Fundamentals of Structural Dynamics

Course “Fundamentals of Structural Dynamics” An-Najah National University April 19 - April 23, 2013 Lecturer: Dr. Alessandro Dazio, UME School

Course “Fundamentals of Structural Dynamics”

April 19 - April 23, 2013

4 Schedule of classes Date

Fundamentals of Structural Dynamics

Day 1 Fri. April 19 2013

1 Course description Aim of the course is that students develop a “feeling for dynamic problems” and acquire the theoretical background and the tools to understand and to solve important problems relevant to the linear and, in part, to the nonlinear dynamic behaviour of structures, especially under seismic excitation. The course will start with the analysis of single-degree-of-freedom (SDoF) systems by discussing: (i) Modelling, (ii) equations of motion, (iii) free vibrations with and without damping, (iv) harmonic, periodic and short excitations, (v) Fourier series, (vi) impacts, (vii) linear and nonlinear time history analysis, and (viii) elastic and inelastic response spectra. Afterwards, multi-degree-of-freedom (MDoF) systems will be considered and the following topics will be discussed: (i) Equation of motion, (ii) free vibrations, (iii) modal analysis, (iv) damping, (v) Rayleigh’s quotient, and (vi) seismic behaviour through response spectrum method and time history analysis. To supplement the suggested reading, handouts with class notes and calculation spreadsheets with selected analysis cases to self-training purposes will be distributed. Lecturer:

Dr. Alessandro Dazio, UME School

Day 2 Sat. April 20 2013

Day 3 Sun. April 21 2013

Day 4 Mon. April 22 2013

2 Suggested reading

Time

Topic

09:00 - 10:30

1. Introduction 2. SDoF systems: Equation of motion and modelling

11:00 - 12:30

3. Free vibrations

14:30 - 16:00

Assignment 1

16:30 - 18:00

Assignment 1

9:00 - 10:30

4. Harmonic excitation

11:00 - 12:30

5. Transfer functions

14:30 - 16:00

6. Forced vibrations (Part 1)

16:30 - 18:00

6. Forced vibrations (Part 2)

09:00 - 10:30

7. Seismic excitation (Part 1)

11:00 - 12:30


14:30 - 16:00

Assignment 2

16:30 - 18:00

Assignment 2

9:00 - 10:30

8. MDoF systems: Equation of motion

11:00 - 12:30

9. Free vibrations

14:30 - 16:00

10. Damping 11. Forced vibrations

[Cho11]

Chopra A., “Dynamics of Structures”, Prentice Hall, Fourth Edition, 2011.

16:30 - 18:00

11. Forced vibrations

[CP03]

Clough R., Penzien J., “Dynamics of Structures”, Second Edition (revised), Computer and Structures Inc., 2003.

09:00 - 10:30


[Hum12] Humar J.L., “Dynamics of Structures”. Third Edition. CRC Press, 2012.

3 Software

Day 5 Tue. April 23 2013

11:00 - 12:30


14:30 - 16:00

Assignment 3

16:30 - 18:00

Assignment 3

In the framework of the course the following software will be used by the lecturer to solve selected examples: [Map10]

Maplesoft: “Maple 14”. User Manual. 2010

[Mic07]

Microsoft: “Excel 2007”. User Manual. 2007

[VN12]

Visual Numerics: “PV Wave”. User Manual. 2012

As an alternative to [VN12] and [Map10] it is recommended that students make use of the following software, or a previous version thereof, to deal with coursework: [Mat12]

MathWorks: “MATLAB 2012”. User Manual. 2012

A. Dazio, April 19, 2013

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Table of Contents


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3.2 Damped free vibrations ....................................................................... 3-6

Table of Contents...................................................................... i

3.2.1

Formulation 3: Exponential Functions ....................................................... 3-6

3.2.2

Formulation 1: Amplitude and phase angle............................................. 3-10

3.3 The logarithmic decrement .............................................................. 3-12 3.4 Friction damping ............................................................................... 3-15

1 Introduction 1.1 Goals of the course .............................................................................. 1-1 1.2 Limitations of the course..................................................................... 1-1 1.3 Topics of the course ............................................................................ 1-2 1.4 References ............................................................................................ 1-3

4 Response to Harmonic Excitation 4.1 Undamped harmonic vibrations ......................................................... 4-3 4.1.1

Interpretation as a beat ................................................................................ 4-5

4.1.2

Resonant excitation (ω = ωn) ....................................................................... 4-8

4.2 Damped harmonic vibration .............................................................. 4-10 4.2.1

2 Single Degree of Freedom Systems 2.1 Formulation of the equation of motion............................................... 2-1 2.1.1

Direct formulation......................................................................................... 2-1

2.1.2

Principle of virtual work............................................................................... 2-3

2.1.3

Energy Formulation...................................................................................... 2-3

2.2 Example “Inverted Pendulum”............................................................ 2-4 2.3 Modelling............................................................................................. 2-10 2.3.1

Structures with concentrated mass.......................................................... 2-10

2.3.2

Structures with distributed mass ............................................................. 2-11

2.3.3

Damping ...................................................................................................... 2-20

3 Free Vibrations

Resonant excitation (ω = ωn) ..................................................................... 4-13

5 Transfer Functions 5.1 Force excitation .................................................................................... 5-1 5.1.1

Comments on the amplification factor V.................................................... 5-4

5.1.2

Steady-state displacement quantities ........................................................ 5-8

5.1.3

Derivating properties of SDoF systems from harmonic vibrations....... 5-10

5.2 Force transmission (vibration isolation) ......................................... 5-12 5.3 Base excitation (vibration isolation)................................................. 5-15 5.3.1

Displacement excitation ........................................................................... 5-15

5.3.2

Acceleration excitation ............................................................................. 5-17

5.3.3

Example transmissibility by base excitation .......................................... 5-20

5.4 Summary Transfer Functions ........................................................... 5-26 3.1 Undamped free vibrations ................................................................... 3-1 3.1.1

Formulation 1: Amplitude and phase angle............................................... 3-1

3.1.2

Formulation 2: Trigonometric functions .................................................... 3-3

3.1.3

Formulation 3: Exponential Functions ....................................................... 3-4

6 Forced Vibrations 6.1 Periodic excitation .............................................................................. 6-1 6.1.1

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Steady state response due to periodic excitation..................................... 6-4

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7.5 Elastic response spectra ................................................................... 7-42

6.1.2

Half-sine ........................................................................................................ 6-5

6.1.3

Example: “Jumping on a reinforced concrete beam”............................... 6-7

7.5.1

Computation of response spectra ............................................................ 7-42

6.2 Short excitation .................................................................................. 6-12

7.5.2

Pseudo response quantities...................................................................... 7-45

Step force .................................................................................................... 6-12

7.5.3

Properties of linear response spectra ..................................................... 7-49

6.2.2

Rectangular pulse force excitation .......................................................... 6-14

7.5.4

Newmark’s elastic design spectra ([Cho11]) ........................................... 7-50

6.2.3

Example “blast action” .............................................................................. 6-21

7.5.5

Elastic design spectra in ADRS-format (e.g. [Faj99]) (Acceleration-Displacement-Response Spectra) .................................... 7-56

6.2.1

7.6 Strength and Ductility ........................................................................ 7-58

7 Seismic Excitation

7.6.1

Illustrative example .................................................................................... 7-58

7.1 Introduction .......................................................................................... 7-1

7.6.2

“Seismic behaviour equation” .................................................................. 7-61

7.2 Time-history analysis of linear SDoF systems ................................. 7-3

7.6.3

Inelastic behaviour of a RC wall during an earthquake ........................ 7-63

Newmark’s method (see [New59]) .............................................................. 7-4

7.6.4

Static-cyclic behaviour of a RC wall ........................................................ 7-64

7.2.2

Implementation of Newmark’s integration scheme within the Excel-Table “SDOF_TH.xls”.................................................................. 7-8

7.6.5

General definition of ductility ................................................................... 7-66

7.6.6

Types of ductilities .................................................................................... 7-67

7.2.3

Alternative formulation of Newmark’s Method. ....................................... 7-10

7.2.1

7.3 Time-history analysis of nonlinear SDoF systems ......................... 7-12

7.7 Inelastic response spectra ............................................................... 7-68 7.7.1

Inelastic design spectra............................................................................. 7-71

7.7.2

Determining the response of an inelastic SDOF system by means of inelastic design spectra in ADRS-format ........................... 7-80

7.3.1

Equation of motion of nonlinear SDoF systems ..................................... 7-13

7.3.2

Hysteretic rules........................................................................................... 7-14

7.3.3

Newmark’s method for inelastic systems ................................................ 7-18

7.7.3

Inelastic design spectra: An important note............................................ 7-87

7.3.4

Example 1: One-storey, one-bay frame ................................................... 7-19

7.7.4

Behaviour factor q according to SIA 261 ................................................. 7-88

7.3.5

Example 2: A 3-storey RC wall .................................................................. 7-23

7.4 Solution algorithms for nonlinear analysis problems .................... 7-26 7.4.1

General equilibrium condition................................................................. 7-26

7.4.2

Nonlinear static analysis ........................................................................... 7-26

7.4.3

The Newton-Raphson Algorithm............................................................... 7-28

7.4.4

Nonlinear dynamic analyses ..................................................................... 7-35

7.4.5

Comments on the solution algorithms for nonlinear analysis problems ..................................................................... 7-38

7.4.6

Simplified iteration procedure for SDoF systems with idealised rule-based force-deformation relationships............................ 7-41

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7.8 Linear equivalent SDOF system (SDOFe) ....................................... 7-89 7.8.1

Elastic design spectra for high damping values ..................................... 7-99

7.8.2

Determining the response of inelastic SDOF systems by means of a linear equivalent SDOF system and elastic design spectra with high damping ............................................. 7-103

7.9 References ........................................................................................ 7-108

8 Multi Degree of Freedom Systems 8.1 Formulation of the equation of motion............................................... 8-1 8.1.1

Equilibrium formulation ............................................................................... 8-1

8.1.2

Stiffness formulation ................................................................................... 8-2

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8.1.3

Flexibility formulation ................................................................................. 8-3

10.3Classical damping matrices .............................................................. 10-5

8.1.4

Principle of virtual work............................................................................... 8-5

10.3.1 Mass proportional damping (MpD) ......................................................... 10-5

8.1.5

Energie formulation...................................................................................... 8-5

10.3.2 Stiffness proportional damping (SpD) .................................................... 10-5

8.1.6

“Direct Stiffness Method”............................................................................ 8-6

10.3.3 Rayleigh damping..................................................................................... 10-6

8.1.7

Change of degrees of freedom.................................................................. 8-11

10.3.4 Example....................................................................................................... 10-7

8.1.8

Systems incorporating rigid elements with distributed mass ............... 8-14

11 Forced Vibrations 9 Free Vibrations

11.1Forced vibrations without damping ................................................. 11-1

9.1 Natural vibrations ................................................................................. 9-1

11.1.1 Introduction ................................................................................................ 11-1

9.2 Example: 2-DoF system ...................................................................... 9-4

11.1.2 Example 1: 2-DoF system.......................................................................... 11-3

9.2.1

Eigenvalues ................................................................................................ 9-4

11.1.3 Example 2: RC beam with Tuned Mass Damper (TMD) without damping... 11-7

9.2.2

Fundamental mode of vibration .................................................................. 9-5

9.2.3

Higher modes of vibration ........................................................................... 9-7

11.2Forced vibrations with damping ..................................................... 11-13

9.2.4

Free vibrations of the 2-DoF system .......................................................... 9-8

11.2.1 Introduction .............................................................................................. 11-13

9.3 Modal matrix and Spectral matrix ..................................................... 9-12

11.3Modal analysis: A summary ............................................................ 11-15

9.4 Properties of the eigenvectors.......................................................... 9-13


9.4.1

Orthogonality of eigenvectors .................................................................. 9-13

9.4.2

Linear independence of the eigenvectors................................................ 9-16

12.1Equation of motion............................................................................. 12-1

9.5 Decoupling of the equation of motion.............................................. 9-17

12.1.1 Introduction................................................................................................. 12-1

9.6 Free vibration response..................................................................... 9-22 9.6.1

Systems without damping ......................................................................... 9-22

9.6.2

Classically damped systems..................................................................... 9-24

12.1.2 Synchronous Ground motion.................................................................... 12-3 12.1.3 Multiple support ground motion ............................................................... 12-8

12.2Time-history of the response of elastic systems .......................... 12-18 12.3Response spectrum method ........................................................... 12-23

10 Damping

12.3.1 Definition and characteristics ................................................................. 12-23

10.1Free vibrations with damping ........................................................... 10-1 10.2Example .............................................................................................. 10-2 10.2.1 Non-classical damping ............................................................................ 10-3 10.2.2 Classical damping .................................................................................... 10-4

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12.3.2 Step-by-step procedure ......................................................................... 12-27

12.4Practical application of the response spectrum method to a 2-DoF system ............................................................................ 12-29 12.4.1 Dynamic properties ................................................................................. 12-29 12.4.2 Free vibrations.......................................................................................... 12-31

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12.4.3 Equation of motion in modal coordinates.............................................. 12-38


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14 Pedestrian Footbridge with TMD

12.4.4 Response spectrum method ................................................................... 12-41 12.4.5 Response spectrum method vs. time-history analysis ........................ 12-50

14.1Test unit and instrumentation ........................................................... 14-1 14.2Parameters .......................................................................................... 14-4

13 Vibration Problems in Structures

14.2.1 Footbridge (Computed, without TMD) ...................................................... 14-4

13.1Introduction ........................................................................................ 13-1 13.1.1 Dynamic action ........................................................................................... 13-2

14.2.2 Tuned Mass Damper (Computed) ............................................................. 14-4

14.3Test programme ................................................................................. 14-5

13.1.2 References .................................................................................................. 13-3

14.4Free decay test with locked TMD ...................................................... 14-6

13.2Vibration limitation ............................................................................. 13-4

14.5Sandbag test ....................................................................................... 14-8

13.2.1 Verification strategies ................................................................................ 13-4

14.5.1 Locked TMD, Excitation at midspan ....................................................... 14-9

13.2.2 Countermeasures ....................................................................................... 13-5

14.5.2 Locked TMD, Excitation at quarter-point of the span .......................... 14-12

13.2.3 Calculation methods .................................................................................. 13-6

14.5.3 Free TMD: Excitation at midspan .......................................................... 14-15

13.3People induced vibrations................................................................. 13-8

14.6One person walking with 3 Hz......................................................... 14-17

13.3.1 Excitation forces......................................................................................... 13-8

14.7One person walking with 2 Hz......................................................... 14-20

13.3.2 Example: Jumping on an RC beam....................................................... 13-15 13.3.3 Footbridges............................................................................................... 13-18 13.3.4 Floors in residential and office buildings .............................................. 13-26 13.3.5 Gyms and dance halls.............................................................................. 13-29

14.7.1 Locked TMD (Measured) .......................................................................... 14-20 14.7.2 Locked TMD (ABAQUS-Simulation) ...................................................... 14-22 14.7.3 Free TMD .................................................................................................. 14-24 14.7.4 Remarks about “One person walking with 2 Hz” .................................. 14-25

13.3.6 Concert halls, stands and diving platforms........................................... 13-30

13.4Machinery induced vibrations......................................................... 13-30

14.8Group walking with 2 Hz .................................................................. 14-26 14.8.1 Locked TMD ............................................................................................ 14-29

13.5Wind induced vibrations.................................................................. 13-31

14.8.2 Free TMD ................................................................................................. 14-30

13.5.1 Possible effects ........................................................................................ 13-31

14.9One person jumping with 2 Hz ........................................................ 14-31

13.6Tuned Mass Dampers (TMD) ........................................................... 13-34

14.9.1 Locked TMD .............................................................................................. 14-31

13.6.1 Introduction............................................................................................... 13-34

14.9.2 Free TMD ................................................................................................. 14-33

13.6.2 2-DoF system ........................................................................................... 13-35

14.9.3 Remarks about “One person jumping with 2 Hz” ................................. 14-34

13.6.3 Optimum TMD parameters....................................................................... 13-39 13.6.4 Important remarks on TMD...................................................................... 13-39

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Table of Contents

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1 Introduction


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1.3 Topics of the course 1) Systems with one degree of freedom

1.1 Goals of the course • Presentation of the theoretical basis and of the relevant tools; • General understanding of phenomena related to structural dynamics;

- Modelling and equation of motion - Free vibrations with and without damping - Harmonic excitation 2) Forced oscillations

• Focus on earthquake engineering; • Development of a “Dynamic Feeling”;

- Periodic excitation, Fourier series, short excitation

• Detection of frequent dynamic problems and application of appropriate solutions.

- Linear and nonlinear time history-analysis - Elastic and inelastic response spectra 3) Systems with many degree of freedom

1.2 Limitations of the course • Only an introduction to the broadly developed field of structural dynamics (due to time constraints);

- Modelling and equation of motion - Modal analysis, consideration of damping

• Only deterministic excitation;

- Forced oscillations,

• No soil-dynamics and no dynamic soil-structure interaction will be treated (this is the topic of another course);

- Seismic response through response spectrum method and time-history analysis

• Numerical methods of structural dynamics are treated only partially (No FE analysis. This is also the topic of another course);

4) Continuous systems

• Recommendation of further readings to solve more advanced problems.

5) Measures against vibrations

1 Introduction

1 Introduction

Page 1-1

- Generalised Systems

- Criteria, frequency tuning, vibration limitation

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1.4 References Blank page Theory [Bat96]

Bathe KJ: “Finite Element Procedures”. Prentice Hall, Upper Saddle River, 1996.

[CF06]

Christopoulos C, Filiatrault A: "Principles of Passive Supplemental Damping and Seismic Isolation". ISBN 88-7358-0378. IUSSPress, 2006.

[Cho11]

Chopra AK: “Dynamics of Structures”. Fourth Edition. Prentice Hall, 2011.

[CP03]

Clough R, Penzien J: “Dynamics of Structures”. Second Edition (Revised). Computer and Structures, 2003. (http://www.csiberkeley.com)

[Den85]

Den Hartog JP: “Mechanical Vibrations”. Reprint of the fourth edition (1956). Dover Publications, 1985.

[Hum12]

Humar JL: “Dynamics of Structures”. Third Edition. CRC Press, 2012.

[Inm01]

Inman D: “Engineering Vibration”. Prentice Hall, 2001.

[Prz85]

Przemieniecki JS: “Theory of Matrix Structural Analysis”. Dover Publications, New York 1985.

[WTY90]

Weawer W, Timoshenko SP, Young DH: “Vibration problems in Engineering”. Fifth Edition. John Wiley & Sons, 1990.

Practical cases (Vibration problems) [Bac+97] Bachmann H et al.: “Vibration Problems in Structures”. Birkhäuser Verlag 1997.

1 Introduction

Page 1-3

1 Introduction

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2 Single Degree of Freedom Systems


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2) D’Alembert principle (2.4)

F+T = 0

2.1 Formulation of the equation of motion

The principle is based on the idea of a fictitious inertia force that is equal to the product of the mass times its acceleration, and acts in the opposite direction as the acceleration The mass is at all times in equilibrium under the resultant force F and the inertia force T = – mu··.

2.1.1 Direct formulation 1) Newton's second law (Action principle) F =

dI = d ( mu· ) = mu·· dt dt

( I = Impulse)

(2.1)

The force corresponds to the change of impulse over time.

y = x ( t ) + l + us + u ( t )

(2.5)

y·· = x·· + u··

(2.6)

T = – my·· = – m ( x·· + u··)

(2.7)

F = – k ( u s + u ) – cu· + mg (2.8) = – ku s – ku – cu· + mg = – ku – cu· F+T = 0

(2.9)

– cu· – ku – mx·· – mu·· = 0 (2.10) – f k ( t ) – f c ( t ) + F ( t ) = mu··( t )

mu·· + cu· + ku = – mx··

Introducing the spring force f k ( t ) = ku ( t ) and the damping force f c ( t ) = cu· ( t ) Equation (2.2) becomes: mu··( t ) + cu· ( t ) + ku ( t ) = F ( t )


(2.11)

(2.2)

(2.3)

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• To derive the equation of motion, the dynamic equilibrium for each force component is formulated. To this purpose, forces, and possibly also moments shall be decomposed into their components according to the coordinate directions. 2 Single Degree of Freedom Systems

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(2.12)

Direct Formulation

• Virtual displacement = imaginary infinitesimal displacement

m

• Should best be kinematically permissible, so that unknown reaction forces do not produce work

δA i = δA a

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2.2 Example “Inverted Pendulum”

2.1.2 Principle of virtual work δu


Fm

(2.13)

a sin(ϕ1)

k

Fp

• Thereby, both inertia forces and damping forces must be considered

l

(2.14)

2.1.3 Energy Formulation

a cos(ϕ1)

( f m + f c + f k )δu = F ( t )δu

Fk

a

• Kinetic energy T (Work, that an external force needs to provide to move a mass)

ϕ1

l


sin(ϕ1) ~ ϕ1 cos(ϕ1) ~ 1

• Deformation energy U (is determined from the work that an external force has to provide in order to generate a deformation)

O

• Potential energy of the external forces V (is determined with respect to the potential energy at the position of equilibrium)

Spring force:

F k = a ⋅ sin ( ϕ 1 ) ⋅ k ≈ a ⋅ ϕ 1 ⋅ k

(2.17)

• Conservation of energy theorem (Conservative systems)

Inertia force:

·· Fm = ϕ1 ⋅ l ⋅ m

(2.18)

External force:

Fp = m ⋅ g

(2.19)

E = T + U + V = T o + U o + V o = cons tan t

(2.15)

dE = 0 dt

(2.16)


l sin(ϕ ϕ1)

Equilibrium

Page 2-3

F k ⋅ a ⋅ cos ( ϕ 1 ) + F m ⋅ l – F p ⋅ l ⋅ sin ( ϕ 1 ) = 0


(2.20)

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2 ·· 2 m ⋅ l ⋅ ϕ1 + ( a ⋅ k – m ⋅ g ⋅ l ) ⋅ ϕ1 = 0

(2.21)

Circular frequency: K -------1 = M1

ω =

2

a ⋅k–m⋅g⋅l ------------------------------------- = 2 m⋅l

2

a ⋅ k- g -----------– --2 l m⋅l

(2.22)

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Spring force:

F k ⋅ cos ( ϕ 1 ) ≈ a ⋅ ϕ 1 ⋅ k

(2.24)

Inertia force:

·· Fm = ϕ1 ⋅ l ⋅ m

(2.25)

External force:

F p ⋅ sin ( ϕ 1 ) ≈ m ⋅ g ⋅ ϕ 1

(2.26)

δu m = δϕ 1 ⋅ l

(2.27)

Virtual displacement: δu k = δϕ 1 ⋅ a ,

The system is stable if: ω > 0:


2

a ⋅k>m⋅g⋅l

(2.23)

Principle of virtual work: ( F k ⋅ cos ( ϕ 1 ) ) ⋅ δu k + ( F m – ( F p ⋅ sin ( ϕ 1 ) ) ) ⋅ δu m = 0

Principle of virtual work formulation

(2.28)

·· ( a ⋅ ϕ 1 ⋅ k ) ⋅ δϕ 1 ⋅ a + ( ϕ 1 ⋅ l ⋅ m – m ⋅ g ⋅ ϕ 1 ) ⋅ δϕ 1 ⋅ l = 0 (2.29)

m

Fm

Fpsin(ϕ1) δum

Fkcos(ϕ1)

k

a

2 ·· 2 m ⋅ l ⋅ ϕ1 + ( a ⋅ k – m ⋅ g ⋅ l ) ⋅ ϕ1 = 0

(2.30)

The equation of motion given by Equation (2.30) corresponds to Equation (2.21).

δuk

l

After cancelling out δϕ 1 the following equation of motion is obtained:

ϕ1 δϕ1 sin(ϕ1) ~ ϕ1

O


cos(ϕ1) ~ 1

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for small angles ϕ 1 we have:

Energy Formulation m

(1-cos(ϕ1)) l ~ 0.5 l ϕ12

2

vm

2

E pot,p = – ( m ⋅ g ⋅ 0.5 ⋅ l ⋅ ϕ 1 )

l

(2.35)

and Equation (2.33) becomes:

Ekin,m

Edef,k

(2.36)

Energy conservation: ϕ1

a

sin(ϕ1) ~ ϕ1 cos(ϕ1) ~ 1 O

1 2 2 1 Spring: E def,k = --- ⋅ k ⋅ [ a ⋅ sin ( ϕ 1 ) ] = --- ⋅ k ⋅ ( a ⋅ ϕ 1 ) (2.31) 2 2 2 2 1 1 · E kin,m = --- ⋅ m ⋅ v m = --- ⋅ m ⋅ ( ϕ 1 ⋅ l ) 2 2

(2.32)

E pot,p = – ( m ⋅ g ) ⋅ ( 1 – cos ( ϕ 1 ) ) ⋅ l

(2.33)

by means of a series development, cos ( ϕ 1 ) can be expressed as: 2

4

2k ϕ1 ϕ1 k x cos ( ϕ 1 ) = 1 – ------ + ------ – … + ( – 1 ) ⋅ ------------- + … (2.34) 2! 4! ( 2k )! 2 Single Degree of Freedom Systems

2

ϕ1 ϕ1 cos ( ϕ 1 ) = 1 – ------ and ------ = 1 – cos ( ϕ 1 ) 2 2

Epot,p a sin(ϕ1)

k

Mass:

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E tot = E def,k + E kin,m + E pot,p = constant

(2.37)

2 2 2 1 ·2 1 E = --- ( m ⋅ l ) ⋅ ϕ 1 + --- ( k ⋅ a – m ⋅ g ⋅ l ) ⋅ ϕ 1 = constant 2 2

(2.38)

Derivative of the energy with respect to time: dE = 0 dt

Derivation rule: ( g • f )' = ( g' • f ) ⋅ f'

2 2 · ·· · ( m ⋅ l ) ⋅ ϕ1 ⋅ ϕ1 + ( k ⋅ a – m ⋅ g ⋅ l ) ⋅ ϕ1 ⋅ ϕ1 = 0

(2.39) (2.40)

· After cancelling out the velocity ϕ 1 : 2 ·· 2 m ⋅ l ⋅ ϕ1 + ( a ⋅ k – m ⋅ g ⋅ l ) ⋅ ϕ1 = 0

(2.41)

The equation of motion given by Equation (2.41) corresponds to Equations (2.21) and (2.30).


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2.3 Modelling

Comparison of the energy maxima 2 1 · KE = --- ⋅ m ⋅ ( ϕ 1,max ⋅ l ) 2

(2.42)

2.3.1 Structures with concentrated mass

2 2 1 1 PE = --- ⋅ k ⋅ ( a ⋅ ϕ 1 ) – --- ⋅ g ⋅ m ⋅ l ⋅ ϕ 1 2 2

(2.43)

Tank: Mass=1000t

Water tank F(t)

F(t)

By equating KE and PE we obtain: RC Walls in the longitudinal direction

§ 2 ⋅ k – m ⋅ g ⋅ l· · ϕ 1,max = ¨ a------------------------------------¸ ⋅ ϕ 1 2 © ¹ m⋅l

(2.44)

· ϕ 1,max = ω ⋅ ϕ 1

(2.45)

• ω is independent of the initial angle ϕ 1

Ground

Tra dir nsver ect se ion

al din itu n ng ctio o L ire d

Bridge in transverse direction

• the greater the deflection, the greater the maximum velocity.

k = …

3EI w k = 2 ----------3 H

F(t) Frame with rigid beam 3EI w k = ----------3 H

F(t)

12EI s k = 2 ------------3 H

mu·· + ku = F ( t ) 2 Single Degree of Freedom Systems

Page 2-9


(2.46) Page 2-10


An-Najah 2013

2.3.2 Structures with distributed mass


δA i =

An-Najah 2013

L

³0 ( EIu'' ⋅ δ [ u'' ] ) dx

(2.53)

• Transformations:

u'' = ψ''U

·· u·· = ψU

and

(2.54)

• The virtual displacement is affine to the selected deformation:

δu = ψδU

δ [ u'' ] = ψ''δU

and

(2.55)

• Using Equations (2.54) and (2.55), the work δA a produced by the external forces is:

u ( x, t ) = ψ ( x )U ( t )

Deformation: External forces:

t ( x, t ) = – mu··( x, t ) f ( x, t )

(2.47)

0

(2.48)

• Principle of virtual work δA i = δA a L

³0

δA a =

(2.49) L

( t ⋅ δu ) dx + ³ ( f ⋅ δu ) dx

(2.50)

(2.56)

0

·· L mψ 2 dx + L fψ dx = δU – U ³ ³ 0

0

• Using Equations (2.54) and (2.55) the work δA i produced by the internal forces is:

δA i =

L

³0

L

( EIψ''U ⋅ ψ''δU ) dx = δU U ³ ( EI ( ψ'' ) 2 ) dx

(2.57)

0

L

0 L

• Equation (2.49) is valid for all virtual displacements, therefore:

0

0

L ·· L mψ 2 dx + L fψ dx U ³ ( EI ( ψ'' ) 2 ) dx = – U ³ ³

(2.58)

* ·· * * m U +k U = F

(2.59)

= – ³ ( mu·· ⋅ δu ) dx + ³ ( f ⋅ δu ) dx δA i =

L ·· ⋅ ψδU ) dx + L ( f ⋅ ψδU ) dx δA a = – ³ ( mψU ³

0

L

³0 ( M ⋅ δϕ ) dx where:

M = EIu''

and


δϕ = δ [ u'' ]

(2.51)

0

0

(2.52) Page 2-11


Page 2-12


An-Najah 2013


An-Najah 2013

• Example No. 1: Cantilever with distributed mass

• Circular frequency L

2 ωn

* ³0 ( EI ( ψ'' ) 2 ) dx k = ------*- = -----------------------------------L 2 m ³ mψ dx

(2.60)

0

-> Rayleigh-Quotient • Choosing the deformation figure - The accuracy of the modelling depends on the assumed deformation figure; - The best results are obtained when the deformation figure fulfills all boundary conditions; - The boundary conditions are automatically satisfied if the deformation figure corresponds to the deformed shape due to an external force; - A possible external force is the weight of the structure acting in the considered direction. • Properties of the Rayleigh-Quotient - The estimated natural frequency is always larger than the exact one (Minimization of the quotient!); - Useful results can be obtained even if the assumed deformation figure is not very realistic.


Page 2-13

πx π 2 ψ'' = § -------· cos § -------· © 2L¹ © 2L¹

πx ψ = 1 – cos § -------· , © 2L¹

*

m =

L

§

§ πx· · dx + ψ 2 ( x = L )M

³0 m © 1 – cos © -----2L¹ ¹

2

πx-· πx πx § 3πx – 8 sin § -----L + 2 cos § -------· sin § -------· L· © 2L¹ © 2L¹ © 2L¹ ¸ 1--- ¨ = m ¨ ----------------------------------------------------------------------------------------------------¸ 2 ¨ π ¸ © ¹

(2.61)

(2.62) L

+M 0

( 3π – 8 ) = -------------------- mL + M = 0.23mL + M 2π


Page 2-14


An-Najah 2013


An-Najah 2013

• Example No. 2: Cantilever with distributed mass * πx 2 π 4 L k = EI § -------· ³ § cos § -------· · dx © 2L¹ ¹ © 2L¹ 0 ©

πx· § πx· · § πx + 2 cos § ------ sin ------- L 4 © ¹ © 2L¹ ¸ ¨ 2L 1 π § · = EI ------- ⋅ --- ¨ ---------------------------------------------------------------¸ © 2L¹ 2 ¨ π ¸ © ¹

(2.63) L

0

4

EI 3EI π EI = ------ ⋅ -----3- = 3.04 ⋅ -----3- ≈ -------3 32 L L L

ω =

πx ψ = 1 – cos § -------· , © 2L¹

3EI ----------------------------------------3 ( 0.23mL + M )L

(2.64)

• Check of the boundary conditions of the deformation figure

πx ψ ( 0 ) = 0 ? -> ψ ( x ) = 1 – cos §© -------·¹ : 2L

ψ ( 0 ) = 0 OK!

π πx ψ' ( 0 ) = 0 ? -> ψ' ( x ) = ------- sin §© -------·¹ : 2L 2L

ψ' ( 0 ) = 0 OK!

π 2 πx ψ'' ( L ) = 0 ? -> ψ'' ( x ) = §© -------·¹ cos §© -------·¹ : 2L 2L

ψ'' ( L ) = 0 OK!

• Calculation of the mass m *

m =

L

§

πx π 2 ψ'' = § -------· cos § -------· © 2L¹ © 2L¹ *

2 § πx· · dx + ψ 2 § x = L ---· M 1 + ψ 2 ( x = L )M 2 © 2¹

³0 m © 1 – cos © -----2L¹ ¹

(2.67)

( 3π – 8 ) 3–2 2 * m = -------------------- mL + § -------------------· ⋅ M 1 + M 2 © 2 ¹ 2π

(2.68)

*

Page 2-15

(2.66)

( 3π – 8 ) π 2 * 2 m = -------------------- mL + § 1 – cos § ---· · ⋅ M 1 + 1 ⋅ M 2 © © 4¹ ¹ 2π

m = 0.23mL + 0.086M 1 + M 2


(2.65)


(2.69)

Page 2-16


• Calculation of the stiffness k

*

* πx 2 π 4 L k = EI § -------· ³ § cos § -------· · dx © 2L¹ ¹ © 2L¹ 0 ©

6

3.04EI ------------------------------------------------------------------------3 ( 0.23mL + 0.086M 1 + M 2 )L

4

(2.70)

(2.71)

(2.72)

3.04EI EI ------------------------------ = 1.673 ----------3 3 ( 1.086M )L ML

3.007EI - = 1.652 ----------EI ----------------------------3 3 ( 1.102M )L ML

(2.73)

(2.74)

As a numerical example, the first natural frequency of a L = 10m tall steel shape HEB360 (bending about the strong axis) featuring two masses M 1 = M 2 = 10t is calculated.


2

2

(2.75) (2.76)

4

EI = 1.673 -----------------------8.638 ×10 - = 4.9170 ω = 1.673 ----------3 3 10 ⋅ 10 ML

(2.77)

1 4.9170 f = ------ ⋅ ω = ---------------- = 0.783Hz 2π 2π

(2.78)

From Equation (2.74): 1.652 EI 1.652 8.638 ×104 f = ------------- ----------= ------------- ------------------------ = 0.773Hz 3 2π ML 3 2π 10 ⋅ 10

The exact first natural circular frequency of a two-mass oscillator with constant stiffness and mass is: ω =

EI = 8.638 ×10 kNm

An-Najah 2013

By means of Equation (2.73) we obtain:

Special case: m = 0 and M 1 = M 2 = M ω =

13

EI = 200000 ⋅ 431.9 ×10 = 8.638 ×10 Nmm

• Calculation of the circular frequency ω ω =


4

EI 3EIπ EI k = ------ ⋅ -----3- = 3.04 ⋅ -----3- ≈ -------3 32 L L L *

An-Najah 2013

Page 2-17

(2.79)

The first natural frequency of such a dynamic system can be calculated using a finite element program (e.g. SAP 2000), and it is equal to: T = 1.2946s , f = 0.772Hz

(2.80)

Equations (2.78), (2.79) and (2.80) are in very good accordance. The representation of the first mode shape and corresponding natural frequency obtained by means of a finite element program is shown in the next figure.


Page 2-18


An-Najah 2013


An-Najah 2013

2.3.3 Damping • Types of damping

M = 10t

Damping

Internal

Material

Hysteretic (Viscous, Friction, Yielding)

M = 10t

External Contact areas within the structure Relative movements between parts of the structure (Bearings, Joints, etc.)

External contact (Non-structural elements, Energy radiation in the ground, etc.)

• Typical values of damping in structures Damping ζ

Material Reinforced concrete (uncraked) Reinforced concrete (craked Reinforced concrete (PT) Reinforced concrete (partially PT) Composite components Steel

HEB 360

0.007 - 0.010 0.010 - 0.040 0.004 - 0.007 0.008 - 0.012 0.002 - 0.003 0.001 - 0.002

Table C.1 from [Bac+97] SAP2000 v8 - File:HEB_360 - Mode 1 Period 1.2946 seconds - KN-m Units


Page 2-19


Page 2-20


An-Najah 2013

• Bearings

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• Dissipators

Source: A. Marioni: “Innovative Anti-seismic Devices for Bridges”. [SIA03]



Page 2-21

Source: A. Marioni: “Innovative Anti-seismic Devices for Bridges”. [SIA03] 2 Single Degree of Freedom Systems

Page 2-22


An-Najah 2013

3 Free Vibrations


An-Najah 2013

• Relationships

“A structure undergoes free vibrations when it is brought out of its static equilibrium, and can then oscillate without any external dynamic excitation”

ωn =

ωn f n = ------ [1/s], [Hz]: Number of revolutions per time 2π

(3.8)

3.1 Undamped free vibrations

2π T n = ------ [s]: ωn

(3.9)

mu··( t ) + ku ( t ) = 0

(3.1)

k ⁄ m [rad/s]: Angular velocity

Time required per revolution

(3.7)

• Transformation of the equation of motion 3.1.1 Formulation 1: Amplitude and phase angle

2 u··( t ) + ω n u ( t ) = 0

• Ansatz:

(3.10)

• Determination of the unknowns A and φ :

u ( t ) = A cos ( ω n t – φ )

(3.2)

2 u··( t ) = – A ω n cos ( ω n t – φ )

(3.3)

The static equilibrium is disturbed by the initial displacement u ( 0 ) = u 0 and the initial velocity u· ( 0 ) = v 0 :

(3.4)

A =

By substituting Equations (3.2) and (3.3) in (3.1): 2

A ( – ω n m + k ) cos ( ω n t – φ ) = 0 2

– ωn m + k = 0

(3.5)

ωn =

(3.6)

k ⁄ m “Natural circular frequency”

3 Free Vibrations

Page 3-1

v0 2 v0 2 u 0 + § ------· , tan φ = ----------© ω n¹ u0 ωn

(3.11)

• Visualization of the solution by means of the Excel file given on the web page of the course (SD_FV_viscous.xlsx)

3 Free Vibrations

Page 3-2


An-Najah 2013

3.1.2 Formulation 2: Trigonometric functions mu··( t ) + ku ( t ) = 0


An-Najah 2013

3.1.3 Formulation 3: Exponential Functions (3.12)

• Ansatz:

mu··( t ) + ku ( t ) = 0

(3.19)

• Ansatz:

u ( t ) = A 1 cos ( ω n t ) + A 2 sin ( ω n t )

(3.13)

u(t) = e

2 2 u··( t ) = – A 1 ω n cos ( ω n t ) – A 2 ω n sin ( ω n t )

(3.14)

2 λt u··( t ) = λ e


2

– ωn m + k = 0

(3.16)

ωn =

(3.17)

k ⁄ m “Natural circular frequency”

• Determination of the unknowns A 1 and A 2 :

2

mλ + k = 0

(3.22)

2 k λ = – ---m

(3.23)

k- = ± iω λ = ± i --n m

(3.24)

u ( t ) = C1 e

iω n t

+ C2 e

–i ωn t

(3.25)

and by means of Euler’s formulas (3.18)

iα

Page 3-3

–i α

iα

–i α

e +e e –e cos α = ----------------------- , sin α = ----------------------2 2i e

3 Free Vibrations

(3.21)

The complete solution of the ODE is:

The static equilibrium is disturbed by the initial displacement u ( 0 ) = u 0 and the initial velocity u· ( 0 ) = v 0 : v0 A 1 = u 0 , A 2 = -----ωn

(3.20)

By substituting Equations (3.20) and (3.21) in (3.19):

A 1 ( – ω n m + k ) cos ( ω n t ) + A 2 ( – ω n m + k ) sin ( ω n t ) = 0 (3.15) 2

λt

iα

= cos ( α ) + i sin ( α ) , e

3 Free Vibrations

–i α

= cos ( α ) – i sin ( α )

(3.26) (3.27)

Page 3-4


An-Najah 2013


An-Najah 2013

3.2 Damped free vibrations

Equation (3.25) can be transformed as follows: u ( t ) = ( C 1 + C 2 ) cos ( ω n t ) + i ( C 1 – C 2 ) sin ( ω n t )

(3.28)

mu··( t ) + cu· ( t ) + ku ( t ) = 0

u ( t ) = A 1 cos ( ω n t ) + A 2 sin ( ω n t )

(3.29)

- In reality vibrations subside

(3.30)

- Damping exists

Equation (3.29) corresponds to (3.13)!

- It is virtually impossible to model damping exactly - From the mathematical point of view viscous damping is easy to treat s Damping constant: c N ⋅ ---m

(3.31)

3.2.1 Formulation 3: Exponential Functions mu··( t ) + cu· ( t ) + ku ( t ) = 0

(3.32)

• Ansatz: u(t) = e

λt

λt 2 λt , u· ( t ) = λe , u··( t ) = λ e

(3.33)

By substituting Equations (3.33) in (3.32): 2

( λ m + λc + k )e

λt

= 0

2

3 Free Vibrations

Page 3-5

(3.34)

λ m + λc + k = 0

(3.35)

2 c 1 λ = – -------- ± -------- c – 4km 2m 2m

(3.36)

3 Free Vibrations

Page 3-6


An-Najah 2013

An-Najah 2013

• Types of vibrations

• Critical damping when: c 2 – 4km = 0 c cr = 2 km = 2ω n m


1

(3.37)

Underdamped vibration Critically damped vibration

• Damping ratio

Overdamped vibration

(3.38)

• Transformation of the equation of motion mu··( t ) + cu· ( t ) + ku ( t ) = 0

(3.39)

c k u··( t ) + ---- u· ( t ) + ---- u ( t ) = 0 m m

(3.40)

2 u··( t ) + 2ζω n u· ( t ) + ω n u ( t ) = 0

(3.41)

u(t)/u0 [-]

c c c ζ = ------ = --------------- = -------------2ω c cr 2 km nm

0.5

0

-0.5

-1

• Types of vibrations:

0.5

1

1.5

2

2.5

3

3.5

4

t/Tn [-]

c ζ = ------ < 1 : c cr

Underdamped free vibrations

c ζ = ------ = 1 : c cr

Critically damped free vibrations

c ζ = ------ > 1 : c cr

Overdamped free vibrations

3 Free Vibrations

0

Page 3-7

3 Free Vibrations

Page 3-8


An-Najah 2013

Underdamped free vibrations ζ < 1


An-Najah 2013

The determination of the unknowns A 1 and A 2 is carried out as usual by means of the initial conditions for displacement ( u ( 0 ) = u 0 ) and velocity ( u· ( 0 ) = v 0 ) obtaining:

By substituting: 2 c c k c ζ = ------ = --------------- = --------------- and ω n = ---2ω n m m c cr 2 km

(3.42)

v 0 + ζω n u 0 A 1 = u 0 , A 2 = --------------------------ωd

(3.51)

in: c 1 c c 2 k 2 λ = – -------- ± -------- c – 4km = – -------- ± § --------· – ---© 2m¹ m 2m 2m 2m

3.2.2 Formulation 1: Amplitude and phase angle (3.43)

Equation (3.50) can be rewritten as “the amplitude and phase angle”:

it is obtained: 2 2

2

2

λ = – ζω n ± ω n ζ – ω n = – ζω n ± ω n ζ – 1 λ = – ζω n ± iω n 1 – ζ

2

u ( t ) = Ae (3.44)

A = ω d = ω n 1 – ζ “damped circular frequency”

(3.46)

λ = – ζω n ± iω d

(3.47)

The complete solution of the ODE is: u ( t ) = C1 e u(t) = e u(t) = e

3 Free Vibrations

( – ζω n + iω d )t

– ζω n t – ζω n t

( C1 e

iω d t

+ C2 e

+ C2 e

( – ζω n – iω d )t

–i ωd t

)

( A 1 cos ( ω d t ) + A 2 sin ( ω d t ) )

cos ( ω d t – φ )

(3.52)

with (3.45)

2

– ζω n t

v 0 + ζω n u 0 v 0 + ζω n u 0 2 2 u 0 + § ---------------------------· , tan φ = --------------------------© ¹ ωd ωd u0

(3.53)

The motion is a sinusoidal vibration with circular frequency ω d and decreasing amplitude Ae

– ζω n t

(3.48) (3.49) (3.50)

Page 3-9

3 Free Vibrations

Page 3-10


An-Najah 2013


3.3 The logarithmic decrement

• Notes - The period of the damped vibration is longer, i.e. the vibration is slower

20

Td u0

15 1

Free vibration

u1

10

0.8 0.7 0.6

ωd = ωn 1 – ζ

0.5

2

Tn T d = -----------------2 1–ζ

0.4 0.3 03

Displacement

0.9

Tn/Td

An-Najah 2013

5 0 -5 -10 -15

0.2

-20

0.1

0

1

2

3

4

5

6

7

8

9

10

Time (s)

0 0

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

1

Damping ratio ζ

• Amplitude of two consecutive cycles

- The envelope of the vibration is represented by the following equation: u ( t ) = Ae

– ζω n t

with A =

v 0 + ζω n u 0 2 2 u 0 + § ---------------------------· © ¹ ωd

(3.54)

- Visualization of the solution by means of the Excel file given on the web page of the course (SD_FV_viscous.xlsx)

3 Free Vibrations

Page 3-11

– ζω t

n u0 cos ( ω d t – φ ) Ae ----- = -------------------------------------------------------------------------------– ζω ( t + T ) d u1 Ae n cos ( ω d ( t + T d ) – φ )

(3.55)

with e

– ζω n ( t + T d )

= e

– ζω n t – ζω n T d

e

cos ( ω d ( t + T d ) – φ ) = cos ( ω d t + ω d T d – φ ) = cos ( ω d t – φ )

3 Free Vibrations

(3.56) (3.57)

Page 3-12


An-Najah 2013

we obtain:

An-Najah 2013

• Evaluation over several cycles

u0 ζω T 1 ----- = ----------------- = e n d – ζω T n d u1 e

(3.58)

• Logarithmic decrement δ u0 2πζ δ = ln § -----· = ζω n T d = ------------------ ≅ 2πζ (if ζ small) © u 1¹ 2 1–ζ

(3.59)

The damping ratio becomes: δ δ ζ = ------------------------- ≅ ------ (if ζ small) 2 2 2π 4π + δ

(3.60)

u u uN – 1 u ζω T N Nζω n T d -----0- = ----0- ⋅ ----1- ⋅ … ⋅ ------------ = (e n d) = e uN u1 u2 uN

(3.61)

u0 1 δ = ---- ln § ------· N © u N¹

(3.62)

• Halving of the amplitude u 1 ---- ln § -----0-· N © u N¹ ζ = ---------------------- = 2π

1 ---- ln ( 2 ) N 1 - --------1------------------ = -----≅ 2π 9N 10N

(3.63)

Useful formula for quick evaluation

10

Exact equation

9

Logarithmic Decrement δ


• Watch out: damping ratio vs. damping constant

Approximation

8

Empty

7

Full

6 5 4 3 2 1 0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Damping ratio ζ

3 Free Vibrations

Page 3-13

m 1, k 1, c 1

m2>m1, k1, c1

c1 ζ 1 = -------------------2 k1 m1

c1 ζ 2 = -------------------- < ζ 1 2 k1 m2

3 Free Vibrations

Page 3-14


An-Najah 2013

3.4 Friction damping


An-Najah 2013

• Free vibrations It is a nonlinear problem! 20 Free vibration Friction force

15

a)

b)

– f k ( t ) – f μ = mu··( t )

– f k ( t ) + f μ = mu··( t )

mu··( t ) + ku ( t ) = – f μ

mu··( t ) + ku ( t ) = f μ

Displacement

10 5 0 -5

-10 -15

• Solution of b) u ( t ) = A 1 cos ( ω n t ) + A 2 sin ( ω n t ) + u μ

fμ with u μ = ---k

u· ( t ) = – ω n A 1 sin ( ω n t ) + ω n A 2 cos ( ω n t )

-20

(3.64)

0

1

2

3

4

5

6

7

8

9

10

Time (s)

Figure: f=0.5 Hz , u0=10 , v0 = 50, uf = 1

(3.65)

• Calculation example: by means of the initial conditions u ( 0 ) = u 0 , u· ( 0 ) = v 0 we obtain the constants:

- Step 1: Initial conditions u ( 0 ) = u 0 , u· ( 0 ) = 0

A1 = u0 – uμ , A2 = v0 ⁄ ωn

A1 = u0 – uμ , A2 = 0

• Solution of a): Similar, with – u μ instead of +u μ

u ( t ) = [ u 0 – u μ ] cos ( ω n t ) + u μ

(3.66) π 0 ≤ t < -----ωn

(3.67)

π End displacement: u § ------· = [ u 0 – u μ ] ( – 1 ) + u μ = – u 0 + 2u μ © ω n¹ 3 Free Vibrations

Page 3-15

3 Free Vibrations

Page 3-16


An-Najah 2013

- Step 2:


An-Najah 2013

• Comparison Viscous damping vs. Friction damping

Initial conditions u ( 0 ) = – u 0 + 2u μ , u· ( 0 ) = 0

Free vibration: f=0.5 Hz , u0=10 , v0 = 50, uf = 1

A 1 = u ( 0 ) + u μ = – u 0 + 2u μ + u μ = – u 0 + 3u μ , A 2 = 0 (3.68) u ( t ) = [ – u 0 + 3u μ ] cos ( ω n t ) – u μ

π 0 ≤ t < -----ωn

Logarithmic decrement: (3.69)

π End displacement: u § ------· = [ – u 0 + 3u μ ] ( – 1 ) – u μ = u 0 – 4u μ © ω n¹

- Step 3:

U0

UN

N

δ

ζ [%]

1

18.35

14.35

1

0.245

3.91

2

18.35

10.35

2

0.286

4.56

3

18.35

6.35

3

0.354

5.63

4

18.35

2.35

4

0.514

8.18

Average

Initial conditions ....

• Important note: The change between case a) and case b) occurs at velocity reversals. In order to avoid the build-up of inaccuracies, the displacement at velocity reversal should be identified with adequate precision (iterate!)

• Characteristics of friction damping - Linear decrease in amplitude by 4u μ at each cycle - The period of the damped and of the undamped oscillator is the same: 2π T n = -----ωn

Comparison: 20 Friction damping Viscous damping

15 10

Displacement

• Visualization of the solution by means of the Excel file given on the web page of the course (SD_FV_friction.xlsx)

5.57

5 0 -5 -10 -15 -20 0

1

2

3

4

5

6

7

8

9

10

Time (s) 3 Free Vibrations

Page 3-17

3 Free Vibrations

Page 3-18


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4 Response to Harmonic Excitation


An-Najah 2013

Linear inhomogeneous differential equation • Particular solution: u p 2 u··p + 2ζω n u· p + ω n u p = f ( t )

(4.4)

• Solution of the homogeneous ODE: u h 2 u··h + 2ζω n u· h + ω n u h = 0

(4.5)

• Complete solution: u = u p + Cu h An harmonic excitation can be described either by means of a sine function (Equation 4.1) or by means of a cosine function (Equation 4.2): mu·· + cu· + ku = F o sin ( ωt )

(4.1)

mu·· + cu· + ku = F o cos ( ωt )

(4.2)

2 u·· + 2ζω n u· + ω n u = f o cos ( ωt )

(4.6)

• Initial conditions u ( 0 ) = u0

,

u· ( 0 ) = v 0

(4.7)

Here we consider Equation (4.2) which after transformation becomes: 2 u·· + 2ζω n u· + ω n u = f o cos ( ωt )

(4.3)

where: ω n : Circular frequency of the SDoF system ω:

Circular frequency of the excitation

f o = F o ⁄ m = ( F o ⁄ k ) ⋅ ω n2


Page 4-1


Page 4-2


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4.1 Undamped harmonic vibrations 2 u·· + ω n u = f o cos ( ωt )

• Ansatz for particular solution (4.9)

2 u··p = – A o ω cos ( ωt )

(4.10)

By substituting (4.9) and (4.10) in (4.8): 2

2

– A o ω cos ( ωt ) + A o ω n cos ( ωt ) = f o cos ( ωt ) 2

Ao ( – ω +

2 ωn )

(4.11)

fo - cos ( ωt ) u = A 1 cos ( ω n t ) + A 2 sin ( ω n t ) + -----------------2 2 ωn – ω

(4.12)

F fo 1 - = -----o ⋅ -----------------------------A o = -----------------2 2 k 1 – ( ω ⁄ ωn )2 ωn – ω

(4.13)

fo - cos ( ωt ) u p = -----------------2 2 ωn – ω

(4.14)

(4.16)

By means of the initial conditions given in Equation (4.7), the constants A 1 and A 2 can be calculated as follows: fo A 1 = u 0 – -----------------2 2 ωn – ω

,

v0 A 2 = -----ωn

(4.17)

• Denominations: - Homogeneous part of the solution:

= fo

- Particular part of the solution:

“transient” “steady-state”

• Visualization of the solution by means of the Excel file given on the web page of the course (SD_HE_cosine_viscous.xlsx)

• Harmonic vibration with sine excitation fo - sin ( ωt ) u = A 1 cos ( ω n t ) + A 2 sin ( ω n t ) + -----------------2 2 ωn – ω

• Ansatz for the solution of the homogeneous ODE (see section on free vibrations) u h = B 1 cos ( ω n t ) + B 2 sin ( ω n t )

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• Complete solution of the ODE: (4.8)

u p = A o cos ( ωt )


(4.15)

By means of the initial conditions given in Equation (3.7), the constants A 1 and A 2 can be calculated as follows: A1 = u0

,

v0 fo ( ω ⁄ ωn ) A 2 = ------ – ----------------------ωn ω2 – ω2 n


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4.1.1 Interpretation as a beat 2 u·· + ω n u = f o cos ( ωt ) with

u ( 0 ) = u· ( 0 ) = 0


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• Case 1: Natural frequency SDoF 0.2 Hz, excitation frequency 0.4 Hz 500

(4.18)

Total response

Envelope

400 300

fo - ⋅ [ cos ( ωt ) – cos ( ω n t ) ] u ( t ) = -----------------2 2 ωn – ω

(4.19)

and using the trigonometric identity α–β α+β cos ( α ) – cos ( β ) = – 2 sin § ------------- t· sin § ------------- t· © 2 ¹ © 2 ¹

Displacement

The solution is:

200 100 0 -100 -200 -300

(4.20)

-400 -500 0

one gets the equation

2

4

6

8

10

12

14

16

18

20

Time [s]

ω–ω ω+ω 2f o - ⋅ sin § ----------------n t· sin § ----------------n- t· u ( t ) = -----------------2 2 © 2 ¹ © 2 ¹ ω – ωn

(4.21)

• Case 2: Natural frequency SDoF 2.0 Hz, excitation frequency 2.2 Hz 80

Total response

Envelope

60

that describes a beat with: Fundamental vibration: f G Envelope:

f – fn f U = -----------2

(4.22)

(4.23)

Displacement

40

f + fn = -----------2

20 0 -20 -40

A beat is always present, but is only evident when the natural frequency of the SDoF system and the excitation frequency are close (see figures on the next page)

-60 -80 0

2

4

6

8

10

12

14

16

18

20

Time [s]


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4.1.2 Resonant excitation (ω = ωn)

• Transition to f = fn

2 u·· + ω n u = f o cos ( ω n t )

f 2.0500 ---- = ---------------fn 2.0000

---f- = 2.0250 ---------------fn 2.0000

(4.24)

• Ansatz for the particular solution u p = A o t sin ( ω n t )

(4.25)

u· p = A o sin ( ω n t ) + A o ω n t cos ( ω n t )

(4.26)

2 u··p = 2A o ω n cos ( ω n t ) – A o ω n t sin ( ω n t )

(4.27)


2

2A o ω n cos ( ω n t ) – A o ω n t sin ( ω n t ) + A o ω n t sin ( ω n t ) = f o cos ( ω n t )

(4.28)

---f- = 2.0125 ---------------fn 2.0000

f 2.0000 ---- = ---------------fn 2.0000 Resonance!


Page 4-7

2A o ω n = f o

(4.29)

fo Fo ωn A o = ---------- = ----- ⋅ -----2ω n k 2

(4.30)

fo u p = ---------- t sin ( ω n t ) 2ω n

(4.31)

• Ansatz for the solution of the homogeneous ODE (see section on free vibrations) u h = B 1 cos ( ω n t ) + B 2 sin ( ω n t )


(4.32)

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(4.33)

By means of the initial conditions given in Equation (4.7), the constants A 1 and A 2 can be calculated as follows: A1 = u0

,

v0 A 2 = -----ωn

(4.34)

• Special case u 0 = v 0 = 0 (The homogeneous part of the solution falls away) fo u = ---------- t sin ( ω n t ) 2ω n

2 u·· + 2ζω n u· + ω n u = f o cos ( ωt )

(4.37)

• Ansatz for particular solution u p = A 3 cos ( ωt ) + A 4 sin ( ωt )

(4.38)

u· p = – A 3 ω sin ( ωt ) + A 4 ω cos ( ωt )

(4.39)

2 2 u··p = – A 3 ω cos ( ωt ) – A 4 ω sin ( ωt )

(4.40)

By substitution Equations (4.38) to (4.40) in (4.37): 2

(4.35)

2

2

2

[ ( ω n – ω )A 3 + 2ζω n ωA 4 ] cos ( ωt ) + [ – 2ζω n ωA 3 + ( ω n – ω )A 4 ] sin ( ωt ) = f o cos ( ωt )

(4.41) Equation (4.41) shall be true for all times t and for all constants A 3 and A 4 , therefore Equations (4.42) and (4.43) can be written as follows:

Is a sinusoidal vibration with amplitude: fo A = ---------- t 2ω n

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4.2 Damped harmonic vibration

• Complete solution of the ODE: fo u = A 1 cos ( ω n t ) + A 2 sin ( ω n t ) + ---------- t sin ( ω n t ) 2ω n


(4.36)

2

2

( ω n – ω )A 3 + 2ζω n ωA 4 = f o

- The amplitude grows linearly with time (see last picture of interpretation “beat”);

– 2ζω n ωA 3 + ( ω n – ω )A 4 = 0

- We have A → ∞ when t → ∞ , i.e. after infinite time the amplitude of the vibration is infinite as well.

The solution of the system [(4.42), (4.43)] allows the calculations of the constants A 3 and A 4 as:


2

2

2

(4.42) (4.43)

2

ωn – ω 2ζω n ω - , A 4 = f o --------------------------------------------------------A 3 = f o --------------------------------------------------------2 2 2 2 2 2 2 2 ( ω n – ω ) + ( 2ζω n ω ) ( ω n – ω ) + ( 2ζω n ω )

(4.44) 4 Response to Harmonic Excitation

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60 Steady-state response

( B 1 cos ( ω d t ) + B 2 sin ( ω d t ) )

(4.45)

with: 2

ω d = ω n 1 – ζ “damped circular frequency”

(4.46)

• Complete solution of the ODE: u = e

– ζω n t

Total response

40

Displacement

– ζω n t

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• Example 1: fn = 1Hz, f = 0.2Hz, ζ = 5%, fo = 1000, u0 = 0, v0 = fo/ωn

• Ansatz for the solution of the homogeneous ODE (see Section 3.2 on damped free vibrations) uh = e


20 0 -20 -40

( A 1 cos ( ω d t ) + A 2 sin ( ω d t ) ) + A 3 cos ( ωt ) + A 4 sin ( ωt )

-60

(4.47)

0

2

4

6

8

10

12

14

16

18

20

Time [s]

By means of the initial conditions of Equation (4.7), the constants A 1 and A 2 can be calculated. The calculation is laborious and should be best carried out with a mathematics program (e.g. Maple).

• Example 2: Like 1 but with F(t) = Fosin(ωt) instead of Focos(ωt) 50 Steady-state response

40

Total response

30

- Homogeneous part of the solution: - Particular part of the solution:

“transient” “steady-state”


Displacement

• Denominations:

20 10 0 -10 -20 20 -30 -40 -50 0

2

4

6

8

10

12

14

16

18

20

Time [s]


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4.2.1 Resonant excitation (ω = ωn) By substituting ω = ω n in Equation (4.44) the constants A 3 and A 4 becomes: fo A 3 = 0 , A 4 = ------------22ζω n

(4.48)

– ζω n t

fo ( A 1 cos ( ω d t ) + A 2 sin ( ω d t ) ) + ------------2- sin ( ω n t ) (4.49) 2ζω n

• Special case u 0 = v 0 = 0 A1 = 0

,

fo fo A 2 = – -------------------------------- = – -------------------2ζω n ω d 2 2 2ζω n 1 – ζ

fo § sin ( ω d t ) – ζω n t· u = ------------2- ¨ sin ( ω n t ) – --------------------- e ¸ 2 ¹ 2ζω n © 1–ζ


- The amplitude is limited, i.e. the maximum displacement of the SDoF system is: fo Fo u st u max = ------------2- = --------- = -----2ζk 2ζ 2ζω n

(4.53)

- For small damping ratios ( ζ ≤ 0.2 ) ω d ≈ ω n and hence Equations (4.51) becomes:

2

1–ζ ≈1

fo – ζω n t – ζω n t u = ------------2- ( 1 – e ) sin ( ω n t ) = u max ( 1 – e ) sin ( ω n t ) (4.54) 2ζω n

It is a sinusoidal vibration with the amplitude: (4.50)

A = u max ( 1 – e

– ζω n t

)

(4.55)

and the magnitude of the amplitude at each maxima j is (4.51)

uj – ζω t ---------- = ( 1 – e n j ) sin ( ω n t j ) u max

(4.56)

Maxima occur when sin ( ω n t ) = – 1 , d.h. when

- After a certain time, the homogeneous part of the solution subsides and what remains is a sinusoidal oscillation of the amplitude: fo A = ------------22ζω n

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where u st = F o ⁄ k is the static displacement.

i.e. if damping is present, the resonant excitation is not a special case any more, and the complete solution of the differential equation is: u = e


Tn t j = ( 4j – 1 ) ⋅ ------ , j = 1…∞ 4 Tn

(4.52)

Page 4-13

(4.57) π

– ζω n ( 4j – 1 ) ⋅ ----– ζ ( 4j – 1 ) ⋅ --uj 4 2 ---------- = 1–e = 1–e u max


(4.58)

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• Magnitude of the amplitude after each cycle: f(ust)

50

50

40

40

ζ = 0.01

abs(uj) / ust

umax / ust

• Dynamic amplification


30

20

10

30

0.02 20

0.05

10

0.10 0

0 0

0.05

0.1

0.15

0.2

0

5

10

15

Damping ratio ζ [-]

20

25

30

35

40

45 0.20 50

Cycle

• Magnitude of the amplitude after each cycle: f(umax) 1

0.2

0.1

0.05 0.02

abs(uj) / umax

0.8

ζ = 0.01 0.6

0.4

0.2

0 0

5

10

15

20

25

30

35

40

45

50

Cycle


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5 Transfer Functions


The maximum dynamic amplitude u max given by Equation (5.5) can be transformed to:

5.1 Force excitation

2

The steady-state displacement of a system due to harmonic excitation is (see Section 4.2 on harmonic excitation): u p = a 1 cos ( ωt ) + a 2 sin ( ωt )

u max =

n

n

n

n

2

2 2

2

( ω n – ω ) + ( 2ζω n ω ) ---------------------------------------------------------------2 2 2 2 2 [ ( ω n – ω ) + ( 2ζω n ω ) ]

(5.8)

2

2

ωn – ω 2ζω n ω - , a 2 = f o --------------------------------------------------------a 1 = f o --------------------------------------------------------2 2 2 2 2 2 2 2 ( ω n – ω ) + ( 2ζω n ω ) ( ω n – ω ) + ( 2ζω n ω )

(5.2)

By means of the trigonometric identity a cos ( α ) + b sin ( α ) =

b a + b ⋅ cos ( α – φ ) where tan φ = --a 2

2

(5.3)

Equation (5.1) can be transformed as follows: u p = u max cos ( ωt – φ )

(5.4)

It is a cosine vibration with the maximum dynamic amplitude u max : 2

2

(5.7) u max = f o

a1 + a2

2

ωn – ω 2ζω n ω § · § ·2 -¸ + ¨ f o ---------------------------------------------------------¸ ¨ f o --------------------------------------------------------© ( ω 2 – ω 2 ) 2 + ( 2ζω ω ) 2¹ © ( ω 2 – ω 2 ) 2 + ( 2ζω ω ) 2¹

(5.1)

with

u max =

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2

(5.5)

1 u max = f o -------------------------------------------------------------2 2 2 2 ( ω n – ω ) + ( 2ζω n ω )

(5.9)

fo 1 u max = -----2- ---------------------------------------------------------------------------------ω n [ 1 – ( ω ⁄ ω ) 2 ] 2 + [ 2ζ ( ω ⁄ ω ) ] 2 n

(5.10)

n

Introducing the maximum static amplitude u o = F o ⁄ k = f o ⁄ ω 2n the dynamic amplification factor V ( ω ) can be defined as: u max 1 V ( ω ) = ----------- = ----------------------------------------------------------------------------------uo 2 2 2 [ 1 – ( ω ⁄ ω n ) ] + [ 2ζ ( ω ⁄ ω n ) ]

(5.11)

The maximum amplification factor V ( ω ) occurs when its derivative, given by Equation (5.12), is equal to zero.

and the phase angle φ obtained from: a2 tan φ = ----a1


(5.6)

Page 5-1


Page 5-2

Course “Fundamentals of Structural Dynamics” 2

2

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2ωω n [ ω – ω n ( 1 – 2ζ ) ] dV = -------------------------------------------------------------------------------4 2 2 2 4 (3 ⁄ 2) dω [ ω – 2 ( 1 – 2ζ )ω ω n + ω n ]

(5.12)

dV 2 = 0 when: ω = 0 , ω = ± ω n 1 – 2ζ dω

(5.13)

The maximum amplification factor V ( ω ) occurs when: 1 2 ω = ω n 1 – 2ζ for ζ < ------- ≈ 0.71 2

2

ω = ω n 1 – 2ζ :

(5.14)

(5.20)

• V ( ω ) ≈ 1 therefore: u max ≈ u o • φ ≈ 0 : Motion and excitation force are in phase

1 V = -----2ζ

(5.15)

1 V = ------------------------2 2ζ 1 – ζ

(5.16)

2ζω n ω 2ζ ( ω ⁄ ω n ) - = -----------------------------tan φ = -----------------2 2 2 ωn – ω 1 – ( ω ⁄ ωn )

(5.17)

ωn 2 • V ( ω ) ≈ § ------· © ω¹

• φ ≈ 180 : Motion and excitation force are opposite

• ( ω ⁄ ω n ) ≈ 1 : (ζ very important)

The phase angle has the following interesting property: 2

2ζ [ 1 + ( ω ⁄ ω n ) ] dφ = --------------------------------------------------------------------------------------------------2 4 2 2 d ( ω ⁄ ωn ) 1 – 2 ( ω ⁄ ω n ) + ( ω ⁄ ω n ) + 4ζ ( ω ⁄ ω n )

• ω ⁄ ω n » 1 : Quick variation of the excitation (ζ not important)

ωn 2 2 • u max ≈ u o ⋅ § ------· = F o ⁄ ( mω ) : Mass controls the behaviour © ω¹

From Equation (5.6), the phase angle φ is:

at ω ⁄ ω n = 1 we have:

1 V ( ω ) = ----------------------------------------------------------------------------------2 2 2 [ 1 – ( ω ⁄ ω n ) ] + [ 2ζ ( ω ⁄ ω n ) ]

• ω ⁄ ω n « 1 : Slow variation of the excitation (ζ not important)

and we have: ω = ωn :

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5.1.1 Comments on the amplification factor V

2

2


(5.18)

1 • V ( ω ) ≈ -----2ζ • u max ≈ u o ⁄ ( 2ζ ) = F o ⁄ ( cω n ) : Damping controls the behaviour

1 180 dφ 1 = --- (= --- ⋅ --------- when φ in deg) d ( ω ⁄ ωn ) ζ ζ π

• φ ≈ 90 : zero displacement when excitation force is maximum

(5.19)


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• Example:

• Amplification factor 10

An excitation produces the static displacement

0.00

9

Amplification factor V(ω)


0.01

8 7

F o cos ( ωt ) u st = -------------------------k

ζ = 0.05

6

(5.21)

5

and its maximum is:

4

0.10

3 2

0.20

Fo u o = ----k

0.50

1

(5.22)

0.70

0 0

0.5

1

1.5

2

ω/ωn

The steady-state dynamic response of the system is: u p = u max cos ( ωt – φ )

• Phase angle

(5.23)

180

therefore:

Phase Angle I

0.00 0.01 ]= 0.05

0.10

0.20

0.50

u st ------ = cos ( ωt ) uo

0.70

,

up ----- = V cos ( ωt – φ ) uo

(5.24)

90

In the next plots the time histories of u st ⁄ u o and u p ⁄ u o are represented and compared.

0 0

0.5

1

1.5

2

The phase angle φ is always positive and because of the minus sign in Equation (5.24) it shows how much the response to the excitation lags behind.

Z/Zn 5 Transfer Functions

Page 5-5


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5.1.2 Steady-state displacement quantities

ω ⁄ ω n = 0.9

• Displacement: Corresponds to Equation (5.4)

u / uo

Frequency of SDoF System fn = 1Hz (ωn = 6.28rad/s), Damping ζ = 0.1 5 4 3 2 1 0 -1 -2 -3 -4 -5 0.00

V

ω = 0.9 ⋅ 6.28 = 5.65 rad/s

Δt

V = 3.82 φ = 43.45° = 0.76 rad

Excitation Steady-state response

0.25

0.50

0.75

1.00

1.25

1.50

1.75

2.00

= 0.14 s

u / uo

Time [s] 5 4 3 2 1 0 -1 -2 -3 -4 -5 0.00

V

ω ⁄ ω n = 1.0 ω = 1.0 ⋅ 6.28 = 6.28 rad/s

Δt

V = 5.00 φ = 90.00° = 1.57 rad

Excitation Steady-state response

0.25

0.50

0.75

1.00

1.25

1.50

1.75

2.00

u / uo

Time [s] 5 4 3 2 1 0 -1 -2 -3 -4 -5 0.00

φ 0.76 Δt = ---- = ---------ω 5.65

1.57 φ Δt = ---- = ---------ω 6.28 = 0.25 s

ω ⁄ ω n = 1.1 V

ω = 1.1 ⋅ 6.28 = 6.91 rad/s

Δt

V = 3.28 φ = 133.66° = 2.33 rad Excitation

φ 2.33 Δt = ---- = ---------ω 6.91

Steady-state response

0.25

0.50

0.75

1.00

Time [s]


1.25

1.50

1.75

2.00

= 0.34 s

Page 5-7

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up -----------= V ( ω ) cos ( ωt – φ ) Fo ⁄ k

(5.25)

• Velocity: Obtained by derivating Equation (5.25) u· p -----------= – V ( ω )ω sin ( ωt – φ ) Fo ⁄ k

(5.26)

u· p ω -----------------------= – V ( ω ) ------ sin ( ωt – φ ) ωn ( F o ⁄ k )ω n

(5.27)

u· p ω --------------------- = – V v ( ω ) sin ( ωt – φ ) with V v ( ω ) = ------ V ( ω ) (5.28) ω F o ⁄ km n • Acceleration: Obtained by derivating Equation (5.26) u··p 2 -----------= – V ( ω )ω cos ( ωt – φ ) Fo ⁄ k

(5.29)

2 u··p ω ----------------------------- cos ( ωt – φ ) = – V ( ω ) 2 2 ( F o ⁄ k )ω n ωn

(5.30)

2 u··p ω -------------- = – V a ( ω ) cos ( ωt – φ ) with V a ( ω ) = ------ V ( ω ) 2 Fo ⁄ m ωn

(5.31)


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• Amplification factors 10

Resonant displacement

0.00 0.01 8 7

10

5

ω = ω n 1 – 2ζ

4

0.10

3 2

0.20 0.70

0 0

0.5

1

1.5

2

ω/ωn 10

Resonant velocity

0.00 Amplificati cation factor Vv(ω)

9

0.01 8 7

ζ = 0.05

6

0 10 0.10

2

0.20

0

0.5

1.5

Resonant acceleration

0.00 0.01 8 7

ζ = 0.05

6

ωn ω = ---------------------2 1 – 2ζ

5

0.10

3

0.20 0.50

1

0.70

0 0.5

1

1.5

2

1 V = ---------------------2ζ 1 – ζ

ω/ωn


2

ζ = 0.05

2ζ ωa

0.5

1

ωb

1.5

2

2

9

0

3

2

ω/ωn

2

4

Condition: 1

10

4

5

ω/ωn

0.70

0

2

6

0

1 V = -----2ζ

0.50

1

V(Resonance)

7

0

4 3

8

1

ω = ωn

5

V(Resonance)

9

2

1 V = ---------------------2ζ 1 – ζ

0.50

1

Amplification factor Va(ω)

• Half-power bandwidth

ζ = 0.05

6

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5.1.3 Derivating properties of SDoF systems from harmonic vibrations

Amp plification factor V(ω)

Amplification Factor V(ω)

9


Page 5-9

V ( ω ⁄ ω n = 1 – 2ζ ) 1 1 V ( ω ) = ----------------------------------------------------- = ------- ⋅ ---------------------2 2 2ζ 1 – ζ 1 1 1- ------------------------------------------------------------------------------------------------------- = -----⋅ 2 2 2 2 2ζ 1 – ζ [ 1 – ( ω ⁄ ω n ) ] + [ 2ζ ( ω ⁄ ω n ) ]

(5.32) (5.33)

ω ·4 ω 2 § ----- – 2 ( 1 – 2ζ 2 ) § ------· + 1 – 8ζ 2 ( 1 – ζ 2 ) = 0 © ω n¹ © ω n¹

(5.34)

ω-· 2 2 2 § ----= 1 – 2ζ ± 2ζ 1 – ζ © ω n¹

(5.35)


Page 5-10


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For small damping, the terms featuring ζ can be neglected: ω----≈ 1 ± 2ζ ≈ 1 ± ζ ωn


5.2 Force transmission (vibration isolation)

(5.36)

This yield the solution for the half-power bandwidth: ωb – ωa 2ζ = ------------------ωn

(5.37)

The mass-spring-damper system, shown here on the right, is excited by the harmonic force F ( t ) = F o cos ( ωt )

• Remarks on the frequency response curve • The natural frequency of the system can be derived from the resonant response. However, it is sometimes problematic to build the whole frequency response curve because at resonance the system could be damaged. For this reason it is often better to determine the properties of a system based on vibration decay tests (see section on free vibration) • The natural frequency ω n can be estimated by varying the Excitation until a 90° phase shift in the response occurs. • Damping can be calculated by means of Equation (5.15) as: 1 uo ζ = --- ⋅ ----------2 u max

What is the reaction force F T ( t ) , which is introduced in the foundation?

The reaction force F T ( t ) results from the sum of the spring force F s and the damper’s force F c F T ( t ) = F s ( t ) + F c ( t ) = ku ( t ) + cu· ( t )

(5.38)

The steady-state deformation of the system due to harmonic excitation F ( t ) is according to Equation (5.4):

However, it is sometimes difficult to determine the static deflection u o , therefore, the definition of half-power bandwidth is used to estimate the damping. • Damping can be determined from the slope of the phase angle curve using Equation (5.19).

Fo u p = u max cos ( ωt – φ ) with u max = u o V ( ω ) = ----- V ( ω ) k

Page 5-11

(5.39)

By substituting Equation (5.39) and its derivative into Equation (5.38) we obtain: Fo F T ( t ) = ----- V ( ω ) [ k cos ( ωt – φ ) – cω sin ( ωt – φ ) ] k


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(5.40) Page 5-12


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with the trigonometric identity from Equation (5.3): Fo 2 2 2 F T ( t ) = ----- V ( ω ) [ k + c ω cos ( ωt – φ ) ] k


• Representation of the transmissibility TR 10

(5.41)

(5.42)

the maximum reaction force becomes: F T,max --------------- = TR ( ω ) Fo

0.01

8

(5.43)

Trrasmissibility TR

ω F T ( t ) = F o V ( ω ) 1 + § 2ζ ------· cos ( ωt – φ ) © ω n¹

0.00

9

and by substituting the identity c = ( 2ζk ) ⁄ ω n : 2

ζ = 0.05

7 6 5

0.10

4 3

0.20

2

where the quantity TR ( ω ) is called Transmissibility and it is equal to:

0.50

1

0.70

0 0

ω TR ( ω ) = V ( ω ) 1 + § 2ζ ------· © ω n¹

2

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0.5

1

2 1.5

2

2.5

3

3.5

4

ω/ωn

(5.44)

- When ω ⁄ ω n > 2 then TR < 1 : Vibration isolation 2

=

1 + [ 2ζ ( ω ⁄ ω n ) ] -----------------------------------------------------------------------------2 2 2 [ 1 – ( ω ⁄ ω n ) ] + [ 2ζ ( ω ⁄ ω n ) ]

- When ω ⁄ ω n > 2 damping has a stiffening effect - High tuning (sub-critical excitation)

Special case:

- Low tuning (super-critical excitation): Pay attention to the starting phase!

2

ω 1 + 4ζ TR § ------ = 1· = ---------------------© ωn ¹ 2ζ


(5.45)

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5.3 Base excitation (vibration isolation)


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The right hand side of the ODE (5.48) can be interpreted as an external excitation force F ( t ) = ky + cy· :

5.3.1 Displacement excitation F ( t ) = ky go cos ( ωt ) – cy go ω sin ( ωt )

(5.49)

ω = ky go cos ( ωt ) – 2ζ ------ sin ( ωt ) ωn ω 2 = ky go 1 + § 2ζ ------· cos ( ωt + φ ) © ω n¹

The external excitation force F ( t ) is harmonic with amplitude: ω 2 F o = ky go 1 + § 2ζ ------· © ω n¹

The mass-spring-damper system, shown here above is excited by the harmonic vertical ground displacement y g ( t ) = y go cos ( ωt )

(5.46)

What is the absolute vertical displacement u ( t ) of the system? The differential equation of the system is: mu·· + c ( u· – y· ) + k ( u – y ) = 0


According to Equations (5.10) and (5.11) the maximum displacement of the system due to such a force is equal to: Fo ω 2 u max = ----- V ( ω ) = y go 1 + § 2ζ ------· V ( ω ) © k ω n¹

(5.51)

By substituting Equation (5.44) we obtain: (5.47)

after rearrangement: mu·· + cu· + ku = ky + cy·

(5.50)

(5.48)

Page 5-15

u max ----------- = TR ( ω ) y go

(5.52)

where again TR ( ω ) is the transmissibility given by Equation (5.44).


Page 5-16


An-Najah 2013

5.3.2 Acceleration excitation This base excitation, like the excitation discussed in the previous Section 5.3.1, is an harmonic excitation and not an arbitrary excitation like e.g. an earthquake (see Section 7).

The mass-spring-damper system, shown above here, is excited by the harmonic vertical ground acceleration. (5.53)

What is the absolute vertical acceleration u··( t ) of the system?

(5.54)

after rearrangement: mu·· + c ( u· – y· ) + k ( u – y ) – my·· = – my··

(5.55)

m ( u·· – y··) + c ( u· – y· ) + k ( u – y ) = – my··

(5.56)

mu··rel + cu· rel + ku rel = – my··g

(5.57)

mu··rel + cu· rel + ku rel = – my··go cos ( ωt )

(5.58)


u rel = a 1 cos ( ωt ) + a 2 sin ( ωt )

Page 5-17

(5.59)

with the constants a 1 and a 2 according to Equation (5.2), and with: Fo – my··go f o = ----- = ---------------- = – y··go m m

(5.60)

By double derivation of Equation (5.59), the relative acceleration u··rel can be calculated as: 2 2 u··rel = – a 1 ω cos ( ωt ) – a 2 ω sin ( ωt )

(5.61)

The desired absolute acceleration is: 2 2 u·· = u··rel + y··g = – a 1 ω cos ( ωt ) – a 2 ω sin ( ωt ) + y··go cos ( ωt )

The differential equation of the system is: mu·· + c ( u· – y· ) + k ( u – y ) = 0

An-Najah 2013

The steady-state relative deformation u rel of the system due to the harmonic ground acceleration y··g is given by Equation (5.1):

Pay attention:

y··g ( t ) = y··go cos ( ωt )


(5.62)

By substituting the constants a 1 , a 2 and f o given by Equations (5.2) and (5.60), and after a long but simple rearrangement, the equations for the maximum absolute vertical acceleration of the system is obtained as: u··max ----------= TR ( ω ) y··go

(5.63)

where again TR ( ω ) is the transmissibility given by Equation (5.44).


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• Additional derivation:


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5.3.3 Example transmissibility by base excitation

The maximum relative displacement given by Equation (5.58) can be easily determined by means of Equations (5.10), (5.11) and (5.60) as: u rel, max

fo – y··go -V(ω) = -----2- V ( ω ) = ------------2 ωn ωn

u rel, max ---------------------- = V(ω) 2 ( y··go ⁄ ω n )

Vertical base excitation:

(5.64)

y··g ( t ) = A 0 cos ( ω 0 t )

(5.65) • Natural frequency SDoF system:

f n = 0.5Hz

• Excitation frequency:

f 0 = 2.0Hz

• Excitation amplitude:

A 0 = 10m ⁄ s 2

Sought is the maximum absolute acceleration u··max of the SDoF system for ζ = 2% and for ζ = 20% . • The steady-state maximum absolute acceleration is: • ζ = 2% , ω 0 ⁄ ω n = 4 :

2 TR = 0.068 and u··max = 0.68m ⁄ s

• ζ = 20% , ω 0 ⁄ ω n = 4 : TR = 0.125 and u··max = 1.25m ⁄ s 2 • Is the steady-state maximum absolute acceleration really the maximum absolute acceleration or at start even larger absolute accelerations may result? • Assumptions: starting time t a = 80s , sinusoidal start function for excitation frequency and excitation amplitude. • Numerical computation using Newmark’s Method (see Section 7) 5 Transfer Functions

Page 5-19


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Excitation freq. [Hz]

2

1

f=fn=0.5Hz 10

20

30

40

50

60

70

80

90

100

10

5

0 0

10

20

30

40

50

60

70

80

90

100

10

Excitation amplitude [m/s2]

0 0

5 0 -5 -10 0

20

30

40

50

60

70

80

90

100

1

f=fn=0.5Hz 0 0

10

20

30

40

50

60

70

80

90

100

10

20

30

40

50

60

70

80

90

100

10

20

30

40

50

60

70

80

90

100

10

20

30

40

50 Time [s]

60

70

80

90

100

10

5

0 0 10 5 0 -5

0 20 Aabs SDoF [m/s2]

Aabs SDoF [m/s2]

2

-10 10

20 10 0 -10 -20 0

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• Case 2: Increase of the damping rate from ζ = 2% to ζ = 20%

Excitation [m/s2]

Excitation [m/s2]



• Case 1: Initial situation with ζ = 2%


10


20

30

40

50 Time [s]

60

70

80

90

100

Page 5-21

10 0 -10 -20 0


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2

1

f=fn=0.5Hz 10

20

30

40

50

60

70

80

90

100

10

5

0 0

10

20

30

40

50

60

70

80

90

100

10


0 0

5 0 -5 -10 0

20

30

40

50

60

70

80

90

100

1

f=fn=0.5Hz 0 0

10

20

30

40

50

60

70

80

90

100

10

20

30

40

50

60

70

80

90

100

10

20

30

40

50

60

70

80

90

100

10

20

30

40

50 Time [s]

60

70

80

90

100

10

5

0 0 10 5 0 -5

0 20 Aabs SDoF [m/s2]

Aabs SDoF [m/s2]

2

-10 10

20 10 0 -10 -20 0

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• Case 4: Change the start function for the amplitude (ζ = 2%)

Excitation [m/s2]

Excitation [m/s2]



• Case 3: Reduction of starting time from ta = 80s to ta = 20s (ζ = 2%)


10


20

30

40

50 Time [s]

60

70

80

90

100

Page 5-23

10 0 -10 -20 0


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5.4 Summary Transfer Functions

• Notes The excitation function in the starting phase has the form: y··g ( t ) = A ( t ) cos ( Ω ( t ) ⋅ t )

(5.66)

The excitation angular frequency varies with time, and is:

ω(t) = d (Ω(t) ⋅ t ) dt

(5.67)

1 V ( ω ) = ----------------------------------------------------------------------------------2 2 2 [ 1 – ( ω ⁄ ω n ) ] + [ 2ζ ( ω ⁄ ω n ) ]

t ω ( t ) = ω 0 ⋅ --- ( 0 ≤ t ≤ t a ) ta

(5.68)

TR ( ω ) =

1 + [ 2ζ ( ω ⁄ ω n ) ] -----------------------------------------------------------------------------2 2 2 [ 1 – ( ω ⁄ ω n ) ] + [ 2ζ ( ω ⁄ ω n ) ]

• Force excitation:

u max ----------- = V ( ω ) uo

• Force transmission:

F T,max --------------- = TR ( ω ) Fo

• Displacement excitation:

u max ----------- = TR ( ω ) y go

• Acceleration excitation

u··max ----------= TR ( ω ) y··go

• Parabolic variation of the excitation circular frequency ω0 2 Ω ( t ) = ------2- ⋅ t : 3t a

t 2 ω ( t ) = ω 0 ⋅ § ---· ( 0 ≤ t ≤ t a ) © t a¹

(5.69)

• Sinusoidal variation of the excitation circular frequency 2ω 0 t a π t π t Ω ( t ) = – -------------- cos § --- ⋅ ---· : ω ( t ) = ω 0 ⋅ sin § --- ⋅ ---· ( 0 ≤ t ≤ t a ) © 2 t a¹ © 2 t a¹ πt

(5.70) • Double-sinusoidal variation of the excitation circular frequency t sin § π ⋅ ---· t a © t a¹ ω0 t Ω ( t ) = ------ 1 – ----------------------------- : ω ( t ) = ω 0 1 – cos § π ⋅ ---· © πt 2 t a¹

(5.71)

(5.72)

2

• Linear variation of the excitation circular frequency ω0 Ω ( t ) = ------- ⋅ t : 2t a

An-Najah 2013

(5.73)

u rel, max ---------------------- = V(ω) 2 ·· ( y go ⁄ ω n ) • For further cases check the literature.

• Visualization of the solution by means of the Excel file given on the web page of the course (SD_HE_Starting_Phase.xlsx) 5 Transfer Functions

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Page 5-26


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6 Forced Vibrations

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with the fundamental frequency 2π ω 0 = -----T0

6.1 Periodic excitation 4.0

(6.3)

Taking into account the orthogonality relations: Half-sine excitation

T0

3.5

T0

³0

3.0

Force F(t) [kN] F


2.5 T0

³0

2.0 1.5

T0

³0

1.0

for n ≠ j 0 sin ( nω 0 t ) sin ( jω 0 t ) dt = ® ¯ T 0 ⁄ 2 for n = j

(6.4)

for n ≠ j 0 cos ( nω 0 t ) cos ( jω 0 t ) dt = ® ¯ T 0 ⁄ 2 for n = j

(6.5)

cos ( nω 0 t ) sin ( jω 0 t ) dt = 0

(6.6)

the Fourier coefficients a n can be computed by multiplying Equation (6.2) by cos ( jω 0 t ) first, and then integrating it over the period T 0 .

0.5 0.0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

• j = 0

Time (s) T0

³0

An excitation is periodic if: F ( t + nT 0 ) = F ( t )

for

n = – ∞, …, – 1, 0, 1, …, ∞ (6.1)

The function F ( t ) can be represented as a sum of several harmonic functions in the form of a Fourier series, namely: ∞

F ( t ) = a0 +

¦ [ an cos ( nω0 t ) + bn sin ( nω0 t ) ]

(6.2)

n=1

6 Forced Vibrations

F ( t ) cos ( jω 0 t ) dt =

Page 6-1

T0

³0

a 0 cos ( jω 0 t ) dt

+

¦ ³ n=1 0

∞

T0

³0

F ( t ) dt =

T0

³0

a 0 dt = a 0 T 0

1 T0 a 0 = ------ ⋅ ³ F ( t ) dt T0 0 6 Forced Vibrations

T0

(6.7) T0

a n cos ( nω 0 t ) cos ( jω 0 t ) dt + ³ b n sin ( nω 0 t ) cos ( jω 0 t ) dt 0

(6.8) (6.9)

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An-Najah 2013

³0

F ( t ) cos ( jω 0 t ) dt =

T0

³0

+

T0

¦ ³0

n=1

T0

(6.10)

a 0 cos ( jω 0 t ) dt ∞

³0

T0

a n cos ( nω 0 t ) cos ( jω 0 t ) dt + ³ b n sin ( nω 0 t ) cos ( jω 0 t ) dt 0

T F ( t ) cos ( nω 0 t ) dt = a n ⋅ -----0 2

2 T0 a n = ------ ⋅ ³ F ( t ) cos ( nω 0 t ) dt T0 0

2 T0 b n = ------ ⋅ ³ F ( t ) sin ( nω 0 t ) dt T0 0

mu·· + cu· + ku = F ( t ) F(t) 2 u·· + 2ζω n u· + ω n u = ---------m F ( t ) = a0 +

∞

¦

(6.14) (6.15)

[ a n cos ( nω 0 t ) + b n sin ( nω 0 t ) ]

(6.16)

n=1

(6.11) (6.12)

Similarly, the Fourier coefficients b n can be computed by first multiplying Equation (6.2) by sin ( jω 0 t ) and then integrating it over the period T 0 . (6.13)

• Static Part ( a 0 ) a0 u 0 ( t ) = ----k

(6.17)

• Harmonic part “cosine” (see harmonic excitation) 2

Co sin e un (t)

nω a n 2ζβ n sin ( nω 0 t ) + ( 1 – β n ) cos ( nω 0 t ) - , β n = ---------0 = ----- ⋅ ---------------------------------------------------------------------------------------2 2 2 k ωn ( 1 – β n ) + ( 2ζβ n )

(6.18) • Harmonic part “sine” (similar as “cosine”) 2

• Notes

Sine un ( t )

- a 0 is the mean value of the function F ( t ) - The integrals can also be calculated over the interval [– T 0 ⁄ 2,T 0 ⁄ 2] - For j = 0 no b-coefficient exists

b n ( 1 – β n ) sin ( nω 0 t ) – 2ζβ n cos ( nω 0 t ) nω - , β n = ---------0 = ----- ⋅ ---------------------------------------------------------------------------------------2 2 2 k ωn ( 1 – β n ) + ( 2ζβ n )

(6.19) • The steady-state response u ( t ) of a damped SDoF system under the periodic excitation force F ( t ) is equal to the sum of the terms of the Fourier series. u ( t ) = u0 ( t ) +

6 Forced Vibrations

An-Najah 2013

6.1.1 Steady state response due to periodic excitation

• j = n T0


Page 6-3

6 Forced Vibrations

∞

Co sin e

¦ un

n=1

(t) +

∞

Sine

¦ un

(t)

(6.20)

n=1

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An-Najah 2013

6.1.2 Half-sine

An-Najah 2013 2

A series of half-sine functions is a good model for the force that is generated by a person jumping. Half-sine excitation

T0

4Aτ cos ( nπτ ) 2A t p πt a n = ------- ⋅ ³ sin § -----· cos ( nω 0 t ) dt = -----------------------------------2 2 © tp ¹ T0 0 π ( 1 – 4n τ )

(6.23)

πt 2A tp 4Aτ sin ( nπτ ) cos ( nπτ )b n = ------- ⋅ ³ sin § -----· sin ( nω 0 t ) dt = -------------------------------------------------------2 2 © tp ¹ T0 0 π ( 1 – 4n τ ) (6.24)

4.0 3.5


tp

Force F(t) [kN] F

3.0

The approximation of the half-sine model for T 0 = 0.5s and t p = 0.16s by means of 6 Fourier terms is as follows:

2.5 2.0

4.0

1.5

Static term (n=0)

T0

First harmonic (n=1)

3.0

1.0

Second harmonic (n=2) Third harmonic (n=3) Total (6 harmonics)

Force F(t) [kN] F

0.5 0.0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time (s)

§ πt-· for 0 ≤ t < t p ° A sin © ---tp ¹ F(t) = ® ° 0 for t p ≤ t < T 0 ¯

2.0 1.0 0.0 -1.0

(6.21) -2.0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time (s)

The Fourier coefficients can be calculated at the best using a mathematics program: A tp πt 2Aτ a 0 = ------ ⋅ ³ sin § -----· dt = ---------© tp ¹ T0 0 π

6 Forced Vibrations

with

tp τ = -----T0

• Note

(6.22)

The static term a 0 = 2Aτ ⁄ π = G corresponds to the weight of the person jumping.

Page 6-5

6 Forced Vibrations

Page 6-6


An-Najah 2013

6.1.3 Example: “Jumping on a reinforced concrete beam” • Beam • Young’s Modulus: E = 23500MPa

Dynamic:

u max = max ( u ( t ) ) with u ( t ) from Equation (6.20)

• Density: ρ = 20.6kN ⁄ m 3

Ratio:

u max V = ----------u st

• Modal mass M n = 0.5M tot • Modal stiffness π 4 EI K n = ----- ⋅ -----32 L

• Excitation (similar to page 186 of [Bac+97]) • Jumping frequency: f 0 = 2Hz

Half-sine excitation

Force F(t) [kN] F

3.0

• Period: T 0 = 0.5s

2.5

• Contact time: t p = 0.16s

2.0 1.5

0.5 0.0 0

0.1

0.2

0.3

0.4

0.5

Time (s)

6 Forced Vibrations

0.6

0.7

0.8

0.9

1

• Investigated cases Length [m]

Frequency fn [Hz]

umax [m]

V [-]

26.80

1

0.003

1.37

19.00

2

0.044

55.94

15.50

3

0.002

3.62

13.42

4

0.012

41.61

12.01

5

0.001

4.20

10.96

6

0.004

25.02

• Notes - When the excitation frequency f 0 is twice as large as the

• Person’s weight: G = 0.70kN

1.0

• Maximum deflections G u st = -----Kn

• Damping rate ζ = 0.017

3.5

An-Najah 2013

Static:

• Bending stiffness: EI = 124741kNm 2

4.0


• Amplitude: A = 3.44kN

natural frequency f n of the beam, the magnification factor V is small. - Taking into account the higher harmonics can be important!

Page 6-7

6 Forced Vibrations

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• Case1: f0 = 2Hz, fn = 1Hz


• Case 3: f0 = 2Hz, fn = 3Hz 0.0020

0.0035

Static term (n=0)

0.0030


0.0015

0.0025

Second harmonic (n=2) Third harmonic (n=3)

0.0020

Displacement [m] D

Displacement [m] D

An-Najah 2013

Static term (n=0) First harmonic (n=1)

0.0015


0.0010

Total (6 harmonics)

0.0005 0 0000 0.0000 -0.0005

Total (6 harmonics)

0.0010 0.0005 0.0000 -0.0005

-0.0010 -0.0015

-0.0010 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0

0.1

0.2

0.3

0.4

Time (s)

0.5

0.6

0.7

0.8

0.9

1

Time (s)

• Case 2: f0 = 2Hz, fn = 2Hz

• Case 4: f0 = 2Hz, fn = 4Hz 0.0150

0.0500


0.0400

Third harmonic (n=3) Total (6 harmonics)

Displacement [m] D

Displacement [m] D

Second harmonic (n=2)

0.0100

0.0300 0.0200 0.0100 0.0000 -0.0100 0 0200 -0.0200

0.0050 0.0000 -0.0050

Static term (n=0)

-0.0300


-0.0400

Third harmonic (n=3)

-0.0100

Second harmonic (n=2)

Total (6 harmonics)

-0.0500

-0.0150 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0

1

Time (s)

6 Forced Vibrations

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time (s)

Page 6-9

6 Forced Vibrations

Page 6-10


An-Najah 2013


An-Najah 2013

6.2 Short excitation

• Case 5: f0 = 2Hz, fn = 5Hz 0.0010

6.2.1 Step force


0.0008


Displacement [m] D

0.0006

The differential equation of an undamped SDoF System loaded with a force F 0 which is applied suddenly at the time t = 0 is:

Total (6 harmonics)

0.0004

mu·· + ku = F 0

0.0002

There is a homogeneous and a particular solution

-0.0002 -0.0004 -0.0006 -0.0008 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time (s)

• Case 6: f0 = 2Hz, fn = 6Hz Static term (n=0) First harmonic (n=1)

0.0040

u h = A 1 cos ( ω n t ) + A 2 sin ( ω n t ) (see free vibrations)

(6.26)

up = F0 ⁄ k

(6.27)

The overall solution u ( t ) = u h + u p is completely defined by the initial conditions u ( 0 ) = u· ( 0 ) = 0 and it is: F0 u ( t ) = ----- [ 1 – cos ( ω n t ) ] k

0.0050

(6.28)


0.0030

Displacement [m] D

(6.25)

0.0000

• Notes

Total (6 harmonics)

• The damped case can be solved in the exact same way. On the web page of the course there is an Excel file to illustrate this excitation.

0.0020 0.0010 0.0000

• The maximum displacement of an undamped SDoF System under a step force is twice the static deflection u st = F 0 ⁄ k .

-0.0010 -0.0020

• The deflection at the time t = ∞ of a damped SDoF System under a step force is equal to the static deflection u st = F 0 ⁄ k .

-0.0030 -0.0040 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time (s)

6 Forced Vibrations

Page 6-11

6 Forced Vibrations

Page 6-12


An-Najah 2013

• Step force: Tn=2s, Fo/k=2, ζ=0


An-Najah 2013

6.2.2 Rectangular pulse force excitation

4.5 Dynamic response Excitation

4

Displacement

3.5 3 2.5 2 1.5

The differential equation of an undamped SDoF system under a rectangular pulse force excitation is:

1 0.5 0 0

1

2

3

4

5

6

7

8

9

10

Time (s)

• Step force: Tn=2s, Fo/k=2, ζ=10% Dynamic response Excitation

3

Displacement

for t ≤ t 1 for t > t 1

(6.29)

Up to time t = t 1 the solution of the ODE corresponds to Equation (6.28). From time t = t 1 onwards, it is a free vibration with the following initial conditions:

4 3.5

·· mu + ku = F 0 ® ·· ¯ mu + ku = 0

2.5 2 1.5

F0 u ( t 1 ) = ----- [ 1 – cos ( ω n t 1 ) ] k

(6.30)

F0 u· ( t 1 ) = ----- ω n sin ( ω n t 1 ) k

(6.31)

The free vibration is described by the following equation:

1

u h = A 1 cos ( ω n ( t – t 1 ) ) + A 2 cos ( ω n ( t – t 1 ) )

0.5 0 0

1

2

3

4

5

6

7

8

9

10

Time (s)

6 Forced Vibrations

Page 6-13

(6.32)

and through the initial conditions (6.30) and (6.31), the constants A 1 and A 2 can be determined. 6 Forced Vibrations

Page 6-14


An-Najah 2013

• Short duration of excitation ( t 1 ⁄ T n is small)


The equation of an undamped free vibration is:

The series expansion of sine and cosine is:

u ( t ) = A cos ( ω n t – φ ) with A =

( ωn t1 )2 2π cos ( ω n t 1 ) = cos § ------ t 1· = 1 – ------------------ + … © Tn ¹ 2

An-Najah 2013

v0 v0 2 u 0 + § ------· and tan φ = -----------© ω n¹ ωn u0

(6.38)

(6.33) therefore, the maximum amplitude of a short excitation is:

( ωn t1 )3 2π sin ( ω n t 1 ) = sin § ------ t 1· = ω n t 1 + ------------------ + … © Tn ¹ 6

(6.34)

and for small t 1 ⁄ T n the expressions simplifies to: cos ( ω n t 1 ) ≅ 1

,

sin ( ω n t 1 ) ≅ ω n t 1

(6.35)

By substituting Equation (6.35) in Equations (6.30) and (6.31) it follows that: u ( t1 ) = 0

,

F0 2 F0 t1 u· ( t 1 ) = ----- ω n t 1 = ---------k m

(6.36)

Equation (6.36) shows, that a short excitation can be interpreted as a free vibration with initial velocity v0 = I ⁄ m

(6.37)

where I is the impulse generated by the force F 0 over the time t 1 . • Rectangular pulse force excitation:

I = F0 t1

• Triangular pulse force excitation:

I = 0.5F 0 t 1

• Arbitrary short excitation:

6 Forced Vibrations

I =

v0 A = -----ωn

(6.39)

• Notes • The damped case can be solved in the exact same way. On the web page of the course there is an Excel file to illustrate this excitation. • Rectangular pulse force excitation: When t 1 > T n ⁄ 2 , the maximum response of the SDoF system is equal to two times the static deflection u st = F 0 ⁄ k . • Rectangular pulse force excitation: When t 1 > T n ⁄ 2 , for some t 1 ⁄ T n ratios (z.B.: 0.5, 1.5, ...) the maximum response of the SDoF system can even be as large as 4F 0 ⁄ k . • Rectangular pulse force excitation: try yourself using the provided Excel spreadsheet. • Short excitation: The shape of the excitation has virtually no effect on the maximum response of the SDoF system. Important is the impulse. • Short excitation: Equation (6.59) is exact only for t 1 ⁄ T n → 0 and ζ = 0 . For all other cases, it is only an approximation, which overestimates the actual maximum deflection.

t1

³0 F ( t ) dt

Page 6-15

6 Forced Vibrations

Page 6-16


An-Najah 2013

• Rectangular pulse: Tn=2s, t1=0.5s (t1/Tn=0.25), Fo/k=2, ζ=0%


An-Najah 2013

• Rectangular pulse: Tn=2s, t1=2s (t1/Tn=1.00), Fo/k=2, ζ=0%

4


3

Dynamic response Excitation

4 3.5

Displacement

Displacement

2 1 0 -1

3 2.5 2 1.5 1

-2 0.5

-3

0

-4

-0.5

0

1

2

3

4

5

6

7

8

9

0

10

1

2

3

4


6

7

8

9

10

• Rectangular pulse: Tn=2s, t1=3s (t1/Tn=1.50), Fo/k=2, ζ=0% 5

5 Dynamic response Excitation

4


4

3

3

2

2

Displacement

Displacement

5

Time (s)

Time (s)

1 0 -1 -2

1 0 -1 -2

-3

-3

-4

-4 -5

-5 0

1

2

3

4

5

6

7

8

9

10

0

1

6 Forced Vibrations

2

3

4

5

6

7

8

9

10

Time (s)

Time (s)

Page 6-17

6 Forced Vibrations

Page 6-18


An-Najah 2013

• Rectangular pulse: Tn=2s, t1=3.5s (t1/Tn=1.75), Fo/k=2, ζ=0%


An-Najah 2013

• Short rectangular pulse: Tn=2s, t1=0.05s, Fo/k=2, ζ=0%

5



4

2

2

Displacement

Displacement

3

1 0 -1

1.5 1 0.5

-2 0

-3 -0.5

-4 0

1

2

3

4

5

6

7

8

9

0

10

1

2

3

4


6

7

8

9

10

• Short rectangular pulse: Tn=2s, t1=0.05s, Fo/k=2, ζ=5% 2.5

4.5



4

2

3.5 3

Displacement

Displacement

5

Time (s)

Time (s)

2.5 2 1.5 1 0.5

1.5 1 0.5 0

0 -0.5

-0.5 0

1

2

3

4

5

6

7

8

9

10

0

1

6 Forced Vibrations

2

3

4

5

6

7

8

9

10

Time (s)

Time (s)

Page 6-19

6 Forced Vibrations

Page 6-20


An-Najah 2013


An-Najah 2013

6.2.3 Example “blast action”

Mass:

m = 3.05 ⋅ 0.276 ⋅ 2.45 = 2.06t ⁄ m

• Test

Concrete

f c' = 41.4MPa

Stiffness

I o = ( 3050 ⋅ 276 ) ⁄ 12 = 5344 ×10 mm

, 3

E c = 5000 ⋅ f c' = 32172MPa 6

4

E c I o = 171.9kNm 2 E c I = 0.30E c I o = 52184kNm 2 (due to cracking!)

- Action t 1 ≈ 0.3ms is by sure much

shorter than the period T n = 64ms of the slab (see Equation (6.51)). Therefore, the excitation can be considered as short.

• Modelling option 1 Within a simplified modelling approach, it is assumed that the slab remains elastic during loading. Sought is the maximum deflection of the slab due to the explosion.

- Equivalent modal SDoF system (see Section “Modelling”)

- Simplified system Cross-Section

Ansatz for the deformed shape: ψ = C1 ⋅ sin ( βx ) + C2 ⋅ cos ( βx ) + C3 ⋅ sin h ( βx ) + C4 ⋅ cos h ( βx )

(6.40) 6 Forced Vibrations

Page 6-21

6 Forced Vibrations

Page 6-22


An-Najah 2013

Boundary conditions: ψ ( 0 ) = 0 , ψ ( L ) = 0 , ψ' ( 0 ) = 0 , ψ'' ( L ) = 0

2

An-Najah 2013

EI ⋅ dx ) = 104.37 ⋅ -----3L

³0 ( EI ⋅ ( ψ'' )

P* =

³L =1.55m ( p ⋅ ψ ⋅ dx )

L 2 =3.45m

(6.45)

= 0.888 ⋅ P tot

(6.46)

1

For this example, the modal properties characterizing the equivalent modal SDoF system are:

[ sin ( βL ) + sin h ( βL ) ] ⋅ [ cos ( βx ) – cos h ( βx ) ] 1.508 ⋅ ψ = sin ( βx ) – sin h ( βx ) + -----------------------------------------------------------------------------------------------------------------– cos ( βL ) – cos h ( βL )

(6.42) with βL = 3.927

L

k* =

(6.41)

By means of the mathematics program “Maple” Equation (6.40) can be solved for the boundary conditions (6.41) and we get:

(6.43)

The shape of the function ψ is: 1 0.8

\ [-]


m * = 0.439 ⋅ 2.06 ⋅ 5 = 4.52t

(6.47)

52184= 43571kN/m k * = 104.37 ⋅ -------------53

(6.48)

P * = 0.888 ⋅ 192000 = 170496kN

(6.49)

ω =

(6.50)

k* ⁄ m* =

43571 ⁄ 4.52 = 98.18rad/s

T n = 2π ⁄ ω = 0.064s

0.6

(6.51)

The maximum elastic deformation of the SDoF system can be calculated using the modal pulse as follows:

0.4 0.2 0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x/L [-]

L

³0 m ⋅ ψ

2

⋅ dx = 0.439mL

(6.44)

25.6 I* v 0 = ------*- = ---------- = 5.66m/s 4.52 m

= 25.6kNs

(6.52)

Page 6-23

(6.53)

The maximum elastic deflection is: Δ m, e = v 0 ⁄ ω = 5.66 ⁄ 98.18 = 0.058m

6 Forced Vibrations

–3

The initial velocity of the free vibration is:

And with the equations given in Section “Modelling”, the modal properties of the equivalent SDoF system are determined: m* =

I * = 0.5 ⋅ P * ⋅ t 0 = 0.5 ⋅ 170496 ⋅ 0.3 ×10

6 Forced Vibrations

(6.54)

Page 6-24


An-Najah 2013


Within a simplified modelling approach, it is assumed that the slab remains elastic during loading. Sought is the maximum deflection of the slab due to the explosion. - Simplified system

ψ [-]

• Modelling option 2

An-Najah 2013

1 0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x/L [-]

And with the equations given in Section “Modelling”, the modal properties of the equivalent SDoF system are determined: L

2

m* =

³0 m ⋅ ψ

k* =

³0 ( EI ⋅ ( ψ'' )

P* =

³L =6.55m ( p ⋅ ψ ⋅ dx )

- Equivalent modal SDoF system (see Section “Modelling”)

⋅ dx = 0.5mL

L

L 2 =8.45m

2

EI 4 EI ⋅ dx ) = 8π ⋅ -----3- = 779.27 ⋅ -----3L L = 0.941 ⋅ P tot

(6.57) (6.58) (6.59)

1

For this example, the modal properties characterizing the equivalent modal SDoF system are:

Ansatz for the deformed shape: 2πx ψ = – sin § ----------· © L ¹

(6.55)

Boundary conditions: ψ ( 0 ) = 0 , ψ ( L ) = 0 , ψ'' ( 0 ) = 0 , ψ'' ( L ) = 0

(6.56)

The shape of the function ψ is: 6 Forced Vibrations

Page 6-25

m * = 0.5 ⋅ 2.06 ⋅ 10 = 10.3t

(6.60)

52184= 40666kN/m k * = 779.27 ⋅ -------------10 3

(6.61)

P * = 0.941 ⋅ 192000 = 180672kN

(6.62)

ω =

(6.63)

k* ⁄ m* =

6 Forced Vibrations

40666 ⁄ 10.3 = 62.83rad/s

Page 6-26


An-Najah 2013

T n = 2π ⁄ ω = 0.10s

(6.64)

The maximum elastic deformation of the SDoF system can be calculated using the modal pulse as follows: I * = 0.5 ⋅ P * ⋅ t 0 = 0.5 ⋅ 180672 ⋅ 0.3 ×10

–3

= 27.1kNs

(6.65)


An-Najah 2013

• Modelling option 3 As a third option, the slab is modelled using the commercial finite element software SAP 2000. - Numerical Model

The initial velocity of the free vibration is: 27.1 I* v 0 = ------*- = ---------- = 2.63m/s 10.3 m

(6.66)

The maximum elastic deflection is: Δ m, e = v 0 ⁄ ω = 2.63 ⁄ 62.83 = 0.042m

(6.67)

The distributed load q is replaced by n = 19 concentrated forces F i : 192000 F i = ------------------ = 10105kN 19

(6.68)

The first period of the system is: T 1 = 0.100s

(6.69)

which corresponds to Equation (6.64).

6 Forced Vibrations

Page 6-27

6 Forced Vibrations

Page 6-28


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An-Najah 2013

And the time-history of the elastic deflection is: Blank Page

0.08

Elastic c deformation [m]

0.06 0.04 0.02 0 -0.02 -0.04 -0.06 -0.08 0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0.45

0.50

Time [s]

The effect of the higher modes can be clearly seen! • Comparison System

6 Forced Vibrations

m* [t]

k* [kN/m]

P* [P]

T [s]

Δ m, e [m]

4.52

43571

0.888

0.064

0.058

10.30

40666

0.941

0.100

0.042

-

-

-

0.100

0.064

Page 6-29

6 Forced Vibrations

Page 6-30

An-Najah 2013

7 Seismic Excitation 7.1 Introduction The equation of motion for a base point excitation through an acceleration time-history u··g ( t ) can be derived from the equilibrium of forces (see Section 2.1.1) as: mu·· + cu· + f s (u,t) = – mu··g


2

(7.1)

where u·· , u· and u are motion quantities relative to the base point of the SDoF system, while f s (u,t) is the spring force of the system that can be linear or nonlinear in function of time and space. The time-history of the motion quantities u·· , u· and u for a given SDoF system are calculated by solving Equation (7.1).

0

Δüg

-1

x [m/s2] ü

2

Ground acceleration

üxg [m/s2]

1

-2

4 2

9.5

1

Δü

0 -1 9.5

-2

9.6 Zeit [s]

-2

0

t

t+Δt

9.6

Time Zeit [s]

-4 0

10

20 Zeit [s] Time [s]

30

40

4 2 x [m/s ] ü

An-Najah 2013

From the previous figure it can be clearly seen that the time-history of an earthquake ground acceleration can not be described by a simple mathematical formula. Time-histories are therefore usually expressed as sequence of discrete sample values and hence Equation (7.1) must therefore be solved numerically.

xg [m/s2] ü


Response of a T=0.5s SDoF system

2 0 -2 -4 0

10

20

30

40

Zeit [s] Time [s] 7 Seismic Excitation

Page 7-1

The sample values of the ground acceleration u··g ( t ) are known from beginning to end of the earthquake at each increment of time Δt (“time step”). The solution strategy assumes that the motion quantities of the SDoF system at the time t are known, and that those at the time t + Δt can be computed. Calculations start at the time t = 0 (at which the SDoF system is subjected to known initial conditions) and are carried out time step after time step until the entire time-history of the motion quantities is computed, like e.g. the acceleration shown in the figure on page 7-1. 7 Seismic Excitation

Page 7-2


An-Najah 2013

7.2 Time-history analysis of linear SDoF systems


An-Najah 2013

7.2.1 Newmark’s method (see [New59]) • Incremental formulation of the equation of motion mΔu·· + cΔu· + kΔu = – mΔu··g t + Δt

t

u = u + Δu ,

t + Δt ·

t u = u· + Δu· ,

(7.4) t + Δt··

t u = u·· + Δu··

(7.5)

• Assumption of the acceleration variation over the time step:

In the case of a linear SDoF system Equation (7.1) becomes: mu·· + cu· + ku = – mu··g

(7.2)

and by introducing the definitions of natural circular frequency ω n = k ⁄ m and of damping ratio ζ = c ⁄ ( 2mω n ) , Equation (7.1) can be rearranged as: u·· + 2ζω n u +

2 ωn u

= – u··g

(7.3)

The response to an arbitrarily time-varying force can be computed using: • Convolution integral ([Cho11] Chapter 4.2)

t + Δt·· t 1 t Δu·· u··( τ ) = --- ( u·· + u ) = u·· + ------2 2 τ

t t Δu·· u· ( τ ) = u + ³ u··( τ ) dτ = u· + § u·· + -------· ( τ – t ) © 2¹ t·

Page 7-3

(7.7)

t τ

τ

t t Δu·· u ( τ ) = u + ³ u· ( τ ) dτ = u + ³ u· + § u·· + -------· ( τ – t ) dτ © 2¹ t

t

t

• Numerical integration of the differential equation of motion ([Cho11] Chapter 5) 7 Seismic Excitation

(7.6)


(7.8)

t

Page 7-4

Course “Fundamentals of Structural Dynamics” 2 t t t Δu·· ( τ – t ) u ( τ ) = u + u· ( τ – t ) + § u·· + -------· ----------------© 2¹ 2

An-Najah 2013

(7.9)

The increments of acceleration, velocity and displacement during the time step are: Δu·· =

t + Δt··

Δu· =

t + Δt ·

t u – u·· = Δu··

(7.10)

t t Δu·· u – u· = § u·· + -------· Δt © 2¹

(7.11)

2 t t Δu·· Δt Δu = u· Δt + § u·· + -------· -------© 2¹ 2

(7.12)

Introducing the parameters γ and β into Equations (7.11) and (7.12) for Δu· and Δu , respectively, can be generalized as follows: t··

Δu· = ( u + γΔu··)Δt

(7.13)

t··

(7.14)

where different values of the parameters γ and β correspond to different assumptions regarding the variation of the acceleration within the time step: Average Acceleration: 1 1 β = --- , γ = --4 2


An-Najah 2013

Linear Acceleration:

Δt---≤ 0.551 T

1 1 β = --- , γ = --6 2

It is important to note that the “average acceleration”-method is unconditionally stable, while the “linear acceleration”-method is only stable if the condition Δt ⁄ T ≤ 0.551 is fulfilled. However, the “linear acceleration”-method is typically more accurate and should be preferred if there are no stability concerns. For a discussion on stability and accuracy of the Newmark’s methods see e.g. [Cho11] and [Bat96]. • Solution of the differential equation: Option 1 Substituting Equations (7.13) and (7.14) into Equation (7.4) gives Equation (7.15), which can be solved for the only remaining variable Δu·· :

2

Δt Δu = u Δt + ( u + 2βΔu··) -------2 t·


Δt ----- ≤ ∞ T

Page 7-5

2

t t t Δt 2 ( m + cγΔt + kβΔt )Δu·· = – mΔu··g – c u··Δt – k § u· Δt + u··--------· © 2 ¹

(7.15)

or in compact form: ˜ m ˜ Δu·· = Δp

(7.16)

Substituting Δu·· into Equations (7.11) and (7.12) gives the increments of the velocity Δu· and of the displacement Δu . In conjunction with Equation (7.5), these increments yield the dynamic response of the SDoF system at the end of the time step t + Δt .


Page 7-6

An-Najah 2013

• Solution of the differential equation: Option 2

Equation (7.15) is implemented in the Excel-Table as follows: Δu··

meq

da

° ° ° ® ° ° ° ¯ ® ¯

2

γ t γΔu γ u Δu· = --------- – ------- + Δt § 1 – ------· u·· © βΔt β 2β¹

(7.18)

Substituting Equations (7.17) and (7.18) into Equation (7.4) gives Equation (7.19), which can be solved for the only remaining variable Δu : t m m γc t γc m γ § k + ----------+ ---------· Δu = – mΔu··g + § --------- + -----· u· + § ------ – Δt § 1 – ------· c· u·· 2 βΔt¹ © βΔt β ¹ © © 2β © ¹ ¹ 2β βΔt

(7.19) or in compact form: kΔu = Δp

(7.20)

–c

u Δt

2

§ tu· Δt + tu··Δt --------· 2 ¹ –k ©

dv

dd

• In the columns C to E the so-called “predictors” dd, dv and da are computed first: t··Δt

t·

2

dd = u Δt + u -------2

(“delta-displacement”)

t dv = u··Δt

(“delta-velocity”)

– mΔu··g – c ⋅ dv – k ⋅ dd da = ------------------------------------------------------ = Δu·· meq

(“delta-acceleration”)

• Afterwards, in the columns F to H the ground motion quantities at the time step t + Δt are computed by means of so-called “correctors: t + Δt··

t u = u·· + da

t + Δt ·

t u = u· + dv + ( da ⋅ γ ⋅ Δt )

For linear systems we have:

Δu

• m ˜ in Equation (7.15), as well as k in Equation (7.20), are also constant and have to be computed only once. Page 7-7

t + Δt

t

2

u = u + dd + ( da ⋅ β ⋅ Δt )

° ° ° ® ° ° ° ¯

• m , c and k are constant throughout the whole time-history.


ΔF ( t )

t··

° ° ® ° ° ¯

Substituting Δu into Equations (7.18) and (7.17) gives the increments of the velocity Δu· and of the acceleration Δx··. In conjunction with Equation (7.5), these increments yield the dynamic response of the SDoF system at the end of the time step t + Δt .

=

– mΔu··g

° ° ® ° ° ¯

( m + cγΔt + kβΔt )

® ¯

(7.17)

and substituting Equation (7.17) into (7.13) we obtain: t·

An-Najah 2013

7.2.2 Implementation of Newmark’s integration scheme within the Excel-Table “SDOF_TH.xls”

Equation (7.14) can be transformed to: t· t·· u u Δu Δu·· = -----------2 – --------- – -----βΔt βΔt 2β


® ¯


Δu 7 Seismic Excitation

Page 7-8


An-Najah 2013

• Finally, in column I the absolute acceleration u··abs at the time step t + Δt is computed as follows: t + Δt·· uabs

=

t + Δt··

u+

t + Δt·· ug

An-Najah 2013

7.2.3 Alternative formulation of Newmark’s Method. The formulation of the Newmark’s Method presented in Section 7.2.1 corresponds to an incremental formulation. It is possible to rearrange the methodology to obtain a total formulation. The equation of motion at the time t + Δt can be written as:

Observations about the use of the Excel-Table • Only the yellow cells should be modified:

m

• The columns A and B contain the time vector and the ground acceleration u··g ( t ) at intervals Δt ; for this ground motion the response of a Single-Degree-of-Freedom (SDoF) system will be computed. To compute the response of the SDoF system for a different ground motion u··g ( t ) , the time and acceleration vector of the new ground motion have to be pasted into columns A and B. • For a given ground motion u·· ( t ) , the response of a linear SDoF g

system is only dependent on its period T = 2π ⁄ ω n and its damping ζ . For this reason, the period T and the damping ζ can be chosen freely in the Excel-Table.

• The mass m is only used to define the actual stiffness of the SDoF 2 system k = m ⋅ ω n and to compute from it the correct spring force f s = k ⋅ u . However, f s is not needed in any of the presented plots, hence the defaults value m = 1 can be kept for all computations. • In the field “Number of periods” (cell V19) one can enter the number of periods T i for which the response of the SDoF is to be computed in order to draw the corresponding response spectra.

• The response spectra are computed if the button “compute response spectra” is pressed. The macro pastes the different periods T i into cell S3, computes the response of the SDoF system, reads the maximum response quantities from the cells F6, G6, H6 and I6 and writes these value into the columns L to P. 7 Seismic Excitation


Page 7-9

t + Δt··

u+c

t + Δt ·

u+k

t + Δt

u = –m

t + Δt·· ug

(7.21)

where t + Δt··

(7.22)

t + Δt ·

(7.23)

t u = u·· + Δu·· t u = u· + Δu·

Using the expressions for Δu·· and Δu· given by Equations (7.17) and (7.18), the acceleration and the velocity at the time t + Δt can be written as: t + Δt··

t t 1 t + Δt 1 t 1 u = -----------2 ( u – u ) – --------- u· – § ------ – 1· u·· © ¹ βΔt 2β βΔt

(7.24)

t γ t + Δt γ t γ t u = --------- ( u – u ) + § 1 + ---· u· + Δt § 1 – ------· u·· © © βΔt β¹ 2β¹

(7.25)

t + Δt ·

Introducing Equations (7.24) and (7.25) into Equation (7.21) and t + Δt solving for the only unknown u we obtain: k⋅

t + Δt

u =

t + Δt

p

(7.26)

where:


Page 7-10


k = k + a1 t + Δt

p = –m

t + Δt·· ug

t t t + a 1 u + a 2 u· + a 3 u··

An-Najah 2013


An-Najah 2013

(7.27)

7.3 Time-history analysis of nonlinear SDoF systems

(7.28)

f el um f y = ------ , μ Δ = -----Ry uy

γc m a 1 = -----------2 + --------βΔt βΔt

(7.29)

m γ a 2 = --------- + § --- – 1· c © βΔt β ¹

(7.30)

γ 1 a 3 = § ------ – 1· m + Δt § ------ – 1· c © 2β ¹ © 2β ¹

(7.31)

This formulation corresponds to the implementation of Newmark’s method presented in [Cho11]. • Strength f y of the nonlinear SDoF system f el k el ⋅ u el f y = ------ = -----------------Ry Ry

(7.32)

• R y = force reduction factor • f el = maximum spring force f s that a linear SDoF system of the same period T and damping ζ would experience if submitted to the same ground motion u··g

• Maximum deformation u m of the nonlinear SDoF system um = μΔ ⋅ uy ,

hence

μΔ = um ⁄ uy

(7.33)

• u y = yield displacement • μ Δ = displacement ductility 7 Seismic Excitation

Page 7-11


Page 7-12


An-Najah 2013


An-Najah 2013

7.3.1 Equation of motion of nonlinear SDoF systems

7.3.2 Hysteretic rules

In the equation of motion for a base point excitation through an acceleration time-history u··g ( t )

The next figure shows typical hysteretic rules (or models) for nonlinear SDoF systems.

mu·· + cu· + f s (u,t) = – mu··g

(7.34)

of a nonlinear SDoF system, the spring force f s (u,t) is no longer constant and varies in function of time and location. Most structural components are characterised by a continuously curved force-deformation relationship like the one shown by means of a thin line on the right of the figure on page 7-12, which however is often approximated by a bilinear curve (thick line in the same figure). When the loading of the nonlinear SDoF system is cyclic, then f s is no longer an unambiguous function of the location u and also for this reason Equation (7.34) shall be solved incrementally. For this reason f s (u,t) must be described in such a way, that starting from a known spring force f s (u,t) at the time t , the still unknown spring force f s (x + Δu,t + Δt) at the time t + Δt can easily be computed. This description of the cyclic force-deformation relationship is known as hysteretic rule. In the literature many different hysteretic rules for nonlinear SDoF systems are available (See e.g. [Saa91]). In the following section a few hysteretic rules are presented and discussed.


Page 7-13


Page 7-14


An-Najah 2013

In the following the often used Takeda hysteretic model is discussed in some detail. The Takeda model was first described in [TSN70] and later modified by various authors. The assumed force-deformation relationship shown in the following figure was derived from the moment-curvature relationship described in [AP88]. Large amplitude cycles + β⋅ξp

k pl =r o ·kel

k u+ = k el ⋅ ( max { μ Δ+ } )

β0 u rev

fs

+ ku

u rev

B+

A+ C+

k pl =r o ·kel

+ ku u rec

k el uy

+ β⋅ξp

A+

B+

A+

fy

+ β⋅ξp

u

u rev + u rec

ku

u

ku

X-

B- X-

xβ⋅ξp

xβ⋅ξp

The initial loading follows the bilinear force-deformation relationship for monotonic loading mentioned in the previous section. The exact definition of this so-called skeleton curve depends on the structural element at hand. For example, in the case of Reinforced Concrete (RC) structural walls the elastic stiffness k el corresponds to 20 to 30% of the uncracked stiffness, while the plastic stiffness k pl = r o ⋅ k el is approximated assuming an hardening factor r o = 0.01…0.05 .


–α

,

k u- = k el ⋅ ( max { μ Δ- } )

–α

(7.35)

Reloading follows a straight line which is defined by the force reversal point (u rev,0) and the point A. The location of point A is determined according to the figure on the previous page as a function of the last reversal point B, the plastic strain ξ p and the damage influence parameter β . The parameter β allows taking into account softening effects occurring during the reloading phase. Again, in the case of RC walls meaningful parameters α and β lay in the following ranges: α = 0.2…0.6 and β = 0.0… – 0.3 .

Cβ>0

β0

100

Spv = 71 cm/s

= pa

. 11 2 cm 10 ] m [c tS en

1

2

/s

em

m

[c

10

pa

d

.S

(7.78)

pl

ac

]

1

0.

• For ζ = 0 :

u(t) = A(t)

• For ζ > 0 :

At u max : u a = A however A < A max Shift of the location of the maxima through damping

Page 7-47

1

1

D

is

Time-history of the pseudo-acceleration A(t)


10

=

cc

u··a ( t ) = – ω 2 u ( t ) – 2ζωu· ( t )

0

cm /s 2 44 7

Sd

0 −a

do

15

10

10 Time [s]

eu

5

10

Ps

0

00

−1.0

10

-1 x Displacement

S

0.0

Pseudo−velocity Spv [cm/s]

0

00

Pseudo-acceleration Pseudo-velocity

15

10

... / ... max [−]

1.0

10

T=1s 0.1


1.0 Period [s]

10.0

Page 7-48


An-Najah 2013

7.5.3 Properties of linear response spectra


An-Najah 2013

7.5.4 Newmark’s elastic design spectra ([Cho11])

100

= = = =

0% 2% 5% 10%

vg = 36.1 cm/s

T [s]

00

dg

10

0

10

= .1 21

10

cm

[-] em

0.

1

1

D

is

10.0

• Response spectra typically show spectral regions where the response is sensitive to different motion quantities, i.e. they show an acceleration sensitive region (small periods), a displacement sensitive region (large periods) and a velocity sensitive region laying in between. 7 Seismic Excitation

d

Spv / vg [-]

pl

ac

[-]

]

1.0 Period [s]

S

g

2

/s

0.1

/d

[c d

tS en

1

3 31 =

/a

g

a

m

[c

10

pa

a

Sp

.S

1

g

m

cc

]

−a

/s 2

do

eu

cm

0

10

10

Ps


0

00

10

Spa / ag [-]

ζ ζ ζ ζ

Page 7-49

T [s] 7 Seismic Excitation

Page 7-50


An-Najah 2013

• Newmark’s elastic design spectra


An-Najah 2013

• Elastic design spectra according to SIA 261 (Art. 16.2.3) 6

ζ = 5%

Zone 3b, ζ = 5%

100 aa g

C

α

0

a

g

10

E

dg

00

B

10 0

Ps

10

10

0

vg 10

F

2

1

[c

0 0.01

tS en

1

2

/s

em

m

[c

10

pa

d

.S

Ground Class A Ground Class B Ground Class C Ground Class D Ground Class E 0.10

1.00

10.00

Period [s]

is

pl

ac

]

1

3

m

cc

A

4

]

−a

do

eu


D dg αd

00

10

αvvg

Pseudo−acceleration Spa [m/s2]

5

0.

TA=1/33s

TB=1/8s

0.1

Damping ζ 2% 5% 10% 20% 7 Seismic Excitation

1

1

D

• Ground class A: firm or soft rock with a maximum soil cover of 5m • Ground class B: deposit of extensive cemented gravel and sand with a thickness >30m.

TE=10s 1.0 Period [s]

Median(50%) αa αv αd 2.74 2.03 1.63 2.12 1.65 1.39 1.64 1.37 1.20 1.17 1.08 1.01

10.0

TF=33s

One sigma (84%) αa αv αd 3.66 2.92 2.42 2.71 2.30 2.01 1.99 1.84 1.69 1.26 1.37 1.38 Page 7-51

• Ground class C: deposits of normally consolidated and uncemented gravel and sand with a thickness >30m. • Ground class D: deposits of unconsolidated fine sand, silt and clay with a thickness >30m. • Ground class E: alluvial surface layer of GC C or D, with a thickness of 5 to 30m above a layer of GC A or B. • Ground class F: deposits of structurally-sensitive and organic deposits with a thickness >10m. 7 Seismic Excitation

Page 7-52


An-Najah 2013

• Displacement elastic design spectra according to SIA 261


An-Najah 2013

• Elastic design spectra according to SIA 261 (linear) 6

25

Displacement Sd [cm]

20

15

Pseudo−acceleration Spa [m/s2]

Zone 3b, ζ = 5% Ground Class A Ground Class B Ground Class C Ground Class D Ground Class E

10

4

3

2

1

Zone 3b, ζ = 5%

5

0 0.01

Ground Class A Ground Class B Ground Class C Ground Class D Ground Class E

5

0 0.00

0.10

1.00

0.50

(7.79)

• Displacement spectra are an important design tool (even within force-based design procedures) because they allow a quick estimate of the expected deformations, hence of the expected damage.

15

Page 7-53

2.50

3.00

10

5

Zone 3b, ζ = 5% 0 0.00


2.00

Ground Class A Ground Class B Ground Class C Ground Class D Ground Class E

20 Displacement Sd [cm]

S pa S d = -------2ω

1.50 Period [s]

25

10.00

Period [s]

• The displacement spectra are computed from the acceleration spectra using equation (7.79)

1.00

0.50


1.00

1.50 Period [s]

2.00

2.50

3.00

Page 7-54


An-Najah 2013

• Elastic design spectra: Newmark vs. SIA 261


7.5.5 Elastic design spectra in ADRS-format (e.g. [Faj99]) (Acceleration-Displacement-Response Spectra)

3.5 Elastic design spectrum according to SIA 261, Ground Type A

S pa = ω 2 S d

3.0

C

B

An-Najah 2013

ζ = 5%

15

0.5

2.5

1.5

0.3 0.2

10

5

0.1 0.0 0

A

1.0

0.5

Sd [cm]

2.0

Spa [g]

Spa / Ag [−]

0.4

Elastic design spectrum according to SIA 261, Ground Type B

1

2 3 Period [s]

0.10

1

2 3 Period [s]

4

5

0.5

1.00

0.4

10.00 Spa [g]

Period [s]

• The SIA 261 spectra, like the spectra of the majority of the standards worldwide, were defined using the same principles as Newmark’s spectra. - SIA 261 takes into account different ground classes; - Different seismic sources were considered; - A larger number of ground motions was considered. • Note: in SIA 261 the corner period T A is not defined.

0.3 0.2

0.0 0

5

10 Sd [cm]

15

Periods T correspond to lines running through the origin of the axes, because: S pa = ω 2 S d

Page 7-55

T = const.

0.1

• However, different ground motion were used:


0 0

5

D

Newmark’s elastic design spectrum

0.0 0.01

4


and after reorganizing: T = 2π S d ⁄ S pa Page 7-56


An-Najah 2013

7.6.1 Illustrative example

Elastic design spectrum according to SIA 261, Ground Type B

Comparison of the time history analyses of an elastic and an inelastic single-degree-of-freedom system (SDOF system):

Elastic design spectrum of an earthquake with equal ag,max

0.4

An-Najah 2013

7.6 Strength and Ductility

• Elastic design spectra in ADRS-format 0.5


T=0.77s: Overestimation Telastic = Tinelastic

T=0.87s: Underestimation

Spa [g]

0.3

ζ = 5%

0.2

0.1

Where: 0.0 0

5

10

15

Sd [cm]

f el R y = ----- : Force reduction factor fy

(7.80)

• Design spectra are defined based on averaged response spectra. For this reason, the spectral values of single response spectra may differ significantly from the design spectra.

f el :

Maximum restoring force that the elastic SDOF system (7.81) reaches over the course of the seismic excitation u··g ( t )

fy :

Yield force of the inelastic SDOF system

• This is a crucial property of design spectra and should be kept in mind during design!

um μ Δ = ------ : Displacement ductility uy


Page 7-57

(7.82) (7.83)

um :

Maximum displacement that the inelastic SDOF system (7.84) reaches over the course of the seismic excitation u··g ( t )

uy :

Yield displacement of the inelastic SDOF system


(7.85)

Page 7-58


An-Najah 2013

• Results

Quantity

0.15

Displacement [m]

T [s]

Elastic SDOF system Inelastic SDOF system (Ry=2) Inelastic SDOF system (Ry=6)

0.10

0.00

Inela. SDOF Ry=2 Inela. SDOF Ry=6 2.0

2.0

134.70

67.35

22.45

Ry [-]

–

2.0

6.0

uy [m]

–

0.068

0.023

um [m]

0.136

0.147

0.126

–

2.16

5.54

μΔ [-]

-0.05

Elastic SDOF

An-Najah 2013

2.0

Fmax [kN]

0.05

• Comments • Both inelastic SDOF systems show a stable seismic response.

-0.10

-0.15 0


5

10

15

20 Time [s]

25

30

35

40

150

100

Force [kN]

50

0

-50

Elastic SDOF system Inelastic SDOF system (Ry=2) Inelastic SDOF system (Ry=6)

-100

-150 -0.15


-0.10

-0.05

0.00 Displacement [m]

0.05

0.10

0.15

Page 7-59


Page 7-60


An-Najah 2013

7.6.2 “Seismic behaviour equation”


An-Najah 2013

• More realistic representation of the decision possibilities

For seismic collapse prevention, the following approximate relationship applies ″quality″ of seismic behaviour ≈ strength × ductility

(7.86)

F

Δ

To survive an earthquake different combination of strength and ductility are possible:

F

Δ

• If the strength of the structure reduces, the stiffness typically reduces too. • If the masses do not change significantly (which is typically the case), the fundamental period T of the softer structure is longer. • Structures with a longer fundamental period T are typically subjected to larger deformations, i.e., the deformation demand is larger.


Page 7-61


Page 7-62


An-Najah 2013

7.6.3 Inelastic behaviour of a RC wall during an earthquake

• Moment-curvature-relationship at the base of the plastic hinge zone.

• The plastic deformation capacity of structures can really be taken into account for seismic design purposes. 7 Seismic Excitation

Moment [kNm]

• Despite reaching and exceeding its elastic limit the wall did not collapse.

100


An-Najah 2013

7.6.4 Static-cyclic behaviour of a RC wall

Wall WDH4

50 0 −50 −100 −30

−20

−10 0 10 Curvature [1/km]

20

30 Wall WSH6 [DWB99]

Page 7-63


Page 7-64


An-Najah 2013

• Hysteretic behaviour of the RC wall under static-cyclic loading


7.6.5 General definition of ductility

500 400

Fy

Wall WSH3

Actuator force [kN]

Ductility demand

…m μ … = --------…y

Ductility capacity:

…u μ … = -------…y

0.75 Fy

300

An-Najah 2013

200 100 0

μΔ=6 μΔ=5 μΔ=4 μΔ=3 μΔ=2 μΔ=1

μΔ=1 μΔ=2 μΔ=3 μΔ=4 μΔ=5 μΔ=6

−100 −200 −300

−0.75 Fy

−400

−Fy

−500 −100

−80

−60

−40

−20 0 20 Top displacement [mm]

40

60

80

100

• Comments • The ductility capacity is a property of the structural member. • The ductility demand is a result of the seismic excitation and also a function of the dynamic properties of the structure. • A structural member survives the earthquake if:

Ductility capacity ≥ Ductility demand • The structural member fractures when locally the deformation capacity of the structural materials (i.e., their strain capacities) are reached. The ductility capacity is therefore exhausted.

Plastic region of test unit WSH6 (left) and close-up of the left boundary region (right). Both photos were taken at displacement ductility 6.


Page 7-65


Page 7-66


An-Najah 2013

f el um R y = ----- , μ Δ = -----fy uy

εu μ ε = ----εy

μΔ ( Ry ) = ?

φu μ φ = ----φy

Sa [m/s2]

10

curvature ductility

ζ = 5%

Sv [m/s]

1.0

θu μ θ = ----θy

0.5

0.0 0.01 0.4

Δu μ Δ = -----Δy

Sd [m]

0.3

displacement ductility

0.2 0.1 0.01


Page 7-67

μΔ=1 μΔ=2 μΔ=4 μΔ=6

5

0 0.01

rotation ductility

An-Najah 2013

7.7 Inelastic response spectra

7.6.6 Types of ductilities

strain ductility



0.10

1.00

10.00

μΔ=1 μΔ=2 μΔ=4 μΔ=6 0.10

ζ = 5%

1.00

10.00

μΔ=1 μΔ=2 μΔ=4 μΔ=6 0.10

100.0

100.0

ζ = 5%

1.00 Period [s]

10.00

100.0

Page 7-68


An-Najah 2013

TD

TE

Ry [−]

μΔ=6 μΔ=4

μΔ=2 Ry=μΔ

Ry=(2μΔ-1)0.5

1

Ry = μΔ


1.0 Period [s]

TA

TF

Page 7-69

TD

TE

TF

Ry=6 Ry=4

Ry=2

1 0.1

2μ Δ – 1

TC

10

10.0

Ry =

TB

ζ = 5%

Displacement ductility [−]

TC

10

0.1

An-Najah 2013

Displacement ductility μ Δ

Force reduction factor Ry TA TB ζ = 5%


1.0 Period [s]

10.0

• In the small period range, already small reductions of the elastic strength of the SDOF system yield very large ductility demands. • If the ductility demand is very large, it can be difficult to provide the structure with a sufficiently large ductility capacity. This problem will be further discussed during the design classes. • Also in the large period range – where the “equal displacement principle” applies – large discrepancies between real and estimated ductility demand can occur. • The “equal displacement principle” and the “equal energy principle” are “historical” Ry-μΔ-Tn relationships. In recent years a lot of research has been done to come up with more accurate formulations (see e.g. works by Krawinkler [KN92], Fajfar [VFF94], Miranda [Mir01], ...) 7 Seismic Excitation

Page 7-70


An-Najah 2013

7.7.1 Inelastic design spectra


An-Najah 2013

• Newmark’s inelastic design spectra [NH82] g

• Inelastic design spectra in combined D-V-A format

ζ = 5%

g

100

V=αvvg

D

vg

B

0 10

.5

D

F

1 m

g

A= a

[c

1

Ay

10 0.

1

1

0.

D

/μ Δ

y

D

[c

]

m

2

]

/s

1

m

10

[c

y

]

1

F’

D

2

/s

A’ 1

]

10

0

m

10

[c

A

E’

g =d

0

Inelastic design spectrum for μΔ=4

B’

D

10

Ay

A/

(2

μ

Δ-

/μ Δ

00

10

D’ 10

10

V/μΔ

1) 0

0 10

C’

10

00

Vy [cm/s]

E

dg

10

Vy [cm/s]

a

g

0

00

10

A= α

aa g

C

dg d =α

100

Elastic design spectrum D


ζ = 5%

0.1 0.1

1.0 Period [s]

10.0

TB=1/8s

1

TA=1/33s

1.0 Period [s]

TE=10s 10.0

TF=33s

• Maximum displacement of the SDOF system

um = μΔ ⋅ Dy

Note the new axes: D y = u y , V y = ω n u y , A y = ω n2 u y

• Yield strength of the SDOF system:

where: u y = yield displacement 7 Seismic Excitation

fy = m ⋅ Ay Page 7-71


Page 7-72


An-Najah 2013

Construction of the spectra using Ry-μΔ-Tn relationships


• Inelastic design spectra according to [NH82] (log. x-axis)

The inelastic design spectra are computed by means of Ry-μΔTn relationships:

μΔ D = S d, inelastic = ------ ⋅ S d, elastic Ry

TA

TB

TD

TE

(7.87) (7.88)

TF μΔ=1 μΔ=2 μΔ=4 μΔ=6

1

It should be noted that: S pa, inelastic ≠ ω 2 ⋅ S d, inelastic

TC

2

Ay / ag [−]

A y = S pa, inelastic

1 = ------ ⋅ S pa, elastic Ry

An-Najah 2013

Equal displacement principle (7.89)

• Ry-μΔ-Tn relationship according to [NH82]

Where:

0 0.01

2

(7.90)

β = log ( T n ⁄ T a ) ⁄ log ( T b ⁄ T a )

(7.91)

T a = 1 ⁄ 33s , T b = 1 ⁄ 8s

(7.92)

T c = Corner period between the constant Spa and the constant Spv regions T c' = Corner period between the constant Spa and the constant Spv regions of the inelastic spectrum 7 Seismic Excitation

0.10

TA

Sd / dg [−]

1 Tn < Ta ° ° ° ( 2μ – 1 ) β ⁄ 2 T < T < T Δ a n b ° ° T b < T n < T c' (EE principle) 2μ Δ – 1 Ry = ® ° T ° -----n- μ Δ T c' < T n < T c ° Tc ° Tn > Tc (ED principle) μΔ ° ¯

Page 7-73

1.00

TB

TC

10.00

TD

TE

100.00

TF


1

Equal displacement principle 0 0.01


0.10

1.00 Period [s]

10.00

100.00

Page 7-74


An-Najah 2013

• Inelastic design spectra according to [NH82] (linear x-axis) 5.0

TB

TC

ζ = 5%

TD


• Ry-μΔ-Tn relationships according to [VFF94] In [VFF94] Ry-μΔ-Tn relationships are defined as follows: cR T n ° -+1 ° c 1 ( μ Δ – 1 ) ⋅ ----T0 Ry = ® ° c ° c1 ( μΔ – 1 ) R + 1 ¯

Ay [m/s2]

4.0


3.0

Where:

2.0

Tn ≤ T0

(7.93)

Tn > T0

T 0 = c 2 ⋅ μ ΔcT ⋅ T c ≤ T c

(7.94)

T c = Corner period between the constant Spa and the constant Spv regions

1.0

0.0 0.0 0.15

An-Najah 2013

0.5

TB

1.0

1.5

TC

2.0

2.5

3.0

ζ = 5%

TD

D [m]

0.10


0.05

The parameters c 1 , c 2 , c R and c T are defined as follows for 5% damping: Model Hysteresis

Damping

c1

cR

c2

cT

Q

Mass

1.0

1.0

0.65

0.30

Q

Tangent stiffness

0.75

1.0

0.65

0.30

Bilinear

Mass

1.35

0.95

0.75

0.20

Bilinear

Tangent stiffness

1.10

0.95

075

0.20

and where the Q-hysteretic rule is a stiffness degrading rule similar to the Takeda-hysteretic rule presented in Section 7.3.2. The table shows the dependency of the Ry-μΔ-Tn relationships both on damping and hysteretic model.

0.00 0.0


0.5

1.0

1.5 Period [s]

2.0

2.5

3.0

Page 7-75


Page 7-76

An-Najah 2013

For the Q-hysteretic model and mass-proportional damping, the Ry-μΔ-Tn relationships by [VFF94] specialise as: T ° ( μ Δ – 1 ) ⋅ -----n- + 1 Ry = ® T0 ° μΔ ¯ Where:

Tn ≤ T0 Tn > T0

T 0 = 0.65 ⋅ μ Δ0.3 ⋅ T c ≤ T c


An-Najah 2013

• Inelastic design spectra according to [VFF94] 5.0

(7.95)

TB

TC

TD

ζ = 5%

4.0


(ED principle) (7.96)

T c = Corner period between the constant Spa and the constant Spv regions

Ay [m/s2]


3.0

2.0

The spectra depicted on the following pages correspond to this case.

1.0

0.0 0.0 0.15

0.5

TB

1.0

1.5

TC

2.0

2.5

3.0

ζ = 5%

TD

D [m]

0.10


0.05

0.00 0.0


Page 7-77


0.5

1.0

1.5 Period [s]

2.0

2.5

3.0

Page 7-78


An-Najah 2013

• Inelastic design spectra in ADRS-format 5.0

TB

TC

ζ = 5%

Spa [m/s2]

4.0

[NH82] μΔ=1 μΔ=2 μΔ=4 μΔ=6

3.0

2.0

TD 1.0


7.7.2 Determining the response of an inelastic SDOF system by means of inelastic design spectra in ADRSformat In this section the response of two example inelastic SDOF systems is determined by means of inelastic design spectra in ADRS-format. • SDOF system 1 with Tn = 0.9 s • SDOF system 2 with Tn = 0.3 s • The spectra according to [VFF94] will be used (see Section 7.7.1)

• Example 1: SDOF system with Tn= 0.9 s m

0.02

TB

0.04

0.06 Sd [m]

0.08

Spa [m/s2]

0.10

0.12

C

TC

ζ = 5%

4.0

[VFF94] μΔ=1 μΔ=2 μΔ=4 μΔ=6

3.0

2.0

TD 1.0

0.0 0.00

fs properties:

0.0 0.00 5.0

An-Najah 2013

0.02


0.04

0.06 Sd [m]

0.08

0.10

0.12

k

m = 100 t fy = 80 kN

kel = 4874 kN/m

uy = 0.016 m kpl = 0 kN/m

fy

ζ = 5%

kpl kel

uy

um

u

• Response of the elastic SDOF system 1: 100- = 0.9s T n = 2π m ---- = 2π ----------k 4874 S pa = 2.62m/s 2 S d = 0.054m f el = 261.7kN

Page 7-79


Page 7-80


An-Najah 2013


An-Najah 2013

• Example 2: SDOF system with Tn= 0.3 s

• Response of the inelastic SDOF system 1: f el 261.7 R y = ----- = ------------- = 3.27 80 fy

m

fs properties:

μ Δ = R y = 3.27 (From Equation (7.95) since T n > T c = 0.5s )

C

u m = u y ⋅ μ Δ = 0.016 ⋅ 3.27 = 0.054m = S d

Representation of the inelastic SDOF system 1 in the inelastic design spectrum in ADRS-format:

k

kel = 43‘865 kN/m kpl = 0 kN/m

5.0

fy,a = 120 kN uy,a = 0.0027 m

m = 100 t

fy,b = 300 kN uy,b = 0.0068 m fy

ζ = 5%

μΔ = 1

kpl kel

uy

um

u

4.0

Spa [m/s2]

• Response of the elastic SDOF system 2 3.0

μΔ = 3.27

Tn = 0.9s Performance point

100 - = 0.3s T n = 2π m ---- = 2π -------------43865 k S pa = 4.71m/s 2 S d = 0.011m

2.0

f el = 471kN 1.0

um = 0.054m 0.0 0.00

0.02

0.04

In this second example two different inelastic SDOF systems will be considered: (a) A SDOF system with a rather low fy and (b) a SDOF system with a rather high fy.

Capacity curve

0.06 Sd [m]

0.08

0.10

0.12

• If the force-deformation relationship of the inelastic SDOF system is divided by its mass m, the “capacity curve” is obtained, which can be plotted on top of the spectrum in ADRS-format. • The capacity curve and the inelastic spectrum intersect in the “performance point”. 7 Seismic Excitation

Page 7-81

• Response of the second inelastic SDOF system 2a f el 471 R y = ----- = --------- = 3.93 120 fy In this case the resulting displacement ductility μ Δ is so large, that Equation (7.96) T 0 = T c = 0.5s results. After rearranging Equation (7.95), the displacement ductility μ Δ can be computed as: 7 Seismic Excitation

Page 7-82


An-Najah 2013

• Response of the inelastic SDOF system 2b

Check that T 0 > T c : T 0 = 0.65 ⋅

⋅ T c = 0.65 ⋅ 5.88

0.3

f el 471 R y = ----- = --------- = 1.57 300 fy

⋅ 0.5 = 0.553s > T c

In this case the displacement ductility μ Δ will be such that Equation (7.96) yields T 0 < T c = 0.5s . To compute μ Δ insert therefore Equation (7.96) in Equation (7.95). This results in following expression:

The maximum displacement response is therefore: u m = u y ⋅ μ Δ = 0.0027 ⋅ 5.88 = 0.016m > S d

Representation of the inelastic SDOF system 2a in the inelastic design spectrum in ADRS-format: 5.0

(7.97)

μ Δ ( T n = 0.3 ,T c = 0.5 ,R y = 1.57 ) = 1.73 Check that T 0 < T c : 0.3

(*) Spa [m/s2]

Tn - + 1 = Ry ( μ Δ – 1 ) ⋅ --------------------------------0.3 0.65 ⋅ μ Δ ⋅ T c Equation (7.97) needs to be solved numerically.

Tn = 0.3s

4.0

T 0 = 0.65 ⋅ μ Δ ⋅ T c = 0.65 ⋅ 1.73

Performance point

3.0

0.3

⋅ 0.5 = 0.383s < T c

Tn 0.3 ( μ Δ – 1 ) ⋅ ------ + 1 = ( 1.73 – 1 ) ⋅ ------------- + 1 = 1.57 = R y T0 0.383 The maximum displacement response is therefore:

2.0

1.0

An-Najah 2013

Now consider the SDOF system 2b:

Tc 0.5 μ Δ = ( R y – 1 ) ⋅ ------ + 1 = ( 3.93 – 1 ) ⋅ ------- + 1 = 5.88 0.3 Tn 0.3 μΔ


μΔ = 1

Capacity curve um = 0.016m

u m = u y ⋅ μ Δ = 0.0068 ⋅ 1.73 = 0.012m > S d

μΔ = 5.88 0.0 0.00

0.02

0.04

0.06 Sd [m]

0.08

0.10

0.12

• Note that the line (*) is no longer vertical as in Example 1, but inclined according to the equation μ Δ = ( R y – 1 ) ⋅ ( T c ⁄ T n ) + 1 .


Page 7-83


Page 7-84


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Representation of the inelastic SDOF system 2b in the inelastic design spectrum in ADRS-format: 5.0

Tn = 0.3s

“Performance Point”

4.0


T ° ( μ Δ – 1 ) ⋅ -----n- + 1 Ry = ® Tc ° μΔ ¯

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Tn ≤ Tc Tn > Tc

(7.98) (ED principle)

This approximation is particularly satisfactory, if the large uncertainties associated with smoothed spectra are considered.

Capacity curve

• Comments

Spa [m/s2]

• A discussion of similar examples can be found in [Faj99]. 3.0

um = 0.012m 2.0

1.0

μΔ = 1

(*)

μΔ = 1.73 0.0 0.00

0.02

0.04

0.06 Sd [m]

0.08

0.10

0.12

• Note that the curve (*) is no longer a straight line as in Examples 1 and 2a. • In Example 2b the curve (*) needs to be computed numerically. • In Example 2a the curve (*) is only an approximation of the curve (*) in Example 2b. As soon as T 0 = T c both curves are identical. In Example 2 this is the case if S pa < 1.6m/s 2 .

• For computing the response of inelastic SDOF systems by means of inelastic design spectra, the Ry-μΔ-Tn relationships in Section 7.7.1 are sufficient. The spectra in ADRS-format are not absolutely necessary, but they illustrate the maximum response of inelastic SDOF systems very well. • Ry-μΔ-Tn relationships should only be used in conjunction with smoothed spectra. They should not be used to derive the inelastic response spectra of a single ground motion • Remember: - Design spectra are very useful tools to design structures for the expected seismic demand. Design spectra represent the average effect of an earthquake with design intensity. - If a single earthquake is considered, the spectra may underestimate the seismic demand for a certain period range (... overestimate ...). - This characteristic of design spectra should be considered when designing structures: The seismic design should aim at structures that are as robust as possible.

• When T 0 < T c (i.e. when S pa > 1.6m/s 2 ) the curve (*)2a predicts larger maximum displacements u m than curve (*)2b. The difference is, however, small. For this reason, in most cases Equation (7.95) can be approximated as:


Page 7-85


Page 7-86


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7.7.3 Inelastic design spectra: An important note


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7.7.4 Behaviour factor q according to SIA 261

The “equal displacement” and the “equal energy” principles represent a strong simplification of the real q = Φo Ry

inelastic behaviour of SDOF systems.

• Design spectra are a powerful tool to design structures to resist the expected seismic action. On average, design spectra are a good representation of the expected peak behaviour of structures.

• However, if single ground motions are considered, then it can easily be the case that design spectra significantly underestimate the expected peak behaviour of structures.

Φ0 = Overstrength

• Design spectra according to SIA 261 4

TB

Ground Class B

TC

TD

Sd / agd [-]

3

• This characteristic of the design spectra shall be taken into account during design by aiming at robust structures.

q=1.5

Elastic design spectrum

2

q=2 1

Φ0=1.5

q=3 q=4

0 0.01

0.10

1.00

10.00

Period [s]


Page 7-87


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7.8 Linear equivalent SDOF system (SDOFe) u .. mu

fs

.. -mug(t)

• Example: Inelastic SDOF system with Takeda-hyst. rule [TNS70]

fs

• • • • •

Elastic (with elastic stiffness kel)

kel

fm fy

.. ug(t)

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The properties of the inelastic SDOF system are:

. cu

ug(t)


kpl kel

uy

Elastic (with „effective stiffness“ or „secant stiffness“ keff)

Inelastic

ζ = 5% (constant, proportional to kel) m = 100t k el = 4874kN/m f y = 80kN Takeda-hysteresis with r o = 0.05 , α = 0.5 , β = 0.0

Damping: Mass: Stiffness: Yield force: Hysteresis: +

+

–α

-

-

–α

k u = k el ( max { μ Δ } ) k u = k el ( max { μ Δ } )

keff

um

u

It is postulated that the maximum response u m of an inelastic SDOF system can be estimated by means of a linear equivalent SDOF system (SDOFe). The properties of the SDOFe are: Stiffness:

k eff = f m ⁄ u m

Damping:

ζe

(7.99) (7.100) The maximum response of the SDOF system when subjected to the NScomponent of the 1940 El Centro Earthquake is:

The differential equation of the SDOFe is: 2 u··( t ) + 2ζ e ω e u· ( t ) + ω e2 u ( t ) = – u··g ( t ) with ω e = k eff ⁄ m (7.101)

The question is how the viscous damping ζ e of the SDOFe can be determined so that max ( u ( t ) ) = u m .

x m = 0.073m , f m = 93.0kN The properties of the corresponding SDOFe are: fm 93.0 kN m- = 2π ----------100- = 1.76s k eff = ------ = ------------- = 1274 ------- , T e = 2π ------0.073 m um k eff 1274 ζ e = 22.89% , the viscous damping ζ e was determined iteratively!


Page 7-89


Page 7-90


An-Najah 2013

Comparison between inelastic SDOF and SDOFe 100

• The damping ζ e is in general larger than the damping ζ , since ζ e of the SDOFe needs to compensate for the hysteretic energy absorption of the inelastic SDOF system.

ζ = 5% ζe = 22.89%

• However, in rare cases it happens that ζ e < ζ . This shows again the difficulties that are associated with the prediction of the seismic response of inelastic SDOF systems.

Force [kN]

• In the example, the viscous damping ζ e was determined iteratively until a value for ζ e was found for which the response of the SDOFe system was equal to the maximum response of the inelastic SDOF system. Hence, if a method was available for estimating the viscous damping ζ e , then the maximum response of the inelastic SDOF system could indeed be estimated by means of the linear equivalent SDOF system.

0

-50

Inelastic SDOF Equivalent SDOF

fm um -0.06

-0.04

-0.02 0.00 0.02 Deformation [m]

0.08

0.04

0.06

0.08

ζ = 5% ζe = 22.89%

0.06

Deformation [m]

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Comments regarding the example:

50

-100 -0.08


• The stiffness k eff and the period T e of the SDOFe system are only known once the maximum response of the inelastic SDOF system are known. Section 7.8.2 shows how the equivalent viscous damping ζ e can be estimated without knowing the stiffness k eff and the period T e of the SDOFe system a priori.

0.04

• Estimating the damping ζ e

0.02

In particular in the sixties significant research has been dedicated to estimating the damping ζ e (see for example [Jac60], [Jen68] and [IG79]). At that time the interest in linear equivalent systems was big because the numerical computation of the response of inelastic systems was extremely expensive. The basic idea behind estimating the damping ζ e was:

0.00 -0.02 -0.04 -0.06 -0.08 0


Inelastic SDOF Equivalent SDOF

um 5

10

15

20 Time [s]

25

30

35

40

Page 7-91

The inelastic SDOF system dissipates energy due to ζ and due to the inelastic deformations, which are a function of its inelastic force-deformation relationship. The equivalent SDOF system, however, dissipates energy solely due to its viscous damping. For this reason the following relationship applies: 7 Seismic Excitation

Page 7-92


ζ e = ζ + ζ eq

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fm = [ 1 + ro ( μΔ – 1 ) ] ⋅ fy

(7.102)

fo = ( 1 – ro ) ⋅ fy

where ζ eq is the viscous damping equivalent to the hysteretic energy absorption of the inelastic system.

u1 f 1 = ----------------- ⋅ f 2 u1 + u3

The simplest method for estimating the equivalent viscous damping is to assume that the inelastic system and the linear equivalent system dissipate the same energy within one displacement cycle. According to this assumption [Cho11] defines the equivalent elastic damping as: 1 Ah ζ eq = ------ ⋅ -----4π A e

f2 = [ 1 + ro ( μΔ – 1 ) ( 1 – β ) ] ⋅ fy u1 = μΔ uy – u2 1 + ro ( μΔ – 1 ) - ⋅ uy u 2 = --------------------------------–α μΔ

(7.103)

Where:

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u3 = [ μΔ – β ( μΔ – 1 ) ] ⋅ uy

The equivalent viscous damping of other important hysteresis rules is:

A h : Energy dissipated by the inelastic SDOF system due to the inelastic deformation of the system. The dissipated energy corresponds to the area of the force-displacement hysteresis of the considered displacement cycle; A e : Potential energy of the equivalent SDOF system at maximum displacement:

Elasto-plastic (EP) rule: 2 μΔ – 1 ζ eq, EP = --- ⋅ --------------π μΔ

(7.105)

2

k eff ⋅ u m A e = -------------------2 The inelastic force-deformation relationship of many structural RC elements can be described by the “Takeda”-hysteresis rule. According to Equation (7.103) the equivalent viscous damping of this hysteresis rule is: 1 ( f m + f o )μ Δ u y + f 1 u 1 – f m u 2 – ( f m + f o )u 3 ζ eq,Tak = ------ ⋅ ----------------------------------------------------------------------------------------------------( fm μΔ uy ) ⁄ 2 4π

2 ( μΔ – 1 ) ( 1 – ro ) ζ eq, BL = --- ⋅ -----------------------------------------π μΔ ( 1 + ro μΔ – ro )

(7.106)

(7.104) Rule according to Clough (Clo) [CP75]:

Where:


Bilinear (BL) rule:

2 3 μΔ – 1 ζ eq, Clo = --- ⋅ ------ ⋅ --------------π 2π μΔ

Page 7-93


(7.107)

Page 7-94


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ζ eq - μ Δ -relationship for these important hysteresis rules: 70

Elasto-plastic

ζeq [-]

50

• The computed value of ζ e was determined iteratively. Eight different inelastic SDOF systems with different periods T n were considered. The strength of each inelastic SDOF system was varied in such a way that seven different displacement ductilities resulted ( μ Δ =2 to 8).

Clough

40 30

Bilinear (ro=5%)

Bilinear (ro=20%)

• The results show that ζ e is not only dependent on μ Δ but also on the period T n of the SDOF system. This effect is not considered by Equations (7.103) and (7.104), respectively.

20

Takeda (α=0.5, β=0.0, ro=5%)

10 0 0

1

2

3

4

5

μΔ [-]

6

7

8

9

10

The next figure compares the theoretical value for ζ e for the Takeda-SDOF (Equ. (7.104), ro=0.05, α=0.5, β=0) with the computed value (for El Centro): 50

11

μΔ,actual [-]

9

ζe [-]

30

20

8 7 6 5

• Similar observations were made when the computation of the inelastic spectra was discussed. • This shows again the difficulties associated with the prediction of the seismic response of inelastic SDOF system.

• Improved estimate for ζ e Over the last years some researchers suggested improved formulas for ζ e by carrying out statistical analyses of time-history responses of inelastic SDOF systems (see [PCK07]). [GBP05] suggest for example the Equation (7.108).

4 3

10

• In some cases the difference between the theoretical value and the computed vale for ζ e is considerable. For this reason there are also considerable differences between μ Δ, t arg et (target ductility) and μ Δ, actual (actual ductility obtained from the time-history analysis of the SDOFe system with the viscous damping ζ e according to Equation (7.104)). • Typically these differences increase as the target ductility increases.

Theory (Eq. (3.46)) Tn=4.00s Tn=2.00s Tn=1.33s Tn=1.00s Tn=0.67s Tn=0.50s Tn=0.33s Tn=0.25s

10 40

2 1

0 0

1

2


3

4

5

6

μΔ,target [-]

7

8

0 0

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Comments regarding the comparison of the theoretical value with the computed value of ζ e for the Takeda-SDOF system when excited by the El Centro earthquake:

2/π

60


1

2

3

4

5

6

μΔ,target [-]

7

8

Page 7-95

ζ e = ζ + ζ eq


where:

§ · 1 ·§ 1 ζ eq = a ¨ 1 – -----b-¸ ¨ 1 + ---------------------d-¸ © ( Te + c ) ¹ μ Δ¹ ©

(7.108)

Page 7-96


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The constants a to d are:

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• Takeda Thin (TT): Hysteretic rule that characterises RC structures which lateral stiffness is provided by walls and columns. • Takeda Fat (TF): Hysteretic rule that characterises RC structures which lateral stiffness is provided by frames. • Ramberg-Osgood (RO): Hysteretic rule that characterises ductile steel structures. • “Flag-Shaped”, β=0.35 (FS): Hysteretic rule that characterises prestressed structures with unbonded tendons. ζ eq - μ Δ -relationships for the most important hysteresis rules according to [GBP05]: 20

a

b

c

d

1) Elasto-Plastic (EPP)

0.224

0.336

-0.002

0.250

2) Bilinear, ro=0.2 (BI)

0.262

0.655

0.813

4.890

3) Takeda Thin (TT)

0.215

0.642

0.824

6.444

4) Takeda Fat (TF)

0.305

0.492

0.790

4.463

5) Ramberg-Osgood (RO)

0.289

0.622

0.856

6.460

6) “Flag-Shaped”, β=0.35 (FS)

0.251

0.148

3.015

0.511 from [GBP05]

The hysteresis rules 1) to 6) were chosen because they can be used to represent the hysteretic behaviour of typical structural types: • Elasto-Plastic (EPP): Hysteretic rule that characterises systems for the seismic isolation of structures (sliding systems that are based on friction). • Bilinear, ro=0.2 (BI): Hysteretic rule that also characterises systems for the seismic isolation of structures. The value of the postyield stiffness rokpl may vary significantly between different systems. 7 Seismic Excitation

Page 7-97

15

ζeq [-]

Hysteresis rule

10

1) EPP 2) BL (ro=0.2) 3) Takeda Thin 4) Takeda Fat 5) FS (beta=0.35) 6) RO

5

Te=2.0s 0 0

1

2

3

μΔ [-]

4

5

6

Important comments: • With these relationships an in a statistical sense improved estimate of the damping ζ e is obtained. • For single systems subjected to a specific ground motion differences between the maximum response of the inelastic system and the maximum response of the equivalent SDOF with ζ e according to these improved ζ eq - μ Δ -relationships can still be significant! 7 Seismic Excitation

Page 7-98


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To compute the response of the equivalent SDOF systems, elastic design spectra can be used. The damping values of equivalent SDOFe systems are in general larger than the typical 5%. For this reason the design spectra needs to be computed for higher damping values. The design spectra for higher values of damping are often obtained by multiplying the design spectra for 5% damping with a correction factor η : S pa (T n,ζ) = η ⋅ S pa (T n,ζ = 5%)

[BE99]:

η =

1.5 ------------------ where 0.05 ≤ ζ ≤ 0.5 1 + 10ζ

η =

1 - where 0.05 ≤ ζ ≤ 0.3 ---------------------0.5 + 10ζ

(7.110) (7.111)

Equation (7.111) corresponds to Equation (29) in the Swiss Code SIA 261 [SIA03]. The correction factors η obtained with Equ.s (7.110) and (7.111) are plotted for different damping values ζ in the next figure:


1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3

[BE99] [TF99]

0.2 0.1 0.0 0.00

0.05

(7.109)

The literature provides different estimates for this correction factor η . Two of these are: [TF99]:

Correction factor η [-]

7.8.1 Elastic design spectra for high damping values

Page 7-99

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0.10

0.15

0.20

0.25

0.30

Damping ζ [-]

0.35

0.40

0.45

0.50

Comments: • A discussion of the different approaches for computing the design spectra for high values of damping can be found in [PCK07]. • Equations (7.110) and (7.111) were derived for ground motions without near-field effects. • Equations (7.110) and (7.111) were derived from the statistical analysis of several response spectra for different ground motions. For this reason Equ.s (7.110) and (7.111) should only be used in conjunction with smoothed response or design spectra. • As for all statistical analyses the resulting design spectra correspond only in average with the true highly damped spectral ordinates. For single periods and ground motions the differences between the highly damped spectral ordinates obtained by Equ.s (7.110)/(7.111) and by time-history analyses of SDOF systems can be significant.


Page 7-100


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• Elastic design spectra according to [BE99] TB

TC

4.0

10%

3.0

20% 30%

5.0

TD

TB

TC

4.0

3.0

ζ = 5% 10% 20%

2.0

0.5

1.0

1.5

2.0

2.5

0.0 0.00

3.0

0.15

TC

5.0

TD

TD

30%

1.0

1.0

TB

[BE99]

ζ = 5%

2.0

0.0 0.0

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• Elastic design spectra in ADRS-format

Spa [m/s2]

Ay [m/s2]

5.0


0.02

TB

0.04

0.06 Sd [m]

0.08

0.10

TC

0.12

[TF99]

ζ = 5% 4.0

D [m]

20% 30%

Spa [m/s2]

10%

0.10

3.0

2.0 0.05

0.00 0.0


0.5

1.0

1.5 Period [s]

2.0

2.5

3.0

Page 7-101

1.0

40 50%

0.0 0.00

0.02


30

ζ = 5% 10% 20%

0.04

0.06 Sd [m]

TD

0.08

0.10

0.12

Page 7-102


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7.8.2 Determining the response of inelastic SDOF systems by means of a linear equivalent SDOF system and elastic design spectra with high damping

• Response of the elastic SDOF system

The computation of the seismic response of inelastic systems by means of linear equivalent systems was studied by Sozen and his co-workers in the seventies (see for example [GS74], [SS76] and [SS81]). Today this approach gains new attention since the “Direct Displacement-Based Design (DDBD)” approach, which was developed by Priestley and his co-workers, is based on the idea of the linear equivalent system ([PCK07]). This sections outlines the procedure for computing the response of an inelastic SDOF system by means of an linear equivalent SDOF system and elastic design spectra with high damping.

fs Properties: C

k

kpl = 244 kN/m ζ = 5%

Step 2: By means of Equ.s (7.102) and (7.104) the nonlinear scale, which represents the damping ζ e as a function of the maximum response of the SDOF system, is plotted along the capacity curve. Step 3: Several spectra for different values of damping are plotted.

• To determine the “Performance Point” exactly, an iterativ approach is typically required.

kpl kel

uy

Step 1: The capacity curve of the SDOF system is plotted on top of the ADRS spectra.

Comments regarding the example:

uy = 0.016 m fy

The maximum response of the inelastic SDOF system will be computed by means of the ADRS-spectra (page 7-106).

S d = 0.065m

fy = 80 kN

kel = 4874 kN/m

S d = 0.054m • Response of the inelastic SDOF system

For the considered example the maxmimum response of the inelastic SDOF system is:

Takeda hysteresis (α=0.5, β=0.0, ro=0.05)

m = 100 t

100- = 0.9s T n = 2π m ---- = 2π ----------4874 k S pa = 2.62m/s 2

Step 4: The “Performance Point” is the point where the spectrum with damping ζ e intersects the capacity curve at the same value of ζe .

• Example: SDOF system with Tn= 0.9 s m

An-Najah 2013

um

u

• The linear equivalent SDOF system is fully defined by the period T e and the damping ζ e . The period T e results from the slope of the line that connects the origin with the “Performance Point”.

For the example the spectra according to [BE99] will be used (Section 7.8.1). 7 Seismic Excitation

Page 7-103


Page 7-104

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• The damping values ζ eq used in the figures on page 7-106 were determined according to Equation (7.104). In the figures on page 7-107 the damping ζ eq was determined using Equation (7.108). The difference is, however, rather small. • It should be noted that in both cases the computed maximum response of the inelastic SDOF system does not comply with the “equal displacement principle”. • The linear equivalent SDOF system leads often to results that do not agree with the “equal displacement principle”. This applies in particular to SDOF systems with long periods or systems with large ductility demands.


An-Najah 2013

Determining the SDOF behaviour by means of elastic ADRS-spectra 5.0

ζ = 5%

4.0

10%

ζ = 18.5%

15% Spa [m/s2]


3.0

Performance point

20%

25%

Capacity curve

2.0

1.0

• A second example is presented on page 7-107. It is a SDOF system with a shorter period and a smaller ductility demand than in Example 1. In this second example the “equal displacement principle” is approximately confirmed.

ζe = 5% 10%

0.0 0.00

0.02

15% 0.04

20% ζe = 18.5% 0.06 Sd [m]

0.08

0.10

0.12

Alternative representation: 5.0

ζ = 5%

Spa [m/s2]

4.0

3.0

ζ = 18.5%

Performance point Tn = 0.9s

2.0

Te = 1.67s 1.0

ζe= 18.5%

μΔ = 3.98 0.0 0.00


Page 7-105

0.02

0.04

uy = 0.016m


0.06 Sd [m]

0.08

0.10

um = 0.065m

0.12

Page 7-106


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Recalculate the example with ζ eq according to [GBP05]: 5.0

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7.9 References

ζ = 5%

[AEM 86] Anderheggen E., Elmer H., Maag H.: “Nichtlineare Finite Element Methoden: Eine Einführung für Ingenieure”. IBK Vorlesungsskript. Institut für Baustatik und Konstruktion (IBK), ETH Zürich. 1986.

4.0

ζ = 17.8% Spa [m/s2]


3.0

Performance point

Tn = 0.90s

[AP88]

Allahabadi R., Powell G.: “Drain-2DX, User Guide”. Report UBC/EERC-88/06. Earthquake Engineering Research Center, University of California, Berkeley 1988.

[Bat96]

Bathe K-J.: Finite Element Procedures. ISBN 0-13-301458-4. Prentice-Hall, Upper Saddle River, New Jersey 1996.

[BE99]

Bommer J.J., Elnashai A.S.: “Displacement spectra for seismic design”. Journal of Earthquake Engineering vol. 3, No. 1, 1999.

[Cho11]

Chopra A.K.: “Dynamics of Structures”. Fourth Edition. Prentice Hall, 2011.

2.0

Te = 1.69s 1.0

0.0 0.00

μΔ = 4.07 0.02

uy = 0.016m

0.04

ζe= 17.8%

0.06 0.08 0.10 Sd [m] um = 0.067m

0.12

Second example with smaller ductility demand: 5.0

ζ = 5%

[DWB99] Dazio A., Wenk T., Bachmann H.: “Versuche an Stahlbetontragwänden unter zyklisch-statischer Einwirkung. (Tests on RC Structural Walls under Cyclic-Static Action)”. IBK-Report No. 239, IBK-ETH Zurich. Birkhäuser Verlag, Basel 1999. (http://e-collection.ethbib.ethz.ch/view/eth:23296)

Performance point

Spa [m/s2]

4.0

ζ = 12.8%

Tn = 0.63s

3.0

Te = 0.88s

[Faj99]

Fajfar P.: “Capacity Spectrum Method Based on Inelastic Demand Spectra”. Earthquake Engineering and Structural Dynamics, 28, 979-993, 1999.

[HHT77]

Hilber H.M., Hughes T.J.R, Taylor R.L.: “Improved Numerical Dissipation for Time Integration Algorithms in Structural Dynamics”. Journal of Earthquake Engineering and Structural Dynamics, Vol. 5, 1977.

[HKS03]

Hibbit Karlsson & Sorensen: “ABAQUS Version 6.4 - User’s Manual”. Pawtucket, RI, 2003.

2.0

1.0

μΔ = 1.97 0.0 0.00

0.02

uy = 0.02m


0.04

um = 0.039m

ζe= 12.8%

0.06 Sd [m]

0.08

0.10

0.12

Page 7-107


Page 7-108


[KN92]

[GBP05]

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Krawinkler H., Nassar A.A.: “Seismic design based on ductility and cumulative damage demands and capacities”. Nonlinear Seismic Analysis and Design of Reinforced Concrete Buildings, eds P. Fajfar and H. Krawinkler. New York: Elsevier Applied Science. 1992. Grant D.N., Blandon C.A., Priestley M.J.N.: “Modelling Inelastic Response in Direct Displacement-Based Design”. Rose School, Report 2005/03, IUSS Press, Pavia, 2005.

[GS74]

Gulkan P., Sozen M.: “Inelastic Responses of Reinforced Concrete Structures to Earthquake Motions”. ACI Journal, Title No. 71-41. December 1974.

[IG79]

Iwan W.D., Gates N.C.: “Estimating Earthquake Response of Simple Hysteretic Structures”. ASCE, Journal of the Engineering Mechanics Division, Vol. 105, EM3, June 1979.

[Jac60]

Jacobsen L.S.: “Damping in composite structures”. Proceedings of the 2nd World Conference in Earthquake Engineering. Vol. 2. Tokio and Kyoto, Japan, 1960.

[Jen68]

Jennings P.C.: “Equivalent Viscous Damping for Yielding Structures”. ASCE, Journal of the Engineering Mechanics Division, Vol. 94, No. EM1, February 1968.

[Mir01]

Miranda E.: “Estimation of inelastic defromation demands of SDOF Systems”. ASCE, Journal of structural engineering, Vol. 127, No. 9, 2001.

[New59]

Newmark N.M.: “A Method of Computation for Structural Dynamics”. ASCE, Journal of the Engineering Mechanics Division, Vol. 85, No. 3, July 1959.

[NH82]

Newmark N.M., Hall W.J.: “Earthquake Spectra and Design”. Earthquake Engineering Research Institute (www.eeri.org), 1982.


Page 7-109


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[PCK07]

Priestley M.J.N., Calvi G.M., Kowalsky M.J.: “DisplacementBased Seismic Design of Structures”. IUSS Press, 2007.

[Saa91]

Saatcioglu M.: “Modeling Hysteretic Force-Deformation Relationships for Reinforced Concrete Elements”. In Publication SP 127 “Earthquake-Resistant Concrete Structures, Inelastic Response and Design”, ACI, Detroit 1991.

[SIA03]

Swiss standard SIA 261: “Actions on structures”. SIA, Zürich 2003.

[SS81]

Saiidi A., Sozen M.: “Simple Nonlinear Seismic Analysis of R/C Structures”. ASCE, Journal of the Structural Division, Vol. 107, No. 5, May 1981.

[SS76]

Shibata A., Sozen M.: “Substitute Structure Method for Seismic Design in R/C”. ASCE, Journal of the Structural Division, Vol. 102, No. 1, January 1976.

[TF99]

Tolis S.V., Faccioli E.: “Displacement design spectra”. Journal of Earthquake Engineering vol. 3, No. 1, 1999.

[TSN70]

Takeda T., Sozen M.A., Nielsen N.N.: “Reinforced Concrete Response to Simulated Earthquakes”. ASCE, Journal of Structural Engineering, Vol. 96, No. 12, December 1970.

[VFF94]

Vidic T., Fajfar P., Fischinger M.: “Consistent inelastic design spectra: strength and displacement”. Earthquake Engineering and Structural Dynamics 23(5) 507-521. 1994.


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8 Multi Degree of Freedom Systems


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8.1.2 Stiffness formulation

8.1 Formulation of the equation of motion The degrees of freedoms are the horizontal displacements u 1 and u 2 at the level of the masses m 1 and m 2

8.1.1 Equilibrium formulation

• Stiffness matrix K =

k 11 k 12 k 21 k 22

=

( k1 + k2 ) –k2 –k2

Unit displacement u 1 = 1 ·· · · · m1 u1 + c1 u1 + k1 u1 = f1 ( t ) + c2 ( u2 – u1 ) + k2 ( u2 – u1 ) ® m 2 u··2 + c 2 ( u· 2 – u· 1 ) + k 2 ( u 2 – u 1 ) = f 2 ( t ) ¯

(8.1)

·· · · m 1 u 1 + ( c 1 + c 2 )u 1 – c 2 u 2 + ( k 1 + k 2 )u 1 – k 2 u 2 = f 1 ( t ) ® m 2 u··2 – c 2 u· 1 + c 2 u· 2 – k 2 u 1 + k 2 u 2 = f 2 ( t ) ¯

(8.2)

m1 0 0 m2

u··1 ( c1 + c2 ) –c2 + ·· u2 –c2 c2

·· + Cu· + Ku = f ( t ) Mu 8 Multi Degree of Freedom Systems

(8.5)

k2

Unit displacement u 2 = 1

u· 1 ( k1 + k2 ) –k2 u1 f (t) = 1 + (8.3) ·u – k u ( t ) k f 2 2 2 2 2

(8.4) Page 8-1


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• Mass matrix M M =

m1 0

(8.6)

0 m2


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By means of the principle of virtual forces the vertical displacement Δ at location d due to a unit force F = 1 acting at location a can be readily determined. FL 3 a d Δ (α,δ) = – α δ ( α 2 + δ 2 – 1 ) ⋅ --------- with α = --- and δ = --L L 6EI

• Equation of motion

(8.9)

The flexibilty matrix consists of the following elemnts: m1 0 0 m2

u··1 ( k1 + k2 ) –k2 u1 + = 0 ·· u2 –k2 k2 u2 0

·· + Ku = 0 Mu

(8.7) (8.8)

8.1.3 Flexibility formulation

(8.10)

u = DF u1 u2

=

d 11 d 12 d 21 d 22

⋅

F1

(8.11)

F2

The d ij factors can be computed by means of Equation (8.9) as follows:

• Flexibility matrix D

4 L3 d 11 = Δ (1 ⁄ 3,2 ⁄ 3) = --------- ⋅ -----243 EI

(8.12)

7 L3 d 12 = Δ (2 ⁄ 3,2 ⁄ 3) = --------- ⋅ -----486 EI

(8.13)

7 L3 d 21 = Δ (1 ⁄ 3,1 ⁄ 3) = --------- ⋅ -----486 EI

(8.14)

4 L3 d 22 = Δ (2 ⁄ 3,1 ⁄ 3) = --------- ⋅ -----243 EI

(8.15)

and the flexibilty matrix D becomes: L3 D = --------------- ⋅ 8 7 486EI 7 8


Page 8-3


(8.16)

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8.1.6 “Direct Stiffness Method”

• Stiffness matrix K 162 EI K = D – 1 = --------- ⋅ -----3- ⋅ 8 – 7 5 L –7 8


• Stiffness matrix of a beam element (8.17)

• Mass matrix M M =

m1 0

(8.18)

0 m2

The stiffness matrix K of a beam element with constant flexural and axial stiffness is well known:

• Equation of motion m1 0 0 m2

u··1 162 EI u + --------- ⋅ -----3- ⋅ 8 – 7 1 = 0 5 L u··2 0 –7 8 u2

·· + Ku = 0 Mu

(8.19)

EA -------L

(8.20) F1

8.1.4 Principle of virtual work

0

F2

• See e.g. [Hum12]

F3 F4

8.1.5 Energie formulation

F5

• See e.g. [Hum12]

(8.21)

F = Ku

F6

0 = EA – -------L 0 0

0

0

12EI----------L3 6EI --------L2

6EI --------L2 4EI --------L

0

0

12EI 6EI - – --------– ----------L2 L3 6EI 2EI --------- --------L L2

EA – -------L 0 0 EA -------L 0 0

0

0

12EI– ----------L3 6EI– -------L2

6EI --------L2 2EI --------L

0

0

12EI 6EI ------------ – --------L3 L2 6EI 4EI - --------– -------L L2

u1 u2 ⋅

u3

(8.22)

u4 u5 u6

If the axial elongation of the beam is not considered, the matrix can be further simplified as follows:


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Equation of motion: mL 0 0 0 0 12

0

6L – 12 6L

2 2 EI K = -----3- ⋅ 6L 4L – 6L 2L L – 12 – 6L 12 – 6L

0 0

0 0

mL 0 -------- 0 2 0 0 0

⋅

u··1 u·· 2

u··3 u·· 4

24

0

– 12 6L

u1

2L 2

u2

8L 2

EI – 6L + -----3- ⋅ 0 ⋅ L u3 – 12 – 6L 12 – 6L 2 2 u4 6L 2L – 6L 4L

0 = 0 (8.25) 0 0

(8.23) mL 0

6L 2L 2 – 6L 4L 2

0

mL 0 -------- 0 2 0 0 0 0 0 0

• Example: Cantilever

0 0 0 0

u··1 u··

24 – 12

0

6L

u1

u··1 u··

24 – 12

0

6L

u1

0 u EI – 12 12 – 6L – 6L ⋅ 3 + -----3- ⋅ ⋅ 3 = 0 (8.26) ·· u2 L u2 0 0 – 6L 8L 2 2L 2 0 u··4 u4 6L – 6L 2L 2 4L 2

Static condensation: mL 0

mL 0 -------- 0 2 0 0 0 0 0 0

Assemblage of the stiffness matrix F1 F2 F3

12 + 12

– 12

6L

2 2 EI = -----3- ⋅ – 6 L + 6L 4L + 4L L – 12 – 6L

– 6L

2L 2

12

– 6L

u3

– 6L

4L 2

u4

6L

2L 2

0 0 0 0

0 u EI ⋅ 3 + -----3- ⋅ – 12 12 – 6L – 6L ⋅ 3 = 0 (8.27) u··2 L u2 0 0 – 6L 8L 2 2L 2 0 u··4 u4 6L – 6L 2L 2 4L 2

u1

– 6 L + 6L

F4

0

⋅

u2

(8.24)

m tt 0 0 0

⋅

·· u t + k tt k t0 ⋅ u t = 0 ·· 0 k 0t k 00 u0 u0

(8.28)

with L = L ⁄ 2


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·· m tt u t + k tt u t + k t0 u 0 = 0 ® k 0t u t + k 00 u 0 = 0 ¯

(8.29)

From the second row of Equation (8.29) the following expression can be derived: u0 =

–1 – k 00 k 0t u t

(8.30)

Substituting Equation (8.30) in the first line of Equation (8.29) we obtain: –1

·· + k u – k k k u = 0 m tt u t tt t t0 00 0t t ·· + ( k – m tt u t tt

–1 k t0 k 00 k 0t )u t

(8.31) (8.32)

= 0


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EI 6 kˆ tt = -----3- ⋅ --- ⋅ 16 – 5 L 7 –5 2

(8.36)

after substituting L = L ⁄ 2 : EI 48 kˆ tt = -----3- ⋅ ------ ⋅ 16 – 5 L 7 –5 2

(8.37)

The final equation of motion of the cantilever is therefore: mL -------- 0 u u·· EI 48 2 ⋅ 1 + -----3- ⋅ ------ ⋅ 16 – 5 ⋅ 1 = 0 0 u··3 L 7 – 5 2 u3 mL 0 -------4

(8.38)

T

and with k t0 = k 0t : T

• Notes –1

·· + ( k – k k k )u = 0 m tt u t tt 0t 00 0t t

(8.33)

• The “Direct Stiffness Method” is often used in the Finite Element Method.

·· + kˆ u = 0 with kˆ = k – k T k – 1 k m tt u t tt 0t 00 0t tt tt t

(8.34)

• The derivation of the stiffness matrix K for a beam element and instructions for assembling the stiffness matrix of entire structures can be found e.g. in the following references:

Where kˆ tt is the condensed stiffness matrix, and in our case it is equal to:

kˆ tt

§ 1 --------¨ 2 EI ¨ 24 – 12 – 0 6L ⋅ 7L = -----3- ⋅ ¨ L 1 –6 L –6 L ¨ – 12 12 – -----------2© 14L

· 1 – -----------2¸ 14L ⋅ 0 – 6 L ¸ ¸ 26L – 6 L ¸ -------¹ 7L 2

[Prz85]

Przemieniecki J.S.: “Theory of Matrix Structural Analysis”. Dover Publications, New York 1985.

[Bat96]

Bathe K-J.: “Finite Element Procedures”. Prentice Hall, Upper Saddle River, 1996.

(8.35) 8 Multi Degree of Freedom Systems

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8.1.7 Change of degrees of freedom


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or:

The equation of motion for free vibration of the 2-DoF system depicted in the following can be immediately set up if the DoFs u 1 and θ 1 are considered.

M⋅

u··1 u +K⋅ 1 = 0 ·· 0 θ1 θ1

(8.40)

As an alternative, the motion of the system can be also expressed in terms of the DoFs u 1 and u 2 . To this purpose, the relationship between the two sets of DoFs can be immediately written as: u1 = u1 ° ® L ° u 2 = --- ⋅ θ 1 2 ¯

(8.41)

which in matricial form yields the following system of equations:

θ1

1 0 0 2⁄L

=

⋅

° ® ° ¯

u1

u1

or u = Au

(8.42)

u2

A Using Equation (8.23), the equation of motion for free vibrations of the system becomes 3m -------2

0 2

mL 0 ----------8

u·· u 12 – 6L EI ⋅ 1 + -----3- ⋅ ⋅ 1 = 0 2 ·· θ1 0 θ1 L – 6L 4L

The matrix A is called coordinate transformation matrix and can be used to transform the mass matrix, the stiffness matrix and the load vector from one set of DoF to the other, i.e. T

K = A KA T

M = A MA T

F = A F 8 Multi Degree of Freedom Systems

(8.43)

(8.39)

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(8.44) (8.45) Page 8-12


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For the example at hand, the stiffness matrix K expressed in the set of DoFs u 1 and u 2 becomes: 12 – 6L EI T ⋅ 1 0 K = A KA = 1 0 ⋅ -----3- ⋅ 2 0 2⁄L L 0 2⁄L – 6L 4L

(8.46)

EI K = -----3- ⋅ 12 – 12 L – 12 16

(8.47)


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8.1.8 Systems incorporating rigid elements with distributed mass The 2-DoF system depicted in the following incorporates a rigid element with distributed mass μ .

while the mass matrix M becomes:

M = A MA = 1 0 ⋅ 0 2⁄L T

3m -------2

0 2

mL 0 ----------8

⋅ 1 0 0 2⁄L

3m -------- 0 M = 2 m 0 ---2

(8.48)

(8.49)

which yields the equation of motion of the 2-DoF systems expressed in terms of the DoFs u 1 and u 2 3m -------- 0 2 ⋅ m 0 ---2

u u··1 EI + -----3- ⋅ 12 – 12 ⋅ 1 = 0 – 12 16 0 u··2 L u2


(8.50)

Page 8-13

The elements of the 2x2 mass matrix can be determined by imparting a unit acceleration üa=1 to one degree of freedom while keeping the acceleration of the other degree of freedom equal to zero (üb=0). The resulting inertia forces are then applied as static forces acting onto the system, and the elements of the mass matrix are computed as the reactions to these static forces.


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In the example at hand, if the DoFs u 1 and u 2 are considered, the elements of the mass matrix can be easily computed as follows: 2 m 11 = --- μL 3

(8.51)

1 m 21 = --- μL 3

(8.52)

1 m 12 = --- μL 3

(8.53)

2 m 22 = --- μL 3

(8.54)


u1 u·· μL EI- 28 – 10 ------- ⋅ 2 1 ⋅ 1 + ----⋅ ⋅ = 0 3 6 ·· – 10 4 1 2 u2 L u2 0

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(8.57)

Hence the mass matrix becomes: μL M = ------- ⋅ 2 1 6 1 2

(8.55)

Due to the fact that the mass is distributed, off-diagonal terms are present and therefore the mass matrix is coupled. The stiffness matrix of the 2-DoF system can be easily computed by means of the methods discussed so far as: EI K = -----3- ⋅ 28 – 10 – 10 4 L

(8.56)

and the equation of motion of the system for free vibration becomes: 8 Multi Degree of Freedom Systems

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9 Free Vibrations

If the matrix A has an inverse A –1 , then Equation (9.7) can be rearranged as follows: (9.1) –1

–1

(9.8)

A Aφ n = A 0

Ansatz: u ( t ) = q n ( t )φ φ n where q n ( t ) = A n cos ( ω n t ) + B n sin ( ω n t )

(9.2)

and therefore φn = 0

The double derivation of Equation (9.2) yields: 2 2 q··n ( t ) = – ω n [ A n cos ( ω n t ) + B n sin ( ω n t ) ] = – ω n q n ( t )

(9.3)

··( t ) = – ω 2 q ( t ) φ u n n n

(9.4)

2 ω n Mφ n

+ Kφ n ]q n ( t ) = 0

The inverse of Matrix A has the form: A

(9.5)

Equation (9.5) is satisfied if q n ( t ) = 0 , which is a trivial solution meaning that there is no movement, because u ( t ) = q n ( t )φφ n = 0 . To obtain a nontrivial solution the term in brackets in Equation (9.5) must be equal to zero, i.e.:

(9.9)

This means that if matrix A has an inverse A –1 , then Equations (9.6) and (9.7) have only the trivial solution given by Equation (9.9).

and by substituting Equations (9.2) and (9.4) in (9.1) we obtain: [–

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Also in the case of Equation (9.7), there is always the trivial solution φ n = 0 , which corresponds to an absence of movement.

9.1 Natural vibrations ·· + Ku = 0 Mu


–1

1 ˆ = ------- A A

(9.10)

If the determinant A is equal to zero, then the matrix is singular and has no inverse. Therefore, Equation (9.6) has a nontrivial solution only if: 2

[–

2 ωn M

+ K ]φ φn = 0

(9.6)

– ωn M + K = 0

(9.11) 2

or: 2

Aφ n = 0 with A = – ω n M + K

9 Free Vibrations

(9.7)

Page 9-1

The determinant yields a polynomial of order N in ω n which is called characteristic equation. The N roots of the characteristic equation are called eigenvalues and allow the calculation of the N natural circular frequencies ω n of the system.

9 Free Vibrations

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As soon as the natural circular frequencies ω n are computed, also the vectors φ n can be computed within a multiplicative constant by means of Equation (9.6). There are N independent Vectors which are called eigenvectors or natural modes of vibration of the system.


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9.2 Example: 2-DoF system

We consider a regular 2-DoF oscillator with m1 = m2 = m

Summary

and

• A MDoF system with N degrees of freedom has N circular frequencies ω n ( n = 1, 2, 3, …, N ) and N eigenvectors. Each eigenvector has N elements. The circular frequencies are arranged in ascending order, i.e.: ω 1 < ω 2 < … < ω n .

k1 = k2 = k

• Natural circular frequencies and eigenvectors are properties of the MDoF system and depends only from its mass and stiffness properties.

The equation of motion of the system corresponds to equation (8.7):

• The index n refers to the numbering of the eigenvectors and the first mode of vibration ( n = 1 ) is commonly referred to as the fundamental mode of vibration.

(9.12)

u·· u m 1 0 1 + k 2 –1 1 = 0 0 1 u··2 –1 1 u2 0

9.2.1 Eigenvalues The eigenvalues are calculated from the determinant: 2

2

K – ωn M =

2k – ω n m

–k

–k

k –ωn m

(9.13)

= 0

2

which gives a quadratic equation in ω 2n 2

2

2

4

2

2

( 2k – ω n m ) ⋅ ( k – ω n m ) – ( – k ) ⋅ ( – k ) = m ω n – 3kmω n + k = 0 (9.14)

9 Free Vibrations

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and both solutions yield the following eigenvalues: 2

2

2


( – 1 + 5 )k ( 1 + 5 ) – kφ 11 + ----------------------------- § --------------------- φ 11· = 0 © ¹ 2 2

2

3km ± 9k m – 4k m 3± 5 k 2 - = ---------------- ⋅ ---ω n = ----------------------------------------------------------2 2 m 2m

(9.15)

– kφ 11 + kφ 11 = 0

For each eigenvalue ω 2n we can now compute an eigenvector and a natural circular frequency.

φ 11 = φ 11

9.2.2 Fundamental mode of vibration 3– 5 k With the smallest eigenvalue ω 21 = ---------------- ⋅ ---- we obtain the 2

1. circular frequency ω 1 =

m

3– 5 k k ---------------- ⋅ ---- = 0.618 --2 m m

(9.16)

2 By substituting this eigenvalue ω 1 into the system of equations

2

[ K – ω 1 M ]φ φ1 =

3– 5 k 2k – § ---------------- ⋅ ----· m © 2 m¹

–k

–k

3– 5 k k – § ---------------- ⋅ ----· m © 2 m¹

• so that the largest element of the eigenvector is equal to 1 • so that one particular element of the eigenvector is equal to 1 • so that the norm of the eigenvector is equal to 1 • ...

Fundamental mode: ⋅

φ 11 φ 21

= 0 0

we obtain two independent equations that can be used to determine the elements of the first eigenvector φ 1 . The first row of the system yield the equation: and

(9.19)

As expected, the eigenvector is determined within a multiplicative constant, and can therefore be arbitrarily normalized as follows:

ω1 =

(9.17)

(-----------------------1 + 5 )k φ 11 – kφ 21 = 0 2

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(1 + 5) φ 21 = --------------------- φ 11 2

“Mode” “Degree of freedom”

φ1 =

3--------------– 5- --k k ⋅ - = 0.618 ---2 m m φ 11 φ 21

2 --------------= 1 + 5 = 0.618 1 1

(9.18)

and by substituting this into the second row we obtain: 9 Free Vibrations

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9.2.3 Higher modes of vibration

9.2.4 Free vibrations of the 2-DoF system

Additionally to the fundamental mode of vibration, the considered 2-DoF system has a second mode of vibration.

According to Equation (9.2), the free vibration of the 2-DoF system is:

The properties of this second mode of vibration can be computed in analogy to the fundamental mode and the following results are obtained:

u = [ C 1 cos ( ω 1 t ) + C 2 sin ( ω 1 t ) ]φ φ 1 + [ C 3 cos ( ω 2 t ) + C 4 sin ( ω 2 t ) ]φ φ2

(9.20) u1 u2

Second mode ω2 =

“Mode” “Degree of freedom”

φ2 =

3--------------+ 5- --k k⋅ - = 1.618 --2 m m

φ 12 φ 22

1

1 = 1– 5 = – 0.618 ---------------2

= [ C 1 cos ( ω 1 t ) + C 2 sin ( ω 1 t ) ]

φ 11 φ 21

+ [ C 3 cos ( ω 2 t ) + C 4 sin ( ω 2 t ) ]

(9.21)

φ 12 φ 22

The still unknown constants C 1 to C 4 can be computed using the initial conditions given by Equation (9.24) and become: φ 22 u1 – φ 12 u2 C 1 = -------------------------------------φ 11 φ 22 – φ 21 φ 12

,

φ 22 v1 – φ 12 v2 C 2 = ------------------------------------------------( φ 11 φ 22 – φ 21 φ 12 )ω 1

(9.22)

φ 11 u2 – φ 21 u1 C 3 = -------------------------------------φ 11 φ 22 – φ 21 φ 12

,

φ 11 v2 – φ 21 v1 C 4 = ------------------------------------------------( φ 11 φ 22 – φ 21 φ 12 )ω 2

(9.23)

u1 ( 0 ) ° ° u2 ( 0 ) Initial conditions: ® · ° u1 ( 0 ) °· ¯u (0) 2

= u1 = u2 = v1

(9.24)

= v2

For an alternative methodology to compute the constants C 1 to C 4 see Section 9.6. 9 Free Vibrations

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• Case 1: u1 = 0.618 , u2 = 1.000 , v1 = v2 = 0 First mode Second mode Total displacement

1.0 0.6 0.4 0.2 0.0 -0.2 -0.4 -0.6

First mode Second mode Total displacement

1.0 0.8

Dis splacement u2 [m]

0.8

0.6 0.4 0.2 0.0 -0.2 -0.4 -0.6

-0.8

-0.8

-1.0

-1.0 -1.2

-1.2 0

5

10

15

0

20

5

1.2

1.2


1.0

15

20


1.0 0.8


0.8

10

Time [s]

Time [s]


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• Case 2: u1 = 1.000 , u2 = – 0.618 , v1 = v2 = 0 1.2

1.2



0.6 0.4 0.2 0.0 -0.2 -0.4 -0.6

0.6 0.4 0.2 0.0 -0.2 -0.4 -0.6

-0.8

-0.8

-1.0

-1.0

-1.2

-1.2 0

5

10

15

20

0

Time [s]

9 Free Vibrations

5

10

15

20

Time [s]

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0.6

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9.3 Modal matrix and Spectral matrix

• Case 3: u1 = 0.618 , u2 = 0.000 , v1 = v2 = 0 0.8


All N eigenvalues and all N eigenvectors can be compactly represented in matricial form: • Modal matrix

0.4 0.2

φ 11 φ 12 … φ 1N

0.0

φ 21 φ 22 … φ 2N

Φ = [ φ jn ] =

-0.2

… … … … φ N1 φ N2 … φ NN

-0.4 -0.6

(9.25)

• Spectral matrix

-0.8 0

5

10

15

20

ω 12 0 … 0

Time [s] 2

Ω =

0.8



0.6

0 ω 22 … 0 … … … … 0

0.4

(9.26)

0 … ω N2

Equation (9.6) can therefore be rearranged as follows:

0.2

Kφ n = Mφ n ω n2

0.0

(9.27)

and it is immediately apparent that the equation for all eigenvalues and all eigenvectors can be expressed in terms of modal and the spectral matrices, as follows:

-0.2 -0.4 -0.6

KΦ = MΦΩ

2

(9.28)

-0.8 0

5

10

15

20

Time [s]

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9.4 Properties of the eigenvectors

vectors are linearly independent and can be chosen so that they are orthogonal (proof complicated).

9.4.1 Orthogonality of eigenvectors

So far we have shown that φ Tn Mφ r = 0 for n ≠ r . By means of Equation (9.32) we can prove also the φ Tn Kφ r = 0 for n ≠ r . We have already seen that for n ≠ r the right hand side of Equation (9.32) is equal to zero. For this reason also the left hand side of Equation (9.32) must be equal to zero, which conclude the verification.

The orthogonality conditions of the eigenvectors are: T

T

φ n Kφ r = 0 and φ n Mφ r = 0 for n ≠ r

(9.29)

and can be proven by means of Equation (9.27). Equation (9.27) is first to be set up for the eigenvector vector n , and then preT multiplied with φ r on both sides: T

T

φ r Kφ n = ω n2 φ r Mφ n

(9.30)

Afterwards, Equation (9.30) shall be transposed making use of T T the symmetry properties of the matrices K = K and M = M : T

T

φ n Kφ r = ω n2 φ n Mφ r

(9.31)

Now, Equation (9.27) shall be set up for the eigenvector vector T r , and then pre-multiplied with φ n on both sides: T φ n Kφ r

=

T ω r2 φ n Mφ r

T

• Relative to the mass matrix 2 (5 + 5) T 2 m 0 ⋅ ---------------- = 2m ----------------------------- ≅ 1.382m φ 1 Mφ 1 = --------------1 ⋅ 1+ 5 2 0 m 1+ 5 (1 + 5) 1

(9.34) 1 T 2 m 0 ⋅ φ 1 Mφ 2 = --------------1 ⋅ 1– 5 = 0 0 m ---------------1+ 5 2

(9.35)

2 T – 5- ⋅ m 0 ⋅ ---------------- = 0 φ 2 Mφ 1 = 1 1--------------1+ 5 0 m 2 1

(9.36)

(9.33)

In the case that the eigenvalues are different, then for n ≠ r we T have ( ω n2 – ω r2 ) ≠ 0 and the expression φ n Mφ r must be zero. In the case that an eigenvalue occurs more than once, the eigen-

9 Free Vibrations

In the following the orthogonality of the eigenvectors of the 2DoF system presented in Section 9.2 is checked:

(9.32)

Equation (9.32) can now be subtracted from Equation (9.31) yielding the following equation: ( ω n2 – ω r2 )φ φ n Mφ r = 0

Example: 2-DoF system

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9.4.2 Linear independence of the eigenvectors

1

m T – 5- ⋅ m 0 ⋅ φ 2 Mφ 2 = 1 1--------------= ---- ( 5 – 5 ) ≈ 1.382m 1 – 5 2 0 m ---------------2 2

The eigenvectors are linearly independent. To prove this, it needs to be shown that if (9.37)

• Relative to the stiffness matrix T φ 1 Kφ 1


α1 φ1 + α2 φ2 + … + αn φn = 0

(9.42)

then all scalars α i must be equal to zero. T

To this purpose, we left-multiply Equation (9.42) by φ i M and we obtain:

2 ---------------2k ( 5 – 5 ) 2 2k – k - ≅ 0.528k ⋅ 1 + 5 = -------------------------= ---------------- 1 ⋅ 2 –k k 1+ 5 (1 + 5) 1

T

(9.38)

T

φ i M ( α 1 φ 1 + α 2 φ 2 + … + α n φ n ) = φ i Mφ i α i = 0

(9.43)

T

1 2 2k – k = ---------------- 1 ⋅ ⋅ 1– 5 = 0 –k k ---------------1+ 5 2

(9.39)

In Section 9.4.1 we have shown that φ i Mφ i ≠ 0 , therefore α i = 0 meaning that the eigenvectors are linearly independent.

2 T 1 – 5 ⋅ 2k – k ⋅ ---------------- = 0 φ 2 Kφ 1 = 1 --------------1+ 5 –k k 2 1

(9.40)

The property that the eigenvectors are linearly independent, is very important because it allows to represent any displacement vector as a linear combination of the eigenvectors.

T φ 1 Kφ 2

1 k T 1 – 5- ⋅ 2k – k ⋅ φ 2 Kφ 2 = 1 --------------= --- ( 5 + 5 ) ≈ 3.618k 1 – 5 2 –k k ---------------2 2

(9.41)

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9.5 Decoupling of the equation of motion The equation of motion for free vibrations is ·· + Ku = 0 Mu

(9.44)

and as a possible solution the displacement vector u(t) =

¦ qi ( t )φφ i

(9.45)

can be assumed, where: linearly independent eigenvectors of the system

qi :

modal coordinates

¦ q··i ( t )φφ i

T

Modal mass:

m n* = φ n Mφ n

Modal stiffness:

k n* = φ n Kφ n

(9.49)

T

(9.50)

and Equation (9.48) can be rewritten as follows: (9.51)

For each n we can set up such an equation, which yields to N decoupled Single Degree of Freedom systems. The total displacement of the system can then be computes as the sum of the contribution of all decoupled SDoF systems, i.e.: N

u(t) =

The displacement vector u ( t ) and its double derivative ··( t ) = u

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m n* q··n ( t ) + k n* q n ( t ) = 0

i

φi :


¦

q i ( t )φ φi

(9.52)

i=1

(9.46)

in matricial form:

i

can be substituted into Equation (9.44), and the latter can be leftT multiplied by φ n yielding the following equation: T T φ n M § ¦ q··i ( t )φ φ i· + φ n K § ¦ q i ( t )φ φ i· = 0 © ¹ © ¹ i

(9.47)

i

Because of the orthogonality properties of the eigenvectors only one term of the summations remains, i.e.: T φ n Mφ n q··n ( t )

+

T φ n Kφ n q n ( t )

= 0

(9.48)

u ( t ) = Φ q ( t ) with q =

q1 ( t )

(9.53)

… qN ( t )

·· + K * q = 0 M*q

(9.54)

with

T

M * = Φ MΦ =

m 1* … 0 … … … and 0 … m n*

where:

T

k 1* … 0

K * = Φ KΦ = … … … 0 … k n*

(9.55) 9 Free Vibrations

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In Equations (9.52) and (9.53) are rewritten q i ( t )φ φ i = Φq ( t )

¦

(9.56)

i=1

the LHS and the RHS of the resulting equation can be pre-mulT tiplied by φ n M and we obtain: N T

φ n Mu ( t ) =

T

¦ φn Mφi qi ( t )

(9.57)

i=1

Because of the orthogonality of the eigenvectors (see Section 9.4.1), Equation can be further simplified to: T

T

φ n Mu ( t ) = φ n Mφ n q n ( t )

(9.58)

which yields the following relationship between q n ( t ) and u ( t ) T

φ n Mu ( t ) q n ( t ) = ---------------------T φ n Mφ n

(9.59)

or introducing the definition of the modal mass given by Equation (9.49) we obtain the equivalent expression qn ( t ) =

T φ n Mu ( t ) ----------------------m n*

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Example: 2-DoF system

N

u(t) =


(9.60)

The modal masses and modal stiffness of the 2-DoF system of Section 9.2 were already checked during the verification of the orthogonality of the eigenvectors (See Equations (9.34), (9.37), (9.38) and (9.41)). They are: 2m ( 5 + 5 ) T ≈ 1.382m m 1* = φ 1 Mφ 1 = ----------------------------2 (1 – 5)

(9.61)

m T m 2* = φ 2 Mφ 2 = ---- ( 5 – 5 ) ≈ 1.382m 2

(9.62)

2k ( 5 – 5 -) T ≅ 0.528k k 1* = φ 1 Kφ 1 = -------------------------2 (1 – 5)

(9.63)

k T k 2* = φ 2 Kφ 2 = --- ( 5 + 5 ) ≈ 3.618k 2

(9.64)

• First modal SDoF system: m 1* q··1 ( t ) + k 1* q 1 ( t ) = 0

(9.65)

1.382mq··1 ( t ) + 0.528kq 1 ( t ) = 0

(9.66)

ω1 =

k* ------1- = m 1*

0.528k- = 0.618 --k----------------1.382m m

(9.67)

The natural frequency corresponds to equation (9.16) of this chapter

Equations (9.59) and (9.60) will be later used to compute the response of MDoF systems. 9 Free Vibrations

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m 2* q··2 ( t ) + k 2* q 2 ( t ) = 0

(9.68)

1.382mq··2 ( t ) + 3.618kq 2 ( t ) = 0

(9.69)

k* ------2- = m 2*

3.618k- = 1.618 --k----------------1.382m m

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9.6 Free vibration response

• Second modal SDoF system:

ω2 =


(9.70)

The natural frequency corresponds to the result shown on page 9-7.

9.6.1 Systems without damping The equation of motion for free vibration of a system without damping is ·· + Ku = 0 Mu

(9.71)

and making use of the possibility of decoupling of the equation of motion, the total deformation u ( t ) under free vibration can be computed as the sum of the contribution of all modes. The equation of motion of the nth decoupled SDoF system is: m n* q··n ( t ) + k n* q n ( t ) = 0

(9.72)

and its solution can be computed as discussed in Chapter 3 for SDoF systems. If we make use of the second formulation with “trigonometric functions” (see Section 3.1.2), the solution is: q n ( t ) = A n cos ( ω n t ) + B n sin ( ω n t )

(9.73)

The the total deformation u ( t ) under free vibration is hence N

u(t) =

¦ i=1

N

φi qi ( t ) =

¦

φ i [ A i cos ( ω i t ) + B i sin ( ω i t ) ]

(9.74)

i=1

The 2 ⋅ N constants A i and B i can be computed by means of the initial conditions u ( 0 ) = u 0 and u· ( 0 ) = v 0 . To this purpose, the vector of the velocity is needed and can easily be computed by deriving Equation (9.74), i.e.:

9 Free Vibrations

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9 Free Vibrations

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Course “Fundamentals of Structural Dynamics” N

u· ( t ) =

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¦ i=1

¦

φ i ω i [ – A i sin ( ω i t ) + B i cos ( ω i t ) ]

(9.75)

i=1

Considering Equations (9.74) and (9.75) at the time t = 0 , we have. N

u(0) =

N

φ i q i ( 0 ) and u· ( 0 ) =

¦ i=1

¦

φ i q· i ( 0 )

(9.76)

i=1

Making use of Equation (9.59) we can now write the equations to compute the initial conditions of the nth decoupled SDoF system as: qn ( 0 ) =

T φ n Mu ( 0 ) -----------------------T φ n Mφ n

q· n ( 0 ) =

T φ n Mu· ( 0 ) -----------------------T φ n Mφ n

(9.77)

(9.78)

In Section 3.1.2 (see Equation 3.18) it as been shown that the constants A n and B n are equal to q n ( 0 ) and q· n ( 0 ) ⁄ ω n , respectively, hence Equation (9.74) can be rewritten as: N

u(t) =

¦ i=1

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9.6.2 Classically damped systems

N

φ i q· i ( t ) =


q· i ( 0 ) φ i q i ( 0 ) cos ( ω i t ) + ------------ sin ( ω i t ) ωi

(9.79)

The equation of motion for free vibration of a system with damping is ·· + Cu· + Ku = 0 Mu

(9.80)

As it will be shown in Chapter 10, if the MDoF system is classically damped, the equation of motion can be decoupled analogously to system without damping, and the total deformation u ( t ) under free vibration can be computed again as the sum of the contribution of all modes. The equation of motion of the nth decoupled SDoF system is: m n* q··n ( t ) + c n* q· n ( t ) + k n* q n ( t ) = 0

(9.81)

2 q··n ( t ) + 2ω n ζ n q n ( t ) + ω n q n ( t ) = 0

(9.82)

or

where c n*

=

T φ n Cφ n

c n* and ζ n = ----------------, respectively. 2m n* ω n

(9.83)

The solution of Equation (9.82) can be computed as discussed in Chapter 3 for SDoF systems. According to Equation (3.50) we have: qn ( t ) = e

– ζω n t

[ A n cos ( ω nd t ) + B n sin ( ω nd t ) ]

(9.84)

where: 9 Free Vibrations

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2

ω nd = ω n 1 – ζ “damped circular frequency of the nth mode”

Blank Page

(9.85) The the total deformation u ( t ) under free vibration is hence N

u(t) =

¦

N

φi qi ( t ) =

i=1

¦

φi e

– ζω n t

[ A i cos ( ω id t ) + B i sin ( ω id t ) ]

(9.86)

i=1

As in the case of the undamped systems, the 2 ⋅ N constants A i and B i can be computed by means of the initial conditions u ( 0 ) = u 0 and u· ( 0 ) = v 0 . For the nth decoupled SDoF system according to Equation (3.51) the constants A n and B n can be expressed in function of the initial conditions of the modal coordinate q as follows: An = qn ( 0 )

(9.87)

q· n ( 0 ) + ζω n q n ( 0 ) B n = -------------------------------------------ω nd

(9.88)

where the initial displacement q n ( 0 ) and the initial velocity q· n ( 0 ) can be computed by means of Equations (9.77) and (9.78). Hence, the total displacement of a classically damped MDoF system under free vibration can be computed as: N

u(t) =

¦ i=1

φi e

– ζω n t

q· i ( 0 ) + ζω i q i ( 0 ) q i ( 0 ) cos ( ω id t ) + ----------------------------------------- sin ( ω id t ) ω id

(9.89)

For nonclassically damped system see e.g. [Cho11], Chapter 14. 9 Free Vibrations

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9 Free Vibrations

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10 Damping


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10.2 Example

10.1 Free vibrations with damping The differential equation to compute the free vibrations of a MDoF system is: ·· + Cu· + Ku = 0 Mu

(10.1)

with the initial conditions: u ( 0 ) = u0

and

u· ( 0 ) = v 0

(10.2)

The displacement vector u ( t ) may be expressed as linear combination of the eigenvectors, i.e. u ( t ) = Φ q ( t ) , and Equation (10.1) becomes: ·· + CΦq· + KΦq = 0 MΦq

(10.3)

Equation (10.3) may be further multiplied by Φ T yielding the following equations: T ·· + Φ T CΦq· + Φ T KΦq = 0 Φ MΦq

(10.4)

·· + C * q· + K * q = 0 M*q

(10.5)

Definition:

The properties of the 2-DoF system are: m 1 = 2m

,

m2 = m

(10.6)

k 1 = 2k

,

k2 = k

(10.7)

while the damping characteristics will be defined later. • Natural frequencies and eigenvectors The natural frequencies and the eigenvectors of the 2-DoF system can be easily computed as: k- , ------2m

Natural frequencies:

ω1 =

Eigenvectors:

φ1 = 1 ⁄ 2 , 1

• A system is classically damped if the matrix C * is diagonal

ω2 =

2k -----m

φ2 = –1 1

(10.8) (10.9)

• A system is non-classically damped if the matrix C * is not diagonal

10 Damping

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10 Damping

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10.2.1 Non-classical damping

10.2.2 Classical damping

The damping characteristics of the 2-DoF system are chosen as:

The damping characteristics of the 2-DoF system are chosen as:

,

c1 = c

(10.10)

c 2 = 4c

The equation of motion of the system can be easily assembled by means of the equilibrium formulation: u u·· u· m 2 0 1 + c 5 –4 1 + 3 –1 1 = 0 –1 1 u2 0 1 u··2 – 4 4 u· 2 0

(10.11)

It is now attempted to decouple the equations by computing the modal properties of the 2-DoF system:

M*

1--1--3--m 0 1 –1 2 2m 0 2 = Φ MΦ = ⋅ ⋅ = 2 0 m 1 1 0 3m –1 1 T

(10.12)

,

c 1 = 4c

The equation of motion of the system can be easily assembled by means of the equilibrium formulation: u·· m 2 0 1 + c 6 –2 0 1 u··2 –2 2

3--1--1--k 0 1 –1 = 4 ⋅ 3k – k ⋅ 2 K * = Φ KΦ = 2 –k k 0 6k –1 1 1 1 1--1--1 –1 2 5c – 4c = C * = Φ CΦ = ⋅ ⋅ 2 – 4c 4c –1 1 1 1 T

5--c 4 7--c 2

(10.13)

7--c 2

(10.14)

u u· 1 + 3 –1 1 = 0 ·u –1 1 u2 0 2

(10.16)

It is now attempted to decouple the equations by computing the modal properties of the 2-DoF system: 3--m 0 M* = 2 0 3m

T

(10.15)

c 2 = 2c

,

3k ------ 0 K* = 4 0 6k

1--1--1 –1 = ⋅ 6c – 2c ⋅ 2 C * = Φ CΦ = 2 – 2c 2c –1 1 1 1 T

(10.17)

3--c 2

0

(10.18)

0 12c

The Matrix C * is diagonal, hence it is possible to decouple the equations!

17c

The Matrix C * is not diagonal, hence it is not possible to decouple the equations! 10 Damping

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10 Damping

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10.3 Classical damping matrices

(10.25)

C = a0 M + a1 K

(10.19)

The damping constant of each mode of vibration is therefore: 2

The damping constant of each mode of vibration is therefore: c n* = a 0 m n*

(10.20)

and the corresponding damping ratio ζ n becomes (see Section 3.2): c n*

a 0 m n*

a0 ζ n = ----------------*- = ----------------*- = --------2ω 2ω n m n 2ω n m n n

(10.21)

(10.22)

The damping constant of each mode of vibration is therefore: 2

c n* = a 1 k n* = a 1 ω n m n*

(10.23)

and the corresponding damping ratio ζ n becomes:

(10.26)

and using the results for MpD and SpD damping ratio ζ n becomes: a0 a1 ζ n = --------- + ----- ω n 2ω n 2

(10.27)

a0 1 a1 ° ----- ⋅ ----- + ----- ⋅ ω i = ζ i ° 2 ωi 2 ® a0 1 a1 ° -------- + ----- ⋅ ω j = ζ j ° 2- ⋅ ω 2 j ¯

(10.28)

In the case that ζ i = ζ j = ζ , coefficients a 0 and a 1 can be computed as follows:

2

c n* a 1 ω n m n* a1 ζ n = ----------------*- = ------------------= ----- ω n * 2 2ω n m n 2ω n m n

c n* = a 0 m n* + a 1 k n* = ( a 0 + a 1 ω n )m n*

The coefficients a 0 and a 1 may be computed for vibration modes i and j by means of equation (10.28):

10.3.2 Stiffness proportional damping (SpD) C = a1 K

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10.3.3 Rayleigh damping

10.3.1 Mass proportional damping (MpD) C = a0 M


(10.24)

Remark

2ω i ω j a 0 = ζ ⋅ ----------------ωi + ωj

2 a 1 = ζ ⋅ ----------------ωi + ωj

(10.29)

Both MpD and SpD, taken alone, are not a good approximation of the behaviour of real structures. Studies have shown that different modes of vibration exhibit similar damping ratios. 10 Damping

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10 Damping

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If we choose values for m , k and ζ so that

10.3.4 Example A damping matrix shall be assembled so that in the case of the 2-DoF system shown on page 10-2 both modes of vibration are characterised by the same damping ratio ζ .

ω 1 = 2rad ⁄ s , ω 2 = 5rad ⁄ s , ζ = 5%

k -------- , 2m

a 0 = 14.287 , ω2 =

2k -----m

(10.30)

(10.34)

the coefficients a 0 and a 1 become:

The natural frequencies are: ω1 =

An-Najah 2013

(10.35)

a 1 = 1.429

and the representation of the damping ratio in function of the natural circular frequency is: 10

hence the coefficients a 0 and a 1 become:

Mass proportional damping

9

4ζ m a 1 = ------ ⋅ -----3 2k

(10.31)

yielding the damping matrix C = a 0 M + a 1 K equal to: k--⋅ 2m + 3k 0–k 4ζ m C = ------ ⋅ ------ ⋅ m 3 2k k--0–k ⋅m+k m

4ζ mk = ------ ⋅ -------- ⋅ 5 – 1 3 2 –1 2

(10.32)

8

Rayleigh Damping

7 6 5 4 3 2 1 0 0

Check:

1

2

3

4

5

6

7

8

9

10

Circular frequency ω T

C * = Φ CΦ =

1 1 --- 1 4ζ --- – 1 mk 5 – 1 mk ----⋅ ⋅ ------⋅ ⋅ = ζ ⋅ -------- ⋅ 3 0 2 2 3 2 2 –1 2 0 12 –1 1 1 1

(10.33) The damping matrix is indeed diagonal.

10 Damping

Dam mping ratio ζn [%]

Stiffness proportional damping

4ζ m k a 0 = ------ ⋅ ------ ⋅ ---- , 3 2k m

Page 10-7

Remarks - If there are more than 2 modes of vibrations, then not all of them will have the same damping ratio. - If more than 2 modes of vibrations should have the same damping, then a different damping modal shall be used. To this purpose see e.g. “Caughey-Damping” in [Cho11]. 10 Damping

Page 10-8

CCourse “Fundamentals of Structural Dynamics”

An-Najah 2013

11 Forced Vibrations 11.1 Forced vibrations without damping


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T ·· + Φ T KΦq = Φ T F ( t ) Φ MΦq

(11.4)

·· + K * q = F * ( t ) M*q

(11.5)

Where 11.1.1 Introduction

- M * : Diagonal matrix of the modal masses m n* - K * : Diagonal matrix of the modal stiffnesses k n* - F * : Vector of the modal forces F n* For the considered 2-DoF system, Equation (11.5) can be rearranged as:

Sought is the response of the 2-DoF system as a result of the external excitation force F ( t ) given by Equation (11.1) F(t) =

F1 ( t )

(11.1)

F2 ( t )

The equation of motion of the system is: ··( t ) + Ku ( t ) = F ( t ) Mu

(11.2)

The displacement vector u ( t ) can be represented as a linear combination of the eigenvectors of the 2-DoF system, u ( t ) = Φ q ( t ) , and Equation (11.2) becomes: ·· + KΦq = F ( t ) MΦq

(11.3)

* ·· * * m1 q1 + k1 q1 = F1 ® * ·· ¯ m 2 q 2 + k 2* q 2 = F 2*

(11.6)

or as alternative: F 1* ·· 2 ° q 1 + ω 1 q 1 = ------*m1 ° ® F 2* ° ·· 2 -----q + ω q = ° 2 2 2 m 2* ¯

(11.7)

The two equations of the system (11.7) are decoupled and can be solved independently. The constants resulting from the solution of the system can be determined by means of the initial conditions u ( 0 ) = u 0 and u· ( 0 ) = v 0 .

T

We can now multiply Equation (11.3) by Φ obtaining:

11 Forced Vibrations

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11.1.2 Example 1: 2-DoF system


F 0 sin ( ωt ) F 0 sin ( ωt ) ° q··1 + ω 21 q 1 = -------------------------- = ------------------------ = f 1 sin ( ωt ) 2 ⋅ ( 3 ⁄ 2 )m 3m ° ® – F 0 sin ( ωt ) – F 0 sin ( ωt ) ° ·· 2 - = --------------------------- = f 2 sin ( ωt ) ° q 2 + ω 2 q 2 = --------------------------3m 3m ¯

An-Najah 2013

(11.13)

with F0 f 1 = -------3m

The properties of the 2-DoF system are: m 1 = 2m

,

m2 = m

(11.8)

k 1 = 2k

,

k2 = k

(11.9)

c1 = 0

,

c2 = 0

(11.10)

The external excitation is: F(t) =

F 0 sin ( ωt )

(11.11)

0

and the modal excitation force is calculated using the modal matrix:

T

F*( t ) = Φ F( t ) =

1 --- 1 2

F 0 sin ( ωt )

–1 1

0

=

F 0 sin ( ωt ) -----------------------2 – F 0 sin ( ωt )

and

F0 f 2 = – -------3m

(11.14)

Each equation of the system (11.13) corresponds to the equation of motion of an undamped SDoF system under an harmonic sine excitation. The complete solution of these differential equations has been discussed in Chapter 4 and it is: fn - sin ( ωt ) q n = A 1 cos ( ω n t ) + A 2 sin ( ω n t ) + -----------------2 2 ωn – ω

(11.15)

The two equations have the following solutions: f1 q = A cos ( ω t ) + A sin ( ω t ) + ------------------ sin ( ωt ) 1 1 2 1 ° 1 2 2 ω – ω ° 1 ® f2 ° - sin ( ωt ) ° q 2 = A 3 cos ( ω 2 t ) + A 4 sin ( ω 2 t ) + -----------------2 2 ω2 – ω ¯

(11.16)

(11.12)

The 4 constants A 1 to A 4 can be easily computed for the initial conditions u ( 0 ) = u· ( 0 ) = 0 by means of the mathematical software “Maple”. They are:

Page 11-3


The system of equations becomes:


Page 11-4


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(11.17)

A1 = 0


u(t) = Φq(t) =

¦ φn qn ( t )

= φ1 q1 ( t ) + φ2 q2 ( t )

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(11.23)

n

ω ⁄ ω1 ω ⁄ ω1 F0 - ⋅ -------- = – ------------------ ⋅ f1 A 2 = – -----------------2 2 2 2 ω 1 – ω 3m ω1 – ω

(11.18)

A3 = 0

(11.19)

ω ⁄ ω2 ω ⁄ ω2 F0 - ⋅ -------- = – ------------------ ⋅ f2 A 4 = -----------------2 3m 2 2 2 ω2 – ω ω2 – ω

(11.20)

The displacements q n becomes:

1 --u = 2 1

§ sin ( ωt ) – ( ω ⁄ ω 1 ) sin ( ω 1 t ) · - ¸+ ¨ f 1 ----------------------------------------------------------------2 2 © ¹ ω1 – ω

(11.24)

sin ( ωt ) – ( ω ⁄ ω 2 ) sin ( ω 1 t ) · – 1 § f ----------------------------------------------------------------- ¸ ¨ 2 2 2 ¹ 1 © ω2 – ω

where:

f1 § ω ⁄ ω1 · ° q = ¨ – ------------------ ⋅ f 1¸ sin ( ω 1 t ) + ------------------ sin ( ωt ) 1 2 2 2 2 ° © ω1 – ω ¹ ω – ω 1 ° ® f2 § ω ⁄ ω2 · ° - ⋅ f 2¸ sin ( ω 2 t ) + ------------------ sin ( ωt ) ° q 2 = ¨ – -----------------2 2 2 2 © ω2 – ω ¹ ° ω2 – ω ¯

(11.21)

°q = f 1 ° 1 ° ® ° ° q2 = f2 ° ¯

(11.22)

F0 F0 f 1 = -------- , f 2 = – -------- , ω 1 = 3m 3m

k -------- , ω 2 = 2m

2k -----m

(11.25)

or sin ( ωt ) – ( ω ⁄ ω 1 ) sin ( ω 1 t ) ----------------------------------------------------------------2 2 ω1 – ω sin ( ωt ) – ( ω ⁄ ω 2 ) sin ( ω 2 t ) ----------------------------------------------------------------2 2 ω2 – ω

Therefore, the total displacement u ( t ) becomes:


Page 11-5


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11.1.3 Example 2: RC beam with Tuned Mass Damper (TMD) without damping


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• RC beam with TMD

• RC Beam • Damping ratio ζ n = 0.0 • Modal mass M n = 5.626t • Modal stiffness K n = 886kN ⁄ m • Natural frequency f n = 2Hz

• Excitation

• TMD (In this case damping is neglected) • Damping ratio ζ T = 0.0 • Mass M T = 0.310t • Stiffness K T = 44kN ⁄ m • Natural frequency f T = 1.90Hz


Page 11-7

As excitation a vertical harmonic sine force acting only on the beam is assumed. F ( t ) = F o sin ( ωt )

with:

(11.26)

ω : excitation frequency F o : static excitation force: F o = 0.8kN

• Solution Both the transient and the steady-state part of the solution are considered. 11 Forced Vibrations

Page 11-8


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• Case 1: K T = 10000kN ⁄ m , excitation frequency f = 2Hz Displacement of the beam

0.03 0.02 0.01 0 -0.01

Amp plification factor A [-]

140

0.04

Disp placement u1(t) [m]

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Displacement of the beam (u1) Displacement of the TMD (u2)

120 100 80 60 40 20

-0.02 0

-0.03

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

1.3

1.4

1.5

Excitation frequency / Natural frequency of the beam [-]

-0.04 0

1

2

3

4

5

6

7

8

9 10 11 12 13 14 15 16 17 18 19 20

Time [s]

• Remarks - The amplification factor A is defined as:

0.04 Displacement of the TMD

A TMD = u 2 ⁄ u st


0.03 0.02

,

A Beam = u 1 ⁄ u st

where u st = F o ⁄ K n

0.01

- The solution was computed by means of the Excel file given on the web page of the course (SD_MDOF_TMD.xlsx)

0 -0.01

- The Tuned Mass Damper (TMD) is blocked

-0.02

- The natural frequency of the beam with TMD is: f n = 1.94Hz

-0.03 -0.04 0

1

2

3

4

5

6

7

8

9 10 11 12 13 14 15 16 17 18 19 20

Time [s]


Page 11-9

- At f = f n resonance occurs. In the diagram above the amplification factor is limited, because the response of the system was only calculated during 60 seconds. 11 Forced Vibrations

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• Case 2: K T = 44kN ⁄ m , excitation frequency f = 2Hz Displacement of the beam

0.006 0.004 0.002 0 -0.002

Amp plification factor A [-]

350

0.008

Disp placement u1(t) [m]

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Displacement of the beam (u1) Displacement of the TMD (u2)

300 250 200 150 100 50

-0.004 0

-0.006

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

1.3

1.4

1.5

Excitation frequency / Natural frequency of the beam [-]

-0.008 0

1

2

3

4

5

6

7

8

9 10 11 12 13 14 15 16 17 18 19 20

Time [s]

• Remarks - The amplification factor A is defined as:

0.04 Displacement of the TMD

A TMD = u 2 ⁄ u st


0.03 0.02

,

A Beam = u 1 ⁄ u st

where u st = F o ⁄ K n

0.01

- The solution was computed by means of the Excel file given on the web page of the course (SD_MDOF_TMD.xlsx)

0 -0.01

- The Tuned Mass Damper (TMD) is free to move

-0.02 -0.03 -0.04 0

1

2

3

4

5

6

7

8

9 10 11 12 13 14 15 16 17 18 19 20

Time [s]


Page 11-11

- No resonance at f = f n occurs. Resonance occurs in correspondence of the first and of the second natural frequencies of the 2-DoF system. In the diagram above the factor A is limited, because the response of the system was only calculated during 60s. 11 Forced Vibrations

Page 11-12


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11.2 Forced vibrations with damping


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Where: - M * : Diagonal matrix of the modal masses m n*

11.2.1 Introduction

- K * : Diagonal matrix of the modal stiffnesses k n* - F * : Vector of the modal forces F n* - C * : Matrix of the modal damping constants. It is diagonal only if the system is classically damped (see Chapter 10). For the considered classically damped 2-DoF system, Equation (11.31) can be rearranged as: Sought is the response of the 2-DoF system as a result of the external excitation force F ( t ) given by Equation (11.27) F(t) =

F1 ( t ) F2 ( t )

(11.27)

The equation of motion of the system is: ··( t ) + Cu· ( t ) + Ku ( t ) = F ( t ) Mu

(11.28)

The displacement vector u ( t ) can be represented as a linear combination of the eigenvectors of the 2-DoF system, u ( t ) = Φ q ( t ) , and Equation (11.28) becomes: ·· + CΦq· + KΦq = F ( t ) MΦq

(11.29)

T We can now multiply Equation (11.29) by Φ obtaining: T ·· + Φ T CΦq· + Φ T KΦq = Φ T F ( t ) Φ MΦq

(11.30)

·· + C * q· + K * q = F * ( t ) M*q

(11.31)


Page 11-13

* ·· *· * * m1 q1 + c1 q1 + k1 q1 = F1 ® * ·· ¯ m 2 q 2 + c 2* q· 2 + k 2* q 2 = F 2*

(11.32)

or as alternative: F 1* ·· 2 · -----q + 2ζ ω q + ω q = ° 1 1 1 1 1 1 m 1* ° ® F 2* ° ·· 2 · -----q + 2ζ ω q + ω q = ° 2 2 2 2 2 2 m 2* ¯

(11.33)

The two equations of the system (11.33) are decoupled and can be solved independently. The constants resulting from the solution of the system can be determined by means of the initial conditions u ( 0 ) = u 0 and u· ( 0 ) = v 0 .


Page 11-14


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11.3 Modal analysis: A summary Blank Page The dynamic response of a Multi-Degree of Freedom (MDoF) system due to an external force F ( t ) can be computed by means of modal analysis. The required steps are: 1) Compute the properties of the MDoF system - Compute the mass matrix M and the stiffness matrix K . - Estimate the modal damping ratios ζ n* 2) Compute the natural circular frequencies ω n and the eigenvectors φ n - Compute the modal properties of the MDoF system ( M * , K* ) 3) Compute the response of every mode of vibration - Set up the equation of motion of the modal SDoF systems F n* q··n + 2ζ n* ω n q· n + ω n2 q n = ------*- and solve it for q n mn - Compute the modal displacements u n ( t ) = φ n q n - Compute the sectional forces by means of the static equivalent forces F n ( t ) = Ku n ( t ) = Kφ n q n = ω n2 Mφ n q n ( t ) 4) Sum up (respectively combine) the contribution from all modes of vibration to obtain the total response of the system.


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The system of equations governing the motion of the system is

12.1 Equation of motion 12.1.1 Introduction In analogy to Section 2.1.1, the equation of motion of the system depicted here can be formulated by means of the d’Alembert principle F + T = 0 applied to each one of the masses.

F1 + T1 = 0 ® ¯ F2 + T2 = 0

(12.1)

·· ·· · · – m 1 ( x + u 1 ) – ( c 1 + c 2 )u 1 + c 2 u 2 – ( k 1 + k 2 )u 1 + k 2 u 2 = 0 ® – m 2 ( x·· + u··2 ) + c 2 u· 1 – c 2 u· 2 + k 2 u 1 – k 2 u 2 = 0 ¯

(12.2)

and in matricial form:

y 1 = x ( t ) + l 1 + u s1 + u 1 ( t ) –

y··1 = x··( t ) + u··1 ( t )

m1 0 0 m2

T 1 = – m 1 y··1 = – m 1 ( x·· + u··1 )

x·· + u··1 c 1 + c 2 – c 2 u· 1 k 1 + k 2 – k 2 u 1 – – = 0 ·· ·· · x + u2 –c2 c2 u2 –k2 k2 u2 0

(12.3)

x·· + u··1 c + c 2 – c 2 u· 1 k + k2 –k2 u1 + 1 + 1 = 0 ·· ·· · x + u2 –c2 c2 u2 –k2 k2 u2 0

(12.4)

or:

F 1 = – k 1 ( u s1 + u 1 ) – c 1 u· 1 + m 1 g + k 2 ( u s2 + u 2 – u 1 ) + c 2 ( u· 2 – u· 1 )

m1 0 0 m2

F 1 = – ( k 1 + k 2 )u 1 + k 2 u 2 – ( c 1 + c 2 )u· 1 + c 2 u· 2

or:

y 2 = x ( t ) + l 1 + u s1 + l 2 + u s2 + u 2 ( t ) m1 0

y··2 = x··( t ) + u··2 ( t )

0 m2

m 0 u··1 c + c 2 – c 2 u· 1 k + k2 –k2 u1 + 1 + 1 = – 1 0 m2 u··2 – c 2 c 2 u· 2 –k2 k2 u2

T 2 = – m 2 y··2 = – m 2 ( x·· + u··2 )

x·· x··

(12.5)

F 2 = – k 2 ( u s2 + u 2 – u 1 ) – c 2 ( u· 2 – u· 1 ) + m 1 g F 2 = k 2 u 1 – k 2 u 2 + c 2 u· 1 – c 2 u· 2


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Page 12-1

which is similar to Equation (8.3) meaning that the base point excitation x ( t ) can be considered equivalent to two external forces f 1 ( t ) = m 1 x··( t ) and f 2 ( t ) = m 2 x··( t ) acting on the masses m 1 and m 2 . This is the same interpretation given in Section 2.1.1 for SDoF systems. 12 Seismic Excitation

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12.1.2 Synchronous Ground motion

Influence vector for some typical cases

As shown in the previous section, the equation of motion of a system subjected to a base excitation is:

• Planar system with translational ground motion (Case 1)

·· + Cu· + Ku = 0 Mu a

In this case all DoFs of the system undergo static displacements u s ( t ) which are equal to the ground displacement u g ( t ) , hence:

(12.6)

·· is vector of the absolute accelerations of the DoFs of where u a the system while u· and u are the vectors of the relative velocities and of the relative displacements of the DoFs of the system, respectively.

1 ι = 1 = 1 … 1

The absolute displacement u a of the system can be expressed as: ua = us + u

where 1 is a vector of order N , i.e. the number of DoFs, with all elements equal to 1.

(12.7)

where u s is displacement of the DoFs due to the static application (i.e. very slow so that no inertia and damping forces are generated) of the ground motion, and u is again the vector of the relative displacements of the DoFs of the system.

• Planar system with translational ground motion (Case 2) The axial flexibility of the elements of the depicted system can be neglected, hence 3 DoFs are defined. In this case DoFs 1 and 2 undergo static displacements which are equal to the ground displacement, while the static displacement of DoF 3 is equal to 0, i.e.:

The “static displacements” u s ( t ) can now be expressed in function of the ground displacement u g ( t ) as follows: us ( t ) = ι ug ( t )

(12.8)

where ι is the so-called influence vector. Equation (12.6) can now be rewritten as: ··) + Cu· + Ku = 0 M ( ιu··g + u

(12.9)

·· + Cu· + Ku = – Mιu·· ( t ) Mu g

(12.10)


Page 12-3

(12.11)

1 ι = 1 0 12 Seismic Excitation

(12.12)

Page 12-4


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• Planar system with rotational ground motion.


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• Spatial system with multiple translational ground motion Consider the spatial frame depicted here:

U6

U5

U4

U3 U2

U1

The depicted system is subjected to a rotational ground motion θ g which generates the following static displacements of the DoFs: h1

h1

u s ( t ) = h θ g ( t ) hence ι = h 2 2 L

(12.13)

L

Excitation in the x-direction ugx(t)

Excitation in the y-direction ugy(t)

Remark If the planar system with rotational ground motion has more than one support and every support is subjected only to the base rotation θ g , then the static application of the base rotations typically create stresses within the system. Such a case must be considered like a multiple support excitation (see Section 12.1.3).


Page 12-5

Picture from: Chaudat T., Pilakoutas K., Papastergiou P., Ciupala M. A. (2006) “Shaking Table Tests on Reinforced Concrete Retrofitted Frame With Carbon Fibre Reinforced Polymers (CFRP),” Proceedings of the First European Conference on Earthquake Engineering and Seismology, Geneva, Switzerland, 3-8 September 2006

The equation of motion of the frame structure for the ground motions u gx ( t ) and u gy ( t ) neglecting damping is: 12 Seismic Excitation

Page 12-6


u··1 u··

u2

u··3 u + K 3 = – Mι x u··gx ( t ) – Mι y u··gy ( t ) ·· u4 u4 u·· u 5

5

u··6

u6

u··1 u··

u1 u2

2

u·· u M 3 +K 3 u··4 u4 u·· u 5

5

u··6

u6

§ ¨ ¨ ¨ = –M ¨ ¨ ¨ ¨ ©

0 1 0 u·· ( t ) + gx 0 1 0

(12.14)

· 1 ¸ 0 ¸ 0 u·· ( t )¸ gy ¸ ¸ 1 ¸ 0 ¸ ¹ 0

(12.15)

u··1 u··

m1 m2

0

2

I3 m4 0

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m5 I6

Structures with a significative spatial extension may be subjected to ground motion time-histories that are different from support to support. A typical example for such structures is the bridge shown in the following figure.

Example of structure where often multiple support excitation is applied: Plan view of the dynamic model for the seismic analysis of a bridge in the transverse direction. The springs represent the piers.

In this case it is distinguished between the DoFs of the structure u a , which are free to move and whose displacements are expressed in absolute coordinates, and those of the ground u g , which undergo the displacements imposed to the support. The vector containing the displacements of all DoFs is hence:

and with

M =


12.1.3 Multiple support ground motion

u1

2

M

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u1 u2

– m 1 u··gy – m u·· 2 gx

u·· u 0 we obtain M 3 + K 3 = ·· u4 u4 – m 4 u··gy u··5 u5 – m 5 u··gx 0 u·· u 6

u =

ua

(12.17)

ug The equation of motion of the system can hence be expressed as (see [Cho11]):

6

(12.16) Remarks

m mg T

m g m gg

·· c c g u· k kg u u 0 a a a (12.18) = + + T T ·· · p ( t ) u ug u g c g c gg g k g k gg g

• For other cases see [Cho11] Sections 9.4 to 9.6. 12 Seismic Excitation

Page 12-7


Page 12-8


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Where p g ( t ) are the forces resulting at the supports when the supports undergo the displacements u g ( t ) . In Equation (12.18) the different matrices … g , … gg are not computed separately, they just result form the partition of the overall system of equations when the DoFs of the structure and of the ground are collected as it is shown in the example of page 12-12. The vector of Equation (12.17) can be rewritten as: u =

ua

=

ug

us ug

+ u 0

(12.19)

where u s is the vector of the displacements of the DoFs of the structure when the ground displacements u g ( t ) are applied statically, and u is the vector of the relative displacements of the DoFs of the structure. The relationship between u s and u g ( t ) is given by the following system of equation: k kg

us

T k g k gg u g

=

0 p g, s


m mg T

m g m gg

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·· + u ·· c c g u· s + u· k kg us + u u 0 s + + = T T ·· · p ug ug ug g(t) c g c gg k g k gg

(12.21) The first line of the system can be rearranged to: ·· + cu· + ku = – ( mu ·· + m u ·· · · mu s g g ) – ( cu s + c g u g ) – ( ku s + k g u g ) (12.22)

According to the first line of Equation (12.20) ku s + k g u g = 0 and hence Equation (12.22) becomes: ·· + cu· + ku = – ( mu ·· + m u ·· · · mu s g g ) – ( cu s + c g u g )

(12.23)

If we now express the vector u s in function of the vector u g as u s = ιu g

(12.24)

the so-called influence matrix ι can be computed, again making use of the first line of Equation (12.20), i.e.: – k g u g = ku s = kιu g

(12.25)

and after rearranging we obtain: (12.20) –1

ι = –k kg

(12.26)

where p g, s is the vector of the support forces needed to impose the displacements u g statically. If the system is statically determinated, p g, s is equal to zero (See example of page 12-12).

The influence matrix ι is a N × N g matrix where N is the number of DoFs of the structure and N g is the number of DoFs of the supports.

By introducing Equation (12.19) into Equation (12.18) we obtain the new system of equations:

By introducing Equation (12.26) into Equation (12.23) the final equation of motion of the system is obtained:



Page 12-9

Page 12-10


·· + cu· + ku = – ( mι + m )u ·· · mu g g – ( cι + c g )u g

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(12.27)

Analogously, with the second line of Equation (12.21), an equation for the computation of the forces at the supports p g ( t ) can be setup and solved.


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Example: 2-DoF system The following 2-DoF system is subjected to multiple support ground motion. Two different ground motions are applied to the degrees of freedom u 1 and u 5 . Sought is the equation of motion of the system:

Remarks In Equation (12.27), the masses associated with the support are often equal to zero, i.e. m g = 0 . If this is the case, Equation (12.27) simplifies to: ·· + cu· + ku = – mιu ·· – ( cι + c )u· mu g g g

(12.28)

And considering that in most cases the damping forces on the LHS of the equation are small (and they are zero if no damping is present) compared to the inertia forces (see [Cho11]), Equation (12.28) can be further simplified to: ·· + cu· + ku = – mιu ·· mu g

u =

ug =

: displacements of the structure

u1

(12.31)

(12.32)

: displacements of the supports (excited, massless)

u5

(12.30)

(12.33)

and with ι = ι1 Equation (12.29) becomes ·· + cu· + ku = – mιu·· mu g

u3 u4

(12.29)

In the case that the movement of the supports is the same at all supports, u g becomes: u g = 1u g

The stiffness matrix of the system is assembled by means of the Direct Stiffness Method and the following degrees of freedom are considered:

u0 =

u2

: displacements of the supports (not excited, massless)

u6

(12.34)

which is the same as Equation (12.10). The stiffness matrix K of the system is:


Page 12-11


Page 12-12


12 6L 6L 4L

2

– 12

6L

– 6L

2L

2

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0

0

u1

0

0

u2

0

6L

2L

2

– 6L 4L

2

24 0

2

6L 4L – 6L 2L

2

0

0

u1

0

0

u2

u EI – 12 – 6L 24 0 – 12 6L ,U = 3 K = -----3- ⋅ 2 2 2 u4 L 6L 2L 0 8L – 6L 2L u5 0 0 – 12 – 6L 12 – 6L 0

2

6L 2L – 6L 4L

0

2

0 8L

u6

6L 2L

24

(12.36)

0

0 8L

u6

– 6L 2L

– 6L 4L

6L 2L

– 12 6L 2

0

u3

u 0 EI – 12 6L 12 6L 0 ,U = 1 K = -----3- ⋅ 2 2 2 u4 L 0 2L 6L 8L – 6L 2L u5 – 12 0 0 – 6L 12 – 6L 6L


0

2

0 2L – 6L 4L

2

0

0 0

0

0

0

u4

,U =

u1

0

12 – 6L

u5

2

u6

– 6L 4L

(12.38)

u2

– 12 – 12 – 6L 6L 2

2

6L – 6L 2L 2L

2 2

6L

u3

2

u4 u1

0

6L

0

12

0

– 6L

u5

0

4L

0

u2

0 – 6L

0

2

4L

,U =

2

(12.39)

u6

By means of static condensation we can now eliminate DoF 2 and 6. We racall that:

u2

0

2

EI – 12 6L 12 K = -----3- ⋅ – 12 – 6L 0 L

By swapping DoF 1 and 3 we obtain: 0

6L 2L – 6L 2L

u3

2

By swapping DoF 2 and 5 we obtain:

6L 2L

2

2

– 12 – 6L 0

with L = L ⁄ 2

24 – 6L – 12

– 12 – 6L – 12 6L 2

EI – 12 6L 12 6L K = -----3- ⋅ 2 2 L – 6L 2L 6L 4L

(12.35) 12 6L – 12 6L

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By swapping DoF 2 and 4 we obtain:

u EI – 12 – 6L 12 + 12 – 6L + 6L – 12 6L K = -----3- ⋅ ,U = 3 2 2 2 2 u4 L 6L 2L – 6L + 6L 4L + 4L – 6L 2L u5 0 0 – 12 – 6L 12 – 6L 0


(12.37)

EI k k EI T –1 ˆ K = -----3- ⋅ tt t0 , k tt = -----3- ⋅ ( k tt – k 0t k 00 k 0t ) k 0t k 00 L L

(12.40)

u6

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By performig the needed calculations we obtain:

–1 k 00

2

1 1 ˆ = ----------- ⋅ k 00 = -----------4- ⋅ 4L 0 k 00 2 16L 0 4L

1 --------- 0 2 = 4L 1 0 --------24L

T

–1

k 0t k 00 k 0t =

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The equation of motion of the system becomes:

(12.41)

M 0 0 0

0 I 0 0

0 0 0 0

u··3 0 ·· EI0 ⋅ u 4 + ----⋅ ·· u1 L3 0 0 u··5

6

0

–3 –3 2

u3

0 6L 3L – 3L ⋅ u 4 u1 – 3 3L 3 0 u5 – 3 – 3L 0 3

0 0 = p1 ( t )

(12.46)

p5 ( t )

We racall that:

– 6L 6L

1 --------- 0 2 2 2 2 2L 2L ⋅ 4L ⋅ – 6L 2L 6L 0 2 1 6L 0 6L 2L 0 – 6L 0 --------24L 0 – 6L


(12.42)

EI k k g ˆ , ι = –k–1 kg k tt = -----3- ⋅ T L k g k gg

(12.47)

By performig the needed calculations we obtain: 18 0 T

–9 –9 2

–1

0 2L 3L 3L – 9 3L 9 0 – 9 3L 0 9

k 0t k 00 k 0t =

§ 24 0 ¨ EI ¨ 0 8L 2 ˆ k tt = -----3- ⋅ ¨ L ¨ – 12 6L ¨ © – 12 – 6L 6

0

(12.43) k

– 9 – 9 ·¸ 6L – 6L – 0 2L 2 3L 3L ¸¸ 12 0 – 9 3L 9 0 ¸ ¸ – 9 3L 0 9 ¹ 0 12

– 12 – 12

–1

1 --- 0 6 1 ˆ L = ------ ⋅ k = ------ ⋅ EI k 1 0 --------26L 3

(12.48)

18 0

(12.44)

1--1 0 --6 EI L –1 – 3 – 3 2 ⋅ -----3- ⋅ = ι = – k k g = – ------ ⋅ EI 1 1 3L 3L 0 --------2- L – ------2L 6L 3

1 --2 1 ------2L

(12.49)

–3 –3 2

EI ˆ k tt = -----3- ⋅ 0 6L 3L – 3L L – 3 3L 3 0 – 3 – 3L 0


(12.45)

3

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Page 12-16


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M ----u·· 2 ⋅ g1 Iu··g5 -----2L

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12.2 Time-history of the response of elastic systems

The vector of the effective forces becomes: M ----2 ·· p eff ( t ) = – ( mι + m g )u g ( t ) = – I – ------2L


(12.50)

As discussed in the previous sections, the equation of motion of a MDoF system under base excitation is: ·· + Cu· + Ku = – Mιu·· ( t ) Mu g

And the equation of motion of the system finally becomes:

(12.52)

Where: M:

·· u EI- 6 0 M 0 ⋅ u 3 + ----⋅ ⋅ 3 3 2 ·· u4 L u4 0 I 0 6L

M ----2 = – I – ------2L

M ----u·· 2 ⋅ g1 I u··g5 ------2L

Mass matrix (symmetric and positive-definite, or diagonal if only lumped masses are present) Stiffness matrix (symmetric and positive-definite) K: Damping matrix (Classical damping: C is typically a C: linear combination of M and K ) ·· u g ( t ) : Ground acceleration ι: Influence vector of order N . In the simplest case of a planar system under translational ground motion ι = 1 .

(12.51)

The following drawings show the interpretation of the elements of the influence matrix ι : if u g1 = 1 then: 1 1 u 3 = --- and u 4 = – ------2 2L

if u g5 = 1 then:

If the damping of the MDoF system is classical, Equation (12.52) can be written in the form of N decoupled modal equations, where N is the number of modes of the system. The modal equations are of the following form:

1 1 u 3 = --- and u 4 = ------2 2L

Remarks:

(12.53)

or:

• See Section 9.7 of [Cho11] for an example with a statically indeterminated system. • In Finite Element analysis, when applying multiple support excitation, support displacements instead of support acceleration are often used. For more details see [Bat96]. 12 Seismic Excitation

T m n* q··n + c n* q· n + k n* q n = – φ n Mιu··g

Page 12-17

T

φ n Mι 2 - u··g q··n + 2ζ n* ω n q· n + ω n q n = – ----------------T φ n Mφ n

(12.54)

The dynamic response of the MDoF system can be written as:


Page 12-18


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N

¦ φn qn ( t )

u(t) =

(12.55)

n=1

nth eigenvector of the MDoF system φn : q n ( t ) : nth modal coordinate of the MDoF system

T

k n* = ωn :

T φn

¦

⋅ K ⋅ φn =

⋅ m n*

(12.57)

nth modal circular frequency of the MDOF system

The modal participation factor Γ n is a measure for the contribution of the n-th mode to the total response of the system. It is defined as follows: T

φ n Mι Γ n = ----------------T φ n Mφ n

(12.58)

In addition the so-called effective modal mass of the nth mode is defined as: 2

m n*, eff = Γ n ⋅ m n*

N

m n*, eff =

n=1

¦ mn

(12.60)

= m tot

n=1

where m tot is the total mass of the dynamic system. *

The effective modal height h n of the nth mode is:

(12.56) 2 ωn

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Unlike the modal mass m n* and the modal participation factor Γ n , the effective modal mass m n*, eff is independent of the normalization of the eigenvectors. The following equation holds: N

Further variables in Equation (12.53) are the modal mass m n* and the modal stiffness k n* of the nth mode. These parameters are defined as follows: m n* = φ n ⋅ M ⋅ φ n


θ

* hn

Ln θ = ------ with L n = Ln

N

¦ hj ⋅ mj ⋅ φjn and Ln

T

= φ n ⋅ Mι (12.61)

j=1

• Significance of the effective modal mass m n*, eff The effective modal mass m n*, eff is the lumped mass of a singlestorey substitute system which is subjected to a base shear force V bn equal to the nth modal base shear force of a multi-storey system. If in addition the height of the single storey substitute system with * the lumped mass m n*, eff equals the modal height h n , the singlestorey system is subjected to a base moment M bn which is equal to the nth modal base moment of the multi-storey system. The following holds:

(12.59) N

V bn =

m n*, eff

⋅ S pa, n =

¦ fjn

(12.62)

j=1


Page 12-19


Page 12-20

¦ fjn ⋅ hj

(12.63)

MDOF System

j=1

where S pa, n is the pseudo-acceleration of the nth mode.

Eigenmodes and equivalent static forces

m3

f31

m2

If the internal forces of the entire system are to be determined, the modal equivalent static forces f jn should be computed first:

h

hs

(12.64)

f21

hs

EI

h2

(12.65)

(12.66)

= Mι

hs m1 hs

EI Mb

Page 12-21

m1*=2.180m

m2

Vb


Mb3

m1*Sa1

hs

(12.67)

n=1

Vb3 Mb2

Set of equivalent SDOF systems and equivalent static forces


m2*=0.646m

h1*=2.50hs

¦ sn

f13

Vb2 Mb1

m3

h

N

f12

Vb1

MDOF System

s n is independent on the normalization of the eigenvector φ n and we have that:

f23

h1

Mb

The excitation vector s n is defined according to equation (12.66) and specifies the distribution of the inertia forces due to excitation of the nth mode:

f22

f11

Vb

s n = Γ n Mφ n

f33

h3 m1

where f n = f 1n f 2n … f nn

f32

hs

• Distribution of the internal forces

f n = s n ⋅ S pa, n

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• MDoF system with eigenmodes and equivalent SDoF systems

N

=


EI1* Vb1

Mb1

m2*Sa2 EI2* Vb2

m3*=0.174m h3*=0.48hs

M bn = m n*, eff ⋅

* S pa, n ⋅ h n

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h2*=0.72hs


Mb2

m3*Sa3 EI3* Vb3

Mb3

Page 12-22


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12.3 Response spectrum method


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• “Absolute Sum (ABSSUM)” Combination Rule N

12.3.1 Definition and characteristics

u i, max ≤

If the maximum response only and not the response to the entire time history according to Equation (12.55) is of interest, the response spectrum method can be applied. The response spectrum can be computed for the considered seismic excitation and the maximum value of the modal coordinate q n, max can be determined as follows: q n, max =

Γ n ⋅ S d (ω n,ζ n*)

1 = Γ n ⋅ -----2- ⋅ S pa (ω n,ζ n*) ωn

(12.68)

Γn : modal participation factor of the n-th mode S d (ω n,ζ n*) : Spectral displacement for the circular eigenfrequency ω n and the modal damping rate ζ n* . S pa (ω n,ζ n*) : Spectral pseudo-acceleration for the circular eigenfrequency ω n and the modal damping rate ζ n* . The contribution of the

φ in ⋅ q n, max

n=1

The assumption that all maxima occur at the same instant and in the same direction yields an upper bound value for the response quantity. This assumption is commonly too conservative. • “Square-Root-of Sum-of-Squares (SRSS)” Combination Rule N

u i, max ≈

¦ ( φin ⋅ qn, max )

(12.71)

• “Complete Quadratic Combination (CQC)” Combination Rule

mode to the total displacement is: (12.69)

The maxima of different modes do not occur at the same instant. An exact computation of the total maximum response on the basis of the maximum modal responses is hence impossible. Different methods have been developed to estimate the total maximum response from the maximum modal responses. 12 Seismic Excitation

2

This rule is often used as the standard combination method and yields very good estimates of the total maximum response if the modes of the system are well separated. If the system has several modes with similar frequencies the SRSS rule might yield estimates which are significantly lower than the actual total maximum response.

N

u n, max = φ n ⋅ q n, max

(12.70)

n=1

where:

nth

¦

Page 12-23

u i, max ≈

N

(j)

(k)

¦ ¦ ui, max ⋅ ρjk ⋅ ui, max

(12.72)

j = 1k = 1

where (j)

(k)

u i, max and u i, max are the max. modal responses of modes j and k ρ jk is the correlation coefficient between nodes j and k : 12 Seismic Excitation

Page 12-24


ρ jk

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ω 8 ζ i ζ k ( ζ i + rζ k )r 3 ⁄ 2 - with r = -----k= --------------------------------------------------------------------------------------------------ωj ( 1 – r 2 ) 2 + 4ζ i ζ k r ( 1 + r 2 ) + 4 ( ζ i2 + ζ k2 )r 2 (12.73)

This method based on random vibration theory gives exact results if the excitation is represented by a white noise. If the frequencies of the modes are well spaced apart, the result converge to those of the SRSS rule. More detailed information on this and other combination rules can be found in [Cho11] Chapter 13.7. • Internal forces The aforementioned combination rules cannot only be applied on displacements but also on internal forces. The maximum modal internal forces can be determined from equivalent static forces F n, max = K ⋅ u n, max ,

(12.74)

which, as a first option, are computed from the equivalent static displacements. Alternatively, the equivalent static forces can be determined from the inertia forces: F n, max = s n ⋅ S pa (ω n,ζ n*) = Γ n Mφ n ⋅ S pa (ω n,ζ n*)


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• Number of modes to be considered. All modes which contribute to the dynamic response of the system should be considered. In practical applications, however, only those modes are considered which contribution to the total response is above a certain threshold. It should be noted that in order to achieve the same accuracy for different response measures (e.g. displacements, shear forces, bending moments, etc.) different numbers of modes might need to be considered in the computation. For a regular building the top displacement can be estimated fairly well on the basis of the fundamental mode only. To estimate the internal forces, however, higher modes need to be considered too. According to Eurocode 8 “Design of Structures for Earthquake Resistance” [CEN04] all modes should be considered (starting from the lowest) until the sum of the effective modal masses m n, eff of all considered modes corresponds to at least 90% of the total mass m tot . As an alternative, Eurocode 8 allows the designer to show that all modes with m n*, eff > 0.05m tot were considered in the computation.

(12.75)

with s n being the excitation vector which represents the distribution of the inertia forces of the nth mode (see Equation 12.67). Attention: It is wrong to compute the maximum internal forces from the maximum displacement of the total response u max . 12 Seismic Excitation

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12.3.2 Step-by-step procedure The maximum response of a N-storey building can be estimated according to the following procedure:


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response spectrum for pseudo-accelerations. (The spectral displacement S d (ω n,ζ n*) should be determined in the same manner) • Computation of the maximum displacements

u n, max = φ n ⋅ Γ n ⋅ S d (ω n,ζ n*)

1) Determine the properties of the MDOF system

• Computation of the maximum equivalent static forces

• Choose DOFs

F n, max = s n ⋅ S pa ( ω n, ζ n ) = Γ n Mφ n ⋅ S pa ( ω n, ζ n )

• Determine mass matrix M and stiffness matrix K . • Estimate modal damping ratios ζ n*

• Computation of the maximum internal forces on the basis of the forces F n, max

2) Carry out modal analysis of the MDOF system • Determine circular eigenfrequencies ω n and eigenvectors φ n 2

( K – ωn M ) ⋅ φn = 0 • Compute the modal properties of the MDOF system ( M * , K * ) T

4) Estimate the total response in terms of displacements and internal forces by means of suitable combination rules. Different combination rules might be applied (ABSSUM, SRSS, CQC). Comment

T

m n* = φ n Mφ n , k n* = φ n Kφ n

In order to consider the non-linear behaviour of the structure the equivalent lateral static forces F n, max can be determined from the spectral ordinate S pa ( ω n, ζ n, q ) of the design spectrum for pseudo-accelerations:

• Compute the modal participation factor Γ n T

φ n Mι Γ n = ----------------T φ n Mφ n

F n, max = s n ⋅ S pa ( ω n, ζ n, q ) = Γ n Mφ n ⋅ S pa ( ω n, ζ n, q ) (12.76)

3) The maximum response of the n-th mode should be determined as described in the following. This should be done for all modes n = 1, 2, …, N which require consideration. • For all periods T n and for the corresponding damping ratios ζ n* , the spectral response S a ( ω n, ζ n ) should be determined from the 12 Seismic Excitation

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12.4 Practical application of the response spectrum method to a 2-DoF system


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• Stiffness The horizontal stiffness of each story is k = 1 , hence the stiffness matrix K is:

12.4.1 Dynamic properties K =

k 11 k 12 k 21 k 22

=

1 –1 –1 2

(12.78)

1. unit displacement u 1 = 1

This 2-DoF system corresponds to the system presented in Chapter 9 with the only difference that the 2 DoFs are swapped.

2. unit displacement u 2 = 1

• Degrees of freedom (DoF) Horizontal displacements u 1 and u 2 in correspondence of the masses m 1 and m 2 • Masses Both story masses have unit value, i.e. m 1 = m 2 = 1 , hence the mass matrix M is: M =

m1 0 0 m2


= 1 0 0 1

(12.77)

Page 12-29

• Damping Damping is small and is neglected, hence the damping matrix C is: C = 0 0 0 0 12 Seismic Excitation

(12.79)

Page 12-30


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12.4.2 Free vibrations

2) Natural modes of vibration and natural frequencies

1) Matrix eigenvalue problem

For each eigenvalue ω a natural mode of vibration and a natural frequency can be computed.

2

(K – ω M) ⋅

φ1 φ2

2

(12.80)

=0 φ1

The nontrivial solution for the eigenvector φ = ≠ 0 exists if φ 2 the determinant is equal to zero: 2

det ( K – ω M ) = 0

(12.81)

1. circular natural frequency ω 1 =

3 – 5- = 0.62 --------------2

(12.86)

2

2

(12.82)

2

2

2

( K – ω1 M ) ⋅

This leads to the quadratic equation in ω : 2

2 3– 5 The smallest eigenvalue ω 1 = ---------------- leads to the 2

When the eigenvalue ω 1 is known, the system

2 –1 = 0 det ( K – ω M ) = det 1 – ω 2 –1 2 – ω

2

• Fundamental mode (first natural mode of vibration)

4

( 1 – ω ) ⋅ ( 2 – ω ) – ( – 1 ) ⋅ ( – 1 ) = 2 – 3ω + ω – 1 = 0 (12.83)

φ1 φ2

(12.87)

=0

φ can be solved for the corresponding vector 11 (fundamental φ 21 mode) to within a multiplicative constant:

or 4

2

ω – 3ω + 1 = 0

(12.84)

The solution of the quadratic equation yield the eigenvalues: − 2 3− + 5+ 9 – 4- = 3 --------------ω = -----------------------2 2

(12.85)

3– 5 1 – ---------------2 –1

–1 3– 5 2 – ---------------2

⋅

φ 11 φ 21

Page 12-31

(12.88)

The first row yields following equation: 2 –( 3 – 5 ) --------------------------- φ 11 – 1φ 21 = 0 2


= 0 0


(12.89)

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Normalizing the largest coordinate of the eigenvector to unity ( φ 11 = 1 ), φ 21 becomes: 2 –( 3 – 5 ) --------------------------- – φ 21 = 0 2


• Higher mode of vibration 3+ 5 2 The largest eigenvalue ω 2 = ---------------- leads to the 2

(12.90)

or 5–1 φ 21 = ---------------- = 0.62 2

(12.91)

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2. circular natural frequency ω 2 =

3+ 5 ---------------- = 1.62 2

(12.93)

In analogy to the fundamental mode, the second mode of vibra2 tion can be computed introducing the second eigenvalue ω 2 into the system of equations:

Hence the first natural mode of vibration is:

3+ 5 1 – ---------------2

–1

–1

3+ 5 2 – ---------------2

⋅

φ 12 φ 22

= 0 0

(12.94)

The first row yields following equation: φ 11 φ 21

1

=

1 5 – 1 = 0.62 ---------------2

2 – ( 3 + 5 )--------------------------φ 12 – 1φ 22 = 0 2

(12.92)

(12.95)

Normalizing the largest coordinate of the eigenvector to unity ( φ 22 = 1 ), φ 12 becomes: 2 –( 3 + 5 ) ---------------------------- φ 12 – 1 = 0 2

(12.96)

–2 1– 5 φ 12 = ---------------- = ---------------- = – 0.62 2 1+ 5

(12.97)

or:


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Hence the second natural mode of vibration is:


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• Orthogonality with respect to the mass matrix The modal mass matrix M* is: 5–1 5 – 1--------------1 – ---------------1 0 2 2 ⋅ ⋅ 0 1 5 – 15–1 --------------1 – ---------------1 2 2 1

T

M* = Φ MΦ = φ 12

1– 5 - = – 0.62 = -----------2 1 1

φ 22

(12.98) 5 – 1--------------2 ⋅

1

= 5–1 – ---------------2

3) Orthogonality of modes

=

5–1 1 + § ----------------· © 2 ¹ 0

In the following the orthogonality of the modes of vibration should be checked. Hence, following matrix of the eigenvectors is needed:

Φ =

φ 11 φ 12 φ 21 φ 22


1 = 5 – 1--------------2

5–1 – ---------------2

1

1

5–1 – ---------------2

5–1 ---------------2

1

2

0 5–1 1 + § ----------------· © 2 ¹

2

= 1.38 0 0 1.38 (12.100)

i.e. the matrix M* is diagonal.

(12.99)

1

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• Orthogonality with respect to the stiffness matrix

5–1 2 2

K* = Φ KΦ =

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k* 22 = 2 + ( 5 – 1 ) + §© ----------------·¹ = 1 + 5 + ---------------------------- = 3.618

The modal stiffness matrix K* is:

T


5–2 5+1 4

(12.105)

5–1 5 – 1--------------1 – ---------------1 – 1 2 2 ⋅ ⋅ –1 2 5 – 15–1 --------------1 – ---------------1 2 2 1

K* = 0.528 0 0 3.618

(12.106)

i.e. the matrix K* is diagonal. =

=

5–1 5–1 1 – ---------------- – 1 + ( 5 – 1 ) 1 – ---------------2 2 ⋅ 5–1 5–1 5–1 ---------------– ---------------- – 1 ---------------- + 2 1 2 2 2 5–1 1 – ( 5 – 1 ) + 2 § ----------------· © 2 ¹

2

5 – 1- § --------------5 – 1-· 2 --------------+ –1 © 2 ¹ 2

12.4.3 Equation of motion in modal coordinates The equation of motion in modal coordinates of a system without damping ( C * = 0 ) is:

5 – 1- § --------------5 – 1-· 2 --------------+ –1 © 2 ¹ 2 5–1 2 + ( 5 – 1 ) + § ----------------· © 2 ¹

·· + K* ⋅ q = – L ⋅ u·· ( t ) M* ⋅ q g

2

where: (12.101) L =

Computation of the single elements of K* : 5–1 2 2

5–2 5+1 2

(12.102) 2 5–2+5–2 5+1–4 5–1 5–1 2 k* 12 = ---------------- + §© ----------------·¹ – 1 = --------------------------------------------------------------- = 0 4 2 2

, and q =

(12.104)

Page 12-37

q1

is the vector of the modal coordinates

q2

Computation of the elements of the vector L : 5–1 ---------------2 L = Φ ⋅M⋅ι = ⋅ 1 0 ⋅ 1 = 1.62 0 1 1 0.382 5–1 – ---------------1 2 (12.108) T

(12.103)


L1 L2

k* 11 = 1 – ( 5 – 1 ) + 2 §© ----------------·¹ = – 5 + ---------------------------- = 0.528

k* 21 = k* 12 = 0

(12.107)


1

Page 12-38


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The influence vector ι represents the displacement of the masses resulting from the static application of a unit ground displacement u g = 1 : ι = 1 1

The modal participation factor Γ n is defined as: Ln Γ n = ---------mn *

(12.114)

and substituting L n and m n * into this definition gives following values for Γ 1 and Γ 2 :

The substitution of L into the equation of motion in modal coordinates (12.110)

leads to: ·· 1.38 0 ⋅ q 1 + 0.528 0 ⋅ q 1 = – 1.62 ⋅ u·· ( t ) g 0 1.38 0 3.618 0.382 q2 q··2 (12.111) Checking the circular natural frequency computed using modal coordinates against the results of Section 12.4.2:

L1 1.62 Γ 1 = ---------- = ---------- = 1.17 m1 * 1.38

(12.115)

L2 0.382 Γ 2 = ---------- = ------------- = 0.28 m2 * 1.38

(12.116)

The effective modal mass is defined as: 2

m n*, eff = Γ n ⋅ m n*

ω2 =

k 11 * -----------= m 11 *

0.528 ------------- = 0.62 OK! 1.38

(12.112)

k 22 * -----------= m 22 *

3.618 ------------- = 1.62 OK! 1.38

(12.113)


m 1*, eff = Γ 1 ⋅ m 1* = 1.17 2 ⋅ 1.38 = 1.894 2

Page 12-39

(12.117)

and substituting Γ n and m n * into this definition gives following values for m 1*, eff and m 2*, eff : 2

ω1 =

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• Additional important modal quantities

(12.109)

·· + K* ⋅ q = – L ⋅ u·· ( t ) M* ⋅ q g


(12.118)

m 2*, eff = Γ 2 ⋅ m 2* = 0.28 2 ⋅ 1.38 = 0.106

(12.119)

m 1, eff + m 2*, eff = 1.894 + 0.106 = 2.000 OK!

(12.120)


Page 12-40


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12.4.4 Response spectrum method

2) Natural modes of vibration and natural frequencies

The 2-DoF system analysed in the previous Sections shall be used to illustrate the response spectrum method. For this reason real masses and stiffnesses shall be assumed. The seismic action on the 2-DoF system is represented by the elastic response spectrum of the “El Centro” earthquake.

The results of the previous Sections computed using unit masses and unit stiffnesses shall be multiplied by the factor:

1) Model

• Fundamental mode Similar to Section 12.4.1, however with a new definition of masses and stiffnesses:

k ---- = m

2

100kg ⁄ s ------------------------ = 1kg

100s

M =

m1 0 0 m2

= 1 0 kg 0 1

The stiffness chosen for each story is k 1 = k 2 = k = 100 N/m and an appropriate units transformation leads to: k = 100 N/m = 100 kgm/s 2 m –1 = 100 kg/s 2

K =

k 11 k 12 k 21 k 22


=

100 – 100 kg/s 2 – 100 200

Natural period:

2π 2π T 1 = ------ = ---------------- = 1.02 s 6.2 Hz ω1

= 6.2 Hz

• Higher vibration mode –1

Natural frequency:

ω 2 = 1.62 ⋅ 100s

Natural period:

2π 2π T 2 = ------ = ------------------- = 0.39 s ω2 16.2 Hz

= 16.2 Hz

The eigenvectors are dimensionless quantities and remain unchanged:

Φ =

φ 11 φ 12 φ 21 φ 22

(12.122)

Page 12-41

–1

ω 1 = 0.62 ⋅ 100s

(12.121)

Hence, the stiffness matrix is:

(12.123)

Natural frequency:

m 1 = m 2 = 1kg Hence, the mass matrix is:

–1


1 = 5 – 1--------------2

5–1 – ---------------2

(12.124)

1

Page 12-42


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3) Modal analysis

4) Peak modal response

Equation of motion in u 1 - u 2 -coordinates (without damping):

The peak modal response of both vibration modes can be computed like in the case of SDoF systems using the spectral value given by the relevant response spectrum.

M ⋅ u·· + K ⋅ u = – M ⋅ 1 ⋅ u··g t

(12.125)

·· 1 0 kg ⋅ u 1 + 100 – 100 kg ⁄ s 2 ⋅ u 1 = – 1 0 kg 1 ⋅ u·· ( t ) g – 100 200 0 1 01 1 u··2 u2 (12.126) Variables transformation in modal coordinates q 1 and q 2 : u = Φ⋅q

(12.127)

where:

L1 q 1, max = ---------- ⋅ S d ( ω 1, ζ 1 ) = Γ 1 ⋅ S d ( ω 1, ζ 1 ) m1 *

(12.130)

where: S d ( ω 1, ζ 1 ) :

spectral displacement for a natural frequency ω 1 and a damping ζ 1 (here ζ 1 = 5% )

If an acceleration instead of a displacement response spectrum is used, then the peak value of the modal coordinate q 1 is:

Φ : Modal Matrix, i.e. the matrix of the eigenvectors The equation of motion in modal coordinates q 1 and q 2 (without damping) is: ·· + K* ⋅ q = – L ⋅ u·· ( t ) M* ⋅ q g

The peak value of the modal coordinate q 1 is:

(12.128)

L1 1 Γ1 q 1, max = ---------- ⋅ -----2- ⋅ S pa ( ω 1, ζ 1 ) = -----2- ⋅ S pa ( ω 1, ζ 1 ) m1 * ω ω1 1

(12.131)

where: ·· 1.38 0 kg ⋅ q 1 + 52.8 0 kg ⁄ s 2 ⋅ q 1 = – 1.62 kg ⋅ u·· ( t ) g q2 0 1.38 0 362 0.382 q··2

S pa ( ω 1, ζ 1 ) : spectral pseudo-acceleration for a natural frequency ω 1 and a damping ζ 1 (here ζ 1 = 5% )

(12.129) yielding the equation of motion in modal coordinates of two independent SDoF systems.


Page 12-43


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The spectral values given by the elastic acceleration response spectrum of the “El Centro” earthquake for the periods T 1 and T 2 are: S pa1 = 4.25 m/s 2 S pa2 = 7.34

and

(12.132)

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5) Inverse transformation The peak deformations and internal forces belonging to each mode of vibration in the original reference system are obtained by multiplying the relevant eigenvector with the corresponding peak value of the modal coordinate.

• Fundamental mode

m/s 2

(12.133) (1)

u max = q 1, max ⋅ φ 1 = 0.130m ⋅

10 Pseudo acceleration [m/s2]


ζ = 5%

1 = 130 mm 0.62 81

(1) (1) f max = K ⋅ u max = 100 – 100 ⋅ 0.130 =

Spa2 = 7.34 m/s2

100 200

0.081

(12.136)

13.0 – 8.1 = 4.9 N – 13.0 + 16.2 3.2

(12.137) Alternatively (allow an approximate consideration of nonlinearities):

2 5 S pa1 = 4.25 m/s

0 0.01

T2 = 0.39 s 0.10

s 1 = Γ 1 Mφ 1 = 1.17 ⋅ 1 0 ⋅ 1 = 1.17 0 1 0.62 0.725

(12.138)

(1) f max = s 1 ⋅ S pa1 = 4.25 1.17 = 4.9 N

(12.139)

0.725

T1 = 1.02 s 1.00

3.2

10.0

Period [s] 1.62kg 1 q 1, max = ----------------- ⋅ ---------------------2- ⋅ 4.25 m/s 2 = 0.130m 1.38kg ( 6.2Hz )

(12.134)

0.38kg 1 q 2, max = ----------------- ⋅ ------------------------2- ⋅ 7.34 m/s 2 = 0.008m 1.38kg ( 16.2Hz )

(12.135)


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• Higher vibration mode

100 200

⋅ – 0.005 =

0.008

(12.140)

– 0.5 – 0.8 = – 1.3 N 0.5 + 1.6 2.1

(12.141)

The total peak response is obtained from the peak response of the single vibration modes using e.g. the SRSS combination rule (SRSS = Square Root of the Sum of Squares). • Peak displacements 2

u 1, max =

Alternatively (allow an approximate consideration of nonlinearities): s 2 = Γ 2 Mφ 2 = 0.28 ⋅ 1 0 ⋅ – 0.62 = – 0.173 01 1 0.28 (1) f max = s 2 ⋅ S pa2 = 7.34 – 0.173 = – 1.3

0.28

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6) Combination

(2) u max = q 2, max ⋅ φ 2 = 0.008m ⋅ – 0.62 = – 5.0 mm 1 8.0

(2) (2) f max = K ⋅ u max = 100 – 100


2.1

) =

2

2

( 130mm ) + ( – 5 mm ) = 130mm

k=1

(12.144) (12.142) 2

u 2, max = N

(n) 2

¦ ( u1

(n) 2

¦ ( u2

) =

2

2

( 81mm ) + ( 8mm ) = 81mm

k=1

(12.143)

(12.145) In this case the total peak displacements are almost identical to the peak displacements of the fundamental mode. The relatively small contributions due to the second vibration mode basically disappear because of the SRSS combination rule.


Page 12-47


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• Peak sectional forces (Shear force V)

12.4.5 Response spectrum method vs. time-history analysis

Upper shear force:

1) Model

V 1, max =

2

2

( 4.9N ) + ( – 1.3N ) = 5.1N (12.146)

Lower shear force: V 2, max =

2

2

( 8.1N ) + ( 0.8N ) = 8.1N (12.147)

Compared to the peak sectional forces due to the fundamental mode, the total peak sectional forces show a slight increase in the upper story of the 2-DoF system. Case study 1 Masses:

Pay attention to following pitfall! It is wrong to compute the total peak sectional forces using the total peak displacements: V 1, max ≠ 100N ⁄ m ⋅ ( 0.130m – 0.081m ) = 4.9N

(12.148)

V 2, max ≠ 100N ⁄ m ⋅ 0.081m = 8.1N

(12.149)

Masses:

m 2 = 1.0kg Stiffnesses:

k 1 = 100N/m k 2 = 100N/m

m 1 = 0.1kg m 2 = 1.0kg

Stiffnesses:

k 1 = 10N/m k 2 = 100N/m

Case study 1 corresponds to the model analysed in Section 12.4.4. Case study 2 represents a dynamic system where the second vibration mode is important.

The sectional forces would be underestimated.


m 1 = 1.0kg

Case study 2

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2) Results


• Demand

• Dynamic properties Case study 1 Periods:

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T 1 = 1.02s

Case study 1

Periods:

T 2 = 0.39s Eigenvectors:

Displacements:

Displacements:

Case study 2

Case study 2

T 1 = 0.74s

1:

Δ = 0.129m

1:

Δ = 0.130m

T 2 = 0.54s

2:

Δ = 0.005m

2:

Δ = 0.072m

Sum:

Δ = 0.134m

Sum:

Δ = 0.202m

Eigenvectors:

1: φ 11 = 1 ,

φ 21 = 0.62

1: φ 11 = 1 ,

φ 21 = 0.27

SRSS:

Δ = 0.130m

SRSS:

Δ = 0.148m

2: φ 12 = 1 ,

φ 22 = – 1.62

2: φ 12 = 1 ,

φ 22 = – 0.37

Time-history:

Δ = 0.130m

Time-history:

Δ = 0.165m

Part. factors:

Γ 1 = 1.17

Part. factors:

Γ 1 = 2.14

Upper shear force:

Upper shear force:

Γ 2 = – 1.14

SRSS:

V = 5.10N

SRSS:

V = 1.36N

Time-history:

V = 5.69N

Time-history:

V = 1.51N

Γ 2 = – 0.17

• Note that in this case the eigenvectors are normalized to yield unit displacement at the top of the second story. Therefore, the eigenvectors and the participation factors of case study 1 differ from the values obtained in previous sections.


Page 12-51

Lower shear force:

Lower shear force:

SRSS:

V = 8.05N

SRSS:

V = 4.40N

Time-history:

V = 8.44N

Time-history:

V = 4.92N


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• Time-histories: Case study 1


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• Time-histories: Case study 2

20

20

First mode Displacement [cm]

Displacement [cm]

First mode 10

0

−10 −20 0

5

10

15

10

0

−10 −20 0

20

20

5

10

Second mode Displacement [cm]

Displacement [cm]

Second mode

0

−10

5

10

15

10

0

−10 −20 0

20

20

5

10

15

Displacement [cm]

Displacement [cm]

Sum

10

0

−10


20

20

Sum

−20 0

20

20

10

−20 0

15

5

10 Time [s]

15

20

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10

0

−10 −20 0


Peak values don’t occur at the same time! 5

10 Time [s]

15

20

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• Time-histories: Summary Blank page

20

Case study 1 Displacement [cm]

10

0

First mode Second mode Sum

−10

−20 0

5

10

15

20

20

Case study 2 Displacement [cm]

10

0

−10

−20 0


First mode Second mode Sum 5

10 Time [s]

15

20

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13 Vibration Problems in Structures


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13.1.1 Dynamic action a) People-induced vibrations

13.1 Introduction

- Pedestrian bridges

There are more and more vibration problems in structures because:

- Floors with walking people

• Higher quality materials with higher exploitation

- Floor with fixed seating and spectator galleries

- slender constructions

- Floors for sport or dance activities - High-diving platforms

- smaller stiffnesses and masses

b) Machinery-induced vibrations

• More intensive dynamic excitations

- Machine foundations and supports

• Increased sensitivity of people

- Bell towers - Structure-borne sound - Ground-transmitted vibrations

Nevertheless vibration sensitive structures are often designed for static loads only

c) Wind-induced vibrations - Buildings - Towers, chimneys and masts

Goal of this chapter • Give an overview of possible causes of vibration problems in buildings and of potential countermeasures • Description of practical cases with vibration rehabilitation

- Bridges - Cantilevered roofs d) Vibrations induced by traffic and construction activity - Roads and bridges - Railways - Construction works

13 Vibration Problems

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13.1.2 References

13.2 Vibration limitation

[Bac+97] Bachmann et al.:”Vibration Problems in Structures”. ISBN 3-7643-5148-9. Birkhäuser Verlag, Basel 1997.

13.2.1 Verification strategies

[BB88]

• Frequency tuning

[SIA06]

[SIA03] [SIA96] [VB87]

Bachmann H., Weber B.: “Tuned Vibration Absorbers for Damping of Lively Structures”. Structural Engineering International, No. 1, 1995. SIA: “Wind – Kommentar zum Kapitel 6 der Normen SIA 261 und 261/1 (2003) Einwirkungen auf Tragwerke”. SIA Dokumentation D0188. Zürich 2006. SIA: “Aktuelle Probleme der Brückendynamik”. SIA Dokumentation D0198. Zürich 2003.

10

0.00

9

0.01

8

Trrasmissibility TR

[BW95]

Baumann K., Bachmann H.: “Durch Menschen verursachte dynamische Lasten und deren Auswirkungen auf Balkentragwerke”. IBK Bericht Nr. 7501-3, 1988.

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ζ = 0.05

7 6 5

0.10

4 3

SIA: “Dynamische Probleme bei Brücken- und Hochbauten”. SIA Dokumentation D0138. Zürich 1996.

2

Vogt R., Bachmann H.: “Dynamische Kräfte beim Klatschen, Fussstampfen und Wippen”. IBK Bericht Nr. 7501-4, 1987.

0

0.20 0.50

1

0.70 0

0.5

1

2 1.5

2

2.5

3

3.5

4

ω/ωn

- High tuning (subcritical excitation) - Low tuning (supercritical excitation) • Amplitude limitation


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13.2.2 Countermeasures

13.2.3 Calculation methods

• Change of the natural frequency

• Computation of the natural frequencies

Strategy

The natural frequencies of structures have to be determined by means of realistic models. Approximate formulas that are often found in design codes and literature shall be checked carefully.

Effects 3 Stiffness: I ∼ bh ½ I ¾ f ∼ ---- ∼ h A Mass: A ∼ bh ¿

• Computation of the Amplitude

2

f a, 1

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π = ------------2- ⋅ EI ------ ( μ = distributed mass) μ 2πL 2

3.93 EI f b, 1 = ------------2- ⋅ ------ = 1.56f a, 1 ( 2.45 ⋅ EI ) μ 2πL

If the frequency of a harmonic of the excitation coincides with a natural frequency of the structure (resonance), the maximum deflection of the structure can be estimated as follows (See Chapter 5): Fo u p = ----- ⋅ V ( ω ) ⋅ cos ( ωt – φ ) k

(13.1)

for ω = ω n we have V ( ω ) = 1 ⁄ ( 2ζ ) and: 2

4.73 EI f c, 1 = ------------2- ⋅ ------ = 2.27f a, 1 ( 5.14 ⋅ EI ) μ 2πL 2

π f d, 1 = ------------------------2- ⋅ EI ------ = 4f a, 1 ( 16 ⋅ EI ) μ 2π ( L ⁄ 2 )

• Increase of the damping - Installation of dampers or absorbers

Fo 1 u max = ----- ⋅ -----k 2ζ

(13.2)

The maximum velocity and the maximum acceleration can be determined from Equation (13.2) as follows: Fo 1 u· max = ω ⋅ u max = ω ⋅ ----- ⋅ -----k 2ζ

(13.3)

2 2 Fo 1 u··max = ω ⋅ u max = ω ⋅ ----- ⋅ -----k 2ζ

(13.4)

- Plastic energy dissipation • Tuned Mass Dampers 13 Vibration Problems

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The amplitude of the nth harmonic component of a force generated by people excitation is proportional to the mass of the person ( F o = G ⋅ α n = g ⋅ M ⋅ α n , see Equation (13.8)). k g ⋅ M ⋅ αn 1 M g ⋅ αn 2 g ⋅ M ⋅ αn 1 u··max = ω ⋅ ----------------------- ⋅ ------ = ---- ⋅ ----------------------- ⋅ ------ = ----- ⋅ ------------k 2ζ m k 2ζ m 2ζ

(13.5)

M g ⋅ αn u··max = ----- ⋅ ------------m 2ζ

(13.6)


13.3 People induced vibrations 13.3.1 Excitation forces In Chapter 6 “Forced Vibrations” it has been already mentioned that excitation due to people, like e.g. walking, running, jumping, and so on, can be represented as Fourier-series: ∞

F ( t ) = a0 +

• Remarks

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[ a n cos ( nω 0 t ) + b n sin ( nω 0 t ) ]

¦

(13.7)

n=1

- A soft structure is more prone to vibration than a rigid one. See Equations (13.2) to (13.4). - The acceleration amplitude is directly proportional on the ratio of the mass of the people to the building mass.

Equation (13.7) can also be represented in a form according to Equation (13.8): N

F(t) = G +

¦

G ⋅ α n ⋅ sin ( n ⋅ 2πf 0 ⋅ t – φ n )

(13.8)

n=1

Where: - G = Weight of the person - α n = Fourier coefficient for the nth harmonic - G ⋅ α n = Amplitude of the nth harmonic of the excitation force - f 0 = Step frequency of the excitation force - φ n = Phase shift of the nth harmonic ( φ 1 = 0 ) - n = Number of the nth harmonic - N = Number of considered harmonics


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The steady-state response of a SDoF system under periodic excitation can be computed in analogy to Chapter 6 as:


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• Jumping (left) and walking (right), see [BB88]

N

u ( t ) = u0 ( t ) +

¦

un ( t )

(13.9)

n=1

Where G u 0 ( t ) = ---- (Static displacement) k

(13.10)

2

G ⋅ α n ( 1 – β n ) sin ( nω 0 t – φ n ) – 2ζβ n cos ( nω 0 t – φ n ) u n ( t ) = --------------- ⋅ --------------------------------------------------------------------------------------------------------------2 2 2 k ( 1 – β n ) + ( 2ζβ n ) ω 0 = 2πf 0

,

nω 0 β n = --------ωn

(13.11) • Clapping, foot stomping and rocking, see [VB87] (13.12)

• Measurement of forces (Example)


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• Walking (see [Bac+97] Figure G.1)

• Jumping (see [Bac+97] Figure G.2)

• Clapping (see [Bac+97] Figure G.3)

• Jumping: Fourier amplitude spectrum (see [Bac+97] Figure G.2)


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Fourier-coefficients for human activities according to [Bac+97] Table G.2

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Characterisation of human activities according to [Bac+97] Table G.1



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• Remarks regarding Table G.2


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• RC Beam

- Coefficients and phase angles represent averages. - Phase angles have strong scattering and therefore, in many cases, it is difficult to provide reasonable values. In such cases (e.g,. running and dancing) in Table G.2 no values are specified. - Decisive are cases in which resonance occurs. In such cases the phase angle no longer plays a role. - Coefficients and phase were checked and discussed internationally. 13.3.2 Example: Jumping on an RC beam

The RC beam has a length of 19 meters. The natural frequency is thus: f n = 2Hz .

(13.13)

• Excitation Here “jumping” is described by means of the Fourier-series given in Table G.2. In Section 6.1 “periodic excitation”, “jumping” was described by means of a half-sine function. Jumping frequency:

f 0 = 2Hz

(13.14)

Contact time:

t p = 0.16s (phase angle computation)

(13.15)

Here the same example as in Section 6.1.3 is considered again: Weight of the person: G = 0.70kN

(13.16)

• Results Excel Table:

u max = 0.043m

(13.17)

Equation (13.2):

F 1 1.8 ⋅ 0.70 1 u max = --- ⋅ ------ = ---------------------- ⋅ --------------------- = 0.042m k 2ζ 886 2 ⋅ 0.017

(13.18) • Remarks - Shape of the excitation “similar” as half-sine - Maximum deflection very close to the solution obtained by means of the half-sine function


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• Excitation

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13.3.3 Footbridges • Frequency tuning

3.5 Static component (n=0)

3.0

First harmonics (n=1)

- Vertical: Avoid natural frequencies between 1.6 and 2.4Hz. In the case of structures with low damping (Steel), avoid also natural frequencies from 3.5 to 4.5Hz.

Second harmonics (n=2)

2.5

Force F(t) [kN]


Third harmonics (n=3) Total (3 harmonics)

2.0 1.5

- Horizontal transverse: Avoid natural frequencies between 0.7 and 1.3Hz (absolutely safe: fht,1 > 3.4Hz).

1.0 0.5

- Horizontal longitudinal: Avoid natural frequencies between 1.6 and 2.4Hz.

0.0 00 -0.5

• Amplitude limitation

-1.0 -1.5 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time (s)

- Calculation of the acceleration maximum amplitude. 2

a max < ca. 0.5m ⁄ s = 5% g

• Response

(13.19)

• Special features of the amplitude limitation 0.0500 Static component (n=0)

- When walking or running, the effectiveness of people is limited, because the forces are not always applied at midspan;

First harmonics (n=1)

0.0400

Second harmonics (n=2)

Displacement D [m]

0.0300

Third harmonics (n=3) Total (3 harmonics)

0.0200

System

0.0100 0.0000 -0.0100 -0.0200 0 0200

Equivalent SDoF system

-0.0300 -0.0400 -0.0500 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time (s)


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An-Najah 2013

- People need a finite number of steps in order to cross the bridge (This limited excitation time may be too short to reach the maximum amplitude) 1

0.2

0.1

0.05 0.02

abs(uj) / umax

0.8

ζ = 0.01 0.6

0.4

0.2

0 0

5

10

15

20

25

30

35

40

45

50

Cycle

- Not all people walk in the step (Exception: Lateral vibrations → Synchronisation effect) To take into account the specificities of the amplitude limitation, sophisticated methods are available. From [Bac+97] the following one is adopted: 2

2

2

a max = 4π ⋅ f ⋅ y ⋅ α ⋅ Φ [ m ⁄ s ]

(13.20)

Where:

The acceleration a max given in Equation (13.20) is the acceleration generated by one person crossing the footbridge. If n people are on the bridge at the same time, the maximum acceleration is typically less than n ⋅ a max because not all people walk in step across the bridge. . The square root of the number of people is often chosen as the multiplication factor, i.e. n ⋅ a max

- y : Static deflection at half the span - α : Fourier coefficient - Φ : dynamic amplification factor 13 Vibration Problems

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Example: “Walking on an RC beam”

• Rough estimate of maximum displacement and acceleration

• Situation

The maximum displacement, and the maximum acceleration can be estimated by means of Equations (13.2) and (13.4):

System

1 G u max,st = ------ = --------- = 0.001128m = 0.11cm 886 Kn

(13.21)

G ⋅ α1 1 1 ⋅ 0.4 1 u max,1 = --------------- ⋅ ------ = --------------- ⋅ --------------------- = 0.0133m = 1.33cm 886 2 ⋅ 0.017 K n 2ζ

(13.22) (13.23)

u max = 0.11 + 1.33 = 1.44cm

Discretisation for FE Analysis

2

2

a max = ω u max,1 = ( 2π ⋅ 2 ) ⋅ 0.0133 = 2.10m ⁄ s

2

(13.24)

• Estimate of the maximum displacement and acceleration using the improved method The maximum acceleration is computed by means of Equation (13.20) as follows: - Stiffness at mid span: K n = 886kN ⁄ m

Walking-velocity:

v = S ⋅ f 0 = 0.7 ⋅ 2 = 1.4m ⁄ s

(13.25)

- Natural frequency: f n = 2Hz

Crossing time:

(13.26)

- Damping: ζ = 0.017

Δt = L ⁄ v = 19 ⁄ 1.4 = 13.57s

Number of cycles:

N = Δt ⋅ f n = 13.57 ⋅ 2 = 27

(13.27)

Amplification factor:

Φ = 23

(13.28)

• Excitation - Walking with f 0 = 2Hz according to Table G.2.

From Equation (13.20):

- Step length: S = 0.70m

2 2 1.00 2 a max = 4π ⋅ 2 ⋅ ---------- ⋅ 0.40 ⋅ 23 = 1.64m ⁄ s 886

- Weight of the Person: G = 1kN


Page 13-21


(13.29)

Page 13-22

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• Computation of displacements and accelerations by means of the FE Programme ABAQUS Displacements and accelerations are computed by means of time-history analysis: - Excitation


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- Time history of the displacement 1.5

Vertical displacement um [cm]


Displacement at mid span

1 0.5 0 -0.5 -1 -1.5 0

5

10

15

20

25

30

35

40

Time [s]

- Time history of the acceleration 1.6

2

Vertical a acceleration am [m/s2]

Forc ce / Static weight [-]

Acceleration at mid span

F1

1.4

F2

1.2

F3

1

F15

0.8

F28

0.6 0.4 0.2 0 0

2

4

6

8

10

12

14

16

Time [s]


1.5 1 0.5 0 -0.5 -1 -1.5 -2 0

5

10

15

20

25

30

35

40

Time [s]

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• Remarks


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13.3.4 Floors in residential and office buildings

- The refined method and the time history calculations show lower values compared to the rough method; - The refined method and the time history calculations are in good agreement;

• Frequency tuning - If the excitation is generated by walking ( f max ≅ 2.4Hz ), the following natural frequencies shall be exceeded:

- The time history of the displacement is not symmetric compared to the time axis, because of the static component of the displacement caused by the weigh of the crossing person;

Damping

Natural frequency [Hz]

Remark

> 5%

>5

Avoid resonance due to the second harmonic

- The time history of the acceleration is symmetric compared to the time axis, because there is no static component of the acceleration.

< 5%

> 7.5

Avoid resonance due to the third harmonic

• Amplitude limitation - Calculation of the acceleration maximum amplitude

• Swinging footbridge on the Internet

2

a max < ca. 0.05m ⁄ s = 0.5% g

http://www.londonmillenniumbridge.com/

(13.30)

- Because of the many non-structural components (wallpaper, furniture, suspended ceilings, technical floors, partitions, ....) it is difficult to estimate the dynamic properties of the floors.

http://www.youtube.com

- Where possible measure the dynamic properties. Response of people to vibrations The sensitivity of people to vibration depends on many parameters:


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- Position (standing, sitting, lying)


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• G2 limit for vibrations parallel to the spinal column

- Direction of the action compared to the spinal column - Activity (resting, walking, running, ...) - Type of vibration - .... Description

Frequency 1 to 10 Hz amax [m/s2]

Frequency 10 to 100 Hz vmax [m/s]

0.034 0.1 0.55 1.8

0.0005 0.0013 0.0068 0.0138

Barely noticeable Clearly noticeable Disturbing Not tolerable

Vertical harmonic vibration action on a standing person. Accepted averages; scatters up to a factor of 2 is possible (from [Bac+97])

• G2 limit for vibrations transverse to the spinal column • ISO 2631 standard a eff =

T 2 --1- ³ a ( t ) dt T 0

(13.31)

Where T is the period of time over which the effective acceleration was measured. 3 limits are defined: - G1: Reduced comfort boundary - G2: Fatigue-decreased proficiency boundary ~ 3 x G1 - G3: Exposure limit ~ 6 x G1


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13.3.5 Gyms and dance halls

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• Amplitude limitation

Due to gymnastics or dancing, very large dynamic forces are generated. This is readily understandable when the Fourier coefficients in Table G2 are considered:

- Calculation of the acceleration maximum amplitude 2

a max < ca. 0.5m ⁄ s = 5% g

(13.32)

- Limits depend on the activity, if e.g. people are sitting in the dance hall, as well, this limit shall be reduced.

- Walking: α 1 = 0.4 , α 2 = 0.1 , α 3 = 0.1 - Running: α 1 = 1.6 , α 2 = 0.7 , α 3 = 0.2 - Jumping: α 1 = 1.9 , α 2 = 1.6 , α 3 = 1.1 - Dancing: α 1 = 0.5 , α 2 = 0.15 , α 3 = 0.1 (however: a - many people moving rhythmically. b - certain dances are very similar to jumping) • Frequency tuning

- Because of the large forces that can be generated through these activities, the dynamic characteristics of the structure shall be estimated as precisely as possible. 13.3.6 Concert halls, stands and diving platforms See [Bac+97].

- If the excitation is generated through jumping ( f max ≅ 3.4Hz ) or dancing ( f max ≅ 3.0Hz ), then the following natural frequencies shall be exceeded: Construction

Gyms Natural frequency [Hz]

Dance halls Natural frequency [Hz]

Reinforced concrete Prestressed concrete Composite structures Steel

> 7.5 > 8.0 > 8.5 > 9.0

> 6.5 > 7.0 > 7.5 > 8.0



Page 13-29

13.4 Machinery induced vibrations It is not possible to carry out here a detailed treatment of machinery induced vibrations. Therefore, reference to [Bac+97] is made.


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13.5 Wind induced vibrations


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• Vortex shedding: Periodic action transversely to the wind direction

Wind-induced vibrations cover a challenging and wide area. It is not possible to carry out here their detailed treatment. Therefore, reference is made to the relevant literature:

- Vortex are not shedded left and right at the same time. If the time-interval of the vortex shedding is equal to the oscillation period of the structure, resonance excitation occurs.

• [Bac+97] • Simiu E., Scanlan R.H.: “Wind Effects on Structures”. Third Edition. John Wiley & Sons, 1996. 13.5.1 Possible effects fe ⋅ d u crit = ----------S

• Gusts: Stochastic effects in wind direction - Turbulent wind with spatially and temporally variable wind speed.

(13.33)

Where u crit :

Critical wind velocity

fe :

Natural frequency of the structure transverse to the wind direction

d:

Diameter of the structure

S:

Strouhal number (about 0.2 for circular cross sections)

• Buffeting: Periodic action in wind direction - Vortex detached from an obstacle hit the structure


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• “Gallopping” and “Flutter”: Unstable interaction between wind flow and structural motion - Gallopping: Motion of the structure transversely to the flow direction. - Flutter: Combined flexural-torsional motion of the structure. Work done by wind forces during flutter


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13.6 Tuned Mass Dampers (TMD) 13.6.1 Introduction When discussing MDoF systems, in Section 11.1.3 a Tuned Mass Damper (TMD) has already been discussed. However, in that case zero damping was assumed for both the structure and the TMD. There it was possible to solve the equation of motion simply by means of modal analysis. Here the theory of the TMD with damping is treated. As we shall see, the damping of the two degrees of freedom is a design parameter, and it shall be possible to chose it freely, therefore: In the case of TMD with damping modal analysis can not be used

Stability curves for bridge cross-sections • References


Page 13-33

[BW95]

Bachmann H., Weber B.: “Tuned Vibration Absorbers for Damping of Lively Structures”. Structural Engineering International, No. 1, 1995.

[Den85]

Den Hartog J.P.: “Mechanical Vibrations”. ISBN 0-48664785-4. Dover Publications,1985. (Reprint of the original fourth edition of 1956)


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13.6.2 2-DoF system


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2 ° [ – ω m H + iω ( c H + c T ) + ( k H + k T ) ]U H + [ – iωc T – k T ]U T = F H ® 2 ° [ – iωc T – k T ]U H + [ – ω m T + iωc T + k T ]U T = 0 ¯

(13.36) To facilitate the solution of the system, some dimensionless parameters are now introduced:

The equations of motion of the 2-DoF system shown above are: ·· · · · mH uH + cH uH + cT ( uH – uT ) + kH uH + kT ( uH – uT ) = F ( t ) ® m T u··T + c T ( u· T – u· H ) + k T ( u T – u H ) = 0 ¯

(13.34)

For an harmonic excitation of the type F ( t ) = F H cos ( ωt ) , a possible ansatz for the steady-state part of the solution is: uH = UH e

iωt

,

uT = UT e

iωt

,

F ( t ) = FH e

iωt

,

(13.35)

Using the complex numbers formulation allows a particularly elegant solution to the problem. The equations of motion become:

γ = mT ⁄ mH :

Mass ratio (TMD Mass/Mass of the structure)

ωT =

kT ⁄ mT :

Natural frequency of the TMD

ωH =

kH ⁄ mH :

Natural frequency of the structure without TMD

β = ωT ⁄ ωH :

Ratio of the natural frequencies

Ω = ω ⁄ ωH :

Ratio of the excitation frequency to the natural frequency of the structure

ζT :

Damping ratio of the TMD

ζH :

Damping ratio of the structure

U H0 = F H ⁄ k H :

Static deformation of the structure

Substituting these dimensionless parameters into Equation (13.36) we obtain: 2 2 2 ° [ – Ω + 2iΩ ( ζ H + βγζ T ) + ( 1 + β γ ) ]U H + [ – 2iΩβγζ T – β γ ]U T = U H0 ® 2 2 2 ° [ – 2iΩβγζ T – β γ ]U H + [ – Ω γ + 2iΩβγζ T + β γ ]U T = 0 ¯

(13.37)


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The system of equation can be easily solved using “Maple”, and we obtain the following expression for the amplification function U H ⁄ U H0 : 2

2

UH ( β – Ω ) + 2iΩβζ T -------- = -----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2 2 2 2 2 2 2 2 2 2 U H0 [ ( β – Ω ) – Ω β ( 1 – γ ) + Ω ( Ω – 4βζ H ζ T ) ] + 2i [ ( β – Ω )ζ H + ( 1 – Ω – Ω γ )βζ T ]

or

U H = U H0 ( x + iy )

(13.39)

The displacement U H has therefore two components: 1) One that is in phase with the displacement U H0 and 2) one with a phase shift equal to π ⁄ 4 . From the vectorial sum of these two components the norm of U H can be computed as: 2

( A + iB ) = U H0 --------------------( C + iD )

(13.41)

2

2 2

2

( β – Ω ) + ( 2Ωβζ T ) --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2 2 2 2 2 2 2 2 2 2 2 2 [ ( β – Ω ) – Ω β ( 1 – γ ) + Ω ( Ω – 4βζ H ζ T ) ] + 4 [ ( β – Ω )ζ H + ( 1 – Ω – Ω γ )βζ T ]

(13.44) A similar procedure can be followed to compute the dynamic amplification function U T ⁄ U H0 . Next figure show a representation of Equation (13.44) in function of Ω for an undamped structure ζ H = 0 . Curves for different values of the parameters β , γ and ζ T are provided. 16

]T=0

(13.40)

Equation (13.38) has however the form UH

UH = -------U H0

]T=inf.

J = 1/20 E=1

12

UH/UH0 [-]

2

U H = U H0 x + y

An-Najah 2013

Thus, the norm of the dynamic amplification function U H ⁄ U H0 can be easily calculated:

(13.38)

The complex expression given in Equation (13.38) shall now be converted into the form: z = x + iy


P 8

]T=0.32

and must be first rearranged as follows: UH

4

( A + iB ) ⋅ ( C – iD ) ( AC + BD ) + i ( BC – AD ) = U H0 ----------------------------------------------- = U H0 ---------------------------------------------------------------2 2 ( C + iD ) ⋅ ( C – iD ) C +D

(13.42) 2

U H = U H0


0.7

0.8

0.9

1

1.1

1.2

1.3

: [-]

(13.43)

Page 13-37

Q

0 0.6

2

A +B ------------------2 2 C +D

]T=0.10


Page 13-38


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13.6.3 Optimum TMD parameters


• Amplification function with TMD: Variation of TMD frequency

Based on observations and consideration at the previous image Den Hartog found optimum TMD parameters for an undamped structure:

ζ T, opt =

3m T ⁄ m H --------------------------------------= 3 8 ( 1 + mT ⁄ mH )

or

1 β opt = ----------1+γ

3γ --------------------3 8(1 + γ)

(13.45)

50

No TMD (ζT=inf.)

γ = 0.01 ζΗ = 0.01

40

UH/UH0 [-]

fH fH f T, opt = ----------------------------- = ----------1 + mT ⁄ mH 1+γ

An-Najah 2013

30

20

(13.46)

fT=0.98fT,opt

fT=1.02f 1.02fT,opt T opt

fT=fT,opt

10

These optimum TMD parameters can be applied also to lowly damped structures providing good response results.

0 0.6

0.7

0.8

0.9

1

1.1

1.2

1.3

Ω [-]

13.6.4 Important remarks on TMD

• Amplification function with TMD: Variation of TMD damping

• The frequency tuning of the TMD shall be quite precise

50

No TMD (ζT=inf.)

• The compliance with the optimum damping is less important

• It is not worth increasing the mass ratio too much

UH/UH0 [-]

• TMDs are most effective when the damping of the structure is low

γ = 0.01 ζΗ = 0.01

40

• Design charts for TMDs shall be computed numerically

30

20

• For large mass ratios, the amplitude of the TMD oscillations reduce

ζT=ζT,opt

10

• Meaningful mass ratios γ are 3-5% • The exact tuning of the TMD occurs experimentally, therefore great care should be paid to construction details.


ζT=0.5ζ 0 5ζT,opt T opt

ζT=1 1.5ζ 5ζT,opt T t

Page 13-39

0 0.6

0.7

0.8

0.9

1

1.1

1.2

1.3

Ω [-]


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• Design charts: Displacement of the structure (from [BW95]) Blank page

• Design charts: Relative TMD displacement (from [BW95])


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14 Pedestrian Footbridge with TMD 14.1 Test unit and instrumentation

Figure 14.1: View of the test setup.

The test unit is a post-tensioned RC beam. The beam is made of lightweight concrete and the post-tensioning is without bond. The dimension of the beam were chosen to make it particularly prone to vibrations induced by pedestrians. A Tuned Mass Damper (TMD) is mounted at midspan.

On the test unit the following quantities are measured: • Displacement at midspan • Acceleration at midspan • Acceleration at quarter point of the span 14 Pedestrian Footbridge with TMD

Page 14-1


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14.2 Parameters

• Characteristics of the TMD A close-up of the TMD is shown in Figure 14.2. We can see: • The 4 springs that define the stiffness K T of the TMD • The 4 viscous dampers that define the damping constant c opt of the TMD • The mass M T , which is made up by a concrete block and two side container filled with lead spheres. The lead spheres are used for the fine-tuning of the TMD.

14.2.1 Footbridge (Computed, without TMD) Modal mass:

M H = 5300kg

Modal stiffness:

K H = 861kN ⁄ m

Natural frequency:

f H = 2.03Hz

(Computed with TMD mass: f = 1.97Hz )

The properties of the TMD are given in Section 14.2.

14.2.2 Tuned Mass Damper (Computed) Mass:

M T = 310kg

Mass ratio:

MT μ = -------- = 0.0585 = 5.85% MH

Natural frequency:

fH f opt = ------------ = 1.92Hz 1+μ

(Measured: f T = 1.91Hz ) 2

Stiffness:

K T = M T ⋅ ( 2πf opt ) = 50.9kN/m

Damping rate

3μ ----------------------- = 0.14 = 14% 8( 1 + μ )3 (Measured: ζ T = 13% )

Damping constant:

c opt = 2ζ opt K T M T = 1.18kNs/m

ζ opt =

Figure 14.2: Close-up of the TMD.


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14.4 Free decay test with locked TMD

Following tests are carried out:

Time history of the displacement at midspan

No. Test

Action location

TMD

1

Free decay

Midspan

Locked

2

Sandbag

Midspan

Locked

3

Sandbag

Quarter-point

Locked

4

Sandbag

Midspan

Free

5

Walking 1 Person 3Hz

Along the beam

Locked

6


Along the beam

Locked

7


Along the beam

Free

8

Walking in group 2Hz

Along the beam

Locked

9

Walking in group 2Hz

Along the beam

Free

10

Jumping 1 Person 2Hz

Midspan

Locked

11

Jumping 1 Person 2Hz

Midspan

Free

Displacement at midspan [mm]

14.3 Test programme

40 30 20 10 0 -10 -20 -30 -40 0

Typical results of the experiments are presented and briefly commented in the following sections.


Page 14-5

5

10 Time [s]

15

20

Figure 14.3: Free decay test with locked TMD: Displacement at midspan.

Evaluation: Logarithmic decrement

Damping ratio

Region 1 Average amplitude: ~30mm

1 41.36 δ = --- ln ------------- = 0.081 8 21.66

0.081 ζ H = ------------- = 1.29% 2π


1 19.91 δ = --- ln ------------- = 0.090 8 9.68

0.090 ζ H = ------------- = 1.43% 2π


1 8.13 δ = --- ln ---------- = 0.092 8 3.90

0.092 ζ H = ------------- = 1.46% 2π


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14.5 Sandbag test

Fourier-spectrum of the displacement at midspan

The sandbag test consists in hanging a 20 kg sandbag 1 meter above the footbridge, letting it fall down and measuring the response of the system.

2.0 "Spectral displacement"


1.5

In order to excite the different modes of vibration of the footbridge, the test is repeated several times changing the position of the impact of the sandbag on the bridge. The considered locations are:

1.0

- at midspan (Section 14.5.1)

0.5

- at quarter-point of the span (Section 14.5.2). 0.0 0.0

0.5

1.0

1.5 2.0 2.5 Frequency [Hz]

3.0

3.5

4.0

Figure 14.4: Free decay test with locked TMD: Fourier-spectrum of the displacement at midspan.

The measured natural frequency of the footbridge with locked TMD is equal to: f = 1.89Hz

(14.1)

This value is less than the value given in Section 14.2.1. This can be explained with the large amplitude of vibration at the start of the test, which causing the opening of cracks in the web of the beam, hence reducing its stiffness. The second peak in the spectrum corresponds to f = 1.98Hz , wich is in good agreement with Section 14.2.1.


Page 14-7

These tests are carried out with locked TMD. In order to investigate the effect of the TMD on the vibrations of the system, the test of Section 14.5.1 is repeated with free TMD (see Section 14.5.3). Remark • The results presented in Section 14.5.1 and those presented in Section 14.5.2 and 14.5.3 belongs to two different series of tests carried out at different point in time. Between these test series the test setup was completely disassembled and reassembled. Slight differences in the assemblage of the test setup (support!) may have led to slightly different natural frequencies of the system.


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1.0

1.0 Acc. at quarter-point [m/s2]

Acceleration an midspan [m/s2]

14.5.1 Locked TMD, Excitation at midspan

0.5 0.0 -0.5 -1.0 0

1

2

3

4

5 6 Time [s]

7

8

9

1

2

3

4

5 6 Time [s]

7

8

9

10

0.040 f1=2.00Hz

"Spectral acceleration"


-0.5

Figure 14.7: Sandbag test with locked TMD: Acceleration at quarterpoint of the span.

0.040 0.030 0.020 0.010 0.000 0

0.0

-1.0 0

10

Figure 14.5: Sandbag test with locked TMD: Acceleration at midspan.

0.5

f3=18.06Hz

5

10 Frequency [Hz]

15

20

Figure 14.6: Sandbag test with locked TMD: Fourier-spectrum of the acceleration at midspan.


Page 14-9

0.030

f1=2.00Hz

0.020 0.010 0.000 0

f3=18.06Hz

5

10 Frequency [Hz]

15

20

Figure 14.8: Sandbag test with locked TMD: Fourier-spectrum of the acceleration at quarter-point of the span.


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14.5.2 Locked TMD, Excitation at quarter-point of the span

• With the sandbag test in principle all frequencies can be excited. Figures 14.5 and 14.7 show a high-frequency vibration, which is superimposed on a fundamental vibration; • The Fourier amplitude spectrum shows prominent peaks at the first and third natural frequencies of the system (Footbridge with locked TMD); • The second mode of vibration of the system is not excited, because the sandbag lands in a node of the second eigenvector. • At midspan, the amplitude of the vibration due to the first mode of vibration is greater than at quarter-point. The amplitude of the vibration due to the third mode of vibration, however, is about the same in both places. This is to be expected, if the shape of the first and third eigenvectors is considered. • The vibration amplitude is relatively small, therefore, the measured first natural frequency f 1 = 2.0Hz in good agreement with the computation provided in Section 14.2.1.

Acceleration at midspan [m/s2]

Remarks


1.0 0.5 0.0 -0.5 -1.0 0

1

2

3

4

5 6 Time [s]

7

8

9

10

Figure 14.9: Sandbag test with locked TMD: Acceleration at midspan.

0.040 "Spectral acceleration"


0.030

f1=2.00Hz

0.020 0.010 0.000 0

f3=18.31Hz

5

10 Frequency [Hz]

15

20

Figure 14.10: Sandbag test with locked TMD: Fourier-spectrum of the acceleration at midspan.


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Remarks • When the sandbag lands at quarter-point of the bridge, the second mode of vibration of the system is strongly excited. Its contribution to the overall vibration at quarter-point of the footbridge is clearly shown in Figures 14.11 and 14.12.

Acc. at quarter-point [m/s2]

1.0 0.5 0.0

• The acceleration sensor located at midspan of the footbridge lays in a node of the second mode of vibration, and as expected in figures 14.9 and 14.10 the contribution of the second mode is vanishingly small.

-0.5 -1.0 0

1

2

3

4

5 6 Time [s]

7

8

9

10

Figure 14.11: Sandbag test with locked TMD: Acceleration at quarterpoint of the span.


0.040 0.030 0.020

f2=8.84Hz f1=2.00Hz

0.010 0.000 0

f3=18.31Hz

5

10 Frequency [Hz]

15

20

Figure 14.12: Sandbag test with locked TMD: Fourier-spectrum of the acceleration at quarter-point of the span.


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14.5.3 Free TMD: Excitation at midspan

• With active (free) TMD the “first” and the “third “natural frequencies of the bridge are excited. As expected, these frequencies are slightly larger than the natural frequencies of the system (bridge with locked TMD), which are given in Figure 14.6. This is because the mass of the TMD is no longer locked and can vibrate freely.

0.5 0.0 -0.5

1

2

3

4

5 6 Time [s]

7

8

9

10

Figure 14.13: Sandbag test with free TMD: Acceleration at midspan.

0.040 "Spectral acceleration"

An-Najah 2013

Remarks

1.0

-1.0 0


0.030

• The effect of the TMD is clearly shown in Figure 14.14. The amplitude of the peak in the “first natural frequency” is much smaller than in Figure 14.6. The amplitude of the peak at the “third natural frequency” is practically the same. The “third mode of vibration” is only marginally damped by the TMD. • In the two comments above, the natural frequencies are mentioned in quotes, because by releasing the TMD number and properties of the natural vibrations of the system change. A direct comparison with the natural vibrations of system with locked TMD is only qualitatively possible.

0.020 “f1=2.05Hz” “f3=19.04Hz”

0.010 0.000 0

5

10 Frequency [Hz]

15

20

Figure 14.14: Sandbag test with free TMD: Fourier-spectrum of the acceleration at midspan.


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14.6 One person walking with 3 Hz

Test results

One 65 kg-heavy person (G = 0.64 kN) crosses the footbridge. He walks with a frequency of about 3 Hz, which is significantly larger than the first natural frequency of the bridge.



• The static deflection of the bridge when the person stands at midspan is: 0.69 G d st = ------- = ---------- = 0.00080m = 0.80mm 861 KH • The maximum measured displacement at midspan of the bridge is about 2 mm (see Figure 14.15), which corresponds to about 2.5 times d st . As expected, the impact of dynamic effects is rather small. • In the Fourier spectrum of the acceleration at midspan of the bridge (see Figure 14.17), the frequencies that are represented the most correspond to the first, the second and the third harmonics of the excitation. However, frequencies corresponding to the natural modes of vibrations of the system are also visible.

2.0 1.0 0.0 -1.0 -2.0 0

5

10

15

20 Time [s]

25

30

35

40

Figure 14.15: One person walking with 3 Hz: Displacement at midspan with locked TMD. Acceleration at midspan [m/s2]

Remarks

An-Najah 2013

1.0 0.5 0.0 -0.5 -1.0 0

5

10

15

20 Time [s]

25

30

35

40

Figure 14.16: One person walking with 3 Hz: Acceleration at midspan with locked TMD.


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0.015

14.7 One person walking with 2 Hz

0.010

One 95 kg-heavy person (G = 0.93 kN) crosses the footbridge. He walks with a frequency of 1.95 Hz, which is approximately equal to the first natural frequency of the bridge. The length of the step is 0.70 m.

0.005

Sought is the response of the bridge under this excitation. A similar problem was solved theoretically in Section 13.3.3.

0.000 0

14.7.1 Locked TMD (Measured) 5

10 Frequency [Hz]

15

20

Figure 14.17: One person walking with 3 Hz: Fourier-Spectrum of the acceleration at midspan with locked TMD.

First the maximum amplitudes are calculated by hand: Static displacement:

0.93 G d st = ------- = ---------- = 0.00108m = 1.08mm 861 KH

(Measured: d st = 1.22mm ) Walking velocity:

v = S ⋅ f 0 = 0.7 ⋅ 1.95 = 1.365m ⁄ s

Crossing time:

Δt = L ⁄ v = 17.40 ⁄ 1.365 = 12.74s

Number of cycles:

N = Δt ⋅ f n = 12.74 ⋅ 1.95 = 25


Φ = 22

Max. acceleration:

a max = 4π ⋅ 1.95 ⋅ 0.00108 ⋅ 0.4 ⋅ 22

(From page 13-20 with 2

ζ H = 1.6%

)

2

= 1.43m ⁄ s 2

(Measured: a max = 1.63m ⁄ s 2 ) Max. dyn. displ.:

d dyn,max = 1.08 ⋅ 0.4 ⋅ 22 = 9.50mm

Max. displacement:

d max = 9.50 + 1.08 = 10.58mm

(Measured: d max = 12.04mm ) 14 Pedestrian Footbridge with TMD

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10 5 0 -5 -10 0

5

10

15

20 Time [s]

25

30

35

40

2.0 1.5 1.0 0.5 0.0 -0.5 -1.0

5

10

15

20 Time [s]

25

30

35

40

Figure 14.19: One person walking with 2 Hz: Acceleration at midspan with locked TMD.


Page 14-21

10 5 0 -5 -10 0

5

10

15

20 Time [s]

25

30

35

40

Figure 14.20: One person walking with 2 Hz: Displacement at midspan with locked TMD. (ABAQUS-Simulation). Acceleration at midspan [m/s2]


Figure 14.18: One person walking with 2 Hz: Displacement at midspan with locked TMD.

-1.5 -2.0 0

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14.7.2 Locked TMD (ABAQUS-Simulation) Displacement at midspan [mm]


Test results


2.0 1.5 1.0 0.5 0.0 -0.5 -1.0 -1.5 -2.0 0

5

10

15

20 Time [s]

25

30

35

40

Figure 14.21: One person walking with 2 Hz: Acceleration at midspan with locked TMD. (ABAQUS-Simulation).


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The curves in Figures 14.20 and 14.21 were computed using the FE program ABAQUS. A similar calculation is described in detail in Section 13.3.3. The input data used in that section were only slightly adjusted here in order to better describe the properties of the test. Maximum vibration amplitude Static displacement:

d st = 1.08mm

(Measured: d st = 1.22mm ) Maximum displacement:

d max = 11.30mm


10 5 0 -5 -10 0

5

10

15

(Measured: d max = 12.04mm )

Maximum acceleration:

20 Time [s]

25

30

35

40

Figure 14.22: One person walking with 2 Hz: Displacement at midspan with free TMD.

d max 11.30 V = ----------- = ------------- = 10.5 1.08 d st a max = 1.68m ⁄ s 2

(Measured: a max = 1.63m ⁄ s 2 ) The maximum amplitudes of the numerical simulation and of the experiment agree quite well and also the time-histories shown in Figures 14.18 and 14.21 look quite similar. Please note that during the first 2 seconds of the experiment, displacements and accelerations are zero, because the person started to walk with a slight delay.



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14.7.3 Free TMD Displacement at midspan [mm]


2.0 1.5 1.0 0.5 0.0 -0.5 -1.0 -1.5 -2.0 0

5

10

15

20 time [s]

25

30

35

40

Figure 14.23: One person walking with 2 Hz: Acceleration at midspan with free TMD.


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Estimate of the maximum vibration amplitude

14.8 Group walking with 2 Hz


about 5.5

Maximum dyn. displ.:

d dyn,max = 1.08 ⋅ 0.4 ⋅ 5.5 = 2.38mm

Maximum displacement:

d max = 2.38 + 1.08 = 3.46mm

All student participating to the test (24 people) cross the footbridge in a continuous flow. A metronome is turned on to ensure that all students walk in the same step and with a frequency of about 2 Hz.

(from page 13-41)

(Measured: d max = 3.27mm ) Maximum acceleration.:

2

2

a max = 4π ⋅ 1.95 ⋅ 0.00108 ⋅ 0.4 ⋅ 5.5 = 0.36m ⁄ s 2

(Measured: a max = 0.34m ⁄ s 2 ) 14.7.4 Remarks about “One person walking with 2 Hz” • The effect of the TMD can be easily seen in Figures 14.22 and 14.23. The maximum acceleration at midspan reduces from 1.63m ⁄ s 2 to 0.34m ⁄ s 2 , which corresponds to a permissible value.

The test is carried out both with locked (Section 14.8.1) and free (Section 14.8.2) TMD. In Figures 14.24 to 14.27 the first 40 seconds of the response of the bridge are shown. Remarks The results of the experiments with several people walking on the bridge are commented by using the results of tests with one person walking (see Section 14.7) as comparison. For this reason the maximum vibration amplitudes shown in Figures 14.18, 14.19, 14.22, 14.23 and 14.24 to 14.27 are summarised in Tables 14.1 and 14.2. Case

Group

1 person

ratio

Maximum acceleration at midspan. Locked TMD

2.05 m/s2 1.63 m/s2

1.26

Maximum acceleration at midspan. Free TMD

0.96 m/s2 0.34 m/s2

2.82

Ratio

2.14

4.79

Table 14.1: Comparison of the accelerations at midspan.


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Case

Group

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1 person

ratio

Maximum displacement at midspan. Locked TMD

20.52 mm 12.04 mm

1.70

Maximum displacement at midspan. Free TMD

12.28 mm 3.27 mm

3.76

Ratio

1.67

3.68

Table 14.2: Comparison of the displacements at midspan.

It is further assumed that only about 16 of the 24 persons are on the footbridge at the same time. The following remarks can thereby be made: • The maximum acceleration measured at midspan of the bridge with locked TMD is only about 1.26-times greater than the acceleration which has been generated by the single person. According to section 13.3.3 we could have expected a larger


An-Najah 2013

of the displacement is larger that the amplification factor of the accelerations, because the static deflection caused by the group is significantly larger than that caused by the single person. • The activation of the TMD results in a reduction of the maximum acceleration caused by the single person by a factor of 4.79. In the case of the group the reduction factor is only 2.14. It should be noted here that when the TMD is active (free), the vibrations are significantly smaller, and therefore it is much easier for the group to walk in step. It is therefore to be assumed that in the case of the free TMD, the action was stronger than in the case of the locked TMD. This could explain the seemingly minor effectiveness of the TMD in the case of the group.

acceleration from the group ( 16 = 4 ). One reason why the maximum acceleration is still relatively small, is the difficulty to walk in the step when the “ground is unsteady.” With a little more practice, the group could probably have achieved much larger accelerations. It is further to note that the person who walked of the bridge for the test presented in Section 14.7 was with his 95 kg probably much heavier than the average of the group. • The maximum displacement measured at midspan of the bridge with locked TMD is 1.70 times larger than the displacement generated by the single person. The amplification factor


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10

0

-10

-20 0

5

10

15

20 Time [s]

25

30

35

40

2.0 1.0 0.0 -1.0 -2.0 5

10

15

20 time [s]

25

30

35

40

Figure 14.25: Group walking with 2 Hz: Acceleration at midspan with locked TMD.


Page 14-29

10

0

-10

-20 0

5

10

15

20 Time [s]

25

30

35

40

Figure 14.26: Group walking with 2 Hz: Displacement at midspan with free TMD. Acceleration at midspan [m/s2]


Figure 14.24: Group walking with 2 Hz: Displacement at midspan with locked TMD.

0

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14.8.2 Free TMD Displacement at midspan [mm]


14.8.1 Locked TMD


2.0 1.0 0.0 -1.0 -2.0 0

5

10

15

20 Time [s]

25

30

35

40

Figure 14.27: Group walking with 2 Hz: Acceleration at midspan with free TMD.


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14.9 One person jumping with 2 Hz

14.9.1 Locked TMD First the maximum amplitudes are calculated by hand: 0.71 G d st = ------- = ---------- = 0.0008m = 0.82mm 861 KH

(Measured: d st = 0.93mm )


1 1 V = ------ = ------------------------- = 31.25 2ζ ( 2 ⋅ 0.016 ) 2

2

a max = 4π ⋅ 1.95 ⋅ 0.0008 ⋅ 1.8 ⋅ 31.25 = 6.92m ⁄ s 2

(Measured: a max = 7.18m ⁄ s 2 ) Max. dyn. displacement: d dyn,max = 0.82 ⋅ 1.8 ⋅ 31.25 = 46.13mm Maximum displacement: d max = 46.13 + 0.82 = 46.95mm (Measured: d max = 51.08mm )



Sought is the response of the bridge under this excitation. A similar problem was solved theoretically in Section 6.1.3.


An-Najah 2013

Test results

One 72 kg-heavy person (G = 0.71 kN) keeps jumping at midspan of the footbridge. He is jumping with a frequency of 1.95 Hz, which is approximately equal to the first natural frequency of the bridge.

Static displacement:


60 40 20 0 -20 -40 -60 0

5

10

15

20 Time [s]

25

30

35

40

Figure 14.28: One person jumping with 2 Hz: Displacement at midspan with locked TMD. Acceleration at midspan [m/s2]


8.0 6.0 4.0 2.0 0.0 -2.0 -4.0 -6.0 -8.0 0

5

10

15

20 Time [s]

25

30

35

40

Figure 14.29: One person jumping with 2 Hz: Acceleration at midspan with locked TMD.

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14.9.2 Free TMD

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Estimate of the maximum vibration amplitude

60


about 5.5

40

Maximum dyn. displ.:

d dyn,max = 0.82 ⋅ 1.8 ⋅ 5.5 = 8.12mm

Maximum displacement:

d max = 8.12 + 0.82 = 8.94mm

20 0 -20


-40 -60 0

2

2

a max = 4π ⋅ 1.95 ⋅ 0.0008 ⋅ 1.8 ⋅ 5.5 = 1.22m ⁄ s 2

5

10

15

20 Time [s]

25

30

35

(Measured: a max = 1.04m ⁄ s 2 )

40

8.0 6.0

14.9.3 Remarks about “One person jumping with 2 Hz” • When jumping, the footbridge can be much strongly excited than when walking. • The achieved acceleration a max = 7.18m ⁄ s 2 = 73% g is very

4.0

large and two jumping people could easily produce the lift-off of the footbridge.

2.0 0.0 -2.0

• The effect of the TMD can be easily seen in Figures 14.30 and 14.31. The maximum acceleration at midspan reduces from

-4.0 -6.0 -8.0 0

(from page 13-41)

(Measured: d max = 8.12mm )

Figure 14.30: One person jumping with 2 Hz: Displacement at midspan with free TMD. Acceleration at midspan [m/s2]


5

10

15

20 Time [s]

25

30

35

40

7.18m ⁄ s 2 to 1.04m ⁄ s 2 , what, however, is still perceived as unpleasant.

Figure 14.31: One person jumping with 2 Hz: Acceleration at midspan with free TMD.


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