x( )

e x( ) Mc* x( ) Mc x( )–=

e x( ) 0= 0 x L≤ ≤Mc

*

e x( )( )2

J 12--- e2 xd

0

L

∫=

mi∂∂J 0= i 1 2 … r, , ,=

am b=

a b m, , a b

aij Ψi x( )Ψj x( ) xd0

L

∫=

i j, 1 2… r,,=

bi Ψi x( )Mc* x( ) xd

0

L

∫=

Mc* x( ) b m

Ψ


obtained by expanding the set of basis functions, and comparing thecorresponding values of J. It should be noted that the exact condition, e(x)=0 for0<x<1, is generally not satisfied by eqn (7.32).

This formulation works with continuous functions, and requires theevaluation of a set of integrals. It is more convenient to work with vectors ratherthan functions, since the computation reduces to matrix operations. Suppose themoment is monitored at n points within the interval . The desiredprestress moment vector is of order n .

(7.35)

Evaluating eqn (7.28) at these observation points leads to

(7.36)

where is of order . The error vector is taken to be the difference between and ,

(7.37)

When and the individual prestress moment fields are linearlyindependent, is non-singular and it is possible to determine an that exactlysatisfies .

(7.38)

When , a least square procedure can be used to establish an approximatesolution for . The error measure is taken as the norm of .

(7.39)

Requiring J to be stationary with respect to leads to an equation having thesame form as eqn (7.32), with and now given by

(7.40)

(7.41)

The following example illustrates the application of the discrete formulation.

0 x L≤ ≤1×

Mc* Mc 1,

* Mc 2,* … Mc n,

*, , , =

Mc Ψm=

Ψ n r×Mc

* Mc

e Mc* Mc– Mc

* Ψm–= =

r n=Ψ m

e 0=

m Ψ 1– Mc*=

r n≠e 0= e

J 12---eTe J m( )= =

ma b

a ΨTΨ=

b ΨTMc*=


Example 7.3: Multiple actuators

Consider the design moment field shown in Fig (1). The longitudinal axisis discretized with 10 equal segments, resulting in 11 (n=11) observation points.Applying the criteria defined by eqn (7.27), and taking leads to thebounding curves for plotted in Fig (2). The problem now consists ofgenerating a prestress moment distribution which lies between these bounds.

Suppose 4 self-equilibrating force system (r=4) that produce bilinearmoment fields are applied at equally spaced interior points. The corresponding

functions are shown in Fig (3a) and the typical field is plotted in Fig (3b). Sincethe prestress moment field is defined in terms of the moments at only 4 fixedpoints (3,5,7,9), and there are 9 interior points, the solution for the case of anarbitrary target distribution will be approximate. Various solutions for aparticular target distribution are plotted in Fig (4). Curve (1) corresponds totaking , , and . Curve (2) is the least squaresolution for the 4 actuator system defined in Fig (3a). Curve (3) is based on the 7actuator system defined in Fig (3c). The 3 additional actuators applied at points2,4, and 6 eliminate the error up to point 6. Incorporating 2 more actuators atpoints 8 and 10 would completely eliminate the error associated with the 4actuator system. It produces a moment field that is fully contained within theallowable zone and has the lowest cost as measured by the sum of the actuatormoment magnitudes, .

Figure 1: Moment due to design loading

M* 1=Mc

*

Ψ

m1 m2 1–= = m3 0= m4 1=

Σmi

Md 3

2

1

-1

-2

1 2 3 4 5 6 7 8 9 10 11


Figure 2: Upper and lower bounds on the prestress moment field

Mc* 3

2

1

-1

-2

1 2 3 4 5 6 7 8 9 10 11

-3

-4

M* 1=upper bound

lower bound


Figure 3: Prestress moment fields for 4 and 7 actuators

1

-1

1 2 3 4 5 6 7 8 9 10 11

1

-1

1 2 3 4 5 6 7 8 9 10 11

1

-1

1 2 3 4 5 6 7 8 9 10 11

1

-1

1 2 3 4 5 6 7 8 9 10 11Ψ1

Ψ2

Ψ3

Ψ4

(a) r=4

1 3 5 7 9 11(b) r=4

m1 m2 m3 m4

1 2 3 4 5 6 7 8 9 10 11(c) r=7

m2 m7

m3m4

m6m1m5


Figure 4: Prestress moment fields

7.3 Quasi-static displacement control of beams

The previous section discussed how active prestressing can be applied tolimit the magnitude of the bending moment in a beam subjected to a quasi-staticloading. In this section, a procedure for controlling the quasi-static displacementprofile of a beam is described. Both prestress and displacement control are basedon the same least square error minimization algorithm which is generalized laterin the next section.

Consider a beam subjected to a loading that produces the displacementfield .Suppose the desired displacement profile is , and the difference

1 2 3 4 5 6 7 8 9 10 11−2.5

−2

−1.5

−1

−0.5

0

0.5

1

Target distribution

4 actuators (1)

4 actuators (2)

7 actuators (3)

2

3

1

o o... ...

uo x( ) u* x( )

7.3 Quasi-static Displacement Control of Beams 507

between and is to be reduced by applying a set of control forcesystems. Let denote the displacement due to the control forces. Theresultant error in displacement is denoted by , and is defined as

(7.42)

Assuming is a function of r control force parameters,

(7.43)

the issue now is how to establish approximate values for the force parameters.The least square procedure works with the square of the displacement errorsummed up over the domain, in this case the length of the beam. Both continuousand discrete forms of the least square error function are considered here.

Continuous least square formulation

In the continuous case, J is taken as

(7.44)

and is expressed as a linear combination of prescribed functions andunknown scalar force parameters,

(7.45)

Substituting for and transforms J to a quadratic form in F,

(7.46)

Introducing the following notation,

uo x( ) u* x( )uc x( )

e x( )

e x( ) uo x( ) uc u* x( )–+=

uc x( )

uc x( ) uc x F1 F2 … Fr, , , ,( )=

J 12--- e2 xd

0

L

∫=

uc x( )

uc x( ) Fiφi x( )i 1=

r

∑=

uc x( ) e

J 12--- uo u*–( )2 x uo u*–( )φi xd

0

L

∫ Fi +

i 1=

r

∑+d0

L

∫=

12--- φiφj xd

0

L

∫ FiFji 1=

r

∑j 1=

r

∑+


(7.47)

reduces eqn (7.46) to a matrix expression,

(7.48)

The first and second differentials of J are

(7.49)

(7.50)

Requiring J to be stationary is enforced by setting the coefficient of equal to 0in eqn (7.49), and leads to a set of r equations for the r force parameters.

(7.51)

The stationary point corresponds to an absolute minimum value when is apositive definite matrix. This property exists when the displacement functions,

, are linearly independent.

Discrete least square formulation

The discrete least square formulation is based on selecting a displacement

F Fi =

b bi =

a aij[ ]=

aij φiφj xd0

L

∫=

bi uo– u*+( )φi xd0

L

∫=

c uo u*–( )2 xd0

L

∫=

J 12---c bTF–

12---FTaF+=

dJFi∂

∂J dFii 1=

r

∑ dFT b– aF+( )= =

d2JFj∂∂

Fi∂∂J dFi

i 1=

r

∑ dFjj 1=

r

∑ dFTadF= =

dFT

dJ 0 for arbitrary dF aF⇒ b= =

a

Ψi x( )


error vector, , and taking J as

(7.52)

One forms by specifying n points along the longitudinal axis and evaluating thedifference between the various displacement components at these points. Usingmatrix notation, is expressed as

(7.53)

where and are displacement vectors of order , and is of order. Substituting for in eqn (7.52) and expressing J in the same form as for the

continuous case

(7.54)

leads to the definition equations for , and .

(7.55)

(7.56)

(7.57)

The stationary requirement for J is still .

Since is now generated by pre-multiplying by its transpose, theproperties of depend on the relative size of n and r. If n > r, will be non-singular provided that the columns of are linearly independent. This conditionis equivalent to requiring the rank of to be equal to the number of its columns.If the rank of is less than r, say r - s, then s control forces do not contribute to theerror and need to be deleted. If n < r, will be singular and a special solutionprocedure needs to be applied. Strang(1993) contains a detailed discussion of thepseudo-inverse procedure for solving when is singular. Thisprocedure is executed by MATLAB with the statement, . Thefollowing example illustrates different scenarios for n vs. r.

e

J 12---eTe=

e

e

e Uo U*– ΦF+=

Uo U* n 1× Φn r× e

J 12---eTe c bTF–

12---FTaF+= =

a b c

b φT Uo– U*+( )=

a ΦTΦ=

c Uo U*–( )T Uo U*–( )=

aF b=

a Φa a

ΦΦ

Φa

aF b= aF pinv a( )*b=


Example 7.4: Cantilever beam - Least square algorithm

Consider the cantilever beam shown in Fig (1). Introducing thedimensionless variable

(1)

and assuming the bending rigidity, , is constant, the expression for the initialand desired displacement fields are written as:

(2)

(3)

(4)

where is a dimensionless parameter.

Figure 1

Suppose a single control force is applied at x = L. The correspondingdisplacement distribution is

x xL---=

DB

uowL4

DB----------g x( )=

g x( ) 124------ x2 6 4x– x2+( )[ ]=

u* Lα---x2=

α

u

x

L/2

L

F2F1

w

F1


(5)

Substituting for the various functions in eqn (7.47) results in

(6)

where

(7)

(8)

Solving for leads to:

(9)

The error function associated with the solution is:

(10)

Table 1 contains the errors at locations and .

uc φ1F1L3

2DB----------- x2 x3

3-----–

F1= =

a11F1 b1=

a11L7

DB2

-------- 0.0262( )=

b1wL8

DB2

---------- 0.0102( )–L5

αDB------------ 0.0722( )+=

F1

F1

DB

L2α---------- 2.7576( ) wL 0.3911( )–=

e x( ) wL4

DB---------- f x( ) L

α---g x( )+=

wL4

DB---------- 0.0545x2 0.1015x3– 0.417x4+ +=

Lα---+ 0.3789x2 0.4596x3–

x 0.5= x 1.0=


Table 1 Comparison of displacement errors

To illustrate the discrete formulation, observation points at andare selected. For this case, n = 2 and r = 1. Using eqns (1), (3), and (5), the

terms defining are:

(11)

(12)

Substituting in eqns (7.55) and (7.56) results in

Case f g f g

1. Continuous leastsquare

0.0035 0.0373 -0.0053 -0.0808

2. Discrete leastsquare and 2

obs. points

0.0048 0.0570 -0.0015 -0.0178

3. Discrete leastsquare and 2

obs. points

-0.0049 0.1292 0.0020 -0.0521

4. Discrete leastsquare and

2 obs. points

0 0 0 0

5. Discrete leastsquare and

1 obs. point

0.0043 0.0703 0 0

x 0.5= x 0.5= x 1.0= x 1.0=

F1

F1

F2

F1 F2,

F1 F2,

x 0.5=x 1.0=

e

Uo U*–wL4

DB----------

17384---------

18---

Lα---

14---

1

–=

Φ L3

DB-------

548------

13---

=


(13)

(14)

and

(15)

This solution approaches the continuous solution, eqn (9), as the number ofobservation points is increased. For example, the solution corresponding to 4observation points located at is

(16)

The location of the point of application of the control force also has aninfluence on the solution. Considering a single force, , applied at , andworking with 2 observation points at and 1.0 generates the followingleast square solution,

(17)

Comparison of the errors listed in Table 1 indicates that applying the force at is more effective for this structure/loading combination.

Applying both and , and taking 2 observation points results inbeing of order 2x2. In this case, one can set and solve

(18)

for . The least square formulation produces the same solution for when issquare. Evaluating eqn (18) results in

a a11L6

DB2

-------- 0.1220( )= =

b b1wL7

DB2

---------- 0.0463–( ) L4

αDB------------ 0.3594( )+= =

F1

DB

αL2---------- 2.9467( ) wL 0.3795( )–=

x 0.25 0.50 0.75 and 1.0, , ,=

F1

DB

αL2---------- 2.8685( ) wL 0.3842( )–=

F2 x 0.5=x 0.5=

F2

DB

αL2---------- 9.0997( ) wL 1.1807( )–=

x 1.0=

F1 F2 Φe 0=

ΦF U* Uo–=

F F Φ


(19)

Suppose control forces are applied at and 1.0, and only oneobservation point is selected. In this case, n = 1 and r = 2. Considering the point tobe at , the corresponding error term defined by eqn (7.53) has thefollowing form:

(20)

Expanding and ,

(21)

(22)

shows that is singular; the second row is 5/16 times the first row. The samerelationship holds for the elements of . Therefore, in this case, there is only oneindependent equation in the set, , generated by requiring to bestationary. Another equation has to be established.

The pseudo inverse solution procedure starts by setting equal to 0.Noting eqn (20),

(23)

An additional equation relating and is generated by requiring the meansquare sum of the force magnitudes to be a minimum, considering eqn (23) to be aconstraint between and . The steps are:

F1

F2

wL0.1964–

0.5714– DB

αL2----------

5.14296.8571–

+=

x 0.5=

x 1.0=

e x 1=( ) wL4

8DB----------- L

α---–

L3

3DB----------- 5

48------ L3

DB-------

F1

F2

+=

e Uo Uc– ΦF+=

a b

a L3

3DB-----------

2 1516------

1 516------ L3

3DB-----------

1 5

16------

516------ 5

16------ 5

16------

= =

b L3

3DB-----------

1516------

wL4

8DB-----------– L

α---+

=

ab

aF b= eTe

e

e x 1=( ) 0= ⇒ F1516------F2+ 3

DB

αL2---------- 3

8---wL–=

F1 F2

F1 F2


a) Form the objective function

(24)

b) Form the first differential

(25)

c) Establish the constraint between the differentials

Noting eqn (23),

(26)

d) Establish the additional equation

Substituting for in terms of in eqn (25),

(27)

and requiring eqn (27) to be satisfied for arbitrary yields

(28)

e) Solve for the primary variable

Using eqn (28), the solution follows form eqn (23).

(29)

Table 1 lists the error at . The error at was taken to be zero at thestart of the solution process. One can obtain the solution defined by eqn (29) withthe MATLAB statement, .

I 12--- F1

2 F22+( )=

dI F1dF1 F2dF2+ 0= =

dF1516------dF2+ 0=

dF2 dF1

F1165------–

F2+ dF1 0=

dF1

F1165------F2=

F11

1 516------

2+

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

3DB

αL2-----------

38---wL–

DB

αL2---------- 2.7331( ) wL 0.3416( )–= =

F2516------F1=

x 0.5= x 1.0=

F pinv a( )*b=


The previous example illustrated the use of concentrated forces to adjustthe deflected shape of a cantilever beam subjected to a uniform transverseloading. The approach involved selecting a set of points at which thedisplacements are monitored, and a set of control forces. Considering thedisplacement field to be continuous corresponds to taking an infinite number ofobservation points. In order to apply the procedure for an arbitrary loading, afinite number of points needs to be specified. The control algorithm is establishedusing the discrete least square formulation. A square nonsingular set of equationsis obtained when the number of observation points, n, is taken to be greater orequal to the number of control forces, r. If n is less than r, the problem is under-specified in the sense that additional equations have to be established. In this case,a pseudo inverse procedure can be applied.

Extended least square formulation

The standard least square formulation works with a weighteddisplacement error, , which represents the difference between the actual andtarget displacement fields. No priority is assigned to a specific location, say ,and also, no limit is placed on the magnitude of the control forces. This approachcan be generalized to include priorities by introducing weighting functions for

and , and employing an extended form of the error functional, J.

Considering first the continuous case, the error at x is weighted with afunction, q(x), and the error measure is taken as the integral of ,

(7.58)

Setting q(x) = constant assigns equal priority in the interval . Taking q(x)as a set of n Dirac delta functions centered at , i=1, 2, ..., n converts eqn(7.58) to the discrete form and allows for different weighting factors,

(7.59)

The magnitudes of the control forces influence J’ indirectly through .Selecting the force magnitudes such that J’ is stationary corresponds to forcing theweighted displacement error to be a minimum. There is a “cost” associated with a

e x( )e xi( )

e x( ) Fi

qe2

J J'⇒ 12--- q x( )e2 xd

0

L

∫=

0 x L≤ ≤x xi=

J' 12--- qiei

2 xi( )i 1=

n

∑=

e x( )


particular choice of the force magnitudes, and this cost is not reflected by thestationary solution for J’. Consequently, as the displacement error is driventoward zero, the magnitude and therefore the “cost” of the forces may becomeexcessive. This situation can be avoided by combining J’ with a force costfunction, and requiring the sum to be stationary. A convenient choice for the forcecost function is the weighted quadratic form,

(7.60)

where is a weight associated with . In general, . Combining this termwith J’ leads to the extended functional, .

(7.61)

The discrete form of is taken as

(7.62)

where e is the displacement error vector, and Q, R are diagonal matrices withnonnegative elements,

(7.63)

(7.64)

Substituting for using eqn (7.53), expands to

(7.65)

where

(7.66)

(7.67)

(7.68)

Force t fuctioncos 12--- r jFj

2

j 1=

r

∑=

r j Fj r j 0>J1

J112--- qe2 x 1

2--- r jFj

2

j 1=

r

∑+do

L

∫=

J1

J112---eTQe 1

2---FTRF+=

Q qiδij[ ]= i j, 1 2 … n, , ,=

R rkδkl[ ]= k l, 1 2 … r, , ,=

e J1

J112---c1 b1

TF–12---FTa1F+=

b1 ΦTQ Uo– U*+( )=

a1 ΦTQΦ R+=

c1 Uo U– *( )TQ Uo U*–( )=


Requiring to be stationary with respect to F leads to a set of r equations for ther force variables,

(7.69)

Although the order is the same as for the conventional formulation thatconsidered only the displacement error, differs from in 2 ways. Firstly, thequadratic term, , is replaced with a weighted form, . Secondly,positive quantities are added to the diagonal elements. The latter operationensures that is non-singular for all values of n vs. r. Holding Q fixed andincreasing the magnitude of R places more emphasis on lowering the cost of thecontrol forces. Consequently, the displacement error will increase. Decreasing Rwith respect to Q has the opposite effect on the displacement error. Ideally, onewants minimum displacement error and minimum force cost, but that goal isimpossible. Generating solutions corresponding to a range of Q and R valuesprovides information concerning the sensitivity of the 2 weighting measureswhich is useful for selecting design values. The following example illustrates thiscomputation.


Example 7.4 considered observation points at and 1.0, and asingle control force applied at . The example is reworked here using theextended least square algorithm. Starting with the basic data contained in eqns(11) and (12) of example 7.4,

(1)

and applying eqns (7.66), (7.67), and (7.69) results in

J1

a1F b1=

a1 aΦTΦ ΦTQQ

a1

x 0.5=x 1.0=

Uo U*–wL4

DB----------

17384---------

18---

Lα---

14---–

1–

+=

Φ L3

3DB-----------

516------

1

=


(2)

(3)

(4)

Suppose the q’s are equal to a common value, . In this case, thecoefficients simplify to

(5)

(6)

The first term on the right hand side of eqn (6) is a measure of the priority for thedisplacement error; the second term reflects the importance of the force cost. Sinceq is present in both and the force depends only on the ratio of to q. Toestablish the appropriate magnitude for , and q can be scaled such thatthe 2 terms on the right hand side of eqn (6) have the same order of magnitude.One possible approach is to take and scale q.

(7)

where is a measure of the relative importance of the force cost vs. thedisplacement error. Using eqn (7) the expression for takes the form

(8)

Equal weighting for displacement and force as assigned when . Increasing

b1wL7

DB2

---------- 124------

85768---------q1– q2–

L4

3DBα--------------- 5

64------q1 q2+

+=

a1L6

9DB2

----------- 516------

2q1 q2+

r1+=

F1 a11– b1=

q

b1wL7

DB2

---------- 85318432---------------–

q L4

DBα------------ 23

64------

q+=

a1L6

DB2

-------- 2812304------------

q r1+=

b1 a1 r1r1 q⁄ r1

r1 1=

r1 1=

qDB

2

L6-------- 2304

281------------

q*=

q*

F1

F1

wl 0.3794–( )DB

L2α---------- 2.9466( )+ q*

1 q*+--------------------------------------------------------------------------------=

q* 1=


beyond 1 places more emphasis on the displacement error, and the solutionapproaches the solution corresponding to no constraint on the control forcemagnitude. Taking ignores the displacement error, and the procedure willminimize the functional corresponding to only the force cost. Consequently, thesolution will be F = 0.

The scaling operation on r and q described in example 7.5 can be applied tothe general formulation represented by eqns (7.66) thru (7.69) in the followingmanner. Firstly, R is taken such that all its elements are between 0 and 1. Thismatrix is denoted as .

(7.70)

where are individual force cost “relative” weighting factors that are now oforder 1. Secondly, Q is scaled with a factor, , and expressed as

(7.71)

The magnitude of can be chosen such that both of the terms comprisinghave essentially the same magnitude. Using these definitions, eqn (7.69) istransformed to

(7.72)

Taking to be equal to the inverse of the maximum element in the matrixproduct, , is a convenient choice since now the elements of can beconsidered to be of order 1.

Sensitivity can be assessed by varying the elements of and . Obviouslycannot be a null matrix, whereas is a limiting case. Setting

and perturbing the elements of is a reasonable strategy for assessing how thesolution varies as the priorities with respect to the displacement error and forcecost are shifted. Increasing places more emphasis on the displacement error.

q*

q* 0≈

R*

R r j*δij[ ] R*≡=

r j*

q

Q qQ*=

q a1

q ΦTQ*Φ( ) R*+( )F qΦTQ* Uo– U*+( )=

qΦTΦ Q*

Q* R*

Q* R* 0= R* I=Q*

Q*

7.4 Quasi-static Control of MDOF Systems 521

7.4 Quasi-static control of MDOF systems

Introduction

The governing equations for the linear static behavior of an n’th ordersystem controlled with r forces are written as:

(7.73)

where contains the n displacement variables, K is the system stiffness matrix, Pis the prescribed load vector, F contains the r control forces, and defines thedistribution of the r control forces in the load vector. For n DOF and r controlforces, is of order n x r. Figure 7.6 illustrates the form of for a particular setof control forces applied to a discrete shear beam model.

Fig. 7.6: Control force distribution matrix

Selecting a control force system involves 2 decisions: firstly, the nature andlocation of the control forces which establishes and secondly, the controlalgorithm which determines the magnitude of the control force parameterscorresponding to the choice of . The decision on depends in turn on theobjective of adding control which usually involves limiting the response of certainvariables. Therefore, the first step is to identify the performance measures. Giventhe performance measures, a set of response quantities which can be “observed”and then used to establish the magnitudes of the performance measures areidentified. Also, once the performance measures are defined, the equilibriumequations are applied to establish the relationship between the “performance”

KU P E f F+=

UE f

E f E f

u1

u2

u3

u4

F1

F2

F2

E f

00

1–

11 00 0

=

E f

E f E f


and the control force measures. Lastly, a mathematical procedure such as the leastsquare method is employed to generate a control algorithm which is used toestablish a quantitative estimate of the control forces.

Selection of measures

By definition performance measures are those quantities whosemagnitudes are to be limited by the application of control forces. In the previoussections dealing with beam type structures, bending moments and displacementswere selected. For a low rise building modelled as a shear beam, limiting thetransverse shear deformation at certain story locations is the objective for motionbased design. Assuming there are np performance measures, the individualmeasures are denoted by and included in the vector .

Fig. 7.7: Deformations as performance measures

Figure 7.7 illustrates the selection process for a single truss modelled as a 3DOF shear beam. The performance measures are taken to be the transverse shearstrains in the first and second stories. They are related to the displacements by

(7.74)

The general matrix form of eqn (7.74) is written as:

(7.75)

vi i 1 2 … np, , ,=( ) v

2u1

u2

u3

vγ1

γ2

=

θ

h

h

h

3

1

v1u1h

------= v2u2 u1–

h------------------=

v BU=


where B can be interpreted as a transformation matrix which convertsdisplacements measures into performance measures.

Observation measures are those quantities which are monitored and usedto establish the magnitude of the performance measures. They may be eitherdeformations or displacements. Let denote the i’th observation measure,the total number of observations, and w the observation vector. The key decisionshere are selecting the number, the nature (i.e., displacement, deformation,...), andthe location of the observations such that the performance measures are uniquelydefined.

Suppose is a particular choice of observation variables. Assuming thebehavior is linear, associated with this choice of w can be expressed as

(7.76)

One evaluates the i’th row of using the definition equation for . If all theelements in row i are equal to 0, cannot be determined by this choice of w.Similarly, if all entries in column j are 0, contributes nothing to v and must bedeleted from . Observability is the term used to characterize the relationshipbetween v and w. A monitoring scheme is said to be observable when the rank of

is equal to , the number of performance measures. In general, a necessarycondition is ; the sufficient condition is for to have at least linearlyindependent columns. Various scenarios are illustrated by the following example.

Example 7.6: Illustrative examples of observability

Consider the physical model defined by Fig 7.7. There are 2 performancemeasures, the shear strains in the first 2 stories. Various cases are consideredbelow. Observability for this example requires the rank of B to be 2.

Case 1.

Applying eqns (7.74) leads to

(1)

wi nob

w*

v

vw* B1w*=

B1 vivi

wjw*

B1 npnob np≥ B1 np

w* u2 u3, =

B11h--- 0 0

1 0=


The second column contains all zero’s since has no effect on either or .

Case 2.

(2)

The rank is 1. The problem here is the incomplete determination of , whichrequires as well as ,

Case 3.

(3)

Here, r(B1)=2. Observability is satisfied.

Case 4. where and are the extensional strains in thediagonal elements located in the first and second stories. Noting the definitionequations for the extensions,

(4)

and substituting for and in eqn (7.74) results in

(5)

In this case, r(B1)=2 for arbitrary finite .

The last group of variables to be defined is the set of control forcesrepresented by the vector F. Once the location of the forces is chosen, isknown, and the equilibrium equations can be solved to determine thedisplacements generated by applying F. Using eqn (7.73), due to F is given by

u3 v1 v2

w* u1 =

B11h--- 1

1–=

γ2u2 u1

w* u1 u2, =

B11h--- 1 0

1– 1=

w* ε1 ε2, = ε1 ε2

e1 u1 θcos hθsin

-----------ε1= =

e2 u2 u1–( ) θcos hθsin

-----------ε2= =

u1 u2

B11

θ θcossin------------------------ 1 0

0 1=

θ 0>

E f

U


(7.77)

Then, substituting for in eqn (7.75) leads to a relation between v and F.

(7.78)

The coefficient matrix, Cf, is of order , and defines the effect of the controlforces on the performance measures. If the elements in the i’th row of Cf are allzero, the choice of control forces has no influence on the i’th performancemeasure, i.e., cannot be controlled by F. Similarly, if all the elements in the j’thcolumn of Cf are zero, contributes nothing and should be deleted from F.

In the usual control scenario, the number of force parameters is less thanthe number of performance parameters. Requiring r(Cf)=r for ensures thatthe columns of Cf are linearly independent and eliminates the situation where Cfhas a zero column. Controllability is defined for an individual performancemeasure, and requires the corresponding row of Cf to have at least one non-zeroelement. The following example illustrates some choices of control force systemsfor example 7.6

Example 7.7: Illustrative examples of controllability

Consider the shear beam model defined in Fig (7.7) and discussed inexample 7.6. There are 2 performance measures, and therefore one would notnormally take more than 2 control forces. The forms of Cf corresponding tovarious force systems are listed below. Case 3 has an excess of control forces. Case4 has a deficiency with respect to controllability; cannot be controlled with thisscheme.

UF

K 1– E f F=

U

vF

BUF

BK 1– E f( )F C f F= = =

np r×

viFj

np r>

v1


Case 1.

Case 2.

Case 3.

F1

E f

001

= C f

1DT1----------

1DT2----------

=

F1

F2

E f

0 10 01 0

= C f

1DT1---------- 1

DT1----------

1DT2---------- 0

=

F1

F3

F2

E f

0 1 00 0 11 0 0

= C f

1DT1---------- 1

DT1---------- 1

DT1----------

1DT2---------- 0 1

DT2----------

=


Case 4.

Least square control algorithms

Suppose the system is subjected to its design loading, and the values of theperformance parameters resulting from this loading are observed with amonitoring scheme. Let denote the “observed” response, and the desiredresponse. The difference between these vectors is the target response for the forcecontrol system. Taking as the response due to the control forces, the netperformance error vector is

(7.79)

The goal is to select F such that e is less than the allowable error. Considering thebehavior to be linear, and noting eqn (7.78), e is related to F by

(7.80)

Equation (7.80) represents equations in r unknowns; Cf is of order .When , there is a unique F which satisfies e = 0. For all other combinationsof and r, the solution for F is not unique and also may not satisfy e = 0.

The least square solution for the case where is generated by formingthe error norm,

(7.81)

and requiring J to be stationary with respect to F. This operation leads to a set of requations, which are written as:

F1

F2

E f

0 1–

1– 11 0

= C f

0 0

0 1DT2----------

=

F1 F2

vo v*

vc

e vo vc v*–+=

e vo C f F v*–+=

np np r×np r=

np

np r>

J 12---eTe=


(7.82)

The coefficient matrix will be non-singular when Cf satisfies the controllabilityrequirement, r(Cf)=r. Since the square error norm is used, the solution of eqn (7.82)does not result in e = 0.


The least square control algorithm is applied to the model considered inexample 7.7. The following values are specified for the performance measures,

(1)

and the transverse shear rigidity is assumed to be constant.

Case 1.

Using eqn (1) of example 7.7, eqn (7.82) reduces to

(2)

The solution for and the corresponding error vector are

(3)

Case 2

Since the number of control forces is equal to the number of performancemeasures, solving eqn (7.82) is equivalent to solving eqn (7.80) with e set equal to

aF b=

a C fTC f=

b C fT v* vo–( )=

vo0.0070.004

= v* 0.0050.005

=

2DT

2--------F1

1DT-------- v1

* v1 0,–( ) v2* v2 0,–( )+[ ]=

F1

F1 0.0005DT≡

e0.00150.0015–

=


0. This result follows by writing eqn (7.82) as

(4)

and noting that the force vector which satisfies

(5)

also satisfies (4). The solution is

(6)

Case 3

In this case, there are 3 force parameters and only 2 performance variables.The problem is undetermined. Setting e = 0 in eqn (7.80) provides 2 equationsrelating , , and

(7)

Taking as the “excess” force parameter, and solving for and leads to

(8)

The additional equation is obtained by requiring the norm of F to be stationarywith respect to . For this example, the norm is taken as the Euclidean norm, I.

(9)

Then

C fT C f F v*– vo+( ) 0=

C f F v* vo–=

F1 0.001DT=

F2 0.003DT–=

e 0=

F1 F2 F3

C f F v* vo–= ⇒

1DT-------- F1 F2 F3+ +( ) 0.002–=

1DT-------- F1 F3+( ) 0.001=

F3 F1 F2

F1 F3– 0.001DT+=

F2 0.003DT–=

F3

I 12---FTF 1

2--- F1

2 F22 F3

2+ +( ) I F3( )= = =


(10)

Finally, the control force magnitudes are

(11)

Other possible solutions that satisfy e = 0 are

and (12)

The pseudo inverse solution given by (11) has the lowest norm, i.e., the smallestvalue of I.

The generalized version of the least square algorithm is based on theextended form of the error norm,

(7.83)

where and are diagonal weighting matrices with elements of order 1, and is a scaling factor defined as

(7.84)

Requiring to be stationary with respect to F leads to r equations for F,

(7.85)

Different priorities are assigned by perturbing the individual elements of and

F3∂∂I 0 F1 F3∂

∂F1 F3+⇒ 0 F1 1–( ) F3+⇒ 0= = =

F1 0.0005( )DT=

F2 0.003–( )DT=

F3 0.0005( )DT=

F 10 3– DT

13–

0

= F 10 3– DT

03–

1

=

J112---qeTQ*e 1

2---FTR*F+ J1 F( )= =

Q* R*

q

q 1 maximum element in C fTC f( )⁄=

J1

a1F b1=

a1 q C fT Q*C f R*+=

b1 q C fT Q* v* vo–( )=

Q*


. If all the elements of are finite, will be non-singular for as wellas for , and it is not necessary to generate additional equations for F.However, since this property is due to including , the equations may be ill-conditioned if and is small. The preferred strategy is to takeand perturb the elements of . The following example illustrates differentscenarios.

Example 7.9: Example 7.7 revisited with an extended least square algorithm

Certain cases considered in example (7.8) are reworked here using theextended algorithm defined by eqn (7.85) and the same performancespecifications as for example (7.8).

Case 1

(1)

(2)

(3)

(4)

(5)

Taking equal weights for the displacement errors and the force cost reduces by 50%.

R* R* a1 r np>r np≤

R*

r np> R* R* I=Q*

C fTC f

2DT

2--------=

q 1maximum element in C f

TC f---------------------------------------------------------------------------

DT2

2--------= =

a1q1

* q2*+

2------------------ r1

*+=

b1 q1* v1

* v1 0,–( ) q2* v2

* v2 0,–( )+[ ]DT2

--------=

0.002–( )q1* 0.001q2

*+[ ]DT2

--------=

F1

DT2

--------0.002–( )q1

* 0.001q2*+

12--- q1

* q2*+( ) r1

*+-----------------------------------------------------=

q1* q2

* r1* 1= = =( ) F1


Case 3

(6)

(7)

(8)

(9)

Note that the elements of are of the same order. When , the thirdrow of is equal to the first row, and is singular. The third row of is alsoequal to the first row, and represents only 2 independent equations inthis case. The third equation relating the 3 control forces can be generated byrequiring the norm, , to be stationary. As mentioned earlier, theMATLAB function, , generates this solution. The standardMATLAB statement, , will also generate a solution for this case, butthe force norm will be greater than the value corresponding to the pseudo inverseprocedure. Results for F and e based on taking , ,and letting range from 1 to are listed in Table 1. The last row contains theerror for the case where no control forces are applied. Taking places allthe emphasis on the error reduction, and results in the maximum control forcemagnitude. Decreasing shifts some of the emphasis to reducing the controlforce.

C fTC f

1DT

2--------

2 1 21 1 12 1 2

=

qDT

2

2--------=

a112---

q1* q2

* 2r1*+ + q1

* q1* q2

*+

q1* q1

* 2r2*+ q1

*

q1* q2

*+ q1* q1

* q2* 2r3

*+ +

=

b1

10 3– DT2

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

2q1*– q2

*+

2q1*–

2q1*– q2

*+

=

a1 r1* r3

* 0= =a1 a1 b1

a1F b1=

1 2⁄( )FTFF pinv a1( )*b1=F a1\b1=

q1* q2

* q*= = r1* r2

* r3* 1= = =

q* ∞q* ∞=

q*


Table 1

0.5 -3.0 0.5 0 0

0 -3.0 1.0 0 0

16 0.3314 -2.3670 0.3314 0.2954 -0.3373

8 0.2264 -1.9623 0.2264 0.4906 -0.5472

4 0.1053 -1.4737 0.1053 0.7363 -0.7895

2 0 -1.0 0 1.0 -1.0

1 -0.0625 -0.625 -0.0625 1.25 -1.125

No control 0 0 0 2.0 -1.0

q* 103F1 DT⁄ 103F2 DT⁄ 103F3 DT⁄ 103e1 103e2

∞pinv a1( )*b1( )

∞a1\b1( )


545

Chapter 8

Dynamic Control Algorithms - Timeinvariant systems

8.1 Introduction

This chapter extends the active control strategy introduced in the previouschapter to deal with time dependent loading. The discussion is restricted to timeinvariant systems, i.e., the case where the system properties and force feedbackalgorithm are constant over the duration of the time response.

The material presented here is organized as follows. Firstly, the state-spaceformulation for a SDOF system is developed, and used to generate the freevibration response. This solution provides the basis for establishing a criterion fordynamic stability of a SDOF system. Secondly, linear negative feedback isintroduced, and the topic of stability is revisited. The primary focus is onassessing the effect of time delay in applying the control force on the stability.Thirdly, the SDOF state-space formulation is specialized to deal with discrete timecontrol, where the feedback forces are computed at discrete time points and heldconstant over time intervals. Stability for discrete time feedback with time delay isexamined in detail, and a numerical procedure for determining the timeincrement corresponding to a stability transition is presented and illustrated withexamples. Fourthly, the choice of the optimal magnitudes of the feedback

546 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

parameters is considered. Optimality is related to the magnitude of a quadraticperformance index (LQR) which is taken as a time integral involving weightedresponse and control force terms. This approach is referred to as the “linearquadratic regulator problem” and leads to a time invariant linear relationshipbetween the control forces and state variables. Fifthly, the state-space formulationis extended to MDOF systems. The modal properties for an arbitrary dampingscheme are derived, and used to generate the governing equations expressed interms of the modal coordinates. The last section deals with optimal feedbackbased on the LQR performance index generalized for MDOF systems. Exampleswhich illustrate the sensitivity of the response and cost parameters to variationsin the location and nature of the control forces, and weighting coefficients arepresented.

8.2 State-space formulation - time invariant SDOF systems

Governing equations

Fig. 8.1: SDOF system.

The dynamic response of the SDOF linear system shown in Fig. 8.1 is governed bythe second order equation,

(8.1)

where is the applied external loading, F is the active force, and m, k, c are systemparameters. Integrating eqn (8.1) in time, and enforcing the initial conditions onand at , one obtains the velocity and displacement as functions of time.These quantities characterize the state of the system in the sense that once and

are specified, the acceleration and internal forces can be determined by backsubstitution.

k

c

m

u ug+

F

ug

p

mu cu ku+ + mag– p F+ +=

pu

u t 0=u

u

8.2 State-space Formulation-Time Invariant SDOF System 547

Rather than working with a second order equation, it is more convenient totransform eqn (8.1) to a set of first order equations involving the state variablesand . The new form is

(8.2)

This form is called the state-space representation. The motivation for the state-space representation is mainly the reduced complexity in generating bothanalytical and numerical solutions.

Matrix notation is convenient for expressing the state-space equations in acompact form. Defining as the state vector,

(8.3)

the matrix equilibrium equation is written as

(8.4)

where the various coefficient matrices are defined below.

(8.5)

(8.6)

(8.7)

The initial conditions at t=0 are denoted by .

uu

uddt------ u=

uddt------ c

m----–

u km----–

u 1–( )ag1m---- p 1

m---- F+ + + +=

X

X uu

X t( )= =

Xddt------- X AX BfF Bgag Bpp+ + += =

A0 1km----– c

m----–

=

Bf Bp

01m----

= =

Bg01–

=

Xo


(8.8)

With this representation, the problem is reduced to solving a first order equationinvolving .

Free vibration uncontrolled response

The free vibration uncontrolled response is governed by a reduced form of eqn(8.4)

(8.9)

When is constant, the general solution has the form

(8.10)

where is an unspecified vector of order 2 and is a scalar. Substituting forresults in

(8.11)

where is the identity matrix. According to eqn (8.11), the eigenvalues ofdefine the frequency and damping characteristics of the free vibration response.

Expanding ,

(8.12)

leads to the characteristic equation

(8.13)

and two eigenvalues

(8.14)

Noting that and , eqn (8.14) is identical to eqn (6.27)which was obtained from the second order equation. Given , eqn (8.11) can besolved for the eigenvectors which define the state-space modes. Since iscomplex, the eigenvectors occur as complex conjugates.

X 0( )u 0( )u 0( )

Xo≡=

X

X AX=

A

X Veλ t=

V λ X

A λI–( )V 0=

I A

A λI– 0=

λ– 1km----– c

m----– λ–

0=

λ2 cm----λ k

m----+ + 0=

λ1 2,12--- c

m----– i 4 k

m---- c

m---- 2

–± λR iλ I±= =

k m⁄ ω2= c m⁄ 2ξω=

λλ


(8.15)

The total free vibration response is obtained by combining the 2 complexsolutions such that the resulting expression is real. Starting with

(8.16)

and taking

(8.17)

where and are real scalars, results in

(8.18)

The constants and are determined by enforcing the initial conditions onat t=0.

(8.19)

Lastly, the solution for u(t) is given by the first scalar equation in eqn (8.18).

(8.20)

V 1 2,1λR

i0λ I

± V R iV I± V1 V1,= = =

X A1eλ1t

V 1 A2eλ1tV 1+=

A112--- AR iAI+( )=

A2 A1=

AR AI

X t( ) eλRt

ARV R AIV I–( ) λ It A– RV I AIV R–( ) λ Isin t+cos =

AR AI X

X 0( ) A1V 1 A1V 1+ ARV R AIV I–= =

⇒

uo

uo

AR

ARλR AIλ I–=

⇒

AR uo=

AI1λ I----- uo λRuo+( )–=

u t( ) eλRt

AR λ It AI λ Itsin–cos( )=


General solution - time invariant systems

The general solution for an arbitrary loading can be expressed as aDuhamel integral involving a specialized form of the free vibration response.Considering first a first order scalar equation,

(8.21)

where is constant, and g is a function of t, the complete solution has the form

(8.22)

A similar form can be generated for the first order matrix equation,

(8.23)

The free vibration solution defined by eqn (8.18), can be expressed as

(8.24)

where is defined by the following series:

(8.25)

This matrix exponential function has the “same” property as the correspondingscalar function.

Using eqn (8.24), the Duhamel integral matrix form of the total solution for eqn(8.4) is

(8.26)

where

(8.27)

The corresponding scalar form of the solution for u(t) is

y ay g+=

a

y t( ) ea t to–( )

yo ea t τ–( )g τ( ) τdto

t

∫+=

X t( ) AX G+=

X t( ) eAtXo=

eAt

eAt I At 12---AAt2 … 1

n!-----An tn …+ + + + +=

tdd eAt( ) AeAt=

X t( ) eA t to–( )

Xo eA t τ–( )G τ( ) τdto

t

∫+=

G t( ) B f F Bgag Bpp+ +=


(8.28)

Equation (8.26) applies for an arbitrary linear time invariant system. It isconvenient for establishing a discrete formulation of the governing equations.This topic is addressed in the next section.

Example 8.1: Equivalence of equations (8.18) and (8.24)

Consider eqn (8.16). The total free vibration response is given by

(1)

Noting eqn (8.11), the and V terms are related by

(2)

Expanding the product, , and using eqn (2), leads to

(3)

It follows that eqn (1) can be written as

(4)

u t( ) eλRt

uo λ It1λ I----- uo λRuo+( ) λ Itsin+cos

+=

1λ I-----e

λR t τ–( ) λ I t τ–( ) ag τ( )– p τ( )m

---------- F τ( )m

-----------+ + sin τd

0

t

∫+

X t( ) A1eλ tV 1 A1eλ tV 1+=

X 0( ) A1V 1 A1V 1+=

λ

AV 1 λV 1=

AV 1 λ V 1=

eλ tV 1

eλ tV 1 V 1 λV 1( ) t λ λV 1( )t2

2---- …+ + +=

I At AAt2

2---- …+ + +

V 1=

eAtV 1=

X t( ) eAt A1V 1 A1V 1+( ) eAtXo= =


Stability criterion

Another advantage of the state-space representation is the ability to relate thestability of the physical system to the eigenvalues of . A system is said to bestable when the motion resulting from some initial disturbance is bounded.Assuming the system state is at time , stability requires

(8.29)

where defines the bound on the perturbation from .

Equation (8.18) defines the general homogeneous solution for a SDOF timeinvariant system. The terms contained inside the brackets depend on the initialconditions and are bounded since the time dependency is harmonic. Therefore, itfollows that the exponential term must be bounded. This requirement is satisfiedwhen the exponent is negative,

(8.30)

In words, the real part of the eigenvalues of must be equal to or less than zero.When , the response is pure harmonic oscillation. A negativeproduces a damped harmonic response.

Plotting in the complex plane provides a geometric interpretation of thestability. For the SDOF case, there are two eigenvalues,

(8.31)

Figure 8.2 shows the corresponding points in the complex plane. These points arereferred to as poles. Undamped motion has poles on the imaginary axis. Holdingstiffness constant and increasing causes the poles to move along the circle ofradius toward the critical damping point, . With further increase indamping, the curves bifurcate with one branch heading in the negative (real axis)

A

Xo t 0=

X t( ) Xo– ε≤ for all t

ε Xo

λR 0≤

AλR 0= λR

λ

λ λR iλ I±=

λRc

2m--------– ξω–= =

λ Ikm----

c2m-------- 2

– ω 1 ξ2–= =

cω ξ 1=


direction, and the other toward the origin. Increasing the stiffness with heldconstant moves the poles in the imaginary direction.

With this terminology, the stability criterion requires all the polescorresponding to the eigenvalues of A to be on or to the left of the imaginary axis,as shown in Fig. 8.3. The uncontrolled SDOF system is, according to thisdefinition, always stable since .

Fig. 8.2: Poles for SDOF system.

Fig. 8.3: Stability condition for SDOF system.

Linear negative feedback

The response of a SDOF time invariant system with negative linear feedback isgoverned by eqn (8.4) with F taken as a linear function of the state variables

c

ξ 0≤

ω

λR

λ I

θc increases

k increases

ξ 1=

ξ 0=

ξ 0=

ω

λR

λ I

stable unstable


(8.32)

Substituting for , the governing equation is transformed to

(8.33)

where

(8.34)

The general form of the free vibration solution of eqn (8.33) is

(8.35)

where and are the eigenvalues and eigenvectors of , the modifiedcoefficient matrix. They are related by

(8.36)

The eigenvalues of are written as

(8.37)

Since is positive for negative feedback (note that the minus sign is incorporatedin the definition equation, eqn (8.32)), the system is stable for arbitrary .Displacement feedback moves the poles in the imaginary direction, andconsequently has no effect on the stability. It follows that increasing the negative

F KfX– kd kvuu

–= =

F

X AcX Bgag Bpp+ +=

Ac A BfKf–0 1

km----–

kdm-----– c

m----–

kvm-----–

= =

X Veλ t=

λ V Ac

A BfKf– λI–[ ]V O=

Ac

λ λR i± λ I=

λR

c kν+

2m--------------– ξeqωeq–= =

λ I ωeq 1 ξeq2

–=

ωeq ω 1kd

ω2m-----------+

1 2⁄=

ξeq ξ ξ a+=

ξa

kν2mωeq----------------- ξ 1 ω

ωeq--------–

–=

kvkv


velocity feedback is a more appropriate mechanism for improving the stability ofa SDOF system.

Noting eqns (8.23) and (8.26), the total solution for time invariant linearfeedback can be expressed in a form similar to eqn (8.26).

(8.38)

The identity established in example 8.1 is also applicable here.

(8.39)

where A is a complex scalar, and are the solution of eqn (8.36). Requiring ensures that is bounded.

The above discussion assumes there is no delay between observing thestate and generating the force. In general, there is some delay and the force at timet is computed using data observed at an earlier time, . The force for linearnegative feedback is taken as

(8.40)

where represents the delay time. Delay introduces additional terms in andunder certain conditions, can cause to become positive and consequently, thesystem becomes unstable. Therefore, although ideal linear negative feedback isunconditionally stable, one needs to examine the potential destabilizing effect ofdelay for the actual control system. One procedure for investigating the effect oftime delay on the stability of a controlled time invariant SDOF system isdescribed in the following section. Additional approaches are discussed in latersections.

Effect of time delay on feedback control

Time delay in feedback control systems in the sum of the times required toexecute the following actions:

• acquire the data from sensors placed at different locations in thestructure;

• process the sensor data and calculate the control force;

X t( ) eAc t to–( )

Xo eAc t τ–( )

Bgag τ( ) Bpp τ( )+( ) τdto

t

∫+=

eAc t to–( )

Xo Aeλ t to–( )

V Aeλ t to–( )

V+≡

λ V,( )λR 0≤ e

Ac t to–( )

t td–

F t( ) KfX t td–( )–=

td λRλR


• transmit the control force signal to the actuator;

• ramp up the actuator to the desired force level.

The resultant time delay affects the synchronization between the control force andthe response of the system and may, under certain conditions, cause a significantdegradation in the performance of the control system which could result toinstability.

In what follows, an analytical procedure for assessing the stability of aSDOF system with time delay is presented. This approach follows the methodproposed by Agrawal et al. (1993). Results based on numerical simulations areincluded here to illustrate the effect of time delay on the response.

Assuming the feedback control algorithm consists of a linear combinationof displacement and velocity terms, and uses data associated with the time, ,the feedback force at time t is written as

(8.41)

Substituting for F in eqn (8.1), the governing equation for this case is

(8.42)

The general form of the homogeneous solution of eqn (8.42) is

(8.43)

where A is an arbitrary constant and satisfies

(8.44)

Letting represent the roots, the stability requirement is .

Expressing and in a convenient form is complicated by the presenceof the exponential terms. A first order approximation can be obtained byintroducing the following expansion

(8.45)

and retaining only the first 2 terms. The result is expressed in the same form aseqn (8.37) with and replaced with modified terms.

t td–

1m----F t( )

kdm-----u t td–( )–

kvm-----u t td–( )– gdu t td–( )– gvu t td–( )–= =

u t( ) 2ξωu t( ) ω2u t( ) gdu t td–( ) gvu t td–( )+ + + + p t( )m

----------=

u Aeλ t=

λ

λ2 2ξω gveλ td–

+ λ ω2 gdeλ td–

++ + 0=

λ λR iλ I±= λR 0≤

λR λ I

eλ td–

1 λ td–12--- λ td( )2 1

6--- λ td( )3– …+ +=

ξeq ωeq


(8.46)

The modified equivalent frequency and damping are related to the time delay by

(8.47)

(8.48)

Equation (8.46) is convenient for identifying behavioral trends.

For no initial damping and no velocity feedback ( ), theapproximation for reduces to

(8.49)

Since is positive, it follows that according to this first order approximation,displacement feedback with time delay produces unstable behavior for aninitially undamped system . The real part of corresponding to purevelocity feedback is estimated as

(8.50)

This result suggests that the response is stable when .

An improved approximation can be obtained by substituting the Pade

λR ξ'eqω'eq–=

λ I ω'eq 1 ξ'eq2

–=

ω'eq ω1 gd ω2⁄+

1 gvtd–--------------------------=

ξ'eq

ξ 12ω------- gv gdtd–( )+

1 gvtd–[ ] 1gd

ω2------+

----------------------------------------------------=

gv 0=λR

λR

gdtd2ω

------------kd td2ωm------------= =

λR

ξ 0=( ) λgd 0=( )

λR ωξ

gv2ω-------+

1 gvtd–--------------------– ω

ξkv

2ωm------------+

1kvtdm

----------–

----------------------–= =

kvtd m⁄( ) 1<


expansion

(8.51)

in eqn (8.44). The resulting expression is now a 3rd degree polynomial in

(8.52)

Introducing the notation for the equivalent frequency and damping definedearlier for the instantaneous feedback case (see eqn (8.37)),

(8.53)


(8.54)

For pure displacement feedback, . For pure velocity feedback, and .

Equation (8.54) is solved numerically for a specific SDOF system with aperiod of , and the two limiting cases of pure displacement and pure velocityfeedback with no initial damping. Figure 8.4 shows the movement of the poles forpure displacement feedback as a function of and . When , the polesmove on the imaginary axis with increasing . As increases, the path shiftsto the right, and when is about , the direction is essentially along thepositive real axis.

eλ td– 1 λ td 2⁄–

1 λ td 2⁄+------------------------- O λ td( )3[ ]+≈

λtd2-----λ3 1 td ξω

gv2

------– + λ2 2ξω gv

td2----- ω2 gd–( )+ + λ ω2 gd+ + + + 0=

ωeq2 ω2 gd+=

ξeqωeq ξω 12---gν+=

ξeq ξ ξ a+=

td2----- λ3 1 2ξω ξeqωeq–( )td+[ ]λ 2

+ +

2ξeqωeq td ω2 ωeq2

2---------–

+ λ ωeq2

+ 0=

ξeqωeq ξω=ωeq ω= ξa kν 2ωm( )⁄=

5s

ωeq td td 0=ωeq td

td 1s


Fig. 8.4: Variation of pole profile with under pure displacement feedback.

The effect of time delay for pure velocity feedback is illustrated by Fig. 8.5.For no time delay, increasing moves the poles further back in the negative realhalf plane until the state of critical damping is reached. As increases, the pathstend to bend toward the positive real half plane, and eventually intersect theimaginary axis. For a given value of , there may be a limiting time delaybeyond which the system is unstable. These observations are based on anapproximation and apply for a particular system, i.e. specific values of and .An exact analysis of the instability problem is presented in the next section.

−0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 0.5−2.5

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

2.5

Real axis, λR

Imag

inar

y ax

is,λ

Iωeq 2.0ω=

ωeq 2.0ω=

ωeq 1.5ω=

ωeq 1.5ω=

ωeq 1.0ω=

ωeq 1.0ω=

T 5s=ξ 0%=

td 0=

td 0= td 0.25=

td 0.25= td 0.5=

td 0.5=

td 1.0=

td 1.0=kv 0=

ωeq2

k kd+

m--------------=

td

kνtd

kν

ω ξ


Fig. 8.5: Variation of pole profile with under pure velocity feedback.

Stability analysis for time delay

Figure 8.5 shows that, for a given system, there is a particular value ofwhich corresponds to a transition in the behavior of . For greater than thisvalue, increases and eventually becomes positive. When , there is atransition from stable to unstable behavior, and the corresponding value ofdefines the stability limit for the system, i.e., for the particular combination of k, c,

, and .

Equation (8.44) is the definition equation for . Noting that the critical timedelay corresponds to , by expressing as

(8.55)

where is a real scalar, one can specialize eqn (8.44) for this case and determinethe critical value of . Substituting for leads to

(8.56)

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4−2.5

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

2.5

Real axis, λR

Imag

inar

y ax

is,λ

IT 5s=ξ 0%=

td 0=

td 0=

td 0.25=

td 0.25=

td 0.5=

td 0.5=

td 1.0=

td 1.0=

ξa 0.4=

ξa 0.4=

0.3

0.3

0.2

0.2 0.1

0.1

ξa

kν2ωm------------=

kd 0=

td

tdλR td

λR λR 0=td

kν kd

λλR 0= λ

λ iΩ=

Ωtd λ

Ω2– i2ξωΩ ω2 gde

itdΩ–igvΩe

itdΩ–+ + + + 0=


Replacing the exponential term by

(8.57)

yields

(8.58)

Equation (8.58) is satisfied when the real and imaginary terms vanish. Then,

(8.59)

(8.60)

It remains to solve these equations for in terms of , , , and .

Squaring both sides and adding the equations eliminates the trigonometricterms and results in a quartic equation for .

(8.61)

The roots of eqn (8.61) are given by

(8.62)

Since the poles correspond to , only the positive value of needs to beconsidered, resulting in two values of .

The next step is to determine . Noting the trigonometric identities,

(8.63)

eitdΩ–

tdΩ( )cos i tdΩ( )sin–=

Ω2– ω2 gd tdΩ( )cos gvΩ tdΩ( )sin+ + +

i 2ξωΩ gd tdΩ( )sin gvΩ tdΩ( )cos+–[ ]+ 0=

gd tdΩ( )sin gv– Ω tdΩ( )cos 2ξωΩ=

gd tdΩ( )cos gvΩ tdΩ( )sin+ Ω2 ω–2

=

td gd gv ω ξ

Ω

Ω4 4ξ2ω2 2ω2– gv

2–( )Ω2 ω4 gd

2–+ + 0=

Ω1 2,4ξ2ω2 2ω2

– gv2

–( )2

-------------------------------------------------–4ξ2ω2 2ω2

– gv2

–( )2

4---------------------------------------------------- ω4 gd

2–( )–+−+−=

iΩ± ΩΩ

tdΩ

tdΩ( )sin2

tdΩ2

----------tan

1 tan2 tdΩ2

----------+

------------------------------------=


(8.64)

eqn (8.59) can be expressed as

(8.65)

The two roots of eqn (8.65) are

(8.66)

Finally, the maximum time delay can be determined from

(8.67)

The minimum positive value of is the maximum allowable time delay.

For the limiting case of pure velocity feedback control of an undampedsystem (i.e. and ), eqn (8.59) reduces to

(8.68)

and it follows that

(8.69)

The expression for can be obtained from either eqn (8.60) or eqn (8.62),

(8.70)

where

(8.71)

Finally, the maximum delay can be expressed in terms of the fundamental period

tdΩ( )cos1 tan2 tdΩ

2----------–

1 tan2 tdΩ2

----------+

------------------------------------=

gvΩ 2ξωΩ–( )tan2 tdΩ2

---------- 2gd

tdΩ2

----------tan gvΩ 2ξωΩ+( )–+ 0=

tdΩ2

----------tangd– gd

2 gv2Ω2 4ξ2ω2Ω2

–++−

gvΩ 2ξωΩ–------------------------------------------------------------------------------=

td max

td max

2Ω----tan 1– gd– gd

2 gv2Ω2 4ξ2ω2Ω2

–++−

gvΩ 2ξωΩ–------------------------------------------------------------------------------

=

td max

gd 0= ξ 0=

gv tdΩ( )cos 0=

td max

π2Ω-------=

Ω

Ωω---- ξa 1 ξa

2++=

ξa

gν2ω-------

kν2ωm------------= =


of the uncontrolled system.

(8.72)

Figures 8.6 through 8.8 illustrate the effect of varying the displacementfeedback, velocity feedback, system damping, and system period on themaximum allowable time delay. Figure 8.6 shows plots of the maximumallowable time delay for an undamped SDOF system as a function of theactive damping ratio (i.e. velocity feedback) for three values of displacementfeedback . The central curve corresponds to . The lower curvecorresponds to , i.e. leading to an increase in the frequency of the controlledsystem, and shows that for no active damping. Furthermore, Fig. 8.6illustrates the effect of underestimating/overestimating the stiffness of the systemon the maximum allowable time delay. If the model used to establish the systemproperties underestimates the stiffness, the actual limit on the time delay will beless than the predicted limit and stability may be a problem.

Figure 8.7 shows the effect of the damping in the system on the maximumallowable time delay. In general, damping increases the allowable time delay. Theeffect of the fundamental period of the system on the maximum allowable timedelay is illustrated by Fig. 8.8. As expected from eqn (8.20), the maximumallowable time delay increases with period and decreases with active damping.Finally, Fig. 8.9 illustrates the degradation in performance, with increasing timedelay, of the SDOF system subjected to seismic excitation. Instability occursbeyond a time delay of which, according to Fig. 8.7, corresponds to themaximum allowable time delay for this level of damping. The instability is due tothe unbounded growth of the homogeneous solution, and will occur for anyarbitrary external excitation.

td maxT

----------------- 1

4 ξa 1 ξa2

++

----------------------------------------=

5sξa

kd kd 0=kd 0>

td 0=

1.1s


Fig. 8.6: Maximum allowable time delay as a function of and .


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.5

1

1.5

2

2.5

t dm

axT 5s=ξ 0%=

ξakv

2ωm------------=

ωeq 0.8ω=

ωeq 1.2ω=

ω

ωeq2 ω2

kdm-----+=

td maxkd kv

0 0.1 0.2 0.3 0.4 0.5 0.60

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t dm

ax

ξakv

2ωm------------=

T 5s=kd 0=

ξ 0%=

ξ 5%= 10%15% 20%25%

td maxkv ξ



Fig. 8.9: Degradation in performance of a SDOF system with time delay.

0 1 2 3 4 5 60

0.5

1

1.5

T - s

ξ 0=kd 0=

ξa 0%=

60%50%40%30%20%10%

70%

t dm

ax

td maxkv T

0 0.5 1 1.50

1

2

3

4

5

6

7

8

9

10

T 5s=ξ 2%=kd 0=ξa 20%=

El CentroTaft

td - s

ut d()

ut d

0=

()

⁄


8.3 Discrete time formulation - SDOF system

Governing equation

The continuous state-space linear formulation considers X and F to becontinuous functions of time that satisfy the following ordinary linear differentialequation and initial conditions,

(8.73)

In the case of a SDOF system, the coefficients are second order matrices involvingthe system properties,

(8.74)

The formulation presented in the previous section was based on the assumptionthat m, k, and c are constant. This restriction allows one to obtain eqn (8.26), theexact analytical solution expressed in terms of a convolution integral. Anumerical integration procedure is required to evaluate the convolution integralwhen the loading is a complex function such as a ground acceleration timehistory. If the system parameters and/or the feedback parameters are timedependent, an exact analytic solution of the equilibrium equation can not beestablished, and one must resort to generating an approximate solution with anumerical procedure such as a finite difference method which works with valuesof the variables at discrete points in time. A discrete time approach is alsonecessary for real time feedback control, since the control force is computed using“observed” values for the response at discrete time points.

X AX B f F Bgag Bpp+ + +=

X t 0=( ) Xo*=

A0 1km----– c

m----–

=

B f Bp

01m----

= =

Bg01–

=

8.3 Discrete Time Formulation- SDOF System 567

One generates a discrete time formulation by subdividing the time domaininto intervals, say , , , , and taking as unknowns thevalues of and at the discrete time points. The notation,

(8.75)

is convenient for representing these discrete variables. Equation (8.73) isapproximated at each time point by an algebraic equation relating the values ofand at that point and neighboring points, and is used to estimate the value ofat a later time. In what follows, the procedure is illustrated using a simpleapproximation for eqn (8.73).

When the system parameters are constant, the solution is given by eqn(8.26) which is listed below for convenience.

(a)

One can use this result to obtain an approximate solution between two timepoints, say and , by introducing assumptions for the variation ofthe force terms during the interval . The simplest model is based on using thevalues at the initial time .

Taking the time limits as and , assuming the force terms areequal to their value for ,

(8.76)

and noting the expansion for transforms the convolution integrals containedin eqn (a) to the following algebraic form,

(8.77)

to t1– t1 t2– . . . tn tf–X t( ) F t( )

X t j( ) X j≡

F t j( ) F j≡

XF X

X t( ) eA t to–( )

X to( ) eA t τ–( )BfF τ( ) τd

to

t

∫ eA t τ–( ) Bpp B+g

ag τ( )( ) τd

to

t

∫+ +=

t j t j 1+ t j ∆t+=∆t

t j

t t j 1+= to t j=t t j=

F τ( ) F t j( ) F j= =

p τ( ) p tj( ) pj= =

ag τ( ) ag t j( ) ag j,= =

eAt

X j 1+ eA∆tX j A 1– eA∆t I–( ) Bgag j, BfF j Bppj+ +[ ]+=


Equation (8.77) provides an estimate for based on data associatedwith the time point, . The first term on the right hand side is the exact freevibration response at , considering to be the initial state at . Theremaining terms represent the contribution of the “constant” loading terms overthe time interval, . Starting at t = 0 which corresponds to j = 0, one specifiesand computes . This process is repeated until the desired time is reached. Thecomputation will be bounded when eqn (8.75) is numerically stable. Stability isdiscussed in more detail later in the section.

This approach can also be applied to an adaptive system. The mass isconsidered to be constant, and therefore and are constant. Stiffness anddamping are assumed to be constant over a time interval, and to vary from oneinterval to another. Discrete values of stiffness and damping are defined for thetime interval, , as follows:

(8.78)

Since A is considered constant over an interval, eqn (a) is still applicable for theinterval. The resulting form of the discrete equilibrium equation is obtained byreplacing A with in eqn (8.77).

(8.79)

Starting at j = 0, one forms and determines . Then, A is updated to andused to compute . This process is continued for successive time points.

Linear negative feedback control

The discrete formulation represented by eqn (8.79) assumes properties andforces are constant over a time interval and are updated at the starting point of theinterval. A feedback law consistent with this assumption is

(8.80)

where is the feedback gain matrix at .

X j 1+t j

t j 1+ X j t j

∆t XoX1

B f Bg

t j t t j 1+<≤

k t( ) k t j( ) kj≡=

c t( ) c t j( ) c j≡=

A t( ) A t j( ) A j≡=

A j

X j 1+ eA j∆t

X j A j1– e

A j∆tI–( ) Bgag j, BfF j Bppj+ +[ ]+=

Ao X1 A1X2

t j t t j 1+<≤

Fj K f j, X j–=

K f j, t t j=


(8.81)

Substituting for F transforms eqn (8.79) to

(8.82)

One can also derive this expression by specializing eqn (8.38). The time historyresponse is generated by starting with , and computing , , ... etc. Keyissues here are the specification of: the time interval; the magnitude and temporaldistribution of the system stiffness and damping; and the feedback parameters,

and .

Stability analysis for time invariant linear feedback control

The numerical stability of the discrete feedback control algorithm isdetermined by examining the nature of the homogeneous solution. The analysispresented here assumes the coefficient matrices are constant, i.e., the system istime invariant. Specializing eqn (8.79) for this case, and considering no externalloading other than feedback leads to the governing equation

(8.83)

For no time delay, is taken as a linear function of . The effect of time delayis to shift the value of used to compute back to . For

, the control force is taken as 0.

(8.84)

Substituting for , the discrete equilibrium equation allowing for time delaytakes the following form

(8.85)

K f j, kd t j( ) kv t j( )=

X j 1+ eA j∆t

A j1– e

A j∆tI–( )B f K f– X j=

A j+ 1– eA j∆t

I–( ) Bgag j, Bppj+[ ]

Xo X1 X2

kd kv

X j 1+ eA∆tX j A 1– eA∆t I–( )B f F j+=

Fj X jX Fj X t j ν∆t–( ) X j ν–≡

j ν<

Fj 0= j ν<

Fj K f X j ν––= j ν≥

Fj

X j 1+ eA∆tX j= 0 j ν<≤

X j 1+ eA∆tX j A 1– eA∆t I–( )B f K f X j ν––= ν j≤


Considering first the case of no time delay, the free vibration response isdetermined with

(8.86)

This equation applies when the control force is updated at the discrete timepoints, and assumed constant during the time intervals. The corresponding formfor the case where the force is assumed to be continuously updated can beobtained by specializing eqn (8.38).

(8.87)

The solution generated with eqn (8.87) is the “exact” damped solution. Accordingto eqn (8.39), this solution is “bounded” for negative feedback applied to aninitially stable system. The error introduced by approximating the control force isrepresented by the difference between the coefficient matrices. It remains todetermine whether this “error” causes eqn (8.86) to became numerically unstablefor particular combinations of and , and therefore to generatehomogeneous solutions which increase rather than decrease with time. In whatfollows, a strategy for determining the numerical stability of first order matrixdifference equations such as eqns (8.86) and (8.87) is presented. The approach isintroduced using a simple scalar equation and is then generalized to handle ann’th order equation.

Consider the first order differential equation,

(8.88)

where a and b are constants and f = f(t). The exact solution is determined usingeqn (8.22).

(8.89)

This solution is bounded when a<0. Specializing the limits for t leads to thedifference form of (8.89)

X j 1+ eA∆t A 1– eA∆t I–( )B f K f–( )X j= j 0 1 2 …, , ,=

X j 1+ eA B f K f–( )∆t

X j=

K f ∆t

x ax bf+=

x t( ) ea t to–( )

x to( ) ea t τ–( )bf τ( ) τdto

t

∫+=


(8.90)

The homogeneous solution is “exact”, and therefore is bounded when the systemis stable, i.e., when a<0.

A similar solution procedure can be followed for the case where the forcingterm, f, is a continuous function of x. Assuming linear negative feedback,

(8.91)


(8.92)

The exact solution has the form

(8.93)

This solution is bounded when bk>a. Normally one starts with an initially stablesystem, (a<0), and adds negative feedback so that “boundness” is always ensured.

Equation (8.93) is based on continuous feedback. The solutioncorresponding to the following discrete feedback

(8.94)

is given by

(8.95)

where

(8.96)

Taking j=0, 1, 2, ... leads to

(8.97)

xj 1+ ea∆t x j ea t j 1+ τ–( )

bf τ( ) τdtj

t j 1+

∫+=

f t( ) kx t( )–= k 0>

x a bk–( )x=

xj 1+ e a bk–( )∆tx j=

f τ( ) kxj–=

t j τ t j 1+≤ ≤

xj 1+ cxj=

c ea∆t bka

----- 1 ea∆t–( )+=

xj c( ) jxo=


For to be bounded, the absolute magnitude of c must be less than 1.

(8.98)

Equation (8.98) represents a constraint on k and .

When the system is initially stable, a is negative,

(8.99)

Specializing eqn (8.98) for this case, and noting that the negative value controls,results in

(8.100)

which is written as:

(8.101)

Equation (8.101) defines the maximum value of k for a particular value of .

The variation of with is shown in Fig 8.10. For small ,the curve tends to as ; for large , the curve is bounded by

. If is less than 1, there is no restriction on the magnitude of. When, is greater than 1, the value of has to satisfy eqn (8.101).

Using Fig (8.10), one determines the limiting value of corresponding to thespecified value of . The range of allowable values for is between zeroand this “limiting” value. For example, suppose is equal to 1.6. Accordingto Fig (8.10), the maximum allowable value of is 1.4, and the range is

.

xj

c 1<

⇒

ea∆t bka

----- 1 ea∆t–( )+ 1<

∆t

a a–=

kba

----- 1 e a ∆t––( ) e a ∆t– 1<–

kba

----- 1 e a ∆t–+

1 e a ∆t––------------------------

kmaxb

a---------------≡<

∆t

kmaxb a⁄ a ∆t a ∆t∞ ∆t 0→ a ∆t

kmaxb a⁄ 1= kb a⁄∆t kb a⁄ ∆t

a ∆tkb a⁄ a ∆t

kb a⁄a ∆t

0 a< ∆t 1.4<


Fig. 8.10: Stability limits for

According to the analysis presented above, the scalar equation,, is numerically stable when . The matrix equation,

, has a similar constraint involving the matrix, c. Observing thatcan be expressed as

(8.102)

it follows that the elements of must be bounded as . A measure foris established by noting that the expression,

(8.103)

where and satisfy the characteristic equation for c,

(8.104)

is a solution of .

Suppose c is of order 2. There are 2 eigensolutions which may be real orcomplex. When the solution is complex, the individual solutions are complexconjugates, and the total real solution is a linear combination of these j solutions,

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

1

2

3

4

5

6

7

8

9

10

kmaxb

a---------------

a ∆t

eqn 8.101( )

xj 1+ cxj=

xj 1+ cxj= c 1<X j 1+ cX j= X j

X j c j Xo=

c j j ∞→ c j

X j ρ jΨ=

ρ Ψ

c ρI–( )Ψ 0=

X j 1+ cX j=


(8.105)

where , are complex constants. These constants are evaluated using theinitial conditions for .

(8.106)

Substituting for A and ,

(8.107)

results in 2 scalar equations for and .

(8.108)

The complex eigenvalue can be expressed as

(8.109)

where is the magnitude of and is the polar angle.

(8.110)

Using the polar form for , the j’th power is given by

(8.111)

Introducing the expressions for , , and , the solution for expands to

(8.112)

Since the terms within the brackets are bounded, it follows that is finite as when the modulus of is less than unity,

(8.113)

X j Aρ jΨ Aρ j Ψ+=

A AXo

Xo AΨ AΨ+=

Ψ

A AR iAI+=

Ψ ΨR iΨI+=

AR AI

2 ARΨR AIΨI–( ) Xo=

ρ ρR iρI+ ρe iθ= =

ρ ρ θ

ρ ρR2 ρI

2+[ ] 1 2⁄=

θtanρIρR------=

ρ

ρ j ρ( ) j ei jθ( )=

A Ψ ρ X j

X j ρ( ) j 2 ARΨR AIΨI–( ) jθ( ) 2 ARΨI AIΨR+( ) jθ( )sin–cos[ ]=

j 1 2 …, ,=

X jj ∞→ ρ

ρ 1<


Extending this analysis to the case where c is of order 2n, there may be npairs of complex conjugates solutions having the same form as eqn (8.105), andthe total solution is given by

(8.114)

Defining as the maximum modulus for the set of eigenvalues of c, thestability requirement is

(8.115)

This requirement also applies when some of the eigenvalues are real. For thiscase, the maximum absolute value must be less than unity.

In general, c depends on the nature of the finite difference approximation,the system parameters, and the magnitude of the time step, . For the case oftime invariant negative linear discrete feedback with no time delay, the form of cfollows from eqn (8.82) specialized for an external loading:

(8.116)

Given , , and , one selects a value for , determines the correspondingeigenvalues of c, and then adjusts the value of if is greater than 1. Thecomplexity of the equation for c necessitates the use of a numerical eigenvalueroutine such as the eig( ) function in MATLAB. The following example illustratescomputational details and presents the stability limit for a SDOF system.

Example 8.2: Stability analysis - SDOF system with no time delay

The various matrices for the SDOF case with negative velocity feedback

X j AlρljΨl Alρl

lΨl+( )

l 1=

n

∑=

ρl( ) j 2 AR l, ΨR l, AI l, ΨI l,–( ) jθl( )cos[

l 1=

n

∑=

2 AR l, ΨI l, AI l, ΨR l,+( ) jθl( ) ]sin–

ρmax

ρmax 1<

∆t

c eA∆t A 1– eA∆t I–( )B f K f–=

A B f K f ∆t∆t ρmax


are:

(1)

where is a dimensionless parameter.

(2)

When , the expansion of can be expressed as

(3)

Using eqn (3), one can obtain analytical expressions for the elements of c. Theresulting form of c is

(4)

Evaluating the eigenvalues of c leads to

(5)

Equation (5) shows that the eigenvalues depend on the dimensionless parameter,

A 0 1

ω2– 2ξω–= A 1–

2ξω------– 1

ω2------–

1 0

=

A∆t 0 ωω----

ωω– 2ξω–

=

B f K f

0 0

0kνm-----

0 00 2ξaω

= =

ω

ω ω∆t 2π∆tT------= =

ξ 0= eA∆t

eA∆t ωI 1ω---- ωsin A+cos=

cωcos 1

ω---- ω 2ξa ω 1–cos( )+sin( )

ω ωsin– ω 2ξa ωsin–cos=

ρ1 2, b1 b12 b2–( )1 2⁄±=

b1 ω ξa ωsin–cos=

b2 1 2ξa ωsin–=


, and the active damping ratio, .

Figure 1 contains plots of the absolute values of the eigenvalues vs.for a representative range of . The curves intersect at , whichcorresponds to a transition in behavior. When , the eigenvalues arecomplex and . For the range , the eigenvalues start out ascomplex quantities, and then shift over to real quantities at a particular value of

which depends on . The curve bifurcates at this value. When , thebifurcation occurs at . The intersection of the upper branch with

defines the limiting value of for the corresponding value of . Thisvalue corresponds to and is related to by .Although for and arbitrary , the modulus is greater than 1 for

and . Therefore, represents the “limit” for anundamped system.

Figure 1

The effect of including damping in the system is to offset the destabilizingeffect of feedback. Figure 2 illustrates this trend. Increasing lowers and

ω 2π ∆t T⁄( )= ξa

∆t T⁄ξa ∆t T⁄ 0.5=

∆t T⁄ 0.5≥ρ 1> 0 ∆t T⁄ 0.5<<

∆t T⁄ ξa ξa 0=∆t T⁄ 0.5=

ρ 1= ∆t T⁄ ξaρ 1–= ξa ω 2ξa ωsin–cos 1–=

ρ 1= ξ 0= ∆t T⁄ξa 0> ∆t T⁄ 0.5> ∆t T⁄ 0.5=

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

ρ

∆t T⁄

ξ 0=

ξa 0.4=

0.00.10.20.3

ξa 0.0=

0.10.20.30.4

kd 0=

ξ ρ


decreases the range of for which is greater than 1. When , c reduces to. The eigenvalues of are related to the eigenvalues of A by

(7)

Noting that , eqn (7) expands to

(8)

and it follows that

(9)

Equation (9) explains the trend for to decrease with increasing . The secondterm in eqn (8) is the source of the transition from real to complex eigenvalues.This shift occurs when .

a)

ω ρ ξa 0=eA∆t eA∆t

ρ eλ∆ t=

λ ξω– i 1 ξ2–( )ω±=

ρ e ξω– e i 1 ξ2–( )1 2⁄ ω±[ ]=

ρ e ξω–=

ρ ω

1 ξ2–( )1 2⁄ ω π=

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

∆t T⁄

ρ

ξ 0.05=

ξa 0.4=0.30.20.10.0

ξa 0.0=

0.10.20.30.4

kd 0=

ξ 0.05=


b)

Figure 2

The limits on depend on both and . As Fig 2 shows, has asignificant effect on , particularly in the region, . When , thisregion is stable even for , which represents significant active damping. Aconservative strategy would be to require to be less than the value forwhich the upper branch first intersects . These values are listed in Table 1.

Table 1: Stability limit for

Figure 3 illustrates the onset of instability for a SDOF system having

0 0.50 0.44 0.38 0.33 0.29

0.05 0.48 0.38 0.33 0.29

0.10 0.39 0.34 0.29

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

ρ

∆t T⁄

ξ 0.1=

ξa 0.4=0.30.20.10.0

ξa 0.0=

0.10.20.3

0.4

kd 0=

ξ 0.1=

∆t T⁄ ξ ξ a ξρ ∆t T⁄ 0.5> ξ 0.1=

ξa 0.4=∆t T⁄

ρ 1=

∆t T⁄

ξ ξa 0= ξa 0.1= ξa 0.2= ξa 0.3= ξa 0.4=

∞

∞ ∞


. The discrete time history response is obtained by evaluating eqn(8.86) for j ranging from 0 to 100. Increasing from 0.33T to 0.34T causes the freevibration response of the initially displaced system to shift from stable to unstablebehavior.

a)

X 0( ) 1 0, =∆t

0 10 20 30 40 50 60 70 80 90 100−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

ξ 0.1=

ξa 0.3=

∆tT------ 0.33=

Dis

plac

emen

tu,

Discrete time point

kd 0=


b)

Figure 3

The stability analysis allowing for time delay starts with eqn (8.86) which iswritten here in a simpler form.

(8.117)

where

(8.118)

(8.119)

The previous analysis corresponds to and . Given and thedamping parameters, and , one needs to establish the limiting value of the

0 10 20 30 40 50 60 70 80 90 100−5

−4

−3

−2

−1

0

1

2

3

4

5

Dis

plac

emen

tu,

Discrete time point

ξ 0.1=ξa 0.3=∆tT------ 0.34=

X j 1+ c1X J c3X j ν–+=

c1 eA∆t=

c3 A 1– eA∆t I–( )B f K f–=

ν 0= c1 c3+ c= νξ ξ a


ratio of the time increment to the fundamental period, . One option is toemploy numerical simulation and solve eqn (8.117) for a range of values of .Figure 8.11 illustrates the stability transition for and a specific set ofparameters. The corresponding results for are shown in Figure 3 ofexample 8.2. In this case, the maximum time step is reduced by about 50%. Theadvantage of numerical simulation is that it allows arbitrary values of , , and

to be considered. The disadvantage is that the search can be inefficient if thestarting point is not reasonably close to the solution. One can start with thesolution for and decrease . Another option is to utilize a simplerformulation based on to generate an estimate of the starting value. Thelatter option is described below.

∆t T⁄∆t T⁄

ν 1=ν 0=

ξ ξaν

ν 0= ∆t T⁄ξ 0=

0 50 100 150 200 250 300 350 400−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

ν 1=

ξ 0.1=

ξa 0.3=

∆tT------ 0.15=

Dis

plac

emen

tu

,

Discrete Time Pointsa)


Fig. 8.11: Illustration of instability due to time delay

The solution of eqn (8.117) can be expressed as

(8.120)

where is a complex scalar, and is an unknown vector. In order for thesolution to be stable, the absolute magnitude of must be less that 1. Substitutingfor , eqn (8.117) takes the following form

(8.121)

When , the coefficient matrix reduces to (see example 8.2)

(8.122)

0 50 100 150 200 250 300 350 400−1500

−1000

−500

0

500

1000

1500

Discrete Time Points

Dis

plac

emen

tu

,

ν 1=

ξ 0.1=

ξa 0.3=

∆tT------ 0.16=

b)

X J ρ jΨ=

ρ Ψρ

X j

c3 ρνc1 ρν 1+ I–+( )Ψ 0=

ξ 0=

ρν ω ρν 1+–cos ρν ωsinω

------------ 2ξaω 1–cosω

----------------------+

ρνω ωsin– ρν ω 2ξa ω ρν 1+–sin–cos


For to have a non-trivial value, the determinant of the coefficient matrix mustvanish. Enforcing this requirement leads to an equation of degree in .

(8.123)

where

(8.124)

Given and , one evaluates the roots of =0 for a representative range of. The following example illustrates this computation.

Example 8.3: Stability analysis -SDOF system with time delay

Figure 1 contains plots of vs. for a range of and . Asexpected, increasing the time delay lowers the magnitude of the time interval atwhich instability occurs. The effect of time delay becomes more significant as theactive damping is increased. Table 1 contains the limiting values of for

and ranging up to 0.4. Both active damping and time delay result in adecrease in the allowable time step. Including passive damping has the oppositeeffect on the time increment, i.e., it increases the allowable time increment. Forexample, the critical time step ratio for , is 0.15 for vs.0.124 for .

Ψν 2+ ρ

p ρ( ) ρν 2+ aρν 1+ ρν bρ b–+ + + 0= =

a 2 ωcos–=

b 2ξa ωsin=

ν ξa p ρ( )ω 2π ∆t T⁄( )=

ρ ∆t T⁄ ξa ν 1 2,=

∆t T⁄ξ 0= ξa

ν 1= ξa 0.3= ξ 0.1=ξ 0.0=


Figure 1

0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.20.6

0.7

0.8

0.9

1

1.1

1.2

1.3

1.4

∆t T⁄

ρ

a) ν 1=

ξ 0=ν 1=

ξa 0.5=

0.4

0.3

0.2

0.1

0.0

0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.20.6

0.7

0.8

0.9

1

1.1

1.2

1.3

1.4

∆t T⁄

ρ

b) ν 2=

ξa 0.5=0.40.3

0.2

0.1

ξ 0=ν 2=


Table 1: Limiting values of for

The actual time increment is governed by the responses of the monitoringand actuator systems. Given data on these response times, one can evaluate theexpected ratio, , for a particular system and determine, using Table 1 as astarting point for numerical simulation, whether stability will be a problem.

8.4 Optimal linear feedback - time invariant SDOF systems

Quadratic performance index

The previous sections were concerned with presenting the state-spaceformulation and numerical procedures for evaluating the time history response ofSDOF systems having linear time invariant feedback. Negative velocity feedbackwas shown to be equivalent to damping, and to result in stable behavior when itis applied in a continuous manner with no time delay. The effect of time delay canbe destabilizing under certain conditions that depend on the magnitude of thefeedback force and the delay time. Discrete time control works with response dataat discrete time points and considers the feedback force to be constant over a timeinterval. Instability can also occur if the time interval exceeds a critical value thatdepends on the level of “feedback” induced damping and the time delay. Given

Stability limit for

0 0.50 0.500 0.5000

0.05 0.158 0.0955

0.10 0.44 0.151 0.0908

0.15 0.144 0.0860

0.20 0.38 0.137 0.0822

0.25 0.131 0.0780

0.30 0.33 0.124 0.0745

0.35 0.118 0.07085

0.40 0.29 0.113 0.0677

∆t T⁄ ξ 0=

∆t T⁄

ξa ν 0= ν 1= ν 2=

∆t T⁄

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 587

the feedback parameters, and , and the delay time, one can determine thelimiting time interval using the procedure described in the previous section. Theremaining issue that needs to be addressed here is the selection of the magnitudeof the feedback parameters. A strategy similar to the approach followed for quasi-static control is introduced in this section. The key step is the formulation of aperformance measure which provides a rational basis for comparing differentchoices for the feedback parameters and establishing the “optimal” solution.

The quasi-static formulation presented in Chapter 7 was concerned withcontrolling a displacement profile, and worked with a quadratic performancemeasure formed by integrating the square of the displacement error over thestructural spatial domain. This measure was extended to reflect the cost of thecontrol forces by incorporating an additional quadratic force term, andintroducing weights for the displacement and force terms. By adjusting theweights, one can assess the effect of assigning more emphasis to either thedisplacement error or the force magnitude.

Dynamic control is concerned with controlling the response of the systemover a specified time period. Starting at some initial state, X(0), the objective is tominimize the deviation between the actual response, X(t), and the desiredresponse, , by applying a control force F(t) over the time interval.Considering the integral of the square of the difference between the actual anddesired responses to be the measure of the “closeness” of the solution, andincluding a cost associated with the control force leads to the following quadraticperformance index for a SDOF system,

(8.125)

where is the time period over which the performance is being measured, andare the weighting functions for the displacement, velocity and force

terms. Specializing F for linear time invariant feedback transforms J to:

(8.126)

where and are constants.

When the system and weighting parameters are specified, J reduces to a

kd kv

X * t( )

J 12--- qd u u*–( )2 qv u u*–( )2 rF2+ + td

0

t*

∫=

t*

qd qv r, ,

J 12--- qd u u*–( )2 qv u u*–( )2 r kdu kvu+( )2+ + td

0

t*

∫=

kd kv


function of the feedback parameters, the external excitation, the desired response,and the time period. The optimal feedback for a specific combination ofexcitation, desired response, and time interval, is defined as the set of values for

and which correspond to a stationary value of J specialized for theseconditions. They are determined with the following equations,

(8.127)

where is J evaluated for specific , , and .

In general the response depends on and , as well as the systemparameters and external excitation. One option is to evaluate numerically anduse simulation to locate the “optimal” values for and . This approach is notvery efficient since it requires ranging over both and . Another option is toapply variational calculus methodology and transform the stationaryrequirement to a differential equation.

In the following section, an example illustrating the first option ispresented. This example deals with the special case where is taken as 0, theresponse is due to an initial velocity at t=0, and the time interval extends out toinfinity. These conditions are typical for the regulator problem which is concernedwith applying control forces to take a system from an arbitrary non-zero initialstate , to a final state at time . The regulator problem is encounteredwhen a system is perturbed from its steady-state equilibrium position and there isa need to eliminate the unwanted perturbation as quickly as possible. Since therequired force increases as the time interval decreases, it is necessary to includethe force cost as well as the response magnitude in the performance index in orderto obtain a realistic solution.

Subsequent sections describe the variational calculus formulation of thestationary requirement for J as a differential equation, and illustrate itsapplication. This derivation is extended to deal with multi-degree of freedomsystems later in the Chapter.

An example - linear quadratic regulator control algorithm

Suppose a single degree of freedom system is subjected to an impulse att=0. Considering only velocity feedback, the response is given by

(8.128)

kd kv

kd∂∂ J* 0=

kv∂∂ J* 0=

J* u u* t*

kd kvJ*

kd kvkd kv

u*

Xo X f t f

uuoλ I-----e

λRt λ Itsin=


where is the initial velocity, and the lambda terms depend on the systemproperties and the velocity feedback parameter.

(8.129)

Interpreting u as a disturbance from the steady-state position, , theobjective is to reduce u to a small quantity, i.e., to regulate the response, in aspecified time interval. The linear quadratic regulator control algorithmspecialized for only velocity feedback is based on the following form of J,

(8.130)

and determines the optimal value of by enforcing the stationary requirementwith respect to ,

(8.131)

Substituting for u in eqn (8.130), considering the weighting functions to beconstant, and expressing in terms of leads to

(8.132)

The form of the terms within the square bracket suggest the following scalefactors for the weights:

(8.133)

With these definitions, all the superscripted weights have the units of (1/velocity)2 and the performance index simplifies to

(8.134)

Finally, enforcing the stationary requirement on J with respect to results in the

uo

λR ξeqω–=

λ I ω 1 ξeq2–[ ] 1 2⁄=

ξeq ξkv

2ωm------------+ ξ ξ a+= =

u* 0=

J 12--- qdu2 qvu2 r kvu( )2+ + td

0

∞

∫=

kvkv

kv∂∂J 0=

kv ξa8ωuo

2------- J 1

ξ ξ a+( )-------------------

qd

ω2------ qv 4m2ω2rξa

2+ +=

qd ω2qd=

qv qv=

r r4m2ω2------------------=

8ωuo

2------- J

qd qv+

ξ ξ a+----------------- r

ξa2

ξ ξ a+--------------+=

ξa


following equation for the optimal active damping ratio,

(8.135)

Figure 8.12 shows the variation of with the weighting parameters. Increasingplaces more emphasis on minimizing the control force, and decreases.

Increasing the displacement and velocity weights places more emphasis ondecreasing the response, and the result is an increase in .

Fig. 8.12: Variation of the optimal velocity feedback with the weightingparameters

The previous analysis considered the weighting functions to be constantover the time period during which the performance index is being evaluated. Thisrestriction fixes the relative priorities on the various terms that contribute to theindex. If the weights are allowed to vary with time, the priorities can be adjusted

ξa optimalξ2

qd qv+

r-----------------+

1 2/ξ–=

ξar ξa

ξa

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

ξa opt

qd qv+

r-----------------

qd ω2qv+

4ω4m2r------------------------=

ξ 0=

ξ 0.1=

ξ 0.2=


as to place more emphasis on a certain term at a particular point in time.

For example, one may want to suppress the response more aggressivelylater in the period. This shift in emphasis can be achieved by taking the weights as

(8.136)

where is a time constant that defines the transition for the priority shift. It isconvenient to express in terms of the fundamental period for the system and ascaling factor, .

(8.137)

The performance index for this case has the following form,

(8.138)

Requiring J to be stationary with respect to results in a cubic polynomial forthe optimal value of .

(8.139)

Figure 8.13 contains plots corresponding to and various values of . Alsoshown is the plot for the uniform (time) weighting case defined by eqn (8.135).Assuming , , and are fixed, the effect of increasing is to reduce theoptimal value of . This trend is due to the decrease in magnitude of theweighted response during the initial phase of the response where the actualresponse magnitude is a maximum. Lowering this term decreases the demand onthe control force, and consequently the required value of decreases.

qd ω2 tT*------qd=

qvt

T*------qv=

r r4m2ω2------------------=

T*

T*

α

T* αT 2παω

----------= =

8ωuo

2------- J 1

4πα---------- 2qd

qd qv+

ξ ξ a+( )2----------------------+

rξa

2

ξ ξ a+--------------+=

ξaξa

ξa ξa ξ+( ) ξa 2ξ+( ) 12πα----------

qd qv+

r-----------------– 0=

ξ 0= α

qd qv r αξa

ξa


a)

b)

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.040

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

ξa opt

qd qv+

r-----------------

ξ 0=

α 14---= α 1

2---=

α 1= α 2=

eqn 8.135( )

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.180

0.05

0.1

0.15

0.2

0.25

0.3

0.35

ξa opt

qd qv+

r-----------------

ξ 0=

α 14---=

α 12---=

α 1=

α 2=

eqn 8.135( )


c)

Fig. 8.13: Variation of the optimal velocity feedback with the time scale factor,

The continuous-time algebraic Riccati equation

The previous example illustrated the application of the linear quadraticregulator algorithm for a particular initial state and velocity feedback. In whatfollows, the performance index is expressed in matrix form, and the analysis isextended to be applicable for both displacement and velocity feedback, andarbitrary initial conditions.

The scalar form of the performance index for the linear quadratic regularproblem is obtained by setting in eqn (8.126) and taking .

(8.140)

0 0.2 0.4 0.6 0.8 1 1.2 1.40

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

ξa opt

qd qv+

r-----------------

ξ 0=

α 14---= α 1

2---=

α 1=

α 2=

eqn 8.135( )

α

u* u* 0= = t* ∞=

J 12--- qdu2 qvu2 r kdu kvu+( )2+ +[ ] td

0

∞

∫=


Introducing the state vector notation, J is expressed as

(8.141)

where Q and R are diagonal weighting matrices, and is the linear statefeedback matrix.

(8.142)

The governing differential equation for free vibration response follows from eqn(8.33).

(8.143)

Solutions of eqn (8.143) satisfy

as (8.144)

when the feedback parameters are selected such that the eigenvalues of have a negative real part.

When the weighting matrices are taken to be constant, an explicitexpression for J can be obtained by expressing the integrand as

(8.145)

where H is a symmetric positive-definite second order matrix. Noting eqn (8.143),the right hand side expands to

(8.146)

and it follows that

(8.147)

With this change in variables, the integral reduces to

J 12--- XT Q K f

TRK f+( )XT td0

∞

∫=

K f

Qqd

qv

=

R r 1[ ]=

K f kd kv=

X A B f K f–( )X AcX= =

X t( ) 0→ t ∞→

kv kd,( )Ac

XT Q K fTRK f+( )X

tdd XTHX( )–=

tdd XTHX( ) XTHX XTHX+ XT Ac

TH H Ac+( )X= =

AcTH H Ac+ Q K f

TRK f+( )–=


(8.148)

Requiring J to be stationary with respect to and noting that isindependent of leads to . The differential of H is generated byoperating on eqn (8.147).

(8.149)

Substituting for ,

(8.150)

and setting , the stationary condition takes the form

(8.151)

For eqn (8.151) to be satisfied for arbitrary , the coefficients must vanish.Then the optimal feedback is related to H by

(8.152)

The last step involves substituting for in eqn (8.147). This operation leads tothe following definition equation for H, which is called the continuous timealgebraic Riccati equation

(8.153)

Given the system parameters , one specifies the weights (Q and R), andsolves for the elements of H. Note that by definition, H is symmetric and positivedefinite. With H known, the feedback matrix is determined with eqn (8.152).

Substituting for A and ,

(8.154)

and expressing H as

J 12---XT 0( )HX 0( )=

K f X 0( )K f δH 0=

AcTδH δH Ac δAc

TH HδAc+ + + δK fTRK f– K f

TRδK f–=

δAc

δAc B f δK f–=

δH 0=

δK fT B f

TH RK f–( ) HB f K fTR–( )δK f+ 0=

δK f

R 1– B fTH K f optimal

=

K f

ATH HA HB f R 1– B fTH–+ Q–=

A B f,( )

B f

A 0 1

ω2– 2ξω–=

B f

01m----

=


(8.155)

in eqn (8.153) yields the following 3 equations for the elements of H

(8.156)

The elements of H are also constrained by the requirement that H must be positivedefinite.

(8.157)

Given H, the elements of are determined with eqn (8.152).

(8.158)

Solving for and , and enforcing the positive definite requirementleads to

(8.159)

The equivalent frequency and damping for the SDOF system with feedbackcorresponding to eqn (8.158) are determined with eqn (8.37) and (8.158).

HH11 H12

H12 H22

=

1rm2---------- H12

2 2ω2H12 qd–+ 0=

1rm2---------- H12H22 H11– 2ξωH12 ω2H22+ + 0=

1rm2---------- H22

2 4ξωH22 2H12– qv–+ 0=

H11 0> H22 0> H11H22 H122 0>–

K f

kd

H12rm----------=

kv

H22rm----------=

H12 H22

H12 rω2m2 1– 1qd

rω4m2-----------------+

1 2/+

=

H22 2rωm2 ξ– ξ2qv

4rω2m2--------------------

H12

2rω2m2--------------------+ +

1 2/

+=


(8.160)

Weighting the displacement introduces additional stiffness and increases thefrequency. Weighting the velocity generates additional damping, but has no effecton the frequency. It follows that pure velocity feedback corresponds to

. The active damping ratio for this case is given by:

(8.161)

This result coincides with eqn (8.135) which was generated by integrating theanalytic solution for the impulse generated response. These solutions mustcoincide since the approach followed earlier is a simplified version of thisprocedure.

The matrix formulation presented above is also applicable to multi-degreeof freedom systems. MATLAB has a function called CARE (Continuous -timeAlgebraic Riccati Equation) which generates H, , and the eigenvalues of

for specified values of A, , Q and R. One option for scaling Q and Ris to use the factors defined by eqn (8.133). The corresponding forms are

(8.162)

where , , and range from 0 to . MATLAB’s default option is to assume. In this case, taking Q as:

ωeq ω 1kd

ω2m-----------+

1 2/ω 1

qd

rω4m2-----------------+

1 4/⇒=

ξeq ξ ωωeq--------

kv2mωeq----------------- ω

ωeq-------- ξ2

qv

4rω2m2--------------------

12--- 1– 1

qd

rω4m2-----------------+

1 2/++ +

1 2/

⇒+=

qd 0 qv 0≠,=

ξa

H22

2rm2ω----------------- ξ– ξ2

qv

4rω2m2--------------------+

1 2/

+= =

K fA B f K f– B f

Qqd

qv

ω2qd 0

0 qv

= =

R r[ ] r4m2ω2------------------= =

qd qv r 1≈R I≡


(8.163)

maintains the same range for and .

The form of the continuous algebraic Riccati solution corresponding to r = 1 and, defined by eqn (8.163) is

(8.164)

A similar scaling strategy can be employed for a multi-degree-of-freedom system.In this case the magnitude of the weighting factor for each of the displacementand velocity variables can be independently assigned.

The discrete time algebraic Riccati equation

The discrete time formulation of the algebraic Riccati equation is based ona performance index that involves a summation of weighted response termsevaluated at discrete times. The discrete form of the performance indexcorresponding to the same assumptions introduced for the continuous timeformulation is

(8.165)

where is determined with eqn (8.86) which is written here as

(8.166)

Taking , eqn (8.166) reduces to

(8.167)

QR I≡

4ω4m2qd 0

0 4m2ω2qv

=

0 1→( ) qd qv

qd qv

kdk----- 1– 1 4qd+[ ] 1 2/+=

kv2ωm------------ ξa ξ– ξ2 qv

kd2k------+ +

1 2/+= =

J 12--- X j

T Q K fTRK f+( )X j

j 0=

∞

∑=

X j

X j 1+ c1X j c2F j+=

c1 eA∆t= c2 A 1– eA∆t I–( )B f=

F j K f X j–=

X j 1+ c1 c2K f–( )X j cX j= =


Finally, eqn (8.165) is expressed as

(8.168)

where H is a symmetrical positive definite matrix. Stability requires thatapproach 0 as . Noting this constraint, the expression for H reduces to (seeproblem 8.11)

(8.169)

Imposing the stationary requirement on J leads to

for arbitrary

(8.170)

Operating on eqns (8.169) and (8.166),

(8.171)

(8.172)

and setting results in the following expression for the optimal feedbackmatrix

(8.173)

The final form of the discrete algebraic Riccati equation is

(8.174)

Given m, k, c, and , and can be evaluated. One specifies theweighting matrices R and Q and then determines H. MATLAB has a functioncalled DARE which computes H, , and the eigenvalues of c specialized for

set equal to the optimal value. The scaling strategies discussed earlier for thecontinuous time case are also applicable for the discrete time formulation. DAREassumes the default value of R to be I and therefore eqn (8.163) can be used todetermine the corresponding value of Q. The following example describes the

J 12---Xo

T c j( )T Q K fTRK f+( )c j

j 0=

∞

∑

Xo12---Xo

THXo= =

c j

j ∞→

H cTHc– Q K fTRK f+=

δJ 12---Xo

TδHXo 0= = δK f

⇒

δH 0=

δH cTδHc– δcTHc– cTHδc– δK fTRK f K f

TRδK f+=

δc c2δK f–=

δH 0=

K f optimalR c2

THc2+( ) 1– c2THc1=

H c1THc1– c1

THc2( ) R c2THc2+( ) 1– c1

THc2( )T+ Q=

∆t c1 c2

K f optK f


solution procedure in considerable detail.

Example 8.4: Solution of the discrete time algebraic Riccati equation for aSDOF system

The undamped case is considered first. Noting the results presented inExample 8.2, the various coefficient matrices are:

(1)

(2)

(3)

(4)

where

(5)

(6)

(7)

c1ωcos ωsin

ω------------

ω ωsin– ωcos

=

c21

mω2----------- 1 ωcos–

ω ωsin=

c1THc2

1mω2-----------

f 1

f 2

=

c2THc2

1m2ω4-------------- f 3=

ω ω∆t 2π∆tT------= =

f 1 H11 ωcos( ) 1 ωcos–( ) H12 ω ωsin( ) 2 ω 1–cos( )+=

H22 ω ωsin( )2–

f 2 H11ω 1 ωcos–( )sin

ω---------------------------------------- H12 2 ωsin( )2 1– ωcos+[ ]+=

H22 ω ω ωsincos( )+


(8)

The optimal feedback matrix is determined using eqn (8.173).

(9)

Finally, the elements of H are obtained by expanding eqn (8.174). This stepleads to the following 3 scalar equations:

(10)

The above equations are reduced to a dimensionless form by scaling thevarious terms according to the following laws

f 3 H11 1 ωcos–[ ] 2 H12 2ω ω 1 ωcos–( )sin[ ]+=

H22 ω ωsin[ ] 2+

K f kd kv1

mω2 rf 3

m2ω4--------------+

----------------------------------------- f 1 f 2= =

ωsin( )2H11 2ω ω ωcossin[ ]H12 ω2 ωsin2[ ]H22–+

qd

f 12

f 3 ω4m2r+------------------------------–=

ω ωcossinω

--------------------------– H11 2 ωsin2[ ]H12 ω ω ωcossin[ ]H22+ +

f 1 f 2–

f 3 ω4m2r+------------------------------=

ωsin2

ω2---------------– H11

2 ω ωcossinω

------------------------------ H12– ωcos2–[ ]H22+

qv

f 22

f 3 ω4m2r+------------------------------–=


(11)

Using eqn (11), the expression for the optimal feedback matrix takes the form

(12)

Equations (10) reduce to

(13)

Lastly, the terms are determined with

(14)

One specifies the relative weights , and the relative time step ratio,, solves eqn (13) for the elements of , computes the optimal

feedback with eqn (12), and lastly checks for stability by evaluating the

f 1 ω4m2r f 1= qd ω4m2rqd=

f 2 ω3m2r f 2= qv ω2m2rqv=

f 3 ω4m2r f 3=

H11 ω4m2rH11=

H12 ω3m2rH12=

H22 ω2m2rH22=

K fω2m

1 f 3+--------------- f 1

1ω---- f 2=

H11 H22–( ) ω 2 ω ωH12cossin+sin2 4qdf 1

2

1 f 3+---------------–=

H11 H22–( ) ω 2 ω ωH12cossin+sin2 4 q– vf 2

2

1 f 3+---------------+=

H11 H22–( ) ω ωcos 2 ωH12sin2–( )sinf 1 f 2

1 f 3+---------------=

f

f 1 H11 ω 1 ωcos–( ) H12 ω 2 ω 1–cos( ) H22 ωsin2–sin+cos=

f 2 H11 ω 1 ωcos–( ) H12 2 ωsin2 1– ωcos+[ ]+sin=

H22 ω ωcossin+

f 3 H11 1 ωcos–( )2 H12 2 ω 1 ωcos–( )sin( )+=

H22 ωsin2+

qd qv,( )ω 2π ∆t T⁄( )= H


eigenvalues of .

As , the solution for the discrete formulation approaches thesolution for the continuous case. This can be shown by introducing second orderapproximations for , and letting approach 0. Assuming , thetrigonometric terms are replaced with

(15)

The corresponding equations are

(16)

and

(17)

Based on eqn (17), the limiting solution is

(18)

Noting eqn (16), the limiting forms for are

(19)

Finally, the feedback terms are

c1 c2K f–

∆t 0→

ωcos ωsin ω ω2<< 1

ω 1≈cosω ω≈sin

f 1 ωH12 O ω2( )+≈

f 2 ωH22 O ω2( )+≈

f 3 ω2H22 O ω3( )+≈

ω2 H11 H22–( ) 2ωH12 4qd ω2H122–≈+

ω2 H11 H22–( ) 2ωH12 4qv– ω2H222+≈+

ω H11 H22–( ) 2ω2H12– ω2H11H22≈

2ωH12 4qd≈

ω2H222

4qv 4qd ω2H122–+≈

4qv 4qd 1 qd–( )+≈

f 1 f 2,

f 1 2qd≈

f 2 2 qv qd 1 qd–( )+ 1 2/≈


(20)

Setting produces pure velocity feedback with

(21)

The corresponding damping ratio is

(22)

This result coincides with eqn (8.161).

In the discrete case, setting does not produce pure velocityfeedback for finite . However, when is small, is small and is usuallyneglected. To determine the behavior of the feedback parameters, DARE is usedto solve the discrete algebraic Riccati equation for a set of damped SDOF systems,taking a range of values for and , with set to 0, scaled according toeqn (11), and r set to 1. The results for are plotted in Fig 1. Figure 2 shows thevariation of .

K f 2ω2 m qd 2ωm qv qd 1 qd

–( )+ 1 2/=

qd 0=

v 2ωm qv( )1 2/=

ξa

kv2ωm------------ qv( )1 2/

qv

4ω2m2r--------------------

1 2/= = =

qd 0=ω ω f 1 kd

qv ∆t T⁄ qd qvkd k⁄

kv 2ωm( )⁄


a)

b)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7−0.35

−0.3

−0.25

−0.2

−0.15

−0.1

−0.05

0

k dk⁄

∆t T⁄

ξ 0.02=qd 0=

qv 4m2ω2qv=

qv 0.01=

qv 0.03=

qv 0.05=

qv 0.07=

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7−0.35

−0.3

−0.25

−0.2

−0.15

−0.1

−0.05

0

k dk⁄

∆t T⁄

ξ 0.05=qd 0=

qv 4m2ω2qv=

qv 0.01=

qv 0.03=

qv 0.05=

qv 0.07=


c)

Figure 1: Displacement feedback predicted by the discrete algebraic Riccatiequation for a SDOF system

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7−0.25

−0.2

−0.15

−0.1

−0.05

0

∆t T⁄

k dk⁄

ξ 0.10=qd 0=

qv 4m2ω2qv=

qv 0.01=

qv 0.03=

qv 0.05=

qv 0.07=


a)

b)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7−0.2

−0.15

−0.1

−0.05

0

0.05

0.1

0.15

0.2

0.25

0.3

∆t T⁄

ξ 0.02=qd 0=

qv 4m2ω2qv=

qv 0.01=

qv 0.03=

qv 0.05=

qv 0.07=

k v2m

ω(

)⁄

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7−0.15

−0.1

−0.05

0

0.05

0.1

0.15

0.2

0.25

∆t T⁄

ξ 0.05=qd 0=

qv 4m2ω2qv=

qv 0.01=

qv 0.03=

qv 0.05=

qv 0.07=

k v2m

ω(

)⁄


c)

Figure 2: Velocity feedback predicted by the discrete algebraic Riccati equation

In general, the feedback parameters decrease with increasing damping in the“uncontrolled” system. As expected, increasing with r held constant placesmore emphasis on lowering the magnitude of the response and consequently themagnitudes of the feedback terms increase. As approaches 0, the solutiontends toward the “continuous” solution given by eqn (8.160). As increases,

decreases, and eventually becomes negative in the region of . Thefeedback stiffness term is also negative. All these solutions are stable, i.e., themaximum modulus of the eigenvalues of is less than unity. The patternof behavior near was also observed in the stability analysis for thediscrete state-space formulation presented in Section 8.3 It should be noted thatthe typical value for is small, on the order of 0.02. For this range, it isreasonable to take . For seismic excitation, the minimum value of iscontrolled by the ground motion data. The typical time interval for acceleration

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7−0.1

−0.05

0

0.05

0.1

0.15

0.2

0.25

∆t T⁄

ξ 0.10=qd 0=

qv 4m2ω2qv=

qv 0.01=

qv 0.03=

qv 0.05=

qv 0.07=

k v2m

ω(

)⁄

qv

∆t T⁄∆t T⁄

kv ∆t T⁄ 0.35≈

c1 c2K f–∆t T⁄ 0.5=

∆t T⁄kd k⁄( ) 0≈ ∆t


time history records is 0.02 sec.

Finite interval discrete time algebraic Riccati equation

The algebraic Riccati equations presented in the previous sections arebased on performance indices which involve integration over an infinite timeinterval. By considering the time interval to be finite, one can generate a spectrumof Riccati type equations corresponding to specific values of the time interval. Theminimum value is equal to the time step used to generate the discrete timesolution; the maximum value is equal to infinity. Eqn (8.174) corresponds to thelatter choice. An equation corresponding to the first choice is derived below.

The discrete quadratic cost function associated with the time intervalbetween and is a convenient choice.

(8.175)

It allows for updating the system properties and weighting parameters at eachdiscrete time and consequently is applicable for adaptive as well as time invariantsystems. The equilibrium equation follows from eqn (8.166).

(8.176)

Requiring to be stationary with respect to ,

(8.177)

and noting that

(8.178)

results in

(8.179)

Finally, substituting for yields the following linear feedback law for

(8.180)

This expression is similar to eqn (8.173) which is based on an infinite time

t j t j 1+

J j j 1+,12--- X j 1+

T Q jX j 1+ F jTR jF j+[ ]=

X j 1+ c1 j, X j c2 j, F j+=

J j j 1+, F j

δJ j j 1+, δX j 1+T Q jX j 1+ δF j

TRF j+ 0= =

δX j 1+ c2 j, δF j=

R jF j c2 j,T Q jX j 1++ 0=

X j 1+ F j

F j R j c2 j,T Q jc2 j,+( ) 1– c2 j,

T Q jc1 j,( )X j– K f j, X j–= =


interval. Replacing H with Q reduces eqn (8.173) to eqn (8.180). Working with thelatter equations is more convenient since it avoids solving the nonlinear Riccatiequation. The following example illustrates the expanded form of for theundamped case.


The various matrices derived in ex 8.4 are applicable for this case when His replaced by Q. The resulting forms are (the subscript j is dropped here tosimplify the notation):

(1)

(2)

(3)

(4)

(5)

(6)

Expressing as

(7)

and using the above equations leads to the dimensionless form of the feedbackparameters

K f

K f1

rf 3

m2ω4--------------+

-----------------------f 1

mω2-----------

f 2

mω2-----------=

qd ω4m2rqd=

qv ω2m2rqv=

f 1

mω2----------- rmω2 qd ω 1 ωcos–( ) qv ωsin2–cos( )=

f 2

mω2----------- rmω qd ω 1 ωcos–( ) qv ω ωcossin+sin( )=

rf 3

m2ω4--------------+ r 1 qd+ 1 ωcos–( )2 qv ωsin2+( )=

K f

K f kd kv=


(8)

(9)

These equations show that depends mainly on , and on . The limitingforms for are:

(10)

(11)

Suppose . The corresponding values of the feedback parameters are:

(12)

(13)

Taking results in an active damping ratio of . The change in stiffnessis on the order of 1.5%, and can be neglected.

Continuous time Riccati differential equation

The algebraic Riccati equation is based on eqn (8.141), which involves anintegral extending from to . A more general formulation considersthe time interval to be finite, and allows for a non-zero final state. Suppose theinitial state at is . Let denote the final time, and the desired finalstate. The generalized form of the performance index is written as

kd

mω2-----------

qd ω 1 ωcos–( ) qv ωsin2–cos

1 qd 1 ωcos–( )2 qv ωsin2+ +--------------------------------------------------------------------------=

kv2mω------------

qd ω 1 ωcos–( ) qv ω ωcossin+sin

2 1 qd 1 ωcos–( )2 qv ωsin2+ +( )-------------------------------------------------------------------------------------=

kd qd kv qvqd 0=

kd

mω2-----------

ωsin2 qv–

1 ωsin2 qv+------------------------------=

kv2mω------------

ω ωqvcossin

2 1 ωsin2 qv+( )--------------------------------------=

∆t T⁄ 0.02=

kd

mω2-----------

0.015775 qv–

1 0.015775 qv+-------------------------------------=

kv2mω------------

0.1243qv2 0.03155qv+----------------------------------=

qv 1= 0.06≈

t 0= t ∞=

t to= Xo t f X f*


(8.181)

where S is a symmetric positive semi-definite matrix which provides a measure ofthe error associated with the boundary conditions at .

Given some initial state the optimal control force F(t) is determined byrequiring J to be stationary. Although the solution procedure for is morecomplicated that for , the optimal feedback law has the same form,

(8.182)

However, H must now satisfy a differential equation rather than an algebraicequation,

(8.183)

and boundary conditions at .

(8.184)

These equations are derived in the following section. In addition to eqn (8.183),the state vector and control force are constrained by

(8.185)

Since the boundary condition on H is at , one has to solve for H backwardin time, i.e., from toward . Given H, one solves for X and F by movingforward in time from . The nature of the solution near is similar to anexponential decay to a steady state solution. Since solving eqn (8.183) is notconvenient, H(t) is approximated by the steady state solution obtained with thealgebraic Riccati equation. This simplification is possible only when A, Q, and Rare constant.

Variational formulation of the continuous time Riccati equation

For completeness, the derivation of the Riccati equation is presented here.The proof is based on the requirement that , , , and are bounded andpiecewise-continuous in the interval . Assuming an optimal control, ,

J 12--- XTQX FTRF+( ) t 1

2--- X f X f

*–( )TS X f X f*–( )+d

to

t f

∫=

t f

t f ∞≠t f ∞=

F t( ) R 1– B f H( )X–=

H ATH H AT HB f R 1– B fTH–+ + Q–=

t t f=

H t f( ) S=

X AX B f F+=

X to( ) Xo=to t t f≤ ≤

t t f=t f to

to t f

A B f Q Rto t tf≤ ≤ F


exists, then an optimal trajectory, , which satisfies

(8.186)

also exists. The corresponding optimal cost functional is given by

(8.187)

The proof proceeds as follows. Let

(8.188)

where is an arbitrary positive or negative scalar and is an arbitrary valuedtime function representing a perturbation of the control force from the optimalvalue. Substituting for , eqn (8.185) expands to

(8.189)

The solution for X can be expressed in terms of the optimal trajectory and aperturbation due to the variation in the control force,

(8.190)

where the state perturbation is related to the control perturbation by

(8.191)

The constraint equation can be incorporated in the cost function byintroducing a Lagrangian multiplier, , and treating as a variable as well as

and (Strang, 1993). The modified cost function is

X

X AX B f F+=

X to( ) Xo=

J F( ) 12--- X f X∗–[ ]

TS X f X∗–[ ] X

TQX F

TRF+[ ] dt

to

t f

∫+

=

F F δF+ F εF+= =

ε F

F

X AX B f F εB f F+ +=

X to( ) Xo=

X X δX+ X εX+= =

X F

X˙

AX B f F+=

X to( ) O=

β t( ) βX F


(8.192)

Noting that

(8.193)

one can write eqn (8.192) as

(8.194)

The cost for optimal control is

(8.195)

Since by definition, and are optimal, the cost difference must be positive,

(8.196)

for all , . Substituting for and in terms of the perturbations,one obtains an expansion for the cost difference in terms of .

J F( ) 12--- X X f–[ ]TS X X f–[ ]( )=

12---XTQX 1

2---FTRF βT AX B f F X–+[ ]+ +

dt

to

t f

∫+

βTXdt

to

t f

∫ βT t f( )X t f( ) βT to( )X to( ) βT

Xdt

to

t f

∫––=

J F( ) 12--- X t f( ) X∗–[ ]T

S X t f( ) X∗–[ ] βT t f( )X t f( ) βT to( )X to( )+–=

12---XTQX βTAX β

TX 1

2---FTRF βTB f F+ + ++ dt

to

tf

∫+

J F( ) 12--- X t f( ) X∗–[ ]

TS X t f( ) X∗–[ ] βT t f( )X t f( ) βT to( )X to( )+–=

12---X

TQX βTAX β

TX 1

2---F

TRF βTB f F+ + ++ dt

to

tf

∫+

F X

J F( ) J F( ) 0≥–

F to t t f≤ ≤ F Xε


(8.197)

A necessary condition for eqn (8.196) to be satisfied is that the first order terms invanish for all , , and . This requirement leads to a first order differential

equation for

(8.198)

and the control law

(8.199)

The sufficient condition for eqn (8.196) is related to the second order terms.

(8.200)

Since and are positive semi-definite and is positive definite, this term isalways positive, and it follows that and represent the solution thatcorresponds to the minimum value of J(F).

Equations (8.186) and (8.198) can be combined into a set of coupled linearvector differential equations involving X and .

J F( ) J F( )– ε X t f( ) X∗–[ ]T

S βT t f( )–( )X t f( ) ε2

2-----X

Tt f( )SX t f( )+=

ε XT

Q βTA βT

+ +[ ] X FT

R βTB f+[ ] F+( ) td

to

t f

∫+

ε2

2----- F

TRF X

TQX+[ ] dt

to

t f

∫+

ε ε X Fβ

XT

Q βTA βT

+ + O=

X t f( ) X∗–[ ]T

S βT t f( )– O=

FT

R βTB f+ O=

⇒

F R 1– B fTβ–=

J F( ) J F( )–ε2

2----- X

Tt f( )SX t f( ) F

TRF X

TQX+[ ] dt

to

tf

∫+

=

S Q RX F

β


(8.201)

The solution can be expressed as

(8.202)

where and are arbitrary times, and is a 2n x 2n state transition matrix ofthe form

(8.203)

Taking and one obtains

(8.204)

Using the boundary condition,

(8.205)

one can solve for in terms of

(8.206)

The above relation is expressed as

(8.207)

Finally, substituting for in eqn (8.199) results in a linear state feedbackcontrol law

(8.208)

The equation for is derived by operating on the governing equations.Differentiating eqn (8.207), one obtains

X

β

A B f R 1– B fT

–

Q– AT–

Xβ

=

X t2( )

β t2( )Φ t2 t1,( )

X t1( )

β t1( )=

t1 t2 Φ

Φ t2 t1,( )Φ11 t2 t1,( ) Φ12 t2 t1,( )

Φ21 t2 t1,( ) Φ22 t2 t1,( )=

t1 t= t2 tf=

X t f( ) Φ11 t f t,( )X t( ) Φ12 t f t,( )β t( )+=

β t f( ) Φ21 t f t,( )X t( ) Φ22 t f t,( )β t( )+=

β t f( ) S X t f( ) X∗–[ ]=

β t( ) X t( )

SΦ12 t f t,( ) Φ22 t f t,( )–[ ]β t( ) Φ21 t f t,( ) SΦ11 t f t,( )–[ ]X t( ) SX *+=

β t( ) H t( )X t( ) G t( )+=

β t( )

F t( ) R 1– B fT HX G+[ ]–=

H


(8.209)

Substituting for , expands to

(8.210)

Finally, introducing and in eqn (8.198),

(8.211)

and setting the coefficient of to zero results in equations for and

(8.212)

(8.213)

The boundary conditions follow from the condition on at and havethe form

(8.214)

For the regulator problem, is taken as zero. Then, and .

As a final point, one can show that the minimum remaining cost at timeis equal to

(8.215)

Example 8.6: Application to scalar case

To illustrate the solution procedure, the scalar form of the LQR formulation isconsidered here. The equations for are :

(1)

β HX HX G+ +=

X β

β H HA HB f R 1– B fTH–+[ ]X G HB f R 1– B f

TG–+=

β β

H HA ATH HB f R 1– B fTH– Q+ + +[ ]X

AT HB f R 1– B fT

–[ ]G G+ O=+

X H G

H HA ATH HB f R 1– B fTH– Q+ + + O=

G AT HB f R 1– B fT

–[ ]G O=+

β t f

H t f( ) S=

G t f( ) SX∗–=

X∗ G O= β HX=

t

J F( ) 12--- XT t( )HX t( )[ ]=

to t tf< <

X AX– BfF– O=


(2)

(3)

At

(4)

At

(5)

Substituting for transforms eqn (1) to

(6)

For stability, .

The homogeneous solution has the form

(7)

(8)

where satisfies

(9)

and the constants are related by

(10)

Then

(11)

β Aβ QX+ + O=

F 1R---- Bfβ–=

t to=

X to( ) Xo=

t tf=

β tf( ) S X tf( ) X∗–[ ]=

F

X AX–Bf

2

R------ β+ 0=

A 0<

X Ceλ t=

β Deλ t=

λ

λ2 A2 QBf2

R-----------+=

C λ A+Q

-------------- D–=

λ1 2, λ±=


(12)

(13)

and the full solution is

(14)

(15)

Imposing the end conditions results in two equations for and .

(16)

(17)

The case of finite is handled by expressing as

(18)

This substitution transforms the solution to

(19)

(20)

The first term applies near and the second term near . Enforcing theboundary conditions leads to

(21)

λ A2 QBf2

R-----------+=

Ci

A λ i+

Q--------------- Di– νiDi= =

β D1eλ t D2e λ t–+=

X ν1D1eλ t ν2D

2e λ t–

+=

D1 D2

ν1D1 ν2D2+ Xo=

1 Sν1–[ ]D1 1 Sν2–[ ]D2e2λ tf–

+ eλ tf–

SX∗–=

tf D1

D1 D1eλ tf–

=

β D1eλ tf t–( )–

D2e λ t–+=

X ν1D1eλ tf t–( )–

ν2D2e λ t–

+=

tf to

D21ν2----- Xo ν1D1e

λ tf––=


(22)

When , the solutions uncouple as follows:

For

(23)

For

(24)

For

D1

SX∗–Xoν2------- 1 Sν2–[ ] e

λ tf––

1 ν1S–[ ]ν1ν2----- 1 ν2S–[ ] e

2λ tf––

--------------------------------------------------------------------------------=

λ tf 3>

0 t 3λ---≤<

β D2e λ t–=

X ν2β=

FBfR----- β–

Bfν2R---------- X–= =

D2

Xoν2-------≈

ν2A λ–

Q-------------– A

Q----–

AQ------- 1

QBf2

A2R-----------++= =

3λ--- t tf

3λ---–< <

β 0=

X 0=

F 0=

tf3λ--- t tf≤<–


(25)

Stability of the controlled system requires . Then, for the stable case

(26)

The feedback is negative near and positive near . Lettingeliminates the solution in the region of .

The algebraic Riccati equation approach for this set of equations starts withthe scalar form of eqn (8.212)

(27)

The roots are

(28)

Enforcing the requirement that the uncontrolled system be stable,

(29)

transforms eqn (28) to

β D1eλ tf t–( )–

=

X ν1β=

FBfR----- β–

Bfν1R---------- X–= =

D1SX∗–

1 ν1S–------------------=

ν1A λ+

Q--------------– A

Q----– A

Q-------– 1

QBf2

A2R-----------+= =

A 0<

ν2 0>

ν1 0<

t 0= t tf= X∗ 0→tf

H2 Bf2

R------– 2AH Q+ + 0=

H1 2,R

Bf2

------ A A 1QBf

2

RA2-----------+±=

A 0<


(30)

Finally, the requirement that eliminates the negative root, and one obtains

(31)

Referring back to eqn (23), the solution near for the case where islarge has the form

(32)

Taking , the expression for becomes

(33)

Forming the ratio of to ,

(34)

and substituting for leads to

(35)

H1 2,R A

Bf2

----------- 1– 1QBf

2

RA2-----------+±=

H 0>

HoptR A

Bf2

----------- 1QBf

2

RA2-----------+ 1–=

t 0= tf

X Xoe λ t–=

βXoν2-------e λ t–

=

ν2AQ----–

AQ------- 1

QBf2

RA2-----------++=

A 0< ν2

ν2AQ------- 1 1

QBf2

RA2-----------++=

β X

βX---- 1

ν2-----=

ν2

1ν2----- Q

A 1 1QBf

2

RA2-----------++

-----------------------------------------------R A

Bf2

----------- 1QBf

2

RA2-----------+ 1–= =

8.5 State-space Formulation for MDOF Systems 623

Comparing eqn (35) with eqn (31) shows that . The negative root ofeqn (30) is equal to , and corresponds to positive feedback in the region of .

8.5 State-space formulation for MDOF systems

Notation and governing equations

The material presented in the previous sections can be readily extended to thecase of a multi-degree-of-freedom system. One has only to generalize thedefinition equations for the various matrices involved in the state-spacerepresentation. In what follows, the essential steps for an nth order linear systemare discussed.

Using the notation defined in Chapter 2, the equations for an nth orderlinear system subjected to seismic excitation and a set of r control forces applied atvarious locations on the system are written as

(8.216)

where is an n x r matrix that defines the location of the control forces withrespect to the degrees of freedom. Figure 8.14 illustrates the notation for the caseof a lumped mass model of a shear beam having four degrees of freedom and twocontrol forces. The initial conditions involve constraints on the displacements andvelocities at time .

(8.217)

Following the approach of section 8.2, the state-space form for eqn (8.216)is taken as

(8.218)

Hopt 1 ν2⁄=1 ν1⁄ tf

MU CU KU+ + MEag– EfF P+ +=

Ef

t 0=

U 0( ) Uo=

U 0( ) Uo=

X AX BfF Bgag BpP+ + +=


Fig. 8.14: 4DOF system with two control forces

where is now a vector of order 2n,

(8.219)

and the coefficient matrices are given by

(2n x 2n) (8.220)

(2n x r) (8.221)

(2n x 1) (8.222)

(2n x 1) (8.223)

m1

m4

m3

m2

u1

u4

u3

u2

F1

F2

U

u1

u2

u3

u4

= E

1111

= Ef

0 00 10 01 0

=

X

X U

U=

AO I

M 1– K– M 1– C–=

BfO

M 1– Ef

=

BgOE–

=

BpO

M 1–=


Free vibration response - time invariant uncontrolled system

Specializing eqn (8.218) for no external forcing leads to a set of 2nhomogeneous first order differential equations.

(8.224)

Considering A to be constant, the general solution has the form

(8.225)

where and V satisfy the following set of 2n homogeneous algebraic equations

(8.226)

Substituting for A, eqn (8.226) expands to

(8.227)

The solution of eqn (8.227) has the form

(8.228)

where and are constrained by n conditions,

(8.229)

For the SDOF case, the complete solution of eqn (8.229) consists of 2solutions involving a pair of complex roots for

(8.230)

Example 3.13 dealt with a 2 DOF system. In this case, there are 4 solutionsinvolving 2 pairs of complex roots:

X AX=

X eλ tV=

λ

A λI–( )V 0=

λI– I

M 1– K– M 1– C– λI–V 0=

VΦλΦ

=

λ Φ

λ2Φ λM 1– CΦ M 1– KΦ+ + 0=

λ

λ λ 1= Φ 1=

λ λ 1= Φ 1=


(8.231)

Then, for an n’th order system, it follows that there are 2n complex solutionsinvolving n pairs of complex roots.

(8.232)

Each term in eqn (8.232) is complex

(8.233)

The complete solution is obtained by combining the 2n solutions usingcomplex constants, and .

(8.234)

Substituting for and ,

(8.235)


λ λ 1 λ1,= Φ Φ1 Φ1,=

and

λ λ 2 λ, 2= Φ Φ2 Φ2,=

λ λ 1 λ1 λ2 λ2 … λn λn, , , , , ,=

Φ Φ1 Φ, 1 Φ2 Φ2 … Φn Φn, , , , ,=

VΦ1

λ1Φ1 Φ1

λ1Φ1

…Φn

λnΦn Φn

λnΦn

, , , ,=

λ j λR j, iλ I j,+=

Φj ΦR j, iΦI j,+=

V j V R j, iV I j,+=

Aj Aj

X Ajeλ jtV j A je

λ jtV j+( )

j 1=

n

∑=

Aj eλ jt

Aj AR j, iAI j,+=

eλ jt e

λR j, t λ It i λ Itsin+cos( )=


(8.236)

One determines and using the initial conditions on X at time t=0. Thiscomputation is discussed in the next section. Stability requires the homogeneoussolutions to be bounded. Noting eqn (8.236), a system having n DOF is stablewhen

(8.237)

The free vibration solution given by eqn (8.234) can also be expressed as

(8.238)

where A is defined by eqn (8.220) and is generated with the following series:

(8.239)

Equation (8.238) is a generalized form of the SDOF solution. One can establishthis result by specializing eqn (8.234) for t = 0,

(8.240)

and noting that

(8.241)

Example 8.7: Free vibration solution for proportional damping

Consider eqn (8.229) premultiplied by M.

X t( ) eλR j, t

AR j, V R j, AI j, V I j,– ) λ I j, tcos([

j 1=

n

∑=

AR j, V I j, AI j, V R j,– ) λ I j,sin t ](–

AR j, AI j,

λR j, 0< j 1 2 … n, , ,=

X t( ) eAtX 0( )=

eAt

eAt I At 12---AAt2 … 1

n!-----Antn …+ + + + +=

X 0( ) AjV j A jV j+( )

j 1=

n

∑=

eAtV j eλ jtV j=

eAtV j eλ jtV j=


(1)

Suppose C is a linear function of K and M

(2)

Substituting for C results in

(3)

When the coefficient of is expressed as

(4)

eqn (3) takes the form of the classical eigenvalue equation.

(5)

There are n eigensolutions to eqn (5)

(6)

All the eigenvectors are real vectors. Also, all the eigenvalues are positive realquantities.

(7)

Given , one solves eqn (4) for . The pair of complex roots are written as

(8)

where the damping ratio is related to and by

(9)

λ2MΦ λCΦ KΦ+ + 0=

C αK βM+=

λ2 β λ+1 αλ+------------------- MΦ KΦ+ 0=

MΦ

λ2 βλ+1 αλ+------------------- ω 2–=

KΦ ω 2MΦ=

ω12 ω2

2 … ωn2, , ,

Φ1 Φ2 … Φn, , ,

ωj2 0>

ωj2 λ j

λ j λ j, ξ j ωj– i ωj 1 ξ j2– 1 2/±=

α β

ξ j

β α ωj2+

2 ωj-------------------------=


In what follows, is considered to be positive, and is set equal to .

The eigenvectors, , are real and satisfy the following orthogonalityrelationships

(10)

(11)

Since is real, the state eigenvector, V, has the following form

(12)

Undamped free vibration response is the special case where .The corresponding solution is

(13)

(14)

Example 8.8: General uncoupled damping

The previous example dealt with the case where the damping matrix is alinear combination of the mass and stiffness matrices. This approach can beextended to deal with a more general form of C. Suppose C satisfies the followingorthogonality conditions,

(1)

where are now considered to be independent parameters. The free vibrationsolution is the same as presented in example 8.7.

ωj ωj ωj

Φj

ΦjTMΦk δjkmk=

ΦjTKΦk δjkωk

2 mk=

Φ

V j

Φj

λ jΦj Φj

λR j, Φj

i0

λ I j, Φj

+= =

α β 0= =

λ j iωj±=

V j

Φj

0

i0

ωjΦj

+=

ΦjTCΦk 2δjkmkωkξk= j 1 2 … r, , ,=

ξk


(2)

The ability to independently specify the modal damping ratio for the first r modesby adjusting C is very convenient since it allows one to deal more effectively withthe problem of suppressing the response of a particular subset of modescontained within this group.

The form of C that satisfies eqn (1) is established by noting the followingidentities:

(3)

A linear combination of these matrices is a candidate solution.

(4)

Taking s = 0 corresponds to Rayleigh damping. Taking allows one to specifyr modal damping ratios. The coefficients for this case are determined with

(5)

If both a and b are used, s = r/2.

The basic problem with this approach is the form of C generated with eqn(3). The coupling between the off-diagonal elements extends beyond adjacentnodes when . This pattern requires a deployment of dampers that involvesconnecting dampers to nodes that are not adjacent, e.g., a damper between floor 1and floor 3. Realistically, the required pattern cannot be achieved.

λ j ξ jωj– iωj 1 ξ j2–[ ] 1 2/±=

V j

Φj

λ jΦj

=

KΦj ωj2MΦj=

ΦjTM M 1– K( )nΦk ωj

2n mjδjk=

ΦjTK M 1– K( )nΦk ωj

2n 2+ mjδjk=

C an M M 1– K( )n[ ] bn K M 1– K( )n

[ ]+( )

n 0=

s

∑=

s r≤

anωj2n 1– bnωj

2n 1++( )n 0=

s

∑ 2ξ j= j 1 2 … r, , ,=

n 1≥


Orthogonality properties of the state eigenvectors

The eigenvector satisfies eqn (8.226).

(8.242)

Since A is not symmetrical, is not orthogonal to .

(8.243)

A set of vectors which are orthogonal to V can be established by considering theeigenvalue problem for the transpose of A:

(8.244)

where , W represent the eigenvalue and corresponding eigenvector for .Eqn (8.244) requires the determinant of the coefficient matrix to vanish.

(8.245)

Noting that the determinant of the transpose of a matrix is equal to thedeterminant of the original matrix,

(8.246)

it follows that

(8.247)

and eqn (8.244) is equivalent to

(8.248)

Starting with

(8.249)

and premultiplying the first equation with and the second with , oneobtains

V j

AV j λ jV j=

V j V k

V jTV k δjk≠

ATW j λ j*W j= j 1 2 … n, , ,=

λ * AT

AT λ *I– 0=

b bT≡

λ j* λ j≡

ATW j λ jW j=

AV j λ jV j=

ATWk λkW k=k j≠

Wk V j


(8.250)

(8.251)

Eqn (8.251) can be written as

(8.252)

Then subtracting eqn (8.250) from (8.252) results in

(8.253)

According to eqn (8.253), and are orthogonal

(8.254)

The result for is written as

(8.255)

One can show that

for all j, k (8.256)

and

(8.257)

Example 8.9: Initial conditions - free vibration response

Consider the general free vibration solution defined by eqn (8.234).

(1)

The integration constants are determined using the initial conditions on X at t = 0.

W kTAV j λ jW k

TV j=

V jTATWk λkV j

TWk=

WkTAV j λkWk

TV j=

0 λk λ j–( )WkTV j=

Wk V j

WkTV j 0 for= j k≠

j k=

W jTV j f j=

WkTV j 0=

WkTV j 0=

W jTV j f j=

X t( ) Ajeλ jtV j A je

λ jtV j+( )

j 1=

n

∑=


(2)

Noting the orthogonality relations, the expressions for the constants are:

(3)

Determination of W and

One establishes W by solving the following eigenvalue problem

(8.258)

Substituting for , the coefficient matrix expands to

(8.259)

Expressing W in partitioned form,

(8.260)

and expanding eqn (8.259) leads to

(8.261)

Solving the first equation for ,

(8.262)

X 0( ) AjV j A jV j+( )

j 1=

n

∑=

Aj1f j-----W j

TX 0( )=

A j1

f j-----W j

TX 0( )=

f j

ATW λW=

AT

λI– KM 1––

I CM 1–– λI–W 0=

WW1

W2

=

λW1– KM 1– W2– 0=

W1 CM 1– W2– λW2– 0=

W2

W2 λMK 1– W1–=


and substituting into the second equation results in

(8.263)

This equation can be transformed to a form similar to the eigen equationfor by expressing as

(8.264)

The result is

(8.265)

Comparing this form with eqn (8.229) shows that

(8.266)

It follows that the forms of and are related by

(8.267)

Using these expressions, the definition equation for f expands to

(8.268)

When C is proportional to either K or M, is real and is given by

(8.269)

The value for no damping is .

General Solution - time invariant system

Following the approach employed for the SDOF case, the general solutionfor an arbitrary loading applied to a time invariant MDOF system can beexpressed as a Duhamel integral involving the free vibration response. Notingeqns (8.238) and (8.239), the complete solution has the form

W1 λCK 1– W1– λ2MK 1– W1–=

Φ W1

W1 KW1=

λ2MW1 λCW1 KW1+ + 0=

W1 Φ≡

V j W j

W j

KΦj

λ jMΦj–

= V j

Φj

λ jΦj

=

f j W jTV j Φj

TKΦj λ j2Φj

TMΦj–= =

2λ j2Φj

TMΦj– λ jΦjTCΦj–=

Φ f j

f j mj ωj2 λ j

2–( )=

f j 2mjωj2=


(8.270)

where the coefficient matrices are defined by eqns (8.220) to(8.223). This equation is similar to the SDOF solution given by eqn (8.26); one hasonly to introduce the appropriate forms of A, , , and . The discrete timeequilibrium equation for the MDOF case follows from eqn (8.77).

(8.271)

Working with the full set of equations for an n’th order system requiresdealing with matrices of order 2n. As n increases, the computational costs for

and become excessive, and one needs to consider an alternativeapproach. In conventional dynamic analysis, this reduction in computationaleffort is achieved by expressing the state vector as a linear combination ofprescribed modal vectors multiplied by unknown coordinate variables which arefunctions of time. Depending upon the nature of the loading, one can obtain areasonably accurate solution by suitably selecting the participating modes so as tominimize the required number of modal coordinate functions, and consequentlyreduce the computational cost. Interpreting the total response as a superpositionof modal responses is also useful from the perspective of active control. A keyissue of active control is the optimal location of “active” forces so as to minimizethe response of a particular mode. In what follows, the formulation of thegoverning equations in terms of modal coordinates is presented. This formulationis used later to establish the form of the LQR algorithm in terms of modalcoordinates.

Modal state space formulation - uncoupled damping

The displacement vector is expressed as a linear combination of a subset ofthe eigenvectors for the undamped system scaled with functions of time.

(8.272)

Assuming the system has n DOF, this expression will produce the exact solutionwhen s is taken to be n. Substituting for in the force equilibrium equation,

X t( ) eA t to–( )

Xo eA t τ–( ) B f F BpP Bgag+ +( ) τdto

t

∫+=

A B f Bp Bg, , ,( )

B f Bp Bg

X j 1+ eA∆tX j A 1– eA∆t I–( ) Bgag j, B f F j BpP j+ +[ ]+=

eA∆t A 1–

U t( ) qj t( )Φj

j 1=

s

∑=

U t( )


premultiplying the result with , and noting the orthogonality conditions for, one obtains s equations having the following form,

(8.273)

These equations uncouple when C is orthogonal to . Assuming this conditionis satisfied, and taking

(8.274)

results in

(8.275)

The initial conditions for are

(8.276)

When F = 0, these equations can be solved separately. The effect of feedback is tointroduce coupling between the modal equations.

In order to deal with feedback, the equations are transformed to thestandard state-space form by defining q as the modal coordinate vector,

(8.277)

ΦkΦk

mkqk ΦkTCΦj( )q j mkωk

2qk+

j 1=

p

∑+

ΦkTMEag– Φk

TP ΦkTE f F+ +=

k 1 2 … s, , ,=

Φj

ΦkTCΦj 2δjkmkωkξk=

qk 2ξkωkqk ωk2qk+ +

k 1 2 … s, , ,=1

mk-------Φk

TMEag–1

mk-------Φk

TP 1mk-------Φk

TE f F+ +=

qk

qk 0( ) 1mk-------Φk

TMU 0( )=

qk 0( ) 1mk-------Φk

TMU 0( )=

q

q1

q2

…qs

= s 1×


and as the modal coordinate state vector,

(8.278)

Note that now contains only 2s modal coordinate terms. The equations areexpressed in the same form as eqn (8.218),

(8.279)

where the “modal” forms of the coefficient matrices along with their sizes are:

(8.280)

(8.281)

(8.282)

(8.283)

(8.284)

(8.285)

(8.286)

(8.287)

Lastly, the initial conditions follow from eqn (8.276)

Xm

Xmq

q

= 2s 1×

Xm

Xm AmXm B fmF BpmP Bgmag+ + +=

m mjδij[ ]= s s×( )

Λ ωj2δij[ ]= s s×( )

Λ1 2ξ jωjδij[ ]= s s×( )

Φ Φ1 Φ2 … Φs= n s×( )

Am0 IΛ– Λ1–

= 2s 2s×( )

B fm0

m 1– ΦTE f= 2s r×( )

Bpm0

m 1– ΦT= 2s r×( )

Bgm0

m 1– ΦTME–= 2s 1×( )


(8.288)

Feedback is introduced by expressing F as a linear function of the modalstate vector

(8.289)

The individual feedback matrices are of order r x s. For pure velocity feedback,is set equal to 0.

Substituting for F in eqn (8.279) leads to the state equation specialized forcontinuous negative linear feedback.

(8.290)

where

(8.291)

The discrete form is similar to eqn (8.271).

(8.292)

For the continuous case, the free vibration solution with feedback is stable whenis stable and is positive. This condition is always satisfied. The discrete

solution is stable provided that the modulus of the largest eigenvalue of is lessthan unity.

(8.293)

(8.294)

Given the selection of the modes and the feedback matrix, , one can establishthe limit on by computing the eigenvalues of for a range of values of .

Xm 0( )q 0( )q 0( ) m 1– ΦTMU 0( )

m 1– ΦTMU 0( )= = 2s 1×( )

F K fmXm– kdq– kvq–= =

kd

Xm Am B fmK fm–( )Xm BpmP Bgmag+ +=

Am B fmK fm–0 I

Λ– m 1– ΦTE f kd– Λ1– m 1– ΦTE f kv–=

Xm j 1+, eAm∆t

Xm j, Am1– e

Am∆tI–( ) B fmF j BpmP j Bgmag+ +[ ]+=

F j K fmXm j,–=

Am kvcm

ρmax cm( ) 1<

cm eAm∆t

Am1– e

Am∆tI–( )B fmK fm–=

K fm∆t cm ∆t


This computation is illustrated in example 8.14

Modal state space formulation - arbitrary damping

When C is an arbitrary symmetric positive definite matrix, the expansionin terms of the eigenvectors for the undamped system and real modal coordinatesdoes not lead to uncoupled modal equations. In this case, one has to work with anexpansion involving complex modal coordinates and complex state eigenvectors.The state space vector is approximated as a linear combination of s modal vectorsand coordinates,

(8.295)

where q and V are complex quantities defined by

(8.296)

(8.297)

The expanded real forms for and are

(8.298)

(8.299)

When damping is uncoupled, . The fully undamped case has ,, and .

The initial conditions for the modal coordinates are determined byspecializing for t = 0.

X t( ) U t( )

U t( )12--- qjV j q jV j+( )

j 1=

s

∑= =

qj t( ) qR j, t( ) iqI j, t( )+=

V jΦj

λ jΦj

=

U t( ) U t( )

U t( ) qR j, ΦR j, qI j, ΦI j,–( )

j 1=

s

∑=

U t( ) qR j, λR j, ΦR j, λ I j, ΦI j,–( ) qI j, λR j, ΦI j, λ I j, ΦR j,+( )–[ ]

j 1=

s

∑=

ΦI j, 0= λR j, 0=λ I j, ωj= ΦI j, 0=

X t( )


(8.300)

Premultiplying by , and noting the orthogonality relation between and, one obtains

(8.301)

Substituting for , the expression for expands to

(8.302)

Introducing the expression for in the state equilibrium equation, andpremultiplying for results in a set of s complex scalar equations,

(8.303)

The terms involving F, , and P can be interpreted as complex modal forcesassociated with the k’th mode. Expanding the matrix products and substitutingthe notation,

(8.304)

reduces the governing modal equations to a simpler form,

(8.305)

X 0( ) 12--- qj 0( )V j q j 0( )V j+( )

j 1=

s

∑=

Wk WkV j

qk 0( ) 2f k-----Wk

TX 0( )=

Wk qk 0( )

qk 0( ) 2f k-----Φk

TK U 0( )

2λkf k

---------ΦkTM

U 0( )–= k 1 2 … s, , ,=

X t( )Wk

qk λkqk2f k-----Wk

TB f F 2

f k-----Wk

TBg ag

2f k-----Wk

TBp P+ + +=

k 1 2 … s, , ,=

ag

b f k,2f k-----Wk

TB f

2λkf k

---------ΦkTE f–= =

bg k,2f k-----W k

TBg

2λkf k

---------ΦkTME= =

bp k,2f k-----Wk

TBp

2λkf k

---------ΦkT–= =

qk λkqk b f k, F bp k, P bg k, ag+ + += k 1 2 … s, , ,=


Since is complex, there are a total of 2s equations.The corresponding set of“real” equations for the k’th mode are:

(8.306)

One establishes the initial conditions for and by expanding eqn(8.302). The resulting expressions are:

(8.307)

(8.308)

Various specialized forms of these equations are presented in the examples listedbelow.

Example 8.10: Modal formulation - undamped case

It is of interest to specialize the general formulation presented in thissection for the case of no damping and compare this result with the solutionobtained with the conventional formulation derived in the previous section.Setting C = 0 reduces the various terms to:

qk

qR k, λR k, qR k, λ I k, qI k,– b f R k, F bpR k, P b+gR k, ag+ +=

qI k, λ I k, qR k, λR k, qI k, b fI k, F bpI k, P b+gI k, ag+ + +=

k=1,2,...,s

qR qI

qR k, 0( ) 2

f R k,2 f I k,

2+---------------------------- f R k, ΦR k,

T f I k, ΦI k,T+ )KU 0( )([=

f R k, λR k, ΦR k,T λ I k, ΦI k,

T–( ) f I k, λR k, ΦI k,T λ I k, ΦR k,

T+( )+ MU 0( ) ]–

qI k, 0( )f I k,f R k,-----------qR k,– 2ΦI k,

T KU 0( )+=

2f R k,----------- λR k, ΦI k,

T λ I k, ΦR k,T+( )MU 0( )–


(1)

Substituting in eqns (8.307) and (8.308), leads to

(2)

(3)

Equation (2) shows that

(4)

Then, eqn (3) can be written as

(5)

This result is identical to eqn (8.225).

Example 8.11: Modal formulation - uncoupled damping

When the damping matrix is proportional to K and M, the eigenvector, ,is real. However, the eigenvalue is complex. The various terms specialized for thiscase are as follows:

ΦI k, 0=

λR k, 0=

λ I k, ωk=

f k 2mkωk2=

b f k,i

mkωk--------------Φk

TE f–=

bp k,i

mkωk--------------Φk

T–=

bg k,i

mkωk--------------Φk

TME=

qR k, ωkqI k,–=

qI k, ωkqR k,1

mkωk--------------Φk

T E f F P MEag–+( )–=

qI k,1ωk------ qR k,–=

qR k, ωk2qR k,+

1mk-------Φk

T E f F P MEag–+( )=

Φ


(1)

Since the forcing terms are pure imaginary, the equations simplify to

(2)

The initial conditions are

(3)

This example shows that involves a combination of and . Theactual displacements are given by

ΦI k, 0=

λR k, ξkωk–=

λ I k, ωk 1 ξk2–[ ] 1 2/=

f k 2mkλ I k, λ I k, iλR k,–( ) 2mkλ Iλ( ) i–= =

λkf k----- i

2mkλ I k,--------------------=

b f k,i

mkλ I k,-----------------Φk

TE f–=

bg k,i

mkλ I k,-----------------Φk

TME–=

bp k,i

mkλ I k,-----------------Φk

T–=

qR k, λR k, qR k, λ I k, qI k,–=

qI k, λ I k, qR k, λR k, qI k,1

mkλ I k,-----------------Φk

T E f F P MEag–+( )–+=

qR k, 0( ) 1mk-------Φk

TMU 0( )=

qI k, 0( ) 1λ I k,---------- λR k, qR k, 0( ) 1

mk-------ΦTMU 0( )–=

qI k, qR k, qR k,


(4)

Example 8.12: Modal parameters - 4DOF system

This example presents modal data for the 4 DOF system shown in Fig. 1.The element stiffness and damping values are listed in Table 1. The elementstiffness factors are selected so that the displacement profile for the first mode isessentially linear. Two damping distributions are considered. Case 1 represents alow damping level for the first mode. The damping coefficients for case 2 areadjusted so that the sum, , is 3 times the corresponding sum for the first case.Since the internodal displacement for the first mode is essentially constant, thisadjustment increases the modal damping ratio for the first mode by a factor of 3.More emphasis is placed on element 1 to exaggerate the non-proportionaldistribution and therefore increase the magnitude of the imaginary part of themodal vector.

Figure 1

U t( ) qR j, Φj

j 1=

s

∑=

U t( ) λR j, qR j, λ I j, qI j,–( )Φj

j 1=

s

∑ qR j, Φj

j 1=

s

∑= =

Σc j

m1

m4

m3

m2

u1

u4

u3

u2

k1 c1,

1000kg

1000kg

1000kg

1000kg

k2 c2,

k3 c3,

k4 c4,


Table 1

Scaled versions of the modal displacement profiles for the 2 dampingdistributions are plotted in Figures 2 and 3. The real parts are essentially identical.Since case 1 represents a low level of damping, the amplitude of thecorresponding imaginary part is negligible in comparison to the real part which isof order 1. Case 2 results show a significant increase in the amplitude for theimaginary part. This shift is due to the large increment in c assigned to element 1.The modal periods and damping ratios are shown in Figures 4 and 5. There isessentially no change in the periods. As expected, all the modal damping ratiosare higher for case 2. Lastly, to illustrate the effect of damping, time historyresponses were generated for the El-Centro ground excitation. The maximumvalues of the inter-nodal displacements for the elements are plotted in Figure 6.Increasing the damping reduces the absolute maximum value from 0.0258 to0.0217, and also decreases the average value.

Elementnumber Element stiffness, k Element damping, c

(kN.s/m)

(kN/m) case 1 case 2

1 1700 4 20

2 1400 3 7

3 1000 2 2

4 700 1 1

Σ 10= Σ 30=


case 1

case 2

Figure 2: Modal displacement profile - real part

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 11

2

3

4

Amplitude

Nod

en

um

ber

Mode 1

Mode 2Mode 3

Mode 4

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 11

2

3

4

Nod

en

um

ber

Amplitude

Mode 1

Mode 2Mode 3

Mode 4


case 1

case 2

Figure 3: Modal displacement profile - imaginary part

−0.02 −0.015 −0.01 −0.005 0 0.005 0.01 0.015 0.02 0.0251

2

3

4

Nod

en

um

ber

Amplitude

Mode 1 Mode 2

Mode 3

Mode 4

−0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.31

2

3

4

Nod

en

um

ber

Amplitude

Mode 1

Mode 2

Mode 3

Mode 4


Figure 4: Modal period - No feedback

Figure 5: Modal damping ratio - No feedback

1 2 3 40.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Mode number

Mod

alpe

riod

sec

()

,

Damping case 1

Damping case 2

1 2 3 40

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

Mode number

Mod

alda

mpi

ng

rati

o

Damping case 1

Damping case 2


Figure 6: Element internodal displacement profile - No feedback

When F = 0, the problem reduces to solving a set of 2 equations for eachmode. Including F couples the modal equations, and now one has to solve 2ssimultaneous equations. In this case, it is convenient to shift over to a state-spacetype formulation.

The state vector is defined in a similar way as for the previousformulations. Firstly, the real and imaginary modal coordinates are groupedseparately,

(8.309)

Secondly, the “modal” state vector is taken as

0.017 0.018 0.019 0.02 0.021 0.022 0.023 0.024 0.025 0.0261

2

3

4

Maximum element internodal displacement, (m)

Ele

men

t num

ber

Damping case 1Damping case 2

El Centro ground motion

qR

qR1

qR2

…qRs

= qI

qI1

qI2

…qIs

=


(8.310)

Lastly, the state equilibrium equation is expressed in the same form as foruncoupled damping

(8.311)

where the corresponding form of the coefficient matrices are

(8.312)

(8.313)

(8.314)

(8.315)

(8.316)

Xm

qR

qI

=

Xm AmXm B fmF BpmP Bgm a+ + +=

λR λR k, δjk[ ]= λ I λ I k, δjk[ ]=

b fR

b fR 1,

b fR 2,

…b fR s,

= b fI

b fI 1,

b fI 2,

…b fI s,

=

bpR

bpR 1,

bpR 2,

…bpR s,

= bpI

bpI 1,

bpI 2,

…bpI s,

=

bgR

bgR 1,

bgR 2,

…bgR s,

= bgI

bgI 1,

bgI 2,

…bgI s,

=

AmλR λ I–

λ I λR

=


(8.317)

The initial conditions for are obtained with eqns (8.307) and (8.308).

Negative linear feedback is taken as

(8.318)

Example 8.10 showed that is a linear function of for no damping.Including damping results in being a function of both and . For a lightlydamped system, it is reasonable to assume depends only on , and thereforeto approximate pure velocity feedback by setting . The feedback term forthis case reduces to:

(8.319)

Example 8.13: Modal response for example 8.12

The state space modal formulation is applied to the 4 DOF system withCase 1 damping defined in example 8.12. Results for the real and imaginary partsof the modal coordinates for the first 2 modes are plotted below. Comparison ofthe plots in Figure (1) shows that the response is dominated by the first mode. Theratio of the amplitudes for the first and second modes is approximately 30, whichindicates that the second mode contributes about 3% to the total response for thiscombination of structure and excitation.

B fmb fR

b fI

= BpmbpR

bpI

= BgmbgR

bgI

=

Xm

F K fmXm– kd kvqR

qI

–= =

qI qRqI qR qR

qI qRkd 0=

B fmF0 b fRkv–

0 b fIkv–

qR

qI

=


a) Mode 1

b) Mode 2

Figure 1: Time history response of the real part of the modal coordinate

0 5 10 15 20 25 30 35 40 45 50−0.08

−0.06

−0.04

−0.02

0

0.02

0.04

0.06

0.08

0.1

Time (sec)

Mod

alco

ordi

nat

eq R

1,

El CentroDamping Case 1

0 5 10 15 20 25 30 35 40 45 50−4

−3

−2

−1

0

1

2

3

4x 10

−3

Mod

alco

ordi

nat

eq R

2,


Time (sec)


a) Mode 1

b) Mode 2

Figure 2: Time history response of the imaginary part of the modal coordinate

0 5 10 15 20 25 30 35 40 45 50−0.1

−0.08

−0.06

−0.04

−0.02

0

0.02

0.04

0.06

0.08

Mod

alco

ordi

nat

eq I1,

Time (sec)


0 5 10 15 20 25 30 35 40 45 50−3

−2

−1

0

1

2

3x 10

−3

Mod

alco

ordi

nat

eq I2,

Time (sec)



Example 8.14: Modal response with feedback for example 8.12

This example illustrates the effect of linear negative velocity feedback onthe modal damping ratios. The modal state space formulation defined by eqns(8.309) through (8.319) is applied to the 4 DOF system with Case 1 dampingdescribed in example 8.12. Two control force systems are considered. The firstsystem is a single force applied at node 4; the second system consists of self-equilibrating sets of nodal forces applied on each element. Figure (1) shows thespatial distribution of the control forces.

Figure 1

Given the feedback matrix, , one determines F using eqn (8.318),

(1)

and then solves for with eqn (8.311).

(2)

The frequency and damping parameters for the system with feedback are relatedto the eigenvalues of , and depend on the control force scheme as

m1

m4

m3

m2

u1

u4

u3

u2

k1 c1,

k2 c2,

k3 c3,

k4 c4,F1

F2F1

F1

F2 F3

F3 F4

Scheme 1 Scheme 2

K fm

F K fmXm–=

Xm

Xm Am B fmK fm–( )Xm BpmP Bgmag+ +=

Am B fmK fm–


well as . Particular forms of obtained with the algebraic Riccatiequation and specialized for pure velocity feedback (see eqn (8.319)) are used hereto generate results for the 2 force schemes. This formulation is discussed in detailin the next section. Our objective here is to illustrate the variation in behaviorassociated with feedback systems.

Figure (2) shows the modal damping ratios for the 4 DOF model with Case1 damping, force scheme 1, and a particular choice for . Feedback results in asignificant increase (a factor of 10) in the damping ratio for the first mode. Sincethis response is dominated by the first mode, there also is a reduction in thedisplacement response. The relevant displacement quantities are plotted in Figs(2b) and (2c). The last plot shows the variation of the control force magnitudewith time.

Results for the control force scheme 2 are plotted in Figure (3). Thiscombination of control forces and weighting produces essentially equal nodaldamping ratios whereas scheme 1 places the priority on the first mode. Since thedamping ratio is less, the displacement response is greater. There also is areduction in the peak magnitudes of the 4 control forces.

a) Modal damping ratio with continuous feedback - no saturation limit

K fm K fm

K fm

1 2 3 40.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

0.22

0.24

Case 1 dampingModel defined in example 8.12

Mode number

Mod

al d

ampi

ng r

atio

wit

h fe

edba

ck


b) Element internodal displacement profile

c) Displacement of node 4

0.0092 0.0094 0.0096 0.0098 0.01 0.0102 0.0104 0.0106 0.0108 0.011 0.01121

2

3

4

El Centro excitation

Maximum element internodal displacement

Ele

men

t num

ber

0 5 10 15 20 25 30−0.05

−0.04

−0.03

−0.02

−0.01

0

0.01

0.02

0.03

0.04

Time (sec)


Nod

al d

ispl

acem

ent (

m)


d) Control force time history

Figure 2: Modal response for control force scheme 1

a) Modal damping ratio with continuous feedback - no saturation limit

0 500 1000 1500−5000

−4000

−3000

−2000

−1000

0

1000

2000

3000


Time step number

Con

trol

forc

e m

agni

tud

e

Max Force = 4860 N

1 2 3 40.11

0.115

0.12

0.125

0.13

0.135

Mode number

Case 1 dampingModel defined in example 8.12

Mod

al d

ampi

ng r

atio

wit

h fe

edba

ck



c) Displacement of node 4

0.0122 0.0124 0.0126 0.0128 0.013 0.0132 0.0134 0.0136 0.0138 0.014 0.01421

2

3

4


Ele

men

t num

ber

0 5 10 15 20 25 30−0.06

−0.04

−0.02

0

0.02

0.04

0.06

Time (sec)

Nod

al d

ispl

acem

ent (

m)



d) Control force time history

Figure 3: Modal response for control force scheme 2

Stability analysis - discrete modal formulation

The discrete form of the state equilibrium equation expressed in terms ofmodal coordinates is used here to evaluate the effect of the time interval and thedelay time on the stability. Allowing for linear feedback, the governing equationfor free vibration response is

(8.320)

Time delay is introduced by taking as

(8.321)

1 2 3 42900

3000

3100

3200

3300

3400

3500

3600

3700

3800

Control force number

Max

imum

con

trol

forc

e m

agni

tud

e (N

ewto

ns) El Centro excitation

Xm j 1+, eAm∆t

Xm j, Am1– e

Am∆tI–( )B fmF j+=

F j

F j 0= 0 j ν<≤

F j K fmXm j ν–,–= j ν≥


With this specification for , the response equations take the following form

(8.322)

where

(8.323)

The stability analysis for the SDOF system showed that time delay loweredthe limiting value of the time interval. Therefore, one can obtain an upper boundestimate for by considering no time delay. For this case, eqn (8.322) simplifiesto

(8.324)

Stability requires the modulus of the largest eigenvalue of to be less than 1.

(8.325)

The complexity of necessitates the use of a numerical computationalprocedure. Example 8.2 illustrates the computational details for a SDOF systemwith negative velocity feedback. In general, one ranges over to determine thevalue corresponding to .

When there is time delay, the eigenvalue problem is of order ,

(8.326)

For the SDOF case, it was possible to expand the determinant of the coefficientmatrix and obtain a polynomial expansion for . However, a numericalprocedure is still required to evaluate the roots. For MDOF systems, this hybridapproach is not feasible and one must resort to a fully numerical approach. Given

and , eqn (8.322) can be used to generate the corresponding free vibrationresponse. By decreasing in increments from the critical value for no time delay,one can converge on the limiting value for the specified time delay. This solutionstrategy is efficient when the number of modal DOF is small, on the order of 5.

F j

Xm j 1+, cm1Xm j,= 0 j ν<≤

Xm j 1+, cm1Xm j, cm3Xm j ν–,+= j ν≥

cm1 eAm∆t

=

cm3 Am1– e

Am∆tI–( )B fmK fm–=

∆t

Xm j 1+, cm1 cm3+( )Xm j, cmXm j,= =

cm

ρmax cm( ) 1<

cm

∆tρmax cm( ) 1=

ν 1+

cm3 ρν cm1 ρν 1+ I–+( )Ψ 0=

ρ

∆t ν∆t


The computational cost for the exponential matrix increases nonlinearly with thenumber of modes and eventually becomes excessive. In this case, one can resort toan approximate formulation which assumes the modal equations are fullyuncoupled. This assumption reduces the stability analysis for a probleminvolving s modes to s individual SDOF stability analyses.

The approximate stability formulation proceeds as follows. Firstly thefeedback force is considered to involve only .

(8.327)

Secondly, is taken as a null matrix in eqn (8.306). These conditionscorrespond to pure velocity feedback for a system with uncoupled damping.Lastly, the feedback terms in the k’th equation are assumed to involve only .With these assumptions, eqn (8.306) takes the form

(8.328)

where

(8.329)

The remaining steps are the same as for the SDOF formulation presented insection 8.3.

Introducing matrix notation, eqn (8.328) is expressed as

(8.330)

where

(8.331)

qI

F kv j, qI j,

j 1=

s

∑=

b fR k,

qI k,

qR k, λR k, qR k, λ I k, qI k,–=

qI k, λ I k, qR k, λR k, qI k, bk qI k,–+=

bk b fI k, kv k,=

qk Akqk Bkqk–=

qk

qR k,

qI k,

=


(8.332)

(8.333)

The discrete form allowing for time delay is written as

(8.334)

where the c matrices are now 2 x 2.

(8.335)

(8.336)

One determines the limiting value of for each k ranging from 1 to s, the totalnumber of modes selected to represent the solution. The lowest value controls thechange of the time interval.

Example 8.15: Stability analysis for example 8.14

This example examines the effect of time delay on the response of the 4DOF system considered in Example 8.14. Modal properties corresponding to the 2control force schemes are listed below in Table 1. The time increment isconsidered to be equal to 0.02 secs. An estimate of the limiting value of for eachforce scheme can be obtained using the data contained in Table 1 of Example 8.3.These results are for , and provide a lower bound estimate, . The limitingvalue of for is higher than . For convenience, the table is listed below asTable 2.

AkλR k, λ I k,–

λ I k, λR k,=

Bk0

bk=

qk j 1+, ck1qk j, ck3qk j ν–,+=

ck1 eAk∆t

=

ck3 Ak1– e

Ak∆tI–( )Bk–=

∆t

ν

ξ 0= νoν ξ 0> νo


Table 1:

Table 2: Limiting values of for (Example 8.3, Table 1)

For force scheme 1, the first and second modes have the larger values ofand therefore control the delay. Assuming , Table 2 indicates that mode 2 isclose to instability for ( 0.102 vs. 0.096). The value for mode 1 iscloser to 3. Since is for modes 3 and 4, they are not critical even though thecorresponding time ratios are higher.

Force scheme 2 produces some damping for modes 3 and 4, and since thesetime ratios are the highest, they control the stability limit. Allowing for someincrease due to initial damping, the critical value for is between 1 and 2 formodes 3, 4, and >2 for modes 1, 2. The least critical mode is mode 1 since it has thelowest time ratio.

Figure 1 contains results generated for control force scheme 1, with

Mode TForce scheme 1 Force scheme 2

1 0.498 0.040 0.013 0.212 0.100

2 0.197 0.102 0.029 0.045 0.096

3 0.129 0.155 0.046 0.005 0.078

4 0.095 0.211 0.071 0.000 0.060

Stability limit for

0 0.50 0.500 0.5000

0.05 0.158 0.0955

0.10 0.44 0.151 0.0908

0.15 0.144 0.0860

0.20 0.38 0.137 0.0822

0.25 0.131 0.0780

0.30 0.33 0.124 0.0745

0.35 0.118 0.07085

0.40 0.29 0.113 0.0677

∆tT------ ξ

ξa ξa

∆t T⁄ ξ 0=

∆t T⁄

ξa ν 0= ν 1= ν 2=

ξaξ 0=

ν 2= ∆t T⁄ =

ξa 0≈

ν

ν


taken equal to 2. They are in close agreement with the solution forpresented in Example 8.14. Increasing to 3 produces the responses plotted inFigure 2. Mode 2 has gone critical, and dominates the response whereas mode 1remains at essentially the same peak level as for . These results are based onlimiting the peak magnitude of the control force to 15,000N, which is 3 times thepeak value for . Limiting F simulates saturation of the force actuators.Comparison of the power input plots shows that shifting from 2 to 3 transformsthe energy dissipation process, resulting in energy being supplied to the systemand a periodic forced vibration response following the seismic induced response.Reducing the peak magnitude down to 5,000N, the value required for (andalso ), leads to the responses plotted in Figure 3. The magnitudes have beenreduced, but the system is still responding as if it were subjected to a periodicforced vibration as well as a seismic excitation. Decreasing further will noteliminate this additional motion, as illustrated by the plot for F = 2,500 and1,000N.

Results for force scheme 2 are plotted in Figs 5 and 6. Mode 4 controls theallowable delay. For , there is a negligible effect on the response. For ,mode 1 is essentially unchanged, while mode 2 shows some additional motiondue to delay. Two out of 4 of the control forces reach the prescribed peak value of15,000N. The response corresponding to = 4,000N shown in Fig 7 exhibitssimilar features.

ν 0=ν

ν 2=

ν 2=ν

ν 0=ν 2=

Fmax

ν 1= ν 2=

Fmax


a) Displacement of node 4

b) Time history response of the real part of the coordinate for mode 2

0 5 10 15 20 25 30 35 40 45 50−0.04

−0.03

−0.02

−0.01

0

0.01

0.02

0.03

0.04

Force scheme 1Damping case 1El Centroν 2=

Time (sec)

Nod

al d

ispl

acem

ent (

m) Fmax = 15000N

0 5 10 15 20 25 30 35 40 45 50−3

−2

−1

0

1

2

3

4x 10

−3


Time (sec)

Mod

al c

oord

inat

e re

spon

se

Fmax = 15000N


c) Time history response of the real part of the coordinate for mode 4

d) Time history response of the magnitude of the control force

0 5 10 15 20 25 30 35 40 45 50−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

2.5x 10

−4


Time (sec)

Fmax = 15000N

Mod

al c

oord

inat

e re

spon

se

0 500 1000 1500 2000 2500−5000

−4000

−3000

−2000

−1000

0

1000

2000

3000

4000

5000

Force scheme 1Damping case 1El Centroν 2=Fmax = 15000N

Time step number

Con

trol

forc

e m

agni

tud

e


e) Time history response of the input power

Figure 1

0 500 1000 1500 2000 2500−3500

−3000

−2500

−2000

−1500

−1000

−500

0

500


Time step number

Pow

er s

uppl

ied

by

the

cont

rol f

orce




0 5 10 15 20 25 30 35 40 45 50−0.2

−0.15

−0.1

−0.05

0

0.05

0.1

0.15


Time (sec)

Nod

al d

ispl

acem

ent (

m)

0 5 10 15 20 25 30 35 40 45 50−0.1

−0.08

−0.06

−0.04

−0.02

0

0.02

0.04

0.06

0.08

0.1


Time (sec)

Mod

al c

oord

inat

e re

spon

se



d) Control force magnitude

0 5 10 15 20 25 30 35 40 45 50−2.5

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

2.5x 10

−4

Time (sec)

Mod

al c

oord

inat

e re

spon

se

0 500 1000 1500 2000 2500−1.5

−1

−0.5

0

0.5

1

1.5x 10

4

Time step number

Con

trol

forc

e m

agni

tud

e



e) Input power

Figure 2

0 500 1000 1500 2000 2500−3

−2

−1

0

1

2

3

4

5x 10

4


Time step number

Pow

er s

uppl

ied

by

the

cont

rol f

orce




0 5 10 15 20 25 30 35 40 45 50−0.06

−0.04

−0.02

0

0.02

0.04

0.06 Force scheme 1Damping case 1El Centroν 3=Fmax = 5000N

Time (sec)

Nod

al d

ispl

acem

ent (

m)

0 5 10 15 20 25 30 35 40 45 50−0.03

−0.02

−0.01

0

0.01

0.02

0.03

Time (sec)

Mod

al c

oord

inat

e re

spon

se




d) Time history response of the input power

Figure 3

0 5 10 15 20 25 30 35 40 45 50−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

2.5x 10

−4


Time (sec)

Mod

al c

oord

inat

e re

spon

se

0 500 1000 1500 2000 2500−6000

−4000

−2000

0

2000

4000

6000

Time step number

Pow

er s

uppl

ied

by

the

cont

rol f

orce




b) Displacement of node 4

Figure 4

0 5 10 15 20 25 30 35 40 45 50−0.06

−0.04

−0.02

0

0.02

0.04

0.06

0.08

Time (sec)

Nod

al d

ispl

acem

ent (

m)


0 5 10 15 20 25 30 35 40 45 50−0.06

−0.04

−0.02

0

0.02

0.04

0.06

0.08

Time (sec)

Nod

al d

ispl

acem

ent (

m)





0 5 10 15 20 25 30 35 40 45 50−0.05

−0.04

−0.03

−0.02

−0.01

0

0.01

0.02

0.03

0.04

0.05Force scheme 2Damping case 1El Centroν 1=Fmax = 15000N

Time (sec)

Nod

al d

ispl

acem

ent (

m)

0 5 10 15 20 25 30 35 40 45 50−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

2.5x 10

−4

Time (sec)

Mod

al c

oord

inat

e re

spon

se



c) Peak values of the control forces

Figure 5

1 2 3 42900

3000

3100

3200

3300

3400

3500

3600

3700

3800


Max

imum

con

trol

forc

e m

agni

tud

e





0 5 10 15 20 25 30 35 40 45 50−0.06

−0.04

−0.02

0

0.02

0.04

0.06

Time (sec)

Nod

al d

ispl

acem

ent (

m)


0 5 10 15 20 25 30 35 40 45 50−3

−2

−1

0

1

2

3x 10

−3

Time (sec)

Mod

al c

oord

inat

e re

spon

se




d) Time history response of the real part of the coordinate for mode 1

0 5 10 15 20 25 30 35 40 45 50−3

−2

−1

0

1

2

3

4x 10

−3

Time (sec)

Mod

al c

oord

inat

e re

spon

se


0 5 10 15 20 25 30 35 40 45 50−0.06

−0.04

−0.02

0

0.02

0.04


Time (sec)

Mod

al c

oord

inat

e re

spon

se


e) Peak values of the control forces

Figure 6

1 2 3 40.5

0.6

0.7

0.8

0.9

1

1.1

1.2

1.3

1.4

1.5x 10

4


Max

imum

con

trol

forc

e m

agni

tud

e



Figure 7: Displacement of node 4

Controllability of a particular modal response

The governing equation for the response of the k’th mode is given by eqn(8.305).

(8.337)

By definition, the j’th column of defines the nodal force patterncorresponding to . Premultiplication by generates the equivalent modalforce for mode k. If is orthogonal to the j’th column of , will have noeffect on the response of the k’th mode. Therefore, for mode k to be controllable,the nodal location of the active control forces must satisfy the following constraint

(8.338)

0 5 10 15 20 25 30 35 40 45 50−0.06

−0.04

−0.02

0

0.02

0.04


Time (sec)

Nod

al d

ispl

acem

ent (

m)

qk λkqk b f k, F+ λkqk

2λkf k

---------ΦkTE f F–= =

E fF j Φk

T

ΦkT E f F j

ΦkTE f 0≠


When damping is uncoupled, real, and this constraint is obvious. One cannotapply a control force at a null point of a mode.

Example 8.16: Controllability analysis for a 20 DOF Model

Controllability is illustrated using a 20 DOF shear beam model havingconstant mass and stiffness. Figure 1 shows the first 4 modal displacementprofiles which are real in this case since there is no damping. The correspondinginternodal displacement profiles are plotted in Fig 2. The latter curves are scaledversions of the element shear deformation profiles.

Figure 1: Modal displacement profiles

Φk

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 10

2

4

6

8

10

12

14

16

18

20

1 2 3 4

Amplitude

Nod

e nu

mbe

r

No feedbackm 1000kg=k 1700000N m⁄=


Figure 2: Scaled modal element shear deformation profiles

Suppose the j’th control force system consists of a single force applied atone of the 20 nodes, say node l. If node l is a null point for a particular mode, thej’th force has no effect on the response of that mode. Considering Fig 1, the firstmode has no null points, and it follows that this mode can be controlled byapplying the force at any of the 20 nodes. The second mode, has a null pointbetween nodes 13 and 14. Therefore, these nodes are ineffective. The optimumlocations for controlling mode 2 are nodes 7 and 20 which correspond to themaximum amplitudes. Mode 3 has null points adjacent to nodes 8 and 16. Nodes4,12, and 20 are the optimal force locations. Lastly, mode 4 has null points adjacentto nodes 6, 12, and 18. The corresponding optimal locations are nodes 3, 9, 15, and20. These results are summarized below in Table 1.

−0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.60

2

4

6

8

10

12

14

16

18

20

1 2 3 4

Amplitude

Ele

men

t num

ber


Table 1

Another option is for the j’th control force system to consist of a set of self-equilibrating force applied at adjacent nodes, say l and l+1. The product of the k’thmodal vector, , and the j’th column of is proportional to the sheardeformation for the element located between nodes l and l+1. Therefore, if thiselement corresponds to a null point for the modal element shear deformationprofile, control force system j has no effect on the response of mode k. It followsthat null points are to be avoided, and the elements corresponding to the peakvalues are the optimal locations for the pair of self-equilibrating control forces.Considering element j to be between nodes j and j-1, and the forces are applied atthe ends of an element, the constraints and optimal locations for this case followfrom Fig 2, and are listed in Table 2.

Table 2

Observability of a particular modal response

Feedback is implemented by observing the response, computing the

ModeNodal location - Single Control Force

Optimal Not allowed

1 20 1

2 7, 20 13, 14

3 4, 12, 20 8, 16

4 3, 9, 15, 20 6, 12, 18

Mode

Element location - Self-equilibratingForce

Optimal Not allowed

1 1 20

2 1, 14 7

3 1, 9, 17 5, 13

4 1, 6, 12, 18 3, 4, 9, 10, 15

Φk E f


control forces with an algorithm and then applying these forces with actuators.When the general modal formulation is used, F is taken as a linear function of themodal coordinates,

(8.339)

Assuming s modes are considered, each q is of order s x 1. If both terms areretained, 2s observations are required to uniquely define the modal response.

Let denote the observations. These observations may bedisplacements, velocities, strains, strain rates, or some other measured that arelinearly related to the state variables. The k’th observation is expressed as

(8.340)

where depends on the nature and location of the observation represented by. One can interpret as defining the nodal variables that is monitoring.

Substituting for the state vector in terms of the full set of n coordinates leads to thefollowing exact expression for

(8.341)

If is orthogonal to , does not contribute to and it follows thatdoes not observe . A similar statement applies for . Therefore, for the j’thmode to be observable, at least one of the set of k D’s cannot be orthogonal to themodal vectors for those modes which are included in the modal approximationfor the solution.

Assuming s modes are retained and N observations are made at aparticular instant in time, eqn (8.341) is approximated by

(8.342)

where y is of order N x 1; are of order s x 1; the j’th row of D contains ;and the k’th column of V contains . Observability for the modal coordinatesselected to approximate the solution requires each column of the products,

F kdqR– kvqI–=

y1 y2 … yN, , ,

yk DkU

UDkX= =

Dkyk Dk yk

yk

yk12--- Dk qjV j q jV j+( )

j 1=

n

∑ Dk V R j, qR j, V I j, qI j,–( )

j 1=

n

∑= =

Dk V R j, qR j, yk ykqR j, qI j,

y DV RqR DV IqI–≈

qR qI, D jV k

DV R


and , to have at least one non-zero term. This constraint can be expressed interms of the rank of the products.

(8.343)

Since DV is of order N x s, it follows that . If N = 2s and the rank constraint issatisfied, the solution for and is unique. Otherwise, a least square typesolution has to be generated.

When damping is uncoupled, the linear velocity feedback law expressed interms of modal coordinates simplifies to

(8.344)

where and are related by

(8.345)

Suppose the observations involve modal velocities. Expressing y as

(8.346)

and substituting for results in

(8.347)

Assuming s modes are retained, observability requires for. This constraint is equivalent to the condition that the observation

points are not located at null points for the j’th modal vector. Requiring that rankof to be equal to s satisfies the observability condition. Since it also requires

, the solution is either unique (N = s), or over-determined.

Example 8.17: Observability analysis for a 20 DOF model

Suppose 4 nodes are retained, and the observed variables are nodalvelocities. At least 4 observations are necessary when damping is uncoupled. Forarbitrary damping, 8 observations are required. If damping is uncoupled, the

DV I

rank DV R( ) s=

rank DV I( ) s=

N s≥qR qI

F kvq–=

q U

U Φq≈

y D' U=

U

y D'Φ( )q=

D'Φj 0≠j 1 2 … s, , ,=

D'ΦN s≥

8.6 Optimal Linear Feedback - MDOF Time Invariant System 685

constraint on the location of the observation points is that they do not coincidewith null of the modal displacement profiles. The most desirable locations arethose which coincide with the maximum modal amplitudes. This choiceeliminates the possibility of an ill-conditioned set of equations for the modalvelocities defined by eqn (8.347). The unacceptable and optimal locations for thevelocity observations are listed below.

Table 1

One needs to select at least 4 observation nodes. Taking 7, 9, 12, and 20, andapplying eqn (8.347) leads to the following set of equations for the first 4 modalvelocities,

(1)

Inverting eqn (1) results in

(2)

Modal responsequantity

Nodal location - Single Control Force

Optimal Not allowed

20 1

7, 20 13, 14

4, 12, 20 8, 16

3, 9, 15, 20 6, 12, 18

q1

q2

q3

q4

u7

u9

u12

u20

0.5114 1.0000 0.4440 0.5757–

0.6367 0.8792 0.3019– 0.9941–

0.7959 0.3741 0.9941– 0.15281.0000 0.9941– 0.9824 0.9650–

q1

q2

q3

q4

=

q1

q2

q3

q4

0.6124 0.6095– 0.8513 0.39730.7397 0.0967– 0.0262 0.3375–

0.8881 0.7186– 0.2170– 0.17610.7767 1.2636– 0.6343 0.0975–

u7

u9

u12

u20

=


8.6 Optimal linear feedback - MDOF time invariant systems

The formulation of the linear quadratic regulator problem developed for atime-invariant SDOF system in Section 8.4 is extended here to MDOF systems.Since matrix notation was used for the SDOF case, the only difference betweenthe SDOF and MDOF formulations is the form of the weighting matrices, Q andR, contained in the quadratic performance index. In what follows, the LQRcontrol algorithm corresponding to the modal state-space formulation is derivedfor both the continuous and discrete time feedback scenarios. The role of Q in thiscase is to assign relative weights to the different modal responses. Weighting thedifferent control forces is achieved by adjusting the elements of R. Examplesillustrating the sensitivity of the modal damping due to feedback as a function ofQ and R are presented. Lastly, various issues involved in “selecting” a controlforce scheme are also discussed.

Continuous time modal formulation

The governing equation for this formulation is eqn (8.311). Forconvenience, the relevant equations are listed below.

(a)

where contains the real and imaginary parts of the modal coordinates,

(b)

and the coefficient matrices depend on the modal properties of the MDOF systemand the nodal distribution of the control forces defined by . Negative linearfeedback is taken as

(c)

An expression for is established by requiring the followingperformance index to be stationary,

(8.348)

Xm AmXm B fmF BpmP Bgm ag+ + +=

Xm

Xm

qR

qI

=

E f

F K fmXm– kd kvqR

qI

–= =

K fm

J 12--- Xm

T QXm FTQF+( ) td0

∞

∫=


where Q and R are taken as diagonal weighting matrices,

(8.349)

When Q and R are constant, the solution is

(8.350)

where H is determined by

(8.351)

Just as for the SDOF case, when is taken as a null matrix. This choiceavoids the potential instability associated with displacement feedback.

Given , , R, and Q, one can determine and the eigenvalues ofwith the MATLAB function CARE. These eigenvalues define the

modal damping ratios and frequencies for the system corresponding to theparticular choice of control force locations, , and weighting matrices. Given

, one can generate a range of modal damping ratio distributions by varying theindividual weighting factors contained in Q and R.

To avoid a potential instability due to displacement feedback, is takenas a null matrix. The form of is a generalized version of eqn (8.163) whichapplies for the SDOF case.

(8.352)

where is the modal mass, is the modal frequency, and is the relativeweighting for the i’th mode. With this scaling law, and are if O(1).Increasing places more emphasis on reducing the response of mode n.Increasing places more emphasis on reducing the “cost” for the m’th controlforce. Both of these perturbations result in changes in the magnitudes of themodal damping ratios.

In addition to these parameters, the modal damping distribution is alsoinfluenced by the nature of the control force scheme. A pair of self-equilibratingcontrol forces is equivalent to “material” damping which tends to produce

QQd 0

0 Qv

=

R riδij[ ]=

K fm R 1– B fmT H kd kv= =

AmT H H Am HB fmR 1– B fm

T H–+ Q=

kd 0≡ Qd

Am B fm K fmAm B fmK fm–

E f( )E f

QdQv

Qv 4mi2ωi

2wiδij[ ]=

mi ωi wiwi r j

wnrm


damping that increases with the modal frequency. A single control force appliedat a node is similar to mass proportional damping, and results in damping thatdecreases with the modal frequency. The force schemes considered in Example8.14 exhibited these behavioral trends.

Continuous time LQR control is useful for establishing an initial estimateof the control force system properties required to achieve certain performanceobjectives such as limiting the peak deformations and accelerations in a structuresubject to constraints on the external power requirement and the peak value of thecontrol forces. This “estimate” is then evaluated using the discrete timeformulation specialized for the particular time step selected, and the design ismodified if necessary. The effect of time delay is also considered at this stage.

Discrete time modal formulation

The discrete time free vibration formulation allowing for feedback is basedon eqn (8.320) which is expressed here as

(8.353)

where

(8.354)

For the discrete time case, the quadratic performance index is taken as

(8.355)

with Q, R defined by eqn (8.349). Expressing as

(8.356)

and requiring J to be stationary with respect to leads to the discrete algebraicRiccati equation,

(8.357)

Xm j 1+, cm1Xm j, cm2F j+=

cm1 eAm∆t

=

cm2 Am1– e

Am∆tI–( )B fm=

J 12--- Xm j,

T QXm j, F jTRF j+( )

j 0=

∞

∑=

F j

F j K fmXm j,–=

K fm

H cm1T Hcm1– cm1

T Hcm2( ) R cm2T Hcm2+( ) 1– cm1

T Hcm2( )T+ Q=


and the following expression for the optimal feedback,

(8.358)

This solution can be generated with the MATLAB function DARE.

Substituting for , the free vibration response with optimal feedback isgoverned by

(8.359)

The frequency and damping properties of the discrete model with feedback arerelated to the eigenvalues of . Suppose s modes are considered. There are spairs of complex conjugates,

(8.360)

The j’th pair is expressed in polar form.

(8.361)

With this notation, the modal periods and corresponding damping ratios aregiven by

(8.362)

These equations are useful for comparing discrete vs. continuous feedback.

Although the actual control system is based on discrete time feedback, it isconvenient to work with the continuous time formulation during the preliminarydesign phase which is concerned with selecting the location and nature of thecontrol forces, and estimating the relative weighting factors in order to satisfy thespecified performance requirements. Shifting from continuous to discrete timefeedback with time delay changes the response characteristics, such as the modaldamping, and can be potentially destabilizing. For this formulation, the freevibration response becomes unstable when , which corresponds to a

K fm optimalR cm2

T Hcm2+( ) 1– cm1T Hcm2( )T=

F j

Xm j 1+, cm1 cm2K fm opt–

X j cmX j= =

cm

ρ ρ1 ρ1 ρ2 ρ2 … ρs ρs, , , , , ,=

ρ jρ j ρ jeiθj±

=

ξ j

δj

δj2 θj

2+[ ] 1 2/-------------------------------=

T j2π∆tθj

-------------=

δj ρ j( )ln–=

ρ 1>


negative value for . Examples illustrating various behavioral aspects of MDOFsystems controlled with the LQR algorithm are presented in the following section.

Application studies - LQR control

The 4 DOF model defined in Fig 8.15 is utilized to illustrate the design ofLQR based control force systems. Versions of this model were used in thepreceding examples to demonstrate instability. The focus here is on modaldamping introduced by feedback. Both global and self-equilibrating forceschemes (Fig 8.15 b, c) are considered.

Fig. 8.15: Definition of model and control force schemes

Case 1: Global forcing

The scheme shown in (b) involves specifying 2 force weighting parameters( and ) in R and 4 modal velocity weighting parameters in .Suppose the design objective is a uniform distribution of the peak internodaldisplacements for a specified dynamic excitation, such as an earthquake. Starting

Node/Element m (kg) k (kN/m) c (kNs/m)

1 1000 1700 4

2 1000 1400 3

3 1000 1000 2

4 1000 700 1

ξ

m1

m4

m3

m2

u1

u4

u3

u2

k1 c1,

k2 c2,

k3 c3,

k4 c4,F1

F2F1

F1

F2 F3

F3 F4

b) c)a)

F2

r1 r2 w1 w4→( ) Qv


with uniform weighting, one can adjust the r’s and w’s separately until thedesired uniform displacement profile is obtained. The strategy followed here isbased on first perturbing the force weights to obtain a reasonable level ofdamping, then adjusting the modal weights to obtain essentially uniform modaldamping, and finally scaling the r’s to shift the average value of the peakinternodal displacement to the desired value.

Results for the first step are shown in Figure 8.16 a). Starting with w and rset to unity, the r’s are reduced to 0.5 and 0.25. The modal damping for the firstmode (period sec) is essentially doubled, while the change in thecorresponding damping ratios for the third and fourth modes is small. There isclose agreement between the continuous and discrete feedback results for all butthe second mode.

a)

0.5≈

0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50.045

0.05

0.055

0.06

0.065

0.07

0.075

0.08

0.085

0.09

Continuous feedbackDiscrete feedback

Modal period (sec)

Mod

al d

ampi

ng r

atio

wit

h fe

edba

ck

w 1=

r1 r2 1= =


b)

c)Fig. 8.16: Sensitivity of the modal damping ratio to the control force weighting

factor, r

0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.550.04

0.05

0.06

0.07

0.08

0.09

0.1

0.11

0.12

0.13


Modal period (sec)

Mod

al d

ampi

ng r

atio

wit

h fe

edba

ck

w 1=

r1 r2 0.5= =

0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.550.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18


Modal period (sec)

Mod

al d

ampi

ng r

atio

wit

h fe

edba

ck

w 1=

r1 r2 0.25= =


Taking r = 0.25 as a first trial value for r, the modal weightings are adjustedto increase the damping ratios for the higher modes. Fig 8.17 shows results for aparticular w distribution which places the primary emphasis on modes 3 and 4.The modal damping is nearly uniform, but note that there is a significantdifference between the continuous and discrete damping ratios. This weightingscheme produces the peak internodal displacement profile plotted in Fig 8.17 b.

a) Modal damping ratio with continuous and discrete feedback - no saturationlimit

0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50.12

0.13

0.14

0.15

0.16

0.17

0.18

0.19

0.2

0.21


w1 1=

w2 4=

w3 70=

w4 50=

r1 r2 0.25= =

Modal period (sec)

Mod

al d

ampi

ng r

atio

wit

h fe

edba

ck


b) Element internodal profileFig. 8.17: Modal damping and peak response for initial weighting scheme

Holding the w values constant and decreasing r generates the resultsshown in Fig 8.18. The average damping in increased, and the average peakdisplacement is decreased. One continues this process until the “design”displacement value is obtained. In addition to motion based design requirements,there are also constraints on the peak values of the control forces and the peakpower required. For this design the peak quantities are:

Optimal design involves a consideration of all of these requirements, andassigning priorities for the multiple objectives.

0.01 0.0105 0.011 0.0115 0.012 0.0125 0.013 0.01351

2

3

4

Maximum displacementAverage displacement

Maximum element internodal displacement (m)

Ele

men

t num

ber

w1 1=

w2 4=

w3 70=

w4 50=

r1 r2 0.25= =


F1 max, 4.234kN=

F2 max, 1.783kN=

Peak Power 3.772kNm s⁄=

Average internodal displacement 0.011m=



b) Element internodal displacement profileFig. 8.18:

0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50.14

0.16

0.18

0.2

0.22

0.24

0.26

0.28


w1 1=

w2 4=

w3 70=

w4 50=

r1 r2 0.125= =

Modal period (sec)

Mod

al d

ampi

ng r

atio

wit

h fe

edba

ck

0.0085 0.009 0.0095 0.01 0.0105 0.011 0.0115 0.0121

2

3

4



Ele

men

t num

ber

w1 1=

w2 4=

w3 70=

w4 50=

r1 r2 0.125= =



Case 2: Self-equilibrating control force

The force scheme defined by Fig 8.15c involves 4 control forces. Startingwith uniform weightings, the scale factors are adjusted until the modal dampingratio distribution is essentially the same as obtained with the final weightingscheme of Case 1. Figure 8.19 contains the final results for Case 2. Since the modaldamping distributions are nearly identical, internodal displacement profiles arealso in close agreement. The peak values of the control forces and power for thiscontrol force scheme are

This scheme requires a larger force, 5.074kN vs. 4.234kN. The peak powerrequired depends on the equivalent damping, and since this quantity isessentially the same, it follows that the power requirements will also be close.

F1 max, 3.818kN=

F2 max, 5.074kN=

F3 max, 4.849kN=

F4 max, 4.449kN=

Peak Power 3.599kNm s⁄=

Average internodal displacement 0.0108m=



b) Element internodal displacement profileFig. 8.19: Modal damping and peak response - Self-equilibrating control force

system

0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.550.14

0.16

0.18

0.2

0.22

0.24

0.26

0.28

0.3

0.32

0.34


Modal period (sec)

Mod

al d

ampi

ng r

atio

wit

h fe

edba

ck

w1 8=

w2 5=

w3 12=

w4 25=

r1 r2 r3 r4 0.35= = = =


Ele

men

t num

ber

w1 8=

w2 5=

w3 12=

w4 25=

r1 r2 r3 r4 0.35= = = =


0.0085 0.009 0.0095 0.01 0.0105 0.011 0.0115 0.0121

2

3

4



Example 8.18: Control force design studies for a 20 DOF shear beam

A 20 DOF shear beam with constant mass (1000kg) and constant elementdamping (10,000Nsec/m) is considered. An estimate for the element stiffnessdistribution is selected such that the element shear deformation profile for thefirst mode is essentially uniform, and the average element relative displacementresponse due to the El-Centro seismic excitation is approximately 0.0125m. Figure1 shows the modal properties and response for this choice of stiffness and nofeedback control.

There is close agreement between the actual and desired deformation fornodes 1 through 13. Beyond this point, the difference increases rapidly andexhibits an exponential type growth pattern, similar to the internodaldisplacement profiles for modes 3 and 4. This result indictates that thecontribution of the higher modes is dominating the response in the upper region.

a) Element shear stiffness distributionFigure 1 Properties for no iteration and no feedback

0 0.5 1 1.5 2 2.5 3 3.5

x 106

0

2

4

6

8

10

12

14

16

18

20

Element stiffness-newtons per meter

Ele

men

t num

ber


b) Modal displacement profile-real part

c) Element internodal modal displacement profile-real part

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 10

2

4

6

8

10

12

14

16

18

20

Nod

e nu

mbe

r

Amplitude

−0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.40

2

4

6

8

10

12

14

16

18

20

Amplitude

Ele

men

t num

ber

Figure 1 Properties for no iteration and no feedback


d) Modal damping ratio without feedback

e) Element internodal displacement profile

0.2 0.4 0.6 0.8 1 1.2 1.4 1.60

0.02

0.04

0.06

0.08

0.1

0.12

0.14

Modal period

Mod

al d

ampi

ng r

atio

0.012 0.014 0.016 0.018 0.02 0.022 0.024 0.026 0.0280

2

4

6

8

10

12

14

16

18

20maximum average design value


Ele

men

t num

ber

Figure 1 Properties for no iteration and no feedback


Various options are possible. One can alter the deformation profiles formodes 3 and 4 by modifying the stiffness in the upper region so that the gradientsare decreased. Another option is to work with the initial stiffness and incorporateaddtional damping with feedback control . These approaches generate themaximum values for stiffness (option 1) and damping (option 2). Combiningthese approaches results in intermediate values for these parameters.

Figure 2 contains the results generated by iterating on the element stiffnessaccording to the following algorithm

(1)

The computation proceeds as follows. Using the initial stiffness, the timehistory reponse due to El-Centro is generated, and the peak value of inter-nodaldisplacement is determined for each element.Equation (1) is applied to updateeach element stiffness, and the complete analysis is then repeated. Covergence isquite rapid for this example. After 2 cycles, the correction process has essentiallyreached the final state. Figure 2c shows the modified element displacementprofiles after 2 iterations. The peak value for element 20, the most critical location,has been reduced by approximately 33%. This correcton results in a significantimprovement in the element reponse profiles plotted in Fig 2d. The “bulge” in theupper region has been eliminated.

knew koldmaximum element displacementdesired element displacement

------------------------------------------------------------------------------------------- =


(a) Two iterations

(b) Four iterations

0 0.5 1 1.5 2 2.5 3 3.5 4

x 106

0

2

4

6

8

10

12

14

16

18

20initial iterated

Iterated element stiffness-newtons per meter

Ele

men

t num

ber

0 0.5 1 1.5 2 2.5 3 3.5 4

x 106

0

2

4

6

8

10

12

14

16

18

20initial iterated


Ele

men

t num

ber

Figure 2 Iterated stiffness with no feedback


(c) Two iterations


−0.4 −0.2 0 0.2 0.4 0.6 0.8 10

2

4

6

8

10

12

14

16

18

20

Ele

men

t num

ber

Amplitude-internodal modal displacement profile (real part)


(d) Two iterations

( e) Four iterations

0.0115 0.012 0.0125 0.0130

2

4

6

8

10

12

14

16

18



Ele

men

t num

ber

0.0118 0.012 0.0122 0.0124 0.0126 0.0128 0.0130

2

4

6

8

10

12

14

16

18



Ele

men

t num

ber



Following the second approach, 4 self-equilibrating pairs of nodal forcesare applied on elements 17,18,19 ,and 20. Figure 3 defines the notation for theseforces. Results based on the initial stiffness and the following set of weightingcoeffcients are plotted in Fig 4.

These coefficients were selected to focus the control mainly on modes 3 and 4.Even though the modal damping for these modes is increased significantly, thesestill is a substantial difference between the actual and desired response in theupper zone.

Fig 3

w1 0 w2 =1 w3 3 w4 5r1 r2 r3 r4 0.125

= = == = = =

F2F1

F2 F3

F3 F4

F120

19

18

17

16

1

F4


Saturation limit


0.2 0.4 0.6 0.8 1 1.2 1.4 1.60

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5continuous discrete−Riccatino feedback

Modal period

Mod

al d

ampi

ng r

atio

0.011 0.012 0.013 0.014 0.015 0.016 0.017 0.018 0.0190

2

4

6

8

10

12

14

16

18



Ele

men

t num

ber

a) Modal damping ration with and without continuous or discrete feedback-no

Figure 4 Initial stiffness and feedback


c) Maximum value of the control force

Figure 4 Initial stiffness and feedback

Results based on iterating once on the stiffness, and then applyingfeedback to the “modifed ” system are plotted in Fig. 5. The weighting coefficientsfor this case are:

Increasing the values reduces the equvialent damping, as shown in Fig 5a.However since the stiffness was corrected, the net effect is a significantlyimproved element displacement profile and lower magnitudes for the controlforces.

1 2 3 43000

3100

3200

3300

3400

3500

3600

3700

3800


Max

imum

con

trol

forc

e m

agni

tud

e

w1 0 w2 =1 w3 3 w4 5r1 r2 r3 r4 1.0

= = == = = =

r

(3)


saturation limit

Element internodal displacement profile

0.2 0.4 0.6 0.8 1 1.2 1.4 1.60

0.02

0.04

0.06

0.08

0.1

0.12


Modal period

Mod

al d

ampi

ng r

atio

n

0.0114 0.0116 0.0118 0.012 0.0122 0.0124 0.0126 0.0128 0.0130

2

4

6

8

10

12

14

16

18



Ele

men

t num

ber

a) Modal damping ration with and without continuous or discrete feeback-no

Figure 5 Single iteration on stiffness and then feedback


c) Maximum value of the control forces

d) Iterated element shear stiffness distribution

1 2 3 41420

1430

1440

1450

1460

1470

1480

1490

1500


Max

imum

con

trol

forc

e m

agni

tud

e

0 0.5 1 1.5 2 2.5 3 3.5 4

x 106

0

2

4

6

8

10

12

14

16

18

20initial iterated


Ele

mem

t num

ber

Figure 5 Single iteration on stiffness and then feedback


Applying a single control force at node 20 produces the results shown inFig 6. The corresponding weights and peak value of the control force are

a) Modal damping ratio with and without continuous or discrete feedback-nosaturation limit

Figure 6 Single iteration on stiffness and a single control force

w1 0 w2 =1 w3 3 w4 5r1 3F1 max 1719N

= = ==

=

0.2 0.4 0.6 0.8 1 1.2 1.4 1.60

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09


Modal period

Ele

men

t num

ber

(4)


b) Element internodal displacement profileFigure 6 Single iteration on stiffness and a single control force

This example illustrates that there is no unique solution. One can vary thestiffness, damping, and active feedback control scheme to adjust the response. Inorder to determine the “optional”solution, cost measures need to be assigned toeach of the paremeters. Simulation studies, as illustrated here, provide the data onsensitivities which allows for a more informed decision as to the final design.

Example 8.19: Alternate choice of response measures

The previous examples are based on the specification of either nodalvelocities or the first derivative of the modal coordinates as the responsemeasures included in the performance index for the LQR control algorithm. Otherresponse measures, such as element shear deformation rate, can also be selected.One needs only to specify the relationship between the alternate measures andthe state vector, and specify weights for the alternate measures. Defining Y as the

0.0114 0.0116 0.0118 0.012 0.0122 0.0124 0.01260

2

4

6

8

10

12

14

16

18



Ele

men

t num

ber

-

(1)


vector containing the alternate measures, the relationship is expressed as either:

or

depending on whether the nodal or modal formulation is used. Taking the indexas and substituting for leads to

The remaining steps are the same as for the standard formulation.

This formulation is applied to the system considered in example 8.18. Thereponse measures are taken as the shear deformation rates for elements 18,19, and20. Uniform weighting ( ) is used for the elementdeformations. The location of the control force is the same as for example 8.18.Figure 1 shows some of the results for this case. These plots correspond to Fig 6 ofexample 8.18. The peak value of the control force is 1200 N for Fig1b vs. 1719 N forFig 6.

a) Modal damping ratio with and without continuous or discrete feedback-

Y DX=

Y DXm=

YTQ0Y

Q DTQ0D=

Y

w 1 w 2 w3 1= = =

0.2 0.4 0.6 0.8 1 1.2 1.4 1.60

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09


Modal period

Mod

al d

ampi

ng r

atio

(2)

(3)


no saturation limit

Figure 1 Solution based on using deformation rates for elements 18,19,20

and a single control force at node 20

.


Fig 1 Solution based on using deformation rates for elements 18,19,20 and a singlecontrol force at node 20.

0.01 0.0105 0.011 0.0115 0.012 0.0125 0.013 0.01350

2

4

6

8

10

12

14

16

18



Ele

men

t num

ber


Problems 715

Problems

Problem 8.1

Refer to eqn (8.28). Consider a SDOF system which is initially at rest( ).

a) Determine for the case of a sinusoidal ground acceleration,.

b) Discuss how you would develop a numerical procedure for evaluating at various times such as t1 , t2 , ... etc.

Problem 8.2

Show that the error in the Pade approximation, (eqn 8.51), is of order( .

Problem 8.3

Consider a single degree-of-freedom system with continuous pure velocityfeedback. Suppose the natural period is 1 second. Use eqn(8.54) to determine themaximum allowable time delay corresponding to the following values for and

:

Problem 8.4

Verify eqn(8.77).

Problem 8.5

Consider eqn(8.73). Integrating this equation between and leads

u0 u0 0= =

u t( )a g a g Ωtsin=

u

λ td( )3

eλ td– 1 1

2--- λ td( )–

1 12--- λ td( )+

-------------------------- Error+=

ξξa

ξ 0.05ξa 0.05 0.10 0.20 0.30, , ,

==

t j t j 1+


to

Suppose the integrand is assumed to vary linearly over the time interval,and the coefficient matrices ,are constant.

a) Derive the expression for corresponding to these conditions.Compare this result with eqn(8.77). Comment on the nature of the error.

b) Specialize a) for negative linear feedback, and compare with eqn(8.82) .

c) Specialize b) for no time delay and free vibration response. Comparewith eqn(8.86). Define the stability requirement for this approximation.

Problem 8.6

A SDOF system has the following properties:

Using the data contained in example 8.2, estimate the limiting time step foreach of the following values of the feedback parameters:

Problem 8.7

Apply numeral simulation to eqn(8.117) and establish an estimate for thestability limit of the following systems:

Xj 1+ Xj– AX BfF Bgag Bp p+ + +( ) td

t j

t j 1+

∫=

A Bf

…, ,

Xj 1+

m 1000 kgk 70 000 N mc

⁄,320 N s⋅ m⁄

===

k d 0kv 0 800 1600 3200 6400 N s⋅ m⁄, , , ,

==

m 1000 kgk 60 000N m⁄c

,750 N s⋅ m⁄

k d 0k v 3000 N s⋅ m⁄υ 0 1 2, ,

===

==

=

Problems 717

Example 8.3 contains data for c=0. The STABILITY option of MOTIONLAB solveseqn (8.117) for given values of the system properties and the time ratio, .

Problem 8.8

Refer to Example 8.3. Extend Table 1 to include . Consider and .

Problem 8.9

Verify eqn(8.158) using the CARE function of MATLAB. Take:

and consider to have the following values: 0,0.5,1.0.

Problem 8.10

Refer to eqn (8.160).Will the LQR algorithm ever produce an unstablesystem?

Problem 8.11

Consider eqn(8.168). Let

Noting the identity,

and the limit condition,

∆t T⁄

ξ 0.05=υ 0 1 2, ,= ξa 0 0.1 0.2 0.3, , ,=

m 1000 kg k 60 000N m⁄ c, 750N ˙ s m⁄

c 1 qd 0 qv 4ω2m2qv˙

= = =

= = =

qv

D Q KfT RKf

Sn

+

Cj T,

DC j

j 0=

n

∑

=

=

CTSnC Sn– D– Cn 1 T,+ DCn 1++=

C j 0 as j ∞→→


derive eqn(8.169)

Problem 8.12

Refer to Figs 1 and 2 of example 8.4. Suppose the time ratio isdetermined by the external loading, and is equal to 0.1. Suggest a value forsuch that is close to 0.2 when .

Problem 8.13

Consider the following system

Suppose . Select the parameters for discrete time feedbackcontrol such that the effective damping ratio is equal to 0.2, Use Figs 1 and 2 ofexample 8.4 to obtain an initial estimate, and the discrete time Riccati equationoption of MOTIONLAB to refine the estimate. Note that the solution tendstoward the continuous time feedback case as approaches 0.

Problem 8.14

Refer to example 8.5. Compare the expressions for andcorresponding to with the continuous time Riccati solution defined byeqn(8.161). Use the discrete time Riccati solution for listed in example8.4 to compare the values of required to produce for 2 timeincrements, and 0.1.

Problem 8.15

Rework Problem 8.13 using the finite interval discrete time algebraicRiccati equation. Note that the weighting factors for the finite intervalformulation are different than the corresponding weighing factors for the discretetime algebraic Riccati equation.

Problem 8.16

Verify eqn(8.236).

∆t T⁄qv

ξa ξ 0.05=

m 1000 kgk 60 000 N m⁄c=1000 N.s m⁄

,=

=

∆t 0.02s=

∆t T⁄

kd kvqd 0=

ξ 0.02=qv kv 2ωm⁄( ) 0.2=

∆t T⁄ 0.02=

Problems 719

Problem 8.17

Refer to eqn(5) of example 8.7. Show that the eigenvetors are real when Kand M are symmetric and the eigenvalues are positive real quantities when K andM are positive definite.

Problem 8.18

Show that for by substituting for W and V using eqn(8.267) and noting the definition equation for , .

Problem 8.19

Refer to eqn(2) of example 8.11. Reduce these equations to a singleequation in terms of and compare with the equation obtained with theconventional modal formulation(e.g, see eqn(8.275). Is there any advantage toworking with the state-space formulation vs the conventional modalformulation?

Problem 8.20

Consider a 20 DOF system having the following constant properties:

a) Determine the modal properties (period, damping, modal displacementprofile, modal intermodal displacement profile) for the first 4 modes.

b) Determine the response for node 20 due to the Kobe ground excitationusing the complete state-space formulation. Take sec.

c) Repeat part b) using the modal state-space formulation and only the first4 modes. Compare the time history and Fourier components for node 20 with thecorresponding results obtained in b). Also compare the peak values of the modalcoordinates.

W jTVk 0= j k≠

φj λ j

λ j2Mφj λ jCφj Kφj+ + 0=

qR

mass 1000 kgelement stiffness 1800 kN m⁄element damping 25 kN s m⁄⋅

===

∆t 0.02=


Problem 8.21

Consider a SDOF system having the following properties:

a) Using the LQR control algorithm, establish values for the weightingparameters such that the effective damping for continuous velocity feedback hasthe following values: .

b) Take seconds. Evaluate the discrete time damping ratios fora).

c) Using the finite time interval control algorithm, establish values for theweighting parameters such that the discrete time damping ratios are the same asfound in b).

d) Using the model properties corresponding to established in a),determine the maximum values of the displacement, control force, and powerassociated with the El-Centro accelerogram. Use .

e) Repeat d) for the Kobe accelerogram.

f) Repeat d) for the Mexico City 1 accelerogram.

Problem 8.22

Consider a 4 DOF system having the following properties:

Suppose a single control force is applied at the top node. Using the LQRalgorithm, select the weighting parameters which result in a value of the dampingratio for discrete feedback equal to 0.2 for the first mode. Take seconds,and apply the following strategies:


1 1000 1700 4

2 1000 1400 3

3 1000 1000 2

4 1000 700 1

m 10 000 kgk

,400 000 N m⁄

c=2500 N s⋅ m⁄,

==

ξeq 0.05 0.1 0.2, ,=

∆t 0.02=

ξeq 0.1=

∆t 0.02=

∆t 0.02=

Problems 721

a) Use the conventional state-space formulation and weight the nodalvelocities uniformly.

b) Use the modal state-space formulation and weight the first derivative ofthe modal coordinates uniformly.

c) Use the conventional state-space formulation and weight the elementdeformation rates uniformly.

d) Repeat c) using the modal state-space formulation.

Problem 8.23

Consider a 5 DOF shear beam with the following constant mass andstiffness properties:

a) Assuming uniform element viscous damping, determine the magnitudeof element damping such that the first mode damping ratio is 0.02.

b) Apply a single control force at mode 5. Assuming all 5 modes areretained, and they are weighted equally, determine the weighting parameterssuch that the equivalent damping for continuous feedback is 0.15 for the firstmode.

c) Determine the corresponding damping ratio for discrete time feedback.Take seconds.

d) Investigate the effect of delay on the free vibration reponse of the modalcoordinates due to an initial displacement. Use seconds and theparameters established in part b.

Problem 8.24

Consider the following 5 DOF systems:


1 1000 2000 5

2 1000 1700 5

m 10 000 kgk

,350 000 N m⁄,

==

∆t 0.02=

∆t 0.02=


a) Suppose control forces are applied at all 5 nodes. Determine the modalcoordinate weighting parameters such that the equivalent damping ratiocorresponding to continuous feedback is equal to 0.15 for the first mode. Assumeuniform weighting.

b) Suppose self-equilibrating sets of control forces are applied on all 5elements and the weighting is applied to the element deformation time rates.Determine the weights such that the first mode damping ratio is 0.15. Assumeuniform weighting.

c) Apply the Northridge earthquake to each system and compare

1. the internode displacement profiles

2. the peak power

3. the peak value of the control forces

Problem 8.25

Consider a 10 DOF shear beam with constant mass, element stiffness, andelement damping. Take m=10000 kg .

a) Determine the stiffness and damping constants such that the propertiesfor the first mode are

b) Select an active control force scheme which provides a damping ratio of0.2 for the first mode.

c) Apply the Kobe ground acceleration to the system defined in b).Examine the responses of the first 3 modes. Generate both the time histories andthe Fourier components.

3 1000 1400 5

4 1000 1000 5

5 1000 700 5


Period 1 ondDamping ratio

sec0.02

==

Problems 723

d) If the design objective is to have uniform peak element sheardeformation throughout the system, what design modifications would yousuggest? Illustrate your strategy for the case where the target value of the relativeinter-nodal displacement is 0.0125m.

Problem 8.26

Consider a 10 DOF shear beam with constant nodal mass equal to 10000kg.

a) Select a parabolic distribution of element stiffness and a constantelement viscous damping so that the period for the first mode is 1 second, andthe modal damping ratio is 0.02.

b) Carry out 2 cycles of iteration on the element stiffness using the El-Centro ground excitation and 0.0125m as the desired value of internodaldisplacement.

c) Incorporate active control in the system obtained in part b. Select theweighting parameters such that the modal damping ratios ( for continuousfeedback ) for the first 3 modes are approximately equal to 0.15. Consider a globalforcing scheme and weight the modal coordinate velocities.

d) Repeat c) using self-equilibrating control force schemes and weight themodal coordinate velocities.

e) Repeat d) using the internodal element displacements as theperformance measures. (See Example 8.19).

Problem 8.27

Consider the bending beam-outrigger system shown below. Assume theoutriggers are inifintely stiff, the beam bending rigidity is constant, and the cablesare initially tensioned to a level of . Suppose the cable tensions can becontinuously adjusted to counteract the effect of lateral load.

T0


a) Take the lateral displacement and rotation at points B, C as the degrees-of-freedom and establish the corresponding matrix form of the equilibriumequations. Work with "lumped" masses, rotatory inertias, and loads.

b) Develop the state-space formulation for a)

c) Describe how you would implement linear velocity feedback control.

d) Suppose the bending rigidity is specified, and the critical dynamicloading is a uniform periodic excitation. Discuss how you would "calibrate" thefeedback parameters for the case where the design objective is to limit themaximum acceleration. Illustrate your strategy.

To ∆T+ To ∆T–

A

B

C

b b

DB,

x

3H/2

m1=a( H)

m1m1

u(x ,t)

ρmH

ρm

Bibliography 735

Bibliography

Akiyama, H. (1985). Earthquake-Resistant Limit-State Design for Buildings,University of Tokyo Press.

Anderson, B.D.O. & Moore, J.B. (1990). Optimal Control: Linear Quadratic Methods,Prentice Hall.

Bachmann, H. & Ammann, W. (1987). Vibrations in Structures. IABSE, Zurich.

Bachmann, H., et al. (1995). Vibration Problems in Structures: Practical Guidelines,Birkauzer Verlag, Basel.

Beskos, D. & Anagnostopoulos, S. (1997). Computer Analysis and Design ofEarthquake Resistant Structures, C.M.P., Boston.

Booth, E. (1998). Seismic Design Practice into the Next Century, Balkema, Rotterdam,Brookfield.

Brogan, W.L. (1991). Modern Control Theory, 3rd Edition, Prentice Hall.

Chopra, A. (1995). Dynamics of Structures, Prentice Hall.

Clark, R.L., Saunders, W.R. & Gibbs, G.P. (1998), Adaptive Structures, Wiley

736 Bibliography

Clough, J.W. & Penzien, J. (1993). Dynamics of Structures, 2nd Edition, McGraw-Hill.

D’Azzo, J.J. & Houpis, C.H. (1995). Linear Control System Analysis and Design:Conventional and Modern, McGraw-Hill.

Den Hartog, J. P. (1956). Mechanical Vibrations, 4th Edition, McGraw-Hill.

DeRusso, P.M., Roy, R.J. & Close, C.M. (1965). State Variables for Engineers, Wiley.

Kelly, J.M. (1993). Earthquake-Resistant Design with Rubber, Springer-Verlag.

Meirovitch, L. (1990). Dynamics and Control of Structures, Wiley.

Soong, T.T. (1990). Active Structural Control: Theory and Practice, Wiley.

Strang, G. (1993). Introduction to Linear Algebra, Wellesley-Cambridge Press.

Suh, N.P. (1990). The Principles of Design, Oxford University Press.

First World Conference on Structural Control (1994). International Association forStructural Control, Pasadena, California.

11th International Conference on Structural Mechanics Reactor Technology (1991).Seismic Isolation and Response Control for Nuclear and Non-Nuclear Structures,Tokyo, Japan.

International Workshop on Recent Developments in Base Isolation Techniques forBuildings (1992). Architectural Institute of Japan, Tokyo.

Introduction to Structural Motion Control Connor

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